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BAER RINGS AND BAER RINGS S K Berberian The University of Texas at Austin Registered US Copyright Of ce March 1988 iii FOREWORD The theory of Baer gtk rings was set down in de nitive form by Irving Kaplansky its creator in 1968 but the subject refuses to stop evolving these notes are an attempt to record the present state of its evolution taking into account especially the simpli cations due to S Maeda SS Holland and D Handelman Also noted are the connections rst explored by J E Roos and G Renault with the theory of regular self injective rings exposed in K R Goodearl7s book Von Neumann regular rings 1979 and with the theory of continuous geometries maximal rings of quotients and von Neumann algebras Kaplansky7s axiomatic approach for studying simultaneously the classical equiv alence relations on projection lattices is developed in detail culminating in the construction of a dimension function in that context The foregoing makes plain that this is less a new venture than it is a con solidation of old debts 1 am especially grateful to Professors Maeda Holland and Handelman for explaining to me several key points in their work their generous help made it possible for me to comprehend and incorporate into these notes substantial portions of their work Each time 1 survey this theory 1 learn something new The most important lessons I learned this time are the following 1 The incisive results of Maeda and Holland on the interrelations of the various axioms greatly simplify and generalize many earlier results 2 The connection with regular right self injective rings puts the theory in a satisfying general algebraic framework the ghost of operator theory surviving only in the involution 3 There is a wealth of new ideas in Handelman7s 1976 Transactions paper it will be awhile before 1 can absorb it all 4 Somehow the touchstone to the whole theory is Kaplansky7s proof of direct niteness for a complete gtk regular ring Annals of Math 1955 That it is an exhiliarating technical tour de force does not diminish one7s yearning for a simpler proof The lack of one suggests to me that this profound result needs to be better digested by ring theory Sterling Berberian Poitiers Spring 1982 Preface to the English version The original French version Anneaua et anneauac de Baer was informally available at the University of Poitiers in the Spring of 1982 Apart from the correction of errors and no doubt the introduction of new iv ones the present version differs from the French only in the addition of a number of footnotes some of them clari cations others citing more recent literature SKB Austin Texas January 1988 Preface to the second English version The rst English version with a cir culation of approximately 40 was produced on a 9 pin dot matrix printer The present version was produced from a TEX le created using LEO ABK Software 1 am most grateful to Margaret Combs the University of Texas Mathematics De partment7s virtuoso TEXperson for gently guiding me through and around the ne points of TEX without insisting that 1 learn it SKB Austin Texas March 1991 Preface to the third English version The diagrams on pages 57 75 and 102 and the block matrices on pages 1197120 have been re coded so as to avoid hand drawn elements SKB Austin Texas April 2003 1 2 mam gt503 5 WDWDWDWDW 5 gt N 11 12 Generalized comparability 14 15 16 17 18 19 20 21 cm 03 TABLE OF CONTENTS FOREWORD Rickart rings Baer rings Corners Center Commutants Equivalence of idempotents gtk Equivalence of projections i i i i i Directly nite idempotents in a Baer ring i Abelian idempotents in a Baer ring type theory Abstract type decomposition of a Baer gtk ring i i Kaplansky7s axioms A H etc a survey of results i i i i Equivalence and gtk equivalence in Baer rings rst properties The parallelogram law Axiom H Polar decomposition Finite and in nite rings Rings of type I Continuous rings i i i i i i Additivity of equivalenc i i Dimension functions in nite rings Continuity of the lattice operations Extending the involution REFERENCES l i INDEX OF TERMINOLOGY INDEX OF NOTATIONS 104 124 126 130 Vi 1 RICKART RINGS BAER RINGS By ring we mean ring with unity a subring B of a ring A is assumed to contain the unity element of A 11 DEFINITION 23 p 510 A Rickart ring is a ring such that the right an nihilator resp left annihilator of each element is the principal right ideal resp left ideal generated by an idempotent 12 Let A be a Rickart ring 1 E A Say ill A1 i 6 06V 1 fA e and f idempotentsl Then y10lt y176ylt y60 20lt 17f22 gtf20 whence 20lt fe0 And 17610117fso 611f 13 If A is a gtk ring that is a ring with involution then righ is enough in De nition 11 one has my T so if y 9A with g idempotent then 1l Agquot 14 DEFINITION 23 p 522 A Rickart ring is a gtk ring such that the right annihilator of each element is the principal right ideal generated by a projection a self adjoint idempotent 15 In a Rickart gtk ring all7s well with left if 1T gA where gquot g 92 then 06F 91W A9 16 The projection in 14 is unique Proof If 6A fA with e and f projections then f6 e and 6f f so 6 6 fe 6 6f f 17 DEFINITION Let A be a Rickart gtk ring 1 E A and write ill A1 6 06V 1 fA 1Foranysubset Sofaring A SLZEA 150 VSES STZEA 310 VsES 2 RlCKART AND BAER RINGS with e and f projections Then e and f are unique 16 and cf 12 y10lt ye0120lt f20 e111f One writes e LP1 f RP1 called the left projection and the right projection of 1 Thus ell A1 14PM gt 06V 1 RPWDA By 15 one has LP1 RP1 RP1 LP1 18 For idernpotents ef in a ring A one writes e S f in case e E fAf that is ef fe e For projections ef in a gtk ring A the following conditions areequivalent e f eef efe eACfA Ae CAf 19 in a Rickart gtk ring A for an element 1 E A and a projection g E A one has g11lt gZLP1 Proof if e LP1 then 1l 17 e so 91 1 gt 17 g1 0 gt 17gEA17elt 17g 17elt e g 110 in a Rickart gtk ring the involution is proper that is 11 0 i 1 0 Proof lf 11 0 then cf 17 0 RP1LP1 LP1LP1 LP1 hence 1 LP1 1 0 111 PROPOSITION 14 Let A be a gtk Ting The following conditions are equivalent a A is a Rickart 7 ing b A is a Rickart ring and Ae Aee for every idempotent e Proof a i b Let e E A be idernpotent Then Ae 1 7 el Ag with g a projection One has eg e and ge g whence ege e and g gquot ge eg eg therefore e ege eege whence eA C eeA But eeA C eA trivially so eA eeA Taking adjoints Aequot Aee applying this to e in place of e one obtains Ae Aee b i a Let e E A be idempotent By b e aee for suitable a Set f aequot Then f aegtlteagt Mew fa whence f is self adjoint and f2 f From f ae one has Af C Ae from e aee fe one has e ef so Aequot C Af thus Ae Af Now let 1 E A be arbitrary Write 1l Ag 9 idernpotent Set e gquot by the preceding there exists a projection f such that Af Aequot Ag 1l thus A is a Rickart gtk ring O 112 EXAMPLE Every von Neurnann regular ring 7 p 1 is a Rickart ring Proof Let A be a regular ring 1 E A Choose y E A with 1 1341 and let e 1y then e is idernpotent 1A eA and 1l 1Al eAl 1 RICKART AND BAER RINGS 3 A17 e Similarly f 341 is idempotent A1 Af and 1T 17 fA Incidentally 1 A17 eT 17 eT eA 1A similarly 1 A1 thus 1A 1A and A1 A1 113 PROPOSITION 27 p 114 Th 45 The following conditions on a gtk ring A are equivalent a A is a regular Rickart gtk ring b A is regular and the involution is proper c for every 1 E A there e1ists a projection e such that 1A eA Proof a i b Every Rickart gtk ring has proper involution 110 b i a By 112 A is a Rickart ring Let 1 E A Since the involution is proper 1y 0 gt 11y 0 for if 11y 0 then 1y1y y11y 0 Thus 1T 11T whence 1 11 in other words proof of 112 A1 A11 In particular Ae Aee for all idempotents e so A is a Rickart gtk ring 111 a i c Let 1 E A Write 1l Af f a projection Then 1 17 fA but 1 1A proof of 112 so 1A 17fA and e 17 f lls the bill c i a Let 1 E A and write 1A eA with e a projection Then e1 1 and e 1y for suitable y whence 1 1y1 thus A is regular And 1l 1Al eAl A1 7 e so A is a Rickart gtk ring Note incidentally that e LP1 by 17 Similarly if f RP1 then A1 Af O 114 DEFINITION A gtk ring satisfying the conditions of 113 is called a regular ring 115 PROPOSITION 16 Lemma 53 The projections of a Rickart gtk ring form a lattice with eUffRPe17f e fe7LPe1fl Proof Write 1 e17 f and let f RP1 Obviously f S 17 f so f f is a projection we are to show that f f serves as supef From 1 1f e17ff ef we have e7ef e e efef ef f thus e S f f so f f majorizes both e and f Suppose also 6 S g and f S g g a prOjection then f f9 9f so 069 617 fg eg17 f e17 f 1 whence f S g and therefore also f f S 9 Thus e U f exists and is equal to f f This establishes the rst formula and the second follows from it by duality e f f exists and 60f1l1fU16l 116Rpl1fel e7RP17fe eiLPl1f6leiLPle if ltgt 4 1 RICKART AND BAER RINGS 116 COROLLARY Let A be a gtk regular ring Then 1 The set of principal right ideals of A is a modular lattice with a canonical complementation I gt gt IT 2 If ef are projections in A then eA fAe fA eAfAeUfA Proof As for any regular ring the set R of principal right ideals of A is a complemented modular lattice for the order relation C with supIJIJ infIJI J 7 p 15 Th 23 Since A is a Rickart gtk ring 114 its projections form a lattice P 115 and since every I E R has the form I eA with e a unique projection 113 the mapping e gt gt eA is a bijection P 7 R clearly an order isomorphism Therefore eUfAeAfA and e fAeA fA for all ef in P Moreover since e gt gt 17 e is an order anti automorphism of P so is eA gt gt 17 eA AeT eAT of R and eA 17eA are obviously complementary in R eA 17eA 0 eA17eAA O 117 PROPOSITION Let A be a Rickart ring R resp the set of all idempotent generated principal right resp left ideals of A ordered by inclusion Then i J gt gt Jl is an order anti isomorphism R 7 with inverse mapping ii If Ja is a family in R that possesses an in mum J in R then iii If Ja is afamily in R that possesses a supremum K in R then K U a lt2 a Proof If e E A is idempotent then eAl el A1 7 e 6 and AeT eT 17 eA E R whence eA eA and Ae Ae ii Suppose there exists J infJa in R Then J C Ja for all 04 so J C Ja Conversely let a E Ja Since A is a Rickart ring my Ae for an idempotent e whence d 17 eA E R For all 04 C Ja so a C Jlof Jet therefore 1 C J and in particular a E J iii Suppose there exists K supJa in R In view of i there exists infJlo in and K infJloT By the dual of ii apply ii in the opposite ring A0 one has infJlo UJal whence K UJalT O 118 COROLLARY 11 Prop 21 Let A be a Rickart gtk ring P its projection lattice 115 ei a family ofprojections in A 1 RICKART AND BAER RINGS 5 i If inf ez eaists in P then inf eiA MelA ii If sup ez eaists in P then sup eiA UeZA EelA In particular for any nite set of projections e1 en in A one has e10 enAe1A enA and e1 UUenA elAUUenA e1AenA Proof As in 117 let R be the set of idempotent generated principal right ideals of A If u E A is idempotent then uA 1 7 uT eA for some projection e clearly e LPu it follows that e gt gt eA is an order isomorphism P a R in particular R is a lattice and the corollary is immediate from 117 O 119 DEFINITION 18 p 3 A Baer ring is a ring A such that for every subset S of A the right annihilator of S is the principal right ideal generated by an idempotent 120 In a Baer ring left annihilators are also idempotent generated hence Baer Rickart Proof One has Sl SM if S eA e idempotent then Sl eAl A17 e 121 PROPOSITION 24 Lemma 13 Let A be a ring R the set of idempotent generated principal right ideals order R by inclusion The following conditions are equivalent a A is a Baer ring b A is a Rickart ring and R is a complete lattice In this case if Jet is any family in R then infJa Ja supJa UJQ Proof In any case note that for every J E R one has J J a i b By hypothesis R is the set of all right ideals ST where S C A As noted in 120 A is a Rickart ring Let Jet be any family in R Then J ltJZgtUJTeR whence Ja obviously serves as infJa in R Let J UJQYT E R Ob viously J 3 Ja for all 04 And if Ja C K E R for all 04 then UJQ C K so J UJQW c K K Thus J serves as supJa in R b i a Let S ma 04 E Q be any subset of A we are to show that ST E R Now sr my 6 1 RICKART AND BAER RINGS Since A is a Rickart ring may 6 R Write Jo may and let J infJa in R which exists by hypothesis By ii of 117 J Ja But Ja air STgt so ST J E R Note It suf ces to assume that A is a Rickart ring and that every family in R has an in mum in O 122 COROLLARY Let A be a regular ring R its lattice of principal right ideals The following conditions are equivalent a A is a Baer ring b R is a complete lattice Proof As noted in the proof of 116 R is a lattice and is the set of idempotent generated right ideals since A is a Rickart ring 112 the corollary is immediate from 121 In particular the lattice operations in R are given by the formulas in 121 O 123 DEFINITION 18 p 27 A Baer ring is a gtk ring A such that for every subset S of A the right annihilator of S is the principal right ideal generated by a projection In view of the formula Sl ST left annihilators are also generated by projections cf 120 124 PROPOSITION Let A be a gtk ring The following conditions are equivalent a A is a Baer gtk ring b A is a Rickart gtk ring whose projection lattice 115 is complete c A is a Rickart gtk ring and a Baer ring In such a ring if S C A then ST 17 eA where e supRPs s E S and if ea is any family of projections in A then inf eaA eaA sup eaA eaA Proof a i c Obvious c i b Let P be the projection lattice of A R the set of idempotent generated principal right ideals of A as noted in the proof of 118 e gt gt eA is an order isomorphism P a R Since R is complete 121 so is P b i a Let S C A and let e supRPs s E S assumed to exist by Then cf 17 STaEA sa0VsES a RPsLPa 0 Vs E S a RPs S 17 LPa Vs E S ae 17LPa a eLPa 0a ea0 er lt176gtA 1 RICKART AND BAER RINGS 7 thus S is a Baer gtk ring The asserted formulas are immediate from those in 121 and the fact that e gt gt eA is an order isomorphism P a R O 125 COROLLARY Let A be a gtk ring The following conditions are equiv alent a A is a regular Boer gtk ring b A is a gtk regular ring whose projection lattice cf 116 is complete c A is a gtk regular Boer ring Proof a i b The involution of A is proper 110 so A is gtk regular 113 114 and its projection lattice is complete by b of 124 b gt c Under either hypothesis A is a Rickart gtk ring so b gt c by 124 c i a Since A is a Rickart gtk ring 114 113 it is a Baer gtk ring by criterion c of 124 0 Because of criterion b such rings are also called complete gtk regular rings 126 EXAMPLE The endomorphism ring of a vector space is a regular Baer ring Proof Let V be a vector space left or right over a division ring D and let A EndDV be the ring of all D linear mappings u V a V Regularity Let u E A we seek v E A With u unit that is for all 1 E V Let W be any supplement of Kern in V V W 69 Kern Since W 0 Kern 0 the restriction of u to W is injective whence an isomorphism no W a Where no has the graph of Consider u61uW WCV and let i E A extend Hal for example take 1 to be 0 on some supplement of Then V y E W 14169 61uoy 2 therefore V y E W WWWD 7111 Thus LL U LL LL on W also uvu0 u on Kern so LL U LL LL on V The Boer property Let S C A say S i E l Then uESr uZu0 gt uV C Kern gt uV C Kerui Let e E A be an idempotent Whose range is Ker ui then u 6 ST gt uV C eV gt eu1 V 1 E V gt u eu lt u E eA 8 1 RICKART AND BAER RINGS thus ST 6A 127 EXAMPLE The algebra of bounded operators on a Hilbert space is a Baer gtk ring Proof Let A LH be the algebra of all continuous linear mappings u H a H where H is a Hilbert space with uquot adjoint of u A is a gtk ring Let S C A say S i E 1 Then N KeruZ is a closed linear subspace of H Let c E A be the projection with range N and kernel NL As in 126 ST 716 A C6H6A 128 Let A be a ring A right ideal I of A is said to be essential in A if 10 J a 0 whenever J is a nonzero right ideal One says that A is right nonsingular if the only 1 E A for which 1T is an essential right ideal is 1 0 in other words if 1 a 0 then there exists a nonzero right ideal J such that 1T 0 J 0 Dually for essential left ideals and left nonsingularity 129 Every Rickart ring in particular every regular ring and every Rickart gtk ring is both right and left nonsingular Proof Let 1 E A 1 a 0 Then 1T 6A with e idempotent e a 1 and J 17 eA is a nonzero right ideal with 1T 0 J 0 130 EXAMPLE If A is a right self injective ring and A is right nonsin gular then A is a Baer ring In 132 we shall see that A is regular Proof The assumptions are that AA is an injective right A module and ii if 1 E A is such that the right ideal 1T is essential then 1 0 Given S C A let 1 ST we seek an idempotent e E A such that l 6A Let f be an injective envelope of l in ModA the category of right A modules that is l E ModA l is injective and l is an essential submodule of 1 19 p 92 Prop 10 Since AA is injective the identity mapping 1 a AA extends to a monomorphism 1 a AA 19 p 91 Lemma 4 thus one can view l as a submodule of AA that is as a right ideal of A Since 1 is injective it is a direct summand of AA say AA lea K K a suitable right ideal of A if e is the component of 1 in 1 for this decomposition one sees easily that e is idempotent and 1 6A thus it will suf ce to show that l Of course 1 C Conversely let u E l we are to show that u E l ST Let s E S to show that su 0 it will suf ce by the hypothesis ii to show that guy is an essential right ideal of A Let luaEA uaEl clearly a right ideal of A since u E 1 and l is essential in f it follows that l u is an essential right ideal of A So it will suf ce to show that guy 3 l If a E l u then ua E l ST so 0 3ua sua thus a 6 guy and we have shown that l u C guy The asterisk signals material mainly functionalanalytic in nature Some of it is needed for 319 dimension function until then it serves only to provide examples and applications of the algebraic development The reader planning to omit dimension can omit the functional analysis altogether 1 RICKART AND BAER RINGS 9 Example 126 is a special case of 130 7 p 11 Cor 123 131 EXAMPLE If A is a right nonsingular ring then the maximal ring of right quotients Q of A is a regular right self injective ring hence is a Baer ring Proof Q is right self injective 19 p 107 Cor of Prop 2 and regular 19 p 106 Prop 2 hence right nonsingular 129 therefore Q is a Baer ring 130 132 EXAMPLE If A is a right self injective ring and A is right nonsin gular then A is a regular Baer ring Cf 141 Proof With Q as in 131 it will suffice to show that A Q Now A C Q and Q is a ring of right quotients of A hence AA is essential in QA 19 p 99 proof of Prop 8 Since AA is injective it is a summand of QA say QA AAG9J Then AA 0 J 0 yields J 0 AA is essential so Q A 133 A gtk ring A is said to be symmetric if for every 1 E A 1 r is invertible in A EXAMPLE Any C algebra With unity 134 LEMMA 18 p 34 Th 26 If A is a symmetric gtk ring then for every idempotent e E A there eacists a projection f E A such that eA fA Proof Let 2 1e 7ee 7e17e7e eequot ee and let t 2 1 since 2 is self adjoint so is t One has e2 eee 2e therefore et te and te et Set f eet teequot then f f and f2 eet eet eeetet eztet eztet eet f thus f is a projection From ef f one has fA C eA and from fe eet e eeet ezt e one has eA C fA Note that the conclusion of the lemma also holds in any Rickart gtk ring A symmetric or not since eA 17 eT O 135 PROPOSITION 18 p 34 Cor Let A be a symmetric gtk ring If A is a Boer ring resp Rickart ring then it is a Boer gtk ring resp Rickart gtk ring Proof Suppose for example that A is a Baer ring and let S C A Write ST eA e idempotent by 134 ST fA With f a projection O 136 EXERCISE Let A be a gtk regular ring 71 a positive integer MnA the gtk ring of n X n matrices over A With gtk transpose as the involution The following conditions are equivalent a MnA is gtk regular b the involution of Mn A is proper c the involution of A is n proper that is TL gala0 implies 1n0 21 Hint By a theorem of von Neumann every full matrix ring over a regular ring is regular 7 p 4 Th 17 cf 113 137 EXERCISE If A is a gtk regular ring Whose involution is 2 proper 11 yy 0 i 1 y 0 then A is symmetric Hint In a Rickart gtk ring With n proper involution one has RP 1 1 RP1 U U RPn 2 p 225 10 1 RICKART AND BAER RINGS 138 An AW algebra is a C algebra A with unity that is also a Baer gtk ring Since every C algebra With unity is symmetric it is the same to say that A is a C algebra and a Baer ring 135 Example A LH H a Hilbert space 127 A Rickart C algebra is a C algebra that is also a Rickart gtk ring equivalently a C algebra that is a Rickart ring For an example of a Rickart C algebra that is not an AW algebra see 2 p 15 Example 2 139 The commutative AW algebras are the algebras CT Where T is a Stonian space a compact space in which the closure of every open set is open 2 p 40 Th 1 The commutative Rickart C algebras are the algebras CT Where T is compact the closed open sets of T are basic for its topology and the closure of the union of any sequence of closed open sets is open 2 p 44 Th 1 140 A commutative C algebra A With unity is an AW algebra if and only if A is the closed linear span of its projections and ii every orthogonal family of projections has a supremum 2 p 43 Exer 1 A commutative C algebra A With unity is a Rickart C algebra if and only if A is the closed linear span of its projections and ii every orthogonal sequence of projections in A has a supremum 2 p 46 Prop 3 141 Every right continuous regular ring is a Baer ring Proof Let A be a regular ring and R its lattice of principal right ideals The hypothesis is that R is upper continuous 7 pp 160 161 in particular R is complete so A is a Baer ring by 122 lncidentally since every regular right self injective ring is right continuous 7 p 162 Cor 135 130 and 131 are special cases of 141 142 Every ring Without divisors of 0 is a Baer ring Whose only idempotents are 0 and 1 ii Conversely if A is a Rickart ring Whose only idempotents are 0 and 1 then A has no divisors of 0 hence is a Baer ring Proof is obvious ii Let 1 E A 1 74 0 Write 1T 6A 6 idem potent By hypothesis 6 0 or e 1 since 1 74 0 necessarily e 0 thus 06V 0 l 143 Every gtk ring Without divisors of 0 is a Baer gtk ring Whose only projec tions are 0 and 1 ii Conversely if A is a Rickart gtk ring Whose only projections are 0 and 1 then A has no divisors of 0 hence is a Baer ring Proof is obvious ii Same proof as in 142 With 6 taken to be a projection 144 Every division ring is a regular Baer ring Whose only idempotents are 0 and 1 ii Conversely if A is a regular ring Whose only idempotents are 0 and 1 then A is a division ring Proof ii If 1 E A 1 74 0 then 1A 1A by the hypothesis thus 1 is right invertible 145 Every involutive division ring is a regular Baer gtk ring Whose only projections are 0 and 1 ii Conversely if A is a gtk regular ring Whose only 1 RICKART AND BAER RINGS 11 projections are 0 and 1 then A is an invoiutive division ring Proo Cf 143 and 144 2 CORNERS 21 If A is a ring and e E A is idempotent the ring eAe with unity ele ment e is called a corner of A Reason If A MAB and e diag1 0 0 then eAe E B is the northwest corner7 of the matrix ring A For example the following proposition says that every corner of a Baer ring is itself a Baer ring 22 PROPOSITION 18 p 6 Th 4 If A is a Boer ring and e E A is idempotent then eAe is a Boer ring Proof Let S C eAe The right annihilator of S in eAe is eAe 0ST Write ST fA f idempotent Since S C eAe one has 17 e 6 ST hence I7 e f17 e f 7 fe left multiplying by e one has 0 ef 7 efe thus ef efe E eAe Write g ef which is an idempotent of eAe 92 efef efef eff ef g It will suf ce to show that eAe 0 ST Q eAe From Sg Sef Sef Sf 0 one has 9 eAe C eAe 0 ST Conversely if 1 E eAe 0 ST then 1 f1 fe1 fe1 91 E Q eAe thus eAe 0 ST C Q eAe O 23 Every corner of a Rickart ring is a Rickart ring Proof If e E A is idempotent and 1 E eAe apply the proof of 22 with S 24 Every corner of a regular ring is regular Proof Let A be regular e E A idempotent 1 E eAe Choose y E A with 1 1341 Then 1y1 1eye1 1eye1 so replacing y by eye one can suppose that y E eAe 25 If A is a Rickart gtk ring and e E A is a projection then eAe is a Rickart gtk ring Proof Since eae eae eAe is a gtk ring Let 1 E eAe and let S Write ST fA f a projection then the idempotent g ef efe see the proof of 22 is self adjoint hence is a projection and the proof continues as in 22 Note incidentally that RP1 1 7 f whereas the right projection of 1 calculated in eAe is e7g 17 but 17e 6 ST fA so 17e S f 17e 17ef f7ef f7g whence 17f e7g thus the right projection of 1 is the same whether calculated in A or in eAe Similarly for the left projection Brie y LPs and RP7s in eAe are unambiguous 26 PROPOSITION 18 p 30 If A is a Boer gtk ring and e E A is a projection then eAe is a Boer gtk ring 12 2 CORNERS 13 Moreover if e2 is any family of projections in eAe then sup ez is the same whether computed in A or in eAe and the some is true of inf ez briefly sups and infs in eAe are unambiguous Proof 1 eAe is a Baer ring 22 and a Rickart gtk ring 25 hence a Baer gtk ring 124 Proof 2 Let S C eAe in the notation of the proof of 22 one can take f to be a projection and then 9 ef efe is a projection Finally let e2 be a family of projections in eAe h sup ez in A Since ez S e for all i one has h S e thus h E eAe clearly serves as supremum of the ez in eAe Similarly for inf ei Alternatively inf ez e 7 supe 7 ei O 27 Let A be a gtk regular ring r E A e LPr f RPr i There exists a unique y E fA such that ry e One calls y the relative inverse of r ii Moreover y E fAe yr f ryr r yry y iii The relative inverse of y is r Proof i ii One has eA rA cf 113 say e ry Right multiplying by e one can suppose y E Ae and e ry rfy rfy so replacing y by fy one can suppose y E fAe 1f also y E fA with ry e then 10721 6 6 0 SO ti 14 0 fy fy 21 21 Moreover 06040671 ryrirf erirf rir 0 so fyr7f 0 whence yr f Finally ryrerr and yryfyy iii From y E fAe one has yA C fA and from yr f one has fA C yA thus yA fA and so f LPy Similarly e RPy and then iii is immediate from 28 PROPOSITION If A is a gtk regular ring and e E A is a projection then eAe is gtk regular More precisely if r E eAe and y is the relative inverse of r in A then yEeAe Proof eAe is regular 24 and its involution induced by that of A is proper hence it is gtk regular 113 Let r E eAe and write f LPr g RPr for the left and right projections of r as calculated in A As noted in 25 fg E eAe thus if y is the relative inverse of r in A then y E gAf C eAe O 29 if A is a regular Baer gtk ring and e E A is a projection then eAe is a regular Baer gtk ring Proof 24 and 26 2 10 Every corner of a regular right self injective ring also has these properties 7 p 98 Prop 98 211 Every corner of a right continuous regular ring also has these properties 7 p 162 Prop 137 212 Every self adjoint corner of a von Neumann algebra is a von Neumann algebra see 414 below 3 CENTER If A is a ring we systematically write Z for the center of A 31 The idempotents of Z form a Boolean algebra that is a complemented distributive lattice with u v 1w qu n1 71w 1 17u u S 1 being de ned by u 1w 32 PROPOSITION 18 p 8 Th 7 The center of a Baer ring is a Baer Ting Proof Let A be a Baer ring with center Z and let S C Z Write ST UA Sl Aw v and w idempotents Since S C Z one has ST Sl thus vA Aw Then 1 E Aw and w E UA so 1 vw w Therefore vA17v Av17v 0 similarly 17 vAv 0 thus for all a E A va17 v 17 vw 0 ya vav w whence v E Z It follows that Z 0 ST UZ thus Z is a Baer ring 0 33 The central idempotents of a Baer ring form a complete Boolean algebra Proof Let nipE1 be any family of idempotents in Z Applying the proof of 32 to S i E I one has ST 0A with v a central idempotent Write u17v For 16A onehas ui0 forall i gt 160A gt oww gt uw 0 In particular if e E A is idempotent then ui S e for all i gt uZ17e 0 for all i gt u17e 0 gt u S e thus it serves as a supremum for the ui in the set of all idempotents of A for the order e S f de ned by e E fAf a fortiori u sup ui in the set of idempotents of Z It then follows that 17 sup17 serves as inf The proof shows 34 If A is a Baer ring a family of central idempotents of A and usupuZ then for w E A one has wu0 gt ML 0 for all i 35 The center of a Rickart ring is a Rickart ring Proof In the proof of 32 let S z where z E Z 36 The center of a regular ring is regular Proof Let A be regular with center Z and let 2 E Z Write 2A UA A2 Aw with v and w idempotents Then 2A A2 yields 1 w E Z as in the proof of 32 Thus 2A 0A with v a central idempotent Write v 22 2 E A replacing 2 by 122 one can suppose 2 E UA Since 222 vz 2 it 14 3 CENTER 15 will suffice to show that 2 E Z Indeed for all a E A one has 12 7 zaz 52 7 221 av 7 pa 0 whence 0 12 7 mpg 12 7 za 12 7 2 37 The center of a regular Baer ring is a regular Baer ring Proof 32 and 36 38 In a Rickart gtk ring every central idempotent is a projection Proof Let A be a Rickart gtk ring with center Z u E Z idempotent we are to show that uquot u As noted in the proof of 118 uA fA with f a projection whence it fu uf f 39 PROPOSITION S Maeda 23 Lemma 21 Let A be a Rickart gtk ring Z its center e E A a projection The following conditions are equivalent a e E Z b effe for all projections f of A c e has a unique complement in the projection lattice of A namely 17e Proof a i b TriVial b i c Let f be a complement of e in the projection lattice of A that is eUf 1 and e f 0 we are to show that f 17 e By hypothesis ef fe therefore e f ef and eUf ef7ef thus ef 0 and 1ef70so f17e c i a Let a E A we are to show that me ea Set a e ea 7 eae Then ea a and ae e it follows that eA aA whence e LPa And a2 aea aea ea a so a is idempotent Write f LP17a as noted in the proof of 118 17 aA fA Mom a17 a 0 we infer that af 0 By the formulas of 118 60fAeA fAaA 17aA0 eUfAeAfA aA17aA AlTA thus e f0 and eUf 1that is f isacomplement of e By the hypothesis Cgt f176gtso 0afa17ea7aea7e thus a e That is e ea 7 eae e whence ea eae similarly ea eae whence 1e eae ea 0 This result is of capital importance from 13 onward cf the proof of 138 310 PROPOSITION 18 p 30 Cor The center of a Baer gtk ring is a Baer gtk ring with unambiguous sups and infs Proof In the proof of 32 one can take 1 to be a projection thus Z is a Baer gtk ring Now suppose uiZEI is a family of projections in Z and let u sup uz as calculated in A 124 Writing S i E I we know that ST 1 7 uA 16 3 CENTER by 124 But as noted in the proof of 32 we have 1711 E Z that is u E Z and Z 0 ST 1 7 uZ therefore it is the supremum of the ui as calculated in Z 124 applied to Z Thus sups in Z are unambiguous by duality so are infs O 311 The center of a gtk regular ring is gtk regular Proof If A is a gtk regular ring with center Z then Z is a gtk ring with proper involution and is regular 36 so it is gtk regular 114 312 The center of a regular Baer gtk ring is a regular Baer gtk ring Proof 36 and 310 313 If A is a right continuous regular ring then its center Z is regular and continuous both right and left Proof A is a regular Baer ring 141 so Z is a regular Baer ring 37 and the idempotents of Z form a complete Boolean algebra B 33 Since Z is regular its lattice of principal ideals is isomorphic to B hence is continuous thus Z is continuous cf 7 pp 160 161 314 If A is a regular right self injective ring then its center Z is regular and self injective right and left Proof 1 p 418 By 129 and 132 A is a regular Baer ring The strategy of the proof is to verify Baer7s criterion for Z Suppose I is an ideal of Z and f l a Z is Z linear we seek to extend f to a Z linear map Z a Z equivalently we seek z E Z such that fy 2y for all y E 1 One can suppose that l is essential in Z for there exists an ideal K of Z with 10 K 0 and l K essential in Z 19 p 60 Lemma 1 and one can extend f to 169 K for example by annihilating K For use later in the proof we now show that the right annihilator 1T of l in A is 0 Write 1T uA u E A idempotent Since 1 C Z one knows that u E Z proof of 32 and so Z 0 1T uZ Writing 0 for annihilator in Z we have 10 Z m V uZ hence loo 17 uZ Then 1010 C100 0 lo 0 since 1 is essential in Z we conclude that 10 0 thus it 0 whence 1T 0 Let J Al 1A be the ideal of A generated by 1 We propose to de ne a right A linear extension f J a A of f by the formula J 214202 Zfltyiai 242 E l a E A 21 21 To see that this is well de ned suppose 2171 yiai 0 Since Z is regular 36 one can write ylZynZUZ with v E Z idempotent 7 p 1 Th 11 Then vyi yi for all i so Z y ai Z finai wawth fU Zyiaz 0 3 CENTER 17 thus f is well de ned and clearly right A linear Since A is right self injective there exists if E A with f1 t1 for all 1 E J In particular V y 61 fy fy 752 so it will suffice to show that t E Z If a E A we wish to show that ta at since IT 0 it will suf ce to show that Ita 7 at 0 Indeed for all y E I one has yt ty fy E Z so yta 7 at yta 7 yat yta 7 1yt 0 315 DEFINITION Let A be a Baer ring 1 E A The central cover of 1 is the central idempotent C1 de ned by the formula C1 infu u1 1 u a central idempotent of A cf 33 An element 1 E A is said to be faithful if C1 1 in other words it 0 is the only central idempotent such that 711 0 316 C1 is the smallest central idempotent it such that 711 1 Proof Let S be the set of all central idempotents it such that 711 1 it clearly suf ces to show that 01 6 S For all u E S one has 117 u 0 so by 34 we have 01sup17u uES 117infu u E S 117 01 whence 1C1 1 as desired 317 PROPOSITION Let A be a Bacr ring 1 E A u a central idempotent of A Then i 711 0 gt uC1 0 ii Cu1 Proof 7110 gt 17u11 gt C1 17u gt uC10 ii Let U From 711 and C1 711 uC11 711 one has 1 S u and v S But u1 vu1 vu1 v1 u7v1 0 so u 7 vC1 0 by i thus uC1 UC1 v 0 318 DEFINITION 19 pp 54 56 A ring A is said to be semiprime if 1A1 0 i 1 0 An equivalent condition is that 0 is the only nilpotent ideal of A it does not matter if one means left right or bilateral ideal It is the same to suppose that I 0 is the only ideal with I2 0 left right or bilateral7it does not matter A ring A is prime if 1Ay 0 i 1 0 or y 0 an equivalent condition is that there are no ideal diVisors of 0 1Note that if 1 is any element of A and if u C1 then 1 is faithful in uA For a the central idempotents of uA are the central idempotents of A that are S u A central idempotent U of uA such that 11 1 must therefore be 2 C1 u hence equal to u 18 3 CENTER 319 Every regular ring is serniprinie Proof Every nonzero left or right ideal contains a nonzero idernpotent 320 Every Baer gtk ring is serniprinie 18 p 31 Proof Suppose 1A1 0 that is 1 E 1AT Since 1 1A is a right ideal If is a bilateral ideal Write 1T uA u a projection Since If is bilateral AuA C uA so 1 7 uAuA 0 1 7 uAu 0 whence on taking adjoints uA1 7 u 0 therefore an uau ua for all a E A so u is central By hypothesis 1 6 If so 711 1 but In 0 in particular 1n 0 whence 1 u1 1n 0 321 PROPOSITION 18 p 15 Th 13 Let A be a semiprime Baer ring i If 1 is a right ideal of A then If is a direct summand of A ii For 11 in A 1Ay0 gt C1Cy0 iii If e E A is idempotent then the central idempotents of the ring eAe are the eu with u a central idempotent of A thus eZ and the center of eAe contain the same idempotents more precisely if f is a central idempotent of eAe then f eCf iv If e E A is an idempotent and 1 E eAe then Ce1 eC1 where Ce denotes central cover relative to the Baer ring eAe Proof Write 1T uA u idernpotent As argued in the proof of 320 one has 17 uAu 0 Setting J uA17 u lT17 n J is a left ideal of A because uA If is such that J2 uA17 uuA17 u 0 since A is serniprinie J 0 thus uA1 7 u 0 As argued in 320 it is central ii Suppose 1Ay 0 Let 1 1A by i If uA with u a central idernpotent By hypothesis y 6 1T thus uy y so 03 3 it But lu 0 so 111 0 hence 01 3 1 7 it therefore C1Cy 0 The reverse irnplication results from the formula 1Ay 1C1AyCy 1AyC1Cy iii Let f beacentralidenipotent of eAe Then e7fAf e7feAef e7fgtlteAegtf7 e7fgtflteAegt 70so clte7fgtcltfgt 70 byltiigtrhen e7fgtcltfgt 7 0 so ecu 7 NO 7 f iv Let 1 E eAe Since 1eC1 1eC1 1C1 1 one has Ce1 S eC1 By iii one can write Ce1 eu with u a central idempotent of A then 1 1Ce1 1eu 171 so 01 3 u whence eC1 3 en 081 0 322 If A is a Baer gtk ring and ei is any family of projections in A then Csup ei sup Cei Proof Write e sup ei ui Cei we are to show that Ce sup 71 One has Ce 2 e Z ei so Ce 2 C e 71 for all i If v is a central projection with u lt v for all i then ez S 71 S v for all i whence e S i so Ce S 1 323 If in a gtk ring ei is a family of projections possessing a supremuni e cf 118 124 then for every central projection 71 the family nei has ue as supremuni brie y usup ei supueZ Similarly if ei has an in rnuni f then uf infueZ for every central projection u 3 CENTER 19 Proof Suppose e sup 6 and u is a central projection From 62 626 one infers uei ueiue thus uei S ue for all 1 Suppose g is a projection with uei S g for all 1 Then 0 reel17 g 62111 7 g so 62 S 17 u17 g for all 1 therefore 6 S 17 u17 g whence eu17 g 0 thus ue S g This proves that ue serves as supremum for the family 116 The second assertion follows from the rst by virtue of the order anti isomorphism g gt gt 1 7 g gtk324 U Sasaki If A is an AW algebra 138 with center Z and if e E A is a projection then the center of 6A6 is eZ Proof 2 p 37 Cor 2 of Prop 4 From 26 one sees that 6A6 is an AW algebra let Z8 be its center also an AW algebra by 310 By 321 iii Z8 and eZ contain the same projections since each is the closed linear span of its projections 140 Z8 6Z See also 334 325 If A is a ring with center Z and if u is a central idempotent of A then uA has center uZ Proof A uA X 17 326 L Jeremy 15 Lemma 02 If A is a regular right self injective ring with center Z and if e E A is idempotent then the center of 6A6 is eZ Proof 4 Let u Ce the central cover of 6 Then 6A6 euAe euAe and eZ 6uZ where uZ is the center of uA 325 dropping down to uA we can suppose that Ce 1 Then A6l 0 by 319 and 321 ii so there exists an isomorphism of rings Lp A 7 End 8A8 A6 where A6 is regarded as a right eAe module in the natural way and for a E A Lpa is left multiplication by a 7 p 98 Prop 98 Let Z8 be the center of 6A6 It is obvious that eZ C Ze Conversely let it 6 ZS De ne 04 A6 7 A6 by 04me 6t since if is central in 6A6 04 is right eAe linear so 04 6 End eAeAe Therefore 04 Lpa for suitable a E A Then for all 1 E A one has 04me Lpae met 116 mi 116 for 1 1 this yields t 16 so it will suf ce to show that a E Z Since 04 is a right multiplication on A6 it commutes with every left multiplication on A6 hence Oat35 Lpbd for all b E A that is 04 is in the center of EndeAeAe since 04 Lpa and Lp is a ring isomorphism a is in the center of A 327 If A is a right continuous regular ring with center Z and if e E A is idempotent then 6A6 has center eZ Proof One has A B X C with B abelian7 all idempotents central and C right self injective 7 p 169 Th 1317 so we may consider these cases separately In view of 326 we need only consider the case that A is abelian but then every idempotent e E A is central and our assertion follows trivially from 325 328 If A is a regular Baer gtk ring with center Z and if e E A is idempotent then 6A6 has center eZ Proof By a theorem of Kaplansky 17 Th 3 or 18 p 117 Th 69 cf 2010 below A satis es the hypotheses of 327 20 3 CENTER 329 DEFINITION 4 We say that a ring A with center Z is compress ible if for every idempotent e E A the center of 6A6 is eZ Examples Every regular right or left self injective ring 326 every right or left continuous reg ular ring 327 every regular Baer gtk ring 328 330 If A is a compressible ring then every corner of A is compressible Proof Let Z be the center of A Let f E A be idempotent and write Zf for the center of fAf by the hypothesis Zf fZ Assuming e E fAf idempotent we are to show that efAfe has center er indeed efAfe 6A6 has by the compressibility of A center eZ efZ er 331 If A is an abelian7 ring all idempotents central then A is compress ible Proof Obvious from 325 332 Every ring isomorphic to a compressible ring is compressible Obvious 333 If A is a Rickart gtk ring with center Z such that for every projec tion f fAf has center fZ then A is compressible Proof Let 6 E A be idempotent Since A is a Rickart gtk ring one has 6A 17 6T fA for a suitable projection f From 6A fA and the idempotence of e and f one infers that e and f are similar 55 below say f tet l Let Lp A a A be the inner automorphism of A induced by t Lpa flat By hypothesis fAf has center fZ hence LpfAf has center wfZ but wfAf wfwAlt2f 6A6 and wfZ wfwZ 6Z thus 6A6 has center eZ as desired gtk334 Every AW algebra is compressible Proof 324 and 333 335 EXAMPLE P Armendariz Not every Baer ring is compressible Proof cf 4 Example 10 Let H be the division ring of real quaternions and let H H Bilto H be the ring of all upper triangular 2 X 2 matrices 06 y 0 z with yz E H Since H is a division ring B is a Baer ring 18 p 16 Exer 2 One readily calculates that the idempotents of B are 0 1 and the matrices 0 y 1 y 0 1 0 0 with y E H Regard IR C C C H in the usual way Let A be the subring of B de ned by C H Ailto Hgtgt 3 CENTER 21 that is A is the set of all matrices of the form 3 Z Where c E C and 11 6 H Since A contains every idempotent of the Baer ring B A is itself a Baer ring for if S C A and ST is the right annihilator of S in B say ST eB With e idempotent then 6 E A and so A ST 6A The center Z of A is the set of all matrices 6 3 With r E lR and in particular Z is one dimensional over R Let 10 eilt0 0gtEA then is commutative hence is its own center and is 2 dimensional over lR Whereas 1R 0 EZ lt0 0 is 1 dimensional over lR The evident relation eAe a eZ shows that A is not compressible In fact eAe is not even isomorphic to eZ 336 If A is a ring such that for some positive integer n MnA is compressible then A is compressible Proof Since A is isomorphic to a corner of MnA this is immediate from 330 and 332 337 PROBLEM If A is compressible is M2A compressible 72 338 PROBLEM ls every regular Baer ring compressible3 339 PROBLEM ls every Baer gtk ring compressible For several other fragmentary results on compressibility see 2Answered in the negative by G M Bergman Carma Algebra 12 1984 18 3My guess is that the answer is no D Castella Comrm Algebra 15 1987 16211635 has proved that if A is a regular Baer ring Without abelian summand cf 827 then A is compressible if and only if the center of A coincides With the center of its maximal ring of right quotients So the task is to construct a regular Baer ring Without abelian summand Whose center fails to satisfy the indicated condition For complements to Castella s paper see E P Armendariz and S K Berberian Carma Algebra 17 1989 17391758 4 COMMUTANTS 41 If S is a subset of a ring A the commutant of S in A is the set S1EAss1forallsES which is a subring of A One also writes S S the bicommutant of S and SW S S One has S C S and ii S C T i S D T It follows that S S For S C S yields S 3 SW whereas S C S by For a subring B of A one has B S for some subset S of A if and only If A is a ring and S is a subset of A s E S i squot E S then S is a subring of A the subrings B of A satisfying B B are the subrings S with S a subset of A If S is a commutative subset of a ring A that is S C S then B S is a commutative subring of A such that S C B and B B Commutativity of B From S C S one infers that S D S B thus B c s s B 42 If S is a subset of a ring A and if B S then B contains inverses b E B invertible i b 1 E B Proof Let b E B be invertible For all s E S one has sb bs b lsbb l b lbsb 1 b ls sb l 43 Let A be a Rickart ring S a subset of A 1 E S Write my eA e idempotent Then for all s E S se ese and ii 17es 17es17e Proof For all s E S one has use sue s 0 0 thus se 6 my eA whence ese se Condition ii is equivalent to condition 44 PROPOSITION Let A be a Rickart gtk Ting S a subset of A If 1 E S then LP1 E S and RP E S thus S is a Rickart i ing with unambiguous LP7s and RP7s Proof Let f RP thus 1 17 fA By ii of 43 fs fsf for all s E S also fs fsf for all s E S because Squot C S whence sf fsf Thus fs fsf sf for all s E S so f E S Writing B S we thus have B 1T 17 fB whence the proposition 0 45 PROPOSITION 18 p 30 Th 20 If A is a Baei 7 ing and S is a subset of A then S is a Baei i ing with unambiguous sups infs LP7s and RP7s 22 4 COMMUTANTS 23 Proof Write B S By 44 1 E B i LP1RP1 E B Let ei be a family of projections in B e supez in A we assert that e E B if s E S then for all i one has es 7 seez esei 7 sez eels 7 sez eis 7 sez 0 whence es 7 see 0 cf 124 thus se ese therefore also se ese because S C S whence es ese se Thus e E S B By 44 B is a Rickart ring and its projection lattice is complete by the preceding therefore B is a Baer ring 124 ltgt 46 There exists a regular Baer ring A with a subset S such that S is not a Rickart ring hence is not regular For the good news see 47 Proof Let F be a eld A M2F the ring of 2 X 2 matrices over F and let S s where s is the matrix LG 1 lt51gtlt33gtZlti2gtlt igtgt 10 bd 7 a ab c d T c cd which says that c 0 and d 1 Thus the ring B S consists of all matrices 8 51 with 11 6 F The only idempotents of B are 01 and B contains nilpotent elements namely the matrices 0 b o o 7 it follows that B is not a Rickart ring 142 However A is a regular Baer ring 126 indeed a regular self injective ring right and left 7 p 11 Cor 123 So far F need not be commutative7it can be any division ring 1f moreover F has no element a such that a21 0 equivalently 12 112 0 i a b 0 then A is regular with transpose as involution 136 hence is a regular Baer ring 125 47 PROPOSITION If A is a regular ring and S is a subset of A then S is a regular ring More precisely if 1 E S and y is the relative inverse of 1 27 then y E S Proof Write B S since A is a Rickart ring 114 so is B 44 Let 1 E B and write e LP1 f RP1 by 44 ef E B Let y be the relative inverse of 1 in A thus y E fAe1y e 341 f Given s E S we are to show that ys sy One has Let us calculate S ys7sy1ys17sy1y1s7sy1fs7sf0 24 4 COMMUTANTS because f E B therefore 0ysisyeyse7syeyesisyeys7sy recall that e E B so se es 0 For an application see 1426 48 COROLLARY If A is a regular Baer ring and S is a subset of A then S is a regular Baer ring with unambiguous sups infs LP7s RP7s and relative inuerses Proof 45 and 47 O 49 The ring A M2F of 46 is self injective thus 46 runs somewhat counter to 314 which says that A Z 2 F is regular and self injective 410 If A is an AW algebra 138 and S is a subset of A then S is an AW algebra with unambiguous sups infs LP7s and RP7s Proof B S is clearly closed in A for the norm topology thus is a C algebra quote 45 It is essential that S be a subset let F C in 46 With conjugate transpose for the involution of A M2 C 411 DEFINITION 6 p 2 Def 1 If H is a Hilbert space and LH is the algebra of all bounded operators on H 127 then a subalgebra A of LH such that A A is called a von Neumann algebra on H thus 41 the von Neumann algebras on H are the algebras S Where S is a subset of 412 Every von Neumann algebra is an AW algebra 138 and 410 But not conversely 413 Let A be a commutative C algebra With unity and write A CT T compact Thus A is an AW algebra if and only if T is Stonian 139 In order that A be isomorphic to a von Neumann algebra on a suitable Hilbert space it is necessary and suf cient that T be hyperstonian Stonian With suf ciently many normal7 measures 414 If A is a von Neumann algebra on a Hilbert space H and if e E A is a projection then eAe may be identi ed via restriction With a von Neumann algebra on the Hilbert space eH 6 p 16 Prop 1 415 If A is a von Neumann algebra on a Hilbert space H and if S is a subset of A then the commutant B of S in A is a von Neumann algebra on H Proof Writing for commutant in LH one has B Ans A S AUS Where A U S is a subset of LH In particular the center Z A f A of A is a von Neumann algebra on H Note incidentally that A f A is the center of both A and A 4 COMMUTANTS 25 416 If A is a von Neumann algebra on the Hilbert space H and A is the commutant of A in LH then A is a von Neumann algebra on H Proof lmmediate from 41 A AW A 417 If A is a von Neumann algebra on a Hilbert space H and if n is a positive integer then Mn A may be identi ed with a von Neumann algebra on the Hilbert space sum nH of 71 copies of H 6 p 23 Lemma 2 5 EQUIVALENCE OF IDEMPOTENTS 51 If A is a ring Viewed as a right A module AA then the direct summands of AA are the principal right ideals eA With e idempotent 52 PROPOSITION For idempotents ef of a ring A the following con ditions are equivalent a eA E fA as right A modules b Ae Af as left A modules c there e1ist 11 in A with 1y e and y1 f d there e1ist 1 E eAf y E fAe with 1y e and y1 f Proof a i d Let Lp eA a fA be an isomorphism of right A modules Set y Lpe E fA and 1 Lp lf E eA then ye Lpee Lpee Lpe y so y E fAe and similarly 1 E eAf For all s E eA W W68 w6s 218 similarly Lp lt 1t for all t E fA Then 1y Lp ly Lp ltpe e similarly y1 f d i c Trivial c i d If 1y e 341 f set 1 e1f E eAf and y fye E fAe Then 1y e1fye e1y1ye e1y1ye e4 e similarly y1 f d i a Let 1y be as in For s E eA one has ys E fAes C fA so a map Lp eA a fA is de ned by Lps ys Lp is right A linear Similarly a right A linear mapping 1 fA a eA is de ned by 1t 1t For all s E eA 1Lp8 1ys es s thus 1 o p 18A similarly p 01 lfA Thus a gt c gt Since c holds for A if and only if it holds for the opposite ring A0 we conclude that c gt O 53 DEFINITION 18 p 22 Idempotents ef of a ring A are said to be equivalent or algebraically equivalent7 in A written e g f if they satisfy the conditions of 52 One also refers to g as ordinary equivalence7 as contrasted With the gtk equivalence7 de ned in the next section for projections of a ring From condition a of 52 it is obVious that g is an equivalence relation in the set of idempotents of A 54 PROPOSITION 18 p 23 Th 15 With notations as in 52 d the mapping s gt gt ys1 is an isomorphism of rings 6 eAe a fAf with inverse mapping t gt gt 1ty For every idempotent g S e one has 9 g 69 26 5 EQUIVALENCE OF IDEMPOTENTS 27 Proof For s E eAe write 6s ysa since y E fA and a E Af one has 6s E fAf The mapping 6 eAe 7 fAf is additive 6e f and for ss in eAe one has 6ss yssa ysesa ysaysa 6s6s thus 6 is a homomorphism of rings 1f 6s 0 then 0 ysa 0 aysay ese s thus 6 is injective and if t E fAf then t ftf yatya 6s where s aty E eAe so 6 is surjective Finally if g S e cf 18 let a gm 2 219 then 062 90619 969 g and 206 2 1906 99 thus 9 3 99 O 55 PROPOSITION 18 p 24 The equivalence of idempotents in a ring A has the following properties 1 e 3 0 i e 0 2 e 3 f i ue 3 uf for all central idempotents u 3 If e 3 f and if e1 en are pairwise orthogonal idempotents with e el en then there eacist pairwise orthogonal idempotents f1 fn such that f1fn and ez 3 fi forall i 4 If e1 en are orthogonal idempotents with sum e if f1 fn are orthogonal idempotents with sum f and if ez 3 fi for all i then e 3 f 5 ef are similar if and only if both e 3 f and 17 e 3 17 f 6 If eAfA then e and f are similar Proof Recall that idempotents ef are said to be orthogonal if ef fe 0 in which case e f is an idempotent with e S e f and f S e f 1 eAO 0Ae 0 so in the notation of 52 d a y 0 whence e my 0 2 is obvious 3 With 6 eAe 7 fAf as in 54 the idempotents fz 6ei clearly ll the bill 4 if a E eiAfi yi E fiAeZ with aiyi ez and yzaz fi then the elements aZnaiEeAf and yZnyiEfAe 21 21 effect an equivalence e 3 f 5 Suppose a E eAf y E fAe with my e ya f and that a E 17eA17f y E 17fA17e with ay 17e ya 17f writing saa tyy wehave stts1 and ftess les 6 1f eA fA then also A17 e eAl fAl A17 f From 52 it is obvious that e 3 f and 17 e 3 17 f so e and f are similar by O 56 in a Rickart ring every idempotent is similar to a projection Proof if e E A is idempotent then eA 1 7 eT fA for a suitable projection f so e and f are similar by 6 of 55 lncidentally A17 f fl el A17 LPe by 17 so f LPe Similarly Ae Ag with gRmei 28 5 EQUIVALENCE OF IDEMPOTENTS 57 If A is a regular ring 1 E A and ef are idempotents With 1A eA and A1Afthen e 3 f Proof Let y E A With 1 1341 and set e 1y f y1 ef are idempotents such that e 3 f As in 112 eA 1A eA thus ee are simi lar 55 likewise Af A1 A so f f are similar Thus e 3 e 3 3 f so e 3 f 58 If A is a regular ring then LP1 3 RP1 for all 1 E A Proof Immediate from 113 and 57 59 In a regular ring eU f 7 f 3 e 7 e f f for all projections e and f Proof Apply 58 to 1 e17 f and cite 115 510 In a regular ring A if e and f are projections such that eAf a 0 then there exist nonzero projections e0 3 e and f0 3 f With e0 3 f0 Proof Choose 1 E eAf With 1 a 0 let e0 LP1 f0 RP1 and cite 581 511 DEFINITION For idempotents ef in a ring one writes e ja f if e 3 f for some idempotent f S f cf 18 We then say that e is dominated by f 512 If ejaf and fjag then ejag Proof Say e 3 f S f and f 3 g S g If 6 fAf 7 gAg is the isomorphism given by 54 then 6 i f 6fgg thus e 3 6f 9 513 e ja 0 i e 0 Clear from 55 514 e ja f i ue ja uf for all central idempotents 11 Clear from 55 515 If u is a central idempotent and e ja u then e S 71 Proof By 514 17 ue ja 17 uu 0 so 17 ue 0 by 513 516 Ina Baer ring ejaf Ce S Cf e 3 f i Ce Cf Proof Immediate from 515 here C denotes central cover de ned in 315 517 LEMMA 18 p 28 Th 18 Let A be a Rickart Ting e and f idempotents of A such that f S e cf 18 Then e 7f 3 RPe 7 RPf 1The analogous remark holds for idempotents in a regular ring With the roles of LP1 and RP1 played by idempotent generators of 1A and A1 of 57 A little fussing is needed to assure that 60 E eAe and f0 E fAf Cf E P Armendariz and S K Berberian 00771771 Algebra 17 1989 17391758 p 1752 75 5 EQUIVALENCE OF IDEMPOTENTS 29 Proof Let g RPe h RPf One knows cf 56 that Ah Af C Ae Ag since h and g are self adjoint this implies that h S 9 Let a e17h yg17f it will su ice to show that aye7f and yag7h lndeed 06y 7 617 h917 f 7 69 7 ehg17 f 676h17f 17h17f e17h7fhfe17h7fh 7617fe76fe7f whereas 917f617h967gf617h g7gf17hg17f17h g17f7hfh917f7hf 7917hg79hg7hltgt 24 518 THEOREM 18 p 42 Th 28 Let A be a Baer rmg e and f projections in A such that e 3 f and eiZEI an orthogonal family ofprojections in A with e supei Then there 6i5t5 an orthogonal family of projections flZEI with f supfi such that ez g fz for all i Proof Say a E eAf y E fAe with my e ya f and let 6 eAe 7 fAf be the ring isomorphism de ned by 65 ysa 54 Let us assume that l is in nite for l nite the argument is an evident simpli cation of what follows We can suppose that l is well ordered with l 04 04 lt D Q a limit ordinal let 9 be the rst ordinal whose cardinality is that of 1 For every 04 E 9 write gasup6pr lt0lt evidently get is an increasing family of projections with supremum e And for each 04 lt D one has go gaea thus ea ga17ga The idempotents ua 6ga of fAf clearly satisfy ua S u S f for 04 S 5 Let ha RPua thus Aua Aha cf 56 by 517 if 04 S 5 then ha 3 h and u 7 ua 3 h 7 ha Set faha17ha dltQ clearly fa is an orthogonal family of projections S f and fa g 471 7 at 6ga17 6got 6gal 7 9a 66a g eat the last equivalence by 54 thus fa 3 ea It remains only to show that sup fa f It is the same to show that sup ha f Let h sup ha then h S f and we are to show that f 7 h 0 Since f 7 h is an idempotent of fAf the element u 6 10 7 h is an idempotent of eAe thus u S e For all 04 we have haf7hhaf7hahha7ha0 30 5 EQUIVALENCE OF IDEMPOTENTS since ha RPua we thus have 0 M0 7 h 9ga9u 6gau so gau 0 since sup go e it follows that eu 0 But it E eAe so u eu 0 that is 6 1f7 h 0 whence f 7110 0 519 If ef are projections in a regular ring and if ef are perspective that is have a common complement then e g f Proof By hypothesis there exists a projection 9 such that eUg ng 1 and e g f g 0 Citing 59 one has ee7e g S eUgiglig and similarly f i 1 7 g whence e g f 520 18 p 48 Exer 4 If ef are orthogonal projections in a Rickart ring A such that e g f then ef are perspective Proof Dropping down to e fAe f which is permissible one can supposethat ef1 Let meeAf yEfAe with mye y1f17e Then 6mfy efeyacfacy 00aceeac thus Ae1 C Afy similarly Afy C Ae1 so Ae1 Afy One sees easily that e and f 14 are idempotents Therefore cf 56 writing 9 RPe 1 we have Ae 1 Ag similarly ARPfyAfyA6Ag so also RPf y Q We assert that eUg 1 Writing h eUg we have f yh fy because h 2 g RPf y and yh y because h 2 e RPy therefore ffyiyeAhAhAz so fgheUg Therefore eUfSeUgbut eUfef1so eUg1 Similarly ng1 We assert that e g0 For let tEAe gi Since 118 Ae gAe AgAe Afy one can write tresfy for suitable 733 in A Then sfreisyEAeAeAe thus sf 6 Af Ae A17e Ae 0 so sf 0 Then 5e 517f sisfs whence 3y seysey s00 Then t3fsy00 thus it 0 and the assertion is proved Similarly f f g 0 thus 9 is a common complement of e and f 6 EQUIVALENCE OF PROJECTIONS 61 DEFINITION Projections ef inagtk ring A are said to be equivalent in A written e i f if there exists 1 E A such that 11 e and 11 f Then e 3 f and one can suppose 1 E eAf 52 cf 62 62 18 p 32 Th 23 Let A be a ring with proper involution aa 0 i a 0 and let 1 E A with 11 e e a projection such an element 1 is called a partial isometry Then 11 is a projection f and 1 E eAf Proof One has 1 7 e11 7 e1 1 7 e11 7 1e 11 7 11e 7 e11 e11e e7e27e2e30 since the involution is proper we conclude that 1 e1 The element f 11 is self adjoint and f2 1e1 11 f thus f is a projection and 1 f1 by the preceding argument thus 1 E eAf 63 In any gtk ring A equivalence is an equivalence relation Proof In question is transitivity Suppose efg are projections with e i f and f i 9 Say 11 e 11 f and 3434 f yy 9 We can suppose 1 E eAf y E ng 61 Then WJXMJV 9599 06f06 11 e and similarly 1y1y 9 thus e i 9 64 PROPOSITION 18 p 33 Th 25 Let A be a ring e and f projections in A such that e i f and let 1 E eAf with 11 e 11 f Then the mapping s gt gt 1s1 is an isomorphism of rings 6 eAe 7 fAf with inverse mapping t gt gt 1t1 For every projection g S e one has 9 i 6 9 Proof To the proof of 54 we need only add the following computations 68 06806 WSW 98 and Q1Q 9119 969 9 WWW 06906 99 O 65 PROPOSITION The equivalence of projections in a ring A has the following properties 31 32 6 gtk EQUIVALENCE OF PROJECTIONS 1 e 5 0 i e 0 2 e 5 f i ue 5 uf for all central projections u cf 38 3 If e 5 f and if e1 en are pairwise orthogonal projections with e el en then there eacist pairwise orthogonal projections f1 fn such that ff1fn and ez 5 fi forall i 4 If e1 en are orthogonal projections with sum e if f1 fn are orthogonal projections with sum f and if ez 5 fz for all i then e 5 f 5 ef are unitarily equivalent if and only if both e 5 f and 17e 5 17f Proof 1 4 See the proof of 55 5 If tet f With t unitary tt tt 1 then tftf tftquot e and tftf fttf f thus e 5 f also t17et tt7tet 17f so 17 e 5 17 f Suppose conversely that a E eAf y E 17 eA17 f with aa e aaf and yyquot 17e yy 17f then tay is unitary and tet f O 66 If A isaBaer ring e and f are projections in A such that e 5 f and eiZE1 is an orthogonal family of projections such that e sup ei then there exists an orthogonal family of projections flZEI such that f sup fz and ez 5 fz for all i Proof lmmediate from 64 67 LEMMA 18 p 44 Th 29 Let A be a Rickart ring e and f projections in A with ef 0 Then e 5 f if and only if there eacists aprojection 9 such that e 2ege f 2fgf g 2geg nyg Proof cf 2 p 99 Prop 2 If such a projection 9 exists then the element a 2egf satis es a e and 1 f thus e 5 f One does not need ef 0 here Conversely suppose a E eAf with aa e and 1 f Set 9 RPe Since e a E e fAe f one has 9136 1 By the de nition of g eaeagegag ii Also eae7ae7eaae7aagtk e7aeae7aa e7007e0 whence ge 7 1 0 gaquot ge thus ageg iii Substituting iii into ii e a 2eg iV EQUlVALENCE OF PROJECTIONS Since 16 afe 0 right multiplication of iV by 6 yields 6 2696 V Since aa afea 0 one has eaaifeaiefaaiaf17007a0 whence ga 7 f 0 thus 906 gf Vi Left multiplying iV by g and citing Vi one has 2geggeggegfg6f 9 the last equality by i thus 9 2969 Vii From ii and iii one has 61 2mg left multiplying by a aeaa 2119 thus a f 2fg Viii Right multiplying Viii by f yields 0 f 2fgf thus f 2f9f 1X Left multiplying Viii by g and recalling that gaquot go see iii one has 2gfggac gfgegfg6f 9 thus 9 2gfg X The equations V Vii ix X establish the lemma 0 68 PROPOSITION 18 p 46 Th 30 Let A be a Baer rmg 6226 an orthogonal family of projections in A with sup 6 e flZEI an orthogonal family of projections in A with sup fz f If e i fz for all i and if 6f 0 then 6 i f Proof Dropping down to e fAe f we can suppose that e f 1 Let ui el fi let S l E l and let T S be the commutant of S in A 41 then T is a Baer ring with unambiguous sups and infs 45 Since S is a commutative set that is S C S T one has T S D S T thus the center of T is T f T T S in particular the ui are orthogonal projections in the center of T With sup ui e f 1 Clearly T contains the 67 the fi and the given partial isometries implementing the equivalences e 3 fi Dropping down further to T we can suppose that the ui el fz are orthogonal central projections in A With sup ui 1 By the lemma there exists 34 6 gtk EQUIVALENCE OF PROJECTIONS for each i E l a projection 92 in A with 62 262926 etc and the proof shows that 92 E uiA Then 92 3 71 so giZE1 is an orthogonal family of projections moreover setting 9 sup 9 it is clear that my 92 for all i For xed i argue that gjuZg 7 92 0 for all j Since the iii are central evidently ui6 7 2696 67 7 2629162 0 for all i whence 6 7 2696 0 Similarly f 2fgf etc therefore 6 5 f by the lemma 0 Under suitable hypotheses on A the conclusion of 68 holds without the re striction 6f 0 cf 1814 69 Under the hypotheses of 68 If moreover a E A with 67 and fi one can show that there exists a partial isometry a E A such that a 6 1 f and 62a az afz for all i 2 p 56 Lemma 3 For other results in this vein see Section 14 610 PROPOSITION 18 p 35 Th 27 Let A be a 7 ing satisfying the following condition for 6116771 1 E A th6i 6 6i5t5 7 E ay th6 bicommutant of 1 such that 1 W7 If 6f are pTOjCCtiOTlS in A such that 6 g f then 6 i f Proof ln 2 p 66 Lemma the condition on A is called the weak square root axiom WSR Let a E 6Af y E fA6 with my 6 ya f Choose 7 E yy with 3434 Mquot W7 since yy is commutative and set w 17 then wwquot arra ayy a ayay 66 6 so it will suffice to show that ww f Now ww raar We assert that 7 commutes with 1 indeed 139yy 1ey figquot WV J f is self adjoint therefore a E yy and since 7 E yy we conclude that 7 commutes with 1 Then a also commutes with 7 so ww raar aarr aayy f and the proof is complete 0 611 The condition in 610 holds for every C algebra by spectral theory one can even take 7 to be positive in which case it is unique in particular in every Rickart C algebra hence in every AW algebra equivalent projections are equivalent 612 Let A bea Rickart ring in which for projections 6 and f 6 3 f i 6 5 f cf 610 Then similar projections are unitarily equivalent 6 gtk EQUIVALENCE OF PROJECTIONS 35 Proof Suppose ef are similar Then 6 3 f and 176 3 17f 55 therefore by hypothesis 6 3 f and 1 7 e 3 1 7 f whence ef are unitariiy equivalent 65 613 Let A be as in 612 Suppose moreover that every isornetry7 in A is unitary that is 1 1 i r 1 Then A is directly nite 3 1 i 341 1 Proof Suppose 1y 1 Then 6 341 is idernpotent and e 3 1 Let f LP6 then 6A fA and f 3 6 of 56 so f 3 1 by transitiVity By the hypothesis on A f 3 1 say wwquot f and ww 1 this implies by supposition that f 1 therefore 6A fA A whence e 1 that is 341 1 7 DIRECTLY FINITE IDEMPOTENTS IN A BAER RING 71 DEFINITION A ring A is directly nite if yr 1 i my 1 If yr 1 then e my is idempotent so direct niteness means the following condition on the equivalence of idempotents e 3 1 i e 1 An idempotent e E A is said to be directly nite if the ring eAe is directly nite by convention 0 is directly nite If A resp e E A is not directly nite it is said to be directly in nite For brevity in this section we say nite and in nite for directly nite and directly in nite 72 The following conditions on a ring A are equivalent a A is directly nite b the right A module AA is not isomorphic to any proper direct summand of itself c the left A module AA is not isomorphic to any proper direct summand of itself Proof 51 and 52 73 If ef are idempotents ofa ring A such that e S f and f is nite then e is also nite Proof If 11 6 eAe with yr e then setting m 1 f 7 e y y f7e one has 13 E fAf with ym f therefore my but myyf7eso mye 74 If ef are idempotents ofaring A with e ja f and f nite then e is nite Proof Say e g e S f By 73 e is nite that is eAe is nite since eAe E eAe 54 it follows that eAe is nite thus e is nite 75 PROPOSITION Let A be a Baer ring a family of central idem potents in A u sup ui cf 33 If every it is nite then so is u Proof Suppose 11 6 uA with yr it Write 12 u yi uZy Then y i uZym uZu ui since uZA is nite 12y ui thus iiimy 7 u 0 for all i whence u1y 7 u 0 34 my u 0 76 DEFINITION A Baer ring A is properly in nite it it contains no nite central idempotent other than 0 An idempotent e E A is said to be properly in nite if the Baer ring eAe is properly in nite by convention 0 is properly in nite Caution In 18 the term purely in nite is used instead following the usage in 6 we shall employ the latter term for another concept 79 36 7 FINITE IDEMPOTENTS 37 77 COROLLARY 18 p 12 Th 10 If A is any Baer ring then there eaists a unique central idempotent it such that uA is nite and 17 uA is properly in nite Proof Let T be the set of all nite central idempotents of A at least 0 E T and let u supT by 75 u E T If v 17u is a nite central idempotent then u E T so u S u whence v 0 thus 1 7 uA is properly in nite If also 1 is a central idempotent such that MA is nite and 1 7 uA is properly in nite then u u17u is nite because u S u and in nite because u S 1 7 1 therefore i 0 whence u S 1 Similarly u S u 0 78 DEFINITION A Baer ring A is said to be semi nite if it possesses a faithful nite idempotent that is there exists a nite idempotent e E A such that Ce 1 cf 315 An idempotent e E A is said to be semi nite if the Baer ring eAe is semi nite by convention 0 is semi nite 79 DEFINITION A Baer ring A is said to be purely in nite or of type III if 0 is the only nite idempotent of A An idempotent e E A is said to be purely in nite if the Baer ring eAe is purely in nite by convention 0 is purely in nite 710 LEMMA 18 pp 12 14 Let A be a Baer ring umE1 a family of pairwise orthogonal central idempotents eiZE1 a family of idempotents such that ez S uz for all i Let S e21 i E I and write ST 17 eA e idempotent Then eie ez and uze eez for all i and Ce sup Cei Proof Remark If the ez are projections in a Baer ring and one takes e to be a projection then e sup ez 124 Since 17e 6 ST one has ei17 e 0 thus ez eie for all i We assert that uze eei For let 12 uz 7 ei One has ezaz eiui 7 ez ez 7 ez 0 and for ja i ejaz ejujuidi 0 thus 12 6 ST 17 eA whence edi 0 that is eui eei Let wz Cei w Ce v sup wz cf 33 we are to show that v ii For all i eiw eiew eiew eie ei so wz S w therefore i S w And ez 3 mi 3 vso ei17v 0 for all i thus 17v 6 ST 17eAso e17v 0 e ev whence w 3 v Incidentally writing fz eez uZe one sees that fz is an idempotent with fiei fz and eifz eieei eiei ei so that Aez Afi When the ez and e are projections in a Baer ring ez fi Note too that cf 121 17 eA 7 ST 7 may 7 g 7 6A 7 17 6A whence Ae S VAei that is Ae is the supremum of the family Aei in the lattice of idempotent generated principal left ideals of A O 711 PROPOSITION 18 p 12 Th 12 If A is any Baer ring there eaists a unique central idempotent u of A such that uA is semi nite and 17 uA is purely in nite 38 7 FINITE IDEMPOTENTS Proof If A has no nite idempotents other than 0 the proof ends with u 0 Otherwise1 let 1161 be a maximal family of pairwise orthogonal nonzero central idempotents such that uZA is semi nite for all i and let u sup 71 For each i 61 let 6 be a nite idempotent in uZA such that e is faithful in uZA7 which means that Cei ui cf 325 Let 6 be the idempotent given by 710 in particular Ce u We assert that e is nite Suppose 11 6 6A6 with ym 6 Set 12 u yi uZy Then y i uZe 662 We have 12 7121 16M eui 1eei whence 6 A6 similarly yi 6 A6 Left multiplying y i 662 by 67 we have eiyimi eieei eiei 67 so 6239 eiy i iyi i1i eiyi i where eiyi i E eiAei since eiAeZ is nite it follows that 621eZyi 67 thus 6239 ii iyi 6239123914239 eiuiy e y Thus 62171y 0 for all i therefore 17mg 6 ST 17eA so 6171y 0 whence e my This completes the proof that e is nite Since it Ce we see that uA is semi nite Suppose f is a nite idempotent with f S 17u we are to show that f 0 Assume to the contrary that f a 0 Then Cf a 0 Cf S 17 u and the maximality of the family is contradicted Suppose also 1 has the properties of u Let 6 be a nite idempotent with Ce 1 Then the idempotent g 61 7 u is nite because 9 S 6 and g S 1 7 u since 1 7 uA is purely in nite necessarily g 0 thus 6 6 u Ce S u Similarly u S 1 O 712 With notations as in 711 if g E A is any idempotent such that gAg is semi nite then 9 S u Proof Let h E gAg be a nite idempotent whose central cover in gAg is 9 Then ChA is semi nitel so Ch S u by the argument in 711 Now gCh is a central idempotent in gAg such that h S gCh therefore gCh 9 because h is faithful in gAg Thus 9 3 C71 but Ch S u so 9 S 713 The argument in 712 shows If g is any idempotent in a Baer ring A and if h E gAg is an idempotent that is faithful in gAg then 9 3 C71 2 714 If A is a Rickart ring and e E A is a nite idempotent then LP6 and RP6 are nite projections Proof Let f LP6 We know 56 that EA fA so 6 3 f indeed e and f are similar by 55 since 6 is nite so is f 74 715 A Baer ring A is a semi nite Baer ring if and only if it contains a faithful nite projection 1If e is a nite idempotent in A then since 5 is faithful in CeA see the footnote for 315 CeA is semi nite that gets the Zorn argument started 2It follows that Cg S But h S 9 so Ch S Cg therefore Ch Cg 7 FINITE IDEMPOTENTS 39 Proof Suppose A is a semi nite Baer ring and let 6 E A be a faithful nite idempotent Let f LP6 As noted in 714 f is a nite projection and f 3 6 therefore 516 Cf Ce 1 716 A Baer ring A is a purely in nite Baer ring if and only if 0 is the only nite projection Proof If 0 is the only nite projection then by 714 it is the only nite idempotent 717 Suppose A is a Rickart ring such that for projections f f g 1 i f 1 Then A is directly nite that is for idempotents e e g 1 i e 1 Proof Suppose e E A is idempotent and e g 1 Let f LP6 then 56 6AfA thus f 3 e g 1 Then f g 1 so by hypothesis f 1 Thus 6A 1AA whence e 1 8 ABELIAN IDEMPOTENTS IN A BAER RING TYPE THEORY 81 DEFINITION 18 p 10 A ring A is said to be abelian if every idem potent of A is central An idempotent e E A is said to be abelian if the ring 6A6 is abelian by convention 0 is abelian 82 Every division ring is an abelian Baer ring Every commutative ring is abelian 83 If A is a semiprime Baer ring and e E A is idempotent then 6 is abelian if and only if f 6Cf for every idempotent f of 6A6 321 84 Every abelian ring resp abelian idempotent is directly nite Proof Suppose A is abelian and 11 6 A with 341 1 Then 6 1y is idempotent by hypothesis central so 6 16 y1e ye1 y1y1 y1y1 1 85 Every suloringl of an abelian ring is abelian Obvious 86 If ef are idempotents ofaring A such that e ja f and f is abelian then 6 is abelian Proof Say 6 g 6 S f Then eAe C fAf so eAe is abelian 85 and 6A6 eAe 54 therefore 6A6 is abelian 87 7 p 26 Th 32 A regular ring A is abelian if and only if 1A A1 for all 1 E A2 Proof Suppose A is abelian and 1 E A Write 1A eA e idempotent By hypothesis 6 is central so 1 e1 16 therefore A1 A16 C A6 6A 1A Similarly 1A C A1 so 1A A1 Conversely suppose this condition holds and e E A is idempotent then 6A A6 so 6A1 7 e 17 eAe 0 whence e1 e16 16 for all 1 E A 88 18 p 17 Exer 5 The following conditions on an idempotent e of a ring A are equivalent a e is in the center of A b e commutes with every idempotent of A Proof b i a Let 1 E A we are to show that 61 16 The element f e 611 7 e is idempotent By hypothesis 6f f6 thus f f6 6 1Here the subring need not contain the unity element of the rin 2A regular ring is abelian if and only if it is reduced no nilpotent elements other than 0 7 p 26 Th 32 40 8 ABELIAN IDEMPOTENTS 41 whence 611 7 e 0 e1 e16 Applying this in the opposite ring A0 we infer that 16 616 so 61 e16 16 89 18 p 10 A ring is abelian if and only if all of its idempotents commute with each other Immediate from 88 810 18 p 37 Exer 2 A Rickart ring is an abelian ring if and only if all of its projections commute with each other Proof Let A be a Rickart ring all of whose projections commute with each other to show that A is abelian it will suf ce by 89 to show that every idempotent of A is a projection Let g E A be idempotent e LPg f RPg As noted in 56 gA 6A and Ag Af thus 696 969 and ffggt 99 By hypothesis 6ff6SO 6f6fgf69fgfgt thus f S e and 6fgefgfege6 so 6 S f Thus 6 f Then 9 eg fg f in particular 9 is a projection 811 18 p 36 A Rickart ring is an abelian ring if and only if every projec tion is central Proof Immediate from 810 812 The following conditions on a Rickart ring A are equivalent a A is an abelian ring b LP1 RP1 for all 1 E A Proof a i b Every projection e is central so 61 1 if and only if 16 1 b i a Assuming b let 9 E A be idempotent we are to show that g is central Let a E A and let 1 ga17 Q then 12 0 so RP1LP1 0 In view of b this means that 1 0 thus gA17 g 0 Similarly 17 gAg 0 thus 9 is central 813 If A is a Rickart ring and g E A is an abelian idempotent then LPg and RPg are abelian projections Proof Write e LPg By 56 EA gA so 6 3 g indeed e and g are similar by 55 since 9 is abelian so is e 86 814 DEFINITION 18 p 11 6 p 123 Th 1 A Baer ring A is of type I or is discrete if it has a faithful abelian idempotent An idempotent e E A is said to be of type I if the Baer ring 6A6 is of type I by convention 0 is of type I 815 DEFINITION 6 p 121 Def 1 and p 123 Th 1 A Baer ring A is continuous if 0 is its only abelian idempotent An idempotent e E A is said to be continuous if the Baer ring 6A6 is continuous by convention 0 is continuous 42 8 ABELIAN IDEMPOTENTS 816 Every Baer ring of type 1 is semi nite cf 78 84 817 A Baer ring A is a Baer ring of type 1 if and only if it has a faithful abelian projection Proof Suppose A is a Baer ring having a faithful abelian idempotent 9 Then e LPg is an abelian projection with e 3 9 cf 813 moreover cit ing 516 Ce Cg 1 818 A Baer ring is a continuous Baer ring if and only if 0 is the only abelian projection Proof Suppose 0 is the only abelian projection in the Baer ring A then 0 is the only abelian idempotent of A 813 so A is continuous 819 LEMMA 18 p 14 Let A be a Baer ring afamily of pairwise orthogonal central idempotents of A such that every uZA is of type 1 and let u sup ui Then uA is of type 1 Proof For each i choose an abelian idempotent ez E uZA with Cei ui By 710 there exists an idempotent e such that Ce u eie ez and uZe eez for all i it will suf ce to show that e is abelian Let fz uZe eei since eie ez and eez fi one has ez 3 fi therefore fi is abelian 86 Given 9 E eAe idempotent we wish to show that g is central in eAe Let a E eAe we are to show that ga ag and it will clearly suffice to show that uZga 7 mg 0 for all i Now uza uZea fla uza uZae 1uZe afi thus uza E fiAfi By the same token uZg E fiAfi Since fiAfZ is abelian the idempotent uZg is central in fiAfi hence commutes with uza 0 i 4905 i 059 O 820 PROPOSITION 18 pp 12 14 If A is any Baer ring there eacists a unique central idempotent it such that uA is of typel and 17uA is continuous Proof cf 711 if A contains no abelian idempotent other than 0 then u 0 lls the bill Otherwise let be a maximal orthogonal family of nonzero central idempotents such that every uZA is of type 1 Let u sup uz cf 33 By 819 uA is of type 1 Suppose f E 1 7 uA is an abelian idempotent Then Cf 17u and CfA is oftype 1 so Cf 0 by maximality whence f 0 Thus 1 7 uA is continuous Suppose also 1 is a central idempotent with uA of type 1 and 1 7 uA continuous Choose abelian idempotents ee with Ce u Ce 1 Then 17ue is abelian because it is S e and is in 17uA therefore 17ue 0 because 17uA is continuous consequently 17uu 0 u S u Similarly uSuthus uultgt 821 With notations as in 820 if g E A is any idempotent such that gAg is of type 1 then 9 S u 8 ABELIAN IDEMPOTENTS 43 Proof cf 712 Let h E A be an abelian idempotent whose central cover in gAg is 9 Then ChA is of type I and C71 S u by the argument in 820 but 9 3 C71 by 713 whence g S 822 DEFINITION 18 p 11 A Baer ring A is said to be of type II if it is semi nite and continuous that is A contains a faithful nite idempotent but no abelian idempotents other than 0 An idempotent e E A is said to be of type II if the Baer ring eAe is of type II by convention 0 is of type II 823 A Baer ring A is a Baer ring of type II if and only if it contains a faithful nite projection but no abelian projections other than 0 Proof If the Baer ring A is a Baer ring of type II then A has a faith ful nite projection 715 but no abelian projections or even idempotents other than 0 The converse is immediate from 813 824 THEOREM 18 p 12 Th 11 Every Baer ring is uniquely the product of Baer rings of types I II and III Proof If A is any Baer ring by 820 one has A B X X with B of type I and X continuous By 711 X C X D with C semi nite and continuous hence of type II and D of type III Then A B X C X D with the required properties The proof of uniqueness is routine cf 921 O 825 DEFINITION 18 p 11 A Baer ring of type I is said to be of type I n if it is directly nite type Iinf if it is properly in nite A Baer ring of type II is said to be of type II n or type IIl if it is directly nite type Iin or type IIOO if it is properly in nite 826 THEOREM 18 p 12 Th 12 Every Baer ring is uniquely the product of Baer rings of types I n linf II n Ilinf and III Proof Routine cf 925 O 827 If A is any Baer ring there exists a unique central idempotent u of A such that uA is abelian and 1 7 uA has no abelian central idempotent other than 0 such rings are called properly nonabelian Proof It is easy to see that if is an orthogonal family of abelian cen tral idempotents then u sup uz is abelian cf the proof of 819 An obvious exhaustion argument completes the proof 828 J M Goursaud and J Valette 9 p 95 Th 14 Let G be a group K a eld such that either K has characteristic p gt 0 or K has characteristic 0 and contains all roots of unity Let A KG be the group algebra of G over K and suppose that A is a regular ring let Q be the maximal ring of right quotients of A by 131 and 132 Q is a regular Baer ring Then the following conditions are equivalent a Q is of type I b Q is of type I n 825 and there exists a nite upper bound on the number of pairwise orthogonal equivalent nonzero abelian idempotents in Q c G has an abelian subgroup of nite index d A satis es a polynomial identity 9 ABSTRACT TYPE DECOMPOSITION OF A BAER RING Throughout this section A denotes a Baer ring with an equivalence relation N de ned on the set of projections of A satisfying the following aaioms A e N 0 i e 0 B e N f i ue N uf for all central projections u D If e1 en resp f1 fn are pairwise orthogonal projections with sum e resp f and if ez N fi for all i then e N f The labelling of the list of axioms follows that of Kaplansky 18 p 41 the list will be rounded out in subsequent sections Axiom A will not be needed until 916 91 EXAMPLES The most important are i and i 55 and 65 respec tively 92 Combining axioms B and D one sees that if ef are projections and u1 un are orthogonal central projections With sum 1 then e N f if and only if uze N uZf for all i So to speak N is compatible With nite direct products 93 If g E A is any projection then the Baer ring gAg also satis es the axioms A B D for the restriction of the equivalence relation N to its projection lattice Proof This is obvious for axioms A and D Suppose ef are projections in gAg With e N f and suppose v is any central projection of gAg Since A is semiprime 320 v ug for some central projection u of Aiu Cv lls the bill 3217and so ve uge ue N uf ugf of 94 If N is the relation g resp 5 on the projection lattice of A and if g E A is a projection then the relation on the projection lattice of gAg induced by N is the relation 3 resp i in the ring gAg Proof Suppose ef are projections in 9A9 and e 3 f relative to A One can choose a E eAf y E fAe With my e and ya f 52 then 11 6 gAg so e g f relative to gAg Similarly for 1 cf 61 95 DEFINITION We say that A is nite relative to the relation N in case for projections e E A e N 1 i e 1 Finiteness relative to g is called direct niteness and means that ya 1 i my 1 71 717 niteness relative to i Will be called niteness and means that u 1 i a 11 Direct 1Every regular ring is nite 13 Ara and P Menal Arch Math 42 1984 126130 44 9 DECOMPOSITION INTO TYPES 45 niteness obviously implies niteness sometimes the converse is true cf 613 A projection g E A is said to be nite relative to N if the Baer ring gAg is nite relative to the restriction of N to its projection lattice by convention 0 is nite If A is not nite it is said to be in nite a projection g E A is said to be in nite if gAg is in nite for the restriction of N to gAg by convention 0 is in nite 96 EXAMPLE If e is nite and e N f one cannot conclude that f is nite this will be true under an extra axiom 0 to be introduced later 153 Counterexample Let A M2 R and declare e N f for all nonzero projections ef The projection e nite 97 Let ef be projections e S f Then e is a nite projection of fAf if and only if it is a nite projection of A Immediate from eAe efAfe 5 8 is nite even minimal and e N 1 so 1 is not 98 If ef are projections of A With f nite and e S f then e is nite Proof Suppose g e gNe By axiom D gf7e Nef7e f so 9 f 7e f by niteness of f whence g e 99 DEFINITION We say that A is properly in nite relative to N if 0 is the only nite central projection A projection e E A is said to be properly in nite if eAe is properly in nite by convention 0 is properly in nite this means cf 320 321 that if u is a central projection of A With ue nite then ue 0 910 LEMMA If is a family of nite central projections then u sup uz is also nite Proof cf 75 Suppose e E uA is a projection With e N u By axiom B uZe N uZu uz for all i since uZA is nite necessarily uZe uz thus uZ17e 0 forall iwhence u17e0 uuee O 911 THEOREM There eacists a unique centralprojection u of A such that uA is nite and 1 7 uA is properly in nite Proof Formally the same as 77 note that 98 is needed for the proof of uniqueness O 912 DEFINITION We say that A is semi nite relative to N if it contains a faithful nite projection A projection e E A is said to be semi nite if eAe is semi nite by convention 0 is semi nite We say that A is of type III or purely in nite relative to N if 0 is the only nite projection of A A projection e E A is said to be of type III if eAe is of type III by convention 0 is of type III 913 LEMMA If ei is a family of nite projections whose central covers are pairwise orthogonal then e sup ez is nite Proof Write uz Cei u sup ui and note that Ce u 322 Note that uZe ez cf 710 For ejuZe 7 e2 ejujuZe 7 ei 0 for j a i and 46 9 DECOMPOSITION INTO TYPES eiuie7eZ eie7ei ei7ei 0 therefore euie7eZ 0 that is uie7ei 0 Suppose f S e and f N e Then uZf 3 me and uZf N uze by axiom B since uze ez is nite necessarily uZf uie thus uZe 7 f 0 for all i therefore ue7f0that is e7f0 O 914 THEOREM There eacists a unique central projection u E A such that uA is semi nite and 1 7 uA is of type III Proof If A contains no nite projections other than 0 then u 0 lls the bill Otherwise let be a maximal orthogonal family of nonzero central projections such that each uZA is semi nite and let u sup iii For each i let ez be a nite projection with Cei iii and let e sup ei Then Ce sup Cei sup ui u and e is nite by 913 thus uA is semi nite If f E 1 7 uA is a nite projection then CfA is semi nite and Cf S 1 7 it therefore Cf 0 by maximality whence f 0 Thus 1 7 uA is of type III Uniqueness is proved exactly as in 711 O 915 With notations as in 914 if g E A is any projection such that gAg is semi nite then 9 S u Proof Formally the same as 712 916 If e Nf then Ce Cf Proof By axiom B 17 Cfe N 17 Cff 0 so 17 Cfe 0 by axiom A whence 17 CfCe 0 Ce S Cf Similarly Cf Remark This is the rst use of axiom A in this section 917 Every abelian projection in A is nite Proof Let e E A be abelian and suppose f S e with f N e By 320 and 83 f eCf but Cf Ce by 916 so f eCe e Remark Only axioms A and B are needed for this 918 If A is of type I then it is semi nite ii If A is of type III then it is continuous Proof Let e E A be a faithful abelian projection By 917 e is nite relative to N so A is semi nite by de nition 912 ii Suppose A is of type III If e E A is abelian then e is nite 917 so e 0 thus A is continuous 815 919 DEFINITION We say that A is of type II relative to N if it is semi nite and continuous that is if A contains a faithful nite projection but no abelian projections other than 0 A projection e E A is said to be of type II if eAe is of type II by convention 0 is of type II 920 For a projection g E A each of the following conditions implies that g 0 g is both type I and type II ii 9 is both type I and type III iii 9 is both type II and type III Proof Assume to the contrary that g a 0 dropping down to gAg we can suppose g 1 Let e E A be a faithful abelian projection Then e 0 A is continuous whence 1 Ce 0 a contradiction ii Let e E A be a faithful abelian projection Then e is nite 917 hence e 0 A is type III 9 DECOMPOSITION INTO TYPES 47 so 1 Ce 0 iii Let e E A be a faithful nite projection Then e 0 A is type III so 1 Ce 0 921 THEOREM The Baer gtk rinq A is uniquely the product of rings of types I II and III relative to the equivalence relation N satisfying aaioms A B and D2 Proof By 820 there exists a central idempotent u a projection by 38 of the Baer ring A such that uA is of type I and IiuA is continuous Apply 914 to 17 uA let v in be orthogonal central projections such that v w 17 u vA is semi nite and wA is of type III Since vA is also continuous because v S 17 u it is of type II Thus A uA gtlt vA gtlt wA with the required properties Suppose also 1 u v w is a central partition of 1 with the indicated properties Then uv 0 by of 920 similarly uw 0 so u u1 uu thus u S 1 Similarly u S u so u 1 Similarly v v w 11 O 922 With notations as in the proof of 921 uA gtlt vA is the semi nite summand of A Proof By de nition vA is semi nite so is uA 918 hence so is uA gtlt vA u vA 923 With notations as in the proof of 921 vA gtlt wA is the continuous summand of A 924 DEFINITION If relative to N A is nite and of type I resp type II we say that A is of type I n resp type II n or Ill If relative to N A is properly in nite and of type I resp type II we say that A is of type Iinf resp type IIinf 925 THEOREM The Baer rinq A is uniquely the product of rings of types I n linf Il n Ilinf and III relative to the equivalence relation N satisfying aaioms A B and D Proof Combine 911 and 921 O 926 All of the above carries through for N an equivalence relation on the partially ordered set of idempotents of a semiprime Baer ring cf 18 Proof Routine notable ingredients being 710 and 321 Here e S f means e E fAf and ef are orthogonal if ef fe 0 927 If A is nite relative to N and LPa N RPa for all a E A then A is directly nite Proof Suppose ya 1 If 12 0 then yaz 0 so 2 0 therefore RPa 1 Let g my which is idempotent and let e LPg then gA eA 56 Now gA ayA C aA whereas a 11 aya aya ga shows that aA C gA thus aA gA eA It follows that e LPa By hypothesis LPa N RPa that is e N 1 since A is nite relative to N e 1 whence g 1 that is my 1 Note that the argument works for a Rickart ring satisfying the hypotheses 2The de nitions of type I7 and continuous are independent of N all other types including type IE 7 and type Iinf depend on the concept of nite therefore on N 10 KAPLANSKY S AXIOMS AH etc A SURVEY OF RESULTS We summarize in this section the principal axioms considered by Kaplansky 18 p 147 for an equivalence relation N on the projection lattice of a Baer ring A e N 0 i e 0 De niteness B e N f i ue N uf for every central projection u Central compatibility C if eiZEI is an orthogonal family of projections With sup ez e and if e N f then there exists an orthogonal family of projections flZEI such that f sup fi and ez N fi for all i Induced partitions 0 Same as C With the index set I assumed to be nite Induced nite partitions D lf e1 en are orthogonal projections With sum e and f1 fn are orthogonal projections With sum f and if ez N fi for all i then e N f Finite additivity E if ef are projections in A With eAf a 0 then there exist nonzero projections e0 3 e f0 3 f With e0 N f0 Partial comparability F lf eiZE1 is an orthogonal family of projections With sup ez e flZEI is an orthogonal family of projections With sup fz f ez N fi for all i E l and ef 0 then e N f Orthogonal additivity G if nip61 is an orthogonal family of central projections With sup ui 1 and if ef are projections such that me N LLif for all i then e N f Central additivity H e U f 7 f N e 7 e f f for every pair of projections ef Parallelogram law There are also axioms J and K 18 p 111 the EP and SR axioms 18 pp 89 90 see below and a spectral axiom 18 p 136 Properties of this sort have their roots in the Murrayivon Neumann theory of operator algebras 1936 for lattice theoretic antecedents of axioms such as A H see Loomis7s memoir 20 and S Maeda7s paper 22 48 10 KAPLANSKY S AXIOMS 49 These axioms are of particular interest for the relations of ordinary equivalence g and equivalence 5 but there are also interesting interactions among the axioms for an abstract relation N1 Work of Loomis 20 p 4 axiom and of S Maeda and S S Holland 26 highlights the following axiom weaker than axiom E E if e f are projections with ef a 0 then there exist nonzero projections e0 e f0 Sf such that e0 Nfo We cite the following results of Maeda and Holland 101 in the presence of axioms A and C the axioms E and E are equivalent this will be proved in 138 Axiom H implies axiom E thus axioms A C and H imply E noted in the proof of 139 This is the key to several of the results mentioned below Other conditions on N contemplated are as follows e j f means that e N e S f for a suitable projection e GC For each pair of projections ef there exists a central projection n such that ue j uf and 1 7 uf j 1 7 ue Generalized comparability Additivity Same as axiom F with the condition ef 0 omitted This is also called complete additivity Continuity of the lattice operations lf ei is an increasingly directed family of projections with supremum e brie y ez T e then ez f f T e f f for every projection f LP N RP For every 1 E A LPa N RPa We now survey some of the principal results to be proved or sometimes merely noted in the sections that follow 102 For 3 39n any Baer ring axioms A D hold see 111 103 For i in any Baer ring axioms A D and F hold see 112 104 Axioms C D F imply the Schroder Bernstein theorem7 e j f and f j e i e N f 18 p 61 Th 41 in particular this holds for i in any Baer ring cf 68 This would seem to be a fundamental result hence enormously important Strangely it seems to be useless At any rate none of the results in these notes make any use of it it has this signi cance writing e for the equivalence class of e under N the relation e S f de ned by e j f is a partial ordering the Schroder Bernstein theorem providing the antisymmetry7 property There is a trivial proof of the Schroder Bernstein theorem in the nite case assuming only axioms C D and niteness in the sense of 95 105 Axioms A G imply complete additivity 18 p 78 Th 52 the proof is sketched in 1816 1Another concrete example is perspeetivity see 107 50 10 KAPLANSKY S AXIOMS 106 if axioms A D F and H hold then GC hence E and complete additivity hence G hold 26 Th 21 This was proved in 18 p 82 Th 54 with axiom E included in the hypothesis For the details see 139 and 1812 107 If relative to N the Baer ring A is nite 95 and satis es axioms A D F and H cf 106 then the lattice operations are continuous and the relation N is the relation of perspectivity that is e N f if and only if there exists a projection 9 such that eUg ng 1 and e g f g 0 18 Ths 69 71 For the proof see 208 108 if A is a regular Baer ring then relative to 3 A is nite ie directly nite and all of the above mentioned conditions hold 17 proof sketched in 2010 leaving out the hardest part2 109 Relative to 5 GC implies complete additivity 2 p 129 Th 1 This is proved in 1814 A square root7 condition on A is important in many applications SR For ever 1 E A there exists r E 11 with r r and r2 11 y Square roots if this condition holds then 3 and i coincide 610 The important conse quence of SR proved by S Maeda is as follows 1010 lf SR holds in A then 3 satis es axiom H 25 Th 2 proved in 1213 The next conditions gure in many applications pertaining to i EP For every 1 E A 1 a 0 there exists r E 11 such that r r and 1quot1r2 is a nonzero projection E1istence of projections Addability of partial isometries lf eiZE1 is an orthogonal family of pro jections with sup ez e flZEI is an orthogonal family of projections with sup fi f ez fz for all i E l and mlLEI is a family of partial isome tries such that wfwz ez and win fz for all i E 1 then there exists a partial isometry in such that ww e wwquot f and we wz fiw for all i E l The term is inelegant but useful for distinguishing this concept from the closely related concept of summability7 see 148 below PD Every 1 E A can be written 1 wr with r E 11 r r r2 11 and w a partial isometry such that ww LP1 ww RP Polar decomposition Obviously PD implies SR and LP 1 RP Here are some further results pertaining to the above three conditions 1011 if A is a Baer ring with no abelian summand and if A satis es GC relative to 3 cf 109 then partial isometries are addable in A 2 p 129 Th 1 in every AW algebra partial isometries are addable same reference 2See the footnote for 95 10 KAPLANSKY S AXIOMS 51 1012 If A is a Baer ring satisfying EP in which partial isornetries are addable then A has PD in particular A satis es SR 14 Th 22 For the details see 1423 and 1429 1013 If A is a Baer ring satisfying EP and satisfying GC relative to 5 then 14131 5 RPQ for all 1 E A For the proof see 1431 1014 If A is a Baer ring satisfying EP and if A is properly in nite relative to i then A has PD For the proof see 1430 11 EQUIVALENCE AND EQUIVALENCE IN BAER RINGS FIRST PROPERTIES 111 PROPOSITION 18 p 47 For 3 every Baer ring satis es the aaioms AiD of 10 Proof The properties A B D are noted in 1 2 4 of 55 property C in 518 O 112 PROPOSITION 18 p 47 For 3 every Baer ring satis es the aaioms AiD and F of 10 Proof The properties A B D are noted in 1 2 4 of 65 property C in 66 property F in 68 O 113 PROPOSITION 18 p 47 For 3 every regular Baer ring satis es the aaioms A F and H of 10 Proof Let A be a regular Baer ring In view of 111 we need only check the properties E F and H For E see 510 for H 59 Property F Suppose as in the statement of F e supei f supfi ez 3 fz forall iand ef0 weareto showthat e 3 f Let uieZfi and let S be the set of all ui Then S is a commutative set so S C S S D S By 48 T S is a regular Baer ring with unambiguous everything and its center is T f T S f S S D S thus the it are central projections in T If a E eiAfi yi E fiAeZ with aiyi ei yzaz fi then clearly any 6 T we may therefore drop down to T and suppose that the it are central in A Then it sup it is central in A dropping down further to uA we can suppose that sup ui 1 By 520 ez and fi are perspective in uZA let 92 E uZA be a projection with ez Ugi fi U 92 it and ez f 92 fz f 92 0 Let g supgi Clearly uig gi uZe ei uif fi By 323 one has 1626 U Q 1626 U 6239 U 9239 239 thus uZ17eUg 0 for all i whence 17eUg 0 eUg 1 Similarly ng 1 e g 0 f g 0 Thus ef are perspective in A Since 3 satis es H it follows that e 3 f 519 O Incidentally LPa 3 RPa for every element a 58 These are the properties of 3 in a regular Baer ring that lie nearest the surface property C being the only one among them that requires a struggle Deeper results are signaled 52 11 EQUIVALENCE IN BAER RINGS 53 in 10 The deepest is surely direct niteness 17 Th 2 1 am unable to improve on Kaplansky7s herculean computation so can only refer the reader to 17 for the details But see 204 for a lattice theoretic approach to this result due to Amemiya and Halperin It may be helpful to signal three minor misprints in 17 on p 529 in the second sentence after Lemma 19 read 6 E ejAeZ and in formula 3 at the bottom of the page read A for A1 on p 530 in the sentence between formulas 5 and 6 the formula should read 61 A2716 on p 532 formula 22 the quanti er 7 gt 0 refers to the particular 7 of the induction hypothesis 114 There exists a Baer ring in which property E fails for N 18 p 43 An example is the ring of 2 X 2 matrices over the eld of three elements cf 18 p 39 Exer 10 2 p 82 Exer Signaled by Kaplansky as open questions are whether axioms E and F hold for 3 in every Baer ring 18 p 47 115 Every AW algebra satis es for 3 cf 611 the axiom G of 10 18 p 75 Proof 2 p 53 Prop 2 Let A be an AW algebra an orthogonal family of central projections with sup ui 1 We rst show that A may be identi ed with a certain subalgebra B of the product algebra C HuiA Let B be the set of all families 1 E C with bounded it is routine to show that B is a subalgebra of C and that B is a C algebra for the norm sup Since the uZA are Baer rings so is their product C and if 67 E C is a projection then every 62 is a projection so 662 622 shows that S 1 for all i whence 67 E B Thus B contains every projection of C hence is itself a Baer ring so B is an AW algebra 138 called the C sum of the family of AW algebras uZA De ne Lp A a B by Lpa uZa clearly a monomorphism of algebras it follows that Lp is isometric 6 p 8 Prop 8 so LpA is norm closed in B If 67 E B is a projection then setting 6 sup 6 in A it is clear that Lpe 67 thus LpA contains every projection of B Since B is the closed linear span of its projections cf 140 and LpA is closed it follows that LpA B Suppose now that for each i one is given projections eifi in uZA with 62 5 fi Let wz E eiAfZ C uZA with win 67 wfwi fi Then S 1 for all i so E B Therefore there exists w E A with Lpw 11 that is uZw wz for all i It follows easily that ww e and wwf where esupeZ fsupr We shall see in subsequent sections that an AW algebra satis es for 5 all properties mentioned in 10 except continuity of the lattice operations which it satis es if and only if the algebra is directly nite see 204 and 2011 12 PARALLELOGRAM LAW AXIOM H In this section A denotes a Rickart ring and N an equivalence relation on the projection lattice of A We review a de nition from 10 121 DEFINITION The relation N is said to satisfy Axiom H or the parallelogram law if e U f 7 f N e 7 e f f for every pair of projections e f in A 122 Every regular ring satis es axiom H for 3 59 123 In the Baer ring of 114 axiom H fails for 5 cf 2 p 75 Exer 1 124 PROPOSITION The following conditions are equivalent a N satis es aaiom H b e 7 e 017 f N f 7 1 7 e f f for every pair ofprojections ef c e N f for every pair ofprojections ef such that e 17f 17e f Proof Let ef be any pair of projections By 115 LP6f LP6l1 1 fl 6 60 1 if therefore RPM 7 LPef 7 Live 7 f 7 f nlt17 e Thus b says that LPef N RPef for every pair of projections whereas cf 115 axiom H says that RPe1 7 fl N LPe1 7 fl for every pair ef Therefore a gt b i c Obvious c i b Let ef be any pair of projections and write e LPef f RPef Then e S e f S f so ef eeff eeff ef thus a LPef e7eo17fgt whence e 17f 0 and fRP ff16 f 54 12 PARALLELOGRAM LAW 55 whence 17 e f f 0 By c e N j that is LPef N RPef in other words b holds 0 125 DEFINITION Projections ef in a Rickart ring are said to be in position p if e 17f 17e f 0 In other words e 17f 0 and eU17f 1 that is e and 17f are complementary equivalently 17e and f are complementary 126 Projections ef are in position p if and only if e LPef and f RPef Clear from the proof of 124 127 Projections in position p are perspective Proof Suppose ef are in position p that is e and 17 f are comple mentary but f 1 7 f are also complementary thus e and f have 1 7 f as a common complement 128 PROPOSITION For every pair of projections e f in a Rickart ring there eacist unique decompositions 66 gt6gt c lgtf such that ef are in position p hence are perspective by 127 and ef e f 0 Necessarily e LPef and f RPef Proof Suppose e ee f ff are decompositions with the indicated properties Since ef are in position p we know 126 that e LPef f RPef But ef e 7e f 7 f ef thus e LPef and f RPef whence uniqueness On the other hand if one de nes e LPef f RPef then as shown in the proof of 124 e and f are in position p also ef eef ef so e7ef 0 and similarly ef7f 0 therefore e e7e and f f7f ll the bill O 129 COROLLARY If N satis es aaiom H then for every pair of projections ef one can write 66ltgt6gt fff with e N and ef e f 0 Proof For the decompositions of 128 one has e N f by condition c Of 124 0 1210 If LPa N RPa holds for all a E A then axiom H holds Immediate from 115 The next results lead up to S Maeda7s theorem that if A satis es the axiom SR of 10 then 3 satis es axiom H 1211 DEFINITION An element s of a ring is called a symmetry if squot s and s2 1 that is s is a self adjoint unitary Note If e is a projection then 1 7 2e is a symmetry Conversely if s is a symmetry and 2 is invertible then 1 7 s is a projection 56 12 PARALLELOGRAM LAW 1212 LEMMA 25 Th 1 The following conditions on d Rickdrt ring A are equivalent a for every pair of projections e f there eacists d symmetry s E A such that sons 7 f6 b for every pair of projections ef in position p there esrists d symmetry s E A such that ses f hence also sfs e one says that e and f are exchanged by the symmetry s These conditions hold if A satis es the daciom SR of 10 Proof a i b Suppose ef are in position p by 126 e LPef f RPef With s as in a the mapping 1 gt gt sds l sdsquot sds is a automorphism of A so ses s LPef s LPsefs LPfe RPfe RPM 7 f b i a Given any pair of projections ef let e LPef f RPef Then ef are in position p 128 so by b there exists a symmetry s ex changing e and j Then sefs sefs sessfs fe ef lter fe Now suppose A satis es SR Let 1 e f 7 1 which is self adjoint By hypothesis there exists r E dd d2 such that r r and r2 d2 Since 1 6 y and r 6 d2 we have rd dr therefore rdr7d r2 7 d2 0 writing 9 RPr 1 we have gr 7 d 0 so gr gd Taking adjoints rg mg Then rdrdgrgdg2rg Since 12effe7e7f1 2 2 one has ed efe d e in particular e 6 y since r 6 12 it follows that er re Similarly f 6 d2 and fr rf Let s 17 29 which is a symmetry Then rsr72rgr7rd by so rs 71 taking adjoints sr 7d whence r 7sd Finally since e 1 17 f by the de nition of 1 one has sefssd17ffs sdfs 7rfs ifmff0 f f1f gt thus a holds 0 1213 THEOREM Si Maeda 25 Th 2 Consider the following conditions on d Rickdrt ring EU For every pair of projections ef there esrists d unitdry it such that uefu fe Existence of unitaries ES For every pair of projections ef there eacists d symmetry s such that sefs fe Existence of symmetries 12 PARALLELOGRAM LAW 57 Then SR ES EU ii EU dacz om H holds for 3 unitary equivalence iii ES dacz om H holds for 3 unitary equivalence by a symmetry In particular SR dacz om H holds for i in other words for 3 610 Proof The rst implication holds by 1212 the second is trivial ii Given any pair of projections e f let u be a unitary such that ue17fu 17 fe Then 115 eUf if RP161 fl LPlU if6l LPue17 fu uLP61fl 39u ue7e fu thus 6 U f 7 f and e 7 6 f f are unitarily equivalent iii Same proof as ii with uquot u 0 We thus have the diagram the arrows signify implication SR l ES EU H for 3 H for 3 l H for i 1214 If A is a Baer ring satisfying the axiom EP of 10 and if A has GC relative to i then A satis es axiom H relative to i and the relations 3 3 coincide Sketch of proof As is shown in 1431 LPd i RPd for all 1 E A there fore A satis es axiom H for 5 1210 If A is abelian then 6 g f implies e f 916 thus both i and 5 coincide with the relation of equality It now suf ces to consider the case that A has no abelian summand 827 Then since A has GC for i partial isometries are addable in A 1011 since moreover A satis es EP it follows that A has polar decomposition 1012 and in partic ular A satis es SR Therefore 3 coincides with i 610 It is the citation of 2 in 1011 that prevents the foregoing from being a complete proof 1215 In a Baer ring satisfying axiom H for N direct niteness and niteness are equivalent conditions 2011 1216 D Handelman has constructed a Baer ring in which axiom H fails for 3 his example is moreover directly nite7even strongly modular7 cf 217 and factorial 0 1 the only central projections 13 13 GENERALIZED COMPARABILITY Throughout this section A is a Baer ring N an equivalence relation on its projection lattice with various axioms added as needed 131 DEFINITION The Baer ring A is said to have generalized com parability GO relative to N if for every pair of projections ef of A there exists a central projection it such that uejuf and 17ufj 17ue A weaker condition is orthogonal GC such a it exists whenever ef 0 132 PROPOSITION Assume adioms B and D of 10 hold Then the fol lowing conditions on A are equivalent a A has GO ID for every pair of projections ef of A there edist orthogonal decomposi tions 66162gt ff1f2 with e1 N f1 and Ce2Cf2 0 where C denotes central cover Proof 2 p 77 Prop 1 a i b Given projections ef let u be a central projection with ue j uf and 17 uf j 17 ue say ueN guf and 17ufNe1 17ue Let elue f17ufthus ei gt er r Set e1 e1e1f1 ff by axiom D e1 N f1 Set e2 e7e1 f2 f 7 f1 then ueg ue 7ue1 e3 7 e3 0 and similarly 17uf2 0 Thus C6217 u and Cfg S u whence Ce2Cf2 0 b i a With notations as in b let u Cfg By axiom B uel N ufl and 17ue1 N 17 uf1 Now ueg Cf2e2 Cf2Ce2e2 0 and 17 uf2 17 uCf2f2 1710qu 0 so ue uel N ufl S uf and 17uf 17uf1 N 17ue1 S 17ue Thus A has GC 0 133 THEOREM 18 p 53 Th 35 Assume adioms B E and F hold Then A has orthogonal GC 13 GENERALIZED COMPARABILITY 59 Proof Let ef be projections in A with ef 0 we seekacentral projection it such that ue j uf and 17uf j 17ue If eAf 0 then CeCf 0 by 320 and 321 and u Cf lls the bill Assume eAf a 0 Let e2 be a maximal pair of orthogonal families of nonzero projections such that ez S e fz S f and ez N fz for all i Zorn7s lemma using axiom E to get started Let g sup ei h sup fi by axiom F g N h recall that ef 0 therefore gh 0 Necessarily e7gAf7h 0 since otherwise an application of axiom E would contradict maximality Therefore Ce 7gCf 7 h 0 Set it Cf 7 h Then ue 7 g 0 so citing axiom B one has ue ug N uh S uf and 17uf7h0so 17uf17uh17ug 17ue O 134 COROLLARY If A is a Baer ring satisfying adiom E for 5 then A has orthogonal GC for i Proof Axioms B and F hold for 5 112 quote 133 0 135 COROLLARY If A is a Baer ring satisfying the EP adiom then A satis es adiom E for i therefore 134 A has orthogonal GC for i Proof Suppose ef are projections with eAf a 0 Choose a E eAf with a a 0 By the EP axiom there exist a nonzero projection g and an element y E dd such that 34 y and 1 y2 9 Then in dy satis es ww yddy 1 y2 g and g y2dd shows that g S RPd S f Set h ww 2411quot then h S LPd S e Since g i h axiom E is veri ed for i Quote 134 0 136 A regular Baer ring has orthogonal GC for g by 113 and 133 since moreover axiom H holds for g in such a ring 59 GC holds by an elementary argument given in the proof of 139 below Another proof of this is given in 1311 The following de nitions and lemmas lead up to S Maeda and S S Holland7s generalization that roughly speaking the parallelogram law implies GC 139 137 DEFINITION Projections ef in A are partially comparable with respect to N if there exist nonzero projections e0 3 e f0 3 f such that e0 N f0 Thus axiom E is equivalent to the condition eAf a 0 i ef are partially comparable If ef are not partially comparable they are said to be unrelated with respect to N Thus axiom E is equivalent to the condition ef unrelated eAf 0 To say that ef are unrelated means that if e0 3 e f0 3 f and e0 N f0 then either e0 0 or f0 0 in the presence of axiom A we can say e0 f0 138 LEMMA 26 Lemma 21 Assume adioms A and 0 hold Then the following conditions are equivalent a adiom E holds b ef unrelated ef 0 c for each projection e there edists a largest projection e unrelated to e and one has ee 0 When these conditions hold the projection e of c is in the center of the ring A 60 13 GENERALIZED COMPARABILITY Proof This is a startling result Axiom E says e f unrelated eAf 0 this is obviously stronger than condition b and one would suppose that it is much stronger Not so says the lemma a i b Obvious b i c Given a projection e E A let S be the set of all projections in A that are unrelated to e at least 0 E S and let e supS We rst show that ee are unrelated Suppose e0 3 e e6 3 e e0 N e6 Note that if f E S then e0f are unrelated because e0 3 e and ef are unrelated Since e6 N e0 it follows that e6f are unrelated For suppose e8 3 e6 f S f e6 N f By axiom C the equivalence e6 N e0 induces an equivalence e8 N g 3 e0 for some 9 thus f N e8 N Q where f S f and g 3 e0 Since fe0 are unrelated it follows that f 0 or g 0 hence axiom A f g 0 and e8 0 By b egf 0 varying f E S ege 0 But e6 3 e so e6 ege 0 This shows that ee are unrelated Consequently e E S thus S has e as largest element moreover ee 0 by Thus c holds c i a Assuming ef unrelated we must show that eAf 0 Choose e as in c thus f S e Apply c to e there exists a largest projection e unrelated to e and ee 0 Since ee are unrelated e S e it will suf ce to show that e is in the center of A for then it will follow that eAf eAef eeAf OAf 0 in view of 39 it will further suffice to show that e has a unique complement namely 17e Assuming g is any complement of e it will suffice to show that g e We rst note that ge are unrelated For suppose go 3 9 e0 3 e go N e0 Since e0 3 e we know that e0e are unrelated since go N e0 it follows from axiom C that goe are unrelated therefore go 3 e Thus go 3 g f e 0 recall that g is a complement of e so go 0 it follows that g S e so we can form the projection e 7 9 Since ee 0 one has e S 1 7 e therefore e 7g 17e 17g17eUg1710 thus 9 e O 139 THEOREM cf 26 Th 21 Let A be a Baer ring N an equivalence relation on its projection lattice satisfying the aaioms A B C D F and especially H of 10 Then A has GC relative to N Proof Let us rst show that axiom E holds Assuming ef are unrelated projections it will suf ce by the lemma to show that ef 0 By axiom H and 124 e7enlt17fgtf7lt17egtm since e f are unrelated either e7e 17f 0 or f 7 1 7e f 0 actually both by axiomA Say f717e f 0 then f 17e f S 17e so ef 0 Since axioms B E F hold A has orthogonal GC 133 By axiom H one can write 666gtfff 13 GENERALIZED COMPARABILITY 61 with e N f and ef e f 0 129 Then er 0 so by orthogonal CC and axiom D one can write e 61 62 f f1f2 with e1 N f1 and Ce2Cf2 0 see the proof of 132 Then 666162gt ffflf2gt where e e1 N f f1 by axiom D and Ce2Cf2 0 whence see the proof of 132 it Cfg satis es ue j uf and 17 uf j 17 ue O This theorem is proved in 18 p 87 Th 57 with axiom E as an added hypothesis redundant as 139 shows 1310 COROLLARY 26 Th 21 If A is a Baer ring satisfying aaiom H for i then A has CC for 5 Proof Immediate from 112 and 139 0 1311 COROLLARY 17 p 534 10 Every regular Baer ring has CC for g Proof Immediate from 113 and 139 0 1312 COROLLARY If A is a Baer ring satisfying the SR aaciom then A has CC for 3 cf 610 Proof Since SR axiom H for L 1213 the corollary is immediate from 1310 Incidentally i coincides with 3 by 610 O 1313 COROLLARY If A is a Baer ring with no abelian summand sat isfying the SR aaciom then partial isometries are addable in A cf 10 Sketch of proof By 1312 A has CC for 5 since moreover A has no abelian summand it follows that partial isometries are addable 2 p 129 Th 1 O This corollary is proved in 18 p 104 Th 64 with the EP axiom as an added hypothesis We remark that in a Baer ring with no abelian summand and satisfying CC for i EP SR 1432 1314 PROPOSITION 18 p 85 Th 55 Let A be a Baer ring satisfying the hypotheses of 139 If ef is any pair of projections in A then there eacists a central projection it such that uejuf and 17u17ej17u17f Proof Write e0 e 017 f f0 17 e f f by axiom H 124 6ieofif0 1 By 139 A has CC applying it to the pair e0 f0 we nd a central projection it such that neojufo 17uf0j17ue0 ii 62 13 GENERALIZED COMPARABILITY From we have ue 7 use N uf 7 ufo which when added to the rst relation of ii yields ue j uf On the other hand the substitutions e gt gt 176 f gt gt 17f transform 60 f0 into f0 60 respectively therefore is transformed into 1 6lif0N1f60 1 From i and the second relation of ii one argues as above that 17 u17 e j 1U17f O The analogue of this propositon is valid for g in a regular right self injective ring 29 Prop 13 14 POLAR DECOMPOSITION 141 DEFINITION A ring A is said to have polar decomposition PD if for every 1 E A one can write a 1m with 7 E 1a 7 7 r2 aa and w a partial isometry such that wwquot LPa ww RPa 142 A ring with PD satis es the SR axiom of 10 as well as LP 3 RP 143 DEFINITION Let A be a Baer ring We say that partial isome tries are addable in A if whenever is a family of partial isometries such that the projections ez wfwi are pairwise orthogonal and the projections fi wiwf are pairwise orthogonal there exists a partial isometry w E A such that ww sup ei wwquot sup fi and we wz fiw for all i In particular equivalence is completely additive in the sense of 10 144 Partial isometries are addable under each of the following hypotheses on a Baer ring A A has no abelian summand and has GC for i 2 p 129 Th 1 ii A any AW algebra same ref iii M2 A is a Baer ring 2 p 131 Exer 1 iv relative to 5 A is properly in nite and satis es axiom E of 10 2 p 131 Exer 3 145 The main result to be proved in this section 1429 is the following theorem of L Herman 14 If A is a Baei i ing such that A satis es the EP aaciom of 10 and ii partial isometries are addable in A then A has PD The idea of the proof is to exhaust on the EP axiom given a E A whose polar decomposition we wish to effect one considers a maximal orthogonal family of nonzero projections ei such that for each i there exists a self adjoint sz 6 ay with 1 s22 ei Replacing sz by eisi one can suppose eisi si Writing wz asi one has wfwz siaasi asf ei one shows easily that the projections fz wiw are pairwise orthogonal and that sup fz LPa sup ez RPa By ii one forms a partial isometry w The trick is to sum up7 the sz more precisely their relative inverses7 n to arrive at an element 7 E aa A machinery for discussing such sums7 was introduced by Loomis 20 p 20ff and elaborated by Herman 14 for the readers convenience we reproduce the details here Herman7s paper has not appeared in print his elegant discussion deserves to be more widely known 146 DEFINITION Elements ab of a ring are said to be orthogonal written a 1 b if abquot ba 0 63 64 14 POLAR DECOMPOSITION Taking gtk one sees that 1 1b gt b1 1 gt 1 1 bquot A family of elements 1 is said to be orthogonal or pairwise orthogonal7 if 12 1 a for i a j this Will be indicated by writing 1 1 Not to be confused With the concept of independent family7 in a lattice for which the same notation is employed 147 For projections ef in a ring 6 1 f means 6f 0 For elements ab of a Rickart ring a 1 b means that RPaLPb RPbLPa 0 that is RPaRPb 0 and LPaLPb 0 In particular 1 for the family in 143 148 DEFINITION An orthogonal family 1 in a ring A is said to be summable to a if there exists an element a E A such that for each index j ai10foralli7 j a1aj1 Expressed in terms of right annihilators this means that for each j any c a 7 ajy i7 j One then writes a 69 called the sum of the 12 In a ring With proper involution the sum is unique 149 In a ring With proper involution if ai a and em b then a b Proof For each j by hypothesis any c a 7 ajy and any c b 7 ajy iii i7 j mphy c a 7 ajy n b 7 a i7 j but a 7b a 7 1 7 b7 1 implies a7 ajr O b 7af C a 7 by therefore air C a 7 by for each index j i7 j By orthogonality and a E air C a 7 by a so a 7 baquot 0 aja 7 b 0 Thus lla 7 b 0 for all i that is a 7 W 6 Hair w 14 POLAR DECOMPOSITION 65 Fix any index j from and we see that a i by E hazy C lilr C a i by 239 a so a 7 ba 7 b 0 Since the involution is proper a 7 b 0 1410 LEMMA 20 p 27 Lemma 47 If a ai in the sense of 148 then the following two conditions hold 1 aa aia aia for all i 2 if an 0 for all i then aa 0 Proof 1 For each j by orthogonality one has a 6 may c a 7 ajy a so a 7 aja 0 aa aja in particular aa is self adjoint so it is equal to its adjoint ajaquot 2 Suppose aza 0 for all i Fix any index j Then 95 E aiir C aiir C a lair 239 a so a7aja0 aaaja0 O 1411 LEMMA 20 p 27 Lemma 48 If in a 7 ing with proper involution a ai and J is a nite set of indices such that baz 0 for all i Z J then ba ZjEJ baj Proof When J 9 this says that the dual of 1410 2 for left annihilators holds Let c b a 7 2a jEJ we are to show that c 0 and it will su ice to show that 00 0 One has i c 7 a 7 2a b jEJ We assert that aicquot 0 for all i Now aicquot aia7E aia b jEJ aia 7 Zaia b by 1410 jEJ aibaZ 7 Zaia Iquot jEJ 66 14 POLAR DECOMPOSITION If i E J then by orthogonality ZjEJaZa aia so ai0 aia 7 aiab 0 whereas if i Z J then bai 0 and ZjEJ aia 0 by orthogonality so and the assertion is proved It then follows from 1410 that a0 0 so 00 b a7Zaj 0 ba0 7Zbaj0 jEJ jEJ b07Zb00 jEJ and the lemma is proved O 1412 LEMMA 20 p 27 Remark Let A be a ring with proper involution ai an orthogonal family in A and a E A Then a ai if and only if for each inded j rail C a 7 ajl i7 j Proof Only if Suppose a E ijail Thus if J then CLZ 0 for all i J By 1411 da daj thus a E a 7 ajl If Suppose the stated condition holds in analogy with 148 let us express this by writing a G9ai By the dual of 1411 if b E A and J is a nite set of indices such that aib 0 for all i Z J then ab Zia aib Letting J we see that the conditions of 148 are ful lled in other words a ai O 1413 In View of 1412 De nition 148 is left right symmetric7 in other words a ai in A if and only if a ai in the opposite ring A0 1414 PROPOSITION 20 p 27 Remark Let A be a ring with proper involution ai an orthogonal family in A and a E A In order that a ai in the sense of 148 it is necessary and su icient that the following two conditions hold 1 aa aia for all i 2 if a 0 for all i then ad 0 One then has aa aia aia for all i Proof Necessity This is 1410 Su iciency Suppose 1 and 2 hold Fix an index j by 1412 it will su ice to show that rail C a 7 ajl i7 j 14 POLAR DECOMPOSITION 67 Suppose aai 0 for all i a j we are to show that aa 7 aj 0 We assert that 1a 7 aaja 0 for all i for if ia j then 1a 7 aaja aaa 7 aaja aaa 7 0 by orthogonality pala by 1 Whereas 1a 7 aaja aaa 7 aja 060110 110 by 1 0 Thus holds in other words aiaa 7 aa 0 for all i then by 2 aaa 7 aa 0 thus 1a 7 aaja 0 aaa aaja aaa aaa 7 a 0 Then 1a 7 aajaa 7 aa 1a 7 aaja 7 aja aaa 7 aja 7 aava 7 aja 01 7 1aja 7 aja 71aja 7 ajaa 0 whence aa 7 1a 0 the involution is proper O 1415 PROPOSITION 20 p 28 In a ring A with proper involution suppose a ai Then i a a ii If a is an element such that the family an is also orthogonal then aZm aa iii If a is an element such that the family 1a is also orthogonal then aai aa iV aa aZa and aa azai V If a and y are elements such that an ya for all i then aa ya Proof If a ai in A then a 69a in the opposite ring A0 because a gt gt aquot is an isomorphism of rings A 7 A0 A0 being equipped With the same involution as A therefore aquot 69a in A 1413 ii Fix j If y E aZay then aiay azay 0 for all i a j whence a 7 ajay 0 by the de nition of a 148 thus y 6 aa 7 a 68 14 POLAR DECOMPOSITION iii if 1a1 then also 1 so a f 1 By and ii i 1quot therefore citing i 1a a a ma iv By 1414 m aa but for i 74 j aa Java aaaja 0 thus aa1 By iii aaquot aa aa Then 11 aIa results on replacing a by 1 V Note that 111 for i 74 j a ha y WMWJ39V 121191 0 aj1a aa 0 So by ii a 11 Also ya a1 so ya ya by iii Thus 106 aac ya ya 0 1416 COROLLARY L6t A b6 6 7 ing with proper involution S C A and a a with a E S for all i Th6n a E 8 Proof Immediate from V of 1415 0 1417 14 Lemma 16 Let A be a ring with proper involution 6 an orthogonal family of projections in A and suppose that there exists 6 696 Then 6 is a projection and 6 sup 6 Proof By 1415 66 6I6 G96 6 thus 6 is a projection Cit ing 1414 66 66 66 6 thus 6 S 6 for all i On the other hand if f is a projection such that 6 3 f for all i then 61 7 f 0 for all i whence 61 7 f 0 by 1414 thus 6 S f This proves that 6 serves as supremum for the family 6 1418 Let A be a Rickart ring 6 an orthogonal family of projections possessing a supremum 6 in the projection lattice of A Then 6 696 Proof 1 One has 66 66 6 66 for all i 2 1f 61 0 for all i then 6LP1 0 6317 LP1 for all i 6 S 17 LP1 6 LP1 0 61 0 By 1414 6 696 1419 Let A be a Rickart ring an orthogonal family of partial isometries in A cf 147 if is summable in the sense of 148 then it is addable in the sense of 143 More precisely if in 6911 then in is a partial isometry and if 6 RPw f LPw then sup6 exists and is equal to ww sup f exists and is equal to wwquot and w6 w fw for all i Proof Suppose w 6911 Then 1415 ww wfw 696 so writing 6 ww from 1417 we know that 6 is a projection and serves as sup 6 Simi larly writing f ww one has f f and f sup f Fix an index j For all i 74 j fjw fjfw 0 so by 1411 one has fjw fjwj 11 Similarly w 911 yields 6jw w w6j w Thus the family is addable in the sense of 143 Conversely 14 POLAR DECOMPOSITION 69 1420 Let A be a Baer ring an orthogonal family of partial isome tries in A cf 147 If is addable in the sense of 143 then it is summable in the sense of 148 Proof Let ez RPwZ fz LPwZ e sup ei f sup fz we are in a Baer ring By hypothesis there exists a partial isometry in with ww e wwquot f and we wz fiw for all i To show that w wi we verify the conditions of 1414 1 For all i ww in eiwf wel 11 wiwf 2 If wza 0 for all i then ez RPwZ S 17 LPa for all i so e S 17 LPa ea 0 therefore wa wea wea 0 1421 14 Lemma 15 In a Rickart ring A suppose a ai in the sense of 148 Let ez RPaZ fz LPaZ Then ei is an orthogonal family of projections having RPa as supremum and is an orthogonal family of projections having LPa as supremum Proof That ei1 and fl1 results from 147 We assert that W may ltgt The inclusion D is 2 of 1414 On the other hand if aa 0 then for all i one has afa aaz dual of 1414 so azaaza aaaia a afaa aaaa 0 whence aza 0 The formula is veri ed This means 17 that 17 RPalA u e RpmHA u 7 enA it follows 118 that inf17 ei exists and is equal to 17 RPa whence sup e2 exists and is equal to RPa Similarly LPa sup fz 1422 DEFINITION L Herman 14 A Loomis ring is a ring with unity satisfying the EP axiom of 10 in which every orthogonal family of partial isometries is summable in the sense of 148 Note The EP axiom trivially assures that the involution is proper 1423 THEOREM 14 Th 17 A ring A is a Loomis ring if and only if it satis es the following three conditions i A is a Baer ring ii A satis es the EP aaciom iii partial isometries are addable in A Proof If From i iii and 1420 one sees that every orthogonal family of partial isometries in A is summable in the sense of 148 thus A is a Loomis ring 1422 Only if Suppose A is a Loomis ring In particular ii holds by the de nition hence the involution is proper 70 14 POLAR DECOMPOSITION claim 1 Every orthogonal family of projections ei has a supremum For by hypothesis ei is summable say e ei By 1417 e is a projec tion and e sup ei claim 2 A is a Rickart ring Let 1 E A we seek a projection e such that 1T 17 eA lf 1 0 then e 0 lls the bill Assume 1 74 0 Let ei be a maximal orthogonal family of nonzero projections such that for each i there exists 34 E 11 with y 34 and 11 y ez Zorn7s lemma get started by EP Let e Gael sup ei by claim 1 it will be shown that 1T 17 eA Suppose 1t 0 Then 11t 0 y 11t 0 thus eit 0 for all i Since e ei it follows that et 0 1414 that is t E 17 eA Thus 1T C 17 eA lf conversely t E 17 eA that is et 0 we are to show that 1t 0 It will su ice to show that 1e 1 for then 1t 1et 1 0 0 Since e ei and ez E 11 for all i one has e E 11 by 1416 therefore 11 17e E 11 We wish to show that 11 7 e 0 Assume to the contrary Then by the EP axiom there exists an element y ezlt17 egtrizlt17 em M17 en c my such that yquot y and 1117 ey2 g g a nonzero projection Replacing y by 17 ey we have y E 11 y y ye 0 and 1quot1y2 9 Obviously ge 0 whence gez 0 for all i contradicting maximality From claims 1 and 2 one sees easily that A is a Baer ring 2 p 20 Prop 1 thus holds Finally iii is immediate from 1419 0 What is striking in this circle of ideas is that the conditions i ii iii imply SR see 1429 below 1424 COROLLARY A C algebra is an AW algebra if and only if it is a Loomis ring Proof If lmmediate from of 1423 and the de nition of AW algebra 138 Only if Every AW algebra is a Baer ring 138 with EP 2 p 43 Cor in which partial isometries are addable 2 p 129 Th 1 thus the conditions i ii iii of 1423 are ful lled O 1425 COROLLARY If A is a Baer ring with no abelian summand satisfying the EP a1iom in which GC holds for i then A is a Loomis ring Proof Condition iii of 1423 holds by 2 p 129 Th 1 whose proof is long and tedious O For various reformulations of this result see 1432 below 1426 COROLLARY If A is a regular Baer ring with no abelian sum mand satisfying the SR a1iom then A is a Loomis ring Proof The SR axiom implies that axiom H holds for L 1213 which in turn implies that GC holds for 3 1310 So by 1425 we need only verify the EP axiom Let 1 E A 1 74 0 By the SR axiom write 11 r2 with r r E 14 POLAR DECOMPOSITION 71 1a and let e RPa RPr Let y be the relative inverse of r 27 in other words y is the inverse of r in eAe Then yquot y because r r y E ay 47 and 1 y2 r2342 e a 0 O 1427 COROLLARY If A is a Baer ring satisfying the EP aaciom properly in nite relative to i then A is a Loomis ring Proof By 135 A satis es axiom E for i since moreover A is properly in nite for 5 it follows that partial isometries are addable in A 2 p 131 Exer 3 Thus the conditions i ii iii of 1423 are satis ed 0 1428 LEMMA In a ring satisfying the EP aaciom if a is a nonzero element then there eacists an element s such that i squot s ii s E 1a iii 1 s2 is a nonzero projection e iv as is a partial isometry with asas e v es s Proof By the EP axiom there exists an element s satisfying i ii iii Moreover asas saas 1as2 e Since e E 1a replacing s by es se we can suppose es s O 1429 THEOREM L Herman 14 Th 22 Every Loomis ring has polar decomposition Proof Let A be a Loomis ring a E A if a 0 then the elements in r 0 meet the requirements of 141 Assume a 74 0 Let ei be a maximal orthogonal family of nonzero projec tions such that for each i there exists sz 6 aa with s si 1 s22 ei eisi sz Zorn7s lemma using 1428 to get started Set wz asi a partial isometry with wfwi ei Write e sup ez which exists by 1423 We assert that e RPa From ez sE a it is clear that ez S RPa for all i therefore e S RPa To show that e Z RPa it will su ce to show that me a that is 11 7 e 0 At any rate since ez E ay for all i one has e E ay 45 whence aa17 e E 1a If on the contrary 11 7 e a 0 by the lemma there exist a nonzero projection g and an element 1 6 Ml 6ll11 all 03051 6H C 051V such that yquot y aa17 e y2 g gy y Evidently ge 0 so gez 0 for all i contradicting maximality Let fz win asfaquot The family of projections is orthogonal if i a j then fifj as a asiaquot as ejaquot asEei ejaquot 0 Let f sup fi From fz asE it is clear that fz S LPa for all i therefore f S LPa We assert that f LPa Let h LPa 7 f For all i one has 72 14 POLAR DECOMPOSITION 1 iffi 0 hence hfz 0 so 0 0a hfza has aa haei thus haei 0 for all i whence 0 hae ha Therefore h LPa 0 but thPaso hhLPa0 LPaf Since ei and are orthogonal families of projections is an or thogonal family of partial isometries 147 By hypothesis is summable in other words 1419 addable and writing in wi we have ww sup ez e wwquot supfz f and wei wz fiw for all i Set r wa Then wr wwa fa a Now w Gem and wfa slaw asi E eiAeZ shows that is an orthogonal family therefore by 1415 one has wa wfa aa si that is r aa si Since 1 sz 6 aa for all i it follows from 1416 that r E 1a Also since the asi are self adjoint by 1415 one has r aa si 1 8239 7 and r2 rr aa aa aa ei aaei ae 11 Summarizing we have a wr with r r E aa r2 aa ww e RPa ww f LPa thus A has PD 141 lncidentally writing rz asi E 1a we have r ri One has rz E eiAei risi siri ei thus rz is the inverse of sz in eiAei Also r3 1 as aa ez ez aa thus rz is a square root7 of aan O 1430 COROLLARY The following classes of Baer rings are Loomis rings hence have polar decomposition AW algebras 1424 ii Baer rings without abelian summand satisfying the EP aaiom and GC for L 1425 iii regular Baer rings without abelian summand satisfying the SR aaciom 1426 iV Baer rings satisfying the EP aaiom and properly in nite for 5 1427 V Baer rings without abelian summand satisfying the EP and SR aacioms Proof V 18 p 104 Th 64 From SR we know that axiom H holds for 3 1213 therefore GC holds for L 1310 So we are in a Loomis ring 1425 whence PD 1429 0 1431 COROLLARY The condition LP 3 RP holds for the following classes of Baer rings AW algebras ii Baer rings satisfying the EP aaiom and GC for 5 14 POLAR DECOMPOSITION 73 iii regular Baer rings satisfying the SR aaiom iv Baer rings satisfying the EP aaiom and properly in nite for 5 V Baer rings satisfying the EP and SR aaioms Proof Since LPa RPa for all a in an abelian ring 812 the corollary is immediate from 827 and 1430 We remark that the assertion for iii is also a triVial consequence of 58 and 610 0 See 2144 for a simpler proof of ii 1432 COROLLARY For a Baer ring A without abelian summand the following conditions are equivalent is a Loomis ring A 97 V gt b A satis es the EP aaiom and has PD c A satis es the EP and SR aacioms d A satis es the EP aaiom and LP 3 RP e A satis es the EP aaiom and aaiom H for i f A satis es the EP aaiom and GC for i Proof a i b 1429 c d triVial e 1213 theorem of S Maeda triVial 115 f 1310 theorem of Maeda and Holland a 1425 ltgt Cf 2144 Vees AAAA H a x 1433 Abelian rings are genuinely pathological in this circle of ideas For example there exists a commutative Baer ring satisfying the EP and SR axioms in which partial isometries are not in general addable 18 p 103 2 p 131 Exer Re ning Kaplansky7s example Herman 14 exhibited a Baer ring satisfying the EP axiom and PD in which partial isometries are not in general addable thus the implication b i a of 1432 fails in general 15 FINITE AND INFINITE RINGS Throughout this section A is a Baer ring and N is an equivalence relation on its projection lattice satisfying the agrioms A B C D F and generalized com parability GC of 10 The precise axioms actually needed for each proof will be noted parenthetically Finiteness and in niteness are de ned relative to N 95 99 151 Axiom E holds Proof using GC and axioms A D Suppose eAf a 0 in other words CeCf a 0 321 we are to show that ef are partially comparable 137 By GC write e e1 e2 f f1f2 with e1 N f1 and Ce2Cf2 0 132 Since CeCf a 0 either e a e2 or f a f2 thus e1 a 0 or f1 a 0 hence axiom A e1 a 0 and f1 a 0 Recall that i always satis es A7D and F 112 thus the present section and following ones is in a sense an exploitation of the consequences of GC Since SR axiomeor i i GC for i by the theorems of Maeda and Holland 1213 and 1310 GC is a natural axiom for many applications 152 If f is a nite projection relative to N see 95 and e S f then e is nite Proof Axiom D This is 98 153 If f is nite and e N f then e is nite 18 p 52 Th 32 Proof Axioms A 0 Suppose g S e g N e By axiom 0 there exists hgf with gNh and e7ng7h Then thNgNeNfthus hoSf since f is nite hfso e7ng7h0thus e7g0 byaxiomA 154 If f is nite and e j f then e is nite Proof Axioms A C D lmmediate from 152 153 155 PROPOSITION 18 p 88 Exer 2 The following conditions on A are equivalent a A is nite b eNf 17eN17f Proof a i b Axioms B D and GC Suppose e N f Let u be a central projection with u17 e j u17 f and 17 u17 f j 17 u17 e Say 74 15 FINITE AND INFINITE RINGS 75 u17e N9 3 u17f and 17u17f N h S 17u17e Since ue Nuf axiom B one has uueu17eNufg u by axiom D since u is nite 152 it follows that uf g u thus 9 u17f whence u17e N u17f Similarly 17u17e N 17u17f and another application of axiom D yields 1 7 e N 1 7 f b i a Axiom A Suppose e N 1 By b 17e N0 so 17e 0 by axiom A O In niteness is characterized in the next proposition 156 LEMMA 18 p 58 Th 38 Let eiZE1 be an in nite family ofpairwise orthogonal projections with ez N ej for all i and j let J be a subset of I with cardJ cardI and let esupeZ tel fsupej jEJ Then e N f Proof Axioms D F Write J as a disjoint union JJ1UJ2 With cardJl carng cardJ cardI Onehas J2 CJ2UI7JI7J1CIso cardI carng 3 cardI 7 J1 S cardI therefore cardI 7 J1 cardI Schroder Bernstein theorem Set f1supeZlEJ1 f2supeZlEJ2 76 15 FINITE AND INFINITE RINGS one has flfg 0 f fl f2 and f2 Nfl by axiom F Set gsupeZiEI7J1 then fig 0 e f1 g and f1 N g by axiom F Then axiom D yields f1f2gfbthatisgt f6 ltgt 157 PROPOSITION 18 p 62 Exer 5 The following conditions are equiv alent a A is in nite relative to N b A contains an in nite family ei of pairwise orthogonal nonzero projec tions with ez N e for all i and j Proof Axioms A C D F b i a Axioms D F Dropping down to a subfamily we can suppose the index set to be I 123 Let J 234 With notations as in 156 e N f S e but f a e thus e is not nite therefore neither is 1 152 a i b Axioms A C By hypothesis there exists a projection f1 a 1 with f1 N 1 Also f1 a 0 axiom A Write e1 17f1 From e1 f1 1N f1 and axiom 0 one obtains an orthogonal decomposition f1 e2 f2 with e1 N eg and f1 N f2 Note that 161f16162f2 in particular 6162 0 From eg f2 f1 N f2 and axiom 0 one obtains an orthogonal decomposition f2 e3 f3 with eg N e3 f2 N f3 Note that 16162f2616263f3gt in particular e1e2e3 are pairwise orthogonal Etc 0 158 LEMMA 18 p 63 Th 42 Absorbing a scrap Let eiZE1 be an in nite family of pairwise orthogonal projections with ez N e for all i and j and let e supei i E I Fiat an indea 1 E I and suppose that f is a projection such that f j el and fe 0 Then e f is the supremum of an orthogonal family hiLEI with h N e1 for all i E I Proof Axioms C D Say f N f1 3 e1 write 91 e1 ifl Then f1 91 e1 N ei so by axiom 0 there is a decomposition ez fi 92 with fz N f1 N f and 92 N 91 Note that the fz are pairwise orthogonal because fz S ei and they are all orthogonal to f because ef 0 thus f U fz i E I is an orthogonal family of pairwise equivalent projections of cardinality cardI Also 92 is an orthogonal family of pairwise equivalent projections and each 92 is orthogonal to every f and to f Since I is in nite there exists a bijection lt7 I 7 1 a I De ne a bijection 6gZ lEIgtfUfZi iEI 15 FINITE AND INFINITE RINGS 77 by 991 fgt 99239 fai for 17 1 For each i 92 and 69 are orthogonal de ne hi gi 6gi 1 61 Explicitly hi Qlfgt higifaifori7 1 The family is clearly orthogonal By axiom D h191fwgif161 and for i a 1 hi gifai Ngifi5i N61 Thus h N e1 for all i E I And suphZ iEIsuphZ 1h1 SUPQifai3 1i91 f Supgi 1 1Supfai 1 191 f supLCIZi i61supfjr i61f SUPQifii i61lf supeZ iEIf 6 f O 159 LEMMA 18 p 64 Th 43 Suppose elkE1 is a maaimal family of pairwise orthogonal nonzero projections in A such that ez N ej for all i and j and suppose that the indea set I is in nite Then there eacist a nonzero central projection u and an orthogonal family hipE1 with u sup h and h N ice for all i E I Proof Fix an index 1 E I Let e sup ez and apply GC to the pair 17e e1 let u be a central projection With u17 e j uel and 17 ue1 j 17 u17e Note that uel a 0 For uel 0 would imply that e1 e17ue1 17ue1 j17u17e 17e brie y e1 j 1 7 e Say e1 N e S 1 7 e then e could be adjoined to the fam ily ei contradicting maximality The ice are pairwise equivalent by axiom B and u1 7 e j nel so 158 provides a decomposition hipE1 of the projection supueZ iEIu17eueu17eu 78 15 FINITE AND INFINITE RINGS meeting the requirements Since 17ee1 are orthogonal all that is really needed is orthogonal GO in View of 133 the proof works assuming B C D E O 1510 LEMMA 18 p 64 Th 44 If A is properly in nite 99 then there eacists an orthogonal sequence en of projections such that e1 N e2 N e3 N and sup en 1 Proof By 157 there exists an in nite family of pairwise orthogonal equiv alent nonzero projections and we can suppose the family to be maximal in these properties Zorn It then follows from 159 that there exist a nonzero central projection u and an orthogonal family flZEI of projections with u sup fi fi N fj for all i and j and I in nite Since N0 cardI cardI one can write I O In 711 with the In pairwise disjoint and cardIn cardI for all n De ning fn supfZ i6 In for n 123 one has f1 Nfg Nfg N by axiomF and the fn are pairwise orthogonal with sup fn u Let uaaEJ be a maximal orthogonal family of nonzero central projections such that for each 04 there exists an in nite sequence of pairwise orthogonal projections eff e3 eg with uasupe n123 and ef NeS Neg N Zorn7s lemma get started by the preceding paragraph De ning en supe 04 EJ for n123 one has e1 N e2 N e3 N by axiom F and sup en sup ua thus it will su ce to show that sup ua 1 Let U 17sup net and assume to the contrary that v 74 0 Since 12A is in nite A is properly in nite an application of the rst paragraph of the proof contradicts maximality Inspecting the proofs of 157 and 159 one sees that the present lemma also holds assuming axioms A B C D E O 1511 THEOREM 18 p 65 Th 45 If A is properly in nite 99 then i there eacists an orthogonal sequence of projections fn with sup fn 1 and fn N1 for all n ii for each positive integer in there eacist orthogonal projections g1 gm with g1gm1 and gkNI for k1m Proof Let I 123 and write 111U12U13U with the In pairwise disjoint and in nite With notations as in 1510 de ne fnsupeZ iEIn then the fn are pairwise orthogonal with sup fn 1 and fn N 1 by 156 15 FINITE AND INFINITE RINGS 79 ii Same as i with just in terms on the right side of The proof is valid assuming axioms A B C D E O a If N means in 1511 ii then A is isomorphic to the matrix ring Mm A and when N means 5 this is a isomorphism with transpose as the involution on the matrix ring cf 2 p 98 Prop 1 1512 COROLLARY Under the hypotheses of 1511 there eacists a projection g with gN1N17g Proof Take in 2 in 1511 ii 0 1513 THEOREM 18 p 86 Th 56 Suppose A is a Baer ring with an equivalence relation on its projection lattice satisfying the aaioms A B C D F and H of 10 If e and f are nite projections in A then the projection e U f is also nite Proof A has GC by the theorem of Maeda and Holland 139 so we are under the hypotheses stated at the beginning of the section One has eUf eUfiffwhere by axiomH eUfifNeie f e thus e U f 7 f j e since e U f 7 f is therefore nite 154 we are reduced to the case that ef 0 Dropping down to e fAe f we can suppose that e f 1 we are to show that A is nite Assume to the contrary that A is in nite Dropping down to a direct sum mand we can suppose that A is properly in nite 911 Let g be a projection with g N 1 N 17 g 1512 Apply 1314 to the pair ge there is a central projection n such that ugjue and17u17gj17u17e1710f From u u 1 N ug j ue we see that u is nite 154 similarly from liu1u1liu1gj1uf we see that Iiu is nite Since A is properly in nite necessarily u Iiu 0 whence 1 0 a contradiction ltgt 16 RINGS OF TYPE I Throughout this section as in 15 A is a Baer ring N is an equivalence relation on its projection lattice satisfying the axioms A B C D F and GC of 10 As noted in 151 axiom E also holds The precise axioms needed for each proof Will be noted parenthetically 161 LEMMA 18 p 53 Th 34 If ef are projections in A with f abelian and e j f then 6 is abelian Proof Axioms A B C In View of 85 it su ices to consider 6 N f Since f is nite 917 so is 6 153 Let g be a projection With g S 6 it Will su ice 83 to show that g 6Cg From 9 S e N f and axiom C one has 9 N h S f for suitable h Since f is abelian h fCh but Cg Ch by 916 and g S e g S Cg therefore 9 S 609 6001 fCh h N 9 thus eCg is equivalent to its subprojection Q But fCh is abelian it is S f hence nite 917 and eCg N fCh so eCg is also nite 153 therefore 9 609 O 162 DEFINITION A is said to be homogeneous if there exists an or thogonal family 6126 of abelian projections in A such that supei 1 and 62 Nej for all i and j 163 Homogeneous i type I Proof Axioms A B With notations as in 162 x an index j One has C62 C61 for all i 916 therefore 322 1 01 cltsupeigt supclteigt ca thus ej is a faithful abelian projection so A is of type I 814 164 If ef are abelian projections With Ce Cf then 6 N f Proof A B D and GC Let u be a central projection With ue j uf and 17 uf j 17 ue Say ue N f S uf Then 317 and 916 ucltegt cltuegt cm Cuf ucm ace 80 16 RINGS OF TYPE I 81 so Cf uCf whence fCf fuCf uf But f is abelian so fCf f 83 therefore uf f Thus ue N f uf Similarly 17 ue N 17 uf therefore e N f by axiom D 165 If ef are projections with e abelian and Ce S Cf then e j f Proof Axioms A B C D and GO Let u be a central projection with ue j uf and 17uf j 17ue Since e is abelian so is 17ue hence 161 so is 1 7 uf moreover cf 916 0107 um 0117 we 7 lt17 ugtcltegt lt17 uCf 7 0107 um whence equality throughout therefore 1 7 ue N 1 7 uf by 164 But also ue j uf therefore e j f 166 DEFINITION A is said to be homogeneous of order N if in 162 the cardinality of the index set I is N We are not af rming that N is uniquely determined by A Problem Is it71 167 DEFINITION A is said to be of type In H a positive integer if i A is nite relative to N and ii A is homogeneous of order n We see in 1611 that n is unique Problem Is condition redundant7 When N means i resp i a ring of type In is isomorphic resp isomorphic to an n X n matrix ring over an abelian ring cf 2 p 98 Prop 1 168 DEFINITION Let N be an in nite cardinal If A is homogeneous of order N it is said to be of type IN Note Axioms D F Such a ring is in nite by 157 169 Suppose A is of type IN If N is in nite then A is in nite 168 if N is nite then A is nite by De nition 167 and N is unique by 1611 below 1610 Suppose A is homogeneous of order N If N is in nite then A is in nite and is of type IN 168 If N is nite we cannot conclude that A is of type IN because we do not know that A is nite In other words it is conceivable that there exists an in nite ring that is homogeneous of order n n an integer Problem Does there The crux of the matter is the following question If A is properly in nite cf 1512 can it be the ring of 2 X 2 matrices over an abelian ring This dif culty disappears if axiom H is added to the hypotheses 1513 1611 PROPOSITION 18 p 68 Th 47 If A is of type In H a positive integer then A cannot contain n 1 pairwise orthogonal equivalent nonzero projections In particular n is unique Proof Axioms A B D and GC Write 1 el en with e1 N N en and the ez abelian 167 Suppose f1 fm are pairwise orthogonal equivalent 1Not necessarily even for AWalgebras M Ozawa Nonuniqueness of the cardinality at tached to homogeneous AWalgebras Proc Amer Math Soc 93 1985 681684 82 16 RINGS OF TYPE I nonzero projections with f1 N N fm and m 2 n we are to show that m n Let u Cfl Then uel 74 0 e1 is faithful Dropping down to uA we can suppose that the fj are also faithful Then ez j fz for 1 S i S n 165 say 5i N S figt SO 1am m mhh since A is nite ff1therefore f1fn1whence nmltgt 1612 LEMMA 18 p 67 Th 46 If A contains a nonzero abelian projection ie A is not continuous in the sense of 815 then A has a nonzero homogeneous direct summand Proof Axioms A B C D and GC Let e E A be a nonzero abelian projection dropping down to CeA we can suppose that A is of type 1 and e is an abelian projection with Ce 1 By Zorn expand e to a maximal family ei of pairwise orthogonal faithful abelian projections By 164 ez N ej for all i and j Set f 17 supei By maximality f contains no faithful abelian projection it follows that Cf 74 1 1f Cf 1 then e j f by 165 say e N e S f then e is a faithful 916 abelian 161 subprojection of f a contradiction Set it 17 Cf 74 0 Then uf 0 that is u17 supei 0 so u sup uei and uA is the desired homogeneous summand O 1613 THEOREM 18 p 67 If A is of type 1 then there eacists an orthogonal family net of nonzero central projections such that sup ua 1 and every uaA is homogeneous Proof Axioms A B C D and GC Obvious exhaustion argument Zorn based on 1612 0 Homogeneous summands of the same order can be combined 1614 LEMMA 2 p 114 Lemma Let uaaEK be an orthogonal family of central projections such that every uaA is homogeneous of the same order N Let u sup ua Then uA is homogeneous of order N Proof Axiom F Let 1 be a set of cardinality N and for each 04 E K let emZEI be a family of pairwise orthogonal equivalent abelian projections with supremum ua For each i E 1 set eisupeM 046K The ez are pairwise orthogonal moreover for each i the Cem 04 E K are orthogonal therefore ez is abelian proof of 819 By axiom F ez N ej for all i and j cf the proof of 1510 Finally sup ez sup ua u 0 1615 THEOREM 2 p 115 Th 3 If A is of type 1 then there eacists an orthogonal family uNNSCMdA of central projections with supremum 1 such that for each N S cardA either uN 0 uNA is homogeneous of order N Proof Axioms A B C D F and GC lmmediate from 1613 and 1614 cardA being trivially an upper bound on the cardinality of orthogonal families of nonzero projections ltgt 16 RINGS OF TYPE I 83 1616 COROLLARY 2 p 115 Th 2 If A is nite and of type 1 then there eacists a unique orthogonal sequence unn21 of central projections with supre mum 1 such that for each n either un 0 or unA is of type In Proof Axioms A B C D F and GC Existence With notations as in 1615 mi 0 for all in nite N 168 And if for a positive integer n un a 0 then unA is of type In by de nition 167 Uniqueness Suppose vn is a sequence with the same properties If umvn a 0 then umvnA is both of type 1m and type In therefore in H 1611 Thus umvn 0 when m a n whence um 3 pm and pm 3 um O 1617 PROPOSITION 2 p 117 Prop 6 If A is of type 1 without abelian summand and if e E A is an abelian projection then e j 1 7 e Proof Axioms A B C D F and GC It will su ice to nd an orthogonal family net of nonzero central projections with sup ua 1 such that uae j ua1 7 e for all 04 for by the orthogonality of e and 1 7 e the equivalences can be added by Axiom F to obtain e j 1 7 e We show rst that there exists a nonzero central projection it such that ue j u17e Since A is not abelian e a 1 so u C17e a 0 Then ue is abelian and 17e is faithful in uA so ue j 17e by 165 in other words ue j u17e Let ua be a maximal orthogonal family of nonzero central projections such that uae j ua17 e for all 04 If sup ua 1 we are done Let u 17 sup net and assume to the contrary that u a 0 Then uA satis es the hypotheses of A and ue is abelian by the preceding paragraph there exists a nonzero central projection v E uA such that vue j vu 7 ue that is ve j 017 e contrary to the maximality of the family O 1618 LEMMA 2 p 81 Prop 9 Let eiZE1 be a family of projections in the Baer ring A such that for every nonzero central projection u the set luiEl ueZa O is in nite Then for each positive integer n there eacist n distinct indices i1 in and nonzero projections gt 3 eiu 1 S 1 S n such that gl N gg N N gn Proof Axioms C E For n 1 any index i1 6 11 will do with gl eil Suppose n 2 2 and assume inductively that i1in11 are distinct indices and f1 fn11 are nonzero projections with fl 3 eiu 1 S 1 S n 7 1 and f1 N N fn11 Consider it Cfl a 0 By hypothesis lu is in nite so we can choose in E lu 7 i1 in11 Then uein a 0 so uCeZn a 0 that is Cf1CeZn a 0 therefore flAeZn a 0 321 By axiom E there exist nonzero projections gl 3 f1 and gn S ein with gl N gn For 1 S 1 S n 71 the equivalence f1 N fl and axiom 0 yield a projection gt 3 fl with gl N 91 Clearly 9192 gn11 gn meet the requirements 0 1619 THEOREM 18 p 69 Th 48 If A is nite and of type 1 and if eiZE1 is any orthogonal family ofprojections in A then there eacists an orthogonal family net of nonzero central projections with sup ua 1 such that for each 04 the set of indices E l uaei a 0 is nite 84 16 RINGS OF TYPE I Proof Axioms A B C D F and GC We rst show that there exists a nonzero central projection it such that the set 1ui61 loci3amp0 is nite Suppose to the contrary that no such it exists By 1612 there exists a nonzero central projection U such that 0A is homogeneous since 0A is nite it is homogeneous of some nite order n 168 hence is of type In For every nonzero central projection u S v the set 1u is by supposition in nite this means that the family of projections 1262261 satis es the hypothesis of 1618 in the Baer ring 0A since 0A is of type In the conclusion of 1618 is clearly contradictory to 1611 Now let uaaEK be a maximal orthogonal family of nonzero central projec tions such that for each 04 the set 1uai61 natal3amp0 is nite Zorn get started by the preceding paragraph It will su ice to show that sup ua 1 Let w 17sup ua and assume to the contrary that w a 0 Then an application of the rst paragraph of the proof to the Baer ring wA contradicts maximality ltgt 17 CONTINUOUS RINGS In this section A is a Baer ring N is an equivalence relation on its projection lattice satisfying the axioms A B C E F of 10 Included are the rings satisfying the common hypotheses of 15 and 16 cf 151 The precise axioms needed for each proof are noted parenthetically 171 LEMMA The following conditions on A are equivalent a A is not abelian b there eacist nonzero projections ef with ef 0 and e N f Proof a i b Axiom E Let g be a projection in A that is not cen tral 811 that is gA17 g a 0 see the proof of 321 By axiom E there exist nonzero projections e S g f S 17 g with e N f b i a Axioms A B With notations as in b one has Ce C by 916 If e and f were both central one would have e Ce Cf f whence 0 ef ee e a contradiction Thus one of ef is noncentral so A is not abelian 81 O 172 The concept of abelian ring is absolute7 ie independent of the axioms of 10 results like 171 and 917 are of interest because they link properties of an absolute7 abelian to a relative7 a postulated equivalence relation If one were to de ne abelian7 in terms of N 171 suggests an appropriate de nition call A abelian relative to N7 if the relations e N f and ef 0 imply e 0 or f 0 173 LEMMA If A is continuous ie has no abelian projections other than 0 then for every positive integer n there eacist n pairwise orthogonal nonzero projections e1 en in A with e1 N e2 N N en Proof Axioms A C E It clearly su ices to consider n 2k k 012 For k 0 e1 1 lls the bill Let k 2 1 m Zk l and suppose inductively that f1 m are pairwise orthogonal equivalent nonzero projections Since A is continuous fl is not abelian therefore by 171 there exist nonzero projections e1 3 f1 e2 3 f1 with 6162 0 and e1 N e2 For i 23 m we see from axiom C and 6162lf15162lf1Nfi that fz contains orthogonal copies7 of e1e2 nonzero by axiom A and the induction is complete 0 85 86 17 CONTINUOUS RINGS 174 THEOREM 18 p 70 Th 49 If A is continuous 818 then for every positive integer n there eacist orthogonalprojections e1 en with e1 en 1 and e1 N626n Proof Axioms A C E F Consider n ples of families of nonzero projections 51iieIgtgt5miel with a common index set I where the e are pairwise orthogonal ehem 0 if A in or and eh N 627 N Nem for each i E I We can suppose that the index set I cannot be enlarged Zorn get started by 173 De ne eAsupegtZ iEl lgAgn The e are pairwise orthogonal and by axiom F one has e1 N e2 N N en so it will suf ce to show that e1 en 1 Let e 17 e1 en and assume to the contrary that e a 0 An application of 173 in the continuous ring eAe contradicts maximality For the case n 2 axioms E F are suf cient by an eVident simpli cation of the above proof 0 175 In 174 if N means 3 then A is isomorphic to the matrix ring Mne1Ae1 and when N means i this is a isomorphism with transpose as the involution on the matrix ring cf 2 p 98 Prop 1 18 ADDITIVITY OF EQUIVALENCE This is the main result 181 THEOREM Let A be a Baer ring N an equivalence relation on its projection lattice satisfying the aaioms AiD and F of 10 Then the following conditions relative to N are equivalent a A has GO ID A satis es aaiom E and equivalence is additive the sense of 10 Stripped to its essentials this says assuming AiF hold GC gt equivalence is additive Still another formulation assuming AiE hold equivalence is additive gt Axiom F and GC hold 182 The proof of b i a is easy same as 133 omitting the hypothesis ef 0 and using complete additivity instead of the weaker axiom One notes that axioms A C D are not needed for this part of the proof Henceforth through 1811 A denotes a Baer ring N an equivalence rela tion on its projection lattice satisfying the aaioms AiD F and GC of 10 Note that axiom E also holds 151 Our objective is to prove that equivalence is additive this Will establish the validity of a i b 183 DEFINITION Let each of eiZE1 and flZEI be an orthogonal family of projections such that ez N fi for all i E I and let e sup ei f sup fi We say that the equivalences are addable if e N f Caution In the case of i to say that a family of partial isometries is addable 143 means something more precise the equivalence e 5 f is required to induce7 the given equivalences ez 5 fi in an appropriate sense Note When I is nite e N f comes free of charge from axiom D when ef 0 it comes free of charge from axiom F 184 LEMMA Let each of eiZE1 and flZEI be an orthogonal family of projections such that eiNfZ foralliEI For each i E I let JZ be an indea set and let 6ijjE i fijjE i be orthogonal decompositions of eifZ respectively such that eij NfZj for all iEI j E Ji 87 88 18 ADDITIVITY OF EQUIVALENCE Then the equivalences are addable if and only if the equivalences are addable Proof supeZj i E l j 6 J2 supiE1supeZj j E JZ supielei similarly sup fij sup fi 0 185 LEMMA 18 p 73 Proof of Th 50 With notations as in 183 if e j 1 7 f then the equivalences are addable that is e N f Proof Axioms C F Say e N f S 17f By axiom C there is an orthogonal decomposition f supfzf with ez N f for all i Then f N ez N fi since ff 0 it follows from axiom F that f N f Thus e N f N f Note Until now the weaker axiom C has suf ced this is the rst use of the full axiom C 186 LEMMA 18 p 73 Th 50 If A is properly in nite then equivalence is additive in A Proof Axioms A7F Let g be a projection with g N 1 N 17 g 1512 Adopt the notations of 183 From e 17 e 1 N 9 we have e N e S g for suitable e Similarly f N f S 1 7 g for suitable f It suf ces to show that e N j invoking axiom C we can suppose e S g and f S 17 9 Then ef 0 and an application of axiom F completes the proof 0 187 LEMMA 18 p 78 proof of Th 52 If A is continuous then equiva lence is additive in A Proof Axioms A7D F and GC note that these imply axiom E by 151 With notations as in 183 write ez e e with e N e 174 The equivalences ez N fi induce decompositions fi f fi with e N f and e N fi whence f N f Set e sup e e sup eg since ee 0 one has e N e axiom Similarly de ning f supfl f supflC one has ff 0 f N f Evidently e e e f f f so by axiom D it will suffice to show that e N f and e N f Let u be a central projection with ue j it and 17 uf j 17ue Then uejufuf u17fu7uf17w whence ue j 1 7 u axiom 0 since ue N ufi for all i it follows from 185 that ue N u Similarly 17ufj17ueN17ue 17u17e 1717ue whence 1 7 uf N 1 7 ue combined with ue N u this yields e N f Similarly e N f O 18 ADDITIVITY OF EQUIVALENCE 89 188 LEMMA If A is abelian then equivalence coincides with equality hence is triuially additive Proof Axioms A B If e N f then citing 916 one has e Ce Cf f 0 Scanning 1867188 we see from structure theory 925 827 that the remain ing case to be considered is type l n without abelian summand This proves to be the most complicated case First a general lemma 189 LEMMA 2 p 38 Prop 5 Let e1 en be any projections in the Baer rinq A Then there eacist orthogonal central projections u1 uT with sum 1 such that for each pair of indices i and V one has either uyei 0 or Cuyei uV Proof No axioms needed Let ul 17 supCeZ 1 S i S Then ulei 0 for all i Dropping down to 1 7 u1A we can therefore suppose that supCeZ Igign1 Let ul ur be orthogonal central projections with sum 1 such that each Cei is the sum of certain of the icy Eg let r 2 and for each n ple 5 51 en E 1 71 de ne Us C61EIC6252 CEn where Ceil Cei and Cei l 17 Cei then let u1ur be any enumeration of the us lf uVCeZ a 0 then ul must be one of the constituents7 of Cei so uVCeZ uV Thus either Cuyei 0 or Cuyei uV O 1810 LEMMA 18 p 84 proof of Th 54 If A is nite and of type 1 then equivalence is additive in A Proof Axioms AiD F and GC i In view of 827 and 188 we can suppose that A has no abelian summand ii Adopt the notations of 183 we are to show that e N f We can suppose the ez hence the fi to be nonzero iii We rst observe that every nonzero projection 9 contains a nonzero abelian projection Proof Let h be a faithful abelian projection 817 Then gAh a 0 321 therefore by axiom E see 151 there exist nonzero projections go 3 9 he 3 h with go N ho Since h is abelian and go j h go is abelian 161 By an obvious Zorn argument it follows that g is in fact the supremum of an orthogonal family of abelian projections In particular for each i one has ez supeZj jE Ji with the eij j 6 J1 an orthogonal family of abelian projections Then ez N f induces an orthogonal decomposition fisupfZj jEJi 90 18 ADDITIVITY OF EQUIVALENCE with eij N fij and therefore fij abelian 161 It suf ces to show that the equivalences eij N fij are addable 184 So changing notation we can suppose that the ez and fi are abelian iv Let uaaEK be an orthogonal family of nonzero central projections with sup ua 1 such that for each 04 the set lai61 uana O is nite 1619 Note that also la 61 uafi a 0 V For each 04 E K there is a nite central partition ua 121 UT of length depending on 04 such that for i 6 la and 1 S 1 S 7 either vVeZ 0 or Cvyei 1 189 Partitioning every ua in this way and replacing the ua by the family of all 07s obtained in this way we can suppose that for each 04 E K the set lai61 uana O is nite and Cuaei ua for all i 6 la vi It follows that for each 04 the nitely many projections uaei 6 la are pairwise equivalent 164 vii If for some 04 la 9 then uaei uafi 0 for all i whence uaeuaf0thus e3 17ua f3 17 Let K0 betheset ofallsuch 04 and let vinf17ua dEKO17supua dEKO Then e S v f S v and dropping down to 0A we can suppose that every la is nonempty viii Let nd cardla thus 1 S nd lt 00 Write each la as a disjoint union 1 1 u 1 u 1 with the following properties a if nd is even then 1301 each have nd2 elements and 1 9 b if nd is odd then 1301 each have no 712 elements and I has one element ix For each 04 de ne e supuan i6 1 e supuan i612 e supuan i612 the sups are actually nite sums and either e 0 or e uaei where I 1 thus e is abelian Since cardli cardli it is clear from vi that 18 ADDITIVITY OF EQUIVALENCE 91 e N e Moreover uae ua supeZ 1 61 supuan i6 1 323 supuan i6 la e e e Similarly de ning fsupuafi i612 for 046K1 t 3 we have naffi f f and f 16 Moreover axiom D eZNf for 046K1 t 3 In particular eiNeiNfiNf forall 046K X For t 123 notations as in ix de ne etsupe 046 K ftsupf 046 Clearly 616263 are pairwise orthogonal as are flf2f3 and esupuae 04 E K sup 6 6 62 sup 6 sup 6 sup 6 61 62 63 similarly f fl f2 f3 xi Since 6162 0 e supo 6 e supo e and e axiom F yields 61 N 62 and similarly f1 N f2 xii Let J di 04 E K i6 la For t 123 de ne 2 l 2 1 aNea forall 04 Jto i 04 6K i612 Since la is partitioned as la 1 U13 U12 it follows that J u u with J1J2J3 pairwise disjoint For t 123 one has supuan 041 6 Jt supuan 04 E K i612 sup supuan i6 1 sup 6 6t 0t 92 18 ADDITIVITY OF EQUIVALENCE and similarly ft supuafZ 041 6 Jt 11 3 where uaei N uafz for all 04 E J xiii Let 11 be a central projection such that uel j 11fl and 1 711 1 j 1 711el Then since f1 N f2 one has uel ufl qu 14171 17W so uel N 11fl by 185 Similarly 1711 l j1711el N 1711e2 S 171117el S 1717ue1 so 17 11 1 N 17 11e1 combining this with uel N 11fl we get e1 N f1 Then e2 N e1 N f1 N f2 so also e2 N f2 Since by x we have e el e2 e3 and f fl f2 f3 it will su ce to show that e3 N f3 xiv Note that e3 is abelian For e3 supe 04 E K where as noted in ix the e are abelian and e S 110 since the 11a are orthogonal e3 is abelian by the argument in the proof of 819 Similarly f3 is abelian For all 04 and 1 one has uaei N uafi therefore Cuaei Cuafi it follows that CeS Csupuan 041 6 J3 supCuan 041 6 J3 322 supCuafZ 041 6 J3 CfS therefore e3 N f3 by 164 We remark that f3 j 17 f3 16171 therefore e3 j 1 7 f3 this is important for the theory of addability of partial isometries cf 2 p 128 proof of Prop O 1811 Completion of proof of 181 Assuming A7D F and GC we have to show that equivalence is additive As observed in 151 axiom E also holds thus A7F and GC are in force The properly in nite case is covered by 186 so we can suppose A is nite 911 The continuous case is covered by 187 so we can suppose that A is type 1 and nite and the proof is nished off by 1810 0 1812 COROLLARY 18 p 82 Th 54 Let A be a Boer 1 1ng N on equ1valence Telat1on on 1ts project1on lattice satisfy1ng the 0311101715 A7D F and H of 10 Then A has GC and equ1valence 1s add1t1ve Proof GC holds by the theorem of Maeda and Holland 139 therefore equiv alence is additive by 181 and axiom E holds Since axiom G is a special case of additivity of equivalence we see that axioms A7H are in force 0 1Since A is nite an alternative proof is available 2 p 118 remark at the end of the proof of Prop 6 18 ADDITIVITY OF EQUIVALENCE 93 1813 COROLLARY 17 p 534 If A is any regular Baer ring then for g A has GC and equivalence is additive Proof Axioms AiF and H hold for g 113 quote 1812 0 1814 THEOREM 2 p 129 Th 1 Let A be a Baer ring i GC holds for i if and only if aaiom E holds for 5 and equivalence is additive ii If GC holds for i and if A has no abelian summand then partial isome tries are addable in A the sense of 143 Proof Axioms AiD and F hold for i in any Baer ring 112 so is immediate from 181 ii can be proved exactly as in 2 20 essentially by the foregoing arguments with minor supplementary remarks 0 1815 If A is a Baer ring such that 1 satis es axiom H in particular if A is any Baer ring satisfying the SR axiom then equivalence is additive in A Proof SR axiom H i GC by 1213 and 1310 quote 1814 1816 THEOREM 18 p 78 Th 52 Let A be a Baer ring N an equiva lence relation on its projection lattice satisfying the aaioms AiG of 10 Then A has GC and equivalence is additive Sketch of proof We recall that axiom G is central additivity of equivalence The properly in nite case is covered by 186 so we can suppose A is nite 911 hence a product of rings of type l n and type ll n 925 Since axioms B E and F hold we have at least orthogonal GC 133 Suppose A is of type I n Adopt the notations of 183 we are to show that e N f There exists an orthogonal family net of central projections with sup ua 1 such that for each 04 the set la E l uaei a 0 is nite For axioms A B C E F are su icient for the proof of 1619 18 p 69 Th 48 in particular one can get by with orthogonal GC By axiom D for every 04 supuaei i 61a N supuafi i 61a that is uae N uaf therefore e N f by axiom G There remains the case of type ll n messier than 187 we refer to Kaplansky7s original proof for the details 18 pp 79 80 0 1817 COROLLARY Let A be a Baer ring N an equivalence relation on its projection lattice satisfying the aaioms AiD and F of 10 cf 112 Then the following conditions are equivalent a A has GC b A satis es aaiom E and equivalence is additive c A satis es aaioms E and G in other words AiG Proof a gt b is 181 b i c is trivial c i b is 1816 0 94 18 ADDITIVITY OF EQUIVALENCE In particular the corollary characterizes the Baer rings satisfying GC for 3 cf 112 1818 Let A be a regular right self injective ring Ad the lattice of principal right ideals2 of A For JJ in write J N J if J J in ModA that is J and J are isomorphic as right A modules Then N is completely additive in in the following sense if each of JibEl KlZEI is an independent family in such that JZ N K for all 1 61 and if J VJi K in then J N K Proof Recall that A is a regular Baer ring 130 so is a complete lattice 122 consequently the indicated suprema exist in 7 p 162 Cor 135 If J is any right ideal of A then J is an essential submodule of J for one knows that J is an essential submodule of a principal right ideal K 7 p 95 Prop 91 e and from J C J C K K we see that J is essential in K but J is a principal right ideal A is a Baer ring hence is an injective right A module whence J K which proves the assertion Suppose now that J are as given in the statement above Write J0 ZJi K0 ZKi by independence these can be viewed as module direct sums J0 Ji K0 Ki therefore the given isomorphisms JZ K2 induce an isomorphism J0 K0 in ModA Now J J3 and K Kg 121 so by the above remark we know that J0 K0 are essential submodules of J K hence the isormorphism JO 2 K0 extends to an isomorphism J E K in ModA in other words J N K as claimed 1819 Let A be a directly nite regular left self injective ring J M Gour saud and L Jeremy have shown that A is right self injective if and only if the analogue of axiom F holds in the lattice Ad of principal right ideals of A with independence playing the role of orthogonality7 8 Th 32 in view of 1818 this says that for a ring A satisfying the initial assumptions complete additivity in is implied by its special case axiom F 2In the notation Ad d abbreviates dexter 19 DIMENSION FUNCTIONS IN FINITE RINGS Let A be a Baer ring N an equivalence relation on its projection lattice satisfying the axioms AiD F and GC of 10 relative to which A is nite 95 Note Axiom E also holds 151 and equivalence is additive 18 Thus Aacioms AiF and GC are in force and A is nite relative to N In particular the results of 15718 are available with these in hand the results to be noted in the present section may be proved verbatim as in 2 Ch ill Thus we shall simply state these results with appropriate references to 191 Let Z be the center of A Let us write PA and PZ for the projection lattices of A and Z In particular PZ is a complete Boolean algebra 33 38 with u v 7 1w as Boolean sum of uv E PZ by M H Stone7s theory it is isomorphic to the Boolean algebra of closed open subsets of a Stonian space X cf 139 One can identify X with the set of maximal ideals of the Boolean ring PZ and u E PZ with the closed open subset of X consisting of the maximal ideals that exclude Let CX be the algebra of continuous complex valued functions on X it is a commutative AW algebra 139 We shall view PZ C CX by identifying u E PZ with the characteristic function of the closed open subset of X to which it corresponds A few other elements of Z can be identi ed with elements of CX cf 2 p 157 but this is not needed for the statement of the following results The set of funtions CfXfeCX0fg1 the positive unit ball7 of CX is a complete lattice Caution The in nite lattice operations in general differ from the pointwise operations 192 THEOREM 2 p 181 Th 1 There eacists a unique function D D3 Du u for all central projections u D4 ef 0 i De f De Df 193 DEFINITION The function D of 192 is called the dimension func tion of A D2 De Z 0 for all e 1The case that N is i that is A is a nite Baer ring satisfying GC relative to i Of 112 95 96 19 DIMENSION FUNCTION 194 THEOREM 2 p 160 Prop 1 p 181 Ths 12 The dimension function has in addition the following properties D5 0 S De S 1 D6 Due uDe when u is a central projection D7 De0 gt e D8 ef7gt DeDf D9 ejf 7gt DltegtsDltfgt D10 If e2 is an increasingly directed family of projections with supre mum e briefly ez T e then De sup Dei in the lattice CfX 2 p 184 D11 If e2 is an orthogonal family of projections with supremum e then De Z Dei the supremum of the family of nite subsums D12 If A is of type II then the range of D is all of CfX 2 p 182 Th 3 195 Along the way one proves the following 2 p 182 Th 4 p 183 Exer 3 Let N be an in nite cardinal If every orthogonal family of nonzero central pro jections has cardinality S N then the same is true for orthogonal families of not necessarily central projections To put it more precisely If there exists an orthog onal family eiZE1 of nonzero projections with cardI gt N then there exists an orthogonal family nipE1 of nonzero central projections In particular if A is a factor7 0 1 the only central projections then every orthogonal family of nonzero projections is countable ie has cardinality S No This is clear for example from D5 and D11 196 One can recover from the properties of dimension the axioms listed at the beginning of the section that permitted its construction We illustrate with i GC ii axiom C and iii niteness i Given any pair of projections ef Consider the open subset of X on which the function De is lt Df form its closure which is closed open and let u be the corresponding central projection Then uDe S uDf and 17uDf S 17uDe citing D6 we have Due S Duf and D17uf S D17 ue so by D9 we have ue j uf and 17 uf j 17 ue ii Let eiZE1 be an orthogonal family with supremum e and let e N f we seek an orthogonal family flZEI with sup fi f and ez N fi for all i We can suppose I to be in nite and well ordered say I 04 04 lt D 9 minimal hence a limit ordinal The case of a nite index set is an obvious simpli cation of the following argument Suppose that 5 lt D and that fa has been de ned for all 04 lt 5 Let e supea 04 lt 5 f supfa 04 lt 5 Then by D11 De Df Clearly ee 0 therefore e S e 7 e so Del we 7 a 7 W 7 W 7 D0 7 W 7 D0 7 D0 7 130 7 gt by D9 D4 D1 whence e j f 7 say e N f S f 7 Thus by trans nite induction we can suppose de ned an orthogonal family f 3ltg with 19 DIMENSION FUNCTION 97 f S f and 6 N f for all 5 Then DU 7 sup f 7 130 7 Wm f5 7 130 7 ZDW 7 136 7 Z D6 7 0 whence f 7 sup f 0 iii Suppose e N 1 Then De D1 De 17 6 De D17 6 therefore D17 e 0 and so 17 e 0 197 The results of this section are valid for the relation 3 in any regular Baer ring Sketch of proof If A is a regular Baer ring then i satis es A7F 113 and GC 1311 by a theorem of Kaplansky 17 Th 1 A is directly nite this is the hard partl 198 The dimension theory on projection lattices sketched here has a parallel for the principal right ideal lattices of directly nite regular right self injective rings 7 Ch 11 20 CONTINUITY OF THE LATTICE OPERATIONS 201 DEFINITION 27 21 A continuous geometry is a lattice L that is complete complemented modular and satis es the following conditions for increasingly directed and decreasingly directed families 10 eiTe ei fTe f forall fEL 2Q eile eiUfleUf forall fEL 202 It is a theorem of I Kaplansky that every orthocomplemented complete modular lattice is a continuous geometry 17 The proof is long and di icult In the application to projection lattices some simpli cations have been achieved as we shall report here but some heavy artillery from lattice theory still has to be called in to get the full results The most di icult ring theoretical argument in 17 is the proof that a regular Baer ring is directly nite 17 Th 1 the following result of I Amemiya and I Halperin offers an alternative path via lattice theory intricate in its own way 203 LEMMA I Amemiya and I Halperin 34 p 516 Th Let L be an orthocomplemented countably complete modular lattice If en is an independent sequence of pairwise perspective elements of L then en 0 for all n 204 PROPOSITION cf 17 Th 1 If A is a Baer ring whose projection lattice is modular then A is directly nite Proof Suppose to the contrary that A is not directly nite then there exists a projection e E A with e a 1 and e g 1 717 By 111 and the proof of a i b of 157 there exists an orthogonal sequence of nonzero projections en such that e1 3 e2 3 the en are pairwise perspective by 520 which contradicts the lemma Remark The proof remains valid for A a Rickart ring whose projection lattice is modular and countably complete for example a Rickart C algebra whose projection lattice is modular O 205 For use in the proof of the main theorem of the section we review some concepts from lattice theory Let L be a lattice with 0 and 1 smallest and largest elements and suppose there exist ordered sets L1 and L2 such that L L1 gtlt L2 as ordered sets L1 gtlt L2 bears the product ordering Then each of L1 L2 is a lattice with 0 and 1 and if Lp L a L1 gtlt L2 is an order isomorphism then w0 00 and wlt1gt 11 98 20 LATTICE CONTINUITY 99 An element 2 E L is said to be central if there exists such an isomorphism Lp L a L1 gtlt L2 with Lpz 10 21 p 27 Def 32 lt is easy to see that 10 has 01 as its only complement in L1 gtlt L2 therefore 2 has a unique complement 2 in L namely 2 Lp 101 Write ZL for the set of all such 2 and call it the center or lattice center7 of L lf 2 E ZL then 2 E ZL let Lp be as above and compose it with the canonical isomorphism L1 gtlt L2 a L2 gtlt L1 consequently z 2 206 S Maeda 24 Th 62 Let A be a Rickart ring L PA its projection lattice 115 Z the center of the ring A and ZL the lattice center of L 205 For it E L the following conditions are equivalent a u E ZL b it has a unique complement in L c u E Z Thus ZPA PZ Proof a i b See 205 b gt c This is 39 c i a The map e gt gt ue 17ue is an order isomorphism L a PuA gtlt 1317 uA under which it gt gt u0 whence it E ZL by de nition 205 207 lf A is a Baer ring and e is a projection in A then its central cover Ce is the smallest projection u in the center of Z such that e S u 38 315 in view of 206 one sees that Ce coincides with the central hull of e7 de ned in 21 p 69 Def 42 208 THEOREM 18 p 117 Th 69 Let A be a Baer ring N an equiv alence relation on its projection lattice relative to which A is nite and satis es aaioms AiD F and H of 10 Then 1 The projection lattice of A is a continuous geometry 2 A is directly nite 3 e Nf gt ef are perspective Proof The projections of A form a complete lattice L 124 with an orthocomplementation e gt gt lie By the theorem of Maeda and Holland axiom E and GC also hold 139 and its proof therefore equivalence is additive in A 181 Since axiom G is a special case of additivity of equivalence we thus see that all of the aaioms AiH are in force Since moreover A is nite relative to N that is e N 1 i e 1 the results of 19 are available A has a dimension function D 1 Modularity Let efg be projections with g 2 e and write h906Uf k6Ug f we are to show that hk Obviously kgh From egkghgeof one sees that eUfkath and from g fgkghg euf oneseesthat g fk fh f Then citing axiom H one has hig fh7h fNhufifka7kaik fkig 100 20 LATTICE CONTINUITY thus h7g ka7g f addingthisto g fg faxiomDyields hNk which combined with k S h and niteness yields k h Continuity of the lattice operations in view of the lattice anti automorphism e gt gt 17e it suf ces to verify 10 of 201 Suppose ez T e and f is any projection setting 9 supez f f we are to show that g e f Clearly g S e f Citing axiom H one has e f7ei fe f7ei e fNe fUei7ei e7ei invoking the properties of the dimension function D 192 194 we therefore have De f7ei f De7eZ whence D62lt136D6mfD6 f3136D9D6 f for all i Since sup DeZ De it follows that 136 S D6 139 i D6 m f whence De f7g 30 Thus De f7g0 e f7g0 2 Immediate from 1 and 204 3 Let g be a common complement of e and f e Ug f Ug 1 e gf g0 Citingaxiom H ee7e gNeUg7g1g similarly f N 1 7 g whence e N f i By 1 the projection lattice L of A is a continuous geometry Given any pair ef in L there exists by the theory of continuous geometries an element u in the lattice center of L7that is 206 a projection u in the center of the ring A7such that u f e is perspective to some e S u f f and 1 7 u f f is perspective to some f S 17 u f e 21 p 87 Satz 11 cf 132 Thus by the implication proved in the preceding paragraph u f e N e S u f f and 17u fo S 17u e inother words ueNe guf and 17ufo 17ue Suppose in addition that e N f Then ue N uf axiom B so e uf by niteness then ue and uf e are perspective in L say with common complement g 6UQufUQ1gt 5 9uf g0 Multiplying through by u it follows cf 323 that ug is a common complement of ue and uf in the projection lattice of uA Similarly 17ue and 17uf 20 LATTICE CONTINUITY 101 have a common complement h in L hence they admit 1 7 uh as a common complement in the projection lattice of 1 7 uA it is then elementary that ug 17 uh is a common complement of e and f in L Thus ef are perspective in L O 209 in a regular ring A modularity of the projection lattice comes free of charge for e gt gt eA is an order isomorphism of the projection lattice onto the lattice of principal right ideals and the principal right ideals of any regular ring form a modular lattice 116 2010 COROLLARY l Kaplansky 17 Th 3 18 p 120 Th 71 Let A be a regular Baer ring i The relation 5 satis es aaioms A7H and niteness and the projection lattice of A is a continuous geometry ii e 5 f gt ef are perspective iii If N is an equivalence relation on the projection lattice of A satisfying aaioms A7D F H and niteness then N coincides with 5 thus withperspectivity Proof The relation 5 satis es A7F and H 113 Since the projection lattice of A is modular 209 A is directly nite by 204 alternatively cite 17 Th 1 that is A is nite relative to 5 71 717 it then follows from 208 that the projection lattice is a continuous geometry Axiom G indeed additivity of equivalence holds by the remarks at the beginning of the proof of 208 ii lmmediate from and 208 iii lmmediate from 208 and ii 0 Cf 213 2011 COROLLARY Let A be a nite 11 1 i aa 1 Baer ring such that e U f 7 f 5 e 7 e f f for all projections ef i The relation 5 satis es aaioms A7H and niteness and the projection lattice of A is a continuous geometry Moreover A is directly nite ii e 5 f gt ef are perspective iii If N is an equivalence relation on the projection lattice of A satisfying aaioms A7D F H and niteness then N coincides with 5 thus withperspectivity Proof For axioms A7D and F see 112 by hypothesis 5 also satis es axiom H and niteness 95 By 208 the projection lattice of A is a continuous geometry and A is directly nite moreover as noted in the proof of 208 axiom E and axiom G indeed additivity of equivalence also hold thus A7H hold ii lmmediate from and 208 iii lmmediate from 208 and ii 0 in particular if A is a Baer ring for which 5 satis es the parallelogram law then A is nite if and only if it is directly nite in view of 1213 this is a generalization of 613 2012 The proof of 208 shows that if N satis es axiom H and niteness and axiom D then the projection lattice is modular We now turn to a criterion for modularity that is independent of axiom H indeed no equivalence relation is postulated 102 20 LATTICE CONTINUITY 2013 LEMMA If ef are projections in a Rickart ring A such that e f f 0 thus e and f are complementary in e U fAe U f then the projection g f 1 7 e U f is a complement of e in A e U Q 1 e f g 0 Proof Since 9 f U 17 eU f it is clear that e U Q 1 Set 1617966Uffeefe17f Citing 115 twice we have e ge7LPae7eie feie0 O The following proposition was communicated to me by David Handelman who attributes it to Israel Halperin 2014 PROPOSITION The following conditions on a Rickart ring A are equivalent a the projection lattice of A is modular b if ef are projections in A such that ef are perspective and e S f then e f Proof a i b Suppose e S f and g is a common complement of e and f Citing modularity ffweUg6Uf g that is f leU0so fe b i a Let efg be projections with e S g and set hwi gkeUU gb the problem is to show that h k Evidently k S h As in the proof of 208 we see that thkaeUf 0 h fk ff g an tfff m eUf h l k f l f g 20 LATTICE CONTINUITY 103 then 71 O f 0 iii for citing ii hofhofof h fo f ng f gf0 Also ka6Uf iV for kw 6UfogUf6Ufongl eUlf gfl6Uf Now kghganSet Eh17euf Ek17euf Then E S h and it will suffice to show that E h in View offbjt will su ice to show that f is a common complement of h and k since k S h it is suf cient to show 7 k U f 1 v E m f 0 Vi Re V Citing iV at the appropriate step EUfkUl16UflUf kaUl1eUfl eUfUl1eUfl 1 Re Vi Let a f b h then 1 Oh 0 by iii so setting 0 b17an it follows from the lemma 2013 that a and c are complementary inparticular 1000 Notethat aneUf For anfUhZfUk eUf Z 1U b Therefore ch1eUfh thus 0a cf hltgt 21 EXTENDING THE INVOLUTION This section is an exposition of results contained in a paper of D Handelman 11 Handelman7s paper contains in addition a wealth of material on matrix rings over Baer rings 211 DEFINITION A ring A will be said to be extendible if its in volution is extendible to an involution of its maximal ring of right quotients For the general theory of rings of quotients which is due to Y Utumi I draw on the exposition of J Lambek 19 437 45 212 PROPOSITION 28 11 Let A be agtk eactendible Baer ring Q its maaimal ring of right quotients Then i Q is also the maaimal ring of left quotients of A ii Q is regular and self injective both right and left iii Q is unit regular hence directly nite hence A is directly nite iv The eatension of the involution of A to Q is unique V The involution of Q is proper vi Q is a regular Baer ring all of whose projections are in A vii A projection of A ie of Q is in the center of A if and only if it is in the center of Q viii The projection lattice of A ie of Q is a continuous geometry in particular it is modular ix For projections e f in A the following conditions are equivalent a e f are perspective in A b ef are perspective in Q c e 3 f in Q Proof The ring isomorphism a gt gt aquot of A onto the opposite ring A0 extends by hypothesis to a ring isomorphism Q a Q0 ii By 129 A is nonsingular right and left so Q is regular and right self injective 19 p 106 Prop 2 and its corollary since Q possesses an involution it is also left self injective or cite iii Immediate from ii and 7 p 105 Th 929 iv Given any two involutions of Q extending that of A their composition is a ring automorphism of Q leaving xed every element of A since Q is a ring of right quotients of A this automorphism must be the identity mapping cf 19 p 99 Prop 8 v Write a gt gt aquot for the involution of Q extending that of A iv For each a E Q the set I A a a E A aa 6 A is a dense right ideal AAAAAA 104 21 EXTENDING THE INVOLUTION 105 of A cf 19 p 96 Lemma 2 1f 11 0 then for all a E 1 one has 1a 6 A and 1a1a 1 11a 0 therefore 1a 0 the involution of A is proper by 110 thus 11 0 so 1 0 by the density of l 19 p 96 Prop 4 See also the proof of 218 below vi By ii and 130 Q is a regular Baer ring since its involution is proper V it is regular 114 hence is a regular Baer ring 125 To complete the proof of vi given any 1 E Q it will su ce to show that LP1 as computed in the Baer ring Q is in A the following argument is taken from 28 Th 33 Let 1 A 1 a E A 1a 6 A which is a dense right ideal of A and let e supLP1a a E l where LP1a is computed in A and the supremum is taken in the projection lattice of A it will su ce to show that 1l Q1 7 e where 1l denotes the left annihilator of 1 in Q cf 17 For all a E 1 one has e Z LP1a so e1a 1a 17 e1a 0 thus 17 e1l 0 therefore 17 e1 0 by the density of 1 This shows that Q1 7 e C On the other hand suppose y E 1l and let J b E A by E A in view of i J is a dense left ideal of A Let b E J for all a E l by1a by1a 0 thus byLP1a 0 for all a E l consequently bye 0 varying b E J we have Jye 0 so ye 0 by the density of J whence y y17 e E Q1 7 e This shows that 1l C Q1 7 e and completes the proof of vi it follows that if 1 E A then LP1 is unambiguous7it is the same whether calculated in A or in Q And of course the lattice operations are unambiguous vii Since A and Q have the same projection lattice vi this is immediate from 39 it is also easy to see from the theory of rings of quotients cf 2130 viii The projection lattice of Q is isomorphic to the lattice of principal right ideals of Q via e gt gt eQ hence is modular cf 116 the continuity of the lattice operations follows from self injectivity 7 p 162 Cor 135 So the requirements of De nition 201 are ful lled One could also cite 2010 but note that in the proof of 2010 one had to cite Kaplansky7s theorem or its generalization by Amemiya and Halperin to get direct niteness whereas in the present context we have an easier route to direct niteness via iii Caution We do not have here an easier proof of Kaplansky7s theorem indeed we do not have here an alternative proof of Kaplansky7s theorem at all ix The equivalence a gt b is trivial since the projection lattices are the same whereas b gt c by unit regularity 7 p 39 Cor 44 One could also cite 2010 for the equivalence of b and c but again this would entail Kaplansky7s theorem 0 213 PROPOSITION 13 Prop 3 Let A be a e1tendible Boer 7 ing Q its ma1imal ring of right quotients and view A as a subi ing of Q 212 Then the following conditions on A relative to 3 written brie y N are equiva lent a A satis es LP N RP 106 21 EXTENDING THE INVOLUTION A satis es aaiom H c A has GC eNf in Q i eNf in A eNf inQlt eNf inA e ef perspective in A i eNf in A e ef perspective in A gt eNf in A Proof a i b Trivial 115 b i d Suppose eNf in Q By ix of212 e and f are perspective in A therefore e N f in A by b see the proof of 208 d gt d The reverse implication in d is trivial d gt e The left sides of d and e are equivalent by ix of 212 e i e Trivial e i a Let a E A and write e LPa f RPa Since Q is regular e N f in Q 58 therefore ef are perspective in A by ix of 212 whence eNf in A by e So far we know that all conditions other than c are equivalent d i c One knows that GC holds for N in the regular Baer ring Q 1311 it is then immediate from d that GC holds for N in A recall that Q and A have the same projection lattices and the same central projections c i d Suppose e N f in Q By c there is a central projection n such that in A one has ueNe guf and 17ufo 317ue for suitable projections ef In Q we have quueNeguf so e uf by the direct niteness of Q thus ue N uf in A Similarly 17ueN17uf inAso eNf in Altgt 214 PROPOSITION 11 Prop 33 13 Let A Q be as in 212 The following conditions are equivalent 1 A satis es LP 5 RP for all a E Q LPa RP in A A satis es aaiom H for i ef perspective in A e 5 f in A ef perspective in A gt e i f in A 5efinQ eifinA 2 3 4 4 JTo put it more succinctly In a extendible Baer ring the conditions LP RP axiom H for i and GO for N are equivalent When these conditions are veri ed the relations in A or in Q it does not matter 5 g f e N f e and f perspective are all equivalent On the other hand let A be the ring of all 2 X 2 matrices over the eld of 3 elements With transpose as involution A is a regular Baer ring trivially extendible A Q in which 5 g f does not imply e i f 18 p 39 Of course axiom H fails for i in this example 2 p 75 Exer 1 21 EXTENDING THE INVOLUTION 107 6 A has GC for 3 Moreover these conditions imply each of the following two 7efinQ efinA 8efinA efinA Proof 1 i 2 Let 1 E Q e LP1 f RP Since Q is regular e g f in Q 58 But from 1 one has afortiori LP g RP in A hence the equivalent conditions of 213 are veri ed in particular condition d of 213yields e g f in A Say r EeAf s EfAe with rs e srf it isthen elementary that e LPr f RPr therefore e 5 f in A by That is LP1 i RP in A 2 i 3 immediate from 115 3 i 4 Same as in the proof of 208 4 gt 4 If e i f in A then ofortiori e 3 f in Q hence ef are perspective in A by ix of 212 Thus the reverse implication in 4 always holds so 4 and 4 say the same thing 4 i 5 Suppose e g f in Q Then ef are perspective in A by ix of 212 so e 5 f in A by 5 i 1 Let r E A e LPr f RPr Then e 3 f in Q by regularity 58so e i f in A by Thus all conditions from 1 through 5 are equivalent 3 i 6 This holds for example by the theorem of Maeda and Holland 1310 Alternatively one could show 5 i 6 using 1311 Or show 5 i 6 using general comparability for regular right self injective rings 7 p 102 Cor 915 6 i 5 Let e 3 f in Q By 6 there exists acentral projection u with ue 5 f guf in A and 1710f eg 17ue in A Afortiori 6 g fSuf in Q but ue g uf in Q so f uf by direct niteness of Q 212 Thus ue uf in A similarly 1 iue i 1710f in A and addition yields e i f in A Summarizing all conditions from 1 to 6 are equivalent The implications 5 i 7 and 5 i 8 are trivial whence the last statement of the proposition 0 215 DEFINITION D Handelman 11 p 7 A ring A is said to be strongly modular if for 1 E A 1 0 i mA is an essential right ideal of A Handelman proved 11 p 12 and Th 23 A Boer ring is eactendible if and only if it is strongly modular We give here an exposition of his proof culminating in 2122 below Observe that in the applications of strong modularity in the results 108 21 EXTENDING THE INVOLUTION leading up to 2122 the implication my 0 i aA essential is cited only for self adjoint a this remark is exploited in 2123 and 2124 216 A strongly modular Rickart ring is directly nite Proof Suppose ya 1 and let e my which is idempotent we are to show that e 1 One has at0 yat0 it0 so my 0 by the hypothesis aA is an essential right ideal But a a1 aya ea so aA C eA therefore eA is also an essential right ideal whence eA A e 1 217 A Rickart ring A is strongly modular if and only if RPa 1 i aA essential in such a ring RPa 1 i LPa 1 Proof 1 17RPaA thus RPa 1 gt my 0 whence the rst assertion On the other hand aA C LPaA so aA essential LPa 1 1 Cf 2126 218 LEMMA Let A be a right nonsingular ring Q its masvimal ring of right quotients a E Q Then as right A modules aA is essential in aQ Proof The assertion is that aAA Ce aQA At any rate one knows that AA Ce QA 19 p 99 proof of Prop 8 Moreover if y E Q and J is an essential right ideal of A with yJ 0 then y 0 For since A is right nonsingular Q is regular and right self injective 19 p 106 Prop 2 and its corollary moreover if 1A is the injective hull of AA then the singular submodule of 1A is 0 But 1A 2 QA 19 p 95 Prop 3 so the singular submodule of QA is also 0 This means that if y E Q and y a 0 then the right ideal 0 y a E A ya 0 of A cannot be essential Now let N be an A submodule of aQA with N f aA 0 we must show N0 Let yEN Since N CaQ 3412 for some zEQ Since AA Ce QA the right ideal I A z a E A 2a 6 A is essential and 21 C A so 3411le 1A But also yl CyACN thus yl CN aA0 3410 By the preceding paragraph y 0 Thus N 0 O 219 LEMMA 11 p 12 If A is a eactendible Baer ring then A is strongly modular Proof Let a E A with RPa 1 we are to show that aA is an essential right ideal of A 217 Let 1 be a right ideal of A with If aA 0 we must show that l 0 1t suf ces to consider the case that l yA for some y E A Then aA yA 0 with RPa 1 and we are to show that y 0 Let Q be the maximal ring of right quotients of A cf 212 Then LPa g RPa 1 in the regular ring Q 58 so LP 1 by direct niteness Now A is right nonsingular 129 so 218 applies aA Ce 1Q and yA Ce yQ as right A modules By elementary module theory aA yA Ce aQ yQ thus 1For a regular ring A strong modularity means A1 A 1A A that is I left invertible I rightinvertible Thus a regular ring known to be niteisee the footnote to 95 is strongly modular if and only if it is directly nite EXTENDING THE lNVOLUTlON 0 Ce 1Q yQ whence 1Q yQ 0 Then by regularity 113 116 0 1Q 0 yQ LPQ0LP11Q Q 0 LPyQ LPyQgt so LPy 0 340 0 To prove the converse we now develop the consequences of strong modularity 2110 LEMMA 11 pp 7 8 In a Rickart ring A if ef are projections with e f0 then RPefeUf Proof Let kRPef Obviously efeUf efso kSeUf But e fT 17 kA so 0 e f17 k and citing 118 we have e17k7f17kEeA fAe fA0 thus e17kf17k0 egk and fgkwhence eUfgkltgt 2111 LEMMA 11 Prop 22 If A is a strongly modular Rickart ring 215 then the projection lattice of A is modular Proof Assuming e f perspective with e S f it will su ce by 2014 to show that ef Let g beacommon complement of e and f eUg ng 1 e gf g0 Since e g 0 by 2110 we have RPeg eUg 1 thus e gT 0 by strong modularity e gA is an essential right ideal So to show that f 7 e 0 it will suf ce to show that f76A0egA0 Suppose t f7ea egb In particular t6 f7eA C fA also eA C fA so t7ebngfA gAf gA0 whence t7ebgb0 Then eb t E eA f7eA0 whence t0 thus is veri ed O 2112 11 p 9 If A is a strongly modular Baer ring then its projection lattice is a continuous geometry Proof In view of 2111 the projections of A form an orthocomplemented complete modular lattice such a lattice must be a continuous geometry by Kaplan sky7s theorem 17 Th 3 We remark that if A satis es axiom H for i then since A is nite even directly nite by 216 a proof of continuity avoiding Kaplansky7s theorem is available via the proof of 2011 2113 The next target 2121 is Handelman7s theorem that if A is a strongly modular Baer ring and l is any right ideal of A then 1 is essential in llT as right A modules The strategy is to prove it rst for a principal right ideal then for a nitely generated right ideal and then infer the general case from the nitely generated case by an application of the continuity of the lattice operations 2112 2114 LEMMA 11 Prop 22 a If A is a strongly modular Rickart ring then for every 1 E A dA is essential in dAlT as right A modules 110 21 EXTENDING THE INVOLUTION Proof Let a E A Set e LPa LPaa Then aAl 1P A1 7 e so aAlT eA thus the problem is to show that aA is essential in eA Set 2 a 1 7 e which is self adjoint claim 1 2T 0 Write 2T 9A 9 a projection Then 0 zg a g 17 eg so aag717egEaA 17eACeA 17eA0 thus a g 17eg 0 Thus 9 eg ge and gaaquot 0 since e LPaa one has ge0 thus gge0 Since A is strongly rnodular we infer from claim 1 that 2A is an essential right ideal of A To show that aA is essential in eA let J be a right ideal of A with J C eA and J aA 0 we are to show that J 0 Let y E J then yA C J so yA aA 0 We must infer that y 0 since 2A is essential it will suffice to establish the following claim 2 zA yA 0 Since 2A 11 17 eA C aA 17 eA it is enough to show that aA17eA yA0 Suppose taa17ebyc with abcEA Then aa7yc717ebEeA 17eA0 so aa7yc17eb0 Then taa17ebaachaA yA0 so t0ltgt 2115 LEMMA If A is a strongly modular Rickart ring then for every pair of projections e1e2 in A elA 62A is essential in el U e2A Proof Write ee1 U62 fel e2 Then f Sei so eiAlteZ7fAeefA 712 1 Note that 61A62A 61 ifA62A For the inclusion D is obvious and the inclusion C follows from e1 e1 7ff and f E egA nally 61 7 62A C 61A 62A 61 62A whereas e1 7 fA C 17 fA whence e1 7 fA 62A 0 From 1 and 2 we have 61A62A 61 ifA62 EXTENDING THE lNVOLUTlON Next we observe that the self adjoint elernent Z1661 f 2ff has right annihilator 0 hence 2A is an essential right ideal by strong modularity For write 2T 9A 9 a projection From 29 0 it follows that 169617f9627fgfg since the right side of is in 6A both sides must be 0 and citing directness of the decomposition 3 we infer 1 6Q61f962f9f90 It follows that 619 fg 0 egg fg 0 Therefore gel 962 0 whence g 6 61A 62Al 61A 62AM 61 U 62Al 6Al A17 6 thus 9 176 But 17eg 0 so 9 S e consequently g 0 thus 2 0 as claimed To prove 61A 62A essential in 6A it will suf ce to show that if 1 6 6A with A 0 61A 62A 0 then 1 0 We rst note that A 0 61A 62A17 6A 0 4 For suppose 1a 61f 620 17 ed Then 1a761b620 17edEeA 17eA0 whence 1a 61b620 E A 61A62A 0 thus 1a 0 proving Now the element 2 de ned earlier obviously belongs to 1 6A61 fA62 fAfA 1 6A61A62Agt therefore A 2A 0 by 4 since 2A is essential A 0 O 2116 In a nonsingular rnodule right say over a ring A if LM are subrnodules with L essential in M brie y L Ce M then L N Ce M N for every subrnodule N Proof Let us rst show that if y E M 2 E N with y z a 0 then yzA LN 0 For let I Ly aEA yaE L since L Ce M l is an essential right ideal of A Since y 2 a 0 yzl a 0 by nonsingularity Choose a 61 with yza a 0 Then ya 6 L and 0 a yza yaza E LN whence yzA LN 3amp0 112 21 EXTENDING THE INVOLUTION Now let K be a nonzero submodule of M N we are to show that K 0 LN 0 Choose a E K 7 0 say dyz y EM ZEN Then y 2A f L N a 0 by the preceding paragraph whence the assertion It follows that if L Ce M2 for i 1n then L1 Ln Ce M1 Mn for example L1 L2 Ce M1 L2 Ce M1 M2 by two applications of the preceding assertion 2117 LEMMA If A is a strongly modular Rickart ring then I Ce 1 for every right ideal I that is generated by two elements Proof Say I dlA d2A Let ez LPlti thus dzA eiA for i 12 and Il 1Al 2Al 7 61 7 62 A1761 1762A1761U62 so I e1 Ue2A By 2114 ZA is essential in eiA therefore I is essential in elA 62A by the remark preceding the lemma the requisite nonsingularity being provided by 129 We thus have 1C8 61A 62A whereas 61A 62A C8 61 U 62A by 2115 therefore I Ce e1 U e2A 1 O 2118 PROPOSITION 11 Lemma 25 If A is a strongly modular Rickart ring then for every nitely generated right ideal I of A I is essential in 1 Proof Say I dlA dnA the proof is by induction on n For n 1 quote 2114 for n 2 2117 Suppose n 2 3 and assume inductively that all7s well with n 7 1 Writing J 11A dnL1A we then have I J dnA with J Ce J and dnA Ce dnAlT whence 2116 1 c J dnAlT i Let ez LPlt1i by the argument in 2117 I e1 U UenA J e1 U U enL1A dnAlT enA in particular J dnAlT e1 U U enL1A enA ii By 2115 e1 U U enL1A enA Ce e1 U U e714 U enA 1 iii Combining i ii iii we get I Ce I O 2119 For submodules of a module suppose L Ce L and M Ce M then L MCe L M and in particular L M0 L M 21 EXTENDING THE INVOLUTION 113 Proof Let K be a submodule of L f M with K 0 L f M 0 Then K L M 0where K L C K C M therefore K L 0 because M Ce M Thus K 0 L 0 where K C L therefore K 0 because L Ce L 2120 LEMMA In a strongly modular Rickart ring if 1 and J are nitely generated right ideals such that 1 0 J 0 then 1 0 J 0 Proof 1 C81 and J Ce J by 2118 quote 2119 0 2121 THEOREM D Handelman 11 Lemma 26 I A is a strongly modular Baer ring then 1 Ce 1 for every right ideal I and J Ce J for every left ideal J Proof Write l Uli with an increasingly directed family of nitely generated right ideals eg consider the set of all nitely generated right ideals C l ordered by inclusion Then I l so I 1n the lattice ofright annihilators 121 Var UW lirrlt 1gtrltU1gtr1r brie y l Vl r Let K be a right ideal with K C l and K 01 0 we are to show that K 0 We can suppose K nitely generated even principal From K 01 0 we have K 0 ll 0 for all i therefore K 01 0 by 2120 Citing continuity of the lattice operations 2112 Klr Ilr Klr A Ilr Klr A ltK A I lt0gt 0 brie y Klf l 0 But K C 1 yields K C 1 l so K Klf l 0 whence K 0 Thus 1 C61 lf J is a left ideal of A then Jquot is a right ideal so by the preceding J Ce Jl7 JTT Jrl gt thus J Ce Jfl as right A modules whence J Ce J as left A modules Alternatively one can show by similar arguments that the ring A0 opposite A is also strongly modular 0 Note that the appeal to Kaplansky7s theorem via 2112 can be avoided if one has an alternate proof of the continuity axioms cf 208 2122 COROLLARY 11 p 12 A Baer ring is eatendible 211 if and only if it is strongly modular 215 Proof Only if This is 219 If Let A be a strongly modular Baer ring and suppose I is a right ideal with ll 0 Then A l so 1 is an essential right ideal of A by 2121 Brie y for right ideals 1 ll 0 i l essential for a nonsingular ring this condition is equivalent to extendibility by a theorem of Y Utumi cf 28 Th 32 O 114 21 EXTENDING THE INVOLUTION 2123 PROPOSITION For a Rickart ring A the following conditions are equivalent a 2 2 2T 0 i 2A essential b 1 Ce 1 for all nitely generated right ideals l c 2A Ce 2A for all 2 E A with 2 2 d dA Ce dA for all a E A e dl 0 i dA essential that is LPd 1 i dA essential Proof a i b Inspecting the proof of 2118 and its preliminaries especially 2111 2114 2115 we see that the condition a is all that is needed the full force of 215 is not used b c Trivial c d Let a E A Then dl 11l because the involution is proper thus dAlT 11A Since dd is self adjoint ddA Ce ddAlT by the hypothesis c thus 11A Ce dAlT but ddA C dA C dA therefore also dA Ce dA d i e If 1P 0 then dA A so dA is an essential right ideal by e i a For 2 2 2T 0 if and only if 2l 0 O i i 2124 COROLLARY For a Baer ring A the following conditions are equivalent 1 A is strongly modular ie 1V 0 i dA essential 2 1 Ce 1 for all right ideals l 3 dA Ce dA for all a E A ie dA Ce LPdA for all a E A 4 dl 0 i dA essential ie LPd 1 i dA essential Proof 1 i 2 This is 2121 2 i 3 i 4 Trivial 4 i 2 Condition 4 is just e of 2123 so we know that a also holds lnspecting the proof of 2121 and its prelirninaries especially 2111 2114 2115 2118 we see that condition a is su icient to establish 2 i 1 By the proof of 2122 A is extendible therefore it is strongly rnodular 219 0 It is tempting to replace the de nition of strongly rnodular by condition 4 but then the easy proof of direct niteness 216 would not work Another de nition one might conternplate 1V 1P 0 i dA and Ad essential2 2125 A ring A is said to have su iciently many projections if for every nonzero element a E A there exists y E A with 1y a nonzero projection It is the same to say that every nonzero right or left ideal contains a nonzero projection For example a ring satisfying the EP axiorn 10 has su iciently rnany projections 2Suppose A has the stated property Then I 0 i 11 I 0 11T 0 so 11A is essential therefore so is 1A thus I 0 1A essential so A is strongly modular by 2124 21 EXTENDING THE INVOLUTION 115 2126 PROPOSITION 11 Prop 210 If A is a Rickart ring with su i ciently many projections then the following conditions are equivalent a A is strongly modular ie RPd 1 i dA is essential b for a E A RPd 1 i LPd 1 Proof a i b See 217 b i a Let a E A with RPd 1 we are to show that dA is essential By hypothesis LPd 1 Suppose to the contrary that there exists a nonzero right ideal J with J 0 dA 0 Since A has suf ciently many projections we can suppose J eA e a projection Thus eA dA 0 Note that d7edT 0 For if d7edy 0 then my edy E dA eA 0 so my 0 since RPd 1 it follows that y 0 Thus RPd7ed 1 by the hypothesis b we have 1 LPd 7 ed LP17 ed S 17 e whence e 0 J eA 0 a contradiction 0 Note that the version of 215 limited to self adjoint a is not suf cient for the proof of 2126 2127 COROLLARY cf 28 Th 52 10 Th 2 11 Prop 210 If A is a nite Rickart ring with su iciently many projections satisfying LP 3 RP cf 1431 1432 then A is strongly modular hence directly nite by 216 Proof If RPd 1 then LPd RPd 1 so LPd 1 by niteness quote 2126 lncidentally it is elementary that a nite Rickart gtk ring3 satisfying LP 5 RP is directly nite 2 p 210 Prop 1 the proof is an easy variation on 216 0 2128 COROLLARY If A is a directly nite Rickart ring with su iciently many projections satisfying LP 3 RP then A is strongly modular Proof Formally the same as 2127 0 Indeed since 2126 makes no reference to a speci c equivalence relation we have the following If A is a Rickart ring with suf ciently many projections and N is a relation on its projection lattice relative to which A is nite e N 1 i e 1 and LP N RP then A is strongly modular hence directly niteI by 216 2129 COROLLARY Let A be a Baer ring satisfying the EP aaciom 10 such that relative to 5 A is nite and satis es CC Then A is strongly modular hence directly nite Proof From EP we know that A has suf ciently many projections by 1431 LP 3 RP and A is nite by hypothesis Quote 2127 0 2130 Let A be a right nonsingular ring Q its maximal ring of right quotients u a central idempotent of A Then it is central in Q and uQ is the maximal ring of right quotients of uA Proof Let a E Q let us show that ud 7 an 0 Let 1 A a a E A da 6 A as noted in the proof of 218 1 is an essential right ideal of A For all a E 1 since da 6 A and u is in the center of A we have um 7 dua uda 7 dua dau 7 dau 0 3For example a regular ring see the footnote to 95 116 21 EXTENDING THE INVOLUTION brie y ua 7 auI 0 since A is right nonsingular it follows that ua 7 an 0 proof of218 Thus it is indeed central in Q Since Q is aring of right quotients of A 19 p 99 Prop 8 it follows easily that uQ is a ring of right quotients of uA and Q uQ X 1 7 uQ shows that uQ is also a right self injective ring consequently it is the maximal ring of right quotients of the right nonsingular ring uA 2131 PROPOSITION Let A be a Baer ring ua an orthogonal family of central projections in A with sup ua 1 In order that A be eactendible it is necessary and su icient that uaA be eactendible for every 04 Proof Let Q be the maximal ring of right quotients of A From 2130 we know that ua is an orthogonal family of central idempotents in Q and that an is the maximal ring of right quotients of uaA The following assertion shows that sup ua 1 in the complete Boolean algebra B of central idempotents of Q cf 33 131 claim 1 If a E Q and uaa0 for all 04 then a0 For let I A a a E A 1a 6 A which is an essential right ideal of A proof of218 Let a E I for all 04 we have uaaa uaaa 0 a 0 therefore 1a 0 because 1a 6 A and sup ua 1 in A Thus aI 0 whence a 0 because I is essential proof of 218 A similar argument shows that if ei is any family of projections in A and e sup ez in A then for a E Q one has ea 0 if and only if eza 0 for all i In particular if is any family of central projections in A and v sup Ui in A then i is also the supremum of the Ui in the complete Boolean algebra B claim 2 1 gt7 uaa is a ring isomorphism Q 7 Han For in view of claim 1 it is a monomorphism of rings Since Q is a regular right self injective ring 131 and since as noted above sup ua 1 in the complete Boolean algebra B claim 2 follows from 7 p 99 Prop 910 Write Lp Q 7 1 1an for the isomorphism Lpa Sufficiency If the uaA are all extendible and if 0a an 7 an is the involution of an extending that of uaA cf 212 then 1 gt7 Lp loauaa de nes an involution of Q extending that of A Necessity The proof is equally straightforward utilizing the following ob servation the ua E A are self adjoint so an involution of Q extending that of A must leave each of the an invariant as sets hence induces an involution of an extending that of uaA O The signi cance of this proposition is that extendibility for a Baer ring A can be reduced via structure theory to proving it for rings of various special types this is illustrated in 2136 below The following proposition can be omitted but in view of 2122 it provides an interesting alternative proof for 2131 2132 Let A be a Baer ring B its complete Boolean algebra of central idempotents 33 ua an orthogonal family in B with sup ua 1 In order 21 EXTENDING THE INVOLUTION 117 that A be strongly modular it is necessary and suf cient that every uaA be strongly modular Proof Suf ciency Let 1 E A with 1T 0 For each 04 it is clear that the right annihilator of ua1 in uaA is also 0 so by hypothesis ua1uaA ua1A is an essential right ideal of uaA Let J be a right ideal of A with J 0 1A 0 we are to show that J 0 For all 04 uaJ is a right ideal of uaA with uaJ ua1ACJ 1A0 whence uaJ 0 Thus for y E J one has nay 0 for all 04 therefore y 0 34 that is J 0 Necessity Assuming A strongly modular and it any central idempotent let us show that uA is strongly modular Let 1 E uA with 0 right annihilator in uA we are to show that 1 uA 1A is an essential right ideal of uA Let y117u lf 26A and yz0then 0yz1z17uzso 12717uzEuA 17uA0 whence 12 17 uz 0 But 0 12 yields uz 0 by the hypothesis on 1 whence 0 1 iuz ziuz 2 Thus yf 0 in A Since A is strongly modular yA is an essential right ideal of A Suppose now J is a right ideal of uA with J 01A 0 J is also a right ideal of A and J yA 0 for if tEJ yAuJ 117uA say if 1 1 7 ua then left multiplication by it yields t ut u1a 1a 6 J 0 1A 0 Since yA is essential J 0 2133 LEMMA 11 p 12 Let A be a e1tendible Baer rinq cf 2122 Q its ma1imal ring of right quotients 212 and n 2 1 an integer Let Mn A MnQ be the rings of n X n matrices i MnQ is the ma1imal ring of right and left quotients of Mn A ii MnQ is unit regular hence directly nite therefore MnA is directly nite iii MnQ is self injective right and left iV Mn A is nonsinqular right and left V each involution of Mn A is uniquely e1tendible to an involution of MnQ Vi If MnA is a Baer rinq relative to some involution not necessarily the natural involution of transposition that it inherits from A then Mn A is a e1tendible Baer rinq hence MnA is a strongly modular ring by 2122 Proof Write B MnA for the ring of n X n matrices over A i By a general theorem of Y Utumi MnQ is the maximal ring of right quotients of B 32 p 5 19 p 101 Exer But Q is also the maximal ring of left quotients of A 212 therefore MnQ is the maximal ring of left quotients of B 118 21 EXTENDING THE INVOLUTION ii Since Q is unit regular 212 so is MnQ 7 p 38 Cor 47 therefore MnQ is directly nite 7 p 50 Prop 52 hence so is its subring B iii iv MnQ is the maximal ring of right quotients of B and is regular therefore B is right nonsingular 19 p 106 Prop 2 and MnQ is right self injective 19 p 107 Cor of Prop 2 Similarly for left V Let be any involution of B Since MnQ is both the left and right maximal ring of quotients of B by i it follows that is uniquely extendible to an involution of MnQ the proof is written out in 28 Th 32 vi Suppose is an involution on B that makes it a Baer ring then extends to its maximal right quotient ring MnQ by v thus B is a extendible Baer ring 211 hence is strongly modular by 2122 Note in cidentally that if e E MnQ is a projection relative to the extension of to MnQ then e E MnA by 212 0 1n the reverse direction we remark that if A is a extendible Baer ring then so is every corner of A Proof Let Q be the maximal ring of right quotients of A and view A as a subring of Q 212 Let e E A be a projection Since A is semiprime 320 and nonsingular 129 eQe is the maximal ring of right quotients of eAe 33 p 135 Prop 02 the proof is completed by the observation that eAe is a subring of eQe 2134 THEOREM D Handelman 11 Prop 29 Let A be a Baer ring in which RPa 1 i LPa 1 in other words my 0 i my 0 and suppose that for some integer n 2 2 A is isomorphic as a ring to the ring of n X n matrices over some ring ie as a ring A has n X 71 matrix units for some n 2 2 Then A is eatendible cf 2122 Proof Let e1 en be orthogonal idempotents in A with 1 el en and e1 g e2 3 g en Writing B elAel we have A MnB as rings On occasion we identify A MnB but we note that while A is a Baer ring for its given involution B is merely a Baer ring 22 thus the involution of A need not arise from an involution of B via transposition Let fLPe1 then elAfA 56so e1 3 f indeed e1f are similar by 55 hence there exists a ring isomorphism Lp elAel a fAf 54 We know that fAf is a Baer ring for the involution of A 26 Let be the unique involution on B elAel that makes Lp a isomorphism then B is a Baer ring but the involution on A MnB that it induces by transposition need not coincide with the given involution on A Let us show that B is astrongly modular ring Suppose b E B with by 0 where the r means right annihilator in B we are to show that bB is an essential right ideal of B Thus if c E B with CB 0 EB 0 it is to be shown that c 0 Since B is a Baer ring there exists an idempotent e E B thus e 3 e1 as idempotents such that eB CT the right annihilator of c in B 21 EXTENDING THE INVOLUTION 119 Let 1 E A MAB be the element given by the matrix b c 0 e 0 61 P 0 61 0 lt0 I 0 0 g 61 Where P Z 6 M2B and I E Mn2B is the identity matrix If n 2 the following argument applies With appropriate simpli cations We assert that my 0 here 7 means right annihilator in A For suppose y E A With 1340 say Q S y ltR T 7 Where Q t E M2B and RST are matrices over B ofthe appropriate sizes Then b7 ct b3 cu PQ PS 0 13 lt R T gt et at R F PS Where R0T0 1380 and brctbscueteu0 Then br7ctEbB cB0 and bs7cuEbB cB0so brctbs cu0 Thus rs br0so rs0and tu creBso tet and ueu But eteu0 by so tu0 Thus Q0 andso 7 os y 00gt Where 1380 Say S sijwhere i 12 and 139 37172 Thus 0pS lt5 C 811 812 sl n2gt 0 5 821 822 51714 i bsll 0821 1812 0522 i 6521 6822 Then for j1n72one has b31j032j0 bslj 705 6 EB cB 0 120 21 EXTENDING THE INVOLUTION so bslj csgj 0 thus slj E by 0 so that the top row of S has all entries 0 Also sgj E CT eB shows that sgj esgj but shows that esgj 0 thus sgj 0 and we conclude that S 0 consequently y 0 Thus 1 0 therefore my 0 by the hypothesis on A Let 2 E A MnB be the element Then 0 0 0 elie 0 0 3310 b 0 e 010 0 1 010 gt thus 2 0 z 6 my 0 z 0 e1 e therefore CT eB elB B whence c 0 This completes the proof that B is a strongly modular ring Therefore the ring fAf isomorphic to B is also strongly modular it follows that the Baer ring fAf with the involution of A is extendible 2122 Now we have ring isomorphisms MnfAf Mne1Ae1 MnB A whence an isomorphism of rings 1 MnfAf a A viewing A as equipped with its original involution let be the involution on Mn fAf that makes 1 a isomorphism Then Mn fAf is also a Baer ring isomorphic to A via iJ since fAf with its original involution is a extendible Baer ring it follows from vi of 2133 that MnfAf is a extendible Baer ring therefore via LJ so is A O 2135 in the wake of 2134 whose proof requires the use of both halves of the equivalence 2122 there is no point in using both of the terms gtk extendible and strongly modular in the context of Baer rings The term strongly modular is attractive because the de nition applies also to non involutive rings and is intrinsic to the ring ie does not require reference to rings of quotients but gtk extendible tells us what we want to know in the Baer ring setting cf 212 214 1 vote for gtk extendible the following results are stated accordingly 2136 COROLLARY Let A be a Baei 7 ing without abelian summand such that RPa 1 i LPa 1 Assume that ordinary equivalence 3 satis es aaiom F and GC cf 10 Then A is eatendible cf 2122 21 EXTENDING THE INVOLUTION 121 Proof Axioms AiD hold by 111 moreover axiom E holds by 151 thus AiF and GC are in force Note that A is directly nite ie nite for g For suppose yr 1 Then 1t 0 i ymt 0 i t 0 so my 0 therefore 1l 0 by the hypothesis but 171mm 17mm 1711 0 so 1711 0 We can write A A1 gtltA2 with A1 of type I n and A2 of type ll n cf 826 or 925 so the two cases can be considered separately case 1 A is of type I n By 1613 there exists an orthogonal family ua of central projections such that every uaA is homogeneous relative to 3 in view of 2131 we are reduced to the case that A is homogeneous say of order N By 157 N is nite this is where axiom F is neededI and N gt 1 because A has no abelian summand Then A is isomorphic as a ring to an N X N matrix ring cf 2 p 98 Prop 1 Thus A is extendible by 2134 case 2 A is of type ll n Since axioms AiF hold 174 applies A is isomorphic as a ring to a 2 X 2 matrix ring Therefore A is extendible by 2134 0 2137 COROLLARY Let A be a directly nite Boer ring without abelian summond and suppose 3 satis es oriom F and LP g RP Then A is eactendible Proof Axioms AiD hold by 111 from LP 3 RP we infer axiom H cf 115 therefore E and GC hold 139 and its proof So axioms AiF and GC are in force Moreover if RP 1 then LP1 g RP 1 so LP1 1 by direct niteness Thus the hypotheses of 2136 are ful lled O 2138 COROLLARY Let A be a Boer ring without abelian summond satisfying GC for equivolence in which RP 1 i LP1 1 Then A is eactendible Proof As noted in the proof of 2136 A is directly nite hence nite ie nite for i By 112 3 satis es axioms AiD and F by 151 it satis es E The proof of 2136 may now be taken over word for word with i replaced by 5 O 2139 Abelian rings are truly exceptional in this circle of ideas 213672138 For example let A be a ring with no divisors of 0 thus A is a Baer ring whose only projections are 0 and 1 hence it is abelian and suppose A is extendible By 212 the maximal ring of right quotients Q of A is a regular Baer ring whose only projections are 0 and 1 thus Q is a division ring 145 it follows that if ab are nonzero elements of A then 1A0 bA 74 0 indeed as noted in the proof of 219 aA bA is an essential right A submodule of IQ bQ Q Q Q Thus A satis es the right Ore condition and because of the involution also the left Ore condition However there exist rings A without divisors of 0 satisfying neither of the Ore conditions which explains why abelian summands were banned from the preceding corollaries For example 35 p 436 Exer 8 let P be the free group with two generators 11 written multiplicatively and let A ZF be the group algebra of F over the ring of integers equipped with the involution a gt gt 1 for 122 21 EXTENDING THE INVOLUTION which aquot a and y y therefore 13quot ya etc For such a ring A Q is a factor of type III by a theorem of J E Roos 31 Prop 1 33 Cor 23 Incidentally the projection lattice of such a ring A7 namely 0 17 is trivially modular and axiom H is trivially veri ed for every equivalence relation N there are only ton thus these conditions are not su icient to assure extendibility 2140 The key to proving strongly modular extendible 2122 is The orem 2121 Whose proof depends on Kaplansky7s proof of continuity of the lattice operations via 2112 If one is Willing to augment strong modularity With ad ditional hypotheses available in many applications such as AW algebras then Kaplansky7s proof can be circumvented by materially simpler arguments this is illustrated in the concluding results of this section 2141 PROPOSITION If A is a strongly modular Baer ring satisfying aaiom H for 5 then A is eactendible Proof We emphasize that for a higher price one can omit axiom H 2122 Strong modularity implies direct niteness 216 hence niteness that is niteness for 5 Therefore the projection lattice of A is a continuous geometry by i of 2011 Whose proof is based on 1 of 208 hence avoids Kaplansky7s theorem One can now repeat the proof of 2121 omitting the reference to 2112 and then the proof of extendibility in the If part of 2122 0 2142 COROLLARY If A is a Baer ring with su iciently many projec tions satisfying aaiom H for 5 such that RPa 1 i LPa 1 then A is eactendible Proof By 2126 A is strongly modular thus the hypotheses of 2141 are ful lled O 2143 COROLLARY I Hafner 10 Th 2 If A is a nite Baer ring with su iciently many projections satisfying LP 5 RP then A is eactendible Proof From niteness and LP 5 RP it is clear that RPa 1 i LPa 1 and 5 satis es axiom H by 115 quote 2142 0 2144 LEMMA If A is a Baer ring satisfying CC for 5 and the EP aaciom then A satis es LP 5 RP Proof Let a E A a a 0 e RPa f LPa Using the EP axiom the rst part of the proof of 1429 yields orthogonal decompositions e sup ei f supfz With ez 5 fz for all i therefore e 5 f by of 1814 Note This is simpler than the proof of the same result in 1431 0 2145 COROLLARY If A is a nite Baer ring satisfying CC for 5 and the EP aaciom then A is eactendible Proof By EP A has su iciently many projections in view of the lemma the hypotheses of 2143 are ful lled O 2146 COROLLARY If A is a nite Baer ring satisfying aaiom H for 5 and the EP aaciom then A is eactendible Proof CC holds for 5 by the theorem of Maeda and Holland 1310 quote 2145 0 21 EXTENDING THE INVOLUTION 123 2147 COROLLARY J E Roos 31 If A is a nite Boer i ing satis fying the EP and SR oacioms then A is eactendible Proof By Maeda7s theorem 1213 SR Axiom H for 5 quote 2146 0 2148 COROLLARY cf 31 2 p 237 Cor If A is o nite AW algebra then A is eactendible Proof Note that A is directly nite by 611 613 By spectral theory A satis es the EP axiom 2 p 43 Cor and the SR axiorn which holds in every C algebra 2 p 70 quote 2147 0 REFERENCES 1 ARMENDARlZ E P and STElNBERG S A Regular self injective rings With a polynomial identity Trans Amer Math Soc 190 1974 4177425 2 BERBERlAN S K Baer Rings Springer Verlag New York 1972 3 The maximal ring of quotients of a nite von Neumann algebra Rocky Mountain J Math 12 1982 1497164 4 The center of a corner of a ring J Algebra 71 1981 5157523 5 DlXMlER J Sur certains espaces consideres par M H Stone Summa Brasil Math 2 1951 1517182 6 Les algebres d7operateurs dans l7espace hilbertien Algebres de Von Neumann 2nd edn Gauthiers Villars Paris 1969 7 GOODEARL K R Von Neumann Regular Rings Pitman London 1979 8 GOURSAUD J M and JEREMY L Sur l7enveloppe injective des anneaux reguliers Comm Algebra 3 1975 7637779 9 GOURSAUD J M and VALETTE J Sur l7enveloppe injective des an neaux de groupes reguliers Bull Soc Math France 103 1975 917102 10 HAFNER 1 The regular ring and the maximal ring of quotients of a nite Baer ring Michigan Math J 21 1974 1537160 11 HANDELMAN D Coordinatization applied to nite Baer rings Trans Amer Math Soc 235 1978 1734 12 The weak parallelogram law in Baer rings unpublished manuscript March 1976 13 Supplement to 12 April 1976 14 HERMAN L A Loomis ring satis es the polar decomposition axiom unpublished manuscript 1971 15 JEREMY L L7arithmetique de von Neumann pour les facteurs injectifs J Algebra 62 1980 1547169 16 KAPLANSKY 1 Projections in Banach algebras Ann of Math 2 53 1951 2354249 124 REFERENCES 125 17 Any orthocomplemented complete modular lattice is a contin uous geometry Ann of Math 2 61 1955 5247541 18 Rings of Operators Benjamin New York 1968 19 LAMBEK J Lectures on Rings and Modules 2nd edn Chelsea New York 1976 20 LOOMlS L H The lattice theoretic background of the dimension theory of operator algebras Memoirs of the Amer Math Soc No 18 Providence R 1 1955 21 MAEDA E Kontinuierliche Geometrien Springer Verlag Berlin 1958 22 MAEDA S Dimension functions on certain general lattices J Sci Hi roshima Univ Ser A 19 1955 2117237 23 On the lattice of projections of a Baer ring J Sci Hiroshima Univ Ser A 22 1958 75788 24 On a ring Whose principal right ideals generated by idempotents form a lattice J Sci Hiroshima Univ Ser A 24 1960 5097525 25 On rings satisfying the square root axiom Proc Amer Math Soc 52 1975 1887190 26 MAEDA S and HOLLAND S 8 Jr Equivalence of projections in Baer rings J Algebra 39 1976 1507159 27 NEUMANN J von Continuous geometry Edited by l Halperin Prince ton Univ Press Princeton N J 1960 28 PYLE E 8 Jr The regular ring and the maximal ring of quotients of a nite Baer ring Trans Amer Math Soc 203 1975 2017213 29 RENAULT G Anneaux reguliers auto injectifs a droite Bull Soc Math France 101 1973 2377254 30 Algebre non commutative Collection Varia Mathematica Gauthiers Villars Paris 1975 31 R008 J E Sur l7anneau maximal de fractions des AW algebres et des anneaux de Baer C R Acad Sci Paris Ser A B 266 1968 A1207A123 32 UTUMl Y On quotient rings Osaka Math J 8 1956 1718 33 CAlLLEAU A and RENAULT G Sur l7enveloppe injective des anneaux semi premiers a ideal singulier nul J Algebra 15 1970 1337141 34 AMEMlYA l and HALPERIN 1 Complemented modular lattices Canad J Math 11 1959 4817520 35 COHN P M Algebra Vol 2 Wiley London 1977 INDEX OF TERMINOLOGY A axiom 9 10 abelian ring or idempotent 81 addability 143 183 additiVity of equivalence 183 1818 annihilators 12 123 axioms AiH 10 AW algebra 138 1424 B axiom 9 10 Baer ring 119 regular 122 Baer ring 123 regular 125 113 212 bicommutant 41 boolean algebra 31 191 C axiom 10 C axiom 10 C sum 1115 center of a lattice 205 of a ring 3 central additiVity 10G central cover 315 hull 207 closed open sets 139 191 commutant 41 comparability generalized 131 151 181 orthogonal 131 partial 137 complete additiVity 10 181 1818 1819 compressible ring 329 continuous geometry 201 212 continuous regular ring left right 141 continuous ring 815 17 idempotent 815 continuity of lattice operations 201 corner of a ring 21 D axiom 9 10 decomposition into types 826 925 126 INDEX 127 dimension function 193 197 198 directly nite ring 71 idempotent 71 directly in nite ring 71 idempotent 71 discrete ring 814 idempotent 814 domination of idempotents 511 E axiom 10 EY axiom 10 endomorphisms of a vector space 126 EP axiom 10 1422 equivalence of idempotents 53 of projections 111 113 equivalence of projections 61 112 essential right ideal 128 submodule 132 218 exchange by a symmetry 1212 extendible 211 F axiom 10 factor factorial ring 195 faithful element 315 nite idempotent 71 projection 95 ring 71 95 15 nite additiVity 10D nite ring 95 G axiom 10 GC axiom 10 generalized comparability 131 group algebra 828 H axiom 10 121 homogeneous Baer ring 162 hyperstonian spaces 413 in nite idempotent 71 projection 95 ring 71 95 15 involutive ring 13 21 involution proper 110 isometry 613 lattice complete 121 continuous 201 128 INDEX modular 116 204 2014 2111 projection 115 left projection 17 Loomis ring 1422 LP 17 LP N RP 10 LP i RP 10 matrix rings 21 1511 167 175 2133 maximal ring of right quotients 131 212 modular lattice 116 2014 modularity strong 215 nonsingular module 2116 ring 128 orthogonal additiVity 10F 112 113 orthogonal comparability 131 operators Hilbert space 127 order of a homogeneous ring 166 orthogonality 146 of idempotents 55 p position 125 parallelogram law 10H 121 partial comparability 10E 137 isometry 62 PD axiom 141 perspectiVity 519 polar decomposition 141 position p 125 prime ring 318 projections 14 central 39 206 equivalent 53 111 113 equivalent 61 112 left right 17 unrelated 137 proper involution 110 136 properly in nite 76 99 nonabelian 827 purely in nite 912 quaternions 335 quotients maximal ring of 131 211 regular ring 112 right left continuous 141 INDEX 129 right self injective 132 1818 regular Baer ring 113 197 212 regular ring 114 95 footnote Baer ring 125 relative inverse 27 Rickart C algebra 138 Rickart ring 11 Rickart ring 14 right continuous regular ring 141 right projection 17 ring 1 RP 17 Schroder Bernstein theorem 104 self injective ring 130 212 semi nite 912 semiprime ring 318 square roots 10 SR axiom 10 Stone spaces 139 191 strongly modular 215 subring 1 subring 41 suf ciently many projections 2125 summable orthogonal family 148 symmetric ring 133 symmetry 1211 type decomposition 826 925 type 1 814 16 I n 924 In 167 linf 924 IN 167 168 11 919 11111 n 924 111 924 111 912 unitarily equivalent 65 1213 unitary 65 unrelated projections 137 von Neumann algebra 411 The Structure of Typed Programming Languages Brief Intro to Haskell 65386L Programming Languages Dr Greg Lavender Department of Computer Sciences The University of Texas at Austin 45m CSSE6L LActuMOQ A Brief Intro to Haskcll Haskell Named after Haskell Curry Haskell is a nonstrict ie lazily evaluated pure non sideeffecting functional programming language referentially transparent lots of usage of recursive functions defined on lists similar syntax to ML but some would say an improvement ML allows side effects both interpreters and compilers are available see wwwhaskellorg Hugs interpreter GHC Glasgow Haskell Compiler i i d H i CSSE6L antuMOQ A Brief Intro to Haskcll iLi FuncTions Func rions are s rrongly Typed fact Integer egt Integer fact fact I n factmel factl Int egt Int factl 0 l factl n n factlmel fact 10 gt 3628800 factl 10 gt 3628800 fact 100 93326215443944152681699238856266700490715968264381621 46859296389521759999322991560894146397615651828625369 7920827223758251185210916864000000000000000000000000 factl 100 gt 0 45m M C53E6L Llama09 A Brief Intro to Haskcll D ifferenf Ways To Express Fu ncTions How many ways can you wri re a fac rorial I recursive fact I if 11 then 1 else I factmel recursive by cases fact 0 l fact I n factmel 01 ingt0nfactnel induc rively recursive fact 0 l fact n1 n1 factn Tail recursive fact I tfact n l where tfact 0 m tfact n m tfact nil nm 45m C53E6L Llama09 A Bricf Intro to Haskcll Lazy evaluation Allows some interesting functions and expressions to be written is the list quotconsquot operator which is right associative ones Iones ones gt IIm gt IIIIIIIIIIm lazy lists partial results are printed asap m n 7 1 n l otherwise 1n gt IIIn gt I22Ingtmgt 12mnH this allows quotinfinitequot lists 0 gt 0l2345678910 45m CSSE6L antuMOQ A Brief Intro to Haskcll A Lazy Fibonacci Two fibonacci functions one consuming time Si space the other space standard recursive fibonacci function fibonacci Int egt Integer fibonacci fibonacci fibonacci fibonaccin72 fibonaccineh infinite fibonacci sequence fibs Intege fibs 0 I zipWith fibs tail fibs fibn Int egt Integer fibn n take n fibs fibonacci function that selects nth element of the sequence fib Int egt Integer fib n fibs ll n 45m C53E6L Llama09 A Bricf Intro to Haskcll Lazy evaluaTion pass argumen rs by name do no r evaua re func rion argumen rs un ril They are needed square Intgt1nt square X X square 5 gt 5 5 square 24 gt 24 24 i i d i i i CSSE6L LlctuMOQ A Brilf Intro to Hasklll iLi LisT comprehensions Ways To genera re lis rs in a se rIike no ra rion sf of pairs ob rained from Two Iis rs xygt i x lt7 x5 yltiysi The higherorder map func rion can be rrivially defined as a Iis r comprehension mapa gt b a gt b mapfxsfxixltxs compare To The recursive and folded versions f xmap f xs map f xxs x ys gt f xys map f xs foldr X5 4504 El l l l CSSE6L Lyman09 A Bricf Intro to Haskcll Lazy Prime Sieve using an infiniTe lisT and a lisT comprehension primes Integer primes sieve 2 W Sieve of Eratosthenes sieve xxs x sieve y l y lt7 xs y rem x0 nthprime n take I primes 45m q l l l C53E6L Llama09 A Bricf Intro to Haskcll Folding over a lisT Fold is a higherorder funcTion ThaT applies a funcTion argumenT successively To The elemenTs of a lisT carrying forward iTs previous resulT requires Three argumenTs The funcTion an iniTial sTarTing value and The lisT iT reTurns a value of The Type of The iniTial argumenT folding from The lefT foldl a gtb gta gta gtb gta foldl 1 ST alazaquotan1 gt f39 ul f39 12 f39 f39un1 f39un folding from The righT foldr 1 ST alazaquotan1 gt 11 f39 oz 1 f39 an1 f39aquot f39g 45m l l El l l l CSSE6L antuMOQ A Brief Intro to Haskcll Foldl vs Foldr whaT are The differences in compuTing n foldr 1 Ln gt 1234n1 foldl 1 Ln gt 1123n noTe ThaT fodr requires The whole expression To be formed before iT can do iTs compuTaTion required On space whaT abouT fod foldl is noT sTricT in iTs second argumenT buT we can force This using The funcTion sTricT agtb gt a gt b sTricT f x seq x f X where seq a gt b gt b is a predefined Haskell funcTion such ThaT seq x y evaluaTes x before reTurning y 45m 1 1 l l El l l l CSSE6L antuMOQ A Brief Intro to Haskcll A STricT Foldl A revised facTorial ThaT used 01 space define a version of fold ThaT uses The sTricT funcTion foldl39 a gt b gt a gt a gt b 7gt a foldl39 f St St foldl39 f st XXs strict foldl39 f f st X XS Then facT becomes fact foldl foldl l n foldl 2 3 n foldl 6 n gt noTe ThaT foldl39 consumes an enTire lisT before giving a resulT so H is useless if applied To infiniTe isTs as H will never produce a resulT 45m q l l l C53E6L Llama09 A Bricf Intro to Haskcll Examples using Foldl summing a lisT of inTeger39s noTe ThaT The sTar Ting value is The addiTive idenTiTy sum Int 7gt Int sum x5 foldl 0 x5 compuTing The lengTh of a lisT we use an anonymous lambda wiTh a don39T car equot 2nd argumenT since we don39T car e whaT is in The lisT length a 7gt Int length x5 foldl n 7 7gt n1 0 x5 NoTe ThaT boTh funcTions can be cur r ied sum foldl 0 length foldl n 7gt n1 0 A UA 7 q l l l C53E6L Llama09 A Bricf Intro to Haskcll Fold is like iTer aTion a facTor39ial funcTion noTe ThaT The sTar Ting value is The mulTiplicaTive idenTiTy also noTe ThaT compuTing facTor ial using a fold is noT recursive buT iTer aTive 1 foldl m 1 2n fact 0 fact n 45m Using fold efficienle how would you naively compuTe The mean of a lisT of elemenTs mean Num a gt a mean x5 sum x5 length x5 7gt Int This approach has The drawback of Traversing The lisT Twice how can we improve on This so ThaT we only Traverse The lisT once noTe ThaT The lengTh of The lisT can be compuTed while we are compuTing The sum 45m El l l l C53E6L Llama09 A Bricf Intro to Haskell A More Eff icienT Mean FuncTion Consider a funcTion mean ThaT compuTes The mean value of a lisT using foldl foldl Takes a funcTion an iniTial value and a lisT as argumenTs noTe ThaT suml is a curried funcTion ThaT reTurns a pair sl where sis The sum and is The lengTh suml Num a gt a 7gt aInt suml foldl 51 x 7gt 5xll 00 mean x5 fst p fromlntegral 5nd m where p suml x5 45m CW iLl CSSE6L antuMOQ A Brief Intro to Haskcll An Even More Efficient Mean Function A strict version of the mean function using a strict version of foldl suml39 Num a gt a 7gt aInt suml39 foldl39 51 x 7gt 5xll 00 mean39 x5 fst p fromlntegral 5nd p where p suml39 x5 45AM 11 l l El l l l C53E6L Llama09 A am mm to Haskcll Homework Assignment Implement Horner39s method for evaluating a polynomialusmg an appropriate verSIon of the fold function Horner39s rule for evaluating a polynomial using recurSIon pnx aux alxn l aniltx an where pnx hnx hum aEl hlx hl1x xal i l2n The horner function type is horner Num a gt a gt a gt a ie takes a numeric value of type althe argument x and a list of coefficients each of 0 type a and computes the value of the polynomial 45m I was 1 MEMORANDUM THE WHITE HOUSE WASHINGTON sscm NODIS XGDS MEMORANDUM OF CONVERSATION PARTICIPANTS Dr Henry A Kissinger Secretary of State and Assistant to the President for National Security Affairs Alan Greenspan Chairman Council of Economic Advisers Charles W Robinson Under Secretary of State for Economic Affairs Lt Gen Brent Scowcroft Deputy Assistant to the President for National Security Affairs DATE AND TIME Monday June 16 1975 PLACE The White House Secretary39s Office SUBJECT Bilateral Oil Agreement with Iran Kissinger The President wanted me to discuss something we have been discussing with Iran This is for you only and is highly sensitive When I saw the Shah in March he complained his liftings were falling and he had 500 000 barrels a day eirgcess I said that under specific conditions we might take it off their hands Chuck Robinson has discussed it further with them and they have now gone up to 700 000 barrels a day They would sell to us for Treasury notes with a forgiveness period The notes could be redeemed for the purchase of American goods I am interested in this idea because number one it breaks the OPEC 7939 front because it shifts the surplus h to mNODIsXGDS if I U CLASSIFIED BY IMW 1 A U r xix EXEMPT FROM GENERAL DECLASSIFICATI ON DECLAsSIFlED 39 SCHEDULE or EXECUTIVE ORDER L165 39 7 I quoti quot9011 5 r r 3 1 x a an gins gal KEMPI IQN LTlx f AJTHORITY Mi 1 z 3i t ICALLY DE Il S FBl on ya Dd By 44 NLE DATE Iauln File scanned from the National Security Adviser39s Memoranda of Conversation Collection at the Gerald R Ford Presidential Library SEGRE BNODISXGDS 2 Greenspan It would have to be at Saudi expense and they wouldn39t like it How long would it be for Kissinger It39s up to us Second it would make it harder to raise the prices Third it39s insurance against another embargo Fourth it puts pressure on the suppliers There are two possible schemes The first one is that we pay the market price for oil The other is to sell at current prices plus an adjustment tied to the wholesale price index Greenspan That breaks the OPEC price structure Kissinger If they give us a forgiveness feature one year gives us 1 dollar discount Greenspan If we could keep Iran at full production it puts severe pressure on the Saudis We would want to insure that Iran would not cut back elsewhere Robinson They could but that obviously is not their scheme They 39 want to keep their revenue up The basic scheme is a barter If they contract way out for oil they need the assurance that the price will go up in line with other goods they would buy By either scheme the OPEC price or the current price plus the wholesale price index 30 long as it didn39t go above OPEC We would give thenyTreausry notes without interest for the first year and that are nonnegotiable for the first year So for the first year we would have 2 billion in our hands This will take imagination and a change in how we operate Greenspan So in effect you have 2 billion in escrow So if they broke the deal or entered an embargo we have the 2 billion Robinson We get a discount We will have to establish a specific arrangement which can go far to break OPEC Greenspan When this gets out the only real issue is price It would be crucial that the price not escalate beyond the OPEC price o sagaE lNODISXGDS 3 0 C 39 1 z g 0 iii339 c9 WT NODISXGDS 3 Robinson I think I got that but I haven t nailed that down yet Greenspan It seems obvious that if you look at l98 at these prices the Saudis are the only ones who will not be a net borrower The others are committing funds at such a rate The 700 000 barrels by itself won t do it But as a symbol The Saudi reaction will be important Kissinger Will the companies give us trouble because the Saudis will be upset Greenspan The idea properly packaged seems very attractive Let me think about it An essential ingredient is not to let the price go above the OPEC price I will think on thenegative elements but I am intrigued Robinson There are two alternatives a government purchase to be auctioned off or we could buy it for the Navy reserve and so on Greenspan We can buy it for the stockpile We are talking about a 1 billion stockpile but we don39t have the Salt Dome capacity Maybe you can solve the company problem by having them take the oil The notes would have to be non negotiable Have you looked at the different interest rates Robinson It is about 1 a barrel atethe current interest rates I am thinking of a fiveyear maturity with no interest if they don39t use it for equipment Kissinger Another optionis to lower the interest rate and have no forgiveness How will it look to our IEA partners It really helps them Robinson But we must be careful how we do it Greenspan It will have a devastating impact on OPEC So you want the maximum apparent price concession Robinson Iran insists that the nonnegotiable aspect be covered by a side letter as well as the price not going above OPEC SEGRH NODISXGDS WNODISXGDS 4 Kissinger I39know we would like the greatest apparent price differential f ihe side letters will look the Saudis will be very upset GreensEan If you then have the same deal with the Saudis I would not favor it Because theh OPEC becomes Saudi Kissinger That is a sepafate issue Let the Saudis worry about that Robinson It would be politically difficult to say we do it with Iran but won39t for the Saudis Greensgan Let me think it through The criticalthing is what happens inA1980 with Iran and with or without the Saudis We may need a total strategy before we move Kissinger I think we should pick up what we can and develop a total strategy after Iran is signed up T he meeting ended e SECRET NODIS XGDS W31M x Ww f4wm W41 W W 19 W4 N WAle hi w 74 fzrrwm Xxwa Jinnil MM 4823 mm MAMr93 exam UAWL 4w chpMQLMURW s WW E 2N2W7327 M 11 W y WW 39 5 I 99 WA 4AM 23 mej C9qu T W4f1AD MM 50 1353520 35 3 GENO 11124msm5 cm eumELmEs 3ka Ewiw 3401 f b 11 M 4 quot BY NARA DATE m lt 02C Gx w A z 9 xCU PWWV xvi w g w ri swxoi 0 m Ya W at V398 J 3 wia aigxmi t r L 351 IvyJXEIIIIs InIIiiaI liliz Ur f9 s Vim gx f eri i 9gt lt 3rb 9 14m Sn eif Mg 9amp5 g 1 If P gfiT lt a x c r Q a 3 93 7 f fi r i m mi m H gNL Vt a BN3 DP 9 f3 ef gy 3 x x y d a mg figkj NRC R m Ev xx manix w w A A w 4 OVERVIEW AND OUTLINE Arithmetic combinatorics is the study of nite subsets of abelian groups and rings For example let AB E Z be nite nonempty subsets of the integers What can we say about the following sets AB abla AbEB A7B a7blaeAbeB AB ablaeAbeB Our main goal is to prove from rst principles a recent theorem of Jean Bourgain Nets Katz and Terry Tao on the growth of subsets in nite elds of prime order see 1 Theorem 01 BKT Let 6 gt 0 be given Then there exist constants c c6 gt 0 and 5 56 gt 0 depending only on 6 such that for any mime p and any subset A E F I ZpZ with lFl6 lt lAl lt lFl1 6 we have maxlA Aly lA Al 2 ClAlHE Since a subring S ofa ring R has the property that lSSl lSl and lSSl lSl we can restate the conclusion of the theorem in the following way In general if A B E Z are subsets of some abelian group then we will let lAl S lBl denote lAl 0lBl ie there is a constant C independent of A and B such that lAl S ClBli Now if Z is a ring7so both addition and multiplication are de ned7we can call A Q Z an approximate subring if lA Al 3 lAl and lA Al 3 The theorem says that nite elds of prime order contain no approximate subringsi As a warmup let s think a little bit about subsets of Z From now on A and B will always denote nite nonempty subsets of the group or ring under consideration unless stated otherwise It may be helpful to observe that translating given sets A and B does not affect lA Bl lA 7 Bl etc For example the sum of the translated sets has the same size as the sum itself lAI Byl lAB Iyl lABl Exercise 01 Let Z Z the ring of integers a Let AB Q Z Show lAl lBl 7 1 S lABl S b Given mn 2 1 and 721 n7 1 S s S mn construct AB E Z such that lAl mlBl n and lABl s A statement similar to a for Z ZpZ where p is prime is called the Cauchy Davenport inequalityi This and a similar lower bound in terms of a multiplicative factor rather than an additive one will play a crucial role in the proof of Theo rem 0 1 and we will prove both inequalities next time One can often think of arithmetic combinatorics as approximate group theory h If H S Z is a nite subgroup1 of an abelian group Z then H is closed under addition and subtraction H H H 7 H Hi In particular lH Hl as we noted above in the case that Z is a ring On the other hand given a subset A E Z such that lA Al S lAl what can we say about lA 7 Al lA A 7 Al etc 1We will always use the H S G notation to distinguish H as a subgroup rather than just a subset of G 1 Exercise 02 Let AB E Z be nite nonempty subsets of an abelian group Z Show that lA Bl lAl if and only if there is a nite subgroup H S Z such that A is the union of cosets of H and B is contained in some coset ofH In particular if A B this exercise says that lA Al lAl if and only if A is the coset of some nite subgroupl It turns out that if A is essentially Closed under addition in the sense that lA Al S lAl then we can say something interesting about the size of sets of the form A i A i i A More generally if A is essentially B invariant ire lA Bl g lAl then lmB 7 nBl g lAl where mB B B B m timesl2 This is known as sumset estimates and it will follow from a more general theorem called Pliinneckels theoreml Since the BKT theorem deals with sizes of subsets we will focus on results from arithmetic combinatorics on cardnality not structure However one can often nd nontrivial information about the structure of say A B given information on i For example a very deep theorem called Friemanls theorem says that if a subset is essentially closed under addition then it is very close to being a generalized arithmetic progressionl We will not require these sorts of results so in order to save time we will not cover theml For more on Friemanls theorem see Lecture 2 of Our strategy for proving the BKT theorem is this after proving the Cauchy Davenport inequality and its re nement we will be able to show that 1 FA51A52A5k for h relatively small depending only on 6 We need a bit more notationl Let P E ZX1 X2 l l Xn be a polynomial with integral coef cients in n variables and let l l l A be its evaluation at the subset A with respect to our notation above For example if PXY X 7 Y then PA A A 7 A Now given the linear surjection77 result as in equation 1 we will deduce the existence of a polynomial P such that PAAHlA Fl However assuming that the conclusion of the theorem is false we will ndiusing standard results such as sumset estimates and the Gowers Balog Szemeredi theoremithat for our polynomial P we have lPA A l l l Al S lAllJrCE for some constant C and any 5 gt 0 Thus we will arrive at a contradiction by choosing a suf ciently small 5 since lAl lt lFl1 6 lt Although the BKT result is only for nite elds of prime order similar results may be obtained for arbitrary nite elds The difference is of course that there are nontrivial sub elds l lt K lt qu when 4 is not prime so in particular lKKl lK Kl lKl contradicting the conclusion of Theorem Olll However this is essentially all that can happen For more on this see Theorem 43 in l as well as Theorem 24 in Our primary resources for this seminar are Taols notes on arithmetic combina torics 3 and the BKT paper 1 in particular with the exception of the proof of Gowers7 result that we will encounter in Week 5 all the results and proofs are from these two sources My sole contributions are reorganization exposition and the occasional correction of a typo I also add some details to Taols proofs in order to aid my understanding of the arguments and I hope these are illuminating rather than distracting Here is my proposed outline for the rest of the course 2I admit that 17m being rather loose with the S notation For now just think of it as meaning small relative to In the next few lectures I hope to be a bit more careful Representation Theory Fall 2004 Lecture 15 16 Fernando Rodriguez Villegas November 9 and 11 2004 1 Tue 119 Recall de nitions GSL2F 3 gtea Nlt1 1gteazrp and if 4p IF a EX is a character7 VW Indgtp Also recall that 0 VW is irreducible if 472 7 1 0 V1 trivial 63 Steinberg We are working on the case where 4p 739 is a quadratic character 7392 1 We claim that VT VTJr 63 V and V are each of dimension p 12 Recall that VT f G a I with fnag 739otfg7 where n E N and alt8 a91gt7a F Recall the de nition of U Ufg ZnEN fw 1ng U VWZVWA is G linear so if 739 7 1 so U VT a V7 then V are the eigenspaces of U Let s verify that U is G linear 91Uf9 Uf991 2 1110471991 nEN U91f9 because 91fw 1ng fw 1ggl Uf 6 V7 does Ulelag TaUf97 Ufn1ag ZnEN fw 1nn1ag ZMEN fw 1n2ag if 712 5 then nga a ml which we call 713 So 2 BEN fw 1angg ZMEN oflw lngw7 since w la a w l we 2W fw 1nsg TaUfg Notes by Kyle Schalm edited by Nick Leger Claim U is an isomorphism ldea Restrict U to a subgroup and diagonalize U Consider VT as a representation of N The characters of N are 1 m 0 1 ewe where 1m is an additive character of 1717 Fix a nonzero 1 then any other character of Pp is of the form 11z 11tm for some t 6 Pp ift 07 11 is the trivial character We will show VT fgjl V where V is the it component of VT7 meaning that if ft 6 Vi ft 75 0731811 ft9n tnft9 A1507 ftn1a9n2 Ta tn2ft9 First7 we claim that ft is determined by its value at 17 w Let s use the Bruhat decomposition G B U BwN either 9 6 B7 in which case 9 na and ftg ftna 739ozf17 or g E BLUN7 in which case 9 mawng and ftg ftn1awn2 Tawtn2ftw Case 257 O ft1 ftn 1 ft1 71 1Jnf17 since 11 is not trivial7 there is an n for which 1171 7 1 so that ft1 0 Hence ft is completely determined by its value at w Normalize by letting ftw 1 1gt gt0 1gt gt1 f0 foo wgt gt1 wgt gt0 Case 25 0 Let These span VT0 which is therefore 2 dimensional To show Vf are stable by U lft 7 O Uft Atft for some At 6 E To compute this At Uftw Atftw At and Uftw ZnEN ftw 1nw7 Where m m a 3E l m m m so fgw lnw t7z 17z 1ftw 117m 1739z 17 since w E N If 35 7g 0 this is A Z akflhw l 637 Z d tm 90 HW71 16F it 2 WWW 7590 H 00 637 3 9H this last sum is a Gauss sum and equal to p7 so g7392 771p and 97 i 771p Gauss i 13 p21 mod 4 7 p E 3 mod 4 39 lft 0 continuing our computation of At Using the fact that V70 is spanned by f0 foo and these are determined by their values at 1 and w7 actually proved 97 Ufo1 ZneN fow 1n 1 ZneN 1 onfow 1 fowlNl as w 1 7w T 1P and with n as before Ufow ZneN f0w 1nw Emery Tfow Emery TM 0 Ufltgtltgtlt1gt ZnEN foow71n pfoo7w T71pfoow 0i Ufoow Zz 07foow foo11since foow 0 So U acts on W as 781 5 with respect to the basis fo foo with eigenvalues ig7 So U diagonalizes as 97 s followed by 7 77s Let V f E V Uf ig739f be the eigenspaces of U G preserves each space V V 63 V each of dimension Summary take any character 4p IF a EX and decompose VW to obtain the principle series Number Space Dim 92 3 g 1 Vw P J 1 w T 2 Vi VF L31 4p 1 2 trivial Steinberg 1p p5 This gives us LES 4 2 representations we need p 4 There is some stuff I don t understand here and then Example with p 3 1G1 3 32 71 24 PSLZUFP SL2lei1 i1 is the center This is simple for p gt 3 For p 3 PSLg A4 Conjugacy classes look like TMU St VJr V 10 lt01 1 3 2 2 0 71 lt 0 71 1 3 72 72 1 1 1 0 11m 11m 0 1 2 2 1 71 1 0 11R 11R 0 1 2 2 71 1 1 0 11m 11m 0 71 2 2 1 1 1 0 11m 11m 0 71 2 2 0 71 lt 1 0 gt 1 71 0 0 0 71 fl 0 gt 1 71 0 0 We should be able to nd some representation of A4 from here The rst two come from PSLZ 01 3 H lt12gtlt34gt 1 064123 lt3 31gtH132 There are still two representations of A4 we haven t found Example p 57 PSL2F5 A5 St 11 1 53 5 33 lt 01 31 gt 5 3 3 lt 1 i gt 0 6 6 lt 1 i gt 0 6 6 lt 31 31 gt 0 6 6 lt 31 31 gt 0 6 6 g1 114 lt 5 if gt62 1 71 0 0 where e 136 E Some correspondences with A5 are ltmwe ltmwe50 wwei lt12gtlt34gt H lt 3 291 PSL2F4 A57 PSLZUFQ A6 PSL4F2 As these are the only isomorphisms between the PSL family and the alternating family 2 Thu 1111 Let H up x L x L7 p gt 2 where L is a vector space over Pp of dimension 2 Let B L x L a Pp be a nondegenerate symmetric bilinear form Let 6 0 B 1 This is a non degenerate skew symmetric form on L x L 01 2 and 01 0 0 B01 1 eZWip y 4p eZWipq b 62MB The whole point is the connection G SL2le H SpU lt E H u H 140 where U L x L SpU In U H U preserves 1 u 1112 m 1112 a f lta11 mm 612 ll and lg are themselves vectors of dimension 2 H 0412 512 7 v12 612 39 a preserves I ie 0412 l ylg 0 l B 0412 l 312 7 i 7 312612 7B 0 VIZ 612 7ltIgtus1ng deta71 We can embed SpU in Aut0H these preserve the center a H 914 H WWW where mu WV2 We have the Schrodinger representation of H pHHgt GLV where V f L H 4 de ned by p0 H C The center acts by multiplication C 6 pp acts by sending f to f o 1 ll0 acts by l H fl 11 0 1 012 acts by l H fl llllg flbllg If a E Aut0H then the composition 0quot p o 0 acts on the center in the same way By Stonevon Neumann Ra 1pRa p for some Ra E GLV Pick such an Ra for each 0 It s well de ned up to a scalar by Schur It follows that C010 2R0 10 2 R01R02 for some CO39l0392 6 Ex R G H PGLV is a projective representation Can we lift it to GLV Plan let s try to understand R for some a E N T and a w and then extend with Bruhat a 0 Case 2 a 0 ail let h 114 6 H for notation Subcase 1 u 110 so an all0 For f E V 71quot fl 0111 hfl fl 11 Subcase 2 u 012 W 041412 71mg ba 1lg mm hfl blglfl afl fal We claim that this works satis es Ra 1pRa p 01 W 30 1hfal 30 1fal11 N all WW Subcase 1 u 110 ua 011 hafl Subcase 2 u 012 ua 7120 hafl To nd R let s de ne a Fourier transform 571 lLl lZ 2 WWW 1 l eL 51711fl7 WW N Z1 fl127 WW W7 12fl and de ne f1l fl ll Then 5711 lLl lZ Z f1l bl 71 l eL lLl lZ Z f1l lbl7l l eL L ilZ Z fllbl7illl WEL 50710570 Now 3 h s 37134 h R915 5H V Case iii a lt 1 Tgtz Ell p Subcase 1 u 110 h 10 u llzll ua llzll ua ua Illl 11 ll zll Bllmll uvu 1117 07 1170 07 so 11014 bllllm2 hamz 1211 law2w xllfl 11 1211 law2w mm 11 WW f1 11 Claim Rafl bzzw2fz siinWWm works RU 1hRU l RU 1hflb17 012 Ra 1fl mm 111 11 z b l 2bl 11z law2m 11 blvllmbll7llm2fl 11 hafW Now use Bruhat to de ne Ra for arbitrary a 39y 7 0 7 04 7 1 04371 0 i39yil 1 Mil If 7 39y 6 0 1 39y 0 0 1 7 de ne 130 Verify if 01 02 0102 E G B then C010 2R0 10 2 R01R02 with 1 I a done a principal series 0 70 i i 1 2 71 H a replace R by 7 R on G B a cusp1dal ser1es Something missing here Consider characters of the circle group K39y F2lN39y1HltEX If A2 7 1 VA is irreducible of dimension p 7 1 If A2 1 VA splits into VAJ V each of dimension These give the missing representations Case I recovers what we did before Arithmetic lattices in SUn 1 D B MCReynolds Arithmetic lattices in SUn 1 Contents Chapter 1 Introduction Acknowledgements Chapter 2 Some basic terminology 1 Commensurability and wide commensurability Lattices in Lie groups Godement s compactness criterion 9 Restriction of scalars 2 Number elds 3 Valuations on number elds 4 Extensions of valuations 5 Local elds 6 Algebraic groups 7 8 Chapter 3 A motivational example 1 The groups Un 1 and SUn 1 2 A family of lattices 3 Real algebraic structure Chapter 4 Hermitian matrices forms and involutions Hermitian and skew Hermitian matrices 2 Hermitian forms and unitary groups 3 Field of de nition and equivalence 4 Diagonalizing forms anisotropic and isotropic factors 5 Involutions associated to Hermitian matrices Chapter 5 Arithmetic lattices of the rst type 1 Subgroups arising from forms 2 Admissible pairs and arithmetic lattices of the rst type 3 Commensurability classes of lattices of rst type Chapter 6 Classifying Hermitian structures over elds 1 Signature and dimension 2 Signatures at other archimedean places 3 Determinant 4 Classifying equivalence classes of Hermitian forms D B McReynolds 4 5 A few examples 26 6 Parity and describing commensurability classes 28 7 All Hermitian forms over Qjd are isotropic 29 8 Local theory for Hermitian forms 30 Chapter 7 Cyclic division algebras 31 1 Central simple algebras 31 2 The Skolem Noether and Structure theorems 32 3 Cyclic algebras 32 4 Splitting elds and degree 33 5 Orders in algebras 35 Chapter 8 lnvolutions and Hermitian elements 37 1 lnvolutions of rst and second kind 37 2 Hermitian elements in unitary algebras and involutions 39 Chapter 9 Arithmetic lattices in SUn 1 41 1 Second type lattices over Q 41 2 Second type lattices over arbitrary number elds 41 3 Mixed type lattices 42 4 Admissible triples and the classi cation theorem 44 5 Tits symbols and a summary of the lattices 44 Chapter 10 Examples for small pairs 47 Lattices of type Ugyz 47 2 Lattices of type U317 47 3 Lattices of type U370 48 4 Lattices of type U371 50 5 Lattices of type U32 51 6 Lattices of type U31 51 Chapter 11 Classifying Hermitian structures over division algebras 53 1 Characteristic polynomials reduced norms and reduced traces 53 2 Invariants 53 3 The classi cation 53 4 A few examples 53 Chapter 12 Producing in nitely many commensurability class 55 1 The Brauer group 55 2 Local invariants 56 3 A local to global theorem 56 4 Local criterion for involutions of the second kind 57 5 Proof of Theorem 121 58 Arithmetic lattices in SUn 1 5 Chapter 13 Unitary algebras over local elds and reductions 61 Chapter 14 Arithmetic lattices in Lie groups a general de nition 63 1 Arithmetic lattices in Lie groups 63 2 kiforms of a Lie group and lattices 63 3 Lkiforms 64 4 Non abelian cohomology and surjectiVity of p 65 5 The classi cation theorem of F iforms of SUn 1 66 Chapter 15 Appendix 69 1 Basic X ihyperbolic geometry 69 2 Invariant trace elds and algebras 72 3 Shimura varieties and other arithmetic moduli spaces 73 Bibliography 75 Index 79 Arithmetic lattices in SUn 1 7 CHAPTER 1 Introduction The purpose of this note is to provide an accessible introduction to arith metic lattices in SUn 1 The mathematical goal is two fold give the clas si cation of arithmetic lattices in SUn 1 and prove that there are in nitely many commensurability classes of lattices of rst and second type see be low for the de nitions of these types of lattices In order to keep the tech nical aspects of this note to a minimum we give the general construction of arithmetic lattices in SUn 1 and only sketch the proof of their totality For the construction of in nitely many commensurability classes for each type some background is needed for instance at the least the Hasse norm theorem and the Hasse invariant for a simple algebra over a number eld Acknowledgements This note is an amalgam of a series of lectures given at the University of Texas and the University of Maryland I would like to thank Bill Goldman Alan Reid David Saltman and Rich Schwartz for conversation before dur ing and after these lectures I would also like to thank John Parker for his interest in these notes Finally I would like to thank Misha Belolipetsky and Matthew Stover for reading a draft of this notes and for pointing out several errors Arithmetic lattices in SUn 1 9 CHAPTER 2 Some basic terminology In this section we present some of the requisite material for the sequel 1 Commensurability and wide commensurability For a group G with subgroups H1H2 lt G we say that H1 and H2 are com mensurable in H1 Hz if a nite index subgroup of H1 and H2 More gen erally if a Giconjugate of H1 is commensurable with H2 we say that H1 and H2 are commensurable in the wide sense 2 Number elds By a number eld k we mean a nite extension of Q and denote the ring of integer of k by 6k This is a nitely generated Zimodule consisting of those elements of k which are roots of a monic polynomial in Zt For a xed number eld k a real embedding 6 of k is an embedding 6 k a R Similarly a complex embedding L39 of k is an embedding T k a C such that 1k is not contained in R Up to post composition by an automorphism of R there are only nitely many real embeddings 616r1 Likewise up to post composition by an automorphism of C there are only nitely many complex embeddings 117172 It is a fundamental fact that see 37 k Q r1 2r2 We say that k is totally real if r2 0 and k is totally imaginary if r1 0 Throughout this note we will be primarily concerned with totally imaginary quadratic extensions E of a totally real number eld F By this we mean that E is totally imaginary with totally real sub eld F such that E F is two The extension E F is necessarily Galois and we denote the Galois group by Ga1EF and its non trivial involution by Note that for a pair of compatible embeddings 6 F a R and L39 E a C such that 1E R 6F gtk is nothing more than the restriction of complex conjugation in C to E In examples we often work with cyclotomic elds By an nth root of unity in a eld k we mean an element 4 E kX such that 4 1 and say that 4 is primitive if 4 is a generator for the nite multiplicative group of all nth D B McReynolds 10 roots of unity Typically we denote a primitive nth root of unity by 4 and call the extension Q n a cyclotomic extension THEOREM 21 Weak Approximation Theorem LetK be a totally real number eld with real embeddings 61 Cy For any p q E N with 17 q s there exists it E K and embeddings SJ1 SJp such that 6koc gt 0 ifand only ifkjfor lp 3 Valuations on number elds By a valuation v on a number eld k we mean a map v k a R such that a vx 2 0 for all x E k b vxy vxvy for all xy E k and c vxy S V06 VO If in addition v satis es the ultra triangle inequality Vltxy S maXVx7Vy we say that v is nonarchimedean and otherwise say v is archimedean Given a valuation v on k we obtain a metric d by setting dvxy vx 7 y We say two valuation v1 and V2 are equivalent if the topologies induced by dv1 and dv2 are the same Alternatively v1 and V2 are equivalent if there exists a positive real number 3 such that v1x V2x3 Given such a 3 it is a simple matter to see that the metric balls in the space kdvl contain and are contained by metric balls in the metric space k V2 Aside from the archimedean valuation given by Q C R for each prime integer p E N we obtain a nonarchimedean valuation on Q as follows For a rational number x write x psy such that neither the numerator nor the denominator of y are divisible by p We de ne the p adic valuation p on Q by le ps One can verify that p is nonarchimedean THEOREM 22 46 Up to equivalence the valuation on Q are p for each prime p E N together with the archimedean valuation 4 Extensions of valuations For any number eld k Q the valuations on Q extend to a family of valua tions on k The archimedean valuation are simply those that come from the embeddings k C R or k C C and the nonarchimedean valuations come from the prime ideal p lt 6k see 30 Speci cally for each prime ideal p over p we obtain an extension of vp to k which we denote by vp We denote the equivalence classes of nonarchimedean valuation on k by Vfk and the archimedean valuations by V For later use we denote the archimedean Arithmetic lattices in SUn 1 11 valuation given by real embeddings and those given by complex embed dings by VR k and VC respectively Finally the equivalence class of all valuations on k is denoted by Vk 5 Local elds Given a valuation v associated to a prime ideal p E 6k we can form the metric completion kdv obtaining a locally compact complete eld kp or kv By a local eld we mean the metric completion of a number eld for a xed nonarchimedean valuation Given a local eld kp with valuation v de ne kypx k vx 1 mkypx k vxlt1 PROPOSITION 23 46 lm is a local principle ideal domain with maximal ideal mkyp As a principle ideal mkyp is generated by 7 E k and we call such an element 7 a uniformizer which is unique up to multiplication by s with ve 1 THEOREM 24 Strong Approximation Theorem Add 6 Algebraic groups Let k be a eld with chark 0 and algebraic closure E By a linear alge braic group G we mean a Zariski closed subgroup of GLn Associated to G is an ideal a lt HT where T denotes an nzituple of indeterminants one for each matrix coef cient The relationship between G and a is simple G is the zero set Va of a Consequently we call a the ideal of vanishing for G It follows from the Hilbert Basis theorem that a is nitely generated We say that G is de ned over a sub eld L C h if there exists a nite set P1T PST E LT which generates a In this case we say that G is an L algebraic group and denote the ideal generated over LT by Q Given an Lialgebraic group G lt GLn and an algebraic embedding p G a GLm de ned over L the groups G GLnL and pG GLmL are isomorphic In fact for any subring R C containing 61 the groups G O GLnR and pG GLmR are commensurable upon view ing G GLnR as a subgroup of GLm via p Thus up to commensu rability the subgroup GR G O GLnR is well de ned and called the R points of G D B McReynolds 12 7 Lattices in Lie groups On a topological group H for h E H we associate the homeomorphism Lh H a H to h where Lhg hg and by R the homeomorphism given by right translation by h A left Haar measure it on H is a regular Borel measure see 16 or 21 such that Lh is uiautomorphism for all h E H It is a fundamental fact see 16 or 21 that every Lie group H can be equipped with a left respectively right Haar measure it and its unique up to scaling For the Lie group SUn 1 this measure can selected to be bi invariant or 2isided A lattice A lt H is a discrete subgroup of H such that u descends to a nite volume measure on the coset space H A Note that since u is A7 equivariant u always descends to a measure on the coset space H A If in addition H A is compact in the quotient topology we say that A is a cocompact lattice The following fundamental result is due to Borel and Harish Chandra 5 THEOREM 25 BoreliHarish Chandra 5 Let G be a semisimple Q algebraic group GR and GZ its group of real and integral points respectively Then GZ is a lattice in GR 8 Godement s compactness criterion In the sequel we also require a compactness criterion known as Godement s compactness criterion This was proved by Borel and Harish Chandra 5 and independently by Mostow and Tamagawa 27 Before stating this the orem we require an additional piece of terminology We say thatA E GLn C is unipotent ifA is conjugate in GLn C into the group of upper triangular matrices with ones along the diagonal THEOREM 26 Godement s compactness criterion5 27 Let G be a semisimple Q algebraic group and A a lattice in G Then A is cocompact if and only if A contains no nontrivial unipotent elements A slightly stronger form of Godement s compactness criterion exists THEOREM 27 Godement s compactness criterion Strong form let A lt G be a lattice in the Q algebraic group G commensurable with GZ Then A is cocompact if and only if GQ contains no nontrivial unipotent elements PROOF If A is noncocompact we simply apply Theorem 26 to ob tain a nontrivial unipotent element x E GZ C GQ For the converse assume that x E GQ is a nontrivial unipotent element To demonstrate Arithmetic lattices in SUn 1 13 the noncocompactness of A it suf ces to construct a noncocompact lattice A1 comrnensurable with A To this end let ZX be the Zariski closure of the cyclic group ltx in G Since x E GQ this is a Qde ned subgroup of G In addition ltx is a lattice in ZX Note the inclusion L ZX a G is a Qide ned injection It follows that there exists a lattice A in GR commensurable with GZ such that Lltx C A see 32 p1657166 for a proof of this Thus by Theorem 26 A is noncocompact D 9 Restriction of scalars As an Rialgebra C can be embedded in M2 R via the map ResCRxiy This extends to GLn C by performing this expansion of each coef cient and produces an embedding ResCR GLnC a GL2nR One way to obtain the map ResCR is as follows Select an Rivector space basis for C say 1 and i For at E C we have an Rilinear map Law oc given by left multiplication This provides us with an injection of Rialgebras C a EndR2 To see that we have constructed precisely the map ResCR above let z x iy be a xed complex number In the basis Li we have taking column vectors may We as asserted Since Li is a Zimodule basis for 600 Zi not surpris ingly ResCRZi ResCRC M2Z For any other imaginary quadratic extension k of Q by selecting instead a Zimodule basis for 6k we can arrange for 69k to be precisely the Z7 points under the induced Rialgebra injection C a EndR2 afforded by More generally for any nite extension F of Q we obtain an injection F a EndRd where d F Q by selecting a Qibasis for F viewed as a didimensional inector space If in addition we select this basis to be a Zimodule basis for 63 6F F MdZ This process is known as restriction of scalars or corestriction Given a kialgebraic group G lt GLn C with k Q d this process can also be performed on G and is denoted by ReskQG Abstractly this provides us with a functor from the category of kialgebraic groups with kialgebraic homomorphisms to the category of Qalgebraic groups with Qialgebraic homomorphisms D B McReynolds 14 A useful take on restriction of scalars is the following Let legal denote the Galois closure of k over Q and GalkgalQ its Galois group For a kialgebraic group G with associated ideal a select a nite generating set P1T PT E kT for a Each element 6 E Galkgalk acts on kgalT yielding a new set of 6k7polynomials GP1T GPT E 6kT The ideal generated by these polynomials is denoted by Ca and produces a new algebraic group 6G By construction 6G is a 6k7algebraic group For two Galois automorphisms 6162 6 GalkgalQ equivalent modulo Galkgal we obtain isomorphic groups 61G and 62G Indeed the im ages of k under the action of Gal legal Q produce all of the embeddings of k a C and it is a simple matter to see that we only need to take a distinct set of coset representatives S of Galkgal Q Galkgal De ne ReskQG H 6G 669 This group is invariant under the action of Galkgal Consequently ReskQG is a Qialgebraic group By construction the groups ReskQGZ and ReskQG k are commensurable Arithmetic lattices in SUn 1 15 CHAPTER 3 A motivational example Before embarking on our rst general construction of lattices in SUn 1 we give a motivational family of lattices whose construction is simplest among the arithmetic lattices in SUn1 These groups are in the same spirit as SLn Z and the Bianchi groups PSL 2 61 1 The groups Un 1 and SUn 1 Let 171 denote the diagonal matrix diag11171 E GLn 1 C As sociated to quot71 is a Hermitian form lt n1 C 1 gtlt C 1 a C given by ltxyn1 y1n71x where gtk denotes the complex transpose of the column vectoryy The group Un 1 is the subgroup of GLn 1C of matrices A such that for all xy 6 CW1 ltAxAyn1 ltxyn1 The subgroup of those elements A of Un 1 such that detA7 1 is denoted by SUn1 The group SUn 1 though embedded in GLn 1C is not a complex algebraic group but instead a real algebraic group In order to apply Theo rem 25 we View SUn 1 as the Ripoints of an algebraic group Using ResCR described in the previous section we obtain an embedding ResCRSUn1 a GL2n 2R such that there exists a Qialgebraic group G lt GL2n 2C for which ResCRSUn1 GR see the proof of Theorem 144 Under the particular description of ResCR in the previous section we have ResCRSUn1Zi GZ In particular by Theorem 25 SUn1Zi is a lattice in SUn1 2 A family of lattices For an imaginary quadratic number eld E Q a selection of a Zibasis for 6 provides us with a Qibasis for E In turn we are afforded an em bedding ResEQ E a M2Q such that ResEQE C M2Z This extends to a Qialgebraic embedding of SUn 1 into GL2n 2R such that SUn 1 6 is precisely the set of Zipoints of SUn 1 under this em bedding THEOREM 31 SUn 1 6 is a noncocompact lattice in SUn1 D B McReynolds 16 PROOF That SUn 1 6 is a lattice follows from Theorem 25 To see that SUn 1 6 is noncocornpact note that SUn 1 6 contains a nontrivial unipotent elernent For example let v 00011 and W 170707 Then explt397vgtnl W7 lt397Wgtnl v is a uniPOtent see 13 p 119 in SUn1 E Thus by Theorem 26 SUn 1 6 is noncompact D 3 Real algebraic structure I should probably say a little more about the real algebraic structure of SUn 1 since this appears on the very last page before the appendix Arithmetic lattices in SUn 1 17 CHAPTER 4 Hermitian matrices forms and involutions In this section we introduce Hermitian matrices from three different angles The rst two are familiar to the reader acquainted with complex hyperbolic nispace or complex projective geometry However it is the third view that is most important for us in the construction of lattices in SUn 1 1 Hermitian and skewHermitian matrices In this section gtk will denote complex transposition on the matrix algebra Mn1C We say that H E Mn1C is Hermitian if H H and skewHermitian if H 7H We denote the set of Hermitian matrices and skew Hermitian matrices by n 1 and 54 11 1 It follows from elementary properties of complex transposition that both n 1 and Y 11 1 are Rivector subspaces of Mn 1C In addition it is easy to verify that n 1 54 n 1 Thus as an Rivector space Mn1C n1 5 n1 In the remainder of this note we denote n 1 simply by and the subset of invertible Hermitian matrices by X 2 Hermitian forms and unitary groups Given H 6 x we can associate to H a non degenerate Hermitian form lt by ltxyH yHx We call lt the associated Hermitianform for H Conversely given a Hermitian form lt we can associate to lt a Hermitian matrix by selecting a Cibasis 61en1 for C 1 and de ne the matrix Hlt7gt to have i joef cient given by lteieJ For a Hermitian matrix H 6 X with associated Hermitian form lt we de ne the signature tuple of H or lt to be the order pair 1711 where p is the number of positive eigenvalues of H and q is the number of negative eigenvalues of H We note that it is a classical result that H can be diago nalized and the coef cients of the associated diagonal matrix are non zero real number D B McReynolds 18 With H as above we de ne the H unitary group to be the subgroup of GLn1C consisting of matricesA such that for all xy 6 CW1 ltAxAygtH ltxygtH We denote this group by UH The special H unitary group SUH consists of those elements A E UH such that detA 1 3 Field of de nition and equivalence We say that H 6 X is de ned over a sub eld k C C if there exists a basis such that in this basis H has coef cients in k The group SUH in turn can be viewed as a kAle ned real algebraic group We say that two Hermitian matrices are isometric if there associated Her mitian pairs on C are isometric More generally we say that two kAle ned Hermitian matrices H1 H2 are equivalent if there exists at E kX R such that ocHl and H2 are isometric EXAMPLE41 Let 1 0 0 0 0 1 0 0 H4374 0 0 1 0 0 0 0 71 and 1 0 0 0 0 1 0 0 H4vzv P 0 0 1 0 0 0 0 7p for p E N For any imaginaiy quadratic number eld E H47271H4727p are EAle ned The reader can check that H4374 and H4727p are equivalent over E if and only if p E NEQEX see Section 3 for the de nition of In Chapter 6 we return to the question of deciding when a pair of forms H1H2 are equivalence over E The importance of equivalence is seen in the following result see 36 THEOREM 41 Let H1H2 be a pair of k de ned Hermitian matrices with associated unitary groups SUH1 and SUH2 Then H1 and H2 are equivalence over k if and only if SUH1 g SUH2 are isomorphic as real k R de ned algebraic groups Arithmetic lattices in SUn 1 19 4 Diagonalizing forms anisotropic and isotropic factors We refer the reader to 12 23 or 29 for the proofs of the results from this section Additional the reader can refer to 47 For a Hermitian matrix H E Mn 1k we say that H is isotropic if there exists v E C 1 such that ltvvgtH 0 and say H is anisotropic other wise Associated to H is a direct sum decomposition Van Bng C 1 such that H Vm is anisotropic and for every v E Visa ltvvgtH 0 The following is a standard application of the Gram Schmidt process THEOREM 42 Every Hermitian matrix H is equivalent to a diagonal form H diagoc1oc1 with ocj 6 69k In fact more can be said about the isotropic factor Hiw THEOREM 43 Up to k isomorphism of associated unitary groups H is equivalent to H H1 BHz where H1 diagoc1 oar is anisotropic and 1 jdimC Vim 0 1 H2 1 0 1 Given H a nondegenerate kide ned Hermitian form and an isotropic k7 vector v we can construct a non trivial unipotent element in SUHk From Theorem 43 there exists akivector v0 such that ltvvogtH 1 ltvvgtH ltv0vogtH 0 Let V0700 denote the Hiorthogonal complement of Cvvo By Theorem 43 H restricted to the space V0700 is Eide ned Select a vec tor w E V0700 such that ltwwgtH E k Finally we complete the set wvvo into an Hiorthogonal basis which is de ned over k Thus for tW expltvgtH w 7 lt wgtHv we obtain the required unipotent element THEOREM 44 If H is isotropic then SUHk contains a nontrivial unipotent element 5 Involutions associated to Hermitian matrices By an involution on Mn 1C we mean an order two map Mn 1C a Mn 1C such that A B A 3 and AB BA For a Hermitian matrix H 6 X we de ne an associated involution H by H pH 0 gtk where 11H denotes conjugation by H The following theorem is fundamental in the classi cation of arithmetic lattices in SUn 1 THEOREM 45 Weil 45 Every involution on Mn 1C is equiv alent to H for some Hermitian matrix D B McReynolds 20 We say that is of rst kind if is the identity on C and of second kind if is complex conjugation on C Finally the reader can verify that 1 SUH A e GLn1C AAH 1 Arithmetic lattices in SUn 1 21 CHAPTER 5 Arithmetic lattices 0f the rst type In this section we de ne arithmetic lattices of rst kind in SUn 1 1 Subgroups arising from forms Let E F be a totally imaginary quadratic extension of a totally real number eld F of degree s over Q Fix a complex embedding 11 of E and a com patible real embedding 61 of F Using these embeddings we can identify E C C and F E R Given an EAle ned Hermitian matrix H of signature 111 we have the associated group SUH For each embedding L39J39 7 11 and compatible real embedding GJ39 7 61 we obtain a new Hermitian matrix by applying L39J39 to the coef cients of H We denote the resulting Hermitian matrix by II H and new associated group SU II H by If SUH In total we obtain an injection SUH a TfSUH j1 Via the diagonal embedding The latter group is nothing more than RespQ SUH which is Qide ned By construction SUH 6E and Resp0SUHZ are commensurable Finally observe that we have a natural projection 7r Resp0SUH a SUH with kernel ker7r H II39SUH j2 D B McReynolds 22 Under the mapping 7 7rResFQSUHZ and SUHE are com mensurable To obtain lattices in SUn 1 we can use the diagram 1 1 11 It SUH RespQSUH SUH 1 We will return to this later see 2 when we discuss kiforms of a Lie group 2 Admissible pairs and arithmetic lattices 0f the rst type The following lemma appears in 24 Lemma 813 and will be used re peatedly LEMMA 51 Let X Y and Z be locally compact topological groups A lt Z be a lattice and 1Yzx1 a split exact sequence Then MA F is a lattice in X if and only if Y is compact In addition if Y is compact then A is cocompact if and only if F is cocompact We are now able to prove the following THEOREM 52 SUH 6 is a lattice in SUH ifand only ifrf SUH is compactfor all j 2 s PROOF From our discussion above we have the short exact sequence 1 112 It SUH 5 ResFQSUH 7 SUH gt1 and this has been constructed so that RespQSUH is a Qireal algebraic group and 7rResFQSUH and SUH 6 are commensurable By Theorem 25 the group RespQSUHZ is a lattice in RespQSUH The theorem now follows from an application of Lemma 51 D The reader can see that this construction is a generalization of the construc tion of the lattices SUn 1 6E where E is an imaginary quadratic number eld In this case since E has only one complex embedding up to complex conjugation the compactness condition was vacuously satis ed With this Arithmetic lattices in SUn 1 23 said we call a pair H E F admissible if II SUH is compact for each j 2 s EXAMPLE 51 Let E Q i F Q and H diag1 1 1 i We as sert that SUH 6 is a lattice in SUn1 To see this rst note that H has signature n 1 Under the other embedding of E we obtain the Hermit ian matrix IH diag111 which has signature n 10 Thus ISUH is compact and so SUH 6 is a lattice in SUH 2 SUn1 COROLLARY 53 IfHEF is admissible then SUH 6 is nonco compact if and only if E is an imaginary quadratic number eld PROOF Since any unipotent in SUH 6 produces a unipotent in the group ker 7 it is enough to prove that compact algebraic groups cannot possess any nontrivial unipotent elements For a compact algebraic group K lt GL n C with associated Lie algebra E and a nontrivial unipotent element x E K we can conjugate K in GLn C such that g lxg is upper triangular with ones along the diagonal For g lxg there exists X E t such that expX x In fact RX maps embeds into K which violates the compactness of K The proof is completed by applying Theorem 26 B Any lattice A lt SUn 1 which is commensurable in the wide sense with a lattice of the form SUH 6 for an admissible pair H E F is called an arithmetic lattice of rst type 3 Commensurability classes of lattices of rst type We conclude this section with the following result THEOREM 54 There exist in nitely many distinct commensurability classes of lattices of the rst type in SUn 1 PROOF LetEF be as above and setH a1oc1 for a1oc1 F By Theorem 21 we can select a1oc1 such that 051 lt 0 ocj gt 0 and for all Cg 7 idF Maj gt 0 for j 1n 1 By Theorem 52 SUH 6 is alattice in SUn1 We assert that iftwo lattices SUH1 6E1 and SUH2 6E2 are commensurable in the wide sense then E1 2 E2 This follows from the commensurability invariance of the invariant trace eld see 2 d f kSUHJ6 Ej 6 Tm11 ye SUHJEj in combination with kSUHJ 6E Ej D Arithmetic lattices in SUn 1 25 CHAPTER 6 Classifying Hermitian structures over elds In this chapter we give a list of invariant which determine the similarity class of a Hermitian form on a vector space V More generally this holds for Hermitian structures on free modules of division algebras see Chapter 11 The main reference for this section is 36 Ch 10 1 Signature and dimension Let V be a nite dimensional E F vector space equipped with a Hermitian pairing H We denote the associated Civector space by VC and the Hermit ian pairing extended to VC by H Above we associated to H a signature pair 1711 A related invariant on the form H is the number 6H 1p 7 11 For two different pairings H1 and H2 we have the following see 12 THEOREM 61 As real algebraic groups SUH1 and SUH2 are real isomorphic if and only if 6H1 6H2 For an Eide ned form H on Vc we have the pair dimE V 6H 2 Signatures at other archimedean places For a different embedding L39J39 of E into C we obtain a new signature for any Hermitian form H on V If V denote the set of inequivalent complex embeddings of E for each v 6 V0 we obtain a signature 6Hv called the visignature 3 Determinant For a Hermitian form H on VC by selecting an Eibasis for V we can associate to H a matrix THWQ The determinant of this matrix is inde pendent not independent of the selection of the basis but is indepen dent viewed as an element of F XNEFEX This produces an invariant detH E F X NE F E X which we call the determinant of H and denote by detH D B McReynolds 26 4 Classifying equivalence classes of Hermitian forms The following result can be found in 36 THEOREM 62 Classi cation of forms Two forms H1 and H2 on V and de ned over E F are equivalent if and only if 6H1 v 6H2v for all V E VDo and detH1 detH2 As a consequence of this result if we x the signatures of at each v E V we see that there are precisely two classes of Hermitian forms For any form H over E F on a Civector space V we form the invariant tu ple dimV 6VHdetH which takes values in N gtlt Nme gtlt FX NEFEX If we denote this tuple by InvH the classi cation theorem states that H1 and H2 are equivalent over E if and only if InvH1 InvH2 5 A few examples 51 Signature 2 1 forms over Qjd For an imaginary quadratic number eld E VNE consists of precisely one complex em bedding up to complex conjugation So long as d 31 1 71 NEQE X According Theorem 62 over E we have the possible invariants 331 3371 301 and 3071 Of interest for us are the latter two since these correspond to Hermitian forms of signature 12 or 21 For 30 1 we take the Hermitian form 71 0 0 11301 0 i1 0 0 0 1 For 30 71 we take the Hermitian form 1 0 0 H3974 0 1 0 0 0 71 By Theorem 62 every form H on C3 de ned over E is equivalent to one of these two forms However note that H3071 7H37071 Thus SUH37071 and SUH37071 are isomorphic Over E Qx 71 we take H3974 again and the form 1 0 0 H303 0 1 0 0 0 i3 Arithmetic lattices in SUn 1 27 Since 73 NOWQ Qi X these two forms represent all the possible classes over Qi On the other hand H37073 is equivalent to 3 0 0 130527 0 3 0 0 0 73 Since H3VOVL27 3H3VOVL1 the groups SUH37073 and SUH37071 are iso morphic THEOREM 63 For any imaginary quadratic number eld E Q and every Hermitian form H of signature 2 1 de ned over E SUH 6 and SU2 1 6 are isomorphic as real Q algebraic groups COROLLARY 64 For E and H as above every arithmetic lattice in SUH is commensurable in the wide sense with SU2 1 6 COROLLARY 65 Every Hermitianform H ofsignature 21 over an imaginary quadratic number eld is isotropic 52 Signature n 1 forms over Qx7d even n ForE Qx7d there are two classes of Hermitian forms over E of signature n 1 or equiv alently 1n As before we have a pair of Hermitian forms 10000 0 1 0 0 0 0 01 0 0 Hn1071 0 0 0 1 0 0 0 0 0 71 and 71 0 0 0 0 0 71 0 0 0 0 0 71 0 0 Hn1olt71gtquot 0 0 0 71 0 0 0 0 0 1 For d 7 1 and n even the forms Hn17071 and Hn17071 represent the two possible classes For E Qx 71 the remaining exceptional case can be handled as before THEOREM 66 Let E be an imaginary quadratic number eld H a Hermitian form over E of signature n1 with n even Then SUH and SUn 1 are isomorphic as real Q algebraic groups D B McReynolds 28 COROLLARY 67 For E and H as above every arithmetic lattice in SUH is commensurable in the wide sense with SUn 1 6 COROLLARY 68 Every Hermitianform H ofsignature n 1 over an imaginary quadratic number eld with n even is isotropic 53 Signature n1 forms over Qjd odd n When n is odd the above forms Hn17071 and HnHkaU represent the same class In this case it follows that detH viewed as an element of Q is negative For xed d 7 1 select aprime integerp such thatpld Thenp Z NQHQQjd X It follows then that any form H with detH 7 p will represent the trivial class since neither p and 71 are norms De ne l 0 0 0 0 0 l 0 0 0 0 0 1 0 0 Hn vov l 0 0 0 1 0 0 0 0 0 7p Then Hn17071 and Hn1707p represent the two possible classes of Her mitian forms over E When d 1 71 is a norm and for any selection of p Hn1707p represents the nontrivial class in QXNQiQQix In this case Hn170771 represents the trivial class THEOREM 69 Let E be an imaginary quadratic number eld and Hn170771 and Hn170 p be as above For any E de ned Hermitian form H of signature n 1 SUH is isomorphic as a real Q algebraic group to exactly one ofthe groups SUn 1 or SUHn1707p COROLLARY 610 For E and H as above every arithmetic lattice in SUH is commensurable in the wide sense to either SUn1E or SUHH1707 P 69E 6 Parity and describing commensurability classes In this section we observe a general reduction of the equivalence classes of Hermitian forms a special case of is Theorem 66 Note that Theorem 611 implies Theorem 69 whose proof was omitted Let E F be a totally imaginary quadratic extension of a totally real number eld with a xed compatible embeddings 61 F a R and 11 E a R As always we now view 61 idF and 11 idE Let H be a Hermitian form such that 661 n 71 and for all GJ39 7 61 66 n 1 According to Theorem 62 there are two equivalence classes of Hermitian forms on E F with this prescribed signature Arithmetic lattices in SUn 1 29 THEOREM 611 Parity theorem a If n is even there is precisely one wide commensurability class for arithmetic lattices of rst type de ned over E F b If n is odd there are precisely two wide commensurability classes of arithmetic lattices of rst type de ned over E F PROOF For a it suf ces to show that any Hermitian form H over EF with EFH admissible then SUH and SUH are isomorphic as real F ialgebraic groups Equivalently we must show that H and H are equivalent For this lnvH and lnvH can differ only on the determinant By Theorem 62 if lnvH lnvH H and H are equivalent Thus we may assume that detH 7 detH viewed as elements of F X NEFE X Without loss of generality assume that detH represents the nontrivial class in F X NEF EX It follows that H and detHH are equivalent On the other hand detdetHH detH 2 Since n is even detH 2 rep resents the trivial class in F X NEF EX Thus by Theorem 62 detHH and H are equivalent and so H and H are equivalent For b let H and H be such that EFH and EFH are admissible and detH and detH represent the trivial and nontrivial class respec tively If H and H are equivalent there exists or E F such that ocH H However detocH an detH Since n is odd an1detH detH in FXNEFEX By assumption an1detH detH which is a con tradiction of our selection of H and H D 7 All Hermitian forms over Q 7d are isotropic In this section we prove the following theorem THEOREM 612 Let E be an imaginary quadratic number eld H an E de ned Hermitian form of signature n 1 over E Then H is isotropic over E The reader can compare this result with the fact that not every QAle ned bilinear form of signature n 1 is isotropic over Q We will return to this in a moment For n even we proved this result above and so it remains to treat the case when n is odd This is achieved instantly for all but n 3 from the following theorem of Kneser see also 36 THEOREM 613 Kneser add reference Every Q de ned bilinear form of signature n 1 with n gt 3 is isotropic over Q D B McReynolds 30 As a result of Theorem 613 the form Hn1yo p and Hn17071 are both isotropic viewed as bilinear forms on Qn so long as n gt 3 The remaining case of n 3 we only need to consider the form H4707p There are several ways to deduce that this form is isotropic over E for instance we could simply nd an isotropic vector As it introduces an im portant tool we introduce the associated quadratic form QH for a Hermitian form H and deduce that H4707p is isotropic from Theorem 613 For any Hermitian form H on V and any v E V H vv E R Viewing V as a real vector space of real dimension 2dimc V we thus obtain a quadratic form QHZ V a R given by QHv Hvv The following result in combination with Theorem 613 implies H4707p is isotropic over E see 36 THEOREM 614 Let H be a Hermitian form over EF and QH be its associated quadratic form over F Then H is isotropic over E if and only if QH is isotropic over F 8 Local theory for Hermitian forms In this section we discuss some local to global results for Hermitian struc tures on free modules over division algebras Additionally we state Witt s theorem Arithmetic lattices in SUn 1 31 CHAPTER 7 Cyclic division algebras In this chapter we introduce the required background needed to generalize the above construction Before introducing a multitude of new algebraic objects required in the con struction we motivate this by reviewing our construction of lattices of rst type in SUn 1 We begin with a Eialgebra Mn 1E together with an involution H de termined by a Hermitian matrix of signature n 1 Using the involution H we obtain a real F ialgebraic group SUH which is isomorphic to SUn 1 In the case E Q is an imaginary quadratic extension of Q we obtain a dis crete subgroup of SUn 1 by taking the discrete subring 6 of C When E is not an imaginary quadratic extension of Q to obtain a discrete subgroup of SUn 1 from the indiscrete subring 6 C C we further insist that the groups If SUH be compact at the other complex embeddings of E 1 Central simple algebras Throughout E will denote a number eld The reader should keep in mind that E will be a totally imaginaly quadratic extension of a totally real num ber eld F for our applications By an E algebra A we mean an Eivector space equipped with an associa tive multiplicative structure We say that an Eialgebra is simple if there are no non trivial two sided ideals in A We say thatA is a central E algebra if the center of A denoted by Z is E Finally if every element of A 0 is invertible in A we call A a division algebra EXAMPLE 71 o E is a central simple Eialgebra o is a central simple Eialgebra 0 Let A be a 441imensional Eivector space with the basis 1xyz such that x2a y2b xyz7 yxxy D B McReynolds 32 The algebraA is a simple central Eialgebra and is called a quater nion algebra Associated to A is the so called Hilbert symbol b 7 o For any central Eialgebra A MrA is a central Eialgebra which encodes all the date necessary to retrieve the algebra 2 The SkolemNoether and Structure theorems We require the following two fundamental results in the sequel We refer the reader to 22 24 or 30 for the proofs THEOREM 71 Skolem Noether theorem LetA be a simple k algebra Then Autk A lnnA A useful corollary of Theorem 71 is COROLLARY 72 LetA be a simple k algebra with simple subalgebra B Then every automorphism of B extends uniquely to an automorphism of A The second result required throughout the remainder of this note is the Wed derburn Structure Theorem THEOREM 73 Wedderburn structure theorem letA be a simple cen tral nite dimensional E algebra Then there exists a central E division algebra D and an integer r such thatA E MrD as E algebras 3 Cyclic algebras Our primary interest is in a special type of algebra known as a cyclic alge bra Given a cyclic extension L E of degree d with Galois group GalLE lt6gt and at 6 Ex we de ne a central simple Eialgebra LE 6 a by LE6otle JXl jeL j0 subject to the relations Xd 0 X6 6 X 3 e L PROPOSITION 74 LE 6 0 is a central simple E algebra ofdimen sion of2 over E as a vector space PROOF That LE 6 a is an Eialgebra of dimension d2 follows im mediately from the de nition of L E 6 a To see that the center of this algebra is precisely E note that the center must be contained in E since Fix6 E One can check using the relations above that E is central which demonstrates the reverse containment To prove that E is simple note that any two sided ideal I C A lifts to an ideal IL in A 8 L However this algebra is isomorphic as an Lialgebra to MdL which is simple D Arithmetic lattices in SUn 1 33 EXAMPLE 72 LetF Q E Qi andLEd Then ixd For at E E LE6oc The following two examples will be important in the sequel see 3 EXAMPLE 73 Let F o E QM and L QltC7 Then 6a 72 A LE 6 is a cyclic Eialgebra EXAMPLE 74 LetF Qm E Km C3 L QltC21 Then 9C21C 13nd A LE 6 4 3 is a cyclic Eialgebra The norm of an element of L over E is de ne by dil I Now H 9W j0 The following theorem is due to Wedderburn see 30 p 2787279 for a proof THEOREM 75 Wedderburn a fad NLELxforj1di 1 then L E 6 0 is a division algebra In addition if d is prime this is necessary as well b or E NLELX ifand only ifLE 6 oz 2 MdE 4 Splitting elds and degree The neXt proposition allows us to View the cyclic algebraA LE 6 as inside MdL PROPOSITION 76 A 8 L g MdL as central simple L algebras PROOF To begin we surely must map the eld E to the scalar matrices Speci cally for at E E a 0 0 0 a Xi gt D B McReynolds 34 To ensure that L C A is not central we can not be so free with how we de ne the homomorphism on L For 3 E L n 0 0 0 0 6 0 0 3H 0 0 62 0 5 5 5 I15d455 Finally for a general element 30 31X d71Xd717 wehave 30 31 32 dil 059l3d71 9l30 9l31 9l3d72 0592l3d72 0592l3d71 92l30 62 d73 aed9lltmgt enemas aed9llt 3gt Q Woo That this is an injective homomorphism is left for the reader D We say that an extension K of E splits A ifA 8 K 2 MmK PROPOSITION 77 Every nite dimensional E algebra splits over a nite extension of E We refer the reader again to 30 p 239 for a proof A stronger result holds and can be useful in practise PROPOSITION 78 Let L C A be a maximal sub eld ofa simple E algebra A Then A splits over L We note in passing that every nite dimensional algebra over E splits over C To see this we apply Proposition 77 Speci cally if a nite dimensional Cialgebra A did not split over C then there would be a nontriVial nite extension L C Using Proposition 77 we de ne the degree of A to be the minimum degree over E of a splitting eld for A and denote this by degE PROPOSITION 79 IfA LE 6 0 is a division algebra then degE A L More generally degEA l L The index indA is the unique positive integer such that indE A2 degEA2 dimE A Arithmetic lattices in SUn 1 35 By Theorem 73 A MmD for some central EAlivision algebra D Alternatively the index and degree of A are given by indE A m and deg A deg PROPOSITION 710 Let A be a simple E algebra LE a nite exten sion and degA L Then L is splits A if and only if A contains a maximal sub eld isomorphic to L 5 Orders in algebras In building lattices in SUn 1 using Hermitian matrices our lattices were constructed up to commensurability as stabilizers of lattices in Hermitian vector spaces Indeed for an admissible pair E FH where H has signa ture n 1 the group SUn 1 6 is precisely the stabilizers of the lattices 631 in C 1 under the action of the Hiunitary group SUH We now in troduce orders in algebras which will play the role of the lattice 6 in the vector space CW1 51 The basics For an Eialgebra A by an E order in A we mean a nitely generated subring of A such that a 6 is a nitely generated Eimodule and b A 6 635 E as an Eimodule EXAMPLE 75 o 6 is a Emrder in E For any nite extension LE 6 1 is a 6E7 order in L o Md 6 is an Emrder in MdE o ForA 71671 Z1ijk is aliorder o ForA LE6oc Big LXj is a Emrder 52 Existence of orders maximal orders In order to build lattices in SUn 1 we require the following existence theorem for orders in a central simple Eialgebra THEOREM 711 Existence of orders 35 For any E algebm A there exists a E order 6 of A In describing commensurability classes the existence of maximal order is quite useful By a maximal order the mean a Eiorder of an Eialgebra A which is not properly contained in any Eiorder of A D B McReynolds 36 THEOREM 712 Existence of maximal orders 35 LetE be a number eld or local eld A an E algebra and 6 a E order of A Then 6 is contained in a maximal E ordex Arithmetic lattices in SUn 1 37 CHAPTER 8 Involutions and Hermitian elements 1 Involutions of rst and second kind 11 The basics In order to relate cyclic central Eialgebras A to the special unitary group SUn1 we require an additional structure on A called an involution of second kind An involution gtk onA is a map 2 A a A such that W x y and of yW We say that gtk is of rst kind if E id and say that gtk is of second kind otherwise Our interest is when E F is a totally imaginary quadratic extension of a totally real number eld F In this case if A is an cyclic Eialgebra we seek an involution of second kind gtk which extends the Galois involution on E F EXAMPLE 81 For a cyclic extension LE with E F as above and or E NLELX the cyclic algebra LE6oc is isomorphic to MdE where d L The algebra MdE admits an involution of second kind which extends the Galois involution on E F Speci cally complex transposition is such an involution Given a cyclic Eialgebra A equipped with an involution of second kind gtk which extends the Galois involution on E in the splitting A 8 L we obtain a new involution gtk on MdL By Theorem 45 gtk H for some Hermitian matrix H 6 X We call gtk standard if in the splitting A 8 C gtllt produces complex transposition The following result is due to Albert see also 22 or 36 PROPOSITION 81 Albert 1 IfA is a simple central E algebra with an involution gtk and L E is any extension then A 8 L also admits an invo lution which extends the involution By a unitary algebra over E F we mean a cyclic Eialgebra A equipped with a standard involution of second kind In the sequel we require some thing more than For each embedding L39J39 of E we insist that gtk inA grim C D B McReynolds 38 be complex transposition In the remainder of this note we call such an involution standard and the pair A a unitary E algebra THEOREM 82 36 Any E algebra A equipped with an involution of second kind admits a standard involution 12 Albert s criterion 0n existence of involutions In order to con struct unitary algebras we require a means of deciding when a given cyclic Eialgebra possesses an involution of second kind We give the rst of two theorems achieving this see 1 p 162 or 22 p 3amp39 Let A LE 6 oz be a cyclic algebra and E F be as before a totally imaginary quadratic extension of a totally real number eld By Proposi tion 81 the Galois involution on E F can be extended to an involution on L L 8 E and we denote the totally real xed eld of this extension by K This yields the lattice of elds 2 L THEOREM 83 Albert 1 LE 6 oz admits an involution ofsecond kind extending the involution gtk E GalEF if and only if there exists 3 E KX such that PROOF Since in application one seeks an explicit involution we prove the converse direction whose proof is constructive see also 15 Given 3 E K such that N K F 3 NEFoz we construction an involution which extends gtk on EF Using 3 we de ne LE6oz a LE6oz by Xj 9l39j 1l3Xj 17 3 ii I3 E L With this the extension to all of LE 6 oz is given by dil I dil I lt2 WW 2 WWW j0 j0 To see that is an involution we must show Y Y for all Y E A For this we need to know XJ 1 Since Xd oz X 1 oz le 1 and more Arithmetic lattices in SUn 1 39 generally XJ39 1 a le j It suf ces to shows that X X Indeed we have WV aileily a 1 6 6d 1 a 1X NKFUDNEFW UX NKF X X NEFW 2 Hermitian elements in unitary algebras and involutions Let Agtk will denote a unitary algebra over E F We say that h E A is Hermitian or 7Hermitian if x x and skewHermitian if x 7x We denote the F ivector subspace of Hermitian and skew Hermitian elements by and 54 respectively PROPOSITION 84 As an F vector space A BY 21 Hermitian involutions Given a Hermitian element h 6 X we associate to h an involution h onA by ac uh o By Theorem 45 applied to A 8 L we obtain an involution h which is equivalent to H for some Hermitian matrix H Since gtk is standard it viewed as an element of MdL is a Hermitian matrix and is precisely the Hermitian matrix which yields the involution h We de ne the signature of h to be the signature of H PROPOSITION 85 The signature of h is independent of the splitting PROOF Let L1 L2 be two minimal splitting elds of A and h E A a Her mitian element We assume in both splittings of A that gtk produces complex transposition Notice rst that if we compute the signature of h in MdLJ or MdLJ Lj C we obtain the same answer Let 171111 and 172112 be the signatures of h computed in MdL1 C and MdL2 8 C These alge bras are both isomorphic to Md C By Theorem 71 this isomorphism can be viewed as an inner automorphism of Md C However the signature of a Hermitian matrix is certainly preserved under conjugation 22 Unitary groups Using 1 as a guide for a Hermitian element h 6 X we de ne the h special unitary group by SUh difpce A E CX xxh 1 Since A 8 CX 2 GLdC ifh has signature n 1 SUh 2 SUn1 More generally SUh SUpq where 17 11 d and it has signature p711 Arithmetic lattices in SUn 1 41 CHAPTER 9 Arithmetic lattices in SUn7 1 In this section we give the most general construction of arithmetic lattices in SUn 1 1 Second type lattices over Q For lattices of rst type in SUn1 when E Q was an imaginary qua dratic extension the groups SUn 1 6 were lattices in SUn 1 How ever when E was a larger extension of Q the indiscreteness of 6 C C complicated the matter considerably This persists in our constructions of lattices in this section For this reason we begin with the case E Q is an imaginary quadratic extension and generalize the construction to arbi trary totally imaginary quadratic extensions of totally real number elds for which this construction is a special case Let A be a unitary division algebra of degree n 1 over an imaginary qua dratic extension E of Q For a Hermitian element h 6 x of signature 111 the group SUh is isomorphic to SUn 1 via the splitting isomor phism A 8 C 2 Mn 1C Given an Eiorder in A we form the group def SUh x 6 6 xxquot 1 THEOREM 91 SUh 6 is a cocompact lattice in SUh PROOF That SUh is a lattice follows from Theorem 25 since SUh can be viewed as the Zipoints of the real Qialgebraic group SUh We demonstrate cocompactness by arguing the contrapositive If SUh is not cocompact by Theorem 26 there exists a nontrivial unipo tent element x E SUh This in turn produces the nontrivial nilpotent element x 7 1 E A which opposes our assumption that A is a division alge bra D 2 Second type lattices over arbitrary number elds Let E F be a totally imaginary quadratic extension of a totally real num ber eld F We denote the distinct complex embeddings of E be 11 L39S and the compatible real embeddings of F by 61 as Via 11 and 61 we D B McReynolds 42 identify E C C and F C R such that E R F Given a cyclic exten sion LE for each embedding L39J39 of E we obtain a family of embeddings 17 itrjyj We select a xed embedding 7LJ for each L39J39 and view L C C via 7L1 Given a cyclic Eialgebra A LE 6 05 via the embedding L C C af forded by 7L1 and the splitting A 8 L 2 MdL for each embedding 7LJ 7 7L1 we obtain a new algebra MA MD1x3le e oi ma Up to an Eialgebra isomorphism this algebra is independent of the selec tion of 7LJ among the family My j ML j We denote this algebra by II A In addition if A is a unitary algebra with Hermitian element h 6 X for each L39J39 7 11 we obtain a new Lie group If SUh SU Tilt As before applying Res F Q we obtain the short exact sequence 1 112 It SUh 4 ResFQSUhb SUh 1 We start with A a unitary division algebra of degree n 1 over E with a Hermitian element h of signature 111 Given an Emrder in A SUh maps to a subgroup of RespQSUh which is commensurable with RespQSUhZ and in the projection 7r RespQSUhZ maps to a subgroup which is commensurable with SUh 6 THEOREM 92 SUh is a cocompact lattice in SUnl ifand only ifrf SUh is compactfor each j 2 s PROOF That SUh is a lattice is identical to the proof of Theo rem 52 For cocompactness we can argue as in the proof of Theorem 91 or Corollary 53 We call the lattices in Theorem 92 arithmetic lattices of second type 3 Mixed type lattices Let nd 6 N rd 11 1 E F a totally imaginary quadratic extension of a totally real number eld and A a unitary algebra over E F of degree d The simple Eialgebra MrA admits an involution of second kind given by 7transposition If L is a splitting eld for A then MrA 8 L MrA 8 L MrMdL MrdL We assume that the involution gtk on MrA is standard in this splitting Arithmetic lattices in SUn 1 43 For a Hermitian element h E MrA with associated twisted involution set SUhA x E MrA xx 1 If in the splitting MrA 8 L h has signature 111 SUhA 8 C 2 SUn1 Given a triple Agtkh above for the pair rd we say that A h is admissible if for all L39J39 7 idE the group If SUhA 8 C is com pact THEOREM 93 Let Agtkh be admissible over E F with associated pair rd Thenfor any E order 6 SUh 6 is a lattice in SUn1 PROOF This follows from Theorem 25 and Lemma 51 D We call the lattices in Theorem 93 arithmetic lattices of mixed type Note that both arithmetic lattices of rst and second type are of mixed type EXAMPLE 91 Let K be as in 2 a1oc E K and h diagoc1oc In the splitting given by Proposition 76 h diagoc16a162a16d 1a1a26d 1oc By Theorem 21 there exists on a E K such that 1 a1lt 0and ocj gt 0 forjgt 1 and 2 6Xj gt 0forall 1d71andj1r Thus A h is admissible and so by Theorem 93 SUh is alattice in SUn 1 for any Eiorder in A THEOREM 94 Ifrd gt 1 and A h is an admissible triple over EF for the associated pair r d then SUh 6 is cocompact for any E order PROOF To begin we note that Theorem 42 Theorem 43 and Theo rem 44 hold in this setting On the free Aimodule A there exists a basis e1 e such that h ha Bhiw and this decomposition coincides with the Aimodule decomposition Ar Ael 7 werim Aerim1z mgr Aan Aisa Since h is non degenerate dimAAlw is even In particular m is even In the splitting afforded by L ha has signature say p 11 It is a simple matter now to see that h has signature p dm2q dm2 Note here that p q d r 7 m and so pqdmdr7mdmdr D B McReynolds 44 as expected However d gt 1 and h n1 which is impossible since dm2 gt 1 and q 2 0 unless h ha By Theorem 26 if SUh is non cocompact SUh contains a nontrivial unipotent element However by Theorem 44 the existence of such an element is equivalent to h 7 ha Therefore by Theorem 26 SUh is cocompact 4 Admissible triples and the classi cation theorem In total we have the following theorem THEOREM 95 Classi cation of arithmetic lattices Let A h be an admissible triple over E F with associated pair rd and 6 an E order in A Then SUh 6 is an lattice in SUn 1 via the injection induced by the isomorphism of SUh with SUn 1 given by the splitting isomorphism MrA 8 C 6 Mn 1 C Moreover ifr 1 andE is not an imaginary quadratic extension of Q or r gt 1 then SUh 6 is cocompact REMARK By Theorem 71 if we change the splitting and hence the isomorphism between SUh and SUn 1 we obtain a new lattice which is commensurable in the wide sense with the original one Thus up to wide commensurability this is independent of the selection of a splitting isomorphism One corollary which the author made use of in 25 is COROLLARY 96 Classi cation of noncocompact arithmetic lattices Let A lt SUn 1 be a noncocompact arithmetic lattice Then there exists an imaginary quadratic extension E Q and an E de ned signature n 1 Hermitian matrix H such that A is commensurable in the wide sense with SUH 6 In combination with Theorem 611 we have an very explicit description of the noncocompact arithmetic lattices in SUn1 5 Tits symbols and a summary of the lattices For an admissible triple Agtkh with pair rd the elements of reduced norm one A 8 C1 is an outer real F form of SLn C In the notation of Tits see 6 these groups are denoted by 2Air71 We associate a Tits 0 symbol to each of the above groups and their lattices namely Unvril Arithmetic lattices in SUn 1 45 Arithmetic lattices in SUn 1 47 CHAPTER 10 Examples for small pairs In this chapter we give several different examples of arithmetic lattices in SUn1 1 Lattices of type Ugyz Our rst class of examples appeared in our discussion of arithmetic lattices of rst type The lattices that arise from Uiz are described as follows For E F and Theorem 611 there is at most one class of Hermitian matrix for each of the 21F 10 possible signatures Without loss of generality select H diagoc1 a2 053 for ocj 6 63 Since Theorem 96 treated the case when F Q we assume that F 7 Q To construct cocompact lattices we apply Theorem 21 Speci cally select 11 a2 053 such that 1 051052 gt 0 and 053 lt 0 and 2 for all 6g 7 idF Maj lt 0 or Maj gt 0 for j 123 EXAMPLE 101 LetF Qgd ENand squarefreeE Qgi SetHdiag117g Then SUH 6 is a cocompact lattice in SU21 2 Lattices of type U317 For EF and a Hermitian form H E Mn 1E up to Eiisomorphism of the associated unitary groups by Theorem 42 we can assume that H diagoc1ocn1 with ocj 6 63 By Theorem 21 select a1ot1 such that 1 an1lt 0 ocj gt 0forj 7 11 1 and 2 for all Cg 7 idF Maj gt 0 or Maj lt 0 for all j Then SUH 6 is a lattice in SUn1 THEOREM 101 Every lattice of type U n is commensurable in the wide 7 sense to SUH Ef0r some pair HEF as above EXAMPLE 102 LetF Qgd ENand squarefreeE Qgi SetHdiag1117g Then SUH 6 is a lattice in SUn1 D B McReynolds 48 3 Lattices of type U370 From several perspectives the lattices of type U7 are well understood The simplest lattices not of this type are of type U370 For E F LE a cyclic Galois extension of degree 3 and K as in 2 let A be a cyclic division algebraA LE 6 oz which admits an involution of second kind For a Hermitian element h E AX we insist that it have signature 21 and at each Tg 7 idE we insist that 61h have signature 30 or 03 Then for any Emrder ofA SUh is a lattice in SU21 THEOREM 102 Every lattice of type U370 is commensurable in the wide sense to SUh for some EF and A h 6 as above Our rst concrete example produces a lattice whose associated arithmetic orbifold is commensurable with Mumford s fake CP2 see 28 We refer the reader to 18 for more on this EXAMPLE 103 LetE F Q L Q 7 and K Qcos27r7 Set A 1 H 05LZ In addition set 6 to be the Galois automorphism 6 L L 6 7 72 The reader can check using Theorem 75 thatA L E 6 oz is a division algebra According to 18 see below 1 1 lnvL A g lnvIA g and any prime which does not divide 2 InvpA 0 Thus by Theo rem 128 A admits an involution of second type Concretely set X EXZ 3 3 3 6L h 7L J 71XZX2 e e1 e x e xz Then SUh is an arithmetic lattice in SU21 By Yau s uniformization theorem the manifold MMum Mumford constructed is a complex hyperbolic 27manifold of minimal Euler characteristic By Wang s theorem 41 MMum is a minimal volume complex hyperbolic 27 manifold Arithmetic lattices in SUn 1 49 THEOREM 103 Kato 18 LetM HSUh 6from 103 Then MMum and M are commensurable More generally Klingler 20 proved that every so called fake CP2 is arith metic of type U370 We refer the reader to 17 33 38 39 and 48 for more on fake CP2 manifolds and piadic uniformization Our next example comes from Amitsur 3 where nite subgroups of divi sion algebras over elds of characteristic 0 are classi ed EXAMPLE 104 LetF Qm E Q7C3LQC21 andKQCOS27F21 ThenA LE 6 4 3 where 6 L a L 6 21 1 is a division alge bra In addition using either Theorem 83 or Theorem 128 one can check that A admits an involution of second kind Finally set it 21 2210 7 Q1 7 E K Having not given the involution gtk on A we can still verify that h is Hermitian since L is nothing more than the non trivial Galois au tomorphism of GalLK To see that it has signature 12 in the splitting A LE given by Proposition 76 h 0 0 h gt gt 0 6 h 0 0 0 62h At the other embedding of F it has signature 30 Thus by Theorem 95 SUh is a lattice in SU2 1 for any Eiorder of A The algebraA in 104 is the group algebra QG2174 where var xy xm 1 y yxy 1xr for any pair m r of positive relatively prime integers and s r71m t ms and n is the multiplicative order of r in Z mZ X More generally we have the algebras 7L Amw ltQltCmgtQltCS7 Cm H C517 2 C55 j0 which generalize 104 Lattices of type U30 and U2 differ greatly see 26 THEOREM 104 a Every lattice of type U2 contains a totally geodesic real hyperbolic 2 orbifold group

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