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# Topics in Analysis 22M 303

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LECTURE NOTES ON GROUP REPRESENTATIONS AND DEFORMATIONS Taken by Jos A V lez Marulanda Graduate Student Depar nent of Mathematics UNIVERSITY OF IOWA January 2008 A mi padre use Lem Velez quotEl crem tscula gris se habit hecha negro y en la invasi n creciente de la sambm la gum del gum rebelde se prayectaba enhiesta en la tiniebla estremecida en la difusi n rajiza de la haguem Home In sambm de un mumll n intacto entre las llamamdas del incendia coma la siluetu delfamll n gigantesca en noche de bOVVQSC butidu pay his alas cubierta par la espuma commute de nukes y myas quot Ibisquot Vargas Vila 1860 1933 INTRODUCTION Part 1 REPRESENTATIONS IN CHARACTERISTIC ZERO Chapter 1 SEMISIMPLICITY Semisimple Rings and Modules 2 Semisimple Group Rings 3 Wedderburn Decomposition Theorem for Semisimple Rings Chapter 2 THE GROUP ALGEBRA 1 Generalities 2 Connection Between Representations of G over F and PG modules 3 Ordinary Characteristic Zero Representation Theory 4 Decomposition of C 5 Integrality Properties of Characters Chapter 3 CHARACTER THEORY Orthogonality Relations for Characters 2 Canonical Decomposition of a CG module Chapter 4 INDUCED REPRESENTATIONS 1 Tensor Products Direct Products Induced Modules The Character of an Induced Representation Mackey s Theorem Mackey s Irreducibility Criterion Chapter 5 EXAMPLES OF INDUCED REPRESENTATIONS 1 Induction from a Normal Subgroup 2 Super Solvable Groups SV P FP CONTENTS iv CONTENTS 3 The Ring RG 4 Grothendieck Groups Chapter 6 A THEOREM BY BRAUER 1 p Regular Elements and p Elementary Subgroups 2 Induced Characters Arising from p Elementary Subgroups 3 Construction of Characters 4 Brauer s Induction Theorem Chapter 7 RATIONALITY QUESTIONS The Rings RKG and Schur Indices Review of Traces over Eields Schur Indices Cont Realizability over Cyclotomic Fields The Rank of RKG Generalization of Brauer s Induction Theorem NEVMU H PNE Part 2 MODULAR REPRESENTATION THEORY INTRODUCTION TO BRAUER THEORY Chapter 8 FUNDAMENTALS 1 Discrete Valuation Rings 2 Completions 3 p Modular Systems 4 Inverse Limits Chapter 9 THE GROUPS RKG RkG AND PkG The Rings RKG and RkG The Groups PRG and PkG Structure of PkG Structure of PRG Self Duality of RKG Duality of PkG and RkG More About Splitting Eields Scalar Extension 9 N U t gt9 l Chapter 10 THE Cde TRIANGLE 1 The Cartan Homomorphism C The Decomposition Homomorphism d The Homomorphism 8 Properties of the Ede Triangle Special Groups Change of Groups Brauer Induction Theorem in the Modular Case N P t gt9 CONTENTS 8 Properties of the Ede Triangle Cont 9 Characterization of the Image of 8 Chapter 11 MODULAR CHARACTERS 1 The Modular Character of a Representation 2 Independence of Irreducible Modular Characters 3 Reformulations 4 A Section of d 5 Example S4 Part 3 INTRODUCTION TO MAZUR S DEEORMATION THEORY Chapter 12 MAZUR S DEEORMATION THEORY 1 Examples of W 2 Back to the General Case 3 Schlessinger s Criteria BIBLIOGRAPHY 105 107 107 108 111 125 INTRODUCTION These are the notes when I took the class 22M2303 Topics in Algebra Group Representations and Deformations with Professor F Bleher during Fall 2006 offered by The Department of Math ematics of THE UNIVERSITY OF IOWA Basically we covered some of the material in 7 and we used some from 2 In the Introduction to Mazur s Deformation Theory we used some mate rial from 6 and 5 The purpose of this manuscript is to be a complement for people reading 7 and for personal interests This is a draft version of this document meaning it stills under corrections I would appreciate any suggestions and any feedback for this document All ob servations can be sent to jose velezmarulandaQui owa edu Iose A VelezeMarulondo Iowa City IA USA January 2008 vii Part 1 REPRESENTATIONS IN CHARACTERISTIC ZERO CHAPTER 1 QEMISIMPLICITY 1 Semisimple Rings and Modules PROPOSITION 11 Let R be u ring and M be u nonzero left Remodulet Then thefollowing state ments are equivalent a Every nonzero submodule is u sum of simple submodules b M is the sum of simple submodules c M is the direct sum of simple submodules d Every submodule T ofM is it direct summund of M ie there is T submodule ofM so that M T 69 TC PROOF u i h This is clear h i C Suppose M Elva S where each Si is a simple R module for i E I 7E 0 Let 3 I Q I XE Sf EI Sf Note that3 is not empty since there exists i0 E I satisfying liege Sf Sio j6i0 Sf thus i0 E 3r Let be partially ordered by inclusion It is easy to see that every chain in 3 has an upper bound take the union By Zorn s Lemma 3 has a maximal element say I0 CLAIM M Eido Si ido S where S Sf 0 withi 7E Otherwise there is i0 E I 7 I0 with Si0 g 21610 sf But Si0 is simple then Si0 m 3610 sf 0 Then 10 u i0 6 5t I0 Q IO U i0 and Eidouyo Si i610Ui0 S This contradicts the maximality of IO U i 01 Suppose M id S for simple submodules Si Let I Q I T 1551 By Zorn s Lemma 3 has a maximal element say I0 CLAIM M T ido Si Otherwise there is i0 E I 7 I0 with So T ielo Si 0 then T lt i610ui0 Si Therefore IO U i0 E 3 and I0 Q IO U i0 This contradicts the maximality of I0 01 3 a CHAPTER 1 SEMlSlMPLICITY 1 Let U 7E 0 be a submodule of M Let V 7E 0 be a submodule of U Then by hypothesis M V 63 Y for some submodule Y of M Then UU MVY U 2 Let U 7E 0 be a submodule of M CLAIM H has a simple submodule PROOF OF CLAIM Let 39U E U 7 Then R1 is finitely generated By Zorn s Lemma R39U has a maximal submodule say X By 1 R1 X 63 S for some submodule S of R1 Since X is maximal then S is simple R module hence S Q R39U Q U D 3 Let U 7E 0 be a submodule of M Let V be the sum of all simple submodules of U Suppose V Q U The by 1 U V 63 Z for some submodule Z of U with Z 7E 0 By 2 Z contains a simple submodule contradicting the definition of V D DEFINITION 12 a A left R module M satisfying the equivalent statements in Propo sition 11 is called semisimple b If M is an R module having simple submodules then the sum of all simple submod ules is the unique maximal semisimple submodule of M called the socle of M written socM c A ring R is called semisimple if RR R as a left module over itself is semisimple REMARK 13 R is called a simple ring if 0 and R are the only 2 sided ideals of R LEMMA 14 a Sums direct sums submodules and quotient modules of semisimple modules are semisimple b A ring R is semisimple if and only if all Remodules ure semisimple c A ring R is semisimple if and only if all Remodules ure projective PROOF a Follows from Proposition 11 b This is clear Assume RR is semisimple Then every free left R module is semisimple Since every left R module is the quotient of a free left R module we get b from u c Every R module is semisimple by b then it is enough to show that every simple module S is projective If S is a simple R module then S is isomorphic to a quotient module of R say 71 R gt S with 71 surjective R module homomorphism Consider the exact sequence 0gtllter7t gtR gtSgt0 Since R is semisimple there is N submodule of R so that R ker7t 63 N Thus splits hence S is direct summand of R Since R is free R module it follows that S is projective Let U 7E 0 be a submodule of R then OaUMRiRUAO w 2 SFMISIMPLE GROUP RINGS 5 is exact with R U R module Then by hypothesis R U is projective Then splits ie there is N submodule of R so that R E U 63 N with RU E N Since U was arbitrary then by Proposition 11 R is semisimple D 2 Semisimple Group Rings DEFINITION 15 Let R be a commutative ring and G be a group we define the group ring RG by RG G gt R fg 7 0 for at mosta finite number ofg E G 11 We define a sum and a product on RG as follows fg 1 01 HEW 12 fng bfbgb 1 l 13 forallfg E RG anda E G Note that by definition we may assume that every element in RG is of the form Egg rgg with Q E RforallgE G DEFINITION 16 a The map 6 RG gt R 14 2 Q8 H 2 7g geG gEG is called the augmentation map of the group ring RG b The ideal IRG kere is called the augmentation ideal of the group ring RG c An RG module M is called trivial if for all g E G and in E M gm in d The trivial submodule of RG is T X E RG gx X for allg E G Note that if G lt 00 then T R EgeGggt and T E R with trivial G action ie gr r for all r E R and g E G The element T7G Egg g is called the trace of G lfG 00 then T 0 LEMMA 17 Let RG he a group ring and 6 its augmentation map Then the augmentation ideal IRG is afree RGemodule with basis g E 1 g E G PROOF Note that for all g E G 6g7 1 1 E 1 0 then gil g E G g kere IRG On the other hand let Egg rgg be arbitrary in IRG Then Egg 7g 0 So we have 2733 Z gg o geG gEG Z gg lt27ggt1 geG gEG ngg71Eg71gE G gEG Thus IRG Q g 7 1 g E G Therefore IRG g 7 1 g E G The result follows E 6 CHAPTER 1 SEMlSlMPLIClTY THEOREM 18 Maschke Let K be afield and G be ufirlite group Then KG is semisimple ifurld only ifG is u urlit ofK iiei ChurK l G In particular CG is semisimplefor unyfirlite group G PROOF ilConsider the augmentation sequence 01KG gtKGgtKgt0 as a short exact sequence of KG modules where K has trivial G action Since KG is semisimple then by Lemma 14 C we have that K is a projective KG module Then the short sequence splits ie there is a KG module section u K gt KG so that eu idK Since G acts trivially on K the image of u is a nonzero trivial submodule of KG Hence Imu K Egg g Note that u1 AzgeGgfor some E K Now 1 6041 e A E g gEG Ae g gEG AG Then G is a unit in K Assume that ChurK i G Let U be a nonzero submodule of KG Consider the short exact sequence OHUmKGLKGUHO M We need to show that has a KG module section We know that has a K Vector space section p KG gt U with pi idu where i denotes the inclusion map Define p KG gt U by A 1 W E E g 1mm 15 gEG for all X E KG Then for allh E G 1 WM flow G g i Z h lg lmghx G geG gghrwghx g6 SI WEDDERBURN DECOMPOSITION THEOREM FOR SEMISIMPLE RINGS 7 Then 3 is a KG module homomorphism Let u E U then 1a u and 7000 u Thus t idu Then splits as a short exact sequence of KG modules It follows that KG U 63 N with N E KGU Therefore KG is semisimple 3 Wedderburn Decomposition Theorem for Semisimple Rings DEFINITION 19 Let R be a ring a An idempotent element is a nonzero element 8 E R with 82 e b Two idempotent elements 8 and f are called orthogonal if ef 0 and f8 0 c Let 8 E R be an idempotent e is called primitive if 8 cannot be written as the sum of two orthogonal idempotents e is called central if e E Z R where ZR denotes the center of the ring R e is called primitive central if e is central and primitive d An equation 1R 1 8 is called a decomposition of 1 if the ei s are pairwise orthog onal idempotents Such a decomposition is called primitive decomposition central decomposition or primitive central decomposition if all of the ei s have this property EXAMPLE 110 1 LetR Mat2C then 1 0 1 0 0 0 lt01ltoolt0139 1 0 1s a primitive decomposition of 1 In R 0 1 Is the only primitive central 1dempo tent in R 2 Let R and R be skew fields ie division rings then 1RgtltS 1R100115 is a primitive central decomposition of 1 3 If 8 7E l is an idempotent in an arbitrary ring R then 1 7 e is an idempotent which is orthogonal to 8 Further I 1 7 e e is a primitive decomposition of l in R 8 CHAPTER 1 SEMISIMPLICITY LEMMA 111 Let R be it ring and 1R 271 e be it decomposition of1 Then R QB Re ds left modules PROOF Since 17 E Rand for allr E Rwe have rr1 re1ek re1rek thenR 251 Rei Let1 g i g kand letx E Rei ltERejgtthen x rie for some r E R and x Ej rjej with r E R Then x riei riei2 xei Eng e 0 z PROPOSITION 112 Let R he a ring a I R v Mi where edch Mi is indecom osdhle ie Mi cdnnot be written as it direct sum o 161 p proper submodules then there is d primitive decomposition of 1 b If RR hds dfinite composition series there is d primitive centrdl decomposition of 1 PROOF dThere is a finite subset I of I and a nonzero e E M for all i E Iwith 1R 28139 ieI Then for all r E R r r1R lid re with re E Re for alli E I Then RR g idRei Thus I IisfiniteandMi Rei for alliE I AssumeI 1k Thus for1 g i g k e ei1R ei2 Zeiej be Since the sum is direct it follows that eiej 0 for all i 7 and e e for all i 1 k Since M is indecomposable for all i we can use Lemma 111 to see that 1R 271 e is a primitive decomposition of 1 11 Since 1R E ZR there is a central decomposition of 1 Suppose that every central decom position of 1 can be refined further then by Lemma 111 RR either has a composition series of arbitrary length or it has no composition series contradicting that RR has a finite composition ser1es COROLLARY 113 Let R be it semisimple ring then there is d primitive decomposition of1 and d primitive centrdl decomposition of 1 PROOF Assume that RR id 5 for simple submodules 5 As in part d in Proposition 112 I has to be finite Then RR has a finite composition series The result follows from part b of Proposition 112 3 WEDDERBURN DECOMPOSITION THEOREM FOR SEMISIMPLE RINGS 9 EXAMPLE 114 Primitive decomposition of 1 may not be unique Take R Mat2C then 1 0 1 0 0 0 lt0 1gtlt0 0gtlt0 1 1 0 1 1 0 71 lt0 1gtlt0 0gtlt0 1 are both primitive decompositions of 1 LEMMA 115 Let R be a ring a If 1 R 271 e 2711 are two primitive central decompositions of 1 then k m and 81 8k f1 In other words primitive central decompositions are unique b If 1 R 271 e is a primitive central decomposition of 1 then k R HeiRe 16 i1 as rings PROOF a Look at eifj for 1 g i g k and 1 g g m These are all central idempotents They are pairwise orthogonal since Eioffoeiiffi Eioeiiffoffi 0 if i0 7E i1 and jo 7E j1 Let 1 g i g k Then e ei1R eifl eifm Since e is primitive central there is a unique E 1m with e eifm Similarly fm 1Rfjv elfw ekfm and since fig is primitive central there is a unique nonzero summand on the right hand side This must be eifw And so fm eifm ei Then e1ek Q f1fm Similarly one proves the inverse inclusion h We know that R Ll Re as left R modules As sets Re eiRei since e is a cen tral idempotent for all 1 lt i lt k Hence we only need to show that eiRe is closed under multiplication and that eiRe hasa multiplicative identity Let rs E R then ereese eirelzse 81397813958139 eizrse eirse E eiRei Thus eiRe is closed under multiplication Note that eevrev eire ereev This e is the identity element in eiRei D THEOREM 116 Wedderburn Let R be a semisimple ring Then R is isomorphic to a finite direct product of fall matrix rings over skew fields More precisely There is a primitive central decomposition 10 CHAPTER 1 SEMISIMPLICITY of 1 say 1R 271 ei which is unique up to permutation and k R HEiREi i1 as rings Moreover 1 Rei y S17 as left Remodules where Si1Si2 Sink are pairwise isomorphic simple R7 modules 2 eiRei MatnxDv as rings where D EndRSv1 g E EndRSv1 thenfg gf with the regular composition on EndRSv1i 3 For all 1 g i g k eiRei is a simple semisimple ring ie it cannot be decomposed further as a ring 4 S11 S21 SM is afull set ofpairwise nonisomorphic simple Remodulesi We prove Theorem 116 by proving several Lemmas LEMMA 117 Let R be a ring a Let 1R 271 ei he a primitive central decompositioni Let S be a simple Remodulei Then there is a unique 1 g i0 g k with elves sfor all s E S and ejS Ofor allj 7E ioi b Suppose RR has a finite composition series Then there is a primitive central decomposition of 1 say 1R 271 81quot Let n he the the number of isomorphism classes of simple Remodulesi Then n 2 k More precisely if S is a composition factor of Rei and T a composition factor of Ref withi jthen S T c Let R be a semisimple ring and let k n e1 ek as in Then k n More precisely 71 REZ39 517 j1 where Si1 Smx are isomorphic simple Remodulesi Then ideals of eiRei are minimal 27sided ideals ofRi PROOF a Let x E S be nonzero Then x 1Rx elx ezx ekx so there is i0 E 1k such that eiox 73 0 Then S Reiox This implies a 11 Let S be a composition factor of Rei and T a composition factor of Ref with i 7 Then there are submodules M Q Rei and N Q Ref with surjections M gt S gt 0 and N gt T gt 0 SinceM Q Rei M Mei eiM ei E ZR Similarly N ejN Then S eiS and T ejT using a we see that HomRST 0 since for f E HomRST satisfies feis eifs eiejfs 0 Then S T c Let U be a simple submodule of Rei Let V DER Ur note that for all r E R Ur is a left R module and Ur E U or Ur 0 Suppose that Rei has a simple submodule not isomorphic to U Let V MSW76R Wr where W is the set of all simple submodules of Rei nonisomorphic to U By assumption V 7E 0 Then V and V are nonzero 2 sided ideals of R and V V 0 since otherwise V V has a simple submodule but the simple submodules of V and V are not isomorphic Therefore Rei V V since all simple submodules of Rei come up in V and V 3 WEDDERBURN DECOMPOSITION THEOREM FOR SEMISIMPLE RINGS 1 1 Then Rei V 63 V as left R modules Then there are uniquef E V and f E Vwith ei f f Therefore for all 390 E V v vei vfvf vf vf E V V 0 Similarly v fv For all 390 E V we obtain 390 f v v f Then eiRei V X V as rings contradicting that ei is a primitive central idempotent Let I Q eiRei be a nonzero 2 sided ideal of R Let U be a simple submodule of I Viewed as a left R module By the above eiRei DER Ur Q I Then I EiREi D LEMMA 118 Let M be a left Remadule M E 911 Hi where U1Ul are isomorphic left Remadules Then EndRM E MatnEndRU1 19 as rings PROOF Let 1 LI gt M the i th inclusion and m M gt U be the i th projection Let 7139 gt U1 be an isomorphism Note that we have the following composition diagram on U1 7 L a 7h 71 U1gtUjgtMgtMgtUi gtUL 110 Letf EndRM gt MatnEndRU1 be defined by f0c Uflmtujtfj y D LEMMA 119 Let R be a ring and M be a left Remadule and e be an central idempotent in R Then a HomRReM E eM as abelian groups b EndRRe E eRe as rings PROOF a Let 39y HomRReM a eM defined by 7a oce 0cee eoce for all X E HomRReM We show that 39y is a group isomorphism as follows Let og E HomRReM then WC 3 0C 98 8 38 NO Yl3 Then 39y is a group homomorphism Let X E HomRReM so that txe 0 It follows that 0cre roce 0 for all r E R Therefore ker39y 0 Thus 7 is injective Let em E eM and define 8 Re gt M so that re rrn for all r E R We show that 8 E HomRReM Let r1rr2e E Re then r1e rze r1 r2e r1 r2rn rlrn rzrn 3r1e 3r2e Further for all rs E R sre srrn s re Thus 8 E HomRReM Now 7 3e e2 e e ern Thus 39y is surjective Hence it is an isomorphism 11 Take 39y from a and note that WC3 DC3W 11038 043032 E38 3000 303048 8 38 D REMARK 120 MatnD gt gt MatnD A E At is an isomorphism of rings 12 CHAPTER 1 SEMlSlMPLICITY LEMMA 121 Let D be a skew 39eld and R MatnD Then R is simple semisimple PROOF Since ZR Aln A E ZD In In is the primitive central decomposition of 1 in R Then by Lemma 117 C the only 2 sided ideals of R are 0 and R Thus R is a simple ring For all 1 g i g 71 let Si 3 g g g 111 where the D s are on the i th column Then S is an R module that is generated by each of its nonzero elements Then S is simple as R module and R is semisimple REMARK 122 The proof of Lemma 121 shows more Suppose R is semisimple and R 111 MatnxDi Let S be a simple R module belonging to the i th component Then S is a simple Mat DD module for all 1 g i g 71 By the proof of Lemma 121 as Div module SigDi Di 71 7copies By Lemma 118 EndeSi E MatnxDi CHAPTER 2 THE GROUP ALGEBRA 1 Generalities DEFINITION 21 Let G be a group and P be a field a A representation of G over F is a group homomorphism p G gt AutpV for some vector space V over F If V is finite dimensional over F say dimpV n then we identify AutpV with GLnP and say that p has degree n p is called fuithfull ifkerp 1g b Given a representation p G gt GLnP the character of p is the map x G gt P defined by Mg Trpg for all g E G In particular 91g n degp Note that x is only a group homomorphism if n 1 since the identity 71 9lg 9lglg x1gxlg n2 is only valid for n 1 c A map y G gt P is called a clussfunctiun if for all gh E G Mg thh l Recall that if AB E MatnP then T7AB T7BA thus T7A TVCAC 1 for all invertible matrix C Hence characters are class functions REMARK 22 We will sometimes extend a representation p G gt GLAP P linearly to get an P algebra homomorphism p P G gt MatnP Since traces are P linear we can also extend the character x of p P linearly to get a map x PG gt P 2 Connection Between Representations of G over F and PGmodules Given an PG module M M is also a vector space over F Hence we get a representation p G gt AutpM defined by for all g E G pgMgtM mHgm 13 14 CHAPTER 2 THE GROUP ALGEBRA Given a representation p G gt AutpV for some vector space V over F V becomes an PG module via the following action for all g E G and v E V 3v Mg v DEFINITION 23 Let p 1p G gt GLnP be representations of G over F of degree n a p and 1p are equivalent if there is A E GLnP so thatpg AlpgA 1 for all g E G In particular equivalent representations have the same character b p is called reducible if p is equivalent to a representation IP with mg 1550 wig gt for all g E G where 1p G gt GLnXP are representations with n g n for i 12 lfp is not reducible then p is called irreducible cpiscalledJ quot ifpis q39 toa I 39 lpwith lag lt 155 1235 for all g E G where 1p G gt GLnXP are representations with n g n for i 12 lfp is not decomposable then p is called indecomposable LEMMA 24 Let p1p G a GLnP be representations of G over F with correspondent PGE modules M and N Then a p and 1p are equivalent if and only if M E N as PGemodules b p is reducible if and only if M has a proper nonzero submodule ie p is irreducible if and only ifM is a simple PGemodule c p is decomposable if and only if M is decomposable ie M M1 69 M2 where M1 and M2 are proper nonzero submodules of M Consequently p is irreducible if and only if M is irreducible REMARK 25 If G is finite and charF i G then reducible is equivalent to decompos able and irreducible is equivalent to indecomposable since every PG module is completely reducible ie the direct sum of simple modules lf charF lG this is not true in general EXAMPLE 26 Let P le G Zp U with p prime Then p Zp A GL2le U gt gt 1 1 0 1 is a representation of G Z p since 3 013 I 3 p is also indecomposable since lt 1 i gt is a Jordan Block and p is reducible 3 ORDINARY CHARACTERISTIC ZERO REPRESENTATION THEORY 15 From now on module meansfinitely generated module 3 Ordinary Characteristic Zero Representation Theory Let G be a finite group and let K be a field with CharK l lGl THEOREM 27 KG has thefollowing properties a KG E 1111 MatnxD whre D is a skewfielafor all 1 g i g m b The number of isomorphism Classes of simple KGimoaules is m c For all 1 g i g m D is afinite dimensional vector space over K and K g ZD IfK is algebraic Closed then D Kfor all 1 g i g m cl If KG 1111 MatnxK then the number of isomorphism Classes of simple KGimoaules is equal to the number of conjugacy Classes of G and m G Eng 21 i1 PROOF a and b follow from Theorems 18 and 116 C Identify K with K1G A1G A E K then KG is a K Vector space of dimension G Let 1G e be the primitive central decomposition of 1 according to a The we identify eKGe MatnxD for all 1 g i g m In particular e corresponds to the n X n identity matrix 1x in MatnxD for 1 g i g m Now Ke A1x A E K E ZKGe Z MatnxD uIx u E ZD Then K Q ZD for all 1 g i g m Thus D is a K Vector space which is isomorphic to a subspace of KG then D is finitely dimensional over K If K is algebraic closed then for all u E D with 1 g i g m there is a l with 1 u u2 u are linear dependent over K Then there is px E Kx 7 0 with pu 0 Thus a E K d The class sums of G form a K basis of ZKG Recall If C is a conjugacy class of G then the correspondent class sum is Egec g Then dimK ZKG of conjugacy classes of G On the other hand at ZKG UZMatK 5 X X K 22 1 1 mifactors Then dimK ZKG m D REMARK 28 Suppose KG 1171 eKGe 1171 Matx D Let S be a simple KG module correspondent to the i th component Then Di EndKGSiO 23 Moreover S is a free D m0dule of rank n then EndeS E MatnxD If D K for all 1 g i g m then dimK S n and if we write KG as a direct sum of simple KG modules then the number of summands isomorphic to S is equal to n Therefore m KG Q9 5 24 i1 16 CHAPTER 2 THE GROUP ALGEBRA as KG modules where S S 63 63 Si nd mitimes REMARK 29 Assume K C and let CG H177 eiCGe H177 MatnxC be the Wedder burn decomposition of CG Let 7 CG gt eiCGe MatnxC be the i th projection Let S be a simple CG module correspond to the i th component with character 9 a 7mg G gt GLnXC is a representation of G over C The character of 7mg is equal to 9 let M be the CG module correspond to mlg then eiM M and ejM 0 for all j 7 i Then M is isomorphic to a MatnxC module of C dimension m Then M is a simple Matnx C module hence a simple eiCGei module thus isomorphic to 5 b View CG as a module over itself The correspondent representation of G pg is called the regular representation of G the correspondent character KG is called the regular character of G Since CG 177 S as CG modules we get m XG Eniin 25 i1 Where 264 T77TilGg LEMMA 210 G ifg 1G 26 XGQ 0 otherwise PROOF pG is given by pG G A AutCCG 27 gH h thVh E G If we take G as C basis of CG and write pgg as a G X G matrix with respect to this basis then if g 7 1G the diagonal elements are zero Then xgg 0 if g 7 lg Also xglg dimC CG G D 4 Decomposition of CG PROPOSITION 211 Fourier Inversion Formula Let 7 CG gt eiCGe MatnxC with 1 g i g m be the ith projection as before Let Lox1 where a E eiCGe MatnxC Let a E Agg 6 CG 28 gEG with 7ta aifor all 1 g i g in Then 1 m 1 Ag E 1quotng u 29 where x Tr7tlg ie xg T77Tilggf07 allg E G 4i DECOMPOSITION OF CG 1 7 PROOF Since the 9 are C linear it is enough to prove this for u h E G ie 1 if g h Ag 0 otherw1se Now i inMg lh iKGQz lh G 11 G where KG is the regular character of G Thus 1 if h of 0 otflerwise D PROPOSITION 212 Let m CG 4 eiCGei MatnxC he the ith prajectianfar 1 g i g m as before Then m sends the center Z CG t0 the center ZMatnx C1711 Hence m induces d C lg br d homomorphism wi ZCG gt C 210 where wiu Xiuf0r all u E ZCG with paw Tr7riu PROOF Letu E ZCG Then 7T u ml for some y E C Thus paw Tr7riu nini It follows that 7T u ixxu nx Therefore wiu nixpmu D COROLLARY 213 The fdmily IA0111 de nes it Cedlgehrd isomorphism wZCGgtCmCgtltgtltC 211 hV a mifdctors u H wzu139i1 PROOF Note that m 71 CG gt HMatMC i1 X H mx ii is a C algebra isomorphism Then nlzm ZG A ZMnXC E C u H7Izu1 1 is a C algebra isomorphism Now use that 71iu wiuInx for all u E ZCG D EXAMPLE 214 1 If G is abelian then CG E C X X C EF Gi mes PROOF Since G is abelian then CG is commutative then all ni in the Wedderburn decomposition have to be 1 MatnC is commutative if and only if n 1 D CHAPTER 2 THE GROUP ALGEBRA 2 Let G g E Zm and g be a primitive m th root of the unity in C For all 1 g i g m pi G gt 02 defined by gt gt U is a group representation and we get m nonequivalent irreducible representations of G If G g X h E Zm gtlt Zfl then we get mfl nonequivalent irreducible representations by Pij2ltggt X01 A 3 gill H E W H a for1 gigmi gjgn me andgne Let G be an arbitrary finite finite group Then the number of nonequivalent irreducible representations of G of degree 1 is equal to G G where G is the commutator group of G PROOF Let p G gt C be a representation of G of degree 1 Then p factors through G G Then p induces a 1 dimensional representation of G G Then the number of nonequivalent1 dimensional representations of G over C is less or equal than G G If l7 GG gt C is a 1 dimensional representation of GG then l7 can be in ated to a group homomorphism p G gt 02 Then the number of nonequivalent 1 dimensional representation of G over C is greater or equal than G G Let N 51 G and let 7r G gt G N be the canonical surjection which can be C linearly extended to a surjective C algebra homomorphism 7T CG A CGN Let 1U 1G 2111 ei be a primitive central decomposition of 1 in CG Then there is Twith T Q 1fl such that 1cGN 1GN Z 7T8l ieT is a primitive central decomposition of 1 in CG N Moreover g HEiCGEi ieT as C algebras ie the Wedderburn decomposition of CG N is contained up to isomorphism in the Wedderburn decomposition of CG SKETCH OF PROOF Let fl E EN Then fl is a central idempotent Thus 1U fl 1H 7 fl is a central decomposition of 1 in CG Thus we can refine it to a primitive central decomposition of 1 Let T be the set of all primitive central idempotents decomposing fl Let ll CGN gt CG be defined by llgN flg and C linear extension Then ll is injective C algebra homomorphism Then CG N E Iml flCGfl as C algebras NOTE Let S be a simple CGN module Then S can be in ated to a CG module ie for all g E G and s E S g 5 gN 5 Moreover S is a simple CG module D 5 lNTEGRALITY PROPERTIES OF CHARACTERS 19 5 Let G S4 Note that G 4 24 We find CG as follows Assume 24 We have that G A4 with G 12 then G G 2 Therefore there are only 2 nonequivalent irreducible representations of degree 1 of CS4 Then n1 n2 1 Thus 22 Ell3 with ni g 4 If 71 4 then 6 2174 Therefore we get more 1 dimensional representations which is a contradiction Thus ni g 3 But 22 i 0 mod 4 then at least one ni 3 say 713 3 Then 13 274 71139 Thus we have another m 3 say 714 3 Then 4 215 Therefore in 5 and n5 2 So we obtain an identification of CS4 as follows CS4 2 C X C X Mat2C gtlt Mat3C gtlt Mat3C 212 DEFINITION 215 Permutation representations Let H g G define GH gH g E G let V be a C vector space with basis GH Since there is a left action of G on GH V becomes a CG module V is called the permutation module denoted by V CGH and its character is called the permutation character If H 1 we get back the regular representation of G EXAMPLE 216 Let G 53 and H 12 Then GH H 123H 132H Let e1 H e2 123H and e3 132H Look at the left action of G on GH Representative of Conjugacy Classes e1 e2 e3 Representation pg 9g 1 0 0 id e1 e2 e3 0 1 0 3 0 0 1 1 0 0 12 81 83 82 0 0 1 1 0 1 0 0 0 1 123 e2 e3 e1 1 0 0 0 0 1 0 This is not an irreducible representation since take X e1 e2 e3 E C53 H then for all g E 53 gx X Then Cx is a trivial submodule OfCSgH Thenx 7 trivial irreducible character is a character which can be shown to be irreducible 5 Integrality Properties of Characters DEFINITION 217 Let R be a commutative ring An element 739 E R is said to be integral over Z if there exists a monic polynomial E ZM with 0 An element 739 E C that is integral over Z is said to be an algebraic integer Note that an algebraic integer in Q is an integer LEMMA 218 Let R be a commutative ring and r E R The following statements are equivalent 20 CHAPTER 2 THE GROUP ALGEBRA a r is integral over Z b the subring Z r of R is finitely generated as a Zernadule c there is a finitely generated Zernadule of R containingZ PROOF a i Let fx xquot an1x a1xa0 E Zx with fr 0 Then rquot EanL1rn E 7 a0 Therefore 1r rquot 1 generates Zr as a Z module h i a Let A Z Zr Zr t 1 g Zr as Z modules Then U A Zr n1 Since Zr is finitely generated there is n 2 1 with An Zr Therefore rquot is a Z linear combination of 1 r rquot 1 h ltgt Use that Z is Noetherian for For i we know that every submodule of a finitely generated Z module is finitely generated D COROLLARY 219 Let R be a commutative ring and r E R Then a If R is finitely generated as a Zernadule then every r E R is integral over the ring Z b The set S r E R r integral over Z is a subring afR PROOF a follows from Lemma 218 c To prove h we know that 0 and 1 are integral over Z thus 01 E S Let rs E S Then Zr and Zs are finitely generated as Z modules Then Zr 292 Z s is finitely generated as a Z module Consider 4 Zr ZZs a R a 29 h gt gt ah Observe that 4 is a Z module homomorphism Then lrnzp Z rs is finitely generated as a Z module Since r i srs E Zrs with Z rs a ring we have that Zr sZr E sZrs g Zrs D LEMMA 220 Let p be a representation of G aver C with character x Then x g is an algebraic integer for all g E G PROOF Observe that 9g Trpg which is equal to the sum of eigenvalues of pg Since pgG is the identity matrix all eigenvalues are roots of the identity D PROPOSITION 221 Let u EgeG Agg E ZCG with Ag an algebraic integerfar all g E G Then a is integral over Z PROOF Let Ci 1 g i g h be the conjugate classes of G and let K Egecx g be the correspondent class sum of C for 1 g i g h Since it E ZCG we have that Ag Ag for all gg E C and this for all 1 g i g h Then it LlhiKi where A Ag with g E C Since the elements in Z CG that are integral over Z form a subring and all A are integral over Z by assumption it is enough to show that K is integral over Z for all 1 g i g h Note that for all 1 lt i lt h K E ZCG ZG ZZG ZK1 63 eBZKh as Z modules 5 lNTEGRALITY PROPERTIES OF CHARACTERS 21 Since ZZG is a ring ZKi Q ZZG for all 1 g i g b By Lemma 218 c K is integral over Zforalllgigh D COROLLARY 222 Let u Egg Agg E ZCG with Ag an algebraic integerfar all g E G Let p be an irreducible representation of G aver C with character x of degree n Then xu i 2 mm 213 gEG is an algebraic integer PROOF Recall that there is a C algebra homomorphism ws 2 Z CG gt C 1 x gt gt 2x00 where S is the simple module associated to p Then ws sends elements that are integral over Z to elements that are integral over Z say if X E ZCG is integral over Z then there is a monic polynomial E Zt with 0 Therefore by Z linearity of ms we have 0 ws f wsX D PROPOSITION 223 Let x be an irreducible character afG aver C afdegree n Then nlGi PROOF Let u Egg xg 1g Since x is a class function u E ZCG We also know 9g 1 is an algebraic integer for all g E G By the Corollary 222 Eg6G xg 1xg is an algebraic integer Further we will see that by Theorem 34 Egg xg 1xg G Then is an algebraic integer hence an integer Thus nlG D CHAPTER 3 CHARACTER THEORY 1 Orthogonality Relations for Characters LEMMA 31 Let KG be the regular Character ofG over C Let CG HZ eiCGei HZ Mat C be the Wedderburn decomposition of CG Let S be a simple CGimodule belonging to the ith compoi nent ie eiSi Si and ejSifori 73 j Let 9 be the Character of Si ofdegree ni Let ei EgeG Aifgg Then 1 1 n 1 Mg 266ng gm 31 PROOF Recall 7 m 7 G if g 1G xGg T gnlxlg T 0 otherwise Then 1 xgeig 1 Did Aiihxdhg l AigG By we also have 2 26681398 1 2711 quot196181398 1 nixig 1 D DEFINITION 32 Let clG be the set of class functions from G to C Then clG is a C algebra Via for all A E C plp E clG and g E G we MW 32 11 Mg Mg Mg 33 WW rpg g 34 The map clG gtlt c1G a c 35 WP H WP 23 24 CHAPTER 3 CHARACTER THEORY where 1 7 7 36 WW G ggmgll gl is a positive defined Hermitian inner product on the C Vector space cl G REMARK 33 If x is a character of G over C then for all g E G g 9g 1 The reason is that if A is a root of unity in C then it A l thus we can define a symmetric bilinear form on clG by 1 Ir1P g E g g 1 37 geG If p and 1p are characters over C then Ir1P 41W 38 THEOREM 34 First Orthogonality Relation Let CG ll1 eiCGei be the Wedderburn def composition ofCG for all 1 g i g m let Si be a simple CGemodule with eiSi Si and ejSi Ofor all i 73 j Let 9 be the Character of Si of degree ni Then 90 xm is an orthonormal basis of the unitary space clG PROOF Recall that ei Egg Airgg where Mg xig 1 Then Mm MW 1 71 i Z xgxg G geG J G quoti 71 71 quotj 71 X39Gg Hm 7141ng G l G J G 2 Ai 1 8 M quot1quot geG On the other hand ier Z M8 2 Mg geG geG lt2 Angelika 16 t Z 2 MW 3 geG geGi g xyeG X lfi 7 then eiej 0 Therefore EgeG AirgilA39rg 0 lfi then eiej ei Therefore 1139 n lj g Aag lALg M16 26416 G From and we conclude that 20 517 39 1 ORTHOGONALITY RELATIONS FOR CHARACTERS 25 It remains to show that dirnC clG m Let C1 Cm be the conjugate classes of G for all 1 g i g mlettxi G gt Cbedefinedby i 1 ifg 6 Ci 1 T 0 otherwise Then 11 0 m is a C basis of clG hence dirnC clG m D PROPOSITION 35 Let V be a CGernadule with Character 4 Let xixm be afull set of irreducible Characters of G aver C corresponding to simple CGemadules 51 Sm T en V2511O ST 310 where ri QOf07 lli 1rn PROOF We know that V is sernisirnple then V 511S7 for certain r1 rm 2 0 Then m 47 Ema i1 Hence 4 271061120 n 1 D COROLLARY 36 Let V and W be CGernadules with correspondent Characters 4 and 1p Then V E W as CGernadules ifiznd only ifzp 1p PROOF If V E W then their representations are equivalent Thus their correspon dent characters are the same 1sz IPthenforall1 g i g Th 47 We By Proposition 35 V E W D COROLLARY 37 Let V 3A 0 be a CGemadule with Character 4 Then 4114 E 2 and V is a simple CGemadule ifiznd only if 4114 1 PROOF Using the notation from Proposition 35 V2511 6969S39 for certain ri 2 0 Then 4 271 rim Therefore 2111 ri2 E 2 since V 7E 0 Moreover 1 ltgtthereexists1 g i g mwithri 1 andrj 0forallj i ltgt there exists 1 g i g rn with 4 xi ltgt 4 is irreducible 26 CHAPTER 3 CHARACTER THEORY DEFINITION 38 Let C1 Cm be the conjugate classes of G let X1 xm be the distinct irreducible characters of G over C also called the ordinary irreducible characters of G For all 1 g ij g in let xCj xig for any representative g E C The m X in matrix M mij with mi xCj is called the ordinary character table of G We say that x is the trivial character of G ifxg 1 for allg E G EXAMPLE 39 Consider G 53 We have the conjugacy classes C1 C2 and C3 with corre spondent representatives id 12 and 123 We know that G A3 then there are 2 dimensional irreducible representations of 53 over C which are the triVial character X1 and the Sign character x2 defined by 07 1 ifUiseven M T 71 ifUisodd Since 6 1 1 4 then 3 has degree 2 Consider the permutation character from Example 216 correspond to the permutation module V CSg and its character p satisfying pC1 3 pC2 1 and pC3 0 Since V has a triVial simple submodule then p is not irreducible but x3 p 7 1 is a character with x3C1 2 x3C2 0 and x3C3 71 Now x3 lxg 22 3 02 2712 1 It follows that 3 is irreducible thus the ordinary character table of 53 is LEMMA 310 Let B by the m X in matrix with 07 bij 759019 311 therl MB Im where M is the ordinary Character table afx PROOF LetC cij MB Then m Cki C17 ZXiCkEXjCk k l iwc xmc x c k1 E XiCkCjCk H Q H 6G 20W 1 ORTHOGONALITY RELATIONS FOR CHARACTERS 27 COROLLARY 311 Second Orthogonality Relation Let st E G then m Cg5 ifs and t are conjugates 39 39 t 312 gxlsxl 0 otherwise PROOF Since MB Im then BM Im thus m Cvi E foCoij 5 k1 Lets E Cf andt E Ci then 1 Ci m 7 1 ifs and t are conjugate 7 t 313 Cgs G gxksxk 0 otherwise D REMARK 312 1 Via the second orthogonality relation the order of the centralizers Cgs for all s E G can be computed from the ordinary character table of G over C 2 Let p be a representation of G over C and let x be its ordinary character Define the kernel of x as kerOc g E G 2 Mg degm xlc geGpgI We can find the kernels of the ordinary irreducible characters of G from the ordinary character table EXAMPLE 313 Let G 53 we already showed that the character table of 53 is kerx1 C1 U C2 U C3 53 kerx2 c1 u 3 123 kerx3 C1 id 3 The ordinary character table does not determine the group up to isomorphism eg D8 and Q8 have the same character table but D8 Q8 28 CHAPTER 3 CHARACTER THEORY 2 Canonical Decomposition of a CGmodule Let CG H177 eiCGei be the Wedderburn decomposition of CG For 1 g i g m let S be a simple CG module correspondent to the i th component with character 9 of degree ni THEOREM 314 Let V be a CGimodale with representation pi Then a V 177 eiV where eiV is the direct sum ofsimple modules all isomorphic to Sifor 1 g i g m This is called the canonical representation of V b Let 7T V a eiV be the ith projection Then n 7 m i Z Xigpg 314 gEG PROOF a Since lCG ei we have that V 31 eiV Since the i s are pairwise orthogonal the sum is direct 11 Let n m i Z xigpg 315 gEG Then 1 V gt V is a C linear It is a C module homomorphism since for all h E G and v E V n 7 Mini i Z Xzgpghv gEG m 7 xvh 1ghghv G g l n h Z xxh lghxh lgmv gEG h lt3 2 xTltzgtltzvgtgt G 26G h lg 2 xTzgtpltzgtltvgtgt 26G hqv Let S be a simple CG module of V with S E Sf We want to show that q 5 is the identity on S ifi j and zero otherwise Since S is a CG submodule of V we have pg 5 S gt S for all g E G Then Mg S gt S By Schur s Lemma there is A E C with Ms Aiids Since S E S n1 ifi 0 otherwise Ainj Trqvs QWXJ39Q g6 Then 1 5 is the identity ofi and zero otherwise Therefore 1 exv is the identity and 1 My 0 forj 7E i Thus 1 m D CHAPTER 4 INDUCED REPRESENTATIONS 1 Tensor Products Let P be an arbitrary field Let M1 and M2 be PG rnodules with representations p1 and p2 over F and characters X1 and x2 over P Then M1 p M2 is an PG rnodule with G action given by Wu m2 gml W2 41 for all g E G and m1 m2 E M1 gtlt M2 with P linear extension The representation of M1 p M2 as a representation of G is denoted by pl 29 p2 and the character is denoted by X1 29 x2 LEMMA 41 For allg E G 261 262 g X18X28 42 PROOF Let b1bm be an P basis for M1 and similarly let C1Cn2 be an F basis forM2Nowfor1 g rgnlandl 5 712 and 712 Then m and 712 30 CHAPTER 4 INDUCED REPRESENTATIONS Therefore for 1 g r g n1 and 1 g s g n2 1 pzgbr Us 91gbr pzgCs n1 n2 Z ZAirBjsgbi Cj i1j1 Then n1 n2 261 2mg Z Z ArrgBssg X18X2g 71 51 2 Direct Products Let G1 and G2 be finite groups Let P be an arbitrary field and for i 12 let V be a PG module with representation p over F and character 9 Then V1 p V2 becomes an PG1 gtlt G2 module Via for all g E G and v E V withi 12 glrgzxvl 712 101 202 43 with P linear extension The representation of V1 29 1 V2 is pl 29 p2 where p1 p2 G1 gtlt G2 A AutpV1 p V2 44 ghgz H 1181 P282 The character of V1 p V2 is denoted by X1 29 x2 Then 261 X2g1182 X18X2g 45 REMARK 42 If G G1 G2 then the restriction of the PG gtlt G rnodule V1 p V2 to the diagonal subgroup AG g g g E G defines an PG rnodule with underlying P Vector space V1 29 1 V2 and diagonal G action 801 7 2 01 gvz Do not confuse the PG gtlt G rnodule V1 p V2 and the PG rnodule V1 p V2 ll Let P C THEOREM 43 Let G1 and G2 be groups a If p is an irreducible representation of G aver C with i 12 then pl 29 p2 is an irreducible representation of G1 29 G2 aver C b Every irreducible representation of G1 gtlt G2 aver C is 0ftbef0rin pl 29 p2f0r irreducible repree sentatians p of G aver C with i 12 PROOF a Let x be the character of pi Then 1 1 x1120 731 Z lxi5il2 seG 2 DIRECT PRODUCTS 31 IfXX1 C2then 1 WC Z msoemszw Gl X G2 slszEG1gtltG2 mlm mlm Then x is irreducible character of G1 gtlt G2 11 Let f be a class function of G1 gtlt G2 over C It is enough to show that if f is orthogonal to all characters of G1 gtlt G2 over C of the form X1 29 x2 where 9 is an irreducible character of G with i 12 then f 0 The reason is the following let A be the span over C of X1 29 x2 xi irreducible character of G with i 12 Then the orthogonal complement of A is zero then A clG1 gtlt G2 lf xl 29 x2 0 for all irreducible character 9 of G with i 12 then f5152m 0 g 51526G1 gtlt G2 for all irreducible characters X1 and 9 of G1 and G2 respectively Fix 9 and define for all 51 E G 12251 2G f s1szxzsz 46 Then 1 is a class function on G1 with fmlxl 0 for all irreducible character X1 of G1 by Therefore fx251 0 for all 51 E G1 Let 51 E G1 be fixed Let gs1 G2 gt C be defined by gslsz f5152 is clear that gs1 is a class function on G2 Since fx251 we get 381lX2 0 for all irreducible character pg of G2 Then gs1 0 for all 51 Then f5152 0 for all 5152 E G1 gtlt G2 D THEOREM 44 Let G be an arbitrary finite group Then the degree of every irreducible representae tiarl afG aver C divides G ZG PROOF Let W be a simple CG module with representation p over C of degree rl Lets E Z G then ps W a W x gt gt 5x is a CG module homomorphism Since for all g E G gs 5g ps is bijective Then by Shur s Lemma there is As E C with ps A5171 Then we get a group homomorphism A ZG gt C s H As Let m 2 0 be an integer Consider the m fold tensor product p m Gm gt AutCW m g1gm Hpg1 pgm 32 CHAPTER 4 INDUCED REPRESENTATIONS If 51sm e ZGm then p mslsm A1 As identity Asl sm identity Let H slsm E ZGm 51 sm 1 51 G H acts trivially on the CGm module WW Then p m induces an irreducible CGmH module of degree degp m nm Then nmlGmH Consider the map p ZGm 1 A H 51sm1 H 51sm1sl sm1 1 It is easy to see that p is an isomorphism of groups Thus From we obtain that nml Zgnml Therefore Gm ZGm 1 G m 1 7 E 72 nZG ZG for all in 2 0 Therefore G C 1 Z nZG ZG 39 with 1652 cyclic so finitely generated Z module By Lemma 218 65 is integral over Z hence an integer since it lies in Q E 3 Induced Modules Let P be an arbitrary field and H leqslilntG Let R be a full set of representations of left cosets of H in G DEFINITION 45 Let V be an PG module and W be an PH submodule We say that V is induced by W if V sW 47 seR REMARK 46 For allg E G s E R and w E W gsw tw for somet E R and w E W Note that gs th for allt E R and Ii E H Then w hw Therefore the action of G on V permutes the sW s LEMMA 47 IfV is an PGimadule and W is an PHisubmadule then V is induced by H ifiznd only if V 2 PG FH W 48 3 INDUCED MODULES 33 PROOF Consider PG 561 sPH as right PH module Then PG FHW 5PH FH W E SW 56R 56R NOTATION we denote the PG module induced by the PH module W to be Indg W Note that PG FH W has as PH submodule the subspace 1 29 W E W as PH modules LEMMA 48 Let V be an PGemadule which is generated as an PGemadule by an PHesubmadule W Then V is induced by W if and only if dimp V G Hd11391 11 W 49 PROOF Assume that V 561 SW then dimp V R dimp W where R G V is generated by W means V Z gW ZSWVS E RsW shWVh E geG 6R Then dimp V g R dimp W G H dimp W Since by assumption this inequality is equality it follows that the sum Egg gW is a direct sum ie V SW seR D REMARK 49 1 Let H lt K g G let W be an PH submodule Then Indg w 2 1nd Ind W 410 PG Hlt PK FH W g PG FH W 2 If V is an PG module and H g G then Resg V is the PH module obtained from V by restricting the action from G to H 3 FROBENIUS RECIPROCITY Let W be an PH module let E be an PG module Then by adjoint associatiVity HompG Indg W E 2 HompH W Res E 411 g w H g3w 93 Indg w 2 PG FH W Furthermore we have that ngGpH is an PG PH bimodule and we get an adjoint pair PG FH HompgPG Then HompgPG FH WE E HompH WH01391 11GPGE 412 since HompgPG E E E as PG modules 34 CHAPTER 4 INDUCED REPRESENTATIONS PROPOSITION 410 Let V be art PGemaalule arid suppose V id Wi where Wi is art Pesuhspace for all i E I Assume that G acts transitively art Wiv61 ie for all ij E I there is g E G with Wi ng Let i0 E larlal put W Wio arid H StahgW g E G gW W Therl Wis art PHesuhmadule afV arid V is induced by W ie V Indg W PROOF By the definition of H W is an PH submodule of V Since G acts transitively on Wii61 for all i E I there is g E G so that Wi gW Moreover sW s W if and only if s 1s E Hif and only ifsH sH Then V id Wi SER sW ifR is a system of left cosets representatives of H in G D EXAMPLE 411 Let P C not really necessary Let H g G and W C be the trivial simple CH module Then Indg w 2 CGH 413 where CGH denotes the permutation module of G with respect to H over C 4 The Character of an Induced Representation DEFINITION 412 Let p H gt C be a class function Define LndICjI p G gt C by 1 G 7 7 71 lt1ndH p s 7 H E pt st 414 t lstEH for all s E G PROPOSITION 413 a Indg p is a Classfurlctiurl afG aver C b If p is the Character of a CH emadule W therl Indg p is the Character aflndg W PROOF a Letgs E G then 1 G 71 7 7 71 71 IndH r g 5g E H E W g Sgt t lg lsgtEH 1 71 7 z 52 H rplt z lszeH ltde 41 s h We use that LndICjI W E 761 rW Let p H gt GLnC be the representation of H over C correspondent to W Let h1 hn be a C basis of W then rh1 rhn is a C basis of rW for all r E RThen lt1nd pgt srhv r pthv where sr r t with r E R and t E H Use that Indg p s is a block matrix If r 74 r this does not contribute to the trace of lt1nd 1pgt If r r ie r lst t E H then the trace of Indgpgt 5 inside the block correspondent to 4 THE CHARACTER OF AN INDUCED REPRESENTATION 35 7W is equal to the trace of ps ie ps Then if Chi is the character of lndg p then mm s 2 110457 76R FlyEH GrhrERhEH 2 pwlst t gitiH We have used that h 1r 1srh pr 1sr for all h E H D DEFINITION 414 Let V1 and V2 be CG modules then V1 V2G dimC HomCG V1 V2 415 LEMMA 415 If V1 and V2 m39 eCGemadules with Characters p1 p2 respectively then V1 V2gtG 111 12 416 PROOF Since G and Homcg are additive in each component we can reduced to simple modules V1 and V2 ie to irreducible characters p1 and p2 Then 1 39f lt II 2gt pl 12 0 otherwise On the other hand by Schur s Lemma C if V1 E V2 H V V 0mCG 1 2 0 otherwise The result follows E PROPOSITION 416 Frobenius Reciprocity Let p H a C and 1p G a C be Classfunctians where H g G Then ltde lt0 we 0 Res 1pm 417 PROOF Since gtG and are bilinear it is enough to prove this for characters p resp IP of a CH module V resp a CG module E But then we have HemCG Indg V E 2 HomCH ltVResg E D LEMMA 417 Let P be afield let H g G let V be a PHemadule and E be an PGemadule Then Indg V F E 21nd V p Resg E 418 as PGemadules 36 CHAPTER 4 INDUCED REPRESENTATIONS PROOF Identify IndICjI V PG FH V and identify V with the PH submodule 1 29 V Then V p ResICjI E is an PH submodule of IndICjI V p E Now dimp 1nd V F E G H dimp Vdimp E dimp Indgw 2 Res 13 It is enough to prove that V p Resg E generates IndICjI V p E as an PG module We use that gv x g E Gv E Vx E E generates IndgV p E Also gv x gv g 1x E gV p Resg E D COROLLARY 418 Let P C let p H a C and 1p G a C be clussfunctitms with H g G Then 1nd p 1p Indg 49 Res 1p 419 LEMMA 419 Let H g G let V be it simple CHemadule with character p let E be it simple CGE module with character 1p Then the number of times V occurs us direct summimd 0f Resg E is equal to the number of times E occurs us it direct summand aflndg V PROOF The number of times V occurs as a direct summand of Resg E is equal to 4A Res gtH 0 Res 1PM By Frobenius Reciprocity Proposition 416 this is equal to Bldg PI le lt Ilndg PlG IndICl PGI which is equal to the number of times that E occurs as a direct summand of Indg V D 5 Mackey s Theorem Let P be an arbitrary field let G be a finite group let HK g G and let S be a full system of K H double cosets in G ie G w KsH 420 565 where 6 denotes disjoint union This is often written as G U KsH 421 sEKGH Let V be an PH module For all s E S we define an PsHs 1 module sV as follows sV V as P Vector spaces and for all h E H shs 1 act on sV as h acts on V ie for all 390 E sV shs 1 v hv If p H gt AutpV is the representation of H correspondent to V then sp sHs 1 lt AutpsV AutV 422 shs l gt gt ph This is an example of transport an structure 5 MACKEYS THEOREM 37 REMARK 420 If V is an FH submodule of an PG module X then 5V g 5V 5v v E V 423 with P Vector space isomorphism 0c 25V 5V 391 H 51 Observe thatocshs 1 v 110w shv shs 1sv shs 1txv EXAMPLE 421 In PG FH V we have 5V E 5 3 V THEOREM 422 Mackey s Theorem Let V be an F H Emadule Then Res Indg V 2 QB md Hs1 K Resjgjil KsV 424 seKGH as PKemadules PROOF For all s E S let Ts be a full system of left cosets representatives of sHs l K in K Then KsH L tETS tsH Therefore G LJ gH LJ Lam ltgEGH seKGH tETs Thus InclgV EB gym K tsv seKGH tETs tET Vs is an PK module since for all k E K kts E KsH Then there are t E T5 11 E H with kts t sh Thus for all 390 E V kstv t shv hv Vi Since 5V E Vs and 5V E 5V as P5Hs 1 K modules we need to show N K sHs 1 VS IndsHs l K RESsHs l K 5V Vs tsV tET hence 5V generates Vs as PK module since Ts Q Now dimp Vs Ts dimp V K sHs l m K dimp V K sHs l m KClil 1 lpSV Then Vs g Indes l K Resi iimx 5V I as PK modules D 38 CHAPTER 4 INDUCED REPRESENTATIONS 6 Mackey s Irreducibility Criterion Let P C and G be a finite group with K g G Let s E G define HS sHs l Let V be a CH rnodule with representation p Then sV is the CsHs l rnodule with representation sp where for all li E H spgtshs 1gt poi 425 with V sV as C Vector spaces Serre s notation p5 Resilpfsi1 sp Let Ressp Resgxp THEOREM 423 Mackey s lrreducibility Criterion X Indg V is a simple CGimodule if and only if a V is simple CHimodule ie p is irreducible and b for all s E G E H p5 and Ressp are disjoint representations of HS ie ifps and Ressp are written as sums of irreducible representations they have no irreducible subrepresentation in common PROOF Since we are working over C X is simple if and only if X XG 1 but X XG V Resg Indg V gt H Proposition 416 V IndH RessHs 1 s V H Theorem 422 lt 69 H3 H3 seHGH E VInd s Res js l 5VgtH seHGH Z Resgs V Resilffsi1 sVH Proposition 416 seHGH Z RESsppsgtHs seHGH lfs E H then RessppsgtHs ppH 2 1 Hence XXG 1 if and only if ppH 1 and for all s E G 7 H RessppsHs 0 if and only ifp is irreducible and for all s E G 7 H Ressp and p5 are disjoint D COROLLARY 424 IfH 31 G then Indg V is a simple CGimodule ifand only if V is a simple CHimodule andfor all s E G E H sV S V CHAPTER 5 EXAMPLES OF INDUCED REPRESENTATIONS 1 Induction from a Normal Subgroup PROPOSITION 51 Let V be a simple CGemodule with representation p Let A 31 G Then either a there is H with A g H g G and a simple CHemodule W with representation a so that p Indg o ie V 2 1nd W 51 b Res V is a direct sum of simple isomorphic CAemodules REMARK 52 If A is abelian h is equivalent to for all a E A there is All E C with pa All identity PROOF OF PROPOSITION Let Resg V 2171 Vi be the canonical decomposition ie if CA 1171 CA lNedderburn decomposition then Vi V In particular each Vi is a direct sum of isomorphic simple CA modules If r 1 we are in case Now assume r gt 1 Note that for all s E G and for all simple CA module T also sT is a simple CsAs 1 module hence since A 51 G a simple CA module Then the action of G on V permutes the Vi Since V is a simple CG module the action of G on V permutes the Vi transitively Let H s E G sV1 V1 ThenA g H g G since r gt 1 H 7 G Therefore V E Indg V1 Since V is a simple CG module V1 has to be s simple CH module See Theorem 423 D THEOREM 53 Let A 31 G where A is abelian Then the degree of every ordinary irreducible representation p of G divides G A Compare with Theorem 44 PROOF By induction on G If G is abelian use Theorem 44 If we are in case a of Proposition 51 there is H with A g H g G and an ordinary irreducible representation 7 of H with p IndICjI 7 By induction degp H A If we are in case b of Proposition 51 then by Remark 52 for all a E A pa All identity for some All E 02 Let G pG A pA and 39 40 CHAPTER 5 EXAMPLES OF INDUCED REPRESENTATIONS p G gt GLnC be inclusion if p G zgt GLAC Then p is irreducible representation of G Moreover A p A g ZGLC Hence A g ZG Then by Theorem 423 n degp divides G ZG Since G ZG HG A we also have tl39lathG A Sincen degp then degpl G A We have a canonical surjection GA A G A gA gt gt pgA Then G A MGzAThusdengGzA D 2 Super Solvable Groups DEFINITION 54 A group G is supersolvable if there exists a finite sequence of subgroups o 1G0ltG1ltltG1ltGnG 52 with G 51 G and GG1 cyclic for 1 g i g n REMARK 55 If G is finite and nilpotent then G is super solvable In general super solvable implies solvable But the converse is not true For example A4 is solvable but not super solvable LEMMA 56 Let G be a noneahelian and superesolvahle group Then there is an abelian normal subgroup A ofG with A g ZG PROOF Since G is super solvable then GZG is super solvable Therefore there is A 51 GZG so that A is non trivial and cyclic Let 7r G gt GZG be the canonical surjection and letA 71 1A D DEFINITION 57 An ordinary representation G is called monomial if it is induced from a degree 1 representation of a subgroup THEOREM 58 Every ordinary irreducible representation of a finite superesolvahle group is monoe mial PROOF By induction on G If G is abelian we are done Let p G gt GLAC be an irreducible representation If kerp 7E 1 then p is an irreducible representation of G kerp and we are done by induction Now assume G is not abelian and p faithful ie kerp By Lemma 56 there is A 51 G abelian with A g ZG Since p is faithful pA g ZpG Thus there is a E A with pa is not a scalar multiple of the identity Therefore Resg p is not a direct sum of equivalent representations Then we are in case a of Proposition 51 Thus there is H with A g H g G and there is an irreducible representation 7 of H with p IndICjI 7 By induction 7 is monomial Since Ind is transitive also p is monomial D 3 The Ring RG Everything can be done for an arbitrary algebraic closed field of characteristic zero instead of C 4 GROTHENDIECK GROUPS 41 Let G be a finite group with h conjugacy classes Let XL xh be the distinct irreducible char acters of G over C A class function f of G over C is a character if and only if is a Z linear combination of X1 xh with non negative coefficients Let RG be the set of all characters of G over C Define RG 2x1 69 69220 53 The elements of RG are called virtual characters of G over C Since the product of two characters is the character of the tensor product RG is a ring In fact RG is a subring of the C algebra clG Serre s notation Pf G clG Since 90 Xh is a C basis of clG clG C 292 RG 54 Let H g G then Indg RH R RG 55 Resg RG R RH 56 Recall that if p E RH and 1p E RG then Ind 11 1P 1116113 11 Res 1P then the image of lndICjI in RG is an ideal of RG By Frobenius Reciprocity Proposition 416 we have Bldg PI 1PgtG ltPl Resg Then IndGRes is an adjoint pair with respect to the symmetric bilinear forms G and H If B is a commutative ring we get bilinear homomorphisms B Zlnd1 2 B ZRH H B Z RG 57 B Z Res B Z RG H B Z RH 58 ANOTHER POINT OF VIEW RG is the Grothendieck group G0CG 4 Grothendieck Groups Let A be a ring not necessary commutative and let C be the category of left unital A modules we assume 1 0 E ObC 2 for all M1M2 E ObC M1 69 M2 6 ObC We also assume that the collection of all isomorphism classes of modules in C forms a set EXAMPLE 59 1 C A 7 Mod the category of all left A modules 2 C A 7 mod the category of all finitely generated left A modules 3 C 73C the category of all finitely generated projective left A modules 42 CHAPTER 5 EXAMPLES OF INDUCED REPRESENTATIONS A short exact sequence of A modules 07LLM gtN7gt0 59 is said to be in C if LMNf and g are in C DEFINITION 510 Let P be the free abelian group generated by symbols M one for each isomorphism class of modules in C Let P0 be the subgroup generated by expressions 7 L 7 N for each short exact sequence in C 07gtL7gtM7gtN7gt0 510 The Gruthendieck Group of C is defined by K0C PPo 511 REMARK 511 Each X E K0C can be written as X E 7 N for MN E ObC This presentation is not unique in general It can happen that N for MN E ObC with M SE N 1 The Gruthendieck Group of the ring A is G0A K0A 7 mod 512 Thus RG G0CG 513 2 The Prujective Class of the ring A is KOltAgt 7 mm 514 REMARK 512 Since every short exact sequence in 73A splits addition in K0A K0 K073A correspond to direct sums in 73A This is not true in general in G0A However ifA is semisimple then A 7 mod 73A So in this case K0A G0A CHAPTER 6 A THEOREM BY BRAUER In this chapter G will denote a finite group and p a prime integer 1 pRegular Elements and pElementary Subgroups DEFINITION 61 Let X E G X is called a pelement or puniputent if the order of X denoted by M is a power of p X is called p element or pregulur if gcdlep 1 REMARK 62 Note that the identity element 1G is p regular for all prime p Let X E G Then X can be written uniquely as X Xu X7 where Xu is p unipotent and X7 is p regular with X1430 nyu Furthermore Xu and X7 are called respectively the p component and the p component of X If lxl yam with gcdpm 1 then there are 75 E Z with tp 5m 1 thus X Xsm and X XW DEFINITION 63 A subgroup H of G is called pelementury if H E C X P where C is a cyclic group of order prime to p and P is a p group REMARK 64 A p elementary subgroup H of G is nilpotent hence super solvable The decomposition H E C X P is unique since C h E H his p regular and P h E H h is p element Let X E G be a p element Define C X and let P be a p Sylow subgroup of CG Then H CP is called a pelementury subgroup associated to X Note that H is unique up to conjugation in CGX 2 Induced Characters Arising from pElementary Subgroups Define V7 to be the subgroup of RG generated by all characters induced from p elementary subgroups Let 90 be the set of all p elementary subgroups of G 43 44 CHAPTER 6 A THEOREM BY BRAUER THEOREM 65 Brauer The index of V7 in RG isfinite and prime to p where V7 is the image of the homomorphism Ind 69 RH A RG 61 H 6x07 fH H E In611in H 6x07 Since for all H g G the image of Indg in RG is an ideal of RG we get that V7 is an ideal of RG CLAIM To prove Theorem 65 it is enough to prove that there is m E 2 with gcdpm 1 and m E V7 means m trivial simple character of G is in Vp PROOF OF CLAIM lfm E V7 then my E V7 for all 1p E RG ThenmRG g V7 g RG Thus RG1VplllRGImRGl mquot where RG 212 D To prove Theorem 65 we need to proof the following Theorem THEOREM 0 Let G 1111 with gcdpl 1 Thenl E Vp Let A where g is a primitive G th root of the unity Then A is finitely generated free Z module of rank 4G where 4 denotes the Euler zp function and A has a Z basis of the form a0 1111 0cc where xi U where C 4G E 1 All elements in A are algebraic integers A Q Z We extend lnd A linearly to get an A module homomorphism A Z Ind 69 ZRH gtA Z RG 62 HEW LEMMA The image of A 292 Ind is A 292 Vp Moreover A 292 Vp RG Vp PROOF The first statement is clear Since V7 g RG and RG is a free abelian group of rank h then V7 is a free abelian group of rank 739 g h Therefore there is a Z basis 1 oh of RG and integers m1 m7 so that m1p1 m7p7 is a Z basis for Vp Then C 7 A 82 Vp 69 ZIX HJTPj i0 j1 where RG lew Therefore A 292 Vp RG 121 Zm pj V7 1x0 1in Z for i 2 1 D COROLLARY To show Theorem 0 it is enough to prove that for all I E A 292 Vp ie for all H E 9p there arefu E RH and ilH E Awith I Z aH IndEUH 64 HEN PROOF Use thatl E RG D 3 CONSTRUCTION OF CHARACTERS 45 3 Construction of Characters LEMMA 1 Let f G gt C be a class function with integer values which are all divisible by G ie f G g ZG Then f is an A linear combination of characters induced from cyclic subgroups of G PROOF We first prove that G has this form Let C g G be cyclic and define 0C to be the class function of C over C with C if C s 9 65 CS 0 otherwise Then 1 G 7 7 71 lndC 9cZ i C tic 9ctzt tzrl ec NOTE Fix 2 E G then for all t E G tzt 1 generates a unique cyclic group of G Then 1 G 66 CltG teG C cyclic lttzt 1gtC Therefore G 2 ME 9C 67 CltG C cyclic Now letf G gt C be a class function such that f G x where x G gt Z is a class function Then f E 111013 9C X C G C cyclic 2 ME ac Resg x C G C cyclic Put KC 0C Resg x To complete the proof it is enough to show that for all C g G with C cyclic KC E A 292 RC For this it is enough to prove that for all p E RC KC pk E A 46 CHAPTER 6 A THEOREM BY BRAUER Indeed if p E RC then KCPk 17 Z xdsws l Z xsps 1 E A seC ltsgtC Note that we use that C s ifC 5 ms 9 lt gt 68 0 otherwise D LEMMA 2 Letf E A 292 RG with integer values ie lt Z Let p be a prime and let X E G with p component X7 Then fX EfX mod p 69 PROOF Using Resay we may assume G Then f 21 mm where a E A and X1xh are the irreducible characters of G X over C Since G X all x are 1 dimensional and thus group homomorphisms Now X X X1 X1 X where X is a p element Then there is a power 1 of p such that X 7 X39l Then xiX 7 xiX 7 for all i Therefore h E Zaxix 7 mod pA i1 EfX 7 mod pA Since Q Z actually fX 7 E fX 7 mod p But for all A E Z A 7 E A mod p Fermat s Little Theorem and q pk for some k then A 7 E A mod p thus we get fX E fXr mod P D LEMMA 3 Let X E G be a p element and let H C P be a p elementary subgroup associated to X ie C X an P a p Sylow of CgX Then there exists 1p E A 292 RH with values in Z such that IP lndICjI 1p satisfies a 1p X E 0 mod p and b lPS 0 for all p elements s of G that are not conjugate to X PROOF Let fc be the class function of C over C with C if u X fCu T 0 otherwise 610 3 CONSTRUCTION OF CHARACTERS 47 The Second Orthogonality Relation Corollary 311 on C is C 39fu x xx 1xu 0 1 th 611 xeRltCgt 0 erw1se xirreducible Then fC Z xx 1x e A Z RCXquot1 e A 612 x6RC xirreducible We in ate fC to a class function IP of H by defining 011 for all u E C and v E P Then 1p E A 292 Take lp Indg 1p Lets be a p regular element then 1 71 Ms 7 tst 613 H t w gt tsrleH If p regular then tsf l E H if and only if tst 1 C en 1 INS fct5t 1 H g tsrlec i E C T H tE tst 1x thus ifs is not conjugate to X then lp s 0 Now 1 1p x 7 C H txrl C mikfcc X CGX P which is relative prime to p Therefore lPX 7 0 mod p D LEMMA 4 There exists 1p E A 292 V7 with values in Z such that 00 0 mod p for all x E G PROOF Let Xi be a full system of conjugates of p regular elements in G For each E A Z V7 and 1509 7 0 mod p and 1509 0forj 11Letlp Eilpg thus 1p E A 292 Vp Let X E G with p component X1 Then X1 is conjugate to a unique Xi By Lemma 2 since all the have integer values 1W 1W1 E IP09 mod P 48 CHAPTER 6 A THEOREM BY BRAUER Therefore W3C E 1PM W99 i 0 mod P D PROOF OF THEOREM 0 Let 1p E A 292 V7 be as in Lemma 4 ie IP has integer values and lpX 7 0 mod p for all X E G Let N Zp Z My where 4 denotes the Euler 1 function Then for all A E Z with ngt p 1 we have AN E 1 mod p Euler 1 function Theorem Therefore for all X E G IPXN E 1 mod p Take the class function 1IPN 7 1 Then for all X E G lleX 7 1 E ZG By Lemma 11IPN 7 1 is an A linear combination of characters induced from cyclic subgroups Since every cyclic group is p elementary for every prime p 11pN 7 1 e A Z V 614 Since V7 is an ideal of RG also A 292 V7 is an ideal of A 292 RG Then lle lle l p E A 292 V7 ii39N J E RG 1y E A I31 Vp Thereforel lle 7 1IPN 7 1 E A 292 V D 4 Brauer s Induction Theorem DEFINITION 66 A subgroup H g G is called elementary if H is p elementary for at least one prime p THEOREM 67 Every Character of G over C is a linear combination with integers coe icients of Characters induced from elementary subgroups PROOF Let V 2 Prime Vp We need to show that RG V Since V7 g V g RG for every prime p then WC IVllWG 2 for all prime p By Theorem 65 gcdRG Vp p 1 Since this is true for all prime p it follows that RG V 1 Thus RG V B Let X be the set of all elementary subgroups of G Then Theorem 67 says that Ind EB RH A RG HEX fH H E InngH HEX is surjective THEOREM 68 Every Character of G over C is a linear combination with integer coe icients of monomial Characters ie Characters induced from 17degree Characters of subgroups PROOF If H is elementary then H is nilpotent hence super solvable By Theorem 58 every character on H is monomial Since induction is transitive the result follows from Theorem 67 REMARK 69 1 Usually the integer coefficients in Theorems 67 and 68 may be pos itive or negative 2 We will use Theorem 68 for rationality questions over cyclotomic fields CHAPTER 7 RATIONALITY QUESTIONS In this chapter G denotes a finite group all modules are supposed to be finitely generated K denotes a field of characteristic zero and C K its algebraic closure 1 The Rings RKG and If V is a KG module then VC C K V E CG K GV as CG modules If p G gt GLMK is the representation correspondent to V then the representation correspondent to VC is given by pdg pg for all g E G with C linear extension Thus the characters of p and pc are the same Define RKG to be the abelian group generated by all characters of G over K Said differently RKG is the Grothendieck group of the category of finitely generated KG modules Then RG is a subring of RCG RG identification PROPOSITION 71 Let V1 V be afall system of representatives ofnoneisomorphic simple KGemodules with Characters T1 T Then T1 T is a Zebasis ofRKG Moreover the T are mutually orthogonal with respect to the symmetric bilinear form Ir1P 2G mam g6 PROOF It is clear that T1 T generate RKG as Z module Now dil l lK HOnglilj dil l lc H0mcclic lt2le By Schur s Lemma 0 ifi A j H VvVv 01391 1KG 1 EndKGUi ifi1 Then Tiffj 0 ifi 7 and my dimKEndKGUi gt 0 ifi In particular T1T are linearly independent over Z D 50 CHAPTER 7 RATIONALITY QUESTIONS DEFINITION 72 a A simple KG module V is called absolutely irreducible if VC is simple C G module b A CG module X resp a representation p of G over C is realizable over K or rational over K if there exists a KG module V resp a representation p of G over K with X E VC r8511 P pc THEOREM 73 Let X be a C Gemadule with Character 9 Then X is realizable aver K if and only if X E PROOF This direction is clear Assume x 2171 rim where m E Z for l g i g r and T1 T7 are the distinct irreducible characters of G over K correspondent to simple KG modules V1 V7 Then for all 1 g i g r 0 S dimc H0mCGXIViC ltX1Tigt quotMETil Since 1317 gt 0 then ni 2 0 for all 1 g i g 7 Let V 121 V7 Vi is a KG module Then VC and X have the same character Therefore X E VC D CONSEQUENCE Assume that H g G Consider Resg RKG E RKH Indg RKH E RKG If X is a CH module realizable over K then X 2 VC for a KH module V Then 1nd X Indg VC E CG CH CH KH VCH V VC g CG KG KG KH V 2 CG KG Indg V g Indg VC Thus Indg VC Indg VC 71 DEFINITION 74 Define to be the subring of RCG RG consisting of all class functions in RCG with values in K Thus MG lt mm 72 2 Schur Indices Fix the following notation Since KG is semisimple then it has a Wedderburn decomposi tion KG KG 2 MatmD 73 gt1 13971 1 L where Di is a skew field and Ii 2 l for l g i g 7 Let Vi be a simple KG module correspond to the i th component Then V1 V7 is a full system of representatives of non isomorphic simple KG modules with Di g EndKGUDO 2 SCHUR INDICES 51 for all 1 g i g r If V is a free Di module with rank 1 then EndeUi E MatxDv 75 foralll g i g r DEFINITION 75 A field L of characteristic zero is called a splittingfield of G if LG is iso morphic to a direct product of full matrix rings over L ie m LG g H Matnx 76 i1 In particular L is a splitting field of G if and only if every simple LG module is absolutely irreducible EXAMPLE 76 L C K is splitting field of G THEOREM 77 Let L be a splittingfield of G with K g L eg L K Let S be a simple LGemadule with representation p and Character x Then a there is a unique i E 1r with fS S and ij Ofar allj 73 i ie S is a simple fiLGfiernadule b K ZD KOO Kxg g E G thefield obtainedfrarn K by adjoining xgf0r g E G PROOF a We have ZKG g ZLG Then 1G gaff is a central decomposition of 1 in LG thus it can be refined to a primitive central decomposition if 1 in LG say 221 ejk is a primitive central decomposition for all 1 g g r Then there is a unique i E 1 r and a unique k0 E 1s39 so that eikOS S and eij 0 for all jk 7e ik0 since S is a simple LG module Then fS 21 eik S eikOS S and for allj 7e ifvS 0 So we obtain a h p induces an injective K algebra homomorphism KG A MatnL 77 where the index i is obtained from the statement a and n dimL Sp can be extended L linearly to an Lalgebra homomorphism p LG a Mat L then pl MM butp is injective since KGf is simple ring and S S i 0 Let B p KG Since is injective it suffices to show that Z09 KOO In 78 where In denotes the n X n identity matrix NOTE By a it follows that for all X E LG W 902 909999 79 Let g E G and Kg be its class sum under conjugation let Cg be the number of G conjugates of g Then by Schur s Lemma since L is splitting field pKg wgl for some wg E L Then Mgquot Trp1ltg Cgxg Thus xg1n p g MfiKgfi E ZB Therefore KOO In E ZB 52 CHAPTER 7 RATIONALITY QUESTIONS Letx Egg Agg E KG Ag E K for all g E G with px E ZB since ZB Q ZMatnL use Schur s Lemma there is A E L with px A In It follows that An Trpx Egg Agxg E Kx Therefore A E Kx Hence ZB Q Kx In D DEFINITION 78 Let L be a splitting field of G containing K and let S be a simple LG module with character x By Theorem 77 there is a unique i E 1 7 with 125 S and ij 0 for allj 7E i and K ZD39 Kx Since KiK is a finite extension C NOTE Since C EEKx D is a simple C algebra there is m E 2 with C k D 2 Math C 710 then dimKx D dimCC EEKx Di We define such m to be the Schur index of x over K written 51x mi Serre calls this the Schur index of V or of KG Ende 3 Review of Traces over Fields Let EK be a finite extension Let X E E Let Ta E gt E be multiplication by X If n E K then Ta can be written as an n X 71 matrix with entries in K with respect to a K basis of E Define TrgK0c TrTa E K 711 Define XE U E gt C Uembedding 71K idK E C K 712 ie XE is the set of all K embeddings of E into C Then TIEK01 Z 0a 713 7625 for all X E E DEFINITION 79 Let 1 g i g 75 E G and pi5 V 4 Vi be the map with pi5v 51 for all 390 E V We take three different types of traces of p5 1 its trace as a K endomorphism of V denoted by 905 this is just the usual K character correspondent to V evaluated at 5 Thus 905 E K 2 its trace as Ki endomorphism of V denoted by p5 this is computed by choosing a K basis of V writing p5 as a matrix with entries in K with respect to this basis and then take its trace thus p5 E K 3 its reduce truce as an element of the simple algebra Ende E Matx Di denoted by 1p5 This is defined as follows Recall that C EEKx D E MatmxC then C K1EndD1Vv g MatmxxC 714 We consider the composition G a EndeVi C x EndDVi MatmlC 5 MS 1 GEMS 9 4 SCHUR INDICES CONT 53 C is Viewed as a Kimodule Via the inclusion I K gt C Then this composition defines an irreducible representation of G over C of degree mili Define 45 to be its character evaluated at 5 By Theorem 77 it follows that K K1pi and hence is E Ki The connection between this traces is 205 TrKKPi5 and MS mi i5 715 REMARK 710 The process of producing an irreducible representation of G over C de scribed in 3 can be used to find all irreducible representations of G over C as follows Let i E 1r Define 2i 2 U Ki gt C UiS anembedding and UlK idK LetU E 2139 Then C can be Viewed as a Ki module Via the embedding U Ki gt C Consider the composition Then this composition defines an irreducible representation of G over C Since we embedded K into C using 7 the character of this representation is 7IPZ39 where IPZ39 is as in we get that forall1 g i 7 Xi TrKKPi Z 0011 2 WWW 716 762 762 Since each irreducible character of G over C has to come up as constituent of exactly one 9 it follows that 01pi 1 g i g 70 E 2 717 is a full system of distinct characters of G over C It follows that for each fixed i GalCK permutes the characters 70 U E 2i transitively since each a E GalCK on restriction to K becomes one of the elements in 2139 4 Schur Indices Cont THEOREM 711 mii 1 g i g r is a Zibasis 0fEG In particular RKG hasfinite index in PROOF Since the 99 s are mutually orthogonal with respect to the symmetric bilinear form we get that mixi with 1 g i g r are linearly independent over Z Let x E R7KG then 7 x Z 2 WWW l1t762 where am E Z Now x E if and only ifx 70 for all 7 E GalCK if and only if for all 1 g i g r divt7 divw for all 70 E 2139 Then 11 762 7 1 Edii midwile E 71Wth i1 mi 762 WEE 54 CHAPTER 7 RATIONALITY QUESTIONS Thusx E I1Z mii D DEFINITION 712 We call KG quasisplit if all of the Di are commutative ie Di K for all1 g i g rifandonlyifrni 1 forall1 g i g r COROLLARY 713 RKG ifana only ifKG is quasiesplit LEMMA 714 Let S be a simple C Gernaaale with Character x Then S is realizable aver K x if and only iszOC 1 PROOF We know thatx 709 for an elementi E 1 r and U E XI The representa tion of G over C correspond to S is therefore equivalent to the composition G EndeV39 I C K1EndD1Vi E MatmxxC 5 715 1 715 This representation is realizable over K then Ki Di If Ki Di then rni 1 therefore 139 71Pi 0pi then this representation is realizable over Ki Since K KOO it follows that 7Ki K This the Schur index of x over K is 1 ie sKOC 1 D 5 Realizability over Cyclotomic Fields DEFINITION 715 Let G be a finite group the exponent rn of G is the least common multiple of all orders of elements of G A field with any characteristic is said to be sufficiently large relative to G if the field contains all rn th roots of the unity THEOREM 716 Let K be a field afcharacteristic zero that is sa iciently large relative to G Then RKG RCG In particular every representation of G aver C is realizable aver K hence K is the splitting field of G PROOF Let x E RCG By Brauer s Induction Theorems Theorems 67 and 68 X anlndg Q E where Hf E I is a collection of subgroups of G nj E Z and Q is the character of Hf over C of degree 1 for all E I Then each Q has values in K Since degCQ 1 this means Q is realizable over K ie Q E RKHj for allj E I Then Ind Q E RKG for allj E I Therefore x E RKG D 6 The Rank of RKG Let K be a field of characteristic zero rn be the exponent of G and w be a primitive rn th root of the unity in K Let L Kw Then LK is Galois with Galois group isomorphic to a subgroup of ZrnZ denote by TK the image of GalLK in ZrnZ Fort E ZrnZ let at E GalLK be defined by 011 wt 6 THE RANK OF RK G 55 DEFINITION 717 Let ss E G We say that s and s are TKconjugate if there is t E PK with st and s are conjugate in G NOTE for all n E Z squot depends only on the congruence class of n mod ls l then it only depends on the congruence class of n mod This defines an equivalence relation in G The classes are called TKclasses of G THEOREM 718 Letf G a L be a classfunction Thenf E K Z RLG ifand only if for alls E Gandt E PK otfs fst PROOF Let p be a representation of G over L with character x Let s E G t E PK If M is the set of eigenvalues of ps then is the set of eigenvalues of pst Then was at 2 2A past Therefore x satisfies Since for all A E K and t E D QM A every f E K 292 RLG satisfies Assume f Z CxX XERLG xirreducible where cC E L We want to show cC E K for all x ie for all t E PK Utcx cx Now 1 7 Ex x E Zf5X5 1 seG Zfltstgtxltswgcdltnmgt 1 seG Z 7tf5 7t265 1INNSj 1151 amp 0r 5 lll M5 for 5 seG MDWOO UtCxr for allt E PK D COROLLARY 719 Let f G a L be a classfunction with values in K ie f E Then f E K 292 RKG ifand only iff is constant on the FKeclasses iefor all s E G andt E PK fs fst In particular the rank ofRKG as afree Zemodule which is equal to dimKK Z RKG is equal to the number of FKeclasses in G PROOF By Theorem 718 for all s E G and t E PK Utf5 fst with Utf5 since fhas values in K Assume f E 9 XERLG x irreducible 56 CHAPTER 7 RATIONALITY QUESTIONS where Ex E L Let s E G and t E PK By assumption fst But we also have that Utf5 fs since fhas values in K Then by Theorem 718 all Cchi are in K As in the proof of Theorem 718 we see that this means Ex x Utft7tx C17xwlll39h f Hf since f is constant on the Lg classes Then Ex Cm for all t E PK This implies that f is a K linear combination of characters of the form 2 M26 TILKOO tELK It is enough to show now that TrLKOC E RKG We can view V the irreducible LG module correspondent to x as a KG module since K C L Then the character of V viewed as a KG module is TrLKOC Then TrLKOC E RKG D 7 Generalization of Brauer s Induction Theorem RECALL Let K be a field of characteristic zero C K and X be the set of all elementary subgroups of G Then H g G is p elementary with p prime if and only if H Z X P where Z is cyclic of order relative prime to p and P is a p group and Ind E RKH A RCG HEX fHHeX H E IndIC lUH H ex is surjective REMARK 720 If we replace C by K this is not always true EXAMPLE 721 Let K lR C C G 53 and let X be the set of elementary subgroups of G ie X id 12 13 23 123 We claim that the map Ind E RKH A RCG HEX fHH6X H E IndIClUH HEX is not surjective PROOF Look at the 2 dimensional lRSg module V with representation p and character x as follows We show that x cannot be written as a Z linear combination of lnd as above We id 12 123 10 01 071 901 10 171 x 2 0 71 first compute all Ind where fH is an irreducible character of H over lR 1 H1 id Let le be the trivial character Then 71 GENERALIZATION OF BRAUERS INDUCTION THEOREM 57 id 12 123 4111ndH1fH1 6 6 6 id 12 123 P21 Indeme 3 1 0 p22 Inng fHZZ 3 71 0 2 H2 12 Let fH21 be the trivial character and fH22 be the sign character Then The subgroups 13 and 23 lead the same induced character as H2 3 H3 123 Let fH31 be the trivial character and fH32 be the character of degree 2 T where T 1p 1p where IP is the 1 dimensional character of H3 with lp123 8 1V is the 1 dimensional character of H3 with 1p 123 87237 Note that IP and IV are not realizable over lR but 1p 1V is realizable over lR Note also that Tid 2 T123 E1 and T132 71 Then id 12 123 p31 Inst pr31 2 0 2 p32 Inng fHSZ 4 0 72 Take any Z linear combination of p p21 p22 p31 and p32 Then if 11 11211122 11315132 E Z 1 1123 21 21123 22 22123 31 31123 32 32123 6 1 2 21 Z zzl which is an even integer But x123 71 This proves our claim Note that over C x Iridij3 1p since mgwam 2 D wawan gmm maw an gewan wwwm 7 7 ir Note that in the above computations we always use the formula 1 InngHXS E E fHt5tTl tsrleH DEFINITION 722 Let H g G is called TKpelementury if H Z gt4 P where Z is a cyclic group of order prime to p and P is a p group such that for all g E P there is t tg E PK with yzy 1 zt for allz E Z H g G is called lquotA p 39 y ifH is lquotA p 39 y for at least one prime p 58 CHAPTER 7 RATIONALITY QUESTIONS THEOREM 723 Generalization of Brauer s Induction Theorem Let XK be the set afall PK 7 peelementm39y subgroups of G Then Ind Z RKH A RCG H EXK fHHeXK H E InngH H EXK is surjectivei PROOF See 7 pgs 99 101 D Part 2 MODULAR REPRESENTATION THEORY INTRODUCTION TO BRAUER THEORY CHAPTER 8 FUNDAMENTALS 1 Discrete Valuation Rings There are basically two ways to talk about discrete valuation rings DEFINITION 81 Discrete Valuation Rings Case Let K ba field A valuation of K is a rnappKgtRU0with i pa 0 if and only ifa 0 ii pab papb for all ab E K iii pa b g pa pb for all ab E K p is called nunarchimedian if iii is replaced by 390 WI H1 S maxf Pb The set V pa a E K g R 81 is called the value group of p If the value group of a valuation p is an infinite cyclic groupie V 7quot n E Z for some real number 739 gt 1 then p is called a discrete valuation Suppose p is non archirnedian then RaEKpa 1 82 is called the valuation ring of p If p is moreover discrete R is called a discrete valuatiun ring DEFINITION 82 Discrete Valuation Rings Case Let K be a field A map 9 K gt Z U 00 is called nunarchimedian discrete valuatiun on K if i 9a 00 if and only ifa 0 ii 9ab 9a 9b for all ab E K iii 9a b 2 rnin9a9b for all ab E K Then R a EK0a 20U0 83 is a ring called the valuation ring of 9 Note that in this case we only talk about discrete valuations so R is a discrete valuatiun ring 61 62 CHAPTER 8 FUNDAMENTALS In both cases and H if R is a discrete valuation ring then R is a local ring with maximal ideal m where In case I m a E K p01 lt 1 84 In case H m a E K2001 gt 1 85 and the units of R are lncase Rgtlt a E K p01 1 86 In case ll Rgtlt a E K2001 0 87 EXAMPLE 83 1 Trivial valuation p K gt lRJr U 0 88 0 gt gt 0 a H 111 73 0 2 llzCgtlRU0 89 a H W is an archimedian valuation 3 Let R be an integral domain and let FracR K where FracR denotes the field of fractions of R Let p be a prime ideal of R satisfying 1 U120 P11 0 ii p1 7E p for alli 7E j iii ifa E p 7 p0 b E p5 7 pl 1 then ab E W 7 NW The we can define a map pp R A 111 u 0 810 0 gt gt 0 a gt gt 8 if E p 7 M where 8 stands for any real number greater than 1 Then we may extend n to K by z My 1 lt5 7 MS 811 Then n is a anon archimedian discrete valuation of K called the pudic valuation of K The discrete valuation ring is RFEV5ERsEp 812 which is the localization of R at the prime ideal p EXAMPLE 84 Let R Z and p p where p is a prime number Then p satisfies 1 7 iii of Example 83 3 Thus we get a p adic valuation of Q 1 DISCRETE VALUATION RINGS 63 DEFINITION 85 Let R be an integral domain and A an extension ring of R ie R Q A The integral closure of R on A is defined by E z a E A there exists a monic polynomial px E RM with pa 0 813 R is integrally closed in A if R H DEFINITION 86 A ring R is said to be a Dedekind domain if R is integral domain Noetherian integrally closed in K FracR and every non zero prime ideal in R is prime Then if R is a Dedekind domain then every maximal ideal m of R satisfies 7 iii of Example 83 3 Therefore we can find m adic valuations EXAMPLE 87 The following are examples of Dedekind domains 0 Every principal ideal domain 0 Let R be a Dedekind domain with K FracR Let L K be a finite field extension Then the integral closure S of R in L ie S E is a Dedekind domain with FracS L 0 Let R Z and K FracZ Q Let U be a n root of unity and L Qw Therefore S Z 4 Suppose p is non archimedian discrete valuation Then R the discrete valuation ring of p is a principal ideal domain with unique maximal ideal p Thus p R71 where p01 is a generator of the value group of p 71 is often called a uniformizer of R or of p All non zero ideals of R have the form R7Ii with i E ZJr U Let R be a complete discrete valuation ring of characteristic zero corresponding to a non archimedian discrete valuation p K gt IRJr U 0 with V rquot n E Z for some real number r gt 1 as in case resp 9 K gt Z U 00 as in case Suppose that the residue field of R ie Rm is a field k of positive characteristic p Then p E R and since p is zero in k p is not a unit in R Hence pp r e resp 91 e for some positive integer e e is called the absolute ramification index of R If e 1 ie p is a uniformizer of R we say that R is absolutely unramified THEOREM 88 For every perfect field k of positive characteristic p there exists a complete discrete valuation ring and only one up to unique isomorphism which is absolutely unramified and has residue field k PROOF See 8 D The ring of Theorem 88 is denoted by Wk and called the ring ofinfinite Witt vectors over k One can show that there is a unique lifting f k gt Wk which commutes with p th powers ie fN7 1 00 and which is multiplicative ie fAu for all A u E k This is also described in 8 Serre calls this the multiplicative system of representatives 64 CHAPTER 8 FUNDAMENTALS 2 Completions Let K be a field p a valuation of K Then K is a metric space via p Namely if a E K then the neighborhoods of a are open spheres x E K px E a lt e 814 where e E IRT We say that two valuations on K are equivalent if they induced the same topology on K A prime of K is an equivalence class of valuations of K Since K is a metric space via p we can form the completion K of that metric space Recall that the elements of K are equivalence classes of Cauchy sequences in K One shows that K is a field and K gt K via the constant Cauchy sequences we can extend p to a valuation on K in a unique way Let C Cnnez be a Cauchy sequence in K then pc EZ is a Cauchy sequence in IR since IR is complete there is limn w En US We set vc K is complete with respect to its adic metric Note that if p is non archimedian then is non archimedian Also the value group of p is the same value group of Hence if p is non archimedian discrete then is non archimedian discrete Let p be non archimedian discrete Let R be the valuation ring of p let R be the valuation ring of Let p be the unique maximal ideal of R and let 5 the unique maximal ideal of R Then p p O R Rp and R pi 2 R A31 815 fori E 24 3 pModular Systems DEFINITION 89 Let p be a prime number A pmodnlur system is a triple K R k where K is a field of characteristic zero R is the valuation ring of a non archimedian discrete valuation of K with unique maximal ideal m and k is the residue field of R ie k Rm such that Chmk p We can complete a p modular system K Rk to KRk k Rm E Rr A p modular system is complete if it is equal to its completion NOTATION G always denote a finite group and p always denote a prime number Also K R k is always a complete p modular system As we state in early chapters any module is assumed to be finitely generated 4 Inverse Limits Let C be the category of groups resp of commutative rings resp of R modules Let I be a partially ordered set and suppose A is an object in C for all i E I Suppose that for indices ij E I withi g i there is a morphism p A gt A in C satisfying yj 39tkj yk for alli S S k and ii p idAHfor alli E I We say that the set Apf61 is an inverse system Let A 01061 E 1161 A pa ig39 a for all i g Then A is an object in C called the inverse limit of the inverse system A p denoted by A liLnA The natural projections from the direct products 1161 A to the A s induce surjective morphisms p A gt A for all i E I The inverse limit satisfies the following 4 INVERSE LIMITS 65 universal property If D is any object in C together with morphisms 71139 D gt Ai for all i E I satisfying 71139 Wyn whenever i g j then there is a unique morphism 7r D gt A such that 171 71139 for all i E I So we obtain the following commutative diagram for all i g in I Ai quotA EXAMPLE 810 1 Let A be a commutative ring with ideal I Then Aquot n E 2 forms an inverse system with morphisms mi AIJ gt A I1 with i g j given by the natural surjections A Aquot is called the I udic completion of A 2 If R belongs to a complete p modular system K R k and m is the unique maximal ideal of R then R is equal to its m adic completion Hence R liLnRmquot 3 Letheagroup LetV N N 51 Gand G N lt 00 Then GN N E N forms an inverse system with morphisms yMN GN gt GM N g M given by the natural surjections Then G GN is called the profinite completion of G CHAPTER 9 THE GROUPS RKG RKG AND PKG 1 The Rings RKG and RkG Let L be a field Then RLG is a Grothendieck group of all finitely generated LG modules This is a commutative ring with multiplication induced by L and multiplicative identity equal to T where T is the trivial simple LG module Recall that if X is an LG module then X denotes its image in RLG Define REG X is an LG module 91 Let 5L be a system of non isomorphic simple LG modules THEOREM 91 RLG is afree Zimadule with basis S E 5L PROOF Let P be a free Z module with basis 5L say P SESL ZS Define 0c P A RLG AskesL H E AslSl SESL Then X is a Z module homomorphism Let S E 5L If X is an LG module let 15X be the number of times that S occurs as a composition factor of X Note that ls is additive with respect to short exact sequences Namely if 0gtX1gtX2gtX3gt0 is exact then lsX2 lsX1 15X3 Hence ls induces a well defined Z module homomor phism 3985 RLG gt Z with 15X for all LG module X and Z linear extension Then we get a Z module homomorphism RLG A P with 5X565L 15X565L and Z linear extension Then we have that tx idp and th idRLG Then RLG E P as Z modules D 67 68 CHAPTER 9 THE GROUPS RKG RKG AND PKG REMARK 92 1 The elements of RKG can be identified with the virtual characters of G over K Then we can use all our previous results about RgG 2 KG is semisimple Therefore if E and E are KG modules then E E E ltgt E E 92 in RKG This is not true ne general for k with Chil7k l G EXAMPLE 93 Let k le G U E 21 with p prime Let V k 69k with trivial G action and let V be the 2 dimensional KG module where 7 acts on V as 1 i Then V 9 V as kG modules but V V in RkG since V and V have the same composition factors 2 The Groups PRG and PkG Let A be a commutative ring Then PAG is the projective class group of AG ie FAG Ko73AG 93 the Grothendieck group of the categories of finitely generated projective AG modules Define P P is finitely projective AG module 94 Note that since KG is semisimple PKG RKG because all KG modules are projective lf Chm39k l G this is not true EXAMPLE 94 Let k B G U E 21 with p prime number Then kG is projective Therefore the trivial simple kG module is not projective LEMMA 95 Let E be a kGemadule let P be a projective kGemadule Then the kGemadule E k P with diagonal Genetian is a projective kGemadule Then PkG is an RkGem0dule PROOF lt suffices to show that E k kG is a free kG module Let E0 be the underlying k vector space of E Each g E G acts on a simple tensor e 29 X E E k kG as ge 29 X ge gX Also each g E G acts on a simple tensor e 29 X E E0 k kG as ge 29 X e 29 gX Define p 2 E0 kkG A E kkG e g gt gt ge 29 g then p is a k vector space isomorphism p is also a G homomorphism since for all h E G 10103 m 118 kg hge hg Mgr g hp8 X g Then E0 k kG E E k kG as kG modules Therefore k ea eak k kG 2 kGdquotmkE0 Wd dimk Eo mes 3 STRUCTURE OF PKG 69 with E0 kkG g kkG dimk Elytimes 3 Structure of PkG DEFINITION 96 Let B be a ring A B module homomorphismf M gt N is called essential iff is surjective and for all M Q M with M submodule 7E N A projective cover of a B module M is a projective module P together with an essential homomorphism e P gt M LEMMA 97 Let A be im artiniim ring Then every nitely generated Aimadule has afinite gener ated projective cover which is unique up to isomorphism PROOF Let M be a finitely generated A module Let X be a finitely generated free A module with short exact sequence 0gtYgtXgtMgt0 ie M g XY For each submodule N of Y consider the canonical epimorphism fN XN gt XY E M Let 3 N N is submodule onwith fN essential Then 3 7E since Y E 3r Order 3 partially by 2 Since all elements of 3 are artinian each chain in 3 has a lower bound By Zorn s Lemma 3 has a minimal element call it N Let P XN and e fN then 6 is essential by minimality of N We need to show that P is projective Let T X gt X N P be the natural projection Let Q minimal among the submodules of X with TlQ surjective Q exists since A is artinian Since X is projective A module since it is free A module we have the following commutative diagram where TlQ39y T By minimality of Q 39y is surjective Let N ker39y then N Q N Consider the following commutative diagram XN fN XYEM HZ E H m TlQ IO lt H quotU 70 CHAPTER 9 THE GROUPS RKG RKG AND PKG where quotQ is essential by minimality of Q Then fN is essential By minimality of N quot in SP N N Then TjQ is an isomorphism Consider the following diagram 0 I N X TtXYEM 0 Q Let X TjQ 1 Then Ta idp Then P is direct summand of X thus P is projective Then P i M is a projective cover Suppose that P gt M is another projective cover Since P is projective we obtain the following commutative diagram P 3 I quot 3 P e gt M 0 0 where e g 6 Since 6 is essential g is surjective Then P E P 63 kerg Since 6 is essential kerg Then P E P DEFINITION 98 Let B a ring and M be a B module A submodule N of M is super uous for any submodule N ofM satisfying N N M then N M LEMMA 99 Let B be a ring Let f M a N be a Bernodule homomorphism Then f is essential if and only if f is surjective and ker f is superfluous PROPOSITION 910 Let I radkG he the Iacohson radical oka ie the maximal nilpotent ideal oka a Let X and P be kGernodules Then P is a projective cover of X if and only if P is projective and PIP g XIX 95 b Let Pi a Mi he a projective coverfor 1 g i g n Then 7t 7t zegPieegMi 96 i1 i1 is a projective cover PROOF We prove first i X and XIX have isomorphic projective covers 3 STRUCTURE OF PKG 71 PROOF Let 6 P gt XIX be a projective cover let T X gt XIX be the natural surjection By Nakayama s Lemma X is super uous hence T is essential Since P is projective we obtain the following commutative diagram P 361 X T XX o 0 where T 6 Then F is surjective with ker Q kere Therefore F is essential Since projective covers are unique up to isomorphism this implies i A Let S be a simple kG module with projective cover P5 Then P5 is indecomposable and P5 P5 2 S PROOF Let 6 P5 gt S be a projective cover Since radkG annihilates ev ery simple kG module 6 factors through E PgIPS gt S Suppose that E is not an isomorphism Then P5 P5 2 S 69 T where S a S is an isomorphism and T 7E 0 since P5 P5 is semisimple Let U XEP52XIPS 65 Then U is a proper submodule of P5 with 6U 611 S contradicting that e is essential Then is an isomorphism and P5 is induced D Now we prove a Let T P gt PIP E XIX be a natural surjection Then kerT P is super uous Thus P is a projective cover of XIX By i P is a projective cover of X Assume that P is a projective cover of X By i P is also a projective cover of XIX Now X X is semisimple say XIX S1Sn where S is simple for all 1 g i g n For all 1 g i g 71 let P5x be a projective cover of S and let Q 321 P51 Then by ii 7t 7t QIQ g Pslps g 635139 XIX i1 i1 Then the natural surjection Q gt QIQ E XIX is essential Then Q is a projective cover of X IX Since projective covers are unique up to isomorphism Q E P Then PIPg QIQ gXIX 72 CHAPTER 9 THE GROUPS RKG RKG AND PKG h This follows from u since 921 Pi is projective and Pil Pigt g Pilpi g MilMi g Mil lt Migt i1 i1 i1 i1 i1 i1 D COROLLARY 911 The Projective Index kGModules PIM are the projective covers of the simple kGimodules Every projective kGimodule is the direct sum of PIM s and this decomposition is unique up to isomorphism PROOF In ii in the proof of Proposition 910 shows that the PlM s are the projective covers of the simple kG modules Clearly every projective kG module is the direct sum of PlM s Suppose P1Pn Q1 Qm 97 where Pi and Qj are PlM s for all 1 g i g n and 1 g g m So we have 69 ENE g 69 QjIQj i1 11 Then by Jordan Holder Theorem m n and after renumbering PiIPlv E QiIQi for all 1 g i g n Since projective covers are unique up to isomorphism it follows that Pi E Q for all 1 g i g n COROLLARY 912 Let 5k be it full system of representations of noniisomorphic simple kGimodules Then ng S E 5k is u Zebusis okaG COROLLARY 913 If P and P are projective kGimodules then P g P gt P P 98 in PkG PROOF If P 356 ngP5 with n5 E Z then P2691335 565k since short exact sequences of projective modules always split D 4 Structure of PRG LEMMA 914 Let A be u commutative ring Let D be u AGimodule Then P is projective us Aimodule ifund only ifthere is u E EndAP so that Z sus 1x x 99 seG for all x E P 4 STRUCTURE OF PRG 73 PROOF Assume that P is projective AG module Then P is the direct summand of AGquot for some n This implies that P is a projective A module Let now P be an arbitrary AG module which is A projective Let P0 be the underlying A module of P and let T Ind P0 AG 2 P0 Then T is a projective AG module and there is a surjective AG module homomorphism a T A P g 29 x gt gt gx for all g E G and X E P with A linearly extension We get that P is projective AG module if and only if there is v E EndAgP T with av idp By Frobenious Reciprocity Proposition 416 we have a surjective map EndAP HomA Resglj 13130 A HomAPAG 2 P0 u gt gt x gt gt Z s us 1x seG Then v has the form vx Z s 29 us 1x seG for some it E EndAP av idp means that for all X E P E sus 1x x seG D LEMMA 915 Let A be a commutative local ring let M be a finitely generated Aemodule Then M is free as Aemodule if and only if M is projective as Aemodule PROOF See 4 D LEMMA 916 Let A be a commutative local ring with maximal ideal mA and residuefield kA A m A a Let P be a AGemodule which is free as a Aemodule Then P is projective as a AGemodule if and only if PPmAP k AP 910 is projective as a kA Gemodule b Two projective AGemodules P and P are isomorphic if and only if P E P as k AGemodules PROOF This is easy to prove since kAG E kA A AG If P is projective as kAG module then by Lemma 914 there is ii E EndAP such that Z sas 1x a 56G 74 CHAPTER 9 THE GROUPS RKG RKG AND PKG for all X E P Since P is projective we obtain the following commutative diagram where 7m n Then u E EndAP and for all x E P E sus 1x E x mod mAP seG Let u 256G sus l then u is a AG module homomorphism with u E idp mod mAP By Nakayama s Lemma u is surjective Since P is finitely generated free A module Lemma 915 this implies that u is a AG module isomorphism Let 39U uu 1 then for all X E P E svs 1x Z suu 1s 1x 56G 56G Z 5u5 1u71x seG u u 1x x Then by Lemma 914 P is a projective AG module l1 This direction is easy Let ID P gt P be a kAG module isomorphism Since P is projective we obtain the following commutative diagram 4 STRUCTURE OF PRG 75 where 71 w 71771 By Nakayama s Lemma w is surjective Since P and P are projective A modules and A is local then P and P are free A modules Lemma 915 Moreover rankA P dimA P dimA P rankA P Then w is also injective D We want to apply Lemma 916 to our complete p modular system K Rk where m is the maximal ideal of R and k Rm PROPOSITION 917 a Let P be on RGemodule Then P is u projective RGemodule ifund only if P is it free Remodule and PmP is u projective kGemodule b If E is u projective kGemodule then there exists it unique projective RGemodule E with E mE E E PROOF u and the uniqueness in h follows from Lemma 916 We prove existence in For all n E 2 let Rn Rmquot Then R is equal to its m adic completition RlltiLl an Since RnG is artinian E when viewed as an RnG module has a projective Rn G cover 71 En gt E Then 71 induces a surjective kG module homomorphism zEnmE A E Since E is projective there is a submodule U of En mEn Q U so that n restricted to UmEn is an isomorphism Then nal E since 71 is essential U En Therefore 7 1 is a kG module isomorphism The En s form an inverse system Let E En then E is an RG module and E is a projective R module since every E is a projective Rn module Since R is local then E is a free R module Moreover EmE E E via 7 1 7 1 Then by u E is a projective RG module D COROLLARY 918 Every projective RGemodule is the direct sum of projective indecomposuble RG7 modules this decomposition is unique up to isomorphism Every projective indecomposuble RGemodule is uniquely determined by its reduction mod m which is u projective indecomposuble kGemodule PROOF Use Proposition 917 and what we know about projective kG modules D COROLLARY 919 Let PQ be two projective RGemodules Then P 2 Q A P Q in PRG COROLLARY 920 Let 5k be ufull system ofnoneisomorphic simple kGemodules Then P5 S E 5k where P5 is u projective RGemodule with P5 mPg u projective cover of S is u Zebusis of PRG COROLLARY 921 Reduction mod m de nes an isomorphism PRGxrighturrow PRG and this isomorphism mups P G into P 76 CHAPTER 9 THE GROUPS RKG RKG AND PKG 5 SelfDuality of RKG The dual of RKG is RKG HomZRKGZ 911 Let E and E be KG modules and define ltEPgtK Clil l lKHOngEP 912 Let p and 1p be the characters of E and E over K respectively We have seen the following Let C K the algebraic closure of K Then E PgtK dil l lc HomcgEcPc ltPI lPgtC ltPI 1PgtKI where EC C K E and E C K E The we get a symmetric bilinear form ltgt ZRKG gtlt RKG A Z Let SK V1 V be a full system of simple KG modules Then 0 if i y V V39 lt JlK di ifij where di dimK EndKGUi Then di 2 1 and di 1 if and only if Vi is an absolutely irreducible KG module Suppose that K is a splitting field of G eg K is sufficiently large relative to G then every simple KG module is absolutely irreducible In this case K is nundegenerate over Z in the sense that it defines an isomorphism between RKG and RE Namely for all 1 g i g 7 let Ti be the character of Vi over K Define RKG gt Z by ltpTigtK Then 51739 since K is a splitting field of G Let X RKG A RIUG Ti gt gt with Z linear extension then X is a Z module isomorphism 6 Duality of PkG and RAG Let P be a projective kG module and let E be a kG module Define P Ek dimk HomkG P E 913 This is additive in each component with respect to short exact sequences namely if 0gtP1gtP2gtP3gt0 is a short exact sequence of projective kG modules then P2Egtk ltP1Egtk ltP31Egtk since P2 E P1 63 P3 Let OaElLHEZLEgHO e 7 MORE ABOUT SPLITTING FIELDS 77 be a short exact sequence of kG modules Without loss of generality assume 1 the inclusion map Since P is projective for all f P 7 E3 we choose and fix vf P 7 E2 with f 7w Define 0c HomkgP E1 63 HomkgPE3 7gt HomkcPE2 a b 7 m 1 h 7 Unh 7111 lt h Note that 7Ih 7 Unh 7th 7 7th 0 for all h E HomkGPE2 Then h 7 vnh P 7 E1 Then 0c is a k vector space isomorphism Then PlEzlk ltPE1gtk PE1gtk Thus we get a bilinear form gtk PkG gtlt RkG 7gt Z 914 Let 5k be a full system of representatives of simple kG modules and for all E E 5k let PE be the projective kG cover of E Note that HomkGPEE E HomkGPEPEE radUcG annihilates E39 g HomkGEE Then 0 if E E in S PEEM Pf k Clll39l lk EndeE 1fE E in 5k As in characteristic zero dimk EndkcE 1 if and only if E is an absolutely irreducible kG module more explanation below Suppose that K is sufficiently large relative to G then k is sufficient large relative to G Therefore below k is a splitting field of G Then all simple kG modules are absolutely irreducible then for all E E E 5k PEHM 6EE 915 Then is non degenerate over Z and the basis PE E E 5k of PkG and E E 5k of RAG are dual to each other with respect to k Then we get a Z module isomorphism 0c RkG A PkG HomZPkGZ 916 El H X H X EM with Z linear extension 7 More About Splitting Fields Let L be a field and G be a finite group DEFINITION 922 a A simple LG module S is called absolutely irreducible if for all L 2 L with L field L L S is a simple L G module 78 CHAPTER 9 THE GROUPS RKG RKG AND PKG b L is a splitting field of G if the Wedderburn decomposition of LG radLG has the form LG radLG MatML 917 i1 Note that since radLG annihilates every simple LG module then simple LG modules can be identified with the simple LG radLG modules Moreover EndLG S EndLG radLG S for all simple LG module S LEMMA 923 Let S be u simple LGemodule Then S is absolutely irreducible ifund only ifEndLG S L PROOF Let C L then SC C L S is a simple CG module and by Schur s Lemma dimL EndLgS dimC Endcgsc 1 Then EndLgS L For all X E LG define ux S gt S Sgt gtXS Let B ux X E LG Then B is an L algebra S can be viewed as a B module Since every subspace of S which is a LG module is also a B submodule of S S is a simple B module S is a faithful B module since if uxs 0 for all s E S then ux 0 This implies that B is a simple L algebra as follows E is artinian since B Q EndLS this has finite L dimension rad B annihilates S since S is faithful this implies rad B 0 Then B is semisimple Therefore Z B HMatD i1 lf 2 2 2 then no simple B module is faithful Therefore 2 1 Hence B E MatnD where D EndBS EndLgS and n rankD S Then B is a simple L algebra Let L 2 L be a field extension We need to show that L L S is a simple L G module it is enough by the above arguments to show that L L S is a simple L L B module since L L B ux X E L G By assumption EndLgS L then EndBS L Therefore B MatnL where n dimL S Thus S is a minimal left ideal of B hence L L S is a left ideal of L L B MatnL Moreover dimLL L S dimL S n Then L L S is a minimal left ideal of of L L B Hence L L S is a simple L L B module D COROLLARY 924 L is u splitting field of G if and only if every simple LGemodule is absolutely irreducible LEMMA 925 Let L be afield with churL p gt 0 Let S be u simple LGemodule then EndLgS is afield which is finitely dimensional and sepuruble over L PROOF Let le be the prime subfield of L Then LG radLG L 1127 le radleG 7 MORE ABOUT SPLITTING FIELDS 79 this is not trivial see 2 pgs 147 149 with m leG radleG E HMatnxNi i1 where all Ni are finitely dimensional rings over le Then all Ni are fields Therefore Nile is finite dimensional and separable for all 1 g i g m Then LG radLG MatMLQ i1 where Li L 1127 Ni for all 1 g i g m Then LiL is finitely dimensional and separable If S is a simple LG module then S is a simple LG radLG module Then it corresponds to an i E 1n and EndLgS Li D LEMMA 926 Let L be afield with CharL p gt 0 Let S be a simple LGemadule Then EndLgS is a field which is nitely dimensional and separable aver L PROOF Let le be the prime subfield of L Then LGradLG L 1Fp leradOFpG and m leG radleG E HMatnxNi i1 where each Ni is a finite skew field then Ni is a finite field Therefore Ni le is finitely dimen sional and separable with 7x L Fp Ni H j1 where each L17 is a field which is finitely dimensional and separable over L Then m 7 LG radLG g Mat L17 i1 j1 Assume that S is simple LG module then S corresponds to a unique i E 1rn and a uniquej E 1 r so that EndLGS g EndLGradLGS g L D PROPOSITION 927 Let L be an arbitraryfield which is su icient large relative to G Then L is a splitting field of G PROOF We know this when CharL 0 see Theorem 716 Suppose now that CharL p gt 0 We have seen in Lemma 926 that if S is a simple LG module then EndLgS is a field which is finitely dimensional and separable over L The the l Schur indices over L are all equal to 1 One shows similarly to the characteristic zero case that this implies RLG 80 CHAPTER 9 THE GROUPS RKG RKG AND PKG shows that the simple LG modules provide a Z basis of Let L L be a field extension Since L is sufficiently large RLG RLG Then every simple L G module is realizable over L then since this is true for every L every simple LG module is absolutely irreducible Then L is a split field of G 8 Scalar Extension Let K R k be a complete p modular system Let K K be a field extension for all KG module M MK K K M is a K G module Then we can get a Z module homomorphism we a Me This homomorphism is in fact injective as can be seen as follows Let SK be a full system of simple KG modules For all E E SK we have that EnngE Dg is a skew field Then 115 K K DE E JHlMatjE E where D is a skew field finitely dimensional over L Note that if K Z Dg then this is actually a simple K algebra If KG E HESSK Mat Dg then 15 KG 9 H H Matnglmwjg EesK jg1 Each D corresponds to a unique simple K G module TjE RKG is 751 1 The our homomorphism is given by and the image of E E RKG in Me a we w 112 El l l Z leLDEl 151 SinceSK E E SK1 g jg g dg and 1 g jg g dg 1 1 g jg g gig for E 7E E in SK Then the homomorphism is injective If KG is quasi split then for all E E SK Dg is commutative therefore I 1 for all 1 g jg g dg Then the homomorphism is a split injection If every simple KG module is absolutely irreducible then dg 1 hence RKG E RKG Let k k be a field extension Then for all kG module N Nk k k N is a k G module If N is projective then Nk is projective Then we get homomorphisms RAG A RkG and PkG A PkG 81 SCALAR EXTENSION 81 Let S E 5k then EndeS L5 is a field which is finitely dimensional and separable over k Then 7 S k 1L3 H LS 918 131 where Lis is a field finite dimensional and separable over k lka radkG E HSESk MatS L5 then 7 k Gradk G H H MatSLs 919 565k 131 Each Lis corresponds to a unique simple k G module X15 and the image of S E RAG in RkG is 21251Xis The our homomorphism is given by RkG A RkG 7s Sl H E Xisl 131 Then this is a split injection similarly PkG Pk1G is a split injection since if Q is a pro jective kG cover of a kG module N then k k Q is a projective k G cover of Nk1 k k N Now let K K be a finite extension let R be the integral closure of R in K Let k be the corre spond residue field of R If P is a projective RG module then P R ER P is a projective R G module and k R P E k RP E k k k R Then we get a commutative diagram PRG quot PRG P1ltGgt PkG Since the bottom row is a split injection the the top row is also a split injection CHAPTER 10 THE C DE TRIANGLE In this chapter K R k denotes a complete p modular system All modules are assumed to be finitely generated We want a commutative diagram FAG C gtRkG m5 1 The Cartan Homomorphism C The homomorphism C PkG A RkG 101 is defined by sending each projective kG module P to its class in RkG and extending Z linearly C is called the Carton homomorphism of G with respect to k Using the canonical bases Pg S E 5k of PkG and S E 5k of RkG we can express C as aS gtlt S matrix C called the Carton Matrix of G with respect to k Namely for all T E 5k PTl Z CSTlSl f in Rle J 102 565k and CST is equal to the number of times S occurs as a composition factor of PT 2 The Decomposition Homomorphism d DEFINITION 101 a An Rluttice is a finitely generated torsion free R module Since R is a principal ideal domain an R lattice is a finitely generated free R module 83 84 CHAPTER 10 THE CDE TRIANGLE b An RGlattice is a finitely generated RG module whose underlying R module is an R lattice c Let V be a finitely generated KG module A full RGlattice in V is an RG lattice M Q V with KM V Note that KM E K R M since M is torsion free LEMMA 102 Every nitely generated KGernadule V contains a full RGelattice PROOF Since we always assume that G is finite V has a finite K basis v1 vn Let M 2111 RGvi then M is an RG lattice and KM V D PROPOSITION 103 Let E be afinitely generated KGernadule and let E1 and E2 be twafull RG7 lattices in E Then E1 E2 in RkG where E EimEi m is the unique maximal ideal afR and m R71 with 71 the uniformizer afm REMARK 104 Proposition 103 only says that E1 and E2 have the same composition factors with multiplicities but E1 E E2 is possible For instance assume Chark 2 G U E Z2 Let E KG then E1 RC is a full RG lattice in E and E1 kG is a indecomposable kG module Also E2 R1 U 63 R1 7 U is a full RG lattice and E2 E k 63 k then E1 E E2 but E1 and E2a both have as composition factors k and k PROOF OF PROPOSITION CASE 1 Assume mE1 E E2 E E1 and let T be the kGmodule E1 E2 So we get an exact sequence 0gtTgtE2gtE1gtTgt0 where T E mE1 mE2 and the homomorphism T gt E2 is obtained by multiplication by 71 Passing to RkG we have Tl 2 1 7 Tl 0 CASE 2 Since kE1 E kE2 and k is the field of fractions of R we can multiply E2 by a scalar r E R 7 0 to be able to assume that E2 is contained in E1 Note that E1 mE2 E rE2mrE2 for all r E R 7 Since all non zero proper ideals of R have the form mquot for some n E Z there exists an integer n 2 0 such that mnEl E E2 E E1 and we proceed by induction on n Let E3 quot 1E1 E2 Then m HE1 g E3 g E1 and mE3 c E2 g E3 By CASE 1 and induction we get which proves the Proposition D Proposition 103 implies that the mapping E gt E1 where E is a KG module E1 is a full RG lattice in E gives a well defined map a RKG E RkG 103 El H E1 4 PROPERTIES OF THE CDE TRIANGLE 85 which is a ring homomorphism called the decomposition homomorphism We can express d as a 5k gtlt Sk matrix D called the decomposition matrix Namely for all E E SK E1 Z DSESin 1mm 104 565k where E1 is a full RG lattice in E and D53 is the number of times S occurs as a composition factor of E1 3 The Homomorphism e We have a homomorphism PR G gt RKG given by sending a projective RG module P to the class K R P in RKG and extending Z linearly Composing this with the isomorphism PRG E PkG we get a homomorphism e PkG A RKG 105 we can express 8 as a 5k gtlt Sk matrix E 4 Properties of the Ede Triangle 1 C deorC DE 2 For allx E PkG and y E RKG KOOWK Xdygtk 106 This means 8 and d are udjoints with respect to these bilinear forms PROOF It is enough to show this for X the class of a projective kG module and y the class of a KG module Then X X for a projective RG module X since PRG E PkG and y Y0 for a KG module Y0 Let Y be a full RG lattice in Y0 then y K R Y Also X and Y are finitely generated free R modules of fi nite rank Then HongX Y is a finitely generated free R module of finite rank say 739 rankR HongXY Then 739 dimKK R HomRGX Y dil l lKH0ngK R XK R Thus 7 exyK Also 7 Clil l lkUC R HongX dimkHomkgk R Xk R Y dimkHongX 17 lterygtk D 3 Let K be sufficiently large relative to G The we have seen that the canonical basis of RKG given by SK is dual to itself with respect to K Also the canonical basis st S E 5k of PkG and S E 5k of RkG are dual with respect to k It follows that 8 can be identified with the transpose of d ie E t D 86 CHAPTER 10 THE CDE TRIANGLE PROOF Let X E PkG and y E RkG be given with respect to the canonical basis Then tXtEy WSW ltex1ygt1lt WWW txDr Since this is true for all X and for all y D tE D Therefore C DE DtD is symmetric 5 Special Groups 51 p Groups THEOREM 105 Let G be a group oforder prirne to p Then a every kGernodule resp every Rfree RGernodule is projective b reduction mod m induces a hijection SK A 5k 107 c ifwe identify SK and 5k Ui h then the matrices C D and E are ll identity matrices In other words the representation theory of G is the some over K as over k PROOF We know by Maschke Theorem Theorem 18 that every kG module is projective Let M be an R free RG module and let P be a free RG module surjecting 71 into M Then we get a short exact sequence 0 a L A P L M a 0 where L ker7t Since M is free as an R module there is an R module homomorphism p2PgtLwithpiidLDefinelpPgtLby 1 71 1p 7 sps 108 G Since p j G G is a unit in R then IP is an RG module homomorphism and W idL So we obtain a Since every simple kG module is projective by a C is the identity matrix To prove h and c let S E 5k Then by a S is projective Let P5 be a projective RG cover of S Then PSmPS E S Let Y5 K K P5 Therefore lm5 Sl Since 01 is additive Y5 is a simple KG module This means 5k A SK S gt gt Y5 is a well defined map which induces an inverse map Then h and c follows E 5 SPECIAL GROUPS 87 52 pGroups Let G be a p group PROPOSITION 106 Let V be a kGemodule Then V has a nonzero suhmodule with trivial Geoction Hence the only simple kGemodule up to isomorphism is the trivial simple kGemodule PROOF Let v E V with v non zero Let X be the additive subgroup of V generated by gv g E G Then EKG oggv 0 g g g p El Since for allz E V pz 0 then X pm Note that X is a G set Let XG X E X gX XVg E G Then X E XG is a disjoint union of non trivial orbits under the action of G on X Since G is a p group it follows that non trivial orbits have as length a positive power of p then X E XG E 0 mod p Therefore XG E X E 0 mod p Thus plXG Since 0 E XG and p 2 2 then there is X E XG with X non zero Then kX is a non zero submodule of V with trivial G action D CONSEQUENCES Assume G pquot By Wedderburn decomposition theorem Theorem 116 we have kG radkG E k Therefore kG is an artinian local ring with residue field k note also that implies that kG is projective indecomposable since k is simple kG module Thus PkG and RkG can be identi fied with Z Therefore C I PkG E Z AZ E RkG is given as multiplication by pquot since ckG dimk kG k pquotk Also d RKG gt RkG is induced by the K dimension since if E is a KG module with full RG lattice E1 then rankR E1 dil l lK KG On the other hand e PkG gt RKG is given as en nKG since e1 eKG K R RG KG 53 C Direct Products of p Groups and pGroups THEOREM 107 Let G H X P where H is a p egroup and P is a pegroup Then kG E kH k kP Moreover a a kGemodule E is semisimple i r P acts trivially on E b a kGemodule E is projective if and only E is isomorphic to M k kP for some kHemodule M c a kGemodule Q is projective ifond only E M R RPfor some Rfree RHEmodule M CONSEQUENCE If P pquot then the Cartan matrix is multiplication by pquot REASON Let S be a simple kG module By a S is a simple kH module in ated to G By 11 the projective kG cover P5 of S is P5 E S 29 kP Then P5 has pquot composition factors which are all isomorphic to S k k E S since k has triVial G action Then ClPsl Pnlsl 88 CHAPTER 10 THE CDE TRIANGLE Thus pquot 0 0 n I C 0 p i pquot identity 3 0 0 0 p71 PROOF OF THEOREM a If P acts trivially on E then E is in ated from a kH module to G Since every kH module is semisimple kG module E is semisimple Let E be a simple kG module Let E X E E 5X XVs E P Then E is akG module since for allg E G and X E E wegetforalls E P 5gx gg 15gx 5 where g lsg E P Since Res E is a kP module it has a non zero submodule with trivial P action see Proposition 106 Thus E 7E 0 Since E is simple E E b If M is an arbitrary kH module then M is a projective kH module Since kH is semisimple then M k kP is a projective kH k kP module hence a projective kG module Moreover M is the maximal quotient with trivial P action Therefore by a M is the maximal semisim ple quotient of M k kP Since every projective kG module is the projective kG cover of its maximal semisimple quotient l1 follows from a since projective covers are unique up to isomorphism C If M is an R free RH module then M is a projective RH module Thus M R RP is a pro jective RG module since RG E RH R RP Now suppose E is a projective RH module Then E EmE is a projective kG module Hence by b E E M k kP for some kH module M Since M is a projective kH module there is a projective RH module M with MmM E M In particular M is an R free RH module therefore M R RP is a projective RG module and a projective RG cover of M k kP E Since E is also an RG cover of E E E M R RP D 6 Change of Groups Let H g G We have seen earlier that for an arbitrary field L and an LH module V lndICjI V E LG LH V Moreover if V is projective LH module then lndICjI V is a projective LG module since LG LG LH E LG We also proved that for an arbitrary LG module E lndV L E E IndgV L Resg El39 with diagonal action of the group G Also Resg E is an LH module If E is a projective LG module then Resg E is a projective LG module then ResICjI E is a projective LH module since ResLG E g6GH LHg E g6GH LH Then we get RKH a RKG RKG a RKH lndgz RkH gt RkG ResICjI RkG gt RkH 109 WW 4 PkG PkG 4 WW We also get that 1nd x Resgygt Indgoc y 1010 7 BRAUER INDUCTION THEOREM IN THE MODULAR CASE 89 for a ifx E RKH andy E RKG tl391enInd1C31xy E RKGor b ifx E RkH andy E RkG then Indxy E RkGor c ifx e RkH andy e PkG then Indgmy e PkG PkG is a module over RkG Note that C d and e commute with induction and restriction 7 Brauer Induction Theorem in the Modular Case RECALL Let m be the exponent of G Let U be a primitive m th root of the unity Then GalKw K E PK where PK g Zm and for all t E PK there is a unique m E GalKw K with 7a wt Aa subgroup H of G is PK 7 l elementary with l a prime number if H Z gt4 P where Z is a cyclic group of orden prime to l and P is a l group and for all y E P there is at E PK with yzy 1 zt for all z E Z Also quotPK elementaryquot means quotPK 7 l elementary for at least one prime 1 THEOREM 108 Let XK be the set afall TKeelementmy subgroups of G Then Indk GB RkH A RkG 1011 HEXK fHHeXK H E IndICjIfH HEXK and Indk 69 PkH A PkG 1012 HEXK are surjecti uei PROOF Let 1K resp 1k be the multiplicative identity in RKG resp in RkG ie the class correspondent to the trivial simple module over K resp over k By Bruer s Induction Theorem over K Theorem 67 1K E mdng K HEX for certain UH E RKH Since d1K 1k and d commutes with induction we get 1k E Ind H HEXK where UH dLIH E RkH Lety E RkG resp y E PkG Then y 1k y E Ind H y HEXK E Indg UH Resgy HEXK where UH Resy E RkH resp UH Res1 y E PkH for all H E XK D 90 CHAPTER 10 THE CDE TRIANGLE COROLLARY 109 IfK is su ficiently large then XK X where X is the set ofizll elementary subgroups of G Hence every y E RkG resp y E PkG has theform y E Imam 1013 HEX for certain fH E RkH resp fH E PkH 8 Properties of the cde Triangle Cont THEOREM 1010 The homomorphism d is injective PROOF For K sufficiently large the general proof is in 7 Chapter 17 we have 11 1de 69 RkH A RkG HEX fHHeX H E InngH HEX is surjective where X is the set of elementary subgroups of G Since d commutes with induc tion it suffices to show that d RKH gt RkH is surjective for all H E X Then reduce the case when G is l elementary for some prime l In this case the result follows from LEMMA 1011 Let G Z X P where Z is a cyclic l igroup and P is on Legroupfor some prime I Then every simple kGimodule E can he lifted to ie is the reduction mod m of on Rfree RGimodule PROOF Let G7 be a p Sylow of G lfl 7E p then G7 lt Z Since is cyclic G7 is unique Then G7 51 G lfl E p then G7 P Therefore cpgc Note that in both cases G E G7 X T where T is a p gr0up If E is a simple kG module then E has trivial Gp action hence E is in ated to G from a simple kT module called again E Therefore E is a projective kT module Thus E can be lifted to an R free RT module E Thus E can be in ated to an R free RG module denoted again by E Moreover E mE E E as LG modules D D THEOREM 1012 The homomorphism e PkG a RKG is a split injection ie there is a Zimodule homomorphism u RKG a PkG with ue idpkG PROOF Suppose first that K is sufficiently large Let D and E be the matrices correspondent to d and e with respect to the usual bases Since K is sufficiently large D tE Since d is surjective for all S E 5k there is X5 E RKG with docs P5 where P5 is the projective cover of S Then there is a 5k gtlt Sk matrix M with DM identity on PkG more precisely the image of PkG under c Then tME tMtD identity on PkG Thus tM defines a Z linear map it RKG gt PkG with ue idpkG Now suppose K is arbitrary Let K K be a finite 8 PROPERTIES OF THE CDE TRIANGLE CONT 91 extension such that K is sufficiently large relative to G Let R be the integral closure of R in K let k be the residue field of R Then we get a commutative diagram PkG 8 RKG W PkG 8 1 RKG where the vertical maps are scalar extensions We have seen that pk is a split injection e is a split injection since K is sufficiently large Then e pk is a split injection Then since the diagram commutes e is a split injection We have actually shown more COROLLARY 1013 IfK K is onyfinite extension the composition PkG L RKG a RKG 1014 is a split injection COROLLARY 1014 Let P and P be two projective RGemodules IfK R P E K R P as KGF modules then P E P as RGemodules PROOF Let P PmP and P P mP Then 8lPl IK R Pl IK R Pl 8IP l since e is injective P P in PkG But PRG E PkG then P P in PRG Therefore P E P as RG modules D THEOREM 1015 Let G pquotm with gcdpm 1 Then every element in RkG that is a multiple ofpquot lies in the image ofc PROOF Assume K sufficiently large By Brauer s Theorem Theorem 108 RkG is gen erated by Indg RkH where H E X with X the set of elementary subgroups of G Since c commutes with induction it suffices to show the Theorem in case G is l elementary for some prime l Assume G Z X P for a cyclic l group Z and an l group P Then exactly as in the proof of surjectivity of d Theorem 1010 it follows that G G7 X T where G7 is a p Sylow and T is a p group In this case we have shown earlier that the Cartan matrix is multiplication by G THEOREM 1016 Let c PkG a RkG is injective and the cokernel ofc is ofinite pegroup PROOF The second statement follows from Theorem 1015 Since PkG and RkG are finitely generated free Z modules of the same rank and since the index of I mc in RkG is fi nite c must be injective otherwise kerc is a free Z module of positive rank Then the cokernel of c is finite 92 CHAPTER 10 THE CDE TRIANGLE COROLLARY 1017 If P and P are two projective kGemodules with the same composition factors then P E P THEOREM 1018 Let K be su ficiently large Then the Cartan matrix C is symmetric and the correspondent quadratic form is positive definitive The determinant of C is is a power of p PROOF Assume K sufficiently large then D tE Therefore C DE DtD is symmetric The correspondent quadratic form is given by PAG A Z x gt gt xcxk where thx xcxk xdexk exexK since K is sufficiently large Since aaK gt 0 for all a 7 o in RKG and e is injective then the quadratic form x gt gtt XCX is positive definitive Therefore detC gt 0 We know c PAG gt RAG is injective and the index of Imc in RAG is a power of p By the fundamental theorem of finitely generated modules over Z there is a Z basis h1 h of RAG and there are a1a E 2 such that p lb1p h is a Z basis of Imc with r Sk Moreover c PAG a Imc is a Z module isomorphism Then there is a Z basis X1 X of PAG with cxi path Then the Cartan matrix C with respect to the basis X1 X7 and h1 11 has the form p l 0 0 0 p 2 i E 0 0 0 pay Then detC p 1 39 D 9 Characterization of the Image of e RECALL An element g E G is called psingular if its order is divisible by p and g is called pregular if its order is no divisible by p Ebery g E G can be written uniquely as g gu gmg gu p component and gyeg p component where gu is an element which order is an power of p and gyeg is p regular Moreover gu gyeg gyeg gu Hence gis regular if and only if gu 7E 1 The elements of RKG can be identified with virtual characters Z linear combinations of characters of G over K THEOREM 1019 The image ofe consists precisely ofthose elements in RKG whose virtual char acters have zero value on all pesingular elements of G PROOF We prove this for K sufficiently large Let first Q be a projective RG module and let x be the character of G over K correspondent to K R Q We want to show that for all s E G with s p singular As 0 Lest s E G be p singular Then we can restrict ourselves to the case G Therefore G H X P where H syeg is a p group and P su is a p group By our results on direct products of p groups and p groups Subsection 53 Q E M R RP for 9 CHARACTERIZATION OF THE IMAGE OF E 93 some R free RH module M Let KM be the character correspond to K R M and let pp be the regular character of P ie the character correspondent to KP P if x 1 0 if x 73 1 Thenx KM eapp Therefore 95 XMSggpp5u 0 Lety E RKG be avirtual character which has value 0 on all p singular elements of G We want to show that y E PRG where we identify PRG with its image of PRG E PkG under 8 By Brauer s induction theorem Theorem 108 since K is sufficiently large 1K Z Ind orH HEX 1170 I for certain fH E RKH where X is the set of elementary subgroups of G Then y 1K y 2 NEW Resy HEX Then yH is zero on all p singular elements of H If we show that yH E PRH for all H E X then y E PRG Then we can reduce to the case that G is l elementary for some prime 1 We have seen before that no mater what the l is G can always be written as G S X T where S is a p group and T is a p group Therefore RKG RKS Z RKT We have that for all s E S 5 7E 1 and for all t E T gst 0 Let p5 be the regular character of S ie S ifx l E S pg 0 ifxy l X Then y pg 29 f for some class function f G gt K Let T be an irreducible character of T over K Then Z 9 W11 TgtK PSUK fTlK ltfITgtK Thus ltfTgtK E Z for all irreducible character T of T Hence f E RKT Since T is a p group f E PR since APR is inclusion On the other hand PS KSl K R Rsl Then p5 E PRSthus y PS f E PRG CHAPTER 11 MODULAR CHARACTERS 1 The Modular Character of a Representation Let K Rk be a complete p modular system and assume K is sufficiently large relative to G Let Gmg be the set of all p regular elements of G Let m be the least common multiple of all the orders of elements of Greg Since K is supposed to be sufficiently large K contains the group pK of all m th roots of the unity in K Similarly the group W of all m th roots of unity in k is contained in k In fact reduction mod m provides an isomorphism pK gt pk Fix this isomorphism For A E pk denote by A the corresponding element in pK DEFINITION 111 Let E be a kG module of k dimension 71 Let pg G gt GLnk be the corresponding representation over k Let s E Greg Since the order of s is relative prime to p and pk E k pgs is diagonalizable in GLnk with eigenvalues A1A E pk Define pgs 21 1 E pK Q R C K Then geta function pg Gmg A R 111 n S gt gt 139 called the modular character or Bruuer character of E The properties of the modular character are 1 PEG n 2 pg is aclass function ie for allt E G and s E Gmg pgtst 1 pgs 3 pg is additive with respect to short exact sequences ie if 0 gt E1 gt E gt E2 gt 0 is a short exact sequence of kG modules then PE 1191 1132 112 95 96 CHAPTER 11 MODULAR CHARACTERS 4 E1 E2 E1 Ezi 5 let g E G with p cornponent 5 Then M91200 11125 114 where the bar represents reduction rnod m PROOF pgg 1s has eigenvalues p th roots of the identity for some X E 2 Since Chmk p all these eigenvalues are 1 Therefore pgg and pgs have the same eigenvalues D 6 Let X be a KG rnodule with representation pg of G over K and character x Let E1 be a full RG lattice in X and let E ElmEl with representation pg of G over k and modular character pg Then PE Maw 115 PROOF It is enough to restrict ourselves to G s where s E Greg Since the composition factors of E and its pg do not depend of the choice of the full RG lattice E1 we can choose E1 to be generated by eigenvalues of pXs Then A1 0 0 0 1 Px5 2 0 0 211 and A1 0 0 0 A PE5 2 0 0 An Then 41195 2121 5w Trpxs X5 D 7 Let P be a projective kG rnodule let I3 be a projective RG rnodule let I3 be a projective RG rnodule with 1311113 E P Let the character corresponding to the KG rnodule K ER 13 be denoted by 912 Let E be a kG rnodule Then E k P is a projective kG rnodule Then we can built g k p Therefore pgs 9125 ifs E Gmg d9 116 E kp 0 otherwise We can then write E 1P 119 117 since d9 is zero outside of Greg 1 THE MODULAR CHARACTER OF A REPRESENTATION 97 PROOF Both d9 and E kp lie in the image of e PkG gt RKG Therefore they have zero values on Greg By 6 for all s E Greg E kP5 E kP5 11195 11175 11195 49175 8 Let P P and E as in Then 1 IDElk E Z P5 1 E5 gEG 171 Egt PROOF PRk clirnk HornkcPE Let H HornkPE This is a kG rnodule by setting for all g E G and f E H gf X gfg 1x for all X E P Then H 2 13 k E with diagonal G action as kG rnodules where P HornkPk Since CG 2 kG P is a projective kG rnodule Then we can lift H to a projective RG rnodule H with H mH E H Therefore clirnk HG rankR HG dil l lKK R HG dil l lKK R HG of times the triVial simple KG rnodule occurs in K R H lt11 HgtK Therefore REM lt11 HgtK l H5 c l H5 c S E p5 g5 by 7with 1245 d9ps 1 s E Gyeg Z ps 1pgs s E Gyeg i G 98 CHAPTER 11 MODULAR CHARACTERS Remember that HG X E H gX XVg E G D 9 Use 8 for E kwith local trivial G action Then 1 r G dimkP 1 ESE ps 118 reg PROOF dimkPG dimkPG dimk HomkG P k ltPrkgtk ltdgtp1gtby 8 D REMARK 112 Since the modular characters are additive with respect to short exact se quences we can define a virtual Brauer character px for every X E RkG Let y E RKG with virtual character Ky and suppose X a By 6 we obtain PX Xyleeg gt xylGreg 2 Independence of Irreducible Modular Characters Let 5k be a full set of non isomorphic simple kG modules Then for all E E 5k the modular character pg is called an irreducible modular character THEOREM 113 Brauer The irreducible modular Characters pg with E E 5k farm a Kebasis 0f the Kevectar space of all Class functions on Gmg with values in K PROOF we first show that the pg with E E 5k are linearly independent over K Suppose X g g 0 EESK for all ag E K not all zero We can multiply this equation by an appropriate element in R to assume that ag E R If all ag E m then since In R71 all ag are divisible by 71 Then since R4 is an integral domain we can divide the equation by 7 Doing this repeatedly if necessary we can assume that at least one a g Then E m E 65k for all s E Gmg where the bar denotes reduction mod m By property 5 if pg is the represen tation of E over 7 Z MPEW 0 5k for all X E G and therefore for all X E kG Since K is sufficiently large k is a splitting field of G then EnderadkGE g EndeE g k 2 INDEPENDENCE OF IRREDUCIBLE MODULAR CHARACTERS 99 for all E E 5k Therefore by Wedderburn decomposition theorem Theorem 116 we have kG radkG E H EndkE E 65k Therefore the map kG A H EndkE EeSk X H PESEesk is an injection Let E E 5k with 7E 6 and choose u E EndkE with trace 1 eg projection of E onto a 1 dimensional quotient space Let X E kG with image u in EndkE and image zero in EndkE for all E 7E E in 5k Then 5 Z T7Psx 565k T u which is a contradiction Now we show that pg with E E 5k generate the k vector space of all class functions on Greg with values in K Let f be a class function on Greg with values in K We extend f to a class functionfon G by setting 0 for all X E Greg Since K is sufficiently large RKG RG and hence h f ZAW i1 with M E K for all 1 g i g Ii and X1xh a full set of representatives of irreducible characters of G over K Then h f fleeg ZAiXileeg 13971 By property 6 xilgmg is the modular character of a kG module hence a Z linear combination of the pg with E E 5k because the modular characters are additive with respect to short exact sequences Theorem 113 is equivalent to THEOREM 114 The map RkG gt Cluss functions on Greg with values in K 1110 x H px induces an isomorphism afKeulgebms K 292 RkG gt Keulgebm 0fthe Clussfunctians on Greg with values in K 1111 COROLLARY 115 Let X and X be two kGemadules with px y Then x x in RkG PROOF Assume that x y by property 3 X and X have the same composition factors Therefore X X in RkG D 100 CHAPTER 11 MODULAR CHARACTERS COROLLARY 116 The kernel ofthe decomposition homomorphism a RKG a RkG consists of all elements y E RKG such that their virtual characters M has zeroevalues on Greg PROOF Let X dy By property 6 x xylem 1112 If X 0 this means xg has zero values on Greg D REMARK 117 This gives a description on RkG as the quotient of RKG mod the Z submodule of all virtual characters of G over K which have zero values on Greg COROLLARY 118 The number of isomorphism classes of simple kgemoalules is equal to the number of conjugacy classes of peregular elements in G Thus K and hence k are su iciently large PROOF The number of isomorphism classes of simple kG modules is equal to Sk By Theorem 114 this is the same as the K dimension of the K algebra of all class functions on Greg with values in K But the latter is equal to the number of conjugacy classes of p regular elements in G 3 Reformulations We have seen that the map x gt gt px gives an isomorphism of K 292 RkG onto the K algebra of all class functions on Greg with values in K We have seen earlier that the image of the injective map e PkG gt RKG consists of all virtual characters of G over K which are zero outside of Greg Then K 29 e identifies K 292 PkG with the K vector space of all class functions of G over K which are zero outside of Greg Therefore tensoring the cale triangle with K over Z gives class functions of G over K K C class functions of G72g which are zero outside of G72g over K 1 113 K 8 K 29 d class functions of G over K where K 29 c and K 29 d are restrictions to Greg and K 29 e is inclusion NOTE c PkG gt RkG is injective and its cokernel is a finite p group Then K 29 c is an isomorphism of K vector spaces QUESTION What do the Cartan matrix C and decomposition matrix D look like Let SK resp 5k be a full set of representatives of non isomorphic simple KG modules resp kG modules If P E SK we denote by xp the correspondent character of G over K If E E 5k we denote by pg its modular character and by PE its up to isomorphism projective 41 A SECTION OF D 101 kG cover Let I3 be a projective RG module with Pmp E PE Let 93 be the character of G over K correspondent to K R P Then XEleg Z DENPE 1114 EeSk Elcswg Z CEUPH 1115 968 Since K is sufficiently large the matrix correspondent to the morphism e is just tD Then I Z DEUCEDEF tng 1116 EESK on G Moreover in property 8 we have seen that with K sufficiently large 5E9 PEH PEI PE 1 ES 1IPES G s Then 5 y 6EE 4 A Section of d We have seen that d I RKG A RkG is surjective Since RkG is a free Z module there is a Z module homomorphism 7 RkG A RKG 1118 with do ideG we want to find 7 Let f be a class function on Gmg with values in K Define fmg to be the class function on G defined by fmgX fX for all X E G where X is the p regular component of X THEOREM 119 a Iff is a modular Character ofG the fmg E RKGi b o RkG a RKG with 0f fmg is a Zemodule homomorphism which is a section ofdi PROOF a By Brauer s induction theorem the class 1K of the trivial simple KG module in RKG satisfies 1K Z IndggH HEX for certain H E RKH where X denotes the set of elementary subgroups of G Then f 2 In ow 2 Indg J39HResfgt H ex H ex If Resf E RKH it follows that f E RKG Hence we can reduce to the case when G is l elementary for some prime 1 No matter what I G E S X P for a p group S and p group P It suffices to take f to be the modular character of a simple kG module E Since G E S X P we 102 CHAPTER 11 MODULAR CHARACTERS know that E is in ated from a simple kS module ie E has trivial P action Since S is a 11 group E can be lifted to a simple KS module E Hence we can in ate E to a simple KG module with trivial P action lf XE is the character of G over K correspondent to E then for all X E G xgx xgx fx Therefore XE frag Thus frag E RKG 11 Obviously U is a Z module homomorphism Let f be a valued modular character Then d0f dfreg fregleg D 5 Example S4 Let G S4 and K R k be a complete 2 modular system with K sufficiently large relative to S4 51 11 Ordinary Character Table of S4 We know that KS4 K X K gtlt Mat2K gtlt Mat3K gtlt Mat3K 1119 Note that S4 E 22 X 22 gt4 S3 S4 N Lwhere N 1d 1234 1324 1423 1120 and L 1d 12 13 23 123 132 1121 RECALL The ordinary character table for S3 is We can in ate 90 6 and 9 to characters of S4 Then we need two more irreducible characters of S4 of degree 3 Consider the natural action of S4 on K4 with basis b1 b2 b3 b4 Consider p the corresponding representation id 12 1 4 123 1234 000HN A H0000 0 0 1 0 01400 H000 000HA H000 0 0 1 0 0 0 1 1 0 0 0 0 quotb 000A 00140 00140 00140 H000 00140 x 4 2 0 1 0 Note that p sends 111 112 113 114 to itself Therefore Kb1 112 113 114 gives a 1 dimensional sub representation of p with trivial S4 action Let 1p X 7 pm We have 1 M1 lt12 1 Wm 1 123 1 IBM 1113111410141 1p1p i62 6 12 3 712 8 02 6 712 1 Then IP is irreducible character of S4 over K Similarly 61p is an irreducible character of S4 over K We have the ordinary character table of S4 5 EXAMPLE 51 103 52 2modular character table of S4 There are precisely 2 conjugacy classes of 2 regular elements namely the class of id and the class of 123 Then 2 isomorphism classes of simple kS4 modules and hence 2 different irreducible 2 modular characters Let p1 be the modular character of the triVial simple kS4 module Suppose the other irreducible 2 modular character has also degree 1 Then there is a 2 modular character T with where w is a primitive 3 rd root of the unity Then 9mg cannot be expressed as a Z linear combination of p1 and T which is a contradiction Then T does not exists Therefore p2 9le28 is an irreducible modular character Then the 2 modular character table of S4 is 1 r1122 53 Decomposition and Cartan matrices For all Chi ordinary character express lewg as sum of p1 and p2 Then 2601628 111 61628 P1 elGreg P2 WGmg 1 21 elPleeg 11 12 We have the decomposition matrix 1 1 0 1 1 D lt0 0 1 1 1 1122 1123 D H HHOHH HHHOO 104 CHAPTER 11 MODULAR CHARACTERS Let Q be the Character of G over K corresponding to the kG Cover of the simple kG module Si with Brauer Character pi let P5x be the kG Cover of 5 R be a projective RG module with Pimpi E P51 thus n is the Character of the KG module K ER 15139 for i 12 Then 1 xoelperp 1124 qgt2 91P61P 1125 We have the Cartan matrix C D tD G 1126 then 11ng 4411 erz 1127 2 G g 2p1 3p2 1128 Part 3 INTRODUCTION TO MAZUR S DEFORMATION THEORY CHAPTER 12 MAZUR S DEFORMATION THEORY SET UP G as always will denote a finite group k an arbitrary field of characteristic p gt 0 and W will denote a complete local commutative Noetherian ring with residue field k ie W is its mw adic completion where mw is the unique maximal ideal of W and k Wmw we denote by C the category of all complete local commutative Noetherian W algebras with residue field k The morphisms in f are W algebra homomorphisms which induce the identity on k mod the maximal ideal 1 Examples of W 1 Let W be k Then objects in C are complete local commutative Noetherian k algebras with reside field k eg k kHtH kHs tH power series algebras 2 Let k be a perfect field Then usually we choose W to be the ring of infinite Witt vec tors over k This is defined as follows Let R be a complete discrete valuation ring of characteristic zero where v K gt lR U 0 is the correspondent valuation Suppose k is the residue field of R in particular chark p gt 0 Since charR 0 Z Q R Then p E R with p is zero in k thus p E R ie p is not a unit in R Therefore 121 lt 1 in our notation Say our value group V va a E K E Z Then V W rt E Z Thus 121 r 5 for some e E 2 and e is called the absolute ramification index of R If e 1 then we say that R is absolutely unramified ie p is a uniformizer of R THEOREM 121 For every perfect field k there is a complete discrete valuation ring arid only one up to isomorphism which is absolutely ariramified arid has residaefield k PROOF See 8 D R is called the ring of infinite Witt vectors over k denoted by wk Given a valuation 9 on K as in 2 we can define an associated valuation v9 K gt lRJr U 0 as in 1 by choosing and fixing an r gt 1 ian and setting v9 a Fa lf W wk then every 107 108 CHAPTER 12 MAZURS DEFORMATION THEORY complete local Noetherian commutative ring A with residue field k has a natural wk algebra structure 2 Back to the General Case Let C be the full subcategory of 6 of Artinian objects and V be a finitely generated kG module DEFINITION 122 Let R E Obf A lift of V over R is a finitely generated RG module M which is free over R together with a kG module isomorphism zpzk RM A V 121 NOTATION Mzp or M but 4 is always kept in mind Two lifts Mzp and M 41 are isomorphic if there is a RG module isomorphism f M gt M so that the following diagram commutes Id k RM l f frk RM 122 A defamation of V over R is an isomorphism class of a lift M 4 of V over R We denote the isomorphism class of M 4 by M Denote by DefcV R the set of all deformations of V over DEFINITION 123 Define a covariant functor F f A Sets 123 by for all R e 0b FR DefGVR o for all X R gt R in define PM DefGVR A DefGVR 124 WI47 H R 8120 MAPl where 4 is the composition mm R my M k R M V 125 where X induces the identity on k F is called the defurmutiunfunctur of V over f Define F to be the restriction of F to C ie F PIC c A Sets 126 Note that Fk Fk Vidv 2 BACK TO THE GENERAL CASE 109 THEOREM 124 Mazur Schlessinger a Suppose that EndeV E k Then F is repre sented by an object RG V E Obf ie F is naturally isomorphic to HomRG V In other words there is a lift UG V to V over RG V so thatfor all R E Obf andfor all lift M ofV over R there is a unique at RG V a R in f with M 2 R ENGV LIGV 127 We call RG V the universal deformation ring of V and the deformation correspondent to UG V the universal deformation ofV over RGV The pair RG V UGV is unique up to a unique isomorphism b Let V arbitrary Then there is RG V E Obf and there is a lift UG V ofV over RG V so thatfor all R E Ob andfor all lift M ofV over R there is a RG V a R in C a not necessary unique with M E R ENGV UGV lfa is not always unique RG V is called the 39oersal deformation ring of V and the deformation correspondent to UG V is called the 39oersal deformation ofV over RG V The par RGV UG is unique up to a nonfunique isomorphism REMARK 125 Since G is finite one can improve a in Theorem 124 to a if EndeV E k then P is representable ie V has a universal deformation ring To prove Theorem 124 we use Schlessinger39s criteria see 6 and 5 DEFINITION 126 Let T C gt Sets be a covariant functor such that Tk is a l element set eg T F our deformation functor a A couple for T is a pair A where A E ObC and g E TA A morphism of couples u A gt A isamorphism u A gt A inCwith g b We can extend T to a functor T f gt Sets by for all R E Obf TR liLnTRm note that R Rm for all X R gt R in C Ttx liLnTan where for all n E 2 0c induces a homomorphism an Rmll gt R m A procouple for T is a pair B where B E Ob and g E TB A morphism of pro couples v B gt BC is a morphism v B gt B in fwith g c Let R be apro couple for T ie g for certain Q E TRm for alln E 2 we say that R prorepresents T if the natural transformation T Hom R gt T defined by T TAA60bC where TA HomcRA A TA 128 u gt gt Tun n since A is artinian there is n E 2 so that u factors through un Rm gt A is a natural isomorphism d Let T C gt Sets be another covariant functor with Tk a l element set A natural transformation 17 T gt T where n 11RR60bc is called smooth if for all surjections 110 CHAPTER 12 MAZURS DEFORMATION THEORY ocBgtAinCthemap TB gt TA XTA T B 129 a H ltTltagtltmglt gtgt is surjective where TA XTA T 09 fg E TA X T B211Af T D g 1210 is the pullback of TA W T B 7 T A 1211 NOTE This implies that the natural transformation 17 T gt T defined by g lgn Bm 1212 is surjective in the sense that 173 TB A 1 09 1213 is surjective for all B E Obf Let Me with 62 0 be the ring of dual numbers Then the tangent space of T is tT Tke 1214 REMARK 127 LetR e 0b Then t mRm me 1215 is called the Zariski cotangent space of R over W t1 is a k Vector space and its k dual is tR Homkt k Moreover tR HomCRke Therefore tR is the tangent space of the functor Home R DEFINITION 128 A pro couple R of T is called a prorepresentable hull of T or just a hull of T if the natural transformation T HomcR gt T defined in part C of the Definition 126 is smooth and quotkm tR gt tT is bijective PROPOSITION 129 Let R and RC be two proerepresentizhle hulls of T Then there is an isomorphism u R a R in f with Tu C Ifmoreover R g and RC proerepresent T then u is unique PROOF We have two natural transformations T Hom R gt T with TA HomcRA A TA u H Tun n 31 SCHLESSINGERS CRITERIA 11 1 where u factors through un Rm gt A and T HomR with quotA HomCRA A TA v H Tvm n where 390 factors through 3907 R hug gt A Therefore 1 and quotE are surjective Thus there is u R a R in f with mom g and there is u R a R in with Toma g If 1 and RC pro represent T then T and T are natural isomorphisms so u and u are unique Since quotkm and Tl k are bijective u and u induces isomorphisms on tangent spaces That is u u R gt R induces an isomorphism on t and uu R gt R induces an isomorphism on t Therefore by Nakayama s Lemma this implies that u u are surjective endomorphisms Since R and R are complete local commutative Noetherian W algebras with residue field k this implies that u u and uu are actually isomorphisms 3 Schlessinger s Criteria Consider diagrams in C of the form A A 1216 where 0dquot AW gt A is a small extension ie adquot is surjection and kertxquot tA for some t E A with mAWt 0 Look at the natural map of pull backs b FA gtltA A A FA gtltFA FA 1217 where A gtltA A OzM E A X A 00 oc a 1218 I F has a pro representable hull if H1 b is always surjective H2 b is bijective ifA k and A Me with 62 0 ring of dual numbers H3 dimkt1 lt 00 where t1 Fke is the tangent space of F which gets a natural k Vector space structure using ADDITION Fltk1e1gt m MM 7 Fkel men F 1219 tFXtF tF 112 CHAPTER 12 MAZURS DEFORMATION THEORY SCALAR MULTIPLICATION Let E k and define mA ke gt ke 1220 x ye gt gt x Aye FkeFkXFkFkel b l FkakelFkel W Mid 1221 where yA denotes multiplication by A II F is pro representable if H1 H3 are satisfied and H4 b is bijective if A A and M 0c WE WILL DO THE FOLLOWING We will check H1 H3 in general for our F and H4 in case EndkcV E k we will also show that F is continuous ie F F meaning for all R E Ob RR liLnFUQm 1222 Then it follows that the pro representable hull in case is the versal deformation ring of V and is unique up to isomorphism In case II F is representable and the representing object is the universal deformation ring of V and is unique up to unique isomorphism MAZUR S IDEA Re write deformations using matrices Fix a k basis v1v2v of V and let p pV G gt GLnk a representation corre sponding to V with respect to v1v2vn Let R E ObC and let ER 7r G gt GLnR 71 is group homomorphism with K1271 p where KR GLR gt GLnk is the natu ral surjection Let GR kerKR g GLnR GR acts on ER by conjugation LEMMA 1210 For all R E ObC there is a bijectian DefGVR L ERGR 1223 PROOF Let M 4 be a lift of 390 over R Since R is commutative local we can lift the k basis v1vn ofv to an R basis m1mn of M so that 411 m v for all 1 g i g n recall 4 k R M i We get the G action on M is given by a group homomorphism 7r G gt GLnR with K1271 p Let MOW E Mzp as lifts of V over R Then there is a 3 SCHLESSINGERS CRITERIA 113 RG module isomorphism g M gt M with zp id 4 k R M k R M k 1224 43 41 V Then g is given by a matrix X E GLAR with X71 7TX thus 71 X7IX 1 Moreover since zp id 4 X E GR then 71 71 in ERGR thus we get a well defined fwhich is a bijection D COROLLARY 1211 Ourfanctar F C 4 Sets is naturally isomorphic to thefanctar which sends each R in C to ER GR We also called the latterfunctar E SCHLESSINGER S CRITERIA Consider 1225 where a is small extension Consider b EA gtltA A GA gtltA A A EA GA gtltEAGA EA GA 1226 H1 h is surjective PROOF Let 71 7t E EA GA gtltEAGA EA GA Then a 7t a 7t in EAG A thus there is X E GA with 071 Xa 7I X 1 1227 Since a is surjective it induces a surjection a GA kerGLnA gt GLnk A GA kerGLnA gt GLnk 1228 Then there is X E GA with a X X Thus a 7t 0c X 7I X 1 Therefore 71 XNHNXH e EA gtltA A and b7127L71VTL7717I7 114 CHAPTER 12 MAZURS DEFORMATION THEORY QUESTION When is b injective Let 7T E HA with 0c 7T 7T Define GnA X e GA X7THXNT1 71 1229 and GnA X e GA X7TX 1 7 1230 LEMMA 1212 b is injective ifthe map LU14 A GnA 1231 induced by 0c is surjectivefar all 7T E EA with oc 7T 7T PROOF Let 7T 75 E EA gtltA A with l lWl MinquotD WWW liq75 where 7T HUTH 7T KKK 7 375 and 7f 3 09 So we get the following commutative diagram A X Then there are X e GA and X e GA with XVIX 1 7 and 2071quotle 7quot let X and X be the images of X and X in GA under M and 0c respectively Let 7T tx 7T 0c 7T and 7 1 0 MOT Then X nX Tl XTHHXTHTL Then Xi lX E GnA By assumption there is Y E G A with Y X FlX Let X X Y then 1 XH XnJYil Xnil 71 Wd H XVHHXVF and 7 gt5 2W XHXHX 22 Then X and X define an element in GA gtltA A and anHXmil 7T7 hence Hm 79 D H2 b is injective ifA k and A Me with 62 0 PROOF Let 7T E HA SinceA k and MOT p then GPA Gpk identity Then the map GnA gt Gpk is surjective Thus we obtain D 3 SCHLESSINGERS CRITERIA 115 H4 lf EndkcV E k then b is injective if A A and 0 a It is enough to prove LEMMA 1213 lendkcV E k thenfar all R E ObC analfar all 71 E ER GAR consists only 0fthe scalar matrices aver R which map to the identity in GLnk PROOF Since R is artinian we prove this by induction on small extensions of the form OHtRlaRlLROHO where R0R1 E ObC t E mR with tmR 0 and tR1 E k By induction assumption for all 71 E ER0 we have that GnR0 consists only of scalar matrices in GR0 Let 711 E ER1 and X1 E Gn1R1 Then kR1X1 identity in GLnk and X1711X1 1 711 Let A011 710 and AX1 X0 Then X0 710K071 710 then X0 E Gn0R0 By induction 710 is a scalar matrix thus there exists 70 E R0 with X0 70 In where In denotes the identity matrix of order 71 Let 71 E R1 with A01 70 then X1 71ln tB1 where B1 E MatnR1 We have 7171391 tB1739L391 X17 L391 7T1X1 7 H171 7T1tB1 Thus tB nl 711tB1 Then tB1 give us an element in EndegV1 where V1 is the RlG module free over R1 corresponding to 711 Observe that lmtB1 g tV1 E V since tR1 E k and k R V1 2 V Therefore tBl E HOleg htvl E HOleg l V g since le annihilates V and E H0mkGVVltV1mR1V1 g mm v1 E V g k Then there exists a1 E R1 with tB1 t lln thus X1 7391 ta1l D Since scalar matrices can always be lifted GnA gt GnA is surjective which implies H4 H3 dil l lkt1 lt 00 IDEA Rewrite tp using group cohomology Let M be a be a right kG module Define Z1G1MfG A M fglgz fglgz fgzVngz E G 1232 and B1GMszAMz mEMwithfgm17g 1233 Z1G M is called the group of l cocycles and B1GM the group of l coboundaries Observe that B1GM g Z1GM Define H1GM Z1GmB1GM 1234 H1GM is called the first cuhumulugy group of G with coefficients in M H1GM is a k vector space Let f1f2 E Z1GM and A E k then W le lfi le 1235 Alfil Afll 1236 116 CHAPTER 12 MAZUR S DEFORMATION THEORY H1GM is a subspace of the k Vector space MapsGM If M is finitely generated kG module then since G is a finite group Maps G M is a finite dimensional k Vector space Thus H1GM is a finite dimensional k Vector space Then H3 follows from CLAIM t E H1GAdV where AdV is the right kG module with underlying k Vector space Matnk and G action as follows for all X E Matnk and for all g E G X g pg 1Xpgl 1237 where p G gt GLnk is the representation correspondent to V with respect to a fixed k basis v1vn PROOF Recall that Eke 71 G a GLnke Kk 71 p where KW GLnke a GLnk is the natural injection and Gke kerKk Let 10 be the so called trivial lift of p over Me This means 10 corresponds to the He G module He k V T0 G A 6141044 1238 g H Mg where pg is Viewed as an element of GLnke Via the embedding k g ke 1239 a gt gt a 0 6 Note that 10 E Eke Let 71 E Eke Since Kk 71 Kk 10 there is a map X G Matnk with 71 101 eX ie for all g E G 71g 10g 1 eXg We want to show that X e Z1GAdVLetg1g2 e G then T0g1821 X8182 182 71g17rg2 T0811 X81 T0821 X82 81g2T082711 X81T0g21 X82 g1g2 1 epg2 1Xg1pg21 6Xg2 10g1g21 elpg2 1Xg1pg2 Xg2l TO TO Then Xg1gz pg2 1Xg1 mm Xg2 Xg1 g2 Xg2 henceX E Z1GAdV Now let 71 101 eX E E with 71 71 in EkeGke Thus there is i E Gke with 71 i lni But i 1 eL for some L E Matnk Thus i l 1 7 6L Therefore 71 Z 171Z ltgt 101 6X 1 E eL101 eX1 6L 101 eX1 eL E eL10 101 eX L E 10p 1eLp 101 eX L E p lel 3 SCHLESSINGERS CRITERIA 11 7 Thus X X L E p le and L E p le is a map f G gt Matnk so that for all g E G fg L E pg 1Lpg L E L g Thus g L1 7g which impliesf E BlGAdV Hence we get a well defined map t1 Pke EkeGke A H1GAdV 1240 which is a k vector space isomorphism D WHAT IS MISSING A We need to prove that F is continuous ie for all R E ObC RR ERRmg EXAMPLE 1214 Assume that k is a perfect field of characteristic p gt 0 let W wk be the ring of infinite Witt vectors over k As before G is a finite group 1 Let V a kG module with dimk V 1 Then RG V E WGZbquot7 where Gabquot7 denotes the maximal abelian p quotient of G PROOF Let p G gt k be a representation of G correspondent to V There is a lifting k gt wk which is multiplicative see 8 Then we get a lift G gt wk ofp over wk W Define T G A WGWV 1241 g H Mg g where 39y G gt Gabquot7 is the natural surjection thus T is a group homomorphism Let m be the maximal ideal of WGZb p Then m is generated by p and the augmentation ideal AugWDZ1W the latter is generated over W by the elements 7g 1 with g E G Since pg Mg pW and 7g E 1 mod AugWG W for all g E G T is a lift of p over WGZb p Now let 7t G gt R be a lift of p over an object R in The we can extend 7t W linearly to aa W algebra homomorphism 7 1 WG gt R Since R is commutative and its residue field has characteristic p 71 factors through the W linear extension 39739 WG gt WGZbquot7 of 39y Then we obtain the following commutative diagram 71 WG v R 1242 WGzlw Let A WGW L WGW be defined by A39yg g 139yg for all g e G and W linear extension Define 0c WGZW7 gt R by X 7 then X is a morphism in 6 since 7 reduces to p and 7 1 reduces to p 1 mod the maximal ideal Moreover we get 118 CHAPTER 12 MAZURS DEFORMATION THEORY Then WGwquot7 is the universal deformation ring of V Let G 53 and Chmk 2 ie p 2 We have the 2 modular character table of 53 Since PV 2 V V is a projective kSg module 7r R WGRbW T since for all g E G Mg 7g MW g 17g Hi Mm g 7g A gWg g E W Tg D 123 Let k be the trivial simple kSg module Then by l RSgk W5 2 W 22 Let V be a 2 dimensional simple kSg module with character p1 LEMMA 1215 RSg V E W PROOF We first determine a representation p 53 gt GL2k corresponding to V For example p Z 53 A 12 3 123Hlt 11 01 CLAIM l V is a projective kSg module 1243 PROOF OF CLAIM 1 Since dimkk 1 and dimk V 2 then 53 E Pk ea PV ea PV where P5 is the projective cover of the simple kSg module S use Wedderburn decom position of k53 radk53 Then dimk k53 6 and dimk Pk 2 1 so dimk Pk 2 2 which implies dimk PV 2 Then PV E V also dimk Pk 2 implies that Pk is uniserial ie Pk has a unique composition series equal to the radical series 0 g rad Pk g Pk D Since W is a complete discrete valuation ring with residue field k then FracW W k is a complete p modular system then PkG E PWG Therefore V can be lifted to a projective WG module V In particular V is a free W module and k w V E V as kG modules Thus V is a lift of V over W CLAIM 2 V is a universal lift of V over W 3 SCHLESSINGERS CRITERIA 119 PROOF OF CLAIM 2 Let R E Obf let M be a lift of V over R Then M is free as an R module and k R M E V as kG modules Since R is commutative local one of our earlier results about projective modules Lemma 916 shows that since V is a projective kG module m is a projective RG module In particular M is a projective cover of V Since R is a W algebra there is a unique 0c W gt R in f giving the W algebra structure Then R wa V is also a projective RG cover of V Then M E R W4 V Which proves the claim D By Claim 2 we have that RS3 V W NOTE Aa representation 7 S3 gt GL2W correspondent to V is given by 12lt1 3 1244 and p123lt 11 01 1245 B Let G S4 and Chmk p 2 So we have the 2 modular character table of S4 123 Let k be the trivial simple kS4 module and let V be a 2 dimensional simple kS4 module with character p1 Both k and V are in ated from the simple kS3 modules k and S3 since W 12 34 13 24 14 23 kSs 54 g 22 X 22 gt4 53 By 1 RS4k 2 wva2 2 w 22 Wtt22t PROPOSITION 1216 RS4V PROOF We can choose a representation of S4 over k corresponding to V p S4 A GL2k 1246 71 71 123 gt gt lt 1 0 gt 0 1 12 gt gt lt1 0 p should be determined by this presentation of S4 s4 75173 152 1rs4 1 1247 120 CHAPTER 12 MAZURS DEFORMATION THEORY REASON The relations of the right hand side group force it to have order 24 On the other hand 739 123 and s 34 satisfy the relations and generate S4 W 9123 Let R E Ob let M be a lift of V over R where V is Viewed as a kS4 module Via p above Let 7r S4 gt GL2R be a representation correspondent to M Let 5312 123 g S4 then ResM is a lift of V over R when V is Viewed as a k53 module Then by 2 ResM E R w V By looking at we see that we can conjugate 71 by an element in kerGlzR gt GL2k to be able to assume that 71 71 nltrgtpltrgt1 0 CLAIM We can conjugate 71 by an element in kerGL2R gt GL2k GR which commutes with 711 31 so as to able to assume that for some X E mR PROOF OF CLAIM We need to find the matrices that commute with 711 x y 71 71 i x y 7x z w 1 0 7 z w 72 71 71 x y 7 7x 7 z 7y 7 w 1 0 z w T x y Thus 2 7y and X y w So the matrices which we are looking for are of the form yw y y w 31 SCHLESSINGERS CRITERIA 121 which lies in GR if y E mR and w E 1 mR Let A 715 We show by induction that mod higher and higher powers of mR we can conjugate A by a product of matrices of the form Y ljyy with y E 11112 so as to be able to assume that 75Alt1ix 1XgtforsomexEmR Ex mod m A lt0 1 T where T E Mat2mR Since A2 I then 10 0 12 0 1 0 1 2 7 1 0 lt10lt10T Tlt1 1 701 W Hi 3 g modmlz2 a 3 46 a Then 1gt1quotT 1 1 modmlzzso 0c ix3 fix mod mg ThenA lt1 73 lid3 mod m Now Y y E1 withy e 1 7 y lt31 mod m Therefore lt gtgtlt1311f ltlt53gtWlt31 31 311 W a 2 3H a 31 1 123 1313 1 lt31 E1 mod mg lt T 0 5 mam3ng mR Then Y l lt 0c 1 0 o 1W 1 0 1 YAY l H 1 oc2y 1By 1737y Ewizy Weneedoc2y ytheny iocEmRThen 17 Z ioc 12 70c 2 YAY 7lt1i2 ilx 72 ilx modmR By induction we assume that we have conjugated A by a product of matrices of the form of Y to get Alt Xi 13 modm g gt mod mfg 1 7 Xi Exi for some Xi E mR 122 CHAPTER 121 MAZURS DEFORMATION THEORY Xi 1 Xi Xi Xi 2 Xi 1Xl39 Xi 1Xl39 Xi 1Xl39 2 7 1 0 1 39 39 lt1ixi X1gtTT 1 39 Xi I T 0 1 V V 0 0 modmgl 0 0 1 0 0 1 11 0 1 11 0 1 1 0 T mod mR 7T 1 0 mod mR Then lt0 1 T T lt0 1 mod m3 which implies mod mg A lt1 gt T for some T E Matzm z Since A2 I then 10 10 V T lt2 5 ix3 71311 mod mg i Xitxi 1Xi3 i1 1 0 1 1 ThusAilt1ixii i xiilx modmR Noin 0 1 2 71 0 V V I withz E mR Then Y Tl E071le lt 11 H 1 0 1 1 X110 1Xil3 i M 1 11 11 YAY Tltlt0 1gtZlti1 0 lixii i xiim Jg 1z 71 0 modmR x1ai Haw3 T 17397 Exiitx 1 1 xi 1Xi Xi 1Xi 1 1 H1 110117 7x11117sz 7391110 1 m dmR Choose 2 7 0 E mR then 23922 E mgl Therefore Xi 0 1 Xi B 7 xi 7 Exi 7 oc wed lt31 M H w a 7lt Xi2 70 1Xi2 70Cgt modmi1 7 R 1 7 YAY i lt1 14943420 4972340 6141 1 Xi1 mod mm 1 X141 XHl R 39 where X141 Xi Z 7 M E mR Since R is complete this process will converge Then the claim follows E Now 7Ir3 I 7TS2 The last condition 7rrs4 I means that the character istic polynomial of 7175 divides the polynomial T4 7 1 m 3 112

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