Top Commutative Algeb
Top Commutative Algeb MATH 681
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68 ALGEBRAIC GROUPS 6 ROOT DATUMS AND REDUCTIVE ALGEBRAIC GROUPS The main goal in this chapter is to give a sketch of how the classi cation of reductive algebraic groups goes I really just want to give you the precise statement and some examples but am going to skip very many details This is a non trivial theorem that would take a substantial amount of work to develop with full proofs 61 Tori and root datums Let us start by talking about tori Recall an n dimensional torus is an algebraic group isomorphic to Gm x x Gm For example the subgroup Dn of GL7 consisting of all diagonal matrices is an n dimensional torus Let T be an n dimensional torus The character group XT HomT Gm HomGm Gm Z An important point is that given any two tori T and T HomTT HomXTXT So any homomorphism f XT a XT of abelian groups induces a unique morphism T a T of algebraic groups and vice versa To be fancy you can view X as a contravariant equivalence of categories between the category of tori and the category of nitely generated free abelian groups All elements of a torus T are semisimple So if V is any nite dimensional representation of T every element of T is diagonalizable in its action on V by the Jordan decomposition Moreover they commute hence we can actually diagonalize V EB VA AEX T where V v E V l 2512 tv for all t E T As before the V s are called the weight spaces of V with respect to the torus T Now let G be an arbitrary connected algebraic group A maximal toms of G is what you d think a closed subgroup T that is maximal subject to being a torus Now start to assume that G is a reductive algebraic group Let T be a maximal torus Let g be the Lie algebra of G We can view g as a representation of T via the adjoint action It turns out moreover 7 using for the rst time that G is reductive 7 that the zero weight space of g with respect to T is exactly the Lie algebra t of T itself So we can decompose g t 63 Ba we where I is the set of all 0 7 Oz 6 XT such that the T weight space g0 7 0 You can already see the root system emerging The difference now however is that the set I of roots is a subset of the free abelian group XT Now using the assumption that G is reductive again you show ALGEBRAIC GROUPS 69 1 Each g0 is one dimensional and Oz 6 Q iff 704 E Q 2 The group W N0TT is a nite group that acts naturally on XT and permutes the subset Q Q XT 3 Let Q be the root lattice the subgroup of XT generated by Q and let E R 8 Q Fix a positive de nite inner product on E that is invariant under the action of W Then E Q is an abstract root system We ve now built out of G a root system E Q and realized the Weyl group W explicitly as the quotient group N0TT Moreover Q is a subset of the character group XT of T I must stress that all these things take work to prove 7 it is usually harder than in the Lie algebra case BUT everything works in arbitrary characteristic If G is semisimple then G is determined up to isomorphism by its root system E Q together with the extra information given by the fundamental group XTQ However this is not the most natural point of view to classify the reductiue not just semisimple groups This is harder since XT will in general be of bigger rank than Q and so there is much more freedom not captured by the fundamental group alone Let s prepare the way to state the classi cation of reductive algebraic groups in general Let G be a reductive algebraic group and let T be a maximal torus Let Q C XT be the root system of G de ned from the decomposition of g as above Let XT HomTGm be the character group of T and let YT HomGm T be the cocharacter group This is also a free abelian group of rank dim T Moreover there is a pairing XT x YT a Z de ned as follows Given E XT and j E YT the composite 0 j is a map Gm a Gm So since AutGm Z A o W 2M for a unique gt E Z For each 04 E Q you prove that there is a unique up to scalars homo morphism ma Ga a G such that tx0ct71 xaoztc for all c E Gmt E T such that the tangent map 1950 LGa gt g0 70 ALGEBRAIC GROUPS is an isomorphism Moreover the mats can be normalized so that there is a homomorphism 10 SL2 a G m 3 izaltcgt a i 2zialtcgt c 0 aVGmgtTaVc alt 0 C71 So 04V 6 YT This is called the coroot associated to the root Oz 6 Q Now we have built a datum XT Q YT QV where QV is the set of all coroots This is the root datum of G with respect to the torus T Actually since all maximal tori in G are conjugate it doesn t depend up to isomor phism on the choice of T The notion of root datum is the appropriate generalization of root system to take care of arbitrary reductive algebraic groups not just the semisimple ones Here is an axiomatic formulation of the notion of root datum a root datum is a quadruple X QY QV where a X characters and Y cocharacters are free abelian groups of nite rank in duality by a pairing X x Y a Z b Q C X roots and QV C Y coroots are nite subsets and there is a given bijection a gt gt 04V from Q to QV such that De ne To record the additional axioms de ne for 04 E Q the endomorphisms 30 3X of X Y respectively by 80M 90 lt90 avgta7 thy y 7 lta M Then we have the axioms RDl For 04 E Q 0404Vgt 2 RD2 For a E Q saQ Q sXQV QV The datum XT Q YT QV built from our algebraic group G earlier is such a gadget There is a notion of morphism of root datum X ltIgt Y M a X ltIgtCYC W a map f X a X that maps Q bijectively onto Q and such that the dual map fV Y a Y maps fOzV to 04V for all 04 E Q Hence there is a notion of isomorphism of root datums Now suppose that G G are reductive algebraic groups with maximal tori T T respectively and corresponding root data XT Q YT QV and the primed version Let f XT a X T be a morphism of root data It induces a dual map f T a T of tori The key step in the proof of the classi cation theorem is to show that f can be extended to a homomorphism f G a G Using it you prove in particular the isomomhism theorem ALGEBRAIC GROUPS 71 Theorem 61 Two reductive algebraic groups G G are isomorphic if and only if their root datums relative to some maximal tori are isomorphic There is also an existence theorem Theorem 62 For every root datum there exists a corresponding reductive algebraic group G Finally one intriguing thing given a root datum X I Y EV there is the dual root datum Y EV X 1 If G is a reductive algebraic group with root datum X I Y EV you see there is a dual group CV with the corresponding dual root datum Note the process of going from G to CV is very clumsy I don t think there is any direct way of constructing the dual group out of the original Example 63 Suppose that G is a semisimple algebraic group Let Q 21 C XT Here Q and XT have the same rank so Q is a lattice in XT and XTQ is a nite group the fundamental group Let P be the dual lattice to Q Fixing a positive de nite W invariant inner product on E R 8 Q we can identify P with the weight lattice of the root system of G and then everything is determined by the relationship between Q Q XT Q P You can formulate the classi cation just of the semisimple algebraic groups in these terms Example 64 Let G be a semisimple algebraic group and suppose that Q Q XT Q P are as in the previous example If XT P then G is called the simply connected group of type I If XT Q then G is called the adjoint group of this type Now let G be the simply connected one Gad be the adjoint one Let G be any other semisimple group of type I Then there is an inclusion XT gt P XTsc This induces a map GM as G Similarly there is always a map G as Gad Example 65 1 Consider the root datum of GL2 Here XT has basis 61 62 these being the characters picking out the diagonal en tries Moreover the positive root is 04 61 7 62 Also YT has basis e 6 the dual basis mapping Gm into each of the diagonal slots The coroot is 04V q 7 6 2 GL2 is its own dual group 3 Consider the root datum of SL2 x Gm Here XT has basis 042 6 YT has the dual basis 04V 6V here 04 is the usual positive root of SL2 4 Consider the root datum of PSLZ x Gm Here XT has basis 04 e YT has the dual basis xv2 6 So PSLZ x Gm is the dual group to SL2 x Gm Exercise 66 7 As an exercise in applying the classi cation you can show that 13 and 4 plus one more the 4 dimensional torus are all the reductive algebraic groups of dimension 4 72 ALGEBRAIC GROUPS 8 The dual group to SLn is PSLn The dual group to Spgn is 02n1 The dual group to PSpgn is Spingn1 The dual group to S02 is S02 The dual group to Spingn is PSOgn For more explicit constructions of root datums7 see Springer7 747 62 Complete varieties and the Borel xed point theorem De nition 67 A variety X is called complete if for all varieties Y the projection pry X X Y gt Y is a closed map This is the analogue of compactness in algebraic geometry Here is an example of a space that is not complete Example 68 A1 is not complete For consider the projection map A1 x A1 a A1 my gt gt y It sends the closed subvariety lzy 1 to A1 7 0 which is not closed Theorem 69 l is complete Proof We need to show for any variety Y that the projection morphism 7r l X Y gt Y is closed It suf ces to deal with the case that Y is af ne and irreducible Put A MY7 S AT07 Tn We can View S as an algebra of functions on k 1 x Y If I is a proper homogeneous ideal in S put VI E l x Y l fmy 0 for all f E I where denotes the point of l de ned by z E k 1 7 You should remember we looked at something like this when we described explicitly the closed sets in lP they were the common zeros of proper homogeneous ideals of MTG 7Tn In the new situation you show similarly o the closed sets in anY are of the form VI for proper homogeneous ideals in S g VI o if and only if To gr 0 VI is irreducible if and only if is a prime ideal Now we have to show that 7139 maps closed sets to closed sets lts enough to show it maps closed irreducible sets to closed irreducible sets since any closed set is a nite union of irreducible closed sets Thus we have to show all the irreducible sets 7rVI are closed for all proper prime homogeneous ideals I in S Let Y0 be the closure of 39n39VI7 also irreducible and af ne Then VI is contained in lexYo and 7139 VI a Y0 is dominant Replacing Y by Y0 this reduces to showing that 7rVI Y for all proper prime homogeneous ideals I in S such that 7139 VI a Y is dominant Note that VI Q l x VA I so 7rVI Q VA O I So the dominance assumption on 7139 means VA I Y7 ie A I ALGEBRAIC GROUPS 73 We are now reduced to showing the following given a proper prime ho mogeneous ideal I in S such that A O I 0 we want to prove for every y E Y that there exists E l with E VI Let M be the max imal ideal of A of all functions vanishing at y Then J MS I is a proper homogeneous ideal in S We have to prove that VJ 7 9 because if zy E VJ then pick i so 7 0 and consider fTi E J for f E M its value on 354 is zero hence fy 0 This is for all f E M which implies y Therefore assume for a contradiction that I is a proper prime homoge neous ideal in S such that A I 0 y E Y is a point with maximal ideal M in A and VJ Q where J MSI This means that T0 Tn so there is some h gt 0 such that Toh T E J Equivalently there is some l gt 0 such that the set S C S of homogeneous polynomials of degree l lies in J Put N SlSl O I This is a nitely generated A module Our assump tions imply that every element of S1 looks like 2 aibi j for ai E M bi E S and j E S O I Hence N MN Hence by Nakayama s lemma there s some element a E A with aN 0 and a 7 1 E M It follows that a I otherwise J S and aSl Q I Since I is a prime ideal we deduce that S Q I ie N But this means that I C S so VI G which is a contradiction Y would be empty D The following is a rather easy exercise Proposition 610 If X is complete and Y is closed in X then Y is complete ii If XY are complete so is X X Y iii If gt X gt Y is a morphism and X is complete then is closed and complete iv If Y is a complete subuarietg of X then Y is closed v IfX is complete and a ne then X is nite For and ii its easy from the de nition For iii consider the graph m gtm which is closed in X x Y You get easily that gtX is closed in Y from this Its also complete which you see by considering j x 1 X x Z a gtX x Z For iv apply iii to the inclusion of Y into X For v let X be irreducible complete and af ne Suppose f X a A1 is a morphism Then its image is irreducible and complete By the example above its not A1 Hence it is a point and f is constant This shows kX k so X is a point We also need the following Lemma 611 Let G be an algebraic group acting transitively on varieties XY Let j X gt Y be a bijectiue G equiuariant morphism If Y is complete then X is complete Proof We know that j X id X x Z a Y x Z is open by 46 Since it is bijective too we deduce that gt id is closed Now prz X x Z a Z factors 74 ALGEBRAIC GROUPS through X x Z a Y x Z a Z where both are closed on the way Hence X is complete D Now we can prove the all important Borel s xed point theorem Theorem 612 Let G be a connected solvable algebraic group and X be a non empty complete G variety Then G has a xed point on X Proof Proceed by induction on dim G the case G 1 being trivial Sup pose then that dim G gt 0 and let H G which is connected solvable of strictly smaller dimension By induction Yz Xlez is non empty It is closed hence complete and G stabilizes Y as H lt1 G So we may as well replace X by Y to assume that H Q Gm for all z E X Since GH is abelian this implies that each Gm 1 G Now choose z so that Gm is of minimal dimension Then Gm is closed hence complete The map GGm a Gm is bijective so we deduce that GGm is complete by the preceeding lemma But GGm is af ne as Gm 1 G So in fact GGm is a point ie G Gm and z is a xed point D Corollary 613 Lie Kolchin theorem Let G be a connected solvable sub group of GLV Then G xes a flag in V Proof Let G act on the ag variety This is projective so G has a xed point D ALGEBRAIC GROUPS 61 5 ROOT SYSTEMS AND SEMISIMPLE LIE ALGEBRAS 51 Characteristic 0 theory Assume in this subsection that chark 0 Let me recall a couple of de nitions made earlier G is called reductiue if it is connected and has no non trivial closed connected unipotent normal sub group G is called semisimple if it is connected and has no non trivial closed connected solvable normal subgroup Semisimple groups are reductive but reductive groups are a little more general including eg the groups CL and D The reductive algebraic groups are the ones for which there is a beautiful structure theory In this section I will discuss the situation for semisimple groups over elds of characteristic 0 Observe right away that a connected G is semisimple if and only if it has no non trivial closed connected abelian normal subgroup Analogously a nite dimensional Lie algebra g is called semisimple if it has no non zero commutative ideal The main theorem 1 want to prove is Theorem 51 Assume chark 0 Then a connected group G is semisimple if and only ifg is semisimple In that case AdG GZG Aut g Note for semisimple G that ZG is nite So the theorem almost classi es the semisimple algebraic groups in characteristic 0 the isomorphism type of GZG at least is classi ed by the isomorphism type of g Since the latter are classi ed by Dynkin diagrams so are the centreless semisimple groups Recall that if g is simple then G is simple over an arbitrary eld But SL7 in characteristic dividing it gave us an example already where G was simple but g was not So the Theorem is completely wrong in positive characteristic For the proof we begin with Lemma 52 Assume chark 0 and G is a connected algebraic group 1 If j G gt H is a morphism of algebraic groups then kerdd Lker gt 2 If A B are closed subgroups of C than a O b LA O B Proof 1 WLOG j is surjective It is automatically separable so we can identify H G ker gt by the rst isomorphism theorem But LG ker gt gLker gt Therefore ker dd Lker gt 2 Let 7r G a GB be the canonical map so kerd7r b Let 71quot A gt GB be the restriction The bres of 71quot are the cosets mA O B for z E A and 71quot is separable automatically Therefore we can identify 71quot with the canonical map A a AA O B hence ker d7r LA O B by But clearly kerdIr u kerd7ru b D Lemma 53 Assume chark 0 Let j G gt GLV be a nite dimen sional representation Then G and g leaue inuariant the same subspaces resp vectors of V Proof We may assume G lt GLV Consider a subspace W of V Let StabGLVW 9 E CLOWW S W StabgiVW X E Elmle E W 62 ALGEBRAIC GROUPS Using dimensions you check easily that LstabGLVW stabgVW in arbitrary characteristicl But stabGW G O stabGLVW stabgW g stabgVW Now using the lemma you get that LstabGW stabgW Finally G stabilizes W if and only if the left hand side is G while g stabilizes W if and only if the right hand side is g The statement about subspaces follows The argument for vectors is similar B Let G be a connected algebraic group The subalgebras of g that are of the form LH for a closed connected subgroup H of G are called the algebraic subalgebras of g Note even in characteristic 0 g may have subalgebras that are not algebraic Theorem 54 Lattice correspondence quot Assume chark 0 and G is con nected Then the map H gt gt E is a 171 inclusion preseruing correspondence between the closed connected subgroups of G and the algebraic subalgebras of g Moreouer normal subgroups correspond to ideals under the correspon dence Proof Suppose that LH We need to show that H K which is false in positive characteristicl But LH O K LH LK by the lemma Hence LH K LH hence dim H K dimH so H K H Similarly H K K Now consider the nal statement about normal subgroups We need to show H is normal in G if and only if E is an ideal ofg We already proved the forward implication in chapter 3 using dlnt AddAd ad Conversely suppose E is an ideal of g Then g stabilizes E acting via ad dAd So by the Lemma G stabilizes E acting via Ad So for z E G Adzh E But Adz E a g is the differential of lntz H a G So the image E Adzh Llnt Lsz 1 by separability Now by the previous paragraph we get that H mHz l because they have the same Lie algebra hence H is normal in G Note this already shows G is simple if and only if g is simple To prove the main theorem above we need one more lemma Lemma 55 Let chark 0 and G be a connected algebraic group For x e G LltCaltzgt0gt cam Proof Recall 00z g E G l mgz l g and cgm X E gl Ade X In general it is obvious that LCGO Q CW For lntz Cazo a Cazo is the identity map so its differential Adz is the identity map on LCaz0 Now suppose for a moment that G GLn Then CC is all invertible matrices that commute with z and cgm is all matrices that commute with ALGEBRAIC GROUPS 63 m Clearly the latter is a principle open subset of the former so we get equality in this case by dimension In general we may assume G is a closed subgroup of GLn Then CW cm g LCGL MG LCGL m G LCG Note we have used chark 0 in this last step D Lemma 56 Let chark 0 and G be a connected algebraic group Then ker Ad ZG Proof We already know ZG Q ker Ad Since LCG ENE we see that if Adz 1 then LCGm g so Caz G by the lattice correspondence ie z E ZG Thus ZG ker Ad D Now we can prove the theorem stated above Theorem 57 Assume chark 0 Then a connected group G is semisimple if and only ifg 239s semisimple In that case AdG GZG Aut g Proof Suppose g is semisimple If N is a closed connected abelian normal subgroup of G then LN is an abelian ideal of g hence LN 0 hence N 1 This shows G is semisimple Conversely suppose G is semisimple Let n be an abelian ideal of g Let H Can0 so E cgn by the previous lemma Moreover E is an ideal in g just check XY Z 0 for all X E gY E E Z 6 m using the Jacobi identity containing 11 in its center By the lattice correspondence H is a normal subgroup of G so Z ZH0 is also normal in G hence zero as G is semisimple By 55 ZH ker Ad so its Lie algebra is ker ad 3h which contains 11 Hence n 0 It remains to prove that AdG Autg0 I m not going to do this D 52 Root systems Hopefully the theorem just proved motivates spending a little time looking at the nite dimensional semisimple Lie algebras over C I want to explain their classi cation The rst step is to introduce the abstract notion of a root system Here is the de nition a root system is a pair E Q where E is a Euclidean space and Q is a nite set of vectors called roots in E such that R1 0 Q spans E R2 04 ca 6 Q implies c i1 R3 Q is invariant under the re ection sa in the hyperplane orthogonal to Oz ie the automorphism B gt gt Bi BaVa where 04V 2aa 04 R4 04 6V 6 Z for all a B E Z Given a root system the Weyl group W is the subgroup of GLE gen erated by the SW for 04 E Q It is a nite group since it acts faithfully on the nite set Q I will give examples in class to give you a geometric way to think about W 64 ALGEBRAIC GROUPS We let H0 B E E l 04 0 be the hyperplane orthogonal to 04 The connected components of E 7 U H0 0 Q are called the Weyl chambers Fix a chamber C which we will call the fundamental chamber Then one can show that the map wi muC is a bijection between W and the set of chambers The choice of C xes several other things We let QJr be the set of all 04 E Q which are in the same half space as C Then Q QJr U 7Q Elements of Q are called positive roots Next let H oz 6 QJr l Ha is one of the walls of C This is called a base for the root system One can show that H is actually a basis for the vector space E and moreover every element of QJr is a non negative integer linear combination of H Elements of H are called simple roots The Weyl group W is actually generated by the sa for 04 E H ie by the re ections in the walls of the fundamental chamber This leads to the idea of the length w of w E W This is de ned as the minimal length of an expression w 501 50 where 041 oz are simple roots Geometrically w is the number of hyperplanes separating wC from C Now let EQ be a root system Let H 041134134 be a base coming from a choice of fundamental chamber Here 6 dimE is the rank of the root system The Cartan matrix A aij1Sjg is the matrix with a 041304 Since all the Weyl chambers are conjugate under the action of W the Cartan matrix is an invariant of the root system up to simultaneous permutation of rowscolumns Here are some basic properties about this matrix C1 a 2 C2 For i 7 j am 6 07172 73 C3 am 0 if and only if a 7 0 Note C2 is not obvious It follows because E on0 together with Q Q O E is a root system of rank 2 The Cartan matrices for the rank two root systems are exactly the following 2 0 2 71 2 72 2 71 A1gtltA1lt02gtA2lt71 2gtB2lt71 2gtG2lt73 Note if am 0 then ai7aiaj7aj Majan E 17 273 So you can work out the ratio of the lengths of the roots 04 04j to each other from the Cartan matrix A root system is called indecomposable if it cannot be partitioned E E1 T E2 Q Q1 U Q2 where E Q are root ALGEBRAIC GROUPS 65 systems For an indecomposable root system you can work out the ratio of lengths of any pair of roots to each other from the Cartan matrix hence you completely recover the form on E up to a scalar from the Cartan matrix You also recover I since the Cartan matrix contains enough information to compute the re ection 30 for each i 1 6 and I WH So an indecomposable root system is completely determined up to isomorphism by its Cartan matrix A convenient shorthand for Cartan matrices is given by the Dynkin di agram This is a graph with vertices labelled by 041 ozg There are algaam edges joining vertices on and 04739 with an arrow pointing towards on if when lt 047047 equivalently a 71aji 7273 Clearly you can recover the Cartan matrix from the Dynkin diagram given properties Cl703 above Now I can state the classi cation of root systems Theorem 58 The Dynkin diagrams of the indecomposable root systems are Exercise 59 5 In class I wrote down the explicit construction of the root system of type A and showed that the length of the longest element we of the Weyl group was 66 12 Do the same thing for the other classical root systems BLCZ and D You will need to look them upll There are many good sources eg Humphreys7 Introduction to Lie algebras and representation theory Bourbaki Groupes et Algebres de Lie Kac ln nite dimensional Lie alge bras Carter Finite groups of Lie type Helgason Differential geometry and symmetric spaces 53 Semisimple Lie algebras Now I sketch how the semisimple Lie al gebras are classi ed by the root systems Of course we need to start with a semisimple Lie algebra and build a root system out of it and vice versa Let us begin with a nite dimensional semisimple Lie algebra g over C Here are some equivalent statements 1 g is semisimple 2 g has no proper abelian ideals 3 g possesses a non degenerate inuaiiant symmetric bilinear form where invariant here means XYZ X Y Note if g is simple there is a unique such form up to a scalar Proving the equivalence of 1 and 2 not at all obvious is the rst major step in developing the theory The experts will know that there is a canoni cal choice of non degenerate form the Killing form but we don t need that here Example 510 Let us consider 5 The bilinear form X Y trXY is non degenerate and invariant So Eta is a semisimple Lie algebra Let em be the ij matrix unit and let E be the diagonal trace zero matrices We can 66 ALGEBRAIC GROUPS decompose 5t h Cemz 7 A basis for E is given by 711 hn1 where hi em 7 6i1i1 Let q E if be the map sending a diagonal matrix to its ith diagonal entry Note 61 en O ie the 61 s are not independent Then leeml 6i EjH5ij7 ie em is a simultaneous eigenvector for F We use the word weight in place of eigenvalue so em is a vector of weight 6 7 6739 Now you recall that the root system of type An1 can be de ned as the real vector subspace of if spanned by 61 en and the roots are 12Ei76j i7 j A base for I is given by by 041 ozn1 where 04 61quot Ei1 Let us nally write g0 Cem if 04 e 7 6739 ie the weight space of g of weight 6 7 6739 Then 21 h 69 EB ga one In other words you see the root system of type An1 when you decompose g into weight spaces with respect to the diagonal matrices Final note the inner product giving the Euclidean space structure is induced by the non degenerate form de ned to start with Indeed if you compute the matrix lily71739 you get back the Cartan matrix of type An71 This example is more or less how things go in general when you start with an arbitrary semisimple Lie algebra g with a non degenerate invariant form The rst step is to develop in g a theory of Jordan decompositions This parallels the Jordan decomposition we proved for algebraic groups You call an element X of g semisimple if the linear map adX g a g is diagonalizable and nilpotent if adX is nilpotent The abstract Jordan decomposition shows that any X E g decomposes uniquely as X XS Xn where XS 6 g is semisimple and Xn E g is nilpotent and X5Xn 0 What is more if you have a representation of g ie a Lie algebra homo morphism p g a g it is true that pX5 oXS and pXn pXn where the semisimple and nilpotent parts on the right hand side are taken just as n x n matrices in g Thus the abstract Jordan decomposition is consistent with all other Jordan decompositions arising from all other rep resentations In particular semisimple elements of g map to diagonalizable matrices under any matrix representation of g For Eta em is nilpotent for i 7 j and semisimple for i j Now you introduce the notion of a maximal toral subalgebra E of g This is a maximal abelian subalgebra all of whose elements are semisimple It turns out that in a semisimple Lie algebra maximal toral subalgebras are non zero and they are all conjugate under automorphisms of g Now x one 7 it doesn t really matter which since they are all conjugate lmportantly ALGEBRAIC GROUPS 67 the restriction of the invariant form on g to E is still non degenerate So we can de ne a map W H h mapping oz 6 if to ta 6 E where ta is the unique element satisfying ta h ah for all h E E Now we can even lift the non degenerate form on E to if by de ning 043 ta t Thus if now has a non degenerate symmetric bilinear form on it too For 04 E if de ne g0 X E g l HX aHX for all H E g Clearly g 046 gm Set I 0 7 oz 6 if lga 7 0 Then you get Cartan decomposition of g g h 69 EB g0 one it is not quite obvious that the right hand side is everything It turns out with some work that each of the g0 spaces are one dimensional Now you can build a root system out of g we ve already constructed the set I Let E be the real vector subspace of if spanned by I The restriction of the form on if to E turns out to be real valued only and makes E into a Euclidean space Now Theorem 511 The pair E 1 just built out ofg starting from a choice of is a root system Moreouer the resulting map from semisimple Lie algebras to Dynkin diagrams giues a bijection between isomorphism classes of semisimple Lie algebras and Dynkin diagrams The decomposition of a semisimple Lie algebra as a direct sum of simples corresponds to the decom position of the Dynkin diagram into connected components For example Eta is the simple Lie algebra corresponding to the Dynkin diagram An1 One other point When you study root systems you x a choice of fundamental chamber giving you the notion IJr of positive roots and H of simple roots This xes in g a triangular decomposition Borel subalgebras Exercise 512 6 Look up or work out the dimensions of the simple Lie algebras of types A1 B1 Cl and D1 In particular check that dim Cl is the same as the dimension of the algebraic group Spy from exercise 50 ALGEBRAIC GROUPS 4 QUOTIENTS 41 More algebraic geometry We will always be concerned with a dom inant morphism j X H Y between irreducible varieties Recall that as j is dominant we have the comorphism w MY H kX which allows us to View kX as a eld extension of If f is an isomorphism then the morphism j is said to be birational Lemma 41 j is birational if and only if there is a non empty open subset U ofX such that gtU is open in Y and j U H gtU is an isomorphism Proof This immediately implies that j is birational by the de nition of function elds a Assume j is birational Without loss of generality7 we can take X and Y to be af ne Then7 f MY H kX as j is dominant7 and the induced map f kY H kX at the level of elds of fractions is an iso morphism So MY Q kX Q kY and we can write kX kYf17 7 fr for f1 71 E Putting the 12 over a common denominator7 we get 0 7 f 6 MY such that all fif E Let U DXf7 aprincipal open sub set of X Then7 gtU Dyf and f kYf H kXf is an isomorphism7 hence U gtU D Two irreducible varieties are said to be birationally equivalent if there exists a birational morphism between them This is an important notion in algebraic geometry birational equivalence is a much weaker thing than being isomorphic7 and it turns out to be much more reasonable to classify varieties up to birational equivalence rather than up to isomorphism Example 42 Let Y z7 y E Azlzz yg be the twisted cubic De ne a morphism j A1 H Y by gtz 3 2 This is obviously a bijective map The inverse is 734 H my for z y 7 07 0 and 00 H O However7 the inverse map is not a morphism of varieties what would its comorphism be So it is not an isomorphism of varieties But 1 is birational consider A1HO which 1 maps to YH07 This is an isomorphism of varieties the inverse map is my H my which Is a morphism of varieties on Y H 07 0 7 you can invert y on this open subset We ve just given an example of a birational7 bijective morphism that is not an isomorphism You might think the problem here is Y it is not a smooth variety since 07 0 is a cusp You would be right Theorem 43 Zariski s main theorem Let 1 X H Y be a morphism of irreducible uarieties that is bijectiue and birational Assume that Y is a smooth uariety Then j is an isomorphism Remark 44 I am not going to prove the theorem We will apply it at a crucial point in the next subsection in our construction of the quotient of ALGEBRAIC GROUPS 51 an algebraic group by a closed subgroup The real statement of Zariski s main theorem replaces the word smooth with the word normal I do not want to get into the de nition and properties of normal varieties so I have stated the weaker result above which will be enough for our purposes Suppose that G is a connected algebraic group acting on a variety X Let x E X l have already told you that dim Gm dim G 7 dim G1 The next result I want to state proves this important fact actually it is a much stronger fact about arbitrary morphisms of varieties Theorem 45 Let X and Y be irreducible uarieties and j X 7gt Y be a dominant morphism Let r dimX 7 dimY Z 0 Then there is a non empty open subset U of X with the following properties i For any uariety Z the restriction of gt X 1 to U X Z de nes an open morphismUX Z7gtYX Z i If Y is an irreducible closed subuariety on and X is an irreducible component of gt71Y intersecting U then dim X dim Y r In particular if y E Y then any irreducible component of jfly inter secting U has dimension r iii If r 0 then for all m E U the number of points in the bre 1 gtm is equal to the degree of the eld extension kX5KY where kX5 is the set of all elements of kX that are separable ouer Roughly speaking the theorem says that a morphism is always well behaved in some open subset It can be quite badly behaved outside of that open set As usual when there is a transitive action of an algebraic group involved we can get a stronger result as a corollary Corollary 46 Let G be a connected algebraic group and XY be uari eties on which G acts transitiuely Suppose j X 7gt Y is a G equiuariant morphism Let r dim X 7 dim Y Then i For any uariety Z the morphism j X 1 X X Z 7gt Y X Z is open ii If Y is an irreducible closed subuariety of Y and X an irreducible component of gt71Y then dim X dim Y r In particular for y E Y all irreducible components of jfly haue dimension r iii j is an isomorphism if and only if it is bijectiue and for some m E X the di erential d gtm TmX 7gt TmY is onto Proof i ii Let U Q X be an open subset as in the theorem Then all translates gU enjoy the the same properties Since these cover X we get and ii iii Suppose that j is bijective and d m is onto Then by ii r 0 and by Theorem 45iii we must have that XS MY ie kX is a purely inseparable extension of On the other hand by the differential 52 ALGEBRAIC GROUPS criterion for separability 316ii etam onto implies that j is a separable morphism ie kX is a separable extension of Hence kX kY and j is birational Finally since Y is an orbit it is smooth so we get that j is an isomorphism by Zariski s main theorem Corollary 47 Let j G H H be a homomorphism of algebraic groups Then i dim G dim im 1 dim ker gt ii j is an isomorphism if and only if it is bijectiue and the di erential d gte is onto Proof This follows easily from Corollary 46 D Corollary 48 Suppose G is a connected algebraic group acting on a variety X Then form 6 X dim Gm dimGH dime Proof Let 1 G H Gmg H gm be the orbit map By Corollary 46ii dim jflm dim G H dim Gm But jflm is the stabilizer Gm D I am not planning on proving Theorem 45 rather I want to give some examples to give you the avor of how things can go wrong in general 7 hopefully that is more useful Example 49 1 Consider the morphism j A2 H A2 z y H I rst want to show that this is a birational morphism that is not an open morphism so the subset U given by the theorem must be a proper subset in this example Let U E A2 lz 7 0 Clearly gtU U Moreover on U we can invert j the inverse map sends zz H 7 this is a morphism because we have localized to invert m Therefore MU U H U is an isomorphism of varieties and j is birational Now consider the image of gt on all of A2 It is AZH 0 y lg 7 0 That is not an open subset of A2 Hence 1 itself is not an open morphism Finally let s compute the irreducible components of the bres of j certainly for z E U the bres are just points So consider z E 142 H U There is only one x to consider z 00 Then gt 100 024 W 6 k which is one dimensional Thus the bres of j are points in the open set U and outside of that they blow up For another example consider 1 A3 H A3 myz H zzyz Let X myz E A3 l y2 1z Let Y gtX I rst want to show that X and Y are irreducible closed of dimension 2 For X we have that kX kT1 T2 TglTZZ H T1 H 1 I claim that kX E kSlSZ The proof uses the fact that the polynomial rings are free the map T1 H S 1T2 H 52 T3 H 53 A to V ALGEBRAIC GROUPS 53 factors through Conversely we can map 52 H T2 53 H E to get the inverse This gives at once that X is irreducible and closed of dimension 2 Consider Y We have that Y mwz lwz z2m3 So it is closed and it is irreducible since its the image of an ir reducible under a morphism Let U zyz E X l x 7 0 which is open in X Then gtU zwz E Y 1 z 7 0 so is open in Y The morphism y U H gtU is an isomorphism the inverse maps z w z to zwz Thus 1 is birational hence dimY dimX 2 too By the theorem we expect for a closed subvariety Y of Y that the generic irreducible components of gt 1Y have the same dimension as Y Consider Y zyz E Yly mz22 17l7z I claim that Y is irreducible of dimension 1 To see this note W17 le17T27TslT22 7 T12 7 T1332 7 M3137 T1 H 1 I claim this is isomorphic to MS the polynomial ring in one variable One way we map S H T3 The inverse is induced by T1 H 52 H 1 T2 H 53 H S T3 H S One has to check that T22 maps to the same thing as T12 717 T13 ie 53 H S2 52 H 12 717 52 H 13 Hence Y A1 Finally consider the irreducible components of zjle lt equals xyz Xlzymz221z So either we have x 022 1y2 1 or z 22 H 1y 2 It follows that providing ChaTk 7 2 the irreducible components of gt 1Y are 0 1 H1 U 0 H1 1 U 22 H 122 12 E k The last of this is E A1 and is the generic case The rst two irreducible components are 0 dimensional so do not meet the well behaved77 open set U 42 Chevalley7s theorem Let G be an algebraic group H a closed sub group We want to give the set GH 9H 1 g E G of cosets the structure of an algebraic variety I am going to assume for simplicity that G is con nected Everything I am saying here can also be done for non connected groups G without very much more work Write g and E for their Lie algebras Recall that g TBG But we identi ed this via a map N TBG H LG with the Lie algebra of all left 54 ALGEBRAIC GROUPS invariant derivations of In particular elements ofg are endomorphisms of MG via N Moreover we have shown that hXengIltHgt 2101 see 320 Lemma 410 Let H be a closed subgroup of C There exists a nite di mensional subspace V of MG and a subspace W of V such that V is stable under all right translations 01 for x E G and Hx GlmeW hX ngWQW Proof Let I Let V be a nite dimensional subspace of MG sta ble under all pm s and containing a linearly independent set of generators f1 fT ofI this is possible by 24 Let W V I Now let us check that the lemma is satis ed by this data Clearly the 01 for x E H stabilize both V and I hence W Suppose x E G stabilizes W Then it stabilizes I since W generates I Therefore for each 1 E I we have that x pmfe 0 as pmf E I This shows that x E H Finally consider E X E g l XI Q I Let p G a GLV be the right regular representation of G on V Let dp g a gV be its differential Then Xf dpXf so all X leave V invariant Combine these two facts D Lemma 411 Let W lt V be nite dimensional uector spaces at dim W Let L be the one dimensional subspace AdW lt AdV Let j GLV a GLd V be the natural morphism 1 For x E GLV xW W if and only if L 2 For X E gV XW Q W if and only if d gtXL Q L Proof Recall gtxu1 A Aud x121 A Axud Also which you can check yourself d gtXu1 A Aud 2211 U1 A A Xvi A w Now you have to do some linear algebra For instance for i pick a basis U1 ud for W and extend to a basis U1 un for V We can to this in such a way that 1411 AAqu os a basis of xW for some l 2 0 Then U1 A Aud is a basis for L and 1411 A A MM is a basis of gtxL lfl gt 0 then these are linearly independent so that x does not stabilize L Hence we must have that l 0 Combining the lemmas gives at once Theorem 412 Let H be a closed subgroup of C There exists a represen tation j G gt GLV and a non zero u E V such that H x E G l E kuh X 6g l d gtXu 6 ho Corollary 413 Let H be a closed subgroup of G Go There exists a quasi projectiue uariety X that G acts transitiuely on and a point x E X such that ALGEBRAIC GROUPS 55 i G H ii The orbit map 1 G gt Xg gt gt gm is separable iii the bres ofiJ are the cosets 9H for g E G Proof Let V and u E V be as in the theorem Take X to be the G orbit Gltugt in lPV This is open in its closure hence it is a quasi projective variety By the Theorem H is the stablizer of z hence using transitivity of the action the bres of the orbit map are exactly the cosets of H Finally to get that the orbit map is separable note that the tangent space to lPV at can be canonically identi ed with The tangent space to X at is therefore a subspace of The differential ti5 maps X E g to So the kernel of the differential is the stabilizer of z ie E So dimkerdiJe dimh dimH dim G7 dimX Hence 1115 is onto by dimension and 11 is separable by 316 D 43 Quotients Continue with G a connected algebraic group H a closed subgroup The assumption G connected can easily be dropped but it makes things slightly simpler De ne a Cheualley quotient of G by H to be a variety X together with a surjective separable morphism 7r G a X such that the bres of 7139 are exactly the cosets zH of H in G We have just shown 413 that Chevalley quotients always exist On the other hand it is far from clear that they are unique even up to isomorphism De ne a categorical quotient of G by H to be a variety X together with a morphism 7r G a X that is constant on all cosets zH of H in G with the following universal property given any other variety Y and morphism j G a Y that is constant on all zH s there is a unique morphism j X a Y such that j j 0 7139 It is obvious that if a categorical quotient exists then it is unique up to canonical isomorphism But it is far from obvious that they should exist at all Goal prove that Chevalley quotients are categorical quotients In par ticular this will give us that categorical quotients exist and that Chevalley quotients are unique up to canonical isomorphism So we need to take a Chevalley quotient X 7139 and check that it has the right universal property Given a morphism j G a Y constant on cosets there is obviously a unique map of sets X a Y factoring 1 since the cosets of 7139 are exactly the cosets But it is almost impossible from this point of view to prove that this map is a morphism of varieties So we need to proceed rather differently Theorem 414 Chevalley quotients are categorical quotients Proof Step one Let us try to construct a categorical quotient not in the category of varieties but in the more general category of ringed spaces De ne GH to be the set gH of cosets of H in G Let 7r G a GH be the map 9 gt gt gH Give GH the structure of a topological space by declaring 56 ALGEBRAIC GROUPS U Q GH to be open if and only if 7r 1U is open This is the coarsest topology on GH so that 7139 is an open map Next de ne a sheaf O of functions on GH if U Q GH is open let 9U consist of all functions f on U such that f 0 7r 6 907r 1U Check the sheaf axioms Now let us check that GH7r as just de ned is a quotient of G by H in the category of ringed spaces So let 1 G a Y be a morphism of ringed spaces constant on cosets of H in G We get induced a unique map as sets zZ GH a YgH gt gt 19 We need to check it is a morphism of ringed spaces For continuity note that for V Q Y open U 1Z 1V is open as w 11Z 1V w lV is open in G Now we show that if maps 1 E OyV to an element of OaHU By de nition we just need to check that 7rzZf E 0011 1V But that is if which does since 1 was a morphism of ringed spaces to start with We are done Okay so now we have constructed GH7r a quotient of G by H in the category of ringed spaces It was quite easy but there is absolutely no guarantee that GH should be a uarletyl If it is then it is also a quotient in the category of varieties and we are done Step two Now let GH 7r be as in step one and let X1J be a Cheval ley quotient Using the universal property of GH we get a unique G equivariant morphism zZ GH a X such that1J 1Z07rie 11g We will prove that zZ is an isomorphism of ringed spaces 7 which will imply that GH is a variety and that X is a categorical quotient to complete the proof We will use Zariski s main theorem and also 46 which tells us For any variety Z the morphism 11 x 1 G x Z a X x Z is open We obviously have that zZ is a continuous bijection If U Q GH is open then 1ZU 117r 1U which is open by Hence 1 is a homeomorphism of topological spaces It remains to prove that if OXU a 01Z 1U is an isomorphism for all open U Q X We need to show for every regular function f on V w lU such that fgh fgVg E V h E H there is a regular function F on U such that F1Ju fuVu E V LetPgfglgEVQVXA1P 11X1PQUXA1SinceP A1 is closed in V x wx1VxA17PUxA17P is open in U x A1 by hence 1quot is closed in U X Al Let A 1quot a U be the morphism induced by the rst projection Clearly A is bijective Note 1 0 pm P a U factors as A o 1 x 1 The former is separable by the de nition of a Chevalley quotient so its differential is onto Hence dA must be onto too S0 A is separable Now we get that A is birational using the separability just proved and 45iii just like we did in the proof of 46iii Finally note that U is open in X and X is smooth So U is smooth Now Zariski s main theorem gives that A is an isomorphism Now look at F prg o A 1 U a A1 which maps u E U to u where 1112 u This is a morphism We are done D ALGEBRAIC GROUPS 57 We have now shown for any closed subgroup H of G the quotient 7r G 7 GH in the category of varieties exists Indeed it is given by any Chevalley quotient In particular GH is a quasi projective variety and the morphism 7139 is separable Also note that THGH g9 canonically Indeed the map 7139 G 7 GH is separable so its differential is onto and has kernel equal to E by dimension So it induces the above isomorphism of tangent spaces Example 415 1 Let G GLV 121 vn be a basis for V and H stabaltv1gt Then X Gltv1gt lPV is a Chevalley quotient of G by H We proved this above in greater generality So GH 71 Similarly let H stabaltv1 vdgt Then Gltv1 vdgt GdV is a Chevalley quotient ie the Grassmann variety GdV is isomor phic to the quotient variety GH Finally let H stabaltv1gt C C v1vkgt C C V ie the stabilizer in G of the standard ag So H is the group of all upper triangular matrices In this case GH V the ag variety because this is the G orbit of the standard ag and the orbit map is separable In all the above examples the quotient variety GH was a projec tive variety Remember by construction it is always at least quasi projective Instead let H stabav1 This time GH G121 which is A 7 0 an open subset of an af ne variety This is nei ther af ne nor projective Let G CL and H 0 g 6 CL l ng 1 assuming the characteristic is not 2 Let S be the set of all n x n symmetric matrices an irreducible af ne variety of dimension nn1 Let Sgtlt be the invertible matrices in S a principal open subset of S hence also af ne of dimension Let G act on S by gm ngg Then Gm On So the orbit map induces by the universal property of the quotient GH a bijec tive function GLnOn 7 Gm By linear algebra all non degenerate symmetric bilinear forms are equivalent to the standard one ie all invertible symmetric matrices can be written as ng for some in vertible matrix 9 Hence we have constructed a bijective morphism GLnOn 7 SX In particular dim CL 7 dim On dim SX ie dim On n2 7 1 71 Moreover to prove that GLnOn Sgtlt we just need to show that the orbit map 9 7 ng is separable Its differential is the map X 7 XT X The tangent space to Sgtlt at I is identi ed with S Clearly any symmetric matrix can be written as XT X except in A to V A DJ V A g V A U V 58 ALGEBRAIC GROUPS characteristic 2 Therefore the orbit map is separable Hence7 the quotient variety GLnOn E SX7 which is an a ne variety Exercise 416 1 Let H1 lt G17H2 lt G2 be closed subgroups of connected algebraic groups Prove that G1 x G2H1 x H2 E GlHl x G2H2 as varieties 2 Prove that GLgnSpgn is isomorphic to the af ne variety of all in vertible 271 x 271 skew symmetric matrices Deduce that dim Spgn n2n 1 ln characteristic 2 a skew symmetric matrix means a symmetric matrix with zeros on the diagonal 44 Normal subgroups Let G be an algebraic group A character of G means a homomorphism X G H Gm of algebraic groups We write XG for the set of all characters of G It has a natural structure of Abelian group X X 9 X9X 9 Example 417 1 Let G SLn7 or any other non abelian simple al gebraic group remember7 that means G has no closed connected normal subgroups other than 1 and itself Then if X G H Gm is a character7 kerX is a closed normal subgroup7 hence is G since G is non abelian Hence7 X is the trivial character 9 H 17 and XG O Let G Ga Consider HomGa7 Gm just as varieties for now It is E HomkTT 17 You get such by mapping T to any invertible element But the invertible elements in MT are just the non zero scalars Therefore the only morphisms of varieties Ga H Gm are the constant ones7 and of these only the one x H 1 is a group homomorphism So XGa O 3 Let G Gm Consider HomGm7 Gm as varieties7 ie HomkTT 1kTT 1 A to They are all the maps sending T to an invertible element The units in MT T4 are the CT for c E kX and n E Z Using in addition that it is a group homomorphism you get 0 1 This shows XGm E Z the map sending n E Z to the character x H z XG x H E XG XH Hence7 recalling Dn E G772 XDn E Z Explicitly7 the n tuple 17 n E Z corresponds to the character diagd1 dn H d1 11quot of D A g V Exercise 418 3 Prove that XG x H E XG XH 4 Prove that XGLn E Z Write down the isomorphism explicitly You should use the fact that SLn is simple Let G be a linear algebraic group and suppose j G H GLV is a representation of G on a nite dimensional vector space V7 ie j is a morphism of algebraic groups For X E XG7 let Vx v E V l gt9v X9UV9 6 G ALGEBRAIC GROUPS 59 This is called the x weight space of V Let me remind of linear independence of characters which is proved in the same way that you prove that different eigenspaces are linearly independent in linear algebra Z v ea X XG X6XG On the other hand it will usually not be the case that V is the sum of all the weight spaces There is one very important case when V is the sum of its weight spaces ie V EB VX X6XG This is when the group G is the group D For the elements of Dn are semisimple hence their images in a representation are semisimple automor phisms of V see 219 Moreover they commute So you can simultaneously diagonalize V with respect to the endomorphisms coming from D The group Dn E Gmm is the algebraic n dimensional torus Why Well perhaps this is convinc ing A torus is a product of 51 s where 51 z E CX 1 7 which is not an algebraic variety The algebraic analogue of 1 is simply all of Ex ie Gm So an algebraic torus is a product of Gm s Characters will play an important role later on For now let us prove Theorem 419 Let G be an algebraic group H a closed normal subgroup Then the variety GH is a ne and the operations 91H92H H glggH and 9H H gilH are morphisms of varieties Proof Let me rst show that 91H 92H H glggH is a morphism The map G x G H GH 9192 H gngH is a morphism that is constant on cosets of H x H Hence by the universal property of the quotients we get induced a unique morphism G x GH x H H GH Now use Exercise 1 above G x GH x H E GH x GH The proof that gH H g lH is a morphism easier So the main thing is to prove that the variety GH is a ne By Cheval ley s theorem we can nd a nite dimensional representation 1 G H GLV and v E V such that H stabGUgt E stabgvgt Let V X6XH VX be the sum of all H weight spaces in V Note 1 E V Moreover H stabilizes each VX and it is a normal subgroup of G hence G must permute the VX amongst themselves In particular V is invariant under G so we may assume replacing V by V that V V Now let W 1 EndVfVX Q VX for all X XH 60 ALGEBRAIC GROUPS Note W is a vector space De ne a morphism of algebraic groups 1 a a GLW7 gf gtltggtf gtltggt 1 6 mm Let us compute kerw if 19 ld then gtg commutes with all f E W hence by Schur s lemma f s gtg acts as scalars on each VX Hence 9 stablizes 12gt so 9 E H So by the universal property of quotients 11 induces a morphism GH a GLW The image of is a closed 7 hence af ne 7 subgroup of GLW It just remains to prove that is an isomorphism onto its image To do this we just need to check that 1115 is onto or equivalently by dimensions that ker dwe Q E Suppose X E kerdzJe Then d115Xf d gt5Xf 7 fd gt5X 0 so d gt5X commutes with all f E W This shows that d gt5 X acts as a scalar on all VX s in particular it stabilizes 12gt hence X E E D Now you can prove things like the First isomorphism theorem77 for alge braic groups Theorem 420 Suppose that j G 7gt H is a separable mmphism of al gebraic groups Le d 5 239s onto Let N ker gt Then j induces an isommphism GN H Note in characteristic 0 separability is equivalent to 1 being onto But in characteristic p you de nitely need the stronger assumption about the differentials Here is an important example Let G H CL Suppose k E Let 1 G a H be the Frobe mus mmphism given by raising matrix entries to the pth power This is a morphism of algebraic groups Moreover it is an isomorphism as abstract groups But the differential d gt5 is the ZERO MAP So 1 is de nitely not an isomorphism of algebraic groups CHAPTER 2 Chevalley groups 1 The main construction Now we ll assume Vk k 8 VZ is the reduction modulo p of some faithful nite dimensional g module via some choice of admissible lattice We wish to study automorphisms of Vk of the form 00 it 0405 exp 25 gtnf for t E k and Oz 6 I Of course the right hand side doesn t make sense until we explain how to interpret it One way is to replace with the element man 1 of Uk which does make sense as an endomorphism of Uk Then 0405 makes perfect sense because all but nitely many 3527 act as zero since Vk is the direct sum of its weight spaces Another way is to consider rst exp Tza as a map from VZ to ZT Z VZ T an indeterminate That makes sense Hence 1 X exp Tza is a map from Vk to k 8 ZT Z VZ Finally composing this map with the evaluation of T at t E k we get a map exp 25 from Vk to Vk Let X0 be the root group mat lt E Clearly matmas 0405 s so X0 is certainly a quotient of the additive group of the eld we ll see soon that actually X E k Let G be the subgroup of GLVk generated by all XDs for 04 E I We call it the Chevalley group associated to g V k and the choice VZ of admissible lattice We will show in a while that G actually only depends on g k and the subgroup A of the weight lattice P generated by HV Recall A for sure includes the root lattice Q That is helpful to know when thinking about examples but remember we didn t prove it yet Now I want to give some trivial but helpful examples taking g 52C Con sider rst V to be the natural 2 dimensional module and VZ 2121 63 2122 Let 04 be the positive root as usual Then 25125 izialttgt 2 and its an easy exercise which I ll do in class to see that these generate the group SL2k This is the case when A P lnstead take V to be the adjoint representation g itself and VZ 2569271692 In this basis we have that 1 722 7252 1 0 0 xat 0 1 t mat it 1 0 0 0 1 7252 2t 1 12 2 CHEVALLEY GROUPS Remembering that Vk 52k these are the same matrices as the linear maps 52k a 52k de ned by conjugating by the 2 x 2 matrices above This is the group PSL2k which by de nition is the image of SL2k under its representation by conjugation on 2 x 2 matrices observe the scalar matrices act as This is the case A Q In fact these are the only two Chevalley groups you can get by starting with g 52C because in this case PQ 22 In general when A P we call G the simply connected or unz39uersal Chevalley group of type g over the eld k When A Q which is always the case eg if V is the adjoint representation then G is the adjoint Chevalley group over k Here is an approximate table giving the possible Chevalley groups you can get by varying g and A P A P intermediate A A Dn n even 22 69 lg n odd 47 Here are some comments about this approximate table First when A P the group G is the simply connected one when A Q the group G is the adjoint one Often you can distinguish these by looking at the center ZG 7 often but not always if the characteristic p of k is too small this is isomorphic to AQ For instance for SLnk when p f n the center of SLnk is the group Inn of nth roots of unity which is cyclic of order n For each subgroup of that you get a central normal subgroup and can take the quotient to get one of the intermediate Chevalley groups between SLnk and PSLnk If you factor out the whole center you get PSLnk Second when k is algebraically closed the table is not approximate it is correct The groups SOnk and Sp2nk in these cases you should know the isometries of a non degenerate symmetric resp skew symmetric bilinear form Probably you ve never seen the right de nition of SOnk when k has characteristic 2 you need both a bilinear form and a quadratic form to get it right The groups PSOnk and PSp2nk are these modulo their centers usually i1 Note when n is odd that the matrix 71 is not of determinant 1 which is why 02n1k has no center there is no P502n1k in the table in this case The Spin groups the simply connected Chevalley groups are a little harder to de ne 7 of course we get them from our construction by taking V to be a spin representation the last fundamental dominant weight Its good enough to know for our purposes that SOnk is usually a double cover of the SOnk Now for the reason the table is approximate Unfortunately it is wrong in the Bn and Dn rows when k is not algebraically closed In that case you need to replace the groups I ve written with their commutator subgroups which might in 2 CHEVALLEY S COMMUTATOR FORMULA 13 these cases be a little bit smaller depending on the eld For example where I ve written 02n1l I should really have written an1k the commutator subgroup of the special orthogonal group of isometries of a 2n 1 dimensional vector space equipped with a non degenerate quadratic form of Witt index n Where I ve written SOgnk I should really have written the commutator subgroup of the special orthogonal group of isometries of a 2n dimensional vector space equipped with a non degenerate quadratic form of Witt index n These are things that we are not going to study here and which are not im portant for what we re going to do so I don t feel too uncomfortable about being a bit crappy in this explanation lts actually rather a technical place My advice ask James In class I will do a couple more examples writing down explicitly the matrices that give some of the root groups for B2 and 02 The point is the Chevalley construction gives some completely concrete matrix groups Its just that apart from some trivial cases the matrices are too big for this to be terribly helpful 2 Chevalley s commutator formula We ll need now to use the following lemma that you ve seen before more than once LEMMA 21 Suppose that A and B are elements of an associatiue algebra let lA be left multiplication by A 734 be right multiplication by A and adA be the map lAirA all are endomorphisms of the giuen associatiue algebra Suppose exp ad A explA exp TA and epr all make sense Then exp ABexp7A exp ad AB PROOF We ve seen this before since lA and TA commute exp adA explA exp7rA lexp ArexpA Now we come to the rst important theorem about Chevalley groups THEOREM 22 Let 043 be roots with 04 B 7 0 Then in the ring of formal power series UZHt u we have the identity exp tza exp uz H exp cijt ujmj where AB ABAilBil the product on the right is ouer all roots ia for ij gt 0 arranged in some xed order and the CM s are integers depending on 043 and the chosen ordering but not on t or u Moreouer cm is the same as the structure constant mmz ir Dag from our original choice of C heualley basis PROOF Sketch ln Ugtu set ftu exp txmexp uz Hexp7cijt u7zm g The product on the right is taken over ij gt 0 such that ia j E I and in the reverse of the xed order The coef cients CM are some complex numbers which 14 2 CHEVALLEY GROUPS are still to be determined In fact l ll treat them like indeterminates to start with then we ll see how to de ne them so that t7 u 17 which is of course what we need to do in order to prove the theorem Note that t exp tma tma exp tza So by the product rule we get that d tam u 7 mm ugt exp tza exp uz 7tmaexp7tza exp7uz H exp7cijtiujmj Zexp tza exp uz H exp7cijtiujmj x M mj gtkal 7ckatkulm1mH x H exp7cijtiujmmj i04j lwtl Now we bring the terms 7tza and 7ckatkulmkam to the front using relations like exp uz 7mg exp ad uz 7tzaexp uz and expadum 7tma 7tza 7 n ytuma a 7 where M9 ma n yza a The result is an expression of the form Aftu with A E Elli Ull Now7 think how you can get ck as a coef cient not inside an exp when you do this You get it from the term but otherwise it only comes when terms to the right of are pulled passed exp7ckltkulzmm These produce coef cients involving ck but of degree gt k l in t and u Thus7 A Z kaJ pkztkulkaw my where pm is a polynomial in the cig s for which i j lt k l At last we can now inductively determine the values of CM 6 Cusing the lexico graphic ordering on pairs i7 j7 to ensure that A 0 Then t t7 u 0 implies that t7 u 07 u 1 and we re almost done It remains to show that the cig s are integers and to compute cm The idea for that is to compute the coef cient of tluj in the commutator formula obtained so far which for sure belongs to UZHt7 It looks like 7cigmmj plus terms coming from exponentials of multiples of zkm with k l lt i j Now you use induction to see that cigzm lies in UZ7 hence CM 6 Z by the basis theorem for U1 COROLLARY 23 Replacing exp txa etc in the theorem with mat then the resulting equation holds in the C heualley group G For example7 from this formula in SL3I 7 taking 04 61 7 62 and B 62 7 63 you get that mt mu 7 mum which you can easily check directly Of course its more complicated for B2 and G2 3 THE BASIC RELATIONS 15 We call a set R of roots closed if 043 6 R 04 B E I implies 04 B E R For example R ltIgt R IJr 04 and R oz 6 ltIgtl hta 2 r are all closed sets of roots We call a subset I of a closed set R an ideal if 04 E I B E R implies 04 B E I The above examples are all ideals in ltIgt LEMMA 24 Let I be an ideal in a closed set R Assume R O 7R G Let XR and XI be the groups generated by all Xa for 04 E R and 04 E I respectively Then X El XR PROOF lmmediate from commutator formula D LEMMA 25 Let R be a closed set of roots such that R O 7R G Then euery element of XR can be written uniquely as HWER ma a for ta 6 k where the product is taken in any xed order PROOF l ll prove this just under the assumption the xed order is compatible with height ie hta lt ht implies 04 lt B The general case can be reduced to this case with a little more work Let 04 be the rst element of B Let I R 04 Its an ideal in B So XR XaXI By induction on R every element of X1 is the product of the z t for B E I with uniqueness of expression Now suppose y 6 X3 written as y zat Hz t Since V was faithful we can nd a weight vector u E Vk of weight such that may 7 0 Then yu u tmau z where u is of weight may is of weight 04 and z is a sum of terms from other weight spaces Hence t is uniquely determined by y The remaining t s are uniquely determined by induction applied to ma7ty Hm t 6 X1 D This lemma has lots of consequences 1 The mapt gt gt zat is an isomorphism from k onto Xa Proof Apply lemma with R a 2 Let U Xqgt Then U Huey Xa with uniqueness of expression where the product is taken in any xed order 3 The subgroup U of G consists of upper uni triangular matrices relative to an appropriately chosen basis of Vk ln particlar every entry of U is a unipotent matrix Proof Pick a basis of weight vectors ordered in some way re ning the dominance ordering 4 For i 2 1 let Ul be the group generated by all Xa for hta 2 i Then Ul is normal in U and U U Q U141 hence U is a nilpotent group 5 If IJr BUS with R and S closed then U XRXS and XR OX5 You should think what these things are saying when G SLnk 3 The basic relations So far we ve de ned elements zat and proved Chevalley s commutator formula explaining how to commute zat with z u when 04 B 7 0 These were all unipotent matrices ie all eigenvalues were 1 Now we introduce some semisimple elements of G ie diagonalizable matrices too For 04 E I and t E kX de ne wt zalttgtzialter1gtzalta 16 2 CHEVALLEY GROUPS and hat watwa1 1 Sometimes I ll write simply wa for wa1 For example if G SL2k and 04 is the positive root then wat lt 731 5 hat lt5 31 1711 refer to these as the monomial elements77 and diagonal elements77 respectively LEMMA 31 For any Toots 043 6 I and t E kx a watz wat 1 ct7lt gtagtm5amgt where c 4043 i1 independent of tk and V and C04 C04 Here sa is the reflection in the hypeiplane Ha b For 1 E Vk of weight a there exists 1 E Vk of weight sum independent 0ft such that watv t M00115 c hat acts on the a weight space of Vk by multiplication by tW D PROOF l ll prove this assuming the ground eld is C In fact everything then follows at the level of power series in UZHtH or as endomorphisms of VZHtH so you get it automatically over an arbitrary eld too We rst show that wDttht 1 sah for each h E E Note l m viewing the left hand side just as some matrices ie endomorphisms of V and the right hand side is the natural action of sa 6 W on b By linearity it suf ces to check this just when h ha because E Chat 63 ker04 and wat commutes with all elements of ker04 But then zat hence wat only involves elements of the three dimensional algebra mungwhat so we can check the thing we re after just by calculations in this algebra Take the usual representation of 5b then we know exactly what 2 x 2 matrices everything is and its easy Now we prove From the de nitions of malt and wat it follows that if U watv UH Z tivi i39eZ for 124 of weight a i04 But acting with h E E and using the previous paragraph hp saphv so 1 lies in the saii weight space Hence the only non zero U4 is when i 7lta04 Now apply b to the adjoint representation using Lemma 21 to get that watm wat 1 ct lt gt gtz5a where c E C is independent oft and V But taking t 1 the left hand side lies in UZ so actually 0 E Z actually 0 i1 since wa1 is invertible Finally Mam wwh w 1 waz w1waz w1 then Ca meat imam so Ca 5 Ca is This praoves a To get c note that wDt 1 wa7t so hat wa7t 1wa71 By b wa7tv it ltt gt gtv and wa71v 71 lt gtD gtv Hence wa7t 1wa71v WWW D COROLLARY 32 For 043 6 lt1 andt e W 3 THE BASIC RELATIONS 17 a won119091031 hmm l b wam tw1 xsa ct with c as in Lemma 31a c halttgtz ltugthalttgt4 WW PROOF For a apply both sides to a vector v E Vk of weight 1 and use the lemma For b we know that waz wgl cz now exponentiate For c we know by c of the lemma applied to the adj oint representation that hatm hat 1 tlt gtagtz Now exponentiate D We have now established all of the following relations R1 zatmau mat I I R2 0405 Hmm Mcmtlul if 04 B 7 0 where the cM s are as in Chevalley s commutator formula wat matma7t 1xat hat watwa1 1 wa wa1 wan119091031 71941305 R7 wam tw1 msg30325 with c as in Lemma 31 R8 halttgtz ltugthalttgt4 tlt gtu Observe all the relations are independent of V So all results proved using R17 R8 will also be independent of V Let H be the subgroup of G generated by all hat s This is an abelian group since all hat s act diagonally on Vk Indeed it is a subgroup of the group of diagonal matrices with respect to any weight basis of Vk Let N be the subgroup of G generated by all wat s Recall also that U is the subgroup generated by all mat s for 04 E ltIgt Let B be the group generated by H and U Keep in mind that if G SLnk then H is the diagonal matrices of determinant 1 U is upper unitriangular matrices B is all upper triangular matrices of determinant 1 and N is the monomial matrices matrices of determinant 1 with exactly one non zero entry in every row and column LEMMA 33 U is normal in B B UH and U O H Hence B is the semidirect product of U by H PROOF The fact that U is normal in B follows by R8 Relative to a suitable basis for Vk any element of U O H is both diagonal and unipotent hence its 1 D LEMMA 34 For a 7 3 X0 7 X PROOF If both 043 are positive roots this follows from Lemma 25 If one is positive and one is negative it follows because we know V has a basis with respect to which X0 is upper unitriangular and X is lower unitriangular For the next lemma we need to use a little information about Weyl groups W is generated by the elements 8a 04 E 1 subject only to the relations 3 1 and 3039504 35am for all 043 6 1 I m not going to prove that here it has nothing to do with Chevalley groups 18 2 CHEVALLEY GROUPS LEMMA 35 H is normal in N and there exists an isommphism j W gt NH such that gtsa Hwat for all Toots oz PROOF By R6 conjugation by wa preserves H and by R4 and R5 wat hatwa Hence H is normal in N Since Hwat Hwa the coset Hwat is independent oft Since wa1wa71 this implies that Hwat2 1 Also waw wgl wam 1z 71z 1w1 which equals by R7 msa cm5a icmsa c wsa c So HwaHw Hwa stam This veri es the relations of W hold so there exists a surjective homomorphism j W a NH Finally suppose w E W lies in ker gt Write w smsaw a product of re ections Then applying 1 we get that 10011 wan1 E H Hence it normalizes each Xa but also it conjugates X0 to Xwa Hence X0 me for all roots 4 By Lemma 34 this implies that wa oz for all a This implies w 1 because it acts faithfully on its re ection representation D 4 The Bruhat decomposition Recall the groups XaGBUHN and W Also we have an isomorphism W E NH l ll identify W with NH via this isomorphism From now on for each w E W l ll x a choice of representative to E N for the coset w E NH We may do this so that 3 0 wa for each 04 E I The goal now is to prove THEOREM 41 Bruhat Chevalley G is the disjoint union of the double cosets BwB for w E W G 7 UwewaB Before I prove that let us discuss the example G SLnk a special case ofthe above theorem Take any 9 E SLnk Left multiplication by a matrix in U lets you add multiples of later rows to earlier rows Right multiplication by a matrix in U lets you add multiples of earlier columns to later columns By repeatedly applying such operations you can transform 9 to an element n E N ie to h w for some h E H and w E W This proves that G UweW BwB We ve at least started to prove the Bruhat decomposition in this case it amounts to Gaussian elimination It is an amusing exercise to nish of this elementary proof So you see 7 like everything else in this subject 7 we are generalizing some classical fact about n x n matrices to an arbitrary representation of an arbitrary semisimple Lie algebra Now let us start to prove the theorem It will take a while LEMMA 42 fa is a simple Toot then B U BwaB is a group PROOF Let S BU BwaB Since B is a group and w1 lies in the same coset of H as wa since its image in W is an involution S is closed under inversion Since S2 Q BB BBwaB U BwaBB U BwaBBwaB Q S O BwanaB we just need to show that wana Q S We rst show X70 Q S If 1 344 y E X70 there exists t E kX such that y xat mat 1wa7t 1mat 1 E BwaB 4 THE BRUHAT DECOMPOSITION 19 Now wanD meHw1 waXaX aw1H XDXltIgtQH since sa leaves the set IJr a invariant This is contained in SB S We re done D LEMMA 43 Ifw E W and 04 is a simple root then a if wa E ltIgt ie if st lw1 then BwBBwaB waaB b in any case BwBBwaB Q waaB U BwB PROOF a Note that anw 1 Q B and wa normalizes Xqgta So BwBBwaB BanXqgtawaB Banw lwwaw1X awaB waaB b If wa E IJr then we re done by a If wa 6 IV set w wsa so uoz E ltIgt By a and the previous lemma BwBBwaB Bw waBBwaB Bw BBwaBBwaB g BuBB u BwaB Bw B u BwwaB D COROLLARY 44 If w E W is written as w 3139 81 reduced expression then BwB BwilBBwi2B BwilB where wi wa for short LEMMA 45 G is generated by mea l 04 E A PROOF This follows since the simple re ections generate W every root is W conjugate to a simple root and waX wgl XSQW Now we can prove the theorem First we show G UweW BwB By the previous lemma the right hand side contains a set of generators for G By the lemma before that the product of two double cosets is another hence the right hand side is closed under multiplication Clearly its closed under inverses So its a group it must be all of G Now suppose BwB BioB for ww E W We show by induction on w that w w For the base case if w 0 then w 1 so it says that w E B Hence u Bu 1 B This implies that w ltIgt ltIgt hence w A A Hence u 1 since W acts simply transitively on bases for I Now say w gt 0 Pick 04 E A so wsa is shorter than w Then Lima 6 Bw BBwaB Q BioB U Bw waB BwB U Bw waB lnduction hypothesis now gives either that wwa w which is a contradiction or that wwa w wa hence The following theorem gives a more precise version of the Bruhat decomposi tion giving a canonical form for elements of G THEOREM 46 For w E W let Uw Uqgtmw71qgti So Uw is the product of all root groups corresponding to positive roots sent to negative roots by w Then any element 9 of G can be written as g uhwu foruniqueu E Uh Hw 6W andu EUw 20 2 CHEVALLEY GROUPS PROOF Take 9 E G By Bruhat decomposition g bwb for unique w E W and some I b E B But B UH HU and w normalizes H Hence h uhwu for u E U h E H and u E U Now U Ui UUw where UL U w71 the product of root groups corresponding to positive roots left positive by w So u Lll ng for ul 6 U and U2 6 Uw and you can commute ul past w to get something still lying in U This shows that g uhwu for u E U h E U and u E Uw Now for uniqueness supposed that ulhlwu l ughgwu z Then hgluglulhl wu 2u 1 1w 1 The left hand side is an upper triangular matrix for a suitable weight basis and the right hand side is strictly lower triangular Hence both sides are 1 This implies u1 Ll2 hl h2 Ll1 D 5 Homomorphisms between Chevalley groups Now we complete our analysis of Chevalley groups by getting the desired in dependence of G on V VZ Recall A is the subgroup of P generated by the weights ofVandQQAQP LEMMA 51 Let G be a group generated by elements 04 E It E k such that the relations R17R8 all hold De ne subgroups U and H as we did for G i Euery element of U can be written in the form Ha6 ma a for ta 6 h product taken in some xed order ii For each w E W x an expression w 3043 as a product of reflections and set w waw where w Then euery element of G can be written as uhwu for u E U h E H w E W and 1 E Uiu PROOF Same as the proof for G 7 this part only depended on the relations D LEMMA 52 Let GG be as above and suppose that j G a G is a group homomorphism such that mat for all at Then uniqueness of expres sion holds in and ii of the previous lemma Moreouer ker gt Q ZG Q H In particular ZG Q H PROOF The uniqueness of expression in G follows easily from the uniqueness already established for G This was something proved for G that depended on more than just the relations R17R8 7 we used the explicit representation of elements of G as matrices which is missing for G For the last sentence let m u h w u lie in ker gt Then 1 gtu gth w gtu From this we get that w 1 so w 1 gtu 1 and gtu 1 so u 1 1 So z h HhMta for some ta e W Now m zbuz 1 tg mu by R8 Applying 1 we get that HQ 2535 1 Hence m commutes with so its central Moreover we have shown that ker gt Q H It just remains to show that ZG Q H If x uhwu E ZG and w 31 1 then there exists oz 6 I such that wa E I Then zza1 ma1z contradicts the uniqueness in the Bruhat decomposition Hence w 1 so z uh Let we be the element sending I to I Then z womgl is both upper and lower triangular hence diagonal Hence z h lies in H 5 HOMOMORPHISMS BETWEEN CHEVALLEY GROUPS 21 COROLLARY 53 The relations R17R8 together with all the relations in H between the hDt s forms a set of de ning relations for G PROOF If the relations in H are imposed on H then the map f in the lemma becomes an isomorphism D COROLLARY 54 If G is constructed as G from g and h but using a di erent V in place of V then there exists a homomorphism from G onto G such that x0t if and only if there exists a homomorphism t9 H gt H such that 0hDt hDt for all 04 and t PROOF Clearly if 1 exists then 9 exists Conversely assume 9 exists Then the relations in H form a subset of the relations in H So we get that 1 exists by the previous corollary So we just need to understand the group H better Recall that hat acts on the a weight space of Vk as the scalar twp LEMMA 55 a For each 04 hDt is multiplicatiue as a function oft b H is an abelian group generated by the hit s where hit hat c 11 hl 1 if and only if 11 Elba 1 for all u e A d The center of G is l l TIMil l HEW 1fOT all 5 E Q i1 i1 In particular ZG is nite PROOF Parts a and c are pretty clear note c is true for all a E A if and only if it is true for all a E HV For b take any 04 E I and write ha ELI nihi Then I claim ha t H Illtm To see this it is enough to check both sides act on the a weight space of V by the same scalar The left hand side acts by twp gt thaw tzmt wi Htmt a Yep Finally for d thlti commutes with x u if and only if Hf1t mgt 1 by R8 I ll leave you to puzzle over why this is a nite group What system of equations have you got to solve COROLLARY 56 If A P then euery h E H can be written uniquely as h Hf1 hl for l e W IfA Q then 2a 1 Finally we can complete the main discussion about Chevalley groups THEOREM 57 Let G and G be C heualley groups constructed from g k and choices V and V of faithful nite dimensional representations with weight groups A and A respectively If A 2 A then there exists a homomorphism f G a G such that gt gt x0t and kerzf Q ZG In particular ifA P then G couers all other C heualley groups de ned from g and k its the uniuersal C heualley group and ifA Q then G is couered by all other C heualley groups de ned from g and k its the adjoint C heualley group 22 2 CHEVALLEY GROUPS PROOF Just note that if A Q A then Hf1taigt 1 for all u E A implies Ufa iffy 1 for all u E A In other words7 by the previous lemma7 all the relations in H are satis ed in H too7 so there is a homomorphism t9 H as H such that ht gt gt Illt for each i This also maps h at to hat for each 04 So there s a j by Corollary 54 D COROLLARY 58 For each 04 E I there is a group homomomhism 10 SL2k a G 5 New 2 New PROOF Apply the theorem to the three dimensional subalgebra of g spanned by ma hut7 ya but keeping the representation V as before You get that the subgroup of SD of G is a quotient of the universal Chevalley group of type Al7 ie SL2k D under which for each 25 E k That concludes the rst look at Chevalley groups In the next chapter I m going to talk about some fun things that should at least make sense after this introduction Here are a couple of exercises you could try 1 If G is of universal type7 then the homomorphism 10 in the last corollary is injective 2 ZG E HomAQkx For example if k C then ZG E AQ 3 The normalizer of U in G is B Assuming gt 37 the normalizer of H in G is N ALGEBRAIC GROUPS 75 63 Borel subgroups Let me state some theorems about maximal tori in connected solvable groups These are proved by induction though it is often quite dif cult Theorem 614 Let G be a connected soluable group Then the set GM of all unipotent elements ofG is a closed connected normal subgroup of G All the maximal tori of G are conjugate and if T is any one of them then G is the semi direct product of T acting on GM As a consequence you show that in an arbitrary connected group G all its maximal tori are conjugate lndeed any maximal torus T of G is contained in a Borel subgroup B If T is another maximal torus contained in a Borel B we can conjugate B to B to assume that T is also contained in B But then T and T are conjugate in B by the theorem Let G be a connected algebraic group De nition 615 A Borel subgroup B of G is a maximal closed connected solvable subgroup of G Example 616 If G is a Chevalley group the subgroup B TU is a Borel subgroup of G Any conjugate of B in G will give another such subgroup ii If G GLn the subgroup B of all upper triangular matrices is a maximal closed connected solvable subgroup by the Lie Kolchin theorem Hence it is a Borel subgroup Any conjugate of this will give another Note in this case that the quotient variety GB is the ag variety so it is a projective variety in particular Theorem 617 For any connected algebraic group G let B be a Borel subgroup Then GB is a projectiue uariety and all other Borel subgroups of G are conjugate to B Proof Let S be a maximal closed connected solvable subgroup of G of max imal dimension Clearly S is a Borel subgroup Apply Chevalley s theorem to construct a representation p G gt GLV and a 1 space L C V such that S stabGL By the LieKolchin theorem S xesa agin VL HenceS xesa agf L f1 C f2 CC fn V in the ag variety Recall this is a projective variety hence it is complete By the choice of L S stabaf Hence the orbit map induces a bijective morphism GS a G C Take any other ag 1quot C Then stabaf is upper triangular in some basis hence it is solvable hence its dimension is dim S This shows that dim Gf dim G7 dim stabaf 2 dim Gf Therefore G is a G orbit in V of minimal dimension hence it is closed This shows that G is also complete so GS is complete too Now GS is complete and its quasi projective hence it is projective Finally let B be another Borel subgroup of G Then B acts on GS so by Borel s xed point theorem B has a xed point gS on GS Therefore 76 ALGEBRAIC GROUPS BgS gS7 ie g lBg Q S By maximality7 we get that g lBg S and this completes the proof D De nition 618 A parabolic subgroup P of G is any closed subgroup of G such that GP is a projective equivalently7 complete variety Theorem 619 Let P be a closed subgroup of G Then P is parabolic if and only if it contains a Borel subgroup In particular P is a Borel subgroup if and only ifP is connected soluable and GP is a projectiue uariety Proof Suppose GP is projective Let B be a Borel subgroup of G It acts on GP with a xed point7 say BgP gP This implies that g lBg Q P7 ie P contains a Borel subgroup Conversely7 suppose P contains a Borel subgroup B The map GB a GP is surjective and GB is complete Hence7 GP is complete too But it is quasi projective too7 so in fact GP is projective D Example 620 Let G CL The subgroups of G containing B upper triangular matrices are exactly the step subgroups7 one for each way of writing n n1 us as a sum of positive integers nl7 7719 There are 2 1 such subgroups ii Let G be an arbitrary Chevalley group Let S be a subset of the simple roots l l7 so there are 2 possibilities for S Let P be the subgroup of G generated by B and all sat for 04 E S Equivalently7 P is the subgroup generated by T and the XDs for 04 E IJr U 7S Then7 P contains B so it is a parabolic subgroup by the theorem7 and GP is a projective variety ln fact7 these P s give allthe parabolic subgroups of G containing the xed choice of Borel subgroup B By the theorem7 all other parabolic subgroups of G are conjugate to one of these There are exactly 2 different conjugacy classes of parabolic subgroup in the Chevalley group G 64 The Bruhat order Let me talk about a very important example7 which gives you a clue of the important role algebraic geometry plays in Lie theory Let G be a Chevalley group once more7 with all the subgroups U7 T7 X07 B7 N7 W NT Recall also that W is generated by the simple re ections 8a l 04 E H For any w 6 W7 we can write w as a product of simple re ections The length of iv was the length of a shortest such expression7 called a reduced expression for W Let me de ne a partial order on W as follows Take w7w E W Let w 31 be a reduced expression for w Declare that u w if and only if u silsij for some 1 i1 lt lt ij r ie if w is a subexpression of w Problem How do you prove this really is a partial order How do you even show that it is well de ned7 ie independent of the choice of the reduced expression of w Answer use algebraic geometry Let me explain By the Bruhat decomposition7 the B orbits on GB are parametrized by the Weyl group W7 ie the orbits are the BwBB s Now7 the closure of an orbit is a union of orbits7 the ones in the boundary being of strictly smaller dimension So there is obviously a partial ordering on ALGEBRAIC GROUPS 77 the orbits of B on GB de ned by 9 g 9 if and only if 9 Q a What l7m going to prove is that this is exactly the partial ordering not de ned in the previous paragraph In other words the ordering in the previous paragraph lS well de ned because there is a geometrically de ned partial ordering that amounts to the combinatorics there Let s proceed with some lemmas Lemma 621 Let G be an algebraic group X a G uariety H g G a closed subgroup and Y Q X a closed H stable subuariety of X If GH is a complete uariety ie ifH is a parabolic subgroup of G then GY is closed in X Proof Let A g z E G x X l g lz E Y which is closed in G x X Let 7r G x X a GH x X be the quotient map Recall this is an open map If g z E A then gh z E A for all h E H since H stabilizes Y Hence 7rA GHx X77rGx X7 A so 7rA is closed in GH x X Since GH is complete the projection prx7rA Q X is also closed This is exactly CY D Lemma 622 Any product of parabolic subgroups of G containing B is closed in G Proof Let P1 PT be parabolic subgroups of G containing B By induc tion P1 P771 is closed in G Consider P1PT1PT Note P1PT1B P1 P1 is closed and B stable Since PTB is complete we get by the lemma that P1 PPLPT is closed too D Theorem 623 Chevalley Let G be a C heualley group Fix w E W and a reduced expression w 31 for w as a product of simple reflections Then BwB U Bw B w where 10 runs ouer all subexpressions Sil n3 of 31 37 Proof Let w 51 s be the xed reduced expression for w Let Pi ltBSigt B U BSZ39B where the last equality comes from the work on Chevalley groups in the preVious chapter We show by induction on r that P1P UBuB w where the union is taken over all subexpressions u of the reduced expression 31 3 The case r 1 is already done Now suppose r gt 1 Then by induction P1P U Bw BB u 353 wl 78 ALGEBRAIC GROUPS where w runs over all subexpressions of 51 371 But that equals U Bw B w as required since Bw sB Q Bw BBSB Q B LUHB U Bw sB by the previ ous chapter By the preceeding lemma P1 PT is closed hence U BwB w union over all subexpressions w of 51 ST is closed So it certainly contains the closure BwB Finally we know dim BwB is equal to dim B the number of positive roots sent to negative roots by w So in fact we must have that U Bw B BwB w by dimension D Now since m is de ned intrinsically independent of any choice of re duced expression of w the relation w w iff w is a subexpression of some xed reduced expression for w is well de ned independent of the choice Moreover it is a partial ordering on W called the Bruhat ordering For w E W the Schubert variety Xw BwB B is a closed subvariety of the ag variety GB Note Xw is no longer an orbit of an algebraic group so it needn t be smooth Schubert varieties are extremely interesting projective varieties with many wonderful properties The Schubert variety XwO is the ag variety itself the Schubert variety X1 is a point We have shown above that in general the lattice of containments of Schubert varieties is isomorphic to the Bruhat order on W ALGEBRAIC GROUPS 17 2 BASIC NOTIONS OF ALGEBRAIC GROUPS Now we are ready to introduce algebraic groups and prove some of their basic properties 21 De nition and rst examples An algebraic group is an af ne va riety G equipped together with morphisms of varieties u G x G a G and i G a G that give the points of G the structure of a group ie u is multiplication and i is inverse satisfying the group axioms In other words7 an algebraic group is both a group and an af ne variety7 so that the group operations are morphisms with respect to the variety structure It is rea sonable to consider algebraic groups that are not necessarily af ne varieties so strictly speaking one should say af ne algebraic group77 for the thing I have just de ned Since we will only ever meet af ne algebraic groups I ll drop the word af ne A closed subgroup H of an algebraic group G is a subgroup that is closed in the Zariski topology Such subgroups are again algebraic groups in their own right A morphism f G a H of algebraic groups is a morphism of varieties that is a group homomorphism too The kernel of a morphism of algebraic groups is a closed7 normal subgroup of G But note we do not yet know that the image fG of a morphism is a closed subgroup of H so we had better not talk about images of morphisms as algebraic groups yet Example 21 1 The additiue group Ga is the group k7 7 ie the af ne variety A1 under addition 2 The multiplicatiue group Gm is the group 182 ie the principal open subset z E A1 l x 7 0 of A1 under multiplication 3 The group CL GLnk is the group of all n x n invertible matri ces over k Obviously7 the set of all n x n matrices can be identi ed with the af ne space Anz Then CL is the principal open subset de ned by the non vanishing of determinant which is a polynomial function in the matrix entries Hence CL is an af ne variety Since the formulae for matrix multiplication and inversion are polynomi als in the matrix entries and 1 det7 the group structure maps are morphisms of varieties 4 The group SLn SLnk is the closed subgroup of CL de ned by the zeros of the function det 71 5 The group Dn of invertible diagonal matrices is a closed subgroup of CL de ned by the zeros of which functions It is isomorphic to the direct product of n copies of Gm 6 The group Un of upper uni triangular matrices in CL is another closed subgroup 7 The group Tn of all upper triangular invertible matrices in CL is yet another It contains Un as a closed normal subgroup7 and Dn as a closed subgroup 18 ALGEBRAIC GROUPS 8 The orthogonal group On z 6 CL lmtz 1 is a closed subgroup of CL But we had better exclude characteristic 2 when discussing this example 9 The special orthogonal group SO On SLn is a normal subgroup of On of index 2 remember the characteristic is not 2 10 The symplectic group szn z E GL2 l mth J where 7 0 In who is another closed subgroup of GL2 Well7 that is a lot of examples Let us go back and look at Ga again lts coordinate ring MGQ is the polynomial ring The comorphisms f and i give us algebra homomorphisms m MT a MT 29 MT f km a MT You can work them out lifT T X 1 1 X T and iT 7T Also let 6 MT a k be the evaluation map at the identity element of Ga7 ie 6T O Then7 M e and i give the comultiplication7 counit and antipode making the algebra MT into a commutative Hopf algebra Big aside de nition of Hopf algebra if you ve never seen it before A coalgebm is a vector space A together with linear maps A A a A X A and e A a F7 called the comultiplication and counit or augmentation respectively7 such that the following diagrams commute A A A lt A A mm TA 1 A A T A k A A gt A k ml H ml lt2gt A Alt A gtA A A A The rst of these diagrams is the coassociatz39ue law If A and B are algebras7 A X B is also naturally an algebra7 with multi plication determined by the formula a1 X b1a2 X b2 a1a2 blbg for L 6 A7 1 E B A bialgebm A is a vector space A that is both an algebra7 with multiplication u and unit 2 and a coalgebra7 with comultiplication A and counit e such that A A a A X A and e A a k are algebra maps ALGEBRAIC GROUPS 19 Finally a Hopf algebra is a bialgebra A together with a linear map S A a A called the antipode such that the following diagrams commute ALAQQA AA A i ei l1 5 i ei i5 1 AA A ALAoA End of big aside In a similar way MGM is the localization of MT at T ie the algebra MT T ll of Laurent polynomials The Hopf algebra structure induced by the group structure of Gm satis es MT T Ti T T 1ET 1 Similarly the coordinate ring of any algebraic group is an a ne Hopf al gebra Indeed the categories of algebraic groups and af ne Hopf algebras are contravariantly equivalent just like the categories of af ne varieties and af ne algebras Exercise 22 8 Consider the algebraic group CL Let TM be the ij coordinate function ie the function on GL7 picking out the ij entry of a matrix Show that the coordinate ring MGLW is the localization of the polynomial algebra l 1 ij n at the function det detTj19jgn Write down explicit formulae for the effect of the comorphisms f and i on each TM Finally what is the counit that makes MGLW into a Hopf algebra 22 Linear algebraic groups A linear algebraic group is a closed sub group of some CL We wish to prove the following theorem which shows that GLn plays the role for algebraic groups that the symmetric group Sn plays in Cayley s theorem about nite groups Theorem 23 Euery a ne algebraic group is linear ie is isomorphic to a closed subgroup of CL To prove the theorem we need to nd a nite dimensional vector space and the place to look is inside the coordinate ring MG of G Given 9 E G there is a linear map Pg 3 lel H lel7 f H ng where pgfh fhg This gives us a group homomorphism p G gt GLkG a homomorphism just as abstract groups GLkG is not an algebraic group since MG is usually in nite dimensional We call p right translation of functions Similarly there is the left translation of functions G a GLkG where where Agfh fg 1h 20 ALGEBRAIC GROUPS Lemma 24 Given any nite dimensional subspace V of MG there is a nite dimensional subspace W of MG containing V that is invariant under all right translations 0m x E G Proof lt suf ces to prove this in the case V is one dimensional spanned by f 6 MG say Let W be the subspace of MG spanned by all pgf for all z E G We need to show that W is nite dimensional Write if 2 f X gt 6 MG X Let X be the nite dimen sional subspace of MG spanned by all flu Now consider z E G We have that pmfh lm pfhz 21fihgiz Hence pmf 21gimfi Hence pmf E X Hence W X and W is nite dimensional D Now to prove the theorem let G be an algebraic group Choose linearly independent generators f1 fn for the coordinate ring Applying the lemma we may assume adding nitely many more generators if necessary that the span E of the 1 is invariant under all right translations Now consider the restriction a GHGLEprmlE of p This is a group homomorphism and GLE is now an algebraic group since E is nite dimensional To see that 11 is a morphism of varieties write if 7 m7 n with the mj s linearly independent As in the proof of the lemma pm Z rimMm Since this lies in E for all z E G we see that each m lies in E So we may assume in fact that m fj for each j ie lift Z fj X 71M 7 Hence pm Z rimMfg which shows that the coordinates of the matrix pmlE E GLE with respect to the basis f1 fn are the n Hence 1 is a morphism of varieties Next we show that 11 is injective lndeed 1390 film fj5mjv SO 27 fjenij 6 then nmlm Lj SO for all The same equation shows that the nm generate MG hence since nm dual the comorphism if kGLE a MC is surjective Hence the image 11G is a closed subgroup of GLE We are done 23 First properties The results in this subsection are absolutely funda mental for everything else that follows Lemma 25 Let G be an algebraic group ALGEBRAIC GROUPS 21 i The identity elemente E G belongs to a unique irreducible component of G ii G0 is a closed normal subgroup of G of nite index and the irre ducible components of G are the cosets of G iii Any closed subgroup of G of nite index contains G0 Proof Let X and Y be two irreducible components containing e Then the closure of XY aX x Y is irreducible too since a is a morphism But W contains both X and Y so X W Y ii The argument in also shows that G0 is closed under multiplica tion It is also closed under the inverse map i since iG0 is an irreducible component containing e Hence G0 is a subgroup of G It is even normal since for g E G gGOg 1 is an irreducible component containing e Finally all cosets zGO of G0 are also irreducible components of G in particular G0 has nite index in G iii Finally let H be a closed subgroup of G of nite index Then H0 is a closed subgroup of nite index of G0 So H0 is both open and closed in G0 But G0 is irreducible hence it is connected so this means that H0 G0 D The lemma shows that for algebraic groups the irreducible components are disjoint hence coincide with the connected components of the topo logical space G Because of this people usually talk just about connected components of an algebraic group not irreducible components The normal subgroup G0 is referred to as the identity component of G Also note at this point that all components of G are isomorphic as varieties to G0 in particular they all have the same dimension Because of this it is reasonable to talk about dim G meaning dim G0 even if G is not connected For example G is nite if and only if dim G O dim GLn n2 Lemma 26 Let U and V be dense open subsets of G Then G UV Proof Let x E G Then zV l and U are dense open subsets Since they are dense they must both intersect G0 so that zV l O G0 and U O G0 are dense open subsets of G0 But G0 is irreducible hence zV l and U in fact intersect each other at some point u E U But then u 171 for some 1 E V hence z uu Lemma 27 Let H be a not necessarily closed subgroup of G i Its closure H is a subgroup of G ii If H contains a non empty open subset of H then H is closed Proof Let x E H Since multiplication by z is a homeomorphism pH E H This shows that HH Q H Now let x E H Then Hz Q H by the previous paragraph hence E Hz Q H Hence H is closed under multiplication Also Hf1 H 1 so we get that H is a subgroup ii Now suppose that H contains a non empty open subset U of H Then H is also open in H since H is a union of translates of U So we get from 26 that H HH H 22 ALGEBRAIC GROUPS Corollary 28 Let j G gt H be a morphism of algebraic groups Then its image is a closed subgroup of H Proof By 128 gtG contains a non empty open subset of its closure Hence it is closed by 27ii Finally we end with a very useful if nasty looking result Theorem 29 Let Xi gtii61 be a family of irreducible uarieties and mor phisms Xi gt G such that e E for all Let H be the smallest subgroup of G containing each Then i H is closed and connected ii H Ya set wise product for some a1an E I and 61En 6 i1 Proof WLOG each of the sets Yfl occur among the Note for each a a1 an E I Ya Y2 Yan is irreducible Hence 7a is irreducible too Obviously YaYb YWb and now arguing as in the proof of 27i you show that 71717 Q YWb Now choose the tuple a such that dimZ is maximal As e 6 Y2 we have for any b that 71 Q Z717 Q 30117 Equality holds by dimension hence 717 Q Z for every b and also 7a is closed under multiplication Now we have shown that Z is a group Since Y contains an open subset of E by 128 we get that Z Yul1 from 26 Therefore H 7a and we have proved and ii D Here are some applications i Assume that Gm61 is a family of closed connected subgroups of G Then the subgroup H generated by them all is closed and connected and moreover H Ga Gan for some a1 an E I The groups Spgn and SO in characteristic 7 2 are connected Incidentally 50 has index two in On hence it is the identity com ponent of On Proof This needs to know a little group theory about symplectic and orthogonal groups which is covered in Kantor s course on Classical groups or in this course but much later on For example in the case of Spgn this group is generated by so called transuection subgroups Let V be the natural 2n dimensional vector space that Spgn acts on with symplectic bilinear form preserved by the group A transvection subgroup is a subgroup consisting of all transformations of the form C V all v gt gt v tvaa for t E h0 a E V The resulting subgroup Ga amt t E h is a closed subgroup of G isomorphic to Ga hence is connected Since these generate all of Spgn which is a theorem in group theory only you get that Spgn is connected by ALGEBRAIC GROUPS 23 iii Let H and K be closed subgroups of G with H connected Then7 the commutator group H7 K generated by all commutators Mk hkh lk l with h 6 H7 k E K is closed and connected Proof Take the index set I in the theorem to be K and the maps 45k H A G to be the maps h gt gt zkz lk l for k 6 1K iv Recall the de nition of the derived series GGlt0gtGlt1gt of a group G CO GIN GU7 CW The group G is then called solvable if G e for some 2 In case G is a connected algebraic group7 each of the derived subgroups are closed7 connected normal subgroups of G Therefore either G Ga 7 and the derived series has stabilized 7 or dim Ga lt dim G Thus we see that for algebraic groups the derived series stabilizes after nitely many terms we will use this later on in particular in developing a nice theory of solvable algebraic groups Similar remarks apply to nilpotent algebraic groups replacing the derived series with the descending central series Exercise 210 9 Show that gtGO ltjgtGO7 for j a morphism of algebraic groups 10 Let G be the group Un of all upper uni triangular matrices Let G0 G and Cl G7 Gi thus de ning the descending central series of G Describe the group Gi for each i 2 0 Hence check directly ie without applying 29 that it is a closed7 connected subgroup of G What is its dimension Now prove that the group Tn of all upper triangular invertible matrices is a solvable algebraic group ALGEBRAIC GROUPS 9 15 Products Now let us work in the category of af ne varieties There is a general notion of product in any category for example a product X x Y of af ne varieties should be an af ne variety together with morphisms p1 X x Y a Xp2 X x Y a Y such that for any other af ne variety Z and maps ql Z a X qg Z gt Y there is a unique morphism r Z a X x Y such that qi p o r Lemma 114 Products exist in the category of a ne varieties Proof We have seen that there is a contravariant equivalence between the category of af ne varieties and the category of af ne algebras Therefore it suf ces to show that coproducts exist in the category of af ne algebras We already know coproducts exist in the category of all algebras the coproduct oanndB isA kB withthemapsAHA kBaHa 1 and B gt A 2 B b gt gt 1 X 1 Therefore we will be done if we can check Let A and B be a ne algebras Then A k B is an a ne algebra To prove this let ELI a I be a nilpotent element of A k B We may assume the b are linearly independent Let f A a k be a morphism Then ELI fab is nilpotent hence ai 0 for all i by the independence of Us This shows that a lies in every maximal ideal of A So Va contains every point of the corresponding algebraic set So Va contains all of the corresponding algebraic set So IVai ai 0 hence ai is 0 as A is reduced D Therefore it makes sense to write X x Y for af ne varieties X and Y It is again an af ne variety with kX x Y kX k Note as a set X x Y is just the Cartesian product and the projection maps 191192 in the de nition of product are just the obvious projections How would you prove this Exercise 115 1 Arguing like in the above lemma show that if A and B are integral domains so is A 2 B Deduce that the product of two irreducible af ne varieties is again irreducible 2 Show that C RC is not an integral domain Therefore the fact that k is algebraically closed is fundamental in the previous exercise 16 Varieties We are nearly ready to give the general de nition of an algebraic variety To start with de ne a preuarz39ety to be a quasi compact ringed space X OX such that every point of X has an open neighbourhood U with the propery that U 9le is an af ne variety Such a U is called an a ne open subset of X Note prevarieties are Noetherian topological spaces A morphism of prevarieties means the same as a morphism of ringed spaces Lemma 116 Products exist in the category of prevarieties Proof Here is the construction of the product X x Y of two prevarieties X and Y As a set X x Y is the Cartesian product of X and Y We need to de ne a topology and a sheaf of functions 10 ALGEBRAIC GROUPS Cover X Ui7Y Uy for af ne opens Ui j Then X x Y is covered by the Ul x These are af ne varieties Now declare that U Q Xx Y is open ifU Ul x is open in Ul x for alli7j Calla function 1 de ned in an open neighbourhood U of z E Ul x regular if its restriction to U Ui x is regular at z in the old sense for af ne varieties This gives the sheaf of functions Now nally call a prevariety X a uariety if the diagonal AX z7 z la 6 X is a closed subset of X x X For example7 any af ne variety is a variety if X is af ne with coordinate ring X7 the diagonal AX is the set of common zeros ofthe ideal I f 217 1 f lf 6 C kX X X7 hence it is closed This last condition 7 that AX is closed 7 is called the separation axiom It is some sort of substitute for the Hausdorff propert when working with the Zariski topology indeed a topological space X is Hausdorff if and only if AX is closed in X x X for the product topology Here are some of the consequences of the separation axiom Lemma 117 Let XY be a uarieties i If gt X gt Y is a morphism then its graph lm E X is closed in X X Y ii If on X gt Y are two morphisms which coincide on a dense subset ofX then gt iJ Proof For ii7 note that z E X l gtz is the inverse image of Ay under the continous map z gt gt Therefore it is closed7 and dense7 therefore is all of X Part is similar7 considering the continuous map X x Y a Y x Ymy gt gt Example 118 Suppose x and gz are polynomials which are equal for in nitely many values of z E k View them as morphisms A1 a A1 then they agree on a dense subset of k so they are equal everywhere Here is an example of a prevariety that is not a variety the af ne line with a point doubled It cannot possibly be a variety because the two maps A1 a X sending A1 to one of the af ne lines or the other are two morphisms that agree on A1 7 07 a dense subset7 but that are not equal iii In a similar way we can make P1 A1 U 00 into a prevariety In this case it is even a variety the cheapest way to prove this is to appeal to the lemma 121 below C V We have one more job to do in this generality we need to understand how to make certain subsets of algebraic varieties into algebraic varieties in their own right We can do this right away for open subsets if X is a variety and U is open7 we have constructed the ringed space U7 9U where 9U OXlU Now ALGEBRAIC GROUPS 11 cover X U2 U by af ne opens Ui Each U O U is open in U therefore can be written as a union U VM for principal open subsets Vi of U Principal open subsets of af ne varieties are af ne so the Vi are af ne Hence U has a nite af ne open cover so it is a prevariety Finally AU is closed as its AX O U x U and the topology on U x U is the subspace topology Exercise 119 3 Consider the open subset X A2 7 0 Give X the induced variety structure Prove that X is not af ne What we would really like is to make closed subsets of a variety into varieties too To start with let X 9X be a ringed space and Y C X an arbitrary closed subset Give Y the subspace topology and for U open in Y de ne 9leU to be all functions on U with the following property there exists an open covering U Q U U0 of U by opens in X and elements f0 6 9XUD for each 04 such that falUmUa flUmUa This makes Y 9le into a ringed space in its own right and generalizes the notion of the restriction of a sheaf of functions to an open subset we had before This is at least the sensible notion for closed subsets of af ne varieties which we already understand as varieties Lemma 120 Let X be an a ne variety with coordinate ring kX and Y VI be a closed subset Then 9le is again an a ne variety with coordinate ring MY kXVI which is of course what we expected Proof The main thing is to show that the restriction map kX 9XX a 9leY is surjective 7 this shows that 9leY MY and given this everything else is routine Let f E 9leY Arguing as in the proof of 110 we can nd ai fi 6 kX such that Y Q Df1U UDfn f on Dfi O Y and aifj ajfi on Y Now X 7 Y is open so we can represent it as Dh1 U U Dhm By the Nullstellensatz kX is generated by the f and his so we can write 161f1 Cnfnd1h1 dmhm Each 71 vanishes on Y so 0 lemm Zcifif 20m 7 l This shows that f is restriction to Y of ciai E kX giving the required surjectivity D Now let X 9X be an arbitrary variety and let Y be a closed subset We want to show that Y 9y is a variety where 9y 9le Cover X Ugl U by af ne opens Then Y U1Y U1 and Y O U is af ne 12 ALGEBRAIC GROUPS since its a closed subset of U use 1201 Hence Y 9y is a prevariety and its obvious that Ay is closed To nish with here is the useful lemma promised in Example above Lemma 121 Suppose that X is a prevariety with an a ne open cover X ULI Ui Then X is a variety if and only if for every pair ij the intersection Ui Uj is a ne open and the images under restriction of OXU and OXU39 in OXUZ39 Uj generate OXUZ39 Uj Proof Suppose that X is a variety and take af ne opens UV We have that AX O U x V is closed in U x V The map i X a AX induces an isomorphism U O V AX O U x V of ringed spaces Hence U O V is af ne as the latter is Moreover the regular functions on AX O U x V are restrictions of regular functions on U x V ie of elements of MU X kV so we are done since the latter algebra is generated by MU X 1 and 1 X Conversely consider for each pair ij the intersection Y AX O U x Uj The map i X a AX again induces an isomorphism U Uj E Y of ringed spaces Hence by assumption Y is an af ne variety and its coordi nate ring is the image under restriction of OXXXUi x Uj X kUj Hence Y is closed in U x Uj see 127i below for the proof Since this is true for all ij this implies that AX is closed in X x X as required D 17 Projective varieties Now we can introduce the most important of the non af ne varieties In fact in this course all varieties we shall ever encounter can be constructed as open subvarieties of projective varieties Let us start by de ning lP As a set this is the set of all lines in kn ie its k 1 7 N where z N y if z cy for a non zero scalar c We will denote the point of l represented by the vector 0 7 z E k 1 by 0m1 zn So 0mn y0yn if and only if z cyi for all i where c is a non zero scalar The x are called homogeneous coordinates for z ForOgign set Ui lmo77l Epnlmi 7amp0 De ne a bijection gti 3 Ul H A73 gtil9007 790nl 9009013 7i71i7i1i7 n nmi Note for i 7 j tiUl Uj is a principal open subset of A de ned by the non vanishing of the jth coordinate function Now make l into a topological space by declaring a subset U is open if and only if U O U is open for all i Check the subspace topology on each U is exactly the Zariski topology on A lifted through in ALGEBRAIC GROUPS 13 Next make l into a ringed space by de ning a function f on an open set U to be regular if UmUi E OUZU Ui for each i where OUZ is the sheaf of functions on Ul obtained by lifting the sheaf of regular functions on A through in Check this de nes a sheaf of functions OHM on lP and moreover Ow lUi OUZ for each i Hence l is a prevariety as the Ui s with the induced structures as ringed spaces are af ne isomorphic to A To check nally that it is a variety it just remains to apply 121 we need to show that OpnUi Uj is generated by the restrictions of functions in OpnUi and OPnUj Writing TiTj for the polynomial function x gt gt zizj on Uj kT0Ti TjTi TnTj kUj MTGTi TiTj TnTj Clearly the restrictions of these generate the localization of at TjTl the localization of kU7 at TiTj which is kU Uj Done De nition A projective variety is a variety that is isomorphic to a closed subvariety of some lP A quasi projective variety is a variety that is isomor phic to an open subvariety of some projective variety We end with a description of the closed sets in lP Let S MTG Tn be the polynomial algebra An ideal I of S is called homogeneous if it is generated by homogeneous polynomials For such ideals we can consider VI z 0 zn E l l x 0 for all homogeneous f E I which makes sense independent of the choice of representative for Lemma 122 The closed sets of l are exactly the VI for homogeneous ideals I in S Proof If I is a homogeneous ideal all VI Ul are closed in Ui hence VI is closed in lP Conversely let U be an open subset of lP ie each U Ul is open in Ui lf 50464 is a family of homogeneous ideals and I DEA Ia then VI aeA VIa Using this and the fact that principal open sets are a base for the Zariski topology on the Ui s you reduce to the special case that U gt1Df for somei and f E So 1 is a polynomial in TOTivHvTrLTi Scaling by a large power of Ti we obtain a polynomial F E S such that F is zero outside Ul and fz 7 0 if and only if x 7 0 for z E Ui Hence U is the complement of Vf D Exercise 123 4 Show that all af ne varieties are quasi projective varieties 5 De ne a map of sets 1 l x Em A llmm by Wm an 240 yml miyjlosi39snnsi39sm 14 ALGEBRAIC GROUPS Show that the image of j is a closed subset of Pmnmn and that j is an isomorphism of varieties between l x l and its image Hence products of projective varieties are projective 18 Dimension Let X be an irreducible variety If X is af ne its coor dinate ring kX is then an integral domain So we can form its eld of fractions denoted Now let 0 7 f E kX and consider the af ne open subvariety Df of X lts coordinate ring is kDf kXf and its eld of fractions kDf is canonically isomorphic to Finally let U be any non empty af ne open subset of X Pick 1 such that Df Q U Then by the preceeding argument applied to U kU the eld of fractions of the coordinate ring of U is canonically isomorphic to kDf E We have shown all af ne open subsets of an irreducible af ne variety have canonically isomorphic elds of fractions Now let X be an arbitrary irreducible variety De ne kX to be the eld of fractions of any non empty af ne open subset U of X By the previous paragraph if V is some other non empty af ne open subset of X then kU E kU O V E kV canonically So kX is well de ned up to canonical isomorphism It is an important invariant of X called the function eld of X Now de ne the dimension of X to be the transcendence degree of the eld extension kX k For example MED E MA E kT1 Tn So dim l dim A n A good rule if your variety X is not irreducible you should not be talking about its dimension but you can talk about the dimensions of its irreducible components We only need for now a couple of lemmas about dimension Lemma 124 IfX is irreducible and Y is a proper irreducible closed subset then dim Y lt dim X Proof Let U be an af ne open subset of X which has non empty intersection with Y Then dim X dim U and dimY dimU O Y In other words replacing X by U and Y by U Y we may assume X and Y are both af ne Let A kX and P Y a non zero prime ideal of X Suppose that dim Y e dim X 1 Let yl ye are algebraically independent elements of MY AP Then their pre images m1 ze in A are also algebraically independent hence e g 1 Now assume e 1 Let f be a non zero element of P There is a relation pf m1 z5 0 where p is a polynomial in e 1 variables T0 T1 Te because 1 m1 ze are dependent WMA by dividing through by To if necessary that p is not divisible by To But then the polynomial q p0T1T5 E kT1T5 is non zero But qy1y5 0 so q is a dependency between the y s which is a contradiction Hence e lt d D Lemma 125 IfX and Y are irreducible varieties then X X Y is again an irreducible variety and dim X X Y dim X dim Y ALGEBRAIC GROUPS 15 Proof X U1 U U UmY V1 U U Vm be af ne open covers Each U x is irreducible af ne since each X is an integral domain Now U1 x V1 U x is a non empty open subset of U x V hence is dense in U x This shows that U1 x V1 2 U x Hence U1 x V1 X x Y Now suppose X x Y Z1 UZZ for closeds Z1 Z2 By irreducibility of U1 x V1 we must have Z 2 U1 x V1 for some i But then Z U1 x V1 X x Y so X x Y is irreducible Now to compute dimX x Y it equals dimU1 x V1 Let 1zd and y1ye be maximal sets of algebraically independent elements of kU1 kV1 respectively Then ml X 1 md 11 y11 ye is a maximal set of algebraically independent elements of kU1 X kV1 Hence dimXxY trdegkU1 x m 15 dim U1dim V1 dim Xdim Y D Exercise 126 6 Give an example of an af ne variety having two irreducible components one of dimension 1 the other of dimension 19 A theorem about morphisms We ll slowly need to know more about algebraic geometry but I ll try to introduce it as we go along from now on But I need one theorem right now about morphisms of varieties which can be very subtle beasts in general Let me call a morphism j X a Y dominant if its image gtX is a dense subset of Y Lemma 127 Let j X gt Y be a mmphism of a ne varieties and let f MY gt kX be its commphism i If f is surjectiue then gtX is a closed subset of Y ii f is injectiue if and only is dominant Proof The statement that gtz y is equivalent to saying evy euz ozf Hence My kerevy kerevm o if Q gt 1Mm In general equality need not hold here the inverse image of a maximal ideal under an algebra homomorphism need not be a maximal ideal However if f is surjective that is true and we see that gtz y if and only if M gt 1Mm Identifying X now with the maximal ideals of kX and Y with the maximal ideals of MY see 18 we see that gtX is the set of all maximal ideals of MY that are inverse images of maximal ideals of kX ie by the correspondence theorem they are the maximal ideals of MY that contain I ker f In other words gtX VI which is closed ii Note I X f 6 MY lf gtm 0 for all x e X f 6 My ltff 0 ker gt 16 ALGEBRAIC GROUPS Hence f is injective if and only if I gtX 0 which is if and only if VltIlt gtXgt TX Y iii We already know the statement about irreducibility see 14 Now consider the restriction 11 X a gt7X of 1 It is dominant so by ii if a kX is injective This obviously implies dim 7 dim X by the de nition of dimension as the maximal number of algebraically independent elements Theorem 128 Let j X gt Y be a morphi of varieties Then gtX contains a non empty open subset of its closure gtX Proof Let us make some reductions 1 We can nd an af ne open V of Y such that 171V is non empty Let U be an af ne open subset of gt 1V It then suf ces to show that gtU contains a non empty open subset of its closure In other words replacing X by U Y by V and j by its restriction we may assume X and Y are af ne 2 Let X1 X9 be the irreducible components of X Then 1 m U U gtX9 Hence we can replace X by one of the X to assume X is irreducible 3 Finally replace Y by gt7X to assume that j is dominant Now we have reduced to the following situation 1 X a Y is a morphism of af ne varieties f MY a kX is injective and kX is an integral domain Now apply the following technical lemma from commutative algebra to B kXA MY see Springer 194 for a proof Let A C B be nitely generated algebras with B an integral domain Then there exists 0 7 a E A such that any homorphism f A gt k with fa 7 0 can be extended to a homomorphism f B gt k We get some element 0 7 a 6 MY with the property that for each of the evaluation homomorphisms evy for y E Y satisfying evya ay 7 0 there exists z E X with egg e1m o f But that is exactly saying that each y E Da is gtz for some z E X ie Da Q gtX We are done D Exercise 129 7 Give an example of a morphism j X a Y of irreducible af ne varieties such that f HY a kX is injective but 1 is not surjective 24 ALGEBRAIC GROUPS 24 Orbits of algebraic groups The reason that algebraic groups are so important is that they act on algebraic varietiesll Let G be an algebraic group and X be a variety not necessarily af nel We say that G acts on X or that X is a G uariety ifwe are given a morphism pGXXgtX of varieties that makes X into a G set in the usual sense I will use all the usual language of group actions in particular I will write gm instead of pgz whenever it is convenient For example G acts on itself in various ways by conjugation by left multiplication The orbit ofz E X is the set Gm gm lg E G The action is transitive if Gm X for z E X The set of xed points ofg resp G on X is denoted X9 resp XG The stabilizer of z E X in G is denoted Gm clearly a subgroup of G For x E X the orbit map fmGgtGmggt gtgx induces a bijection 2 between the set GGm of cosets of Gm in G and the orbit Gz Of course it would be nice if both GGm and the orbit Gm were va rieties in some natural way so that this map 1 was a morphism of varieties we will see that is the case eventually More generally for a subset Y Q X let COO g E G l gy y for all y EY and N00 g E G l gY Q Y If you take Y to be a subgroup of G and the action of G on itself by conjugation these are the usual notions of centralizer and normalizer of a subgroup of G Finally the center of G is the normal subgroup ZG z E G l 92 29 for all g E G which is just the set of xed points of G on itself for the conjugation action Lemma 211 Let G act on X Let Y Z be subsets ofX with Z closed i The set 9 E G l gY Q Z is closed in particular N0Z is closed ii For each m E X the stabilizer GE is a closed subgroup of G in particular 000 is closed iii The xed point set X9 ofg E G is closed in X in particular XG is closed Proof For each y E X the orbit map fy G a X g gt gt gy is a morphism So f1Z is closed in G So the intersection yey fy 1Z is closed in G which is g E Gng Q Z ii Observe Gm g E G l gz Q m COO yey Gy and apply iii To see that X9 is closed consider 1 X a X x X de ned by z gt gt zgm The xed point set X9 is the inverse image under 1 of the diagonal which is closed since X is a variety The lemma shows things like centralizers of subsets normalizers of closed subsets xed point sets are closed However orbits themselves will not be closed in general In fact the structure of orbits of an algebraic group on a variety can be extremely interesting ALGEBRAIC GROUPS 25 Theorem 212 Let G be a connected algebraic group acting on X Each orbit Gm is irreducible open in its closure hence is an irreducible uariety in its own right and its boundary E 7 Gm is a union of orbits of strictly smaller dimension In particular orbits of minimal dimension are closed so closed orbits exist Proof Let 0 Gm It is the image of C under the orbit map fm hence contains a non empty open subset U of its closure But G acts transitively on 0 so 0 U960 gU is open in its closure too Therefore 57 O is closed so all of its irreducible components have strictly smaller dimension than 0 itself It is G stable so it is a union of orbits B Let us now pause to give some examples of algebraic group actions and their orbits Example 213 1 Let G CL GLV where V k viewed as af ne n space Note the coordinate ring MV of V is the polynomial algebra kT1 Tn where T is the ith coordinate function lnci dentally if you like the basis free approach then MV is the algebra of all functions on V which is just SV the symmetric algebra of the dual space What are the orbits of G on V There are just two the point 0 a closed orbit of dimension 0 and the rest V 7 0 an open orbit of dimension n 1 The stabilizer of a point in the open orbit is conjugate to the closed subgroup Note dim G n2 dim H nn71 and dim G7dim H n which is the dimension of the open orbit This is true in general dim Gm dimG 7 dim Gm for all z E X though we are not quite ready to prove this yet Note V V 7 0 is an open subset of V hence a variety in its own right Let lPV E EW l be the projective variety consisting of all 1 subspaces of V There is an obvious function j V 7gt lP Vm gt gt This is actually a morphism To prove this we will use the following simple lemma that reduces the check to af ne open subsets A to V Lemma 214 Let j X 7gt Y be a function Suppose there is an a ne open couering Y U1 U U Un and open subsets V1 Vn ofX such that i M g U ii 1 0 j E wheneuer f E OyUi A g V ALGEBRAIC GROUPS Then j is a momhism of varieties In our case we take U Mil the af ne open subset of P We take V to be gt 1Ui ie the vectors in V that have non zero i th coordinate So V is the principal open subset in the original af ne space V So we have that kT1 TnT71 Now you can easily see that gtkUil is the subalgebra kT1Ti TnTi of This checks the condition in the lemma so 1 is a morphism By a similar argument to 2 the map 0 i G X WV H WV 97 ltvgt H lt9vgt is a morphism of varieties This is just the natural action of G on 1 spaces in V This action is transitive and the stabilizer of a point is conjugate to This time dim G n2 dimH nn 1 1 and dim G 7 dimH n 7 1 dim le l as we expected Now I want to re ne this example It is time to introduce the Grass mann varieties GdV Here V is an n dimensional vector space and 0 g d g it As a set GdV is the set of all d dimensional subspaces of V For example G1V lPV E le l We want to give GdV the structure of a variety There is an injective map 1 1i am a lPVltv1vdgt Hm Avd So to endow GdV with the structure of a projective variety it suf ces to show that the image of it is closed We will do this by showing its intersection with the af ne open subsets of lPdV are all closed Fix an ordered basis v1 vn for V Then AdV has basis Ui1AAvidl1 i1ltltid7n Without loss of generality lets consider the af ne open subset U C lPdV consisting of the span of all vectors whose 121 A A Ud coordinate is non zero Let V0 v1vdgt Clearly for a d dimensional subspace W of V iMW E U if and only ifthe projection 7139 of V onto V0 along the basis maps W isomorphically onto V0 Then 11 v z where w 7139 n 90139 E am39va39 jd1 ALGEBRAIC GROUPS 27 for some abs Conversely given any choice of the dn 7 d scalars cum7s there is a unique subspace W with zMW E U So we have parametrized 11 1U by d x n 7 d matrices Now let us think about the coordinates of11W in U The image of W under 1 is U1AAvdZaMU1AAUjAAUd jgtd 12739 in the ith position where consists of wedges with two or more from 121 vd replaced by higher vk s Obviously the coef cients in 5 will look like some polynomial functions in the LLs Hence the coordinates of11W with respect to the standard coordinates of U is of the form aijfkaij where the fk are some polynomial functions in the LLs As W varies we get all such points for arbitrary a E Adm d This shows that GdV O U 3571095 35 E Adltn7d where f is a morphism from Adm d to some other af ne space But the graph of a morphism is always closed 117i Hence we can identify GdV with its image under 1 to give it the structure of a projective algebraic variety as a closed subvariety of some Now consider the natural action of G GLV on GdV By 3 it is just the restriction of the action of G on lPdV to GdV hence the action is a morphism Moreover it is obviously transitive This shows that GdV is an irreducible variety and you can compute its dimension by computing the stabilizer of a point to get dim GdV dn 7 1 As a nal example we have the flag variety This is the following closed subvariety of G1V x G2V x x GMV FV 00171027 wfn l fi 6 GiV7 fi C 1241 Clearly GLV acts morphically on FV and the action is transi tive Hence FV is an irreducible projective variety Consider the standard flag f4r f1 f where f 121 vgt lts stabilizer on G is the subgroup Tn of all upper triangular matrices Hence dim FV dim CL 7 dim Tn nn 712 A U V A a V Exercise 215 1 Prove the lemma in Example 3 above 2 Use the lemma to verify that the map p in Example 4 above really is a morphism of varieties as was claimed 28 ALGEBRAIC GROUPS 25 The Jordan decomposition Let us review some linear algebra Let V be a nite dimensional vector space and a E EndV Then a is called semisimple if there is a basis for V with respect to which a is a diagonal matrix a is called nilpotent if a9 0 for some 3 2 1 equivalently all eigenvalues of a are zero and a is called unipotent if a 7 1 is nilpotent equivalently all eigenvalues of a are 1 Given a set S of pairwise commuting elements of EndV you can always nd a basis for V with respect to which all elements of S are upper triangu lar matrices If in addition all elements of S are semisimple you can nd a basis for V with respect to which all elements of S are diagonal It follows easily from this that the product of two commuting semisimple resp nilpo tent resp unipotent matrices is again semisimple resp nilpotent resp unipotent This is de nitely false if you remove the word commuting Lemma 216 Let a E EndV i There are unique elements asan E EndV such that as is semisim ple an is nilpotent asan anas and a as an ii There are polynomials P Q without constant term so that as Paan Qa Hence as and an leaue inuariant any subspace W of V that a stabilizes and alWs aslWalWn an w Similarly for the induced maps on VW iii Let j V 7gt W be a linear map and b E EndW be such that gtoabo gt Then gtoasbso gt gtoanbno gt Proof Let HT 7 im be the characteristic polynomial of a where the i are the distinct eigenvalues of a Let Viu Vla7imu0 a non zero a stable subspace of V By CRT there exists P 6 MT such that PT E 0 mod T PT E i mod T 7 for all i Set as Pa Then as stabilizes all V and the restriction of as to V is scalar multiplication by i Hence as is semisimple and a 7 as has all eigenvalues zero so is nilpotent Everthing in and ii follow pretty easily once we have checked the uniqueness For uniqueness suppose that a bs bn is another such decomposition Since as and an are polynomials in a bs and bn commute with as and an Hence as 7 bs bn 7 an is both semisimple and nilpotent hence zero D The lemma is known as the additive Jordan decomposition As a corollary we get the multiplicatiue Jordan decomposition Corollary 217 Let a E GLV There exist unique asau E GLV such that as is semisimple an is unipotent and a asau auas They have properties similar to ii and iii in the theorem too Proof Let a as an be the additive Jordan decomposition Since a is invertible none of its eigenvalues are 0 Hence taking a basis with respect ALGEBRAIC GROUPS 29 to which both as and an are upper triangular matrices you see that none of the eigenvalues of as are 0 Hence as is invertible Now set an 1 a 1 5 anv and check as and au do the job We can also think about a 1 and a X b if a E EndV b E EndW So aeab is the endomorphism of VQBW such that abuw aubw and a b is the endomorphism of V W such that a bu w au bw You can check easily that if a b are semisimple resp nilpotent resp unipotent then so are aeab and a b It then follows from the uniqueness in the Jordan decompositions that Lemma 218 Let a asau and b bsbu be the Jordan decompositions of a E GLVb E GLW Then a 691 as eabs an 691 is the Jordan decomposition of a EB 12 E GLV EB W and a X b as X bsau X bu is the Jordan decompostion of a X b E GLV X The theory just developed generalizes to in nite dimensional vector spaces V providing we restrict our attention to locally nite endommphisms a ie endomorphisms such that any 1 E V belongs to a nite dimensional a invariant subspace We call a locally nite endomorphism a of V semisimple if its restriction to every nite dimensional a invariant subspace is semisim ple and de ne nilpotent unipotent similarly For a general locally nite a E EndV we have its Jordan decompo sitions a as an and a asau with all the properties of the nite dimensional case holding To de ne as take 1 E V nd a nite dimen sional a invariant subspace W containing 1 and de ne MU alwsv The fact that this is well de ned independent ofthe choice of W follows from the uniqueness statement in the nite dimensional Jordan decomposition The elements an and au are de ned similarly and properties like ii and iii of the lemma hold Now let G be an algebraic group Let g E G We know by 24 that the right translation of functions pg MG a MC is a locally nite endomor phism of the vector space Therefore we have a Jordan decomposition p9 P9sp9u Theorem 219 For any 9 E G there are unique elements gsgu E G such that p9s pgs p9u p9u and g gsgu guys Moreover Z39f gt a a H 239s a momhism of algebraic groups ms gtltggts gtltgugt My Proof Let m MG X MG a MC be the algebra multiplication We have that m 0 099 109 p9 O m Hence by 216iii and 218 m 0 p9s P9s mm 0 m 30 ALGEBRAIC GROUPS Hence pg5 is an automorphism of MG so 5 f gt gt pgsfe is an algebra homomorphism MG a k ie there is a point 95 E G such that E is evaluation at gs Now we compare pg5 and 999 note z and pg commute for all z E G Hence z commutes with ogS too by 216iii p9sf 1P9sf6 P9s 1f6 96 1fgs aws Pgsf pltggts pus ln a similar way you get a gu E G such that pgu gu But the representation p G a GLkG is faithful so now we see that gsgu guys 9 since We know pgsp9u p9up9s p9 Now let 1 G a H be a homomorphism of algebraic groups Then we have f a Take 9 E G and set h gtg You check that w opHlthgt my ow Hence w opHlthgts poo ow Hence w o pm PG99 o w For any a e lel pHhsae PHhsa6 Wis This equals Pags gta6 ags gtgs a Since this is true for all a we get that hs gs The argument for unipotent parts is similar D In the case G CL gs is just the semisimple part of 9 Viewed as an automorphism of V km and gu is its unipotent part To see this let 0 7 f E V For 1 E V de ne u 6 MG by fv9 f9v This gives an injective map f V a MC and satis es gv P9fv Hence gsv p9sfv pgsfv gsv where the gs on the left is the semisimple part of g E GLV in the old sense and the 99 s on the right hand side are the semisimple part of g in the abstract Jordan decomposition For an arbitrary G we can embed G as a closed subgroup of some CL Again the Jordan decomposition g 959 of g as an element of the abstract group G and as an automorphism of V coincide Exercise 220 3 Show that the set Gu gu lg E G of all unipotent elements of G is closed ALGEBRAIC GROUPS 31 26 Unipotent algebraic groups An algebraic group G is called unipo tent if all of its elements are unipotent Theorem 221 Let G be a unipotent closed subgroup of CL Then there is g 6 CL such that gGg 1 Q U Proof Let V k We need to show that G xes some ag in V Proceed by induction on n If there is a proper subspace W of V stablized by G7 then by induction G xes a ag in W and in VW and we can glue them together So we need to consider the case that C does not stabilize and subspace of V7 ie G acts irreducibly on the vector space V Then by Wedderburn s the orem7 the elements of G span the vector space EndV of all endomorphisms of G Since G is unipotent7 all elements of C have trace n Hence7 trVh 7 gh trV17 gh 0 for all 971 6 G7 hence for all g E G and h E EndV Taking It to be the various matrix units7 you now get that 1 7 g 07 1e 9 1 Hence G e D Corollary 222 Unipotent algebraic groups are nilpotent Here is a nice application of the theory so far Theorem 223 Rosenlicht Let G be a unipotent algebraic group acting on an a ne uariety X Then all orbits of G on X are closed Proof Let 0 be an orbit Replacing X by 5 we may assume that O is open dense in X Let Y be its complement Consider the action of G on kX by translation of functions This action is locally nite7 by the same proof as 24 Moreover7 G stabilizes Y hence it leaves Y lt kX invariant By the above theorem7 we can nd a non zero function f E Y xed by G But then f is constant on 07 so since 0 is dense7 f is constant on all of X This shows 1 is a non zero scalar7 hence Y kX and Y 0 Thus 0 X B Let me nally make an important de nition Lemma 224 Let X and Y be closed subgroups of G IfX normalizes Y then XY is a closed subgroup of G Proof It is immediate that XY is a subgroup It is the image of X x Y under a morphism of varieties7 so it contains a non empty open subset of its closure by 128 Therefore XY is closed by 27ii Now let G be an arbitrary connected algebraic group Suppose that X and Y are two closed connected normal solvable subgroups Then X Y is again a closed normal subgroup of G by the lemma7 and it is connected and solvable exercise Hence X Y is another closed connected normal solvable 32 ALGEBRAIC GROUPS subgroup It follows that G contains a unique closed connected normal solvable subgroup of maximal dimension This is called the radical of G denoted RG An algebraic group is called semisimple if its radical RG 5 Similarly you de ne the unipotent radicalof G denoted RAG This is the unique closed connected normal unipotent subgroup of maximal dimension An algebraic group is called reductive if is unipotent radical RAG 5 Since RAG is unipotent it is nilpotent hence solvable hence RAG RG So semisimple groups are reductive For example SLn is a semisimple group but CL has a one dimensional radical consisting of the scalar matrices Thus CL is not semisimple but since scalar matrices are semisimple its unipotent radical is trivial so CL is reductive Later on we shall restrict our attention to the study of reductive algebraic groups for which there is a beautiful structure theory and classi cation ALGEBRAIC GROUPS Disclaimer There are millions of errors in these notes 1 SOME ALGEBRAIC GEOMETRY The subject of algebraic groups depends on the interaction between alge braic geometry and group theory To get going we therefore need to learn some basic algebraic geometry I ll try to keep it to a minimum Note I am following the book by Springer very closely here 11 Algebraic sets Let k be an algebraically closed eld ALWAYS We can think of the elements of the polynomial algebra S kT1 T2 Tn MT for short as k valued functions on the space k if v v1 vn E k and f fT1Tn then u fv1vn We call u E k a zero of f if u 0 Given an ideal I C S let VI denote the set of all common zeros of all functions in I ie VI v E k l fv 0 for all f E I Given a subset X C k let IX denote the ideal of all functions f E S vanishing on all of X ie IX f Slfv Ofor all u EX Note that X g V1Xi I g awn But neither one need be equality To understand the problem recall the radical of an ideal I is the ideal fESlfneIforsomenzl It is obvious that the ideal IX is a radical ideal ie equals its radical Now we need Hilbert s Nullstellensatz Nullstellensatz i If I is a proper ideal of S ie I 7 S then VI is non empty ii For any ideal I of S IVI By an algebraic set in k we mean one of the form VI for some ideal I of S Using the Nullstellensatz you check that the operators V and I set up a 171 inclusion reversing correspondence between the radical ideals of S and the algebraic sets in k The function I gt gt VI has the following properties a V0 kni WS 0 2 ALGEBRAIC GROUPS b VI J VI U VJ c If 50464 is a family of ideals and I ZOEKID then VI aeA VUDrl39 These are all easy to prove except perhaps for For this take z E VI U VJ wlog z E VI Then everything in I vanishes on x So everything in I O J certainly vanishes on m ie z E VI O J Conversely take z VI U VJ Then we can nd some functions f E I and g E J such that x 7 09z 7 0 Then fgz 7 0 But fg E I O J hence z VI Hence Lemma 11 The algebraic sets in k form the closed sets of a topology on k called the Zariski topology Example 12 The closed sets in k other than k itself are the nite ones ii Given X Q h let X denote its closure in the Zariski topology Then X VIX iii On C the closed sets in the Zariski topology are closed sets in the Euclidean topology But there are MANY closed sets in the Euclidean topology that are not Zariski closed The Zariki topology has very few closed sets for example it is not Hausdorff By the way the points in k are closed eg v v1 vn is the zero set of the maximal ideal of S generated by T1 7 v1 Tn 7 vn Indeed by the Nullstellensatz the points of k parametrize the maximal ideals in S in this way Now recall by the Hilbert basis theorem that S is a Noetherian ring ie it has ACC on ideals or equivalently any nonempty collection of ideals of S has a maximal element relative to inclusion Hence a k has DCC on closed sets b Any non empty family of closed sets in k has a minimal one Note a topological space satisfying either of the equivalent properties a or b is called a Noetherian topological space Lemma 13 k is quasi compact ie every open cover has a nite open subcover I reserve the term compact for Hausdor spaces which k is not Proof In terms of closed sets this says that if IQWEA is a family of ideals of S such that aeA VIa 0 there is a nite subset A0 of A such that uer VIa Q We know VZDEA Ia 0 Hence by the Nullstellensatz DEA ID S Hence there are nitely many of the ID such that 1 lies in their sum ie 0er ID S for some nite subset A0 C A But then uer VIa VS Q and we are done Finally note that if X is any algebraic set in k with the subspace topol ogy it is also Noetherian and quasi compact ALGEBRAIC GROUPS 3 12 Irreducible topological spaces Let X be a non empty topological space Then X is called reducible if it is the union of two proper closed subsets Otherwise X is called irreducible A subset A C X is called irreducible if it is irreducible in the induced topology Exercises 14 Let X be a topological space 1 A C X is irreducible if and only ifA is 2 Let f X a Y be a continuous map of topological spaces If X is irreducible then so is fX Lemma 15 Let X be a Noetherian topological space Then X has nitely many maximal irreducible subsets Moreouer these are closed and couer X Proof By 14 maximal irreducible subsets are closed Now suppose that X cannot be written as a union of nitely many ir reducible closed subsets Using the Noetherian property we can nd a minimal non empty closed subset A of X which is not a nite union of irre ducible closed subsets Clearly A is reducible so it is a union of two proper closed subsets But by the minimality of A each of these is a nite union of irreducible closed subsets hence A is too a contradiction Therefore we can write X X1 U U XS where the Xi are irreducible and closed We may assume that there are no inclusions amongst them Now suppose that Y is an irreducible subset of X Then Y Y X1 U U Y O X9 so by de nition of irreducibility we get that Y Q Xi for some i Hence any irreducible subset of X is contained in one of the Xi Hence the X are the maximal irreducible subsets of X The maximal irreducible subsets of X are called the irreducible compo nents of X We want to apply this theory to study algebraic sets in k recalling that the Zariski topology is Noetherian What does irreducibility mean in this context Lemma 16 A closed subset X of k is irreducible if and only if X is a prime ideal Proof Let X be irreducible To show X is a prime ideal suppose fg E S and fg E X Then X Q Vf U Vg ie X Xm w u Xm vltggtgt Since X is irreducible we must therefore have wlog that X Q Vf So 1 E X and X is prime Conversely suppose X is a prime ideal but that X is reducible Sup pose X VI U VJ VI O J with VI Q X for radical ideals IJ Then X Q I so we can pick 1 E 71X But fg E X for all g E J hence by primeness g E X This shows that J Q X whence X Q VJ and X is irreducible D 4 ALGEBRAIC GROUPS So now we know that any closed set X in k can be decomposed uniquely into irreducible components as XX1UUXm Moreover each X VP for some prime ideal in S You can rephrase this purely in terms of algebra if you like any radical ideal I of S is an intersection of prime ideals Moreover assuming there are no inclusions amongst them these prime ideals are uniquely determined Exercises 17 3 Describe all irreducible Hausdorfftopological spaces 4 Let X z y E k2 l my 0 Show that X is a closed connected subset of k2 What are its irreducible components 5 Show that the Zariski topology on k2 is NOT the same as the product topology on k x k arising from the Zariski topology on each copy of k 13 A ine algebras So far we can consider the Zariski topology on a closed subset X of k We would like a more intrinsic description of the topology on X that does not depend on its embedding in h To start with let X be a closed set in k Let lel SHX called the coordinate ring of X Clearly elements of kX can be viewed as functions on X in a well de ned way lndeed restriction of functions from k to X determines a natural surjection S a kX with kernel IX Now kX has the following properties a kX is nitely generated as a k algebra b kX is reduced 0 is its only nilpotent element We will call a k algebra with these properties an a ne algebra Clearly every af ne algebra arises as the coordinate ring of some algebraic set X Moreover we can completely recover the topological space X from its coordinate ring Lemma 18 Let X be an algebraic set with coordinate ring i For each pointx E X let Mm denote the set of allf E kX vanishing at m The map m gt gt M1 is a bijection between the points of X and the set maXkX of maximal ideals of ii For each ideal I ofkX let VI denote the set of all common zeros of all functions in I Then the closed sets ofX are the VI as I runs over all ideals of Proof Everything follows because the ideals of kX are in 171 correspon dence with the maximal ideals of S containing IX The point of the lemma is that you can completely recover X and its Zariski topology from the algebra By the way X is irreducible if and only if its coordinate ring kX is an integral domain by 16 ALGEBRAIC GROUPS 5 So now let X be an algebraic set with coordinate ring Note I no longer need to talk about the embedding X gt knl Let us now introduce some special open subsets of X for f E kX let DU le7 0XV By the way7 Df9 Df D97 DU Df The Df are called principal open subsets of X Lemma 19 If g 6 kX and Q Dg then f E g for some n 2 1 ii The principal open sets form a base for the topology on X Proof Both statements are equivalent to 1 Q ii Equivalently7 every closed set is an intersection of Vf s But VI Vf1 Vfn if 1371 7 so this is obvious D 14 A ine algebraic varieties To start with let X be an algebraic set with coordinate ring We understand the polynomials in kX which make sense as functions de ned on all of X But if we are only interested in functions de ned on a principal open set Df7 more functions make sense any rational function77 of the form gf makes sense as a function on Df since 1 7 0 there In other words7 elements of the localization kXf of kX at the multiplicative set 171 71 27 make sense as functions on Df Let us try to make this more precise we want to work out exactly which functions make sense on an open subset U of X R1 Start with a point z E X A k valued function 1 de ned in a neigh bourhood of z is called regular at x if there are 97 h E kX and an open neighbourhood V Q U Dh such that fy for all y E X So a regular function at z is one that looks like gh near 1 R2 Now let U be a non empty open subset of X A k valued function 1 de ned on U is called regular if it is regular at all points of U We denote by OXU or 9U the algebra of all regular functions on U Notice the following properties are obvious A If U and V are non empty open subsets of X with U C V7 restriction de nes a k algebra homomorphism 0V a OU B Let U UaeA Ua be an open covering of an open set U Suppose we are given fa E 0Ua such that whenever Ua O U is non empty7 fa and f restrict to the same element of 0Ua U Then there is f E 9U whose restriction to each Ua is equal to fa We have now constructed an example OX of a sheaf of functions on the topological space X You can probably guess the general de nition for an arbitrary topological space X7 a sheaf OX of functions on X determines for each open subset U of X a subalgebra OXU of the algebra of all k valued 6 ALGEBRAIC GROUPS functions on U such that the properties A and B above hold The pair X OX is then called a ringed space Let me record a few other generalities to do with ringed spaces G Let X OX be a ringed space and let U be an open subset of X Then we have a new ringed space U 9le where U is given the subspace topology and 9le is the restriction of OX to U de ned on an open V Q U simply by OXlUV OXV Let X OX be a ringed space and let x E X be a point We denote by OXm or 9x the algebra Lin OXU U31 E Thus elements of 9m consist of equivalence classes of pairs U f where U is an open neighbourhood of z and f E OXU two such pairs U f and V 9 being equivalent if there is an open neighbour hood W Q U O V such that W ng Finally let X OX and Y 9y be two ringed spaces Let 1 X a Y be a continuous map For each open set V Q Y and f E OyV we can consider ff f o if as a function on gt 1V This need not belong to OX gt 1V If it does for all V and 1 then 1 is called a morphism of ringed spaces So a morphism of ringed spaces is a continuous map 1 X a Y such that gtOyV Q OX gt 1V for all open subsets V of Y E Now for an important de nition an a ne algebraic variety is a ringed space X OX that is isomorphic as a ringed space to an algebraic set equipped with its sheaf of regular functions as de ned in R2 above In particular the ringed space arising from the algebraic set X k itself is the af ne algebraic variety denoted A a ne n space You probably think the de nition of af ne variety is a little stupid After all the pair X OX was built from the pair X In turn X was built from kX as the set of all maximal ideals So there is no more information in the ringed space X OX than there was in the original af ne algebra kX we started with though the point of view is very different Actually there might be much less information in X OX than there was in That is the point of the next theorem we can recover kX out of the sheaf OX Let X be an algebraic set with coordinate ring kX and let X OX be the corresponding af ne algebraic variety By the de nitions there is a natural map 1 kX a OXX Theorem 110 1 is an isomorphism Proof By the Nullstellensatz the only function in kX which is zero on all of X is 0 This shows that j is injective The problem is surjectivity Let f E OXX For each x E X we can nd an open neighbourhood Um of z and gm 71m 6 kX such that 711 is non zero on all of Um and for each ALGEBRAIC GROUPS 7 y E Um y 9zyhzy This is just the de nition of the functions in OXX Since the principal open sets form a base for the Zariski topology we may assume Um Dam for some am 6 So Dam Q Dhm hence there exists h E kX and an integer nm 2 1 with a hmhm Note 1 gmh mazx on Um So since Dam Da we may as well replace gm by gmh m and 711 by a3 In other words we may assume that Uz Dhz Now X is quasi compact so nitely many of the Dhm cover X Say X Dh1 U U Dhn Now we have picked h1hn and 9 E kX such that the restriction of f to Dhi equals gihfl Now gihgl and gjhgl coincide on Dhi Dhj and hihj vanishes out side of this set So hihjgihj igjhi 0 Since the Dhi cover X the ideal generated by 71 713 is So there exist I E kX with n 1 212713 i1 Finally let x E Dhj Then n n 7171902 Z bi9ihi Z bihi2hj9j i1 i1 71719097190 hj2f This shows that f ELI bigihi giving surjectivity D Exercise 111 6 Let Df be a principal open subset of X Copy the proof of the above theorem to show that there is a natural isomorphism kXf a OXDf Remember kXf denotes the localization of kX at 112 gt 7 Le X OX be an af ne algebraic variety For a point z E X we have de ned D the algebra Om of regular functions at m Let Mm f E kX l fz 0 Prove that Om is isomorphic to the localization kXMx of kX at the multiplicative set kX 7 Mm Hence Om is a local ring with unique maximal ideal mm U f l fz 0 We have de ned in the notion of morphism of ringed space Now let X and Y be af ne varieties Then according to the de nition E we should call a map 1 X a Y a momhism of varieties if it is a continuous map such that gtOyV Q OX gt 1V for each open subset V of Y In particular this means by 110 that 1 de nes a map f MY a kX of the associated coordinate rings called the comomhism of j 8 ALGEBRAIC GROUPS Conversely suppose we are just given a map 1 HY a kX of algebras For x E X let evm be the homomorphism kX a k determined by evalua tion at m Then there is a unique map 1 X a Y such that evwm 511m 01 for all z E X see 18i You should check for yourself that this map is a mmphz39sm of ringed spaces it is obvious that f 1 Exercise 112 8 Show that a morphism j X a Y of af ne varieties is an isomorphism ie has a two sided inverse if and only if f MY a kX is an isomorphism of algebras 9 Suppose that k is of characteristic p gt 0 Consider the map 1 A a A 121 v gt gt 12f v Show that j is a bijective morphism of af ne varieties but is not an isomorphism of af ne varieties 10 Let X be an af ne algebraic variety and f E k X Show that DU OXlDf is again an af ne algebraic variety with coordinate ring kX f Now let X and Y be two af ne varieties So X and Y are topological spaces equipped with sheaves OX and 9y of regular functions We also have their coordinate rings kX OXX and MY OXY We under stand the notion of a morphism j X a Y a continuous map such that gtOyV Q OX gt 1V for every open in V But we have shown that these properties are entirely equivalent to saying simply that the comorphism f associated to the function 1 maps MY into Example 113 Let X A3 and Y VT1T2 71 C A2 S0 kX kT1T2T3 and kT1T2T1T2 7 kSlSZ where is the image of Ti i What do morphisms g5 X a Y look like Well to make sense just as a function we must have that gtv gtv1 122123 1102 gt2v where gt1v gt2v 1 Now what is the comorphism f on 5 Well gtS U1U2U3 U 120 Thus 11 has to belong to kX ie be a polynomial in T1T2 and T3 Similarly 12 has to be a polynomial Thus gtv 1102 gt2v for polynomials 1 Moreover gt1 gt2 1 so they both have to be invertible polynomials so 11 c 12 0 1 for a non zero scalar 0 So the only 1 are the constant ones with gtv 0 0 1 ie HomX Y E kX What do morphisms j Y a X look like We must have gtw gtw1 102 gt1w gt2w gt3w What is gtT Its the function on Y with gtTw jgw So each coordinate function 1 has to belong to M51 52 But that is all The 1 can be chosen arbitrarily lh kSl C V The moral morphisms between af ne varieties are functions whose coordi nates are polynomials ALGEBRAIC GROUPS 43 35 Separable morphisms Recall that a morphism j X a Y of irre ducible varieties is called dominant if its image is dense in Y In the case of af ne varieties that is equivalent to f HY a kX being injective Hence even for arbitrary irreducible X Y the map f kY a kX in duced by a dominant morphism is injective So we can view kX as an extension eld of The morphism j is called separable if kX is a separable extension of Go back to the case that XY are af ne The composite of f and 1X kX a x is a derivation MY a x So by the universal property of differentials we get induced a kY module map 13 Qy gt 9X such that 1X 0 f 13 o dy Let x E X y The kX module km viewed as a kY module via f is kg We used this before to see that f induced the linear map 1 TAX DerkkX km a my DerkkY kg D H D 0 Equivalently we can view ddm as the map 1 t H0mkXQX7 km A Homkmmw kyle H 0 O 5 Applying adjointness of tensor and hom we can even view ddm as a linear map Homkmxwt k H Homkmiyt k Theorem 315 Let j X gt Y be a morphism of irreducible uarieties i Assume that m E X and y E Y are simple points and that d gtm is surjectiue Then j is a dominant separable morphism ii Assume j is a dominant separable morphism Then the points m E X with the property of form a non empty open subset of X Proof We may assume X and Y are af ne and x 9y are free kX resp kY modules of rank at dimX resp e dim Y In particular X and Y are smooth The map 13 fly a x of kY modules induces a homomorphism of free kX modules 1 C ky Qy H Qx We can represent 1 as a d x e matrix A with entries in kX xing bases for Six 9y Suppose that ddm is surjective Then Az which represents the dual map 14 Qyy a Qmz is injective hence a matrix of rank e Hence the rank of A is at least e hence equal to e since rank cannot be more than the number of columns This shows that 11 is injective Hence 13 is injective too Since x and fly are free modules this implies that f MY a kX must be injective so 1 is dominant Moreover injectivity of 1 implies injectivity of MX kY QY A MX kX 9X 44 ALGEBRAIC GROUPS This is the map 04 in the exact sequence 19 MX kY QkYk i QkXk QkXkY 0 Hence kX is a separable extension of MY by the differential criterion for separability The proof of ii is similar to the proof of ii in 312 D Corollary 316 Let G be a connected algebraic group i If X is a variety on which G acts transitively then X is irreducible and smooth In particular G is smooth ii Let j X gt Y be a G equivariant morphism between two varieties on which G acts transitively Then j is separable if and only if d gtm is surjective for some m E X which is if and only if d gtm is surjective for all m E X iii Let j G gt H be a surjective homomorphism of algebraic groups Then j is separable if and only if d gt5 is surjective Proof Take z E X The orbit map 9 gt gt gm is onto so X is irreducible as G is Since at least one point of z is simple and the action is transitive we see that all points of z are simple so X is smooth ii This follows at once from the theorem iii Apply ii to X G Y G D 36 Lie algebras Notation if G is an algebraic group let g denote the tangent space T5G to G at the identity At the moment g is just a vector space of dimension dim G we will see in a while that it has additional structure as a Lie algebra For example the tangent spaces to CL SLn Spgn SO at the identity will be denoted gln 5 532 50 At the moment these are all just vector spaces Example 317 1 First CL Then gin DermaLni k5 is the n2 dimensional vector space on basis em em is the point derivation 1 ij n where 8f f H 8TH e Here TM is the ij coordinate function and MGLW is the localization of the polynomial ring at determinant I will always identify the vector space gn with the vector space of n x n matrices over k so that em is identi ed with the ij matrix unit 2 5L Since SLn Vdet71 inside of CL 5b is a canonically embedded subspace of gln Indeed it will be all matrices X 2 Ma a m ALGEBRAIC GROUPS 45 such that Xdet 71 0 But calculate Z aijeij Z sgnwT1w1Tnwn 7 1 2am L we i So 5 is all matrices in g of trace zero 3 Spgn Recall that Spgn z 6 CL l mth J where 7 0 1 rho as a closed subgroup of GL2 Let T be the 2n x 2n matrix with ij entry TM the ij coordinate function on GL2 Then Spgn VTtKT7J 4n2 polynomial equations written as one matrix equa tionl Then 5332 is the X E g2 such that XTtJT 7 J 0 You calculate this is exactly the condition that X J JX 0 Now you can calculate dim Spgn by linear algebra 4 Similarly 50 in characteristic 7 2 is all X E g satisfying XT 0 5 Also recall that Tn is all upper triangular invertible matrices The tangent space tn will be all upper triangular matrices embedded into g Similarly you can compute the tangent spaces of D U Okay now let s introduce on g the structure of a Lie algebra It is con venient to go via an intermediate LG This is de ned to be the vector space LG D e DerkkG kG l Dom Mpg for all f e kGx e G Recall Amfg fm 1g We call LG the left invariant derivations because they are the derivations commuting with the left regular action of G on Now DerkkGkG is a Lie algebra with operation being the com mutator DD D o D 7 D o D Obviously if D and D are left invariant derivations so is D D Therefore LG is a Lie subalgebra of DerkkGkG Lemma 318 Let G be a connected algebraic group g T5G The maps LGgtgDgt gteveoD and g gt LG X H X where X MG gt MG is the derivation with X971f for all g E G are mutually inverse isomorphisms 46 ALGEBRAIC GROUPS Proof We d better rst make sure the maps make sense ie that eve o D is a point derivation and that X is a left invariant derivation Only the last thing is tricky Xgfh Xh71gf Xh7lgfXf9 1h9Xfh Now let s compute eve o X eve o XW ifWe Xf for all f 6 MG hence eve o X X Finally let s compute eve o D for a left invariant derivation D ammoeveoDAg71f DAg71feAg71DfeDf9 Hence em D D Because of the lemma we get on g an induced structure as a Lie algebra induced by the Lie algebra structure on LG XY eve o X Example 319 Let s compute the Lie algebra structure on g We need to work out Well for g E G atnag 7 etaAgata 7 x Twpmus 7 awaylt9 Therefore MTPg 6MT Now you can compute the commutator in le it acts on each Tm in the same way as 6jk il 7 zigzag Hence we ve worked out leij75kll jk5il 6ilekj The right hand side is just the commutator emek 7 ewe of the matrix units So in general X Y XY 7 YX for X Y E g 7 the usual commutator as matricesl Lemma 320 Let H be a closed subgroup of G and set I Then hX ngIOX ngI I In particular E is a Lie subalgebra of g Proof That hX ng10 is just the de nition of tangent space of a closed subvariety lf XI 0 consider Xf for f E I Evaluating at h E H Kmz 7 Xwaf which is zero as Ah71f is still in I as IIis a subgroupl So Xf vanishes on H ie is containing in I This shows XI Q I The opposite namely that if XI Q I then XI 0 is easy ALGEBRAIC GROUPS 47 Finally observe that if DD E LG satisfy D Q I D I Q I then the commutator D D also does This implies that E is a subalgebra ofg D Now we get the Lie algebra structure in all the above examples 5n50n since they are all just Lie subalgebras of gln where the Lie bracket is just the commutator as matrices The next lemma I m going to leave to you to supply the proof hint its easier to rephase things in terms of LG and Lemma 321 Let j G H H be a momhlsm of connected algebraic groups Then d gt5 g H E is a Lie algebra homomomhlsm 37 Some differential calculations 1 now want to compute differentials of various natural morphisms In all cases since an arbitrary algebraic group can be embedded as a closed subgroup of some GLn it is enough to make the computation in the case of GLn when we can be very explicit Example 3392239 1 Let M 3 G x G H G be multiplication Then diam g ea g H g is addition Proof Suppose G GLn Recall geag Daniela we kw the isomorphism mapping X Y to the map f X g H Xfg5 f6Yg By de nition dme em 5klTrs 6M7 Sam Tm Tm 51375 kink t 5ij l 5klTrs Hence dlu5eeijekj em em ie elm is addition Let i G H G be the inverse map Then die g H g is the map X H HX Proof Consider the composite G H G x G H Gg H glg H gig e The composite is a constant function so its differential is zero But the differential of a composite is the composite of the differentials so appying 1 0 dide dl5 The differential of the identity map is the identity map so we are done Fix z E G Let lntz G H G be the automorphism of algebraic groups 9 H zgm l The differential dlnt ze g H g is a Lie algebra automorphism It is usually denoted Adz g H g For example suppose G GLn Then for a matrix X E gln Ad sz l ie Adz is just the Lie algebra automorphism given by conjugation by z Hence for H and closed subgroup of G and z E H Adz E H E is just conjugation by z too 7 it leaves the subspace E of g invariant Proof Let us compute lntzTj InthTmMg Tij9 1 if WTijigi71 Z TikTkl9Tlj7139 kl A to V A DJ V g V A U V A a V ALGEBRAIC GROUPS Hence lnt xTM Z mmTkJW lhj k l The ij entry of Ad is Ad XTij Z ikXTkl 1ij M which is the ij entry of zXz l Here is a consequence of For each x E G Adz g a g is an invertible linear map even a Lie algebra automorphism So you can think of Ad as a group homomorphism from G to GLg or even to Autg which is a closed subgroup of GLg I claim that Ad G a GLg is a morphism of algebraic groups Proof Embed G as a closed subgroup of some CL Then by 3 Adz is conjugation by m which is clearly a morphism of varieties 7 since it is given by matrix multiplication and inversion which are polynomial operations The image of Ad G a GLg is a closed connected subgroup of Autg denoted Ad G It is interesting to consider the kernel of Ad a closed normal subgroup of G Obviously every element of ZG belongs to ker Ad ie ZG Q ker Ad I warn you that equality need not hold here However it usually does for example it always does in characteristic 0 For example if G CL kerAd ZG check directly ker Ad is the invertible matrices which commute with all other matrices the scalar matrices In this case AdG is the group known as PGLn As an abstract group PGLn E GLnscalars Similarly if G SLn ker Ad is the scalar matrices of determinant one ie the matrices w 0 0 0 0 0 0 w where w runs over all nth roots of unity The group Ad SL7 is known as PSLn As an abstract group PSLn E SLnscalars Warning ln characteristic pln ZSLn is trivial So PSLn E SL7 as an abstract group the isomorphism being the map 9 gt gt Adg However SL7 is not isomorphic to PSLn as an algebraic group the problem is that Ad is a bijective morphism that is NOT an isomorphism of varieties ie its inverse is not a morphism of varieties Because of 4 we have a morphism Ad G a GLg So we can consider its differential again dAd g a gg ALGEBRAIC GROUPS 49 The map dAd is usually denoted ad I claim that adX E gg is the map Y H XY in particular adX is a derivation of g so ad is a Lie algebra homomorphism g H Derg In other words The di erentz39al of Ad is ad The proof is so nasty I m not going to type it in Exercise 323 8 Here is an example to show that ker Ad may be larger than ZG if Charis p gt 0 Let G be the two dimensional closed subgroup of GL3 consisting of all matrices of the form a 0 0 0 ap b 0 0 1 where a 7 0 and b are arbitrary elements of is Describe the Lie algebra g explicitly as a subspace of g3 Now compute ZG and ker Ad and show that they are not equal 9 Consider the morphism Ad SL2 H PSLZ Both SL2 and PSLZ are three dimensional algebraic groups so their Lie algebras are three dimensional too Consider the differential ad 52 H p52 Using Example 8 above compute the kernel of ad an ideal in the Lie algebra 52 Deduce by 316 that the morphism Ad is separable if and only if Charis 7 2 So in characteristic 2 7 when Ad is a bijective morphism 7 it is inseparable so it cannot be an isomorphism 10 Let Charis p gt 0 Take X E g DerkisGis5 Recall g is isomorphic to LG the left invariant derivations of isG via the map X H Show that X is a left invariant derivation of Hence there is a unique element lel E g with lei for This gives an extra operation on g in characteristic p the map X H lel which makes g into what is known as a restricted Lie algebra Describe this map explicitly in the one dimensional cases G Ga and G Gm Lie algebras Examples sheet 2 g denotes a nite dimensional7 semisimple Lie algebra over C unless other wise stated Representations 1 If V and W are g modules7 we made homCV7 W into a g module by set ting zfv 7 fmv for all z E g f E homCV7 Wv E V Verify directly that this gives a well de ned g module structure on homCV7 2 Using the fact that the Lie algebra g 5LnC is simple7 show that the Killing form x y trgadzady is related to the form x y trmy by any 2749071 Representations of s2C In these exercises7 Lm denotes the irreducible 52C module of dimension m l 3 Embed 52C into 53C in its upper left hand 2 x 2 position The restriction of the adjoint representation of 53C de nes an 8 dimensional 52C module V Show that V E L0 L1 L1 L2 4 Let V L1 denote the natural 52C module7 with its usual basis m1 m2 Let P C172 be the polynomial algebra in two variables7 and extend the action of g on Cu 63 Cmg lt P to all of P by the rule 2f9 2f9 29 for all z E g g 6 P Show that this makes P into an in nite dimensional g module7 and that the subspace Pm lt P of homogeneous polynomials of degree m is a g submodule of P Show P Lm 5 Let 0 g m g n Prove the Clebsch Gordan formula Lm Ln Ln7mLn7m2Lnm72Lnm The Jordan decomposition 6 Show that g is nilpotent if and only if every element x ofg is nilpotent ie adgm is a nilpotent endomorphism ofg Give an example to show that the statement g is semisimple if and only if every element x ofg is semisimple77 ie adgm is diagonalisable is false 7 Let g be the Lie algebra C with multiplication 0 for all 71 6 C Every element of g is both semisimple and nilpotent Verify that the following maps g a g2C are representations of g meg 8ltbgtw8 3ltcgtw3 Show that in a every non zero element of the image of g is semisimple but not nilpotent and that in b every non zero element of the image of g is nilpotent but not semisimple ln c7 show that the semisimple and nilpotent parts of every non zero element of the image of g are not even elements of the image ofg Thus7 the Jordan decomposition is false in general ifg is not a semisimple Lie algebra 8 Let g lt g be two semisimple Lie algebras For x E g show that its abstract Jordan decomposition regarded as an element of g agrees with its abstract Jordan decomposition regarded as an element of g The Cartan decomposition 9 Compute explicitly the Cartan decomposition of the Lie algebra 5nC taking E to be the set of all diagonal matrices ie verify all the details from the lectures This is the most important example on this sheetll 10 Compute the restriction of the Killing form on 5LnC to the set E of all diagonal matrices directly without using question 4 Hence verify directly that the restriction of the Killing form to E is non degenerate 2Dr 11 Calculate explicitly the Cartan integers for 5C the numbers for all 043 in the root system I they should all be 07 2 or i1 12 If g is semisimple7 E a maximal toral subalgebra7 prove that E 151 13 Prove that every maximal toral subalgebra of52C is one dimensional 14 Prove that every three dimensional semisimple Lie algebra is isomorphic to 52C 15 Using just the Cartan decomposition prove that no 47 5 or 7 dimensional semisimple Lie algebras exist Jon Brundan7 10142005
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