14 Semisimple algebraic groups and Lie algebras

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1 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS Semisimple algebraic groups and Lie algebras Semisimple algebraic groups Recall (11.30) that a connected algebraic group G 1 is said to be semisimple if its radical RG = 1. In other words, it is semisimple if it has no normal connected solvable subgroup 1. PROPOSITION 14.1 A connected algebraic group 1 is semisimple if and only if it has no normal connected commutative algebraic subgroup 1. PROOF. Obviously, a semisimple algebraic group has no such subgroup. For the converse, we use that, for any algebraic group H, RH and DH are characteristic subgroups, i.e., every isomorphism H H maps RH onto RH and DH onto DH. This is obvious from their definitions: RH is the largest connected normal solvable algebraic subgroup and DH is the smallest normal algebraic subgroup such that H/DH is commutative. Therefore the chain G RG D(RG) D 2 (RG) D r (RG) 1 is preserved by every automorphism of G. In particular, the groups are normal in G. If RG 1, then the last nonzero term in its derived series is a normal connected commutative algebraic subgroup of G. Semisimple Lie algebras The derived series of a Lie algebra g is g g = [g, g] g = [g, g ]. A Lie algebra is said to be solvable if the derived series terminates with 0. Every Lie algebra contains a largest solvable ideal, called its radical r(g). A nonzero Lie algebra g is semisimple if r(g) = 0, i.e., if g has no nonzero solvable ideal. Similarly to the case of algebraic groups, this is equivalent to g having no nonzero commutative ideal. Semisimple Lie algebras and algebraic groups THEOREM 14.2 Let G be a connected algebraic group. If Lie(G) is semisimple, then G is semisimple, and the converse is true when char(k) = 0. PROOF. Suppose Lie(G) is semisimple, and let N be a normal connected commutative subgroup of G we have to prove N = 1. But Lie(N) is a commutative ideal in Lie(G) (p93), and so is zero. Hence N = 1 (see 13.9). Conversely, suppose G is semisimple, and let n be a commutative ideal in g we have to prove n = 0. Let G act on g through the adjoint representation Ad: G GL g, and let H be the subgroup of G whose elements fix those of n (see 13.17). Then (ibid.), the Lie algebra of H is h = {x g [x, n] = 0}, which contains n. Because n is an ideal, so also is h: [[x, y], n] = [x, [y, n]] [y, [x, n]]

2 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 100 which lies in n if x h and n n. Therefore, H is normal in G (13.18), and so its centre Z(H ) is normal in G. Because G is semisimple, Z(H ) = 1, and so z(h) = 0 (13.19). But z(h) n, which must therefore be zero. COROLLARY 14.3 Assume char(k) = 0. For a connected algebraic group G, Lie(R(G)) = r(g). PROOF. From the exact sequence we get an exact sequence 1 RG G G/RG 1 1 Lie(RG) g Lie(G/RG) 1 in which Lie(RG) is solvable (obvious) and Lie(G/RG) is semisimple (14.2). Therefore Lie RG is the largest solvable ideal in g. The map ad For a k-vector space with a k-bilinear pairing a, b ab: C C C, we write Der k (C) for the space of k-derivations C C, i.e., k-linear maps δ : C C satisfying the Leibniz rule δ(ab) = aδ(b) + δ(a)b. If δ and δ are k-derivations, then δ δ need not be, but δ δ δ δ is, so Der k (C) is a subalgebra of gl(c), not End k-lin (C). For a Lie algebra g, the Jacobi identity says that the map ad(x) = (y [x, y]) is a derivation of g: [x, [y, z]] = [y, [z, x]] [z, [x, y]] = [y, [x, z]] + [[x, y], z]. Thus, ad: g End k-lin (g) maps into Der k (g). The kernel of ad is the centre of g. THEOREM 14.4 Let k be of characteristic zero. If g is semisimple, then ad: g Der k (g) is surjective (and hence an isomorphism). The derivations of g of the form ad(x) are often said to be inner (by analogy with the automorphisms of G of the form inn(g)). Thus the theorem says that all derivations of a semisimple Lie algebra are inner. We discuss the proof of the theorem below (see Humphreys 1972, 5.3). The Lie algebra of Aut k (C) Again, let C be a finite-dimensional k-vector space with a k-bilinear pairing C C C. PROPOSITION 14.5 The functor is an algebraic subgroup of GL C. R Aut k-alg (R k C)

3 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 101 PROOF. Choose a basis for C. Then an element of Aut k-lin (R k C) is represented by a matrix, and the condition that it preserve the algebra product is a polynomial condition on the matrix entries. [Of course, to be rigorous, one should write this out in terms of the bi-algebra.] Denote this algebraic group by Aut C, so Aut C (R) = Aut k-alg (R k C). PROPOSITION 14.6 The Lie algebra of Aut C is gl(c) Der k (C). PROOF. Let id +εα Lie(GL C ), and let a+a ε, b+b ε be elements of C k k[ε] C Cε. When we first apply id +εα to the two elements and then multiply them, we get ab + ε(ab + a b + aα(b) + α(a)b); when we first multiply them, and then apply id +εα we get ab + ε(ab + a b + α(ab)). These are equal if and only if α satisfies the Leibniz rule. The map Ad Let G be a connected algebraic group. Recall (p89) that there is a homomorphism Ad: G GL g. Specifically, g G(R) acts on g k R G(R[ε]) as inn(g), x gxg 1. On applying Lie, we get a homomorphism ad: Lie(G) Lie(GL g ) End(g), and we defined [x, y] = ad(x)(y). LEMMA 14.7 The homomorphism Ad has image in Aut g ; in other words, for all g G(R), the automorphism Ad(g) of g k R preserves the bracket. Therefore, ad maps into Der k (g). PROOF. Because of (3.8), it suffices to prove this for G = GL n. But A GL(R) acts on g k R = M n (R) as X AXA 1. Now A[X, Y ]A 1 = A(XY Y X)A 1 = AXA 1 AY A 1 AY A 1 AXA 1 = [AXA 1, AY A 1 ].

4 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 102 LEMMA 14.8 Let g G(k). The functor C G (g) R {g G(R) gg g 1 = g } is an algebraic subgroup of G with Lie algebra c g (g) = {x g Ad(g)(x) = x}. PROOF. Embed G in GL n. If we can prove the statement for GL n, then we obtain it for G, because C G (g) = C GLn (g) G and c g (g) = c gln (g) g. Let A GL n (k). Then C GLn (A)(R) = {B GL n (R) AB = BA}. Clearly this is a polynomial (even linear) condition on the entries of B. Moreover, Lie(C GLn (A)) = {I + Bε Lie(GL n ) A(I + Bε)A 1 = (I + Bε)} {B M n ABA 1 = B}. PROPOSITION 14.9 For a connected algebraic group G over a field k of characteristic zero, the kernel of Ad is the centre Z(G) of G. PROOF. Clearly Z N = Ker(Ad). It suffices 43 to prove Z = N when k = k. If g N(k), then c g (g) = g, and so C G (g) = G (by 14.8). Therefore g Z(k). We have shown that Z(k) = N(k), and this implies 44 that Z = N. THEOREM For a semisimple algebraic group G over a field of characteristic zero, the sequence 1 Z(G) G Aut g 1 is exact. PROOF. On applying Lie to Ad: G Aut g, we get ad: g Lie(Aut g ) Der(g). But, according to (14.4), the map g Der(g) is surjective, which shows that ad: g Lie(Aut g ) is surjective, and implies that Ad: G Aut g is a quotient map (13.7). Recall that two semisimple groups G 1, G 2 are said to be isogenous if G 1 /Z(G 1 ) G 2 /Z(G 2 ). The theorem gives an inclusion {semisimple algebraic groups}/isogeny {semisimple Lie algebras}/isomorphism. In Humphreys 1972, there is a complete classification of the semisimple Lie algebras up to isomorphism over an algebraically closed field of characteristic zero, and all of them arise from algebraic groups. Thus this gives a complete classification of the semisimple algebraic groups up to isogeny. We will follow a slightly different approach which gives more information about the algebraic groups. For the remainder of this section, k is of characteristic zero. 43 Let Q = N/Z; if Q k = 0, then Q = The map k[n] k[z] is surjective let a be its kernel. Since m = 0 in k[n], if a 0, then there exists a maximal ideal m of k[n] not containing a. Because k = k, k[n]/m k (AG 2.7), and the homomorphism k[n] k[n]/m k is an element of N(k) Z(k).

5 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 103 Interlude on semisimple Lie algebras Let g be a Lie algebra. A bilinear form B : g g k on g is said to be associative if B([x, y], z) = B(x, [y, z]), all x, y, z g. LEMMA The orthogonal complement a of an ideal a in g with respect to an associative form is again an ideal. PROOF. By definition a = {x g B(a, x) = 0 for all a a} = {x g B(a, x) = 0}. Let a a and g g. Then, for a a, B(a, [g, a ]) = B(a, [a, g]) = B([a, a ], x) = 0 and so [g, a ] a. The Killing form on g is κ(x, y) = Tr g (ad(x) ad(y)). That is, κ(x, y) is the trace of the k-linear map z [x, [y, z]]: g g. LEMMA The form κ(x, y) = Tr g (ad(x) ad(y)) is associative and symmetric. PROOF. It is symmetric because for matrices A = (a ij ) and B = (b ij ), Tr(AB) = a ijb ji = Tr(BA). i,j By tradition, checking the associativity is left to the reader. EXAMPLE The Lie algebra sl 2 consists of the 2 2 matrices with trace zero. It has as basis the elements ( ) ( ) ( ) x =, h =, y =, and Then [x, y] = h, [h, x] = 2x, [h, y] = 2y adx = 0 0 1, adh = 0 0 0, ady =

6 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 104 and so the top row (κ(x, x), κ(x, h), κ(x, y)) of the matrix of κ consists of the traces of , , In fact, κ has matrix 0 8 0, which has determinant Note that, for sl n, the matrix of κ is n 2 1 n 2 1, and so this is not something one would like to compute. LEMMA Let a be an ideal in g. The Killing form on g restricts to the Killing form on a, i.e., κ g (x, y) = κ a (x, y) all x, y a. PROOF. Let α be an endomorphism of a vector space V such that α(v ) W ; then Tr V (α) = Tr W (α W ), because when we choose a basis for W and extend it to a basis for V, the matrix for α takes the form ( ) A B 0 0 where A is the matrix of α W. If x, y a, then adx ady is an endomorphism of g mapping g into a, and so its trace (on g), κ(x, y), equals Tr a (adx ady a) = Tr a (ad a x ad a y) = κ a (x, y). PROPOSITION (Cartan s Criterion). Tr V (x y) = 0 for all x [g, g] and y g. A Lie subalgebra g of gl(v ) is solvable if PROOF. If g is solvable, then an analogue of the Lie-Kolchin theorem shows that, for some choice of a basis for V, g t n. Then [g, g] u n and [[g, g], g] u n, which implies the traces are zero. For the converse, which is what we ll need, see Humphreys 1972, 4.5, p20 (the proof is quite elementary, involving only linear algebra). 45 COROLLARY If κ([g, g], g) = 0, then g is solvable; in particular, if κ(g, g) = 0, then g is solvable. PROOF. The map ad: g gl(v ) has kernel the centre z(g) of g, and the condition implies that its image is solvable. Therefore g is solvable. THEOREM (Cartan-Killing criterion). A nonzero Lie algebra g is semisimple if and only if its Killing form is nondegenerate, i.e., the orthogonal complement of g is zero. PROOF. = : Let a be the orthogonal complement of g, a = {x g κ(g, x) = 0}. It is an ideal (14.11), and certainly κ(a, a) = 0 45 In Humphreys 1972, this is proved only for algebraically closed fields k, but this condition is obviously unnecessary since the statement is true over k if and only if it is true over k.

7 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 105 and so it is solvable by (14.14) and (14.16). Hence, a = 0 if g is semisimple. = : Let a be a commutative ideal of g. Let a a and g g. Then g adg g ada a adg a ada 0. Therefore, (ada adg) 2 = 0, and so 46 Tr(ada adg) = 0. In other words, κ(a, g) = 0, and so a = 0 if κ is nondegenerate. A Lie algebra g is said to be a direct sum of ideals a 1,..., a r if it is a direct sum of them as subspaces, in which case we write g = a 1 a r. Then [a i, a j ] a i a j = 0 for i j, and so g is a direct product of the Lie subalgebras a i. A nonzero Lie algebra is simple if it is not commutative and has no proper nonzero ideals. In a semisimple Lie algebra, the minimal nonzero ideals are exactly the ideals that are simple as Lie subalgebras (but a simple Lie subalgebra need not be an ideal). THEOREM Every semisimple Lie algebra is a direct sum g = a 1 a r of its minimal nonzero ideals. In particular, there are only finitely many such ideals. Every ideal in a is a direct sum of certain of the a i. PROOF. Let a be an ideal in g. Then the orthogonal complement a of a is also an ideal (14.11, 14.12), and so a a is an ideal. By Cartan s criterion (14.16), it is solvable, and hence zero. Therefore, g = a a. If g is not simple, then it has a nonzero proper ideal a. Write g = a a. If and a and a are not simple (as Lie subalgebras) we can decompose them again. Eventually, g = a 1 a r with the a i simple (hence minimal) ideals. Let a be a minimal nonzero ideal in g. Then [a, g] is an ideal contained in a, and it is nonzero because z(g) = 0, and so [a, g]= a. On the other hand, [a, g] = [a, a 1 ] [a, a r ], and so a = [a, a i ] for exactly one i. Then a a i, and so a = a i (simplicity of a i ). This shows that {a 1,... a r } is a complete set of minimal nonzero ideals in g. Let a be an ideal in g. The same argument shows that a is the direct sum of the minimal nonzero ideals contained in a. COROLLARY All nonzero ideals and quotients of a semisimple Lie algebra are semisimple. PROOF. Obvious from the theorem. COROLLARY If g is semisimple, then [g, g] = g. PROOF. If g is simple, then certainly [g, g] = g, and so this is also true for direct sums of simple algebras. 46 If α 2 = 0, the minimum polynomial of α divides X 2, and so the eigenvalues of α are zero.

8 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 106 REMARK The theorem is surprisingly strong: a finite-dimensional vector space is a sum of its minimal subspaces but is far from being a direct sum (and so the theorem fails for commutative Lie algebras). Similarly, it fails for commutative groups: for example, if C 9 denotes a cyclic group of order 9, then C 9 C 9 = {(x, x) C 9 C 9 } {(x, x) C 9 C 9 }. If a is a simple Lie algebra, one might expect that a embedded diagonally would be another simple ideal in a a. It is a simple Lie subalgebra, but it is not an ideal. LEMMA For any Lie algebra g, the space {ad(x) x g} of inner derivations of g is an ideal in Der k (g). PROOF. Recall that Der k (g) is the space of k-linear endomorphisms of g satisfying the Leibniz condition; it is made into a Lie algebra by [δ, δ ] = δ δ δ δ. For a derivation δ of g and x, y g, [δ, adx](y) = (δ ad(x) ad(x) δ)(y) = δ([x, y]) [x, δ(y)] = [δ(x), y] + [x, δ(y)] [x, δ(y)] = [δ(x), y]. Thus, is inner. [δ, ad(x)] = ad(δx) (44) THEOREM If g is semisimple, then ad: g Der(g) is a bijection: every derivation of g is inner. PROOF. Let adg denote the (isomorphic) image of g in Der(g). It suffices to show that the orthogonal complement (adg) of adg in D for κ D is zero. Because adg and (adg) are ideals in Der(g) (see 14.22, 14.11), [adg, (adg) ] adg (adg). Because κ D adg = κ adg is nondegenerate (14.17), Let δ (adg). For x g, adg (adg) = 0. ad(δx) (44) = [δ, ad(x)] = 0. As ad: g Der(g) is injective, this shows that δx = 0. Since this is true for all x g, δ = 0.

9 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 107 Semisimple algebraic groups A connected algebraic group G is simple if it is noncommutative and has no normal algebraic subgroup except G and 1, and it is almost simple if it is noncommutative and has no proper normal algebraic subgroup of dimension > 0. Thus, for n > 1, SL n is almost simple and PSL n = df SL n /µ n is simple. An algebraic group G is said to be the almost direct product of its algebraic subgroups G 1,..., G n if the map (g 1,..., g n ) g 1 g n : G 1 G n G is a quotient map (in particular, a homomorphism) with finite kernel. In particular, this means that the G i commute and each G i is normal. THEOREM Every semisimple group G is an almost direct product G 1 G r G of its minimal connected normal algebraic subgroups of dimension > 0. In particular, there are only finitely many such subgroups. Every connected normal algebraic subgroup of G is a product of those G i that it contains, and is centralized by the remaining ones. PROOF. Write Lie(G) = g 1 g r with the g i simple ideals. Let G 1 be the identity component of C G (g 2 g r ) (notation as in 13.17). Then Lie(G 1 ) = c g (g 2 g r ) = g 1, and so it is normal in G (13.18). If G 1 had a proper normal connected algebraic subgroup of dimension > 0, then g 1 would have an ideal other than g 1 and 0, contradicting its simplicity. Therefore G 1 is almost simple. Construct G 2,..., G r similarly. Then [g i, g j ] = 0 implies that G i and G j commute. The subgroup G 1 G r of G has Lie algebra g, and so equals G (13.6). Finally, Lie(G 1... G r ) = g 1... g r = 0 and so G 1... G r is étale (13.9). Let H be a connected algebraic subgroup of G. If H is normal, then Lie H is an ideal, and so is a direct sum of those g i it contains and centralizes the remainder. This that H is a product of those G i it contains, and is centralized by the remaining ones. COROLLARY All nontrivial connected normal subgroups and quotients of a semisimple algebraic group are semisimple. PROOF. Obvious from the theorem. COROLLARY If G is semisimple, then DG = G, i.e., a semisimple group has no commutative quotients. PROOF. This is obvious for simple groups, and the theorem then implies it for semisimple groups.

10 15 REDUCTIVE ALGEBRAIC GROUPS Reductive algebraic groups Throughout this section, k has characteristic zero. Structure of reductive groups THEOREM 15.1 If G is reductive, then the derived group G der of G is semisimple, the connected centre Z(G) of G is a torus, and Z(G) G der is finite; moreover, Z(G) G der = G. PROOF. It suffices to prove this with k = k. By definition, (RG) u = 0, and so (11.28) shows that RG is a torus T. Rigidity (9.15) implies that the action of G on RG by inner automorphisms is trivial, and so RG Z(G). Since the reverse inclusion always holds, this shows that R(G) = Z(G) = torus. We next show that Z(G) G der is finite. Choose an embedding G GL V, and write V as a direct sum V = V 1 V r of eigenspaces (see 9.14). When we choose bases for the V i, then Z(G) (k) consists of the matrices A A r with each A i nonzero and scalar, 47 and so its centralizer in GL V consists of the matrices of this shape with the A i arbitrary. Since G der (k) consists of commutators, it consists of such matrices with determinant 1. As SL(V i ) contains only finitely many scalar matrices, this shows that Z(G) G der is finite. Note that Z(G) G der is a normal algebraic subgroup of G such that G/(Z(G) G der ) is commutative (being a quotient of G/G der ) and semisimple (being a quotient of G/R(G)). Now (14.26) shows that G = Z(G) G der. Therefore G der G/R(G) is surjective with finite kernel. As G/R(G) is semisimple, so also is G der. REMARK 15.2 Among the semisimple algebraic groups in an isogeny class, there is one (said to be simply connected) having all others as quotients. Let G be a reductive algebraic group and let G G der be the simply connected covering group of G der. The centre Z(G) of G is a group of multiplicative type, and G defines a homomorphism Z( G) Z(G der ) Z(G). Thus, from G we get a triple ( G, Z, Z( G) Z) with G a simply connected semisimple group, Z a group of multiplicative type, and Z( G) Z a homomorphism. Conversely, every such triple ( G, Z, ϕ) gives rise to a reductive group, namely, the quotient (ϕ,inclusion) Z( G) Z G G 0. (45) 47 That is, of the form diag(a,..., a) with a 0.

11 15 REDUCTIVE ALGEBRAIC GROUPS 109 Generalities on semisimple modules Let k be a field, and let A be a k-algebra (not necessarily commutative). An A-module is simple if it does not contain a nonzero proper submodule. PROPOSITION 15.3 The following conditions on an A-module M of finite dimension 48 over k are equivalent: (a) M is a sum of simple modules; (b) M is a direct sum of simple modules; (c) for any submodule N of M, there exists a submodule N such that M = N N. PROOF. Assume (a), and let N be a submodule of M. Let (S i ) i I be the set of simple modules of M. For J I, let N(J) = j J S j. Let J be maximal among the subsets of I for which (i) the sum j J S j is direct and (ii) N(J) N = 0. I claim that M is the direct sum of N(J) and N. To prove this, it suffices to show that each S i N + N(J). Because S i is simple, S i (N + N(J)) equals S i or 0. In the first case, S i N + N(J), and in the second J {i} has the properties (i) and (ii). Thus (a) implies (b) and (c), and it is obvious that (b) and (c) each implies (a). DEFINITION 15.4 An A-module is semisimple if it satisfies the equivalent conditions of the proposition. Representations of reductive groups Throughout this subsection, k is algebraically closed. Representations are always on finitedimensional k-vector spaces. We shall sometimes refer to a vector space with a representation of G on it as a G-module. The definition and result of the last subsection carry over to G-modules. Our starting point is the following result. THEOREM 15.5 If g is semisimple, then all g-modules are semisimple. PROOF. Omitted see Humphreys 1972, pp25 28 (the proof is quite elementary but a little complicated). THEOREM 15.6 Let G be an algebraic group. All representations of G are semisimple if and only if G is reductive. LEMMA 15.7 The restriction to any normal algebraic subgroup of a semisimple representation is again semisimple. PROOF. Let G GL V be a representation of G, which we may assume to be simple. Let S be a simple N-submodule of V. For any g G(k), gs is a simple N-submodule, and V is a sum of the gs. LEMMA 15.8 All representations of G are semisimple if and only if all representations of G are semisimple 48 I assume this only to avoid using Zorn s lemma in the proof.

12 15 REDUCTIVE ALGEBRAIC GROUPS 110 PROOF. = : Since G is a normal algebraic subgroup of G (8.13), this follows from the preceding lemma. = : Let V be a G-module, and let W be a sub G-module (i.e., a subspace stable under G). Then W is also stable under G, and so V = W W for some G -stable subspace W. Let p be the projection map V W it is a G -equivariant 49 map whose restriction to W is id W. Define q : V W, q = 1 n g gpg 1, where n = (G(k): G (k)) and g runs over a set of coset representatives for G (k) in G(k). One checks directly that q has the following properties: (a) it is independent of the choice of the coset representatives; (b) for all w W, q(w) = w; (c) it is G-equivariant. Now (b) implies that V = W W, where W = Ker(q), and (c) implies that W is stable under G. REMARK 15.9 The lemma implies that the representations of a finite group are semisimple. This would fail if we allowed the characteristic to divide the order of the finite group. LEMMA Every representation of a semisimple algebraic group is semisimple. PROOF. From a representation G GL V of G on V we get a representation g gl V of g on V, and a subspace W of V is stable under G if and only if it is stable under g (see 13.15). Therefore, the statement follows from (15.5). Proof of Theorem 15.6 Lemma 15.8 allows us to assume G is connected. = : Let G GL V be a faithful semisimple representation of G, and let N be the unipotent radical of G. Lemma 15.7 shows V is semisimple as an N-module, say V = V i with V i simple. Because N is solvable, the Lie-Kolchin theorem (11.23) shows that the elements of N have a common eigenvector in V i (cf. the proof of the theorem) and so V i has dimension 1, and because N is unipotent it must act trivially on V i. Therefore, N acts trivially on V, but we chose V to be faithful. Hence N = 0. = : If G is reductive, then G = Z G where Z is the connected centre of G (a torus) and G is the derived group of G (a semisimple group) see (15.1). Let G GL V be a representation of G. Then V = i V i where V i is the subspace of V on which Z acts through a character χ i (see 9.14). Because Z and G commute, each space V i is stable under G, and because G is semisimple, V i = j V ij with each V ij simple as a G -module (15.10). Now V = i,j V ij is a decomposition of V into a direct sum of simple G-modules. REMARK It is not necessary to assume k is algebraically closed. In fact, for an algebraic group G over k of characteristic zero, all representations of G are semisimple if and only if all representations of G k are semisimple (Deligne and Milne 1982, 2.25) 50. However, as noted earlier (11.33), it is necessary to assume that k has characteristic zero, even when G is connected. 49 That is, it is a homomorphism of G -representations. 50 Deligne, P., and Milne, J., Tannakian Categories. In Hodge Cycles, Motives, and Shimura Varieties, Lecture Notes in Math. 900 (1982), Springer, Heidelberg,

13 15 REDUCTIVE ALGEBRAIC GROUPS 111 REMARK Classically, the proof was based on the following two results: Every semisimple algebraic group G over C has a (unique) model G 0 over R such that G(R) is compact. For example, SL n = (G 0 ) C where G 0 is the special unitary group (see p91). Every representation of an algebraic group G over R such that G(R) is compact is semisimple. To prove this, let, be a positive definite form on V. Then, 0 = G(R) x, y dg is a G(R)-invariant positive definite form on V. For any G-stable subspace W, the orthogonal complement of W is a G-stable complement. A criterion to be reductive There is an isomorphism of algebraic groups GL n GL n sending an invertible matrix A to the transpose (A 1 ) t of its inverse. The image of an algebraic subgroup H of GL n under this map is the algebraic subgroup H t of GL n such that H t (R) = {A t A H(R)} for all k-algebras R. Now consider GL V. The choice of a basis for V determines an isomorphism GL V GL n and hence a transpose map on GL V, which depends on the choice of the basis. PROPOSITION Every connected algebraic subgroup G of GL V such that G = G t for all choices of a basis for V is reductive. PROOF. We have to show that (RG) u = 0. It suffices to check this after passing to the algebraic closure 51 k of k. Recall that the radical of G is the largest connected normal solvable subgroup of G. It follows from (11.29c) that RG is contained in every maximal connected solvable subgroup of G. Let B be such a subgroup, and choose a basis for V such that B T n (Lie-Kolchin theorem 11.23). Then B t is also a maximal connected solvable subgroup of G, and so RG B B t = D n. This proves that RG is diagonalizable. EXAMPLE The group GL V itself is reductive. EXAMPLE Since the transpose of a matrix of determinant 1 has determinant 1, SL V is reductive. It is possible to verify that SO n and Sp n are reductive using this criterion (to be added; cf. Humphreys 1972, Exercise 1-12, p6). They are semisimple because their centres are finite (this can be verified directly, or by studying their roots see below). 51 More precisely, one can prove that R(G k ) = (RG) k and similarly for the unipotent radial (provided k is perfect).

14 16 SPLIT REDUCTIVE GROUPS: THE PROGRAM Split reductive groups: the program In this, and all later sections, k is of characteristic zero. Split tori Recall that a split torus is a connected diagonalizable group. Equivalently, it is an algebraic group isomorphic to a product of copies of G m. A torus over k is an algebraic group that becomes isomorphic to a split torus over k. A torus in GL V is split if and only if it is contained in D n for some basis of k. Consider for example {( ) } a b T = a 2 + b 2 0. b a The characteristic polynomial of such a matrix is and so its eigenvalues are X 2 2aX + a 2 + b 2 = (X a) 2 + b 2 λ = a ± b 1. It is easy to see that T is split (i.e., diagonalizable over k) if and only if 1 is a square in k. Recall ( 9) that End(G m ) Z: the only homomorphisms G m G m are the maps t t n for n Z. For a split torus T, we set There is a pairing Thus X (T ) = Hom(T, G m ) = group of characters of T, X (T ) = Hom(G m, T ) = group of cocharacters of T., : X (T ) X (T ) End(G m ) Z, χ, λ = χ λ. χ(λ(t)) = t χ,λ for t G m (R) = R. Both X (T ) and X (T ) are free abelian groups of rank equal to the dimension of T, and the pairing, realizes each as the dual of the other. For example, let a 1 0 T = D n = a n Then X (T ) has basis χ 1,..., χ n, where and X (T ) has basis λ 1,..., λ n, where χ i (diag(a 1,..., a n )) = a i, λ i (t) = diag(1,..., i t,..., 1).

15 16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 113 Note that i.e., χ j, λ i = χ j (λ i (t)) = { 1 if i = j 0 if i j, { t = t 1 if i = j 1 = t 0 if i j. Some confusion is caused by the fact that we write X (T ) and X (T ) as additive groups. For example, (5χ 2 + 7χ 3 )diag(a 1, a 2, a 3 ) = χ 2 (a 2 ) 5 χ 3 (a 3 ) 7. For this reason, some authors use an exponentional notation χ(a) = a χ. With this notation, the preceding equation becomes Split reductive groups a 5χ 2+7χ 3 = χ 2 (a 2 ) 5 χ 3 (a 3 ) 7, a = diag(a 1, a 2, a 3 ). Let G be an algebraic group over a field k. When k = k, a torus T G is maximal if it is not properly contained in any other torus. For example, D n is a maximal torus in GL n because it is equal to own centralizer in GL n. In general, T G is said to be maximal if T k is maximal in G k. A reductive group is split if it contains a split maximal torus. Let G a reductive group over k. Since all tori over k are split, G is automatically split. It is known that there exists a split reductive group G 0 over k, unique up to isomorphism, such that G 0k G. EXAMPLE 16.1 The group GL n is a split reductive group (over any field) with split maximal torus D n. On the other hand, let H be the quaternion algebra over R. As an R-vector space, H has basis 1, i, j, k, and the multiplication is determined by i 2 = 1, j 2 = 1, ij = k = ji. It is a division algebra with centre R. There is an algebraic group G over R such that G(R) = (R k H). In particular, G(R) = H. As C R H M 2 (C), G C GL 2, but G is not split. EXAMPLE 16.2 The group SL n is split reductive (in fact, semisimple) group, with split maximal torus the diagonal matrices of determinant 1. EXAMPLE 16.3 Let (V, q) be a nondegenerate quadratic space (see 5), i.e., V is a finitedimensional vector space and q is a nondegenerate quadratic form on V with associated symmetric form φ. Recall (5.5) that the Witt index of (V, q) is the maximum dimension of an isotropic subspace of V. If the Witt index is r, then it follows from (5.2) that V is an orthogonal sum V = H 1... H r V 1 (Witt decomposition)

16 16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 114 where each H i is a hyperbolic plane 52 and V 1 is anisotropic. It can be shown that the associated algebraic group SO(q) is split if and only if its Witt index is as large as possible. (a) Case dim V = n is even. When the Witt index ( is as) large as possible, n = 2r, and 0 I there is a basis for which the matrix 53 of the form is, and so I 0 Note that the subspace of vectors q(x 1,..., x n ) = x 1 x r x r x 2r. (,..., r, 0,..., 0) is totally isotropic. The algebraic subgroup consisting of the diagonal matrices of the form diag(a 1,..., a r, a 1 1,..., a 1 r ) is a split maximal torus in SO(q). (b) Case dim V = n is odd. When the Witt index is as large as possible, n = 2r + 1, and there is a basis for which the matrix of the form is 0 0 I, and so 0 I 0 q(x 0, x 1,..., x n ) = x x 1 x r x r x 2r. The algebraic subgroup consisting of the diagonal matrices of the form diag(1, a 1,..., a r, a 1 1,..., a 1 r ) is a split maximal torus in SO(q). Henceforth, by SO n, I ll mean one of the above two groups. EXAMPLE 16.4 Let V = k 2n, and let ψ be the skew-symmetric form with matrix so ψ( x, y) = x 1 y n x n y 2n x n+1 y 1 x 2n y n. ( ) 0 I, I 0 The corresponding symplectic group Sp n is split, and the algebraic subgroup consisting of the diagonal matrices of the form is a split maximal torus in Sp n. diag(a 1,..., a r, a 1 1,..., a 1 r ) That is, has matrix relative a suitable basis, and so q = xy + yx Moreover, SO(q) consists of the automorphs of this matrix with determinant 1, i.e., SO(q)(R) consists of 0 I 0 I the n n matrices A with entries in R and determinant 1 such that A t A =. I 0 I 0

17 16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 115 Program Let G be a split reductive group over k. Then any two split maximal tori are conjugate by an element of G(k). Rather than working with split reductive groups G, it turns out to be better to work with pairs (G, T ) with T a split maximal torus in G To each pair (G, T ) consisting of a split reductive group and a maximal torus, we associate a more elementary object, namely, its root datum Ψ(G, T ). The root datum Ψ(G, T ) determines (G, T ) up to isomorphism, and every root datum arises from a pair (G, T ) Classify the root data Since knowing the root datum of (G, T ) is equivalent to knowing (G, T ), we should be able to read off information about the structure of G and its representations from the root datum. This is true The root data have nothing to do with the field! In particular, we see that for each reductive group G over k, there is (up to isomorphism) exactly one split reductive group over k that becomes isomorphic to G over k. However, there will in general be many nonsplit groups, and so we are left with the problem of understanding them. In linear algebra and the theory of algebraic groups, one often needs the ground field to be algebraically closed in order to have enough eigenvalues (and eigenvectors). By requiring that the group contains a split maximal torus, we are ensuring that there are enough eigenvalues without requiring the ground field to be algebraically closed. Example: the forms of GL 2. What are the groups G over a field k such that G k GL 2? For any a, b k, define H(a, b) to be the algebra over k with basis 1, i, j, ij as a k-vector space, and with the multiplication given by i 2 = a, j 2 = b, ij = ji. This is a k-algebra with centre k, and it is either a division algebra or is isomorphic to M 2 (k). For example, H(1, 1) M 2 (k) and H( 1, 1) is the usual quaternion algebra when k = R. Each algebra H(a, b) defines an algebraic group G = G(a, b) with G(R) = (R H(a, b)). These are exactly the algebraic groups over k becoming isomorphic to GL 2 over k, and G(a, b) G(a, b ) H(a, b) H(a, b ). Over R, every H is isomorphic to H( 1, 1) or M 2 (R), and so there are exactly two forms of GL 2 over R. Over Q, the isomorphism classes of H s are classified 54 by the subsets of {2, 3, 5, 7, 11, 13,..., } having a finite even number of elements. In particular, there are infinitely many forms of GL 2 over Q, exactly one of which, GL 2, is split. 54 The proof of this uses the quadratic reciprocity law in number theory.

18 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP The root datum of a split reductive group Recall that k has characteristic zero. Roots Let G be a split reductive group and T a split maximal torus. Then G acts on g = Lie(G) via the adjoint representation Ad: G GL g. In particular, T acts on g, and so it decomposes g = g 0 g χ where g 0 is the subspace where T acts trivially, and g χ is the subspace on which T acts through the nontrivial character χ (see 9.14). The nonzero χ occurring in this decomposition are called the roots of (G, T ). They form a finite subset Φ of X (T ). Example: GL 2 Here {( ) } x1 0 T = x 0 x 2 1 x 2 0, ( ) X (T ) = Zχ 1 Zχ 2, x1 0 χi 0 x 2 xi, g = M 2 (k), 55 and T acts on g by conjugation: ( ) ( ) ( ) ( x1 0 a b x a x 1 0 x 2 c d 0 x 1 = 2 d x 2 x 1 c ) x 2 b Write E ij for the matrix with a 1 in the ij th -position, and zeros elsewhere. Then T acts trivially on g 0 = E 11, E 22, through the character α = χ 1 χ 2 on g α = E 12, and through the character α = χ 2 χ 1 on g α = E 21. Thus, Φ = {α, α} where α = χ 1 χ 2. When we use χ 1 and χ 2 to identify X (T ) with Z Z, Φ becomes identified with {(1, 1), ( 1, 1)}.. Example: SL 2 Here 55 Recall that, for example, {( )} x 0 T = 0 x 1, ( ) X (T ) = Zχ, x 0 χ 0 x x, 1 g = { ( a b c d ) M2 (k) a + d = 0}. x x 2 3χ1 +5χ 2 x 3 1 x 5 2 x x 2 3χ1 5χ 2 x 3 1 /x 5 2.

19 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 117 Again T acts on g by conjugation: ( ) ( ) ( ) ( ) x 0 a b x 1 0 a x 0 x 1 = 2 b c a 0 x x 2 c a Therefore, the roots are α = 2χ and α = 2χ. When we use χ to identify X (T ) with Z, Φ becomes identified with {2, 2}. Example: PGL 2 Recall that this is the quotient of GL 2 by its centre: PGL 2 = GL 2 /G m. One can prove that for all rings R, PGL 2 (R) = GL 2 (R)/R. Here {( ) }/ T = x1 0 0 x x1 2 x 2 0 {( x 0 x 0 ) x 0}, ( ) X (T ) = Zχ, x1 0 χ 0 x 2 x 1, x 2 g = M 2 (k)/{ai} (quotient as a vector space). and T acts on g by conjugation: ( ) ( ) ( ) ( x1 0 a b x a x 1 0 x 2 c d 0 x 1 = 2 d x 2 x 1 c ) x 2 b Therefore, the roots are α = χ and α = χ. When we use χ to identify X (T ) with Z, Φ becomes identified with {1, 1}.. Example: GL n Here {( x1 ) } 0 T =... x 1 x n 0, 0 x n X (T ) = ( x1 ) 0 Zχ χ i,. 1 i n.. i xi, 0 x n g = M n (k), and T acts on g by conjugation: ( x x n ) a 11 a 1n. a ij.. a n1. a nn x = 0 x 1 n a 11.. x 1 xn a 1n x i a x ij j. xn x1 a n1 a nn Write E ij for the matrix with a 1 in the ij th -position, and zeros elsewhere. Then T acts trivially on g 0 = E 11,..., E nn and through the character α ij = χ i χ j on g αij = E ij, and so Φ = {α ij 1 i, j n, i j}. When we use the χ i to identify X (T ) with Z n, then Φ becomes identified with {e i e j i j} where e 1,..., e n is the standard basis for Z n...

20 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 118 Definition of a root datum DEFINITION 17.1 A root datum is a quadruple Ψ = (X, Φ, X, Φ ) where X, X are free Z-modules of finite rank in duality by a pairing, : X X Z, Φ, Φ are finite subsets of X and X in bijection by a map α α, 56 satisfying the following conditions rd1 α, α = 2, rd2 s α (Φ) Φ where s α is the homomorphism X X defined by s α (x) = x x, α α, for x X, α Φ, rd3 the group W = W (Ψ) of automorphisms of X generated by the s α is finite. Note that (rd1) implies that s α (α) = α, and that the converse holds if α 0. Moreover, if s α (α) = α, then 57 s 2 α = 1. Clearly, also s α (x) = x if x, α = 0. Thus, s α should be considered an abstract reflection in the hyperplane orthogonal to α. The elements of Φ and Φ are called the roots and coroots of the root datum (and α is the coroot of α), and W (Ψ) is called the Weyl group of the root datum. We want to attach to each pair split reductive group G and split maximal torus T, a root datum Ψ(G, T ) with X = X (T ), Φ = roots, X = X (T ) with the pairing X (T ) X (T ) Z in 16, Φ = coroots (to be defined). First examples of root data EXAMPLE 17.2 Let G = SL 2. Here X = X (T ) = Zχ, X = X (T ) = Zλ, Φ = {α, α}, ( ) x 0 χ 0 x 1 t α = 2χ λ Φ = {α, α }, α = λ. 56 Thus, a root datum is really an ordered sextuple, but everyone says quadruple. 57 Because X, X,,, Φ, Φ, Φ Φ, s α(s α(x)) = s α(x x, α α) x ( ) t 0 0 t 1 = (x x, α α) x, α s α(α) = x.

21 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 119 Note that and so As always, t ( ) λ t 0 2χ 0 t t 2 1 s α (α) = α, α, α = 2. s α ( α) = α etc., and so s ±α (Φ) Φ. Finally, W (Ψ) = {1, s α } is finite, and so Ψ(SL 2, T ) is a root system, isomorphic to (Z, {2, 2}, Z, {1, 1}) (with the canonical pairing x, y = xy and bijection 2 1, 2 1). EXAMPLE 17.3 Let G = PGL 2. Here Φ = {α, α }, α = 2λ. In this case Ψ(PGL 2, T ) is a root system, isomorphic to (Z, {1, 1}, Z, {2, 2}). REMARK 17.4 If α is a root, so also is α, and there exists an α such that α, α = 2. It follows immediately, that the above are the only two root data with X = Z and Φ nonempty. There is also the root datum (Z,, Z, ), which is the root datum of the reductive group G m. EXAMPLE 17.5 Let G = GL n. Here X = X (D n ) = i Zχ i, diag(x 1,..., x n ) χ i x i Note that and so X = X (D n ) = i Zλ i, Φ = {α ij i j}, α ij = χ i χ j Φ = {αij i j}, αij = λ i λ j. t λ i λ j t λ i diag(1,..., 1, t, i 1,..., 1) diag(1,..., t, i..., t 1,...) χ i χ j t 2 j α ij, α ij = 2. Moreover, s α (Φ) Φ for all α Φ. We have, for example, s αij (α ij ) = α ij s αij (α ik ) = α ik α ik, α ij α ij = α ik χ i, λ i α ij (if k i, j) = χ i χ k (χ i χ j ) = α jk s αij (α kl ) = α kl (if k i, j, l i, j).

22 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 120 Finally, let E(ij) be the permutation matrix in which the i th and j th rows have been swapped. The action A E(ij) A E(ij) 1 of E ij on GL n by inner automorphisms stabilizes T and swaps x i and x j. Therefore, it acts on X = X (T ) as s αij. This shows that the group generated by the s αij is isomorphic to the subgroup of GL n generated by the E(ij), which is isomorphic to S n. In particular, W is finite. Therefore, Ψ(GL n, D n ) is a root datum, isomorphic to (Z n, {e i e j i j}, Z n, {e i e j i j} where e i = (0,..., i 1,..., 0), the pairing is the standard one e i, e j = δ ij, and (e i e j ) = e i e j. In the above examples we wrote down the coroots without giving any idea of how to find (or even define) them. Before defining them, we need to review some general results on reductive groups. Semisimple groups of rank 0 or 1 The rank of a reductive group is the dimension of a maximal torus, i.e., it is the largest r such that G k contains a subgroup isomorphic to G r m. Since all maximal tori in G k are conjugate (see below), the rank is well-defined. THEOREM 17.6 (a) Every semisimple group of rank 0 is trivial. (b) Every semisimple group of rank 1 is isomorphic to SL 2 or PGL 2. PROOF. (SKETCH) (a) Take k = k. If all the elements of G(k) are unipotent, then G is solvable (11.25), hence trivial. Otherwise, G(k) contains a semisimple element (10.1). The smallest algebraic subgroup H containing the element is commutative, and therefore decomposes into H s H u (see 11.8). If all semisimple elements of G(k) are of finite order, then G is finite (hence trivial, being connected). If G(k) contains a semisimple element of infinite order, H s is a nontrivial torus, and so G is not of rank 1. (b) One shows that G contains a solvable subgroup B such that G/B P 1. From this one gets a nontrivial homomorphism G Aut(P 1 ) PGL 2. Centralizers and normalizers Let T be a torus in an algebraic group G. The centralizer of T in G is the algebraic subgroup C = C G (T ) of G such that, for all k-algebras R, C(R) = {g G(R) gt = tg for all t T (R)}. Similarly, the normalizer of T in G is the algebraic subgroup N = N G (T ) of G such that, for all k-algebras R, N(R) = {g G(R) gtg 1 T (R) for all t T (R)}. Of course, one has to show that these subfunctors are in fact representable.

23 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 121 THEOREM 17.7 Let T be a torus in a reductive group G. (a) The centralizer C G (T ) of T in G is a reductive group; in particular, it is connected. (b) The identity component of the normalizer N G (T ) of T in G is C G (T ); in particular, N G (T )/C G (T ) is a finite étale group. (c) The torus T is maximal if and only if T = C G (T ). PROOF. (a) Omitted. (When k = k, the statement is proved in Humphreys 1975, 26.2.) (b) Certainly N G (T ) C G (T ) = C G (T ). But N G (T ) /C G (T ) acts faithfully on T, and so is trivial by rigidity (9.15). For the second statement, see 8. (c) Certainly, if C G (T ) = T, then T is maximal because any torus containing T is contained in C G (T ). Conversely, C G (T ) is reductive group containing T, and so C G (T ) = D(C G (T )) Z(C G (T )) (see 15.1). It follows that Z(C G (T )) = T and D(C G (T )) is a semisimple group of rank 0, and hence is trivial. Thus C G (T ) = Z(C G (T )) = T. The quotient W (G, T ) = N G (T )/C G (T ) is called the Weyl group of (G, T ). It is a constant étale algebraic group 58 when T is split, and so may be regarded simply as a finite group. Definition of the coroots LEMMA 17.8 Let G be a split reductive group with split maximal torus T. The action of W (G, T ) on X (T ) stabilizes Φ. PROOF. Take k = k. Let s normalize T (and so represent an element of W ). Then s acts on X (T ) (on the left) by (sχ)(t) = χ(s 1 ts). Let α be a root. Then, for x g α and t T (k), t(sx) = s(s 1 ts)x = s(α(s 1 ts)x) = α(s 1 ts)sx, and so T acts on sg α through the character sα, which must therefore be a root. For a root α of (G, T ), let T α = Ker(α), and let G α be centralizer of T α. THEOREM 17.9 Let G be a split reductive group with split maximal torus T. (a) For each α Φ, there is a unique s α 1 in W (G α, T ) and a unique α X (T ) such that s α (x) = x x, α α, for all x X (T ). (46) Moreover, α, α = 2. (b) The system (X (T ), Φ, X (T ), Φ ) with Φ = {α α Φ} and the map α α : Φ Φ is a root datum. PROOF. (SKETCH) (a) The key point is that C Gα (T α ) is a reductive group of rank one with T as a maximal torus. Thus, we are essentially in the case of SL 2 or PGL 2, where everything is obvious (see below). Note that the uniqueness of α follows from that of s α. (b) We noted in (a) that (rd1) holds. The s α attached to α lies in W (G α, T ) W (G, T ), and so stabilizes Φ by the lemma. Finally, all s α lie in the Weyl group W (G, T ), and so they generate a finite group (in fact, the generate exactly W (G, T )). 58 That is, W (R) is the same finite group for all integral domains R.

24 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 122 EXAMPLE Let G = SL 2, and let α be the root 2χ. Then T α = T and G α = G. The unique s 1 in W (G, T ) is represented by and the unique α for which (46) holds is λ. ( ) 0 1, 1 0 EXAMPLE Let G = GL n, and let α = α 12 = χ 1 χ 2. Then T α = {diag(x, x, x 3,..., x n ) xx 3... x n 1} and G α consists of the invertible matrices of the form Clearly n α = represents the unique nontrivial element s α of W (G α, T α ). It acts on T by For x = m i χ i, diag(x 1, x 2, x 3,..., x n ) diag(x 2, x 1, x 3,..., x n ). s α x = m 2 χ 1 + m 1 χ 2 + m 3 χ m n χ n = x x, λ 1 λ 2 (χ 1 χ 2 ). Thus (46) holds if and only if α is taken to be λ 1 λ 2. Computing the centre PROPOSITION Every maximal torus T in a reductive algebraic group G contains the centre Z = Z(G) of G. PROOF. Clearly Z C G (T ), but (see 17.7), C G (T ) = T. Recall (14.9) that the kernel of the adjoint map Ad: G GL g is Z(G), and so the kernel of Ad: T GL g is Z(G) T = Z(G). Therefore Z(G) = Ker(Ad T ) = α Φ Ker(α).

25 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 123 We can use this to compute the centres of groups. For example, Z(GL n ) = i j Ker(χ i χ j ) = {( x {( ) Z(SL 2 ) = Ker(2χ) = x 0 0 x x 2 = 0 1 Z(PGL 2 ) = Ker(χ) = 1. On applying X to the exact sequence we get (see 9.11) an exact sequence 0 x } n ) = µ 2, x 1 = x 2 = = x n 0 0 Z(G) T t (...,α(t),...) α Φ G m (47) α Φ Z (...,mα,...) mαα X (T ) X (Z(G)) 0, }, and so For example, X (Z(G)) = X (T )/{subgroup generated by Φ}. (48) X (Z(GL n )) Z n / e i e j i j (a 1,...,a n) a i Z, X (Z(SL 2 )) Z/(2), X (Z(PGL 2 )) Z/Z = 0. Semisimple and toral root data DEFINITION A root datum is semisimple if Φ generates a subgroup of finite index in X. PROPOSITION A split reductive group is semisimple if and only if its root datum is semisimple. PROOF. A reductive group is semisimple if and only if its centre is finite, and so this follows from (48). DEFINITION A root datum is toral if Φ is empty. PROPOSITION A split reductive group is a torus if and only if its root datum is toral. PROOF. If the root datum is toral, then (48) shows that Z(G) = T. Hence DG has rank 0, and so is trivial. It follows that G = T. Conversely, if G is a torus, the adjoint representation is trivial and so g = g 0.

26 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 124 The main theorems. From (G, T ) we get a root datum Ψ(G, T ). THEOREM Let T, T be split maximal tori in G. Then there exists a g G(k) such that T = gt g 1 (i.e., inn(g)(t ) = T ). PROOF. Omitted for the present. EXAMPLE Let G = GL V, and let T be a split torus. A split torus is (by definition) diagonalizable, i.e., there exists a basis for V such that T D n. Since T is maximal, it equals D n. This proves the theorem for GL V. It follows that the root datum attached to (G, T ) depends only on G (up to isomorphism). THEOREM (ISOMORPHISM) Every isomorphism Ψ(G, T ) Ψ(G, T ) of root data arises from an isomorphism ϕ: G G such that ϕ(t ) = T ; moreover, ϕ is unique up to composition by a homomorphism inn(t) with t T (k) and α(t) k for all α. PROOF. Springer 1998, THEOREM (EXISTENCE) Every reduced root datum arises from a split reductive group. PROOF. Springer 1998, A root datum is reduced if the only multiples of a root α that can also be a root are ±α. Examples We now work out the root datum attached to each of the classical split semisimple groups. In each case the strategy is the same. We work with a convenient form of the group G in GL n. We first compute the weights of the split maximal torus on gl n, and then check that each nonzero weight occurs in g (in fact, with multiplicity 1). Then for each α we find a natural copy of SL 2 (or PGL 2 ) centralizing T α, and use it to find the coroot α. Example (A n ): SL n+1. Let G be SL n+1 and let T be the algebraic subgroup of diagonal matrices: {diag(t 1,..., t n+1 ) t 1 t n+1 = 1}. Then X (T ) = Zχ i / Zχ, { X (T ) = { a i λ i a i = 0}, diag(t 1,..., t n+1 ) χ i t i χ = χ i t χ i diag(t a 1,..., t an ), a i Z, with the obvious pairing,. Write χ i for the class of χ i in X (T ). Then all the characters χ i χ j, i j, occur as roots, and their coroots are respectively λ i λ j, i j. This follows easily from the calculation of the root datum of GL n.

27 17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 125 Example (B n ): SO 2n+1. Consider the symmetric bilinear form φ on k 2n+1, φ( x, y) = 2x 0 y 0 + x 1 y n+1 + x n+1 y x n y 2n + x 2n y n Then SO 2n+1 = SO(φ) consists of the 2n + 1 2n + 1 matrices A of determinant 1 such that φ(a x, A y) = φ( x, y), i.e., such that A t 0 0 I A = 0 0 I. 0 I 0 0 I 0 The Lie algebra of SO 2n+1 consists of the 2n + 1 2n + 1 matrices A of trace 0 such that i.e., such that φ(a x, y) = φ( x, A y), A t 0 0 I = 0 0 I A. 0 I 0 0 I 0 Take T to be the maximal torus of diagonal matrices Then diag(1, t 1,..., t n, t 1 1,..., t 1 n ) X (T ) = 1 i n Zχ i, X (T ) = 1 i n Zλ i, diag(1, t 1,..., t n, t 1 1,..., t 1 n ) χ i t i t λ i diag(1,..., t, i..., 1) with the obvious pairing,. All the characters ±χ i, ±χ i ± χ j, i j occur as roots, and their coroots are, respectively, ±2λ i, ±λ i ± λ j, i j. Example (C n ): Sp 2n. Consider the skew symmetric bilinear form k 2n k 2n k, φ( x, y) = x 1 y n+1 x n+1 y x n y 2n x 2n y n. Then Sp 2n consists of the 2n 2n matrices A such that φ(a x, A y) = φ( x, y), i.e., such that ( ) A t 0 I A = I 0 ( ) 0 I. I 0