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14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 99 14 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]]
42

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Page 1: 14 Semisimple algebraic groups and Lie algebrasjmilne/aag140.pdf · 2005-04-20 · 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 101 PROOF.Choose a basis for C. Then an element

14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 99

14 Semisimple algebraic groups and Lie algebras

Semisimple algebraic groups

Recall (11.30) that a connected algebraic group G 6= 1 is said to be semisimple if its radicalRG = 1. In other words, it is semisimple if it has no normal connected solvable subgroup6= 1.

PROPOSITION 14.1 A connected algebraic group 6= 1 is semisimple if and only if it hasno normal connected commutative algebraic subgroup 6= 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 fromtheir definitions: RH is the largest connected normal solvable algebraic subgroup and DHis the smallest normal algebraic subgroup such that H/DH is commutative. Therefore thechain

G ⊃ RG ⊃ D(RG) ⊃ D2(RG) ⊃ · · · ⊃ Dr(RG) ⊃ 1

is preserved by every automorphism of G. In particular, the groups are normal in G. IfRG 6= 1, then the last nonzero term in its derived series is a normal connected commutativealgebraic subgroup of G. 2

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 Liealgebra 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 ofalgebraic 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 issemisimple, and the converse is true when char(k) = 0.

PROOF. Suppose Lie(G) is semisimple, and let N be a normal connected commutativesubgroup 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 haveto prove n = 0. Let G act on g through the adjoint representation Ad: G → GLg, and letH be the subgroup of G whose elements fix those of n (see 13.17). Then (ibid.), the Liealgebra 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]]

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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 centreZ(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. 2

COROLLARY 14.3 Assume char(k) = 0. For a connected algebraic groupG, Lie(R(G)) =r(g).

PROOF. From the exact sequence

1→ RG→ G→ G/RG→ 1

we get an exact sequence

1→ Lie(RG)→ g→ Lie(G/RG)→ 1

in which Lie(RG) is solvable (obvious) and Lie(G/RG) is semisimple (14.2). ThereforeLieRG is the largest solvable ideal in g. 2

The map ad

For a k-vector space with a k-bilinear pairing

a, b 7→ ab : C × C → C,

we write Derk(C) for the space of k-derivations C → C, i.e., k-linear maps δ : C → Csatisfying 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 Derk(C) is asubalgebra of gl(C), not Endk-lin(C).

For a Lie algebra g, the Jacobi identity says that the map ad(x) = (y 7→ [x, y]) is aderivation of g:

[x, [y, z]] = −[y, [z, x]]− [z, [x, y]] = [y, [x, z]] + [[x, y], z].

Thus, ad: g→ Endk-lin(g) maps into Derk(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→ Derk(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 theautomorphisms of G of the form inn(g)). Thus the theorem says that all derivations of asemisimple Lie algebra are inner.

We discuss the proof of the theorem below (see Humphreys 1972, 5.3).

The Lie algebra of Autk(C)

Again, let C be a finite-dimensional k-vector space with a k-bilinear pairing C × C → C.

PROPOSITION 14.5 The functor

R 7→ Autk-alg(R⊗k C)

is an algebraic subgroup of GLC .

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14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 101

PROOF. Choose a basis for C. Then an element of Autk-lin(R ⊗k C) is represented bya matrix, and the condition that it preserve the algebra product is a polynomial conditionon the matrix entries. [Of course, to be rigorous, one should write this out in terms of thebi-algebra.] 2

Denote this algebraic group by AutC , so AutC(R) = Autk-alg(R⊗k C).

PROPOSITION 14.6 The Lie algebra of AutC is gl(C) ∩Derk(C).

PROOF. Let id+εα ∈ Lie(GLC), and let a+a′ε, b+b′ε be elements ofC⊗kk[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. 2

The map Ad

Let G be a connected algebraic group. Recall (p89) that there is a homomorphism

Ad: G→ GLg .

Specifically, g ∈ G(R) acts on g⊗k R ⊂ G(R[ε]) as inn(g),

x 7→ gxg−1.

On applying Lie, we get a homomorphism

ad: Lie(G)→ Lie(GLg) ' End(g),

and we defined[x, y] = ad(x)(y).

LEMMA 14.7 The homomorphism Ad has image in Autg; in other words, for all g ∈G(R), the automorphism Ad(g) of g⊗k R preserves the bracket. Therefore, ad maps intoDerk(g).

PROOF. Because of (3.8), it suffices to prove this for G = GLn. But A ∈ GL(R) acts ong⊗k R = Mn(R) as

X 7→ AXA−1.

Now

A[X,Y ]A−1 = A(XY − Y X)A−1

= AXA−1AY A−1 −AY A−1AXA−1

= [AXA−1, AY A−1]. 2

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14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 102

LEMMA 14.8 Let g ∈ G(k). The functor CG(g)

R 7→ {g′ ∈ G(R) | gg′g−1 = g′}

is an algebraic subgroup of G with Lie algebra

cg(g) = {x ∈ g | Ad(g)(x) = x}.

PROOF. Embed G in GLn. If we can prove the statement for GLn, then we obtain it for G,because CG(g) = CGLn(g) ∩G and cg(g) = cgln(g) ∩ g.

Let A ∈ GLn(k). Then

CGLn(A)(R) = {B ∈ GLn(R) | AB = BA}.

Clearly this is a polynomial (even linear) condition on the entries of B. Moreover,

Lie(CGLn(A)) = {I +Bε ∈ Lie(GLn) | A(I +Bε)A−1 = (I +Bε)}' {B ∈Mn | ABA−1 = B}. 2

PROPOSITION 14.9 For a connected algebraic groupG 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 suffices43 to prove Z = N when k = k. Ifg ∈ N(k), then cg(g) = g, and so CG(g) = G (by 14.8). Therefore g ∈ Z(k). We haveshown that Z(k) = N(k), and this implies44 that Z = N . 2

THEOREM 14.10 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→ Autg, we get

ad: g→ Lie(Autg) ⊂ Der(g).

But, according to (14.4), the map g → Der(g) is surjective, which shows that ad: g →Lie(Autg) is surjective, and implies that Ad: G→ Aut◦g is a quotient map (13.7). 2

Recall that two semisimple groups G1, G2 are said to be isogenous if G1/Z(G1) ≈G2/Z(G2). 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 upto isomorphism over an algebraically closed field of characteristic zero, and all of themarise from algebraic groups. Thus this gives a complete classification of the semisimplealgebraic groups up to isogeny. We will follow a slightly different approach which givesmore information about the algebraic groups.

For the remainder of this section, k is of characteristic zero.43Let Q = N/Z; if Q

k= 0, then Q = 0.

44The map k[N ]→ k[Z] is surjective — let a be its kernel. Since∩m = 0 in k[N ], if a 6= 0, then there existsa maximal ideal m of k[N ] not containing a. Because k = k, k[N ]/m ' k (AG 2.7), and the homomorphismk[N ]→ k[N ]/m→ k is an element of N(k) r Z(k).

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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 14.11 The orthogonal complement a⊥ of an ideal a in g with respect to an asso-ciative 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⊥. 2

The Killing form on g is

κ(x, y) = Trg(ad(x) ◦ ad(y)).

That is, κ(x, y) is the trace of the k-linear map

z 7→ [x, [y, z]] : g→ g.

LEMMA 14.12 The form

κ(x, y) = Trg(ad(x) ◦ ad(y))

is associative and symmetric.

PROOF. It is symmetric because for matrices A = (aij) and B = (bij),

Tr(AB) =∑

i,jaijbji = Tr(BA).

By tradition, checking the associativity is left to the reader. 2

EXAMPLE 14.13 The Lie algebra sl2 consists of the 2× 2 matrices with trace zero. It hasas basis the elements

x =

(0 10 0

), h =

(1 00 −1

), y =

(0 01 0

),

and[x, y] = h, [h, x] = 2x, [h, y] = −2y.

Then

adx =

0 −2 00 0 10 0 0

, adh =

2 0 00 0 00 0 −2

, ady =

0 0 0−1 0 00 2 0

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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 of0 0 −20 0 00 0 0

,0 0 0

0 0 −20 0 0

,2 0 0

0 2 00 0 0

.

In fact, κ has matrix

0 0 40 8 04 0 0

, which has determinant −128.

Note that, for sln, the matrix of κ is n2 − 1× n2 − 1, and so this is not something onewould like to compute.

LEMMA 14.14 Let a be an ideal in g. The Killing form on g restricts to the Killing formon a, i.e.,

κg(x, y) = κa(x, y) all x, y ∈ a.

PROOF. Let α be an endomorphism of a vector space V such that α(V ) ⊂ W ; thenTrV (α) = TrW (α|W ), because when we choose a basis for W and extend it to a basisfor V , the matrix for α takes the form

(A B0 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

Tra(adx ◦ ady|a) = Tra(adax ◦ aday) = κa(x, y). 2

PROPOSITION 14.15 (Cartan’s Criterion). A Lie subalgebra g of gl(V ) is solvable ifTrV (x ◦ y) = 0 for all x ∈ [g, g] and y ∈ g.

PROOF. If g is solvable, then an analogue of the Lie-Kolchin theorem shows that, for somechoice of a basis for V , g ⊂ tn. Then [g, g] ⊂ un and [[g, g], g] ⊂ un, which implies thetraces 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

2

COROLLARY 14.16 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 impliesthat its image is solvable. Therefore g is solvable. 2

THEOREM 14.17 (Cartan-Killing criterion). A nonzero Lie algebra g is semisimple if andonly 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 certainlyκ(a, a) = 0

45In Humphreys 1972, this is proved only for algebraically closed fields k, but this condition is obviouslyunnecessary since the statement is true over k if and only if it is true over k.

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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

gadg−→ g

ada−→ aadg−→ a

ada−→ 0.

Therefore, (ada ◦ adg)2 = 0, and so46 Tr(ada ◦ adg) = 0. In other words, κ(a, g) = 0,and so a = 0 if κ is nondegenerate. 2

A Lie algebra g is said to be a direct sum of ideals a1, . . . , ar if it is a direct sum ofthem as subspaces, in which case we write g = a1 ⊕ · · · ⊕ ar. Then [ai, aj ] ⊂ ai ∩ aj = 0for i 6= j, and so g is a direct product of the Lie subalgebras ai. A nonzero Lie algebra issimple 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 aresimple as Lie subalgebras (but a simple Lie subalgebra need not be an ideal).

THEOREM 14.18 Every semisimple Lie algebra is a direct sum

g = a1 ⊕ · · · ⊕ ar

of its minimal nonzero ideals. In particular, there are only finitely many such ideals. Everyideal in a is a direct sum of certain of the ai.

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, andhence 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 anda⊥ are not simple (as Lie subalgebras) we can decompose them again. Eventually,

g = a1 ⊕ · · · ⊕ ar

with the ai 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, a1]⊕ · · · ⊕ [a, ar],

and so a = [a, ai] for exactly one i. Then a ⊂ ai, and so a = ai (simplicity of ai). Thisshows that {a1, . . . ar} 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 minimalnonzero ideals contained in a. 2

COROLLARY 14.19 All nonzero ideals and quotients of a semisimple Lie algebra are semi-simple.

PROOF. Obvious from the theorem. 2

COROLLARY 14.20 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 ofsimple algebras. 2

46If α2 = 0, the minimum polynomial of α divides X2, and so the eigenvalues of α are zero.

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14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 106

REMARK 14.21 The theorem is surprisingly strong: a finite-dimensional vector space is asum of its minimal subspaces but is far from being a direct sum (and so the theorem failsfor commutative Lie algebras). Similarly, it fails for commutative groups: for example, ifC9 denotes a cyclic group of order 9, then

C9 × C9 = {(x, x) ∈ C9 × C9} × {(x,−x) ∈ C9 × C9}.

If a is a simple Lie algebra, one might expect that a embedded diagonally would be anothersimple ideal in a⊕ a. It is a simple Lie subalgebra, but it is not an ideal.

LEMMA 14.22 For any Lie algebra g, the space {ad(x) | x ∈ g} of inner derivations of g

is an ideal in Derk(g).

PROOF. Recall that Derk(g) is the space of k-linear endomorphisms of g satisfying theLeibniz condition; it is made into a Lie algebra by [δ, δ′] = δ ◦ δ′ − δ′ ◦ δ. For a derivationδ 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,[δ, ad(x)] = ad(δx) (44)

is inner. 2

THEOREM 14.23 If g is semisimple, then ad: g→ Der(g) is a bijection: every derivationof g is inner.

PROOF. Let adg denote the (isomorphic) image of g in Der(g). It suffices to show that theorthogonal complement (adg)⊥ of adg in D for κD is zero.

Because adg and (adg)⊥ are ideals in Der(g) (see 14.22, 14.11),

[adg, (adg)⊥] ⊂ adg ∩ (adg)⊥.

Because κD|adg = κadg is nondegenerate (14.17),

adg ∩ (adg)⊥ = 0.

Let δ ∈ (adg)⊥. For x ∈ g,

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. 2

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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 al-gebraic subgroup except G and 1, and it is almost simple if it is noncommutative and hasno proper normal algebraic subgroup of dimension > 0. Thus, for n > 1, SLn is almostsimple and PSLn =df SLn /µn is simple. An algebraic group G is said to be the almostdirect product of its algebraic subgroups G1, . . . , Gn if the map

(g1, . . . , gn) 7→ g1 · · · gn : G1 × · · · ×Gn → G

is a quotient map (in particular, a homomorphism) with finite kernel. In particular, thismeans that the Gi commute and each Gi is normal.

THEOREM 14.24 Every semisimple group G is an almost direct product

G1 × · · · ×Gr → G

of its minimal connected normal algebraic subgroups of dimension > 0. In particular, thereare only finitely many such subgroups. Every connected normal algebraic subgroup of G isa product of those Gi that it contains, and is centralized by the remaining ones.

PROOF. WriteLie(G) = g1 ⊕ · · · ⊕ gr

with the gi simple ideals. LetG1 be the identity component ofCG(g2⊕· · ·⊕gr) (notation asin 13.17). Then Lie(G1)

13.17= cg(g2⊕· · ·⊕gr) = g1, and so it is normal inG (13.18). IfG1

had a proper normal connected algebraic subgroup of dimension > 0, then g1would havean ideal other than g1 and 0, contradicting its simplicity. Therefore G1 is almost simple.Construct G2, . . . , Gr similarly. Then [gi, gj ] = 0 implies that Gi and Gj commute. Thesubgroup G1 · · ·Gr of G has Lie algebra g, and so equals G (13.6). Finally,

Lie(G1 ∩ . . . ∩Gr)12.24= g1 ∩ . . . ∩ gr = 0

and so G1 ∩ . . . ∩Gr is etale (13.9).Let H be a connected algebraic subgroup of G. If H is normal, then LieH is an ideal,

and so is a direct sum of those gi it contains and centralizes the remainder. This that H is aproduct of those Gi it contains, and is centralized by the remaining ones. 2

COROLLARY 14.25 All nontrivial connected normal subgroups and quotients of a semi-simple algebraic group are semisimple.

PROOF. Obvious from the theorem. 2

COROLLARY 14.26 If G is semisimple, then DG = G, i.e., a semisimple group has nocommutative quotients.

PROOF. This is obvious for simple groups, and the theorem then implies it for semisimplegroups. 2

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15 REDUCTIVE ALGEBRAIC GROUPS 108

15 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 Gder of G is semisimple, theconnected centreZ(G)◦ ofG is a torus, andZ(G)∩Gder is finite; moreover, Z(G)◦·Gder =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 innerautomorphisms 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)◦ ∩Gder is finite. Choose an embedding G ↪→ GLV , and writeV as a direct sum

V = V1 ⊕ · · · ⊕ Vr

of eigenspaces (see 9.14). When we choose bases for the Vi, then Z(G)◦(k) consists of thematrices A1 0 0

0. . . 0

0 0 Ar

with each Ai nonzero and scalar,47 and so its centralizer in GLV consists of the matrices ofthis shape with the Ai arbitrary. Since Gder(k) consists of commutators, it consists of suchmatrices with determinant 1. As SL(Vi) contains only finitely many scalar matrices, thisshows that Z(G)◦ ∩Gder is finite.

Note that Z(G) ·Gder is a normal algebraic subgroup ofG such thatG/(Z(G) ·Gder) iscommutative (being a quotient of G/Gder) and semisimple (being a quotient of G/R(G)).Now (14.26) shows that

G = Z(G) ·Gder.

ThereforeGder → G/R(G)

is surjective with finite kernel. As G/R(G) is semisimple, so also is Gder. 2

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 alge-braic group and let G→ Gder be the simply connected covering group of Gder. The centreZ(G) of G is a group of multiplicative type, and G defines a homomorphism Z(G) →Z(Gder) → Z(G). Thus, from G we get a triple (G, Z, Z(G) → Z) with G a simplyconnected semisimple group, Z a group of multiplicative type, and Z(G) → Z a homo-morphism. Conversely, every such triple (G, Z, ϕ) gives rise to a reductive group, namely,the quotient

Z(G)(ϕ,inclusion)−→ Z × G→ G→ 0. (45)

47That is, of the form diag(a, . . . , a) with a 6= 0.

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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 issimple if it does not contain a nonzero proper submodule.

PROPOSITION 15.3 The following conditions on an A-module M of finite dimension48

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 (Si)i∈I be the set of simplemodules of M . For J ⊂ I , let N(J) =

∑j∈J Sj . Let J be maximal among the subsets of

I for which(i) the sum

∑j∈J Sj 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 eachSi ⊂ N +N(J). Because Si is simple, Si ∩ (N +N(J)) equals Si or 0. In the first case,Si ⊂ 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). 2

DEFINITION 15.4 An A-module is semisimple if it satisfies the equivalent conditions ofthe proposition.

Representations of reductive groups

Throughout this subsection, k is algebraically closed. Representations are always on finite-dimensional k-vector spaces. We shall sometimes refer to a vector space with a representa-tion of G on it as a G-module. The definition and result of the last subsection carry over toG-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 alittle complicated). 2

THEOREM 15.6 Let G be an algebraic group. All representations of G are semisimple ifand only if G◦ is reductive.

LEMMA 15.7 The restriction to any normal algebraic subgroup of a semisimple represen-tation is again semisimple.

PROOF. Let G→ GLV be a representation of G, which we may assume to be simple. LetS be a simple N -submodule of V . For any g ∈ G(k), gS is a simple N -submodule, and Vis a sum of the gS. 2

LEMMA 15.8 All representations of G are semisimple if and only if all representations ofG◦ are semisimple

48I assume this only to avoid using Zorn’s lemma in the proof.

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15 REDUCTIVE ALGEBRAIC GROUPS 110

PROOF. =⇒ : Since G◦ is a normal algebraic subgroup of G (8.13), this follows from thepreceding 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◦-stablesubspace W ′. Let p be the projection map V → W — it is a G◦-equivariant49 map whoserestriction to W is idW . Define

q : V →W, q =1n

∑ggpg−1,

where n = (G(k) : G◦(k)) and g runs over a set of coset representatives forG◦(k) inG(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 ′′ isstable under G. 2

REMARK 15.9 The lemma implies that the representations of a finite group are semisim-ple. This would fail if we allowed the characteristic to divide the order of the finite group.

LEMMA 15.10 Every representation of a semisimple algebraic group is semisimple.

PROOF. From a representation G → GLV of G on V we get a representation g → glV ofg on V , and a subspace W of V is stable under G if and only if it is stable under g (see13.15). Therefore, the statement follows from (15.5). 2

Proof of Theorem 15.6

Lemma 15.8 allows us to assume G is connected.=⇒ : Let G → GLV 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, sayV =

⊕Vi with Vi simple. Because N is solvable, the Lie-Kolchin theorem (11.23) shows

that the elements of N have a common eigenvector in Vi (cf. the proof of the theorem) andso Vi has dimension 1, and because N is unipotent it must act trivially on Vi. Therefore, Nacts 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→ GLV

be a representation of G. Then V =⊕

i Vi where Vi is the subspace of V on which Z◦

acts through a character χi (see 9.14). Because Z◦ and G′ commute, each space Vi is stableunder G′, and because G′ is semisimple, Vi =

⊕j Vij with each Vij simple as a G′-module

(15.10). Now V =⊕

i,j Vij is a decomposition of V into a direct sum of simpleG-modules.

REMARK 15.11 It is not necessary to assume k is algebraically closed. In fact, for analgebraic group G over k of characteristic zero, all representations of G are semisimpleif and only if all representations of Gk 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.

49That is, it is a homomorphism of G◦-representations.50Deligne, P., and Milne, J., Tannakian Categories. In Hodge Cycles, Motives, and Shimura Varieties, Lec-

ture Notes in Math. 900 (1982), Springer, Heidelberg, 101-228.

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15 REDUCTIVE ALGEBRAIC GROUPS 111

REMARK 15.12 Classically, the proof was based on the following two results:Every semisimple algebraic group G over C has a (unique) model G0 over R such that

G(R) is compact.For example, SLn = (G0)C where G0 is the special unitary group (see p91).

Every representation of an algebraic group G over R such that G(R) is compact issemisimple.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 orthogonalcomplement of W is a G-stable complement.

A criterion to be reductive

There is an isomorphism of algebraic groups GLn → GLn sending an invertible matrix Ato the transpose (A−1)t of its inverse. The image of an algebraic subgroupH of GLn underthis map is the algebraic subgroup Ht of GLn such that Ht(R) = {At | A ∈ H(R)} forall k-algebras R.

Now consider GLV . The choice of a basis for V determines an isomorphism GLV ≈GLn and hence a transpose map on GLV , which depends on the choice of the basis.

PROPOSITION 15.13 Every connected algebraic subgroup G of GLV such that G = Gt

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 thealgebraic closure51 k of k. Recall that the radical of G is the largest connected normalsolvable subgroup of G. It follows from (11.29c) that RG is contained in every maximalconnected solvable subgroup of G. Let B be such a subgroup, and choose a basis for Vsuch that B ⊂ Tn (Lie-Kolchin theorem 11.23). Then Bt is also a maximal connectedsolvable subgroup of G, and so

RG ⊂ B ∩Bt = Dn.

This proves that RG is diagonalizable. 2

EXAMPLE 15.14 The group GLV itself is reductive.

EXAMPLE 15.15 Since the transpose of a matrix of determinant 1 has determinant 1, SLV

is reductive.

It is possible to verify that SOn and Spn are reductive using this criterion (to be added;cf. Humphreys 1972, Exercise 1-12, p6). They are semisimple because their centres arefinite (this can be verified directly, or by studying their roots — see below).

51More precisely, one can prove that R(Gk) = (RG)

kand similarly for the unipotent radial (provided k is

perfect).

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16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 112

16 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 algebraicgroup isomorphic to a product of copies of Gm. A torus over k is an algebraic group thatbecomes isomorphic to a split torus over k. A torus in GLV is split if and only if it iscontained in Dn for some basis of k.

Consider for example

T =

{(a b−b a

)| a2 + b2 6= 0

}.

The characteristic polynomial of such a matrix is

X2 − 2aX + a2 + b2 = (X − a)2 + b2

and so its eigenvalues areλ = 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(Gm) ' Z: the only homomorphisms Gm → Gm are the maps

t 7→ tn for n ∈ Z. For a split torus T , we set

X∗(T ) = Hom(T,Gm) = group of characters of T,

X∗(T ) = Hom(Gm, T ) = group of cocharacters of T.

There is a pairing

〈 , 〉 : X∗(T )×X∗(T )→ End(Gm) ' Z, 〈χ, λ〉 = χ ◦ λ.

Thusχ(λ(t)) = t〈χ,λ〉 for t ∈ Gm(R) = R×.

Both X∗(T ) and X∗(T ) are free abelian groups of rank equal to the dimension of T , andthe pairing 〈 , 〉 realizes each as the dual of the other.

For example, let

T = Dn =

a1 0

. . .0 an

.

Then X∗(T ) has basis χ1, . . . , χn, where

χi(diag(a1, . . . , an)) = ai,

and X∗(T ) has basis λ1, . . . , λn, where

λi(t) = diag(1, . . . ,it, . . . , 1).

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16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 113

Note that

〈χj , λi〉 =

{1 if i = j0 if i 6= j

,

i.e.,

χj(λi(t)) =

{t = t1 if i = j1 = t0 if i 6= j

.

Some confusion is caused by the fact that we write X∗(T ) and X∗(T ) as additivegroups. For example,

(5χ2 + 7χ3)diag(a1, a2, a3) = χ2(a2)5χ3(a3)7.

For this reason, some authors use an exponentional notation χ(a) = aχ. With this notation,the preceding equation becomes

a5χ2+7χ3 = χ2(a2)5χ3(a3)7, a = diag(a1, a2, a3).

Split reductive groups

Let G be an algebraic group over a field k. When k = k, a torus T ⊂ G is maximal if itis not properly contained in any other torus. For example, Dn is a maximal torus in GLn

because it is equal to own centralizer in GLn. In general, T ⊂ G is said to be maximal ifTk is maximal in Gk. 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 G0 over k, unique up to isomorphism,such that G0k ≈ G.

EXAMPLE 16.1 The group GLn is a split reductive group (over any field) with split max-imal torus Dn. On the other hand, let H be the quaternion algebra over R. As an R-vectorspace, H has basis 1, i, j, k, and the multiplication is determined by

i2 = −1, j2 = −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 ≈M2(C), GC ≈ GL2, but G is not split.

EXAMPLE 16.2 The group SLn is split reductive (in fact, semisimple) group, with splitmaximal 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 finite-dimensional vector space and q is a nondegenerate quadratic form on V with associatedsymmetric form φ. Recall (5.5) that the Witt index of (V, q) is the maximum dimension ofan isotropic subspace of V . If the Witt index is r, then it follows from (5.2) that V is anorthogonal sum

V = H1 ⊥ . . . ⊥ Hr ⊥ V1 (Witt decomposition)

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16 SPLIT REDUCTIVE GROUPS: THE PROGRAM 114

where each Hi is a hyperbolic plane52 and V1 is anisotropic. It can be shown that theassociated algebraic group SO(q) is split if and only if its Witt index is as large as possible.

(a) Case dimV = n is even. When the Witt index is as large as possible, n = 2r, and

there is a basis for which the matrix53 of the form is

(0 II 0

), and so

q(x1, . . . , xn) = x1xr+1 + · · ·+ xrx2r.

Note that the subspace of vectors

(∗, . . . , r∗, 0, . . . , 0)

is totally isotropic. The algebraic subgroup consisting of the diagonal matrices of the form

diag(a1, . . . , ar, a−11 , . . . , a−1

r )

is a split maximal torus in SO(q).(b) Case dimV = 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

1 0 00 0 I0 I 0

, and so

q(x0, x1, . . . , xn) = x20 + x1xr+1 + · · ·+ xrx2r.

The algebraic subgroup consisting of the diagonal matrices of the form

diag(1, a1, . . . , ar, a−11 , . . . , a−1

r )

is a split maximal torus in SO(q).Henceforth, by SOn, I’ll mean one of the above two groups.

EXAMPLE 16.4 Let V = k2n, and letψ be the skew-symmetric form with matrix

(0 I−I 0

),

soψ(~x, ~y) = x1yn+1 + · · ·+ xny2n − xn+1y1 − · · · − x2nyn.

The corresponding symplectic group Spn is split, and the algebraic subgroup consisting ofthe diagonal matrices of the form

diag(a1, . . . , ar, a−11 , . . . , a−1

r )

is a split maximal torus in Spn.

52That is, has matrix�

0 11 0

�relative a suitable basis, and so q = xy + yx.

53Moreover, SO(q) consists of the automorphs of this matrix with determinant 1, i.e., SO(q)(R) consists of

the n× n matrices A with entries in R and determinant 1 such that At

�0 II 0

�A =

�0 II 0

�.

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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 byan element of G(k). Rather than working with split reductive groups G, it turns out to bebetter to work with pairs (G,T ) with T a split maximal torus in G.

16.5 To each pair (G,T ) consisting of a split reductive group and a maximal torus, we as-sociate 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 ).

16.6 Classify the root data.

16.7 Since knowing the root datum of (G,T ) is equivalent to knowing (G,T ), we shouldbe able to read off information about the structure ofG and its representations from the rootdatum. This is true.

16.8 The root data have nothing to do with the field! In particular, we see that for eachreductive group G over k, there is (up to isomorphism) exactly one split reductive groupover k that becomes isomorphic to G over k. However, there will in general be manynonsplit 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 fieldto be algebraically closed in order to have enough eigenvalues (and eigenvectors). By re-quiring that the group contains a split maximal torus, we are ensuring that there are enougheigenvalues without requiring the ground field to be algebraically closed.

Example: the forms of GL2. What are the groups G over a field k such that Gk ≈ GL2?For any a, b ∈ k×, define H(a, b) to be the algebra over k with basis 1, i, j, ij as a k-vectorspace, and with the multiplication given by

i2 = a, j2 = b, ij = −ji.

This is a k-algebra with centre k, and it is either a division algebra or is isomorphic toM2(k). For example, H(1, 1) ≈ M2(k) and H(−1,−1) is the usual quaternion algebrawhen 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 GL2

over k, andG(a, b) ≈ G(a′, b′) ⇐⇒ H(a, b) ≈ H(a′, b′).

Over R, every H is isomorphic to H(−1,−1) or M2(R), and so there are exactly twoforms of GL2 over R.

Over Q, the isomorphism classes of H’s are classified54 by the subsets of

{2, 3, 5, 7, 11, 13, . . . ,∞}

having a finite even number of elements. In particular, there are infinitely many forms ofGL2 over Q, exactly one of which, GL2, is split.

54The proof of this uses the quadratic reciprocity law in number theory.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 116

17 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→ GLg .

In particular, T acts on g, and so it decomposes

g = g0 ⊕⊕

where g0 is the subspace where T acts trivially, and gχ is the subspace on which T actsthrough the nontrivial character χ (see 9.14). The nonzero χ occurring in this decomposi-tion are called the roots of (G,T ). They form a finite subset Φ of X∗(T ).

Example: GL2

Here

T =

{(x1 00 x2

) ∣∣∣∣∣ x1x2 6= 0

},

X∗(T ) = Zχ1 ⊕ Zχ2,(

x1 00 x2

)χi7−→ xi,

g = M2(k),55and T acts on g by conjugation:(

x1 00 x2

)(a bc d

)(x−1

1 00 x−1

2

)=

(a x1

x2b

x2x1c d

).

Write Eij for the matrix with a 1 in the ijth-position, and zeros elsewhere. Then T actstrivially on g0 = 〈E11, E22〉, through the character α = χ1 − χ2 on gα = 〈E12〉, andthrough the character −α = χ2 − χ1 on g−α = 〈E21〉.

Thus, Φ = {α,−α} where α = χ1 − χ2. When we use χ1 and χ2 to identify X∗(T )with Z⊕ Z, Φ becomes identified with {(1,−1), (−1, 1)}.

Example: SL2

Here

T =

{(x 00 x−1

)},

X∗(T ) = Zχ,(

x 00 x−1

)χ7−→ x,

g = {(

a bc d

)∈M2(k) | a+ d = 0}.

55Recall that, for example, �x1 00 x2

� 3χ1+5χ27−→ x31x

52�

x1 00 x2

� 3χ1−5χ27−→ x31/x5

2.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 117

Again T acts on g by conjugation:(x 00 x−1

)(a bc −a

)(x−1 00 x

)=

(a x2b

x−2c −a

)

Therefore, the roots are α = 2χ and −α = −2χ. When we use χ to identify X∗(T ) withZ, Φ becomes identified with {2,−2}.

Example: PGL2

Recall that this is the quotient of GL2 by its centre: PGL2 = GL2 /Gm. One can provethat for all rings R, PGL2(R) = GL2(R)/R×. Here

T ={(

x1 00 x2

) ∣∣∣ x1x2 6= 0}/{( x 0

0 x ) | x 6= 0} ,

X∗(T ) = Zχ,(

x1 00 x2

)χ7−→ x1

x2,

g = M2(k)/{aI} (quotient as a vector space).

and T acts on g by conjugation:(x1 00 x2

)(a bc d

)(x−1

1 00 x−1

2

)=

(a x1

x2b

x2x1c d

).

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: GLn

Here

T =

{(x1 0

. . .0 xn

) ∣∣∣∣∣ x1 · · ·xn 6= 0

},

X∗(T ) =⊕

1≤i≤nZχi,

(x1 0

. . .0 xn

)χi7−→ xi,

g = Mn(k),

and T acts on g by conjugation:

(x1 0

. . .0 xn

)a11 ··· ··· a1n

... aij

......

...an1 ··· ··· ann

x−1

1 0

. . .0 x−1

n

=

a11 ··· ··· x1

xna1n

... xixj

aij

......

...xnx1

an1 ··· ··· ann

.

Write Eij for the matrix with a 1 in the ijth-position, and zeros elsewhere. Then T actstrivially on g0 = 〈E11, . . . , Enn〉 and through the character αij = χi−χj on gαij = 〈Eij〉,and so

Φ = {αij | 1 ≤ i, j ≤ n, i 6= j}.

When we use the χi to identify X∗(T ) with Zn, then Φ becomes identified with {ei − ej |i 6= j} where e1, . . . , en is the standard basis for Zn.

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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∨,

56satisfying the following conditionsrd1 〈α, α∨〉 = 2,rd2 sα(Φ) ⊂ Φ where sα is the homomorphism X → X defined by

sα(x) = x− 〈x, α∨〉α, for x ∈ X , α ∈ Φ,

rd3 the group W = W (Ψ) of automorphisms of X generated by the sα is finite.

Note that (rd1) implies thatsα(α) = −α,

and that the converse holds if α 6= 0. Moreover, if sα(α) = −α, then57

s2α = 1.

Clearly, also sα(x) = x if 〈x, α∨〉 = 0. Thus, sα should be considered an “abstractreflection 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 = SL2. Here

X = X∗(T ) = Zχ,(

x 00 x−1

)χ7−→ x

X∨ = X∗(T ) = Zλ, tλ7−→(

t 00 t−1

)Φ = {α,−α}, α = 2χ

Φ∨ = {α∨,−α∨}, α∨ = λ.

56Thus, a root datum is really an ordered sextuple,

X, X∨, 〈 , 〉, Φ, Φ∨, Φ→ Φ∨,

but everyone says quadruple.57Because

sα(sα(x)) = sα(x− 〈x, α∨〉α)

= (x− 〈x, α∨〉α)− 〈x, α∨〉sα(α)

= x.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 119

Note thatt

λ7−→(

t 00 t−1

)2χ7−→ t2

and so〈α, α∨〉 = 2.

As always,sα(α) = −α, sα(−α) = α

etc., and so s±α(Φ) ⊂ Φ. Finally, W (Ψ) = {1, sα} is finite, and so Ψ(SL2, T ) is a rootsystem, 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 = PGL2. Here

Φ∨ = {α∨,−α∨}, α∨ = 2λ.

In this case Ψ(PGL2, 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. Itfollows immediately, that the above are the only two root data withX = Z and Φ nonempty.There is also the root datum

(Z, ∅,Z, ∅),

which is the root datum of the reductive group Gm.

EXAMPLE 17.5 Let G = GLn. Here

X = X∗(Dn) =⊕

iZχi, diag(x1, . . . , xn)

χi7−→ xi

X∨ = X∗(Dn) =⊕

iZλi, t

λi7−→ diag(1, . . . , 1,it, 1, . . . , 1)

Φ = {αij | i 6= j}, αij = χi − χj

Φ∨ = {α∨ij | i 6= j}, α∨ij = λi − λj .

Note that

tλi−λj7−→ diag(1, . . . ,

it, . . . ,

j

t−1, . . .)χi−χj7−→ t2

and so〈αij , α

∨ij〉 = 2.

Moreover, sα(Φ) ⊂ Φ for all α ∈ Φ. We have, for example,

sαij (αij) = −αij

sαij (αik) = αik − 〈αik, α∨ij〉αij

= αik − 〈χi, λi〉αij (if k 6= i, j)

= χi − χk − (χi − χj)= αjk

sαij (αkl) = αkl (if k 6= i, j, l 6= i, j).

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 120

Finally, let E(ij) be the permutation matrix in which the ith and jth rows have beenswapped. The action

A 7→ E(ij) ·A · E(ij)−1

of Eij on GLn by inner automorphisms stabilizes T and swaps xi and xj . Therefore, it actson X = X∗(T ) as sαij . This shows that the group generated by the sαij is isomorphic tothe subgroup of GLn generated by the E(ij), which is isomorphic to Sn. In particular, Wis finite.

Therefore, Ψ(GLn,Dn) is a root datum, isomorphic to

(Zn, {ei − ej | i 6= j},Zn, {ei − ej | i 6= j}

where ei = (0, . . . ,i1, . . . , 0), the pairing is the standard one 〈ei, ej〉 = δij , and (ei−ej)∨ =

ei − ej .

In the above examples we wrote down the coroots without giving any idea of how tofind (or even define) them. Before defining them, we need to review some general resultson 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 largestr such that Gk contains a subgroup isomorphic to Gr

m. Since all maximal tori in Gk areconjugate (see 17.17 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 SL2 or PGL2.

PROOF. (SKETCH) (a) Take k = k. If all the elements of G(k) are unipotent, then Gis 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 thereforedecomposes intoHs×Hu (see 11.8). If all semisimple elements ofG(k) are of finite order,then G is finite (hence trivial, being connected). If G(k) contains a semisimple element ofinfinite 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 ≈ P1. From this

one gets a nontrivial homomorphism G→ Aut(P1) ' PGL2. 2

Centralizers and normalizers

Let T be a torus in an algebraic groupG. The centralizer of T inG is the algebraic subgroupC = CG(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 = NG(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.

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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 CG(T ) of T in G is a reductive group; in particular, it is connected.(b) The identity component of the normalizer NG(T ) of T in G is CG(T ); in particular,

NG(T )/CG(T ) is a finite etale group.(c) The torus T is maximal if and only if T = CG(T ).

PROOF. (a) Omitted. (When k = k, the statement is proved in Humphreys 1975, 26.2.)(b) Certainly NG(T )◦ ⊃ CG(T )◦ = CG(T ). But NG(T )◦/CG(T ) acts faithfully on T ,

and so is trivial by rigidity (9.15). For the second statement, see §8.(c) Certainly, if CG(T ) = T , then T is maximal because any torus containing T is

contained in CG(T ). Conversely, CG(T ) is reductive group containing T , and so CG(T ) =D(CG(T )) · Z(CG(T ))◦ (see 15.1). It follows that Z(CG(T ))◦ = T and D(CG(T )) is asemisimple group of rank 0, and hence is trivial. Thus CG(T ) = Z(CG(T ))◦ = T . 2

The quotient W (G,T ) = NG(T )/CG(T ) is called the Weyl group of (G,T ). It is aconstant etale algebraic group58 when T is split, and so may be regarded simply as a finitegroup.

Definition of the coroots

LEMMA 17.8 Let G be a split reductive group with split maximal torus T . The action ofW (G,T ) on X∗(T ) stabilizes Φ.

PROOF. Take k = k. Let s normalize T (and so represent an element of W ). Then s actson X∗(T ) (on the left) by

(sχ)(t) = χ(s−1ts).

Let α be a root. Then, for x ∈ gα and t ∈ T (k),

t(sx) = s(s−1ts)x = s(α(s−1ts)x) = α(s−1ts)sx,

and so T acts on sgα through the character sα, which must therefore be a root. 2

For a root α of (G,T ), let Tα = Ker(α)◦, and let Gα be centralizer of Tα.

THEOREM 17.9 Let G be a split reductive group with split maximal torus T .(a) For each α ∈ Φ, there is a unique sα 6= 1 in W (Gα, T ) and a unique α∨ ∈ X∗(T )

such thatsα(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 7→

α∨ : Φ→ Φ∨ is a root datum.

PROOF. (SKETCH) (a) The key point is that CGα(Tα) is a reductive group of rank onewith T as a maximal torus. Thus, we are essentially in the case of SL2 or PGL2, whereeverything is obvious (see below). Note that the uniqueness of α∨ follows from that of sα.

(b) We noted in (a) that (rd1) holds. The sα attached to α lies inW (Gα, T ) ⊂W (G,T ),and so stabilizes Φ by the lemma. Finally, all sα lie in the Weyl group W (G,T ), and sothey generate a finite group (in fact, the generate exactly W (G,T )). 2

58That is, W (R) is the same finite group for all integral domains R.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 122

EXAMPLE 17.10 Let G = SL2, and let α be the root 2χ. Then Tα = T and Gα = G. Theunique s 6= 1 in W (G,T ) is represented by(

0 1−1 0

),

and the unique α∨ for which (46) holds is λ.

EXAMPLE 17.11 Let G = GLn, and let α = α12 = χ1 − χ2. Then

Tα = {diag(x, x, x3, . . . , xn) | xx3 . . . xn 6= 1}

and Gα consists of the invertible matrices of the form∗ ∗ 0 0∗ ∗ 0 00 0 ∗ 0

. . ....

0 0 0 · · · ∗

.

Clearly

nα =

0 1 0 01 0 0 00 0 1 0

. . ....

0 0 0 · · · 1

represents the unique nontrivial element sα of W (Gα, Tα). It acts on T by

diag(x1, x2, x3, . . . , xn) 7−→ diag(x2, x1, x3, . . . , xn).

For x =∑miχi,

sαx = m2χ1 +m1χ2 +m3χ3 + · · ·+mnχn

= x− 〈x, λ1 − λ2〉(χ1 − χ2).

Thus (46) holds if and only if α∨ is taken to be λ1 − λ2.

Computing the centre

PROPOSITION 17.12 Every maximal torus T in a reductive algebraic groupG contains thecentre Z = Z(G) of G.

PROOF. Clearly Z ⊂ CG(T ), but (see 17.7), CG(T ) = T . 2

Recall (14.9) that the kernel of the adjoint map Ad: G → GLg is Z(G), and so thekernel of Ad: T → GLg is Z(G) ∩ T = Z(G). Therefore

Z(G) = Ker(Ad |T ) =⋂

α∈ΦKer(α).

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 123

We can use this to compute the centres of groups. For example,

Z(GLn) =⋂

i6=jKer(χi − χj) =

{(x1 0

. . .0 xn

) ∣∣∣∣∣ x1 = x2 = · · · = xn 6= 0

},

Z(SL2) = Ker(2χ) ={(

x 00 x−1

)| x2 = 0

}= µ2,

Z(PGL2) = Ker(χ) = 1.

On applying X∗ to the exact sequence

0→ Z(G)→ Tt7→(...,α(t),...)−−−−−−−−→

∏α∈Φ

Gm (47)

we get (see 9.11) an exact sequence

⊕α∈Φ

Z(...,mα,...) 7→

∑mαα

−−−−−−−−−−−−→ X∗(T )→ X∗(Z(G))→ 0,

and soX∗(Z(G)) = X∗(T )/{subgroup generated by Φ}. (48)

For example,

X∗(Z(GLn)) ' Zn/〈ei − ej | i 6= j〉(a1,...,an) 7→

∑ai−−−−−−−−−−→

'Z,

X∗(Z(SL2)) ' Z/(2),X∗(Z(PGL2)) ' Z/Z = 0.

Semisimple and toral root data

DEFINITION 17.13 A root datum is semisimple if Φ generates a subgroup of finite indexin X .

PROPOSITION 17.14 A split reductive group is semisimple if and only if its root datum issemisimple.

PROOF. A reductive group is semisimple if and only if its centre is finite, and so this followsfrom (48). 2

DEFINITION 17.15 A root datum is toral if Φ is empty.

PROPOSITION 17.16 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 thatG = T . Conversely, ifG is a torus, the adjoint representationis trivial and so g = g0. 2

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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 17.17 Let T, T ′ be split maximal tori in G. Then there exists a g ∈ G(k) suchthat T ′ = gTg−1 (i.e., inn(g)(T ) = T ′).

PROOF. Omitted for the present. 2

EXAMPLE 17.18 Let G = GLV , 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 ⊂ Dn. Since T is maximal, itequals Dn. This proves the theorem for GLV .

It follows that the root datum attached to (G,T ) depends only on G (up to isomor-phism).

THEOREM 17.19 (ISOMORPHISM) Every isomorphism Ψ(G,T ) → Ψ(G′, T ′) of rootdata arises from an isomorphism ϕ : G → G′ such that ϕ(T ) = T ′; moreover, ϕ is uniqueup to composition by a homomorphism inn(t) with t ∈ T (k) and α(t) ∈ k for all α.

PROOF. Springer 1998, 16.3.2. 2

THEOREM 17.20 (EXISTENCE) Every reduced root datum arises from a split reductivegroup.

PROOF. Springer 1998, 16.5. 2

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 inGLn. We first compute the weights of the split maximal torus on gln, and then check thateach nonzero weight occurs in g (in fact, with multiplicity 1). Then for each α we find anatural copy of SL2 (or PGL2) centralizing Tα, and use it to find the coroot α∨.

Example (An): SLn+1.

Let G be SLn+1 and let T be the algebraic subgroup of diagonal matrices:

{diag(t1, . . . , tn+1) | t1 · · · tn+1 = 1}.

Then

X∗(T ) =⊕

Zχi

/Zχ,

{diag(t1, . . . , tn+1)

χi7−→ tiχ =

∑χi

X∗(T ) = {∑

aiλi |∑

ai = 0}, tχi7−→ diag(ta1 , . . . , tan), ai ∈ Z,

with the obvious pairing 〈 , 〉. Write χi for the class of χi inX∗(T ). Then all the charactersχi−χj , i 6= j, occur as roots, and their coroots are respectively λi−λj , i 6= j. This followseasily from the calculation of the root datum of GLn.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 125

Example (Bn): SO2n+1 .

Consider the symmetric bilinear form φ on k2n+1,

φ(~x, ~y) = 2x0y0 + x1yn+1 + xn+1y1 + · · ·+ xny2n + x2nyn

Then SO2n+1 = SO(φ) consists of the 2n+ 1× 2n+ 1 matrices A of determinant 1 suchthat

φ(A~x,A~y) = φ(~x, ~y),

i.e., such that

At

1 0 00 0 I0 I 0

A =

1 0 00 0 I0 I 0

.The Lie algebra of SO2n+1 consists of the 2n+ 1× 2n+ 1 matrices A of trace 0 such that

φ(A~x, ~y) = φ(~x,A~y),

i.e., such that

At

1 0 00 0 I0 I 0

=

1 0 00 0 I0 I 0

A.Take T to be the maximal torus of diagonal matrices

diag(1, t1, . . . , tn, t−11 , . . . , t−1

n )

Then

X∗(T ) =⊕

1≤i≤nZχi, diag(1, t1, . . . , tn, t−1

1 , . . . , t−1n )

χi7−→ ti

X∗(T ) =⊕

1≤i≤nZλi, t

λi7−→ diag(1, . . . ,it, . . . , 1)

with the obvious pairing 〈 , 〉. All the characters

±χi, ±χi ± χj , i 6= j

occur as roots, and their coroots are, respectively,

±2λi, ±λi ± λj , i 6= j.

Example (Cn): Sp2n .

Consider the skew symmetric bilinear form k2n × k2n → k,

φ(~x, ~y) = x1yn+1 − xn+1y1 + · · ·+ xny2n − x2nyn.

Then Sp2n consists of the 2n× 2n matrices A such that

φ(A~x,A~y) = φ(~x, ~y),

i.e., such that

At

(0 I−I 0

)A =

(0 I−I 0

).

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 126

The Lie algebra of Spn consists of the 2n× 2n matrices A such that

φ(A~x, ~y) = φ(~x,A~y),

i.e., such that

At

(0 I−I 0

)=

(0 I−I 0

)A.

Take T to be the maximal torus of diagonal matrices

diag(t1, . . . , tn, t−11 , . . . , t−1

n ).

Then

X∗(T ) =⊕

1≤i≤nZχi, diag(t1, . . . , tn, t−1

1 , . . . , t−1n )

χi7−→ ti

X∗(T ) =⊕

1≤i≤nZλi, t

λi7−→ diag(1, . . . ,it, . . . , 1)

with the obvious pairing 〈 , 〉. All the characters

±2χi, ±χi ± χj , i 6= j

occur as roots, and their coroots are, respectively,

±λi, ±λi ± λj , i 6= j.

Example (Dn): SO2n .

Consider the symmetric bilinear form k2n × k2n → k,

φ(~x, ~y) = x1yn+1 + xn+1y1 + · · ·+ xny2n + x2ny2n.

Then SOn = SO(φ) consists of the n× n matrices A of determinant 1 such that

φ(A~x,A~y) = φ(~x, ~y),

i.e., such that

At

(0 II 0

)A =

(0 II 0

).

The Lie algebra of SOn consists of the n× n matrices A of trace 0 such that

φ(A~x, ~y) = φ(~x,A~y),

i.e., such that

At

(0 II 0

)=

(0 II 0

)A.

When we write the matrix as

(A BC D

), then this last condition becomes

A = Dt, C = Ct, B = Bt.

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17 THE ROOT DATUM OF A SPLIT REDUCTIVE GROUP 127

Take T to be the maximal torus of matrices

diag(t1, . . . , tn, t−11 , . . . , t−1

n )

and let χi, 1 ≤ i ≤ r, be the character

diag(t1, . . . , tn, t−11 , . . . , t−1

n ) 7→ ti.

All the characters±χi ± χj , i 6= j

occur, and their coroots are, respectively,

±λi ± λj , i 6= j.

REMARK 17.21 The subscript on An, Bn, Cn, Dn denotes the rank of the group, i.e., thedimension of a maximal torus.

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18 GENERALITIES ON ROOT DATA 128

18 Generalities on root data

Definition

The following is the standard definition.

DEFINITION 18.1 A root datum is an ordered 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 correspondence α↔ α∨,

satisfying the following conditionsRD1 〈α, α∨〉 = 2,RD2 sα(Φ) ⊂ Φ, s∨α(Φ∨) ⊂ Φ∨, where

sα(x) = x− 〈x, α∨〉α, for x ∈ X , α ∈ Φ,s∨α(y) = y − 〈α, y〉α∨, for y ∈ X∨, α ∈ Φ.

Recall that RD1 implies that sα(α) = −α and s2α = 1.Set

Q = ZΦ ⊂ X Q∨ = ZΦ∨ ⊂ X∨

V = Q⊗Z Q V ∨ = Q⊗Z Q∨.

X0 = {x ∈ X | 〈x,Φ∨〉 = 0}By ZΦ we mean the Z-submodule of X generated by the α ∈ Φ.

LEMMA 18.2 For α ∈ Φ, x ∈ X , and y ∈ X∨,

〈sα(x), y〉 = 〈x, s∨α(y)〉, (49)

and so〈sα(x), s∨α(y)〉 = 〈x, y〉. (50)

PROOF. We have

〈sα(x), y〉 = 〈x− 〈x, α∨〉α, y〉 = 〈x, y〉 − 〈x, α∨〉〈α, y〉〈x, s∨α(y)〉 = 〈x, y − 〈α, y〉α∨〉 = 〈x, y〉 − 〈x, α∨〉〈α, y〉,

which gives the first formula, and the second is obtained from the first by replacing y withs∨α(y). 2

In other words, as the notation suggests, s∨α (which is sometimes denoted sα∨) is theendomorphism of X∨ defined by sα.

LEMMA 18.3 The following hold for the mapping

p : X → X∨, p(x) =∑α∈Φ

〈x, α∨〉α∨.

(a) For all x ∈ X ,〈x, p(x)〉 =

∑α∈Φ〈x, α∨〉2 ≥ 0, (51)

with strict inequality holding if x ∈ Φ.(b) For all x ∈ X and w ∈W ,

〈wx, p(wx)〉 = 〈x, p(x)〉. (52)

(c) For all α ∈ Φ,〈α, p(α)〉α∨ = 2p(α), all α ∈ Φ. (53)

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18 GENERALITIES ON ROOT DATA 129

PROOF. (b) It suffices to check this for w = sα, but

〈sαx, α∨〉 = 〈x, α∨〉 − 〈x, α∨〉〈α, α∨〉 = −〈x, α∨〉

and so each term on the right of (51) is unchanged if x with replaced with sαx.(c) Recall that, for y ∈ X∨,

s∨α(y) = y − 〈α, y〉α∨.

On multiplying this by 〈α, y〉 and re-arranging, we find that

〈α, y〉2α∨ = 〈α, y〉y − 〈α, y〉s∨α(y).

But

−〈α, y〉 = 〈sα(α), y〉(49)= 〈α, s∨α(y)〉

and so〈α, y〉2α∨ = 〈α, y〉y + 〈α, s∨α(y)〉s∨α(y).

As y runs through the elements of Φ∨, so also does s∨α(y), and so when we sum overy ∈ Φ∨, we obtain (53). 2

REMARK 18.4 Suppose mα is also a root. On replacing α with mα in (53) and using thatp is a homomorphism of Z-modules, we find that

m〈α, p(α)〉(mα)∨ = 2p(α), all α ∈ Φ.

Therefore,(mα)∨ = m−1α∨. (54)

In particular,(−α)∨ = −(α∨). (55)

LEMMA 18.5 The map p : X → X∨ defines an isomorphism

1⊗ p : V → V ∨.

In particular, dimV = dimV ∨.

PROOF. As 〈α, p(α)〉 6= 0, (53) shows that p(Q) has finite index in Q∨. Therefore, whenwe tensor p : Q → Q∨ with Q, we get a surjective map 1 ⊗ p : V → V ∨; in particu-lar, dimV ≥ dimV ∨. The definition of a root datum is symmetric between (X,Φ) and(X∨,Φ∨), and so the symmetric argument shows that dimV ∨ ≤ dimV . Hence

dimV = dimV ∨,

and 1⊗ p : V → V ∨ is an isomorphism. 2

LEMMA 18.6 The kernel of p : X → X∨ is X0.

PROOF. Clearly, X0 ⊂ Ker(p), but (51) proves the reverse inclusion. 2

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18 GENERALITIES ON ROOT DATA 130

PROPOSITION 18.7 We have

Q ∩X0 = 0Q+X0 is of finite index in X.

Thus, there is an exact sequence

0→ Q⊕X0(q,x) 7→q+x−−−−−−−→ X → finite group→ 0.

PROOF. The map1⊗ p : Q⊗X → V ∨

has kernel Q⊗X0 (see 18.6) and maps the subspace V of Q⊗X isomorphically onto V ∨

(see 18.5). This implies that

(Q⊗Z X0)⊕ V ' Q⊗X,

from which the proposition follows. 2

LEMMA 18.8 The bilinear form 〈 , 〉 defines a nondegenerate pairing V × V ∨ → Q.

PROOF. Let x ∈ X . If 〈x, α∨〉 = 0 for all a∨ ∈ Φ∨, then x ∈ Ker(p) = X0. 2

LEMMA 18.9 For any x ∈ X and w ∈W , w(x)− x ∈ Q.

PROOF. From (RD2),sα(x)− x = −〈x, α∨〉α ∈ Q.

Now(sα1 ◦ sα2)(x)− x = sα1(sα2(x)− x) + sα1(x)− x ∈ Q,

and so on. 2

Recall that the Weyl group W = W (Ψ) of Ψ is the subgroup of Aut(X) generated bythe sα, α ∈ Φ. We let w ∈W act on X∨ as (w∨)−1, i.e., so that

〈wx,wy〉 = 〈x, y〉, all w ∈W , x ∈ X , y ∈ X∨.

Note that this makes sα act on X∨ as (s∨α)−1 = s∨α (see 49).

PROPOSITION 18.10 The Weyl group W acts faithfully on Φ (and so is finite).

PROOF. By symmetry, it is equivalent to show that W acts faithfully on Φ∨. Let w be anelement of W such that w(α) = α for all α ∈ Φ∨. For any x ∈ X ,

〈w(x)− x, α∨〉 = 〈w(x), α∨〉 − 〈x, α∨〉= 〈x,w−1(α∨)〉 − 〈x, α∨〉= 0.

Thus w(x) − x is orthogonal to Φ∨. As it lies in Q (see 18.9), this implies that it is zero(18.8), and so w = 1. 2

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18 GENERALITIES ON ROOT DATA 131

Thus, a root datum in the sense of (18.1) is a root datum in the sense of (17.1), and thenext proposition proves the converse.

PROPOSITION 18.11 Let Ψ = (X,Φ, X∨,Φ∨) be a system satisfying the conditions (rd1),(rd2), (rd3) of (17.1). Then Ψ is a root datum.

PROOF. We have to show that

s∨α(Φ∨) ⊂ Φ∨ where s∨α(y) = y − 〈α, y〉α∨.

As in Lemma 18.2, 〈sα(x), s∨α(y)〉 = 〈x, y〉.Let α, β ∈ Φ, and let t = ssα(β)sαsβsα. An easy calculation59 shows that

t(x) = x+ (〈x, s∨α(β∨)〉 − 〈x, sα(β)∨〉)sα(β), all x ∈ X.

Since

〈sα(β), s∨α(β∨)〉 − 〈sα(β), sα(β)∨〉 = 〈β, β∨〉 − 〈sα(β), sα(β)∨〉 = 2− 2 = 0,

we see that t(sa(β)) = sα(β). Thus,

(t− 1)2 = 0,

and so the minimum polyonomial of t acting on Q ⊗Z X divides (T − 1)2. On the otherhand, since t lies in a finite group, it has finite order, say tm = 1. Thus, the minimumpolynomial also divides Tm − 1, and so it divides

gcd(Tm − 1, (T − 1)2) = T − 1.

This shows that t = 1, and so

〈x, s∨α(β∨)〉 − 〈x, sα(β)∨〉 = 0 for all x ∈ X.

Hences∨α(β∨) = sα(β)∨ ∈ Φ∨. 2

REMARK 18.12 To give a root datum amounts to giving a triple (X,Φ, f) where— X is a free abelian group of finite rank,— Φ is a finite subset of X , and— f is an injective map α 7→ α∨ from Φ into the dual X∨ of X

satisfying the conditions (rd1), (rd2), (rd3) of (17.1).

59Or so it is stated in Springer 1979, 1.4 (Corvallis).

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 132

19 Classification of semisimple root data

Throughout this section, F is a field of characteristic zero, for example F = Q, R, or C.An inner product on a real vector space is a positive-definite symmetric bilinear form.

Generalities on symmetries

Let V be a finite-dimensional vector space over F , and let α be a nonzero element of V . Asymmetry with vector α is an automorphism of V such that

— s(α) = −α, and— the set of vectors fixed by s is a hyperplane H .

Then V = H ⊕ 〈α〉 with s acting as 1⊕−1, and so s2 = 1.Let V ∨ be the dual vector space HomQ-lin(V, F ) of V , and write 〈x, f〉 for f(x). The

compositeV → V/H

α+H 7→2−→ F

is the unique element α∨ of V ∨ such that α(H) = 0 and 〈α, α∨〉 = 2; moreover,

s(x) = x− 〈x, α∨〉α all x ∈ V. (56)

In this way, symmetries with vector α are in one-to-one correspondence with vectors α∨

such that 〈α, α∨〉 = 2.

LEMMA 19.1 Let Φ be a finite subset of V that spans V . Then, for any nonzero vector αin V , there exists at most one symmetry s with vector α such that α(Φ) ⊂ Φ.

PROOF. Let s, s′ be such symmetries, and let t = ss′. Then t defines the identity map onboth Fα and on V/Fα, and so

(t− 1)2V ⊂ (t− 1)Fα = 0.

Thus the minimum polynomial of t divides (T −1)2. On the other hand, because Φ is finite,there exists an integer m ≥ 1 such that tm(x) = x for all x ∈ Φ and hence for all x ∈ V .Therefore the minimum polyomial of t divides Tm − 1, and hence also

gcd((T − 1)2, Tm − 1) = T − 1.

This shows that t = 1. 2

LEMMA 19.2 Let ( , ) be an inner product on a real vector space V . Then, for any nonzerovector α in V , there exists a unique symmetry s with vector α that is orthogonal for ( , ),i.e., such that (sx, sy) = (x, y) for all x, y ∈ V , namely

s(x) = x− 2(x, α)(α, α)

α. (57)

PROOF. Certainly, (57) does define an orthogonal symmetry with vector α. Conversely,suppose s′ is a second such symmetry. Let H = 〈α〉⊥. Then H is stable under s′, and mapsisomorphically on V/〈α〉. Therefore s′ acts as 1 on H . As V = H ⊕ 〈α〉 and s′ acts as −1on 〈α〉, it must coincide with s. 2

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 133

Generalities on lattices

In this subsection V is a finite-dimensional vector space over F .

DEFINITION 19.3 A subgroup of V is a lattice in V if it can be generated (as a Z-module)by a basis for V . Equivalently, a subgroup X is a lattice if the natural map F ⊗Z X → Vis an isomorphism.

REMARK 19.4 When F = Q, every finitely generated subgroup of V that spans V is alattice, but this is not true for F = R or C. For example, Z1 + Z

√2 is not a lattice in R.

DEFINITION 19.5 A perfect pairing of free Z-modules of finite rank is one that realizeseach as the dual of the other. Equivalently, it is a pairing into Z with discriminant ±1.

PROPOSITION 19.6 Let〈 , 〉 : V × V ∨ → k

be a nondegenerate bilinear pairing, and let X be a lattice in V . Then

Y = {y ∈ V ∨ | 〈X, y〉 ⊂ Z }

is the unique lattice in V ∨ such that 〈 , 〉 restricts to a perfect pairing

X × Y → Z.

PROOF. Let e1, . . . , en be a basis for V generating X , and let e′1, . . . , e′n be the dual basis.

ThenY = Ze′1 + · · ·+ Ze′n,

and so it is a lattice, and it is clear that 〈 , 〉 restricts to a perfect pairing X × Y → Z.Let Y ′ be a second lattice in V ∨ such that 〈x, y〉 ∈ Z for all x ∈ X , y ∈ Y ′. Then

Y ′ ⊂ Y , and an easy argument shows that the discriminant of the pairing X × Y ′ → Z is±(Y : Y ′), and so the pairing on X × Y ′ is perfect if and only if Y ′ = Y . 2

Root systems

DEFINITION 19.7 A root system is a pair (V,Φ) with V a finite-dimensional vector spaceover F and Φ a finite subset of V such thatRS1 Φ spans V and does not contain 0;RS2 for each α ∈ Φ, there exists a symmetry sα with vector α such that sα(Φ) ⊂ Φ;RS3 for all α, β ∈ Φ, 〈β, α∨〉 ∈ Z.

In (RS3), α∨ is the element of V ∨ corresponding to sα. Note that (19.1) shows that sα

(hence also α∨) is uniquely determined by α.The elements of Φ are called the roots of the root system. If α is a root, then sα(α) =

−α is also a root. If tα is also a root, then (RS3) shows that t = 12 or 2. A root system

(V,Φ) is reduced if no multiple of a root except its negative is a root.The Weyl group W = W (Φ) of (V,Φ) is the subgroup of GL(V ) generated by the

symmetries sα for α ∈ Φ. Because Φ spans V , W acts faithfully on Φ; in particular, it isfinite.

PROPOSITION 19.8 Let (V,Φ) be a root system over F , and let V0 be the Q-vector spacegenerated by Φ. Then

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 134

(a) the natural map F ⊗Q V0 → V is an isomorphism;(b) the pair (V0,Φ) is a root system over Q.

PROOF. The example in (19.4) shows that (a) does require proof. For a proof of the propo-sition, see Serre 1987, p42. 2

Thus, to give a root system over R or C amounts to giving a root system over Q.

Root systems and semisimple root data

Compare (18.12; 19.7)Semisimple root datum Root system (over Q)X,Φ, α 7→ α∨ : Φ ↪→ X∨ V,ΦΦ is finite Φ is finite(X : ZΦ) finite Φ spans V

0 /∈ Φ〈α, α∨〉 = 2, sα(Φ) ⊂ Φ ∃sα such that sα(Φ) ⊂ Φ

〈β, α∨〉 ∈ Z, all α, β ∈ ΦWeyl group finite

For a root system (V,Φ), let Q = ZΦ be the Z-submodule of V generated by Φ and letQ∨ be the Z-submodule of V ∨ generated by the α∨, α ∈ Φ. Then, Q and Q∨ are lattices60

in V and V ∨, and we letP = {x ∈ V | 〈x,Q∨〉 ⊂ Z}.

Then P is a lattice in V (see 19.6), and because of (RS3),

Q ⊂ P . (58)

PROPOSITION 19.9 If (X,Φ, α 7→ α∨) is a semisimple root datum, then (Q⊗Z X,Φ) is aroot system over Q. Conversely, if (V,Φ) is root system over Q, then for any choice X ofa lattice in V such that

Q ⊂ X ⊂ P (59)

(X,Φ, α 7→ α∨) is a semisimple root datum.

PROOF. If (X,Φ, α 7→ α∨) is a semisimple root datum, then 0 /∈ Φ because 〈α, α∨〉 = 2,and 〈β, α∨〉 ∈ Z because α∨ ∈ X∨. Therefore (Q⊗Z X,Φ) is a root system.

Conversely, let (V,Φ) be a root system. LetX satisfy (59), and letX∨ denote the latticein V ∨ in duality with X (see 19.6). For each α ∈ Φ, there exists an α∨ ∈ V ∨ such that〈α, α∨〉 = 2 and sα(Φ) ⊂ Φ (because (V,Φ) is a root datum), and (19.1) shows that it isunique. Therefore, we have a function α 7→ α∨ : Φ → V ∨ which takes its values in X∨

(because X ⊂ P implies X∨ ⊃ Φ∨), and is injective. The Weyl group of (X,Φ, α 7→ α∨)is the Weyl group of (V,Φ), which, as we noted above, is finite. Therefore (X,Φ, α 7→ α∨)is a semisimple root datum. 2

60They are finitely generated, and Φ∨ spans V ∨ by Serre 1987, p28.

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 135

The big picture

Take the ground field k to be of characteristic zero.

Split reductive groups ↔ Reduced root data...

...Split semisimple groups ↔ Reduced semisimple root data

↓ ↓Lie algebras k=k↔ Reduced root systems

19.10 As we discussed in (§17), the reduced root data classify the split reductive groupsover k.

19.11 As we discussed in (15.1), from a reductive group G, we get semisimple groupsDG and G/Z(G) together with an isogeny DG → G/Z(G). Conversely, every reductivegroup G can be built up from a semisimple group and a torus (15.2).

19.12 As we discuss in the next section, the relation between reduced root data and re-duced semisimple root data is the same as that between split reductive groups and splitsemisimple groups. It follows that to show that the reduced root data classify split reduc-tive groups, it suffices to show that reduced semisimple root data classify split semisimplegroups.

19.13 From a semisimple group G we get a semisimple Lie algebra Lie(G) (see 14.2),and from Lie(G) we can recover G/Z(G) (see 14.10). Passing from G to Lie(G) amountsto forgetting the centre of G.

19.14 From a semisimple root datum (X,Φ, α 7→ α∨), we get a root system (V = Q⊗ZX,Φ). Passing from the semisimple root datum to the root system amounts to forgettingthe lattice X in V .

19.15 Take k = k, and let g be a semisimple Lie algebra over k. A Cartan subalgebrah of g is a commutative subalgebra that is equal to its own centralizer. For example, thealgebra of diagonal matrices of trace zero in sln is a Cartan subalgebra. Then h acts on g

via the adjoint map ad: h→ End(g), i.e., for h ∈ h, x ∈ g, ad(h)(x) = [h, x]. One showsthat g decomposes as a sum

g = g0 ⊕⊕

α∈h∨gα

where g0 is the subspace on which h acts trivially, and hence equals h, and gα is the subspaceon which h acts through the linear form α : h→ k, i.e., for h ∈ h, x ∈ gα, [h, x] = α(h)x.The nonzero α occurring in the above decomposition form a reduced root system Φ in h∨

(and hence in the Q-subspace of h∨ spanned by Φ — see 19.8). In this way, the semisimpleLie algebras over k are classified by the reduced root systems (see Serre 1987, VI).

Classification of the reduced root system

After (19.8), we may as well work with root systems over R.

PROPOSITION 19.16 For any root system (V,Φ), there exists an inner product ( , ) on Vsuch that the sα act as orthogonal transformations, i.e., such that

(sαx, sαy) = (x, y), all α ∈ Φ, x, y ∈ V.

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 136

PROOF. Let ( , )′ be any inner product V × V → R, and define

(x, y) =∑

w∈W(wx,wy)′.

Then ( , ) is again symmetric and bilinear, and

(x, x) =∑

w∈W(wx,wx)′ > 0

if x 6= 0, and so ( , ) is positive-definite. On the other hand, for w0 ∈W,

(w0x,w0y) =∑

w∈W(ww0x,ww0y)′

= (x, y)

because as w runs through W , so also does ww0. 2

REMARK 19.17 There is in fact a canonical inner product on V , namely, the form inducedby x, y 7→ (x, p(x)) (see 18.3).

Thus, we may as well equip V with an inner product ( , ) as in the proposition. Oncomparing (57) with (56)

sα(x) = x− 2(x, α)(α, α)

α,

sα(x) = x− 〈x, α∨〉α,

we see that

〈x, α∨〉 = 2(x, α)(α, α)

.

Thus (RS3) becomes the condition:

2(β, α)(α, α)

∈ Z, all α, β ∈ Φ.

Study of two roots

Let α, β ∈ Φ, and let n(β, α) = 2 (β,α)(α,α) . We wish to examine the significance of the

condition n(β, α) ∈ Z. Write

n(β, α) = 2|β||α|

cosφ

where | · | denotes the length of a vector and φ is the angle between α and β. Then

n(β, α) · n(α, β) = 4 cos2 φ ∈ Z. (60)

Excluding the possibility that β is a multiple of α, there are only the following possibilities(in the table, we have chosen β to be the longer root).

n(β, α) · n(α, β) n(α, β) n(β, α) φ |β|/|α|0 0 0 π/2

11−1

1−1

π/32π/3

1

21−1

2−2

π/43π/4

√2

31−1

3−3

π/65π/6

√3

The proof of this is an exercise for the reader, who should also draw the appropriate pictures.

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 137

REMARK 19.18 Let α and β be roots with neither a multiple of the other. Clearly, n(α, β)and n(β, α) are either both positive or both negative. From the table, we see that in the firstcase at least one of n(α, β) or n(β, α) equals 1. If it is, say, n(β, α), then

sα(β) = β − n(β, α)α = β − α,

and so ±(α− β) are roots.

Bases

DEFINITION 19.19 A base for Φ is a subset S such that(a) S is a basis for V (as an R-vector space), and(b) when we express a root β as a linear combination of elements of S,

β =∑

α∈Smαα,

the mα are integers of the same sign (i.e., either all mα ≥ 0 or all mα ≤ 0).

The elements of a (fixed) base S are often called the simple roots (for the base).

PROPOSITION 19.20 There exists a base S for Φ.

PROOF. Serre 1987, V 8. The idea of the proof is the following. Choose a vector t in thedual vector space V ∨ such that, for all α ∈ Φ, 〈α, t〉 6= 0, and set

Φ+ = {α | 〈α, t〉 > 0}Φ− = {α | 〈α, t〉 < 0}

(so Φ = Φ− t Φ+). Say that an α ∈ Φ+ is decomposable if it can be written as a sumα = β + γ with β, γ ∈ Φ+, and otherwise is indecomposable. One shows that the inde-composable elements form base. 2

REMARK 19.21 Let α and β be simple roots, and suppose n(α, β) and n(β, α) are positive(i.e., the angle between α and β is acute). Then (see 19.18), both of α − β and β − α areroots, and one of them, say, α− β, will be in Φ+. But then α = (α− β) + β, contradictingthe simplicity of α. We conclude that n(β, α) and n(α, β) are negative.

EXAMPLE 19.22 Consider the root system of type An, i.e., that attached to SLn+1 (seep124). We can take V to be the subspace61 of Rn+1 of n + 1-tuples such that

∑xi = 0

with the usual inner product, and Φ = {ei − ej | i 6= j} with e1, . . . , en+1 the standardbasis of Rn+1. When we choose t = ne1 + · · ·+ en,

Φ+ = {ei − ej | i > j}.

For i > j + 1,ei − ej = (ei − ei−1) + · · ·+ (ej+1 − ej)

is decomposable, and so the indecomposable elements are e1 − e2, . . . , en − en+1. Theyobviously form a base.

61The naturally occurring space is Rn+1 modulo the line R(e1 + · · · + en+1), but V is the hyperplaneorthogonal to this line and contains the roots, so gives an isomorphic root system. Alternatively, it is naturallythe dual Φ∨.

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 138

Action of the Weyl group

Let W be the subgroup of GL(V ) generated by {sα | α ∈ Φ}.

PROPOSITION 19.23 Let S be a base for Φ. Then(a) W is generated by the sα for α ∈ S;(b) W · S = Φ;(c) if S′ is a second base for Φ, then S′ = wS for some w ∈W .

PROOF. Serre 1987, V 10. 2

EXAMPLE 19.24 For the root system An,

sαij (~x) = ~x− 2(~x, αij)

(αij , αij)αij , αij = ei − ej ,

= ~x+ (0, . . . , 0,i

xj − xi, 0, . . . , 0,j

xi − xj , 0, . . . , 0)

= (x1, . . . ,ixj , . . . ,

jxi, . . . , xn+1).

Thus, sαij switches the ith and jth coordinates. It follows thatW has a natural identificationwith the symmetric group Sn+1, and it is certainly generated by the sαii+1 . Moreover,W · S = Φ.

Cartan matrix

For a choice S of a base, the Cartan matrix is (n(α, β))α,β∈S . Thus, its diagonal termsequal 2 and its off-diagonal terms are negative (or zero).

PROPOSITION 19.25 The Cartan matrix doesn’t depend on the choice of S, and it deter-mines the root system up to isomorphism.

PROOF. The first assertion follows from (19.23c). For the second, let (V,Φ) and (V ′,Φ′)be root systems such that for some bases S and S′ there is a bijection α 7→ α′ : S → S′

such that n(α, β) = n(α′, β′). Then α 7→ α′ extends uniquely to an isomorphism x 7→x′ : V → V ′. Because

sα(β) = β − n(β, α)α,

this isomorphism sends sα to sα′ for α ∈ S. Because of (19.23a), it also maps W onto W ′,which (by 19.23b) implies that it maps Φ onto Φ′. 2

EXAMPLE 19.26 For the root system An and the obvious base S, the Cartan matrix is

2 −1 0 0 0−1 2 −1 0 00 −1 2 0 0

. . .0 0 0 2 −10 0 0 −1 2

because 2 (ei−ei+1,ei+1−ei+2)

(ei−ei+1,ei−ei+1) = −1, for example.

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19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 139

The Coxeter graph

This is the graph with nodes indexed by the elements of a base S for Φ and two nodes joinedby n(α, β) · n(β, α) edges.

We can define the direct sum of two root systems

(V,Φ) = (V1,Φ1)⊕ (V2,Φ2)

by taking V = V1 ⊕ V2 (as vector spaces with inner product) and by taking Φ = Φ1 ∪ Φ2.A root system is indecomposable if it can’t be written as a direct sum of two nonzero rootsystems.

PROPOSITION 19.27 A root system is indecomposable if and only if its Coxeter graph isconnected.

PROOF. One shows that a root system is decomposable if and only if Φ can be written as adisjoint union Φ = Φ1 tΦ2 with each root in Φ1 orthogonal to each root in Φ2. Since rootsα, β are orthogonal if and only n(α, β) · n(β, α) = 4 cos2 φ = 0, this is equivalent to theCoxeter graph being disconnected. 2

Clearly, it suffices to classify the indecomposable root systems.

The Dynkin diagram

The Coxeter graph doesn’t determine the root system because for any two base roots α, β,it only gives the number n(α, β) · n(β, α). However, for each value of n(α, β) · n(β, α)there is only one possibility for the unordered pair

{n(α, β), n(β, α)} = {2 |α||β|

cosφ, 2|β||α|

cosφ}.

Thus, if we know addition which is the longer root, then we know the ordered pair. TheDynkin diagram is the Coxeter graph with an arrow added pointing towards the shorterroot (if the roots have different lengths). It determines the Cartan matrix and hence theroot system. Specifically, to compute the Cartan matrix from the Dynkin diagram, numberthe simple roots α1, . . . , αn, and let aij = n(αi, βj) be the ijth coefficient of the Cartanmatrix; then

for all i, aii = 2;if αi and αj are not joined by an edge, then aij = 0 = aji;if αi and αj are joined by an edge and |αi| ≤ |αj |, then aij = −1;if αi and αj are joined by r edges and |αi| > |αj |, then aij = −r.

THEOREM 19.28 The Dynkin diagrams arising from reduced irreducible root systems areexactly those listed below.

PROOF. See Humphreys 1979, 11.4, pp 60–62. 2

Page 42: 14 Semisimple algebraic groups and Lie algebrasjmilne/aag140.pdf · 2005-04-20 · 14 SEMISIMPLE ALGEBRAIC GROUPS AND LIE ALGEBRAS 101 PROOF.Choose a basis for C. Then an element

19 CLASSIFICATION OF SEMISIMPLE ROOT DATA 140

An: ◦ ◦ · · · ◦ ◦ n ≥ 1

α1 α2 αn−1 αn

Bn: ◦ ◦ · · · ◦=====⇒◦ (n ≥ 2)

α1 α2 αn−1 αn

Cn: ◦ ◦ · · · ◦⇐=====◦ (n ≥ 3)

α1 α2 αn−1 αn

αn−1

Dn: ◦ ◦ · · · ◦ (n ≥ 4)

α1 α2 αn−2

αn

◦ α2

E6: ◦ ◦ ◦ ◦ ◦

α1 α3 α4 α5 α6

◦ α2

E7: ◦ ◦ ◦ ◦ ◦ ◦

α1 α3 α4 α5 α6 α7

◦ α2

E8: ◦ ◦ ◦ ◦ ◦ ◦ ◦

α1 α3 α4 α5 α6 α7 α8

F4: ◦ ◦=====⇒◦ ◦

α1 α2 α3 α4

G2:Omitted for the present.