Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 22 July 2013

§13. Representations continued

In this section the irreducible representations of characteristic p (the characteristic of the base field k) of the finite Chevalley groups and their twisted analogues will be considered. The main result is as follows.

Theorem 43: Let G be a finite universal Chevalley group or one of its twisted analogues constructed as in §11 as the set of fixed points of an automorphism of the form xα(t)xρα(±tq(α)). Then the Παsimpleq(α) irreducible polynomial representations of the including algebraic group (got by extending the base field k to its algebraic closure) for which the highest weights λ satisfy 0λ,αq(α)-1 for all simple α remain irreducible and distinct on restriction to G and form a complete set.

By Theorem 41 we also have a tensor product theorem with the product Π0 suitably truncated, for example to Π0n-1 if G is a Chevalley group over a field of pn elements.

Exercise: Deduce from Theorem 43 the nature of the truncations for the various twisted groups.

Instead of proving the above results (see Nagoya Math. J 22 (1963)), which would take too long, we shall give an a priori development, similar to the one of the last section.

We start with a group of twisted rank 1 (i.e. of type A1, A22, C22, or G22, the degenerate cases A1(2), A1(3), not being excluded). The subscript σ on Gσ, Wσ, will henceforth be omitted. We write X,Y,w,K for 𝔛a (a>0), 𝔛-a, the nontrivial element of the Weyl group realized in G, and an algebraically closed field of characteristic p. We observe that U=X and U-=Y in the present case. In addition, X will denote the sum of the elements of X in KG.

Definition: An element v in a KG-module V is said to be a highest weight vector if it is nonzero and satisfies

(a) xv=v for all xU.
(b) hv=λ(h)v for all hH and some character λ on H.
(c) Xwv=μv for some μK.
The couple (λ,μ) is called the corresponding weight.

Remark: The refinement (c) of the usual definition is due to C. W. Curtis (I11. J. Math. 7(1963) and J. für Math. 219 (1965)). That such a refinement is needed is already seen in the simplest case G=SL2(p). Here there are p representations realized on the spaces of homogeneous polynomials of degrees i=0,1,,p-1, with the highest weight in the usual sense being iλ1. Since the group H is cyclic of order p-1, the weights 0 and (p-1)λ1 are identical on H, hence do not distinguish the corresponding representations from each other.

Theorem 44: Let G (of rank 1) and the notations be as above.

(a) Every nonzero KG-module V contains a highest weight vector v. For such a v we have KGv=KYv =KU-v.
(b) If V is irreducible, then it determines Kv as the unique line of V fixed by U, hence it also determines the corresponding highest weight.
(c) Two irreducible KG-modules are isomorphic if and only if their highest weights are equal.

The proof depends on the following two lemmas.

Lemma 80: Let Y and K be as above, or, more generally, let Y be any finite p-group and K any field of characteristic p.

(a) Every nonzero KY-module V contains nonzero vectors invariant under Y.
(b) Every irreducible KY-module is trivial.
(c) RadKY= {Σc(y)y|Σc(y)=0} is the unique maximal (one-sided or two-sided) ideal of KY. It is nilpotent.
(d) KY is the unique minimal ideal of KY.


(a) By induction on |Y|. Assume |Y|>1. Since Y is a p-group, it has a normal subgroup Y1 of index p, and the subspace V1 of invariants of Y1 on V is nonzero by the inductive assumption. Choose yY to generate Y/Y1, and v in V1 and nonzero. Then (1-y)pv=0. Now choose r maximal so that (1-y)rv0. The resulting vector is fixed by Y, whence (a).

(b) By (a).

(c) By inspection RadKY is an ideal, maximal because its codimension in KY is 1. Bring the kernel of the trivial representation, which is the only irreducible one by (b), it is the unique maximal left ideal; and similarly for right ideals. On each factor of a composition series for the left regular representation of KY on itself Y acts trivially by (b), hence RadKY acts as 0, so that RadKY is nilpotent.

(d) By (b).

Lemma 81: For xX-1, write wxw=f(x)h(x)wg(x) with f(x),g(x)X and h(x)H.

(a) f and g are permutations of X-1.
(b) (Xw)2= Xw2+ ΣxX-1 h(x)Xw g(x).


(a) If f(x)=1, we get the contradiction xwB, while if f(x)=f(x), we see that w-1x-1xwB, so that x=x. Similarly for g.

(b) (Xw)2= Xw2+ ΣxX-1 Xf(x) h(x)wg(x).

Here f(x) gets absorbed in X and h(x) normalizes X, whence (b).

Proof of Theorem 44.

(a) By Lemma 80(b) the space of fixed points of U=X on V is nonzero. Since H normalizes U and is Abelian, that space contains a nonzero vector v such that (a) and (b) of the definition of highest weight vector hold. Let Xwv=v1. If v1=0, then (c) holds with μ=0. If not, we replace v by v1. Then (a) and (b) of the definition still hold with wλ in place of λ, and by Lemma 81(b) so does (c) with μ=Σλ(h(x)). Now to prove that KGv=KYv it is enough, because of the decomposition G=YBwB, to prove that wvKYv. By the two parts of Lemma 81 we may write Xwv=μv, after some simplification, in the form

(*) wv+ΣxX-1 λ(wh-1w)y (x)v=μv,

with y(x)=wxw-1Y, whence our assertion.

(b) Let V=Kv and V=RadKY·v. It follows from Lemma 80(c) that the sum V=V+V is direct. Now assume there exists some v1V, v1V, fixed by X. We may assume that v1V and also that v1 is an eigenvector for H since H is Abelian. We have Xwv1=wYv1=0; since Y(1-y)=0 for any yY. Thus v1 is a highest weight vector. By (a), V=KYv1KYV=V, a contradiction, whence (b).

(c) By (b) an irreducible KG-module determines its highest weight (λ,μ) uniquely. Conversely, assume that V1 and V2 are irreducible KG-modules with highest weight vectors v1 and v2 of the same weight (λ,μ). Set v=v1+v2V1+V2 and then V=KGv=KYv. Now v2V, since otherwise we could write v2=cv+v with cK and vRadKY·v and then projecting on V1 and V2 get that c=0 and c=1, a contradiction. Thus we may complete the proof as in the proof of Theorem 39(e).

Theorem 45: Let (λ,μ) be the nighest weight of an irreducible KG-module V.

(a) If λ1, then μ=0. If λ=1, then μ=0 or -1.
(b) Every weight as in (a) can be realized. Thus the number of possibilities is |H|+1.


(1) Proof of (b). In KG let Hλ=ΣhHλ(h-1)h, then u=XHλwX and v=XHλ. As in the proof of Theorem 44(a), asimple calculation yields Xw(u+cv)= ΣxX-1 λ(h(x))u + cXwHλX. Here H acts, from the left, according to the characters wλ,λ,wλ on the respective terms. Thus if wλλ, we may realize the weight (λ,0) by taking c=0. If wλ=λ, we take c=-Σλ(h(x)) instead. Finally if λ=1, then Σλ(h(x))=-1 and we get μ=-1 by taking c=0. To achieve (λ,μ) in an irreducible module we simply take KG(u+cv) modulo a maximal submodule.

(2) If dimV=1, then V is trivial and (λ,μ)=(1,0). Since X and Y are p-groups, they act trivially by Lemma 80(b), whence (2).

(3) If dimV1, then λ determines μ. Write V=V+V as in the proof of Theorem 44(b). We have V0. Since Y ? fixes V it fixes some line in it, uniquely determined in V, by Theorem 44(b) with Y in place of X. Since Y clearly fixes wv, we conclude that wvV. Projecting (*) of the proof of Theorem 44(a) onto V, we get

(**) Σλ (wh-1w) =μ,

whence (3).

(4) Proof of (a). Combine (1), (2) and (3).

Corollaries to Theorem 44 and 45:

(a) If λ1, then ΣxX-1 λ(h(x)) =0. The number of solutions n(h) of h(x)=h with h given is, modulo p, independent of h, in particular for each h is at least 1 (cf. Lemma 64, Step (1)).
(b) The irreducible representation of weight (λ,μ) can be realized in the left ideal generated by XHλwX+cXHλ with c=1 for the trivial representation (1,0), c=0 otherwise.
(c) If LK is a splitting field for H, it is one for G.
(d) DimV=|X| if (λ,μ)=(1,-1), dimV<|X| if not.
(e) The number of p-regular conjugacy classes of G is |H|+1.


(a) If λ1, then μ=0 by Theorem 45 (a), so that Σλ(h(x))=0 by (**) above applied with λ replaced by wλ. Then Σn(h)λ(h)=0 for every λ1. By the orthogonality relations for the characters on H (which are valid since p|h|), we conclude that n(h), as an element of K, is independent of h. If n(h) were 0 for some h, we would get |H|=Σn(h)=0 (modp), a contradiction.

(b) Let V be an irreducible module whose dual V* has the highest weight (λ,μ). We consider, as in Theorem 40 the isomorphism φ of V* into the induced representation space of functions a:GK such that a(yb)=a(y)λ*(b) for all yG,bB defined by (φf)(x)= f(xv+) with v+ a highest weight vector for V. Using the decomposition V=Kwv++RadKX·wv+, we may define f+ as in Lemma 73, prove that it is a highest weight vector, and that a+=φf+ is given by the equations of Lemma 74, with λ* in place of λ. Coverting functions on G to elements of KG in the usual way, aΣa(x)x, we see that a+ becomes the element of (b), whence (b). At the same time we see that V may be realized in the induced module Bλ*G, as the unique irreducible submodule in case λ1, as one of two in case λ=1.

(c) By (b).

(d) If μ=0, then Xwv=0, whence Yv=0 and dimV<|X| by Theorem 44(a). Conversely if dimV<|X|, then the annihilator of v in KY contains Y by Lemma 80(d), so that μ=0.

(e) By a classical theorem of Brauer and Nesbitt (University of Toronto Studies, 1937) the number in question equals the number of irreducible KG-modules, hence equals |H|+1 by Theorem 45(b).

Example: G=SL2(q). Here |H|=q-1, so that |H|+1=q.


(a) We see that the extra condition Xv=μv serves two purposes. First it distinguishes the smallest module (1,0) from the largest (1,-1). Secondly, in the proof of the key relation KGv=KU-v it takes the place of the density argument (U-B dense in G) used in the infinite case.
(b) The preceding development applies to a wide class of doubly transitive permutation groups (with B the stabilizer of a point, H of two points), since it depends only on the facts that H is Abelian and has in B a normal complement U which is a p-Sylow subgroup of G.

Now we consider groups of arbitrary rank. W (=Wσ) will be given the structure of reflection group as in Theorem 32 with Σ/R (see p. 177), projected into Vσ and scaled down to a set of unit vectors, the corresponding root system. For each simple root a, we write Ya for X-a and choose wa in Xa,Ya to represent wa in W. If wW is arbitrary, we choose a minimal expression w=wawb as a product of simple reflections, and set w=wawb. Then w is independent of the minimal expression chosen. We postpone the proof of this fact, which could (and probably should) have been given much earlier, to the end of the section so as not to interrupt the present development. As a consequence we have:

Lemma 82: If w=wawb is any minimal expression, then Xw= Xawa ·Xbwb.


Since w=wawb this easily follows by induction on N(w) or by Appendix II.25.

We extend the earlier definition of highest weight vector by the new requirement:

(c) Xawav =μav (μaK) for every simple root a.

Theorem 46: Let G be a (perhaps twisted) finite Chevalley group (of arbitrary rank).

(a), (b), (c) Same as (a), (b), (c) of Theorem 44.
(d) Let HaXa,Ya. If (λ,μa) is the highest weight of some irreducible module then μa=0 if λ|Ha1, and μa=0 or -1 if λ|Ha=1.
(e) Every weight as in (a) can be realized on some irreducible KG-module.


We shall prove this theorem in several steps.

(a1) There exists in V a nonzero eigenvector v for B.

This is proved as in Theorem 44(a).

(a2) If v is as in (a1), then so is v1=Xawav (a simple), unless it is 0.


Let x be any element of U. Write x=xaxa with xaXa and xaXa, the subgroup of elements of U whose Xa components are 1. We recall that Xa and wa normalize Xa (see Appendix I.11). Thus xv1=xaXawav=v1, since Uv=v. Since H normalizes Xa and is normalized by wa, we see that v1 is also an eigenvector for H.

(a3) Choose v as in (a1), then wW so that N(w) is maximal subject to v1=Xwwv0. Then v1 is a highest weight vector.


By Lemma 82 and (a2), v1 is an eigenvector for B. Let a be any simple root. If w-1a>0, then N(waw)=N(w)+1 by Appendix II.19, so that Xwaw waw = Xawa Xww Lemma 82, and Xawav1=0 by the choice of w. If w-1a<0, then we may choose a minimal expression w=wawb starting with wa. Then

Xa wa v1 = (Xawa)2 Xb wbv by Lemma 82 = μXa wa Xb wbv, withμKby Lemma 81(b) = μv1.

By (a1) and (a3) we have the first statement in (a).

(a4) KGv= KU-v.


We have G=w0GWwU-B by Theorem 4. Thus it is enough to show each w fixes KU-v, and for this we may assume w=wa with a simple. Assume yU-. Write y=yaya as above, but using negative roots instead. Then ya and wa normalize Ya, so that wayv= waya yav U-wa yav KU-v by Theorem 44(a) applied to Xa,Ya.

(b1) If V=V+V with V=Kv and V=RadKU-v as before, then V is fixed by every Xawa.


Write y=yaya as before.

Then Xawayv= ΣxXa y(x)xwa v with y(x)U-.

Thus Xawa(y-1)v= ΣxXa (y(x)-1)x wav V.

(b2) Proof of (b). If this is false, there exists v1V, v10, v1 fixed by X. As usual we may choose v1 as an eigenvector for H, and then by (b1) and (a3) also an eigenvector for each Xawa. Then, as before, V=KGv1= KU-v1 V, a contradiction.

(c) Same proof as for Theorem 44(c).

(d) By Theorem 45(a) applied to Xa,Ya.

(e) Let π be the set of simple roots a such that μa=0, and Wπ the corresponding subgroup of W. The reader should have no trouble in proving that the left ideal of KG generated by ΣwWπw0 UHλw Xw is an irreducible KG-module whose highest weight is (λ,μa).


(a) A splitting field for H is also one for G.
(b) Let V be irreducible, of highest weight (λ,μa).
Then dimV=|U| if λ=1 and all μa=-1, dimV<|U| if not.
(c) For each set π of simple roots, let Hπ be the group generated by all Ha (aπ). Then the number of irreducible KG-modules, or, equivalently, of p-regular conjugacy classes of G is Σπ|H/Hπ|.


(a) Clear.

(b) Write Uw0v= Xw0w0v= Xawa Xbwb v, with w0=wawb as in Lemma 82, and then proceed as in the proof of Cor. (d) to Theorems 44 and 45.

(c) The given sum counts the number of possible weights (λ,μa) according to the set π of simple roots a such that μa=-1.

Exercise: If G is universal, the above number is Πasimple (|Ha|+1) = Παsimple q(α), in the notation of Theorem 43. If in addition G is not twisted, then the number is q.

It remains to prove the following result used (inessentially) in the proof of Lemma 82.

Lemma 83:

(a) If wW, then any two minimal expressions for w as a product of simple reflections can be transformed into each other by the relations (*) wawbwa= wbwawb (n terms on each side, n= order wawb, with a and b distinct simple roots).
(b) Assume that for each simple root a the corresponding element wa of G (any Chevalley group) is chosen to lie in Xa,X-a. Let w=wawb be a minimal expression for wW. Then w=wawb is independent of the minimal expression chosen.


(a) This is a refinement of Appendix IV.38 since the relations wα2=1 are not required. It is an easy exercise to convert the proof of the latter result into a proof of the former, which we shall leave to the reader.

(b) Because of (a) we only have to prove (b) when w has the form of the two sides of (*). For this we can refer to the proof of Lemma 56 since the extra restrictions there, that G is untwisted and that wa=wa(1) for each a, are not essential for the proof.

Remark: It would be nice if someone could incorporate in the elementary development just given the tensor product theorem mentioned after Theorem 43 or at least a proof that every irreducible K-module for G can be extended to the including algebraic group, hence also to the other finite Chevalley groups contained in the latter group.

Notes and References

This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract ARO-D-336-8230-31-43033.

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