## Lectures on Chevalley groups

Last update: 22 July 2013

## §13. Representations continued

In this section the irreducible representations of characteristic $p$ (the characteristic of the base field $k\text{)}$ 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 ${x}_{\alpha }\left(t\right)\to {x}_{\rho \alpha }\left(±{t}^{q\left(\alpha \right)}\right)\text{.}$ Then the $\underset{\alpha \text{simple}}{\Pi }q\left(\alpha \right)$ irreducible polynomial representations of the including algebraic group (got by extending the base field $k$ to its algebraic closure) for which the highest weights $\lambda$ satisfy $0\le ⟨\lambda ,\alpha ⟩\le q\left(\alpha \right)-1$ for all simple $\alpha$ 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 $\underset{0}{\overset{\infty }{\Pi }}$ suitably truncated, for example to $\underset{0}{\overset{n-1}{\Pi }}$ if $G$ is a Chevalley group over a field of ${p}^{n}$ 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 ${A}_{1},$ ${}^{2}A_{2},$ ${}^{2}C_{2},$ or ${}^{2}G_{2},$ the degenerate cases ${A}_{1}\left(2\right),$ ${A}_{1}\left(3\right),\dots$ not being excluded). The subscript $\sigma$ on ${G}_{\sigma },$ ${W}_{\sigma },\dots$ will henceforth be omitted. We write $X,Y,w,K$ for ${𝔛}_{a}$ $\left(a>0\right),$ ${𝔛}_{-a},$ the nontrivial element of the Weyl group realized in $G,$ and an algebraically closed field of characteristic $p\text{.}$ We observe that $U=X$ and ${U}^{-}=Y$ in the present case. In addition, $\stackrel{‾}{X}$ will denote the sum of the elements of $X$ in $KG\text{.}$

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

 (a) $xv=v$ for all $x\in U\text{.}$ (b) $hv=\lambda \left(h\right)v$ for all $h\in H$ and some character $\lambda$ on $H\text{.}$ (c) $\stackrel{‾}{X}wv=\mu v$ for some $\mu \in K\text{.}$
The couple $\left(\lambda ,\mu \right)$ 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 $G={SL}_{2}\left(p\right)\text{.}$ Here there are $p$ representations realized on the spaces of homogeneous polynomials of degrees $i=0,1,\dots ,p-1,$ with the highest weight in the usual sense being $i{\lambda }_{1}\text{.}$ Since the group $H$ is cyclic of order $p-1,$ the weights $0$ and $\left(p-1\right){\lambda }_{1}$ 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 $KG\text{-module}$ $V$ contains a highest weight vector $v\text{.}$ For such a $v$ we have $KGv=KYv=K{U}^{-}v\text{.}$ (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 $KG\text{-modules}$ 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 $p\text{-group}$ and $K$ any field of characteristic $p\text{.}$

 (a) Every nonzero $KY\text{-module}$ $V$ contains nonzero vectors invariant under $Y\text{.}$ (b) Every irreducible $KY\text{-module}$ is trivial. (c) $\text{Rad} KY=\left\{\Sigma c\left(y\right)y | \Sigma c\left(y\right)=0\right\}$ is the unique maximal (one-sided or two-sided) ideal of $KY\text{.}$ It is nilpotent. (d) $K\stackrel{‾}{Y}$ is the unique minimal ideal of $KY\text{.}$

 Proof. (a) By induction on $|Y|\text{.}$ Assume $|Y|>1\text{.}$ Since $Y$ is a $p\text{-group,}$ it has a normal subgroup ${Y}_{1}$ of index $p,$ and the subspace ${V}_{1}$ of invariants of ${Y}_{1}$ on $V$ is nonzero by the inductive assumption. Choose $y\in Y$ to generate $Y/{Y}_{1},$ and $v$ in ${V}_{1}$ and nonzero. Then ${\left(1-y\right)}^{p}v=0\text{.}$ Now choose $r$ maximal so that ${\left(1-y\right)}^{r}v\ne 0\text{.}$ The resulting vector is fixed by $Y,$ whence (a). (b) By (a). (c) By inspection $\text{Rad} KY$ is an ideal, maximal because its codimension in $KY$ is $1\text{.}$ 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 $\text{Rad} KY$ acts as $0,$ so that $\text{Rad} KY$ is nilpotent. (d) By (b). $\square$

Lemma 81: For $x\in X-1,$ write $wxw=f\left(x\right)h\left(x\right)wg\left(x\right)$ with $f\left(x\right),g\left(x\right)\in X$ and $h\left(x\right)\in H\text{.}$

 (a) $f$ and $g$ are permutations of $X-1\text{.}$ (b) ${\left(\stackrel{‾}{X}w\right)}^{2}=\stackrel{‾}{X}{w}^{2}+\underset{x\in X-1}{\Sigma }h\left(x\right)\stackrel{‾}{X}wg\left(x\right)\text{.}$

 Proof. (a) If $f\left(x\right)=1,$ we get the contradiction $xw\in B,$ while if $f\left(x\prime \right)=f\left(x\right),$ we see that ${w}^{-1}{x}^{-1}x\prime w\in B,$ so that $x\prime =x\text{.}$ Similarly for $g\text{.}$ (b) ${\left(\stackrel{‾}{X}w\right)}^{2}=\stackrel{‾}{X}{w}^{2}+\underset{x\in X-1}{\Sigma }\stackrel{‾}{X}f\left(x\right)h\left(x\right)wg\left(x\right)\text{.}$ Here $f\left(x\right)$ gets absorbed in $\stackrel{‾}{X}$ and $h\left(x\right)$ normalizes $\stackrel{‾}{X},$ whence (b). $\square$

 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 $\stackrel{‾}{X}wv={v}_{1}\text{.}$ If ${v}_{1}=0,$ then (c) holds with $\mu =0\text{.}$ If not, we replace $v$ by ${v}_{1}\text{.}$ Then (a) and (b) of the definition still hold with $w\lambda$ in place of $\lambda ,$ and by Lemma 81(b) so does (c) with $\mu =\Sigma \lambda \left(h\left(x\right)\right)\text{.}$ Now to prove that $KGv=KYv$ it is enough, because of the decomposition $G=YB\cup wB,$ to prove that $wv\in KYv\text{.}$ By the two parts of Lemma 81 we may write $\stackrel{‾}{X}wv=\mu v,$ after some simplification, in the form $(*) wv+Σx∈X-1 λ(wh-1w)y (x)v=μv,$ with $y\left(x\right)=wx{w}^{-1}\in Y,$ whence our assertion. (b) Let $V\prime =Kv$ and ${V}^{\prime \prime }=\text{Rad} KY·v\text{.}$ It follows from Lemma 80(c) that the sum $V=V\prime +{V}^{\prime \prime }$ is direct. Now assume there exists some ${v}_{1}\in V,$ ${v}_{1}\notin V\prime ,$ fixed by $X\text{.}$ We may assume that ${v}_{1}\in {V}^{\prime \prime }$ and also that ${v}_{1}$ is an eigenvector for $H$ since $H$ is Abelian. We have $\stackrel{‾}{X}w{v}_{1}=w\stackrel{‾}{Y}{v}_{1}=0\text{;}$ since $\stackrel{‾}{Y}\left(1-y\right)=0$ for any $y\in Y\text{.}$ Thus ${v}_{1}$ is a highest weight vector. By (a), $V=KY{v}_{1}\subseteq KY{V}^{\prime \prime }={V}^{\prime \prime },$ a contradiction, whence (b). (c) By (b) an irreducible $KG\text{-module}$ determines its highest weight $\left(\lambda ,\mu \right)$ uniquely. Conversely, assume that ${V}_{1}$ and ${V}_{2}$ are irreducible $KG\text{-modules}$ with highest weight vectors ${v}_{1}$ and ${v}_{2}$ of the same weight $\left(\lambda ,\mu \right)\text{.}$ Set $v={v}_{1}+{v}_{2}\in {V}_{1}+{V}_{2}$ and then $V=KGv=KYv\text{.}$ Now ${v}_{2}\notin V,$ since otherwise we could write ${v}_{2}=cv+{v}^{\prime \prime }$ with $c\in K$ and ${v}^{\prime \prime }\in \text{Rad} KY·v$ and then projecting on ${V}_{1}$ and ${V}_{2}$ get that $c=0$ and $c=1,$ a contradiction. Thus we may complete the proof as in the proof of Theorem 39(e). $\square$

Theorem 45: Let $\left(\lambda ,\mu \right)$ be the nighest weight of an irreducible $KG\text{-module}$ $V\text{.}$

 (a) If $\lambda \ne 1,$ then $\mu =0\text{.}$ If $\lambda =1,$ then $\mu =0$ or $-1\text{.}$ (b) Every weight as in (a) can be realized. Thus the number of possibilities is $|H|+1\text{.}$

 Proof. (1) Proof of (b). In $KG$ let ${H}_{\lambda }=\underset{h\in H}{\Sigma }\lambda \left({h}^{-1}\right)h,$ then $u=\stackrel{‾}{X}{H}_{\lambda }w\stackrel{‾}{X}$ and $v=\stackrel{‾}{X}{H}_{\lambda }\text{.}$ As in the proof of Theorem 44(a), asimple calculation yields $\stackrel{‾}{X}w\left(u+cv\right)=\underset{x\in X-1}{\Sigma }\lambda \left(h\left(x\right)\right)u+c\stackrel{‾}{X}w{H}_{\lambda }\stackrel{‾}{X}\text{.}$ Here $H$ acts, from the left, according to the characters $w\lambda ,\lambda ,w\lambda$ on the respective terms. Thus if $w\lambda \ne \lambda ,$ we may realize the weight $\left(\lambda ,0\right)$ by taking $c=0\text{.}$ If $w\lambda =\lambda ,$ we take $c=-\Sigma \lambda \left(h\left(x\right)\right)$ instead. Finally if $\lambda =1,$ then $\Sigma \lambda \left(h\left(x\right)\right)=-1$ and we get $\mu =-1$ by taking $c=0\text{.}$ To achieve $\left(\lambda ,\mu \right)$ in an irreducible module we simply take $KG\left(u+cv\right)$ modulo a maximal submodule. (2) If $\text{dim} V=1,$ then $V$ is trivial and $\left(\lambda ,\mu \right)=\left(1,0\right)\text{.}$ Since $X$ and $Y$ are $p\text{-groups,}$ they act trivially by Lemma 80(b), whence (2). (3) If $\text{dim} V\ne 1,$ then $\lambda$ determines $\mu \text{.}$ Write $V=V\prime +{V}^{\prime \prime }$ as in the proof of Theorem 44(b). We have ${V}^{\prime \prime }\ne 0\text{.}$ Since $Y$ ? fixes ${V}^{\prime \prime }$ it fixes some line in it, uniquely determined in $V,$ by Theorem 44(b) with $Y$ in place of $X\text{.}$ Since $Y$ clearly fixes $wv,$ we conclude that $wv\in {V}^{\prime \prime }\text{.}$ Projecting $\left(*\right)$ of the proof of Theorem 44(a) onto $V\prime ,$ we get $(**) Σλ (wh-1w) =μ,$ whence (3). (4) Proof of (a). Combine (1), (2) and (3). $\square$

Corollaries to Theorem 44 and 45:

 (a) If $\lambda \ne 1,$ then $\underset{x\in X-1}{\Sigma }\lambda \left(h\left(x\right)\right)=0\text{.}$ The number of solutions $n\left(h\right)$ of $h\left(x\right)=h$ 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 $\left(\lambda ,\mu \right)$ can be realized in the left ideal generated by $\stackrel{‾}{X}{H}_{\lambda }w\stackrel{‾}{X}+c\stackrel{‾}{X}{H}_{\lambda }$ with $c=1$ for the trivial representation $\left(1,0\right),$ $c=0$ otherwise. (c) If $L\subset K$ is a splitting field for $H,$ it is one for $G\text{.}$ (d) $\text{Dim} V=|X|$ if $\left(\lambda ,\mu \right)=\left(1,-1\right),$ $\text{dim} V<|X|$ if not. (e) The number of $p\text{-regular}$ conjugacy classes of $G$ is $|H|+1\text{.}$

 Proof. (a) If $\lambda \ne 1,$ then $\mu =0$ by Theorem 45 (a), so that $\Sigma \lambda \left(h\left(x\right)\right)=0$ by $\left(**\right)$ above applied with $\lambda$ replaced by $w\lambda \text{.}$ Then $\Sigma n\left(h\right)\lambda \left(h\right)=0$ for every $\lambda \ne 1\text{.}$ By the orthogonality relations for the characters on $H$ (which are valid since $p\nmid |h|\text{),}$ we conclude that $n\left(h\right),$ as an element of $K,$ is independent of $h\text{.}$ If $n\left(h\right)$ were $0$ for some $h,$ we would get $|H|=\Sigma n\left(h\right)=0$ $\text{(mod} p\text{),}$ a contradiction. (b) Let $V$ be an irreducible module whose dual ${V}^{*}$ has the highest weight $\left(\lambda ,\mu \right)\text{.}$ We consider, as in Theorem 40 the isomorphism $\phi$ of ${V}^{*}$ into the induced representation space of functions $a:G\to K$ such that $a\left(yb\right)=a\left(y\right){\lambda }^{*}\left(b\right)$ for all $y\in G,b\in B$ defined by $\left(\phi f\right)\left(x\right)=f\left(x{v}^{+}\right)$ with ${v}^{+}$ a highest weight vector for $V\text{.}$ Using the decomposition $V=Kw{v}^{+}+\text{Rad} KX·w{v}^{+},$ we may define ${f}^{+}$ as in Lemma 73, prove that it is a highest weight vector, and that ${a}^{+}=\phi {f}^{+}$ is given by the equations of Lemma 74, with ${\lambda }^{*}$ in place of $\lambda \text{.}$ Coverting functions on $G$ to elements of $KG$ in the usual way, $a\sim \Sigma a\left(x\right)x,$ 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 ${B}_{{\lambda }^{*}}\to G,$ as the unique irreducible submodule in case $\lambda \ne 1,$ as one of two in case $\lambda =1\text{.}$ (c) By (b). (d) If $\mu =0,$ then $\stackrel{‾}{X}wv=0,$ whence $\stackrel{‾}{Y}v=0$ and $\text{dim} V<|X|$ by Theorem 44(a). Conversely if $\text{dim} V<|X|,$ then the annihilator of $v$ in $KY$ contains $\stackrel{‾}{Y}$ by Lemma 80(d), so that $\mu =0\text{.}$ (e) By a classical theorem of Brauer and Nesbitt (University of Toronto Studies, 1937) the number in question equals the number of irreducible $KG\text{-modules,}$ hence equals $|H|+1$ by Theorem 45(b). $\square$

Example: $G={SL}_{2}\left(q\right)\text{.}$ Here $|H|=q-1,$ so that $|H|+1=q\text{.}$

Remarks.

 (a) We see that the extra condition $\stackrel{‾}{X}v=\mu v$ serves two purposes. First it distinguishes the smallest module $\sim \left(1,0\right)$ from the largest $\left(1,-1\right)\text{.}$ Secondly, in the proof of the key relation $KGv=K{U}^{-}v$ it takes the place of the density argument $\text{(}{U}^{-}B$ dense in $G\text{)}$ 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 $p\text{-Sylow}$ subgroup of $G\text{.}$

Now we consider groups of arbitrary rank. $W$ $\left(={W}_{\sigma }\right)$ will be given the structure of reflection group as in Theorem 32 with $\Sigma /R$ (see p. 177), projected into ${V}_{\sigma }$ and scaled down to a set of unit vectors, the corresponding root system. For each simple root $a,$ we write ${Y}_{a}$ for ${X}_{-a}$ and choose ${\stackrel{‾}{w}}_{a}$ in $⟨{X}_{a},{Y}_{a}⟩$ to represent ${w}_{a}$ in $W\text{.}$ If $w\in W$ is arbitrary, we choose a minimal expression $w={w}_{a}{w}_{b}\dots$ as a product of simple reflections, and set $\stackrel{‾}{w}={\stackrel{‾}{w}}_{a}{\stackrel{‾}{w}}_{b}\dots \text{.}$ Then $\stackrel{‾}{w}$ 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 $w={w}_{a}{w}_{b}\dots$ is any minimal expression, then ${\stackrel{‾}{X}}_{w}={\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}·{\stackrel{‾}{X}}_{b}{\stackrel{‾}{w}}_{b}\dots \text{.}$

 Proof. Since $\stackrel{‾}{w}={\stackrel{‾}{w}}_{a}{\stackrel{‾}{w}}_{b}\dots$ this easily follows by induction on $N\left(w\right)$ or by Appendix II.25. $\square$

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

 (c) ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}v={\mu }_{a}v$ $\left({\mu }_{a}\in K\right)$ for every simple root $a\text{.}$

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 ${H}_{a}\cap ⟨{X}_{a},{Y}_{a}⟩\text{.}$ If $\left(\lambda ,{\mu }_{a}\right)$ is the highest weight of some irreducible module then ${\mu }_{a}=0$ if $\lambda |{H}_{a}\ne 1,$ and ${\mu }_{a}=0$ or $-1$ if $\lambda |{H}_{a}=1\text{.}$ (e) Every weight as in (a) can be realized on some irreducible $KG\text{-module.}$

Proof.

We shall prove this theorem in several steps.

 (a1) There exists in $V$ a nonzero eigenvector $v$ for $B\text{.}$

This is proved as in Theorem 44(a).

 (a2) If $v$ is as in (a1), then so is ${v}_{1}={\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}v$ $\text{(}a$ simple), unless it is $0\text{.}$

 Proof. Let $x$ be any element of $U\text{.}$ Write $x={x}_{a}^{\prime }{x}_{a}$ with ${x}_{a}\in {X}_{a}$ and ${x}_{a}^{\prime }\in {X}_{a}^{\prime },$ the subgroup of elements of $U$ whose ${X}_{a}$ components are $1\text{.}$ We recall that ${X}_{a}$ and ${\stackrel{‾}{w}}_{a}$ normalize ${X}_{a}^{\prime }$ (see Appendix I.11). Thus $x{v}_{1}={x}_{a}^{\prime }{\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}v={v}_{1},$ since $Uv=v\text{.}$ Since $H$ normalizes ${X}_{a}$ and is normalized by ${\stackrel{‾}{w}}_{a},$ we see that ${v}_{1}$ is also an eigenvector for $H\text{.}$ $\square$

 (a3) Choose $v$ as in (a1), then $w\in W$ so that $N\left(w\right)$ is maximal subject to ${v}_{1}={\stackrel{‾}{X}}_{w}\stackrel{‾}{w}v\ne 0\text{.}$ Then ${v}_{1}$ is a highest weight vector.

 Proof. By Lemma 82 and (a2), ${v}_{1}$ is an eigenvector for $B\text{.}$ Let $a$ be any simple root. If ${w}^{-1}a>0,$ then $N\left({w}_{a}w\right)=N\left(w\right)+1$ by Appendix II.19, so that ${\stackrel{‾}{X}}_{{w}_{a}w}\stackrel{‾}{{w}_{a}w}={\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}{\stackrel{‾}{X}}_{w}\stackrel{‾}{w}$ Lemma 82, and ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}{v}_{1}=0$ by the choice of $w\text{.}$ If ${w}^{-1}a<0,$ then we may choose a minimal expression $w={w}_{a}{w}_{b}\dots$ starting with ${w}_{a}\text{.}$ Then $X‾a w‾a v1 = (X‾aw‾a)2 X‾b w‾b…v by Lemma 82 = μX‾a w‾a X‾b w‾b…v, with μ∈K by Lemma 81(b) = μv1.$ $\square$

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

 (a4) $KGv=K{U}^{-}v\text{.}$

 Proof. We have $G={\stackrel{‾}{w}}_{0}G\subseteq \underset{W}{\cup }\stackrel{‾}{w}{U}^{-}B$ by Theorem 4. Thus it is enough to show each $\stackrel{‾}{w}$ fixes $K{U}^{-}v,$ and for this we may assume $w={w}_{a}$ with $a$ simple. Assume $y\in {U}^{-}\text{.}$ Write $y={y}_{a}{y}_{a}^{\prime }$ as above, but using negative roots instead. Then ${y}_{a}$ and ${\stackrel{‾}{w}}_{a}$ normalize ${Y}_{a}^{\prime },$ so that ${\stackrel{‾}{w}}_{a}yv={\stackrel{‾}{w}}_{a}{y}_{a}{y}_{a}^{\prime }v\in {U}^{-}{\stackrel{‾}{w}}_{a}{y}_{a}v\subseteq K{U}^{-}v$ by Theorem 44(a) applied to $⟨{X}_{a},{Y}_{a}⟩\text{.}$ $\square$

 (b1) If $V=V\prime +{V}^{\prime \prime }$ with $V\prime =Kv$ and ${V}^{\prime \prime }=\text{Rad} K{U}^{-}v$ as before, then ${V}^{\prime \prime }$ is fixed by every ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}\text{.}$

 Proof. Write $y={y}_{a}{y}_{a}^{\prime }$ as before. Then ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}yv=\underset{x\in {X}_{a}}{\Sigma }y\left(x\right)x{\stackrel{‾}{w}}_{a}v$ with $y\left(x\right)\in {U}^{-}\text{.}$ Thus ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}\left(y-1\right)v=\underset{x\in {X}_{a}}{\Sigma }\left(y\left(x\right)-1\right)x{\stackrel{‾}{w}}_{a}v\in {V}^{\prime \prime }\text{.}$ $\square$

 (b2) Proof of (b). If this is false, there exists ${v}_{1}\in {V}^{\prime \prime },$ ${v}_{1}\ne 0,$ ${v}_{1}$ fixed by $X\text{.}$ As usual we may choose ${v}_{1}$ as an eigenvector for $H,$ and then by (b1) and (a3) also an eigenvector for each ${\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}\text{.}$ Then, as before, $V=KG{v}_{1}=K{U}^{-}{v}_{1}\subseteq {V}^{\prime \prime },$ a contradiction.

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

(d) By Theorem 45(a) applied to $⟨{X}_{a},{Y}_{a}⟩\text{.}$

(e) Let $\pi$ be the set of simple roots $a$ such that ${\mu }_{a}=0,$ and ${W}_{\pi }$ the corresponding subgroup of $W\text{.}$ The reader should have no trouble in proving that the left ideal of $KG$ generated by $\underset{w\in {W}_{\pi }{w}_{0}}{\Sigma }\stackrel{‾}{U}{H}_{\lambda }\stackrel{‾}{w}{\stackrel{‾}{X}}_{w}$ is an irreducible $KG\text{-module}$ whose highest weight is $\left(\lambda ,{\mu }_{a}\right)\text{.}$

$\square$

Corollary:

 (a) A splitting field for $H$ is also one for $G\text{.}$ (b) Let $V$ be irreducible, of highest weight $\left(\lambda ,{\mu }_{a}\right)\text{.}$
Then $\text{dim} V=|U|$ if $\lambda =1$ and all ${\mu }_{a}=-1,$ $\text{dim} V<|U|$ if not.
 (c) For each set $\pi$ of simple roots, let ${H}_{\pi }$ be the group generated by all ${H}_{a}$ $\left(a\in \pi \right)\text{.}$ Then the number of irreducible $KG\text{-modules,}$ or, equivalently, of $p\text{-regular}$ conjugacy classes of $G$ is $\underset{\pi }{\Sigma }|H/{H}_{\pi }|\text{.}$

 Proof. (a) Clear. (b) Write $\stackrel{‾}{U}{\stackrel{‾}{w}}_{0}v={\stackrel{‾}{X}}_{{w}_{0}}{\stackrel{‾}{w}}_{0}v={\stackrel{‾}{X}}_{a}{\stackrel{‾}{w}}_{a}{\stackrel{‾}{X}}_{b}{\stackrel{‾}{w}}_{b}\dots v,$ with ${w}_{0}={w}_{a}{w}_{b}\dots$ 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 $\left(\lambda ,{\mu }_{a}\right)$ according to the set $\pi$ of simple roots $a$ such that ${\mu }_{a}=-1\text{.}$ $\square$

Exercise: If $G$ is universal, the above number is $\underset{a \text{simple}}{\Pi }\left(|{H}_{a}|+1\right)=\underset{\alpha \text{simple}}{\Pi }q\left(\alpha \right),$ in the notation of Theorem 43. If in addition $G$ is not twisted, then the number is ${q}^{\ell }\text{.}$

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

Lemma 83:

 (a) If $w\in W,$ then any two minimal expressions for $w$ as a product of simple reflections can be transformed into each other by the relations $\left(*\right) wawbwa…= wbwawb…$ $\text{(}n$ terms on each side, $n=$ order ${w}_{a}{w}_{b},$ with $a$ and $b$ distinct simple roots). (b) Assume that for each simple root $a$ the corresponding element ${\stackrel{‾}{w}}_{a}$ of $G$ (any Chevalley group) is chosen to lie in $⟨{X}_{a},{X}_{-a}⟩\text{.}$ Let $w={w}_{a}{w}_{b}\dots$ be a minimal expression for $w\in W\text{.}$ Then $\stackrel{‾}{w}={\stackrel{‾}{w}}_{a}{\stackrel{‾}{w}}_{b}\dots$ is independent of the minimal expression chosen.

 Proof. (a) This is a refinement of Appendix IV.38 since the relations ${w}_{\alpha }^{2}=1$ 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 $\left(*\right)\text{.}$ For this we can refer to the proof of Lemma 56 since the extra restrictions there, that $G$ is untwisted and that ${\stackrel{‾}{w}}_{a}={w}_{a}\left(1\right)$ for each $a,$ are not essential for the proof. $\square$

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 $K\text{-module}$ 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.