Representations concluded - Lectures on Chevalley groups

Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 24 July 2013

§14. Representations concluded

Now we turn to the complex representations of the groups just considered. Here the theory is in poor shape. Only ${G L}_{n}$ (Green, T.A.M.S. 1955) and a few groups of low rank have been worked out completely, then only in terms of the characters. Here we shall consider a few general results which may lead to a general theory.

Henceforth $K$ will denote the complex field. Given a (one-dimensional) character $λ$ on a subgroup $B$ of a group $G,$ realized on a space $V_{λ},$ we shall write $V_{λ}^{G}$ for the induced module for $G .$ This may be defined by $V_{λ}^{G} = K G \otimes_{K B} V_{λ}$ (this differs from our earlier version in that we have not switched to a space of functions), and may be realized in $K G$ in the left ideal generated by $B_{λ} = \underset{b \in B}{Σ} λ (b^{- 1}) b$ (and will be used in this form). Its dimension is $| G / B | .$

Exercise: Check these assertions.

Lemma 84: Let $B, C$ be subgroups of a finite group $G,$ let $λ, μ$ be characters on $B, C,$ and let $V_{λ}^{G}, V_{μ}^{G}$ be the corresponding modules for $G .$

(a)	If $x \in G,$ then $B_{λ} x C_{μ}$ in $K G$ is determined up to multiplication by a nonzero scalar by the $(B, C)$ double coset to which $x$ belongs.
(b)	${Hom}_{G} (V_{λ}^{G}, V_{μ}^{G})$ is isomorphic as a $K -space$ to the one generated by all $B_{λ} x C_{μ} .$
(c)	If $B = C$ and $λ = μ,$ then the isomorphism in (b) is one of algebras.
(d)	The dimension of ${Hom}_{G} (V_{λ}^{G}, V_{μ}^{G})$ is the number of $(B, C)$ double cosets $D$ such that $B_{λ} x C_{μ} \neq 0$ for some, hence for every $x$ in $D,$ or, equivalently, such that the restrictions of $λ$ and $x μ$ to $B \cap x C x^{- 1}$ are equal.

Proof.

(a) This is clear.

(b) Assume $T \in {Hom}_{G} (V_{λ}^{G}, V_{μ}^{G}) .$ Since $B_{λ}$ generates $V_{λ}^{G}$ as a $K G -module,$ $T B_{λ}$ determines $T .$ Let $T B_{λ} = \underset{x \in G / C}{Σ} c_{x} x C_{μ} .$ Since $b B_{λ} = λ (b) B_{λ},$ we get by averaging over $B$ that $T B_{λ} = \underset{x \in B \ G / C}{Σ} c_{x} B_{λ} x C_{μ} .$ Thus $T$ is realized on $B_{λ},$ hence on all of $V_{λ}^{G},$ by right multiplication by ${| B |}^{- 1} Σ c_{x} B_{λ} x C_{μ} .$ Conversely, any such right multiplication yields a homomorphism, which proves (b).

(d) The first statement follows from (a) and (b). Let $B_{1} = B \cap x C x^{- 1}$ and $C_{1} = x^{- 1} B x \cap C,$ and ${y_{i}}$ and ${z_{j}}$ systems of representatives for $B_{1} \ B$ and $C / C_{1} .$ Set $B_{λ}^{'} = Σ λ (y_{i}^{- 1}) y_{i}$ and $C_{μ}^{'} = Σ μ (z_{j}^{- 1}) z_{j} .$ Then $B_{λ} x C_{μ} = B_{λ}^{'} B_{1 λ} B_{1, x μ} x C_{μ}^{'}$ with $(x μ) (x c x^{- 1}) = μ (c)$ since $x C_{1} x^{- 1} = B_{1} .$ If $λ \neq x μ$ on $B_{1},$ then $B_{1 λ} B_{1, x μ} = 0 .$ If $λ = x μ,$ this product is $| B_{1} | B_{1 λ},$ and then $B_{λ} x C_{μ} \neq 0$ since the elements $y_{i} b_{1} x z_{j}$ are all distinct.

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Remarks:

(a)	This is a special case of a theorem of G. Mackey. (See, e.g., Feit's notes.)
(b)	The algebra of (c) is also called the commuting algebra since it consists of all endomorphisms of $V_{λ}^{G}$ that commute with the action of $G .$

Theorem 47: Let $G$ be a (perhaps twisted) finite Chevalley group.

(a)	If $λ$ is a character on $H$ extended to B in the usual way, then $V_{λ}^{G}$ is irreducible if and only if $w λ \neq λ$ for every $w \in W$ such that $w \neq 1 .$
(b)	If $λ, μ$ both satisfy the conditions of (a), then $V_{λ}^{G}$ is isomorphic to $V_{μ}^{G}$ if and only if $λ = w μ$ for some $w \in W .$

Proof.

(a) $V_{λ}^{G}$ is irreducible if and only if its commuting algebra is one-dimensional (Schur's Lemma), i.e., by (c) and (d) of Lemma 84, if and only if $λ$ and $w λ$ agree on $B \cap w B w^{- 1},$ hence on $H,$ for exactly one $w \in W,$ i.e. for only $w = 1 .$

(b) Since $V_{λ}^{G}$ and $V_{μ}^{G}$ irreducible, they are isomorphic if and only if $dim {Hom}_{G} (V_{λ}^{G}, V_{μ}^{G}) = 1,$ which, as above, holds exactly when $λ = w μ$ for some (hence for exactly one) $w \in W .$

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Exercise:

(a)	$dim {Hom}_{G} (V_{λ}^{G}, V_{μ}^{G}) = \| W_{λ} \|$ if $λ = w μ$ for some $w,$ $dim {Hom}_{G} (V_{λ}^{G}, V_{μ}^{G}) = 0$ otherwise.
(b)	In Theorem 47 the conclusion in (b) holds even if the condition in (a) doesn't.

Here $W_{λ}$ is the stabilizer of $λ$ in $W .$ We see, in particular, that if $λ = 1$ then the commuting algebra of $V_{λ}^{G}$ is $| W | -dimensional.$ But more is true.

Theorem 48: Let $V$ be the $K G -module$ induced by the trivial one-dimensional $K B -module.$ Then the commuting algebra ${End}_{G} (V)$ is isomorphic to the group algebra $K W .$

Reformulations:

(a)	The multiplicities of the irreducible components of $V$ are just the degrees of the irreducible $K W -modules.$
(b)	The subalgebra, call it $A,$ of $K G$ spanned by the double coset sums $\overline{B} w {\overline{X}}_{w}$ is isomorphic to $K W .$
(c)	The algebra of functions $f : G \to K$ biinvariant under $B$ $(f (b x b') = f (x)$ for all $b, b' \in B)$ with convolution as multiplication is isomorphic to $K W .$

Proof.

The theorem is equivalent to (a) by Schur's Lemma and to (b) by Lemma 84, while (b) and (c) are clearly equivalent. We shall give a proof of (b), due to J. Tits, $\hat{w}$ will denote the average in $K G$ of the elements of the double coset $B w B .$ The elements $\hat{w}$ form a basis of $A$ and $\hat{1}$ is the unit element. If $a$ is a simple root, $c_{a}$ will denote ${| X_{a} |}^{- 1} .$

(1) $A$ is generated as an algebra by ${{\hat{w}}_{a} | a simple}$ subject to the relations

$(α)$	${\hat{w}}_{a}^{2} = c_{a} \hat{1} + (1 - c_{a}) {\hat{w}}_{a}$ for all $a .$
$(β)$	${\hat{w}}_{a} {\hat{w}}_{b} {\hat{w}}_{a} \dots = {\hat{w}}_{b} {\hat{w}}_{a} {\hat{w}}_{b} \dots,$ as in Lemma 83(a).

Proof.

We observe that if each $c_{a}$ is replaced by $1$ then these relations go over into a defining set for $K W,$ by Appendix II.38. Since $B \cup B w_{a} X_{a}$ is a group, we have ${(\overline{B} w_{a} {\overline{X}}_{a})}^{2} = r \overline{B} + s \overline{B} w_{a} {\overline{X}}_{a}$ with $r, s \in K .$ Since $B w_{a} X$ contains with each of its elements its inverse, we get $r = | B w_{a} X_{a} |,$ and then from the total coefficient $s = | B w_{a} X_{a} | - | B | .$ Thus $(α)$ holds in $A,$ and so does $(β)$ by Lemma 25,Cor., which also shows that the ${\hat{w}}_{a}$ generate $A .$ Conversely, let $A_{1}$ be the abstract associative algebra (with $\hat{1})$ generated by symbols ${\hat{w}}_{a}$ subject to $(α)$ and $(β) .$ For each $w \in W$ choose a minimal expression $w = w_{a} w_{b} \dots$ and set $\hat{w} = {\hat{w}}_{a} {\hat{w}}_{b} \dots .$ By Lemma 83(a) and $(β)$ this is independent of the expression chosen. By $(α)$ it follows that

\begin{matrix} {\hat{w}}_{a} \hat{w} & = & \hat{w_{a} w} & if w^{- 1} a > 0, \\ = & c_{a} \hat{w_{a} w} + (1 - c_{a}) \hat{w} & if w^{- 1} a < 0 . \end{matrix}

Thus the $\hat{w}$ form a basis for $A_{1},$ which, having the same dimension as $A,$ is therefore isomorphic to it.

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(2) There exists a positive number $c = c (G)$ such that. $c_{a} = c^{n_{a}}$ with $n_{a}$ a positive integer depending only on the type of $G .$ The multiplication table of $A$ in terms of the basis ${\hat{w}}$ is given by polynomials in $c$ depending only on the type.

Proof.

Consider ${}^{2}A_{4} (q^{2}),$ for example. Here the two possibilities for $| X_{a} |$ are $q^{2}$ and $q^{3}$ by Lemma 63(c). If we set $c = q^{- 1},$ then the corresponding values of $n_{a}$ are 2 and 3, which depend only on the type. For each of the other types the verification is similar. From the first statement of (2) and the equations of the proof of (1) the second statement follows.

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(3) An associative algebra $A_{c}$ with multiplication table given by the polynomials of (2) exists for every complex number $c .$ In particular $A_{c (G)} = A$ and $A_{1} = K W .$

Proof.

Since the type of the group $G$ contains an infinite number of members, the multiplication table is associative for an infinite set of values of $c,$ hence for all values.

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(4) $A_{c}$ is semisimple for $c = c (G),$ for $c = 1,$ and for all but a finite number of values of $c .$

Proof.

$A_{c}$ is for $c = c (G)$ the commuting algebra of a $K G -module$ and for $c = 1$ a group algebra $K W,$ hence semisimple in both cases. The discriminant of $A_{c}$ is a polynomial in $c,$ nonzero at $c = 1$ since then $A_{c}$ is semisimple, hence nonzero for all but a finite number of values of $c .$

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(5) Completion of proof. Since $A$ is semisimple and $K$ is an algebraically closed field, $A$ is a direct sum of complete matric algebras, of certain degrees over $K$ (see, e.g., Jacobson's Structure of Rings or Feit's notes), and similarly for $K W .$ We have to show that the degrees are the same in the two cases. If $𝒜$ is any finite-dimensional associative algebra which is separable, i.e. which is semisimple when the base field is extended to its algebraic closure, we define the numerical invariants of $𝒜$ to be the degrees of the resulting matric algebras. The proof of Theorem 48 will be completed by the following lemma.

Lemma 85: Let $R$ be an integral domain, $F$ its field of quotients, and $f$ a homomorphism of $R$ onto a field $K .$ Let $𝒜$ be a finite—dimensional associative algebra over $R,$ and $𝒜_{F}$ and $𝒜_{K}$ the resulting algebras over $F$ and $K .$ If $𝒜_{F}$ and $𝒜_{K}$ are separable, then they have the same numerical invariants.

In fact, from (3) and the lemma with $R = K [c],$ $𝒜 = A_{c},$ and first $f : c \to c (G)$ and then $f : c \to 1,$ it follows that $A$ and $K W$ have the same numerical invariants, hence that they are isomorphic.

Proof of the lemma.

(a) Assume that $ℬ$ is a finite-dimensional semisimple associative algebra over an algebraically closed field $L,$ that $b_{1}, b_{2}, \dots, b_{n}$ form a basis for $ℬ / L,$ that $x_{1}, x_{2}, \dots, x_{n}$ are independent indeterminates over $L,$ that $b = Σ x_{i} b_{i},$ and that $P (t)$ is the characteristic polynomial of $b$ acting from the left on $ℬ_{L (x_{1}, \dots, x_{n})},$ written as $P (t) = Π P_{i} {(t)}^{p_{i}}$ with the $P_{i}$ distinct monic polynomials irreducible over $L (x_{1}, \dots, x_{n}) .$ Then:

(a1)	The $p_{i}$ are the numerical invariants of $ℬ .$
(a2)	$p_{i} = d g_{t} P_{i}$ for each $i .$
(a3)	If $P (t) = Π Q_{j} {(t)}^{q_{j}}$ is any factorization over $L (x_{1}, \dots, x_{n})$ such that $q_{j} = d g_{t} Q_{j}$ for each $j,$ then it agrees with the one above so that the $q_{j}$ are the numerical invariants of $ℬ .$

Proof.

For (a1) and (a2) we may assume that $ℬ$ is the complete matric algebra $End L^{p}$ and that $b = Σ x_{i j} E_{i j}$ in terms of the matric units $E_{i j} .$ If $X = [x_{i j}],$ then $P (t) = det {(t I - X)}^{p},$ so that we have to show that $det (t I - X)$ is irreducible over $L (x_{i j}) .$ This is so since specialization to the set of companion matrices

[\begin{matrix} 1 \\ ⋱ \\ 1 \\ x_{p 1} & x_{p 2} & \dots & x_{p p} \end{matrix}]

yields the general equation of degree $p .$ In (a3) if some $Q_{j}$ were reducible or equal to some $Q_{k}$ with $k \neq j,$ then any irreducible factor $P_{i}$ of $Q_{j}$ would violate (a2).

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(b) Let $R^{*}$ be the integral closure of $R$ in $\overline{F}$ (consisting of all elements satisfying monic polynomial equations over $R),$ and $x_{1}, x_{2}, \dots, x_{n}$ indeterminates over $\overline{F} .$ Then $R^{*} [x_{1}, \dots, x_{n}]$ is the integral closure of $R [x_{1}, \dots, x_{n}]$ in $\overline{F} (x_{1}, \dots, x_{n}) .$

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Proof.

See, e.g., Bourbaki, Commutative Algebra, Chapter V, Prop. 13.

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(c) If $R^{*}$ is as in (b), then any homomorphism of $R$ into $K$ can be extended to one of $R^{*}$ into $\overline{K} .$

Proof.

By Zorn's lemma this can be reduced to the case $R^{*} = R [α],$ where it is almost immediate since $\overline{K}$ is algebraically closed.

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(d) Completion of proof. Let ${a_{i}}$ be a basis for $𝒜 / R,$ hence also for $𝒜_{F} / F,$ and ${x_{i}}$ independent indeterminates over $\overline{F}$ and also over $\overline{K} .$ The given homomorphism $f : R \to K$ defines a homomorphism $f : 𝒜 \to 𝒜_{K} .$ By (c) it extends to a homomorphism of $R^{*}$ into $\overline{K}$ and then naturally to one of $R^{*} [x_{1}, \dots, x_{n}]$ into $\overline{K} [x_{1}, \dots, x_{n}] .$ If $a = Σ x_{i} a_{i}$ and $P (t) = Π P_{i} {(t)}^{p_{i}}$ is its characteristic polynomial, factored over $\overline{F} (x_{1}, \dots, x_{n})$ as before, then the coefficients of each $P_{i}$ are integral polynomials in its roots, hence integral over the coefficients of $P,$ hence integral over $R [x_{1}, x_{2}, \dots, x_{n}],$ hence belong to $R^{*} [x_{1}, \dots, x_{n}]$ by (b). Thus if $f (a) = Σ x_{i} f (a_{i}),$ then its character polynomial has a corresponding factorization $P_{f} (t) = Π P_{i f} {(t)}^{p_{i}}$ over $\overline{K} (x_{1}, \dots, x_{n}) .$ By (a1) the $p_{i}$ are the numerical invariants of $𝒜_{F},$ and by (a2) they satisfy $p_{i} = d g_{t} P_{i} = d g_{t} P_{i f},$ so that by (a3) with $ℬ = 𝒜_{\overline{K}}$ they are also the numerical invariants of $𝒜_{K},$ which proves the lemma.

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Exercise: If $λ$ is a character on $H$ extended to $B$ in the usual way, then ${End}_{G} (V_{λ}^{G})$ is isomorphic to $K W_{λ} .$ (Observe that this result includes both Theorem 47 and Theorem 48.)

Remark: Although $A$ is isomorphic to $K W$ there does not seem to be any natural isomorphism and no one has succeeded in decomposing the module $V$ of Theorem 48 into its irreducible components, except for some groups of low rank. We may obtain some partial results, in terms of characters, by inducing from the parabolic subgroups and using the following simple facts.

Lemma 86: Let $π$ be a set of simple roots, $W_{π}$ and $G_{π}$ the corresponding subgroups of $W$ and $G$ (see Lemma 30), and $V_{π}^{W}$ and $V_{π}^{G}$ the corresponding trivial modules induced to $W$ and $G;$ and similarly for $π' .$

(a)	A system, of representatives in $W$ for the $(W_{π}, W_{π'})$ double cosets becomes in $G$ a system of representatives for $(G_{π}, G_{π'})$ double cosets.
(b)	$dim {Hom}_{G} (V_{π}^{G}, V_{π'}^{G}) = dim {Hom}_{W} (V_{π}^{W}, V_{π'}^{W}) .$

Proof.

(a) Exercise.

(b) By Lemma 84(d) and (a).

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Corollary 1: Let $χ_{π}^{G}$ denote the character of $V_{π}^{G}$ and similarly for $W .$ If ${n_{π}}$ is a set of integers such that $χ^{W} = Σ n_{π} χ_{π}^{W}$ is an irreducible character of $W,$ then $χ^{G} = Σ n_{π} χ_{π}^{G}$ is, up to sign, one of $G .$

Proof.

Let ${(χ, ψ)}_{G}$ denote the average of $χ \overline{ψ}$ over $G .$ We have ${(χ_{π}^{G}, χ_{π'}^{G})}_{G} = dim {Hom}_{G} (V_{π}^{G}, V_{π'}^{G}),$ and similarly for $W .$ Since $χ^{W}$ is irreducible, ${(χ^{W}, χ^{W})}_{W} = 1,$ it follows that ${(χ^{G}, χ^{G})}_{G} = 1,$ so that $\pm χ^{G}$ is irreducible, by the orthogonality relations for finite group characters.

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Remarks:

(a)	In a beautiful paper in Berliner Sitzungsberichte, 1900, Frobenius has constructed a complete set of irreducible characters for the symmetric group $S_{n},$ i.e. the Weyl group of type $A_{n - 1},$ as a set of integral combinations of the characters $χ_{π}^{W} .$ Using his method and the preceding corollary one can decompose the character of $V$ in Theorem 48 in case $G$ is of type $A_{n - 1} .$ (See R. Steinberg T.A.M.S. 1951).
(b)	The situation of (a) does not hold in general. Consider, for example, the group $W$ of type $B_{2},$ i.e. the dihedral group of order 8. It has five irreducible modules (of dimensions 1,1,1,1,2), while there are only four $χ_{π}^{W}'s$ to work with.
(c)	A result of a general nature is as follows.

Corollary 2: If the notation is as above and ${(- 1)}^{π}$ is as in Lemma 66(d), then $χ^{G} = Σ {(- 1)}^{π} χ_{π}^{G}$ is an irreducible character of $G$ and its degree in $| U | .$

Proof.

Consider $χ^{W} = Σ {(- 1)}^{π} χ_{π}^{W} .$ By (8) on p. 142, extended to twisted groups (check this using the hints given in the proof of Lemma 66), $χ^{W} = det,$ an irreducible character. Hence $\pm χ^{G}$ is also one by Cor. 1 above. We have $χ^{G} (1) = Σ {(- 1)}^{π} | G / G_{π} | .$ If $G$ is untwisted and the base field has $q$ elements, then by Theorem 4' applied to $G$ and to $G_{π}$ this can be continued $Σ {(- 1)}^{π} W (q) / W_{π} (q) = q^{N} = | U |,$ as in (4) of the proof of Theorem 26. If $G$ is twisted, the proof is similar.

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We continue with some remarks on the algebra $A$ of Theorem 48(b).

Lemma 87: The homomorphisms of $A$ onto $K$ are given by: $f ({\hat{w}}_{a}) = 1$ or $- c_{a}$ for each simple root $a,$ subject to the condition that $f ({\hat{w}}_{a})$ is constant on each $W -orbit.$

Proof.

For $a$ and $b$ simple, let $n (a, b)$ denote the order of $w_{a} w_{b}$ in $W .$ We claim that $(*)$ $a$ and $b$ belong to the same $W -orbit$ if and only if there exists a sequence of simple roots $a = a_{0}, a_{1}, \dots, a_{r} = b$ such that $n (a_{i}, a_{i + 1})$ is odd for every $i .$ The equation ${(w_{a_{i}} w_{a_{i + 1}})}^{n} = 1$ with $n$ odd can be rewritten to show that $w_{a_{i}}$ and $w_{a_{i + 1}}$ are conjugate, so that if the sequence exists then $a$ and $b$ are conjugate. If $a$ and $b$ are conjugate, then so are $w_{a}$ and $w_{b},$ and this remains true when we project into the reflection group obtained by imposing on $W$ the additional relations: ${(w_{c} w_{d})}^{2} = 1$ whenever $n (c, d)$ is even. In this new group $w_{a}$ and $w_{b}$ must belong to the same component, so that the required sequence exists. By $(*)$ the condition of the lemma holds exactly when $f ({\hat{w}}_{a}) = f ({\hat{w}}_{b})$ whenever $n (a, b)$ is odd, i.e. exactly when $f$ preserves the relations $(β)$ (of the proof of Theorem 48). Since $f ({\hat{w}}_{a}) = 1$ or $- c_{a}$ exactly when $f$ preserves $(α)$ (solve the quadratic), we have the lemma.

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Remark: By finding the annihilator in $A$ of the kernel of each of the homomorphisms of Lemma 87, we get a one-dimensional ideal $I$ in $A .$ This corresponds to an irreducible submodule of multiplicity 1 in $V,$ realized in the left ideal $K G I$ of $K G .$ By working out the corresponding idempotent, the degree of the submodule can be found.

Exercise:

(a)	If $f ({\hat{w}}_{a}) = 1$ for all $a,$ show that $I = K Σ q_{w} \hat{w}$ with $q_{w} = \| X_{W} \|,$ and that the corresponding $K G -module$ is the trivial one. (Hint: by writing $W$ as a union of right cosets relative to ${1, w_{a}}$ and writing $(α)$ in the form $({\hat{w}}_{a} - 1) ({\hat{w}}_{a} + c_{a}) = 0,$ show that $I$ is as indicated.)
(b)	If $f ({\hat{w}}_{a}) = - c_{a}$ for all $a,$ show that $I = K Σ (det w) \hat{w},$ and that in this case the dimension is $\| U \| .$ (Hint: if $e$ is the given sum and $c_{w} = q_{w}^{- 1},$ show that $e^{2} = m e$ with $m = Σ c_{w} = \| G / B \| / \| U \| .)$
(c)	For $G = B_{2} (q)$ work out all (four) cases of Lemma 87, hence obtain the degrees of all (five) irreducible components of $V .$
(d)	Same for $A_{2}$ and for $G_{2} .$

Remark: It can be shown that the module of (b) is isomorphic to the one with character $χ^{G}$ as in Lemma 86, Cor.2. If we reduce the element $Σ (det w) | U / X_{W} | \overline{B} w {\overline{X}}_{W}$ (which is $| B | | U | e)$ of $I$ mod $p,$ we see from the proof of Theorem 46(e) that the given module reduces to the one of that theorem for which $λ = 1$ and every $μ_{a} = - 1 .$ This latter module can itself be shown to be isomorphic to the one in Theorem 43 for which $⟨ λ, α ⟩ = q (α) - 1$ for every $α .$ From these facts and Weyl's formula the character of the original module can be found up to sign and the result used to prove that the number of $p -elements$ (of order a power of $p)$ of $G$ is ${| U |}^{2} .$ For the details see R. Steinberg, Endomorphisms of linear algebraic groups, to appear.

Finally, we should mention that the algebra $A$ admits an involution given by ${\hat{w}}_{a} \to 1 - c_{a} - {\hat{w}}_{a}$ for all $a$ (which in case $c_{a} \to 1$ and $A \to K W$ reduces to $w \to (det w) w).$

The preceding discussion points up the following
Problem: Develop a representation theory for finite reflection groups and use it to decompose the module $V$ (or the algebra $A)$ of Theorem 48.

It is natural that in studying the complex representations of $G$ we have considered first those induced by characters on $B$ since for representations of characteristic $p$ this leads to a complete set. In characteristic 0, however, this is not the case, as even the simplest case $G = {S L}_{2}$ shows. One must delve deeper. Therefore, we shall consider representations of $G$ induced by (one-dimensional) characters $λ$ on $U .$ We can not expect such a representation ever to be irreducible since its degree $| G / U |$ is too large (larger than ${| G |}^{\frac{1}{2}}),$ but what we shall show is that if $λ$ is sufficiently general then at least it is multiplicity-free. In other words, ${End}_{G} (V_{λ}^{G})$ is Abelian, hence a direct sum of fields. (If $λ$ is not sufficiently general, we can expect the Weyl group to play a role, as in Theorem 48.)

Before stating the theorem, we prove two lemmas.

Lemma 88: Let $k$ be a finite field and $λ$ a nontrivial character from the additive group of $k$ into $K^{*} .$ Then every character can be written uniquely $λ_{c} : t \to λ (c t)$ for some $c \in k .$

Proof.

The map $c \to λ_{c}$ is a homomorphism of $k$ into its dual, and its kernel is clearly 0.

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Lemma 89: For $w \in W$ the following conditions are equivalent.

(a)	If $a$ and $w a$ are positive roots and one of them is simple, then so is the other.
(b)	If $a$ is simple and $w a$ is positive, then $w a$ is simple.
(c)	$w = w_{0} w_{π}$ for some set $π$ of simple roots, with $w_{0}$ as usual and $w_{π}$ the corresponding object of $W_{π} .$

There are

2^{ℓ}

possibilities for

w .

Proof.

(c) $\Rightarrow$ (a) Because $w_{π}$ maps $π$ onto $- π$ and $(*)$ permutes the positive roots with support not in $π$ (same proof as for Appendix I.11).

(a) $\Rightarrow$ (b) Obvious.

(b) $\Rightarrow$ (c) Let $π$ be the set of simple roots kept positive, hence simple, by $w .$ We claim: $(* *)$ if $a > 0$ and $supp a ⊈ π,$ then $w a < 0 .$ Write $a = b + c$ with $supp b \subseteq π,$ $supp c \subseteq Π - π .$ Then $w a = w b + w c .$ Here $w c < 0$ by the choice of $π,$ and $supp w c ⊈ w π \supseteq supp w b .$ Thus $w a < 0 .$ If $a$ is a simple root not in $π$ then $w w_{π} a < 0$ by $(*)$ and $(* *),$ while if $a$ is in $π$ this holds by the definition of $π .$ Thus $w w_{π} = w_{0},$ whence (c).

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Theorem 49: Let $G$ be a finite, perhaps twisted, Chevalley group and $λ : U \to K^{*}$ a character such that $λ | X_{a} \neq 1$ if $a$ is simple, $λ | X_{a} = 1$ if $a$ is positive but not simple. Then $V_{λ}^{G}$ is multiplicity-free. In other words, ${End}_{G} (V_{λ}^{G})$ is Abelian, or, equivalently, the subalgebra $A$ of $K G$ spanned by the elements $U_{λ} h \overline{w} U_{λ}$ $(h \in H, w \in W)$ is Abelian.

Here $U_{λ} = \underset{u \in U}{Σ} λ (u^{- 1}) u,$ and we assume that the $\overline{w}$ are chosen as in Lemma 83(b).

Remarks:

(a)	If $a$ is not simple, then usually $X_{a} \subseteq 𝒟 U,$ so that the assumption $λ \| X_{a} = 1$ is superfluous, but this is not always the case, e.g. for $B_{2}$ or $F_{4}$ with $\| k \| = 2$ or for $G_{2}$ with $\| k \| = 3 .$ In these latter groups, there are other possibilities, which because of their special nature will not be gone into here.
(b)	The proof to follow is suggested by that of Gelfand and Graev, Doklady, 1963 who have given a proof for ${S L}_{n}$ and announced the general result for the untwisted groups. T. Yokonuma, C. Rendues, Paris, 1967, has also given a proof for these latter groups, but his details are unnecessarily complicated.

Proof of Theorem 49.

The fact that $A$ is Abelian will follow from the existence of an (involutory) antiautomorphism $f$ of $G$ such that

(a) $f U = U .$

(b) $λ f = λ$ on $U .$

(c) For each double coset $U n U$ such that $U_{λ} n U_{λ} \neq 0,$ we have $f_{n} = n$ (here $n \in N = Σ H \overline{w}).$

For since $f$ extended to $K G$ and then restricted to $A$ is an antiautomorphlsm and at the same time the identity (by (a), (b), (c)) it is clear that $A$ is Abelian. The existence of $f$ will be proved in several steps.

(1) If $U_{λ} n U_{λ} \neq 0$ and $n \in H \overline{w},$ then $w = w_{0} w_{π}$ for some set $π$ of simple roots.

Proof.

By Lemma 89 we need only prove that if $a$ is simple and $w a$ positive then $w a$ is simple. Writing the first $U_{λ}$ above with the $X_{w a}$ component on the right, and the second with the $X_{a}$ component on the left, we get $X_{w a, λ} n X_{a, λ} n^{- 1} \neq 0 .$ Since $λ$ is nontrivial on $X_{a}$ it is also so on $X_{w a},$ whence $w a$ is simple by the assumptions on $λ,$ which proves (1).

$□$

The condition in (1) essentially forces the correct definition of $f .$ We set $a^{*} = - w_{0} a .$ If $a$ is simple, so is $a^{*} .$ In order to simplify the discussion in one or two spots we assume henceforth that $G$ (i.e. its root system) is indecomposable. If $G$ is untwisted, we start with the graph automorphism corresponding to $*$ (see the Corollary on p. 156), compose it with the inversion $x \to x^{- 1},$ and finally with a diagonal automorphism so that the result $f$ satisfies, not only $f U = U$ but also $λ f = λ$ on $U .$ This is possible because of Lemma 89 and the assumptions on $λ$ in the theorem. If $G$ is twisted, then we may omit the graph automorphism, (because $*$ is then the identity), and use the explicit isomorphism $X_{a} / 𝒟 X_{a} ≅ k$ of (2) of the proof of Theorem 36 in combination with Lemma 89 to achieve the second condition. We see that

(2) $f$ is an involutory antiautomorphism which satisfies the required conditions (a) and (b). We must prove that it also satisfies (c). As consequences of the construction we have:

(3) $f h = {\overline{w}}_{0} h {\overline{w}}_{0}^{- 1}$ for every $h \in H .$

(4) If $a^{*} = a,$ then $f$ is the identity on $X_{a} / 𝒟 X_{a} .$

(5) If $a^{*} = a,$ there exists a nontrivial element of $X_{a}$ fixed by $f .$

Proof.

For $X_{a}$ of type $A_{1}$ this follows from (4). For $X_{a}$ of type ${}^{2}A_{2}$ we choose the element $(t, u)$ of Lemma 63(c) with $t = 2,$ $u = 2$ if $p \neq 2,$ and $t = 0,$ $u = 1$ if $p = 2,$ since $f (t, u) = (t, - t t^{θ} u)$ (check this, referring to the construction of $f).$ For types ${}^{2}C_{2}$ and ${}^{2}G_{2}$ we may choose $(0, 1)$ and $(0, 1, 0)$ since $f$ is the identity on ${(0, u)}$ and ${(0, u, 0)} .$

$□$

(6) The elements ${\overline{w}}_{a} \in G$ may be so chosen that:

(6a)	$f {\overline{w}}_{a} = {\overline{w}}_{a^{*}}$ for every simple root $a .$
(6b)	If $a$ and $b$ are simple and $n \in N$ is such that $n X_{a} n^{- 1} = X_{b}$ and $λ (n x n^{- 1}) = λ (x)$ for all $x \in X_{a},$ then $n {\overline{w}}_{a} n^{- 1} = {\overline{w}}_{b} .$

Proof.

Under the action of $f$ and the inner automorphisms $i_{n}$ by elements $n$ as in (6b) the $X_{a}$ $(a$ simple) form orbits. From each orbit we select an element $X_{a} .$ If $a^{*} = a,$ we choose $x_{a} \in X_{a}$ as in (5), write it as $(*)$ $x_{a} = x_{1} {\overline{w}}_{a} x_{2}$ with $x_{1}, x_{2} \in X_{- a},$ and choose ${\overline{w}}_{a}$ accordingly. Since $f$ is an antiautomorphism and fixes $X_{- a},$ it also fixes ${\overline{w}}_{a}$ by the uniqueness of the above form. If $a^{*} \neq a,$ we choose $x_{a} \in X_{a},$ $x_{a} \neq 1,$ arbitrarily. We then use the equations $f {\overline{w}}_{a} = {\overline{w}}_{a^{*}}$ and $i_{n} {\overline{w}}_{a} = {\overline{w}}_{b}$ of (6a) and (6b) to extend the definition of $\overline{w}$ to the orbit of $a .$ We must show this can be done consistently, that we always return to the same value. Let $a_{1}, a_{2}, \dots, a_{n}$ be a sequence of simple roots such that $a_{1} = a_{n} = a$ and for each $j$ either $a_{j + 1} = a_{j}^{*}$ or else there exists $n_{j}$ such that the assumptions in (6b) holds with $a_{j}, a_{j + 1}, n_{j}$ in place of $a, b, n .$ Let $g$ denote the product of the corresponding sequence of $f's$ and $i_{n_{j}}'s.$ We must show that $g$ fixes ${\overline{w}}_{a} .$ We have $g X_{a} = X_{a},$ $g X_{- a} = X_{- a},$ and in fact $g$ acts on $X_{a} / 𝒟 X_{a},$ identified with $k,$ by multiplication by a scalar $c$ as follows from the definition of $f$ and the usual formulas for $i_{n} .$ Since $λ g = λ$ by the corresponding condition on $f$ and each $i_{n},$ it follows from Lemma 89 that $c = 1,$ so that $g$ is the identity on $X_{a} / 𝒟 X_{a} .$ If $𝒟 X_{a} = 0,$ then $g$ fixes the element $x_{a}$ above, hence also ${\overline{w}}_{a}$ by $(*),$ whether $g$ is an automorphism or an antiautomorphism. If $𝒟 X_{a} \neq 0,$ then $G$ is twisted so that $a^{*} = a .$ If $g$ is an automorphism, then by the proof of Theorem 36 from (5) on its restriction to $⟨ X_{a}, X_{- a} ⟩$ is the identity so that it fixes ${\overline{w}}_{a},$ while if $g$ is an antiautomorphism then by the same result its restriction coincides with that of $f$ so that it fixes ${\overline{w}}_{a}$ by the choice of ${\overline{w}}_{a} .$

$□$

Remark: If $G$ is untwisted, the above proof is quite simple.

We assume henceforth that the ${\overline{w}}_{a}$ are as in (6).

(7) If $\overline{w} = {\overline{w}}_{a} {\overline{w}}_{b} \dots$ as in Lemma 83(b) and $w^{*} = w_{0} w^{- 1} w_{0}^{- 1},$ then $f \overline{w} = {\overline{w}}^{*} .$

Proof.

Since $w_{a} w_{b} \dots$ is minimal, $\dots w_{b^{*}} w_{a^{*}}$ is also. (Check this.) Since $f$ is an antiautomorphlsm it follows from (6a) that $f \overline{w} = \dots {\overline{w}}_{b^{*}} {\overline{w}}_{a^{*}} = \overline{\dots w_{b^{*}} w_{a^{*}}} = {\overline{w}}^{*} .$

$□$

(8) If $w$ is as in (1) then $f \overline{w} = {\overline{w}}^{*} .$

Proof.

$w^{*} = w$ in this case (see (7)).

$□$

(9) If $n$ is as in (1) then $f n = n .$

Proof.

By (1), $n \in \overline{w} H$ with $w = w_{0} w_{π} .$ Assume $a \in π .$ Then $w a$ is simple and $λ (n x n^{- 1}) = λ (x)$ for all $x \in X_{a},$ by the inequality in the proof of (1). Thus $n {\overline{w}}_{a} n^{- 1} = {\overline{w}}_{w a}$ by (6b), from which we get, on picking a minimal expression for $w_{π},$ that $(*)$ $n {\overline{w}}_{π} n^{- 1} = \overline{w_{π^{*}}} .$ Since $N (w) = N (w_{0} w_{π}) = N (w_{0}) - N (w_{π}),$ it follows that if we put together minimal expressions for $w$ and $w_{π}$ we will get one for $w_{0} .$ Thus ${\overline{w}}_{0} = \overline{w} {\overline{w}}_{π}$ by Lemma 83(b), and similarly ${\overline{w}}_{0} = {\overline{w}}_{π^{*}} \overline{w},$ so that $(* *)$ $\overline{w} {\overline{w}}_{π} {\overline{w}}^{- 1} = {\overline{w}}_{π^{*}} .$ If now we write $n = \overline{w} h,$ then $h$ commutes with ${\overline{w}}_{π}$ by $(*)$ and $(* *) .$ Hence

\begin{matrix} f n & = & f h \cdot f \overline{w} & since f is an antiautomorphism \\ = & {\overline{w}}_{0} h {\overline{w}}_{0}^{- 1} w & by (3) and (8) \\ = & \overline{w} {\overline{w}}_{π} h {\overline{w}}_{π}^{- 1} & since {\overline{w}}_{0} = \overline{w} {\overline{w}}_{π} \\ = & \overline{w} h & since h commutes with w_{π} \\ = & n . \end{matrix}

$□$

Thus $f$ satisfies condition (c) and the proof of Theorem 49 is complete.

$□$

Exercise:

(a)	Prove that if ${{\overline{w}}_{a}}$ is as in (6) and $w$ as in (1), then $U_{λ} \overline{w} U_{λ} \neq 0 .$
(b)	Deduce that if $H_{π}$ denotes the kernel of the set of simple roots $π$ then the dimension of $A,$ hence the number of irreducible components of $V_{λ}^{G},$ in Theorem 49 is $Σ \| H_{π} \| .$

Remark: The natural group for the preceding theorem seems to be the adjoint group extended by the diagonal automorphisms, a group of the same order as the universal group, but with something extra at the top instead of at the bottom. For this group, $G',$ prove that the dimension above is just $Π (| H_{a} | + 1) = Π q (α)$ in the notation of the exercise just before Lemma 83. Prove also that in this case $V_{λ}^{G}$ is independent of $λ .$

Remark: The problem now is to decompose the algebra $A$ of Theorem 49 into its simple (one-dimensional) components. If this were done, it would be a major step towards a representation theory for $G .$ As far as we know this has been done only for the group $A_{1}$ (see Gelfand and Graev, Doklady, 1962). It would not, however, be the complete story. For not every irreducible $G -module$ is contained in one induced by a character on $U,$ i.e., by Frobenius reciprocity, contains a one-dimensional $U -module,$ as the following, our final, example, due to M. Kneser, shows (although it is for some types of groups such as $A_{n}).$

As remarked earlier, reduction mod 3 yields an isomorphism of the subgroup $W^{+}$ of elements of determinant 1 of the Weyl group $W$ of type $E_{6}$ onto the group $G = {S O}_{5} (3),$ the adjoint group of type $B_{2} (3) .$ If we reverse this isomorphism and extend the scalars we obtain a representation of $G$ on a complex space $V .$ The assertion is that $U,$ i.e. a 3-Sylow subgroup of $G,$ fixes no line of $V .$ Consider the following diagram.

This is the Dynkin diagram of $E_{6}$ with the lowest root $α_{7}$ adjoined $(α_{7}$ is the unique root in $- D$ (see Appendix III.33), unique because all roots are conjugate in the present case. It is connected as shown because of symmetry and the fact that each proper subdiagram must represent a finite reflection group.) We choose as a basis for $V$ the $α's$ with $α_{3}$ omitted, a union of three bases of mutually orthogonal planes. $w_{1} w_{2}$ acts as a rotation of $120^{\circ}$ in the plane $⟨ α_{1}, α_{2} ⟩$ and as the identity in the other two planes, and similarly for $w_{4} w_{5}$ and $w_{6} w_{7} .$ The group $W^{+}$ also contains an element permuting the three planes cyclically as shown, because of the conjugacy of simple systems and the uniqueness of lowest roots, and the four elements generate a 3-subgroup of $W .$ It is now a simple matter to prove that this subgroup fixes no line of $V .$

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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