Representations - 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: 21 July 2013

§12. Representations

In this section we consider the irreducible representations of the infinite Chevalley groups. As we shall see, here the theory is quite complete. All representations are assumed to be finite-dimensional and the standard terminology is used. In particular 1 must act as the identity, and the trivial $0 -dimensional$ (but not the trivial $1 -dimensional)$ representation is excluded from the list of irreducible representations. We start with a general lemma.

Lemma 68: Let $K$ be an algebraically closed field, $B$ and $C$ associative algebras with $1$ over $K,$ and $A = B \otimes C .$

(a)	If $(β, V)$ and $(γ, W)$ are (finite-dimensional) irreducible modules for $B$ and $C,$ then $(α, U) = (β \otimes γ, V \otimes W)$ is one for $A .$
(b)	Conversely, every irreducible $A -module$ $(α, U)$ is realizable, uniquely, as a tensor product as in (a).

Proof.

(a) By Burnside's Theorem (see, e.g., Jacobson, Lectures in Abstract Algebra, Vol. 2), $β B = End V$ and $γ C = End W,$ whence $α A = End U$ and $(α, U)$ is irreducible.

(b) Let $V$ be an irreducible $B -submodule$ of $U .$ Such exist since $U$ is finite-dimensional. Let $L$ be the space of $B -homomorphisms$ of $V$ into $U .$ This is nonzero and is a $C -module$ under the rule $c ℓ = α (c) \circ ℓ .$ (Check this.) Let $(γ, W)$ be an irreducible submodule. The map $φ : V \otimes W \to U$ defined by $v \otimes f \to f (v)$ is easily checked to be an $A -homomorphism.$ $V \otimes W$ is irreducible by (a), and $U$ is by assumption. Hence by Schur's Lemma (see loc. cit.) $φ$ is an isomorphism. If $α = β' \circ γ'$ is a second decomposition of the required form, then restriction to $B$ yields $β \otimes 1 ≅ β' \otimes 1,$ i.e. multiples of $β$ and $β'$ are isomorphic, so that by the Jordan-Holder or Krull-Schmidt theorems $β$ and $β'$ are also. Similarly $γ$ and $γ'$ are isomorphic, which proves the uniqueness in (b).

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Corollary:

(a)	If $K$ is an algebraically closed field and $G = Π G_{i}$ is a direct product of a finite number of groups, then the tensor product $V$ of irreducible $K G_{i} -modules$ $V_{i}$ is an irreducible $K G -module,$ and every irreducible $K G -module$ is uniquely realizable in this way.
(b)	Similarly for a direct sum $ℒ = Σ ℒ_{i}$ of Lie algebras over $K .$

Proof.

We apply Lemma 68, extended to several factors, in (a) to group algebras, in (b) to enveloping algebras.

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Exercise: If the direct product above is one of algebraic groups over $K$ (of topological groups, of Lie groups,...), then $V$ is rational (continuous, analytic,...) if and only if each $V_{i}$ is.

Remark: If we are interested in the irreducible representations of a Chevalley group $G,$ we may as well assume it is universal. The corollary then implies that we may as well also assume that $G$ (i.e. that $Σ)$ is indecomposable. This we will do whenever it is convenient.

Now we take up the study of rational representations for Chevalley groups over algebraically closed fields viewed as algebraic groups. In such representations the coordinates of the representative matrix are required to be rational functions of the original coordinates. Whether this requirement is to be taken locally (e.g. as in the proof of Theorem 7, Cor. 1) or globally is immaterial, in view of the following result.

Lemma 69: Let $G$ be a Chevalley group viewed as an algebraic group as above, and $f : G \to k$ a function. Then the following conditions are equivalent.

(a)	$f$ is expressible as a rational function locally.
(b)	$f$ is expressible as a rational function globally.
(c)	$f$ is expressible as a polynomial.

Proof.

It will be enough to show that (a) implies (c). Let A be the algebra of polynomial functions on $G .$ By assumption there exists an open covering ${U_{i}}$ of $G,$ which may be taken finite by the maximal condition on the open subsets of $G$ (which holds by Hilbert's basis theorem in $A),$ and elements $g_{i}, h_{i}$ in $A$ such that $f = g_{i} / h_{i}$ and $h_{i} \neq 0$ on $U_{i}$ for all $i .$ Since the $h_{i}$ don't all vanish together, by Hilbert's Nullstellensatz there exist elements $a_{i}$ in $A$ such that $1 = Σ a_{i} h_{i}$ on $G .$ Let $U_{0} = \cap U_{i};$ it is nonempty, in fact dense, since $G$ is irreducible. On $U_{0}$ we have $f = Σ a_{i} f h_{i} = Σ a_{i} g_{i},$ a polynomial, hence by density also on each $U_{i}$ and on $G,$ as required.

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Presently we will need the following result.

Lemma 70: The algebra $A$ of polynomial functions on $G$ is integrally closed (in its quotient field).

Proof.

We observe first that $A$ is an integral domain (since $G$ is irreducible as an algebraic set, the polynomial ideal defining it is prime), so that it really has a quotient field. Assume $f = p_{1} / p_{2}$ $(p_{i} \in A)$ is integral over $A : f^{n} + a_{1} f^{n - 1} + \dots + a_{n} = 0$ for some $a_{i} \in A .$ On restriction to the open set $U^{-} H U$ of $G$ the $p_{i}$ and $a_{i}$ become, by Theorem 7(b), polynomials in the coordinates ${t_{α}, t_{i}, t_{i}^{- 1}} .$ Since such polynomials form a unique factorization domain, we see by the above equation that $f$ itself is such a polynomial, on $U^{-} H U .$ The same being true on each of the translates of $U^{-} H U$ by elements of $G,$ we conclude that $f$ is a polynomial on $G,$ i.e. $f$ is in $A,$ by Lemma 69.

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Two more lemmas and then the main theorem.

Lemma 71: The rational characters of $H$ (homomorphisms into $k^{*})$ are just the elements of the lattice $L$ generated by the global weights of the representation defining $G .$

Proof.

Let $λ$ be a character. Then it is a polynomial in the diagonal elements of $H$ (written as a group of diagonal matrices), i.e. a linear combination of elements of $L .$ Being multiplicative, it equals some element of $L .$ (Prove this.) Conversely, if $λ \in L,$ then $λ$ is a power product of weights in the representation defining $G,$ and all exponents may be taken positive since the product of the latter weights (as functions) is $1,$ so that $λ$ is a polynomial on $H .$

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Now for any rational $G -module$ $V$ we may define the weights $λ$ and the corresponding weight spaces $V_{λ},$ relative to $H,$ in the obvious way.

Lemma 72: Let $V$ be a rational $G -module,$ $λ$ a weight, $v$ an element of $V_{λ},$ and $α$ a root. Then there exist vectors $v_{i} \in V_{λ + i α}$ $(i = 1, 2, \dots)$ so that $x_{α} (t) v = v + Σ t^{i} v_{i}$ for all $t \in k .$

Proof.

Since $V$ is rational and $t \to x_{α} (t)$ is an isomorphism, $x_{α} (t)$ is a polynomial in $t : x_{α} (t) v = Σ t^{i} v_{i} .$ If we apply $h$ to this equation and compare the result with the equation got by replacing $x_{α} (t)$ by $h x_{α} (t) h^{- 1} = x_{α} (α (h) t),$ we get $v_{i} \in V_{λ + i α} .$ Setting $t = 0,$ we get $v = v_{0},$ whence the lemma.

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Theorem 39 (Compare with Theorem 3): Let $G$ be a Chevalley group over an algebraically closed field $k$ (i.e. a semisimple algebraic group over $k),$ and assume the notations as above.

(a)	Every nonzero rational $G -module$ $V$ contains a nonzero element $v^{+}$ which belongs to some weight $λ \in L$ and is fixed by all $x \in U .$
(b)	Assume $V = k G v^{+}$ with $v^{+}$ as in (a). Then $V = k U^{-} v^{+} .$ Further $dim V_{λ} = 1,$ every weight $μ$ on $V$ has the form $λ - Σ α$ $(α$ positive root), and $V = Σ V_{μ} .$
(c)	In (a) $⟨ λ, α ⟩ \in ℤ^{+}$ for every positive root $α .$
(d)	If $V$ is irreducible, then the weight $λ$ (the "highest weight") and the line $k v^{+}$ of (a) are uniquely determined.
(e)	Given any character $λ$ on $H$ satisfying (c), there exists a unique irreducible rational $G -module$ $V$ in which $λ$ is realized as in (a).

Proof.

(a) The proof is the same as that of Theorem 3(a) with Lemma 72 in place of Lemma 11.

(b) Since $U^{-} B$ is dense in $G$ (Theorem 7) and $V$ is rational, any linear function on $V$ which vanishes on $U^{-} B v^{+}$ also vanishes on $G v^{+} .$ Thus $V = k U^{-} B v^{+} = k U^{-} v^{+} .$ The other assertions of (b) follow from this equation and Lemma 72.

(c) $w_{α} λ$ is a weight on $V$ (with $w_{α} v^{+}$ a corresponding weight vector). Since $w_{α} λ = λ - ⟨ λ, α ⟩ α,$ it follows from (b) that $⟨ λ, α ⟩ \in ℤ^{+} .$

(d) This follows from the second and third parts of (b).

(e) We will use the correspondence between local weights (on $ℒ)$ and global weights (on $G)$ (see p. 60). Let $λ$ be as in (e). By Lemma 71, $λ \in L .$ Let $λ$ also denote the corresponding weight on $ℒ,$ so that $λ (H_{α}) = ⟨ λ, α ⟩ \in ℤ^{+}$ for all $α > 0 .$ Let $(ρ, V')$ be an irreducible $ℒ -module$ with $λ$ as its highest weight, $v^{+}$ a corresponding weight vector, and $G'$ a corresponding Chevalley group over $k$ (constructed from $ρ ℒ$ and some choice of the lattice $M$ in $V').$ Since $λ$ is in the lattice generated by the weights of the representation of $ℒ$ used to construct $G,$ it follows (Theorem 7, Cor. 1) that there exists a rational homomorphism $φ : G \to G'$ such that $x_{α} (t) \to x_{α}^{'} (t)$ or 1 for all $α$ and $t,$ in the usual notation. The resulting representation of $G$ on $V'$ need not be irreducible (and its representation class may vary with the choice of $M),$ but at least it contains the vector $v^{+}$ which is of weight $λ$ and is fixed by every $x \in U .$ Let $V^{''}$ be the submodule of $V'$ generated by $v^{+},$ and $V^{'''}$ a maximal submodule of $V^{''} .$ $(V^{'''}$ is in fact unique as follows from the equation $V^{''} = k U^{-} v^{+}$ of (b). Check this.) The $G -module$ $V = V^{''} / V^{'''}$ meets the existence requirements of (e). For the uniqueness, let $V_{i}, v_{i}^{+}$ $(i = 1, 2)$ satisfy the conditions on $V, v^{+}$ in (e). Let $v^{+} = v_{1}^{+} + v_{2}^{+} \in V_{1} + V_{2},$ and $V = k G v^{+} .$ Then $V_{λ} = k v^{+}$ by (b), so that $v_{2}^{+} \notin V .$ Consider the $G -homomorphism$ $p_{1} : V \to V_{1},$ projection on the first factor. Since $v_{1}^{+}$ generates $V_{1}, p_{1}$ is onto. Since also $ker p_{1} \subseteq V_{2} \cap V,$ which is 0 because $V_{2}$ is irreducible and $v_{2}^{+} \notin V,$ it follows that $p_{1}$ is an isomorphism. Thus $V$ is isomorphic to $V_{1},$ and similarly to $V_{2},$ so that $V_{1}$ and $V_{2}$ are isomorphic, as required.

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A complement: If $char k = 0,$ then in the existence proof above $V'$ is itself irreducible, i.e. $V' = V^{''}$ and $V^{'''} = 0 .$ In other words, if the Chevalley group $G$ is constructed from an irreducible $ℒ -module$ $V$ and a field $k$ of characteristic $0,$ then as a linear group it is irreducible.

Proof.

Recall that $V$ was originally an $ℒ_{ℂ} -module$ (irreducible by assumption), and that a lattice $M$ as in Theorem 2, Cor. 1 was then used to shift the coefficients to $k .$ Clearly $V_{ℚ}$ is irreducible relative to $ℒ_{ℚ} .$ It follows that $V_{k}$ is irreducible relative to $ℒ_{k} :$ otherwise there would be a proper invariant subspace $V_{1},$ excluding $k v^{+}$ since $k v^{+} \cap V_{ℚ} \neq 0,$ then some nonzero $v \in V_{1}$ such that $X_{α} v = 0$ for all $α > 0,$ so that writing $v = Σ t_{i} m_{i}$ $(m_{i} \in M, t_{i} \in k$ and linearly independent over $ℚ)$ and choosing $α$ so that $X_{α} m_{i} \neq 0$ for some $i,$ we would arrive at the contradiction $Σ t_{i} X_{α} m_{i} = 0 .$ Since we can recover each $X_{α}$ from $G$ by using $x_{α} (t) = 1 + t X_{α} + \dots$ for several values of $t$ and the $X_{α}'s$ generate $ℒ,$ we conclude that $V_{k}$ is irreducible for $G .$

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In contrast to the case just considered, if $char k \neq 0,$ then $V' \neq V^{''}$ and $V^{'''} \neq 0$ in general and the exact situation is not at all understood, except in a few scattered cases (types $A_{1},$ $A_{2},$ $B_{2}$ or when $char k$ is large "compared" to $λ).$ However, the following is true.

Exercise:

(a)	For the lattice $M$ of Theorem 2, Cor. 1 (with $V$ there assumed to be irreducible) assume that $ℂ v^{+} \cap M$ is prescribed. Prove that there is a unique minimal choice for $M$ (contained in all others) and a unique maximal choice.

Assume now as in the complement, except that

char k \neq 0 .

(b)	If $M$ is maximal, then $V^{'''} = 0,$ i.e. $V^{''}$ is irreducible.
(c)	If $M$ is minimal, then $V' = V^{''} .$

Example: If $ℒ$ is of type $A_{1},$ $char k = 2,$ and the adjoint representation is used, then (b) holds for $M_{max} = ⟨ X, H / 2, Y ⟩$ and (c) holds for $M_{min} = ⟨ X, H, Y ⟩,$ but not vice versa.

The proof given above for the existence in Theorem 39(e) brings out the connection between the representations of $G$ and those of $ℒ$ and shows that every irreducible rational representation of a Chevalley group in characteristic $p \neq 0$ can be constructed by the reduction mod $p$ of a corresponding representation of a group in characteristic $0 .$ It depends, however, on the existence of representations of $ℒ,$ which we have not proved here, thus in its entirety is very long. We shall now develop an alternate, more intrinsic, proof.

We start with the connection between a $G -module$ $V$ and its dual $V^{*},$ on which $G$ acts by the rule $(x f) (v) = f (x^{- 1} v)$ for all $x \in G, f \in V^{*},$ and $v \in V .$ We recall that $w_{0}$ is the element of the Weyl group which makes all positive roots negative.

Lemma 73: Let $V$ be an irreducible rational $G -module,$ $λ$ its highest weight, $v^{+}$ a corresponding weight vector, $λ^{*} = - w_{0} λ,$ and $f^{+}$ the element of $V^{*}$ defined thus: if we write $v^{-} = w_{0} v^{+}$ and $v \in V$ as $v = c v^{-} +$ terms of other (hence higher) weights, then $f^{+} (v) = c .$ Then $λ^{*}$ and $f^{+}$ are highest weight and highest weight vector for $V^{*} .$

Proof.

In the definition of $f^{+}$ we have used the fact that $dim V_{w_{0} λ} = dim V_{λ} = 1 .$ Here, and also in similar situations later, we extend $λ$ to $B$ by the rule $λ (b) = λ (h)$ if $b = u h$ $(u \in U, h \in H),$ and similarly for $λ^{*} .$ If we write $v \in V$ as in the lemma and use Lemma 72, we see that $b v = c (w_{0} λ) (h) v +$ higher terms. Since $c = f^{+} (v)$ and $(w_{0} λ) (h) = λ^{*} (b^{- 1}),$ we have $f^{+} (b v) = λ^{*} (b^{- 1}) f^{+} (v) .$ On replacing $b$ by $b^{- 1}$ we get $b f^{+} = λ^{*} (b) f^{+},$ as required.

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Theorem 40: For $λ \in L$ let $A_{λ}$ be the space of polynomial functions $a$ on $G$ such that $a (y b) = a (y) λ (b)$ for all $y \in G, b \in B,$ made into a $G -module$ in the obvious way.

(a)	If $V, λ, v^{+}$ are as in Lemma 73, then the map $φ : V^{} \to A_{λ}$ defined by $(φ f) (x) = f (x v^{+})$ for $f \in V^{}$ and $x \in G$ is a $G -isomorphism$ into.
(b)	Conversely, if $λ$ is such that $A_{λ} \neq 0,$ then $A_{λ}$ contains a unique irreducible $G -submodule.$ The latter is finite-dimensional and rational and its highest weight is $λ^{*} .$

Proof.

(a) The points to be checked here will be left as an exercise.

(b) We observe first that as a $G -module$ $A_{λ}$ is locally finite-dimensional (in fact, it is finite-dimensional, but we shall not prove this), since the set of polynomials of a given degree is. Thus there exist irreducible submodules and all of them are finite-dimensional and rational. Let $μ$ be the highest weight of any one of them and $a^{+}$ a corresponding nonzero weight vector. We have $(*)$ $a^{+} (b x b') = μ (b^{- 1}) a^{+} (x) λ (b')$ for all $x \in G, b, b' \in B .$ Since $B w_{0} B$ is dense in $G,$ $a^{+} (w_{0}) \neq 0 .$ Since also $a^{+} (b w_{0}) = a^{+} (w_{0} \cdot w_{0}^{- 1} b w_{0}),$ we get from the above equation that $μ (b^{- 1}) = λ (w_{0}^{- 1} b w_{0}),$ so that $μ = λ^{*} .$ Since $μ$ is uniquely determined by $λ,$ the function $a^{+}$ is determined by its value at $w_{0}$ by $(*)$ with $x = w_{0}$ and the density of $B w_{0} B$ in $G,$ proving the uniqueness in (b).

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Remarks:

(a)	In characteristic $0$ it easily follows from the theorem of complete reducibility that $A_{λ}$ itself is irreducible.
(b)	The representation of $G$ on $A_{λ}$ is, in the context of polynomial representations, the one induced by the character $λ$ on $B .$ The fact that it contains a representation of highest weight $λ^{*},$ is, in view of Theorem 39(a), a form of Frobenius reciprocity.

Lemma 74: Let $f^{+}$ be as in Lemma 73 and $a^{+} = φ f^{+}$ with $φ$ as in Theorem 40(a) so that $x v^{+} = a^{+} (x) w_{0} v^{+} +$ higher terms. Let $W_{λ}$ be the stabilizer of $λ$ in $W,$ and for $w \in W_{λ}$ assume that the corresponding representative $w \in G$ has been chosen so that $w v^{+} = v^{+} .$ Then if $x \in G$ is written $u h w_{0} w u_{1}$ (see Theorem 4') we have $a^{+} (x) = λ^{*} (h^{- 1})$ if $w \in W_{λ},$ $a^{+} (x) = 0$ otherwise.

Proof.

A choice for $w \in G$ as above is always possible: if $λ = 0,$ then $V$ is trivial since $G = 𝒟 G,$ while if $λ \neq 0,$ then $w v^{+}$ has weight $w λ = λ,$ hence is a multiple of $v^{+},$ so that by modifying it by a suitable element of $H$ we can achieve $w v^{+} = v^{+} .$ From the definitions and Lemma 72 we have $a^{+} (x) w_{0} v^{+} = h w_{0} w v^{+} .$ If $w \in W_{λ},$ then $a^{+} (x) = λ^{*} (h^{- 1})$ by the choice of $w,$ while if $a^{+} (x) \neq 0,$ then $w$ fixes $k v^{+}$ by the equation so that $w \in W_{λ} .$

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This brings us to the

Second proof of the existential part of Theorem 39(e):

Proof.

Let $λ$ be as in Theorem 39(c). It will be enough to prove that the function defined by the last equations of Lemma 74 is rational on $G .$ The existence will then follow from Theorem 40(b) with $λ^{*}$ in place of $λ .$ By Lemma 70 any power of this function will do, so that by Lemma 74 it will be enough to construct an irreducible representation whose highest weight is some positive power (positive multiple if we write characters on $H$ additively) of $λ .$ This we will do, using the following interesting result.

Lemma 75 (Chevalley): Let $G$ be a linear algebraic group and $P$ a closed subgroup. Then there exists a rational $G -module$ $V$ and a line $L$ in $V$ whose stabilizer in $G$ is $P .$

Proof.

Let $A$ be the algebra of polynomials in the matrix entries and $I$ the ideal defining $P .$ By Hilbert's basis theorem $I$ is generated by a finite number of its elements, so that there exists a finite-dimensional $G -invariant$ subspace $B$ of $A$ such that $B \cap I,$ say $C,$ generates $I .$ For $x \in G$ we have the following equivalent conditions: $x \in P;$ $f (x^{- 1} y) = 0$ for all $f \in I, y \in P;$ $x I \subseteq I;$ $x C \subseteq C .$ If now $c = dim C,$ $V = Λ^{c} B,$ $v$ is the product in $V$ of a basis for $C,$ and $L = k v,$ it follows that the stabilizer of $L$ is exactly $P .$

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We resume the proof of existence. Let $Π = {α_{1}, α_{2}, \dots, α_{ℓ}}$ be the set of simple roots. For $i = 1, 2, \dots, ℓ$ let $P_{i}$ be the parabolic subgroup of $G$ corresponding to $Π - {α_{i}}$ (see Lemma 30), $L_{i} = k v_{i}$ the corresponding line of Lemma 75, $μ_{i}$ the corresponding rational character on $P_{i},$ hence also on $B$ and $H,$ and $V_{i} = k G v_{i} .$ If $j \neq i,$ then $w_{j}$ is represented in $P_{i},$ so that $w_{j} μ_{i} = μ_{i}$ and $⟨ μ_{i}, α_{j} ⟩ = 0 .$ Since $w_{i}$ does not fix $L_{i},$ by choice, it follows from parts (b) and (c) of Theorem 39 applied to $V_{i}$ that $⟨ μ_{i}, α_{i} ⟩$ is a positive integer, say $d_{i} .$ If now $λ$ is as before so that $⟨ λ, α_{i} ⟩ = c_{i} \in ℤ^{+},$ it follows that $d λ = Σ e_{i} μ_{i}$ with $d = Π d_{i}$ and $e_{i} = c_{i} d / d_{i} .$ If we form the tensor product $Π V_{i}^{e_{i}},$ then $Π v_{i}^{e_{i}}$ is a vector of weight $d λ$ for $B,$ so that we may extract an irreducible component whose highest weight is $d λ,$ and thus complete our second existence proof.

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Remark: We are indebted to G. D. Mostow for the proof just given.

The extra problems that arise when $char k \neq 0$ are compensated for by the fact that only a finite number of representations has to considered in this case, as we shall now see.

Lemma 76: Assume $char k = p \neq 0 .$ Let Fr (for Frobenius) denote the operation of replacing the matrix entries of the elements of $G$ by their $p^{th}$ powers. If $ρ$ is an irreducible rational representation of $G,$ then so is $ρ \circ Fr.$ If the highest weight of $ρ$ is $λ,$ that of $ρ \circ Fr$ is $p λ .$

Proof.

Exercise.

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Theorem 41: Assume that $G$ above is universal (i.e. $G$ is a simply connected algebraic group), and that $char k = p \neq 0 .$ Let $ℛ$ be the set of $p^{ℓ}$ irreducible rational representations of $G$ for which the highest weight $λ$ satisfies $0 \leq ⟨ λ, α_{i} ⟩ \leq p - 1$ $(α_{i}$ simple). Then every irreducible rational representation of $G$ can be written uniquely $\otimes_{j = 0}^{\infty} ρ_{j} \circ {Fr}^{j}$ $(ρ_{j} \in ℛ) .$

Sketch of proof.

We observe first that since $G$ is universal $L = L_{1},$ so that all $λ's$ with all $⟨ λ, α_{i} ⟩ \in ℤ^{+}$ occur as highest weights, in particular those used to define $ℛ .$ Consider $ρ = \otimes ρ_{j} \circ {Fr}^{j} .$ Let $λ_{j}$ be the highest weight of $ρ_{j} .$ The product of the corresponding weight vectors yields for $ρ$ a highest weight vector of weight $λ = Σ p^{j} λ_{j},$ by Lemma 76. If we vary the $ρ_{j}'s$ in $ℛ,$ we obtain, in view of the uniqueness of the expansion of a number in the scale of $p,$ each possible highest weight $λ$ exactly once. Thus to prove the theorem we need only show that each $ρ$ above is irreducible. The proof of this fact depends eventually on the linear independence of the distinct automorphisms ${Fr}^{j}$ $(j = 0, 1, \dots)$ of $k .$ We omit the details, referring the reader to R. Steinberg, Nagoya Math. J. 22 (1963), or to P. Cartier, Sém. Bourbaki 255 (1963).

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Corollary: Assume that one of the special situations of Theorem 28 holds. Let $ℛ_{ℓ}$ (resp. $ℛ_{s})$ be the subsets of $ℛ$ defined by $⟨ λ, α_{i} ⟩ = 0$ for all $i$ such that $α_{i}$ is long (resp. short). Then every element of $ℛ$ can be written uniquely $ρ_{ℓ} \otimes ρ_{s}$ with $ρ_{ℓ} \in ℛ_{ℓ}$ and $ρ_{s} \in ℛ_{s} .$

Proof.

Given $ρ \in ℛ,$ write the corresponding highest weight $λ$ as $λ_{ℓ} + λ_{s}$ so that the corresponding irreducible representations $ρ_{ℓ}$ and $ρ_{s}$ are in $ℛ_{ℓ}$ and $ℛ_{s} .$ We have to show that $ρ_{ℓ} \otimes ρ_{s}$ is irreducible. If we define $φ$ as in Theorem 28 but with $G$ and $G^{*}$ interchanged, and set $ρ_{ℓ}^{*} = φ \circ ρ_{ℓ},$ $ρ_{s}^{*} = φ \circ ρ_{s},$ we have to show that the representation $ρ_{ℓ}^{*} \otimes ρ_{s}^{*}$ of $G^{*}$ is irreducible. Since the corresponding highest weights satisfy

\begin{matrix} ⟨ λ_{ℓ}^{*}, α^{*} ⟩ & = & ⟨ λ_{ℓ}, α ⟩ & if α is short \\ = & 0 & if not \\ ⟨ λ_{s}^{*}, α^{*} ⟩ & = & p ⟨ λ_{s}, α ⟩ & if α is long \\ = & 0 & if not, \end{matrix}

we see that $λ_{ℓ}^{*}$ and $λ_{s}^{*} / p$ correspond to elements of $ℛ^{*},$ so that the corollary follows from Theorem 41 applied to $G^{*} .$

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Examples:

(a)	${S L}_{2} .$ Here there are $p$ representations $ρ_{i}$ $(i = 0, 1, \dots, p - 1)$ in $ℛ,$ the $i^{th}$ being realized on the space of polynomials homogeneous of degree $i$ over $k^{2} .$
(b)	${S p}_{4}, p = 2 .$ Here there are 4 representations in $ℛ .$ If $φ$ is the graph automorphism of Theorem 28 then $ℛ_{ℓ} \circ φ = ℛ_{s}$ so that by the above corollary, these 4 are, in terms of the defining representation $ρ,$ just 1 (trivial), $ρ, ρ \circ φ,$ and $ρ \otimes (ρ \circ φ) .$

The results we have obtained can easily be extended to the case that $k$ is infinite (but perhaps not algebraically closed). We consider representations on vector spaces over $K,$ some algebraically closed field containing $k,$ and call them rational if the coordinates of the image are polynomial functions over $K$ in the coordinates of the source. The preceding theory is then applicable almost word for word because of the following two facts both coming from the denseness of $G$ in $G_{K}$ (this is $G$ with $k$ extended to $K).$

(a)	Every irreducible polynomial representation of $G$ extends uniquely to one of $G_{K} .$
(b)	On restriction to $G$ every irreducible rational representation of $G_{K}$ remains irreducible.

Exercise: Prove (a) and (b).

The structure of arbitrary irreducible representations is given in terms of the polynomial ones by the following general theorem. Given an isomorphism $φ$ of $k$ into $K,$ we shall also write $φ$ for the natural isomorphism of $G$ onto the group $φ G$ obtained from $G$ by replacing $k$ by $φ k .$

Theorem 42 (Borel, Tits): Let $G$ be an indecomposable universal Chevalley group over an infinite field $k,$ and let $σ$ be an arbitrary (not necessarily rational) irreducible representation of $G$ on a finite-dimensional vector space $V$ over an algebraically closed field $K .$ Assume that $σ$ is nontrivial. Then there exist finitely-many isomorphisms $φ_{i}$ of $k$ into $K$ and corresponding irreducible rational representations $ρ_{i}$ of $φ_{i} G$ over $K$ such that $σ = \underset{i}{\otimes} ρ_{i} \circ φ_{i} .$

Remarks:

(a)	As a corollary we see that $k$ is necessarily imbeddable as a subfield of $K .$ In other words, if $k$ and $K$ are such that no such imbedding exists, e.g. if $char k \neq char K,$ then every irreducible representation of $G$ on a finite-dimensional vector space over $K$ is necessarily trivial. (Deduce that the same is true even if the representation is not irreducible.) If $k$ is finite, these statements are, of course, false.
(b)	The theorem can be completed by statements concerning the uniqueness of the decomposition and the condition for irreducibility if the factors are prescribed. Since these statements are a bit complicated we shall omit them.
(c)	The theorem was conjectured by us in Nagoya Math. J. 22 (1963). The proof to follow is based on an as yet unpublished paper by A. Borel and J. Tits in which results of a more general character are considered.

Lemma 77: Let $G, G'$ be indecomposable Chevalley groups over fields $k, k'$ with $k$ infinite and $k'$ algebraically closed, and $σ : G \to G'$ a homomorphism such that $σ G$ is dense in $G' .$

(a)	There exists an isomorphism $φ$ of $k$ into $k'$ and a rational homomorphism $ρ$ of $φ G$ into $G'$ such that $σ = ρ \circ φ .$
(b)	If $G$ is universal, then $ρ$ can be lifted, uniquely to the universal covering group of $G' .$

Proof.

(a) If the reader will examine the proof of Theorem 31 he will observe that what is shown there is that $σ$ can be normalized so that $σ x_{α} (t) = x_{α'} (ε_{α} φ {(t)}^{q (α)})$ with $α \to α'$ an angle-preserving map of $Σ$ on $Σ',$ $ε_{α} = \pm 1,$ $φ$ an isomorphism of $k$ onto $k',$ and $q (α) = 1$ on $p .$ Since we are assuming only that $σ G$ is dense in $G',$ not that $σ G = G',$ the proof of the corresponding result in the present case is somewhat harder. However, the main ideas are quite similar. We omit the proof. From the above equations and the corresponding ones on $H,$ it follows from Theorem 7 that $σ$ has the form of (a).

(b) From these equations we see also, e.g. by considering the relations (A), (B), (C) of Theorem 8, that $ρ$ can be lifted to any covering of $G',$ uniquely since $G = 𝒟 G .$

$□$

Proof of Theorem 42.

Let $A = \overline{σ G},$ the smallest algebraic subgroup of $G L (V)$ containing $σ G .$ We claim $A$ is a connected semisimple group, hence a Chevalley group. As in the proof of Theorem 30, step (12), $\overline{σ U}$ is connected, and similarly for $\overline{σ U^{-}},$ so that $A,$ being generated by these groups, is also. Let $R$ be a connected solvable normal subgroup of $A .$ By the Lie-Kolchin theorem $R$ has weights on $V,$ finite in number. $A$ permutes the corresponding weight spaces, and, being connected, fixes them all. Since $V$ is irreducible, there is only one such space and it is all of $V,$ so that $R$ consists of scalars, of determinant 1 since $A = 𝒟 A,$ so that $R$ is finite. Since $R$ is connected, $R = 1,$ so that $A$ is semisimple, as claimed. Let $A_{1} = Π A_{i 1}$ be the universal covering group of $A$ written as a product of its indecomposable components, $A_{0} = Π A_{i 0}$ the corresponding factorization of the adjoint group, and $α, β, γ = Π γ_{i}$ the corresponding natural maps as shown:

\begin{matrix} A_{1} & = & Π A_{i 1} \\ α \begin{matrix} ↓ \end{matrix} & ↓ \\ G & \overset{σ}{⟶} & A & γ = Π γ_{i} \\ β \begin{matrix} ↓ \end{matrix} \\ A_{0} & = & Π A_{i 0} \end{matrix}

By Lemma 77 we can lift $β σ$ componentwise to get a homomorphism $δ : G \to A_{1}$ of the form $δ (x) = Π ε_{i} φ_{i} (x)$ with each $φ_{i}$ an isomorphism of $k$ into $K$ and $ε_{i}$ a rational homomorphism of $φ_{i} G$ into $A_{i 1} .$ We have $α δ = σ$ since otherwise we would have a homomorphism of $G$ into the center of $A .$ By Lemma 68, Cor. (a), $α,$ interpreted as an irreducible rational representation of $A_{1},$ may be factored $\underset{i}{\otimes} α_{i}$ with $α_{i}$ an irreducible rational representation of $A_{i 1} .$ On setting $ρ_{i} = α_{i} ε_{i},$ we see that $σ = α δ = \otimes α_{i} ε_{i} φ_{i} = \underset{i}{\otimes} ρ_{i} φ_{i},$ as required.

$□$

Corollary:

(a)	Every absolutely irreducible real representation of a real Chevalley group $G$ is rational.
(b)	Every holomorphic irreducible representation of a semisimple complex Lie group is rational.
(c)	Every continuous irreducible representation of a simply connected semisimple complex Lie group is the tensor product of a holomorphic one and an antiholomorphic one.

Proof.

(a) If $G$ is universal, this follows from the theorem and the fact that the only isomorphism of $ℝ$ into $ℝ$ is id. The transition to the nonuniversal case is an easy exercise.

(b) The proof is similar to that of (a).

$□$

Exercise: Prove that the word "absolutely" in (a) and the words "simply connected" in (c) may not be removed.

Now we shall touch briefly on some additional results.

Characters. As is customary in representation theory, the characters (i.e. the traces of the representative matrices) play a vital role. We state the principal results in the form of an exercise.

Exercise:

(a)

Prove that two irreducible rational

G -modules

are isomorphic if and only if their characters are equal. (Consider the characters on

H .)

(b)

Assume that

char k = 0

and that the theorem of complete reducibility has been proved in this case. Prove (a) for representations which need not be irreducible.

(c)

Assume

char k = 0 .

Prove Weyl's formulas: Let

V, λ

be as in Theorem 39(e),

χ

the corresponding character, and

δ

one-half the sum of the positive roots, a character on

H .

Set

S_{λ} = \underset{w \in W}{Σ} det w \cdot w (λ + δ),

a sum of functions on

H .

Then

(1)	$χ (h) = S_{λ} (h) / S_{0} (h)$ at all $h \in H$ where $S_{0} (h) \neq 0 .$
(2)	$dim V = \underset{α > 0}{Π} ⟨ λ + δ, α ⟩ / ⟨ δ, α ⟩ .$

(Hint: use the corresponding formulas for Lie algebras (see, e.g., Jacobson, Lie Algebras) and the complement to Theorem 39).

Remark: The formula (1) determines $χ$ uniquely since it turns out that the elements of $G$ which are conjugate to those elements of $H$ for which $S_{0} \neq 0$ form a dense open set in $G .$

The unitarian trick. The basic results about the irreducible complex representations of a compact semisimple Lie group $K,$ i.e. a maximal compact subgroup of a complex Chevalley group $G$ as in §8, can be deduced from those of $G$ because of the following important fact: $(*)$ $K$ is Zariski-dense in $G .$ Because of Lemmas 43(b) and 45 $(K$ is generated by the groups $φ_{2} {S U}_{2})$ this comes down to the fact that ${S U}_{2}$ is Zariski-dense in ${S L}_{2} (ℂ),$ whose proof is an easy exercise. By $(*)$ the rational irreducible representations of $G$ remain distinct and irreducible on restriction to $K .$ That a complete set of continuous representations of $K$ is so obtained then follows from the fact that the corresponding characters form a complete set of continuous class functions on $K .$ The proof of this uses the formula for Haar measure on $K$ and the orthogonality and completeness properties of complex exponentials, and yields as a by-product Weyl's character formula itself. This is how Weyl proved his formula in Math. Zeit. 24 (1926) and it is still the best way. The theorem of complete reducibility can be proved as follows. Given any rational representation space $V$ for $G$ and an invariant subspace $V',$ we can, by averaging over $K,$ relative to Haar measure, any projection of $V$ onto $V'$ and taking the kernel of the result, get a complementary subspace invariant under $K,$ thus also invariant under $G$ because of $(*) .$ It is then not difficult to replace the complex field by any field of characteristic $0 .$

Invariant bilinear forms. $G$ denotes an indecomposable infinite Chevalley group, $V$ an irreducible rational $G -module,$ and $λ$ its highest weight.

Lemma 78: The following conditions are equivalent.

(a)	There exists on $V$ a (nonzero) invariant bilinear form.
(b)	$V$ and its dual $V^{*}$ are isomorphic.
(c)	$- w_{0} λ = λ .$

Proof.

Exercise (see Lemma 73).

$□$

Exercise: Prove that $- w_{0}$ is the identity for all simple types except $A_{n}$ $(n \geq 2),$ $D_{2 n + 1},$ $E_{6},$ and for these types it comes from involutory automorphism of the Dynkin diagram. (Hint: for all of the unlisted cases except for $D_{2 n}$ the Dynkin diagram has no symmetry.)

Exercise: If there exists an invariant bilinear form on $V,$ then it is unique up to multiplication by a scalar and is either symmetric or skew-symmetric. (Hint: use Schur's Lemma.)

Lemma 79: Let $h = Π h_{α} (- 1),$ the product over the positive roots.

(a)	$h$ is in the center of $G$ and $h^{2} = 1 .$
(b)	If $V$ possesses an invariant bilinear form then it is symmetric if $λ (h) = 1,$ skew-symmetric if $λ (h) = - 1 .$

Proof.

(a) Since $h_{α} (- 1) = h_{- α} (- 1)$ (check this), $h$ is fixed by all elements of $W .$ This implies that $h$ is in the center, as easily follows from Theorem 4'. Since $h_{α} {(- 1)}^{2} = h_{α} (1) = 1,$ we have $h^{2} = 1 .$

(b) We have an isomorphism $φ : V \to V^{*},$ $v^{+} \to f^{+}$ with $v^{+}$ and $f^{+}$ as in Lemma 73; the corresponding bilinear form on $V$ is given by $(v, v') = (φ v) (v') .$ It follows that $(x v^{+}, y v^{+}) = f^{+} (x^{- 1} y v^{+})$ for all $x, y \in G .$ Thus $(v^{+}, w_{0} v^{+}) = f^{+} (w_{0} v^{+}) \neq 0$ by the definition of $f^{+},$ and $(w_{0} v^{+}, v^{+}) = f (w_{0}^{- 1} v^{+}) .$ If $w_{0} = w_{α} w_{β} w γ \dots$ is a minimal product of simple reflections in $W,$ then for definiteness we pick $w_{0} = w_{α} (1) w_{β} (1) \dots$ in $G,$ so that $w_{0}^{- 1} = \dots w γ (- 1) w_{β} (- 1) w_{α} (- 1) .$ We have $w_{α} (- 1) = w_{α} (1) h_{α} (1),$ and similarly for $β, γ, \dots .$ Substituting into the expression for $w_{0}^{- 1}$ and bringing all the $h's$ to the right, by repeated conjugation by $w's,$ we get to the right $h$ by Appendix II (25) and to the left $\dots w γ (1) w_{β} (1) w_{α} (1)$ which is just $w_{0}$ by a lemma to be proved in the last section. Thus $w_{0}^{- 1} = w_{0} h,$ and $(w_{0} v^{+}, v^{+})$ becomes $λ (h) f^{+} (w_{0} v^{+}) = λ (h) (v^{+}, w_{0} v^{+}),$ as required.

$□$

Observation: $h$ as in Lemma 79 is 1 in each of the following cases, since the center is of odd order.

(a)	$G$ is adjoint.
(b)	$Char k = 2 .$
(c)	$G$ is of type $A_{2 n}, E_{6}, E_{8}, F_{4}, G_{2} .$

Exercise: In the remaining cases find $h,$ as a product $Π h_{α} {(- 1)}^{n_{α}}$ over the simple roots.

Example: ${S L}_{2} .$ For every $V$ there is an invariant bilinear form. Assume $char k = 0,$ so that for each $i = 1, 2, 3, \dots$ there is exactly one $V$ of dimension $i,$ viz. the space of polynomials homogeneous of degree $i - 1 .$ Then the invariant form is symmetric if $i$ is odd, skew-symmetric if $i$ is even.

Invariant Hermitean forms. Assume now that $G$ is complex, $σ$ is the automorphism of Theorem 16, $K = G_{σ}$ is the corresponding maximal compact subgroup, $V$ and $v^{+}$ are as before, and $f : G \to ℂ$ is defined by $x v^{+} = f (x) v^{+} +$ terms of other weights.

(a)	Prove that $f (σ x^{- 1}) = \overline{f (x)} .$ (First prove it on $U^{-} H U,$ then use the density of $U^{-} H U$ in $G .)$
(b)	Prove that there exists a unique form $(,)$ from $V \times V$ to $ℂ$ which is linear in the second position, conjugate linear in the first, and satisfies $(x v^{+}, y v^{+}) = f (σ x^{- 1} y),$ and that this form is Hermitean.
(c)	Prove that $(,)$ is positive definite and invariant under $K .$

Dimensions. Assume now that $G$ is a Chevalley group over an infinite field $k,$ that $V$ and $λ$ are as before, and that $𝒰_{ℤ}$ is the universal algebra of Theorem 2, written in the form $𝒰_{ℤ}^{-} 𝒰_{ℤ}^{0} 𝒰_{ℤ}^{+}$ of page 16.

(a)	Prove that there exists an antiautomorphism $σ$ of $𝒰_{ℤ}$ such that $σ X_{α} = X_{- α}$ and $σ H_{α} = H_{α}$ for all $α .$
(b)	Define a bilinear form $(u, u')$ from $𝒰_{ℤ}$ to $𝒰_{ℤ}^{0}$ thus: write $σ u \cdot u'$ in the above form and then set every $X_{α} = 0 .$ Prove that this form is symmetric.
(c)	Now define a bilinear form from $𝒰_{ℤ}$ to $ℤ$ thus: ${(,)}_{λ} = λ \circ (,)$ (interpreting $λ$ as a linear form on $ℋ$ such that $λ (H_{α}) \in ℤ^{+}$ for all $α > 0).$ Assuming now that this form is reduced modulo the characteristic of $k,$ prove that its rank is just the dimension 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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