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Finite-group invariants are exact when the group order is invertible

Statement

Let $k$ be a field, let $U$ be a finite group, and assume that the order $∣ U ∣$ is invertible in $k$ (equivalently, that $char k$ does not divide $∣ U ∣$ ). For a $k$ -linear $U$ -module $X$ (An $R$ -linear action of $G$ on a left $R$ -module, and a $G$ -module over $R$ ) write $X^{U} : = {x \in X : u \cdot x = x for every u \in U}, π_{X} : = \frac{1}{∣ U ∣} \sum_{u \in U} u \in {End}_{k} (X)$ for the invariants and the averaging (Reynolds) operator. Then:

Projector. $π_{X}$ is $k$ -linear and idempotent, $π_{X} (X) = X^{U}$ , $π_{X}$ restricts to the identity of $X^{U}$ , and $X = X^{U} \oplus \ker π_{X}$ ; in particular $X^{U}$ is a direct summand of $X$ as a $k$ -vector space.
Exactness. For every short exact sequence of $k$ -linear $U$ -modules $0 ⟶ X \overset{f}{\to} Y \overset{g}{\to} Z ⟶ 0$ (Exact sequences and short exact sequences of modules) the sequence of invariants $0 ⟶ X^{U} \overset{f ∣_{X^{U}}}{\to} Y^{U} \overset{g ∣_{Y^{U}}}{\to} Z^{U} ⟶ 0$ is again exact; equivalently, the invariants functor $X \mapsto X^{U}$ is additive and preserves kernels, images and cokernels, hence is exact.
Unipotent radicals. For $k = C$ and every standard parabolic $P_{α} = L_{α} ⋉ U_{α}$ of $G = {GL}_{n} (F_{q})$ (Block Levi decomposition of standard parabolics) the group $U_{α}$ is finite and $∣ U_{α} ∣$ is invertible in $C$ , so with the Harish–Chandra restriction $^{*} ⁣ R_{L_{α}}^{G} (X) = X^{U_{α}}$ (Harish-Chandra induction and restriction for finite general linear groups) claims 1 and 2 apply to $U = U_{α}$ : the functor $^{*} ⁣ R_{L_{α}}^{G}$ is exact.

Facts & Assumptions

Given: A field $k$ , a finite group $U$ with $∣ U ∣$ invertible in $k$ , a short exact sequence $0 \to X \overset{f}{\to} Y \overset{g}{\to} Z \to 0$ of $k$ -linear $U$ -modules, and the averaging operators $π_{X}, π_{Y}, π_{Z}$ .

[L1]

A $k$ -linear $U$ -module structure makes every $u \in U$ act as a $k$ -linear map, so sums and scalar multiples of such operators are again $k$ -linear, and the action is associative: $(u v) \cdot x = u \cdot (v \cdot x)$ (An $R$ -linear action of $G$ on a left $R$ -module, and a $G$ -module over $R$ ).

[L2]

A short exact sequence $0 \to X \overset{f}{\to} Y \overset{g}{\to} Z \to 0$ of modules is one that is exact at $X$ , $Y$ and $Z$ ; equivalently $f$ is injective, $g$ is surjective, and $im f = \ker g$ (Exact sequences and short exact sequences of modules, Module homomorphism and isomorphism, kernel, image and cokernel).

[L3]

$∣ U ∣$ invertible in $k$ means $∣ U ∣ \cdot 1_{k} \in k^{\times}$ , so the element $∣ U ∣^{- 1} \in k$ exists and $∣ U ∣ \cdot ∣ U ∣^{- 1} = 1_{k}$ (Field, Standard subgroups of finite general linear groups).

[L4]

For a composition $α$ of $n$ the standard parabolic subgroup of $G = {GL}_{n} (F_{q})$ is $P_{α} = L_{α} ⋉ U_{α}$ with $U_{α}$ the unipotent radical, a normal subgroup of $P_{α}$ (Block Levi decomposition of standard parabolics, Compositions, partial flags, and standard parabolics).

[L5]

In the finite general linear group setting, $G = {GL}_{n} (F_{q})$ is a subset of the finite set of $n \times n$ matrices over $F_{q}$ , where $∣ F_{q} ∣ = q$ . This implies that its subgroups, including each $U_{α}$ , are finite (Standard subgroups of finite general linear groups).

[L6]

$char C = 0$ , and $0$ does not divide the order of any finite group (Standing hypotheses for ordinary character theory: $G$ finite, $k = C$ , and every representation finite-dimensional).

Proof

technique · direct

1.1

$π_{X}$ is a $k$ -linear idempotent. Since each $u$ acts $k$ -linearly by [L1], the sum $\sum_{u \in U} u$ and its scalar multiple $π_{X}$ are $k$ -linear. For the composite, associativity of the action gives $(\sum_{u \in U} u) (\sum_{v \in U} v) = \sum_{u, v \in U} u v = \sum_{w \in U} c_{w} w$ with $c_{w} = # {(u, v) \in U \times U : u v = w} = ∣ U ∣$ for every $w$ , because for fixed $w$ the pairs are $(u, u^{- 1} w)$ with $u \in U$ arbitrary; hence $π_{X}^{2} = ∣ U ∣^{- 2} ∣ U ∣ \sum_{w \in U} w = π_{X}$ .

L1L3

1.2

Kernels are preserved. Let $f : X \to Y$ be a $U$ -linear map, i.e. $f (u \cdot x) = u \cdot f (x)$ for all $u \in U$ , $x \in X$ . Then $f (X^{U}) \subseteq Y^{U}$ , because $u \cdot f (x) = f (u \cdot x) = f (x)$ for $x \in X^{U}$ ; and $\ker (f ∣_{X^{U}}) = (\ker f)^{U},$ because $x \in X^{U}$ lies in $\ker (f ∣_{X^{U}})$ exactly when $f (x) = 0$ , which is exactly $x \in (\ker f)^{U}$ . In particular, if $f$ is injective then $f ∣_{X^{U}}$ is injective.

L1L2

2.1

Image and kernel of $π_{X}$ . For every $u_{0} \in U$ one has $u_{0} π_{X} = ∣ U ∣^{- 1} \sum_{u} u_{0} u = ∣ U ∣^{- 1} \sum_{u^{'}} u^{'} = π_{X}$ by [L1], so every value $π_{X} x$ is fixed by every $u_{0}$ , that is $π_{X} (X) \subseteq X^{U}$ ; conversely $π_{X} x = x$ for $x \in X^{U}$ because every $u$ fixes $x$ and the sum has $∣ U ∣$ terms, so $X^{U} \subseteq π_{X} (X)$ . Hence $π_{X} (X) = X^{U}$ and $π_{X}$ restricts to the identity on $X^{U}$ . Finally, every $x \in X$ is $x = π_{X} x + (x - π_{X} x)$ with $π_{X} x \in X^{U}$ and $π_{X} (x - π_{X} x) = π_{X} x - π_{X}^{2} x = 0$ ; and if $x \in X^{U} \cap \ker π_{X}$ then $x = π_{X} x = 0$ . Hence $X = X^{U} \oplus \ker π_{X}$ .

step 1.1L1L3

3.1

Images and cokernels are preserved. Let $f : X \to Y$ and $g : Y \to Z$ be $U$ -linear. The averaging operator is natural: since $g$ is $U$ -linear and additive, $g (π_{Y} (y)) = ∣ U ∣^{- 1} \sum_{u} g (u \cdot y) = ∣ U ∣^{- 1} \sum_{u} u \cdot g (y) = π_{Z} (g (y))$ for every $y \in Y$ , that is $g \circ π_{Y} = π_{Z} \circ g$ . If now $g$ is surjective and $z \in Z^{U}$ , choose $y \in Y$ with $g (y) = z$ by [L2]; then $π_{Y} (y) \in Y^{U}$ by step 2.1 and $g (π_{Y} (y)) = π_{Z} (g (y)) = π_{Z} (z) = z$ because $z$ is fixed; hence $g ∣_{Y^{U}}$ is surjective. The same naturality computation gives $f \circ π_{X} = π_{Y} \circ f$ . Thus $f (X^{U}) = f (X) \cap Y^{U}$ : inclusion from left to right follows from step 1.2, while if $y = f (x) \in Y^{U}$ then $y = π_{Y} (y) = f (π_{X} (x))$ with $π_{X} (x) \in X^{U}$ by step 2.1. Combining this with $im f = \ker g$ of [L2], we get $im (f ∣_{X^{U}}) = f (X^{U}) = f (X) \cap Y^{U} = \ker g \cap Y^{U} = \ker (g ∣_{Y^{U}})$ , the last step because $g (y) = 0$ for $y \in Y^{U}$ exactly when $y \in \ker g$ .

step 2.1step 1.2L1L2

4.1

Exactness and the invariant functor. In a short exact sequence as in [L2], $f$ is injective and $g$ surjective, so by steps 1.2 and 3.1 the map $f ∣_{X^{U}}$ is injective, the map $g ∣_{Y^{U}}$ is surjective and its kernel is exactly the image of $f ∣_{X^{U}}$ ; hence $0 \to X^{U} \to Y^{U} \to Z^{U} \to 0$ is exact by [L2]. The assignments $X \mapsto X^{U}$ and $f \mapsto f ∣_{X^{U}}$ define a functor that is additive — restriction of $k$ -linear maps is $k$ -linear, so $(f + f^{'}) ∣_{X^{U}} = f ∣_{X^{U}} + f^{'} ∣_{X^{U}}$ — and the computations of steps 1.2 and 3.1 show that it preserves kernels, images and cokernels, which is the exactness of claim 2.

step 1.2step 3.1L1L2

5.1

Unipotent radicals of standard parabolics. Let $α$ be a composition of $n$ and $P_{α} = L_{α} ⋉ U_{α}$ as in [L4]. The group $U_{α}$ is a subgroup of $G = {GL}_{n} (F_{q})$ , whose elements are matrices with entries in the finite field $F_{q}$ ; hence $U_{α}$ is finite, so $∣ U_{α} ∣ \in N$ is a positive integer, and by [L6] the characteristic $0$ of $C$ does not divide $∣ U_{α} ∣$ , so $∣ U_{α} ∣$ is invertible in $C$ by [L3]. Therefore steps 1.1–4.1 apply with $U = U_{α}$ and $k = C$ : for every short exact sequence of $C$ -linear $U_{α}$ -modules the $U_{α}$ -invariants form a short exact sequence, that is, the Harish–Chandra restriction $^{*} ⁣ R_{L_{α}}^{G} (X) = X^{U_{α}}$ is exact. ∎

step 1.1step 4.1L3L4L5L6

Remark. The averaging operator is the diagonal action of the idempotent $e_{U} = ∣ U ∣^{- 1} \sum_{u \in U} u$ , and its existence is exactly the point at which the hypothesis on $char k$ enters; over a field of characteristic dividing $∣ U ∣$ the invariants functor need not be exact. For a finite group $U$ the operator is available for every field $k$ with $char k ∤ ∣ U ∣$ , so the lemma applies verbatim to the unipotent radicals $U_{α}$ of the standard parabolics of ${GL}_{n} (F_{q})$ , whose orders are powers of the prime $p$ with $q = p^{m}$ . No choice principle is used: $U$ is finite and the average is a finite sum.

Depends on

Used by

Existence and uniqueness of cuspidal support Theorem

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Sources

Olivier Dudas and Jean Michel, Lectures on Finite Reductive Groups and Their Representations - Proposition 9.4(ii), printed p. 35 (standard reference, not scraped)
Jay Taylor, Finite Reductive Groups - Definition 5.2, printed p. 42 (standard reference, not scraped)