代数学笔记——有限群的表示论(一)

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这个专题主要介绍有限群的线性表示。粗略地说,我们把有限群中的元素看作是某个有限维线性空间上的(可逆)线性变换,从而来研究群的性质。本文主要介绍群表示论的最基础的知识,并介绍 Maschke 定理,这个定理在群表示论的教材中往往一开始就会碰到,虽然证明并不难,但它在有限群的表示论中起着至关重要的作用。

Let V be a finite dimensional vector space over a field k. Let G be a finite group. A linear representation of G in V is a group homomorphism \rho:G\to\operatorname{GL}(V).

Define the group algebra k[G]=\{\sum_{g\in G}a_g g:a_g\in k\} with obvious addition and multiplication. Then linear representations of G are the same things as k[G]-modules, since any group homomorphism G\to\operatorname{GL}(V) can be linearly extended to a k-algebra homomorphism k[G]\to\operatorname{End}_k (V).

Let \rho:G\to\operatorname{GL}(V), \rho':G\to\operatorname{GL}(V') be two representations of G. We say a linear map \varphi:V\to V' is a morphism of representations of G if the diagram

commutes for all g\in G. This exactly means that \varphi is a k[G]-module homomorphism.

Remark

In fact a representation of G can be viewed as a functor \rho from the category \mathcal{C} consisting of a single object X whose morphisms are elements of the group G, to the category of k-vector spaces, sending a group element g to \rho(g)\in\operatorname{GL}(V). A natural transformation from \rho to \rho' is exactly a linear map \varphi:V\to V' such that the diagram above commutes, i.e. a morphism of representations is exactly a natural transformation. Then we say two representations \rho and \rho' of G are isomorphic if they are naturally isomorphic as functors.

Definition

We say that a representation \rho:G\to\operatorname{GL}(V) is simple (or irreducible) if V has no k[G]-submodules except 0,V.

Note that there are only finitely many irreducible representations up to isomorphism. This is because any irreducible k[G]-module is isomorphic to a quotient module of V, thus is a composition factor of V, and V has finitely many composition factors.

If \rho_1:G\to \operatorname{GL}(V_1), \rho_2:G\to \operatorname{GL}(V_2) are two representations of G, then we can define their

(1) direct sum

\rho_1\oplus\rho_2:G\to\operatorname{GL}(V_1\oplus V_2), (\rho_1\oplus\rho_2)(g)(v_1,v_2)= (\rho_1(g)v_1,\rho_2(g)v_2).

(2) tensor product

\rho_1\otimes\rho_2:G\to\operatorname{GL}(V_1\otimes V_2), (\rho_1\otimes\rho_2)(g)(v_1\otimes v_2) = \rho_1(g)v_1\otimes \rho_2(g)v_2

Suppose that \rho_1:G\to \operatorname{GL}(V_1), \rho_2:G\to \operatorname{GL}(V_2) are two representations of G. For any linear map f:V_1\to V_2, define its trace to be

\operatorname{Tr}_G(f):V_1\to V_2, \quad v\mapsto\sum_{g\in G}\rho_2(g)\circ f\circ \rho_1(g^{-1})v.

Then \operatorname{Tr}_G(f) is a morphism of k[G]-modules.

Suppose \rho_1, \rho_2 are irreducible representations. By Schur's lemma, if \rho_1 and \rho_2 are not isomorphic, then \operatorname{Tr}_G(f)=0. If k=\mathbb{C}, and \rho_1=\rho_2:G\to\operatorname{GL}(V), then \operatorname{Tr}_G(f)=\lambda\cdot\operatorname{id}_V with \lambda=\frac{|G|}{\dim V}\operatorname{tr}(f) (by taking trace).

Theorem[Maschke]

The group algebra k[G] is semisimple if and only if \operatorname{char} (k)\nmid |G|.

Proof

Suppose \operatorname{char}(k)\nmid |G|. Then for any representation \rho:G\to \operatorname{GL}(V), and any k[G]-submodule W of V, let W' be any subspace of V such that V=W\oplus W'. Let \pi:V\to W be the projection map with respect to this composition. Then \psi:=\frac{1}{|G|}\operatorname{Tr}_G(\pi) is a morphism of k[G]-modules and we have \psi|_W=\operatorname{id}_W. Thus the exact sequence of k[G]-modules

splits. Hence V\cong W\oplus\ker\psi as k[G]-modules. Therefore, k[G] is semisimple.

On the other hand, if \operatorname{char}(k)\mid |G|, then (\sum_{g\in G} g)^2=|G|\sum_{g\in G} g=0. Hence \sum_{g\in G} g is contained in the nilradical of k[G]. Thus the Jacobson radical is not 0. So k[G] is not semisimple.
\square

Remark

Note that if k[G] is semisimple, then \rho is irreducible if and only if \rho is indecomposable. In this case, every representation of G is a direct sum of irreducible representations, by induction on \dim V.

编辑于 2020-10-13

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