# Representation Theory III

Let’s set up some notation first. Recall that if $\phi: G\to GL(V)$ is a representation, then it makes V into a kG-module. Let’s denote this module by $V^\phi$. Now we want to prove that given two representations into GL(V), that $V^\phi \cong V^\sigma$ if and only if there is an invertible linear transformation $T: V \to V$ such that $T(\phi(x))=\sigma(T(x))$ for every $x\in G$.

The proof of this is basically unwinding definitions: Let $T: V^\phi \to V^\sigma$ be a kG-module isomorphism. Then for free we get $T(xv)=xT(v)$ for $x\in G$ and $v\in V$ is vector space iso. Now note that the multiplication in $V^\phi$ is $xv=\phi(x)(v)$ and in $V^\sigma$ it is $xv=\sigma(x)(v)$. So $T(xv)=xT(v)\Rightarrow T(\phi(x)(v))=\sigma(x)(T(v))$. Which is what we needed to show. The converse is even easier. Just check that the T is a kG-module iso by checking it preserves scalar multiplication.

This should look really familiar (especially if you are picking a basis and thinking in terms of matrices). We’ll say that T intertwines $\phi$ and $\sigma$. Essentially this is the same notion as similar matrices.

Now we will define some more concepts. Let $\phi: G\to GL(V)$ be a representation. Then if $W\subset V$ is a subspace, then it is “$\phi$-invariant” if $\phi(x)(W)\subset W$ for all $x\in G$. If the only $\phi$-invariant subspaces are 0 and V, then we say $\phi$ is irreducible.

Let’s look at what happens if $\phi$ is reducible. Let W be a proper non-trivial $\phi$-invariant subspace.Then we can take a basis for W and extend it to a basis for V such that the matrix $\phi(x)=\left(\begin{matrix} A(x) & C(x) \\ 0 & B(x) \end{matrix}\right)$
and $A(x)$ and $B(x)$ are matrix representations of G (the degrees being dim W and dim(V/W) respectively).

In fact, given a representation on V, $\phi$ and a representation on W, $\psi$, we have a representation on $V\oplus W$, $\phi \oplus \psi$ given in the obvious way: $(\phi \oplus \psi)(x) : (v, w)\mapsto (\phi(x)v, \psi(x)w)$. The matrix representation in the basis $\{(v_i, 0)\}\cup \{(0, w_j)\}$ is just $\left(\begin{matrix}\phi(x) & 0 \\ 0 & \psi(x)\end{matrix}\right)$ (hence is reducible since it has both $V\oplus 0$ and $0\oplus W$ as invariant subspaces).

I’m going to continue with representation theory, but I’ll start titling more appropriately now that the basics have sort of been laid out.

### Author: hilbertthm90

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### 2 thoughts on “Representation Theory III”

1. I’m not sure, but I think there should be something more general here. Let’s say we have a bifunctor F, from Vect x Vect -> Vect. Say covariant in both variables, such as the tensor product or direct sum. Then F should induce a bifunctor from Mod-G x Mod-G into Mod-G, I think. Indeed, an element of Mod-G is just an element of the category Vect with a group homomorphism G -> Aut(V). So, given two spaces V, W and maps G-> Aut(V), Aut(W), we get a map G->End( F(V,W)) which must land inside the automorphism group since G itself is a group.

For functors which are partially or fully contravariant, there should be something similar except you might have to multiply by g^{-1} for the contravariant part, as you do with Hom.

It would be nice if this were a special case of an even more general phenomenon. It shouldn’t work for algebras in general, but perhaps for semigroups (except Aut wouldn’t be valid any more)?