Posted by: hilbertthm90 | July 7, 2009

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.

Posted in algebra | Tags: , , , ,

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Responses

  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)?

    By: Akhil Mathew on July 8, 2009
    at 8:39 am

  2. [...] The decomposition is also unique up to isomorphism. (Note that the direct sum of representations is a representation itself.) [...]

    By: Basics of group representation theory « Delta Epsilons on July 10, 2009
    at 1:58 pm


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