## The Group Algebra (II)

We continue our discussion of the group algebra.

## Constructing K[G]-modules

Recall that such a module V is also called a representation of G over K, and corresponds to a group homomorphism $\rho : G \to GL_K(V).$

(i) Given a K[G]-module V, a submodule W of V is precisely a vector subspace such that g(W) ⊆ W for all g in G.

(ii) Given K[G]-modules V and W we have the direct sum V ⊕ W.

(iii) Given K[G]-modules V and W we have the tensor product V ⊗K W. This is a K[G]-module since for each gG, the map $V \times W \to V\otimes_K W$ taking $(v, w) \mapsto gv\otimes gw$ is K-bilinear, so it induces a linear map $V\otimes_K W \to V\otimes_K W$ taking $v\otimes w \mapsto gv \otimes gw.$

[ Note: if you’re not familiar with tensor product, we will discuss this in a more general setting later. ]

(iv) Given a K[G]-module V, its dual (as a K-vector space) is given by $V^* := \text{Hom}_K(V, K).$ It is given the structure of a K[G]-module as follows:

$f : V\to K, g\in G \ \mapsto \ (g\cdot f) : V \to K, v \mapsto f(g{^-1}v).$

Again, the inverse is required to ensure group action occurs in the right order. The dual gives us a representation with the same dimension.

(v) More generally, given K[G]-modules VW, let $\text{Hom}_K(V, W)$ be the space of K-linear maps V → W. It is given the structure of a K[G]-module as follows:

$f:V \to W, g\in G \ \mapsto \ (g\cdot f): V\to W, v\mapsto g\cdot f(g^{-1}v).$

This clearly generalises (iv), if we assume G acts trivially on K.

Note : from basic linear algebra, $\text{Hom}_K(V, W) \cong V^* \otimes_K W$ when V and W are finite-dimensional. The LHS has a K[G]-module structure via (iv) while the RHS has a K[G]-module structure via (ii) and (iii). It is easy to check that they are consistent. ]

## Character Theory

The character of a K[G]-module V is defined to be:

$\chi_V : G\to K, \quad \chi_V(g) := \text{tr}(g : V\to V),$

where tr is the trace as a K-linear map.

Note that $\chi_V(hgh^{-1}) = \chi_V(g)$ since $\text{tr}(B\cdot AB^{-1}) = \text{tr}(AB^{-1}\cdot B) = \text{tr}(A)$ for any square matrices A and B with B invertible. Generally, a function χ : G → K is said to be a class function if $\chi(hgh^{-1}) = \chi(g)$ for any gh in GThus characters are class functions.

Note : the set of class functions is a vector space over K, whose dimension is the number of conjugancy classes of G.

Elementary linear algebra tells us it’s not hard to compute the characters of the direct sum, tensor product, and dual from the characters of the individual modules:

\begin{aligned}\chi_{V\oplus W}(g) &= \chi_V(g) + \chi_W(g), \\ \chi_{V\otimes W}(g) &= \chi_V(g) \times \chi_W(g)\\ \chi_{V^*}(g), &= \chi_V(g^{-1}).\end{aligned}

As a result, we have

$\chi_{\text{Hom}(V,W)}(g) = \chi_{V^* \otimes W}(g) = \chi_{V^*}(g) \chi_W(g) = \chi_V(g^{-1})\chi_W(g).$

The next lemma is simple yet critical.

Lemma. If V is a K[G]-module, let $V^G := \{v\in V: gv = v \forall g\in G\}$ be the space of G-invariant vectors. Then:

$\dim_K V^G = \frac 1 {|G|} \sum_{g\in G} \chi_V(g).$

Proof

Let pV → V be the K-linear map $p(v) := \frac 1{|G|}\sum_{g\in G} g\cdot v$. We have:

1. Image of p lies in VG : indeed for any h in G, $h\cdot p(v) = \frac 1{|G|}\sum_{g\in G} hg\cdot v \stackrel{x=hg}{=} \frac 1{|G|}\sum_{x\in G} x\cdot v = p(v).$
2. If v ∈ VG, then p(v) = v : this follows from gvv for all g.
3. p2p : for any v in V, property 1 says p(v) ∈ VG; then property 2 says p(p(v)) = p(v).

Thus, p is a projection map onto VG, and so its trace is precisely dim(VG). But that trace is precisely $\frac 1{|G|} \sum_{g\in G} \text{tr}(g:V\to V)$ which proves our lemma. ♦

## Orthogonality of Irreducible Characters

Let VW be K[G]-modules; we now apply the above lemma on the K[G]-module $U:=\text{Hom}_K(V, W).$ The lemma says that the dimension of the space UG is:

$\frac 1{|G|} \sum_{g\in G} \chi_U(g) = \frac 1{|G|}\sum_{g\in G} \chi_V(g^{-1})\chi_W(g).$

On the other hand, what is UG ? An element f of U gives a linear map VW, and g acts on it via $(g\cdot f):v \mapsto g\cdot f(g^{-1} v)$; thus f is G-invariant if and only if $(f(g\cdot v) = g\cdot f(v)$ for all g in Gv in VIn short, the space of G-invariant elements of U is precisely the space of intertwining operators V→W, or equivalently, K[G]-module homomorphisms V→W.

But we recall Schur’s lemma: if V and W are simple K[G]-modules, then:

• V non-isomorphic to W ⇒ HomK[G](VW) = 0.
• V isomorphic to W ⇒ HomK[G](VW) is a division ring containing K, so its dimension ≥ 1.

Let us summmarise everything here.

Summary. Define an inner product on the space of class functions as follows: given class functions χ and ψ, let:

$\left<\chi, \psi\right> := \frac 1 {|G|}\sum_{g\in G} \chi(g^{-1}) \psi(g).$

This is a bilinear map onto K. If χ and ψ are characters of simple K[G]-modules, then they are orthogonal if the modules are not isomorphic. [ We call such characters irreducible characters. ]

So the collection of irreducible characters forms an orthogonal set in the space of class functions. In particular, this means they are linearly independent! In general, however, they do not span the full space of class functions.

Example

Let G = {egg2} be the finite group of order 3 and KQ. The group ring Q[G] is easily seen to be isomorphic to Q[T]/(T3-1) (via mapping g to T). This is isomorphic to Q × Q(√-3), so it has only two simple modules up to isomorphism. Yet it has 3 conjugancy classes, so the characters do not span the full space of class functions.

Note that if we had picked KC instead, there would be no problem since (T3-1) is completely factorisable in C. This hints that the case where K is algebraically closed is nice.

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