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I have got the next question that I am pondering the answer to.

Let $\tau$ be a completely reducible representation of finite dimension of a group $G$, and let $\pi$ be another irreducible representation of finite dimension of $G$. Suppose these representations are over a closed algebraic field, $K$.

I want to show that $\pi$'s multiplicity in $\tau$ equals $dim_K(Hom_G(\pi,\tau))$, where $T\in Hom_G(\pi,\tau) \Leftrightarrow \forall g \in G\ T\pi(g) = \tau(g) T$.

Do you have any insight or good reference for this question?

Basically, if $\tau \cong n_1 \pi_1 \oplus \cdots \oplus n_k \pi_k$, and $\pi \cong m_i\pi_i$ for some $1\leq i\leq k$, then I want to show that $$m_i=dim_K(Hom_G(\pi,\tau))$$

I am not sure how to show this? any good refernece or better yet hint to the right way?

Thanks in advance.

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Hint: Schur's lemma. – user27126 Dec 4 '12 at 18:29
The title of your question is the reverse of your actual question, i.e., you're asking about the multiplicity of the irreducible in the reducible representation. – Keenan Kidwell Dec 4 '12 at 18:45
$Keenan, sorry for that. :-( – MathematicalPhysicist Dec 5 '12 at 11:24
Actually, $\dim_K\hom_G(\pi,\tau)=n_im_i$ as you have it. – anon Dec 5 '12 at 20:25
up vote 2 down vote accepted

I leave it to you to parse Schur's lemma, that if $V$ and $W$ are non-isomorphic irreducible representations of a group $G$, then $\hom_G(V,W)=0$, i.e. contains only the trivial homomorphism.

Furthermore, for vector spaces $A,B,C,X,Y,Z$ we have $\hom$ -$\oplus$ distributivity:

$$\begin{cases}\hom(A\oplus B,C)\cong\hom(A,C)\oplus\hom(B,C); \\ \hom(X,Y\oplus Z)\cong\hom(X,Y)\oplus\hom(X,Z).\end{cases} \tag{$\bullet$}$$

Note that we view the homs as vector spaces with pointwise addition and scalar multiplication. If our group $G$ linearly acts on $V$ and $W$, then it also acts linearly and from the left on $\hom(V,W)$ by each $g\in G$ sending the linear map $x\mapsto f(x)$ to the linear map $x\mapsto gf(g^{-1}x)$. Convince yourself that this is indeed a left action and is linear. Importantly, $g\cdot f=f$ as functions if and only if $f$ is an equivariant map, i.e. iff it intertwines with the $G$-action as $f(gv)=gf(v)\,\forall v\in V\,\forall g\in G$.

Thus the $G$-invariant subspace $\hom(\cdot,\cdot)^G$ of $\hom$ is precisely $\hom_G$; taking $G$-invariants distributes through direct sums (i.e. $(V\oplus W)^G=V^G\oplus W^G$), so each $\hom$ in $(\bullet)$ may be given the subscript and interpreted as homomorphisms between representations instead of just vector spaces.

Writing $V^{\oplus n}$ for the direct sum of $n$ copies of $V$, we have (using a higher-arity version of $(\bullet)$)

$$\begin{array}{ll} \hom_G\left(\bigoplus_i V_i^{\large\oplus e_i},~\bigoplus_jV_j^{\large\oplus s_j}\right) & \cong\bigoplus_{i,~j}\hom_G\big(V_i,\,V_j\big)^{\large \oplus(e_i\cdot s_j)} \\ & \cong\bigoplus \mathrm{End}_G(V_i)^{\large \oplus e_is_i} \\ & \cong\Bbb C^{\large\sum\limits_i e_is_i}.\end{array}\tag{$\circ$}$$

We may replace $\Bbb C$ as necessary, so long as Schur's holds. The dimension $\dim_{\,\Bbb C}\hom_G(V,W)$ then is a kind of inner product of the vectors of multiplicities of irreducible representations in $V$ and $W$.

In particular, if $V$ is irreducible and has multiplicity $m$ in $W$'s decomposition, a quick computation shows that $\dim_{\Bbb C}\hom_G(V,W)=(\dots,0,1,0,\dots)\cdot(\dots,m,\dots)=\ldots+0+1\cdot m+0+\ldots=m$.

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