The Unapologetic Mathematician

Dimensions of Hom Spaces

Now that we know that hom spaces are additive, we’re all set to make a high-level approach to generalizing last week’s efforts. We’re not just going to deal with endomorphism algebras, but with all the $\hom$-spaces.

Given $G$-modules $V$ and $W$, Maschke’s theorem tells us that we can decompose our representations as \displaystyle\begin{aligned}V&\cong m_1V^{(1)}\oplus\dots\oplus m_kV^{(k)}\\W&\cong n_1V^{(1)}\oplus\dots\oplus n_kV^{(k)}\end{aligned}

where the $V^{(i)}$ are pairwise-inequivalent irreducible $G$-modules with degrees $d_i$. I’m including all the irreps that show up in either decomposition, so some of the coefficients $m_i$ or $n_i$ may well be zero. This is not a problem, since it just means direct-summing on a trivial module.

So let’s use additivity! We find \displaystyle\begin{aligned}\hom_G(V,W)&\cong\hom_G\left(\bigoplus\limits_{i=1}^km_iV^{(i)},\bigoplus\limits_{j=1}^kn_jV^{(j)}\right)\\&=\bigoplus\limits_{i=1}^k\bigoplus\limits_{j=1}^k\hom(m_iV^{(i)},n_jV^{(j)})\end{aligned}

Now to calculate these summands, we can pick a basis for $V^{(i)}$ and $V^{(j)}$ and use the same sorts of methods we did to calculate commutant algebras. We find that if $i\neq j$ $V^{(i)}\not\cong V^{(j)}$ — then there are no $G$-morphisms at all, even if we include multiplicities. On the other hand, if $i=j$ we find that an intertwinor between $m_iV^{(i)}$ and $n_iV^{(i)}$ has the form $M_{m_i,n_i}\boxtimes I_{d_i}$, where $M_{m_i,n_i}$ is an $m_i\times n_i$ complex matrix. That is, as a vector space it’s isomorphic to the space of $m_i\times n_i$ matrices.

We conclude $\displaystyle\hom_G(V,W)\cong\bigoplus\limits_{i=1}^k\mathrm{Mat}_{m_i,n_i}(\mathbb{C})$

and its dimension is $\displaystyle\dim\hom_G(V,W)=\sum\limits_{i=1}^k\dim\mathrm{Mat}_{m_i,n_i}(\mathbb{C})=\sum\limits_{i=1}^km_in_i$

Notice that any $i$ for which $m_i=0$ or $n_i=0$ doesn’t count for anything.

As a special case, we consider the endomorphism algebra $\mathrm{End}_G(V)=\hom_G(V,V)$. This time we assume that none of the $m_i$ are zero. We find: $\displaystyle\mathrm{End}_G(V)\cong\bigoplus\limits_{i=1}^k\mathrm{Mat}_{m_i}(\mathbb{C})$

with dimension $\displaystyle\dim\mathrm{End}_G(V)=\sum\limits_{i=1}^k\dim\mathrm{Mat}_{m_i}(\mathbb{C})=\sum\limits_{i=1}^km_i^2$

Just like before, we can calculate the center, which goes summand-by-summand. Each summand is (isomorphic to) a complete matrix algebra, so we know that its center is isomorphic to $\mathbb{C}$. Thus we find that the center of $\mathrm{End}_G(V)$ is the direct sum of $k$ copies of $\mathbb{C}$, and so has dimension $k$.

As one last corollary, let $V=V^{(1)}$ be irreducible and let $W$ be any representation. Then we calculate the dimension of the $\hom$-space: \displaystyle\begin{aligned}\dim\hom_G(V,W)&=\dim\bigoplus\limits_{i=1}^k\hom_G(V^{(1)},n_iV^{(i)})\\&=\sum\limits_{i=1}^k\dim\hom_G(V^{(1)},n_iV^{(i)})\\&=n_1\end{aligned}

That is, the dimension of the space of intertwinors is exactly the multiplicity of $V^{(1)}$ in the representation $W$.

October 12, 2010