# The Unapologetic Mathematician

## Hom Space Duals

Again, sorry for the delay but I was eager to polish something up for my real job this morning.

There’s something interesting to notice in our formulæ for the dimensions of spaces of intertwinors: they’re symmetric between the two representations involved. Indeed, let’s take two $G$-modules: \displaystyle\begin{aligned}V&\cong\bigoplus\limits_{i=1}^km_iV^{(i)}\\W&\cong\bigoplus\limits_{j=1}^kn_jV^{(j)}\end{aligned}

where the $V^{(i)}$ are pairwise-inequivalent irreducible $G$-modules with degrees $d_i$. We calculate the dimensions of the $\hom$-spaces going each way: \displaystyle\begin{aligned}\dim\hom_G(V,W)&=\sum\limits_{i=1}^km_in_i\\\dim\hom_G(W,V)&=\sum\limits_{i=1}^kn_im_i\end{aligned}

but these are equal! So does this mean these spaces are isomorphic?

Well, yes. Any two vector spaces having the same dimension are isomorphic, but they’re not “naturally” isomorphic. Roughly, there’s no universal method of giving an explicit isomorphism, and so it’s regarded as sort of coincidental. But there’s something else around that’s not coincidental.

It turns out that these spaces are naturally isomorphic to each other’s dual spaces. That is, for any $G$-modules $V$ and $W$ we have an isomorphism $\displaystyle\hom_G(W,V)\cong\hom_G(V,W)^*$

Luckily, we already know that their dimensions are equal, so the rank-nullity theorem tells us all we need is to find an injective linear map from one to the other.

So, let’s take an intertwinor $h:W\to V$ and use it to build a linear functional $\lambda_h$ on $\hom_G(V,W)$. For any intertwinor $f:V\to W$ we define $\displaystyle\lambda_h(f)=\mathrm{Tr}_V(h\circ f)$

Where $\mathrm{Tr}$ is the trace of an endomorphism. Given a matrix, it’s the sum of the diagonal entries. Since the composition of linear maps is linear in each variable, and the trace is a linear function, this is a linear functional as desired. It should also be clear that the construction $h\mapsto\lambda_h$ is itself a linear map.

Now, we must show that this map is injective. That is, for no $h$ to we find $\lambda_h=0$. This will follow if we can find for every nonzero $h:W\to V$ at least one $f:V\to W$ so that $\mathrm{Tr}_V(h\circ f)\neq0$. To do so, we pick a basis for each irreducible representation that shows up in either $V$ or $W$ so we can replace $V$ and $W$ with matrix representations. Now we can write $\displaystyle h=\bigoplus\limits_{i=1}^kM_i\boxtimes I_{d_i}$

where $M_i\in\mathrm{Mat}_{n_i,m_i}(\mathbb{C})$ is an $n_i\times m_i$ complex matrix. To construct our $f$, we simply take the conjugate transpose of each of these matrices: $\displaystyle f=\bigoplus\limits_{i=1}^kM_i^\dagger\boxtimes I_{d_i}$

where now $M_i^\dagger\in\mathrm{Mat}_{m_i,n_i}(\mathbb{C})$ is an $m_i\times n_i$ complex matrix, as desired. We multiply the two matrices: $\displaystyle hf=\bigoplus\limits_{i=1}^k(M_iM_i^\dagger)\boxtimes I_{d_i}$

and find that each $M_iM_i^\dagger\in\mathrm{Mat}_{m_i,m_i}(\mathbb{C})$ is a $m_i\times m_i$ square matrix. Thus the trace of this composition is the sum of their traces.

We’ve already seen that the composition of a linear transformation and its adjoint is self-adjoint and positive-definite. In terms of complex matrices, this tells us that the product of a matrix and its conjugate transpose is conjugate-symmetric and positive-definite. This means that it’s diagonalizable with all nonnegative real eigenvalues down the diagonal. And thus its trace is a nonnegative real number, and it can only be zero if the original matrix was zero.

The upshot, if you didn’t follow that, is that if $h\neq0$ we have an $f$ so that $\lambda_h(f)=\mathrm{Tr}(h\circ f)\neq0$. And thus the map $h\mapsto\lambda_h$ is injective, as we asserted. Proving naturality is similarly easy to proving it for additivity of $\hom$-spaces, and you can work it out if you’re interested.

October 13, 2010 -