# The Unapologetic Mathematician

## Intertwiner Spaces

Another grading day, another straightforward post. It should come as no surprise that the collection of intertwining maps between any two representations forms a vector space.

Let’s fix representations $\rho:A\rightarrow\mathrm{End}(V)$ and $\sigma:A\rightarrow\mathrm{End}(W)$. We already know that $\hom_\mathbb{F}(V,W)$ is a vector space. We also know that an intertwiner $f\in\hom_{\mathbf{Rep}_\mathbb{F}(A)}(\rho,\sigma)$ can be identified with a linear map $f\in\hom_\mathbb{F}(V,W)$. What I’m asserting is that $f\in\hom_{\mathbf{Rep}_\mathbb{F}(A)}(\rho,\sigma)$ is actually a subspace of $f\in\hom_\mathbb{F}(V,W)$ under this identification.

Indeed, all we really need to check is that this subset is closed under additions and under scalar multiplications. For the latter, let’s say that $f$ is an intertwiner. That is, $\left[\sigma(a)\right]\left(f(v)\right)=f\left(\left[\rho(a)\right](v)\right)$. Then given a constant $c\in\mathbb{F}$ we consider the linear map $cf$ we calculate

\displaystyle\begin{aligned}\left[\sigma(a)\right]\left(\left[cf\right](v)\right)=\left[\sigma(a)\right]\left(cf(v)\right)\\=c\left[\sigma(a)\right]\left(f(v)\right)\\=cf\left(\left[\rho(a)\right](v)\right)\\=\left[cf\right]\left(\left[\rho(a)\right](v)\right)\end{aligned}

And so $cf$ is an intertwiner as well.

Now if $f$ and $g$ are both intertwiners, satisfying conditions like the one above, we consider their sum $f+g$ and calculate

\displaystyle\begin{aligned}\left[\sigma(a)\right]\left(\left[f+g\right](v)\right)=\left[\sigma(a)\right]\left(f(v)+g(v)\right)\\=\left[\sigma(a)\right]\left(f(v)\right)+\left[\sigma(a)\right]\left(g(v)\right)\\=f\left(\left[\rho(a)\right](v)\right)+g\left(\left[\rho(a)\right](v)\right)\\=\left[f+g\right]\left(\left[\rho(a)\right](v)\right)\end{aligned}

Which shows that $f+g$ is again an intertwiner.

Since composition of intertwiners is the same as composing their linear maps, it’s also bilinear. It immediately follows that the category $\mathbf{Rep}_\mathbb{F}(A)$ is enriched over $\mathbf{Vect}_\mathbb{F}$.