The Unapologetic Mathematician

Kernels and Images of Intertwiners

The next obvious things to consider are the kernel and the image of an intertwining map. So let’s say we’ve got a representation $\rho:A\rightarrow\mathrm{End}(V)$, a representation $\sigma:A\rightarrow\mathrm{End}(W)$, and an intertwiner $f:\rho\rightarrow\sigma$ defined by the linear map $f:V\rightarrow W$ which satisfies $\left[\sigma(a)\right]\left(f(v)\right)=f\left(\left[\rho(a)\right](v)\right)$ for all $v\in V$.

Now the linear map $f$ immediately gives us two subspaces: the kernel $\mathrm{Ker}(f)\subseteq V$ and the image $\mathrm{Im}(f)\subseteq W$. And it turns out that each of these is actually a subrepresentation. Showing this isn’t difficult. A subrepresentation is just a subspace that gets sent to itself under the action on the whole space, so we just have to check that $\rho(a)$ always sends vectors in $\mathrm{Ker}(f)$ back to this subspace, and that $\sigma(a)$ always sends vectors in $\mathrm{Im}(f)$ back into this subspace.

First off, $v\in V$ is in the kernel of $f$ if $f(v)=0$. Then we calculate

\displaystyle\begin{aligned}f\left(\left[\rho(a)\right](v)\right)=\left[\sigma(a)\right]\left(f(v)\right)\\=\left[\sigma(a)\right](0)=0\end{aligned}

which shoes that $\left[\rho(a)\right](v)$ is also in the kernel of $f$.

On the other hand, if $w\in W$ is in the image of $f$, then there is some $v\in V$ so that $w=f(v)$. We calculate

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

And so $\left[\sigma(a)\right](w)$ is also in the image of $f$.

So we’ve seen that the image and kernel of an intertwining map have well-defined actions of $A$, and so we have subrepresentations. Immediately we can conclude that the coimage $\mathrm{Coim}(f)=V/\mathrm{Ker}(f)$ and the cokernel $\mathrm{Cok}(f)=W/\mathrm{Im}(f)$ are quotient representations.