The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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December 12, 2008 Posted by | Algebra, Representation Theory | 2 Comments