The Unapologetic Mathematician

Mathematics for the interested outsider

Do Short Exact Sequences of Representations Split?

We’ve seen that the category of representations is abelian, so we have all we need to talk about exact sequences. And we know that some of the most important exact sequences are short exact sequences. We also saw that every short exact sequence of vector spaces splits. So does the same hold for representations? It turns out that no, they don’t always, and I’ll give an example to show what can happen.

Consider the group \mathbb{Z} of integers and the two dimensional representation \rho:\mathbb{Z}\rightarrow\mathrm{GL}\left(\mathbb{F}^2\right) defined by:

\displaystyle\rho(n)=\begin{pmatrix}1&n\\ 0&1\end{pmatrix}

Verify for yourself that this actually does define a representation of the group of integers.

Now it’s straightforward to see that all these linear transformations send every vector of the form \begin{pmatrix}x\\ 0\end{pmatrix} to itself. This defines a one-dimensional subspace fixed by the representation — a subrepresentation \tau:\mathbb{Z}\rightarrow\mathrm{GL}\left(\mathbb{F}^1\right) defined by:

\displaystyle\tau(n)=\begin{pmatrix}1\end{pmatrix}

Then there must be a quotient representation \sigma=\rho/\tau, and we can arrange them into a short exact sequence: \mathbf{0}\rightarrow\tau\rightarrow\rho\rightarrow\sigma\rightarrow\mathbf{0}. The question, then, is whether this is isomorphic to the split exact sequence \mathbf{0}\rightarrow\tau\rightarrow\tau\oplus\sigma\rightarrow\sigma\rightarrow\mathbf{0}. That is, can we find an isomorphism \rho\cong\tau\oplus\sigma compatible with the the inclusion map from \tau and the projection map onto \sigma?

First off, let’s write the direct sum representation a little more explicitly. The direct sum \tau\oplus\sigma acts on pairs of field elements by \tau on the first and \sigma on the second, with no interaction between them. That is, we can write the representation as

\displaystyle\left[\tau\oplus\sigma\right](n)=\begin{pmatrix}\tau(n)&0\\ 0&\sigma(n)\end{pmatrix}=\begin{pmatrix}1&0\\ 0&\sigma(n)\end{pmatrix}

And we’re looking for some isomorphism so that for every n\in\mathbb{Z} we get from the matrix \rho(n) to the matrix \left[\tau\oplus\sigma\right](n) by conjugation. Explicitly, we’ll need a matrix \begin{pmatrix}a&b\\c&d\end{pmatrix}. But we also need to make sure that \tau as a subrepresentation of \rho is sent to \tau as a subrepresentation of \tau\oplus\sigma. That is we must satisfy

\displaystyle\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}1\\ 0\end{pmatrix}

Thus a=1 and c=0 right off the bat! Now the intertwining condition (equivalent to the conjugation) is that

\displaystyle\begin{pmatrix}1&b\\ 0&d\end{pmatrix}\begin{pmatrix}1&n\\ 0&1\end{pmatrix}=\begin{pmatrix}1&0\\ 0&\sigma(n)\end{pmatrix}\begin{pmatrix}1&b\\ 0&d\end{pmatrix}

\displaystyle\begin{pmatrix}1&n+b\\ 0&d\end{pmatrix}=\begin{pmatrix}1&b\\ 0&\sigma(n)d\end{pmatrix}

But this says that n+b=b for all n\in\mathbb{Z}, and this is clearly impossible!

So here’s an example where a short exact sequence of representations can not be split. At some point later we’ll see that in many cases we’re interested in they do split, but for now it’s good to see that they don’t always work out so nicely.

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December 17, 2008 - Posted by | Algebra, Representation Theory

3 Comments »

  1. Every time I read one of your posts a tiny bit of the algebra I once was sorta-kinda proficient at comes back (ever so slightly).

    Comment by jrshipley | December 17, 2008 | Reply

  2. That’s good! And there’s plenty more to come!

    Comment by John Armstrong | December 17, 2008 | Reply

  3. […] decomposable. Indeed, in categorical terms this is the statement that for some groups there are short exact sequences which do not split. To chase this down a little further, our work yesterday showed that even in the reducible case we […]

    Pingback by Decomposability « The Unapologetic Mathematician | September 24, 2010 | Reply


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