Some funny representations

01 Oct 2016

As part of a discussion in our Galois representations course John Bergdall challenged us to come up with a representation that is irreducible but not absolutely semi-simple. I found this a pretty fun thing to think about so I thought I’d write up my progress and what the next steps are.

First things first a reminder of the definitions: an irreducible representation is one with no fixed subspace, and a semisimple representation is one which can be written as a direct sum of irreducible representations. Adding the word absolutely just means that we require the property to be true for the representation acting on the vector spaces over algebraic closure of the base field. By allowing more general coefficients things that are irreducible to start with can easily become reducible.

To start our search lets work with the most simple type of potentially interesting representations I can think of, these look something like

$$\rho\colon \mathbf{Z} \to \operatorname{GL}_2(K)$$

which are entirely determined by the image of 1, lets try and find an appropriate matrix to make the example we want work.

In my head non-semi-simple things look something like this

$$\begin{pmatrix} *&*\\0&* \end{pmatrix}$$

So in order to find an irreducible non absolutely semisimple representation we want to find a matrix over a field which has no eigenvalues, but which over the algebraic closure has repeated eigenvalues. This is not possible over $\mathbf{C}$ for example, as the trace would have to be twice the eigenvalue. This would then give that the eigenvalues were themselves in the ground field, and therefore the eigenvalues were defined over $K$ in the first place.

This sort of weird multiple roots of irreducible polynomials stuff happens only for non-perfect fields (by definition), of which the most quoted example is $\mathbf{F}_p((t))$. Here the polynomial $x^p - t$ is irreducible but has repeated roots over the algebraic closure as it factors as $(x-\sqrt[p]{t})^p$. As we are dealing with 2-dimensional representations here we should look for a matrix over $\mathbf{F}_2((t))$ with characteristic polynomial $x^2 - t$, one simple example of this is the matrix

$$M = \begin{pmatrix} 0&t\\1&0 \end{pmatrix}.$$

So we have a matrix that has no eigenvalues over the base field but repeated eigenvalues over the algebraic closure. We now need to check that it only has a one dimensional eigenspace, to be sure that we have a non absolutely semisimple representation. This eigenspace is given by the kernel of

$$M - \sqrt{t}I = \begin{pmatrix} -\sqrt{t}&t\\1&-\sqrt{t} \end{pmatrix}$$

which is indeed one dimensional and so we are done.

One property of this example is that if we consider the restriction of the representation to the subgroup $2\mathbf{Z}$ we get something semisimple as the square of our matrix $M$ is diagonal. The new improved challenge is therefore to find a representation for which this doesn’t happen either; more explicitly to find an irreducible not absolutely semisimple representation which remains that way on all finite index subgroups. I don’t think the example I have right now can be extended to this case and it might be necessary to look at representations of more exotic groups than simply $\mathbf{Z}$ for this.

Thoughts welcome!