# Dimension of subspace fixed by subgroup representation.

If $G$ is an abelian group with cyclic subgroup $H$ and $(\rho,V)$ is a (permutation) representation of $G$. Then I can form a representation of $H$ by considering the composition $\lambda=\rho\circ\iota:H\to GL(V)$ where $\iota:H\to G$ is the inclusion map. Now if I let $V^{H}\subset V$ be the subspace of $V$ fixed by $H$ (the image of the projection map $\frac{1}{|H|}\sum_{h\in H}\lambda(h)$) then what can I say about the dimension of $V^{H}$? Can it be understood in terms of the dimension of $V$, the order of $H$, and the order of $G$?

This seems intimately related to the following question. Are there versions of the orbit-stabilizer theorem and/or Burnside lemma in which dimensions (rather than cardinalities) appear and that can be applied in the theory of linear representations?

I'm particularly interested in the case that $(\rho,V)$ is the regular representation modulo the trivial representation.

As to the broader context of this question. Aside from being a purely academic question that I'd like to know the answer to, I was motivated to ask it when attempting to understand an answer provided to the following question on MathOverlow:

Subject to some conditions, is it possible to conclude a subfield of an abelian extension generated by a unit is a cyclic extension

-
If $\rho$ is the regular representation of $G$, then $\lambda$ is the direct sum of $|G|/|H|$ copies of the regular representation of $H$. Does this not imply that if $V$ is the regular representation of $G$ modulo the trivial 1-d module, then $$\dim V^H=|G|/|H|-1?$$ – Jyrki Lahtonen Apr 15 '12 at 18:14
Thanks! This certainly seems reasonable. I guess the bit of information I was lacking was that if $\rho$ is the regular representation of $G$ then $\lambda$ is the direct sum of $|G|/|H|$ copies of the regular representation of $H$. I guess this is pretty obvious is hindsight and I feel silly for even having asked the question. I'll need to give some thought as to how this relates to #2 at here. Any thoughts on this matter may be helpful. – Paul Apr 15 '12 at 18:44

## 1 Answer

$\def\triv{\boldsymbol{1}}$ In general, it is often helpful to write $\mathrm{dim}V^H$ as $\langle \mathrm{Res}_{G/H}V,\triv_H\rangle$, where $\langle-,-\rangle$ denotes the usual inner product of characters. If $V$ is induced from a subgroup, you can use Mackey, Frobenius reciprocity, etc. This essentially gives you the answer to the question what is the analogue of orbit-stabiliser for representations. For example if $U$ is a subgroup and $V=\mathrm{Ind}_{G/U}\triv$ (the regular representation being the special case $U=1$), then $$\begin{array}{c c l} \mathrm{dim}V^H & = &\langle \mathrm{Res}_{G/H}\mathrm{Ind}_{G/U}\triv,\triv\rangle\\ & = & \langle\bigoplus_{U\backslash G/H}\mathrm{Ind}_{H/H\cap U^g}\triv,\triv\rangle\\ & = &\langle \bigoplus_{U\backslash G/H}1_{H\cap U^g},1_{H\cap U^g}\rangle\\ & = & \#U\backslash G/H. \end{array}$$ This recovers Jyrki's comment in the case that $U=1$, but is true in general. Of course this calculation works just as well, with the necessary modifications, if $V$ is induced from any other representation, not necessarily the trivial one.

-