# If $G$ is compact, $H \leq G$ open, $V$ an irreducible $H$-rep, is $\text{Ind}_H^G$ semisimple?

Let $G$ be a compact group, $H$ a normal open subgroup, and $K$ a $p$-adic field (so that not all $G$-reps with coefficients in $K$ are semisimple). Let $V$ be a finite-dimensional topological $K$-vector space with continuous action of $H$, such that $V$ is irreducible as an $H$-representation. Is the induced representation $\text{Ind}_H^G(V)$ semisimple?

In this specific context, I'm thinking of $G$ as the Galois group of a field, if that's helpful.

• Is $V$ is a topological vector space over $K$? Finite-dimensional? Is the $G$-action continuous? Commented Jul 23, 2014 at 19:25
• @Keenan: Yes, yes, and yes. Edited to reflect. Commented Jul 23, 2014 at 20:07
• The continuity of the action does not imply that V is finite diml. Commented Jul 24, 2014 at 6:22
• @Laurent: thanks, you're right. Somehow after my first edit, the 'irreducible' got lost in the shuffle. Commented Jul 24, 2014 at 12:32
• I don't see why irreducible implies finite dimensional either. This is only true for complex irreps of compact groups. Commented Jul 24, 2014 at 12:43

The answer is yes. I think it should be an exercise in any book on representation theory. Since $H$ has finite index in $G$, and $V$ is finite dimensional, so is the representation $W=Ind _H^G (V)$ induced to $G$. In characteristic zero, this is equivalent to proving that the Zariski closure of $G$ in $GL(W)$ is reductive.
The connected component ${\mathcal G}^0$ of the Zariski closure ${\mathcal G}$ of $G$ in $GL(W)$ is unchanged if we replace $G$ by any finite index subgroup $K$. We take $K$ to be the intersection $\cap gHg^{-1}$ as $g$ varies over $G$; this is a finite intersection since $G/H$ is finite.
Restricted to $K$, the representation is semi-simple since $W$ restricted to $K$ is the span of the restriction to $K$ of the semi-simple $gV$ as $g$ varies over $G/H$. Therefore, the unipotent radical of ${\mathcal G}^0$ acts trivially on $gV$ for every $g$ and hence on $W$. Thus, ${\mathcal G}^0$ has no unipotent radical.
 I should have said that I am assuming $Char (K)=0$.