Let $\rho$ be irreducible representation of group $G$. How one can characterize all subgroups $H< G$ such that $\rho$ can be embedded into permutation representation $F^X$, where $X=G/H$.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Minor edits. I guess F is the underlying field. Note that the characteristic of the field and whether G is finite might influence the answer or the methods used. Permutation representations most often come up for finite groups. $\endgroup$– Jim HumphreysCommented Sep 7, 2010 at 17:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
There is the following adjointness (a form of Frobenius reciprocity):
$Hom_G(\rho,F^X) = Hom_H(\rho,trivial).$
Thus $\rho$ embeds in $F^X$ if and only if $\rho$ admits a non-trivial $H$-fixed quotient.
(If $H$ is finite and $F$ has characteristic zero, or at least prime to the order of $H$, so that $\rho$ is semi-simple as an $H$-representation, then this is equivalent to requiring that $\rho$ have a non-trivial $H$-fixed subrepresentation.)
(Note also that a non-zero $G$-equivariant map out of $\rho$ is automatically injective, because $\rho$ is irreducible.)
-
2$\begingroup$ Note also that, if $\chi$ is the character of $\rho$, then $Hom_H(\rho, trivial)$ has dimension $1/|H| \sum_{h \in H} \chi(h)$. $\endgroup$ Commented Sep 7, 2010 at 17:48
-
$\begingroup$ Sometimes I feel like I use this fact 5000 times a day. $\endgroup$– JSECommented Sep 7, 2010 at 18:26