1
$\begingroup$

Frobenius reciprocity states that if $G$ is a group with representation $V$ and $H$ a subgroup of $G$ with representation $W$, then

$\langle \chi_{\mathrm{Ind}_H^G W}, \chi_V \rangle_G = \langle \chi_W, \mathrm{Res}_H^G\chi_V\rangle_H $

According to my notes, it follows that $\mathrm{Hom}_G(\mathrm{Ind}_H^G W, V) \cong \mathrm{Hom}_H(W, \mathrm{Res}_H^G V)$. I assume this means isomorphic as representations of $G$. Why is this true? $\langle \chi_V, \chi_W \rangle_G = \mathrm{dim}\ \mathrm{Hom}_G(V,W)$ in general, so I can see why the above are isomorphic as vector spaces.

Thanks

$\endgroup$

1 Answer 1

2
$\begingroup$

It means isomorphic as vector spaces. They're not representations of $G$ (or, well, they're trivial representations).

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .