4
$\begingroup$

(Remark: This has previously been posted on math.stackexchange, but I believe it might be suitable for this site as well. https://math.stackexchange.com/questions/1428350/smooth-admissible-representations-hom-tensor-and-extension-of-scalars )

Let $G$ be a locally profinite group, and consider $V$ and $W$ smooth admissible representations of $G$ over some field $F$ (or char. $0$). Let $E/F$ be any field extension.

I'd like to find conditions to ensure that the natural map $$ Hom_{F[G]}(V,W)\otimes_{F}E\to Hom_{E[G]}(V_E,W_E) $$

is an isomorphism. Here $V_E=V\otimes_FE$, $W_E=W\otimes_FE$ and $Hom_{F[G]}$ (resp. $Hom_{E[G]}$) are the $F$-linear (resp. $E$-linear) $G$-equivariant maps.

If $V$ is generated by $V^K$ as a $G$-space, for some compact open $K\subset G$ (for ex. if it's irreducible), then it's easy to see that the above map is injective by restricting to $V^K$. But I don't know about surjectivity. I don't want to assume too much on $W$ if possible.

Does anyone have any ideas about this?

The question might be related to try to find conditions under which the map $$ \widetilde V\otimes_FW\to Hom_{F}(V,W)_{0} $$ is an isomorphism ($\widetilde V$ is the smooth dual and the subindex $0$ means the smooth vectors), at least when restricting to the subspaces of $G$-fixed vectors.

$\endgroup$
3
  • 1
    $\begingroup$ If $V$ is a finitely generated representation, then the map is an isomorphism for any field extension. If $E/F$ is finite, it's an isomorphism for all $V$ and $W$. The map is always injective. This is all independent of the characteristic of $F$. $\endgroup$ Sep 10, 2015 at 14:04
  • 1
    $\begingroup$ I should say the above holds for an arbitrary group with $V$ and $W$ arbitrary $F[G]$-modules. I don't know what more is true if one supposes $G$ is locally profinite and the representations are (admissible) smooth. $\endgroup$ Sep 10, 2015 at 14:16
  • $\begingroup$ Thanks Keenan! Would you post it as an answer so I can accept it? $\endgroup$
    – dbluesk
    Sep 14, 2015 at 14:39

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.