Smooth versus classical induction for an open subgroup of a profinite group

Question

Let $G$ be a locally profinite group and $H \leq G$ an subgroup. If $(\sigma, W)$ is a smooth representation of $G$ then its classical induction is the pair $(\Sigma, \text{IND}_H^G(W))$ where

$$\text{IND}_H^G(W)= \{ f: G \rightarrow W: f(hg) = \sigma(h)f(g) \text{ for } g \in G, h \in H \}, $$

and $\Sigma: G \rightarrow GL(\text{IND}_H^G(W))$ is the group homomorphism defined by $(\Sigma(g)(f))(x)= f(xg)$ for $x \in G$. Taking the smooth part of this representation yields the smooth induction $(\Sigma, \text{Ind}_H^G(W))$ of $(\sigma, W)$ . Here, $\text{Ind}_H^G(W)$ consists of the vectors $f \in \text{IND}_H^G(W)$ such that there exists a compact open subgroup $K_f \leq G$ with $f(gx)=f(g)$ for all $g \in G, x \in K$.

If $G$ is profinite and $H$ is open, so finite index, do we have that $\text{Ind}_H^G(W)=\text{IND}_H^G(W)$?

I have been trying to answer this question in the affirmative. My hope was that, since any $f \in \text{IND}_H^G(W)$ is determined by its values on a right transversal $\{ g_1,...,g_n \}$ and $f(g_i) \in W$ is stabilised by a compact open subgroup $K_i \leq H$ by assumption then I had hoped that $f$ would be stabilised by (at least something related to) $\cap_{i=1}^n K_i$. (If $H$ were not of finite index then this argument breaks as the intersection of infinitely many opens need not be open). However, I can't get the computations to work and now I am unsure what I am trying to prove is even true.

I have also tried thinking about this using Frobenius reciprocity but I think this was a bad approach to take because the relevant adjoint pairs all live in different categories. I can elaborate on this if this is not clear.

Any help would be hugely appreciated. Thank you in advance!

Tom

Answer 1

This is true, and your guess of $\cap_{i = 1}^n K_i$ is almost correct: instead you should consider $$ K = \cap_{i = 1}^n g_i^{-1} K_i g_i. $$ Indeed, suppose that $k \in K$, and that $f \in \text{IND}_H^G(W)$. We want to show that $k*f = f$, or in other words, that $$ f(xk) = f(x) $$ for all $x \in G$. An arbitrary element of $G$ is of the form $x = hg_i$ for some $i$. Because $k \in K$, $k = g_i^{-1} t g_i$, for some $t \in K_i$, and therefore: $$ f(xk) = f(hg_i g_i^{-1} t g_i) = f (ht g_i) = \sigma(ht) f(g_i) = \sigma(h)f(g_i) = f(hg_i) = f(x), $$ and we're done.