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

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

• I now think this will be true: H\G is a finite discrete set, since H is open in G, and hence is compact. Thus, the image of any element f of the classical induction will have compact support and thus belongs to the compact induction $\text{c)-Ind_H^G(W)$. It now follows from the argument in Ngo's notes on 'Representation theory of p-adic reductive groups' (argument just after (5.22)) that such a vector must be smooth. I'm still thinking this through but it seems plausible right now! Commented Feb 29 at 14:04
• Ah, I think this requires functions to be locally constant Commented Apr 30 at 15:33

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.