compactly induction of smooth modules over Hecke algebras

Let $$G$$ be a locally profinite group and $$H$$ its closed subgroup. It is well-known that a smooth representation of $$G$$ is identified with a smooth module over $$\mathcal{H}(G)$$, where $$\mathcal{H}(G)$$ is the Hecke algebra of $$G$$. That is, $${\rm Rep}(G)\simeq \mathcal{H}(G){\text -}{\rm mod}.$$ Now let us consider the compactly induction functor $$c{\text -}{\rm Ind}_{H}^{G}\colon {\rm Rep}(H)\longrightarrow {\rm Rep}(G).$$ For any object $$(\sigma,W)$$ of $${\rm Rep}(H)$$, we have $$c{\text -}{\rm Ind}_{H}^{G}\sigma =(\Sigma,X_{c}).$$ The space $$X_{c}$$ is $$X_{c}=\{f\colon G\longrightarrow W\mathrel{\vert} (1),(2),(3)\},$$

• (1). for any $$h\in H,g\in G$$, $$f(hg)=\sigma(h)f(g),$$
• (2). there exists a open compact subgroup $$K_{f}$$ such that $$f(gk)=f(g)$$ for any $$k\in K_{f}$$
• (3). the support of $$f$$, $${\rm supp}(f)$$ is compact modulo $$H$$.

And $$\Sigma$$ is $$\Sigma\colon G\longrightarrow {\rm Aut}(X_{c});g\longmapsto \left(f\longmapsto (x\longmapsto f(xg))\right).$$

I want to know about the functor $$c{\text -}{\rm Ind}_{H}^{G}\colon \mathcal{H}(H){\text -}{\rm mod}\longrightarrow \mathcal{H}(G){\text -}{\rm mod}$$. Probably this is given as a thensor product. Let $$\mu_{G}$$ and $$\mu_{H}$$ be left Haar measures on $$G,H$$ respectively. For any $$\phi\in \mathcal{H}(H)$$ and $$f\in \mathcal{H}(G)$$, we define $$R(\phi)f\in \mathcal{H}(G)$$ to be $$g\longmapsto \int_{H}\phi(h)f(hg)d\mu_{H}(h).$$ Then $$R$$ provides $$\mathcal{H}(G)$$ with a structure of right smooth module over $$\mathcal{H}(H)$$. Perhaps there exists a canonical ismorphism $$c{\text -}{\rm Ind}_{H}^{G}\sigma \simeq \mathcal{H}(G)\otimes_{\mathcal{H}(H)}\sigma$$. But I dont know the explicit correspondence of this isomorphism.

• You will also need the notion of sufficiently many idempotents in a ring (such as these full Hecke algebras), and correct formation of tensor products over them. And pay attention to the modular functions (regarding left-versus-right Haar measures...) Apr 23, 2021 at 18:17