Let $G$ be a locally profinite (i.e., locally compact, Hausdorff, and totally disconnected) topological group, $H \le G$ a closed subgroup, $(\pi, V)$ a smooth representation of $G$, and $(\sigma, W)$ a smooth representation of $H$. Here smoothness of $\pi$ means that for every $v \in V$ there exists a compact open subgroup $K \le G$ such that $\pi(k)v = v$ for every $k \in K$, and likewise for $\sigma$. Also, we take it that $\pi$ and $\sigma$ are $R$-linear representations where $R$ is a commutative ring; that is, $V$ and $W$ are $R$-modules and the actions are $R$-linear. If it makes a difference, feel free to assume below that $R = \mathbb{C}$.

The smooth induction $\mathrm{Ind}_H^G \sigma$ is the $R$-module of all functions $f\colon G \rightarrow W$ such that $f(hg) = \sigma(h) f(g)$ for all $h \in H$ and $g \in G$ and that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$ for some compact open subgroup $K \le G$ that depends on $f$. The action of $g\in G$ on $\mathrm{Ind}_H^G \sigma$ is by $f \mapsto f(*g)$, so $\mathrm{Ind}_H^G \sigma$ is a smooth $R$-linear representation of $G$. My question is: is the "projection formula map" $$ \pi \otimes_R \mathrm{Ind}_H^G \sigma \rightarrow \mathrm{Ind}_H^G(\pi|_H \otimes_R \sigma) $$ that sends $\sum v_i \otimes f_i$ to $g \mapsto \sum \pi(g)v_i \otimes f_i(g)$ an isomorphism?

I see that it is a well-defined $G$-homomorphism, but I don't see whether it is injective or is surjective. Nor do I see how to write down an inverse.


The map is not surjective in general. The basic idea is the induction is something like a product and tensor products and products don't commute.

Example. $H$ and $\sigma$ trivial, $R$ a field and $\pi$ is an infinite dimensional $R$ vector space with trivial $G$-action. Then if $\psi$ lies in the image of your map then the span set $\{\psi(g): g\in G\}$ is finite dimensional $R$-vector space.

It is enough to construct a function $\psi$ that doesn't satisfy it. Let $\{v_1, v_2, \ldots\}$ be an infinite set of linear independent vectors in $\pi$. Assume that $G$ is not compact choose an open compact subgroup $K$ of $G$. Then can choose infinitely many distinct cosets $\{ g_1K, g_2 K, \ldots\}$. Let $\psi: G\rightarrow \pi$ be the function with support $\bigcup_{n\ge 1} g_n K$, such that $\psi(g)=v_n$ for all $g\in g_n K$. This function is fixed by $K$, hence smooth.

  • $\begingroup$ Thanks, that's very nice. Apparently there's a 24 hour lower bound on awarding bounty, so I'll have to wait before awarding you the +100. $\endgroup$ – Question Mark Oct 23 '14 at 3:25
  • $\begingroup$ But is in general injective? $\endgroup$ – João Dias Sep 27 '19 at 15:56

The projection formula does hold if you assume that $H$ is open and replace induction by compact induction. Indeed if $H$ is open, compact induction (denoted by ${\rm ind}$) is left adjoint to restriction functor. With your notation, let us prove that $\pi\otimes_R {\rm ind}_H^G \sigma \simeq {\rm ind}_H^G (\pi\otimes_R \sigma )$. To see this, let $\tau$ be any smooth $G$-module. Then we have the following isomorphisms of $R$-modules, all functorial in $\tau$:

${\rm Hom}_G ( {\rm ind}_H^G (\pi\otimes_R\sigma ), \tau) \simeq {\rm Hom}_H (\pi\otimes_R\sigma ,\tau )$

$\simeq {\rm Hom}_H (\sigma ,{\rm Hom}_R (\pi ,\tau )) $ (adjoinction property of $\otimes$ and ${\rm Hom}$)

$\simeq {\rm Hom}_G ({\rm ind}_H^G \sigma , {\rm Hom}_R (\pi ,\tau ))$ (I use the fact that $ {\rm Hom}_R (\pi ,\tau )$ is naturally a $G$-module)

$\simeq {\rm Hom}_G (\pi\otimes_R {\rm ind}_H^G \sigma , \tau) $

Now your result follows by a standard application of Yoneda Lemma (I learnt this kind of trick from Guy Henniart).

  • $\begingroup$ Thanks, this is very nice and helpful: I was debating whether I should include the compact induction as an additional question or not. I wonder if any trouble in the chain of isomorphisms is caused at all by the possible nonsmoothness of $\mathrm{Hom}_R(\pi, \tau)$ (do we fall out of the category in which the adjunction holds?). Could you clarify this? $\endgroup$ – Question Mark Oct 22 '14 at 22:36
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    $\begingroup$ @Question Mark. I think this is not a problem for you may replace ${\rm Hom}_R (\pi ,\tau )$ by its smooth part. $\endgroup$ – Paul Broussous Oct 23 '14 at 6:45
  • $\begingroup$ I see. And I guess it doesn't matter whether it is the $G$-smooth or $H$-smooth part being formed, because $H$ is open. $\endgroup$ – Question Mark Oct 23 '14 at 15:14

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