3
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
+100
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Oct 23, 2014 at 3:25
  • $\begingroup$ But is in general injective? $\endgroup$
    – João Dias
    Sep 27, 2019 at 15:56
2
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Oct 22, 2014 at 22:36
  • 1
    $\begingroup$ @Question Mark. I think this is not a problem for you may replace ${\rm Hom}_R (\pi ,\tau )$ by its smooth part. $\endgroup$ Oct 23, 2014 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. $\endgroup$ Oct 23, 2014 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.