Let $G$ be a locally profinite (i.e., locally compact Hausdorff and totally disconnected) topological group, $H \le G$ a closed subgroup, and $(W, \sigma)$ a representation of $H$ over $\mathbb{C}$ such that $W$ is smooth, i.e., every $w \in W$ is fixed by a compact open subgroup $K \le H$.

One may form compact induction $\mathrm{ind}_H^G W$ as the $\mathbb{C}$-vector space of all functions $f\colon G \rightarrow W$ such that $f(hg) = \sigma(h) f(g)$ for all $h \in H$ and $g \in G$, that there is a compact open subgroup $K \le H$ such that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$, and that the support of $f$ has compact image in $H\setminus G$. This is a smooth representation of $G$ by setting $(g\cdot f)(x) = f(xg)$.

On the other hand, one may form the smooth $G$-representation $(\mathbb{C}G \otimes_{\mathbb{C}H} (\delta_G/\delta_H) W)^{\infty}$, where the superscript indicates the subspace of smooth vectors and $\delta_G$ (resp., $\delta_H$) is the modulus character of $G$ (resp., $H$) defined by $\delta_G(g) = [gKg^{-1} : K]$ (fractional index) for (no matter which) compact open subgroup $K \le G$ (resp., analogous thing for $H$).

My question is: are these representations isomorphic, i.e., is $$ \mathrm{ind}_H^G W \cong (\mathbb{C}G \otimes_{\mathbb{C}H} (\delta_G/\delta_H) W)^{\infty} $$ as smooth $G$-representations?

  • $\begingroup$ What is the definition of $RG$ and $RH$ here? $\endgroup$ Oct 24, 2014 at 4:55
  • $\begingroup$ These are group rings: I meant $R = \mathbb{C}$ but forgot this, having the more general setup in mind. (The question makes sense for more general coefficient rings $R$, namely, for commutative rings $R$ such that $G$ has a compact open subgroup of invertible pro-order in $R$, so that the modular characters are $R^*$-valued.) $\endgroup$ Oct 24, 2014 at 5:54
  • $\begingroup$ I think you need to replace the group algebras by the Hecke algebras as PL suggests, but also to assume that $H$ is open (see my comment to PL's answer). $\endgroup$ Oct 24, 2014 at 14:25
  • $\begingroup$ The philosophy is that in the framework of $p$-adic group compact induction from on open subgroup looks very much like induction in the framework of finite groups $\endgroup$ Oct 24, 2014 at 14:39

2 Answers 2


No. Take $H$ the trivial subgroup, $W$ the trivial rep. The left hand side is all smooth functions on $G$ with compact support. The right hand side is zero. The reason for this is that elements of $\mathbb C G$ are linear combinations of finitely many elements of $G$, but it is easy to see that such element cannot be fixed by an open subgroup.


Not quite. Rather than work with the group algebra, you want to work with the Hecke algebra. This is the $\Bbb{C}$-algebra of locally constant, compactly supported functions $f:G\rightarrow\Bbb{C}$ under convolution, so it's a subalgebra of $\Bbb{C} G$. The category of smooth representations of $G$ is then equivalent to the category of modules over the Hecke algebra (the equivalence isn't yet natural, but that can be fixed by treating the Hecke algebra as an algebra of distribution instead; choice of Haar measure give the isomorphism with the definition I've given).

Making free use of this equivalence, I'll abuse notation a bit and go backwards and forwards over this equivalence without explicitly saying it. You get two functors $\mathrm{Rep}(H)\rightarrow \mathrm{Rep}(G)$, defined by


where the subscript $G$ denotes the operation of taking the subspace of smooth vectors, and

$$P_H^G:W\mapsto G\otimes_H W.$$

These turn out to be left- and right- adjoint, respectively, to the restriction functor, so they coincide with smooth induction and compact induction, respectively. This is all worked out in David Renard's Representations des groupes reductifs p-adiques.

P.S. Having just written this, I read your definition of compact induction as a space of functions more carefully and realised that your tensor product definition is actually consistent with what your function space definition. Compact induction is actually the subspace of the space you describe given by locally constant, compactly supported functions, just as the Hecke algebra is given by the same subspace of the group algebra.

  • $\begingroup$ Don't you need that $H$ is open in $G$ ? I think that for taking the tensor product of ${\mathcal H}(G)$ over ${\mathcal H}(H)$ (the Hecke algebras of $G$ and $H$ respectively), you need an inclusion ${\mathcal H}(H)\subset {\mathcal H}(G)$ which exists only when $H$ is open in $G$ ... $\endgroup$ Oct 24, 2014 at 14:22
  • $\begingroup$ Regarding your P.S., isn't it more natural to require the support be compact modulo $H$ because the support will in any event be a union of right cosets of $H$? $\endgroup$ Oct 24, 2014 at 16:58

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