The structure of $\textrm{c-Ind}_H^G(\sigma)$

Question

Let $G$ be a group of td type, and $H$ a closed subgroup. Let $(\sigma,W)$ be a a smooth representation of $H$. I have always had trouble thinking about the compactly induced representation $\textrm{c-Ind}_H^G(\sigma)$. Is there a nice way to think about this space besides the definition?

Answer 1

You can at least describe the $K$-fixed vectors $\textrm{c-Ind}_H^G(\sigma)^K$ in a nice way for an open compact subgroup $K$. Let $f \in \textrm{c-Ind}_H^G(\sigma)$ be a $K$-fixed vector. Then $f$ vanishes off a finite union of double cosets $HgK$. Let $\mathcal G$ be a set of representatives for $H \backslash G /K$. Then $f$ is completely determined by where it sends the elements of $\mathcal G$, since $f(hgk) = \sigma(h)f(g)$.

Let $g_1, ... , g_t$ be elements of $\mathcal G$ such that the support of $f$ is contained in $\bigcup\limits_{i=1}^t Hg_iK$. Define $f_i: G \rightarrow W$ to be $f_i(hg_ik) = f(hg_ik)$, and $f_i(x) = 0$ for $x$ outside $Hg_iK$. Then $f_i$ lies in $\textrm{c-Ind}_H^G(\sigma)^K$, and is supported in $Hg_iK$, and $f = f_1 + \cdots + f_t$.

Also $f(g_i)$ lies in $W^{H \cap g_iKg_i^{-1}}$, since for $h = g_ikg_i^{-1}$ with $h \in H$ and $k \in K$, we have

$$\sigma(h)f(g_i) = f(g_ikg_i^{-1}g_i) = f(g_ik) = f(g_i)$$

Conversely, if $w \in W^{H \cap g_iKg_i^{-1}}$, there is a unique $f_i \in \textrm{c-Ind}_H^G(\sigma)$, supported in $Hg_iK$, such that $f_i(g_i) = w$.

Conclusion: the $K$-fixed vectors of $\textrm{c-Ind}_H^G(\sigma)$ is, as a vector space, nothing more than the direct sum

$$\bigoplus\limits_{g \in \mathcal G} W^{H \cap gKg^{-1}}$$

In particular, if $\pi$ is admissible, then the summands are finite dimensional, and if $H \backslash G$ is compact, then the direct sum is finite, so $\pi$ admissible, $H \backslash G$ compact implies $\textrm{c-Ind}_H^G(\sigma) = \textrm{c-Ind}_H^G$ is admissible.