Equivalent definitions of compactly supported in the case of induced representation

Question

Let $G$ be a topological group of totally disconnected (td) type, meaning every neighborhood of $1_G$ contains a compact open subgroup. Let $H$ be a closed subgroup of $G$. Then $H$ is also of td type.

Let $(\sigma,W)$ be a smooth representation of $H$. So $W$ is a complex vector space, $\sigma: H \rightarrow \textrm{GL}(W)$ is an abstract representation of $H$, and the stabilizer in $H$ of any vector is an open subgroup. Equivalently, every vector is stabilized by an open compact subgroup of $H$.

Let $\textrm{Ind}_H^G(\sigma)$ be the complex vector space of functions $f: G \rightarrow W$ with the following properties:

1 . There is an open compact subgroup $K$ of $G$ (depending on $f$) such that $f$ is right $K$-invariant ($f(gk) = f(g)$ for all $g \in G, k \in K$).

2 . For all $h \in H$ and $g \in G$, $f(hg) = \sigma(h)f(g)$.

Let $\textrm{c-Ind}_H^G(\sigma)$ be the subrepresentation of $\textrm{Ind}_H^G(\sigma)$ of functions $f$ satisfying the addition property:

3 . $f$ is compactly supported modulo $H$. This means that there is a compact set $Z$ (depending on $f$) such that $\{ x \in G : f(x) \neq 0 \}$ is contained in the product set $H.Z$. Equivalently, we can take closures and say that the support of $f$ is contained in $H.Z$. This is because $H.Z$ is closed, see e.g. here.

I have seen people interchange this definition with saying that the image of the support of $f$ is $H \backslash G$ is compact. But looking at the comments of this question, I am a bit confused on what is and isn't true. Compactly supported modulo a subgroup doesn't appear to always coincide with having the image in the coset space compact. Can someone clarify what, if anything, the condition (3) is equivalent to?

Answer 1

Let $K$ be a compact open subgroup of $G$, and $f \in \textrm{Ind}_H^G(\sigma)$. Let $S$ be the support of $f$. Then $S$ is stabilized by $H$ on the left, and stabilized on the right by $K$. Hence $S$ is a union of double cosets $HgK$. The following are equivalent:

(i): $f$ vanishes off a set of the form $H.Z$ for $Z \subseteq G$ compact.

(ii): $S$ is contained in a set of the form $H.Z$ for $Z$ compact.

(iii): $S = H.Z$ for a compact set $Z \subseteq G$.

(iv): The image of $S$ in $H \setminus G$ is compact.

(v): $S$ is a finite union of double cosets $HgK$.

Proof: Since the nonvanishing set of $f$ is right $K$-stable (resp. left $H$-stable), so is its closure $S$, which is the support of $f$. Hence $S$ is a union of right cosets $xK$, and in particular $S$ is both open and closed. Moreover $S$ is a union of left cosets $Hx$, and being left $H$ and right $K$-stable is a union of double cosets.

(iii) $\Rightarrow$ (ii) $\Rightarrow$ (i) is clear, and (i) $\Rightarrow $ (ii) is obtained by taking closures, and using the fact that $H.Z$ is closed (since $H$ is closed and $Z$ is compact).

For (iv) $\Rightarrow$ (iii), suppose the image of $\pi(S)$ in $H \backslash G$ is compact. There is a compact set $Z$ of $G$ such that $\pi(Z) = \pi(S)$ e.g. Lemma 5.2 of these [notes][2]. Since $S$ is left $H$-stable, the preimage of $\pi(S)$ is $S$. And the preimage of $\pi(Z)$ is $H.Z$.

For (ii) $\Rightarrow$ (iv), the image of $S$ in $H \backslash G$ is closed, since $S = \pi^{-1}\pi(S)$ is a saturated closed set. Being a closed subset of the compact set $\pi(H.Z) = \pi(Z)$, it must be compact.

(v) $\Rightarrow$ (iv): We have $S = Hg_1K \cup \cdots Hg_tK = H(g_1K \cup \cdots g_tK)$ with $g_1K \cup \cdots g_tK$ compact.

(iv) $\Rightarrow$ (v): $S$ is a union of various right cosets $xK$. The image of these cosets are open in $H \backslash G$, so the image of $S$ is a union of finitely many images $\pi(xK)$, say $\pi(S) = \pi(x_1K) \cup \cdots \cup \pi(x_tK)$. Taking preimages, we get

$$S = \pi^{-1}\pi(S) = \pi^{-1}\pi(x_1K) \cup \cdots \cup \pi^{-1}\pi(x_tK)$$

with $\pi^{-1} \pi(xK) = HxK$.

Answer 2

Lemma: Let $G$ be a locally compact Hausdorff topological group, and $H$ a closed subgroup of $G$. Let $\pi: G \rightarrow G/H$ be the canonical projection. If $\Omega \subset G/H$ is compact, then there exists a compact set $Z \subset G$ such that $\pi(Z) = \Omega$.

Proof: if we let $E = \pi^{-1}(\Omega)$, then $E$ is closed, but not necessarily compact. Since $G$ is locally compact, we may cover $E$ by open sets $U_i$ whose closures are compact. Since $\pi$ is an open mapping, and $\Omega$ is compact, we can write $\Omega \subseteq \pi(U_1) \cup \cdots \cup \pi(U_n)$ for a finite set of indices $1 \leq i \leq n$.

Now let $Z = \Big( \overline{U_1} \cup \cdots \cup \overline{U_t} \Big) \cap E$. This is a compact set in $G$, and we claim

$$\pi(Z) = \Omega.$$

It is clear that $\pi(Z) \subseteq \Omega$. Conversely, supposing $x \in \Omega$, we know that there exists $1 \leq i \leq n$ and a $g \in U_i$ such that $x = \pi(g)$. Clearly $g \in E$ and therefore $g \in Z$. Thus every $x \in \Omega$ is equal to $\pi(g)$ for some $g \in Z$, which means that $\Omega \subseteq \pi(Z)$.