# support compact modulo subgroup

I am studying (co)-induced representations of topological groups and I came across the following situation:

$G$ is a topological group, $H$ a closed subgroup and $f\colon G\to W$ a set-theoretic map, where $W$ is a (complex) vector space (in fact, it is a vector space on which $H$ acts but for the moment let us ignore this). We say that $f$ is compactly supported modulo $H$ if $supp(f) \subseteq HC$ for some compact set $C$ in $G$.

My question: is this equivalent with saying that the image of $supp(f)$ is compact in $H\backslash G$ (where the quotient space is endowed with the usual topology)?

Now I add some further assumptions: $G$ is a locally profinite group (= locally compact + totally disconnected or equivalently: Hausdorff + the unit $1\in G$ has a fundamental system of compact open subgroups), $H$ a closed subgroup of $G$ acting on $W$, and $f$ a map satisfying

$f(hg) = \sigma(h)f(g)$

for all $h\in H,g\in G$ (where $\sigma$ is the map describing the action of $H$). Is the previous statement true now? Thanks in advance!

AYK

Yes this is true. I guess you know this, but to give the question context for anyone else reading it, let me point out that if $(\sigma,W)$ is a representation of a closed subgroup $H$ of $G$, then the compactly induced represenation $cInd_H^G \sigma$ of $G$ is defined to be the set of functions $f:G\rightarrow W$ such that, for all $h\in H$, $g\in G$, $f(hg)=\sigma(h)f(g)$ and $supp(f)$ is compact in $H\backslash G$, equipped with the action of $G$ by left translation.
So our claim is that $supp(f)$ is compact in $H\backslash G$ if and only if there is a compact set $C\subset G$ such that $supp(f)\subset HC$.
The "if" direction is clear. For the "only if" direction, suppose $supp(f)$ is compact in $H\backslash G$, and take its preimage $X$ in $G$, which is a disjoint union of $H$ cosets. Let $\{g\}$ denote a set of representatives of these cosets, reduced in the sense that each coset is represented exactly once; then the projection $G\rightarrow H\backslash G$ homeomorphically maps $\{g\}$ to $supp(f)$. Write $C=\{g\}$. Then $C$ is compact and one certainly has $supp(f)\subset HC$.
• Thanks, I missed the fact that $\{g\}\to supp(f)$ is indeed a homeomorphism. – AYK Jul 5 '15 at 16:32
• @PL. in other words I don't understand why if $A \subseteq \{g\}$ and $A = U\cap\{g\}$ where $U \subseteq G$ is open then $AH \subseteq G$ is open. (Note that $UH$ can be bigger than $AH$) – darkl Jun 12 '16 at 22:40