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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

AYK
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1 Answers1

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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 $supp(f)\subset HC$.

PL.
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  • Thanks, I missed the fact that ${g}\to supp(f)$ is indeed a homeomorphism. – AYK Jul 05 '15 at 16:32
  • @PL. Why is the quotient map a homeomorphism in this case? I tried to show this using the fact that the quotient map from G is open, but got stuck. – darkl Jun 11 '16 at 16:37
  • The quotient map might not be a homeomorphism; all that's true is that when you restrict it to the system of coset representatives it gives a homeomorphism. That much is immediate from the quotient map being open. – PL. Jun 12 '16 at 15:02
  • @PL. How do I do show it? I want to show that an open subset of the representatives maps to an open subset. I want to use the fact that the quotient map (without restriction) is open. I thought about writing the open subset of representatives as an intersection of an open subset and the set of representatives, but this open subset might have a different image under the quotient map. This is where I'm stuck. – darkl Jun 12 '16 at 20:44
  • @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
  • I was asked by @darkl to look at this question, and I agree with his/her doubts. The equivalence fails in general case. – Alex Ravsky Oct 12 '16 at 06:10
  • In the complicated special case, consider here, it may holds, for instance taking into account that the map $f$ satisfies $f(hg) = \sigma(h)f(g)$ for all $h\in H,g\in G$ (which may imply that $supp(f)$ is the union of the cosets) or that the group $G$ is locally profinite, but I don’t see the proof of this. – Alex Ravsky Oct 12 '16 at 06:10
  • The set $C$ of representatives may be descriptively very bad and the restriction of a quotient map onto it is not open in general. For instance, consider $G=\Bbb R^2$, $H={0}\times\Bbb R$. As sets of representatives of a compact subset $[0,1]$ of $H\setminus G$, pick the set $C_1=[0;1/2]\times{0}\cup (1/2;1]\times{1}$ and $C_2={(0,0)}\cup{(x,1/x):0<x\le 1}$. Then $C_1$ is not closed in $G$, $C_2$ is closed in $G$, but no one of the restriction of the quotient map onto $C_i$ is open. – Alex Ravsky Oct 12 '16 at 06:11
  • Also it looks strange that PL. spokes on $supp(f)$ as a subset of $H\setminus G$ despite that $f$ is a function on the group $G$ and AYK explicitly states that $H\setminus G$ contains an image of $supp(f)$. – Alex Ravsky Oct 12 '16 at 06:11