# Inverting the integration along a subgroup

Given a locally compact group $G$ and a closed subgroup $H$, one often uses an operator of the form $$P: C_c(G) \rightarrow C_c(H \backslash G), \qquad Pf(Hg) = \int_H f(hg) d_H h,$$ where $d_H h$ denotes a Haar measure on $H$. This map is surjective. Is there an explicit form for the right inverse $D: C_c(H \backslash G) \rightarrow C_c(G)$ of $P$?

Consider $\pi: G \mapsto H \backslash G$:

See comments of Daniel Litt for two nice solutions in the extremal cases:

1. If $H$ is compact, $D \phi = \phi \circ \pi$.

2. If $H$ is cocompact, then choose $q$ in the preimage of $1$ under $P$, and $D\phi = q \cdot (f \circ \phi)$.

So what about the groups inbetween?

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There are of course many left inverses; if $H$ is compact, there is an obvious one which sends a function $f$ on $H\G$ to a function on $G$ which is constant on each coset of $H$. (Namely, $f\mapsto (f\circ \pi)/\operatorname{Vol}(H)$ where $\pi$ is the projection $\pi: G\to H\G$.) Is there a particular left inverse you are interested in? – Daniel Litt Jul 27 '11 at 7:18
But I need $P^{-1} :C_c(H\backslash G) \rightarrow C_c(G)$ in full generality, and your construction applies iff $H$ is compact. But this is a nice observation. – Marc Palm Jul 27 '11 at 8:13
I'll write with left cosets since the TeX seems to be screwed up in my last comment. Here's a construction that works in general. Choose any function $g: G\to \mathbb{R}$ such that $P(g)$ vanishes nowhere, where $g$ is chosen so that $P(g)$ makes sense (e.g. $g$ is integrable over each coset of $H$). Then given $f: G/H\to \mathbb{R}$, we send $f$ to $S(f):=g\cdot ((f\circ \pi)/(P(g)\circ \pi))$. The factor in parentheses is constant on cosets of $H$, so we may pull it out of the integral and so $P(S(f))=f$. (I think you mean left inverse everywhere, not right inverse, by the way.) – Daniel Litt Jul 27 '11 at 9:06
*Rather, right inverse, not left inverse. – Daniel Litt Jul 27 '11 at 9:06
For right cosets you have to use \backslash. Yes, right inverse=) I will check your construction... thx – Marc Palm Jul 27 '11 at 10:03

I have a bit of time, so I'll be explicit. This construction assumes $\pi: G\to G/H$ is a fiber bundle, and that $G/H$ is paracompact Hausdorff (and so admits partitions of unity). This should cover many examples that occur "in nature." My comments below the original question also do the case where $H$ is compact, which I think need not be contained in this case.
Lemma. Under the given assumptions, there is a continuous function $s: G\to \mathbb{R}$ such that $s$ has compact support when restricted to each fiber of $\pi$, and such that for each $g\in G$, $\int_H s(gh) d_Hh=1$. Proof. If $\pi: G\to G/H$ is trivial as a principle $H$-bundle (that is, it admits a section $t: G/H\to G$), this is easy; namely, pick any continuous function $s': H\to \mathbb{R}$ with compact support, satisfying $\int_H s'(h) d_Hh=1$. Let $s(x)=s'(t(\pi(x))^{-1}x)$.
Now if $\pi: G\to G/H$ is a fiber bundle, we may cover $G/H$ by open $U_i$ such that the bundle is trivial over each $U_i$. By the previous paragraph, we may choose $s_i: \pi^{-1}(U_i)\to \mathbb{R}$ with compact support on each fiber of $\pi$, and whose integral over each fiber equals $1$. Now by assumption we may choose a partition of unity $\{\phi_j\}$ subordinate to the cover $\{U_i\}$. Let $s=\sum_{i,j} s_i\cdot (\phi_j\circ \pi)$. $\Box$
We now construct a right inverse $D$ to $P$. Namely, for $f$ a compactly supported continuous function on $G/H$, let $D(f)=s\cdot (f\circ \pi)$. It is clear that $P(D(f))=f$; one need only check that $s\cdot (f\circ \pi)$ has compact support, which I leave as an easy exercise (again using that $\pi$ is a fiber bundle).
I guess that $C_c(G)$ has a $C^*$ algebra completion and $P$ is $*$ algebra homomorphism and that gives $C_c(G/H)$ a $C^*$ algebra structure, and every $C^*$ algebra has an approximate identity, which would be a suitable replacement for $G/H$ posessing a partition of unity, I guess. But I will have to diggest you contruction a little bit. Thanks for your time. – Marc Palm Jul 28 '11 at 9:50