The coarea formula for topological groups The classical coarea formula provides us a possibility to reduce the integration over some set to the integration over the slices of this set. For example, we can reduce the integration over a unit ball to the integration over spheres:
$$
   \int\limits_{B(0,1)} f(x) \, dx = \int\limits_{0}^{1} \int\limits_{S(0,t)} f(x)\,dS \, dt.
$$
I'm looking for the generalization of this result to the case of topological groups. For example, is there some similar formula that allows us to reduce the integral over $GL(n)$ to an integral over $O(n)$ and etc?
I'm also looking for such decomposition of integral over the additive group of symmetric matrices. I want to reduce the integration to the integration over the set of eigenvalues.
 A: Such an analogous result indeed exists and is called Weil's formula.
If $G$ is a locally-compact group and $H$ is a closed normal subgroup, then $G/H$ is naturally a locally-compact group. Let $\lambda$ and $\mu$ be Haar measures on $G$ and $H$, respectively (notice that in general the Haar measure on a group does not push-forward to a Haar measure on a subgroup, therefore the measure on $H$ must be chosen). Then there exists a unique measure $\lambda _\mu$ on $G/H$ and for any continuous $\varphi :G \to \Bbb C$ with compact support we have
$$\int \limits _G \varphi (g) \ \Bbb d \lambda (g) = \int \limits _{G/H} \left( \int \limits _H \varphi (g h) \ \Bbb d \mu (h) \right) \Bbb d \lambda_\mu (\hat g) .$$
To see that the value of the inner integral indeed depends only on $\hat g$ and not on $g$ (as it may seem at a first look), notice that for $h' \in H$
$$\int \limits _H \varphi \big( (g h') h \big) \ \Bbb d \mu (h) = \int \limits _H \varphi \big( g (h' h) \big) \ \Bbb d \mu (h) = \int \limits _H \varphi \big( g \underbrace{(h' h)} _{h''} \big) \ \Bbb d \mu \big( h'^{-1} \underbrace{(h'h)} _{h''} \big) = \\
\int \limits _H \varphi (g h'') \ \Bbb d \mu \big( h' ^{-1} h'' \big) = \int \limits _H \varphi (g h'') \ \Bbb d \mu (h'') = \int \limits _H \varphi (g h) \ \Bbb d \mu (h) \ ,$$
so indeed the inner integral evaluated in $g$ has the same value as when evaluated in $gh'$, which shows that it depends only on $\hat g$.
To make the analogy clear with the coarea formula, here the map is the natural projection $G \ni g \mapsto \pi(g) = \hat g \in G/H$, which can be viewed as a fibration with the total space being $G$, the base being $G/H$ and the fibers being $\pi^{-1} (\hat g)$ which all "look like" the "generic" fibre $H$.
Notice the important thing that given any two out of the three measures $\lambda$, $\mu$ and $\lambda_\mu$, the third one is automatically fixed by those two.
You can find more information in "Introduction to Harmonic Analysis and Generalized Gelfand Pairs" by Gerrit van Dijk.
