Graduate Decomposition into irreps of compact Lie group

[Orodruin]

When decomposing a representation $\rho$ of a finite group $G$ into irreducible representations, we can find the number of times the representation contains a particular irrep $\rho_0$ through the character inner product
$$
\langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$
where $\chi$ and $\chi_0$ are the characters of $\rho$ and $\rho_0$, respectively. Since all group elements in the same conjugacy class have the same characters, this may be rewritten as
$$
\langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{C \in [G]} n(C) \chi(C) \chi_0(C)^*,
$$
where $[G]$ is the set of conjugacy classes of $G$ and $n(C)$ is the number of elements in $C$.

This should generalise to compact Lie groups with the character inner product given by
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{g\in G} \chi(g) \chi_0(g)^* d\mu_G
$$
where $V_G = \int_{g\in G}d\mu_G$ and $d\mu_G$ is the Haar measure on $G$. The corresponding expression on the conjugacy class level should be
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{C\in [G]} \chi(C) \chi_0(C)^* d\mu_{[G]}
$$
with $d\mu_{[G]}$ being a measure on the conjugacy classes determined by integrating the Haar measure over each conjugacy class. Right?

My particular problem is doing this for $SO(3)$. More explicitly, the conjugacy classes of $SO(3)$ should be parametrised by the rotation angle $\theta$ (with the conjugate elements being rotations around different axis by the same $\theta$). I am especially looking for the function $f(\theta)$ such that
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{0}^\pi \chi(\theta) \chi_0(\theta)^* f(\theta) d\theta.$$

Any insights?

 

[martinbn]

May be the Weyl integration formula is what you need.

https://en.wikipedia.org/wiki/Weyl_integration_formula

 

[fresh_42]

I have found the formula
$$
\int_{\operatorname{SU}(2)}f(\theta)\,d\theta = \dfrac{2}{\pi}\int_0^\pi (f\circ e)(t)\sin^2(t)\,dt
$$
including a proof, but in German. Maybe it helps anyway.

Source: https://wwwold.mathematik.tu-dortmund.de/~lschwach/SS10/Bachelor-Seminar/Averim.pdf

 

[fresh_42]

May be the Weyl integration formula is what you need.

https://en.wikipedia.org/wiki/Weyl_integration_formula

The German version of this page notes the transformation function explicitly.

 

[Orodruin]

I have found the formula
$$
\int_{\operatorname{SU}(2)}f(\theta)\,d\theta = \dfrac{2}{\pi}\int_0^\pi (f\circ e)(t)\sin^2(t)\,dt
$$
including a proof, but in German. Maybe it helps anyway.

Source: https://wwwold.mathematik.tu-dortmund.de/~lschwach/SS10/Bachelor-Seminar/Averim.pdf

I think I found my issue. I was trying $\sin^2(\theta)$, but when mapping from SU(2) to SO(3), the $t$ here is actually $\theta/2$?

Using $\sin^2(\theta)$ I obtained that the fundamental representation should contain the trivial one once, which would be absurd. Using $\sin^2(\theta/2)$ gives zero as expected.

It also gives the correct result (2) for the representation $\operatorname{Sym}^2(\operatorname{Sym}^2 V)$ (with $V$ being the fundamental representation), which is what I was really looking for.