Calculating the Haar integral on $SU(2)$ in practice

Question

I'm trying to calculate the Haar integral on $SU(2)$ of a given function $f: SU(2) \to \mathbb{R}$.

For this particular function, I know the value of it on the subgroup

$$T = \left\{ \begin{pmatrix} z & 0 \\ 0 & z^{-1} \end{pmatrix}: z \in \mathbb{C}, |z|=1 \right\} \leq SU(2)$$

and that every element of $SU(2)$ is conjugate to something in $T$.

I happen to also know that $f$ is a class function, i.e. $f(A) = f(XAX^{-1}) \forall X$, indeed, I only have an explicit formula for $f$ on $T$, say $f\left( \bigl( \begin{smallmatrix} z & 0 \\ 0 & z^{-1} \end{smallmatrix}\bigr) \right) = \overline{f}(z)$.

So if I were to calculate the Haar integral $\int_{SU(2)} f(A)dA$, I'd ideally want to equate it to an integral over $T$ in some way, that is

$$\int_{SU(2)} f(A)dA = \int_{T} f(A') \omega(A') dA' = \int_{0}^{2 \pi} \overline{f}(z) \overline{\omega}(z)dz$$

Where $\omega$ is some kind of weight function quantifiying how common $A' \in T$ is a conjugate of $A \in SU(2)$ over all of $SU(2)$. That is, something similar to how the Jacobian acts during change of variables in multivariate integration. $\overline{\omega}$ is defined like how $\overline{f}$ is.

But I'm struggling to figure out how I might begin calculating $\omega$ (all I really need is to calculate $\overline{\omega}$) or if this is even the ideal approach to calculate the Haar integral.

Answer 1

You want the Weyl Integration Formula in the simple case of SU(2). This set of lecture notes seems to work it out nicely https://www.math.utah.edu/~milicic/Math_6260/weyl_character.pdf

Also the best elementary introduction to Haar measures on compact groups I know is the one by Joel Feldman https://personal.math.ubc.ca/~feldman/m606/haar.pdf the formula for SU(2) is in Example 5, ii.

Another approach which is suitable for integrating polynomial functions is explained in https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality/255853#255853 It is formulated for $SU(d)$ but it considerably simplifies if you plug $d=2$.

Answer 2

Abdelmalek Abdesselam's answer is good, especially for more general groups like $SU(d)$ or more general Lie groups.

I want to share another approach in case it helps anyone:

In my above question, the function $\overline{\omega}$ is $\overline{\omega}(e^{i \theta}) = \frac{\sin^2 \theta}{\pi}$.

That is because of the following. Denote by $\mathcal{O_{\theta}}$ the conjugacy class of $SU(2)$ of elements which are conjugate to $\bigl( \begin{smallmatrix} e^{i \theta} & 0 \\ 0 & e^{-i \theta} \end{smallmatrix}\bigr) \in T$:

$$\begin{aligned} \int_{SU(2)} f(g)dg &= \frac{1}{2 \pi^2} \int_0^{\pi} \left[ \int_{\mathcal{O_{\theta}}} \overline{f}(e^{i \theta}) d\psi \right] d\theta \\ &= \frac{1}{2 \pi^2} \int_0^{\pi} 4 \pi \sin^2 \theta \overline{f}(e^{i \theta}) d \theta \\ &= \frac{1}{\pi} \int_0^{2 \pi} \overline{f}(e^{i \theta}) \sin^2 \theta d \theta \\ &= \int_0^{2 \pi} \overline{f}(z) \overline{\omega}(z) dz \end{aligned}$$

for the given $\overline{\omega}$ in this answer.

On the first and second lines, we use the fact that $\overline{f}$ is a class function, i.e. constant on $\mathcal{O_{\theta}}$.

Note in particular $\overline{\omega}(\pm 1)=0$, which is to be expected, as the matrices $\pm I$ are only conjugate to themselves.