# Integrals of the type $\delta(g^{n})$ on $\mathrm{SU}(2)$

I posted this question previously to MathSE. However, I have still not solved it, so lets try to ask it here. When doing some calculations with spin-foam models for 3d quantum gravity for some research project, I quite generically stumble over integrals of the following form

$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g^{n})$$

where $$n\in\mathbb{N}_{>0}$$, $$\mathrm{d}g$$ denotes the normalized Haar measure and where $$\delta(g)$$ denotes the $$\mathrm{SU}(2)$$-delta function, i.e. the distribution defined via

$$\langle\delta,f\rangle:=\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(gh^{-1})f(g)=f(h).$$

Now I was told that it is possible to calculate these integrals explicitely and what one gets is formally something like the sum over the $$n$$-roots of unity, which is also what I would expect, since $$\delta(g^{n})$$ in the end just says that $$g^{n}=1$$. So, my approach to this question was the following: It is well-known, using the Theorem of Peter-Weyl, that the delta-function can formally be written as the sum

$$\delta(g)=\sum_{j\in\mathbb{N}/2}(2j+1)\chi^{j}(g),$$

where $$\chi^{j}$$ are the characters of the spin-$$j$$-representation, i.e. the unique (up to unitary equivalence) irreducible unitary representation of dimension $$(2j+1)$$ of $$\mathrm{SU}(2)$$. Now, as a next step, I parametrize the $$\mathrm{SU}(2)$$-group elements as $$g=e^{i\varphi\vec{n}\cdot\vec{\sigma}}$$ where $$\vec{n}$$ is a unit vector and $$\varphi\in [0,2\pi]$$. In this parametrization, the Haar measure is given by $$\mathrm{d}g=\frac{1}{\pi}\mathrm{sin}(\varphi)^{2}\mathrm{d}\varphi\mathrm{d}^{3}\vec{n}.$$ Since $$\chi^{j}$$ are class functions, we can always rotate to the $$z$$-direction, which means that we have that $$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\chi^{j}(g)=\frac{1}{\pi}\int_{0}^{2\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,\chi^{j}(e^{i\varphi\sigma_{z}})=\frac{2}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,\chi^{j}(e^{i\varphi\sigma_{z}}).$$ Hence, let us look at integrals of the type $$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\chi^{j}(g^{n})=\frac{2}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,\chi^{j}(e^{in\varphi\sigma_{z}})=\frac{2}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,\frac{\mathrm{sin}((2j+1)n\varphi)}{\mathrm{sin}(n\varphi)}$$ where in the last step we used that the characters are explicitly given by $$\chi^{j}(g(\varphi))=\frac{\mathrm{sin}((2j+1)\varphi)}{\mathrm{sin}(\varphi)}.$$

Now, using Mathematica, I got the following results:

Lets consider the following integral for all $$j\in\mathbb{N}/2$$ and for all $$n\in\mathbb{N}$$ $$\mathcal{I}_{j,n}=\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\chi^{j}(g^{n})=\frac{2}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,\frac{\mathrm{sin}((2j+1)n\varphi)}{\mathrm{sin}(n\varphi)}$$ For $$n=1$$, the integral is given by $$\mathcal{I}_{j,1}=\delta_{j0}=\begin{cases}1 &\text{if j=0}\\0 > &\text{if j\neq 0}\end{cases}$$ and for $$n=2$$ by $$\mathcal{I}_{j,2}=(-1)^{2j}=\begin{cases}1 &\text{if > j\in\mathbb{N}}\\-1 &\text{if > j\in\mathbb{N}_{\mathrm{odd}}/2}\end{cases}$$ and for $$n=3$$ by $$\mathcal{I}_{j,3}=\frac{1+(-1)^{2j}}{2}=\begin{cases}1 &\text{if > j\in\mathbb{N}}\\0 &\text{if > j\in\mathbb{N}_{\mathrm{odd}}/2}\end{cases}$$

We could continue with higher $$n$$'s, but let us stick to the cases $$n=1,2,3$$.

Now, we are interested in integrals of the following type:

$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g^{n})=\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\sum_{j\in\mathbb{N}_{0}/2}(2j+1)\chi^{j}(g^{n})=\frac{2}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\sum_{j\in\mathbb{N}_{0}/2}(2j+1)\chi^{j}(e^{in\varphi\sigma_{z}})=\sum_{\mathbb{N}_{0}/2}(2j+1)\mathcal{I}_{j,n}$$

As a quick consistency check, let us see what happens for the case $$n=1$$:

$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g)=\sum_{\mathbb{N}_{0}/2}(2j+1)\mathcal{I}_{j,1}=\sum_{\mathbb{N}_{0}/2}(2j+1)\delta_{0j}=1$$

as it should. However, for $$n=2,3$$, it seems that we get an undefined or divergent series, as can be seen in the following calculation:

$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g^2)=\sum_{\mathbb{N}_{0}/2}(2j+1)\mathcal{I}_{j,2}=\sum_{\mathbb{N}_{0}/2}(2j+1)(-1)^{2j}$$ $$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g^3)=\sum_{\mathbb{N}_{0}/2}(2j+1)\mathcal{I}_{j,3}=\sum_{\mathbb{N}_{0}/2}(2j+1)\frac{1+(-1)^{2j}}{2}$$

Of course, since the delta-function should more precisely be viewed as a distribution, we should also add a test function. However, decomposing the test function also using the Theorem of Peter-Weyl, this should not make any difference.

Does anyone know where my error is? Or maybe some alternative method to calculate the integrals?

For the special case of $$SU_2$$ it is often useful to think of this as the group of unit quaternions. So $$g^n=1$$ really is the question for $$n$$-th roots of unity, only your asking for quaternion solutions.
in $$\mathbb{H}$$ one can write every number as $$g=a+bx$$ with $$a,b\in\mathbb{R}$$ and $$x$$ a purely imaginary unit quaternion so $$x=\alpha i + \beta j + \gamma k$$ with $$\alpha,\beta,\gamma\in\mathbb{R}$$ and $$\alpha^2+\beta^2+\gamma^2=1$$. In matrix language, this is the sphere $$\{A\in SU_2 \mid \operatorname{tr}(A) = 0\}$$. And then $$g^n$$ can be computed just like in the complex numbers. Only now we have a whole 2-sphere of choices for the imaginary unit $$x$$ whereas in $$\mathbb{C}$$ there are only two discrete imaginary units, $$+i$$ and $$-i$$.
In other words, $$g^n=1$$ means that $$a=\cos(k/2n\pi)$$ and $$b=\sin(k/2n\pi)$$ for some $$k\in\{0,1,\ldots,n-1\}$$. Therefore one should expect the integral you're asking about to be equal (up to some constant) to a sum of $$n$$ integrals. For odd $$n$$, these would be $$n-1$$ integrals over a 2-sphere (corresponding to $$k=1,\ldots,n-1$$) and one is just the evaluation the identity (corresponding to $$k=0$$), i.e. another $$\delta$$. For even $$n$$, the case $$k=n/2$$ also corresponds to a single point, namely $$-1$$.
• The case $k=0$ degenerates to a point, so you have $n-1$ spheres and a point. See my answer for a slightly different way to see this. Dec 22, 2021 at 8:15
• I realized that what I wrote was not quite right and I don't have time to correct it now. So I removed my answer. The idea was to describe the elements by looking at the quotient $\mathrm{SO}(3)\simeq\mathrm{SU}(2)/\{\pm 1\}$. Dec 22, 2021 at 13:16