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


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


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


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:


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


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?


1 Answer 1


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 $-1$.

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 22, 2021 at 8:15
  • $\begingroup$ You're right. I'll correct my post. $\endgroup$ Dec 22, 2021 at 9:53
  • $\begingroup$ Yes, and now I see that I was also wrong so I will correct mine. $\endgroup$ Dec 22, 2021 at 10:14
  • $\begingroup$ 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 $\mathrm{SO}(3)\simeq\mathrm{SU}(2)/\{\pm 1\}$. $\endgroup$ Dec 22, 2021 at 13:16
  • $\begingroup$ There is some double counting here. The points corresponding to k and n-k are on the same sphere. $\endgroup$
    – Tom Mrowka
    Dec 22, 2021 at 13:40

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