# Symmetry Properties of Wigner's Matrices

I have an expression of the form

$$S=\sum_{m,n=-j}^{j}(-1)^{m-n}D^{j}_{mn}(g)D^{j}_{mn}(g)$$

This is the end result of a long calculation, from which I am pretty confident that it is correct. For a number of reasons, I expect that it should be possible to write the result in terms of the characters of SU(2). However, I can't see how this is possible here.

In the end, I have to get somehow the trace. First of all, we would need to change the order of indiced of one of the Wigner matrices. This can be done by taking the transpose, i.e.

$$D^{j}(g)_{mn}=D^{j}(g)^{T}_{nm}$$

However, we still need to get rid of the sign factor and the only way I can think of is to use the formula

$$D^{j}_{mn}(g)=(-1)^{m-n}D^{j}_{-m-n}(g)^{\ast}$$

Combining this, I get something like

$$S=\sum_{m,n=-j}^{j}{D^{j}(g)^{\dagger}_{-n-m}}D^{j}_{mn}(g)$$

However, this does not seem to help. Do I miss some property of the Wigner matrices?

• Have you tried a simple example, e.g. with j=1/2 or 1? Commented Dec 14, 2021 at 15:35

$$(-1)^{M'-M}D^{J}_{MM'}(\alpha,\beta,\gamma)=D^{J}_{M'M}(\gamma,\beta,\alpha)=D^{J}_{M'M}(\bar{g})$$
Thus your sum $$\sum_{m,n}(-1)^{m-n}D^{j}_{mn}(g)D^{j}_{mn}(g)=\sum_{m,n}D^{j}_{nm}(\bar g)D^{j}_{mn}(g)=\sum_{n}D^{j}_{nn}(\bar g g)$$ is indeed a character, but of $$\bar g g$$, not of $$g$$.