I am trying to calculate the same question that got asked a few years back.
Identities related to Schur's orthogonality relations.
But in the comments one guy gives a way to try and compute it with the polarization identity. But i get confused why he get $\frac{1}{8}$ out, when we writes that formula. I would think he got $\frac{1}{16}$ out instead?. And in the last expression the last terms isn't squared???
And why would it not be smart to write\begin{align*} (u,u')\overline{(v,v')} = (u,u')(v',v) \end{align*} and then try and multiply the equation out? \begin{align*} (u,u')(v',v) = \frac{1}{16} \bigg( &\big( \|u+u'\|^2 - \|u-u'\|^2+i\|u+iu\|^2 - i\|u-iu\|^2 \big) \\ &\big( \|v'+v\|^2 - \|v'-v\|^2+i\|v'+iv\|^2 - i\|v'-iv\|^2 \big) \bigg) \end{align*} But if i multiply all this out i dont seem to get anything "nice". Am i thinking this wrong, and any suggestions to reach the desired result would be appreciated :)
Let us focus on the first term. Multiplying out, we get \begin{align*}\|u+u'\|^2 &\|v'+v\|^2 - \|u+u'\|^2\|v'-v\|^2\\+i \|u+u'\|^2&\|v'+iv\|^2 - i \|u+u'\|^2\|v'-iv\|^2\\ &= \int_{G}\big(\vert\langle{\pi(g)(u+u’),v’+v}\rangle\vert^{2}-\vert\langle{\pi(g)(u+u’),v’-v}\rangle\vert^{2}\\&\quad+i \vert\langle{\pi(g)(u+u’),v’+iv}\rangle\vert^{2}-i \vert\langle{\pi(g)(u+u’),v’-iv}\rangle\vert^{2}\big)\, \mu(dg). \end{align*} Now, the expressions $$ \vert\langle{\pi(g)u,v}\rangle\vert^{2} $$ will have their own polarization identity, and so you can go back basically following the same way backwards.