Integral of product of Schur functions

Question

Schur functions are irreducible characters of the unitary group $\mathcal{U}(N)$. This implies the integration formulae $$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)dU=\frac{|s_\lambda(A)|^2}{s_\lambda(1)}$$ and $$ \int_{\mathcal{U}(N)}s_\lambda(AU)\overline{s_\mu(A U)}dU=\frac{\delta_{\lambda\mu}s_\lambda(AA^{\dagger})}{s_\lambda(1)},$$ where the overline means complex conjugation.

My question is whether we can compute the integral $$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)\overline{s_\mu(AUA^\dagger U^\dagger)}dU$$

Answer 1

As a first step towards a solution, using the identity $$\int_{\mathcal{U}(N)} U_{\alpha a}U_{\alpha' a'}\bar{U}_{\beta b}\bar{U}_{\beta' b'}\,dU=\frac{1}{N^{2}-1}\bigl( \delta_{\alpha\beta}\delta_{ab}\delta_{\alpha'\beta'}\delta_{a'b'}+ \delta_{\alpha\beta'}\delta_{ab'}\delta_{\alpha'\beta}\delta_{a'b}\bigr)$$ $$\qquad\qquad\mbox{}-\frac{1}{N(N^{2}-1)}\bigl( \delta_{\alpha\beta}\delta_{ab'}\delta_{\alpha'\beta'}\delta_{a'b}+ \delta_{\alpha\beta'}\delta_{ab}\delta_{\alpha'\beta}\delta_{a'b'}\bigr),$$

I computed this integral for the trace, $$\int_{\mathcal{U}(N)}{\rm tr}\, (AUA^\dagger U^\dagger)\,\overline{{\rm tr}\,(AUA^\dagger U^\dagger)}\,dU=\frac{1}{N^2-1}\left(|{\rm tr}\, A|^4+|{\rm tr}\,AA^\dagger|^2\right)-\frac{2}{N(N^2-1)}|{\rm tr}\,A|^2({\rm tr}\,AA^\dagger).$$

_{As a check, take $A=1$ and see that the right-hand-side reduces to $(N^4+N^2)/(N^2-1)-2N^3/(N^3-N)=N^2$.}