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Jun 26, 2020 at 20:04 comment added Marcel Thank you for your input. The idea of working with explicit Euler angles seems scary. Let me point out that it is easy to derive the Weingarten function for $O(N)$ using the theory of zonal polynomials, as I show here: arxiv.org/abs/1406.2182. Zonal polynomials appear because of the relation $\int_{O(N)}s_{2\lambda}(Au)du=Z_\lambda(A^TA)/Z_\lambda(1)$, where $s$ is a Schur function. So a Weingarten calculus for $SO(N)$ might follow from the integral $\int_{SO(N)}s_{\lambda}(Au)du$. Is this integral known?
Jun 26, 2020 at 19:44 history edited Abdelmalek Abdesselam CC BY-SA 4.0
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Jun 26, 2020 at 19:22 history answered Abdelmalek Abdesselam CC BY-SA 4.0