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Mar 14, 2023 at 18:04 comment added Giuseppe Negro The opposite extreme is when $[G, G]=\{1\}$, for in that case the orbits are singletons, and this argument places no obstructions on eigenfunctions. Your answer is a really nice observation, thank you for it. I also discovered that $[SO(3), SO(3)]=SO(3)$. Actually, $SO(3)$ has no nontrivial normal subgroups, something that I had never seriously thought about before. By the way, this is exactly what is suggested in comments to the main question (through the fact that the abelianization of $SO(3)$ is trivial).
Mar 8, 2023 at 18:34 history answered Will Sawin CC BY-SA 4.0