My question might be a silly one. I think that the only free action of $S^1$ (thought as a lie group) on itself (up to equivariant diffeomorphisms) is the complex multiplication, but I don't know how to prove/disprove this. Am I correct?
2026-03-30 08:15:17.1774858517
Free actions of the circle on itself
65 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in LIE-GROUPS
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