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Aug 3, 2018 at 22:33 comment added mme For the special case of the torus $T^n$, the fibration $\Omega T^n \to LT^n \to T^n$ is trivial: the fiber is contractible to $\Bbb Z^n$, so $LT^n$ is homotopy equivalent to a covering space, and each element of the fiber belongs to a distinct component of $LT^n$ (this uses that $\Bbb Z^n$ is an abelian group, so every conjugacy class is one element), so the covering space is trivial. So $LT^n \simeq \Bbb Z^n \times T^n$. I imagine the $\text{Diff}(S^1)$-equivariant homotopy type is more complicated.
Aug 3, 2018 at 20:23 vote accept Yining Zhang
Aug 3, 2018 at 20:00 answer added Manuel Rivera timeline score: 11
Aug 3, 2018 at 15:35 history asked Yining Zhang CC BY-SA 4.0