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Apr 2, 2015 at 22:21 comment added Ryan Budney Also, the higher homotopy groups of $Emb(S^1,S^3)$, with the exception of the unknot component, are the same as the higher homotopy groups of $S^3$. I don't think the embedding calculus offers much insights into that -- from this perspective the homotopy groups of spheres are building blocks rather than things to gain insight into.
Apr 2, 2015 at 22:14 comment added Ryan Budney It's quite unclear what the Goodwillie calculus sees of the homotopy-groups of $Emb(S^1,S^3)$. In contrast, it would appear to see "quite a bit" of the homology. But in homotopy, it looks like the calculus sees relatively little. For example, it would be very difficult to extract much of the non-abelian nature of $\pi_1 Emb(S^1,S^3)$ using the calculus. In higher dimensions $Emb(S^1,S^n)$ for $n > 3$ all those troubles vanish, though.
S Jun 6, 2014 at 16:26 history answered Niles CC BY-SA 3.0
S Jun 6, 2014 at 16:26 history made wiki Post Made Community Wiki by Niles