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Jun 22, 2022 at 7:16 history edited CommunityBot
replaced http://front.math.ucdavis.edu/ with https://arxiv.org/abs/
Apr 21, 2010 at 8:15 comment added Craig Westerland More: I think you can adapt your argument to get a homotopy equivalence from $S^3 \setminus \sqcup_k S^1$ to a bouquet of $k$ circles and $k-1$ 2-spheres. Unfortunately, however, look at the universal covering of this space: it's an infinite $k$-valent tree with $k-1$ $S^2$'s wedged on at every vertex. So $\pi_2$ of this cover (and hence the space itself) is infinitely genereated!
Apr 21, 2010 at 4:21 comment added Craig Westerland Good point! So perhaps it embeds in $$\prod_n Aut(\pi_n(S^3 \setminus \sqcup_k S^1))$$ That will be nonzero for infinitely many values of $n$, but I bet that your argument indicates that it is determined by $n=1, 2$.
Apr 21, 2010 at 2:40 comment added Tom Church @Craig: I started to write down that argument, but I don't think $S^3-\sqcup_k S^1$ is aspherical (it's not a $K(G,1)$) for unlinked circles. For each circle, add back in all but one point: this gives a map $S^3-\sqcup_k S^1\to S^3-\sqcup_k \ast$. The latter is equivalent to a bouquet of 2-spheres, and this map seems to be surjective on $\pi_2$.
Apr 21, 2010 at 0:24 comment added Craig Westerland About your question -- since $S^3 - \sqcup_k S^1$ is a $K(F_k, 1)$ (for the free group $F_k$), every nontrivial self-map is determined by what it induces in $\pi_1$. So if homotopy = isotopy in this dimension (not obvious to me), then $pi_0(Diff(S^3 - \sqcup_k S^1))$ will embed in $Aut(F_k)$.
Apr 19, 2010 at 0:08 comment added Craig Westerland Tom, do you know anything about the stability of the corresponding group as the dimension of the knots increases? Note that n-spheres are automatically unlinked for n>1. Thanks.
Apr 18, 2010 at 16:40 history answered Tom Church CC BY-SA 2.5