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Let $K$ be a knot in $S^3$, and let $M=S^3/N(K)$ be its knot complement, where $N(K)$ is a tubular neighborhood of $K$. $K$ is given for example by a its projection onto the plane.

The question is then how many spin structures there are in $M$, for a given knot $K$? Since $M$ has a boundary $\partial M=T^2$, and since $T^2$ has four spin structures, I think I could refine this question by asking, "how many spin structures are there in $M$, which extends a fixed spin structure (one of the four) on $\partial M$?

I believe this is related to the computation of $H^1(M, \partial M; \mathbb{Z}_2)$, but do not know how to compute this.

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migrated from mathoverflow.net Nov 28 '15 at 6:17

This question came from our site for professional mathematicians.

  • $\begingroup$ Have a look at section 3.9 of the following notes www3.nd.edu/~lnicolae/Torsion.pdf $\endgroup$ – Liviu Nicolaescu Nov 27 '15 at 23:52
  • $\begingroup$ $H_1(-, \mathbb{Z})$ of a knot complement is $\mathbb{Z}$ (e.g. by Alexander duality), so $H^1(-, \mathbb{Z}_2)$ is $\mathbb{Z}_2$. So there are two spin structures. $\endgroup$ – Qiaochu Yuan Nov 28 '15 at 0:07
  • $\begingroup$ Compute the co-homology group you're interested via excision, relating the pair $(M,\partial M)$ to $(S^3, K)$. I've voted to migrate to MSE. $\endgroup$ – Ryan Budney Nov 28 '15 at 0:48

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