There are a variety of characterizations of spin structures on the tangent bundle of a manifold. Two facts about them:

  1. Spin structures on $TM$ are an affine space over $H^1(M; \mathbb{Z}/2\mathbb{Z})$, but in general there's no canonical way to identify them with $H^1$.
  2. Spin structures on $TM$ are the same as trivializations of $TM$ along the $1$-skeleton that extend to trivializations on the $2$-skeleton.

I'm interested in the case where $M = S^3 \setminus L$ is a knot or link complement. (I'm not completely clear whether I want to think about $M$ as compact with torus boundary or non-compact, but I guess my proof below uses ideal triangulations.) I am suspicious that the following is true:

Theorem: Spin structures on $S^3 \setminus L$ are in one-to-one correspondence with orientations of $L$, hence with $H^1(S^3 \setminus L; \mathbb{Z}/2\mathbb{Z})$.

Proof idea: From an oriented diagram of $L$ we can use the octahedral decomposition to get an ideal triangulation of $S^3 \setminus L$. For such a triangulation it's possible to orient everything so that the obvious trivialization of the $1$-skeleton (coming from its orientation) extends over the $2$-skeleton.

This is pretty vague, but I think the "compatible orientations" are a branching, in the sense of [1]. I believe [1] works out the details, since they have a "spin calculus" for triangulations of spin $3$-manifolds.

My question: Is the theorem true? If so, is there a more direct proof than what I've given? It seems like there should be a more direct path than choosing some special triangulation.

[1] R. Benedetti, C. Petronio, Branched Standard Spines of 3-manifolds, Lect. Notes Math. 1653, Springer (1997)

EDIT: Some motivation for the question is the "well known" fact that spin structures on a hyperbolic knot complement are in natural 1-1 correspondence with lifts of the holonomy representation $\rho : \pi_1(S^3 \setminus K) \to \operatorname{PSL}_2(\mathbb C)$ to $\operatorname{SL}_2(\mathbb C)$. (The argument has to do with identifying $\mathbb H^3$ as $\operatorname{PSL}_2(\mathbb C)/\operatorname{SO}_3$.) Then for a hyperbolic knot it's straightforward to pick a canonical lift/spin structure, because in one lift the trace of the meridian is $2$ and in the other it's $-2$. It doesn't matter which meridian you pick, because $\mathfrak m$ and $\mathfrak m^{-1}$ have the same trace.

However, upon further thought we can't identify this choice of lift with a choice of meridian ($\mathfrak m$ versus $\mathfrak m^{-1}$), which is the same as an orientation of the knot. This agrees with the answers that say that orientations are not naturally the same as spin structures.

  • $\begingroup$ Perhaps the simplest way to see spin structures can't be identified (naturally) with link orientations would be to consider the Hopf link case. The exterior is $S^1 x S^1 x I$, so you can understand the spin structures and the mapping class group in fairly elementary terms. $\endgroup$ Sep 14 at 19:48

Is the theorem true? There is an non-natural bijection. There is no natural bijection.

A link exterior is homotopy-equivalent to a $2$-complex, so a trivialization of the tangent bundle over the $2$-skeleton is simply a map:

$$S^3 \setminus L \to SO_3.$$

This is because there is a canonical trivialization of $TS^3$ coming from, say, a left-invariant framing.

Similarly, if you had a trivialization over the $1$-skeleton that admits an extension to the $2$-skeleton, any two extensions $S^3 \setminus L \to SO_3$ have to be homotopic, due to $\pi_2 SO_3$ being trivial.

So spin structures on link exteriors are canonically in bijection with

$$\pi_0 Maps(S^3 \setminus L, SO_3).$$

$SO_3$ is diffeomorphic to $\mathbb RP^3$, which is the $3$-skeleton of $\mathbb RP^\infty$, which is a $K(\mathbb Z_2,1)$-space.

So you have a natural map between the spin structures on the link exteriors and

$$\pi_0 Maps(S^3 \setminus L, K(\mathbb Z_2,1)) \equiv H^1(S^3 \setminus L, \mathbb Z_2).$$

You can argue this map is injective. The argument that comes to mind is that $\pi_1 \mathbb RP^3 \to \pi_1 \mathbb RP^\infty$ is an isomorphism.

I think the relation to Kevin Walker's observation is that diffeomorphisms of the link exterior act trivially on some of the spin structures, but not on all. A diffeomorphism that reverses a link component does not affect the spin structure. The only way a diffeomorphism can act non-trivially on spin structures is if they permute components of the link. Similarly the diffeomorphism group of a knot exterior acts trivially on the spin structures, even for invertible knots (where diffeomorphisms can reverse the knot). The action of the diffeomorphism group on the spin structures factors through the permutation representation, i.e. the map $\Sigma_n \to GL_n \mathbb Z_2$ given by the $n \times n$ matrices that permute the rows and columns, and that permutation representation is the permutation of the link components.

Technically to get the above factorization you need to know that the link has no split Hopf link components. When you have a split Hopf link you get a $GL_2(\mathbb Z)$ in the mapping class group and this action would not belong to the above representation.

  • $\begingroup$ @ryan_budney -- Upon reading your answer I realize that I forgot to specify that I was focusing on the case where $L$ is the unknot in my answer. Oops. $\endgroup$ Sep 14 at 20:29

The unique spin structure on $S^3$ restricts to a spin structure on $S^3 \setminus L$, and so provides a basepoint to the affine space of all spin structures on the latter.

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    $\begingroup$ How does this concern orientations of the link, which is what the OP asked? $\endgroup$ Sep 14 at 18:56
  • $\begingroup$ Spin structures are an affine space over $H^1(S^3\setminus L; \mathbb{Z}/2)$ (mentioned in the OP). The base point mentioned in this post means they are actually canonically identified with $H^1(S^3\setminus L; \mathbb{Z}/2)$. The OP already mentioned that this latter is orientations of L, which I guess is just a form of Alexander duality. $\endgroup$ Sep 15 at 13:20
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    $\begingroup$ I wrote "the unique spin structure", but really what is unique is a spin structure once an orientation is selected. These are what is a torsor for $H^1$. I suspect that the choice of orientation will make a difference. In any case, I was taking for granted the OP's claim that $H^1(S^3 \setminus L; \mathbb{Z}/2)$ classified orientations of $L$ --- I didn't think it through myself. $\endgroup$ Sep 15 at 19:46

I don't think there can be a natural 1-1 correspondence between orientations of $L$ and spin structures on $S^2\setminus L$. Consider the case where L is the unknot. [Note: the previous sentence was inadvertently omitted from the original version of this answer.] There are diffeomorphisms of $(S^3, L)$ which reverse the orientation of $L$. These same diffeomorphisms act trivially on the spin structures on $S^2\setminus L$.


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