There are a variety of characterizations of spin structures on the tangent bundle of a manifold. Two facts about them:
- 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 $H^1$.
- 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.