Let's consider two n-dimensional closed Riemannan manifolds N and M that are cobordant to an undetermined (n+1) dimensional manifold W (That is: N and M are the boundary of W). If all considered manifolds admit spin structures, then does each class of inequivalent spin structures on N,M determine a different cobordism (and hence W)? An answer to this general question may be to long so I'll be more spedific:

An example in 1-d is easiest. Choosing two circles as our 1-dimensional N and M, We have two spin structures (the connected or unconnected double cover over each). Trivially the connected double cover gives N and M each as the boundary of a disk $D^{2}$ (see for example here ). Then our W is the disjoint union of two 2-disks.

Similarly, choosing the unconnected double cover gives N and M as the boundary of a cylinder.

(Note here, that the connected sum of these surfaces is the topological sphere, does a similar result happen in higher dimensions...?)

How much does this generalize? If I have g different spin structures on my boundaries, does each choice of spin structure correspond to a different W? If so, what is the relation between these different W, are they homeomorphic (or diffeomorphic for $n\geq3$)?

My specific case:

I am working with closed three-manifolds of genus g which each admit $2^{g}$ spin structures. How do I know which spin structures are a subset of a 4-manifold resulting from a cobordism of these manifolds? The last question is what I'm trying to answer, though an answer covering the general case would be amazing!

  • $\begingroup$ There is no spin cobordism from $S^1$ with the connected double cover to $S^1$ with the unconnected double cover. This seems to answer the question negatively if I understand it. $\endgroup$ Oct 22, 2021 at 19:58
  • 1
    $\begingroup$ The spin bordism group in dimension 3 is trivial, so yes in dimension 3 there is always a spin bordism between two spin structures on $M$. Its uniqueness up to bordism is measured by the 4 dimensional spin bordism group which is $\mathbb{Z}$. $\endgroup$ Oct 22, 2021 at 20:04
  • $\begingroup$ @Connor_Malin thanks! I had been trying to wrap my head around the meaning of the spin bordism group! I generally study physics, and have found my knowledge in this area lacking. $\endgroup$
    – R. Rankin
    Oct 23, 2021 at 1:21
  • $\begingroup$ @Connor_Malin regarding your first comment, since the connected sum of the surfaces is the 2-sphere, would the sum of the two spin structures somehow be related to the spin structure on $S^{2}$? Thanks again $\endgroup$
    – R. Rankin
    Oct 23, 2021 at 1:40
  • $\begingroup$ Almost, but instead of comparing different spin structures on the manifold you should be considering different null bordisms of the same spin structure. So for two copies of the circle with the trivial spin structure there is a cylinder realizing the null bordism, but also just two disks. Gluing these together along their boundaries give $S^2$. The reason you can’t compare different spin structures on this way is because they won’t glue together to give a spin structure on the resulting manifold. $\endgroup$ Oct 23, 2021 at 16:01


You must log in to answer this question.