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We know that all compact orientable manifolds of dimension 3 are spin. In 4 dimensions, $CP^2$ is not spin. I would like to ask if all 4-dimensional compact orientable mapping tori are spin?

See also a related question Spin structure on mapping torus

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  • $\begingroup$ No, for example $\mathbb RP^2 \times S^1 \times S^1$ is not orientable, so it is not spin. Do you mean to ask a different question? The question you cite seems to answer your question here. $\endgroup$
    – Ryan Budney
    Commented May 8, 2014 at 17:07
  • $\begingroup$ I am sorry, I mean orientable ones. I update the question. Thanks! $\endgroup$
    – Xiao-Gang Wen
    Commented May 8, 2014 at 17:08
  • $\begingroup$ The question I cited is related, but it did not directly answer if all 4-dimensional compact orientable mapping tori are spin. $\endgroup$
    – Xiao-Gang Wen
    Commented May 8, 2014 at 17:10
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    $\begingroup$ I think the answer is the mapping torus admits a spin structure if and only if the monodromy of the 3-manifold fixes a spin structure. You should be able to construct a diffeomorphism of $(S^1)^3$ that does not fix any spin structures. $\endgroup$
    – Ryan Budney
    Commented May 8, 2014 at 17:26
  • $\begingroup$ @RyanBudney : won't a (orientation-preserving?) diffeomorphism of the 3-torus always fix the trivial spin structure, due to the fact that this the only one which has a non-trivial kernel for its associated Dirac operator? $\endgroup$
    – Jan Jitse Venselaar
    Commented May 8, 2014 at 21:51

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The answer is no. As was described in the thread: Spin structure on mapping torus

a mapping torus has a spin structure if and only if the monodromy of the bundle (over $S^1$) fixes a spin structure. The mapping torus I'm going to describe is a bundle over $S^1$ with fibre $S^1 \times S^2$.

The mapping class group of $S^1 \times S^2$ is of order $8$. If you pass to the subgroup that preserves the fundamental classes of the factors, you get a subgroup of order $2$, the generator can be thought of as the diffeomorphism that twists the $S^2$ factor by $2 \pi$ as one walks around the circle factor.

This automorphism does not preserve any spin structure on $S^1 \times S^2$, it acts as an involution on the spin structures, with no fixed spin structures.

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  • $\begingroup$ Do you have the ref for The mapping class group of $S^1\times S^2$ is of order 8. $\endgroup$
    – Xiao-Gang Wen
    Commented May 13, 2014 at 13:01
  • $\begingroup$ @Xiao-GangWen See Theorem 5.1 of Gluck's 1962 paper "The embedding of two-spheres in the four-sphere" jstor.org/stable/1993581 $\endgroup$
    – j.c.
    Commented May 13, 2014 at 16:55
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    $\begingroup$ @Xiao-GangWen: j.c.'s reference is probably as good as any. I'm not sure when it was originally proved. But the basic argument is fairly simple. The action of the mapping class group on $H_i(S^1 \times S^2)$ gives an epi-morphism to $\mathbb Z_2^2$. A little cut-and-paste topology shows you that diffeomorphisms in the kernel must preserve fibres of the bundle $S^1 \times S^2 \to S^1$ (projection onto the first factor), moreover, they must preserve orientation of $S^2$. So by Smale's theorem on $Diff(S^2)$, such an element is in $\pi_1 Diff^+(S^2) = \pi_1 SO_2 = \mathbb Z_2$. $\endgroup$
    – Ryan Budney
    Commented May 13, 2014 at 20:18
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    $\begingroup$ That last $SO_2$ should have been $SO_3$. There is a homotopy-equivalence $O_3 \simeq Diff(S^2)$ and $SO_3 \simeq Diff^+(S^2)$. $\endgroup$
    – Ryan Budney
    Commented May 13, 2014 at 21:46

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