Every closed-orientable 3-manifold admits an $su(2)$ spin structure. I'm working with manifolds that are of the form:

$$M=\#_{k}\left(S^{2}\times S^{1}\right)$$

For a Riemann surface I know the genus $g$ regulates the number of inequivalent spin structures. I'm wondering if there's something similar for the above class of manifolds? I get that the number of inequivalent spin structures is given by $H_{1}\left(M,\mathbb{Z}_{2}\right)$ but I honestly have literally no idea how to calculate that, and searching the internet has yielded nothing of use thus far


1 Answer 1


If $M$ admits a spin structure, then the set of isomorphism classes of spin structures on $M$ is in bijection with $H^1(M; \mathbb{Z}_2)$. One way to calculate $H^1(M; \mathbb{Z}_2)$ is to use the isomorphisms

$$H^1(M; \mathbb{Z}_2) \cong \operatorname{Hom}(H_1(M; \mathbb{Z}), \mathbb{Z}_2) \cong \operatorname{Hom}(\pi_1(M), \mathbb{Z}_2).$$

The first isomorphism follows from the Universal Coefficient Theorem, while the second follows from the Hurewicz Theorem, namely that $H_1(M; \mathbb{Z}) \cong \pi_1(M)^{\text{ab}}$, together with the fact that $\mathbb{Z}_2$ is abelian.

Another fact which is useful in this case is that if $\dim M_1 = \dim M_2 > 1$, then $$H^1(M_1\# M_2; \mathbb{Z}_2) \cong H^1(M_1; \mathbb{Z}_2)\oplus H^1(M_2; \mathbb{Z}_2),$$

which is proved by using the Mayer-Vietoris sequence (twice). For example, a genus $g$ Riemann surface has $H^1(\Sigma_g; \mathbb{Z}_2) \cong H^1(T^2; \mathbb{Z}_2)^g \cong \mathbb{Z}_2^{2g}$, so there are $2^{2g}$ isomorphism classes of spin structures.

You can use the Seifert-van Kampen Theorem to show that if $\dim M_1 = \dim M_2 \geq 3$, then $\pi_1(M_1\# M_2) \cong \pi_1(M_1)\ast\pi_1(M_2)$. It follows that $\pi_1(k(S^2\times S^1)) \cong F_k$, the free group on $k$ generators. Using this, or the connected sum fact, you can deduce that there are $2^k$ isomorphism classes of spin structures on $k(S^2\times S^1)$.

  • $\begingroup$ Thank you! I'm not 100% on the fundamental group for the above manifold, but I'll see if I can work it out. (: $\endgroup$
    – R. Rankin
    Jul 29, 2021 at 4:41
  • 1
    $\begingroup$ I have added some more details to my answer in case you weren't able to figure it out. $\endgroup$ Jul 29, 2021 at 12:16

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