# Number of inequivalent spin structures on $\#_{k} (S^2 \times S^1)$?

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

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 $$k(S^2\times S^1)$$.

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