pin$^{\pm}$ structure under connected sum

Question

A manifold (orientable or unorientable) may or may not be pin$^+$. If we know the pin$^+$ property of $M$ and $N$, can we determine the pin$^+$ property of $M\# N$? Similarly, if we know the pin$^-$ property of $M$ and $N$, can we determine the pin$^-$ property of $M\# N$?

Answer 1

pin$^+$: A manifold $M$ admits a pin$^+$ structure if and only if $w_2(M) = 0$.

If $M$ is a zero-dimensional, one-dimensional, or a non-closed two-dimensional manifold, then $M$ admits pin$^+$ structures as $w_2(M) \in H^2(M; \mathbb{Z}_2) = 0$.

On a closed two-dimensional manifold, the second Stiefel-Whitney class is zero if and only if the second Stiefel-Whitney number is zero. As the top Stiefel-Whitney number of a closed manifold is the mod $2$ reduction of its Euler characteristic, we see that a closed surface admits pin$^+$ structures if and only if it has even Euler characteristic. As $\chi(M\# N) = \chi(M) + \chi(N) - 2$ for closed surfaces $M$ and $N$, we see that if $M$ and $N$ admit pin$^+$ structures, then so does $M\# N$. Note that the converse doesn't hold. For example, $M = N = \mathbb{RP}^2$ doesn't admit pin$^+$ structures as $\chi(\mathbb{RP}^2) = 1$, but $K = \mathbb{RP}^2\#\mathbb{RP}^2$ does admit pin$^+$ structures as $\chi(K) = 0$.

If $\dim M = \dim N > 2$, then $H^2(M\# N; \mathbb{Z}_2) \cong H^2(M; \mathbb{Z}_2) \oplus H^2(N; \mathbb{Z}_2)$ and under this isomorphism, $w_2(M\# N)$ is identified with $w_2(M) + w_2(N)$. Therefore, $M\# N$ admits a pin$^+$ structure if and only if $M$ and $N$ admit pin$^+$ structures.

Therefore, we have the answer to your first question:

If $M$ and $N$ admit pin$^+$ structures, then so does $M\# N$.

Furthermore, if $\dim M = \dim N \neq 2$, the converse is also true.

pin$^-$: A manifold $M$ admits a pin$^-$ structure if and only if $w_1(M)^2 + w_2(M) = 0$.

If $M$ is a zero-dimensional, one-dimensional, or a non-closed two-dimensional manifold, then $M$ admits pin$^-$ structures as $w_1(M)^2 + w_2(M) \in H^2(M; \mathbb{Z}_2) = 0$.

Furthermore, every closed manifold $M$ of dimension $n = 2$ or $3$ admits pin$^-$ structures. To see this, first recall that $w_1^2(M) + w_2(M)$ is the second Wu class $\nu_2(M)$ and that $\operatorname{Sq}^2 : H^{n-2}(M; \mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$ is given by $\operatorname{Sq}^2(x) = \nu_2(M)\cup x$. But for any $x \in H^{n-2}(M; \mathbb{Z}_2)$, $\operatorname{Sq}^2(x) = 0$ as $\operatorname{deg} x = n - 2 \leq 1 < 2$, so by Poincaré duality, $\nu_2(M) = 0$ and hence $M$ admits pin$^-$ structures.

If $\dim M = \dim N > 2$, then $H^i(M\# N; \mathbb{Z}_2) \cong H^i(M; \mathbb{Z}_2) \oplus H^i(N; \mathbb{Z}_2)$ for $i = 1, 2$ and under these isomorphisms, $w_i(M\# N)$ is identified with $w_i(M) + w_i(N)$. Furthermore, the ring structure of $H^*(M\# N)$ is such that if $x \in H^k(M)$ and $y \in H^l(N)$, then their product (as classes in $H^*(M\# N)$) is zero, so $w_1(M\# N)^2 = (w_1(M) + w_1(N))^2 = w_1(M)^2 + w_1(N)^2$. Therefore $w_1(M\# N)^2 + w_2(M\# N) = (w_1(M)^2 + w_2(M)) + (w_1(N)^2 + w_2(N))$, and hence $M\# N$ admits pin$^-$ structures if and only if $M$ and $N$ admit pin$^-$ structures.

Therefore, we have the answer to your second question:

If $M$ and $N$ admit pin$^-$ structures, then so does $M\# N$.

Furthermore, the converse is also true.