# Does $(S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$ admit a symplectic form?

This is a crosspost from this MSE question from a year ago.

Consider the smooth four-manifold $$M = (S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$$. Does $$M$$ admit a symplectic form?

If $$\omega$$ is a symplectic form, then the real cohomology class $$[\omega]$$ satisfies $$[\omega]^2 = [\omega^2] \neq 0$$. Note that $$H^*(M; \mathbb{R})$$ has such classes.

Recall that a symplectic manifold admits an almost complex structure. In general, a closed four-manifold $$N$$ admits an almost complex structure if and only if there is $$c \in H^2(N; \mathbb{Z})$$ such that $$c \equiv w_2(N) \bmod 2$$ and $$c^2 = 2\chi(N) + 3\sigma(N)$$. As $$M$$ is spin (i.e. $$w_2(M) = 0$$), and $$\chi(M) = \sigma(M) = 0$$, the class $$c = 0$$ satisfies the two conditions so $$M$$ admits an almost complex structure.

In addition to the above, the vanishing of $$w_2(M)$$, $$\chi(M)$$, and $$\sigma(M)$$ implies that $$M$$ is parallelisable by the Dold-Whitney Theorem. Moreover, it follows from our knowledge of complex surfaces that $$M$$ does not admit a complex structure, i.e. $$M$$ admits almost complex structures, but none of them are integrable.

One special aspect of symplectic manifolds in dimension four is that they have a non-trivial Seiberg-Witten invariant by a theorem of Taubes. When $$b^+ \geq 2$$, a non-trivial Seiberg-Witten invariant implies that no metric of positive scalar curvature exists, but in this case $$b^+(M) = 1$$ and $$M$$ does admit a metric of positive scalar curvature. I'm not sure if a deeper understanding of Seiberg-Witten theory could be used to provide a negative answer to the above question. In particular, I don't know if $$M$$ has a non-trivial Seiberg-Witten invariant.

• M does have nontrivial Seiberg-Witten invariants using the "loop constraints" version (here the moduli space of SW solutions has positive dimension and characterized by holonomy around the circle factors). I think that after compensating for possible wall-crossing, this would contradict the manifold being symplectic (the hypothetical symplectic chamber would need to have dimension zero). Apr 15 at 4:44
• @ChrisGerig: My understanding of Seiberg-Witten theory is fairly limited; in particular, I am not aware of the "loop constraints" version. If you have time to sketch some details of the argument, I'd be very interested to learn more. Apr 15 at 6:50
• The complete SW invariants take values in $\Lambda^*H_1(M)$, where the "usual" SW invariant is in $\Lambda^0H_1(M)\cong\mathbb{Z}$. Here, when the SW dimension is $2d>0$, you can either use the $\mu^d$ class (from the SW configuration space) to integrate over the SW moduli (giving the "usual" SW invariant), or you can use classes coming from $H_1(M)$ (which again are part of the cohomology of the SW configuration space), such as $\mu^{d-1}\gamma_1\gamma_2$. Apr 15 at 20:48

No, $$M$$ is not symplectic. Consider a double cover $$\tilde{M}$$ of $$M$$ along one of the $$S^1$$ components. Then it is not hard to prove that $$\tilde{M}$$ is diffeomorphic with $$(S^1\times S^3)\#2(S^1\times S^3)\# 2 (S^2\times S^2)$$. Now if $$M$$ were symplectic then you could pull back the symplectic structure on $$\tilde{M}$$. But notice that $$\tilde M$$ has a 3-sphere separating it into two copies with $$b_2^+ = 1$$, so $$SW(\tilde M) =0$$, so this in contradiction with the fact that $$\tilde M$$ is symplectic.