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.

  • $\begingroup$ 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). $\endgroup$ Apr 15, 2022 at 4:44
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    $\begingroup$ @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. $\endgroup$ Apr 15, 2022 at 6:50
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    $\begingroup$ 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 $\mu^{d-1}\gamma_1\gamma_2$. $\endgroup$ Apr 15, 2022 at 20:48

1 Answer 1


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.

  • $\begingroup$ I should have thought of this. Looking at such a covering is how you show it doesn't admit a complex structure. Thanks! $\endgroup$ Apr 15, 2022 at 0:23

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