*This is a crosspost from [this MSE question](https://math.stackexchange.com/q/4395420/39599) from a year ago.*




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> 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?
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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.



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


