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I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So:

Is $\Bbb S^2 \times \Bbb S^4$ symplectic?


Edit: after some time I came back to those calculations. I had missed the obvious case $(n_1,n_2) = (2,2)$. The proof given in the answers can be adapted to show that products of the form $\Bbb S^2 \times \Bbb S^{n_2}$ for even $n_2>2$ are not symplectic. The conclusion of what happened here is the

Theorem: Let $1 \leq n_1 \leq n_2$ be natural numbers. Then $\Bbb S^{n_1}\times \Bbb S^{n_2}$ is symplectic if and only if $n_1=n_2=1$ or $n_1=n_2=2$.

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    $\begingroup$ For what it's worth: All oriented surfaces are symplectic and all products of symplectic manifolds are symplectic. So any product of tori and 2-spheres is symplectic; the content of the Kunneth calculations here are that this is all that's possible: a product of spheres which is symplectic must be even-dimensional and have no factors of dimension at least 3. $\endgroup$ – Mike Miller Nov 2 '18 at 15:58
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No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ is a symplectic form on $S^2\times S^4$. Then $[\omega]=c[\pi^*\alpha]$ for some $c\in\mathbb R^\times$. Then $[\omega^3]=c^3[\pi^*\alpha^3]=0$, contradicting the requirement that $\omega^3$ is everywhere nondegenerate.

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  • $\begingroup$ Piotr Hajlasz beat me to it. $\endgroup$ – Fan Zheng Jul 13 '18 at 4:36
  • $\begingroup$ What do you mean? $\endgroup$ – Piotr Hajlasz Jul 13 '18 at 4:38
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    $\begingroup$ @PiotrHajlasz I am just saying that you answered 1 min faster than me, with essentially the same proof :) $\endgroup$ – Fan Zheng Jul 13 '18 at 4:41
  • $\begingroup$ @Piotr Probably because you answered a couple of seconds before. The reference is very interesting, since I'm still learning the subject. I'm accepting Fan's proof since it's very straightforward. Thanks a lot you both. :) $\endgroup$ – Ivo Terek Jul 13 '18 at 4:42
  • $\begingroup$ OK. But your answer was accepted, not mine :( and upvoted by me :) $\endgroup$ – Piotr Hajlasz Jul 13 '18 at 4:42
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There is no symplectic form on $\mathbb{S}^2\times\mathbb{S}^4$. More generally, there is no symplectic form on $M\times \mathbb{S}^{2n}$, if $n>1$ and $M$ is compact, see Symplectic structures on $M\times \mathbb{S}^{2n}$ .

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