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:



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



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> **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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Is $\Bbb S^2 \times \Bbb S^4$ symplectic?
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.


