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

Question

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?

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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

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 $n_1=n_2=2$.

Answer 1

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.

Answer 2

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 $M\times \mathbb{S}^{2n}$ .