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

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 $$n_1=n_2=2$$.

• 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.
– mme
Nov 2, 2018 at 15:58

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