I was trying to prove that $S^3 \times S^1$ is not a symplectic manifold but got stuck. I consider reasoning by contradiction as follows:
Suppose that there is a symplectic form $\omega$ on $S^3 \times S^1$. That is, $\omega$ is a 2 form such that $\mathrm{d}\omega = 0$ and $\omega \wedge \omega \neq 0$. Now, I figure since $\omega \wedge \omega$ is a non-zero 4-form on $S^3 \times S^1$, it is a volume form, but I don't know how proceed, I was thinking I could try to do something along the lines of Proof that $S^4$ is not a symplectic manifold, by computing the de Rham cohomology using the Kunneth formula, but this didn't seem to simplify anything since it seems that that $H_{dR}^4(S^3 \times S^1) \neq \{ 0\} $ (If I have computed this properly). Are any of you able to provide me with a hint? Thanks.
Edit: (Question regarding comment) I think that the Kunneth formula states that: $$H^k(S^3 \times S^1) \cong \sum_{k=i+j}H^i(S^3) \otimes H^j(S^1) $$ Using: $$H^i(S^n) \cong \begin{cases} \mathbb{R} & i=0,n \\ 0 & \mathrm{else} \end{cases} $$ I get: $$H^2(S^3 \times S^1) \cong (\mathbb{R} \otimes 0) \oplus (0 \otimes \mathbb{R}) \oplus( 0 \otimes \mathbb{R}) $$ I this correct?
