I'm trying to verify whether or not $M:=\mathbb{T}^3\times \mathbb{S}^3$ is a symplectic manifold.
Here is my attempt: first, I've noticed that $\mathbb{S}^1\times \mathbb{S}^3$ is not symplectic, since $H^2(\mathbb{S}^1\times \mathbb{S}^3)=0$. Second, $\mathbb{T}^2$ is symplectic and $M=\mathbb{T}^2\times (\mathbb{S}^1\times\mathbb{S}^3)$.
I don't know if this is possible, but I'd like to argue that if both $M$ and $\mathbb{T}^2$ are symplectic, then $\mathbb{S}^1\times\mathbb{S}^3$ must be symplectic (which is absurd), so $M$ cannot be symplectic.
What I have in mind is the following fact from symplectic linear algebra: if $(V,\omega)$ is a symplectic vector space and $W$ is a symplectic subspace (i.e., $(W,\omega|_W$) is symplectic), then $V=W\oplus W^\omega$ and $W^\omega$ is also symplectic.
In this case, fixing $p\in\mathbb{T}^2$ and $q\in\mathbb{S}^1\times\mathbb{S}^3$, we have $T_{(p,q)}M=T_p\mathbb{T}^2\oplus T_q(\mathbb{S}^1\times\mathbb{S}^3)$. Since $\mathbb{T}^2$ is symplectic, I feel like I'm getting close, but I don't know how to finalize it.
Does this strategy work? Any suggestion is welcome, thanks!
Suppose that it is symplectic, by Kunneth, $H^2(T^3\times S^3)$ is isomorphic to $H^2(T^3)$. Thus a symplectic form $w$ verifies $w^3=0$ contradiction.