# Vanishing differential forms in cohomology

Let $$X$$ be a smooth differentiable manifold. Consider on $$X$$ a closed $$p$$-form $$\eta$$ and a closed $$q$$-form $$\omega$$, which have associated cohomology classes $$[\eta] \in H^p(X)$$ and $$[\omega] \in H^q(X)$$.

Now assume their wedge product is zero in cohomology $$[ \eta \wedge \omega ] = 0 \in H^{p+q}(X)$$. My question is:

Is it always possible to find cohomologically equivalent elements $$\eta' \in [\eta]$$ and $$\omega' \in [\omega]$$ such that $$\eta' \wedge \omega' = 0$$ (i.e. such that the wedge product is genuinely zero, not only in cohomology)?

Naively one needs to determine whether the exact form $$\mathrm{d}\xi$$ making $$\eta \wedge \omega + \mathrm{d}\xi = 0$$ can always be written in the form $$(\eta + \mathrm{d}\alpha) \wedge (\omega + \mathrm{d}\beta) - \eta \wedge \omega$$ for some $$\alpha$$ and $$\beta$$. But this seems a difficult question, so I am wondering if there is a better argument.

Let me describe the basic setting. Assume we are given three cohomology classes $$[x] \in H^k(X), [y] \in H^l(X), [z] \in H^m(X)$$ such that $$[x \wedge y] = [y \wedge z] = 0$$. Then one can define a cohomology class $$\left< [x], [y], [z] \right> \in H^{k+l+m-1}(X) / \left( [x] \cdot H^{l+m-1}(X) + [z] \cdot H^{k+l-1}(X) \right)$$ by the following construction: Choose $$\lambda$$ such that $$x \wedge y = d \lambda$$ and $$\mu$$ such that $$y \wedge z = d \mu$$ and set $$\left< [x], [y], [z] \right> = \left[ \lambda \wedge z - (-1)^k x \wedge \mu \right].$$
This operation is called the Massey triple product and one can check that it is independent of the representatives chosen for the cohomology classes $$[x],[y],[z]$$.
Now assume that $$[\eta] \in H^1(X), [\omega] \in H^1(X)$$ such that $$[\eta \wedge \omega] = 0$$. Then we also have $$[\eta \wedge \eta] = 0$$ (as $$\eta$$ is a one-form) so the Massey triple product $$\left< [\eta],[\eta],[\omega] \right>$$ is well-defined. If one can choose representatives $$\eta' \in [\eta], \omega' \in [\omega]$$ with $$\eta' \wedge \omega' = 0$$ then working with $$\eta', \omega'$$ instead of $$\eta,\omega$$, we can take $$\lambda = \mu = 0$$ and then $$\eta' \wedge \eta' = 0 = d\lambda, \eta' \wedge \omega' = 0 = d\mu$$ and so $$\left< [\eta], [\eta], [\omega] \right> = \left< [\eta'], [\eta'], [\omega'] \right> = \left[ 0 \wedge \omega' + \eta' \wedge 0 \right] = [0].$$
Hence, any manifold $$X$$ in which there exists $$[\eta], [\omega] \in H^1(X)$$ with $$[\eta \wedge \omega] = 0$$ and $$\left< [\eta], [\eta], [\omega] \right> \neq 0$$ will give you a counterexample. For example, one can take $$X$$ to be an $$S^1$$-bundle over $$\mathbb{T}^2$$ whose Euler class is one. For the explicit construction, calculation and more details about the triple Massey product, I refer you to pages 136-137 in Morita's "Geometry of Differential Forms".