# 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.

No, this is not always possible. One can use differential forms to define higher order cohomology operations called Massey products and if they don't vanish, then you have an obstruction for the possibility to choose representatives with zero wedge product.

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".