This looks elementary, but somehow I am stuck, please bear with me:


Let $H$ be a differential 3-form, nowhere vanishing, but not necessarily closed. What is a sufficient condition that the sequence of forming wedge products with $H$ is exact, i.e. that


$$
 \mathrm{ker}(H \wedge(-))\,/\,\mathrm{im}(H \wedge(-)) \;=\; 0
$$


?


A necessary condition is clearly that $H \neq \alpha\_1 \wedge \beta\_2$ for a 1-form $\alpha\_1$. When is this sufficient?


If $H$ is closed, this is asking for vanishing "$H$-cohomology" in the terminology of [arXiv:0501406](https://arxiv.org/abs/math/0501406).


I am really interested in the variant where $H$ is not a differential form, but an element of bi-degree $(3,\mathrm{even})$ in the Chevalley-Eilenberg algebra of a super Lie algebra, specifically the element in (28) of [hep-th/0406020](https://arxiv.org/abs/hep-th/0406020). 


But that shouldn't make make much of a difference of principle. More generally I could ask this question for any free graded-commutative algebra in characteristic zero.



Vanishing of H-cohomology
I have not thought about the variant problem, but in the ordinary differential form case, I do not believe it is possible to have vanishing $H$-cohomology on a (finite-type) smooth manifold $M$, regardless of any conditions on $H$ such as nowhere vanishing, closed, etc.


We proceed by contradiction, supposing that the $H$-cohomology does vanish on $M$. Note that as a differential form, one can write $H$ as a finite sum $H = \sum\_{i=1}^{k}\beta\_i \wedge \alpha\_i$ (one establishes finiteness via a partition of unity argument on a finite covering by coordinate patches). Our key claim is the following:


**Claim:** For any sequece $1 \leq i\_1 < \ldots < i\_{\ell} \leq k$, there exist differential forms $\gamma\_{i\_1\cdots i\_{\ell}} \in \Omega^{k-3-\ell}(M)$ (in particular they are zero if $\ell > k-3$) such that the differential forms $\alpha\_{\widehat{i\_1\cdots i\_{\ell}}} := \alpha\_1 \wedge \cdots \wedge \widehat{\alpha\_{i\_1}} \wedge \cdots \wedge \widehat{\alpha\_{i\_{\ell}}} \wedge \cdots \wedge \alpha\_k$ satisfy $$\alpha\_{\widehat{i\_1\cdots i\_{\ell}}} = (-1)^{i\_1+\cdots+i\_{\ell}+\ell} \left(H \wedge \gamma\_{i\_1\cdots i\_{\ell}} + \sum\_{j=1}^{\ell}(-1)^{j+1}\beta\_{i\_j}\wedge \gamma\_{i\_1\cdots\widehat{i\_j}\cdots i\_{\ell}}\right).$$


**Remark on Notation:** If $\ell = 0$, then we just take $\alpha\_{\widehat{\emptyset}} = \alpha\_1 \wedge \cdots \wedge \alpha\_k$. Similarly, we take $1 = \alpha\_{\widehat{1,2,\ldots,k}}$.


**Claim implies result:** For the case of $1 = \alpha\_{\widehat{1,2,\ldots,k}}$, we see that the right-hand side of the equation is just $0$ since the the corresponding $\gamma$ terms are in negative degree. This yields a contradiction. (One could also have taken $\ell = k-1$ instead of $k$ and assumed that the decomposition of $H$ had minimal $k$.)


**Proof of claim:** We proceed by induction on $\ell$, beginning with the case of $\ell = 0$. We see that $$H \wedge (\alpha\_1 \wedge \cdots \wedge \alpha\_k) = 0,$$ and so by vanishing of $H$-cohomology, indeed we find $\alpha\_{\widehat{\emptyset}} = H \wedge \gamma\_{\emptyset}$.


Now suppose that we have that the formula holds for all $\ell < \ell\_0$. Then
\begin{eqnarray\*}
H \wedge \alpha\_{\widehat{i\_1\cdots i\_{\ell\_0}}} &=& \sum\_{i=1}^{k} \beta\_i \wedge \alpha\_i \wedge \alpha\_{\widehat{i\_1\cdots i\_{\ell\_0}}} \\
&=& \sum\_{j=1}^{\ell\_0} (-1)^{i\_j+j}\beta\_{i\_j} \wedge \alpha\_{\widehat{i\_1\cdots \widehat{i\_j}\cdots i\_{\ell\_0}}} \\
&=& \sum\_{j=1}^{\ell\_0} (-1)^{i\_1 + \cdots + i\_{\ell\_0}+\ell\_0+j+1}\beta\_{i\_j} \wedge \\ && \left(H \wedge \gamma\_{i\_1\cdots \widehat{i\_j}\cdots i\_{\ell\_0}}+\sum\_{k=1}^{j-1}(-1)^{k+1}\beta\_{i\_k}\wedge \gamma\_{i\_1\cdots \widehat{i\_k} \cdots \widehat{i\_j} \cdots i\_{\ell\_0}} + \sum\_{k=j+1}^{\ell\_0}(-1)^k\beta\_{i\_k}\wedge \gamma\_{i\_1\cdots \widehat{i\_j} \cdots \widehat{i\_k} \cdots i\_{\ell\_0}}\right) \\
&=&(-1)^{i\_1 + \cdots + i\_{\ell\_0}+\ell\_0}\sum\_{j=1}^{\ell\_0} H \wedge (-1)^{j+1}\beta\_{i\_j} \wedge \gamma\_{i\_1\cdots \widehat{i\_j}\cdots i\_{\ell\_0}}
\end{eqnarray\*}


By vanishing of $H$-cohomology, the inductive step now follows.


