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.

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.

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.


1 Answer 1


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.


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