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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.