5
$\begingroup$

Let $M$ be a (compact) manifold and $\lambda$ a closed 1-form on $M$. Under which condition on $\lambda$ does there exists a non-trivial (complex) function $F$ such that

$$dF = F\lambda \quad \quad ?$$

A couple of remarks :

  • If the equation above has a solution $f$ and $g$ is some function on $M$, then $e^g f$ is a solution to the equation

$$dF = F(\lambda+dg),$$

so any characterization only depends on the cohomology class of $\lambda$.

  • Since this equation can be written locally as $d \ln (F) = \lambda$, the condition that $\lambda$ be closed is natural.
$\endgroup$
1
$\begingroup$

Your question is about what is called Morse-Novikov cohomology. Namely, if you have a manifold $M$ and a closed one-form $\lambda\in\Omega^{1}(M)$, then the following operator is a differential: $$ d_{\lambda}:\Omega^{k}(M)\rightarrow\Omega^{k+1}(M):\beta\mapsto d\beta+\lambda\wedge\beta, $$ i.e. $d_\lambda \circ d_\lambda=0$. The associated cohomology groups $H_{\lambda}^{\bullet}(M)$ are called Morse-Novikov cohomology groups. They arise for instance in the context of locally conformal symplectic (lcs) structures: the form $\lambda$ is then the Lee form, which is defined in terms of the lcs structure.

So you are basically asking what is known about the zeroth Morse-Novikov cohomology group $H_{-\lambda}^{0}(M)$, since $$ dF=F\lambda \Leftrightarrow F\in H_{-\lambda}^{0}(M). $$

I know of 2 interesting statements about this group:

1) Related to your first remark: the groups $H_{\lambda}^{k}(M)$ only depend on the cohomology class $[\lambda]\in H^{1}(M)$. Indeed, if $\lambda'=\lambda+dg$, then we get an isomorphism $$ H_{\lambda'}^{k}(M)\rightarrow H_{\lambda}^{k}(M):[\beta]\mapsto [e^{g}\beta]. $$ In particular, if $\lambda=dg$ is exact then $H_{\lambda}^{0}(M)\cong H^{0}(M)$ as $$ H_{\lambda}^{0}(M)=\{f/e^{g}:f\ \text{constant on connected components of}\ M\}. $$

2) If $\lambda$ is not exact and $M$ is connected, then $H^{0}_{\lambda}(M)=0$. This can be done by showing that, if $F\in H^{0}_{\lambda}(M)$ then $F^{-1}(0)$ is nonempty and both open and closed. It is clearly closed, and it is nonempty since $\lambda$ is not exact (as demonstrated in your second remark). To show that it is open, choose $x\in F^{-1}(0)$ and let $U$ be a neighborhood of $x$ on which $\lambda=dg$ is exact. Setting $h:=e^{g}$, we then have $$ 0=dF|_{U}+F|_{U}dg=dF|_{U}+F|_{U}d(\ln(h))=dF|_{U}+F|_{U}\frac{1}{h}dh. $$ Multiplying by $h$, we get that $d(F|_{U}h)=0$, i.e. $F|_{U}h$ is constant on $U$. But $F(x)=0$ so that $F|_{U}h\equiv 0$. As $h$ is nowhere zero, this implies that $F|_{U}\equiv 0$. So $U$ is an open neighborhood of $x$ that is contained in $F^{-1}(0)$, showing that $F^{-1}(0)$ is open.

These statements, and much more, can be found in the literature. For instance, see Section 2 of https://arxiv.org/pdf/1904.09759.pdf

$\endgroup$
  • $\begingroup$ Thank you very much for the references ; with the right keywords, I'll surely find out what I want ! A small caveat: the vanishing of $H^0_\lambda$ holds for real-valued $\lambda$, and is no longer true for complex-valued $1$-form (takes for $F$ trigonometric monomials on a torus). $\endgroup$ – D. Thomine Nov 8 at 18:57

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.