# Global existence of solutions to a differential equation

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.

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 $$dF=F\lambda \Leftrightarrow F\in H_{-\lambda}^{0}(M).$$

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.
• 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). – D. Thomine Nov 8 at 18:57