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I'm reading this paper and there is a step in theorem $3.1$ on pages $4$ and $5$ that I can't understand. Basically, my doubt is how to ensure that there is a potential function for the vector field $X$, i.e., a smooth function $f: M \longrightarrow \mathbb{R}$ such that $X = \nabla f$? I heard once that Frobenius' theorem enable find this potential function and I read the Frobenius' theorem in "An introduction to smooth manifolds" by John Lee, but I can't see how the Frobenius' theorem provide a potential function $f$ for the vector field $X$.

Another approach that I thought is by Poincaré's lemma: "my" answer in this topic gives details about the definition of a conformal vector field according the author of the paper that I linked in the initial of this topic. Thus, $\mathscr{L}_X g(Y,Z) = (\nabla_Y \omega) (Z) + (\nabla_Z \omega) (Y) = 2\nabla_Y \omega(Z)$, where the last equality remains because the author restrict the vector field $X$ to vector fields such that $\nabla_i X_j = \nabla_j X_i$, so $\nabla_i X_j = 2\nabla_{\partial_i} \omega(\partial_j) = 2 \lambda g_{ij}$ for some smooth function $\lambda: M \longrightarrow \mathbb{R}$. This structure and the hypothesis that the vector field is defined on a simply connected Riemannian manifold on theorem $3.1$ led me to think that I need to apply Poincaré's lemma, but, in fact, I can't apply this lemma because $\nabla_Y \omega (Z)$ is a $2$-form. I would like to know if I need to apply the Poincaré's lemma in this theorem? If I need of the Poincaré's lemma, then can you say how I can do this?

Thanks in advance!

George
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In the paper they include also the following:

Since $\nabla_k X_j$ is symmetric in $k$ and $j$ we can integrate $X_i$ locally, i.e. on any simply connected domain $\Omega \subset N$ we can find a smooth function $f$ such that $X_i = \nabla_i f$.

This is indeed how one proves the Poincare lemma for 1 one form. Indeed you are always dealing with the one form $X = X_i dx^i$. Since

$$ (dX)_{ij} = \partial _i X_j - \partial _j X_i = \nabla _j X_i - \nabla _i X_j = 0,$$

$X$ is a closed one form and thus $X = df$ for some smooth function $f$.

Arctic Char
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