# Tangential harmonic $1$-forms are pullbacks of harmonic functions

This question has also been posted on MSE, but maybe here is the right place to obtain an answer.

Let $$(M^3,g)$$ be a compact connected oriented Riemannian $$3$$-manifold with nonempty boundary. The Hodge-de Rham Theorem says that there is an isomorphism between $$H^1_{dR}(M)$$, the first de Rham cohomology group of $$M$$, and the space $$\mathcal{H}_N^1(M)$$ of tangential harmonic $$1$$-forms on $$M$$, given by

$$\mathcal{H}_N^1(M) = \{ \omega \in \Omega^1(M) : d \omega = 0,\, d^\ast \omega=0 \text{ and } \omega(\nu) = 0 \text{ on }\partial M\},$$

where $$\nu$$ is a unit normal for $$\partial M$$.

Suppose that $$H^1(M; \mathbb{Z}) \neq 0$$. Then, since $$\mathbb{S}^1$$ is a $$K(\mathbb{Z},1)$$, there is a bijection

$$\Phi : [M, \mathbb{S}^1] \to H^1(M; \mathbb{Z})$$

given by $$\Phi([u]) = [u^\ast(d\theta)]$$, where $$d\theta \in \Omega^1(\mathbb{S}^1)$$ is the volume element of $$\mathbb{S}^1$$. Now, de Rham Theorem and the Universal Coefficient Theorem give isomorphisms

$$H_{dR}^1(M) \cong H^1(M; \mathbb{R}) \cong H^1(M;\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R}$$

How do I conclude from this information that there exists a smooth map $$u : M \to \mathbb{S}^1$$ such that $$u$$ is harmonic with Neumann condition and $$u^\ast(d\theta)\in \Omega^1(M)$$ is tangential and harmonic?

## 1 Answer

Any constant map satisfies the requirements of the final question :). More seriously, if you want to find tangential harmonic form of this type, which represents a given $$[u]\in [M,S^1]\cong H^1(M,\mathbb{Z})$$ in de Rham cohomology, then you can proceed as follows (unless I am missing something):

Pick a smooth map $$v:M\rightarrow S^1$$ representing $$[u]\in [M,S^1]$$. Then $$\sigma:=v^*d\theta$$ is in the right de Rham cohomology class, if $$\int_{S^1}d\theta=1$$. By the relative Hodge-de Rham theorem, there is an exact form $$df\in \Omega^1(M)$$ such that $$\tau:=\sigma+df$$ is tangential and harmonic. Since we can write $$df$$ as $$df=(p\circ f)^*d\theta$$ (where $$p:\mathbb{R}\rightarrow S^1$$ is a covering map with $$p^*d\theta=dt$$, with $$t$$ as the standard coordinate on $$\mathbb{R}$$), we find that $$\tau=(v+p\circ f)^*d\theta$$ has the desired form. Moreover $$v+p\circ f$$ is harmonic since $$\tau$$ is $$d^*$$-closed and it satisfies the von Neumann condition since $$\tau$$ is tangential.

Edit: As requested in the comments, here are some more details on how one concludes that $$v+p\circ f$$ is harmonic:
A map $$w:M\rightarrow S^1$$ is harmonic if and only if for all local isometries $$L:S^1\supset U\rightarrow \mathbb{R}$$ the function $$L\circ w:w^{-1}(U)\rightarrow \mathbb{R}$$ is harmonic. A real valued function $$g$$ is harmonic if and only if $$dg=g^*dt$$ is $$d^*$$-closed (as the Laplacian reduces on functions to $$d^*d$$). If $$g=L\circ w$$, then $$(L\circ w)^*dt= c w^*d\theta$$ for a constant $$c$$. Combining these equivalences with the fact that $$w^*d\theta$$ is $$d^*$$-closed for $$w=v+p\circ f$$ we find that $$v+p\circ f:M\rightarrow S^1$$ is harmonic as desired.

• Nice! Could you further explain why $v + p \circ f$ is harmonic, please? Mar 11, 2021 at 19:32
• @EduardoLonga I explained it a bit more in an edit; let me know, if it remains unclear. Mar 12, 2021 at 12:33