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 1


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.

  • $\begingroup$ Nice! Could you further explain why $v + p \circ f$ is harmonic, please? $\endgroup$ Mar 11, 2021 at 19:32
  • 1
    $\begingroup$ @EduardoLonga I explained it a bit more in an edit; let me know, if it remains unclear. $\endgroup$
    – user_1789
    Mar 12, 2021 at 12:33

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