2
$\begingroup$

Let $(M^3,g)$ be a compact, connected and oriented Riemannian $3$-manifold with boundary. For a harmonic map $u : M \to \mathbb{S}^1$ satisfying Neumann condition along $\partial M$, let $h = u^*(d \theta)$, so that $d h = 0$, $\delta h = 0$ and $h(\nu) = 0$, where $\nu$ is the unit exterior normal to $\partial M$. Let $i : \partial M \to M$ be the inclusion.

Denote by $J$ the complex structure of $\partial M$. Let $J(i^\ast h)$ denote the $1$-form dual to the vector field $JX$, where $X$ is the vector field dual in $\partial M$ dual to $i^\ast h$.

My question is: does it hold that $J(i^\ast h)$ is a closed $1$-form in $\partial M$? Is it harmonic?

Why is it relevant? If $S$ denotes a connected component of a regular level set of $u$, then the vector field $JX$ as above is tangent to $\partial S$ and nonzero along this set. So,

$$ \int_{\partial S} J(i^\ast h) \neq 0,$$

thus $\partial S$ represents a nontrivial class in $H_1(\partial M)$ provided that $J(i^\ast h)$ is a closed $1$-form.

$\endgroup$
1
  • 1
    $\begingroup$ I'm not sure but if I remember correctly, an isometric immersion pulls harmonic $1$-forms back to harmonic $1$-forms iff it is minimal. If that is the case for $\partial M$, and $J$ is compatible with $g$, then it seems that $J(i^*h)$ is harmonic, since $i^*h$ is harmonic and $J$ preserves the space of harmonic forms. (Also hello Eduardo, hope everything is fine at IME) $\endgroup$
    – Caramello
    Commented May 10, 2020 at 22:56

0

You must log in to answer this question.