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$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$

Let $\M,\N$ be closed $d$-dimensional oriented Riemannian manifolds. Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative.

Let $1 \le k \le d$ be fixed. Consider the following two properties $f$ can have:

  1. $\delta^{\N} \omega=0 \Rightarrow \delta^{\M}(f^*\omega)=0$ for every $k$-form $ \omega \in \Omega^k(\N)$.

$\,\,\,\,$ 2. $\omega\in \Omega^k(\N)$ is harmonic $\Rightarrow f^*\omega$ is harmonic.

Question: Does property 2 implies property 1?

(Property 1 certainly implies property 2, since a form is harmonic if and only if it's closed and co-closed, and closedness of forms is preserevd automatically, by any smooth map).

Comment: A map which satisfies property $2$ for $k=1$, is called harmonic morphism.

I suspect the answer is negative, since the space of harmonic forms is finite-dimensional. Thus property $2$ gives us a "finite-dimensional" information about $f$, while the requirement of property $1$ is on the much larger space $\text{ker} \delta$, which is infinite dimensional.

Asaf Shachar
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  • Just to make sure - are $\mathcal{M}$ and $\mathcal{N}$ boundaryless? – Amitai Yuval May 02 '18 at 15:01
  • To be honest, I didn't think about that. By the adjoint here I refer to $\delta=\pm \star d \star$ This is indeed the "true" adjoint when the manifolds are compact and boundaryless. (Otherwise, we probably should restrict to test the "adjointness" to pairs of compactly supported forms, were at least one is zero on the boundary, I guess). – Asaf Shachar May 02 '18 at 15:11
  • I don't care so much about the terminology. What bothers me more is that on a manifold with boundary (or on a non-compact manifold, actually) it is not true that a harmonic form is closed and coclosed. – Amitai Yuval May 02 '18 at 15:13
  • Hmmm... Yeah you are probably right. Then let's assume that the manifolds are closed. (I am curious about your last remark, but we can probably discuss this at a more convenient place). – Asaf Shachar May 02 '18 at 15:26

1 Answers1

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Since you've phrased the question in terms of global forms, the answer is definitely no. For example, let $M=N=\mathbb S^2$. Then by the Hodge theorem, there are no nontrivial harmonic $1$-forms, so every smooth map $f\colon \mathbb S^2\to\mathbb S^2$ satisfies property 2 in the case $k=1$. But not every such $f$ satisfies property 1. (I was thinking that $f\colon \mathbb S^2\to\mathbb S^2$ satisfies property 1 for $k=1$ if and only if it's holomorphic or antiholomorphic, but I don't believe that's true.)

Jack Lee
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  • Thanks. Can you say something (even heuristic) for why do you think property $(1)$ (which is $\delta^{\mathbb{S}^2} \omega=0 \Rightarrow \delta^{\mathbb{S}^2}(f^*\omega)=0$ for every one-form $\omega$) implies $f$ being holomorphic or anti-holomorphic? (I know that this is equivalent to $f$ being conformal, and that conformal maps do satisfy property $(1)$ in general, but I wondered if you had an idea for the converse direction). – Asaf Shachar May 09 '18 at 08:24
  • I take it back -- I no longer believe it's true. What I was vaguely thinking was that your condition 1 was equivalent to preserving the Hodge star operator up to sign, and for maps between open subsets of $\mathbb C$, that's equivalent to being conformal. But your condition is weaker than that, and it's not even clear that the stronger condition implies conformality on a non-flat Riemann surface. Sorry to lead you astray. – Jack Lee May 09 '18 at 22:01
  • Thanks. By the way, the preservation of the Hodge star does imply conformality, even in the non-flat case. (This is in fact a pointwise linear algebra statement- you can see here. But you are right that the stronger condition does not imply Hodge star preservation). – Asaf Shachar May 10 '18 at 08:23
  • Ah, right. Thanks. I see that I was making a stupid calculation error. – Jack Lee May 10 '18 at 17:15