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Consider $(S^1 \times \Sigma^2, g)$, where $g$ is any Riemannian metric on the compact and closed $3$-manifold $S^1 \times \Sigma^2$.

Question: Does there always exist a nowhere vanishing harmonic $1$-form on $S^1 \times \Sigma^2$? If the answer to this question is No, how about the generalisation to $k$-parameter families of metrics?

So far I tried to find an example of a harmonic $1$-form on $T^3=S^1 \times S^1 \times S^1$ that does have a zero but did not succeed.

I have cross-posted this question to: https://mathoverflow.net/questions/407340/nowhere-vanishing-harmonic-1-forms-on-3-manifolds.

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    $\begingroup$ By non-vanishing you mean nowhere-vanishing? $\endgroup$ Sep 28 '21 at 23:31
  • $\begingroup$ @TedShifrin Yes. I edited the question to reflect this. $\endgroup$
    – user505117
    Sep 29 '21 at 8:55
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    $\begingroup$ I made a bit of progress: using Calabi's characterisation of harmonic $1$-forms from mathoverflow.net/questions/319107/…, I can get harmonic $1$-forms with zeros on some manifolds. Take a some 3-manifold and a Morse function $f$ with exactly one minimum and maximum. Attach a 3-handle connecting the minimum and maximum. $f$ cannot be extended to the new manifold, but $df$ can and is intrinsically harmonic by Calabi's theorem. If $f$ had other critical points (which may or may not be the case), then $df$ has zeros. $\endgroup$
    – user505117
    Oct 16 '21 at 10:09
  • $\begingroup$ The answer looks to be no. A closed nowhere vanishing 1-form (you could try to generalize the Tishler's theorem) will give you a fiberation to S^1 and the 1-form will be the pull-back of the S^1 factor. It looks to me that for genus>1 Riemann surface, the fiberation is unique. Thus, you could choose any Riemannian metric to make the S^1 factor closed 1-form not harmonic, then for these metric, it doesn't have the thing you want. $\endgroup$
    – Siqi He
    Dec 20 '21 at 20:09
  • $\begingroup$ @SiqiHe I don't understand how to complete your argument. (1) The closed 1-form coming from the $S^1$-fibration may not be harmonic, but its cohomology class contains a harmonic representative. How can I say anything about its zeros? (2) if the Riemann surface has genus$>1$, there are other harmonic forms not in the homology class of the closed 1-form coming from the $S^1$-fibration. I don't know how to say anything about their zeros. $\endgroup$
    – user505117
    Dec 23 '21 at 13:53
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This is not a full answer, but I just wanted to add holonomy to what you already found and it became to long for a comment.

The Bochner Theorems say, that for a Riemann manifold $(M,g)$ with $Ric \geq 0$ any harmonic $1$-form is parallel.

In this case, if $\omega$ is harmonic and there exists $p \in M$ with $\omega_p = 0$ then $\omega = 0$ globally. That is why you cannot find non-trivial harmonic $1$-forms on $T^3$ having zeroes.

If $Ric = 0$, then $1$-forms are harmonic iff they are parallel. Then finding a nowhere vanishing harmonic $1$-form is the same as finding a nontrivial parallel vector field, which exists if and only if $$ Hol((M,g)) \subset SO(m-1) \subset SO(m).$$

So the metrics constructed in Calabi's Theorem you linked for nontrivial transitive $1$-forms with zeroes, cannot have $Ric \geq 0$ globally.

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  • $\begingroup$ Great observation! I think that in the case of $M=T^3$, we can't even have $Scal \geq 0$ globally, as that would contradict the Schoen-Yau Torus Rigidity theorem. I found this statement as Theorem 2.1 in "Conjectures on Convergence and Scalar Curvature" (arxiv.org/pdf/2103.10093.pdf) but didn't find the original reference. $\endgroup$
    – user505117
    Nov 16 '21 at 14:48

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