# Nowhere vanishing harmonic 1-forms on 3-manifolds

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.

• By non-vanishing you mean nowhere-vanishing? Sep 28, 2021 at 23:31
• @TedShifrin Yes. I edited the question to reflect this. Sep 29, 2021 at 8:55
• 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. Oct 16, 2021 at 10:09
• 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. Dec 20, 2021 at 20:09
• @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. Dec 23, 2021 at 13:53

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.
• 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. Nov 16, 2021 at 14:48