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Does anyone know an example of a closed spin hyperbolic manifold of dimension 3 or greater such that the kernel of the Dirac operator is non-trivial?

I'm mainly interested in the 3-dimensional case but would be happy to hear about higher dimensions as well. Note that Hitchin showed that for a particular choice of spin structure on a surface, the Dirac operator necessarily has kernel. There are strong results of Bär (The Dirac Operator on Hyperbolic Manifolds of Finite Volume, J. Diff. Geom. 54 (2000), 439--488) for finite-volume hyperbolic manifolds but I haven't found anything for closed hyperbolic manifolds.

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  • $\begingroup$ By the way, in an earlier paper of Baer, he shows that every 3-manifold has a metric which has harmonic spinors. Of course, it is not clear that the hyperbolic metric is it... $\endgroup$
    – Igor Rivin
    Commented Apr 17, 2015 at 17:46
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    $\begingroup$ @MatthiasLudewig As far as I recall Pontrjagin classes of a hyperbolic manifold are trivial in real cohomology, using Chern-Weil theory. (I don't have a handy reference, but compare Igor Belgradek's comments in mathoverflow.net/questions/107716/…). This implies that the A-hat class is zero, so the index of the Dirac operator is 0. That's kind of the point of the question, to find examples where the kernel is non-zero in a setting where the A-S index theorem doesn't help. $\endgroup$
    – Danny Ruberman
    Commented May 18, 2015 at 14:54
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    $\begingroup$ I spent a couple of minutes looking for a reference, and found that it comes from Chern's paper, On Curvature and Characteristic Classes of a Riemann Manifold. Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg October 1955, Volume 20, Issue 1-2, pp 117-126. Of course, this predates the Dirac operator and the index theorem, but the characteristic class computation is there. $\endgroup$
    – Danny Ruberman
    Commented May 19, 2015 at 16:54
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    $\begingroup$ I know nothing about harmonic spinors, but my suggestion is to talk to people who study automorphic forms. There is a neat trick (using theta-series and congruence subgroups) for constructing compact hyperbolic manifolds supporting nonvanishing automorphic forms of certain kind, which may work in your setting. I learned this trick from Gordan Savin years ago, see Proposition on page 204 of his paper "Cusp forms", Israel Math Journal, 1992. You have to get lucky for this trick to work with harmonic spinors since it requires absolute convergence, but it's worth asking. $\endgroup$
    – Misha
    Commented Nov 24, 2015 at 18:58
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    $\begingroup$ I have a feeling that you can construct examples, and even an idea for a construction. I tried to get a student to follow up on this plan, but, well, kids these days... I'd be happy to share my idea off-line if you're interested. I also think that the suggestion from @Misha is a good one. In any event the question is still open. $\endgroup$
    – Danny Ruberman
    Commented Apr 12, 2016 at 20:29

2 Answers 2

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I'm happy to be able to answer my own question! John Ratcliffe, Steven Tschantz and I showed that the Dirac operator on the Davis manifold (a closed hyperbolic 4-manifold constructed by Mike Davis) has non-zero kernel. This is in the preprint Harmonic spinors on the Davis hyperbolic 4-manifold that we posted on the arxiv today.

The method is to find a symmetry of the manifold that respects a particular spin structure, and use the G-spin theorem to show that the G-spinor index is non-zero. This can only happen if the kernel of the Dirac operator is non-zero. The method also works in dimension 2 (where the result is of course not new) and in principle in any even dimension. It doesn't (to my knowledge) work in odd dimensions, although Misha's suggestions in the comments above seem promising there.

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  • $\begingroup$ Is the Davis manifold another non-trivial contractible manifold but not homeomorphic to $\mathbb{R}^{4}$, like the Whitehead manifold? I feel kind of amazed. $\endgroup$
    – Bombyx mori
    Commented Mar 21, 2018 at 3:20
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    $\begingroup$ @Bombyxmori No, it is nothing like the Whitehead manifold - look at the link in Danny's answeer. $\endgroup$
    – Igor Rivin
    Commented Mar 21, 2018 at 3:36
  • $\begingroup$ @IgorRivin: I see. I was told by Tam it is something similar. $\endgroup$
    – Bombyx mori
    Commented Mar 21, 2018 at 3:54
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    $\begingroup$ @Bombyxmori: Same Davis, different manifold. The ones you are thinking of are closed aspherical manifolds of dimension $\geq 4$ whose universal covers are not simply connected at $\infty$. Davis, Michael W. Groups generated by reflections and aspherical manifolds not covered by Euclidean space. Ann. of Math. (2) 117 (1983), no. 2, 293-324. $\endgroup$
    – Danny Ruberman
    Commented Mar 21, 2018 at 13:06
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According to Rula Tabbash in this paper, all negatively curved closed manifolds admit harmonic spinors (and no positively curved ones do). I am not 100% certain, since this is a physics paper, and so it is not clear if he is assuming something unstated...

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    $\begingroup$ I'm not sure if it's polite to comment negatively on specific papers, but I feel skeptical of the one you cite. There is no argument for the existence of harmonic spinors in negative curvature, other than the statement of the well-known theorem of Lichnerowicz (the converse of his claim!) that positive scalar curvature obstructs harmonic spinors. Even for surfaces, the claim that all negatively curved closed manifolds admit harmonic spinors is false, by results of Hitchin and Ammann-Dahl-Humbert. $\endgroup$
    – Danny Ruberman
    Commented Apr 17, 2015 at 14:52
  • $\begingroup$ @DannyRuberman Yes, precisely. I didn't know about the results you mention, but certainly the "Bochner" obstruction does not seem to show a positive result. $\endgroup$
    – Igor Rivin
    Commented Apr 17, 2015 at 14:54

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