# What are some geometric reasons why a Dirac operator would have a gap in its spectrum?

My question is motivated by the following well-known computation. Let $M$ be an even dimensional Riemannian spin manifold and let $D$ be the spinor Dirac operator on $M$. Lichnerowicz showed that $D^2 = \nabla^* \nabla + \kappa/4$ where $\nabla$ is the spin connection on the spinor bundle and $\kappa$ is the scalar curvature of $M$. It is not hard to show that $\nabla^* \nabla$ is a positive operator, and thus if $\kappa > 0$ there is an interval containing $0$ in the real line which avoids the spectrum of $D^2$ (and therefore $D$). A corollary is the well-known fact that the Fredholm index of $D$ vanishes if $M$ has positive scalar curvature.

This example is a starting point for the entire theory of positive scalar curvature obstructions. The machinery involved gets more sophisticated, but in the end all one really needs about positive scalar curvature metrics on spin manifolds is the fact that they create a gap around 0 in the spectrum of the spinor Dirac operator.

So I am wondering if there are other geometric causes for gaps in the spectrum of a Dirac operator. Note that I do not want to restrict my attention to the spinor Dirac operator; I think the question is particularly interesting for the signature operator or the Dolbeault operator, for example. I am aware that elliptic operator theory of this sort helps produce a proof of the Weyl character formula, so conceivably the sort of answer that I'm looking for could come from representation theory.

Any ideas?

-
Somehow you restricted this discussion to first-order operators of Dirac type, but maybe you are interested in the "zero-in-the-spectrum"-question of Gromov? I also think that a result by Gromov/Lawson says that the gap exists even if the scalar curvature is only uniformly positive outside of a compact subset. That's in their "long article". –  Daniel Pape Feb 28 '11 at 1:29
Hi Daniel! I wasn't thinking about the zero in the spectrum conjecture when I wrote the question, but I agree that it sounds quite related to the sort of thing that I'm asking about. I seem to recall that there are counterexamples to the conjecture (and therefore spectral gaps) if one asks about general complete Riemannian manifolds rather than the universal cover of an aspherical manifold. Do you know anything further about this subject? –  Paul Siegel Feb 28 '11 at 22:33
Regarding the other part of your comment, it is my understanding that most of the non-compact elaborations on spectral gaps due to PSC can still be reduced to the Lichnerowicz formula by introducing large-scale (coarse) geometry, including the one that you mentioned. Still, it would be interesting if there are other mechanisms other than the Lichnerowicz formula which explain the spectral gap. –  Paul Siegel Feb 28 '11 at 22:39
zero in the spectrum: Farber-Weinberger ams.org/mathscinet-getitem?mr=1847591 One familiar case is for the signature operator (or the DeRham) operator twisted by a flat (unitary) connection then the kernel is identified via the Hodge theorem with cohomology, so if the corresponding cohomology vanishes, there is a gap in the spectrum near zero for any riemannian metric. –  Paul Mar 1 '11 at 1:52

## 1 Answer

The area in dimension $2$ for the sphere (with abitrary metric): In C. Bär: Lower eigenvalue estimates for Dirac operators (Math. Ann. 293) it was shown that $$\lambda^2 area\geq 4\pi.$$ This can be generalized to higher genus surfaces if one restricts to eigenvalues with higher multiplicity.

-