For a (compact) spin manifold, we know that the eigenvalues $\lambda_n$ of the Dirac operator are countable, with finite multiplicity, and satisfy $$ |\lambda_n| \to \infty, ~~~ \text{ as } n \to \infty. $$ This can be concluded, for example, from the fact that they have compact resolvent, as established in Friedrich's book on Dirac operators in Chapter 4.2.

I am wondering for the gap between the eigenvalues, as we tend to infinity, will it become as large as we want, or at least is there a minimum distance between succesive eigenvalues.

(Honestly, I care most about Hermitian manifolds that are spin, so if it is easier in this case, please let me know!)

  • $\begingroup$ Welcome to MathOverflow! Could you be a bit more explicit concerning your statement "the eigenvalues [...] tend to infinity since it is an elliptic differential operator"? As this is currently worded it seems to be a rule of thumb rather than a mathematical theorem. Could you, maybe, add a reference or a detailed statement of the result that you use here? $\endgroup$ Jul 21, 2019 at 16:24
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    $\begingroup$ Thanks for suggestions. I have edited to make more clear. $\endgroup$ Jul 21, 2019 at 17:25
  • $\begingroup$ In the case of a round sphere, the eigenvalues of the Dirac operator have distance 1, which means that the gaps for the Laplace operator (between $\lambda^2$ and $(\lambda+1)^2$) indeed tend to infinity. But to the best of my knowledge, this is very far from the generic case. $\endgroup$ Jul 21, 2019 at 17:27
  • $\begingroup$ But for the easier question, in general, will there a minimal bound between eigenvalues? Or is there an example where eigenvalues can be as close as we want? $\endgroup$ Jul 21, 2019 at 17:48
  • $\begingroup$ If for some $n \in \mathbb{N}$ the $n$-the power of the resolvent of the Laplace operator maps $L^2$ into $L^\infty$, then we can conclude that the $n$-th power of the resolvent is a Hilbert-Schmidt operator - which gives you some information on how far the eigenvalues tend to $\infty$ (but I don't know if the resolvent behaves that smoothly in your setting). $\endgroup$ Jul 21, 2019 at 18:07

1 Answer 1


Edit: In Eigenvalues of the dirac operator, M. Atiyah, 1984, the following results are stated and proved. Let $M$ be a $d$-dimensional closed spin manifold, and $D$ the Dirac operator associated to a Hermitian bundle $V$ with a connection over $M$. Let $\lambda_k$ be its eigenvalues, ordered by absolute value and increasing. The following results are named after their version in the above paper.

Theorem 2.

There exists a constant $C$ independent of $V$ such that $$|\lambda_k|\leq Ck^{1/d}.$$

Theorem 1*. If $d$ is odd, there exists a constant $C$ independent of $V$ such that every interval of length $C$ contains an eigenvalue.

Theorem 2*. If $d$ is odd, there exists a constant $C$ independent of $V$ such that every interval of length $Ck^{1/d}$ contains at least $k$ eigenvalues.

About the case of even-dimensional manifolds, Atiyah gives an argument showing that such a constant would have to depend on $V$, but I don't think it rules out similar theorems for fixed $V$.

Out of transparency, here is my previous answer, which in the odd case is of course way worse. It seems classical, although I cannot seem to be able to find a reference with a complete proof (Atiyah writes a sentence about it in the above article), that the Dirac operator on a closed spin manifold $M$ of dimension $d$ satisfies a Weyl type law: $$\frac{N(\lambda)}{|\lambda|^d}\to C$$ for some known constant $C>0$ ($C_d\mathrm{Vol}(M)$, for $C_d$ a constant depending only on $d$), where $N(\lambda)$ is the number of eigenvalues with absolute value less than $\lambda$ (counted with multiplicity). I claimed before that it implied that the gap between two consecutive eigenvalues cannot be bounded below. In fact, we get $$\liminf_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{k^{1/d-1}}<\infty$$ so for $d\geq2$ the gaps cannot be bounded below at least on one side.

However it does not rule out large gaps; it only rules out relatively large gaps in absolute value in the sense that $$\limsup_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{\lambda_k} =\limsup_{k\to\infty}\frac{|\lambda_{k+1}|-|\lambda_k|}{k^{1/d}}=0.$$ Again, I don't know if there exists Dirac operators on even-dimensional manifolds with arbitrarily large eigengaps.

A few references: Weyl laws on open manifolds (S. Moroianu, 2003), claims that this law is well-known in the case of closed manifolds, and that the case of compact manifolds with boundary is treated in Spectral Asymmetry and Riemannian Geometry. I (M. F. Atiyah, V. K. Patodi and I. M. Singer, 1975). It is also possible that the result of Moroianu holds trivially in the compact case, see Theorem 3. The more recent work The Dirac spectrum on manifolds with gradient conformal vector fields (A. Moroianu and S. Moroianu, 2007) states in the introduction that

[The eigenvalues of the Dirac operator on a closed spin manifold] grow at a certain speed determined by the volume of the manifold and its dimension.

which I read as “the Weyl law holds for the Dirac operator on compact spin manifolds.” I bet someone with more experience with Dirac operators than myself will be able to find the missing pieces.


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