Question: If M is a spin manifold. Condider the dirac operator on a spinor bundle. Can the eigenvalue of this operater tend to negative infinity? If it can, can we choose a riemann matrix such that under this matrix, the eigenvalue of dirac operator does not tend to negative infinity??
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2 Answers
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Given any first order, graded, self-adjoint, elliptic operator on a manifold the spectrum is unbounded in both the positive and negative directions. There are probably elementary ways to see this, but I'll give an argument which is overly sophisticated but a bit more quantitative.
Given $D$ as above, consider the eta function associated to $D$:
$$\eta(s) = \sum_{\lambda} sign(\lambda) |\lambda|^{-s}$$
(the sum ranges over the eigenvalues of $D$)
By a variation on Weyl's asymptotic formula for the eigenvalues of the Laplacian together with some basic complex analysis, this function extends to a meromorphic function on the complex plane. It is a very deep topological fact that the eta function is actually bounded at $0$; in the case of Dirac-type operators this follows from the sort of heat kernel analysis used in the local proof of the Atiyah-Singer index theorem, and for general elliptic operators this follows from the Dirac case together with some cobordism theory.
In any event, $\eta(0)$ can be regarded as a sort of "renormalized difference" between the number of positive eigenvalues and the number of negative eigenvalues, and the fact that it is bounded means that there are infinitely many positive and negative eigenvalues.
As I said above this machinery is probably overkill as far as your actual question is concerned, but it proves something stronger: it makes precise the assertion that the cardinality of the positive part of the spectrum agrees with the cardinality of the negative part "up to a finite error".
answered Oct 29, 2011 at 17:19
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1$\begingroup$ "There are probably elementary ways". I think the following should work: It suffices to show that the Rayleigh quotient $\langle \phi, D\phi\rangle/|\phi|^2$ is doubly unbounded. Take $\phi$ to be supported in a ball where $D$ is well approximated by a constant coefficient operator. In a much smaller domain, take $\phi$ to be an eigenvector $\psi$ for the latter operator thought of as an operator with periodic boundary condition (i.e. an operator on a torus). $\endgroup$ Oct 29, 2011 at 22:56
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1$\begingroup$ In the larger ball, define $\phi$ by cutting off $\psi$, but slowly, so that the behaviour of $\psi$ dominates. This reduces the problem to the case of constant coefficient operators over the torus, for which Fourier series are available. $\endgroup$ Oct 29, 2011 at 22:57
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$\begingroup$ BTW, a good account of these facts about the eta function is given by Ken Richardson in his note "Introduction to the eta invariant" ncatlab.org/nlab/show/eta+invariant#Richardson $\endgroup$ Aug 27, 2014 at 18:52
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Can the eigenvalue of this operater tend to negative infinity?
Yes -- actually this question is closely tied to the physical origins of the Dirac operator. Very roughly speaking negative eigenvalues correspond to negative-energy eigenstates; the Dirac equation predicts that particles with positive energy (e.g. electrons, which can have arbitrarily large positive energy) there is an antiparticle "twin" with negative energy (positrons in this case).
For instance, if $M$ is the unit sphere in $\mathbb{R}^3$ then Dirac eigenfunctions give you the angular part of the electron wave function; the corresponding eigenvalues are simply all the integers $n \in \mathbb{Z}$ which show up with multiplicity $n+1$.
