3
$\begingroup$

It is said that if we take the spacetime manifold to be a sphere $S^d$ of large volume so that the spectrum of Dirac operator $$i\gamma^\mu D_{\mu}$$ is discrete.

For example at least for $d=4$, this has been discussed in p.2 of Witten paper from this Physics Letters B, Volume 117, Issue 5, 18 November 1982, Pages 324-328 Physics Letters B, 117(5), 324–328

My question is simple that

  1. why the spectrum of Dirac operator $i\gamma^\mu D_{\mu}$ is discrete? in any $d$? Can the spectrum of Dirac operator be continuum?

In what setup can it be discrete?

In what setup can it be continuum?

I thought that in a finite volume, the eigenvalues may have discrete. In an infinite volume, the eigenvalues may have a chance to have a part to be continuum?

  1. under what conditions do we have zero eigenvalues for Dirac operator $i\gamma^\mu D_{\mu}$?

p.s. I know this is related to the Atiyah Singer index theorem. But I am trying to ask for an explicit user-friendly down-to-earth answer. I am not asking for a reference only. (The paper I quoted is a reference itself.)

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ So, the Dirac operator on a $1$-dimensional flat Riemannian manifold is, up to constant rescaling, just the momentum operator $\hat{P} := -\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t}$ with various boundary conditions. On the line $\mathbb{R}$, so that your boundary conditions consist of vanishing at infinity, the spectrum of $\hat{P}$ is $\mathbb{R}$, since for every $k \in \mathbb{R}$, the function $(t \mapsto \mathrm{e}^{2\pi\mathrm{i}kt}$ is a distributional eigenvector for $k$. $\endgroup$
    – Branimir Ćaćić
    Commented Apr 11, 2021 at 18:31
  • 3
    $\begingroup$ On the flat circle $\mathbb{S}^1$ (with the trivial spin structure), which is just, say, $[0,1]$ with the periodic boundary condition, you have discrete spectrum $\{2\pi n \mid n \in \mathbb{Z}\}$ by the usual theory of ODE. The difference is that $\mathbb{R}$ is non-compact while $\mathbb{S}^1$ is compact—it’s compactness that implies that self-adjoint elliptic operators on compact Riemannian manifolds, e.g., Laplace and Dirac operators, necessarily have discrete spectrum with finite multiplicities. $\endgroup$
    – Branimir Ćaćić
    Commented Apr 11, 2021 at 18:35
  • 1
    $\begingroup$ The standard method to establish discrete spectrum in such situations is to show that the resolvent is compact. $\endgroup$
    – Christian Remling
    Commented Apr 12, 2021 at 14:43

0

Reset to default

You must log in to answer this question.