The spectral growth of the Dirac operator

Since $\Dslash^2 = \Dl^S + \quarter s$, and $\Dl^S$ is closely related to the Laplacian $\Dl$ on the (compact, boundaryless) Riemannian manifold $(M, g)$, the general features of $\spec(\Dslash)$ may be deduced from those of $\spec(\Dl)$.

We require two main properties of Laplacians on compact Riemannian manifolds without boundary. Recall that

$$\Dl = -\Tr_g(\nb^{T^*M\ox T^*M}\circ\nb^{T^*M}) = - g^{ij} (\del_i \del_j - \Ga_{ij}^k \del_k)$$

is the local expression for the Laplacian (which depends on $g$ through the Levi-Civita connection and $g^{ij}$). Thus $\Dl$ is a second order differential operator on $\Coo(M)$.

To make $\Dl$ selfadjoint, we must complete $\Coo(M)$ to a larger domain, by defining

$$\snorm{f}^2 := \braket{f}{(1 + \Dl)f} = \Vert f\Vert^2 + g^{ij} \braket{\del_i f}{\del_j f},$$

where $\braket{f}{f} := \int_M \vert f\vert^2 \nu_g$. Taking $\Dom\Dl := \set{f \in L^2(M,\nu_g) : \snorm{f} < \infty}$, $\Dl$ becomes selfadjoint and $(1 + \Dl)^{-1}\: L^2(M,\nu_g) \to (\Dom\Dl, \snorm{\cdot})$ is bounded. Then one shows that the inclusion $(\Dom\Dl, \snorm{\cdot}) \hookto L^2(M,\nu_g)$ is a compact operator (by Rellich's theorem); and $(1 + \Dl)^{-1}$, as a bounded operator on $L^2(M,\nu_g)$, is then the composition of these two, so it is also compact.

As a convention, when $A$ is a compact positive operator, we write $\la_k(A)$ to denote the $k$-th eigenvalue of $A$ in decreasing order (with multiplicity): $\la_0(A) \geq \la_1(A) \geq\cdots$; on the other hand, if $A$ is an unbounded positive selfadjoint operator with compact inverse, we write the eigenvalues in increasing order, as we did for $\Dl$.

We shall not prove Weyl's theorem, in particular why the number of eigenvalues (up to $\la$) is proportional to $\Vol(M)$, but we shall compute the constant by considering an example. For a simple and clear exposition of the proof, we recommend Higson's ICTP lectures [Hig].

For the spinor Laplacian $\Dl^S$, a similar estimate holds, but with $C_n$ replaced by $2^m C_n$ (recall that in the flat torus case with untwisted spin structure, $\sS \isom \Coo(\bT^n) \ox \bC^{2^m}$). Now by Lichnerowicz' formula, $\Dslash^2$ differs from $\Dl^S$ by a bounded multiplication operator $\quarter s$, thus $N_{\Dslash^2}(\la) \sim N_{\Dl^S}(\la)$ as $\la\to\infty$, hence

$$N_{\Dslash^2}(\la) \sim \frac{2^m \Omega_n}{n(2\pi)^n} \Vol(M) \la^{n/2}, \as \la \to \infty.$$

Consider the positive operator $\vert\Dslash\vert := (\Dslash^2)^{1/2}$; remember that $\mu$ is an eigenvalue for $\vert\Dslash\vert$ if and only if $\mu^2$ is an eigenvalue for $\Dslash^2$ (with the same multiplicity). We arrive at the following estimate.