# Why is this operator compact?

Let $D$ be the Dirac-Operator on $\mathbb{R}^n$ or more generally the Dirac spinor bundle $\mathcal{S}\to M$ of a (semi-)Riemannian spin manifold $M$. Then we consider $D$ as an unbouded Operator on $\mathcal{H}=L^2(\mathbb{R}^n)$ with domain $C^\infty_c(\mathbb{R}^n,\mathbb{C}^N)$. Then it is said that the operator $f\langle D\rangle^{-n}$ is compact, where $f\in C^\infty_c(\mathbb{R}^n,\mathbb{C})$ is considered as a multiplication operator on $\mathcal{H}$ and $\langle D\rangle:=\sqrt{D^\dagger D+ DD^\dagger}$.

Since I am not really a crack in functional analysis, it is not even obvious for my how exactly $\langle D\rangle$ works. I suspect that the Operator $D^\dagger D+DD^\dagger$ is (essentially) self-adjoint and then the spectral theorem is used for defining $\langle D\rangle$ and its powers $\langle D\rangle^{-n}$.

But what is even more mysterious to me is the claim that $f\langle D\rangle^{-n}$ is actually compact (note that $f$ has compact support, however). Why is this true?

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I think this is a Hilbert-Schmid-Operator : It's an integral operator with kernel f(x)g(x-y) where g is the fourier transform of $\langle D\rangle^{-n}$ . Hope g is good enough such that this argument works. –  jjcale Jul 3 '13 at 19:20
Can you explain what you mean by $D^\dagger$? It seems to me that in order for $\langle D \rangle^{-n}$ to make sense, $\langle D \rangle$ has to have a gap around $0$ in its spectrum. If this happens to be the case then I can answer your question. –  Paul Siegel Jul 3 '13 at 20:20
In "Smoothness and locality for nonunital spectral triples" by Adam Rennie (which doesn't seem to available online anymore anywhere, what's up with that?), they give a reference to Higson and Roe's book "Analytic K-Homology" for this precise fact. –  Jan Jitse Venselaar Jul 3 '13 at 20:31
Upon closer reading, the claim in that article is that in fact $f(1+D^2)^{(-p/2)}$ is compact. Maybe this solves Paul's objection. One needs to be very careful here, many standard texts are somewhat sloppy when working with these noncompact cases. Rennie also gives a reference to "Summability for nonunital spectral triples and the local index theorem" by himself (luckily still available online), which also proves a similar result. –  Jan Jitse Venselaar Jul 3 '13 at 20:46
For the pseudo-Riemannian case, all I can find is Section 3 (particularly Proposition 3.8) of "Pseudo-Riemannian spectral triples and the harmonic oscillator" by Van den Dungen--Paschke--Rennie, arxiv.org/abs/1207.2112. –  Branimir Ćaćić Jul 4 '13 at 8:37

I think it is useful to ask the simpler question, why $f\cdot (1-\Delta)^{-1}$ is compact, on $\mathbb R^n$, when $f$ is a test function. Part of the point is that $\Delta$ itself (nevermind the Dirac operator) does not have compact resolvent on $\mathbb R^n$, essentially because Fourier inversion shows that the spectrum is purely continuous. An even nicer case of positive outcome is the Schrodinger Hamiltonian $-\Delta+|x|^2$ on $\mathbb R^n$, which provably has discrete spectrum (without looking at specific formulaic aspects), for general geometric reasons.

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As discussed in the comments, the statement probably needs to be modified in order for $\langle D \rangle^{-n}$ to be defined. I'm guessing that the correct statement should fit into the following framework:

Proposition: Let $D$ be an essentially self-adjoint first order elliptic operator on a possibly non-compact manifold $M$, let $f \in C_c^\infty(M)$, and let $g \in C_0(\mathbb{R})$. Then $m(f) g(D)$ is compact where $m(f)$ is the multiplication operator by $f$.

I'm not going to be able to drudge up all of the gory details, but in the end the proof can be pieced together using standard elliptic analysis. First consider the resolvent function $g(t) = (i + t)^{-1}$. Let $K$ denote the support of $f$ and let $v$ be a vector in the domain of the closure $\overline{D}$ of $D$, so that $v' = m(f)v$ is in the Sobolev space $L_1^2(K)$. Garding's inequality estimates the Sobolev $1$-norm of $v'$ in terms of the $L^2$ norm of $v'$ and of $\overline{D}v'$; from this it follows that $m(f)(i + \overline{D})^{-1}$ maps $L^2(M)$ continuously into $L_1^2(K)$. But the Rellich lemma asserts that the inclusion of $L_1^2(K)$ into $L^2(M)$ is compact, so the result is proved for the specific $g$ above. Now, the set of all $g$ for which the result is true is closed under linear combinations, pointwise multiplication, complex conjugation, and uniform limits, so by the Stone-Weierstrass theorem the result is true for any $g \in C_0(\mathbb{R})$. Notice, however, that the proposition does not apply to $g(t) = t^{-n}$, hence my concerns in the comments.

Now, the Dirac operator on a complete Riemannian manifold is essentially self adjoint and therefore fits into the proposition above. This ultimately follows from the fact that its symbol is the Clifford multiplication endomorphism which is not only invertible (away from the $0$ section) but bounded in norm on the unit cosphere bundle. In other words, it has "finite propagation speed". I'm not quite sure how this works out in the semi-Riemannian case, but my guess is that it does as long as you have some counterpart of the completeness assumption: note that the Dirac operator on $(0,1)$ is not essentially self-adjoint.

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Would you happen to know, by any chance, when $0$ can be an isolated point in the spectrum of the Dirac operator on a complete but non-compact spin manifold? Because then one could still use the spectral triples convention of replacing $g(t) = |t|^{-n}$ with some $\tilde{g} \in C_0(\mathbb{R})$ with $\tilde{g}(t) = g(t) = |t|^{-n}$ for $|t|>\epsilon$ and $\tilde{g}(0) = 0$, where $\sigma(D) \cap [-\epsilon,\epsilon] = \{0\}$. Otherwise, I suppose one would have no choice but too use, rather, something like $g(t) = (1+|t|)^{-n}$ to make things work? –  Branimir Ćaćić Jul 4 '13 at 18:11

What is it that you find mysterious exactly? On a compact manifold $M$, the eigenvalues of $\langle D\rangle$ tend to infinity, hence the eigenvalues of $\langle D\rangle^{-1}$ tend to zero, so that we have a compact operator, namely the image of unit ball is compact. Multiplication by $f$ is a bounded operator, therefore composing with it will respect the compactness of the image of the unit ball, so that the composition is a compact operator, as well.

Note 1. I see that I may have misread the OP's question as he seems to be mostly interested in the noncompact case. Paul Garrett's answer seems to suggest that the claim may be problematic in the non-compact case.

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Why do the eigenvalues of $\langle D\rangle$ tend to infinity? Why does this imply that the eigenvalues of $\langle D\rangle^{-1}$ tend to zero? And finally, why does this imply that $\langle D\rangle^{-1}$ is compact. Do not forget that we are dealing with unbounded operators, so the spectrum may have continuous parts as well. –  Robert Rauch Jun 25 '13 at 16:50
I am particularly interested in those little details. Also, a generic selfadjoint operator may of course have continuous spectrum. For example, the multiplication operator of the identity map $t\mapsto t$ on $L^2([0,1])$ has purely-continuous spectrum $\sigma=[0,1]$. –  Robert Rauch Jun 26 '13 at 20:07
Maybe I should have pointed out more clearly, that I am mainly interested in non-compact semi-riemannian manifolds, so I'm not sure if Lawson/Michelson is a good starting point to answer my question. –  Robert Rauch Jun 27 '13 at 23:23
also you seem to.imply that the spectrum.is discrete. it isn't for Rn. –  Marc Palm Jul 3 '13 at 20:19
The claim's true in the non-compact but complete Riemannian case, and even in the complete, bounded geometry pseudo-Riemannian case. –  Branimir Ćaćić Jul 4 '13 at 8:21