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Let $X$ be a complete Riemann surface with a smooth metric, and $L$ a line bundle on it also equipped with a smooth metric; associated to this data there is a Laplace-Beltrami operator $D_L$ acting on sections of $L$.

We know that if $X$ is compact the operator $D_L$ has only discrete spectrum.

In the specific situation I'm considering $X$ is not compact but the metrics are such that $D_L$ is uniformly elliptic, and not just elliptic as in the general non-compact case. I would like to conclude from this that $D_L$ is discrete, viewing the uniform ellipticity as a "replacement" for the compactness of the surface.

Does it work? If it does not, is it possible to add some more hypothesis to deduce such a result or the facts

  • $D_L$ is uniformly elliptic on a complete surface
  • $D_L$ is discrete

are completely unrelated?

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    $\begingroup$ Isn't the usual Laplacian on $\mathbb R^n$ "uniformly elliptic", but with purely continuous spectrum? $\endgroup$ Commented Oct 14, 2014 at 13:32
  • $\begingroup$ Yes, and it is also complete; I didn't think about it, thanks! Now I have to think about what could help me to still get some similar statement. $\endgroup$ Commented Oct 14, 2014 at 13:44

1 Answer 1

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I tend to consider such things from the perspective of Dirac operators. Let $D$ be a Dirac operator on a complete Riemann manifold, acting on a vector bundle $E$. There is a Weitzenboeck type formula $D^2 = \nabla^{\ast} \nabla + R$. The Weitzenboeck remainder $R$ is of order zero and symmetric. We write $\nabla^* \nabla =: \Delta$, the Laplace operator. The Laplace you are interested in is $D^2$, not $\Delta$. It is a standard result that $D^2$, $D$ and $\Delta$ are essentially selfadjoint. If you take the operator $D=d+d^{*}$ on exterior forms, suitably twisted with a line bundle, you get the situation you were takling about in the question. You want to know a criterion for $D$ or equivalently $D^2$ to have a discrete spectrum. I believe that the following results are true, and indicate a proof (potentially flawed). The basic result is the following extension of the Fredholm property on noncompact manifolds.

THEOREM 1: ''Let $\Delta =\nabla^{\ast} \nabla$. Let $P$ be an essentially selfadjoint elliptic operator of order $2$. Assume that there exists $c>0$ and a compact set $K \subset M$ such that for all compactly supported sections $u$ with support disjoint from $K$, one has $$\langle Pu,u \rangle \geq \langle \Delta u , u \rangle + c^2 \langle u, u\rangle. $$ Then $P$ is Fredholm.''

I defer the proof to the end of this post, as it is technical, and discuss some applications.

COROLLARY 1: ''Let $D$ be a Dirac operator on a complete manifold $M$. Assume that there exists a compact $K$ and a $c>0$ such that the Weitzenboeck remainder $R$ is estimated from below by $c^2$ over $M-K$. Then $spec (D) \cap (-c,c)$ is discrete.''

One could phrase this by saying that an estimate of the Weitzenboeck remainder from below enforces an essential spectral gap, at least as large as $c$.

PROOF: Let $\lambda^2 < c^2$. Let $Q := D^2-\lambda^2$. Over $M-K$, we find that $Q = \Delta + R-\lambda^2 \geq \Delta + (c^2-\lambda^2)$. By Theorem 1, $Q$ is Fredholm. Thus $\lambda^2$ is an isolated point of the spectrum of $D^2$, and hence $\pm \lambda$ are isolated points of $spec (D)$. QED

Note that in Corollary 1, the size of $K$ was inessential and could be increased, which allows the possibility for a stronger estimate on the complement of $K$ and hence to enlarge the essential spectral gap. This is the content of the next result.

THEOREM 2: ''Let $D$ be a Dirac operator on $M$ and $f: M \to \mathbb{R}$ be a proper function that is bounded from below. If the remainder term $R$ in the Weitzenboeck formula $D^2 = \Delta + R$ is bounded below by $f$, then the spectrum of $D$ (and hence that of $D^2$) is discrete.''

PROOF: For each $C>0$, we find a compact set $K \subset M$ and an estimate $R \geq C^2$ on $M-K$. By Corollary 1, $spec (D) \cap (-C,C)$ is discrete. As $C$ was arbitrary, $spec (D) \subset \mathbb{R}$ is discrete. QED.

PROOF OF THEOREM 1: The first step is a strengthening of Gardings inequality. Namely, we prove that there is a constant $C$ with

$$ (!!!) \; \|u \|_2 \leq C (\| u|_K \|_0 + \| Pu \|_0). $$

Without loss of generality $c \leq 1$. Let $u$ be a section with compact support $supp (u)\subset M-K$. I write $\| u \|_r$, $r=0,1,2$, for the Sobolev norms. Estimate using the condition on $P$:

$$ \| u \|_2^2 = \|u \|_0^2 + \|\Delta u \|_0^2 \leq \frac{1}{c^2} \|P u\|_0^2 $$

If $K \subset U$ is a relatively compact neighborhood, and $u$ supported in $U$, the usual Garding inequality shows

$$ \| u \|_2^2 \leq C(\| u\|_0^2 + \| Pu\|_0^2) $$

To patch both estimates together, choose $a$ supported in $U$, $b$ supported in $M-K$ such that $a + b =1$. For an arbitrary compactly supported section $u$, we find that

$$ \| u \|_2 \leq \| au \|_2+\| bu \|_2 \leq C' (\| au\|_0^2 + \| P a u\|_0^2 + \| Pbu\|_0^2) $$ The first term is bounded by $\| u|_K \|_0 $. Furthermore $\| P a u\|_0 \leq \| [P,a] u \|_0 + \| a P u \|_0 \leq \| [P,a] u \|_0 + C\| P u \|_0$. But $[P,a]$ has order $1$ and is supported in $U$. Therefore, there is a function $h$ with $[P,a]h=[P,a]$ and hence $\| [P,a] u \|_0 \leq C \| hu \|_1 $. By the Peter-Paul inequality, there is for $\epsilon>0$ a constand $C(\epsilon)$ such that $\| hu \|_1 \leq \epsilon \| hu \|_2 + C(\epsilon) \| hu \|_0\leq C\epsilon \| u \|_2 + \| u|_K\|_0$. Picking $\epsilon$ small enough, we can swap the first summand to the left hand side of the estimate. Similarly, the term $\|Pbu\|$ is estimated, and we have proved the strengthened inequality (!!!).

Why did I do all this stuff? Well, the operator $W^2 \to L^2 \stackrel{|_K}{\to} L^2$ is compact by Rellichs theorem. It is a general fact (e.g. Proposition 3.7.2 in my lecture notes http://wwwmath.uni-muenster.de/u/jeber_02/skripten/mainfile.pdf) that this implies that $P$ has closed range and finite-dimensional kernel. Since $P$ is essentially selfadjoint, the cokernel is the kernel of $P$ and so finite-dimensional as well. QED (Theorem 1).

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  • $\begingroup$ I forgot to mention that the Dirac operator is uniformly elliptic. It seems that the conditions in my post concern the order 0 part, while uniform ellipticity concerns the order 1 part. $\endgroup$ Commented Oct 15, 2014 at 17:43
  • $\begingroup$ Thanks for the answer, +1! But to accept it I still need to understand a few things. The Laplacian I'm interested into is $D^2$ where $D = \bar\partial + \bar\partial^*$ and $\bar\partial : A^{p,q}(X,L) \rightarrow A^{p,q+1}(X,L)$ is the extension of the Cauchy-Riemann operator to sections of $L$, this is what Griffit and Harris call the (\bar\partial)-Laplacian. For me $\nabla_L:A^{n}(X,L) \rightarrow A^{n+1}(X,L)$ is the unique hermitian holomorphic connection on $L$, from it we get the "connection" Laplacian $\nabla_L^* \nabla_L$. Is this your notation as well? $\endgroup$ Commented Oct 18, 2014 at 11:37
  • $\begingroup$ In this case, is it clear that the Weitzenboeck remainder lives in order 0? Also, is not uniform ellipticity an order 2 condition? Thank you again! $\endgroup$ Commented Oct 18, 2014 at 11:38

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