# Pólya's conjecture on the spectra of the Laplacians

Recently I've learned something about the spectra of the Laplacians. Given a bounded domain $\Omega \subset \mathbb{R}^n$ with $\partial \Omega$ smooth, we can consider eigenfunctions of Dirichlet type, i.e. $u \in C^2(\Omega)\cap C(\partial \Omega)$ s.t. $-\triangle u=\lambda u$ and $u|_{\partial \Omega}=0$. By standard results in functional analysis, $-\triangle$ has a discrete spectrum $0<\lambda_1 \leq \lambda_2 \leq \cdots$ with $\lambda_k \to +\infty$.

A well-known asymptotic formula by Weyl says $\displaystyle \lambda_k \sim W_n (\frac{k}{V(\Omega)})^{2/n}$. We refer $W_n$ as the Weyl constant. And Pólya conjectured that $\displaystyle \lambda_k \geq W_n (\frac{k}{V(\Omega)})^{2/n}$ holds.

As far as I know, the best known result is due to Li and Yau. They proved the conjecture in the sense of "average": $\displaystyle \sum_{j=1}^k \lambda_j \geq \frac{nW_n}{n+2}{k}^{(n+2)/n}{V(\Omega)}^{-n/2}$.

I find their argument is elementary, only employing some standard Fourier tricks. And the big picture is quite clear if put into the quantum framework. My question is in some sense "soft", but it does make me feel absurd: what makes it so difficult to estimate the eigenvalues one by one while their average is so well understood? Does anyone work on this problem by carrying further Li & Yau's analysis? I do know one instance: Kröger has transplanted their proof to Neumann settings, but how about the original Dirichlet problem?

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Fourier analysis comes with an uncertainity principle, which allows at most counting eigenvalues in a certain range, but disallows us to isolate them. At least, that is what one encounters in the spectral analysis of hyperbolic manifolds. – Marc Palm Apr 15 '12 at 21:22
Some reference would be extremely appreciated, but thanks any way Mrc Plm! – Zhang Xiao Apr 16 '12 at 2:47

Since your question is "soft", I think it is okay if I give a soft answer in the case of a hyperbolic compact Riemann surface.

The Selberg trace formula describes the spectrum of the Beltrami-Laplace operator pretty well. You get an identity $$\sum\limits_{\lambda} f(\lambda) = \sum\limits_{\gamma} \widehat{f( \log \gamma)} + \dots ,$$ where $\lambda$ is asymptotic to the square of an eigenvalue and $\gamma$ to the lengths of closed geodesics.

The best known error term for the Weyl law her is $O(\sqrt{T} / \log T)$, so approxiamtely the square root of the main term.

Laplace-Beltrami Operator on Surfaces

Eigenvalues of Laplacian-Beltrami operator

Now why can we not do better? One of the main problems, is that you are only allowed to plugin holomorphic $f$, so it will be hard to estimate single objects. On the other hand, if you would be allowed to pluggin something compactly supported, then $\widehat{f}$ becomes entire, and is not compactly supported.

The fact that the support of $f$ and $\widehat{f}$ can not be simultaneously small, is a first instance of the uncertainity principle of Fourier analysis. The name is certainly derived from the Heisenberg uncertainity prinicple, eigenvalues are here "waves" and length of closed geodesics here "particles".

Similar things are happening for prime numbers and zeros of the Riemann zeta function (Weil's excplicit formula) or in quantum chaos (Gutzwiller trace formula). Already in finite group theory, if you try to compare traces of irreducible representations with conjugacy classes.

Perhaps Paley-Wiener theorem's give you a good flavor for first instances of Fourier uncertainity. Stein-Shakarchi "Complex Analysis" has a good treatment, chapter 4, I guess.

Best, Marc.

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Perhaps you will like "Spectra of Hyperbolic Surfaces" from Sarnak. It's freely available. – Marc Palm Apr 16 '12 at 17:55
I am reading the fascinating Sarnak now. Best regards Plm:) – Zhang Xiao Apr 17 '12 at 17:15

You might want to check out Laptev's notes. -- he gives some background and explanations of methods involved.

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To be frank I don't think the above notes make an attempt to attack the problem by detailed analysis on the "quantum spectrum". But still many thanks! – Zhang Xiao Apr 16 '12 at 2:49