# Lower bound on the first eigenvalue of the Laplace-Beltrami on a closed Riemannian manifold

It has been proved by Li-Yau and Zhong-Yang that if $M$ is a closed Riemannian manifold of dimension $n$ with nonnegative Ricci curvature, then the first nonzero eigenvalue $\lambda_1(M)$ of the (positive) Laplace-Beltrami operator $\Delta$ satisfies

\begin{equation} \lambda_1(M) \geq \frac{\pi^2}{d^2}, \tag{1} \end{equation}

where $d$ is the diameter of $M$. Another well known result is the Lichnerowicz theorem: if $\mathrm{Ric} \geq (n-1)K >0$ then

$$\lambda_1 \geq n K.$$

All these lower bounds require some assumption on the curvature.

1. Are there universal (i.e. curvature-independent) lower bounds for $\lambda_1$ on a closed Riemannian manifold?
2. More precisely, does the inequality (1) hold even with no assumption on the Ricci curvature?
• If not, are there counter-examples?
• If yes, is this the best bound one can achieve in this sense (clearly (1) is not sharp as $\lambda_1(\mathbb{S}^{n}) = n$)?
• You may be interested in the following generalization of the Lichnerowicz theorem ams.org/mathscinet-getitem?mr=690651 by Berard and Meyer. They assume some control on the isoperimetric profile, rather than any curvature conditions. – Otis Chodosh May 28 '15 at 21:27
• Related: I remember it was shown that the equality case in $(1)$ is attained precisely when $M = \mathbb S^1$. (I can't remember the name of the author though) – Arctic Char May 29 '15 at 17:12

To see this consider two (2D) sphere's of constant curvature 1 and a (flat) cylinder of unit length and radius $\epsilon$. Cut out a disk from each sphere and glue in the cylinder (and smooth everything out in a small neighborhood of the gluing). By Gauss Bonnet this must introduce negative curvature.

Now take a function which is identically 1 on one sphere and identically -1 on the other sphere and which linearly interpolates between the two on the cylinder. It's clear that this can be chosen to be orthogonal to the constants. Moreover, this function has $L^2$ norm which is (up to a small error) independent of $\epsilon$ (and positive) while the Dirichlet energy is on the order of $\epsilon$. In particular, the variational characterization of $\lambda_1$ implies that $\lambda_1$ is smaller than $C\epsilon$ for some fixed constant $C$ -- i.e. as small as one likes.

• It seems to me that, in this example, the diameter goes to $\infty$ for $\epsilon \to 0$, as you are introducing a "region" of large negative curvature (that scales roughly with $-1/\epsilon^2$). – Raziel May 28 '15 at 18:13
• The diameter is roughly bounded by $2\pi+1$ (the distance between antipodal points on the two spheres via the cylinder). The smoothed out metric is $C^0$ close to the glued metric so this won't effect diameter. I'm not sure I follow what you are getting at with your comment about a region of large negative curvature. – foliations May 28 '15 at 18:40

A well-known result of Y. Colin de Verdière states that given any compact connected manifold $M$, with $\dim M\geq 3$, and any finite sequence $0<a_1\leq\dots\leq a_k$, there exists a Riemannian metric on $M$ such that the first eigenvalues in the spectrum of its Laplacian are $0< a_1\leq\dots\leq a_k$. The original paper can be found here. Therefore, there is no hope for a "universal" lower bound on $\lambda_1(M)$.

• Hey Renato, nice answer! Two remarks: (1) there should be strict inequality for $a_1$. (2) An interesting extension of this, which I find very surprising, is due to Lohkamp which says that you can prescribe the first $k$ eigenvalues while forcing $Ric < - \alpha^2$ for any $\alpha$. ams.org/mathscinet-getitem?mr=1356779 Corollary 2.7 – Otis Chodosh May 28 '15 at 21:26
• @Otis: Thanks for catching that typo, I've just fixed it. One curiosity I always had is what happens to these metrics as $k\nearrow +\infty$. I don't remember the details, but I think I somehow concluded that the volume of these metrics must go to zero (or the diameter has to blow up, depending what you want to normalize). The statement in Lohkamp's paper says one can arrange these metrics to also have Ricci bounded above by an arbitrarily negative number -- I wonder what the GH limit is, maybe some sort of graph? Do you know anything about that? – Renato G. Bettiol May 28 '15 at 22:08
• For some reason I remember that you should also be able to force the volume to be 1, but I couldn't find that statement in Lohkamp's paper. I think that Lohkamp is perturbing Colin de Verdière's metrics in $C^0$, so its probably just an issue of understanding the limit of the ones you mention. I guess it isn't even obvious without looking into Colin de Verdière's paper that there's a GH limit! – Otis Chodosh May 28 '15 at 23:34
• @Otis: Good point; now that I'm thinking about it, I'm not so sure there's a GH limit either... – Renato G. Bettiol May 29 '15 at 2:07