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 $\lambda\_1(\mathbb{S}^{n}) = n$)?



Lower bound on the first eigenvalue of the Laplace-Beltrami on a closed Riemannian manifold
The answer is no.


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. 


