# The smallest positive eigenvalue and the length of the shortest geodesic

I'm confused about some things concerning lengths of geodesics on Riemann surfaces and positive eigenvalues of the Laplacian. Moreover, I'm also interested in the relation between these two.

Let $X$ be a compact (connected) Riemann surface of genus $g\geq 2$.

Since the complex upper half plane $\mathfrak{h}$ is the universal cover of $X$, we have that $X$ inherits the structure of a Riemannian manifold from $\mathfrak{h}$. The length of the shortest geodesic with respect to the smooth volume form on $X$ induced by the hyperbolic metric, denoted by $\ell_X$, is well-defined in this case. Let $\lambda_X$ be the smallest positive eigenvalue of the Laplace operator on $L^2(X)$.

Question. What's the relation between $\ell_X$ and $\lambda_X$? Is there some kind of correspondance?

Now, let's suppose that $b_1,\ldots,b_n$ are points in $X$. Then $X$ is the compactification of a quotient $G\backslash \mathfrak{h} = X\backslash \{b_1,\ldots,b_n\}$ by adding the cusps'' $b_1,\ldots,b_n$. (Note that $G \backslash \mathfrak{h}$ inherits the structure of Riemannian manifold from $\mathfrak{h}$.) In this case there is no shortest geodesic on $X$ (due to the existence of cusps). Let $\lambda_G$ be the smallest positive eigenvalue of the Laplace operator on $L^2(G\backslash \mathfrak{h}$.

Question. Does $\lambda_X$ equal $\lambda_G$?

You can have a sequence surfaces $X_i$ with arbitrarily short geodesics but $\lambda_{X_i}$ bounded away from zero. The geodesics are non-separating, so when they shrink to length zero, the Cheeger constant stays bounded away from zero, and therefore so does $\lambda$. You may see this explicitly using Fenchel-Nielsen coordinates.
On the other hand, there exist surfaces $X_i$ with $\lambda_{X_i}\to 0$ but the shortest geodesic is bounded away from zero. Just take a surface whose shortest geodesic is $\geq N$, and take a large cyclic cover to make $\lambda_X$ arbitrarily small.
As for filling in cusps, $\lambda_X \neq \lambda_G$ (if I understand your statement correctly). With further geometric information, you can find some relation between them. Robert Brooks has investigated this.
There is some relation between the lengths of geodesics and the eigenvalues of the Laplacian. In principle, Selberg's trace formula can determine one from the other. If there is a very short geodesic, then one obtains many eigenvalues near $\frac14$, because the spectrum is approaching the continuous spectrum of a cusped surface.
• Thanks alot. This is perfect! If I may ask another question, there's a remark (Remark 5.11) in an article of Jorgenson and Kramer on canonical Green's functions which states that for an arbitrary cover of $Y\rightarrow X$ of compact Riemann surfaces, we have that $1/\lambda_Y = O_{X}(g_Y^2)$. Assuming I have correctly quoted this result from their paper, do you know if something similar holds or can hold for $1/\ell_Y$? That is, do we have that $1/\ell_Y = O_X(g_Y^a)$ for some integer $a$? Here I used the Big-O notation and $g_Y$ denotes the genus of $Y$. – Ariyan Javanpeykar Sep 18 '11 at 10:50
• The behavior of $\lambda_Y$ and $l_Y$ for covers is quite different. In covers, $\lambda_Y$ can approach zero, but the inequality you state shows that the growth is at most quadratic in the index of the cover (I think one may see this from Cheeger's inequality). On the other hand, the systole $l_Y$ can only grow in covers, and the fastest rate of growth possible would be like $g_Y^{\frac12}$ by Gromov's systolic inequality. en.wikipedia.org/wiki/… – Ian Agol Sep 18 '11 at 15:45