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 $\lambda\_G$?



The smallest positive eigenvalue and the length of the shortest geodesic
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](http://en.wikipedia.org/wiki/Cheeger_constant#Cheeger.27s_inequality) stays bounded away from zero, and therefore so does $\lambda$. You may see this explicitly using [Fenchel-Nielsen coordinates](http://en.wikipedia.org/wiki/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](http://www.ams.org/mathscinet-getitem?mr=1677565). 


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. 


