A friend in physics asked this question, and I didn't know the answer.
Are there lower bounds on the first eigenvalue of the Laplacian of a Riemann surface equipped with a metric of constant negative curvature?
A friend in physics asked this question, and I didn't know the answer. Are there lower bounds on the first eigenvalue of the Laplacian of a Riemann surface equipped with a metric of constant negative curvature? 


I guess you mean constant curvature $=1$; otherwise you get an example by rescaling. On a sphere with two handles there are metrics with curvature $\equiv1$ which look roughly as a long neck attached to two tori. Such metric has as small eigenvalue as you want. On the other hand upper diameter + lower curvature bound (in particular curvature $\equiv1$) imply a lower bound on the eigenvalue. 


All hyperbolic Riemann surfaces are quotients of the upper halfplane ${\mathbb H}$ by a Fuchsian group $\Gamma$. When $\Gamma$ is a congruence subgroup, Selberg's eigenvalue conjecture says that the smallest eigenvalue is at least $1/4$, and, in fact, he proved it was at least $3/16$ (for $SL_2({\mathbb Z})$, a better lower bound is known: $3\pi^2/2$). This lower bound has been improved somewhat since then (I think the current record is something like 0.228 due to Kim and Shahidi, via functoriality for the symmetric cube). For general hyperbolic Riemann surfaces, the lower bound is zero. See this paper. Apparently, this, too, but I don't have access to it, so I can't confirm. 

