Compact surfaces with arbitrary gaps in spectrum Consider a sequence of positive numbers $a_n$. My question is, can we select a closed Riemann surface whose spectrum $\lambda_i$ satisfies the condition that $\lambda_{i + 1} - \lambda_i > a_i$? Of course, this implies that the eigenvalues do not repeat. If this is impossible, what are some obstructions, and what is the best that can be done? The literature on spectra of Riemann surfaces is really really huge, I could use some help where to get started. Thanks in advance.
 A: If you are allowed to choose a metric, together with the Riemann surface, this is related to the well-known result of Colin de Verdiere (you can choose the bottom of the spectrum as you like):
MR0932800 (90d:58156) Reviewed 
Colin de Verdière, Yves(F-GREN-F)
Construction de laplaciens dont une partie finie du spectre est donnée.   (French) [Construction of Laplacians for which a finite subset of the spectrum  is given] 
Ann. Sci. École Norm. Sup. (4) 20 (1987), no. 4, 599–615. 
58G25 

This paper of Koji Fujiwara is also related:
MR1257106 (95j:58171) Reviewed 
Fujiwara, Koji(J-KEIOE)
Eigenvalues of Laplacians on a closed Riemannian manifold and its nets. (English summary) 
Proc. Amer. Math. Soc. 123 (1995), no. 8, 2585–2594. 
58G25 (58G99) 

A: You seem to be talking about the eigenvalues of the laplacian associated with the surface and to be tacitly assuming compactness (otherwise the spectrum would not be discrete).  If this is so, then Weyl's asymptotic formula would preclude a result of this type.  This is valid for any compact manifold and states the the sequence $(\lambda_n)$ of eigenvalues is asymptotically like one of the form $(n^k)$, the power $k$ depending on the dimension of the manifold.
A: As indicated by estrelha Weyls estimate states that, on any compact Riemann manifold of dimension $m$ the number of $N(r)$ eigenvalues $\leq R^2$  asymptotically equal to   ${\rm vol}\;(B^m_R)$ as $R\to\infty$, where $B^m_R$ denotes the $m$-dimensional Euclidean ball of radius $R$.  Set $\omega_m:={\rm vol}\; (B_1^m)$. 
This leads to a constraint of the  type
$$\sum_{j\leq \omega_m R^m} a_j\geq  R^2,\;\;\forall R\gg 1, $$
$$ \sum_{k\le K}  a_k\geq \left(\frac{K}{\omega_m}\right)^{2/m}, \;K\gg 1. $$
In your case $m=2$ and  $\omega_2=\pi$.
