# Teichmuller distance between isospectral riemann surfaces

Let $S$ be a surface of negative Euler characteristic (for simplicity let's assume $S$ to be closed), and let $\mathcal{M}(S)$ denote the moduli space of hyperbolic surfaces homeomorphic to $S$.

Given $X \in \mathcal{M}(S)$, let $L(X)= l_{1} \leq l_{2} \leq l_{3} \leq...$ denote the increasing sequence of lengths of all closed geodesics on $X$. This is the so-called "unmarked length spectrum". It is known that $L(X)$ does not determine $X$ up to isometry; indeed, if $\mathcal{V} \subset \mathcal{M}(S)$ denotes the set of all surfaces $X$ for which there exists $Z \neq X$ with $L(X) = L(Z)$ (in this case $X$ and $Z$ are called $\textit{iso-spectral}$), then it is a theorem of Wolpert that $\mathcal{V}$ is a real-analytic subvariety of $\mathcal{M}(S)$ with positive dimension. On the other hand, $\mathcal{V}$ also has positive codimension, so "generically" $L(X)$ does determine the metric.

McKean proved that any subset $W \subset \mathcal{M}(S)$ consisting of pairwise iso-spectral surfaces must be finite, and Buser gave an explicit upper bound on $|W|$ which depends only on the genus of $S$.

Let $d_{T}( , )$ denote the projection of the Teichmuller metric to $\mathcal{M}(S)$, and similarly $d_{WP}( , )$ for the Weil-Petersson metric.

$\textbf{Question 1}:$ Let $D_{T}(S):= \sup \left\{ d_{T}(X,Y) : \hspace{2 mm} X,Y \in \mathcal{M}(S), X \hspace{2 mm} \mbox{iso-spectral to} \hspace{2 mm} Y \right\}$, and define $D_{WP}(S)$ similarly. Is $D_{T} < \infty$? How about $D_{WP}$? (of course if $D_{T}$ is finite then so is $D_{WP}$).

$\textbf{Question 2:}$ Let $LS(X)$ denote the subsequence of $L(X)$ consisting only of lengths corresponding to $\textbf{simple}$ closed geodesics. Is it the case that $LS(X)= LS(Y)$ implies $L(X)= L(Y)$?

$\textbf{Regarding Question 1:}$ I suspect that the answer is "no" for both $D_{T}$ and $D_{WP}$, but I don't have an argument in mind, nor am I very familiar with what is known about iso-spectrality in general. I'll remark that there exists fairly general/algebraic methods for constructing iso-spectral surfaces with arbitrarily large genus, such as the following due to Sunada:

Let $\phi: \pi_{1}(X) \rightarrow G$ be a surjective homomorphism on to a finite group $G$, and let $H_{1}, H_{2}$ be $\textit{almost conjugate}$ subgroups of $G$, meaning that

$$| [g] \cap H_{1}| = | [g] \cap H_{2} |, \forall g \in G,$$

where $[g]$ denotes the conjugacy class of $g$. Then the covering spaces of $X$ associated to the subgroups $\phi^{-1}(H_{1})$ and $\phi^{-1}(H_{2})$ will be iso-spectral. Perhaps, one can construct iso-spectral surfaces that are very far apart in $\mathcal{M}(S)$ by carefully choosing $G$, $\phi$ and $X$.

$\textbf{Regarding question$2$:}$ It is known that the converse is false. That is, $L(X)= L(Y)$ does not imply that $LS(X)= LS(Y)$ (see the following preprint of Maungchang: http://arxiv.org/abs/1111.4317). The proof uses Sunada's construction described above.

Thank you very much for reading.

• I think you are correct, regarding $D_T$ -- Sunada's construction will give a negative answer to Question 1. Fix $S$ the closed surface of genus two. Fix data $G$, $\phi$, $H_1$ and $H_2$ as required by Sunada. Vary $X \in M(S)$ as follows. Fix a pants decomposition of $S$. For any $n > 0$... – Sam Nead Apr 5 '15 at 12:50
• let $X_n$ to be a hyperbolic genus two surface where all cuffs have length $1/n$. Let $Y_n$ and $Z_n$ be the two covers. These will also have many short curves, but in different (non-homeomorphic) configurations. Thus $d_T(Y_n,Z_n)$ goes to infinity with $n$. --- You might be able to shorten the proof by factoring through the map to $F_2$ (free group of rank two) that kills the conjugacy classes of the cuffs. Then $Y_n$ and $Z_n$ come to us with distinct pants decompositions. – Sam Nead Apr 5 '15 at 12:51

As a partial answer to question one: The diameter of $M(S)$ with respect to the Weil--Petersson metric is finite. So $D_{WP}$ is finite.
For $D_T:$ if you have a bound on $\lambda_1,$ then, by Cheeger's inequality you have a (lower) bound on the shortest geodesic. An estimate then follows from the known estimate on the diameter of the $\epsilon$-thick part of moduli space. (due to, who else, Kasra Rafi).
EDIT 1 As pointed out by Ian in the comments, that is not quite true ($\lambda_1$ gives a bound on the shortest separating multicurve, not the shortest curve).
• A lower bound on $\lambda_1$ doesn't necessarily give a lower bound on the shortest geodesic, if the geodesic is non-separating (maybe if $\lambda_1 >1/4$). – Ian Agol Apr 5 '15 at 18:38