Lengths of generators of surface group

Question

Let $\Sigma$ be a closed genus $g\geq 2$ Riemann surface, which we equip with its unique constant curvature $-1$ hyperbolic metric. Let $\pi_1(\Sigma)$ be its fundamental group with respect to some base point. For each $\gamma\in \pi_1(\Sigma)$, there is a unique closed geodesic on $\Sigma$ in the free homotopy class of $\gamma$; let the length of this geodesic be $\ell(\gamma)$, which depends on the complex structure of $\Sigma$, of course.

Question: does there exist an $\epsilon =\epsilon(g) >0$, independent of the complex structure of $\Sigma$, such that for any $N\geq 2$ and elements $\gamma_1, \cdots, \gamma_N$ generating $ \pi_1(\Sigma)$, there is a $k$ with $\ell(\gamma_k) \geq \epsilon$?

Note that for each fixed $\Sigma$, and $\gamma$ non-trivial, $\ell(\gamma)$ is bounded below by the systole, so the question is really about a possible uniform bound over all choices of $\Sigma$. Any references or pointers would also be much appreciated!

Answer 1

In order to remove this question from the "unanswered list." Let $\epsilon>0$ be the Margulis constant for the hyperbolic plane (with curvature $-1$). Then for every complete hyperbolic surface $S$, if $c_1, c_2$ are closed (nonconstant) geodesics of length $\le \epsilon$, then either $c_1, c_2$ are disjoint or have the same image. Moreover, the image of each $c_i$ is diffeomorphic to $S^1$. Suppose $S$ is a compact connected hyperbolic surface and $\gamma_1,...,\gamma_N$ are elements of $\pi_1(S)$ with $\ell(\gamma_i)\le \epsilon$ for all $i$; let $c_i, i=1,...,N$ denote the (oriented) closed geodesics in $S$ freely homotopic to the loops corresponding to $\gamma_i, i=1,...,N$. Then $c_i$'s are pairwise disjoint (or equal). Hence, $\{[c_1],...,[c_N]\}$ does not generate $H_1(S)$. In particular, $\{\gamma_1,...,\gamma_N\}$ does not generate $\pi_1(S)$.