We are interested in the following ´relative´ version of residual finiteness for fundamental groups of surfaces. Similar discussions where given in this question: [injectivity radius of hyperbolic surface](https://mathoverflow.net/questions/126352/injectivity-radius-of-hyperbolic-surface) (in particular with [this answer](https://mathoverflow.net/questions/126352/injectivity-radius-of-hyperbolic-surface/126372#126372)). 


The question is the following: 


Given a closed hyperbolic surface $S$ with a simple closed geodesic of length less than $\ell$ and a value $K>0$, does there exist a finite cover $p: \hat S \to S$ such that $\hat S$ has a unique simple closed geodesic of length less than $\ell$ and every other simple closed geodesic has length $\geq K$. 


(Notice that such a cover cannot be a [normal covering](https://en.wikipedia.org/wiki/Covering_space#Deck_transformation_group.2C_regular_covers) as there cannot be an isometry of $\hat S$ sending $\gamma$ (the curve of length $\leq \ell$) to other preimages of $p(\gamma)$. ) 



Finite covers of hyperbolic surfaces and the `second systole´
[17/4/17: edited to correct proof.]


This is true. First, we need a lemma which builds a related cover. Throughout, $\alpha$ is a simple closed geodesic of length $\ell$, and $\beta\_1,\ldots,\beta\_n$ are the finitely many (not necessarily simple) closed geodesics on $S$ of length at most $K$ which are not equal to $\alpha$. Note that we may assume that $\alpha$ is non-separating, by passing to a double cover if necessary.



> 
> **Lemma 1:** Let $S$ be a closed, hyperbolic, orientable surface and $\alpha$ a simple closed curve. For every $K>0$ there exists a finite-sheeted cover $\widetilde{S}\to S$ so that $\alpha$ lifts to $\widetilde{S}$ and every simple closed curve in $\widetilde{S}$ of length less than $K$ is a preimage of $\alpha$.
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*Proof:*
Let $S\_0$ be the result of cutting $S$ along $\alpha$. 


Consider the result of killing $\alpha^k$, for sufficiently large $k$. There are various ways of thinking about this; my preferred way is to think of this as an orbispace $\Sigma$, obtained as follows. First, construct an orbifold $\Sigma\_0$ from $S\_0$ by replacing the two boundary components with cone points of degree $k$; then glue the two cone points together to obtain $\Sigma$.


Note that $\Sigma\_0$ is still a hyperbolic orbifold; in particular, its fundamental group is Fuchsian and therefore residually finite. Furthermore, it's a classical fact that the fundamental group of a graph of residually finite groups with finite edge groups is itself residually finite; in particular, $\pi\_1\Sigma=\pi\_1S/\langle\langle\alpha^k\rangle\rangle$ is residually finite.


We next consider the images $\bar{\beta}\_i$ of the $\beta\_i$ in the orbispace $\Sigma$. For sufficiently large $k$, the images $\bar{\beta}\_i$ still have infinite order in $\pi\_1\Sigma$. This follows from the combinatorial Dehn filling machinery of Osin/Groves--Manning, but one can certainly give a more elementary proof in this context.


Since $\pi\_1\Sigma$ is residually finite, we may find a finite quotient
$$
\pi\_1S\stackrel{\pi}{\to}\pi\_1\Sigma\stackrel{\eta}{\to} Q
$$ 
so that $\eta(\bar{\beta}\_i)$ has order greater than $k$, for all $i$, and $\eta(\bar{\alpha})$ has order $k$.


Finally, let $q=\eta\circ\pi$ and let $\widetilde{S}\to S$ be the covering space corresponding to the (usually not normal) subgroup $H=\langle \alpha\rangle\ker q$, which has finite index in $\pi\_1S$. Since $\alpha\in H$, we see that $\alpha$ lifts to $\widetilde{S}$. On the other hand, any closed geodesic $\tilde{\gamma}$ on $\widetilde{S}$ of length less than $K$ which doesn't map to a power of $\alpha$ maps to one of the $\beta\_i$, which leads to a contradiction since the $\beta\_i$ unwrap when they lift to $\widetilde{S}$. *QED*


At this point, we are only concerned about the components of the preimages of $\alpha$ on $\widetilde{S}$; let $\tilde{\alpha}\_0,\ldots,\tilde{\alpha}\_m$ denote these, where without loss of generality $\tilde{\alpha}\_0$ is a lift of $\alpha$. Note that they form a disjoint set of non-separating curves on $\widetilde{S}$, and the complementary regions are covers of $S\_0$, hence have genus at least one. 



> 
> **Lemma 2:** Fix an integer $N>0$. There is a finite-sheeted cover $\widehat{S}\to\widetilde{S}$ so that $\tilde{\alpha}\_0$ has a unique lift to $\widehat{S}$, and the remaining components of the preimages of the $\tilde{\alpha}\_i$ all unwrap at least $N$ times.
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*Proof:* By pinching a non-separating curve that separates a punctured torus containing $\tilde{\alpha}\_0$ from the other $\tilde{\alpha}\_i$, and collapsing the resulting torus to a circle, we obtain a surjection
$$
\pi\_1\widetilde{S}\to\langle\tilde{\alpha}\_0\rangle\*\pi\_1S'
$$
where $S'$ is a closed surface and the set $\{\tilde{\alpha}\_i\mid 1\leq i\leq n\}$ maps to a disjoint collection of non-separating simple closed curves on $\pi\_1S'$. Since the $\tilde{\alpha}\_i$ are all non-zero in $H\_1(S')$, it follows that we may find a surjection
$$
\pi\_1\widetilde{S}\to F\_2=\langle\tilde{\alpha}\_0\rangle\*\langle b\rangle
$$
so that the other $\tilde{\alpha}\_i$ all map to non-zero powers of $b$.


We now pick a large prime $p$, and map $F\_2\to \mathbb{F}^\times\_p\ltimes \mathbb{F}\_p$ in such a way that $\tilde{\alpha}\_0$ maps to a generator of the Frobenius complement $\mathbb{F}^\times\_p$, and $b$ maps to a generator of the Frobenius kernel $\mathbb{F}\_p$. This gives a surjection
$$
r: \pi\_1\widetilde{S}\to \mathbb{F}^\times\_p\ltimes \mathbb{F}\_p
$$
with the property that, for every $g\in\pi\_1\widetilde{S}$, $\langle r(\tilde{\alpha}\_0)\rangle\cap \langle r(\tilde{\alpha}\_0^g)\rangle=1$ unless $r(g)\in \langle r(\tilde{\alpha}\_0)\rangle$ (see [here](https://en.wikipedia.org/wiki/Frobenius_group)). For large enough $p$, we also have that every $r(\tilde{\alpha}\_i)$ has order greater than $N$, for every $i$. 


We now define $\widehat{S}\to\widetilde{S}$ to be the cover corresponding to the subgroup $\langle \tilde{\alpha}\_0\rangle\ker r=r^{-1}(\mathbb{F}\_p^\times)$. Clearly $\tilde{\alpha}\_0$ lifts to $\widehat{S}$.


On the other hand, covering space theory shows that any other component of the preimage of $\tilde{\alpha}\_i$ in $\widehat{S}$ corresponds to a non-trivial double coset $\langle r(\tilde{\alpha}\_0)\rangle r(g) \langle r(\tilde{\alpha}\_i)\rangle$, and the degree of unwrapping corresponds to the minimal power of $r(g\tilde{\alpha}\_ig^{-1})$ contained in $\langle r(\tilde{\alpha}\_0)\rangle$. By construction, this degree of unwrapping is at least $N$. *QED*


Choosing $\ell N>K$, we immediately obtain the cover we were looking for.



> 
> **Theorem:** For any $K>0$ there is a finite-sheeted cover $\widehat{S}\to S$ so that $\alpha$ lifts to $S$ and every other closed curve on $\widehat{S}$ is of length greater than $K$.
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