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 $\widehat{S}$$\widetilde{S}$ and every simple closed curve in $\widehat{S}$$\widetilde{S}$ of length less than $K$ is a preimage of $\alpha$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a preimage of $\alpha$.
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 $\alpha$.
[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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a liftpreimage of $\alpha$.
Consider the result of killing $\alpha^2$$\alpha^k$, for sufficiently large $k$. There are various ways of thinking about this; my preferred way is to think of this as an orbispace $\Sigma$, obtained as follows. First, construct an orbifold $\Sigma_0$ from $S_0$ by replacing the two boundary components with cone points of degree 2;$k$; then glue the two cone points together to obtain $\Sigma$.
Note that $\Sigma_0$ is still a hyperbolic orbifold; in particular, its fundamental group is Fuchsian and therefore residually finite. Furthermore, it's a classical fact that the fundamental group of a graph of residually finite groups with finite edge groups is itself residually finite; in particular, $\pi_1\Sigma=\pi_1S/\langle\langle\alpha^2\rangle\rangle$$\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 $\Sigma$. Since a curve on a cone-type orbifold is torsion if and only if it bounds a cone point, and since the $\beta_i$ were in minimal position with respect toFor sufficiently large $\alpha$$k$, it follows that the images $\bar{\beta}_i$ still have infinite order in $\pi_1\Sigma$. This follows from the combinatorial Dehn filling machinery of Osin/Groves--Manning, but one can certainly give a more elementary proof in this context.
Since $\pi_1\Sigma$ is residually finite, we may find a finite quotient $$ \pi_1S\stackrel{\pi}{\to}\pi_1\Sigma\stackrel{\eta}{\to} Q $$ so that $\eta(\bar{\beta}_i)$ has order at least 3greater than $k$, for all $i$, and $\eta(\bar{\alpha})$ has order 2$k$.
To construct $\widehat{S}$ as required by the questionAt this point, we now modifyare only concerned about the constructioncomponents of the preimages of $\alpha$ on $\widetilde{S}$ slightly; let $\tilde{\alpha}_0,\ldots,\tilde{\alpha}_m$ denote these, using the following lemma.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for each $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle $$ whenever $r_N(g)\notin\langle r_N(\alpha)\rangle$.
Proof: Sincewhere without loss of generality $\pi_1S$ maps to$\tilde{\alpha}_0$ is a free group sendinglift of $\alpha$ to a basis element this follows from the fact that there are such elements in finite groups. Frobenius groups, for instance, provide examples. Note that they form a disjoint set of non-separating curves on QED
To complete the construction$\widetilde{S}$, we just modifyand the map $q$ usingcomplementary regions are covers of $r_N$$S_0$, for large enough $N$hence have genus at least one.
TheoremLemma 2: For any Fix an integer $K>0$ there$N>0$. There is a finite-sheeted cover $\widehat{S}\to S$$\widehat{S}\to\widetilde{S}$ so that $\alpha$ lifts$\tilde{\alpha}_0$ has a unique lift to $S$ and every other closed curve on $\widehat{S}$ is, and the remaining components of length greater thanthe preimages of the $K$$\tilde{\alpha}_i$ all unwrap at least $N$ times.
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 Let$$ \pi_1\widetilde{S}\to\langle\tilde{\alpha}_0\rangle*\pi_1S' $$ where $q:\pi_1S\to Q$ be as$S'$ 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 proofother $\tilde{\alpha}_i$ all map to non-zero powers of Lemma 1$b$. Let
We now pick a large prime $N$ be$p$, and map $F_2\to \mathbb{F}^\times_p\ltimes \mathbb{F}_p$ in such a way that $\ell N>K$$\tilde{\alpha}_0$ maps to a generator of the Frobenius complement $\mathbb{F}^\times_p$, and let $\widehat{S}\to S$ be the covering corresponding$b$ maps to a generator of the subgroupFrobenius kernel $\langle \alpha\rangle \ker(q\times r_N)$$\mathbb{F}_p$. For notational convenience This gives a surjection $$ r: \pi_1\widetilde{S}\to \mathbb{F}^\times_p\ltimes \mathbb{F}_p $$ with the property that, let's writefor every $p=q\times r_N$$g\in\pi_1\widetilde{S}$, $\langle r(\tilde{\alpha}_0)\rangle\cap \langle r(\tilde{\alpha}_0^g)\rangle=1$ unless $r(g)\in \langle r(\tilde{\alpha}_0)\rangle$ (see here). For large enough $p$, we also have that every $r(\tilde{\alpha}_i)$ has order greater than $N$, for every $i$.
ClearlyWe now define $\widehat{S}$ also satisfies$\widehat{S}\to\widetilde{S}$ to be the conclusion of Lemma 1, andcover corresponding to the subgroup $\alpha$$\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 $\alpha$$\tilde{\alpha}_i$ in $\widehat{S}$ corresponds to a non-trivial double coset $\langle p(\alpha)\rangle p(g) \langle p(\alpha)\rangle$$\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 $p(\alpha^g)$$r(g\tilde{\alpha}_ig^{-1})$ contained in $\langle p(\alpha)\rangle$$\langle r(\tilde{\alpha}_0)\rangle$. By Lemma 2construction, this degree of unwrapping is at least $N$, and hence every other component of the preimage of $\alpha$ in $\widehat{S}$ has length at least $\ell N>K$, as required. 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 $K$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a lift of $\alpha$.
Consider the result of killing $\alpha^2$. 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 2; 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, $\pi_1\Sigma=\pi_1S/\langle\langle\alpha^2\rangle\rangle$ is residually finite.
We next consider the images $\bar{\beta}_i$ of the $\beta_i$ in the orbispace $\Sigma$. Since a curve on a cone-type orbifold is torsion if and only if it bounds a cone point, and since the $\beta_i$ were in minimal position with respect to $\alpha$, it follows that the images $\bar{\beta}_i$ still have infinite order in $\pi_1\Sigma$.
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 at least 3, for all $i$, and $\eta(\bar{\alpha})$ has order 2.
To construct $\widehat{S}$ as required by the question, we now modify the construction of $\widetilde{S}$ slightly, using the following lemma.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for each $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle $$ whenever $r_N(g)\notin\langle r_N(\alpha)\rangle$.
Proof: Since $\pi_1S$ maps to a free group sending $\alpha$ to a basis element this follows from the fact that there are such elements in finite groups. Frobenius groups, for instance, provide examples. QED
To complete the construction, we just modify the map $q$ using $r_N$, for large enough $N$.
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 $K$.
Proof: Let $q:\pi_1S\to Q$ be as in the proof of Lemma 1. Let $N$ be such that $\ell N>K$, and let $\widehat{S}\to S$ be the covering corresponding to the subgroup $\langle \alpha\rangle \ker(q\times r_N)$. For notational convenience, let's write $p=q\times r_N$.
Clearly $\widehat{S}$ also satisfies the conclusion of Lemma 1, and $\alpha$ lifts to $\widehat{S}$.
On the other hand, covering space theory shows that any other component of the preimage of $\alpha$ in $\widehat{S}$ corresponds to a non-trivial double coset $\langle p(\alpha)\rangle p(g) \langle p(\alpha)\rangle$, and the degree of unwrapping corresponds to the minimal power of $p(\alpha^g)$ contained in $\langle p(\alpha)\rangle$. By Lemma 2, this degree of unwrapping is at least $N$, and hence every other component of the preimage of $\alpha$ in $\widehat{S}$ has length at least $\ell N>K$, as required. QED
[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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a preimage of $\alpha$.
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 $k$.
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.
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). 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 $K$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a lift of $\alpha$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a lift of $\alpha$.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for each $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle $$ whenever $r_N(g)\notin\langle r_N(\alpha)\rangle$.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for all $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle~. $$ Proof: Since $\pi_1S$ maps to a free group sending $\alpha$ to a basis element this follows from the fact that there are such elements in finite groups. Frobenius groups, for instance, provide examples. QED
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 $K$.
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 $K$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a lift of $\alpha$.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for all $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle~. $$ Proof: Since $\pi_1S$ maps to a free group sending $\alpha$ to a basis element this follows from the fact that there are such elements in finite groups. QED
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 $K$.
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 $\widehat{S}$ and every simple closed curve in $\widehat{S}$ of length less than $K$ is a lift of $\alpha$.
Lemma 2: For any $N$ there is a homomorphism $r_N:\pi_1S\to R_N$, for $R_N$ a finite group, so that, for each $g\in\pi_1S$, $$ \langle r_N(\alpha^g)\rangle\cap\langle r_N(\alpha)\rangle\subseteq \langle r_N(\alpha)^N\rangle $$ whenever $r_N(g)\notin\langle r_N(\alpha)\rangle$.
Proof: Since $\pi_1S$ maps to a free group sending $\alpha$ to a basis element this follows from the fact that there are such elements in finite groups. Frobenius groups, for instance, provide examples. QED
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 $K$.