[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 $\alpha$.

*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.

*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 $K$.

uniquegeodesic of length $\leq \ell,$ then what exactly would cover it? Either the cover is connected (so you lose $\leq \ell,$ or not, so you lose uniqueness. Am I missing something? $\endgroup$regularcovers. $\endgroup$1more comment