12
$\begingroup$

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 (in particular with this answer).

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 as there cannot be an isometry of $\hat S$ sending $\gamma$ (the curve of length $\leq \ell$) to other preimages of $p(\gamma)$. )

$\endgroup$
  • $\begingroup$ I am a little confused. If there is a unique geodesic 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? $\endgroup$ – Igor Rivin Mar 13 '17 at 23:37
  • $\begingroup$ @IgorRivin The unique (simple closed) geodesic of length $\leq \ell$ is in $\hat S$, so it projects to some curve $\gamma$ of length $\leq \ell$ in $S$. Clearly, the preimage of $\gamma$ by $p$ has many other preimages which are longer than $K$ in addition to the 'short' one. For other curves of length $\leq K$ in $S$ different from $\gamma$, \textit{all} the preimages in $\hat S$ must have length $\geq K$. Did I understand right? $\endgroup$ – rpotrie Mar 14 '17 at 0:34
  • 2
    $\begingroup$ Yes, that's right, I was asleep, BUT I believe that what I said makes sense for regular covers. $\endgroup$ – Igor Rivin Mar 14 '17 at 0:49
  • $\begingroup$ Correct. I just edited to emphasise that. Thanks!. $\endgroup$ – rpotrie Mar 14 '17 at 0:55
  • 1
    $\begingroup$ I can construct a cover with the property that only lifts of your favouite curve have length $\ell$, and all other sccs have length greater than any fixed $K$. It should be relatively easy to get from there to what you want. I'll try to work out the details in the morning. $\endgroup$ – HJRW Mar 14 '17 at 22:38
5
$\begingroup$

[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 $K$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ The general principle is that if you can kill (a power of) the element that you're interested in and obtain something that's still residually finite, then you are in good shape. In this case, we can do it explicitly because the curve you are interested in is simple and so the quotient is an orbispace. Wise's Malnormal Special Quotient Theorem can be used in much more general situations. $\endgroup$ – HJRW Mar 15 '17 at 14:58
  • 1
    $\begingroup$ @HJRW: Very good. The first step can be a bit simplified (depending on one's taste) using the fact that the quotient of $\pi_1(S)$ by $<<\gamma^n>>$, where $\gamma$ is a simple loop, is the fundamental group of a 3-dimensional hyperbolic orbifold and, hence, is residually finite. I do not know if a similar thing is true for nonsimple loops (where instead of 3-d hyperbolic orbifolds one uses higher-dimensional ones). I think, it is true if n is large enough. $\endgroup$ – Misha Mar 15 '17 at 15:48
  • 1
    $\begingroup$ @HJRW: I'm confused about Lemma 2. If $r_N(g)$ centralizes $\alpha$, then this won't be true (assuming $\langle r_N(\alpha)\rangle \neq \langle r_N(\alpha)\rangle^N$, which you seem to be assuming given the later argument). So it seems to me that you need to make some extra assumption here, or maybe I am misunderstanding the quantifiers. Anyway, a reference to this fact about finite groups would be helpful. $\endgroup$ – Ian Agol Mar 15 '17 at 19:45
  • 1
    $\begingroup$ @IanAgol, sorry, yes, it only needs to be true for $r_N(g)$ not a power of $r_N(\alpha)$. I'll edit in the morning. For specific examples, Frobenius complements, for instance, can be used and are overkill. $\endgroup$ – HJRW Mar 15 '17 at 21:16
  • 1
    $\begingroup$ @Misha, I would say that the orbifold theorem is a much bigger hammer than the residual finiteness gluing result I quoted! But I would be very interested in a reference for the fact that killing high powers of nonsimple loops gives orbifolds. (In this case, if they're higher dimensional, is there any reason to think that they're residually finite? In fact, they are residually finite by Agol--Wise, but a proof avoiding cubical machinery would be interesting.) $\endgroup$ – HJRW Mar 16 '17 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.