# Curves on surfaces lifting to embedded curves in finite covers

Let $$S$$ be a orientable closed surface with genus $$g \geq 1$$ and let $$\gamma \subset S$$ be an immersed curve. Does there exist a finite cover of $$S$$ where $$\gamma$$ lifts to a curve that is homotopic to an embedded curve?

If so, and $$\gamma$$ has $$k$$ double points, is there an explicit bound on the degree of the finite cover that we need to take in order to lift $$\gamma$$ to a curve homotopic to an embedded curve?

• Doesn't this depend on the topology of $\gamma$? What if the curve crosses itself in a tiny disk in $S$? – Ted Shifrin Dec 21 '18 at 17:10
• @TedShifrin: cant you just slightly perturb $\gamma$ so that it's self-transverse? – Hempelicious Dec 21 '18 at 22:38
• I was in fact assuming it was self-transverse, but even so you still won't lift a neighborhood of the crossing point to a smooth curve. @Hempelicious – Ted Shifrin Dec 21 '18 at 22:43
• @TedShifrin I will edit things to be more clear - I am really only concerned in my curves up to homotopy. – user101010 Dec 21 '18 at 23:37
• Aha, @user101010, so what is going on topologically? When can you homotop an immersion to an embedding in the first place, and when can you not? – Ted Shifrin Dec 21 '18 at 23:54

The links in Lee's answer give you a part of the story but not the whole story. There is a missing step for going from residual finiteness to the "lifting property". To explain the missing part I will need two definitions.

Definition 1. A subgroup $$H$$ of a group $$G$$ is called "separable" (or "closed in profinite topology") if for every finite subset $$F\subset G-H$$ there exists a finite index subgroup $$G_0< G$$ which contains $$H$$ and is disjoint from $$F$$.

Thus, residual finiteness of $$G$$ amounts to separability of the trivial subgroup of $$G$$. In general, residual finiteness of $$G$$ does not imply separability of cyclic subgroups.

Example. Let $$G=BS(p,1)=$$ for some $$p>1$$. This group is residually finite, even linear: It embeds in the subgroup of upper triangular matrices in $$SL(2,{\mathbb Q})$$. However, the cyclic subgroup $$H=$$ is not separable. Indeed, $$a\notin H$$. On the other hand, for every homomorphism $$\phi: G\to F$$ to a finite group, the subgroups $$<\phi(a)>$$ and $$<\phi(a^p)>$$ are conjugate in $$F$$, hence, have the same order. Since $$<\phi(a^p)>\subset <\phi(a)>$$ it follows that these cyclic groups are equal. Hence, $$\phi(a)\in \phi(H)$$ for every homomorphism $$\phi$$ from $$G$$ to a finite group and, thus, $$H$$ is not a separable subgroup of $$G$$.

Definition 2. An abelian subgroup $$H< G$$ is "self-centralizing" if the centralizer of $$H$$ in $$G$$ equals $$H$$.

This condition clearly fails for the cyclic subgroup $$H=$$ in the example.

Lemma 1. If $$H$$ is finitely generated, abelian, self-centralizing and $$G$$ is residually finite, then $$H$$ is separable.

Proof. Let $$F\subset G-H$$ be a finite subset. Let $$h_1,...,h_n$$ be a finite generating set of $$H$$. Define the finite subset $$C\subset G$$ to be the set of commutators $$[h_i,f], f\in F$$. Since $$G$$ is residually finite, there exists a finite group $$K$$ and a homomorphism $$\phi: G\to K$$ such that $$\phi(C)$$ does not contain the neutral element $$e\in K$$. Since $$H$$ is self-centralizing, for each $$f\in F$$, $$\phi(f)\notin \phi(H)$$ (otherwise, $$\phi(f)$$ commutes with all the generators of $$\phi(H)$$). Then let $$G_0:= \phi^{-1}(\phi(H)).$$ This is clearly a finite index subgroup containing $$H$$, but disjoint from $$F$$ (as noted above). qed

As a corollary we obtain that maximal cyclic subgroups of surface groups are separable. In fact, all finitely generated subgroups of surface groups are separable, but this is a much harder result (due to Peter Scott).

Lemma 2. Let $$G=\pi_1(S)$$ be a hyperbolic surface group, $$A< G$$ is a maximal cyclic subgroup. Then there exists a finite covering $$S_0\to S$$ such that $$A$$ is realized by the fundamental group of a simple loop in $$S_0$$.

Proof. Let $${\mathbb H}^2\to S$$ be the universal covering of $$S$$. Let $$L\subset {\mathbb H}^2$$ be the unique geodesic invariant under the action of $$A$$. Let $$D=pq\subset L$$ be a subinterval which is a fundamental domain for the action of $$A$$ on $$L$$, i.e. if $$a$$ is a generator of $$A$$, then $$a(p)=q$$. Since $$D$$ is compact and the action of $$G$$ on $${\mathbb H}^2$$ is properly discontinuous, there is a finite subset $$F\subset G$$ such that for all $$g\in G-F$$, either $$g(D)\cap D=\emptyset$$ or $$g=e$$, the neutral element. By Lemma 1, there exists a finite index subgroup $$G_0< G$$ disjoint from $$F$$ and containing $$A$$. Then, the projection of $$L$$ (which is the same as the projection of $$D$$) to $$S_0;{\mathbb H}^2/G_0$$ is a simple loop. qed

From this proof you can see that residual finiteness of $$G$$ is not enough here, one needs separability of $$A$$ in $$G$$. You also get an answer to the question asked in a comment to Lee's answer. The subgroup separability (even for surface subgroups) of hyperbolic 3-manifold groups was unknown before the work of Agol and Wise. You cannot use the "commutator trick" from Lemma 1 since surface subgroups are non-abelian. A version of this trick works for maximal surface subgroups represented by immersed totally geodesic subsurfaces, it is due to D.Long, see here. But this case is too special.

We are not done yet, as we still have to deal with non-maximal cyclic subgroups.

Lemma 3. Given a maximal cyclic subgroup $$A< G$$ as above, there is a finite covering $$S_1\to S$$ such that $$A$$ is represented by a simple nonseparating loop in $$S_1$$.

I will skip the proof, for instance, you again can use the separability of $$A$$.

Lemma 4. The conclusion of Lemma 2 holds for all infinite cyclic subgroups of $$G$$.

Proof. Let $$C< G$$ be an infinite cyclic subgroup and $$C< A$$ is a maximal cyclic subgroup of $$G$$ containing $$C$$. Let $$i$$ denote the index of $$C$$ in $$A$$. By lemma 3, we can assume that $$A$$ is realized by a simple nonseparating loop $$\alpha$$ on our hyperbolic surface $$S$$. Let $$\sigma\in H^1(S; {\mathbb Z})$$ be the cohomology class Poincare dual to the homology class represented by $$\alpha$$. Then $$\sigma$$ lifts to a homomorphism $$\psi: G\to H_1(S; {\mathbb Z})\to {\mathbb Z}$$. Let $$G_2:= \psi^{-1}(i{\mathbb Z})$$; it is an index $$i$$ subgroup of $$G$$ containing $$C$$. The loop $$\alpha^i$$ then lifts to a simple loop on $$S_2:= {\mathbb H}^2/G_2$$. qed

This proof actually works in much greater generality and provides a topological interpretation of subgroup separability due to Peter Scott:

Lemma 5. A subgroup $$H< G$$ of a countable group $$G$$ is separable if and only if the following holds: For every locally compact Hausdorff topological space $$X$$ (say, a manifold), a properly discontinuous action $$G\times X\to X$$ and a compact subset $$K\subset X/H$$, there exists a finite index subgroup $$G_0 containing $$H$$ such that the projection of the covering map $$p: X/H\to X/G_0$$ restricts to a one-to-one map on $$K$$.

In Lemma 4, the space $$X$$ is the universal covering of $$S$$, the action of $$G$$ on $$X$$ is the covering action and the compact subset $$K$$ is a simple loop which carries the fundamental group of the cylinder $$X/H$$.

The proof of Lemma 4 of course, will not give you an estimate on the degree of a finite covering of $$S$$ to which the given loop $$\gamma$$ in $$S$$ lifts to a loop homotopic to a simple loop. Getting such an estimate is an interesting project worth pursuing. If you are interested: The first step is to get such an estimate in the case when $$S$$ is replaced with a finite graph $${\mathcal G}$$. The construction of a finite covering $${\mathcal G}_0\to {\mathcal G}$$ to which given "tight" loop $$\gamma$$ in $${\mathcal G}$$ lifts to a simple loop is pretty explicit, you can find it for instance in the paper

P.Scott, T.Wall "Topological methods in group theory".

From this proof you can extract an estimate on the degree of the covering. After that, if you are still interested in the problem, you can read the proof of subgroup separability of surface groups which is due to Wise, his argument is combinatorial, inspired by the one by Scott and Wall, and you can get an explicit degree estimate.

• Thanks, this looks wonderful and will take me a while to get through. I was just in the process of looking at the references that LeeMosher mentioned. From his reference I saw how I could take any loop and lift it to some cover where it would be an embedding - but not a loop! This seems to follow from residual finiteness now that I have some idea what is going on. I guess your argument is fixing this issue. Is that a correct summary? – user101010 Dec 23 '18 at 20:22
• So it was me that asked the question in the comments, not OP, but thank you for answering! From my limited searching since then, it seems like Wise showed certain subgroups of RAAGs are LERF, and Agol proved $\pi_1(M)$ for a closed hyperbolic 3-manifold are always (virtually) such subgroups. Part of my perpetual confusion on this was why the full strength of Kahn-Markovic was needed- it seemed like a surface subgroup + LERF is enough. But now I understand (slightly) that Kahn-Markovic gets you the cube complex. – Hempelicious Dec 24 '18 at 2:06
• @user101010: Right. – Moishe Kohan Dec 24 '18 at 12:03
• @Hempelicious: It is more complicated. Wise proves that a "special" CAT(0) cube complex gives subgroup separability for quasiconvex subgroups of hyperbolic groups. Kahn-Markovic (plus Sageev) give you a CAT(0) cube complex. Agol proves that CAT(0) cube complex implies virtually special cube complex. – Moishe Kohan Dec 24 '18 at 12:06

The existence of the finite cover you are asking for is a well known consequence of the residual finiteness of surface groups, a theorem with a long history.

For a discussion of that theorem, which starts out with the proof that answers your first question, see this post and its followup.

For your second question I don't have a good answer. Also, I'm not sure what you would count as an "explicit" bound. However, I know exactly how I would go about finding an "explicit" bound in the case of a free group, where the proof of residual finiteness is much easier. I'm guessing that the surface group case can also be done by working through the proof of residual finiteness.

Comment: As pointed out in the answer of @MoisheCohen, my answer is incomplete. Essentially what I have written produces a cover of $$S$$ such that some iterate $$\gamma^n$$ lifts to an embedded curve. The additional requirement that $$n=1$$ needs subgroup separability ideas as expalined in the other answer, and as applied to the cyclic subgroup generated by $$\gamma$$.

• can I ask an off-topic but related question? What goes wrong in the 3-dimensional case? If we have a 3-manifold $M$ and a surface subgroup $S\le \pi_1(M)$, why can't we use residual finiteness to lift this to an embedding (of a different genus surface) in a finite cover of $M$? From the very limited I know, Agol had to prove the stronger RFRS condition to get there. [With the goal of having an incompressible surface to show a finite cover was Haken.] – Hempelicious Dec 22 '18 at 17:39
• That is a good question, but I would suggest that you ask it in a new question. As a comment to my answer, it's kind of invisible to anyone but you and me. – Lee Mosher Dec 23 '18 at 1:16