# Fundamental groups of noncompact surfaces

I got fantastic answers to my previous question (about modern references for the fact that surfaces can be triangulated), so I thought I'd ask a related question. A basic fact about surface topology is that if $S$ is a noncompact connected surface, then $\pi_1(S)$ is a free group (possibly trivial or $\mathbb{Z}$). I've had a lot of people ask me for references for this fact. I know of two such references:

1) In section 44A of Ahlfors's book on Riemann surfaces, he gives a very complicated combinatorial proof of this fact.

2) This isn't a reference, but a high-powered 2-line proof. Introducing a conformal structure, the uniformization theorem shows that the universal cover of $S$ is contractible. In other words, $S$ is a $K(\pi,1)$ for $\pi=\pi_1(S)$. Next, since $S$ is a noncompact $2$-manifold, its integral homology groups vanish in dimensions greater than or equal to $2$. We conclude that $\pi_1(S)$ is a group of cohomological dimension $1$, so a deep theorem of Stallings and Swan says that $\pi_1(S)$ is free.

There should be a proof of this that you can present in a first course in topology! Does anyone know a reference for one?

• If you accept the polygonal representation of a closed surface on the plane, then a punctured surface will be represented by the same polygon with punctures easily managable to be in the interior, but now this picture clearly homotopes to a wedge of circles hence $\pi_1$ is free. – Maharana Mar 17 '10 at 6:11
• Andy, one does not need uniformization to show that $S$ is Eilenberg-MacLane. Namely, the universal cover of $S$ is acyclic (it does not have homology above dimension $1$ by Poincare duality and it does not have first homology since it is simply-connected). Now Whitehead theorem gives you that the inclusion of any point into the universal cover is a homotopy equivalence. – Igor Belegradek Mar 17 '10 at 13:04
• Igor - Excellent observation! Thanks! – Andy Putman Mar 17 '10 at 13:42
• Maharana - I'm not talking about punctured surfaces. I'm talking about general noncompact surfaces, like a closed surface minus a Cantor set or the boundary of a regular neighborhood of an infinite graph embedded in R^3. – Andy Putman Mar 17 '10 at 13:44
• ah, got it. Was thinking of a very special and easy case! – Maharana Mar 18 '10 at 3:22

I'm reluctant to advertise, but since no one else has answered yet, I'll mention the proof on pp. 142--144 of my book Classical Topology and Combinatorial Group Theory.

• Very nice! Another nice feature is that you include a reference to a paper of Johansson from 1931 that (I guess) is the origin of this result. – Andy Putman Mar 17 '10 at 2:57
• Yes, I'm confident that the result is due to Johansson. – John Stillwell Mar 17 '10 at 2:58
• If I had been faster I would have proposed the same reference so you wouldn't have had to advertise your own book. I'm a fan :-) – Alon Amit Mar 17 '10 at 16:52
• That book is surely in the list of books that should be advertised loud and often! :) – Mariano Suárez-Álvarez Mar 31 '12 at 0:19
• @Andy I think checkmarks only move an answer to the top when it's not the original poster's answer. – Mike Miller May 25 '18 at 15:53

If you assume the existence of smooth structure on a noncompact surface then it is easy to show the existence of a proper morse function with no local maximum.This shows that the surface is homotopic to a one dim CW complex. This is the smooth version of Igor's answer.

EDIT BY ANDY PUTMAN: Mohan isn't registered and thus isn't able to comment, but he sent me an email with more details. The result is true in all dimensions : any noncompact smooth n-manifold is homotopy equivalent to an n-1 complex. The key is to construct a strictly subharmonic morse exhaustion function. The subharmonicity prevents the function from having local maxima. Details of this can be found in his paper "Elementary Construction of Exhausting Subsolutions of Ellitpic Operators", which was joint with Napier and was published in L’Enseignement Math´ematique, t. 50 (2004), p. 1–24.

• Is there a topological version? – Mariano Suárez-Álvarez Mar 17 '10 at 15:17
• Is this a general fact about noncompact manifolds, or is this special to surfaces? I haven't managed to prove it, but that doesn't mean that it isn't as easy as you claim it is... – Andy Putman Mar 17 '10 at 15:34
• It would be nice to have a good source of information on proper Morse functions; I could not find any reasonable account, and I suspect there may be some subtle issues there. But I agree with Mohan that Morse theory argument should work for the problem at hand, and I think it should work in any dimension. – Igor Belegradek Mar 17 '10 at 16:05
• I was not aware of this interesting result, and I cannot immediately picture it visually for $\mathbb{R}P^2$ or for Klein bottle. Could you refer me to an explicit construction for either? – Michael May 25 '18 at 16:35
• @Michael: Since both the real projective plane and the Klein bottle are compact, the theorem doesn't apply to them. – Andy Putman May 25 '18 at 20:25

I just ran across this question, and thought I would give a precise version of the proof Ilya suggested. I believe I learned this proof in Richie Miller's topology course, Michigan State University, 1977 or so.

Choose a triangulation of the surface $S$, equipped with the simplicial metric. Choose a maximal one-ended subtree $T$ of the dual 1-skeleton $S^{(1)}$. The subtree $T$ contains every dual $0$-cell, that is, the barycenter of every 2-simplex. Also, $T$ contains dual 1-cells crossing certain $1$-simplices. Let $U$ be the union of the open 2-simplices and open 1-simplices that contain a point of $T$. The metric completion of $U$, denoted $\bar U$, is a closed disc with one boundary point removed, and so there is a deformation retraction from $\bar U$ onto its boundary $\partial \bar U$. Attaching $\bar U$ to $S - U$ in the obvious way to form the surface $S$, the deformation retraction $\bar U \to \partial\bar U$ induces a deformation retraction of $S$ onto $S-U$, wnich is a subcomplex of the 1-skeleton.

By the way, the subtree $T \subset S^{(1)}$ can be constructed by an explicit process. Enumerate the dual $0$-cells $v_1,v_2,\ldots \in S^{(1)}$. Construct one-ended subtrees $T_1,T_2,\ldots \subset S^{(1)}$ as follows. $T_1$ is any proper ray based at $v_1$. If $v_n \in T_{n-1}$ then $T_n = T_{n-1}$. If $v_n \not\in T_{n-1}$, let $T_n$ be the union of $T_{n-1}$ with any arc $\alpha \subset S^{(1)}$ having one endpoint at $v_n$ and intersecting $T_{n-1}$ in its opposite endpoint. Each $T_n$ is a one-ended tree by induction, and since the radius $r$ neighborhood of $v_1$ in $T_n$ stabilizes as $n \to \infty$, it follows that $T = \cup_n T_n$ is a one-ended subtree of $S^{(1)}$, and it is maximal because it contains each $v_i$.

I think this proof generalizes to any dimension, to give the theorem that Igor Belegradek refers to.

--- Edited to simplify and clarify the argument ---

• This is a beautiful proof. Thanks Lee! – Andy Putman Mar 30 '12 at 23:46
• Sorry to raise an issue in an ancient answer, but I was explaining this to one of my graduate students and I realized that I don't understand your argument for why you can choose a single one-ended tree. It is true that the radius $r$ neighborhood of $v_1$ stabilizes, but I don't think that means the union is one-ended. While I don't have a counterexample for triangulated surfaces, if you apply your argument to $\mathbb{R}^1$ with its usual triangulation, the tree you end up with has two ends. What I think you should do instead is construct a forest in the dual triangulation each – Andy Putman May 23 '18 at 18:39
• component of which is $1$-ended. You then collapse each component. This is, by the way, what is done in the paper of Whitehead that Igor references in his answer. – Andy Putman May 23 '18 at 18:39
• @user_1789 It's not possible for the map from the components of the forest to the ends of $S$ to be onto. The surface can have uncountably many ends (for example a sphere minus a Cantor set), but there's only countably many vertices hence only countably many components to the forest. – Lee Mosher May 26 '18 at 12:56
• @AndyPutman Thanks for pointing this out Andy. Looking back on it, I can see I was overly optimistic in trying for a connected forest. I had hoped to avoid Zorn's Lemma, but maybe that's not realistic........ It does make me want to ask the logicians if this might be another avatar of Zorn's Lemma, i.e. another equivalent property ... – Lee Mosher May 26 '18 at 12:58

I don't have a reference but here is one way to prove what you want. It is a basic result of PL-topology that any open PL-manifold deformations retracts to a subcomplex of lower dimension. Thus you are reduced to showing that the fundamental group of a graph is free, so collapse a maximal subtree to a point, get a wedge of circles and apply Van Kampen.

• Do you (or anyone else) know a reference for this "basic fact in PL-topology". I just skimmed through Rourke-Sanderson and didn't find it. – Andy Putman Mar 17 '10 at 3:19
• Andy, this fact was mentioned in Brown's book "Cohomology of groups" in the proof of Proposition 8.1 (Chapter 8, Section 8). Brown does not give a reference. I do not know how the proof goes, and also would be interested in a reference. – Igor Belegradek Mar 17 '10 at 12:07
• @Andy: This doesn't seem to be so hard if we can assume that the surface is a compact triangulated surface with some points removed. This is the same as removing the interiors of a few triangles. Then, whenever a see an edge that bounds a triangle on only one side, we remove both the edge and the interior of that triangle. This is a deform. retraction. The only way that this process can stop is when there are no filled-in triangles (assuming the surface was connected), so we have something 1-dimensional left. Did I miss something? Or do we need some generalization of this in your case? – Ilya Grigoriev Mar 17 '10 at 17:37
• @Ilya - that is the trivial case! There are a whole zoo of non-compact surfaces that are nowhere close to these examples. For instance, take a compact surface and remove a cantor set. I want arguments that handle those cases as well. – Andy Putman Mar 17 '10 at 18:06

Not to take anything away from the other answers, but I believe that the result (the more general n-dimensional version mentioned by @Igor Belegradek) is actually due to J. H. C. Whitehead: The immersion of an open 3-manifold in euclidean space, Proc. London Math. Soc 11 1961, 81-90., lemma 2.1 (JHC was a little modest calling this a lemma).

• @Igor: That's a nice reference! Actually, Whitehead's proof is essentially the same as the one outlined by Lee. – Misha Mar 31 '12 at 3:10
• @Misha: Thanks! I did notice that this was the same argument -- I guess Whitehead found the canonical argument... – Igor Rivin Mar 31 '12 at 4:30

In case anyone is interested, I wrote up a detailed account synthesizing the various answers here and correcting some issues I ran into. It is entitled "Spines of manifolds and the freeness of fundamental groups of noncompact surfaces" and can be downloaded from my page of notes here.

Related to Mohan's answer, one can give an overkill proof using the fact that non-compact Riemann surfaces are Stein, and every complex $n$-dimensional Stein manifold is homotopy equivalent to an n-dimensional CW complex. This is Theorem 7.2 in Milnor's book on Morse theory.

• That’s a hilariously awesome way of doing this! – Andy Putman Jul 24 '18 at 3:50

A new approach to spines is available via mass transport theory and Kantorovich duality. This is developed in my PhD thesis.

The idea is elementary: consider the retract $$x\mapsto x/|x|$$ from the closed unit ball $$B:=\{x \in \mathbb{R}^{N} | ~~ |x| \leq 1\}$$ to its boundary sphere $$\partial B$$. The retract has locus-of-discontinuity $$Z=\{pt\}$$ equal to a point, namely $$x=0$$. Observe that the inclusion $$\{pt\} \hookrightarrow B$$ is a homotopy-equivalence. Claim: this is general principle which follows from Kantorovich duality and the optimal transport theory.

For example, let $$S$$ be closed hyperbolic surface with metric $$d$$, and $$C\hookrightarrow S$$ an embedded Cantor set. Let $$X:=S-C$$ be the Cantor-punctured surface, and let $$\sigma$$ be the Hausdorff measure on $$X$$. Similarly let $$\tau$$ be the Hausdorff measure of $$C$$ viewed as a subset of $$(S,d)$$. Now consider the function $$c: X \times C \to (0,\infty)$$ defined by the rule $$c(x,y_0):= [\int_C d(x,y)^{-2} d\tau(y) ] - \frac{1}{2} d(x,y_0)^{-2}.$$ We view $$c(x,y_0)$$ as the cost of transporting a unit mass from the source $$x\in X$$ to target $$y_0\in C$$. If $$\int_X \sigma > \int_C \tau$$, then there exist semicoupling measures $$\pi$$ on $$X\times C$$ with the property $$proj_X \# \pi \leq \sigma, ~~~\text{and}~~~~proj_C \# \pi = \tau.$$ In otherwords, $$\pi$$ is a transference plan from the abundant source $$\sigma$$ to the prescribed target $$tau$$. (Such measures are called "Semicouplings"). It is standard result of optimal transport that there exists a unique $$c$$-optimal semicoupling $$\pi_*$$ which minimizes the total cost $$c[\pi]:=\int_{X\times C} c(x,y) d\pi(x,y).$$

Now imagine we rescale the target measure $$\tau\mapsto \lambda \tau$$ for scalar $$\lambda>0$$. If $$\lambda \int_C \tau$$ is sufficiently close to $$\int_X \sigma$$, then the $$c$$-optimal semicoupling $$\pi_*$$ will have a "locus-of-discontinuity" $$Z \hookrightarrow X$$ such that $$Z$$ is a strong-deformation retract of $$X$$ and $$Z$$ will be codimension-one (i.e., the "singularity" is the spine).

More specifically, the $$c$$-optimal semicouplings $$\pi_*$$ are characterized by the existence of a $$c$$-concave potentials $$\psi: C \to \mathbb{R}$$ satisfying $$(\psi^c)^c=\psi$$. This is Kantorovich's duality theory. The $$c$$-optimal transport has the form $$x\mapsto \partial^c \psi^c (x)$$ for every $$x\in X$$. Here $$\partial^c \psi^c(x)$$ is a subset of $$C$$, namely the $$c$$-subdifferential of $$\psi^c$$ at $$x\in X$$. The "locus-of-discontinuity" is more precisely described as the set of $$x\in X$$ where $$\# \partial^c \psi^c (x) \geq 2$$, i.e. where the $$c$$-convex potential $$\psi^c$$ is not uniquely differentiable. The locus-of-discontinuity $$Z$$, where $$\psi^c$$ is finite and not uniquely differentiable, is a closed lipschitz subvariety of $$X$$. And Kantorovich duality shows $$Z \hookrightarrow X$$ is a deformation-retract. The existence of this retract is probably not obvious, unless you are well-studied in the mass-transport theory...

But all the details are in my thesis, including applications to spines for the Teichmueller spaces and symmetric spaces of arithmetic-groups. I'd be happy to share the details, since my supervisor has absolutely zero interest in topological applications, and is quite plainly indifferent to algebraic topology.

• I have trouble extracting what is going on in this answer, but would be delighted to look at your thesis if you emailed it to me. – Andy Putman Nov 10 '18 at 2:40

Let $M$ be a non-compact connected surface. One can assume that $\partial M = \varnothing$. It follows from Lemma 2.2 this paper

• Epstein, D. B. A. Curves on 2-manifolds and isotopies. Acta Math. 115 1966 83–107.

that

1) $\pi_1 M$ is locally free, i.e. every finitely generated subgroup $G$ of $\pi_1 M$ is free;

2) $\pi_1 M$ is a union of an increasing countable sequence of finitely generated free subgroups.

However there are locally free groups that are not free.

Recall that a connected subsurface $N \subset M$ is incompressible if the homomorphism $\pi_1 N \to \pi_1 M$ is injective, so one can regard $\pi_1 N$ as a subgroup of $\pi_1 M$.

Lemma 2.2 of Epstein's paper. Let $X \subset M$ be a compact subset and $G$ be a finitely generated subgroup of $\pi_1 M$. Then there is a compact incompressible subsurface $N \subset M$ such that

• $X \subset int(N)$

• $G \subset \pi_1 N \subset \pi_1 M$.

The proof is elementary and is based on Jordan curve theorem and properties of covering spaces.

Proof that $\pi_1 M$ is locally free. Indeed, since $N$ is compact and has non-empty boundary, $N$ can be deformed onto a finite graph, and therefore $\pi_1 N$ is free and contains $G$. Hence, by Nielsen–Schreier theorem, $G$ is free as well.

Proof that $\pi_1 M$ is a union of an increasing countable sequence of finitely generated free subgroups. Represent $M$ as a countable union of compact subsets $X_1 \subset X_2 \subset \cdots$ such that $M = \cup_i X_i$. Let also $G_0 = 1$ be the unit subgroup of $\pi_1 M$, and $N_0 \subset M$ be an incompressible subsurface such that

• $X_0 \subset int(N_0)$ and $G_0 \subset \pi_1 N_0 \subset \pi_1 M$.

Denote $G_1 = \pi_1 N_0$, and let $N_1 \subset M$ be an incompressible subsurface such that

• $X_1 \subset int(N_1)$ and $G_1 \subset \pi_1 N_1 \subset \pi_1 M$,

repeating this process we will obtain that an increasing sequence of incompressible compact subsurfaces $N_0 \subset N_1 \cdots$ such that $M = \cup_i N_i$.

Since every loop in $M$ is contained in some compact subset and therefore in some $N_i$, it follows that $\pi_1 M$ is a union of its finitely generated free subgroups $\pi_1 N_0 \subset \pi_1 N_1 \subset \cdots$