I know a sketch of the proof.

M. A. Armstrong 's Basic Topology says that

Suppose $X$ has a universal covering space, and denote it by $\tilde{X}$. Then the covering transformations form a group isomorphic to the fundamental group of $X$. Given any subgroup $H$ of $\pi_1(X)$, it acts on $\tilde{X}$ and the associated orbit space $\tilde{X}/H$ is a covering space of $X$ whose fundamental group is isomorphic to $H$.

$\mathbb{R}^2$ is the universal covering space of $T$. The fundamental group of a torus $T=S^1\times S^1$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}=\mathbb{Z}^2$. Any subgroup of $\mathbb{Z}^2$ is isomorphic either to the trivial group, $\mathbb{Z}$ or $\mathbb{Z}^2$.

Since $\mathbb{R}^2/\mathbb{Z}$ can be $S^1\times\mathbb{R}$, $\mathbb{R}^2/\mathbb{Z}^2$ can be $T$, we get the conclusion we want to prove.

However, Basic Topology also says:

A given group may act in many different ways on the same space.

Indeed, it gives an example that $T/\mathbb{Z}^2$ can be sphere, torus and Klein bottle.

So my problem is: Is it possible that $\mathbb{R}^2/\mathbb{Z}$ or $\mathbb{R}^2/\mathbb{Z}^2$ can be other spaces?

Can you please help? Thank you.

  • 11
    $\begingroup$ Note that you are not only given the universal covering space but actually also the universal cover $p: \mathbb R^2 \to T$. This also specifies how $\mathbb Z^2$ acts on $\mathbb R^2$. Now how do subgroups act on $\mathbb R^2$? What are the resulting quotient spaces? $\endgroup$ Commented Apr 21, 2011 at 6:14
  • $\begingroup$ @Alexander: Thanks. I have re-read my textbook and found that it is actually the subgroup of the covering transformations group, which is isomorphic to the fundamental group, that acts on $\mathbb{R}^2$. Does that mean it is not enough to only know the fundamental group is isomorphic to $\mathbb{Z}^2$ and I should actually find out the group of covering transformations? Is the group of covering transformations what you mentioned as "how $\mathbb{Z}^2$ acts on $\mathbb{R}^2$?" Thank you. $\endgroup$
    – Roun
    Commented Apr 21, 2011 at 10:50
  • $\begingroup$ @Alexander: And if I know how $\mathbb{Z}^2$ acts on $\mathbb{R}^2$ to get a torus $T$, say $(x,y)\to(x+m,y+n)$, can I get the group of covering transformations directly? $\endgroup$
    – Roun
    Commented Apr 21, 2011 at 11:01

1 Answer 1


To expand on what Alexander said, we aren't looking at group actions of $\pi_1(X)$ on $\widetilde{X}$, but we're looking at the group actions that are compatible with the covering map. Let's go over the construction to see what is happening.

If we have a covering map $p:E\to B$ of spaces, where $b\in B$ is a fixed basepoint, then for any path $\gamma:[0,1] \to B$, with $\gamma(0)=\gamma(1)=b$, $x\in p^{-1}(b)$, there is a unique lift $\gamma_x:[0,1]\to E$ of $\gamma$ with $\gamma_x(0)=x$. Note that by a lift of $\gamma$, we mean that $p(\gamma_x(t))=\gamma(t)$. Let us assume that $\gamma$ is a loop, that is $\gamma(0)=\gamma(1)=b$,

We assert the following facts: (1) The map $x\mapsto \gamma_x(1)$ is in fact an automorphism of $p^{-1}(b)$. (2) This map depends only on the class of $\gamma$ in $\pi_1(B)$. (3) This map $\pi_1(B) \to \operatorname{Aut}(p^{-1}(b))$ is actually a group homomorphism.

Given an automorphism of $\phi$ of $p^{-1}(b)$, there is at most one automorphism $\psi$ of $E$ which both restricts to $\phi$ and satisfies $p(\psi(e))=p(e)$ for all $e\in E$. Of course, not every automorphism lifts. For example, if $\mathbb{R}\to S^1$ is the standard covering map, the automorphism of $\mathbb{Z}$ (viewed as a space, not a group) defined by $z\mapsto (-1)^{z} z$ doesn't extend to a continuous automorphism of $\mathbb{R}$. However, by using the path lifting property, we can show that all the automorphisms we've generated actually do extend (prove this!).

So what we have shown is that there is a unique action of $\pi_1(B)$ on $E$ which is compatible with the projection map $p$, in the sense that the action comes from path lifting. There are probably other characterizations of the action in terms of an action of $\pi_1(B)$ on $\pi_1(E)$.

Of course, there are lots of other actions. For example, given any homomorphism from $\pi_1(B)$ to itself we can compose these with our given action to get a new one

$\pi_1(B) \to \pi_1(B) \to \operatorname{Aut}(E)$

For example, composing with the trivial group homomorphism, we get the trivial action (all loops act trivially). That doesn't take away from the fact that there is a canonical action.

In the case of the universal covering space, there is a very simple description of the action. One of the constructions of $\widetilde{X}$ is as the space of homotopy classes (preserving the endpoints) of paths in $X$ which start at the basepoint. The covering map sends a path to its endpoint. The action of $\pi_1(X)$ takes a loop $\gamma$ and a path $\rho$ and gives the path obtained by following $\gamma$ and then following $\rho$.

The point is that all of this fits together very nicely: For every covering of a space $X$, we have a canonical action of $\pi_1(X)$, and if we have a map of two covering spaces of $X$ (a continuous map which is compatible with the projection maps), then this map is compatible with the action of $\pi_1(X)$. If $X$ is nice (so that the universal exists), and if we look only at connected covering spaces, then there is always a unique map of covering space from the universal covering to any other given cover. This map is the unique map which has the property you stated in your question.

To go back to the original question, we have a natural action of $\mathbb{Z}^2$ on $\mathbb{R}^2$. Namely, we view both as groups under addition, and the action comes from the natural inclusion.

Every subgroup of $\mathbb{Z}^2$ is either isomorphic to $\{0\}$, $\mathbb{Z}$, or $\mathbb{Z}^2$, although there are many different subgroups of the latter two types, which up to automorphism correspond to orbits of points or pairs of points in $\mathbb{Z}^2$ under the action of $\operatorname{SL}_2(\mathbb{Z})$. While the quotient of any two different subgroups gives a different covering map, we have that (in this special case, with these particular actions) the quotient of $\mathbb{R}^2$, up to homeomorphism, by the subgroup depends only on the isomorphism class of the subgroup. For isomorphic but non-equivalent subgroups, even though the space on top is the same, the action of $\mathbb{Z}^2$ is different. So in fact, this gives an example of how a group can act on a space in multiple ways.

But what if we don't want to use this action? What spaces can you get? Well, then depends on what kind of action you want to take. If you act by $(m,n).(x,y)=(-1)^m x, (-1)^n y)$, you will something that looks like the first quadrant of the plane. If you act by $\mathbb{Z}$ with $n.(a,b)=(2^n a,2^n b)$, the quotient is almost the torus: every point other than the original can be moved inside the the annulus between the unit circle and the circle of radius $2$, and so the quotient of the action on $\mathbb{R}^2\setminus \{0\}$ is the torus, but then you have an extra point. You can get all sorts of interesting things if you are okay with the action not treating every point equally. I am not sure what the classification of such quotients looks like, or even what familiar things you can get from such quotients.

For a nice action, where each point has the same stabilizer, and modulo this stabalizer the action is free, you will get some surface whose fundamental group is a quotient of $\mathbb{Z}^2$, and whose universal cover is $\mathbb{R}^2$. The first condition rules out getting things like higher genus surfaces and the Klein bottle, whose fundamental groups are not abelian.. The second condition rules out the projective plane, whose universal cover is $S^2$.

Update in response to the clarification request:

When you have a group acting on a space, quotienting out by a subgroup will generally depend on the specific subgroup, and not just the isomorphism type of the subgroup. This is easy to see for non-covering-space examples, where the group doesn't act in a uniform way: consider $\mathbb{Z}^2$ acting on $\mathbb{R}$ where the second copy of $\mathbb{Z}$ acts trivially. Quotienting out by one copy of $\mathbb{Z}$ does nothing, and by the other gives you the circle.

However, covering maps are nice (the action of the fundamental group of the base is uniform). Let's just consider the case of the universal cover.

Suppose that $H_1,H_2\subset G=\pi_1(X,x)$, are subgroups of the fundamental group which are abstractly isomorphic, but which are not mapped into each other by any automorphism of $G$. For example, we could have $4\mathbb{Z}\subset 2\mathbb{Z} \subset \mathbb{Z}$. The bundles $\widetilde{X}/H_i \to X$, $i=1,2$ will usually be non-isomorphic. For example, if $H_i$ has finite index, then $\widetilde{X}/H_i \to X$ will be a finite covering of degree equal to that index. For the example $4\mathbb{Z}\subset 2\mathbb{Z} \subset \mathbb{Z}$, the two subgroups have index $2$ and $4$.

However, what if we ignore the bundle structure? I believe that $\widetilde{X}/H$ spaces need not be determined by the isomorphism type of $H$. Unfortunately, in one or two dimensions, we can't get any interesting counterexamples: $\widetilde{X}/H$ has fundamental group isomorphic go $H$, and surfaces are determined by their fundamental groups. Additionally, up to homotopy, there are no counterexamples if $\widetilde{X}$ is contractible, as we would have an Eilenberg-Maclane space, which is up to homotopy determined by its fundamental group. See this wikipedia article for more information on EM-spaces.

So is there a conterexample? Probably, but I do not know it. It will require at least $3$ dimensions and a covering space which is not contactable. Still, I have no good a priori reason to believe that quotienting the universal cover by a subgroup depends only on the isomorphism type of the subgroup, and you shouldn't bu lulled into believing this is true in general until you see a theorem asserting it. I don't even know where to begin constructing a counterexample, as the obvious algebraic invariant that might help, the quotient $G/H$, is lost when you forget the bundle structure.

  • $\begingroup$ Wow! Thank you very much! I really appreciate your answer. You've really help me a lot on understanding this topic. I still have a little question (to check whether I understand it right or not). You mentioned that "in this special case, with these particular actions." Does that mean I need to know the specific form of the groups which are isomorphic to some particular group (say $\mathbb{Z}$) when dealing with similar problems? Thanks again. $\endgroup$
    – Roun
    Commented Apr 21, 2011 at 12:15
  • 2
    $\begingroup$ Here is a counterexample for you in dimension 3. Consider $S^2\times S^1$. The group $\mathbb{Z}/2\mathbb{Z} + \mathbb{Z}/2\mathbb{Z}$ acts on this where each $\mathbb{Z}/2\mathbb{Z}$ summand is generated by the antipodal map on $S^2$ or $S^1$. Of course, quotienting by the first $\mathbb{Z}/2\mathbb{Z}$ gives $\mathbb{R}P^2 \times S^1$ while quotienting by the second gives $S^2\times \S^1$. Since the first is nonorientable and the second is orientable, the two manifolds are not not even homotopy equivalent. $\endgroup$ Commented Apr 22, 2011 at 3:31
  • 1
    $\begingroup$ @Jason: How is a möbius strip not homotopy equivalent to a circle? $\endgroup$ Commented Apr 22, 2011 at 8:45
  • 1
    $\begingroup$ @Jason: You have claimed, that two manifolds cannot be homotopy equivalent, because one is oriented and the other is not. $\endgroup$ Commented Apr 22, 2011 at 17:36
  • 2
    $\begingroup$ @Alexander: You're right. However, for closed manifolds, orientability is a homotopy invariant - I should have been clearer. @user9863: My actions are honest to goodness covering actions in the sense that the projection maps onto the orbit spaces are covering maps (which is precisely what you're looking for, except that my spaces weren't simply connected). Modifying my examples to $S^2 \times S^n$ for $n\geq 3$ gives a counterexample for any dimension 5 or greater. $\endgroup$ Commented Apr 23, 2011 at 0:52

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