Examples: (Symmetric) Quotients of Euclidean Space

Question

What are some examples of $n$-dimensional topological manifolds $M$ (possibly with boundary), which can be written as a quotient $\mathbb{R}^d/G$ for some topological group acting on $\mathbb{R}^d$?

Answer 1

I will interpret your question as asking about manifolds which are quotients of $R^d$ by proper continuous actions of topological groups.

First of all, if $M$ is a connected manifold and $p: \tilde M\to M$ is the universal covering then the discrete group $G=\pi_1(M)$ acts properly on $\tilde M$ by deck-transformations of $p$ so that $M= \tilde M/G$.

Thus, you obtain some easy examples: Manifolds whose universal cover is homeomorphic to ${\mathbb R}^d$, e.g. tori $T^d$. Also, if $S$ is a connected surface not homeomorphic to $S^2$ or ${\mathbb R} P^2$, then the universal cover of $S$ is homeomorphic to ${\mathbb R}^2$: Once can derive this from the Uniformization Theorem. This gives you more examples by taking products of surfaces and circles. With a bit more work, you also prove that if $$ F\to M\to B $$ is a fiber bundle, where the fiber and base ($F$ and $B$) have universal covers homeomorphic to ${\mathbb R}^k, {\mathbb R}^n$ respectively, then the universal cover $\tilde M$ of $M$ is homeomorphic to ${\mathbb R}^{n+k}$: This is because the fiber bundle above lifts to a fiber bundle of universal covers: $$ \tilde F\to \tilde M\to \tilde B $$ and the latter is trivial since $\tilde B$ is contractible.

Conjecture. Every connected topological manifold $M^m$ (possibly with boundary) is homeomorphic to the quotient of some ${\mathbb R}^d$ ($d=m$ or $d=m+1$) by a proper continuous group action of a topological group.

I will prove this conjecture (when $d=m$) in the case of (arbitrary) surfaces and compact 3-manifolds ($m=3$). There is a strategy for proving the conjecture in higher dimensions, assuming that $M$ is compact and triangulated (the proof I have in mind may need $m+1$ instead of $m$ and a nondiscrete topological group $G$). I am not sure about topological manifolds.

Proof. In order to keep the discussion reasonably short, I will assume that $\partial M=\emptyset$.

If $M$ is a noncompact surface (without boundary), the universal cover of $M$ is homeomorphic to ${\mathbb R}^2$ and we are done. The same for closed (compact, with empty boundary) surfaces different from $S^2$ and ${\mathbb R} P^2$.

To represent $S^2$ as the quotient of the plane, start with a square $Q\subset {\mathbb R}^2$ and let $\Gamma$ denote the group of Euclidean isometries generated by reflections $\tau_1,...,\tau_4$ in the consecutive edges of $Q$. Then ${\mathbb R}^2/\Gamma$ is naturally homeomorphic to $Q$ (since every $\Gamma$-orbit in ${\mathbb R}^2$ intersects $Q$ exactly once). Let $G_1< \Gamma$ denote the index 2 subgroup consisting of orientation-preserving isometries. Then ${\mathbb R}^2/G_1$ is homeomorphic to $S^2$: You see this by taking the union $D$ of $Q$ and $Q'=\tau_1(Q)$ and checking the identification pattern on the boundary of $D$ given by the rotations $\tau_1 \circ \tau_i, i=1,2,3$. The key fact to keep in mind is that while $\Gamma$ (and, hence, its subgroups) acts properly on the plane, it does not act freely on the plane, so the action of $G_1$ is not by covering transformations (the universal cover of $S^2$, of course, is not ${\mathbb R}^2$). In order to get ${\mathbb R}P^2$ as the quotient, let $\sigma$ denote the unique reflection $\ne \tau_1$ and preserving $D$. Define the group $G_2$ generated by $G_1$ and the composition $\tau_1\circ \tau_2\circ \sigma$. The group $G_2$ contains $G_1$ as an index 2 subgroup and one verifies that $R^2/G_2$ is homeomorphic to ${\mathbb R}P^2$.

In order to understand what is really going on here, it is better to use the language of orbifolds. The best reference I know is:

Scott, Peter, The geometries of 3-manifolds, Bull. Lond. Math. Soc. 15, 401-487 (1983). ZBL0561.57001.

What I am using is that both $S^2$ and ${\mathbb R}P^2$ appear as underlying spaces or parabolic (Euclidean) orbifolds: For $S^2$ I am placing four elliptic points of order 2 on the sphere and for ${\mathbb R}P^2$ I am placing two elliptic/cone points of order 2 on ${\mathbb R}P^2$. This creates, respectively, parabolic orbifolds ${\mathcal O}_1$ and ${\mathcal O}_2$ with respective fundamental groups isomorphic to $G_1, G_2$: The Euclidean plane on which $G_1, G_2$ are acting serves as the universal covering space (in the orbifold sense!) of both orbifolds. The groups $G_1, G_2$ are acting on the plane as the groups of covering transformations (again, in the orbifolds sense!), hence, fixed points are allowed (and even required). I could have used different parabolic or even hyperbolic orbifold structures with the underlying spaces $S^2$ and ${\mathbb R}P^2$ to the same effect. In the latter case, the universal covering space (as a metric space) would be the hyperbolic plane.

The latter generalizes to dimension 3. I will use a theorem mostly due to Thurston:

Theorem. Let $M$ be a closed oriented connected 3-manifold. Then there exists a link (even a knot) $L\subset M$ such that placing a suitable orbifold data at $L$, one obtains a hyperbolic orbifold ${\mathcal O}$ with the underlying space $M$ and the singular locus $L$.

You can find proofs for instance in

Myers, Robert, Simple knots in compact, orientable 3-manifolds, Trans. Am. Math. Soc. 273, 75-91 (1982). ZBL0508.57008.

Brooks, Robert, A construction of metrics of negative Ricci curvature, J. Differ. Geom. 29, No. 1, 85-94 (1989). ZBL0628.53043.

The proof in the second paper has a chance to generalize in higher dimensions, but one needs to weaken negative curvature to nonpositive curvature (in suitable sense).

The result also holds for compact manifolds with boundary (one needs to put suitable "boundary reflectors" on $\partial M$) and non-orientable 3-manifolds.

Now, given this, let $G=\pi_1({\mathcal O})$ be the orbifold-fundamental group of ${\mathcal O}$. The property that ${\mathcal O}$ is hyperbolic means that $G$ acts isometrically and properly on the hyperbolic 3-space ${\mathbb H}^3$ so that the quotient-orbifold ${\mathbb H}^3/G$ is isomorphic (as an orbifold) to ${\mathcal O}$. In particular, this quotient, regarded as a topological space, is homeomorphic to $M$.