Here are some examples of compact homogeneous 3 manifolds for different Thurston geometries:

(1) Spherical: $\mathbb{S}^3 \cong \mathrm{SU}_2$ modulo any finite subgroup

(2) Euclidean: 3 torus $\mathbb{R}^3/\mathbb{Z}^3$

(3) $\mathbb{S}^2 \times \mathbb{R}$: 2 sphere times a circle $ S^2\times S^1 \cong (SO_3/SO_2) \times SO_2 $ or projective plane times a circle $ \mathbb{RP}^2 \times S^1 \cong (SO_3/O_2) \times SO_2 $

(4) Nil: Take the three dimensional real Heisenberg group, which is just the group of upper triangular $3\times 3$ real matrices with all 1s on the diagonal, and mod out by the subgroup with integer entries. This quotient is called the "Heisenberg nilmanifold." This is an iterated principal circle bundle like all compact nilmanifolds in any dimension (https://arxiv.org/abs/1805.06585).

(5) $\widetilde{\mathrm{SL}_2(\mathbb{R})}$: $ SL_2(\mathbb{R}) $ mod a Fuchsian surface group gives a unit tangent bundle (a type of circle bundle) over a hyperbolic surface.

Now I want to find examples for the other 3 geometries of a matrix group $ G $ mod some cocompact subgroup $ H $ whose quotient is a 3 manifold with the desired geometry.

(6) Sol: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact 3 dimensional manifold with Sol geometry? (EDIT: this is answered here https://math.stackexchange.com/questions/4317139/bianchi-classification-of-solvable-lie-groups-and-cocompact-subgroups )

(7) $ \mathbb{H}^2 \times R $: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact 3 manifold with $ \mathbb{H}^2 \times R $ geometry ?

(8) Hyperbolic $ \mathbb{H}^3 $: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact hyperbolic 3 manifold?

(EDIT: As Moishe Kohan states in his answer every compact homogeneous 3 manifold is either $ G/H $ for $ G $ 3 dimensional and $ H $ discrete (which covers my examples here for all geometries except $ S^2 \times R $) or if it is not of this form then it must be a fiber bundle of the following sort: either a bundle of circles over a sphere/projective plane/torus/klein bottle or a bundle of sphere/projective plane/torus/klein bottle over a circle)

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    $\begingroup$ The Cartesian product of a Klein bottle with a circle has Euclidean geometry, since it is double covered by $T^3$. The standard example of Sol geometry is a torus bundle over the circle with Anosov monodromy, eg $[2,1;1,1]$ $\endgroup$
    – Josh Howie
    Nov 20, 2021 at 0:15
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    $\begingroup$ I am not sure what you are asking exactly, but solmanifolds are quotients of the group Sol by a lattice, and closed hyperbolic manifolds are quotients of the hyperbolic space by uniform torsion free lattices in $PSL(2,\mathbb C)$ $\endgroup$
    – Nicolast
    Nov 20, 2021 at 20:18
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    $\begingroup$ For an example of a solmanifold, I don't think you will do better than the example on Wikipedia (the mapping torus of the automorphism of the two-torus defined by the matrix $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, but you can express it as the group of $4 \times 4 $ matrices of the form $ \begin{pmatrix} x & y & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & z \\ 0 & 0 & 0 & x \end{pmatrix}$ with $x>0$, or something like that. $\endgroup$
    – Will Sawin
    Nov 22, 2021 at 23:25
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    $\begingroup$ The point is just that the action of the $2 \times 2$ matrix on the tangent space of the two-torus has two eigenvalues which are both positive and inverse to each other. So we need a group with a one-parameter subgroup that acts by conjugation on a two-parameter subgroup, with eigenvalues positive and inverse to each other. This we obtain with the one-parameter subgroup $y=z=0$ and the two-parameter $x=0$. $\endgroup$
    – Will Sawin
    Nov 25, 2021 at 3:32
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    $\begingroup$ So your example would correspond to the lattice in this group where $x$ is an integer power of the eignvalue $ 3+ 2 \sqrt{2} $ and $(y,z)$ are coordinates of an integer vector in the eigenbasis of your matrix, i.e. are integer combinations of $(1 + \sqrt{2}, 1- \sqrt{2})$ and $(1,1)$ or something like that. $\endgroup$
    – Will Sawin
    Nov 25, 2021 at 3:45

3 Answers 3


This is an answer to questions 7 and 8 (I have to say, having 8 questions in one post is way too much for my taste):

Suppose that $M$ is a finite-volume quotient of $H^3$ or a compact quotient of $H^2\times {\mathbb R}$ by a discrete group of isometries. Then $M$ cannot be homogeneous (in the non-Riemannian sense!), meaning that there is no (finite-dimensional) connected Lie group $G$ and its closed subgroup $H$ such that $M$ is diffeomorphic to $G/H$.

This can be derived, for instance, from Theorem C in

G. D. Mostow, A structure theorem for homogeneous spaces. Geom. Dedicata 114 (2005), 87–102.

Mostow's theorem provides the most complete, to my knowledge, description of smooth connected homogeneous manifolds: Mostow describes them as iterated fiber bundles with homogeneous fibers where, with exception of one of the fibrations, the fibers of the bundles are diffeomorphic to quotients of connected Lie groups by discrete subgroups. When you translate this into the setting of 3-dimensional homogeneous manifolds, this becomes the statement that a homogeneous connected 3-manifold $M$ is either diffeomorphic to $G/\Gamma$ where $G$ is a connected Lie group and $\Gamma< G$ is discrete, or $M$ is a fiber bundle whose base and fiber are either circle and a surface of Euler characteristic $\ge 0$ or vice-versa. In the case of hyperbolic 3-manifolds (of finite volume) and compact $H^2\times {\mathbb R}$-manifolds neither type is possible. For instance, a fibration of one of the above types is ruled out by the fact that $\pi_1(M)$ is not (virtually) solvable. To rule out a representation of $M$ as a quotient $G/\Gamma$, where $G$ is a 3-dimensional connected Lie group and $\Gamma< G$ is discrete, one observes that, for the same algebraic reason, $G$ has to be locally isomorphic to $SL(2, {\mathbb R})$. But in this case, $M$ is a manifold which admits the $\widetilde{SL}(2, {\mathbb R})$-geometry, which is known to be incompatible with a hyperbolic structure of finite volume and with a $H^2\times {\mathbb R}$-structure in the case of compact manifolds.

For the sake of completeness:

  1. The classes of noncompact manifolds of finite volume of the types $H^2\times {\mathbb R}$ and $\widetilde{SL}(2, {\mathbb R})$ are indistinguishable topologically.)

  2. "Most" quotients of ${\mathbb H}^3$ by torsion-free discrete subgroups cannot be diffeomorphic to quotients of $\widetilde{SL}(2, {\mathbb R})$ by discrete subgroups. The exceptions all have free fundamental groups or fundamental groups isomorphic to ${\mathbb Z}^2$ or to the fundamental group of the Klein bottle.


Here is a cool fact about $\mathrm{SL}(2, \mathbb{R}) / \mathrm{SL}(2, \mathbb{Z})$; it is homeomorphic to the complement of the trefoil knot in the three-sphere. Apparently this was first proved by Quillen - see page 84 of Introduction to algebraic K-theory by Milnor. For a discussion with further links, see here: Why S^3-K and SL(2,R)/SL(2,Z) are diffeomorphic? Here K is a trefoil in S^3.

And here is a cool fact about $M = \mathrm{SO}(3) \backslash \mathrm{PSL}(2, \mathbb{C}) / \Gamma = \mathbb{H}^3 / \Gamma$, where $\Gamma$ is an index twelve subgroup of the Bianchi group $\mathrm{PSL}(2, O_3)$: this manifold is the complement of the figure eight knot in the three-sphere. Unfortunately, this is the only "arithmetic knot". See the book The arithmetic of hyperbolic 3-manifolds by Maclachlan and Reid. They also discuss the cocompact case, but a quick search (the book is 472 pages!) does not find a cocompact example as beautiful as this.

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    $\begingroup$ This is an awesome contribution and thanks for adding even more. I'm specifically looking for compact and homogeneous so just to be clear I'm not going to accept this as an answer. But the noncompact case is interesting and I'm a huge lover of double coset spaces and arithmetic manifold in particular! So feel free to keep the fun facts coming Sam Nead! $\endgroup$ Nov 21, 2021 at 15:30

These days, many popular quotients of ("real") hyperbolic three-space are obtained as $\Gamma\backslash SL_2(\mathbb C)/SU(2)$, where $\Gamma$ is the group of units in a quaternion division algebra over a complex-quadratic field. More generally, with a field $k$ with exactly one complex archimedean completion, and the rest real, a quaternion algebra which splits at the complex place, and is the Hamiltonian quaternions at all the real places, gives a compact quotient.

The $SL_2(\mathbb R)$ analogues of this are sometimes called "Shimura curves".

  • $\begingroup$ Hey this sounds good could you expand on this example? Is Q(i) a complex quadratic field? Is curly C a quaternion division algebra over a complex quadratic field(are the rational quaternions an example?)? $\endgroup$ Nov 20, 2021 at 20:41
  • $\begingroup$ Ah, first, that was meant to be "\mathbb C", not "\mathfrak". Changed it just now. And, yes, $\mathbb Q(i)$ is a complex quadratic field. Since its unique archimedean completion is algebraically closed, locally at that archimedean place the only quaternion algebra is the two-by-two matrices. But we can construct (by "arithmetic" means) quaternion division algebras that are not split at some finite (even) number of primes, over any number field. $\endgroup$ Nov 20, 2021 at 20:46
  • $\begingroup$ Ok great this sounds like exactly the sort of example I want. So If I understand you correctly you are saying that $\Gamma$ is a cocompact closed subgroup of real dimension 3. Could you pick a specific $\Gamma $ and describe it more as a subgroup of $ SL_2(\mathbb{R})$? $\endgroup$ Nov 20, 2021 at 23:49
  • $\begingroup$ Ah, no, the relevant (co-compact!) subgroups of $SL_2(\mathbb C)$ are not inside the subgroup $SL_2(\mathbb R)$, for many reasons. $\endgroup$ Nov 23, 2021 at 23:51
  • $\begingroup$ Again a very interesting contribution but I'm looking for a compact 3 manifold that is a coset spaces ( admits a transitive lie group action) not a double coset space $\endgroup$ Nov 25, 2021 at 14:52

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