It is a well known fact that there is a correspondence between complete hyperbolic $n$-manifolds up to isometry and discrete subgroups of isometries of the hyperbolic space $\mathbb{H}^n$ that act freely on $\mathbb{H}^n$ up to conjugation.

The correspondece is given by $\Gamma < Isom(\mathbb{H}^n)\mapsto \mathbb{H}^n/\Gamma$ and the inverse is given by the map $M\mapsto \pi_1(M)\hookrightarrow Isom(\tilde{M})$ where $\tilde{M}\simeq \mathbb{H}^n$ is the fundamental cover of $M$. The requirement that $\Gamma$ acts freely on $\mathbb{H}^n$ is equivalent to requiring that there are no elliptic isometries in $\Gamma$ or equivalently if every element in $\Gamma$ has infinite order.

In particular any parabolic or hyperbolic element in $Isom(\mathbb{H}^n)$ generates an infinite cyclic subgroup. This will correspond to manifolds with $\pi_1 M \simeq \mathbb{Z}$.

What are the complete hyperbolic manifolds with fundamental group $\mathbb{Z}$? Can we say something at least in the case of $3$-manifolds?

  • 2
    $\begingroup$ Yes, in dim 3, it consists in classifying conjugacy classes of non-elliptic elements (modulo inversion) in the isometry group. The isometry group can be viewed as those $z\mapsto\frac{az+b}{cz+b}$ and $z\mapsto\frac{a\bar{z}+b}{c\bar{z}+d}$. If I'm correct, up to conjugacy and inversion, we get the loxodromics: $z\mapsto az$, with $|a|>1$ of nonnegative imaginary part, and $z\mapsto a\bar{z}$ with $a$ positive real. And the horocyclic (aka parabolic): $z\mapsto z+1$ and $z\mapsto\bar{z}+1$. $\endgroup$
    – YCor
    Mar 22, 2020 at 18:35
  • $\begingroup$ In dimension 2: identifying the isometry group with $\mathrm{PGL}_2(\mathbf{R})$, we get $x\mapsto ax$ for $|a|>1$, $x\mapsto x+1$. Converted into an action on the upper half-plane, this yields $z\mapsto az$ and $z\mapsto -a\bar{z}$ for $a>1$, $z\mapsto z+1$. $\endgroup$
    – YCor
    Mar 22, 2020 at 19:11

1 Answer 1


This consists in classifying non-elliptic elements of the Lie group $\mathrm{Isom}(\mathbf{H}^n)\simeq\mathrm{PO}(n,1)$ up to conjugacy and inversion.

One can do separately loxodromics and horocyclics ("parabolics").

Loxodromics: they have two invariants: the translation length (a positive real number), and the transverse isometry, namely an isometry of $\mathbf{H}^{n-1}$ fixing a point (up to conjugation fixing this point), and this is classified by an element of $\mathrm{O}(n-1)$ up to conjugation [and inversion].

Horocyclics: they are classified by their action on the horosphere (which is a non-geodesic copy of the Euclidean space $\mathbf{R}^{n-1}$, modulo conjugation [and inversion] by the whole group of similarities. Hence, by a non-elliptic isometry of $\mathbf{R}^{n-1}$, modulo conjugation by similarities. In general, the horocyclic is orthogonal direct sum of a nontrivial translation and an element of $\mathrm{O}(n-2)$. Hence horocyclics are classified by conjugacy classes of $\mathrm{O}(n-2)$.

The horocyclic case corresponds to the existence of a cusp in the quotient manifold. If one sticks to orientable manifolds, one should restrict to $\mathrm{SO}(n-1)$ in the loxodromic case and $\mathrm{SO}(n-2)$ in the horocyclic case.

["And inversion" will not play any role since it follows that all isometries are conjugate to their inverse, since this holds in $\mathrm{O}(k)$ for every $k$.]

Let's specify in small dimension:

$n=2$: loxodromics are classified by a positive real number, and a sign (preserving or not the orientation). There's a single horocyclic (orientation-preserving).

$n=3$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(2)$ up to conjugation (hence, either a rotation of angle in $[0,\pi]$, or a reflection). Horocyclics: it can be a translation, or a glide reflection.

$n=4$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(3)$ up to conjugation (hence a rotation or antirotation of angle in $[0,\pi]$). Horocyclics: classified by some conjugacy class of $\mathrm{O}(2)$.

Topological classification:

actually, in the orientable case, the quotient manifold is analytically diffeomorphic to $\mathbf{R}^{n-1}\times (\mathbf{R}/\mathbf{Z})$, and in the non-orientable case, it is analytically diffeomorphic to $\mathbf{R}^{n-2}\times (\text{Möbius})$.

Indeed, in both case one sees that the isometry is analytically conjugate to a non-elliptic isometry of the Euclidean space $\mathbf{R}^n$. Such an isometry can be conjugated to have the form $f:(t,y)\mapsto (t+1,Sy)$ with $t\in\mathbf{R}$, $y\in\mathbf{R}^{n-1}$ and $S\in\mathrm{O}(n-1)$. If $S\in\mathrm{SO}(n-1)$, there is a 1-parameter subgroup $(S^t)$ with $S^1=S$, and conjugating $f$ by the analytic self-diffeomorphism $(t,y)\mapsto (t,S^ty)$ yields a translation. If $S\notin\mathrm{SO}(n-1)$, write $f$ as $(t,u,z)\mapsto (t+1,-u,Tz)$ with $z\in\mathbf{R}^{n-2}$ and $T\in\mathrm{SO}(n-2)$. Then conjugating as above only on the last variable conjugates to $(t,u,z)\mapsto t+1,-u,z)$ and this yields the requested description.

  • $\begingroup$ Thank you YCor. I'm interested in the topological type of these object. For example, in 3D do we get solid tori and solid klein bottles or what else? It's not immediate to understand the quotient by these actions. $\endgroup$ Mar 22, 2020 at 21:44
  • $\begingroup$ @WarlockofFiretopMountain It somewhat trivializes (the homeomorphism type is determined by the dimension and orientability). I added a paragraph. $\endgroup$
    – YCor
    Mar 22, 2020 at 22:05

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