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 $\pi\_1 M \simeq \mathbb{Z}$.



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> What are the complete hyperbolic manifolds with fundamental group $\mathbb{Z}$? Can we say something at least in the case of $3$-manifolds?
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Hyperbolic manifolds with infinite cyclic fundamental group
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 $\mathrm{O}(2)$.




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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.


