Seifert manifolds and Fuchsian group

A Fuchsian group is a discrete subgroup of $PSL(2, \mathbb{R})$.

Let $M$ be a Seifert manifold (maybe with boundary) and $t \in \pi_1(M)$ the class of a regular Seifert fiber.

Hempel claims in his books on 3-Manifolds that the group $G = \pi_1(M)/ \langle t \rangle$ is a Fuchsian group.

Q: How can one see this? I can kind of imagine this when the Seifert surface is a hyperbolic surface.

To answer this question it might be helpful to know that the group $G$ has the following representation. $$G = \langle a_1, b_1,\ldots, a_g, b_g, c_1, \ldots, c_q, d_1, \ldots, d_k : c_i^{n_i} = 1, [a_1, b_1]\ldots[a_g,b_g]c_1\ldots c_q d_1 \ldots d_k =1 \rangle$$

Edit: One definitly has to make the claim a bit more precise. I definitly don't see this with M = S^3.

• First, the trivial group is a Fuchsian group! Second, a Seifert-fibred space is just a circle bundle over a surface (or orbifold). Thus the fundamental group is an extension of $\pi_1(\text{circle})$ by $\pi_1(\text{surface})$. The generator of the circle's fundamental group is $t$. So if your surface (or orbifold) is hyperbolic, the quotient you describe (which is just $\pi_1(\text{surface})$) is a Fuchsian group. But this is not in general true, as a torus times a circle shows.
– user641
Jul 25, 2013 at 4:10
• Thank you. I somehow still had cocompact in mind. I see that $\pi_1(M)$ is an extension of $\pi_1(surface)$ by $\pi_1(circle)$. If $G$ is the fundamental group of a hyperbolic surface $\Sigma$ I can take the universal covering and obtain an embedding of $\pi_1(\Sigma)$ in $PSL(2,\mathbb{R})$ by the monodromy action. How does this work with orbifolds?
– mna
Jul 25, 2013 at 7:42
• If an orbifold has negative Euler characteristic, it is "good" in the sense that it is (finitely) covered by a manifold, and hence its universal cover is $\mathbb{H}^2$. Everything works almost exactly the same as in the surface case. The only difference being the singularities of the orbifold introduce torsion in the monodromy.
– user641
Jul 29, 2013 at 18:31
• As an example, consider an orbifold which is a sphere with three cone points. If the cone angles are less than $2\pi$, then this orbifold has negative Euler characteristic (in fact, it double covers a hyperbolic triangle). The associated Seifert fibre space $M$ has three singular fibres coming from the cone points. It's fundamental group is given by the presentation $\langle x,y,z\mid x^p=y^q=z^r=xyz=1\rangle$, which is also known as a (hyperbolic) triangle group. These embed (very nicely) in $PSL(2,\mathbb{R})$, gottten by tiling $\mathbb{H}^2$ with the triangle mentioned above.
– user641
Jul 29, 2013 at 18:33

This claim is, strictly speaking, false, my guess is that you are missing some hypothesis. All Seifert manifolds fall into 3 main classes: The ones with spherical base-orbifold, ones with Euclidean base-orbifolds and ones with hyperbolic base, orbifolds. You can find a detailed discussion of this in Peter Scott "Geometries of 3-manifolds". In all cases, the quotient group $G$ in your question is the fundamental group of the base-orbifold. If the base is hyperbolic, then $G$ embeds as a discrete subgroup in $SO(2,1)$, the full isometry group of the hyperbolic plane. The group $PSL(2,R)$ is surely not enough since it preserves the orientation on the hyperbolic plane, but you can just take a circle bundle over a nonorientable hyperbolic surface, as your Seifert manifold.
In the case of Euclidean base, the group $G$ is never Fuchsian as it contains a rank 2 free abelian subgroup. In the case of spherical base, sometimes the group $G$ is Fuchsian, i.e., when it is cyclic or dihedral. But in some cases it is not Fuchsian, for instance, when it is isomorphic to $A_4$, the alternating group on 4 letters. The reason is that all finite subgroups of $SO(2,1)$ are cyclic or dihedral and $A_4$ is not.
The correct statement is that the group $G$ acts as a discrete isometry group either on the 2-dimensional sphere or on the Euclidean plane or on the hyperbolic plane, take a look at Scott's paper.