# Lattice generated by parabolics

Let $$G$$ be a semisimple Lie group of split-rank one and let $$\Gamma$$ be a non-cocompact lattice which is torsion-free. For the group $$G=\mathrm{SL}_2(\mathbb{R})$$ it then follows that $$\Gamma$$ is generated by parabolic elements.

My question is, whether this is always the case, so for groups like $$G=\mathrm{SO}(n,1)$$ or $$G=\mathrm{SU}(n,1)$$, is every torsion-free, non-uniform lattice generated by parabolic elements?

• Why is this true for $SL_2(\mathbb R)$? If $\Gamma$ is a congruence lattice in $SL_2(\mathbb Z)$ then $\Gamma$ is the fundamental group of some modular curve $X(n)$. The modular curve $X(n)$ has some compacification $Y(n)$, and we have a surjection $\pi_1(X(n)) \to \pi_1(Y(n))$ whose kernel contains all the parabolic elements as these are loops around the cusps. But $\pi_1(Y(n))$ need not be trivial, as $Y(n)$ can have genus arbitrarily large. Commented Jan 15 at 12:14
• @WillSawin in this case, is the index of the subgroup generated by parabolics still finite?
– YCor
Commented Jan 15 at 13:12
• @YCor: It’s infinite-index. The quotient is a closed surface group. Commented Jan 15 at 13:33
• Interestingly, if $G$ has rank $\ge 2$ and $\Gamma$ is a uniform and irreducible lattice in $G$, then a finite index subgroup in $\Gamma$ is generated by (finitely many) unipotent elements. Commented Jan 15 at 14:52
• @MoisheKohan you mean "non-uniform", not "uniform".
– YCor
Commented Jan 15 at 19:45

Here is an example (similar but perhaps simpler to the one given in the comments) showing that parabolics need not generate, even when $$G = \mathrm{SL}(2, \mathbb{R})$$.
Let $$r = 1/\sqrt{2}$$. Take $$A = \begin{pmatrix} 3r & r \\ r & r \end{pmatrix}$$ and $$B = \begin{pmatrix} -3r & r \\ r & -r \end{pmatrix}$$. Let $$F = \langle A, B \rangle$$ be the subgroup of $$\mathrm{SL}(2, \mathbb{R})$$ generated by $$A$$ and $$B$$. The group $$F$$ acts, in the usual way, on the upper half-plane model of $$\mathbb{H}^2$$. The quotient $$\mathbb{H}^2 / F$$ is a once-punctured torus. To see this, consider the tiling of $$\mathbb{H}^2$$ by ideal squares, where we start with the one with vertices at $$-1$$, $$0$$, $$1$$, and $$\infty$$ in the upper half plane model and generate the others by reflections. The squares are fundamental domains for the action. We deduce that $$F$$ is a rank-two free group.
The loop about the puncture is given by the element $$[A, B]$$, which is parabolic. We find that $$[A, B]$$ normally generates $$F'$$, the commutator subgroup. Also, all parabolics in $$F$$ are conjugates of powers of $$[A, B]$$. So the subgroup generated by parabolics (in $$F$$) is exactly $$F'$$. Finally note that $$F / F'$$ is isomorphic to $$\mathbb{Z}^2$$. We deduce that the parabolics do not generate all of $$F$$.
Interestingly, for the non-uniform lattices in $$\mathrm{SL}(2, \mathbb{C})$$ coming from knot complements in the three-sphere, the parabolic elements do generate. (This can be proven using the Wirtinger presentation.) However this is very special to knot complements. It is easy to give examples (say the SnapPy manifold m003) where the fundamental group of the boundary does not generate all of the homology of the manifold. Thus the parabolics do not generate the entire fundamental group.