What is $T^1(\mathbb H^2/PSL_2(\mathbb Z))$?

Question

Let $\mathbb H^2$ be the upper-half plane. The group $PSL_2(Z)$ acts on $\mathbb H^2$ by isometries, and hence we get an action on $T^1(\mathbb H^2)$. This action is free, smooth, and proper, and thus $X=T^1(\mathbb H^2)/PSL_2(Z)$ is a smooth manifold (with unique smooth structure such that the projection $T^1(\mathbb H^2)\to T^1(\mathbb H^2)/PSL_2(Z)$ is a submersion).

In these notes, on page 3, the first line reads that: Note that $X$ is a unit tangent bundle of non-compact hyperbolic surface (with two "singular points").

I am unable to make sense of this statement. Can somebody please explain how we can see $X$ as the unit tangent bundle of a Riemannian manifold?

Also, what is meant by 'singular points' here?

Thank you.

Edit. Definition of $T^1$: Let $(M, g)$ be a Riemannian manifold. Then $T^1M$, the unit tangent bundle of $M$, is the collection of all the members of $TM$ which are of unit length. Now since $\mathbb H^2$ is a Riemannian manifold (the metric being $(dx^2+dy^2)/y^2$), we can talk about $T^1(\mathbb H^2)$.

Answer 1

The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).

Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C \times S$.

More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $\ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.

I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.

In the case of $T^1(\mathbb H^2 / PSL_2(\mathbb Z))$, the base orbifold $\mathbb H^2 / PSL_2(\mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(\mathbb H^2 / PSL_2(\mathbb Z))$ over those two singularities, by examining, for each $p \in \mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(\mathbb Z)$, how that subgroup acts on $T^1_p(\mathbb H^2)$. That way you will have completely worked out the classification of $T^1(\mathbb H^2 / PSL_2(\mathbb Z))$ in Siefert fiber space language.

But, if you want to know some familiar example which is homeomorphic to $T^1(\mathbb H^2 / PSL_2(\mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(\mathbb H^2 / PSL_2(\mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.

Answer 2

If $\Gamma$ is a torsion free lattice in $G:=PSL_2(\mathbb{R})$, thus acting freely and discontinuously on $\mathbb{H}$, one can unambiguously make sense of $T^1(\Gamma\backslash\mathbb{H})$ since $\Gamma\backslash\mathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $\Gamma\backslash G$ with $T^1(\Gamma\backslash\mathbb{H})$ and one can see that the $\mathbb{R}$-action of diagonal matrices with entries $\lbrace e^{t/2},e^{-t/2}\rbrace$ is intertwined with the geodesic action on $T^1(\Gamma\backslash\mathbb{H})$ under this identification.

But in general when $\Gamma$ is has some fixed points, the notation $T^1(\Gamma\backslash\mathbb{H})$ becomes ambiguous\innacurate as you have noticed. However, there is still the space $\Gamma\backslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $\Gamma\backslash\mathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).

As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).