What is the relationship between the Liouvelle measure on $T^1(\mathbb{H})$, and the Haar measure $\mu$ on $PSL(2, \mathbb{R})$?

Question

Let $\frac{dx dx d\theta}{y^2}$ be the Liouvelle measure on $T^1(\mathbb{H}) \cong \mathbb{H} \times \mathbb{S}^1$. Let $\mu$ be the Haar measure on $PSL(2, \mathbb{R})$.

What is the relationship between the Liouvelle and Haar measures, knowing that we can identify $PSL(2, \mathbb{R})$ and $T^1(\mathbb{H})$?

Answer 1

As long as you are careful about the identification $f : PSL(2,\mathbb{R}) \to T^1(\mathbb{H})$, the answer is that the two measures are the same. Let me avoid actual calculations and list the properties that you need $f$ to verify:

• $f$ is smooth.
• $f$ is an equivariant map with respect to the two natural actions of $PSL(2,\mathbb{R})$: the natural (left) action on the domain $PSL(2,\mathbb{R})$ which is just left multiplication; and the natural (right) action on the range $T^1(\mathbb{H})$ which is just the adjoint of the derivative.

The equivariance condition is that if $\omega = f(\psi)$ then $\omega \cdot \phi = f(\phi^{-1} \cdot \psi)$ (I think I have that right; mixing left and right actions is always treacherous). The notation here is $\phi,\psi \in PSL(2,\mathbb{R})$ and $\omega \in T^1(\mathbb{H})$.

As long as these properties hold, which you can verify for the usual identification map $f$ by a calculation, then using the fact that Liouville is a right invariant Borel measure on $T^1(\mathbb{H})$ you can prove very quickly that $f^*(\text{Liouville})$ is a left invariant Borel measure on $PSL(2,\mathbb{R})$, which by uniqueness of Haar measure must be equal to a constant multiple of Haar measure.