Consider the action of $SL_2(\mathbb R)$ on real projective space $P^1(\mathbb R)$; given $A \in SL_2(\mathbb R)$ and $\alpha \in P^1(\mathbb R)$ we write $A . \alpha \in P^1(\mathbb R)$ for this action. Let $\mu$ be a probability measure on $SL_2(\mathbb R)$ and $\nu$ a probability on $P^1(\mathbb R)$. The **convolution** $\mu * \nu$ is then defined.

Let us call $\mu$ **smoothing** if it has the property that for any $\alpha \in P^1(\mathbb R)$, we have that $\mu * \delta_\alpha$ is absolutely continuous w.r.t. Lebesgue measure on $P^1(\mathbb R)$ (here, $\delta_\alpha$ is the unit Dirac mass at $\alpha$). Clearly, if $\mu$ is absolutely continuous on $SL_2(\mathbb R)$ then $\mu$ is smoothing. However, the converse is not true: for instance, take $\mu$ to be the law of rotation matrices by an angle $\theta$ chosen uniformly from $[0,2 \pi)$.

Given $\mu$, let us write $\bar \mu$ for the law of $A^{-1}$ when $A$ is distributed with law $\mu$.

**Question:** If $\mu$ is smoothing, does it follow that $\bar \mu$ is also smoothing?

**Edit (5-17-2018):** I thought of a possible direction and would be happy to hear comments about it.

Parametrizing $P^1(\mathbb R)$ by the extended real line $\bar {\mathbb R} = \mathbb R \cup \{ \infty\}$, the action of $SL(2, \mathbb R)$ is given by Mobius transformations. To wit, if $$ A = \left( \begin{array}{c c} a & b \\ c & d \end{array} \right) $$ and $x \in P^1(\mathbb R) \cong \bar{\mathbb R}$, then $A$ maps $$ x \mapsto \frac{a x + b}{c x + d} \, . $$ This action preserves the cross ratio $$ (z_1, z_2; z_3, z_4) = \frac{z_1 - z_3}{z_1 - z_4} \frac{z_2 - z_4}{z_2 - z_3} $$ Plugging in $z = z_1, z_2 = 1, z_3 = 0, z_4 = \infty$ and letting $w_i = A . z_i, i = 2,3,4$ and $w = A.z$, we have $$ z = (w, w_2; w_3, w_4) \, . $$ That is, $z = A^{-1} . w$ is essentially determined by the random variables $w_i, i = 2,3,4$. In this way we can perfectly reconstruct the law $\bar \eta$ of $A^{-1}$ using the joint law of $(w_2, w_3, w_4)$.

**Punchline:** Under the smoothing property, the law of each $w_i$ is smooth on $\bar {\mathbb R}$. If the joint law of $(w_2, w_3, w_4)$ were smooth on $\bar {\mathbb R}^3$, then $\bar \eta$ is smoothing.

However, smoothness of the joint law does not follow from the smoothing property for $\eta$: a simple example is that $(w_2, w_3, w_4)$ is supported on some 1D curve in $\bar {\mathbb R}^3$. In fact, there are plenty of possible and quite ugly distributions for the joint law $(w_2, w_3, w_4)$ and this makes me wonder about a counterexample.