Manifold structure of $\mathrm{PSL}(2,\mathbb{R})$ I am trying to prove that $\mathrm{PSL}(2,\mathbb{R})$ is a 3-dimensional connected non-compact Lie group, where
$$\mathrm{PSL}(2,\mathbb{R}) = \left\{\left(\begin{array}{cc}a&b\\c&d\end{array}\right) : ad-bc=1 \right\}\bigg/\{\pm I_2\}$$
I've already shown that it's a group, and we can see that it's a smooth 3-manifold by using the charts
$$\varphi_a\colon U_a\to\mathbb{R}^3$$
$$\varphi_a\colon \left[\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\right]\mapsto\left(\begin{array}{c}a\\b\\c\end{array}\right)$$
where $U_a=\{a\neq0\}$ and we pick a representative matrix such that $a>0$, and
$$\varphi_b\colon U_b\to\mathbb{R}^3$$
$$\varphi_b\colon\left[\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\right]\mapsto\left(\begin{array}{c}a\\b\\d\end{array}\right)$$
where $U_b=\{b\neq0\}$ and we pick a representative matrix such that $b>0$.
Then $U_a\cong\{(x,y,z)\in\mathbb{R}\mid x>0\}$ and $U_b\cong\{(x,y,z)\in\mathbb{R}\mid y>0\}$, but I'm struggling to see how we can glue these two subsets of $\mathbb{R}^3$ together to see what $\mathrm{PSL}(2,\mathbb{R})$ looks like as a smooth $3$-manifold.
So, my questions:


*

*Is all of the above correct?

*Can we draw some pictures of how to glue together these two patches to find a diffeomorphic copy of $\mathrm{PSL}(2,\mathbb{R})$ inside $\mathbb{R}^3$?

*Is there another (maybe simpler) way of proving connectedness?

 A: As $ad-bc=1$ the glueing happens by $U_u\supseteq U_a\cap U_b\to U_b$, $\begin{pmatrix}a\\b\\c\end{pmatrix}\mapsto \begin{pmatrix}a\\b\\\frac{1+bc}{a}\end{pmatrix}$, which works because $a\ne0$ in $U_a$.
A: Well, your maps aren't really very well defined, because of the "P" in PSL. You might instead define
$$
\varphi_a\colon \left[\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\right]\mapsto\left(\begin{array}{c}b/a\\c/a\\d/a\end{array}\right)
$$
which is invariant with respect to scaling, and well defined because $a \ne 0$. 
As for connected-ness, you might instead prove path-connectedness: show that for any matrix in your group, there's a path (in the group) to the identity matrix. To do so, you can take the SVD of the matrix, writing
$$
A = U^t D V
$$
where $D$ is diagonal with determinant 1. By negating columns of $V$ if necessary, we can assume that $D$'s entries are all positive. Let $H(t) = (1-t)D + tI$. Now
$$
A_t = U^t H(t) V(t)
$$
is a path from $A$ to a pure rotation, or rotation-and-reflection; since the det is positive, you know there's no reflection. And each rotation can surely be connected to the identity through a path in the rotation group (which is just $S^1$). 
Finally: you probably aren't going to be able to glue together the two images to put PSL(2,R) into 3-space. Most 3-manifolds don't embed in 3-space. I guess that SL(2, R) can be embedded as a kind of solid torus ($S^1 \times B$, where $B$ is an open ball) because of the QR decomposition, but I suspect that PSL is probably messier. (But don't trust me here -- I really know very little about it.)
