TITLE: Lie group. How is manifold defined?
QUESTION [8 upvotes]: If I have the Lie group $SL(2,\mathbb{R})$. Then how is the manifold structure on this algebraic group defined, could anybody explain this to me? I mean this is the group of matrices that determinant one, but where comes the manifold structure from?

REPLY [2 votes]: A slightly different take: you can show that $SL_2\mathbb{R}$ is diffeomorphic to $S^1 \times \mathbb{H}$, where $\mathbb{H}$ is the upper half-plane.
The idea is as follows. $SL_2\mathbb{R}$ acts on $\mathbb{C}$ via fractional linear transformations. That is, via
$$
\begin{pmatrix}a & b \\ c & d \end{pmatrix} \cdot z = \frac{az + b}{cz + d}
$$
Since all entries of the matrix are real, it in particular preserves the real line in $\mathbb{C}$. Moreover, since the determinant is 1 (and not -1), it preserves the upper half-plane. So we have an action (actually---this is an action of $PSL_2\mathbb{R}$, and we are really showing that $PSL_2\mathbb{R} \cong S^1 \times \mathbb{H}$---but since $SL_2\mathbb{R}$ is a double cover...).
Essentially, we now pick our favorite point in the upper half plane (say, $\sqrt{-1}$), and note that its stabilizer is just rotations around that point i.e. a copy of $S^1$. Since the group acts transitively on the upper half-plane, it follows then that $PSL_2\mathbb{R}\cong S^1 \times \mathbb{H}$ as desired.
Anyhow, this is a super sketchy argument, but I always think it's a nice one, and it brings the group action nicely into the picture.