So $\mathfrak{sl}_2(\mathbb{R})$ is generated by the following three traceless matrices.
\begin{equation}
X_1 = \begin{bmatrix} 0&1\\1&0\end{bmatrix},\quad X_2 = \begin{bmatrix} -1&0\\0&1\end{bmatrix}, \quad X_3 = \begin{bmatrix} 0&-1\\1&0\end{bmatrix}
\end{equation}
The exponential map goes from the lie algebra to the lie group such that for $X\in \mathfrak{sl}_2(\mathbb{R})$, $e^{tX} = \gamma_X(t)$, i.e the integral curve generated by $X$. So applying this map to the three generators gives;
\begin{equation}
e^{tX_1} = \begin{bmatrix}\cosh(t)&\sinh(t)\\ \sinh(t)&\cosh(t)\end{bmatrix}, \quad e^{tX_2} = \begin{bmatrix}e^{-t}&0\\0&e^t\end{bmatrix}, \quad e^{tX_3} = \begin{bmatrix}\cos(t)&-\sin(t)\\ \sin(t) &\cos(t)\end{bmatrix}
\end{equation}
So my question is, do these three global flows generate $SL_2(\mathbb{R})$? I don't see how the first matrix does since it is only trivially a shear mapping. I don't really understand what information we can get about the group if we know the algebra.
Can someone explicitly show that these three matrices generate $SL(2,\mathbb{R})$? Because people are saying it does but I really don't see it.
EDIT: So the reason I am having trouble seeing this is because I was thinking about 'generating' $SL(2,\mathbb{R})$ from the Iwasawa decomposition. The Iwasawa decomposition says that $SL(2,\mathbb{R}) = KAN$ such that every $g\in SL(2,\mathbb{R})$ has a unique representation $g = kan$ where $k \in K, a\in A, n \in N$. Where
\begin{equation}
K=\left\{\left(\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right)\right\}, \quad A=\left\{\left(\begin{array}{cc}
r & 0 \\
0 & 1 / r
\end{array}\right): r>0\right\}, \quad N=\left\{\left(\begin{array}{ll}
1 & x \\
0 & 1
\end{array}\right)\right\}
\end{equation}
So I was having trouble showing that these three matrices generate $SL(2,\mathbb{R})$ since I didn't know how to show that $e^{tX_1}$ and/or some combination of $e^{tX_2}$ and $e^{tX_3}$ gives you $N$.
However, the Cartan decomposition seems to imply that any $g \in SL(2,\mathbb{R})$ can be represented as $g = k_1 a k_2$ for some $k_1,k_2 \in K$ and $a \in A$. Does this mean that we don't even need $e^{tX_1}$ to generate $SL(2,\mathbb{R})$? My understanding is that the Cartan decomposition is not unique for each element. So does the Iwasawa decomposition need $N$ in order to have this uniqueness but it is otherwise not necessary to generate the whole group?
Best Answer
To conclude the discussion: the Cartan decomposition (aka the SVD) implies that already the 1-parameter subgroups of positive diagonal matrices and of rotations generate $SL(2, {\mathbb R})$. In particular, you do not need the subgroup $\exp({\mathbb R} X_1)$, the other two subgroups suffice.