On the page 4 of this paper, it indicates that $\text{SL}(2, \mathbb R)$ is generated by one-parameter unipotent subgroups. I wonder how to prove this fact. By definition, a unipotent matrix is the one whose only eigenvalues are $1$'s and a unipotent subgroup is the image of a map $t\to u_t$ where $u_t$ is unipotent for all $t\in \mathbb R$.
I remember that $\text{SL}(2, \mathbb R)$ is generated by three types of elements $$\begin{bmatrix} a & 0\\ 0 & a^{-1} \end{bmatrix}, \begin{bmatrix} 1 & b\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}.$$
The last two types of elements are clearly unipotent by definition but the first one is not necessarily unipotent.
So how to see that $\text{SL}(2, \mathbb R)$ is generated by one-parameter unipotent subgroups. Or did I misunderstand the paper?
Write $a=1+x$, then $$\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}=\begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&-\frac{x}{a}\\0&1\end{pmatrix}\begin{pmatrix}1&0\\-a&1\end{pmatrix}$$