I am confused about some facts on $SL(2,\mathbb R)$.
The Lie algebra of $SL(2,\mathbb R)$ is $\mathfrak{sl}(2,\mathbb R)$. However, the map $$ \exp:\mathfrak{sl}(2,\mathbb R)\ \rightarrow SL(2,\mathbb R) $$ has image $$ \{g\in SL(2,\mathbb R),\mathrm{Tr}(g)>-2\}\cup\{\pm I\}. $$ That is to say, the elements $\begin{pmatrix}-1&t\\&-1\end{pmatrix}$ with $t\neq 0$ are not in the image.
Now, Let $G=SL(3,\mathbb{R})$ and $\mathfrak{g}=\mathfrak{sl}(3,\mathbb R)$. The analytic subgroup of $G$ with Lie algebra $$ \left\{\begin{pmatrix} a&*&0\\ *&-a&0\\ 0&0&0 \end{pmatrix}\right\} $$ is of the form $$ \left\{\begin{pmatrix} H&\\ &1 \end{pmatrix}\right\} $$ So, what is $H$?
$SL(2,\mathbb R)$, $PSL(2,\mathbb R)$ or $\exp(\mathfrak{sl}(2,\mathbb R))$?