# Question on $\mathfrak{sl}(2,\mathbb R)$

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 $\exp(\mathfrak{sl}(2,\mathbb R))$?

• $SL(2,\mathbb{R})$, in my opinion. A Lie group is in general not equal to the image of the exponential of its Lie algebra, as you see. It is just generated by the image. Often, the image of the exponential is not a group at all. – Peter Franek Jun 17 '14 at 15:15
• I understand. Thanks a lot ! – Qinghua Pi Jun 17 '14 at 16:08
• @PeterFranek, could you remind me why it's not a group? I thought it was a subgroup (since it has the identity element and the Baker-Campbell-Hausdorff formula suggests it is closed under multiplication, right?) but rarely is it the full group...am I mistaken? – Alex Nelson Jun 17 '14 at 17:19
• @AlexNelson: you are probably right, sorry for the confusion. I think it is a subgroup but not a Lie subgroup; the differential of $\exp$ is not a local diffeomorphism in the neighborhood of some points -- for example in the neighborhood of $\mathrm{diag}(i\pi,-i\pi)$; the image might be topologically very wild.. But this is already slippery slope for me :) – Peter Franek Jun 17 '14 at 17:42
• As I said, only near the identity. – Moishe Kohan Jun 17 '14 at 20:10