How to compute the Lie group of this Lie algebra?

Question

I have this set, let's say $A$, $$ A=\left\lbrace \begin{pmatrix} 0 & a & b \\ a & 0 & c \\ b & -c & 0 \end{pmatrix}; a,b,c\in\mathbb{R} \right\rbrace, $$ which is a sub algebra of the $SL(3,\mathbb{R})$-Lie algebra (because it has trace zero and is closed by brackets). By Lie's third theorem we know that there exists a Lie group, namely $G$, having $A$ as Lie algebra.

My question is, does anyone know what group is that? How to compute $G$? *I've tried to compute it by exponentiation, but I've got stuck in computations.

It is easy to see what the Lie algebra is, see this site for a classification in dimension $3$. Then it is not hard to say which Lie groups we can take. — Dietrich Burde, Nov 26 '21 at 17:59
Good opportunity to add the link of my own lecture notes on the classification. It is section $1.7$. — Dietrich Burde, Nov 26 '21 at 19:20

Dietrich Burde · Answer 1 · 2021-11-27T19:19:12.903

Denoting the Lie algebra basis of matrices by $\{x_1,x_2,x_3\}$, according to chosing $a,b,c$ equal to one and the other entries equal to zero, we obtain the following commutating relations: $$ [x_1,x_2]=x_3,\; [x_1,x_3]=x_2,\;[x_2,x_3]=-x_1. $$ So we have $[L,L]=L$ and $L=A$ must be isomorphic to either $\mathfrak{so}_3(\Bbb R)$ or $\mathfrak{sl}_2(\Bbb R)$. The Killing form of $L$ is negative definite, so it is the orthogonal Lie algebra. So we have $G\cong SO(3,\Bbb R)$.

On the other hand, the Lie brackets are very similar to the ones of $\mathfrak{so}_{2,1}(\Bbb R)$, see this post, which is isomorphic to $\mathfrak{sl}_2(\Bbb R)$.

How to compute the Lie group of this Lie algebra?

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