The isomorphism with SO(2,1)

Asked Sep 17 '23 at 14:50

Active Sep 17 '23 at 15:55

Viewed 41 times

It is written in that Paper that

The commutator table for the Lie algebra is $$ \begin{aligned} & {\left[L_p, L_q\right]=(C-A) L_q ; \quad\left[L_p, L_r\right]=(A-C) L_r ;} \\ & {\left[L_q, L_r\right]=-L_p ;\left[\begin{array}{c} L_p, L_s \\ q \\ r \end{array}\right]=0 .} \end{aligned} $$ It is easy to show that this group is isomorphic to $\mathrm{SO}(2,1)$ when $A-C \neq 0$.

where, $A$ and $C$ are real numbers. Moreover, $$[L_{p},L_{s}]=[L_{q},L_{s}]=[L_{r},L_{s}]=0.$$

I know that

$$SO(2,1)=\{X\in \operatorname{Mat}_3(\mathbb{R}) \,|\, X^t\eta X=\eta, \det(X)=1\},$$ where $$\eta = \left ( \begin{array}{ccc} 1&0&0\\0&1&0\\0&0&-1\end{array}\right ) .$$

But my question is how to prove that the Lie algebra of ${L_{p}, L_{q}, L_{r}}$ and the lie algebra of SO(2,1) are isomorphic?

Any help with that is appreciated.

edited Sep 17 '23 at 15:55

asked Sep 17 '23 at 14:50

Math Lover

@user10354138 Actually not, I need to prove the isomorphism between the two Lie algebras – Math Lover Sep 17 '23 at 15:02
Why not? Just check the obvious Lie brackets... – user10354138 Sep 17 '23 at 15:04
@user10354138 Could you kindly provide me with additional elaboration, as my expertise lies somewhat outside the realm of Algebra and Lie groups? – Math Lover Sep 17 '23 at 15:14
Please do not use pictures of symbols for critical portions of your post. Pictures may not be legible, cannot be searched, and are not viewable to some, such as those who use screen readers. – Shaun Sep 17 '23 at 15:23
@Shaun I edited my post – Math Lover Sep 17 '23 at 15:56
Thank you, @MathLover. – Shaun Sep 17 '23 at 17:42

The isomorphism with SO(2,1)

0 Answers0