- #1
geoduck
- 258
- 2
I'm trying to derive the SO(2,1) algebra from the SO(3) algebra.
The generators for SO(3) are given by:
[tex]M^{ij}_{ab}=i*(\delta_{a}^{i}\delta_{b}^{j}-\delta_{b}^{i}\delta_{a}^{j}) [/tex]
where "a" represents the row, and "b" represents the column, and "ij" represents the generator (12=spin in 3 direction, 23=spin in 1 direction, etc.).
To get SO(2,1), I thought you just raised the "a" term by using the metric tensor:
[tex]M^{(ij)a}_{b}=i*(g^{ai}\delta_{b}^{j}-\delta_{b}^{i}g^{aj}) [/tex]
and took the commutators of the Mij with each other to derive the Lie Algebra.
But when I do this I don't seem to get the Lie algebra of SO(2,1). I have some notes that say the Lie Algebra of SO(2,1) is given by:
[D,H]=-iH
[K,D]=-iK
[H,K]=2iD
where D, H, and K are the generators. I get nothing like that when I take the commutators of M12, M23, and M31.
The generators for SO(3) are given by:
[tex]M^{ij}_{ab}=i*(\delta_{a}^{i}\delta_{b}^{j}-\delta_{b}^{i}\delta_{a}^{j}) [/tex]
where "a" represents the row, and "b" represents the column, and "ij" represents the generator (12=spin in 3 direction, 23=spin in 1 direction, etc.).
To get SO(2,1), I thought you just raised the "a" term by using the metric tensor:
[tex]M^{(ij)a}_{b}=i*(g^{ai}\delta_{b}^{j}-\delta_{b}^{i}g^{aj}) [/tex]
and took the commutators of the Mij with each other to derive the Lie Algebra.
But when I do this I don't seem to get the Lie algebra of SO(2,1). I have some notes that say the Lie Algebra of SO(2,1) is given by:
[D,H]=-iH
[K,D]=-iK
[H,K]=2iD
where D, H, and K are the generators. I get nothing like that when I take the commutators of M12, M23, and M31.