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SO(2,1) Lie Algebra Derivation

  1. May 10, 2012 #1
    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:


    where D, H, and K are the generators. I get nothing like that when I take the commutators of M12, M23, and M31.
  2. jcsd
  3. May 10, 2012 #2


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    The algebra above is written in a completely different basis from the angular momentum algebra you wrote before. This is a representation appropriate to the conformal algebra acting on a particle in 1d:

    [tex]\begin{split}& H = -\frac{\partial}{\partial x} ,&~~~ \mathrm{translation} \\
    &D = -i x \frac{\partial}{\partial x}, & ~~~\mathrm{dilatation} \\
    &K = - x^2 \frac{\partial}{\partial x} , & ~~~\mathrm{special ~conformal}. \end{split}[/tex]

    If you want to write this in terms of angular momentum and boosts, it looks like you can write
    [tex]\begin{split}& J = D, \\
    &B_1 = \frac{i}{2} (H+K), \\
    &B_2 = \frac{i}{2} (-H+K). \end{split}[/tex]
    [tex]\begin{split}& [J,B_1] = iB_2, \\
    & [J,B_2] = -iB_1, \\
    & [B_1,B_2] = -i J. \end{split}[/tex]

    The algebra you want is obtained by identifying

    [tex] M_{23} = J,~~M_{12}=B_1,~~M_{31}=B_2.[/tex]

    I could have signs wrong, but you should be able to get the general idea by working through the algebra.
  4. May 10, 2012 #3

    It seems then that different Lie Algebras can describe the same group?

    The conformal algebra, when taking a linear combination of the generators, gives you the 3-dimensional Lorentz algebra: this must mean that you can use either set of generators and exponentiate to produce a representation of the group.

    Does this mean that you can perform a dilation, translation, and special conformal transformation and achieve the same result with rotations and boosts?

    I'm a little confused about the terminology: there's the conformal algebra, and the Lorentz algebra, but is the group that they are describing called SO(2,1)?
  5. May 10, 2012 #4


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    What we have are two different physical systems. The conformal symmetry of a particle in 1d and the Lorentz symmetry of 3d Minkowski space. In both cases, it turns out that the associated Lie algebra is SO(2,1). So, even though we started with different physics, the mathematical symmetry is the same. If my calculations above are correct, I've given an isomorphism between the natural generators for the two physical systems.

    More generally, it can be the case that two different Lie algebras describe the same group. For example, the Lie algebras of SO(3) and SU(2) are the same, even though the groups are different. Though, they're not that different: SU(2) is mathematically a double cover of SO(3) as a manifold.
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