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Lie-Algebra of the SO(3,1)

  1. Jan 25, 2012 #1
    Cheers everybody,

    I have a question regarding the Lie-Algebra of the SO(3,1), i.e. the Lorentz group. The generators are denoted by [itex] M_{\mu \nu} [/itex] with [itex] M_{\mu \nu} = - M_{\nu \mu} [/itex] and fulfill the following commutation relations
    [itex] [M_{\mu \nu}, M_{\rho \sigma}] = g_{\mu \rho} M_{\nu \rho} + g_{\nu \sigma} M_{\mu \rho} - g_{\mu \sigma} M_{\nu \rho} - g_{\nu \rho} M_{\mu \sigma}[/itex],
    where [itex] g_{\mu \nu} [/itex] are metric coefficients.

    I can now construct from these generators the operators
    [itex] \mathbf{J} = (M_{23},M_{31},M_{12}),\ \mathbf{K} = (M_{01},M_{02},M_{03})[/itex],
    which fulfill the relations
    [itex] [J_i,J_j] = i \epsilon_{ijk} J_k,\ [K_i,K_j]= -i \epsilon_{ijk} J_k,\ [J_i,K_j]=i \epsilon_{ijk} K_k [/itex].

    In a final step, I define
    [itex] \mathbf{J}^\prime = \frac{1}{2} ( \mathbf{J} + i \mathbf{K}),\ \mathbf{K}^\prime = \frac{1}{2} (\mathbf{J} - i \mathbf{K})[/itex]
    with the commutation relations
    [itex] [J_i^\prime,J_j^\prime] = i \epsilon_{ijk} J_k^\prime,\ [K_i^\prime,K_j^\prime]= i \epsilon_{ijk} K_k^\prime,\ [J_i^\prime, K_j^\prime]=0[/itex].

    This means that I can construct generators that fulfill exactly the commutation relations of the Lie-Algebra of the SU(2). What does this mean for the group SO(3,1)? Can I deduce from this something like SU(2) x SU(2) is a representation of SO(3,1)?

    Greetings, Syrius
  2. jcsd
  3. Jan 25, 2012 #2


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    Okay, as always, I find the physicist take on these matters to be confusing. What do you mean by SO(3,1), exactly? Since you're concerned with its Lie algebra, I assume you're referring to the identity component of SO(3,1) (i.e. what some people call the "restricted" Lorentz group), since it's all that matters here.

    You're noticing the commutation relations of (the Lie algebra of) SU(2) because:
    (1) SL(2,C) is the complexification of SU(2), and
    (1) SL(2,C) covers (the identity component of) SO(3,1) in the topological sense,
    (2) so the (complexified) Lie algebras of SU(2) (which is the complex Lie algebra of SL(2,C)) and (the identity component of) SO(3,1) are the same.

    Generally, if a Lie group G covers a Lie group H (both assumed to be connected), then the Lie algebras will be isomorphic, and in fact the differential of the covering map at the identity will implement the isomorphism.

    What you can say about representations is a bit delicate. First of all, you really need to clarify whether you're looking for representations of the group or of its Lie algebra.
    Last edited: Jan 25, 2012
  4. Jan 26, 2012 #3
    Hello morphism,

    Yes, I mean the identity component.

    Thank you, that helps a lot. So in my own words, since the Lie algebra of SL(2,C) is nothing else than the direct sum of the SU(2) Lie algebra, and SL(2,C) is covering the identity component of SO(3,1), I am getting 6 generators which are equivalent to two times the generators of the SU(2).

    Maybe it is a dumb question, but if we find that the Lie Algebra of SL(2,C) is just a direct sum of the Lie algebra of SU(2), what does that mean for the underlying groups. Is the SL(2,C) group then also just the direct sum of SU(2)'s?

    And another question: How can I see that SL(2,C) is the complexification of SU(2)?

    I mean here representations of the group.

    Many thanks in advance,
  5. Jan 26, 2012 #4


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    No. An isomorphism of Lie algebras does not guarantee an isomorphism of the underlying Lie groups; all you get is a cover.

    In this specific example it's easy to see that SL(2,C) isn't a direct sum of SU(2)'s: SL(2,C) isn't compact but SU(2) (and hence a direct sum of two SU(2)'s) is.

    There are a bunch of different definitions of "complexification", but here all we need to know is that SU(2) is a maximal compact subgroup of SL(2,C) and that at the Lie algebra level [itex]\mathfrak{sl}(2,\mathbb C) = \mathbb C \otimes \mathfrak{su}(2)= \mathfrak{su}(2) \oplus i \mathfrak{su}(2)[/itex].

    In that case, you generally can't "lift" representations from the Lie algebra to the group - not unless the group is simply connected. Generally you will get a representation of some covering group (you can always guarantee a representation of the universal covering group, but sometimes a "smaller" cover might suffice). Do you have a specific question you want answered? (I'm having trouble following what you said towards the end of your original post.)
    Last edited: Jan 26, 2012
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