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Generators of Lorentz Lie Algebra being complex?

  1. Jun 5, 2015 #1
    I was wondering about the following
    T are the generators and Λ a continuous LT transformation, thus it is real. Therefore T needs to be imaginary.
    And we can find two sets one being the generators for SO(3) J_i and the other for boosts K_i, which are both imaginary.
    Now I am wondering about introducing the new basis of generators to make the Lie algebra look more like the one from angular momentum
    How can this be a generator as it is complex and would make Λ complex therefore in the first expression?
    What am I overlooking or misunderstanding here?
    Thank you for your help.
  2. jcsd
  3. Jun 5, 2015 #2


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    The point is that you first investigate the Lie algebra of the Lorentz group, i.e., the "infinitesimal transformations" close to the group's unit element (mathematically it's the trangent space, inheriting the Lie algebra from the Lie group).

    To analyze the possible representations of this Lie algebra, it is convenient to complexify it first and then use the fact that by introducing the pseudo-angular-momentum operators $$S_i^{\pm}=\frac{1}{2}(J_i \pm \mathrm{i} k_i),$$
    which fulfill the commutator relations of two independent angular-momentum operators.

    Now all finite-dimensional representations of su(2) and thus also ##\mathrm{su}(2) \oplus \mathrm{su}(2)## are equivalent to a unitary representation, and you know them from quantum theory. So you use these and then go back to the original rotation and boost transformations. As it turns out, you cannot find a finite-dimensional unitary representation of the Lorentz Lie algebra and the corresponding Lorentz group (except the trivial one).

    Nevertheless there are purely real representations like the "fundamental" one, acting on Minkowski four-vectors. Of course, these are not orthogonal transformations (which is the special case of a unitary transformation on real vector spaces) but ##\mathrm{SO}(1,3)## transformations, which leave the Minkowski product and not the Euclidean product on ##\mathbb{R}^4## invariant. This fundamental representation is given by the (1/2,1/2) representation of the Lorentz group, i.e., the two su(2) algebras are both represented by the s=1/2 representation.

    A very good book covering all this in a really very comprehensible way and delivers also some interesting physics on the way is

    Sexl, Roman U., Urbandtke, Helmuth K.: Relativity, Groups, Particles, Springer, 2001
  4. Jul 1, 2015 #3
    I have a somewhat mathematical and abstract answer, I hope you find it useful. The Lorentz Group SO(1,3; R) is a real, simple, and non-compact group. Since it is a simple group, it cannot be written as the direct product of two other Lie Groups. However, all of that changes when you complexify the algebra to SO(1, 3; C) algebra. The complex algebra is isomorphic to SU(2; C) X SU(2; C). (This isomorphim is an accidental one. For example if you were to complexify SO(1, 11; R) to SO(1, 11; C), it does not decompose into a direct product.)

    If you check the Dynkin diagrams (for complexified algebras) you can tell when an accidental isomorphism will take place. All these so-called accidental isomorphisms take place for small Lie groups. For groups large enough, there are no such cases.
    Last edited: Jul 1, 2015
  5. Jul 1, 2015 #4


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    If you choose the rotations ##J_i## to be real and the boosts ##K_i## to be imaginary, then

    $$ \mathcal{J}^\pm_i = \frac{1}{2} ( J_i \pm i K_i)$$

    are real.
  6. Jul 2, 2015 #5


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    There's no nontrivial finite-dimensional unitary representation of the Lie algebra sl(2,C), which is the Lie algebra of the covering group of the Lorentz group SL(2,C). The generators ##J_i## for the rotation subgroup SU(2) can be made hermitian, because SU(2) is compact and semisimple, and this is the usual choice for the representations in physics. The ##K_i## are then necessarily not hermitian, but the ##J_i^{\pm}## are, they represent the Lie algebra ##\mathrm{su}(2) \oplus \mathrm{su}(2)## which is thus equivalent to the Lie algebra sl(2,C). Note, however that the exponential mapping from ##\mathrm{su}(2) \oplus \mathrm{su}(2)## to SL(2,C) is not surjective. For details see Appendix B of my QFT notes:

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