Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About the Lie algebra of our Lorentz group

  1. Mar 25, 2013 #1
    Hello!

    I'm currently reading Ryder - Quantum Field Theory and am a bit confused about his discussion on the correpsondence between Lorentz transformations and SL(2,C) transformations on 2-spinor.
    He writes that the Lie algebra of Lorentz transformations can be satisfied by setting
    [tex]\vec{K} =\pm \frac{i \vec{\sigma}}{2}. [/tex]
    Here it seems as if the dimensions are mixed up. The Pauli matrices are 2 times 2 while the Loretnz generators are 4 times 4.

    Secondly he argues that the Lorentzgroup can be ''factorized'' into [tex]SU(2) \times SU(2)[/tex] but how come this goes along with the fact that the Loretnz group is non-compact.
    It seems as if we take the product group of two compact group the resulting group is compact?
    Am I wrong about this?
     
  2. jcsd
  3. Mar 25, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    He means K and σ as abstract operators which satisfy a certain set of commutation relations, not a particular matrix representation of those operators.
    To go from one set of generators to the other you have to consider the complexified Lie algebra, A = (J + i K)/2 and B = (J − i K)/2. This does not preserve group compactness.
     
  4. Mar 25, 2013 #3
    Ah cool!
    Maybe I'm a bit under the level that this book is written on.. I come back with future confusions. :)
    Thanks Bill!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: About the Lie algebra of our Lorentz group
Loading...