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Is the vectorial representation of the Lorentzian Group unitary?

  1. Aug 10, 2011 #1
    I am 99% sure it is not, but I would like to hear that from someone else to be more serene.
  2. jcsd
  3. Aug 10, 2011 #2
  4. Aug 10, 2011 #3
    Thanks I already was thinking with that argument, I did the calculation of [itex] \Lambda \Lambda ^\dagger[/itex] for some random Lorentz Matrix. Was just looking for a confirmation.
    Last edited: Aug 10, 2011
  5. Aug 10, 2011 #4


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    It's because the Lorentz group (or its component connected to the identity) is non-compact. It's a classical result, a proof of which can be found in Cornwell's compendium.
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