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If one generator commutes with all other,why must it be generator of U(1)?

  1. Dec 30, 2011 #1
    Please teach me this:

    If one generator of a Lie algebra commutes with all other generator,then why must this generator be the generator of U(1) trasformation?(I know that generator of U(1) commutes with all other generators).

    By the way,compact simple Lie algebras belong to 3 families called classical groups,with only 5 exceptions.What are the five exceptions?(QFT of Peskin and Schroeder)

    Thank you very much in advance
     
  2. jcsd
  3. Dec 30, 2011 #2

    strangerep

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    Science Advisor

    This question needs more context to be answered properly, but I guess you're dealing with a case where the representation of the other generators is irreducible. If so, I suspect Schur's lemma applies.

    http://en.wikipedia.org/wiki/Schur's_lemma

    This leads to the generator in question being a multiple of the identity, which generates U(1) when represented on a Hilbert space (i.e., when exponentiated to become a unitary operator).

    Well (surprise!) they are the "exceptional groups".

    See the section "Exceptional cases" in this Wiki page:
    http://en.wikipedia.org/wiki/Exceptional_group
    which also has links to details of these groups.
     
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