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I Adjoint representation of SU(3)

  1. Jul 24, 2016 #1
    Not sure if this is the correct forum but here goes.

    I am trying to prove [Ta,Tb] = ifabcTc

    Where (Ta)bc = -ifabc and fabcare the structure constants for SU(3).

    I picked f123 and generated the three 8 x 8 matrices .. T1, T2 and T3.
    The matrices components are all 0 except for,

    (T1)23 = -i
    (T(1)47 = -i/2
    (T1)56 = i/2

    (T2)46 = -i/2
    (T2)57 = -i/2

    (T3)45 = -i/2
    (T3)67= i/2

    When I compute [T1,T2] I get 0. What am I missing?
  2. jcsd
  3. Jul 24, 2016 #2


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    All your generators ##T^i## seem to be nilpotent which can't be. They have to be regular, i.e. invertible.
  4. Jul 24, 2016 #3
    Yes, so how does one build the correct generators? This worked fine for SU(2) so something else is missing. I suspect the Jacobi identities play a role but I'm not sure how to proceed. Thanks
  5. Jul 24, 2016 #4


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    Have a look on the Gell-Mann matrices: (try ##\lim_{t→0} \exp(t \lambda_i)##, or simply ##1+\lambda_i##)


    Remember that the adjoint representation ##Ad## is a group homomorphism ##SU(3) \longrightarrow GL(\mathfrak{su}(3))## with ##Ad(\exp(\lambda_i)) = \exp(ad (\lambda_i)).##
    Last edited: Jul 24, 2016
  6. Jul 24, 2016 #5
    Thanks. I am familiar with the G-M matrices. I am after the 8 x 8 adjoint versions of the generators. I think I am making progress. When I add indeces:




    into the mix I almost get the right answer. I am wondering if interchanging the second and third indeces changes the sign of the constant.

    Forgive me if I am making this hard work but I no expert in GT.
  7. Aug 2, 2016 #6
    Adjoint representation of SU(3) has 8 matrices, not 3.
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