Adjoint transformation of gauge fields

  1. A gauge field [itex] W_\mu [/itex] is known to transform as
    [tex]
    W_\mu\to W'_\mu=UW_\mu U^{-1} +(\partial_\mu U)U^{-1}
    [/tex]
    under a gauge transformation [itex]U[/itex], where the first term [itex]UW_\mu U^{-1}[/itex] means it transforms under the adjoint representation. Can anyone explain to me why it means a transformation under the adjoint representation? all I know is the definition of the adjoint representation
    [tex] [T_i]_{jk}=-if_{ijk} [/tex]
     
    Last edited: Aug 19, 2014
  2. jcsd
  3. ChrisVer

    ChrisVer 2,285
    Gold Member

    Because the gauge field [itex]W_{\mu}[/itex] belongs to the adjoint representation of the gauge group.
     
  4. How can you see that?
     
  5. ChrisVer

    ChrisVer 2,285
    Gold Member

    Well in fact you need to check the adjoint representation. The definition of the adj. repr is that of an automorphism.
    http://en.wikipedia.org/wiki/Adjoint_representation
    In particular it's because the transformation [itex]W \rightarrow U^{-1} W U[/itex] preserves the Lie Bracket.
    The above transformation in practice means [itex] W^{a}_{\mu} T^{a}_{ij} \rightarrow W_{\mu} (U^{-1} T^{a} U)_{ij} [/itex].

    Actually are we talking in particular for SU(2)?
    In the SU(2) case you have the (dim) reprs:
    [itex] 2 \equiv \bar{2}, 2 \otimes \bar{2} = 3 \oplus 1 [/itex]
    the [itex]3[/itex] is where the gauge bosons (spin=1) belong. and that's the adjoint repr.
     
    Last edited: Aug 19, 2014
  6. I understand. But I don't see how this transformation rule is consistent with the definition I know of the adjoint rep: Is it possible to assume that T transforms as [itex] GT_a G^{-1} [/itex] and then prove that it is given by the adjoint representation [itex] [T_a]_{bc}=-if_{abc} [/itex]?
    where [itex] f_{abc} [/itex] are determined from:
    [tex]
    [T_a,T_b]=if_{abc}T_c
    [/tex]
    and I don't look specifically for SU(2) reps. also SU(3), and generally any Lie group.
     
  7. An infinitesimal symmetry transformation can be parametrized by some numbers ##\omega^a##, where ##a## runs over the generators of the symmetry group. Then an object ##A_i## is said to transform in the representation ##R## if, under an infinitesimal transformation,

    ##A_i \to A_i + i \omega^a (T^a_R)_{ij} A_j##.

    where the ##T_R^a##'s are the representations of the generators in the representation ##R##.

    Let's look at how the vector potential ##W_\mu^a## transforms under a global gauge transformation. I'll drop the Lorentz index ##\mu## because it's irrelevant. We have

    ##W \to U W U^{-1}##

    where ##U## is the gauge transformation matrix (we will look at a global transformation, so ##\partial_\mu U = 0##). For an infinitesimal gauge transformation ##U## can be written

    ##U = 1 + i \omega^a T^a_F##

    where the ##T^a_F## are the generators in the fundamental representation. Similarly ##W## can be written in terms of the fundamental generators:

    ##W = W^a T^a_F##.

    So we can rewrite the transformation rule, for an infinitesimal gauge transformation, as

    ##W^a T^a_F \to (1 + i \omega^a T^a_F) W^b T^b_F (1 - i \omega^c T^c_F)##

    or, dropping negligible terms of order ##\omega^2##,

    ##W^a T^a_F \to W^a T^a_F + i \omega^a W^b [T^a_F, T^b_F]##.

    But we know from the commutation rules that ##[T^a_F, T^b_F] = i f^{abc} T^c_F##. So the transformation rule becomes

    ##W^a T^a_F \to W^a T^a_F - \omega^a f^{abc} W^b T^c_F##

    By renaming indices this can be rewritten

    ##W^a T^a_F \to (W^a + \omega^c f^{cab} W^b) T^a_F##

    or just

    ##W^a \to W^a + \omega^c f^{cab} W^b##

    Looking back at the first equation, this is the transformation rule for an object that lives in a representation ##R## where the generators are given by

    ##(T_R^c)^{ab} = -i f^{cab}##.

    This is exactly the adjoint representation.
     
  8. Hi. Thank you very much! This is exactly what I was looking for. the explanation is very clear. I only have one question (that I think does not affect the proof): Why did you assume that [itex]U[/itex] is given in the fundamental representation [tex] U=1+i\omega^aT_F^a [/tex]
     
  9. haushofer

    haushofer 999
    Science Advisor

    I guess because we are talking about matrix Lie groups (their elements are matrices), and their algebra elements correspond to the fundamental representation.
     
  10. It doesn't matter; you can pick any representation ##R## and think of ##W## as the matrix ##W^a T_R^a## and ##U## as the matrix ##1 + i \omega^a T_R^a##.
     
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