Adjoint transformation of gauge fields

  1. Aug 19, 2014 #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. Aug 19, 2014 #2

    ChrisVer

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    Gold Member

    Because the gauge field [itex]W_{\mu}[/itex] belongs to the adjoint representation of the gauge group.
     
  4. Aug 19, 2014 #3
    How can you see that?
     
  5. Aug 19, 2014 #4

    ChrisVer

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    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. Aug 19, 2014 #5
    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. Aug 19, 2014 #6
    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. Aug 19, 2014 #7
    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. Aug 20, 2014 #8

    haushofer

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    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. Aug 20, 2014 #9
    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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