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Why are gauge fields in the adjoint rep?

  1. Mar 11, 2007 #1

    nrqed

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    Does anyone know a deep reason why we always put the gauge fields in the adjoint representation of the group? I am not sure if there is a deep reason or it's just that it "happens" to work for SU(2) and SU(3).

    Just wondering.
     
  2. jcsd
  3. Mar 11, 2007 #2

    Kea

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    Good question. The answer requires stepping away from gauge theory into the proper version of the Standard Model, where symmetry is NOT fundamental and groups tend to arise as automorphism groups.

    http://en.wikipedia.org/wiki/Adjoint_representation

    Cheers
    :smile:
     
  4. Mar 11, 2007 #3

    mjsd

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    note that a rep with dimension equal to the number of generators always exists (may be constructed from the structure constants), and this is your adjoint rep.
     
  5. Mar 21, 2007 #4

    George Jones

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    It is general.

    Equation (11.58) of Griffiths is

    [tex]\mathbf{\tau} \cdot \mathbf{A}'_\mu = S \left( \mathbf{\tau} \cdot \mathbf{A}_\mu \right) S^{-1} + i \left( \frac{\hbar c}{q} \right) \left( \partial_\mu S \right) S^{-1}. [/tex]

    Here, [itex]S[/itex] is a member of the Lie (gauge) group and [itex]\mathbd{\tau}[/itex] is a basis for its Lie algebra.

    If [itex]S[/itex] is a rigid (independent of spacetime position), then only the first term on right survives, and this is just the definition of the adjoint representation of the Lie group on its Lie algebra.

    I am not sure if this answers your question.

    For mathematicians, [itex]\mathbf{\tau} \cdot \mathbf{A}_\mu[/itex] is a Lie algebra-valued one-form. Since the the spacetime index [itex]\mu[/itex] is downstairs, the [itex]A_\mu[/itex] are the components of a standard one-form. Evaluating this one-form at a 4-vector and using the result in [itex]\mathbf{\tau} \cdot \mathbf{A}_\mu[/itex] gives the sum of a bunch of scalars times the basis elements of the Lie algebra, which is just an element of Lie algebra. Hence, the name Lie algebra-valued (instead of real-valued) one-form.
     
    Last edited: Mar 21, 2007
  6. Dec 3, 2007 #5

    blechman

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    Hey.
    I was just perusing old posts and found this one, and thought I might add something. One way you can see that the gauge fields must be in the adjoint is to remember that they are connection coefficients. The way that you define the covariant derivative (I don't know if this is the "fundamental" way to think of it, but it's the way I learned it) is that you want the covariant derivative of a field to transform the same way that the field transforms. In other words:

    [tex]\psi\rightarrow U\psi\Rightarrow D_\mu\psi\rightarrow UD_\mu\psi[/tex]

    This can be thought of as an operator tranforming as:

    [tex]D_\mu\rightarrow UD_\mu U^{\dagger}[/tex]

    If [itex]U[/itex] is a fundamental rep transformation, this is how the adjoint transforms.

    Conclusion: the covariant derivative transforms as an adjoint, and therefore the gauge fields are in the adjoint rep (up to the inhomogeneous term that is cancelled by the ordinary derivative).

    Now here's a cute follow-up question: what if the quarks/leptons/higgs are NOT in the fundamental? So let's imagine that there is a colored object that is in the 6 of SU(3). Applying the same rule as above, the covariant derivative could transform as any rep of the [itex]6\otimes\bar{6}=27\oplus 8\oplus 1[/itex]. In particular, I see no reason why the gauge fields cannot be in the 27(!)

    Now, it should be clear that if there is EVEN **ONE** fundamental, then the covariant derivative of that field forces the gauge fields to be in the adjoint. So this is all irrelevant for the SM. However, if there is some more complicated symmetry at a higher scale, and no fundamentals of that symmetry, then perhaps we can have more exotic gauge fields!

    I don't know if this is right or not. What do you think?
     
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