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Nonabelian gauge field

  1. Jun 3, 2005 #1
    Dear friends
    I am here with mathematical physics question:
    we know tha if i have a compact Lie group G with g its Lie algebra, and a connection A on the fibre,
    For nonabelain Lie algebra
    The relation between covariant derivative and the curvature of A is
    Code (Text):
    [ tex ]\begin{equation*}[D_{m},D_{n}]F_{ab}=[F_{mn},F_{ab}]\end{equation*}[ /tex ]
    for any representation of g the Lie algebra
    D is the covariant derivative
    F the curvature of the connection A
    my problem:
    I will be so grateful if someone could help me to prove that
    Code (Text):
    [ tex ][D_{m},D_{n}]F_{ab}=[F_{mn},F_{ab}][ /tex ]
    is valid for any representation of the Lie algebra g especially for the fundamental (defining) representation, because i already did it for the adjoint representation of g.
    thank you in advance wissam
  2. jcsd
  3. Jun 3, 2005 #2


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    You got me all lost here.Why doesn't the curvature have 4 suffixes...?What are the gauge fields (potentials) and the field tensors...?

  4. Jun 3, 2005 #3
    the gauge field is the yang-mills

    Dear daniel
    Thank u for your interest in my question, the story here, physically talking means that the gauge field is the non-abelain gauge field "A_{mu}, and the field strength is the F_{mu nu}.
    the gauge field is not the one of the gravidation
    PS: please make difference between abelian( Maxwell theory), and non abelian (yang-Mills).
    thank u
  5. Jun 3, 2005 #4


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    Do you know how to build any finite representation of the gauge group...?

    In the case u do,u'll find that your problem reduces to checking the validity of that equality only in the case of the 2 fundamental (possibly inequivalent) irreps of the gauge group:the fundamental and its contragradient one.

  6. Jun 3, 2005 #5
    Dear daniel
    we know that
    F=dA+1/2[A,A]=dA+A^A (differential forms)
    F_{mu nu}=[D_{mu},D_{nu}] in any local basis.
    F here is g-valued
    F_{mu nu}=F^a T_{a} where T_{a} is the generator of g.

    In the irre. adjoint representation of g gives C_{ab}^{c}=(T_{a})^{c}_{b}
    where C is the structure constant of g,
    in the adjoint irr.rep the covariant derivative is

    So using these facts, the equality in question is feasible.

    my problem once again is:
    i know that this relation is valid in any irr.rep of the algebra,
    i will be so grateful if i can see the proof at least for the irr. fundemantal representation.
    In any arbitrary representation, the covariant derivative written as
    D=d+A without the commutator
    A, F are G-valued
    In components language for any field Q the covariant derivative is
    thank you
  7. Jun 26, 2005 #6
    [tex]\begin{equation*}[D_{m},D_{n}]F_{ab}=[F_{mn},F_{ab}]\end{equation*}[/tex] Sorry, I can't view this code, so I posted it to see what it looks like
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