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I Product rule for exterior covariant derivative

  1. May 15, 2017 #1
    It is well known that the product rule for the exterior derivative reads
    [tex]d(a\wedge b)=(da)\wedge b +(-1)^p a\wedge (db),[/tex]where a is a p-form.
    In gauge theory we then introduce the exterior covariant derivative [tex]D=d+A\wedge.[/tex] What is then D(a ∧ b) and how do you prove it?

    I obtain
    [tex]D(a\wedge b)=d(a\wedge b)+A\wedge a \wedge b=(da)\wedge b +(-1)^p a\wedge (db)+A\wedge a \wedge b,[/tex]
    which is neither (Da) ∧ b +(-1)p a ∧ (Db) nor (Da)∧ b + a∧ (Db). I have, however, seen the latter been used without proof.
     
  2. jcsd
  3. May 19, 2017 #2

    martinbn

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    May be the definition is for 1-forms and then you extend it to satisfy the product rule.
     
  4. May 25, 2017 #3

    Ben Niehoff

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    This statement is your error. The correct statement is

    $$D = d + \rho(A) \wedge$$
    where ##\rho : G \to GL(n)## is the homomorphism which defines the representation of the gauge group you need. If you have two 1-forms ##\alpha## and ##\beta##, each in the fundamental representation, then the 2-form ##\alpha \wedge \beta## is no longer in the fundamental representation! It will be in the antisymmetrized product of two fundamental representations (which is usually the adjoint rep, I think). Higher-degree wedge products will give you antisymmetrized products of more fundamental reps.
     
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