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I Commutator of covariant derivatives

  1. Nov 19, 2016 #1

    naima

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    Hi there
    I came across this paper.
    the author defines a covariant derivative in (1.3)
    ##D_\mu = \partial_\mu - ig A_\mu##
    He defines in (1.6)
    ##F_{jk} = i/g [D_j,D_k]##
    Why is it equal to ##\partial_j A_k - \partial_k A_j - ig [A_j, A_k]##?
    I suppose that it comes from a property of Lie derivatives.
    Thanks
     
  2. jcsd
  3. Nov 19, 2016 #2

    ShayanJ

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    Just put the definition of ## D_\mu ## in ## [D_\mu,D_\nu] ## and calculate!
    You can apply the above operator on a scalar field, i.e. calculate ## [D_\mu,D_\nu]\phi ## to avoid confusion.
     
  4. Nov 19, 2016 #3

    naima

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    Why does ##A_j \partial_k## disappear?
     
  5. Nov 19, 2016 #4

    Orodruin

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    What you have is a commutator of operators, in order to compute the commutator - which is another operator - you need to consider the action of the commutator.

    I suggest you start with the following example with two operators acting on functions on the real line:

    Operator ##A## is such that ##Af(x) = A(x)f(x)##, where ##A(x)## is a function.
    Operator ##B## is the derivative operator such that ##Bf(x) = f'(x)##.

    The commutator ##[A,B]## is the operator such that ##[A,B]f(x) = (AB - BA)f(x)## for all ##f##. What is this operator?
     
  6. Nov 21, 2016 #5

    naima

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    Thank you

    This shows that when you have product of operators D and B acting on a vector V, the notation DBV may be considered as a compositiion law
    D(BV) or else (DB) V if you can give sense to (DB)
    IF D and B are patrices the internal law of multiplication gives
    (DB)V = D(BV)
    here (DB) = D(B .)
    If D is a derivative we have D(BV) = (DB)V + B (DV)
    This gives a sense to (DB)
    here (DB) = DB - BD

    So the equality in the paper comes from an implicit choice for the product of operators.
    Unless it is based on a more intrinsic way to mix covariant derivatives.
     
  7. Nov 21, 2016 #6

    Orodruin

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    It does not matter whether D and B are matrices or linear operators on a general function space (including derivatives). The commutator [D,B] is defined as the linear operator such that [D,B]f = D(Bf) - B(Df) for all f in the vector space that the operators act on.

    If D is a derivative and B is multiplication by a function b, you have D(Bf) = D(bf) = f (Db) + b(Df) = f(Db) + B(Df) and therefore D(Bf) - B(Df) = [D,B]f = f (Db), i.e., the commutator [D,B] is multiplication by the function Db. Note that I have used capital and lower case Bs to distinguish DB (operator product) from Db (a function).
     
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