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Connections and tensor fields

  1. Feb 23, 2015 #1

    CAF123

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    Gold Member

    1. The problem statement, all variables and given/known data
    Let ##T## be a ##(1, 1)## tensor field, ##\lambda## a covector field and ##X, Y## vector fields. We may define ##\nabla_X T## by requiring the ‘inner’ Leibniz rule, $$\nabla_X[T(\lambda, Y )] = (\nabla_XT)(\lambda, Y ) + T(\nabla_X \lambda, Y ) + T(\lambda, \nabla_X Y ) . $$

    (a) Prove that ##\nabla_XT## defines a ##(1, 1)## tensor field.
    (b) Prove that the components of the ##(1, 2)## tensor ##\nabla T## are, $$\nabla_cT^{a}_{\,\,\,b} = e_c(T^{a}_{b}) + \Gamma^{a}_{\,\,dc}T^{d}_{b} − \Gamma^{d}_{ bc}T^{a}_{d}$$
    (c) Deduce that the Kronecker delta tensor is covariantly constant ##\nabla \delta = 0.##
    2. Relevant equations

    Tensor fields and linearity

    3. The attempt at a solution

    Really just need to check what I am going to do is correct. For a), ##\nabla_X T## defines a (1,1) tensor if it is linear in the arguments ##X## and ##Y##? Linear in ##X## by definition of a connection and for ##Y##;

    $$\nabla_X (T (\lambda, fY)) = (\nabla_X T)(\lambda fY) + T(\nabla_X \lambda, fY) + T(\lambda, \nabla_X (fY)) $$ which can be written using the axioms of a connection $$ f(\nabla_X T)(\lambda, Y) + fT(\nabla_X \lambda, Y) + fT(\lambda, f \nabla_X Y) + T(\lambda, X(f)Y)$$ It doesn't seem to be linear in Y at all?

    Thanks!
     
  2. jcsd
  3. Feb 23, 2015 #2

    stevendaryl

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    Well, you started with the equation:

    [itex]\nabla_X (T(\lambda, Y)) = (\nabla_X T)(\lambda, Y) + T(\nabla_X \lambda, Y) + T(\lambda, \nabla_X Y)[/itex]

    You can rewrite this to get:

    [itex](\nabla_X T)(\lambda, Y) = \nabla_X (T(\lambda, Y)) - T(\nabla_X \lambda, Y) - T(\lambda, \nabla_X Y)[/itex]

    So it's the combination of all three terms on the right-hand side that must be linear in [itex]\lambda[/itex] and [itex]Y[/itex]. The individual terms don't have to be linear. To prove it's linear in [itex]Y[/itex], replace [itex]Y[/itex] by [itex]f Y[/itex]:

    [itex](\nabla_X T)(\lambda, f Y) = \nabla_X (T(\lambda, f Y)) - T(\nabla_X \lambda, f Y) - T(\lambda, \nabla_X (f Y))[/itex]

    Then you have to prove that the right-hand side is equal to [itex]f (\nabla_X (T(\lambda, Y)) - T(\nabla_X \lambda, Y) - T(\lambda, \nabla_X Y))[/itex]

    To prove this, you have to use the linearity of [itex]T[/itex] and the Leibniz rule to rewrite the right-hand side so that hopefully you can get [itex]f[/itex] on the outside. What that means is that derivatives of [itex]f[/itex] have to cancel.
     
  4. Feb 24, 2015 #3

    CAF123

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    Thanks! Ok so I proved the linearity in ##Y##, can I use the same method to find the linearity in ##\lambda##? So let ##\lambda \rightarrow f \lambda##?

    For part b), I am not sure how much detail is required for an answer. The definition I have in my notes is that 'Given an ##(r,s)## tensor field ##T##, the covariant derivative ##\nabla T## is an ##(r,s+1)## tensor with components $$\nabla_c T^{a_1 .... a_r}_{\,\,\,\,\,\,b_1 ....b_s} = e_c T^{a_1....a_r}_{\,\,\,\,b_1...b_s} + \Gamma^{a_1}_{dc} T^{d.....a_r}_{\,\,\,\,b_1....b_s} + .... + \Gamma^{a_r}_{dc}T^{a_1 ....d}_{\,\,\,\,b_1 ..... b_s} - \Gamma^d_{b_1 c} T^{a_1....a_r}_{\,\,\,\,d....b_s} - .... - \Gamma^{d}_{b_s c} T^{a_1...a_r}_{\,\,\,\,\,b_1....d}$$ So applying this gives the result immediately, as far as I can see we are evaluating all the possible contractions of the covariant and contravariant components of the tensor. Do you think this is fine? And is this result a definition or does it follow from something? Thanks!
     
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