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Covariant exterior derivative vs regular exterior derivative

  1. Apr 21, 2012 #1

    Matterwave

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    Quick question.

    Suppose we have a manifold with a metric and a metric compatible symmetric connection.

    Suppose further that we have a smooth vector field V on this manifold.

    I see two ways to take the derivative of this vector field.

    I can regard my vector field as a vector-valued 0-form and take the covariant exterior derivative (which, in this simple case, is simply the covariant derivative) and obtain ∇V.

    Alternatively, I can use the musical isomorphisms induced by the metric to change my vector field into a one-form field and then take the regular exterior derivative, and then use the metric again to "raise the second index" to turn my 2-form into a (1,1) tensor (like ∇V above).

    This operation would give me: [itex](dV^\flat)^\sharp[/itex]
    Where the sharp is implied to act on the "second index".

    The question is: are these two operations equivalent? It seems to me that they would not be, but it also seems to me that it's plausible that they are since both operations involved using the metric, the first in the definition of the covariant derivative, and the second explicitely in raising and lowering indices.

    Are there any relation between the two? Thanks.
     
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  3. Apr 21, 2012 #2

    quasar987

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    Strangely, I compute that the difference between the two, when evaluated on (X,Y), is [itex]g(\nabla_YV,X)[/itex].
     
  4. Apr 22, 2012 #3

    Matterwave

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    Hmm interesting...I don't see any intuitive reason why that should be...
     
  5. Apr 22, 2012 #4

    quasar987

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    I did not express myself very well. What I did is I said ok , [itex]\nabla V=(d(V^{\flat}))^{\sharp}[/itex] iff [itex] (\nabla V)^{\flat}=d(V^{\flat})[/itex] and then I computed both sides when evaluated on (X,Y). If you want to check my computations, here is what I did: for the RHS, I first used the invariant formula for the exterior derivative of a 1-form, which yields 3 terms. Then I used the metric compatibility condition on the first 2 terms, and the symetry condition on the third one. Some terms cancel out, and then LHS-RHS = [itex]g(\nabla_YV,X)[/itex].
     
  6. Apr 22, 2012 #5

    Matterwave

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    Yea, I figured that's what you did (the first part) since that was the only option that made sense based on your answer. I don't doubt your result. It also makes sense in the sense that all the pertinent variables are accounted for - the metric, and the vector fields V, X and Y.

    I just don't have any intuitive explanation for such a result. I guess what I'm saying is, the result doesn't make my understanding of the 2 concepts much deeper (of course, this is not your fault).
     
  7. Apr 23, 2012 #6
    I have doubts about this operation (most likely due to my lack of knowledge). I don't understand well enough how do you make a an antisymmetric two form into a symmetric vector-valued one-form just by raising an index with the metric. I can see how that can be done with a regular covariant rank two tensor to get a mixed tensor that is also a vector-valued one-form, but then both objects are symmetric. Could you clarify this?
     
  8. Apr 23, 2012 #7

    Matterwave

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    What do you mean symmetric? I'm just raising one of the indices. I'm not symmetrizing anything.

    A 2-form is also a (0,2) tensor is it not? I can use the metric to turn that into a (1,1) tensor.
     
  9. Apr 23, 2012 #8
    Sure, I thought you were implying that what you obtained was a vector valued one-form like the ∇V you obtained the other way.
     
  10. Apr 23, 2012 #9

    Ben Niehoff

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    No, they are not. They are both tensors (i.e., covariant objects) of the same type, however, so they must differ by a tensor.
     
  11. Apr 23, 2012 #10

    Matterwave

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    Is a (1,1) tensor not always a vector valued one-form? My understanding of these concepts is not very complete.

    However, both of the object are at least (1,1) tensors, and so should be comparable.
     
  12. Apr 23, 2012 #11
    I would say not always, or at least in this special case when we are dealing with rank two tensors so they are either completely antisymmetric (the usual differential 2-forms) or completely symmetric (like say the Ricci tensor), but Ben is the master on these things, hopefully he'll say something if I'm wrong or to better clarify it.
    This much is true, as Ben said.
     
  13. Apr 23, 2012 #12

    Matterwave

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    Why do rank 2 tensors need to be either symmetric or anti-symmetric? They certainly should be able to be neither...(i.e. a sum of a symmetric and an anti-symmetric part).
     
  14. Apr 23, 2012 #13

    quasar987

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    TM-valued 1-forms, (1,1)-tensors, [itex]C^{\infty}(M)[/itex]-bilinear mappings [itex]\Gamma(TM)\times\Gamma(T^*M)\rightarrow C^{\infty}(M)[/itex] and bundle maps [itex]TM\rightarrow TM[/itex] are all the same thing (up to a canonical isomorphism).
     
    Last edited: Apr 23, 2012
  15. Apr 23, 2012 #14
    Oh, ok, and when you exchange indices you get neither the same tensor nor a change of sign right? Sorry, I didn't consider that option.
     
  16. Apr 23, 2012 #15
    So then all mixed tensors can be considered vector(or tensor) valued forms?
     
  17. Apr 23, 2012 #16

    quasar987

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    yep. (to both questions)
     
  18. Apr 23, 2012 #17

    lavinia

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    If your vector field is the gradient of a function then the 1 1 tensor is identically zero because the exterior derivative of the differential of the function is zero.

    But the covariant derivative of the field is not generally zero.

    An arbitrary vector field can be modified by a gradient without changing the exterior derivative of its musical dual but the covariant derivative will change.
     
    Last edited: Apr 23, 2012
  19. Apr 23, 2012 #18
    Yes, d(df)=0, actually matterwave didn't specify if he wanted his vector field to be the gradient of a function. I guess if it was he would have gotten zero (so no two-form) on one hand and a symmetric tensor on the other (hessian: [itex]\nabla[/itex]df), the difference is a (symmetric) tensor (the Hessian).
    If it was not the gradient of a function then a 1,1 vector valued one form would be antisymmetric (from the 2-form) and the other would be part symmetric and part antisymmetric but their difference is also a (symmetric) tensor. Like in the other case I guess the one quasar987 computed [itex](\nabla_Y V,X)[/itex]
     
  20. Apr 25, 2012 #19

    lavinia

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    You can also take the covariant derivative of the one form to get a (0,2) tensor.

    If the vector field is a gradient and if the connection is symmetric then this tensor is symmetric.
     
  21. Apr 25, 2012 #20

    Matterwave

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    ok thanks for the input. =]
     
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