Covariant derivative of an anti-symmetric tensor

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  • #1
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Given an antisymmetric tensor

[itex]T^{ab}=-T^{ab}[/itex]

show that

[itex]T_{ab;c} + T_{ca;b} + T_{bc;a} = 0[/itex]

If I explicitly write out the covariant derivative, all terms with Christoffel symbols cancel pair-wise, and I'm left to demonstrate that

[itex]T_{ab,c} + T_{ca,b} + T_{bc,a} = 0[/itex]

and this I have no idea how to do. Could anybody put me on the right track please?
 

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  • #2
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OK I just came up with an idea. Assuming (under what conditions?!) that I can express

[itex] T_{ab} = T_{a,b}-T_{b,a} [/itex]

so as to get antisymmetry, and doing the same for all three terms, they all cancel pair-wise, on the condition that ordinary partial derivatives commute. So my updated question now is, (i) can I always regard a generic tensor T_ab as a partial derivative of some other tensor T_a, and (ii) do the partial derivatives always commute?
 
  • #3
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Have you presented your problem correctly? You're second equation presented in post #1 doesn't seem to be generally true.

In any case, I don't see any obvious tricks. Just expand your covariant derivatives into partial derivatives and connection terms using the general rule for tensors with lower indices that you've been given.

What you have in post #2 is not even close.
 
  • #4
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I've been given this question as an assignment. Anyway, if the proposition is not true, could you suggest a counter example such that

[itex] T_{ab;c} + T_{ca;b} + T_{bc;a} \neq 0 [/itex]

when [itex] T_{ab} = -T_{ba}[/itex]?
 
  • #5
dextercioby
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T is a 2-form. dT is a 3-form which is generally non zero, unless T is exact. The question probably assumes working on a manifold with trivial de Rham cohomology, thus T is assumed to be exact.
 
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  • #6
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I've been given this question as an assignment. Anyway, if the proposition is not true, could you suggest a counter example such that

[itex] T_{ab;c} + T_{ca;b} + T_{bc;a} \neq 0 [/itex]

when [itex] T_{ab} = -T_{ba}[/itex]?

Sure. For antisymmetric tensors with lower indeces

[tex]T_{ab;c} + T_{ca;b} + T_{bc;a} = T_{ab,c} + T_{ca,b} + T_{bc,a} [/tex]

is always true for a Christoffel connection (and I think, any connection). Can you demonstrate this by expansion?
 
  • #7
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Yes, I have demonstrated it. What remains to show is that the right-hand-side is identically equal to zero (or give a counter example).
 
  • #8
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Yes, I have demonstrated it. What remains to show is that the right-hand-side is identically equal to zero (or give a counter example).

OK, so you no longer need to involve the Christoffel connection but just use the covariant derivatives

This defines the exterior derivative of an asymmetric tensor with two lower indices:

[tex]\partial_{[a} T_{bc]} \equiv T_{ab,c} + T_{ca,b} + T_{bc,a}[/tex]

a, b, and c are cyclically permuted in each term.

Hint: The exterior derivative is not in general zero, as bigubau has pointed out.

P.S. I was a lot sleepier than I thought when I first posted this, so left out a constant. For T antisymmetric,

[tex]\partial_{[a} T_{bc]} \equiv 6 \left[ T_{ab,c} + T_{ca,b} + T_{bc,a} \right][/tex]
 
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  • #9
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Shouldn't this be in HOMEWORK????
 
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  • #10
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Shouldn't this be in HOMEWORK????

Are you offended?
 
  • #11
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maybe I should be ..having been severely chastised in the past for once simply posting a technical question in the wrong techncial section.....
 

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