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Covariant Derivative Identity

I am trying to solve an exercise from MTW Gravitation and the following issue has come up:

Let D denote uppercase delta (covariant derivative operator)
[ _ , _ ] denotes the commutator
f is a scalar field, and A and B are vector fields

Question:
Is it true that
[D_A,D_B]f = D_[A,B]f
?
 

dextercioby

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Using a more familiar notation, i'll write

[tex] \left[\nabla_{\mu},\nabla_{\nu}\right]_{-}f(x) \ ? \ \nabla_{\left[\mu,\nu\right]} f(x) [/tex]

A remark. I've never seen the operator in the RHS. I'm sure [tex] ? \ \leftrightarrow \ \neq [/tex].

Daniel.
 
Well the original problem is to show that if [tex]f(P)[/tex] is a scalar field such that [tex]f(P_0)=1[/tex], and [tex]A,B,C[/tex] are vector fields, then
[tex][\nabla_A,\nabla_B](f(P)C(P))-[\nabla_A,\nabla_B{A}](C(P))=[\nabla_{[A,B]}(C(P))[/tex]
Unless I am doing something wrong, this immediately reduces to the above identity, which doesn't look true to me either...
 
Last edited:
Hurkyl -- your (deleted) post was helpful. I see now that I was having trouble with the notation in which a vector times a scalar field denotes the directional derivative in that direction. Then the identity does follow, as you pointed out.
Thanks
 

Hurkyl

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Well, the identity in your original post seems clear, unless I'm making a dumb mistake (which is easily possible): according to Spivak, [itex]\nabla_X f = X f[/itex], so we have:

[tex]
[\nabla_A, \nabla_B] f = (\nabla_A \nabla_B - \nabla_B \nabla_A) f = (AB - BA) f = [A, B] f = \nabla_{[A, B]} f
[/tex]

Your second post boggles me, because it looks like you're trying to take X(Y) where X and Y are both vector fields.
 
Whoops I wrote my second post incorrectly, it's editted.
 

dextercioby

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Ouch, i guess you both realized that i hadn't read the "A and B are vector fields" part and mistakenly took them as subscripts.


Daniel.
 

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