Vector Field Commutator Identity in Covariant Derivative

[schulmannerism]

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]

Using a more familiar notation, i'll write

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

A remark. I've never seen the operator in the RHS. I'm sure $$? \ \leftrightarrow \ \neq$$.

Daniel.

 

[schulmannerism]

Well the original problem is to show that if $$f(P)$$ is a scalar field such that $$f(P_0)=1$$, and $$A,B,C$$ are vector fields, then
$$[\nabla_A,\nabla_B](f(P)C(P))-[\nabla_A,\nabla_B{A}](C(P))=[\nabla_{[A,B]}(C(P))$$
Unless I am doing something wrong, this immediately reduces to the above identity, which doesn't look true to me either...

 

[schulmannerism]

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]

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

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

Your second post boggles me, because it looks like you're trying to take X(Y) where X and Y are both vector fields.

 

[schulmannerism]

Whoops I wrote my second post incorrectly, it's editted.

 

[dextercioby]

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.