Covariant Derivative Identity

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
?

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dextercioby

Homework Helper
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...

Last edited:

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

Staff Emeritus
Gold Member
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

Homework Helper
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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