# 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
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.

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:
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.

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

dextercioby