- #1

schulmannerism

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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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- Thread starter schulmannerism
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- #1

schulmannerism

- 4

- 0

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

?

- #2

- 13,256

- 1,290

[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.

- #3

schulmannerism

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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...

[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:

- #4

schulmannerism

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Thanks

- #5

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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[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.

- #6

schulmannerism

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

- #7

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Daniel.

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