Covariant Derivative Commutation

  • Thread starter PLuz
  • Start date
  • #1
64
0

Main Question or Discussion Point

Hello,

Can anyone tell me the general formula for commuting covariant derivatives, I mean, given a (r,s)-tensor field what is the formula to commute covariant derivatives?

I found a formula here page 25, Eq.6.18 but it doesn't seem right, since for a vector field one would write:

[itex](\nabla_{\alpha} \nabla_{\beta}- \nabla_{\beta}\nabla_{\alpha})U^{\gamma}=R^{\gamma}\hspace{.5 mm}_{\delta \alpha \beta}U^{\delta}[/itex]

And according to the formula in the link it would be, for a vector field


[itex](\nabla_{\alpha} \nabla_{\beta}- \nabla_{\beta}\nabla_{\alpha})U^{\gamma}=-R^{\gamma}\hspace{.5 mm}_{\delta \alpha \beta}U^{\delta}[/itex]

Thank you
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
I see only a difference in minus signs between the two expressions which can be accounted for in the convention used to define the Riemann tensor. I believe this formula is correct (at least for a coordinate basis, I cannot be sure if there are more terms for a non-coordinate basis).
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
12,986
541
For non-coordinate (aka anholonomic) basis, compute the Riemann tensor (it doesn't matter which sign convention you use) with the connection coefficients as provided in MTW page 210, formula 8.24b.
 

Related Threads on Covariant Derivative Commutation

Replies
5
Views
4K
Replies
1
Views
3K
Replies
3
Views
3K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
13
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
8
Views
2K
Replies
0
Views
2K
Top