Commutator of two covariant derivatives

In summary: In the second term, the Christoffel symbol should have a different index from the first term to account for the contraction with the covariant index of the vector field. In the third term, the Christoffel symbol should have a different index from the first term to account for the contraction with the contravariant index of the vector field.
  • #1
ibazulic
5
0
Hello all,

I'm trying to calculate a commutator of two covariant derivatives, as it was done in Caroll, on page 122. The problem is, I don't get the terms he does :-/

If ##\nabla_{\mu}, \nabla_{\nu}## denote two covariant derivatives and ##V^{\rho}## is a vector field, i need to compute ##[\nabla_{\mu}, \nabla_{\nu}]V^{\rho}##. Covariant derivative is defined as

[tex]\nabla_{\mu} = \partial_{\mu} + \Gamma^{\nu}_{\mu\lambda}[/tex]

Putting it into the definition of the commutator, one can write

[tex]
\begin{align}
[\nabla_{\mu}, \nabla_{\nu}]V^{\rho} &= \nabla_{\mu} \nabla_{\nu} V^{\rho} - \nabla_{\nu}\nabla_{\mu} V^{\rho} \nonumber \\ &=\partial_{\mu} (\nabla_{\nu} V^{\rho})+\Gamma^{\rho}_{\mu\sigma}(\nabla_{\nu}V^{\sigma})-\Gamma^{\lambda}_{\mu\nu}\nabla_{\lambda}V^{\rho}+(\mu \leftrightarrow \nu) \nonumber \\ &=\ldots \nonumber
\end{align}
[/tex]

What gives me problems is the 3rd term in the 2nd row. I don't know where this third term comes from. The expansion is even more problematic, Caroll expands these three terms into 7:

[tex]\ldots\partial_{\mu}\partial_{\nu}V^{\rho}+(\partial_{\mu}\Gamma^{\rho}_{\nu\sigma})V^{\sigma}+\Gamma^{\rho}{\gamma\sigma}\partial_{\mu}V^{\rho}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda}V^{\rho}-\Gamma^{\lambda}_{\mu\nu}\Gamma^{\rho}_{\lambda\sigma}V^{\sigma}+\Gamma^{\rho}_{\mu\sigma}\partial_{\nu}V^{\sigma}+\Gamma^{\rho}_{\mu\sigma}V^{\lambda}-(\mu \leftrightarrow \nu)\ldots[/tex]

Any ideas would be very much appreciative.

:-)
 
Physics news on Phys.org
  • #2
ibazulic said:
What gives me problems is the 3rd term in the 2nd row. I don't know where this third term comes from.
The action of the first covariant derivative is on a type (1,1) tensor. As such, you must include one term with a Christoffel symbol for both the covariant and the contravariant index of that tensor.

ibazulic said:
. Covariant derivative is defined as
This seems to be part of your problem. Your definition of the contravariant derivative both contains too many indices on one side and seems like it is coming from the action on a contravariant vector only. You need to look up the general definition.
 
  • #3
Caroll defines the covariant derivative as follows:

[tex]\nabla_\mu V^{\nu} = \partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu\sigma}V^{\sigma}[/tex]

(formula 3.2 on page 93)

:when i wrote the formula in my first post, i omitted the vector field, just gave a definition on what ##\nabla_{\mu}## is. that's why i don't really understand your comment. can you clarify a bit further?
 
  • #4
ibazulic said:
Caroll defines the covariant derivative as follows:
You will note that in this definition the indices are contracted between the vector field and the Christoffel symbol. It is therefore not possible to simply remove the vector field from the definition.

ibazulic said:
that's why i don't really understand your comment. can you clarify a bit further?
Which part do you have problems with? The fact that ##\nabla_\mu V^\nu## is a type (1,1) tensor or how the covariant derivative acts on arbitrary tensors? The latter should be defined a little bit later in Carroll.
 
  • #5
Orodruin said:
You will note that in this definition the indices are contracted between the vector field and the Christoffel symbol. It is therefore not possible to simply remove the vector field from the definition.

true, my mistake about that one.

Which part do you have problems with? The fact that ##\nabla_\mu V^\nu## is a type (1,1) tensor or how the covariant derivative acts on arbitrary tensors? The latter should be defined a little bit later in Carroll.

now when i reread the first comment again, i see your point :-) thanks. one more question: are indicies in the 2nd and 3rd term correct? now that i look at it, it seems they aren't but I'm not sure.
 
  • #6
ibazulic said:
are indicies in the 2nd and 3rd term correct?
Yes.
 

FAQ: Commutator of two covariant derivatives

What is the commutator of two covariant derivatives?

The commutator of two covariant derivatives is a mathematical operation that measures the failure of two derivatives to commute with each other. It is denoted by [Dμ, Dν], where Dμ and Dν are two covariant derivatives.

What is the significance of the commutator of covariant derivatives?

The commutator of covariant derivatives is an important quantity in differential geometry, as it is related to the curvature of a manifold. It describes the failure of parallel transport to be path-independent, which is a fundamental concept in the study of curved spaces.

How is the commutator of covariant derivatives calculated?

The commutator of covariant derivatives is calculated by taking the difference of two covariant derivatives applied to a function or tensor. This quantity is then contracted with the appropriate metric tensor to obtain the final result. The specific calculation can vary depending on the type of manifold and covariant derivatives used.

Can the commutator of covariant derivatives be zero?

Yes, it is possible for the commutator of covariant derivatives to be zero. This would indicate that the manifold is flat and the covariant derivatives commute with each other. However, in most cases, the commutator is non-zero and describes the curvature of the manifold.

How is the commutator of covariant derivatives related to the Lie bracket?

The commutator of covariant derivatives is closely related to the Lie bracket, which is a mathematical operation that measures the failure of two vector fields to commute with each other. In fact, the commutator of covariant derivatives can be expressed in terms of the Lie bracket, making it a useful tool in the study of differential geometry and curved spaces.

Similar threads

Back
Top