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Covariant derivative of connection coefficients?

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pellman
#1
Feb17-12, 06:43 AM
P: 582
The connection [tex]\nabla[/tex] is defined in terms of its action on tensor fields. For example, acting on a vector field Y with respect to another vector field X we get

[tex]\nabla_X Y = X^\mu ({Y^\alpha}_{,\mu} + Y^\nu {\Gamma^\alpha}_{\mu\nu})e_\alpha
= X^\mu {Y^\alpha}_{;\mu}e_\alpha[/tex]

and we call [tex]{Y^\alpha}_{;\mu}={Y^\alpha}_{,\mu} + Y^\nu {\Gamma^\alpha}_{\mu\nu}[/tex] the covariant derivative of the components of Y. We can similarly form the covariant derivative of the components of any rank tensor, by including other appropriate terms with the connection coefficients.

So what does it mean to take the covariant derivative of the connection coefficients themselves? They are not components of a tensor? I have just come across a reference to [tex]{\Gamma^\alpha}_{\mu\nu;\lambda}[/tex] and don't know what to do with it.
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pellman
#2
Feb17-12, 09:46 AM
P: 582
I figured this out. Apparently, its covariant derivative does have the same form as the covariant derivative of the components of a (1,2) tensor. But if someone can confirm this result is correct, I would appreciate it.
Ben Niehoff
#3
Feb17-12, 11:41 AM
Sci Advisor
P: 1,594
That's a very bastard notation, and whoever wrote it down should explain what they mean. As you say, the connection coefficients are not a covariant object, so it is not sensible to talk about their covariant derivatives.

My guess is someone probably noticed they could write down the formula for the Riemann tensor in a kind of shorthand. It is technically incorrect.

Ben Niehoff
#4
Feb17-12, 11:51 AM
Sci Advisor
P: 1,594
Covariant derivative of connection coefficients?

By the way, I'm not sure of your level of knowledge, but if you're still learning this stuff, I would say to avoid getting in the habit of using "comma, semicolon" notation, for two reasons:

1. Since covariant derivatives do not commute, it is unclear what is meant by objects such as

[tex]A^\mu{}_{;\nu\rho} = \nabla_\nu \nabla_\rho A^\mu \quad \text{or} \quad \nabla_\rho \nabla_\nu A^\mu \; \text{?}[/tex]
2. On the printed page, little marks like commas and semicolons can be hard to see, especially in photocopies.

Whoever invented the notation thought they were being clever by saving space, but seems to have forgotten that the main purpose of scientific papers is to communicate...
pellman
#5
Feb18-12, 06:17 AM
P: 582
Thanks, guys. Yeah, I never liked the semi-colon notation either.
ianhoolihan
#6
May23-12, 06:26 PM
P: 145
Quote Quote by pellman View Post
I figured this out. Apparently, its covariant derivative does have the same form as the covariant derivative of the components of a (1,2) tensor. But if someone can confirm this result is correct, I would appreciate it.
Sorry to drag this up, but in trying to verify the formula for the components of the Riemann tensor in a non--coordinate basis, I need to know how to take the covariant derivative of the connection coefficients. Pellman, can you let me know the resource that confirmed that

[tex]\nabla_a \Gamma^b{}_{cd} = \partial_a \Gamma^b{}_{cd} + \Gamma^b{}_{ma}\Gamma^m{}_{cd} - \Gamma^m{}_{ca}\Gamma^b{}_{md}- \Gamma^m{}_{da}\Gamma^b{}_{cm}
[/tex]

Cheers
ianhoolihan
#7
May23-12, 06:39 PM
P: 145
Ah, working backward from the definition of the Riemann tensor, it would appear that

[tex]
\nabla_d \Gamma^a{}_{bc} = \partial_d \Gamma^a{}_{bc}
[/tex]

...?
elfmotat
#8
May23-12, 07:02 PM
elfmotat's Avatar
P: 260
Since the connection coefficients aren't a tensor, taking a covariant derivative of them doesn't really make sense.
ianhoolihan
#9
May23-12, 07:11 PM
P: 145
Quote Quote by elfmotat View Post
Since the connection coefficients aren't a tensor, taking a covariant derivative of them doesn't really make sense.
OK, but [itex]\Gamma^a{}_{bc}e^b[/itex] is a vector, so it makes sense to take its covariant derivative.
ApplePion
#10
May24-12, 07:50 AM
P: 205
A covariant derivative is the covariant analogue of a regular derivative. But if you use the affine connection as the thing to operate on, even if it has a form looking like a covariant derivative, it still will not be--it will not be a tensor.

You can do the operation anyway, and if it has physical usefulness it will still have physical usefulness even though it is not covariant.
pellman
#11
May24-12, 08:28 AM
P: 582
Quote Quote by ianhoolihan View Post
Sorry to drag this up, but in trying to verify the formula for the components of the Riemann tensor in a non--coordinate basis, I need to know how to take the covariant derivative of the connection coefficients. Pellman, can you let me know the resource that confirmed that

[tex]\nabla_a \Gamma^b{}_{cd} = \partial_a \Gamma^b{}_{cd} + \Gamma^b{}_{ma}\Gamma^m{}_{cd} - \Gamma^m{}_{ca}\Gamma^b{}_{md}- \Gamma^m{}_{da}\Gamma^b{}_{cm}
[/tex]

Cheers
Sorry. I never found an independent confirmation. I just proved it to my own satisfaction. I don't recall the details now either.
ianhoolihan
#12
May24-12, 03:18 PM
P: 145
Quote Quote by ApplePion View Post
A covariant derivative is the covariant analogue of a regular derivative. But if you use the affine connection as the thing to operate on, even if it has a form looking like a covariant derivative, it still will not be--it will not be a tensor.

You can do the operation anyway, and if it has physical usefulness it will still have physical usefulness even though it is not covariant.
I realise the connection coefficients are not the components of a tensor. However, in the case of the vector [itex]\nabla_c e_b = \Gamma^a{}_{bc}e_a[/itex] I'm pretty sure you can just treat the connection coefficient as the component of the vector: [itex]\Gamma^a{}_{bc} = [\nabla_c e_b]^a[/itex]. Hence

[tex]\nabla_d (\nabla_c e_b) = \partial_d \Gamma^a{}_{bc} + \Gamma^a{}_{fd}\Gamma^f{}_{bc}[/tex].

Quote Quote by pellman View Post
Sorry. I never found an independent confirmation. I just proved it to my own satisfaction. I don't recall the details now either.
I guess my case is different to yours.
ianhoolihan
#13
May24-12, 03:19 PM
P: 145
Quote Quote by ianhoolihan View Post
OK, but [itex]\Gamma^a{}_{bc}e^b[/itex] is a vector, so it makes sense to take its covariant derivative.
Sorry, I meant [itex]\Gamma^a{}_{bc}e_a[/itex] in this case, as in the post above.
dextercioby
#14
May24-12, 03:48 PM
Sci Advisor
HW Helper
P: 11,927
The connection coefficients are not the components of any tensor. The covariant derivative, if applied onto this set of components, would lose their meaning and purpose as a derivative. I haven't seen any source in geometry defining a covariant derivative to the connection coefficients.
ianhoolihan
#15
May24-12, 05:22 PM
P: 145
Quote Quote by dextercioby View Post
The connection coefficients are not the components of any tensor. The covariant derivative, if applied onto this set of components, would lose their meaning and purpose as a derivative. I haven't seen any source in geometry defining a covariant derivative to the connection coefficients.
Yes, as before, I understand this. However, as in the previous post, [itex]\nabla_d (\nabla_c e_b) = \nabla_d(\Gamma^a{}_{bc}e_a)[/itex] is a valid equation. Is my previous result correct?


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