Covariant derivative of connection coefficients?by pellman Tags: coefficients, connection, covariant, derivative 

#1
Feb1712, 06:43 AM

P: 565

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. 



#2
Feb1712, 09:46 AM

P: 565

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.




#3
Feb1712, 11:41 AM

Sci Advisor
P: 1,563

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. 



#4
Feb1712, 11:51 AM

Sci Advisor
P: 1,563

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... 



#5
Feb1812, 06:17 AM

P: 565

Thanks, guys. Yeah, I never liked the semicolon notation either.




#6
May2312, 06:26 PM

P: 145

[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 



#7
May2312, 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] ...? 



#8
May2312, 07:02 PM

P: 260

Since the connection coefficients aren't a tensor, taking a covariant derivative of them doesn't really make sense.




#9
May2312, 07:11 PM

P: 145





#10
May2412, 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 beit 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. 



#11
May2412, 08:28 AM

P: 565





#12
May2412, 03:18 PM

P: 145

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



#13
May2412, 03:19 PM

P: 145





#14
May2412, 03:48 PM

Sci Advisor
HW Helper
P: 11,866

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.




#15
May2412, 05:22 PM

P: 145




Register to reply 
Related Discussions  
Table of Connection Coefficients  Special & General Relativity  6  
Covariant derivative vs Gauge Covariant derivative  Differential Geometry  3  
Covariant Derivative and Gauge Covariant Derivative  General Math  0  
Spherical connection coefficients  Advanced Physics Homework  2  
Covariant derivative repect to connection?  Differential Geometry  3 