Subtraction of Christoffel symbol

  • A
  • Thread starter mertcan
  • Start date
  • #1
340
6
hi, How do we prove the torsion tensor is a tensor using the subtraction of christoffel symbols?? I am aware of the fact that subtraction of christoffel symbols equals the torsion. How can we use this fact to prove the tensor?? Could you please give the proof or share the link which prove it??
 

Answers and Replies

  • #2
wrobel
Science Advisor
Insights Author
741
452
use the law of transformation of the Christoffel symbols under the change of local coordinates
 
  • #3
haushofer
Science Advisor
Insights Author
2,474
864
Carroll shows this in great detail. If you write down the transformation of the difference between two connections, the inhomogeneous term drops out. This is because the inhomogeneous term is symmetric in the lower indices. As such this difference transforms as a tensor.

You're asking different questions, so maybe it's a good idea to tell us what your background and literature is. This should help if you have any further questions.
 
  • #4
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,496
2,632
hi, How do we prove the torsion tensor is a tensor using the subtraction of christoffel symbols?? I am aware of the fact that subtraction of christoffel symbols equals the torsion. How can we use this fact to prove the tensor?? Could you please give the proof or share the link which prove it??
A general rule of thumb is that if you can define an operation without mentioning a particular coordinate system, then it is a coordinate-free definition, and linear operations with coordinate-free definitions are (almost?) always tensors. So a way to show that [itex]T[/itex] is a tensor is to show:
  1. There is a coordinate-free way to define [itex]T(U,V)[/itex]
  2. [itex]T(U,V)[/itex] is linear in both [itex]U[/itex] and [itex]V[/itex]
The coordinate-free definition of [itex]T(U,V)[/itex] is this:

[itex]T(U,V) = \nabla_U V - \nabla_V U - [U,V][/itex]

where [itex]\nabla_U[/itex] means the covariant derivative, or directional derivative along [itex]U[/itex], and where [itex][U,V][/itex] is the vector field defined by its action on a scalar field:

[itex]\nabla_{[U,V]} \phi \equiv \nabla_U (\nabla_V \phi) - \nabla_V (\nabla_U \phi) [/itex]

So [itex]T(U,V)[/itex] has a coordinate-free definition. Now, you have to show that it is linear in [itex]U[/itex] and [itex]V[/itex]. Since [itex]T(U,V) = -T(V,U)[/itex], it's enough to show that [itex]T(U,V)[/itex] is linear in [itex]U[/itex]. So let [itex]\bar{U} \equiv \phi U + W[/itex], where [itex]\phi[/itex] is a scalar field, and [itex]W[/itex] is a vector field. Then we need to show that [itex]T(\bar{U}, V) = \phi T(U,V) + T(W,V)[/itex]

[itex]T(\bar{U},V) = \nabla_{\bar{U}} V - \nabla_V \bar{U} - [\bar{U},V][/itex]

Using properties of covariant derivatives, we have:

[itex]\nabla_{(\phi U + W)} V = \phi \nabla_U V + \nabla_W V[/itex]
[itex]\nabla_V (\phi U + W) = (\nabla_V \phi) U + \phi (\nabla_V U) + \nabla_V W[/itex]
[itex][\phi U + W, V] = \phi [U,V] - (\nabla_V \phi) U + [W,V][/itex]

So we have:

[itex]T(\bar{U},V) = \phi \nabla_U V + \nabla_W V - (\nabla_V \phi) U - \phi (\nabla_V U) - \nabla_V W - \phi [U,V] + (\nabla_V \phi) U - [W,V][/itex]

The two occurrences of [itex]\nabla_V \phi[/itex] cancel out, giving:

[itex]T(\bar{U},V) = \phi \nabla_U V + \nabla_W V + \phi (\nabla_V U) - \nabla_V W - \phi [U,V] - [W,V][/itex]
[itex] = \phi (\nabla_U V - \nabla_V U - [U,V]) + \nabla_W V - \nabla_V W - [W,V][/itex]
[itex] = \phi T(U,V) + T(W,V)[/itex]
 
  • #5
340
6
A general rule of thumb is that if you can define an operation without mentioning a particular coordinate system, then it is a coordinate-free definition, and linear operations with coordinate-free definitions are (almost?) always tensors. So a way to show that [itex]T[/itex] is a tensor is to show:
  1. There is a coordinate-free way to define [itex]T(U,V)[/itex]
  2. [itex]T(U,V)[/itex] is linear in both [itex]U[/itex] and [itex]V[/itex]
The coordinate-free definition of [itex]T(U,V)[/itex] is this:

[itex]T(U,V) = \nabla_U V - \nabla_V U - [U,V][/itex]

where [itex]\nabla_U[/itex] means the covariant derivative, or directional derivative along [itex]U[/itex], and where [itex][U,V][/itex] is the vector field defined by its action on a scalar field:

[itex]\nabla_{[U,V]} \phi \equiv \nabla_U (\nabla_V \phi) - \nabla_V (\nabla_U \phi) [/itex]

So [itex]T(U,V)[/itex] has a coordinate-free definition. Now, you have to show that it is linear in [itex]U[/itex] and [itex]V[/itex]. Since [itex]T(U,V) = -T(V,U)[/itex], it's enough to show that [itex]T(U,V)[/itex] is linear in [itex]U[/itex]. So let [itex]\bar{U} \equiv \phi U + W[/itex], where [itex]\phi[/itex] is a scalar field, and [itex]W[/itex] is a vector field. Then we need to show that [itex]T(\bar{U}, V) = \phi T(U,V) + T(W,V)[/itex]

[itex]T(\bar{U},V) = \nabla_{\bar{U}} V - \nabla_V \bar{U} - [\bar{U},V][/itex]

Using properties of covariant derivatives, we have:

[itex]\nabla_{(\phi U + W)} V = \phi \nabla_U V + \nabla_W V[/itex]
[itex]\nabla_V (\phi U + W) = (\nabla_V \phi) U + \phi (\nabla_V U) + \nabla_V W[/itex]
[itex][\phi U + W, V] = \phi [U,V] - (\nabla_V \phi) U + [W,V][/itex]

So we have:

[itex]T(\bar{U},V) = \phi \nabla_U V + \nabla_W V - (\nabla_V \phi) U - \phi (\nabla_V U) - \nabla_V W - \phi [U,V] + (\nabla_V \phi) U - [W,V][/itex]

The two occurrences of [itex]\nabla_V \phi[/itex] cancel out, giving:

[itex]T(\bar{U},V) = \phi \nabla_U V + \nabla_W V + \phi (\nabla_V U) - \nabla_V W - \phi [U,V] - [W,V][/itex]
[itex] = \phi (\nabla_U V - \nabla_V U - [U,V]) + \nabla_W V - \nabla_V W - [W,V][/itex]
[itex] = \phi T(U,V) + T(W,V)[/itex]
Thank you "stevendarly", I appreciate your endeavor. The things you have written down have remarkable and striking structure. They are based on a nice structure.
 
  • #6
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,496
2,632
Thank you "stevendarly", I appreciate your endeavor. The things you have written down have remarkable and striking structure. They are based on a nice structure.
You are welcome.
 
  • #7
340
6
Hi, I would like to ask whether or not linear operations with coordinate-free definitions are always tensors, Because "stevendarly" said that linear operations with coordinate-free definitions were (almost?) always tensors. Is there a proof or derivation that just tensors are related to linear operations with coordinate-free definitions ???
 
  • #8
340
6
Guys, ?????
 
  • #9
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,496
2,632
Hi, I would like to ask whether or not linear operations with coordinate-free definitions are always tensors, Because "stevendarly" said that linear operations with coordinate-free definitions were (almost?) always tensors. Is there a proof or derivation that just tensors are related to linear operations with coordinate-free definitions ???
A tensor is just a multilinear function on vectors and covectors. So if you can prove that [itex]T(u,v)[/itex] is linear in both its arguments, then it follows that it is a tensor. The point about coordinate-free definitions is that such definitions reveal the functional dependency of the object in a way that the component definition can obscure. For a counter-example, the connection coefficients [itex]\Gamma^\mu_{\nu \lambda}[/itex] look like components of a tensor, but there is no coordinate-free way to write it as a function [itex]\Gamma(u,v)[/itex] that takes a pair of vectors and returns a vector. In contrast, the directional derivative [itex]\nabla_u v[/itex] does have a coordinate-free definition, but it isn't linear in the second argument ([itex]v[/itex]).
 
  • Like
Likes vanhees71 and mertcan

Related Threads on Subtraction of Christoffel symbol

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
10
Views
8K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
14
Views
2K
Replies
4
Views
762
Top