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- Thread starter mertcan
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wrobel

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use the law of transformation of the Christoffel symbols under the change of local coordinates

- #3

haushofer

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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

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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:

- There is a coordinate-free way to define [itex]T(U,V)[/itex]
- [itex]T(U,V)[/itex] is linear in both [itex]U[/itex] and [itex]V[/itex]

[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

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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.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:

The coordinate-free definition of [itex]T(U,V)[/itex] is this:

- There is a coordinate-free way to define [itex]T(U,V)[/itex]
- [itex]T(U,V)[/itex] is linear in both [itex]U[/itex] and [itex]V[/itex]

[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]

- #6

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You are welcome.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.

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- #8

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Guys, ?????

- #9

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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]).

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