Why can we use metric tensors to lower index of Christoffel symbol

[ZealScience]

I haven't learned much of advanced mathematics. It seems that we can use metric tensors to lower or raise index of christoffel symbols. But isn't christoffel symbols made of metric tensors and derivatives of metric tensors? How can we contract indices of a derivative directly with metric tensors while they are non constants.

 

[WannabeNewton]

Remember that the metric tensor in and of itself is such that g_{p}:T_{p}(M) \times T_{p}(M) \mapsto \mathbb{R} for a p\in M. But let's say, instead of g(\mathbf{U}, \mathbf{V}), (\mathbf{U, V})\in T_{p}(M)\times T_{p}(M) we simply had g(\mathbf{U}, ), \mathbf{U}\in T_{p}(M). What this statement is equivalent to is that we essentially have a function of a single vector of the tangent space at that point p on M. We know that one - forms are mappings of vectors to the reals so what we can say is g(\mathbf{U}, ) = \tilde{U}, \tilde{U}\in T^{*}_{p}(M) where the tilde denotes a one - form. This is essentially the gist of contracting with the metric tensor. So when you have g_{\alpha \beta }\Gamma ^{\beta }_{\gamma \delta } = \Gamma _{\alpha \gamma \delta } what you are doing is taking the vector component (which is the upper index) and summing it over a single index of the metric tensor which is equivalent to taking the metric as a mapping but with one missing argument. This will give you the corresponding one - form which is the lower index. The situation with the inverse metric is, of course, analogous to this. You can contract partial derivatives because remember that, for p\in M for the n - manifold M in question, the partial derivatives \partial _{\alpha } at p are the n directional derivatives at p and these \partial _{\alpha } are actually the basis vectors for T_{p}(M). As you know, the metric with a single vector argument will give you the corresponding one form so contracting the index of a partial derivative is fine.

 

[ZealScience]

So you mean that metric tensors are not functions of those basis. Then the ∂α and gαβ commute?

 

[ZealScience]

Remember that the metric tensor in and of itself is such that g_{p}:T_{p}(M) \times T_{p}(M) \mapsto \mathbb{R} for a p\in M. But let's say, instead of g(\mathbf{U}, \mathbf{V}), (\mathbf{U, V})\in T_{p}(M)\times T_{p}(M) we simply had g(\mathbf{U}, ), \mathbf{U}\in T_{p}(M). What this statement is equivalent to is that we essentially have a function of a single vector of the tangent space at that point p on M. We know that one - forms are mappings of vectors to the reals so what we can say is g(\mathbf{U}, ) = \tilde{U}, \tilde{U}\in T^{*}_{p}(M) where the tilde denotes a one - form. This is essentially the gist of contracting with the metric tensor. So when you have g_{\alpha \beta }\Gamma ^{\beta }_{\gamma \delta } = \Gamma _{\alpha \gamma \delta } what you are doing is taking the vector component (which is the upper index) and summing it over a single index of the metric tensor which is equivalent to taking the metric as a mapping but with one missing argument. This will give you the corresponding one - form which is the lower index. The situation with the inverse metric is, of course, analogous to this. You can contract partial derivatives because remember that, for p\in M for the n - manifold M in question, the partial derivatives \partial _{\alpha } at p are the n directional derivatives at p and these \partial _{\alpha } are actually the basis vectors for T_{p}(M). As you know, the metric with a single vector argument will give you the corresponding one form so contracting the index of a partial derivative is fine.

What I am looking for is whether the partial differentiation and metric tensor commute, and if yes, why do they commute?

 

[WannabeNewton]

No I'm saying the metric tensor is. You simply have the metric with an empty argument and the other argument as a basis vector of the tangent space in question (which here is the corresponding partial derivative) and, since we know that a one - form is a function of single vector, the metric with one basis vector and an empty argument will give you the corresponding basis one - form for the cotangent space to the tangent space in question. And yes, g^{\alpha \beta }\partial _{\alpha } = \partial _{\alpha }g^{\alpha \beta } = \partial ^{\beta } because we are just summing over \alpha.

 

[ZealScience]

No I'm saying the metric tensor is. You simply have the metric with an empty argument and the other argument as a basis vector of the tangent space in question (which here is the corresponding partial derivative) and, since we know that a one - form is a function of single vector, the metric with one basis vector and an empty argument will give you the corresponding basis one - form for the cotangent space to the tangent space in question. And yes, g^{\alpha \beta }\partial _{\alpha } = \partial _{\alpha }g^{\alpha \beta } = \partial ^{\beta } because we are just summing over \alpha.

Thanks. I think if they commute, we can just do it. Short fire way of doing that.

 

[atyy]

I haven't learned much of advanced mathematics. It seems that we can use metric tensors to lower or raise index of christoffel symbols. But isn't christoffel symbols made of metric tensors and derivatives of metric tensors? How can we contract indices of a derivative directly with metric tensors while they are non constants.

It's a definition.

http://www.mathematics.thetangentbundle.net/wiki/Differential_geometry/Christoffel_symbol

 

[ZealScience]

It's a definition.

http://www.mathematics.thetangentbundle.net/wiki/Differential_geometry/Christoffel_symbol

Thank you for replying. I know that lowering index of C.S. is the definition of transformation of C.S. of 1st kind and C.S. of the second. I am just wondering whether partial derivatives of metric tensor (which is involved in C.S.) commute with metric tensors (as they are not constant over space). I don't want to go through the whole bunch of calculations. Thus I posted it.

 

[WannabeNewton]

Thank you for replying. I know that lowering index of C.S. is the definition of transformation of C.S. of 1st kind and C.S. of the second. I am just wondering whether partial derivatives of metric tensor (which is involved in C.S.) commute with metric tensors (as they are not constant over space). I don't want to go through the whole bunch of calculations. Thus I posted it.

Oh I see what you are saying now. I shouldn't have said they commuted in the way I posted above. g^{\alpha \beta }\partial _{\alpha } is not the same as \partial _{\alpha }g^{\alpha \beta } when you treat the \partial _{\alpha } as an operator. But in the definition of the CS you have something like g^{\gamma \delta }\partial _{\gamma }g_{\alpha \beta } which is fine.

 

[ZealScience]

Oh I see what you are saying now. I shouldn't have said they commuted in the way I posted above. g^{\alpha \beta }\partial _{\alpha } is not the same as \partial _{\alpha }g^{\alpha \beta } when you treat the \partial _{\alpha } as an operator. But in the definition of the CS you have something like g^{\gamma \delta }\partial _{\gamma }g_{\alpha \beta } which is fine.

Thanks. But Christoffel Symbol of the1st kind only contains partial derivative of metric tensors. I think what you have posted is components in the christoffel symbol of the 2nd kind. And it's just the metric tensor in the front contract indices with the partial derivatives to form the C.S. with only three indices.

 

[atyy]

Thank you for replying. I know that lowering index of C.S. is the definition of transformation of C.S. of 1st kind and C.S. of the second. I am just wondering whether partial derivatives of metric tensor (which is involved in C.S.) commute with metric tensors (as they are not constant over space). I don't want to go through the whole bunch of calculations. Thus I posted it.

They don't commute.

Defining the connection to be metric-compatible and torsion free results in the CS of the 2nd kind being given by Eq 3.21 of http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html .

Then the CS of the 1st kind is just another definition given by Eq 1.39 of http://www.blau.itp.unibe.ch/lecturesGR.pdf .

(I think Carroll gets through his whole textbook without CS of the 1st kind?)

 

[WannabeNewton]

(I think Carroll gets through his whole textbook without CS of the 1st kind?)

Yeah he doesn't use them in the text itself and they really only come up in calculations that you do yourself because its easier to get the CS of the 2nd kind from the 1st kind and just contracting after that instead of using the definition of the CS of the 2nd kind in its entirety.

 

[ZealScience]

They don't commute.

Defining the connection to be metric-compatible and torsion free results in the CS of the 2nd kind being given by Eq 3.21 of http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html .

Then the CS of the 1st kind is just another definition given by Eq 1.39 of http://www.blau.itp.unibe.ch/lecturesGR.pdf .

(I think Carroll gets through his whole textbook without CS of the 1st kind?)

Thank you for these notes. I am reading Dirac's lecture notes which is a little bit too concise.

 

[atyy]

Yeah he doesn't use them in the text itself and they really only come up in calculations that you do yourself because its easier to get the CS of the 2nd kind from the 1st kind and just contracting after that instead of using the definition of the CS of the 2nd kind in its entirety.

Good, so we don't need to know them for the final?

Thank you for these notes. I am reading Dirac's lecture notes which is a little bit too concise.

! Only a little?

 

[WannabeNewton]

Good, so we don't need to know them for the final?

Beats me, if there isn't a wolfram app on the iPhone that has it then I would be screwed if it was on a final =p.

 

[atyy]

BTW, another presentation that may be a useful supplement to Dirac is http://casa.colorado.edu/~ajsh/phys5770_10/grbook.pdf , see Eq 2.54 - 2.55.