# How to get metric field using a path dependent parallel propagator

• I
• Jianbing_Shao
Jianbing_Shao
TL;DR Summary
metric compatible equation, parallel propagator, path dependence
If a vector ##V(x)## being transported down a path ##l##, The vector field is described with equation:

$$\partial_\mu V(x)=\Gamma_\mu V(x)$$

The solution of the equation can be described with parallel propagator ##P(x, x_0)##(in mathematics it is also called product integration):

$$V(x)=P(x, x_0)V(x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right)V(x_0)$$

If there exists a matrix function ##A(x)## which satisfies the equation:

$$\partial_\mu A(x)=\Gamma_\mu A(x)$$

Then##\Gamma_\mu## can be written as ##\partial_\mu A(x) A^{-1}(x)##. and the parallel propagator constructed with such a connection is:
$$P(x, x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right) =A(x)A^{-1}(x_0)$$

So the parallel propagator is path independent, the curvature of ##\Gamma_\mu## is zero.

$$R_{\mu\nu}=\partial_\mu \Gamma_\nu-\partial_\nu \Gamma_\mu+[\Gamma_\mu \ \Gamma_\nu]=0$$

For a matrix function the change of the matrix between two points is path independent. It is very natural get a zero curvature connection from it.

If we use parallel propagator to construct a metric field. From metric compatible equation:

$$∇_ρ g_{μν}=g_{μν,ρ} -Γ_{ρμ}^λ g_{λν}-Γ_{ρν}^λ g_{μλ}=0$$

If the metric ##g_{μν}## moves along a path ##l## between points ##x_0## and ##x##, then the metric can be described with parallel propagator:

$$g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$$

If ##Γ_\mu(x)## the connection compatible with metric ##g(x)## whose curvature is not zero. then parallel propagator ##P(x,x_0)## is path dependent. but because ##g(x)## is a field, So ##P(x,x_0)g(x_0)P^T(x,x_0)## must be path independent. The only possible parallel propagator should take such a form:

$$P(x,x_0)=A(x)P'(x,x_0)=A(x)\mathcal{P}\exp\int^x_{x_0}\Gamma'_\mu(x)dx^\mu$$

Here ##A(x)## is a matrix function. We demand the curvature of ##\Gamma'_\mu(x)## is not zero. ##R(\Gamma'_\mu)\neq 0## and:

$$P'(x,x_0)g(x_0)P'^T(x,x_0)=g(x_0)$$

Then we can get a path dependent parallel propagator and a metric field at the same time. The connection is:##\partial_\mu A(x) A^{-1}(x)+A(x)\Gamma’_\mu(x) A^{-1}(x)##

The metric is only determined by the matrix function ##A(x)##, ##g(x)= A(x) g(x_0)A^T(x)## and has nothing to do with the path dependent part ##P'(x,x_0)##. Then I have some quertions:

1. It seems that for any metric field there exist connections with zero curvature compatible with it. is it true?

2. Although there exist connections compatible with a metric field whose curvature is not 0. But in the definition of the metric field, the path dependent part ##P'(x,x_0)## was canceled. So we can get a metric field (path independent) at the same time a path dependent parallel propagator. Then how can we get a non-zero curvature connection purely from a metric field?

3. For a connection with zero curvature, it can be described with a matrix function, so we need ##n^2## parameters to described it. An arbitrary connection has ##n^3## parameters, and the zero curvature condition reduces the number of parameters from ##n^3## to ##n^2##. A Christoffel connection can be described with metric which only has ##\frac{1}{2}n(n+1)## parameters. Then what kind of conditions the Christoffel connection should satisfy then the parameters was reduced from ##n^3## to ##\frac{1}{2}n(n+1)##?

1. Yes, at least locally. Simply choose a basis of orthonormal vector fields and declare them to be parallel as the definition of your connection. This fails globally if you cannot find a full set of global orthonormal vector fields, such as for the sphere.

2. You can use a path independent propagator to construct a metric. There is no guarantee this metric will be any particular metric. There is no contradiction between a metric being path independent and a propagator being path dependent. Why would there be? Take the Levi-Civita connection as example.

3. The Levi-Civita connection is the unique torsion free metric compatible connection.

Orodruin said:
1. Yes, at least locally. Simply choose a basis of orthonormal vector fields and declare them to be parallel as the definition of your connection.
The matrix function ##A(x)## and the parallel propagator are all defined globlly. So I am confused how to relate an arbitrary matrix function ##A(x)## to a basis of orthonormal vector fields? Can the change of the matric function ##A(x)## between two neighbouring points can always be described by the relation of two orthonormal basis at the two points?

Orodruin said:
There is no contradiction between a metric being path independent and a propagator being path dependent. Why would there be?
For a differential equations likes $$\partial_\mu A(x)=\Gamma_\mu A(x)$$
We can draw a colclusion: if the curvature ##R(\Gamma_\mu)\neq 0##(it means the parallel propagator generated from ##\Gamma_\mu## is path dependent ), then we can not find a global matrix function ##A(x)## which can satisfy the equation.

Then to metric compatible equation, we can say for some connections with non-zero curvature, we can not find a metric field which can satisfy the equation. So at least not all path dependent movement of a vector can be explained using metric field? isn't it?

Orodruin said:
The Levi-Civita connection is the unique torsion free metric compatible connection.
Independent parameters of a Levi-Civita connection is not more than ##\frac{1}{2}n(n+1)##(it can not be more than the parameters of a metric field). Then how many independent parameters we need to describe a metric compatible connection? and how many independent parameters can be cancelled from torsion free condition? So at last can we find that the parameter of a Levi-Civita connection is ##\frac{1}{2}n(n+1)##?

Jianbing_Shao said:
$$P(x, x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right) =A(x)A^{-1}(x_0)$$
Can you post your proof that ##\mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)=A(x)A^{-1}(x_0)##?

renormalize said:
Can you post your proof that ##\mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)=A(x)A^{-1}(x_0)##?
The proof is trivial, suppose there exists a matrix function ##A(x)## fulfill the equation:
$$\partial_\mu A(x)=\Gamma_\mu(x) A(x)$$
Then ##A(x)## from a point ##x_0## to point ##x## along a particular path, then
$$A(x)=P(x, x_0)V(x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right)A(x_0)$$
At the same time, ##\Gamma_\mu## can be expressed using matrix function:
$$\Gamma_\mu(x)=\partial_\mu A(x)A^{-1}(x)$$
So we can get:
$$\mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)=A(x)A^{-1}(x_0)$$

Last edited:
Orodruin said:
You can use a path independent propagator to construct a metric. There is no guarantee this metric will be any particular metric. There is no contradiction between a metric being path independent and a propagator being path dependent. Why would there be? Take the Levi-Civita connection as example.
From the study on the differential equations l $$\partial_\mu A(x)=\Gamma_\mu A(x)$$
We can draw a colclusion: if the curvature ##R(\Gamma_\mu)\neq 0##, then we can not find a global matrix function ##A(x)## which can satisfy the equation. only when ##R(\Gamma_\mu)= 0## there exist global matrix function ##A(x)## which can satisfy the equation.

So from the metric compatible equations can we draw similar conclusion.s？

If the curvature of the connection in equation is zero, we always can find a global metrc field compatible the connection. and for all metric fields there exist zero curvature connections compatible with them.
If the curvature of the connections is not zero, then in most cases we can not find a global metric field compatible with the connection. For those have corresponding compatible metric, to get a metric field,
the path dependent part of the parallel propagator should be cancelled in the definition of the metric field?

Jianbing_Shao said:
If the curvature of the connections is not zero, then in most cases we can not find a global metric field compatible with the connection.
This appears to contradict basic GR, in which we have spacetime manifolds with connections with nonzero curvature that have global metric fields on them.

Jianbing_Shao said:
The proof is trivial, suppose there exists a matrix function ##A(x)## fulfill the equation:
$$\partial_\mu A(x)=\Gamma_\mu(x) A(x)$$
Then ##A(x)## from a point ##x_0## to point ##x## along a particular path, then
$$A(x)=P(x, x_0)V(x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right)A(x_0)$$
At the same time, ##\Gamma_\mu## can be expressed using matrix function:
$$\Gamma_\mu(x)=\partial_\mu A(x)A^{-1}(x)$$
So we can get:
$$\mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)=A(x)A^{-1}(x_0)$$
Since you haven't explicitly displayed the indices on your matrix, I can only guess that they refer to a coordinate basis, so that the matrix ##A(x)## is actually the (assumed invertible) rank-2 tensor field ##A_{\alpha\beta}(x).## In that case, your first two matrix equations are wrong; they should read:$$\partial_{\mu}A\left(x\right)=\Gamma_{\mu}\left(x\right)A\left(x\right)+A\left(x\right)\Gamma_{\mu}^{T}\left(x\right)$$$$A\left(x\right)=P\left(x,x_{0}\right)A\left(x_{0}\right)P^{T}\left(x,x_{0}\right)$$(This of course mirrors parallel-transport for the metric tensor ##g##, except for no assumption of symmetry for ##A##.) Using these corrected forms, can you now demonstrate how you write ##\Gamma## and ##P## in terms of ##A## and its first derivative?

PeterDonis said:
This appears to contradict basic GR, in which we have spacetime manifolds with connections with nonzero curvature that have global metric fields on them.
1. From the metric compatible equations, there exist non-zero connections compatible with a metric field.

2. To a particular metric field, there exist many connections compatible with it. especially there exist zero curvature connections compatible with it, and all metric fields have corresponding compatible zero curvature connections.

3.I only doubt that if we can get a non zero curvature connection from a metric field, Just like in calculas, we can get a zero curl vector field from a scalar field(a field just means the scalar moves path independently), The path integral of a vector field with non-zero curl is path dependent.
So from a matrix function ##A(x)##, we can get a zero curvature connection, and parallel propagator is path independent, only the curvature is not zero, then the parallel propagator generated with the connection is path dependent.

So if we claim that we can get a non-zero curvature connection, then we should using parallel propagator to show that if we can recover the metric field and why the path dependence disppeared in a metric field.

Jianbing_Shao said:
... all metric fields have corresponding compatible zero curvature connections.
You certainly haven't demonstrated this as far as I can tell. Please address my question in post #8.

Jianbing_Shao said:
I only doubt that if we can get a non zero curvature connection from a metric field
And, as I have already said, any curved spacetime in GR will give you a nonzero curvature connection from a metric field. So I cannot understand why you doubt that such a thing exists.

@Jianbing_Shao you keep repeating the same statements without responding to questions that I and others are asking. If you cannot respond to questions, this thread will be closed as it is pointless for you to keep repeating yourself.

renormalize said:
Since you haven't explicitly displayed the indices on your matrix, I can only guess that they refer to a coordinate basis,
At first, I follow the way in the discussion about product integration, so I treat##\Gamma_\mu(x)## as a matrix function. Actually the indice ##\mu## refer to a coordinate basis. In fact I try to avoid to write the indices. the reason is:
if we define metric field with a basis ##g_{\mu\nu}=e_\mu \cdot e_\nu##, and ##e_\mu## fulfill the relation:
$$\partial_\mu e_\nu=\Gamma_{\mu\nu}^\rho e_\rho$$
Here Partial derivative operator ##\partial_\mu## is determined by the basis ##e_\mu##, As an operator, at each point the meaning of ##\partial_\mu## is Immutable, this needs the basis ##e_\mu## should be a field in the space. so we can ralate the basis at different point using parallel propagator generated with ##\Gamma_{\mu\nu}##.
$$e(x)=\mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right) e(x_0)$$
If we demand that the basis ##e_\mu## should be a field, then parallel propagator should path independent, then the curvature of the connection should be zero. if the curvature is not zero, then how can we define derivative operator ##\partial_\mu##.
In fact even the basis ##e_\mu## is a field is not enough. if the basis field is a non-coordinate basis, then the Infinitesimal parallelogram we need to define curvature is not closed. then the definition of curvature need some amendments.
renormalize said:
so that the matrix A(x) is actually the (assumed invertible) rank-2 tensor field Aαβ(x). In that case, your first two matrix equations are wrong; they should read:
If we only treat the matrix ##A(x)## as a matrix. then the equation $$\partial_\mu A(x)=\Gamma_\mu A(x)$$ just the derivative of the matrix is codetermined by ##\Gamma_\mu## and matrix ##A(x)## itself. This is a defferential equations.
Of course you can treat it as a tensor, just like a metric field. So the differential equations will different from this one.
renormalize said:
∂μA(x)=Γμ(x)A(x)+A(x)ΓμT(x)A(x)=P(x,x0)A(x0)PT(x,x0)(This of course mirrors parallel-transport for the metric tensor g, except for no assumption of symmetry for A.) Using these corrected forms, can you now demonstrate how you write Γ and P in terms of A and its first derivative?
If we try to write Γ and P in terms of A and its first derivative, then first we must admit that because ##A(x)## is a field, then this just means that parallel propagator ##P(x,x_0)## should be path independent( we neglect the symmetry of ##A##). then this put a ristrict on the connections, because the connection with zero curvature always can be expressed with a matrix function, so zero curvature condition will reduce the number of independent parameters of a connection from ##n^3## to ##n^2##, then the conditon of ##\Gamma_{\mu\nu}=\Gamma_{\nu\mu}## will reduce the parameter to ##\frac{1}{2}n(n+1)##, it is the same as the number of a metric field, then we can use the method of linear equations to get the result just as we do in general ralativity.

Now there is a problem if ##P(x,x_0)## is path dependent, then ##A(x)## is not a field(we have neglect the symmetry of ##A##) then can we express connection field ##\Gamma## using a path dependent ##A(x)##?

Jianbing_Shao said:
If we demand that the basis ##e_\mu## should be a field, then parallel propagator should path independent, then the curvature of the connection should be zero.
Wrong. Again, every curved spacetime in GR can be given a coordinate basis that defines a set of 4 vector fields on the spacetime, and still have parallel transport be path dependent and the connection have nonzero curvature.

Jianbing_Shao said:
if the curvature is not zero, then how can we define derivative operator ##\partial_\mu##.
The same way you always did; but now that derivative operator, by itself, does not define parallel transport. Parallel transport is defined by the covariant derivative ##\nabla_\mu##.

Jianbing_Shao said:
In fact even the basis ##e_\mu## is a field is not enough. if the basis field is a non-coordinate basis, then the Infinitesimal parallelogram we need to define curvature is not closed. then the definition of curvature need some amendments.
Not "amendments", just recognition of how curvature is properly defined and interpreted. Most GR textbooks discuss this. See, for example, Chapter 11 of Misner, Thorne & Wheeler, and in particular Fig. 11.2, showing the extra fifth segment to "close the quadrilateral", and the accompanying discussion.

PeterDonis said:
And, as I have already said, any curved spacetime in GR will give you a nonzero curvature connection from a metric field. So I cannot understand why you doubt that such a thing exists.
What I am discussing is just the property of differential equations like parallel transport equations whose solution can be expressed with parallel propagator. so there exist problems you can point out.
renormalize said:
You certainly haven't demonstrated this as far as I can tell. Please address my question in post #8.
Of course I have demonstrated it in my question. I can repeat it:
If we define a connection ##\Gamma_\mu=\partial_\mu A(x) A^{-1}(x)##, the parallel propagator defined with the connection is path independent. so the curvature of the connenction if zero. and if we define the metric ##g(x)=A(x)A^T(x)##, then the two will satisfy the metric compatible equation.

So I only claim that we can get a metric field(a field just means path independence) using a path independent parallel propagator(generated with a zero-curvature connection).

If this is not true, it seems that we can only get a metric field with a path dependent parallel propagator? it is impossible to define a metric with a path independent parallel propagator?

Then from the metric compatible equation. if given a metric field can you point out all possible connections compatible to the metric field and point out that their connections are impossible to be zero?

Last edited:
PeterDonis said:
Wrong. Again, every curved spacetime in GR can be given a coordinate basis that defines a set of 4 vector fields on the spacetime, and still have parallel transport be path dependent and the connection have nonzero curvature.
Do you mean that the metric ##g(x)## is define using a coordinate basis ##e_\mu(x)##, and the connection defined $$\partial_\mu e_\mu(x)=\Gamma_{\mu\nu}^rho e_\rho$$ should be a connection with non-zero curvature?
Then the solution of the of the euqation(if we choose a particular path) is:
$$e(x)=\mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right) e(x_0)=P(x,x_0)e(x_0)$$
Then if we demand that ##e(x)## to be a field then it is path imdependent, then parallel propagator ##P(x,x_0)## should be path dependent, then do the two demands contradict with each other?

Jianbing_Shao said:
What I am discussing is just the property of differential equations
What you are discussing, as I have already pointed out several times now, contradicts basic GR. So you are making a mistake somewhere. You have not addressed this point at all.

Jianbing_Shao said:
Do you mean that the metric ##g(x)## is define using a coordinate basis ##e_\mu(x)##, and the connection defined $$\partial_\mu e_\mu(x)=\Gamma_{\mu\nu}^rho e_\rho$$ should be a connection with non-zero curvature?
Have you read any basic GR textbooks? Any of them will tell you how to (a) write down a metric in the form of a line element given a choice of coordinates, (b) compute the connection coefficients in terms of the metric and its derivatives, and (c) compute the Riemann curvature tensor from these. In fact you don't even need to do most of the work by hand now since computer programs like Maxima will do much of it for you.

The same textbooks will show you multiple examples of metrics with nonzero curvature.

I do not know where you are getting the equation ##\partial_\mu e_\mu(x)=\Gamma_{\mu\nu}^rho e_\rho ## for the connection, but it does not look like what I see in GR textbooks.

I suspect that you are confusing yourself by using bad notation, but I can't make enough sense of your posts to be sure. I strongly advise you to check some GR textbooks.

Jianbing_Shao said:
and the connection defined $$\partial_\mu e_\mu(x)=\Gamma_{\mu\nu}^rho e_\rho$$
This is nonsensical. Apart from the index mismatch and missing \, you cannot apply a regular partial differential operator to a vector field. It is simply not defined. What is defined is the connection
$$\nabla_\mu e_\nu = \Gamma^\rho_{\mu\nu} e_\rho$$

The rest of the post therefore does not follow.

Last edited:
Orodruin said:
This is nonsensical. Apart from the index mismatch and missing \, you cannot apply a regular partial differential operator to a vector field. It is simply not defined. What is defined is the connection
$$\nabla_\mu e_\nu = \Gamma^\rho_{\mu\nu} e_\rho$$

The rest of the post therefore does not follow.
From my expression, we can use parallel propagator to describe the relation between the basis field ant the connection.
Then from you foumula, can you also point out the relation between basis field and connection in a similar way?

Jianbing_Shao said:
From my expression, we can use parallel propagator to describe the relation between the basis field ant the connection.
You really cannot because, as mentioned, it is nonsensical.

Jianbing_Shao said:
Then from you foumula, can you also point out the relation between basis field and connection in a similar way?
The connection coefficients, expressed in whatever basis, define the connection, not the basis. Generally, the basis is not parallel because ##\nabla_\mu e_\nu## is not generally zero.

PeterDonis said:
What you are discussing, as I have already pointed out several times now, contradicts basic GR. So you are making a mistake somewhere. You have not addressed this point at all.

Have you read any basic GR textbooks? Any of them will tell you how to (a) write down a metric in the form of a line element given a choice of coordinates, (b) compute the connection coefficients in terms of the metric and its derivatives, and (c) compute the Riemann curvature tensor from these. In fact you don't even need to do most of the work by hand now since computer programs like Maxima will do much of it for you.

The same textbooks will show you multiple examples of metrics with nonzero curvature.

I do not know where you are getting the equation ##\partial_\mu e_\mu(x)=\Gamma_{\mu\nu}^rho e_\rho ## for the connection, but it does not look like what I see in GR textbooks.

I suspect that you are confusing yourself by using bad notation, but I can't make enough sense of your posts to be sure. I strongly advise you to check some GR textbooks.
I only use the method of parallel propagator to investigate the metric compatible equation. if you can point out that it is wrong to use it on metric compatible equation and tell me why, or simply you point out the mistakes I made in my analysis, then I'd like to accept your advice.

You can find the formula in John Dirk Walecka' book "Introduction to general relativity' P39,3.64, I found the formula there. perhaps it is not a popular textbook. but I do not think it is wrong.

I don't like play with comcept. So I sinserely ask a quertion. From the metric compatible equation, if given a connection field, then can we express the metric using parallel propagator generated from the connection field.

Last edited:
Orodruin said:
You really cannot because, as mentioned, it is nonsensical.
In John Dirk's book He start from two formula:$$g_{ij}=e_i \cdot e_j$$ and$$\partial_i e_j= \Gamma_{ij}^k e_k$$
Then get metric compatible equation, then if the formula ##\partial_i e_j= \Gamma_{ij}^k e_k## is nonsensical, then also the metric compatible equation?
Orodruin said:
The connection coefficients, expressed in whatever basis, define the connection, not the basis. Generally, the basis is not parallel because ∇μeν is not generally zero.
Then return the discussion about metric compatible equation. how the change of basis affect the parallel propagator generated with connection coefficients? then affect the metric compatible with the connection?

Last edited:
Jianbing_Shao said:
You can find the formula in John Dirk Walecka' book "Introduction to general relativity' P39,3.64
Equation 3.64 in that textbook is not what you wrote. It is (if we substitute Equation 3.65):

$$d \mathbf{e}_i = \Gamma^k_{ij} \mathbf{e}_k dq^i$$

PeterDonis said:
Equation 3.64 in that textbook is not what you wrote. It is (if we substitute Equation 3.65):

$$d \mathbf{e}_i = \Gamma^k_{ij} \mathbf{e}_k dq^i$$
Then the same question: if the basis moves along a path in space, then can it be described with parallel propagator?
$$e_i(x)=\left[P(x, x_0)\right]^j_ie_j(x_0) = \left[\mathcal{P} \exp\left(\int_{x_0}^x \Gamma_k(x){\rm dx^k }\right)\right]^j_ie_j(x_0)$$
If it is, then I think the difference is trivial.
Because the connection field has ##n^3## components, but a metric field only has ##\frac{1}{2}n(n+1)##. The difference is too big. Can I say the metric compatible equation can generate a map from the set of all metric field to set of connection fields. The calculation of Levi-civita connection is only a one to one map. so the set of all Levi-civita connections is only a subset of the set of all connections. Even regardless of the torsion free condition. all the connections we can get from a metric field is a subset of all connection fields set.

Those connections has no corresponding compatible metric fields also can generate a path dependent
parallel propagator, then what is the difference between the two types of path dependence? If we can not tell the difference, then when we encounter a path dependent movement of a vector, then we can not assert it is am effect of a metric field.

Last edited:
weirdoguy
Jianbing_Shao said:
if the basis moves along a path in space, then can it be described with parallel propagator?
Have you even read the reference you gave? Do you understand what it is describing in the text around the equation you referenced?

I ask because I am extremely confused as to why you would even be asking the question quoted above if you had read the book. The book is telling you that the connection coefficients ##\Gamma^k_{ij}## describe how the basis vectors change as you move from coordinates ##q^i## to coordinates ##q^i + dq^i##. So you are thinking of it backwards: it's not that parallel transport defines how the basis vectors change, it's that how the basis vectors change defines parallel transport. Once you know the connection, i.e., how the basis vectors change, you know the covariant derivative operator ##\nabla_\mu##, and that tells you what parallel transport is.

Jianbing_Shao said:
$$e_i(x)=\left[P(x, x_0)\right]^j_ie_j(x_0) = \left[\mathcal{P} \exp\left(\int_{x_0}^x \Gamma_k(x){\rm dx^k }\right)\right]^j_ie_j(x_0)$$
You keep conjuring up these equations out of thin air. You really should be looking at what the book you referenced actually says instead.

Jianbing_Shao said:
At this point? Quite frankly, no. At least not in this thread, because I am now closing it. Take some time and actually read the reference you gave, and then if you have another question, you can ask it in a new thread. But I strongly suggest that you read your reference carefully before asking a question, because, as I said above, it really seems like you are not doing so, which means the questions you are asking are just wasting other people's time pointing out to you things that are already there in the book you reference, which you could have read there for yourself.

weirdoguy

• Special and General Relativity
Replies
62
Views
4K
• Special and General Relativity
Replies
35
Views
3K
• Special and General Relativity
Replies
5
Views
1K
• Special and General Relativity
Replies
9
Views
1K
• Special and General Relativity
Replies
5
Views
216
• Special and General Relativity
Replies
4
Views
168
• Special and General Relativity
Replies
28
Views
2K
• Special and General Relativity
Replies
3
Views
2K
• Special and General Relativity
Replies
1
Views
1K
• Special and General Relativity
Replies
1
Views
157