Metric defined with a non-coordinate basis

[Jianbing_Shao]

TL;DR Summary

For a non-coordinate basis, because there is no corresponding coordinate system, then the definition of partial derivative and a closed path is in question.

We always can define a metric with a basis field $g_{\mu\nu}=e_\mu \cdot e_\nu$, For a basis field $e_\mu$, it can belong to a coordinate basis, then there is a corresponding coordinate system$\{x^\mu\}$,then we can have $e_\mu=\frac{\partial}{\partial x^\mu}$, and $[e_\mu , e_\nu]=0$ ,but for a non-coordinate system, there is no corresponding coordinate system. then we will find that:

(1) If the metric is defined with a non-coordinate system, then there is no cooresponding coordinate system, then it seems that we will have trouble defining the partial derivative $\partial_\mu$:

(2) To define curvature we should parallel a vector along a closed path in the space. In a coordinate system, it is very easy to describe a closed path. For example we can define a parallelogram which contains four infinitesimal segments: $\epsilon e_a , \epsilon e_b , -\epsilon e_a , -\epsilon e_b$. But in a space only equipped with a non-coordinate basis, because $[ e_\mu ,e_\nu]\neq 0$, so the path $\epsilon e_\mu , \epsilon e_\nu , -\epsilon e_\mu , -\epsilon e_\nu$ is not a closed path, so how to define a closed path?

 

[andrewkirk]

We can always find a local coordinate basis, and that is sufficient to define curvature. Any Riemannian manifold can be covered by an atlas of open neighbourhoods, each of which is homeomorphic to Euclidean space and hence can have a local coordinate basis derived from that homeomorphism. To measure curvature at any point, we choose such an open neighbourhood containing it, choose a homeomorphism from that neighbourhood to Euclidean space, then choose a parallelogram that follows the lines of the coordinate system derived from that homeomorphism, and is small enough to lie entirely within the coordinate patch.

However, coordinates are not needed to define the Riemann curvature tensor, and its formal definition does not use coordinates, eg see here. All you need are two vectors at a and b to define the directions of your parallelogram, and the covariant derivative (Christoffel symbols) to define parallel transport. You parallel transport those vectors around the parallelogram so you always know which direction to head after each vertex. You follow geodesics to get the sides of the parallelogram, and you alternate the directions at each vertex as a, b, a, b. As you point out, you may not end up back where you started if all sides are the same length. To fix that, adjust the length of the third side so that the fourth side runs through the starting point. That should always be possible because the manifold is smooth.

 

[Jianbing_Shao]

We can always find a local coordinate basis, and that is sufficient to define curvature. Any Riemannian manifold can be covered by an atlas of open neighbourhoods, each of which is homeomorphic to Euclidean space and hence can have a local coordinate basis derived from that homeomorphism. To measure curvature at any point, we choose such an open neighbourhood containing it, choose a homeomorphism from that neighbourhood to Euclidean space, then choose a parallelogram that follows the lines of the coordinate system derived from that homeomorphism, and is small enough to lie entirely within the coordinate patch.

The difference between a coordinate basis and non-coordinate basis is just if the commutator $[ e_\mu(x), e_\nu(x)]$ is zero.it is not determined by the basis at point $x$,$e_\mu(x)$,but it’s derivative. So you can define a basis at a point and regard it as a coordinate basis, but it can not describe the properties of a non-coordinate basis.
The partial derivative $e_\mu=\frac{\partial}{\partial x^\mu}$ corresponding to a coordinate system is a global operator, Of course I am not doubt that locally I can find a coordinate system to define a locall partial derivative operators, but I doubt the exisence of a global operator $e_\mu=\frac{\partial}{\partial x^\mu}$ if we only have a non-coordinate basis. of course, with the help of a coordinate coordinate basis $e_a$, we can define a global operator $e_\mu=e_\mu^a \frac{\partial}{\partial x^a}$, but $e_\mu^a$ can not be expressed with a coordinate transformation.

However, coordinates are not needed to define the Riemann curvature tensor, and its formal definition does not use coordinates, eg see here. All you need are two vectors a and b to define the directions of your parallelogram, and the covariant derivative (Christoffel symbols) to define parallel transport. You parallel transport those vectors around the parallelogram so you always know which direction to head after each vertex. You follow geodesics to get the sides of the parallelogram, and you alternate the directions at each vertex as a, b, a, b.

Here if only with a non-coordinate basis, the parallelogram which contains four infinitesimal segments: $\epsilon e_\mu , \epsilon e_\nu , -\epsilon e_\mu , -\epsilon e_\nu$ is just not a closed path. It is not a closed path even locally, It is not the same as the situations in the coordinate system, So if a space equipped with a metric defined with a non-coordinate basis. Because we can not have a global coordinate system, so it is not easy to define a closed path, and if you choose a local coordinate system in stead of the non-coordinate basis system, How can you be sure that the non-commutative of basis vector do not affect the calculation of the curvature. In fact if the path we choose is not closed, we obviously will have a non-zero curvature, but in fact it is not the real curvature we understand.

 

[PeterDonis]

you can define a basis at a point and regard it as a coordinate basis, but it can not describe the properties of a non-coordinate basis

This is nonsense. You can describe the properties of any vectors and their derivatives, including those that make up a non-coordinate basis, using a valid coordinate chart and its accompanying coordinate basis. If this were not possible, doing GR using standard differential geometry as it is taught in most of the major textbooks on the subject would be impossible.

The partial derivative corresponding to a coordinate system is a global operator

No, it isn't, it's a local operator in the tangent space at a point.

I think you need to learn some basic differential geometry.

 

[Jianbing_Shao]

This is nonsense. You can describe the properties of any vectors and their derivatives, including those that make up a non-coordinate basis, using a valid coordinate chart and its accompanying coordinate basis. If this were not possible, doing GR using standard differential geometry as it is taught in most of the major textbooks on the subject would be impossible.

Now in a space defined with a basis field, then at each point there is a basis, it can be regarded as a basis of a local coordinate system, then the transformation group relate two basis at two neighboring points can be seemed as the transformation group between two local coordinate systems. So the language of basis field does not contradict with the language of chart in general relativity. I think they are equivalent.

For a coordinate basis field it can be transformed to an orthonormal basis using a coordinate transformation, but a non-coordinate basis can’t, it can be transformed to an orthonormal basis field by a frame transformation but can't by a coordinate transformation. Then what is the difference?

No, it isn't, it's a local operator in the tangent space at a point.

I think you need to learn some basic differential geometry.

If I can say the orthonormal basis of a Cartesian coordinate system is a global operator, then a global coordinate transformation can change the orthonormal basis field to a coordinate basis field, then perhaps I can say a coordinate basis field is a global operator. If I understand the meaning of global correctly.

 

[PeterDonis]

in a space defined with a basis field

There is no such thing. Spacetime as a manifold with its geometric properties, including curvature, exists independently of any choice of coordinates or any choice of basis. The basis does not define the spacetime and is not necessary to define the spacetime.

at each point there is a basis

At each point there are an infinite number of possible sets of basis vectors, none of which are privileged over any other.

the language of basis field does not contradict with the language of chart in general relativity

I never said it did. You said a coordinate chart cannot describe the properties of a non-coordinate basis, which is false.

For a coordinate basis field it can be transformed to an orthonormal basis using a coordinate transformation

You don't transform a basis into another basis. A basis is just a set of 4 linearly independent vectors. A coordinate basis is such a set with a particular property, that all of the commutators are zero. An orthonormal basis is such a set with a different property, that the vectors are all mutually orthogonal and of unit norm. Talk of "transforming" one into the other is meaningless; they're just two different sets of 4 linearly independent vectors.

a non-coordinate basis can’t, it can be transformed to an orthonormal basis field by a frame transformation but can't by a coordinate transformation

This is nonsense. See above.

If I can say the orthonormal basis of a Cartesian coordinate system is a global operator

You can't. It's wrong.

Again, I think you need to learn some basic differential geometry. You appear to have a number of fundamental misconceptions.

 

[Jianbing_Shao]

There is no such thing. Spacetime as a manifold with its geometric properties, including curvature, exists independently of any choice of coordinates or any choice of basis. The basis does not define the spacetime and is not necessary to define the spacetime.

Do you think I can't even define a vector field in a space? if I can, why can't I define an object contains N vector fields?

 

[PeterDonis]

Do you think I can't even define a vector field in a space?

Of course you can. But that is not "a space defined with a basis field". The vector field is not required to define the space. You have to define the space first, before you can even figure out what vector fields exist in the space.

 

[Jianbing_Shao]

Of course you can. But that is not "a space defined with a basis field". The vector field is not required to define the space. You have to define the space first, before you can even figure out what vector fields exist in the space.

I wonder something needed to define the space first is independent of all the comecptions such as coordinate system, basis field?

So I can, but call what doesn't matter, but with such objects, we obviously can define a metric. and can expand a vector using this as a base. or we also can get a connection with such objects.

If I can't say "a space defined with a basis field", then can I say "a space can be equivallently described with a basis field" , Of course the choose of basis field is not unique.

For example: I can define a coordinate basis field in a Cartesian coordinate system. and the coordinate system can be get under a global coordinate transformation, then I am sure the coordinate basis must relate to a curvilinear coordinate system. and using such a basis field we can define a metric. we also can define the corresponding connection, and I am sure that the metric can be changed to zero and the connection can be changed to zero under coordinate transformation.

 

[PeterDonis]

I wonder something needed to define the space first is independent of all the comecptions such as coordinate system, basis field?

Yes. The definition of the space with its geometry is independent of those things, as I've already said.

with such objects, we obviously can define a metric

No, you can't. You have to have the space and its metric first, or you don't know if a set of vectors are linearly independent or orthogonal or have unit norm.

can I say "a space can be equivallently described with a basis field"

No.

I can define a coordinate basis field in a Cartesian coordinate system

A Cartesian coordinate system, and a space that admits one, have many special properties that will mislead you if you try to use this as your example case. The most important such property is that any space which admits a Cartesian coordinate chart at all must be flat, i.e., zero curvature. Pretty much all of the intuitive properties of Cartesian coordinates that you are relying on depend on the space being flat; in a curved space they no longer hold.

You really, really need to learn some basic differential geometry. You keep on asking questions based on obvious misconceptions. It would save time if you would learn the basics and correct your misconceptions all at once, instead of dragging the process out question after question.

 

[Jianbing_Shao]

Yes. The definition of the space with its geometry is independent of those things, as I've already said.
No, you can't. You have to have the space and its metric first, or you don't know if a set of vectors are linearly independent or orthogonal or have unit norm.
No.
A Cartesian coordinate system, and a space that admits one, have many special properties that will mislead you if you try to use this as your example case. The most important such property is that any space which admits a Cartesian coordinate chart at all must be flat, i.e., zero curvature. Pretty much all of the intuitive properties of Cartesian coordinates that you are relying on depend on the space being flat; in a curved space they no longer hold.

You really, really need to learn some basic differential geometry. You keep on asking questions based on obvious misconceptions. It would save time if you would learn the basics and correct your misconceptions all at once, instead of dragging the process out question after question.

Then ,do you mean that the definition of coordinate basis and non- coordinate basis is wrong?

 

[PeterDonis]

do you mean that the definition of coordinate basis and non- coordinate basis is wrong?

Read post #6.

 

[pervect]

Physicists generally define a coordinate system by giving a metric, $g_{uv}$. See for instance Misner, "Precis of General Realtivity", https://arxiv.org/abs/gr-qc/9508043. This coordinate system only needs to be local, not global. One may need several coordinate systems to cover a specific geometry, which exists independently of any coordinates, as others have mentioned.

A basis field, such as an orthonormal basis, is typically described by physicisits in terms of coordinate basis vectors, associated with some particular coordinate choic, but that's just a human choice, there's no physical significance to a particular coordinate choice.

This is a bit terse, I have to run.

Some more mathematical types have raised eyebrows at the idea that the metric defines a coordinate system, so I can't say for sure what mathematicians think. But I'll point to Misner as an example of a physicists saying that a metric defines a local coordinate system. I say local, because the very definition of a manifold admits for the need for multiple charts to cover a particular geometry. Again, the geometry exists independent of the coordinate choice.

 

[Jianbing_Shao]

A Cartesian coordinate system, and a space that admits one, have many special properties that will mislead you if you try to use this as your example case. The most important such property is that any space which admits a Cartesian coordinate chart at all must be flat, i.e., zero curvature. Pretty much all of the intuitive properties of Cartesian coordinates that you are relying on depend on the space being flat; in a curved space they no longer hold.

You really, really need to learn some basic differential geometry. You keep on asking questions based on obvious misconceptions. It would save time if you would learn the basics and correct your misconceptions all at once, instead of dragging the process out question after question.

Of course a Cartesian coordinate system is a flat, But does any principle forbid us to define an object contains N vector fields in the Cartesian coordinate system? and can these vector field changes under coordinate transformation?

Read post #6.

I know what do you mean, In my opinion, the coordinate of the space is only the label of the point in the space, In such a space, if we define a orthonormal basis, then the length of between point $x$ and $x+dx$ is $\delta_{\mu\nu}dx^\mu dx^\nu$, and if we define a different basis(corresponding metric is $g_{\mu\nu}$), then the length will be $g_{\mu\nu} dx^\mu dx^\nu$, this is a different space,

In your opinion, you firstly demand that the length between two points is invariant, then the metric and the coordinate change accordingly under coordinate transformation, so the length keeps invariant, if you want to get a new space, you need change the length between the two points, then need a new set of metrics and corresponding coordinate.

It seems the difference between us is that I think we can define coordinate system and basis(metric) independently, and once we define them, they will change accordingly under coordinate transformation and keep the length between two points is invariant. But you seem to think that length between two points is firstly determined by the space, so the metric and the coordinate system is firstly determined, the invariance of length under coordinate transformation demand the metric and coordinate system changes accordingly under coordinate transformation.

In fact, I really can’t find https://www.physicsforums.com/javascript%3A; between the two.

 

[Jianbing_Shao]

Physicists generally define a coordinate system by giving a metric, $g_{uv}$. See for instance Misner, "Precis of General Realtivity", https://arxiv.org/abs/gr-qc/9508043. This coordinate system only needs to be local, not global. One may need several coordinate systems to cover a specific geometry, which exists independently of any coordinates, as others have mentioned.

A basis field, such as an orthonormal basis, is typically described by physicisits in terms of coordinate basis vectors, associated with some particular coordinate choic, but that's just a human choice, there's no physical significance to a particular coordinate choice.

This is a bit terse, I have to run.

Some more mathematical types have raised eyebrows at the idea that the metric defines a coordinate system, so I can't say for sure what mathematicians think. But I'll point to Misner as an example of a physicists saying that a metric defines a local coordinate system. I say local, because the very definition of a manifold admits for the need for multiple charts to cover a particular geometry. Again, the geometry exists independent of the coordinate choice.

Thanks :
I have a question, if a theory is coordinate independent , then does it means that in fact we can choose any arbitrary coordinate system in our calculation, and the effection of choose of a partilular coordinate will disspear in the process of our calculation?
So for a specific problem, we always can choose a particular coordinate system for convenience?

 

[PeterDonis]

Of course a Cartesian coordinate system is a flat

No, the coordinate system is not what's flat. The underlying geometry is flat. That geometry can be described by an infinite number of different coordinate systems, one of which is Cartesian.

does any principle forbid us to define an object contains N vector fields in the Cartesian coordinate system?

Of course not.

can these vector field changes under coordinate transformation?

Vector fields themselves never change under coordinate transformation. They are coordinate independent geometric objects.

In my opinion, the coordinate of the space is only the label of the point in the space

This is correct: coordinate charts are simply one-to-one mappings between points in the space and n-tuples of real numbers, with some requirements regarding continuity, differentiability, etc.

In such a space, if we define a orthonormal basis, then the length of between point $x$ and $x + dx$ is $\delta_{\mu \nu} dx^\mu dx^\nu$

More precisely, the length of the line element $ds^2$ between the two points will be $ds^2 = \delta_{\mu \nu} dx^\mu dx^\nu$. Assuming Cartesian coordinates (and a Riemannian space, not a pseudo-Riemannian spacetime), so the metric is $\delta_{\mu \nu}$ when expressed in those coordinates.

However, the key property that makes the basis you've defined a coordinate basis--which is an unstated assumption you've made, that the basis in which the metric is $\delta_{\mu \nu}$ is a coordinate basis--is not that the basis is orthonormal, but that the basis vectors are all mutually commuting. It just so happens that the coordinate basis vectors in Cartesian coordinates (which are only possible in a flat space, as I've said) have both properties. But that is by no means true of the coordinate basis in all coordinates. And for a non-coordinate basis, you cannot express the metric in the form you give. There are other ways to express it, but not that way.

and if we define a different basis (corresponding metric is $g_{\mu \nu}$), then the length will be $g_{\mu \nu} dx^\mu dx^\nu$,

More precisely, if we choose a different coordinate chart, such that the metric when expressed in this chart is $g_{\mu \nu}$, then if we have the same points as above, but now labeled with new transformed coordinates $x'$ and $x' + dx'$, then we can express the same line element length $ds^2$ as above in the new coordinates as $ds^2 = g_{\mu \nu} dx'^\mu dx'^\nu$.

this is a different space

No, it is not. As I emphasized above, it is the same pair of points and the same line element between them, in the same underlying geometry, in both cases. All that has changed is the choice of coordinates and the corresponding expression for the metric in those coordinates.

In your opinion, you firstly demand that the length between two points is invariant

That's not a "demand" I'm making. It's a basic fact about the geometry. Anyone who starts an "I" level thread on this topic should already know that.

 

[PeterDonis]

if a theory is coordinate independent , then does it means that in fact we can choose any arbitrary coordinate system in our calculation, and the effection of choose of a partilular coordinate will disspear in the process of our calculation?
So for a specific problem, we always can choose a particular coordinate system for convenience?

Yes to both questions. Anyone who starts an "I" level thread on this topic should already know that.

 

[PeterDonis]

you seem to think that length between two points is firstly determined by the space

Of course it is. That's part of what being a "geometry" means.

so the metric and the coordinate system is firstly determined

No, the geometry is determined. But how that geometry is described in coordinate form depends on the choice of coordinates. Different choices of coordinates will lead to different expressions for the metric. But invariants like the distance between two points will be the same no matter what coordinates you calculate them in.

 

[Jianbing_Shao]

No, the coordinate system is not what's flat. The underlying geometry is flat. That geometry can be described by an infinite number of different coordinate systems, one of which is Cartesian.
Of course not.
Vector fields themselves never change under coordinate transformation. They are coordinate independent geometric objects.
This is correct: coordinate charts are simply one-to-one mappings between points in the space and n-tuples of real numbers, with some requirements regarding continuity, differentiability, etc.
More precisely, the length of the line element $ds^2$ between the two points will be $ds^2 = \delta_{\mu \nu} dx^\mu dx^\nu$. Assuming Cartesian coordinates (and a Riemannian space, not a pseudo-Riemannian spacetime), so the metric is $\delta_{\mu \nu}$ when expressed in those coordinates.

However, the key property that makes the basis you've defined a coordinate basis--which is an unstated assumption you've made, that the basis in which the metric is $\delta_{\mu \nu}$ is a coordinate basis--is not that the basis is orthonormal, but that the basis vectors are all mutually commuting. It just so happens that the coordinate basis vectors in Cartesian coordinates (which are only possible in a flat space, as I've said) have both properties. But that is by no means true of the coordinate basis in all coordinates. And for a non-coordinate basis, you cannot express the metric in the form you give. There are other ways to express it, but not that way.
More precisely, if we choose a different coordinate chart, such that the metric when expressed in this chart is $g_{\mu \nu}$, then if we have the same points as above, but now labeled with new transformed coordinates $x'$ and $x' + dx'$, then we can express the same line element length $ds^2$ as above in the new coordinates as $ds^2 = g_{\mu \nu} dx'^\mu dx'^\nu$.

I see, the difference between us is just I think that the coordinate and the basis field can change independently, So what I talking about is not confined to a particular space. but you think in a particluar space they should change accordingly, so we can keep the length of line element invariant,

Then the problem is very simple, if we assume that the two can change independently, then if the coordinate does not change, but the basis(metric) changes independently a little, then the length of line element will different from the original one, so the geometry of the space changes, and if we change the basis, and at the same time we demand the coordinate change accordingly to keep the geometry of the space invariant.

Am I righe to understand in such a way?

 

[PeterDonis]

I think that the coordinate and the basis field can change independently

The coordinate chart and the coordinate basis cannot change independently. You can choose a non coordinate basis if you like, which doesn't have to change when you change coordinates; but if you use a non-coordinate basis, you have to use different mathematical machinery to describe things like the invariant lengths of line elements.

if the coordinate does not change, but the basis(metric) changes independently a little

This is not possible. You can't use the "metric" mathematical machinery you are using with a non-coordinate basis, as I've already said.

the length of line element will different from the original one, so the geometry of the space changes

This makes no sense. There is no way to pick out "corresponding" points in spaces with different geometries, so there is no way to say "the line element between these two points has changed because I changed the geometry of the space".

Am I righe to understand in such a way?

No. See above.

 

[pervect]

Thanks :
I have a question, if a theory is coordinate independent , then does it means that in fact we can choose any arbitrary coordinate system in our calculation, and the effection of choose of a partilular coordinate will disspear in the process of our calculation?
So for a specific problem, we always can choose a particular coordinate system for convenience?

Yes. This is also known as "general covariance". Theories expressed in tensor form are automatically covariant, due to the tensor transformation properties.

To prove the coordinate independence of tensor methods, you'll need some familiarity with the machinery of tensors. This includes the notion of basis vectors which is vital to understanding tensors. There may be other routes to a proof, but I think that knowing that coordinates have an associated coordinate basis, which is the set of basis vectors are represented as partial derivative operators in those coordinates, is the most straightforwards. You'll also need to understand the mechanics of a change of basis.

Note that is is required that different coordinate systems have a 1:1 correspondence, a bijection, to use this approach. A set of coordinates must specify a specific unique point, and a specific point must have a unique set of coordinates, which can be regarded as "labels". Then a change in coordinate systems corresponds to a change in the labels associated with "the same" point. There are some other conditions required as well, that involve some advanced notions of "smoothness" of transformations. At an elementary level, we talk about infinitely differentiable transformations, at a more advanced level you may run into a more advanced notion of what it means for a transformation to be "smooth". The basic point is that the basis vectors can be identified with partial derivative operators, thus a coordinate basis is mathematically based on the partial derivative operators with respect to the coordinates. Hence to define a new coordinate basis from a new set of coordinates requires that the partial derivative operators exist in the new coordinates. As long as they do exist, there is a straightforwards linear relationship between the coordinate basis vectors expressed in one basis to another.

 

[Jianbing_Shao]

The coordinate chart and the coordinate basis cannot change independently. You can choose a non coordinate basis if you like, which doesn't have to change when you change coordinates; but if you use a non-coordinate basis, you have to use different mathematical machinery to describe things like the invariant lengths of line elements.
This is not possible. You can't use the "metric" mathematical machinery you are using with a non-coordinate basis, as I've already said.

If the geometry of the space itself is changing? then the the basis also can't be regarded as can chang indenpendently?

Ok, I think assuming them to change indenpendently does not mean I don't approve your definition. I think if I demand that they change accordingly, then the situation I discussed will degenerate to yours, and because parameters of coordinate and different from the basis', so it is possible that they change independemtly, and perhaps there are something new can appear if we assume they can change independently. in fact, I have found some studies in mathetics. the situations they discussed is just similar to the situations I have mentioned. there they try to resolve the differential equations similar to the parallel transport equation.

By the way, although parallel transport of a vector is an important comcept in general relativity, but it seems very few people are interested in the problem of how to get the vector field if we parallel transport a vector from a point along a particular path in the space which equipped with a connection.

 

[PeterDonis]

If the geometry of the space itself is changing?

There is no such thing. The geometry of spacetime does not change. Spacetime is a 4-dimensional geometry that already includes all the information about what we normally think of as "change" over the entire history of the universe. What we call "change" is not the geometry of spacetime changing; it is just looking at different segments of our worldlines that lie in different parts of the unchanging spacetime geometry.

Also, as I have already said, there is no way to identify "corresponding points" in different geometries, so the conceptual picture you are implicitly using when you think of "geometry changing" is not well-defined.

 

[PeterDonis]

it seems very few people are interested in the problem of how to get the vector field if we parallel transport a vector from a point along a particular path in the space which equipped with a connection.

You must be joking. This very subject takes up multiple chapters in multiple GR textbooks.

 

[Jianbing_Shao]

You must be joking. This very subject takes up multiple chapters in multiple GR textbooks.

Sorry, I think the way to solve differential equations similar to parallel transport equations is the method of integral production, and in my opinion, this is the the only way to solve such problem, and the calculation is too complicated to get an exact solution, because the solution can be expressed as the Dyson expansion.and only in a book I find someone to solve the problem in such a way. And if the curvature of the space is not zero(the parallel of a vector is path dependent), we can not get an exact solution the parallel transport of a vector along a path.

So I think the most difficult in general relativity is just how to solve the parallel transport equation.

 

[Jianbing_Shao]

There is no such thing. The geometry of spacetime does not change. Spacetime is a 4-dimensional geometry that already includes all the information about what we normally think of as "change" over the entire history of the universe. What we call "change" is not the geometry of spacetime changing; it is just looking at different segments of our worldlines that lie in different parts of the unchanging spacetime geometry.

Also, as I have already said, there is no way to identify "corresponding points" in different geometries, so the conceptual picture you are implicitly using when you think of "geometry changing" is not well-defined.

Whatever you use to describe the geometry of the spacetime, I really find difficulties to comprehend they can contain all the infomation of the past of the universe.

 

[PeterDonis]

I think the way to solve differential equations similar to parallel transport equations is the method of integral production, and in my opinion, this is the the only way to solve such problem, and the calculation is too complicated to get an exact solution, because the solution can be expressed as the Dyson expansion.and only in a book I find someone to solve the problem in such a way.

I don't know what books you're looking at, but they don't seem to be GR textbooks. Try looking at some GR textbooks.

if the curvature of the space is not zero(the parallel of a vector is path dependent), we can not get an exact solution the parallel transport of a vector along a path.

Certainly you can. Parallel transport along a path is uniquely defined in a curved space. What is not well-defined is parallel transport between two points without specifying a path. But if you specify a path you specify a unique result for parallel transport.

 

[PeterDonis]

Whatever you use to describe the geometry of the spacetime, I really find difficulties to comprehend they can contain all the infomation of the past of the universe.

Then you need to spend some time learning GR from GR textbooks.

 

[PeterDonis]

The OP question has been answered. Thread closed.