A Global solution to parallel transport equation?

1. Nov 15, 2017

Jianbing_Shao

In general relativity, a vector parallel along a curve on a manifold M with a connection field Γ can be expressed:
∂v+Γv=0
We know that if the curvature corresponding to Γ is non-zero, which means if we parallel transport a vector along different paths between two points, we will get different result, so we can say we can not get a globally parallel vector field on the manifold.
Does this conclusion means that to the parallel transport equation, if curvature of Γ is not zero, then we can not find a global vector field on manifold M to satisfy the parallel equation?

2. Nov 16, 2017

andrewkirk

The equation that defines parallel transport is
$$\forall s\in[0,1]:\ \nabla_{\dot \gamma(s)}\mathbf X=0$$
for parallel transport along a curve $\gamma:[0,1]\to M$ where $M$ is a manifold and $\mathbf X$ is a vector field on $M$.

So a solution to the equation needs to relate to a specific curve.

Is your question perhaps asking whether there is some vector field $\mathbf X$ on $M$ such that for every curve $\gamma$ the equation is satisfied? If so then I'm pretty sure the answer is No, because by parallel transporting a vector around a nontrivial loop it will be changed, so that the mapping via $X$ from the start point to the vector bundle is multivalued and hence not a function.

3. Nov 16, 2017

Orodruin

Staff Emeritus
It means you cannot find a set of global vector fields that are parallel and form a complete basis of the tangent space at each point. A priori, there might exist a smaller number of global parallel fields.

4. Nov 16, 2017

Jianbing_Shao

Yes ,I also think so, from the property of Riemman curvature, I think only when the Rieman curvature is zero, then we can find a global solution (a global vector field v(x)) satisfy the equation. So I can say the existence of a global solution of the parallel transport equation is determined by the Riemman curvature.
Ok then to the similar equations such as metric compatibility condition \nabla g=0, If given a connection firstly, then perhaps a global metric field satisfy the equation also does not exist?

5. Nov 16, 2017

Jianbing_Shao

A globally parallel transport vector field does not exist, but we can define local parallel transport of a vector.

6. Nov 16, 2017

Orodruin

Staff Emeritus
It is unclear what you are trying to say. A parallel field $X$ is a field such that $\nabla_Y X = 0$ for all vector fields $Y$. Such a field may exist even if the curvature is non-zero, but it depends on the connection and the manifold. What cannot exist is a set of parallel fields that form a basis at every point.

It is determined by the curvature, yes, but the curvature need not be identically zero for a single such field to exist.

7. Nov 16, 2017

Jianbing_Shao

I mean that if we parallel a vector along different paths from a point to another, then we will get the same result, so the parallel transport is independent of choice of path, From the definition of curvature, the curvature should be zero. so perhaps I can not understand the if the curvature is non-zero, then such a vector field can exist. Would you mind giving me an example?

' What cannot exist is a set of parallel fields that form a basis at every point' I also think this is right if curvature is non-zero.

8. Nov 16, 2017

Orodruin

Staff Emeritus
This is the point. You can only draw this conclusion if the above is true for all vectors. For a single vector field, it is sufficient that the corresponding components of the curvature is zero. You cannot draw the conclusion that $R(X,Y)Z = 0$ just because you know that $R(X,Y)W = 0$, where $Z$ and $W$ are different vector fields.

Take $\mathbb R^3$ in spherical coordinates and exclude the origin. Equip it with a connection for which the radial $\partial_r$ is parallel and $\partial_\theta$ and $\partial_\varphi$ behave in the same way as they would on the corresponding sphere. This connection has a non-zero curvature, but a parallel field $\partial_r$ exists.

9. Nov 19, 2017

Jianbing_Shao

Yes,,from the differential equations I have given above $\partial_\mu V^\nu +\Gamma_{\mu\lambda}^\nu V^\lambda=0$, and if a vector field$V^\nu$ can satisfy the equation above, then it must fulfill the parallel transport equation along an arbitrary path $\gamma$:$\nabla_\gamma V=0$. So if such a vector field $V^\mu$exists, then we can get it by parallel transport a vector（but not a vector field）along different path on the manifold and get a vector field. so we can draw a conclusion that the curvature should be zero, the reason is if curvature is nonzero, then:
because by parallel transporting a vector around a nontrivial loop it will be changed, so that the mapping via $X$ from the start point to the vector bundle is multivalued and hence not a function.

Last edited by a moderator: Nov 19, 2017
10. Nov 20, 2017

Jianbing_Shao

11. Nov 20, 2017

Orodruin

Staff Emeritus
You seem to be having a problem of nomenclature here. A parallel vector field by definition is a vector field $X$ that satisfies $\nabla_Y X = 0$ everywhere for all $Y$. The reason we introduce parallel transport is that it is generally not possible to construct parallel fields.

Also, you are wrong that you can draw the conclusion that the curvature is zero from the existence of just one parallel field. I suggest you try to work out the example I gave you. All you can do is conclude that some components of the curvature are zero. This is generally not sufficient to conclude that the entire tensor is zero.

Let us say that we have a vector field $V$ for which it is (locally) true that a parallel transport around an infinitesimal loop gives back the same vector up to corrections of third order and higher. It is then true that $R(X,Y)V = 0$ at that point for all $X$ and $Y$. The curvature tensor generally has $N^2(N^2-1)/12$ independent components, where $N$ is the dimension of the manifold. However, since you only have access to $N^3$ relations (actually fewer due to the anti-symmetries of the tensor) on the form $R(X,Y)V = 0$, you simply cannot compute all of your curvature components from this knowledge.

Now, in two dimensions, the curvature tensor has only one component and so it should come as no surprise to you that my example was three-dimensional.