Talk:Parallel transport

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Generality of this article?

Latest comment: 1 year ago4 comments3 people in discussion

There are various levels of generality at which one can discuss parallel transport:

• for vector fields;
• for sections of arbitrary vector bundles;
• for sections of arbitrary geometric bundles (e.g. frame bundles and their associated bundles);
• for sections of arbitrary fiber bundles (e.g. principal bundles and their associated bundles);

At what level should this article employ most of its energies? Geometry guy 16:08, 12 March 2007 (UTC)Reply

Why not all of them? But if I had to pick one, I believe it should focus primarily on vector fields since that is likely to be the area of interest to most people looking in the general community. At least, it should start out with vector fields. The other levels of generality are also important, though, for the article holonomy (which also needs a lot of work). Silly rabbit 14:11, 13 April 2007 (UTC)Reply

I tend to agree: work with vector fields, then explain that it all generalizes to each of the above levels (with fewer details in the more general cases). Geometry guy 14:41, 13 April 2007 (UTC)Reply

This is certainly the correct answer, but as of 2024 this vision is still yet to be realized. I attempted to put some work on it though. Mathwriter2718 (talk) 14:19, 3 July 2024 (UTC)Reply

Another question

Latest comment: 18 years ago5 comments2 people in discussion

In the earlier part of the 20th century, connections were defined in terms of parallel transport satisfying certain properties (see e.g. the Springer Encyclopedia articles on connections). Should we present this approach here? Geometry guy 22:21, 20 March 2007 (UTC)Reply

I'm all for it, since PT really is the unifying theme behind all the connection concepts. So organizing Category:connection (mathematics) around this idea was one of my original goals in this bulk revision. That said, I don't think it can be done in a way that will be at all familiar to readers who have just recently encountered the covariant derivative (*) and are looking for some further reading on the subject. So, if it is to be done, it shouldn't be the primary approach: perhaps a section later on in the article like Defining the covariant derivative in terms of parallel transport.

(*) Ok, anywhere except Kobayashi-Nomizu.

Silly rabbit 14:20, 13 April 2007 (UTC)Reply

I agree with your conclusion here, especially as this is an encyclopedia, not a new book on connections ;) ! Also, I would not say PT is the unifying theme, since there are several points of view, such as Cartan's infinitesimal one. And parallel transport is quite hard to use as a primitive definition (see again the Encyclopedia of Mathematics) because one has to capture the idea that it is not arbitrary, but only depends on 1-jets. This involves taking an awkward derivative at time zero. I feel some of the infinitesimal intuition gets lost when PT is overemphasised. I hope with both of us working on this we can achieve a good balance between the different points of view ;) Geometry guy 14:41, 13 April 2007 (UTC)Reply

For future reference, some starting material for this point of view is buried in Talk:Connection (vector bundle). Silly rabbit 19:23, 13 April 2007 (UTC)Reply

Yes, I put it there, to save it from being buried even deeper in the history archives. Geometry guy 19:28, 13 April 2007 (UTC)Reply

Transporting tensors on Lie groups

Latest comment: 15 years ago1 comment1 person in discussion

I'm trying to understand how in general one transports tensors in the tangent space of a point in a Lie group to some other point in a Lie group, particularly using matrix representation. For example, the exponential map on SO(3) lets you represent tangent vectors in SO(3) as skew-symmetric matrices, and there is a 1:1 mapping between 3x3 skew-symmetric matrices and 3-vectors, however they transform differently... If I have an axis-angle vector, ${\displaystyle \omega }$ , I can rotate it (transport it) using a rotation matrix: ${\displaystyle R\omega }$ . However, if I have the skew-symmetric representation ${\displaystyle \omega }$ , ${\displaystyle [\omega ]_{\times }}$ , I believe it transforms like a matrix; that is,

${\displaystyle [R\omega ]_{\times }=R[\omega ]_{\times }R^{\top }}$ .

Two questions:

1. In SO(3), it seems like the "matrix" representation of a "second-order" tensor would be a (3x3)x(3x3) matrix. How would that transform? Presumably you'd need to transform both 3x3 portions, so it would look something like ${\displaystyle R_{ij}R_{nk}T_{jk\ell m}R_{\ell p}R_{qm}}$ , which seems pretty kludgy.
2. Outside of SO(3), the exponential map on a matrix gives you tangent "vectors" that are matrices with no obvious "simpler" representation like an axis angle. There must be a wholistic way of transporting tangent vectors and tensors. Is it simpler than constructing some high-order object and transforming along each index as above? —Ben FrantzDale (talk) 02:42, 5 November 2009 (UTC)Reply

Equation for parallel transport in general relativity

Latest comment: 7 years ago1 comment1 person in discussion

It would be useful to include the equation for parallel transport of a vector as used in general relativity:

${\displaystyle {\frac {dv^{\mu }}{ds}}+\Gamma _{\alpha \beta }^{\mu }{\frac {dx^{\alpha }}{ds}}v^{\beta }=0}$ 

But where to put it in the article? 193.198.162.14 (talk) 09:49, 6 February 2018 (UTC)Reply

Illustration of Parallel Transport

Latest comment: 5 years ago1 comment1 person in discussion

It appears that the vectors illustrating the parallel transport are viewed form a point above the equatorial plane. The vector at B is shown properly. The direction of the vector between A and B is not correct. It should be tangent to the equator, i.e. showing downward from the equator, just like the vector at B. Also, the vector at A showing to the N-S line is in error. It should be tangent to the equator, i.e. directed downward from the equator. It is absolutely misleading to end it at that line. It has nothing to do to the N-S line.178.48.163.158 (talk) 10:56, 20 August 2020 (UTC)lbReply

For whom is this written?

This article only raises questions. It explains parallel transport on a level only experts understand. It confirms a well-known fact that academics can't explain things, because they don't have any empathy.

Problems with this article

Latest comment: 1 year ago5 comments2 people in discussion

This article has some serious problems.

1. There is not even a single example.

2. It includes the case of a general vector bundle and even further generalizations beyond that (as it should), but it barely discusses the case of the tangent bundle. Most readers of this page probably only care about this case.

3. Then there is a section called "Parallel transport in Riemannian geometry", even though the previous section was also about Riemannian geometry.

I will try to make some first steps towards fixing these problems. Mathwriter2718 (talk) 13:06, 3 July 2024 (UTC)Reply

I went ahead and removed the section "Parallel transport in Riemannian geometry" and added a new section "Parallel transport of tangent vectors" that I think is written for a more appropriate audience. This new section is spliced together from 1. other parts of this page that I moved, 2. content I recently wrote for Riemannian manifold, 3. the part about parallel transport from Affine connection. This section and the article as a whole still needs significant work. Mathwriter2718 (talk) 14:17, 3 July 2024 (UTC)Reply

Looks good, except I think the definition of the linear parallel transport map ${\displaystyle \Gamma _{s}^{t}}$  should be properly introduced in the tangent vectors section. (It is referred to in the Metric connections part before the introduction later in the vector bundles section.) Tito Omburo (talk) 15:38, 3 July 2024 (UTC)Reply

Certainly true. I went ahead and did this. Mathwriter2718 (talk) 18:59, 3 July 2024 (UTC)Reply

Wonderful! Tito Omburo (talk) 14:20, 4 July 2024 (UTC)Reply

Smooth curves

Latest comment: 16 days ago10 comments3 people in discussion

Tito Omburo, I'm puzzled by this edit. I understand the need for a piecewise C¹ curve, or at least I cannot immediately give a weaker sufficient condition for parallel transport. However, smoothness generally refers to C^∞, which is clearly overkill, since parallel transport makes sense on jagged curves, and even the illustration has points of non-differentiability. —Quondum 22:02, 12 August 2025 (UTC)Reply

Just being a "curve" is not sufficient. Details are given in the article. Sobolev conditions (in addition to continuity) may be sufficient for parallel transport, but I haven't found any sources emphasizing this viewpoint. For geodesics, you need C^2, for parallel transport, most sources require C^1. "Smooth" seems completely appropriate for the lede, because generically one expects smoothness, and simple continuity is not enough. Tito Omburo (talk) 22:27, 12 August 2025 (UTC)Reply

I agree that simple continuity is not enough; your "piecewise differentiable curve" seems perfect. I'm not convinced that a lead portion needs to give such conditions in general, since these will be in the body and would introduce clutter on the lead. Placing conditions in the lead that will serve more to confuse than help introduce the topic seems counterproductive. I expect that some sources will require smoothness, but that is likely what I would call laziness: addressing only a convenient subset of the ___domain of applicability for the purposes of pedagogy or economy of exposition. I would argue to simply remove the word "smooth" from the lead, and keep the more precise qualification where you put it in § Precise definition. If you really want a simple qualifier, I would think that "differentiable" is far more suitable than "smooth" is. —Quondum 22:44, 12 August 2025 (UTC)Reply

This is a view taken in many textbooks, to their detriment in my opinion. For example, Marsden and Tromba repeatedly refer to "curve" in casual discussion, without saying that curves satisfy at least some smoothness condition. Fortunately "smooth curve" covers continuity and differentiability, in a way that is not likely to mire the reader of the lede into technicalities, without telling lies to children. Tito Omburo (talk) 22:55, 12 August 2025 (UTC)Reply

And "differentiable"? It does the same, is more precise while being no more verbose, and is a simpler concept to boot. —Quondum 23:08, 12 August 2025 (UTC)Reply

Differentiable is both more technical and less correct. Parallel transport exists along smooth curves, but not differentiable ones. Tito Omburo (talk) 23:22, 12 August 2025 (UTC)Reply

I suppose I need to brush up on the definitions used in differential geometry. I thought smoothness was defined in terms of, and inherently included, differentiability. —Quondum 23:25, 12 August 2025 (UTC)Reply

Fair enough. Interested what you find out. Tito Omburo (talk) 23:36, 12 August 2025 (UTC)Reply

I don't know if it is on topic for this article, but there are notions of parallel transport in discrete differential geometry, e.g., [1], [2] that have applications in computer graphics. But for traditional parallel transport, I've only ever seen ${\displaystyle C^{\infty }}$  requirements mentioned.--{{u|Mark viking}} {Talk} 00:10, 13 August 2025 (UTC)Reply

I do not put a lot of credence in textbooks' presentations. When an author is interested in presenting a topic in their area, they will typically restrict their definitions accordingly, e.g. they might define a vector space, algebra or other very general structure as being over R or C without comment that the term is actually used more generally. Smoothness is often assumed from the start just to avoid the need to call out edge cases in the general treatment. The unfortunate result is that we, as WP editors, have difficulty teasing out the definitions and constraints that apply at the level of generality that we are targeting. —Quondum 00:42, 13 August 2025 (UTC)Reply