• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,345 answers , 22,720 comments
1,470 users with positive rep
818 active unimported users
More ...

  Question about the inverse of parallel transport on principal bundle.

+ 2 like - 0 dislike

Let \(\tilde{\gamma}\) be a horizontal lift of the curve \(\gamma : [0,1] \to M\) on a principal bundle \(P(M,G)\), with the projection \( \pi : P \to M\). The parallel transport is defined by the map:

\(\Gamma(\tilde{\gamma}^{-1}) : \pi^{-1}(\gamma(1)) \to \pi^{-1} (\gamma(0)) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \tag{1}\)

where \(\tilde{\gamma}^{-1}(t) = \tilde{\gamma}(1-t) \). Apparently, this means that \(\Gamma(\tilde{\gamma}^{-1}) = \Gamma (\tilde{\gamma})^{-1}\), but I don't see how this is true. As far as I understand, equation (1) implies that:

\(\Gamma(\tilde{\gamma}) : \pi^{-1}(\gamma(1)) \to \pi^{-1} (\gamma(0))\)

because \(\Gamma(\tilde{\gamma})\) sends the fibre at \(t=0\) (i.e.\(\gamma(1-0) = \gamma(1)\)) to the fibre at \(t=1\) (i.e. \(\gamma(1-1)=\gamma(0)\)). But this is clearly wrong because it implies that \(\Gamma(\tilde{\gamma}^{-1}) = \Gamma (\tilde{\gamma}) \implies \Gamma(\tilde{\gamma}^{-1}) \neq \Gamma (\tilde{\gamma})^{-1}\). Anybody knows where I'm making a mistake?

asked Jun 2, 2014 in Mathematics by Hunter (520 points) [ revision history ]
edited Jun 3, 2014 by Hunter

So I'm starting to get the feeling that book means (with abuse of notation):

\(\Gamma(\tilde{\gamma}(t)^{-1}) = \Gamma(\tilde{\gamma}(t))^{-1}\)

whereas I thought they meant:

\(\Gamma(\tilde{\gamma}(t)^{-1}) = \Gamma(\tilde{\gamma}(1-t))^{-1}\)

But I'm not sure though.

Btw, the notation $\Gamma(\tilde\gamma)$ doesn't make much sense, because each vector you want to transport comes with its own $\tilde\gamma$, ie $$ \Gamma(\gamma):G_{\gamma(0)}\to G_{\gamma(1)},X=\tilde\gamma(X,0)\mapsto\tilde\gamma(X,1) $$

...where 'vector' in this case means $X\in G_{\gamma(0)}$, which in case of connections on principal bundles (instead of vector bundles) is generally no such thing

Hmm interesting, the author does change notation when he discussed loops (i.e. closed path), because then he will write someting like:

\(\tau_\gamma : \pi^{-1}(p) \to \pi^{-1}(p)\)

for \(p \in M\). Does this resemble the notation that you prefer to use?

Since I'm only studying from one book (except for the few occasions that I try to look something up on the internet), I am only used to the authors notation.

1 Answer

+ 2 like - 0 dislike

The problem is the formal notation, which is stupid, and makes the content of the statements much, much, less clear, because you focus on pedantic nonsense, like the distinction between $\gamma$ (the curve on the manifold) and $\tilde{\gamma}$, the curve on the bundle.

The notation $\gamma^{-1}$ means the curve oriented backwards. The book's statement is just that the parallel transport going backwards on a curve undoes the parallel transport going forward. It's obvious from the definition of parallel transport. $\Gamma(\gamma^{-1}) = \Gamma(\gamma)^{-1}$, by definition, it seems from the comment that you confused yourself by reversing the curve twice, once by reversing parametrization, and again with the -1 superscript.

answered Jun 3, 2014 by Ron Maimon (7,720 points) [ no revision ]

Thanks for the reply and confirming that my confusion indeed comes from reversing the curse twice (the notation did confuse me a lot, but should be clear now).

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights