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Question about the inverse of parallel transport on principal bundle.

+ 2 like - 0 dislike
282 views

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 3, 2014 in Mathematics by Hunter (510 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 4, 2014 by Ron Maimon (7,535 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).

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