Question about the inverse of parallel transport on principal bundle.

Question

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?

Comments

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.

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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 $$ \Gamma(\gamma):G_{\gamma(0)}\to G_{\gamma(1)},X=\tilde\gamma(X,0)\mapsto\tilde\gamma(X,1) $$

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...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

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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.

Answer 1

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.

Comments

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).