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Consider a principal bundle [itex]\pi:E\rightarrow M[/itex] with structure group G. Let [itex]I=(a-\epsilon,b+\epsilon)[/itex]. We can use a curve [itex]\alpha:I\rightarrow M[/itex] to construct a pullback bundle [itex]Pr_1:IE\rightarrow I[/itex]. I'm calling the total space of the pullback bundle IE because that reminds me that it's a subspace of [itex]I\times E[/itex]. To be more specific, we have

[tex]IE=\{(t,p)\in I\times E|\alpha(t)=\pi(p)\}[/tex]

This pullback bundle is a principal bundle too, and [itex]Pr_2[/itex] is a principal bundle isomorphism. If [itex]\omega[/itex] is a connection 1-form on the original principal bundle, then [itex]Pr_2^*\omega[/itex] is a connection 1-form on the pullback bundle. (I was able to prove that, so that's not what I'm going to ask about).Now let D be the derivative operator that takes a real-valued function on I to its derivative, i.e. Df=f' and D

[tex]IE=\{(t,p)\in I\times E|\alpha(t)=\pi(p)\}[/tex]

This pullback bundle is a principal bundle too, and [itex]Pr_2[/itex] is a principal bundle isomorphism. If [itex]\omega[/itex] is a connection 1-form on the original principal bundle, then [itex]Pr_2^*\omega[/itex] is a connection 1-form on the pullback bundle. (I was able to prove that, so that's not what I'm going to ask about).Now let D be the derivative operator that takes a real-valued function on I to its derivative, i.e. Df=f' and D

_{t}f=f'(t). This D is a vector field on I. The "horizontal lift" of this vector field is defined as the unique horizontal vector field [itex]D^H[/itex] that satisfies the condition [itex]\pi_*D^H=D[/itex]. "Horizontal" means that it's taken to 0 by the connection 1-form, so we have [itex](Pr_2^*\omega)_{(t,p)}D^H_{(t,p)}=0[/itex].I understand all of the above, but there's another detail that looks like it should be easy to prove, and yet I can't figure it out. Define [itex]\gamma[/itex] to be the integral curve of [itex]D^H[/itex] trough (a,p) (where a is the real number mentioned in the definition of I, and p is some arbitrary point in E). This [itex]\gamma[/itex] is a curve in IE, so we can write [itex]\gamma(t)=(\gamma_I(t),\gamma_E(t))[/itex], but why is [itex]\gamma_I(t)=t[/itex]?I'm stuck at a "clearly we have" statement in the book I'm reading, and that statement is only true if [itex]\gamma_I(t)=t[/itex], so apparently the author of this book (Isham) felt that this is so trivial that he didn't even need to include a "clearly we have" statement about*that*. I'm hoping that someone can tell me why the integral curve [itex]\gamma[/itex] must also be a section.This stuff is on page 264 of "Modern differential geometry for physicists" by Chris Isham, but for some reason it isn't possible to view page 264 either at Google books or at Amazon. If you happen to own a copy of the book, you might get confused by the fact that my notation is different from his.
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