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Horizontal lift, or parallel transport

  1. Nov 13, 2012 #1
    Hello, everyone!

    I'm studying Nakahara's book, Geometry, Topology and Physics and now studying the connection theory. I come across a problem. Please look at the two attachments.

    In the attachment Screenshot-1.jpg , Nakahara said we could use the similar method in the attachment Screenshot.jpg to get [itex]\tilde X[/itex], but why does the first term have [itex]g_i(t)^{-1}[/itex]. According to the first figure, the first term should have the following form


    Since [itex]R_{g*}X=Xg[/itex], it becomes


    So there shouldn't be [itex]g_i(t)^{-1}[/itex]. But why did the author put it there? Thank you!
  2. jcsd
  3. Nov 14, 2012 #2
    When I read this chapter I found this strange. I think it's a mistake. The next equation however is right by (10.3b')
  4. Nov 14, 2012 #3
    Hey, thanks for your reply! But what is "the next equation", the second equation in the first attachment? If so, I cannot agree with you, because there is no a pullback of a right action.
  5. Nov 15, 2012 #4
    I have to read this chapter again. Here is what I wrote for me:

  6. Nov 15, 2012 #5
    [tex] 0=\omega(\tilde{X})=\omega(R_{g_{i}*}(\sigma_{i*}X))+\omega([g_{i}^{-1}dg_{i}(X)]^{\sharp})=R_{g_{i}}^{*}\omega(\sigma_{i*}X)+g_{i}^{-1}dg_{i}(X)=g_{i}^{-1}\omega(\sigma_{i*}X)g_{i}+g_{i}^{-1}\frac{dg_{i}}{dt} [/tex]
  7. Nov 15, 2012 #6
    Great! Thank you very much!
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