Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [itex]R_{g_i(t)*}\sigma_{i*}X[/itex]

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

    [itex](\sigma_{i*}X)g_i(t)[/itex]

    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
    Hello.
    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:

    [tex]0=\omega\left(\tilde{X}\right)=\omega\left(R_{g_{i}*}\left(\sigma_{i*}X\right)\right)+\omega\left(\left[g_{i}^{-1}dg_{i}\left(X\right)\right]^{\#}\right)=R_{g_{i}}^{*}\omega\left(\sigma_{i*}X\right)+g_{i}^{-1}dg_{i}\left(X\right)=g_{i}^{-1}\omega\left(\sigma_{i*}X\right)g_{i}+g_{i}^{-1}\frac{dg_{i}}{dt}[/tex]
     
  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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Horizontal lift, or parallel transport
  1. Parallel Transport (Replies: 20)

  2. Parallel transport (Replies: 14)

Loading...