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Horizontal lift, or parallel transport

by qinglong.1397
Tags: horizontal, lift, parallel, transport
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qinglong.1397
#1
Nov13-12, 12:21 PM
P: 107
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 Click image for larger version

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ID:	52916, Nakahara said we could use the similar method in the attachment Click image for larger version

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ID:	52915 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!
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bloby
#2
Nov14-12, 05:06 AM
P: 112
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')
qinglong.1397
#3
Nov14-12, 05:44 PM
P: 107
Quote Quote by bloby View Post
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')
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.

bloby
#4
Nov15-12, 03:07 AM
P: 112
Horizontal lift, or parallel transport

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]
bloby
#5
Nov15-12, 03:16 AM
P: 112
[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]
qinglong.1397
#6
Nov15-12, 07:34 AM
P: 107
Quote Quote by bloby View Post
[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]
Great! Thank you very much!


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