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Horizontal lift, or parallel transport 
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#1
Nov1312, 12:21 PM

#2
Nov1412, 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') 


#3
Nov1412, 05:44 PM

P: 107




#4
Nov1512, 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] 


#5
Nov1512, 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]



#6
Nov1512, 07:34 AM

P: 107




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