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Lie transport vs. parallel transport 
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#1
Mar2504, 09:20 PM

P: 33

I learnt that there exists a difference between Lie transport and parallel transport and what that difference is in differential geometry, but I'm getting all confused again when I read the explanation given in the 'intro to differential forms' thread (below).
How do we do Lie transport of a vector without embedding the manifold in some higher dimensional manifold? Thanks! 


#2
Mar2604, 11:21 AM

P: 1,308

Good question.
Is a parallel transported vector always form the same angle with respect to the manifold or curve or something? 


#3
Mar2704, 07:39 AM

P: 657




#4
Mar2704, 08:59 PM

P: 1,308

Lie transport vs. parallel transport
Could someone explain parallel transport, please?



#5
Mar2904, 08:14 PM

P: 1,308

Found this at:
http://planetmath.org/encyclopedia/Connection.html "when we integrate a connection we get parallel transport, and when we take the derivative of parallel transport we get a connection." Could someone elaborate? I think I'd completely understand if I could see that integral of a connection. Thanks. 


#6
Mar3004, 01:54 PM

P: 1,308

Thanks. 


#7
Mar3004, 07:39 PM

P: 33

Maybe I'll try to give my understanding of the explanation given in the 'intro to differential forms' sticky here. If I got it wrong, then please correct me. Consider FLAT R2 and ANY closed curve. If we attach a vector at its tail to this curve and drag it along the curve such that the direction of the vector is everywhere parallel to the initial vector, the vector will coincide with the initial vector when it returns to its initial position. Lie transport produces the same effect. However, if we consider the surface of a 2sphere embedded in R3 and ANY closed curve (not a maximal geodesic or maybe a piecewise curve consisting of segments of maximal geodesics), the parallelly transported vector will not coincide with the initial vector when it returns to its initial position. The Lie transported vector would. If we consider a closed curve that is a maximal geodesic on this surface (the greatest circle of the sphere), then the parallelly transported vector will coincide with the initial vector once it returns to its initial position. The Lie transported vector would too. 


#8
Mar3004, 08:11 PM

P: 33




#9
Mar3004, 09:13 PM

P: 33

(sorry, I don't know how to post equations here) From what I learnt, the connection is introduced as a multiplicative factor when we want to do a covariant derivative of a vector. Let's say we have a point p and a vector with its tail attached to p on an arbitrary manifold. At another point q an infinitesimal distance delx from p, we want to build a vector that is "parallel" to the vector at p on this manifold. This "parallel" vector must be the vector at p plus a term that is linear with respect to the vector at p and the distance delx between points p and q. The term is the linear product of the multiplicative factor with the vector at p and the distance delx. 


#10
Mar3104, 11:28 AM

P: 1,308




#11
Mar3104, 08:51 PM

P: 33




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