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Lie transport vs. parallel transport

by maddy
Tags: parallel, transport
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maddy
#1
Mar25-04, 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).

Intrinsic curvature is defined by using the fairly easy to understand idea of "parallel transport". Imagine some closed curve on a flat surface with the tail of a vector placed on a point of this curve. Now push the tail around the curve in such a way that in moving it between infinitessimally separated points on the curve, the vector is kept parallel to itself. When the tail returns to the starting point the vector will be pointing in the same direction as it was initially. However, in performing the same exercise on a curved surface, the final and initial orientations of the vector will in general differ.
(posted by jeff)

How do we do Lie transport of a vector without embedding the manifold in some higher dimensional manifold?
Thanks!
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Mike2
#2
Mar26-04, 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?
lethe
#3
Mar27-04, 07:39 AM
lethe's Avatar
P: 657
Quote Quote by maddy

How do we do Lie transport of a vector without embedding the manifold in some higher dimensional manifold?
Thanks!
perhaps i am not sure where you are coming from, but why would you think you need an embedding? Lie derivatives are intrinsically defined.

Mike2
#4
Mar27-04, 08:59 PM
P: 1,308
Lie transport vs. parallel transport

Could someone explain parallel transport, please?
Mike2
#5
Mar29-04, 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.
Mike2
#6
Mar30-04, 01:54 PM
P: 1,308
Quote Quote by Mike2
"when we integrate a connection we get parallel transport, and when we take the derivative of parallel transport we get a connection."
Does this mean that, for a surface, the vector of interest always forms the same angle with the connection vector,... the dot product is always the same as you move that vector along a curve? Or does that mean that to parallel transport you must move a vector in the direction of the connection vector? Is the differential change in a parallel transported vector alway the connection vector associated with that change on a curve?

Thanks.
maddy
#7
Mar30-04, 07:39 PM
P: 33
Quote Quote by lethe
perhaps i am not sure where you are coming from, but why would you think you need an embedding? Lie derivatives are intrinsically defined.
Yes, thanks. Was a bit fixed on vectors living in the higher dimensional manifold. My reference is J. A. Thorpe's Elementary Topics in Differential Geometry. In addition to the mathematics, the author provides some intuitive pictures. I have gone through the mathematics for both the Lie and covariant derivatives in Ray D'Inverno's Introducing Einstein's Relativity, but couldn't grasp an intuitive picture (am a physics student ) of Lie and parallel transports.

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 2-sphere 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.
maddy
#8
Mar30-04, 08:11 PM
P: 33
Quote Quote by Mike2
Could someone explain parallel transport, please?
Mmm...maybe you can also view it with respect to a geodesic (I find it easier to understand and picture intuitively). Parallelly transporting a tangent vector produces a trajectory on an arbitrary manifold, and this trajectory is a geodesic on the manifold. When we have an arbitrary vector and want to parallely transport it, we can do so "using the geodesic as a frame of reference", that is, the parallely transported vector must always "form the same angle with the tangent vector of some geodesic in the same specific congruence/family" (sorry using very vague language here). In the case of a flat R2, geodesics are just straight lines passing through every point in it.
maddy
#9
Mar30-04, 09:13 PM
P: 33
Quote Quote by Mike2
Does this mean that, for a surface, the vector of interest always forms the same angle with the connection vector,... the dot product is always the same as you move that vector along a curve? Or does that mean that to parallel transport you must move a vector in the direction of the connection vector? Is the differential change in a parallel transported vector alway the connection vector associated with that change on a curve?

Thanks.
The connection is not a vector, and it's also not a tensor coz it's not independent of coordinates.

(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.
Mike2
#10
Mar31-04, 11:28 AM
P: 1,308
Quote Quote by maddy
The connection is not a vector, and it's also not a tensor coz it's not independent of coordinates.

(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.
Thanks for the effort. But it leaves me with my origianl question: Is the differential change in a parallel transported vector alway the connection vector associated with that change on a curve?
maddy
#11
Mar31-04, 08:51 PM
P: 33
Quote Quote by Mike2
Thanks for the effort. But it leaves me with my origianl question: Is the differential change in a parallel transported vector alway the connection vector associated with that change on a curve?
I'd say no.


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