# About the parallel transport and choice of connection

## Thought Experiment

Consider a 2-sphere, $S^2$, and let $p$ be a point at the equator.

Case 1

Let us parallel transport a vector, $V$ from $p$ using the recipe:

• Move one unit of length East.
• Move one unit of length North.
• Move one unit of length West.
• Move one unit of length South.

Case 2

Let us parallel transport a vector, $V$ from $p$ using the recipe:

• Move one unit of arc East.
• Move one unit of arc North.
• Move one unit of arc West.
• Move one unit of arc South.

## Differences

It's clear that the first case draw a path which is not closed, while the second path closed on itself. And moreover, the vectors might change, in the usual way, under the action of an element of the Holonomy group.

Questions

• Why does that difference appear?
• Am I misunderstanding the notion of parallel transport? (In that case which one is the correct one)
• It is well known that one might choose different connections in order to define the parallel transport. Is the above difference due to a choice of connection?
• Are these point of view the corresponding to Levi-Civita and Weitzenboeck connections?

I'd thank any help to clarify my doubts. Cheers.

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What do you mean by moving East? Do you move along latitude? In any case, your cases 1 and 2 look identical. –  Misha Mar 8 '13 at 18:41
Perhaps you should also explain the difference between a "unit of length" and a "unit of arc". –  Lee Mosher Mar 8 '13 at 20:35
@Dox: I think, it is not really a question about parallel transport, but about arc-length and distance calculation of 2-dimensional sphere. As such, it is not appropriate for MO, but you can ask it at math.stackexchange. –  Misha Mar 8 '13 at 21:15
@misha @lee-mosher Hello. In order to explain myself, moving East is the same as in real life, moving to the right side along the equatorial line. If one move 1cm along the geodesic (in this case maximum circles), it is obvious that the path is not close... since the sphere has "curvature". In the second case, if you move 1sec of arc along the azimuthal angle, and then polar angle, you in fact will obtain a closed path (not a square). –  Dox Mar 9 '13 at 0:36