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Parallel transport around a loop

  1. Feb 15, 2014 #1
    From the Wikipedia article on the Riemann curvature tensor:
    The last sentence assumes that ##\tau_{sX}^{-1}\tau_{tY}^{-1}\tau_{sX}\tau_{tY}## returns Z back to x0. It isn't obvious to me that this should be true. My guess is that this is true because X and Y commute, but I can't think of a proof for this claim.
  2. jcsd
  3. Feb 15, 2014 #2

    Ben Niehoff

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    X and Y being commuting fields is equivalent to the quadrilateral being closed (provided there is no torsion).

    In the presence of torsion, quadrilaterals of commuting vectors fail to close, and the torsion tensor measures the amount by which they fail to close:

    [tex]T(X,Y) \equiv \nabla_X Y - \nabla_Y X - [X,Y][/tex]
    Here ##Z = T(X,Y)## is the vector that closes the gap left after taking into account the possibility that X and Y fail to commute.
  4. Feb 15, 2014 #3
    That's what I suspected. Do you have a proof for this?
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