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Parallel Vector Fields

  1. Sep 26, 2008 #1
    Let M be a Riemmanian manifold. Prove that a parallel vector field along a curve c(t) preserves the length of the parallel transported vector.

    Furthermore if M is an oriented manifold, prove that P preserves the orientation.

    So I want to prove that d/dt <P, P> = 0, so <P, P> = constant.

    d/dt <P, P> = 2<dP/dt, P>

    But dP/dt is not an element of the tangent space at c(t) for any t. DP/dt is however always an element of the tangent space, but since DP/dt is 0, we get our desired result.

    But am I allowed to do say that dP/dt is in general not an element of the tangent space? DP/dt is always an element of the tangent space as we can see when we use a local expression for it. However, I feel like I am using an exterior argument by essentially saying that:

    dP/dt = orth(dP/dt) + tang(dP/dt) i.e. I'm splitting it up into tangential and orthogonal components. Then when we "dot" an orthogonal part with the purely tangential part, it drops out and we are left with:

    <DP/dt, P> = 0.

    Is there another way to do this in terms of local coordinates?

    And if M is an oriented manifold, what does it mean exactly for P to preserve orientation?

    Thanks guys!
     
  2. jcsd
  3. Sep 26, 2008 #2

    Dick

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    <P,P> is a scalar function along the curve. So d/dt along the curve is the same as the covariant derivative of <P,P> along the curve with respect to the vector c'(t). Nuff said? Another form of this problem is in Misner, Wheeler and Thorne and they warn you that it's the easiest problem in the whole book. For good reason.
     
  4. Oct 19, 2008 #3
    Let M be a Riemmanian manifold. let P be a parallel vector field along a curve c(t) preserves the length of the parallel transported vector.

    I have shown that P is a linear isometry between tangent spaces along c(t).

    However, if M is oriented, why does P preserve the orientation? Let c(t_0) be any point on the curve, and consider the tangent space at that point c(t_0); let P_1, P_2,...,P_n be an orthonormal set of basis vectors of the tangent space to c(t_0).

    Well we can parallel transport the basis vectors to form an orthonormal basis for the tangent spaces along the curve c.

    Now since P must preserve the length of vectors and we are dealing with an orthonormal basis, P must be an unitary matrix, i.e. P*P^(T), i.e. P times it's transpose is the identity mapping, where P = the matrix whose rows are the orthonormal basis vectors, and this is true along the curve c.

    In particular, P we have that P must have determinant 1 or -1. If P has determinant +1, then it does preserve the orientation.

    I don't see where I use that M is oriented. Am I doing this correctly?
     
  5. Oct 19, 2008 #4

    Dick

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    If P starts out with determinant 1, parallel transport is a continuous operation, so it can't suddenly flip to -1, right? This issue for an orientable manifold is whether transport along ANY path preserves orientation. Along any single path you never flip orientation. You don't need to use that it's orientable to prove that. The point is just that if you start in one orientation you end up in the same one. So if the manifold is orientable and you start in the designated orientation you must wind up in the same one.
     
  6. Oct 20, 2008 #5
    Got it, thank you both so much for your help!
     
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