Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel transport analog of Stoke's theorem

  1. Apr 20, 2009 #1
    In a Stokes theorem, the integral of all curls of a vector field enclosed in some region is equal to the line integral around the boundary.

    I'm wondering if a similar theorem exists for parallel transport. The Riemann curvature tensor gives a change in a vector when parallel transported around a small loop. If we were to parallel transport a vector around a much larger loop, would the change in the vector be proportional to the integral of the Riemannian tensor around small loops inside the larger loop?
     
  2. jcsd
  3. Apr 20, 2009 #2
    You'll have to be more specific, since as it stands, nothing is well-defined. Integration gives a number. Unless you're in dimension two (where the curvature tensor is ~ Gaussian curvature, which is a function), you'll have to input specific vectors into the curvature tensor to even get a number to come out when you integrate.

    On the other side of the equation, you have a vector. I'm not sure how one would make the necessary changes so that the sides would come out apples to apples. The closest thing (in dimension two) would be the Gauss-Bonnet theorem, and in higher dimensions, the Chern-Gauss-Bonnet theorem. However, it doesn't seem to do exactly what you want it to do. One immediate problem is that you would have a very difficult time "integrating" over the "interior" of a closed loop, if you're in dimension greater than two - it's just not possible to do it with the volume element on a Riemannian manifold. You could attempt to generalize to boundaries of submanifolds, but then you're not working with vectors and loops anymore.

    In a broader sense, curvature does measure the non-commutativity of mixed covariant derivatives, so it seems like you're barking up the right tree. But I just don't think it's going to work the way you'd like.
     
  4. Apr 21, 2009 #3
    yea basically the idea is to integrate non-commutativity (or non-commutativite density) in a 2D closed region on a plane spanned in some nth dimensional manifold, and hoping it would equal to the change of a vector parallel transported around the boundary.

    I guess that's not possible, thanks for clarifying
     
  5. Apr 21, 2009 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Something's fishy here; what have I done wrong? (Or am I right?)

    Let's suppose our n-dimensional manifold M can be covered by a single coordinate chart. The chart provides an isomorphism TM --> R^n. In these coordinates, parallel transport is a function P that takes any curve c and returns an element P(c) of GL(n).

    Parallel transport is (?) differentiable, so given any tangent vector v to a point p, the derivative of P gives us an element P(v) of gl(n).

    In other words, there is a gl(n)-valued one-form A such that P(c) is computed by integrating A along c. (A can be written as n^2 ordinary one-forms)

    Now, by Stoke's theorem, if a closed loop c is the boundary of an (oriented) surface S in M, then P(c) can be computed by integrating dA over all of S. In particular, it doesn't matter what surface you use, they all give the same value.

    Isn't that essentially what waht is trying to think of?
     
  6. Apr 21, 2009 #5
    Could dA can be the riemann curvature tensor?

    If I understand the tensor correctly, you give it two vectors to define a plane, and a third vector v to be parallel transported around an infinitesimal closed loop on the plane; the result will be a change in the vector v after completing the circuit.

    By integrating the tensor in a region inside a closed curve c on the plane, by (stoke's theorem or any other theorem?), the result should be the same as if doing the parallel transport of the vector v around the curve c; or integrating the riemann tensor along c?
     
  7. Apr 21, 2009 #6
    I'm not quite sure what you're trying to do with that, but it certainly wouldn't be GL(n)-valued - parallel transport is trivial in R^n.
     
  8. Apr 21, 2009 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not using the standard connection on Rn. I'm using the (coordinate representation of the) connection on M.

    More explicitly: parallel transport along c is a linear isomorphism

    P(c) : Tc(0)M --> Tc(1)M

    My coordinate chart yields a basis for TM, and thus linear isomorphisms

    Rn --> Tc(0)M
    Rn --> Tc(1)M

    And thus the coordinate representation of P(c) is the composite

    Rn --> Tc(0)M --> Tc(1)M --> Rn

    which is an element of GL(n).
     
  9. Apr 21, 2009 #8
    It ought to work if you take the standard connection on R^n, right? You're taking the derivative of the identity, which you claimed was an element of GL(n). That's what I'm disagreeing with.
     
  10. Apr 21, 2009 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ah, I see.

    Well, I can definitely take the derivative, and reconstruct the identity by integrating zero along the curve, so that part works. And 0 is an element of gl(n).

    But I suppose that the derivative wouldn't need to be in gl(n). (?) I know there's some good Lie group/algebra stuff going on, but I can never manipulate that as adeptly as I would like; I suppose there should really be an exponential or logarithm in there to do things 'right'.
     
  11. Apr 21, 2009 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ah right.

    Let P(c,t) denote parallel transport along the subcurve of c with domain [0,t]. The derivative wrt t is going to be in the tangent space to GL(n) at P(c,t), but we really do want to "normalize" it to an element of gl(n), and so we need to translate back to the identity on GL(n).

    By P being differentiable, what I mean is that there is a gl(n)-valued one-form A (whose definition doesn't depend on the choice of c) satisfying

    (d/dt) P(c,t) = A(dc/dt) P(c,t)

    But recovering P(c) is no longer simply the integral of a one-form, and that was my mistake.


    Oh, but in a two-dimensional surface, if we actually had a metric and chose an orthonormal basis, SO(2) is commutative, so the equation above becomes

    (d/dt) log P(c,t) = A(dc/dt)

    and so the logarithm of P(c) can be recovered by integrating a one-form, and thus we can use Stoke's theorem and everything works out!


    You could still do what I suggested, but each different curve c would have yielded an entirely different A, and therefore wouldn't be all that useful!


    Sorry to go off on a tangent (heh heh). At least it was useful for me....
     
    Last edited: Apr 21, 2009
  12. Apr 21, 2009 #11
    can we integrate a riemann tensor over a region inside a closed curve?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Parallel transport analog of Stoke's theorem
  1. Parallel transport (Replies: 14)

Loading...