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Parallel Translation

  1. Oct 10, 2006 #1
    I've been wrestling with this all day, and it's starting to drive me crazy; if you can help me out, please take just a few minutes to read and answer.:smile:

    I'm looking at S^2 in R^3 (the two dimensional unit sphere in space), and at a large circle going through the north pole as my closed curve.

    I've proved that parallel translation in T_p(S^2) is simply the reduction of parallel translation in T_p(R^3) [1]. I'll write how in a second, but my problem is that this gives me a stupid contradiction:

    Let's look at the tangent vector (0,0,1) at the north pole. Parallel translation of that vector in R^3 along my curve gives me the constant vector field (0,0,1). But that can't be the parallel translation in S^2, seeing as this vector field has tangent vectors that aren't in T_p(S^2) (almost all of them, actually)!


    Now, for the relatively easy proof (but obviously wrong) of [1]. Covariant derivative is the same in both, since an isometry phi respects the covariant derivative up to phi_*, and for phi=id, phi_*=id. Now if we look at parallel translation in S^2, since by the previous line it's also parallel in R^3, then by uniqueness it must be the parallel translation in R^3!


    Am I going crazy? What did I do wrong?
     
  2. jcsd
  3. Oct 10, 2006 #2
    parallel translation in S^2 is not the same as that in R^3. The parallel translate of a vector along a great circle is found simply by moving the vector along the great circle by the rotation about (0,0,0) with axis of rotation given by the line through (0,0,0) and perpendicular to the plane of the great circle.
     
  4. Oct 11, 2006 #3
    I figured as much, but what's wrong with my 'proof'?
     
  5. Oct 11, 2006 #4
    i think the problem lies in the very first line. i'm having a hard time understanding what you mean. there is no isometry between S^2 and R^3 for obvious reasons. But the inclusion of S^2 into R^3 (which is the identity on S^2) is isometric. Is that the map you're talking about there?
     
  6. Oct 11, 2006 #5
    That's the one. It is an isometry (in my book isometry doesn't mean onto), right?

    So am I confusing something?

    I'm starting to think maybe its * isn't the inclusion after all...
     
  7. Oct 11, 2006 #6

    HallsofIvy

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    An isometry "between" A and B means an isometry that maps A one-to-one onto B. That's not what you have here. What you have is an "embedding" of S2 in R3.
     
  8. Oct 11, 2006 #7
    Actually, I used the term "isometric" (adjective) rather than "isometry" (noun) very carefully. A mapping (in particular, an embedding) can be isometric (i.e. preserves the Riemannian metric) without being an honest-to-Goddess isometry (i.e. a diffeomorphism that preserves the Riemannian metric). See, for example, Chapter 6 of Do Carmo, called "Isometric Immersions."

    In other words, the inclusion map S^2 into R^3 is the isometric immersion to which I referred.

    Nevertheless, I still don't understand your "proof." Covariant differentiation is decidedly *not* the same on S^2 as it is on R^3. In particular, in S^2, you need to eliminate the normal part of the covariant derivative. In fact, I think that's the key: the covariant derivative of two vector fields on S^2 is the R^3 covariant derivative of the vector fields (with domain expanded to an open set of R^3) projected onto the tangent space of S^2.
     
  9. Oct 12, 2006 #8
    Don't get me wrong, I understand what you're saying, and I'm also now calculating with brute force in spherical coordinates to see what I'm gonna get. It's just bothering me that this theorem that my book states seems to require the isometry to be onto, but doesn't explicitely mention it.

    Since the proof is left as an exercise, I haven't checked yet if onto is mandatory. I guess I'll have to prove that theorem then...

    Thanks a lot for your guidance.
     
  10. Oct 12, 2006 #9

    George Jones

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    Yes an isometry is onto - see Doodle Bob's previous post.

    There can't be an isometry between S^2 and R^3 because, e.g., S^2 is compact and R^3 is not.
     
  11. Oct 12, 2006 #10
    It's a matter of convention. We defined it to be one to one, but not onto. Call it an isometric mapping then.

    I've calculated the thing brute force, yet my equations tell me that parallel translation, in spherical coordinates, keeps the 'theta' component constant, while the 'phi' one behaves like sinus.

    If anyone would be willing to go over my (brief) calculations, I'd send them the word document by mail. It would really help me understand this whole subject...
     
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