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Dense geodesic

  1. Nov 21, 2008 #1
    Are there surfaces that have a geodesic curve which completely covers the surface, or (if that's not possible) is dense in the surface?

    In other words, if you were standing on the surface and started walking in a straight line, eventually you would walk over (or arbitrarily close to) every point on the surface.

    My current thinking is of a torus where you start walking at a wonky angle so that your path never repeats, like so, but I'm not sure if it quite works:
    [​IMG]
     
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  3. Nov 21, 2008 #2

    Hurkyl

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    It's actually pretty easy to see that it does work. Pick any horizontal line: where does the geodesic pierce it?
     
  4. Nov 21, 2008 #3
    Ahh yes so simple, thanks.

    Now actually, thinking about it a bit, this is a very roundabout way of showing that the cardinality of RxR is equal to the cardinality of R. This torus construction is directly analogous to the diagonal counting used when you show ZxZ = Z. EDIT: Scratch that, I can only show it is dense...
     
  5. Nov 21, 2008 #4
    While a geodesic on a torus might not do it, you can construct a continuous, surjective function from [0, 1] to [0, 1]2, so that |R| = |[0, 1]| ≥ |[0, 1]2| = |R2|.
     
  6. Nov 21, 2008 #5
    I question whether the cardinality of RxR is equal to the cardinality of R in terms of a "space filling" curve (geodesic or not) because the fractal dimension is always less than 2. The curve always has the cardinality of R, but not R^2. The fractal dimension of the curve (C) lies on the interval 1< C < 2 , so the interval is closed at the lower bound, but open on the upper bound.
     
    Last edited: Nov 21, 2008
  7. Nov 21, 2008 #6
    The idea of the space-filling curve is that you can construct a sequence of continuous functions fn: I → I2 (with I = [0, 1]) that converges uniformly to another, also continuous function f: I → I2 that is surjective (which can be proven by showing that every point in I2 is in the closure of f(I), which is just f(I) because I is compact). Munkres' Topology (§44 in the second edition) describes it better than I.

    (If you're still not convinced that |R| = |R2|, it's easy to find a surjection [0, 1) → [0, 1)2 using decimal expansions.)
     
  8. Nov 21, 2008 #7

    gel

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    you can't have a geodesic completely covering a surface though (not if its a differentiable manifold), just dense. The geodesic would have zero area.
     
  9. Nov 21, 2008 #8
    Could you elaborate a little bit? Maybe I'm just being dense (ha ha), but I can't see it.
     
  10. Nov 21, 2008 #9

    gel

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    take a coordinate patch on the manifold and a segment of the geodesic lying in this patch. Using the coordinates, this gives a differentiable curve in R2, which must have zero area. Adding up all the segments of the geodesic lying in this coord patch will give a set with zero area and can't cover the whole coordinate patch.
     
  11. Nov 21, 2008 #10
    So the patch contains only countably many segments, each with zero measure, and so the union of all segments has zero measure. OK
     
  12. Nov 22, 2008 #11

    Hurkyl

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    That's not true.
     
  13. Nov 22, 2008 #12
    For some reason, I am not convinced that there would countably many segments. But a non-measure theoretic argument that the geodesic will not fill the torus is simply finding a point that won't be on the geodesic: if this is the standard unit square torus and the geodesic is given as a line through (0,0) with an irrational slope, then any point of the form (rational, rational) except (0,0) will not be on the geodesic.
     
  14. Nov 22, 2008 #13

    gel

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    The point of the measure theoretic argument is that it applies to any manifold, not just the torus which was used as an example. You could also use the Baire category theorem.

    and a geodesic will intersect a coordinate patch in countably many segments simply because any open subset of the real line is a union of countably many connected segments
     
  15. Nov 22, 2008 #14
    Hurkyl

    This quote is in response to my statement that the fractal dimension is always less than two in this particular example.

    Yes, the Peano curve passes through every point in the plane, but how do you get there? This is the limiting condition. How many iterations of the relevant (non intersecting) fractal generator does it take to get there from some arbitrary starting point? The fractal dimension is a real number (n), in this case on the interval D1 < n < D2 where D1 and D2 are whole number fractal dimensions. The upper bound of the interval is open. In what way can it be closed? The Peano curve (sometimes called a 'monster') is either a limit value or identical with all the points on a plane by definition.
     
    Last edited: Nov 22, 2008
  16. Nov 22, 2008 #15

    Hurkyl

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    My original comment was under the assumption you were talking about fractal dimensions of cuves in general.

    If not, then what exactly are you trying to say? The Peano curve, being space-filling, is obviously of fractal dimension 2. Each of the chosen approximations to it, being a polygonal chain, are obviously of fractal dimension 1.

    (This is not a problem, because fractal dimension is not continuous)
     
  17. Nov 23, 2008 #16
    I'm questioning Maze's assertion that the cardinality of RxR is equal to R by virtue of a space filling curve described in his first post where he walks arbitrarily close to every point on a surface. This suggests approximations to the Peano curve (P) not P itself. All approximations to P would have a fractal dimension less than 'two'. P itself has fractal dimension 'two' (but zero area). The cardinality of all approximations to P is the cardinality of R (or aleph 0).

    What is the cardinality of P itself?

    What is the cardinality of R^2?

    Why is the fractal dimension of approximations to P 'one' instead of a real number 'Dn' (actually a fraction) on the interval D1 < Dn < D2? Is P necessarily a polygonal chain?,
     
    Last edited: Nov 23, 2008
  18. Nov 23, 2008 #17
    I think you might getting a bit confused between the two curves discussed: the dense geodesic (which is not a space-filling curve and does not cover the entire surface, but is only dense), and the Peano curve (which does cover the entire unit square). The geodesic, as gel said, has zero area, while the Peano curve has area 1.

    The cardinality of the Peano curve P, being an image of [0, 1], has cardinality at most that of [0, 1], which is the same as that of R. But P is the entire unit square, so it has the cardinality of [0, 1]2, which is that of R2. Thus, |P| = |R| = |R2|.

    P is the limit of a (uniformly converging) sequence of polygonal chains, but is not itself a polygonal chain. (Indeed, if it were, it would be the union of countably many lines, but each line has measure 0, so the entire curve must have measure 0. But P has measure 1.)

    Minor note: You say "The cardinality of all approximations to P is the cardinality of R (or aleph 0)", but [itex]\aleph_0[/itex] is the cardinality of the natural numbers, which is different from the cardinality of R, which is [itex]2^{\aleph_0}[/itex].
     
  19. Nov 23, 2008 #18

    Hurkyl

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    How are you defining area? If you do it in what seems to me to be the most obvious way by taking the area of its image, P would have area equal to the square it fills.

    The thing a space-filling curve does is, by definition, give a surjective function
    [0,1] --> [0,1] x [0,1]​
    (at least in the case where we consider curves that merely fill the unit square), thus proving
    |RxR| <= |R|​
     
  20. Nov 23, 2008 #19

    Hurkyl

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    P is not. But the usually-used sequences of approximations to P are. For example, this image shows the first three approximations in one particular sequence.
     
    Last edited: Nov 23, 2008
  21. Nov 23, 2008 #20
    Thanks Hurkyl, adriank, maze et al. My remaining question is if |RxR|= |R|, then why confine P to the plane? Can |RxRx....xR| = |R|? If so, what do the transfinite numbers represent?
     
    Last edited: Nov 23, 2008
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