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Definition of arc length on manifolds without parametrization

  1. Jun 3, 2008 #1

    mma

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    Curves are functions from an interval of the real numbers to a differentiable manifold.
    Given a metric on the manifold, arc length is a property of the image of the curves, not of the curves itself. In other word, it is independent of the parametrization of the curve. In the case of the Euclidean space, arc length can be defined without using any parametrization of the curve, namely by the limit of the length of the approximating polygons. My question is whether could we give similar definition of the arc length in the case of manifolds? The standard definition is [tex]l(\gamma)=\int_{0}^{1} {|\dot\gamma(t)|} dt\ ,[/tex] and unlike in the case of the definition with polygons, this definition involves the parametrization of the curve, in spite that all parametrizations give the same result. How could we eliminate the unnecessary parametrization from the definition?
     
    Last edited: Jun 3, 2008
  2. jcsd
  3. Jun 3, 2008 #2
    But how do you define a "curve" that doesn't have a parametrization? For instance, if you say a "curve" is a connected compact 1-dimensional submanifold with boundary, it's easy to show by a gluing argument that this "curve" is just a piecewise regular curve in the ordinary sense, and you can use your formula to get the arc length.
     
    Last edited: Jun 3, 2008
  4. Jun 4, 2008 #3

    mma

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    I say an analogy of what I mean.
    The definition of the tangent vector of a curve also uses the parametrization of the curve (and what is more, depends on in). The tangent space over a point of the manifold consinsts of the tangent vectors of the curves at this point. Even so, tangent space can be defined without mentioning curves: it is the space of derivations. Derivation is a linear and Leibniz-rule compliant real-valued map from the space of the smooth functions around the point.
     
    Last edited: Jun 4, 2008
  5. Jun 4, 2008 #4
    the answer to your question is "no, there is no canonical method of defining the arc length of a curve on a general manifold." The reason is quite simple: defining arc length on a curve amounts to locally defining a (not necessarily Euclidean) arc length of a curve in R^n, i.e., on the coordinate patches of the manifold. But the local coordinate systems of the manifold are related to each other by diffeomorphisms, which in general will not preserve a distance-function or arc length function.

    In order for your specific scheme to work, the transition mappings of the local coordinate systems of the manifold would have to be elements of SO(n,R), i.e., it would be an affine manifold. And, in general, for any method of assigning arc length to curves, the transition mappings would have to somehow preserve this method.

    I imagine, though, that there are plenty of particular manifolds with particular diffeomorphic structures on which an arc length function can be defined parametrization-free on its smooth curves.
     
  6. Jun 4, 2008 #5
    Why not? The coordinate charts don't have to be isometric to R^n with the standard inner product.

    It seems like you can generalize the argument given by the OP for standard Euclidean space. Obviously you need a metric (but Euclidean space has one too). Then cover your curve in finitely many totally normal neighborhoods of radius < epsilon, and construct a piecewise-geodesic curve such that each piece is a minimizing geodesic of arc length < epsilon, and each point between two pieces lies on your original curve. Take epsilon -> 0.
     
  7. Jun 4, 2008 #6
    I somewhat misunderstood the original question. I didn't realize that he was assuming a fixed metric, in which case, your method works fine. Of course, in order for this to be easier than the standard parametrization method of calculating arc length, we need a manifold for which it is relatively easy to calculate the distance between two given points, e.g., the hyperbolic plane.

    Off-hand, this is a nice way to intuitively envision arc length.
     
  8. Jun 4, 2008 #7

    mma

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    I am afraid that I don't understand this. It doesn't look like a definition of arc length.
     
  9. Jun 5, 2008 #8
    It essentially amounts to the polygonal situation on the plane that you described.

    You're given a smooth curve whose arc length you want to determine from point a to point b. For every positive integer n, choose n points on the curve between points a and b such that any two consecutive points on the curve are connected by a unique shortest-length geodesic (we are assuming that a and b are among these points). Then connect the points by those geodesics and sum up their lengths. If we do this so that the longest length of the geodesics for each partitioning corresponding to n goes to zero as n goes to infinity, then the length of the arc will be the limit of the sums of the lengths of the geodesics as n goes to infinity.

    Of course, even in the Euclidean case, this only works with a smooth curve (at least C^1), since we all know that the coast of England is infinitely long.
     
  10. Jun 5, 2008 #9

    mma

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    OK, but then how is the length of a geodesic defined?
     
  11. Jun 6, 2008 #10
    Well, actually, through the use of parametrizations. I'm afraid there's just no getting around it. See Do Carmo's Riemannian Geometry for details.
     
  12. Jun 7, 2008 #11

    mma

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    It's clear that parametrization is superfluous in the definition because the value of the arc length does not depend on it. But I don't know how can we get rid of it.:(
     
  13. Jun 7, 2008 #12
    On a Lie group can you use the inverse of the exponential map to locally compute the arc length of geodesics?
     
  14. Jun 8, 2008 #13

    mma

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    Do you mean of curves t->exp(tv) where |v|=1?. On the one hand, such curves are only very special ones among all curves on the Lie-group, and on the other hand this works only for Lie-groups. Otherwise good idea.
     
  15. Jun 8, 2008 #14
    Yes, they are the geodesics, but then you can find the arc length of any arbitrary curve by using the argument above.

    You're right about it only working on Lie groups, of course.
     
  16. Jun 8, 2008 #15

    mma

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    I am not too familiar in Lie groups, so I don't know how are geodesics deined on Lie groups. According your remark it seems to me that all t-> exp(tv)g is a geodesic where g is an arbitrary elemt of the group (perhaps this is the definition itself), because the above mentioned t-> exp(tv) curves are only those curves that pass across the unit element of the group; and of course we need geodesics across arbitrary points in arbitrary direction if we want do define the arc length of any curves by them. And it means of course that Lie groups are always equipped with this unique metric determined by the group itself. Am I right?
     
  17. Jun 8, 2008 #16
    Actually, that's not very clear at all. Just because the arc length doesn't depend on it, does not mean that it's superfluous. There might be other ways to define arc length, but in order to find them you should first understand why parametrization is used in the definition in the first place:

    If you're given two points and a smooth curve connecting them, this is an immersed submanifold of the overall manifold. A parametrization of this curve can be then thought of as a generalized coordinate patch of this submanifold. The Riemannian metric on the ambient space induces a Riemannian metric onto the curve, which in turn induces a "volume form" on the curve. Arc length is then the measure of the entire curve with respect to this volume form, which is in fact a 1-form on the curve. If you can find some parametrization-free way to redefine the integral of this 1-form along the curve, then you've done it.

    The idea of splitting the curve up into partitions and taking the limit as the number of points go to infinity and the distances between these points go to 0, etc., is essentially calculating some particular Riemann sums of the integral described above. Conceptually a nice idea, but calculationwise quite prohibitive in specific cases. For example, try using this method to calculate the length of a lemniscate on the plane.
     
  18. Jun 9, 2008 #17

    mma

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    Yes, integration of forms on manifolds also suffers from the same thing. Perhaps it is unavoidable to use sometimes some inert objects in the definitions.
     
  19. Jun 10, 2008 #18

    gel

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    How about the following. If C(M) is the space of smooth functions M->R, then you can define a distance between points as
    [tex]
    d(P,Q)=\sup\left\{ |f(P)-f(Q)| : f\in C(M), \Vert \nabla f \Vert \le 1\right\}
    [/tex]

    You can then define the length of a curve (even a non-smooth one) in the same way as for Rn. I've seen this idea used in noncommutative geometry, where you don't have coordinate charts.
     
    Last edited: Jun 10, 2008
  20. Jun 12, 2008 #19

    mma

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    What does here [tex] \Vert \nabla f \Vert [/tex] mean?
     
  21. Jun 15, 2008 #20

    gel

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    By [itex]\nabla f[/itex] I meant the differential of f, maybe df would be better notation. and ||df|| is the norm under the metric, [itex]\sqrt{g(df,df)}[/itex].
     
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