Results 1 to 5 of 5

Math Help - asymptotic curve on a complete surface.

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    131

    asymptotic curve on a complete surface.

    Hello,

    We have a 2-dimensional Riemannian Manifold immersed in \mathbb{R}^3, which is complete and has constant negative curvature.
    Then the statement is, that for every point p\in M, there is an asymptotic curve c:\mathbb{R}\rightarrow M, parametrized by arclength, s.t. c(0)=p.

    I don't know how i can prove this. I mean the Surface M is complete, that is we can extend every geodesic c: (-a,a)\rightarrow M into a geodesic on the whole real line c:\mathbb{R}\rightarrow M.
    And we know, locally there exists always a geodesic. But a geodesic is not an asymptotic curve. Since for an asymptotic curve we have that c''(t)\in T_p M. But for a geodesic we have that c''(t) is perpendicular to T_p M.

    So we have to search for another idea:
    I could show, that for any point p, we have locally an asymptotic curve, parametrized by arclength.

    But i have no idea, why we can extend such an asymptotic curve to the real line.....
    I hope you can help me with this problem.

    Regards
    Last edited by Sogan; December 26th 2011 at 12:45 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23

    Re: asymptotic curve on a complete surface.

    Didn't look in detail yet but your first statement if false since according to Hilbert theorem, there is no complete regular surface of constant negative gaussian curvature immersed in R^3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    131

    Re: asymptotic curve on a complete surface.

    Yes you are right. I want to prove this theorem. And the author first assumes, that there is such an immersion of M, in the proof. (His strategy is proof by contradiction)
    Then he says, that there must be such an asymptotic curve to each point.

    Maybe we can do it like this:
    We take an local asymptotic curve c_1 : [-a,a]\rightarrow M with c_1 (0)=p. (as restriction of a regular curve on (-a-\epsilon , a+\epsilon)

    Then we take the point c(a)\in M and do the same process again with some function c_2 : [-b,b]\rightarrow M.
    Now we define: c : [-a,a+b]\rightarrow M by c(x)=c_2 (x-a) if x>=b and c(x):=c_1 (x) otherwise.

    and so on.... Is this the right idea?

    Regards
    Last edited by Sogan; December 27th 2011 at 12:27 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23

    Re: asymptotic curve on a complete surface.

    Since the Gaussian curvature is negative, the two principal curvatures k1 and k2, are of different signs. Let's say k1 > 0 and k2 < 0, then according to Euler's theorem, k = k_1 cos^2\theta + k_2 sin^2\theta, this is a continuous function of \theta with k(0)=k_1 > 0 and k(\pi / 2)=k_2 < 0 , so there must be a value of \theta \in (0, \pi/2) so that k=0. This is the asymptotic direction.

    So this defines a unit vector field X on the surface. At each point p, X(p) is the unit vector at the asymptotic direction. Since k_1, k_2 and the principal directions are smooth on the surface, the above construction of X is smooth on the surface. Then at each p, there is locally a unique integral curve of X. Since M is complete, we can extend this integral curve indefinitely.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    Posts
    131

    Re: asymptotic curve on a complete surface.

    Hello,

    thank you so much for your help. But i do not unterstand your last point.

    "Since M is complete, we can extend this integral curve indefinitely."

    is every integral curve extendable to the real line, if the surface is complete?
    I mean the definition of complete surfaces just says, that every geodesic is extendable in such a way.

    Regards
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Showing that a curve is normal to a surface
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 6th 2010, 09:03 AM
  2. [SOLVED] Volume generated by curve in one complete rotation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 13th 2010, 06:26 PM
  3. [SOLVED] Finding the Surface Area of the curve?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 26th 2010, 01:34 AM
  4. Replies: 1
    Last Post: May 13th 2010, 06:46 AM
  5. area of surface by rotating the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2010, 05:36 PM

Search Tags


/mathhelpforum @mathhelpforum