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Math Help - Hyperbolic point

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Hyperbolic point

    Let  f : U \to \mathbb{R}^3 be a surface element. Suppose that the image of  f lies in the region  \{(x, y, z) | z > 0 \} , and the principal curvatures of  f at  f(0) satisfy  \kappa_1 \kappa_2 < 0 . Show the tangent plane of  f at that point is not parallel to the  xy axis.
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  2. #2
    Super Member Rebesques's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Let  f : U \to \mathbb{R}^3 be a surface element. Suppose that the image of  f lies in the region  \{(x, y, z) | z > 0 \} , and the principal curvatures of  f at  f(0) satisfy  \kappa_1 \kappa_2 < 0 . Show the tangent plane of  f at that point is not parallel to the  xy axis.

    Clearly f(0)=0 is minimum and unique.
    We use Taylor's expansion in a neighbourhood B(0,r) of 0, to obtain
    f(x,y)=1/2(f_{xx}(0)x^2+2f_{xy}(0)xy+f_{yy}y^2)+0(x^2+y^2)
    or
    f(x,y)=\langle (x,y),D^2f(0)(x,y)\rangle+0(x^2+y^2)=Q(x,y)+0(x^2+  y^2)
    with the obvious notation.
    We also have (k_1k_2)(0)=(f_{xx}f_{yy}-f_{xy}^2)(0)={\rm det}D^2f(0)<0. One of the principal curvatures at 0 must be negative, say k_1<0. Since this is an eigenvalue of D^2f(0)
    there is a direction w on the plane, of length |w|<r, such that Q(w)=k_1|w|^2<0. Along this direction, we will have
    f(sw)=s^2|w|^2k_1+0(s^2|w|^2)=s^2(|w|^2k_1+sC(s)), \ s>0, \ C(s) a smooth function.
    Therefore, there is an s_0>0 such that f(s_0w)<0, a contradiction.
    Last edited by Rebesques; December 29th 2010 at 03:23 PM. Reason: "a c ontradiction". sounds like a new meme
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