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Thread: Hyperbolic point

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

    Let $\displaystyle f : U \to \mathbb{R}^3 $ be a surface element. Suppose that the image of $\displaystyle f $ lies in the region $\displaystyle \{(x, y, z) | z > 0 \} $, and the principal curvatures of $\displaystyle f $ at $\displaystyle f(0) $ satisfy $\displaystyle \kappa_1 \kappa_2 < 0 $. Show the tangent plane of $\displaystyle f $ at that point is not parallel to the $\displaystyle 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 $\displaystyle f : U \to \mathbb{R}^3 $ be a surface element. Suppose that the image of $\displaystyle f $ lies in the region $\displaystyle \{(x, y, z) | z > 0 \} $, and the principal curvatures of $\displaystyle f $ at $\displaystyle f(0) $ satisfy $\displaystyle \kappa_1 \kappa_2 < 0 $. Show the tangent plane of $\displaystyle f $ at that point is not parallel to the $\displaystyle xy $ axis.

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