# Hyperbolic point

• Oct 6th 2009, 02:50 PM
chiph588@
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.
• Dec 29th 2010, 08:56 AM
Rebesques
Quote:

Originally Posted by chiph588@
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.