Hyperbolic point

**chiph588@** · October 6th 2009, 02:50 PM

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.

**Rebesques** · December 29th 2010, 08:56 AM

Originally Posted by chiph588@

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.

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