[Math] Gaussian curvature

calculusdifferential-geometry

Let us take a surface $z =x^3-3x^2+y^2$. If you calculate the critical points on the surface it will come as $(0,0)$ and $(2,0)$. Moreover one can find the
local behavior of the critical points. It will turn out that $(0,0)$ corresponds to a saddle point, and $(2,0)$ corresponds to local min. Up to here, I am ok.

I take this surface as a Monge Patch $f(u,v) = (u,v,u^3-3u^2+v^2)$ and write the first fundamental form, second fundamental form and compute Gaussian curvature via
$K = (LN-M^2)/(EG-F^2)$ (standard notation can found in Pressley's "Elementary Differential Geometry").

The problem comes here:

$K(0,0)=-12<0$ saddle ok, BUT

$K(2,0)=12>0$ !

I get confused here. **If it is a local min, then Gaussian curvature is positive?

Why is this notion bothering? If a surface has local min, isn't it true that at every local min you have negative Gaussian curvature?

Can anyone please explain with some some examples and references?

Best Answer

For a surface of the form $z = f(x,y)$, the Gaussian curvature is

$$K = \frac{f_{xx}f_{yy} - f_{xy}^2}{(1 + f_x^2 + f_y^2)^2}.$$

(This is Exercise 8.1.1 in the second edition of Pressley's book.)

By the Second Derivative Test, we see that:

  • If $K(x,y) < 0$, then $(x,y)$ is a saddle point.
  • If $K(x,y) > 0$, then $(x,y)$ is a local maximum or a local minimum (depending on whether $f_{xx}(x,y)$ is negative or positive).

Example: Consider the elliptic paraboloids $z = x^2 + y^2$ and $z = -x^2 - y^2$. Both have positive Gaussian curvature $K(0,0) = 4 > 0$ at the origin. However, the first one has a minimum there, and the second has a maximum.

Example: The hyperbolic paraboloid $z = x^2 - y^2$ has negative Gaussian curvature $K(0,0) = -4 < 0$ at the origin, and has a saddle point there.

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