Find a surface with the Gaussian curvature is zero at only one point but nonzero at others.

Question

Are there any smooth surface $S$ such that there is only one point $p \in S$ with $K(p) = 0$ but $K(q) \neq 0$ for all $q \in S$ with $q \neq p$, where $K$ is the Gaussian Curvature.

Answer 1

With Ted Shifrin's answer, I tried to give an example. Consider a graph of $h$ : $$ h(x,y)=(x^2+ y^2)^2 $$

So $$ h_{xx}= 12x^2+4y^2,\ h_{yy}=12y^2+4x^2,\ h_{xy}=8xy $$

$$K=\frac{ h_{xx} h_{yy}- h_{xy}^2}{(1+ h_x^2+ h_y^2)^2} =\frac{48 (x^2+y^2)^2}{ (1+16(x^2+y^2)^2 )^2 }$$

Answer 2

Sure. Try a surface of revolution obtained by rotating a curve with a point of zero curvature (not necessarily an inflection point) on the axis of rotation.

Answer 3

There can be many. If the handy surfaces of revolution are taken, locally straight and flat/inflection point should be on axis of rotation as Gauss curvature should vanish.

For example take Cornu's Spirals. $ \kappa = s$ or $ \kappa = s^2 $ with the inflection / flat point at origin. Rotate it about an arbitrary axis in $\mathbb R^3 $ through the origin. The surface of revolution so formed has $K=0 $ at origin.

One such simple rotationally symmetric example is:

$$ z(x,y) =(x^2 + y^2)^{\frac32}.$$

Finding a nice non-axisymmetric example as an answer to your question would, I believe be interesting.