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Premise: I'm a bit unsure about the terminology most frequently used, so I'll try to be as clear as possible, please tell me if something isn't clear.

Talking about regular smooth manifolds, by which I mean function $P: E \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $P_u \times P_v \not\equiv 0 \; \forall (u,v)\in E$ (by "$\times$" I mean vector product) and $P \in C^{\infty}(E)$, we can define something which I'd like to call "cartesian" manifolds, manifolds with parametrization that looks like $(u,v,f(u,v))$, which coincide with the graph of some function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$.

My question is: is it true that if any umbilical point exists on the manifold, then it must be a maximum or a minimum of the function $f$? If this condition is too strong, would it suffice for the point to be a stationary point (namely, the gradient of $f$ vanishes)?

As for the viceversa, would it be sufficient for a point to be umbilical on a cartesian manifold that the gradient of $f$ vanished in that point, and that such point was not flat? (A flat point is a point where the matrix of the second fundamental form is the zero matrix)

My question arises in the context of my geometry class (where we covered both algebraic and differential geometry from an introductory point of view)...my professor probably said something about this (he showed us an example of a elliptic paraboloid which has its "vertex" as a umbilical point), but I can't remember well. Thanks in advance for any insight.

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up vote 1 down vote accepted

Generically, an umbilic is an isolated point on the surface $S$ where $K=H^2$, $K$ and $H$ being the Gaussian and the mean curvature, respectively. What this implies for the partial derivatives of your height function $f$ you have to find out by yourself by doing the calculations.

In any case, for an umbilic it is not necessary that the tangent plane to $S$ is horizontal there. See it this way: If $p$ is an umbilic on an arbitrary surface $S\subset{\mathbb R}^3$ and if the tangent plane to $S$ at $p$ is not parallel to the $z$-axis then you can always parametrize $S$ in a neighbourhood of $p$ in the form $(u,v)\to(u,v, f(u,v))$ for a suitable $f$.

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