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I am trying to answer the following problem:

  • Show that the two-sheeted cone, with its vertex at the origin, that is, the set $\{(x,y,z)∈R^3 :x^2+y^2−z^2=0\},$ is not a regular surface.

I am trying the following: I rewrite it as $(x,y,\pm \sqrt{x^2+y^2})$ and then compute the differential:

$$\begin{pmatrix} {1}&{0}\\ {0}&{1}\\ {\frac{x}{\sqrt{x^2+y^2}}}&{\frac{y}{\sqrt{x^2+y^2}}} \end{pmatrix}$$

In the definition of regular surface, we ask that: For each $q$, the differential $d\textbf{x}_q$ is one to one.

I think I can use that. The trouble for me is: The differential does not exist at $(0,0,0)$, does this also means that the surface is non regular?

Red Banana
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1 Answers1

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I'm guessing you're reading Do Carmo's book on curves and surfaces. Example $5$ in the preceding section, showing that the one-sheeted cone is not a regular surface, is the example to look at here.

Using Proposition $3$, if the two sheeted cone (call it $C_1$) were a regular surface, then in a neighbourhood of $(0,0,0) \in C_1$ it would be the graph of a differentiable function having one of three forms: $y=h(x,z)$, $x=g(y,z)$ or $z=f(x,y)$. As in the case of the one-sheeted cone, the first two forms can be dismissed, since the projections of $C_1$ over the $xz$ and $yz$ planes are not one-to-one. And in this case (the case of the two-sheeted cone) we can use the same reasoning with regard to the other form $z=f(x,y).$ If $\pi: \mathbb R^3 \to \left\{(x,y,z): z=0 \right\}$ is the projection onto the $xy$ plane, then clearly we have $\pi(x,y,\sqrt{x^2+y^2}) = \pi(x,y,-\sqrt{x^2+y^2}),$ ie. the projection onto the $xy$ plane is also not one-to-one.

Elliot Herrington
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