Let $C$ be a figure $‘‘8"$ in the $xy$ plane and let $S$ be the cylinder surface over $C$; that is, $$S=\{(x,y,z)\in\mathbb{R^3}:(x,y) \in C \}$$ Is the set $S$ a regular surface?
I know that the answer is no and the reason is because this surface have self-intersections in the levels of the intersection point of $‘‘8"$, but how can I prove this more mathematically?
Best Answer
From the implicit function theorem it follows that if $S$ is a regular surface and if $p \in S$ then there is a well-defined 2-dimensional tangent plane $T_p S$ with the property that for any differentiable curve $\gamma$ whose image is in $S$ and any $t$ such that $\gamma(t)=p$ then the vector $D\gamma(t)$ is parallel to the plane $T_p S$. In particular the maximum number of linearly independent tangent vectors at $p$ is 2.
But for your figure 8 curve $C$ in the $xy$-plane, letting $X \in C$ be the crossing point, at the point $X$ the curve $C$ has a linearly independent pair of tangent vectors that I'll denote $V_1,V_2$. And so for any point $p$ in $(x,y,z)$-space lying over $X$ you can show that $S$ has three linearly independent tangent vectors at $p$:
This contradiction shows that $S$ cannot be a regular surface.