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Let $S = \{(u\cos v, u\sin v, v): 0<v<2\pi, -\infty<u<+\infty\}.$ Show that S is a regular surface. $\newcommand{\D}{\mathrm{d}}$

I'm using DoCarmo's book, Differential Geometry of Curves and Surfaces. For $S \subset \mathbb{R}^3$ to be a regular surface, we would need for each $p \in S$, there exists a neighborhood $V$ in $\mathbb{R}^3$ and a map $\mathbb{x}: \to V \cap S$ of an open set $U \subset \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that $\mathbb{x}$ is differentiable (i.e., its component functions have continuous partial derivatives of all orders in $U$), $\mathbb{x}$ is a homeomorphism (i.e., it has a continuous inverse), and for each $q \in U$, the differential $\mathrm{d}\mathbb{x}_q: \mathbb{R}^2 \to \mathbb{R}^3$ is one-to-one.

Is this definition equivalent to showing that the vectors $\mathbb{x}_u$ and $\mathbb{x}_v$, where $\mathbb{x}(u,v) = (u\cos v, u \sin v, v)$ are linearly independent? If so, isn't that quite trivial?

If not, can someone show me why not, and explain what we can do?

Pom pom
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  • It is indeed pretty trivial for this particular surface. As you probably suspect, you just note that $\sin v, \cos v$ are never simultaneously $0$ (easy) and then check the last coordinate of each of the derivative vectors to see they're independent. – Adam Hughes Feb 27 '15 at 19:13

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Showing that $\mathbf x_u$ and $\mathbf x_v$ are linearly independent everywhere is not quite enough; if you look closely at the definition of regular surface, you'll see that you also have to show that $\mathbf x$ is a homeomorphism onto its image. (In this case, for all points in $S$ we can take $U=\mathbb R^2$ and $V=\mathbb R^3$, so $V\cap S = S$.)

To show that $\mathbb x$ is a homeomorphism onto its image, you can construct a continuous inverse explicitly. Define $\phi\colon S\to \mathbb R^2$ by $$ \phi(x,y,z) = \begin{cases} \left(\dfrac{x}{\cos z},z\right), & \cos z \ne 0,\\[2ex] \left(\dfrac{y}{\sin z},z\right), & \sin z \ne 0. \end{cases} $$ The two formulas for $\phi$ agree where they overlap, and at every point of $S$ either $\cos z$ or $\sin z$ is nonzero, so $\phi$ is globally defined and continuous. You can show by computation that $\phi\circ \mathbf x(u,v) = (u,v)$ for all $(u,v)\in\mathbb R^2$, and $\mathbf x\circ\phi(x,y,z) = (x,y,z)$ for all $(x,y,z)\in S$. Thus $\phi$ is a continuous inverse for $\mathbf x$, hence $\mathbf x$ is a homeomorphism onto its image.

Jack Lee
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