Are there non-parametrizable surfaces?

Question

Are there any surfaces that cannot be parameterized? (I'm in multivariable calc and we were talking about parametrizing surfaces for Stokes' Theorem so I was wondering if there are any surfaces that cannot be parameterized.) Please don't make the answer overly complicated. Thanks in advance!

Answer 1

One possible definition of a surface is that each point has a neighborhood $N$ that "looks" two-dimensional. Rigorously, this means there is an open set $U$ in $R^2$ and a smooth, one-to-one, and onto function $f$ that maps $U\mapsto N$. If this is how you think of a surface, then every surface is, by definition, parametrizable. This certainly does not necessarily yield a simple, algebraic form of the parametrization, however.

Answer 2

The notion of a parameterization of a surface $S$ in $R^3$ is notoriously sloppy. Here are some definitions and it is pretty much up to you (or your professor) which one you accept as valid. In all these definitions, smooth means _infinitely differentiable (one can require less differentiability).

1. Local parameterization, which is what Mark McClure has in mind in his answer: For every point $p\in S$ one chooses a local chart $f: U\to S$, where $U\subset R^2$ is an open subset, say, the open unit disk centered at $0$, $f$ is a smooth one-to-one mapping whose derivative at each point has rank 2, such that $f(0)=p$ (the image of $f$ is a neighborhood $N$ of $p$ in $S$).

2. Global parameterization: A smooth mapping $f: U\to S$ which is onto and whose derivative has rank 2 at each point of $U$; here $U$ is again an open subset of $R^2$.

Note that in Definition 2 the map $f$ is by no means one-to-one, points in $S$ are allowed to have infinitely many preimages under $f$. This is not good if you are trying to define, say, integrals, over $S$ of some functions. The last definition is meant to handle this problem:

3. $f: U\to S$ is as in 2, but, in addition, one is given a domain $D\subset U$ with piecewise-smooth boundary, such that $f(D)=S$ and $f$ is one-to-one on the interior of $D$, i.e., on $D$ with the boundary curve removed.

It is a nice exercise to see that 3->2->1, meaning that a parameterization in the sense 2 yields a parameterization in the sense 1, etc.

Each (smooth) surface admits parameterizations in the sense 3 (the strongest), but it requires some work to prove it and is, typically, not done in multivariable calculus classes.