Understanding of the Regularity Condition in the definition of regular surfaces

Question

I am having some difficulty in understanding the meaning/motivation of the regularity condition in the definition of regular surfaces.

The definition (restricted to $\mathbb{R}^2$ and $\mathbb{R}^3$):

A subset $S \subset \mathbb{R}^3$ is a regular surface if for all $p \in S$, there exists a neighborhood $V$ in $\mathbb{R}^3$ and a map ${\textbf x}: U \to V \cap S$ of an open set $U \subset \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that:

1. ${\textbf x}$ is smooth.
2. ${\textbf x}$ is a homemorphism.
3. (Regularity Condition): for each $q \in U$, the differential $d{\textbf x}_q : \mathbb{R}^2 \to \mathbb{R}^3$ is one-to-one.

Would appreciate any kind of assistance with it.

Thanks.

Answer 1

The differential $ d{\textbf x}_q : \mathbb{R}^2 \to \mathbb{R}^3 $ is a presentation of the tangent map $T_q {\textbf x} : T_q \mathbb{R}^2 \to T_q \mathbb{R}^3$ with respect to the identifications $T_q \mathbb{R}^2 \cong \mathbb{R}^2$ and $T_q \mathbb{R}^3 \cong \mathbb{R}^3$, and also the map $T_q {\textbf x}$ is identified with $d{\textbf x}_q \in T^{*}_q \mathbb{R}^2 \otimes {\textbf x}^* T_q \mathbb{R}^3$. Usually, in elementary courses of differential geometry these identifications are made tacitly, but in W.P.A. Klingenberg, A course if differential geometry, Graduate texts in mathematics 51, Springer 1978, for instance, a bit of work is done to explain them carefully (without going into the tensor notation).

A fancier way of stating the regularity condition is to say that the map ${\textbf x} : U \to \mathbb{R}^3$ is an immersion, that is the map $T_q {\textbf x}$ is injective for each $p \in U$. This condition allows us to pullback the metric from $\mathbb{R}^3$ onto the surface along ${\textbf x}$. Usually, the metric in $\mathbb{R}^3$ is assumed to be Euclidean (that is the usual dot product). This way, the map ${\textbf x} : U \to \mathbb{R}^3$ is turned into an isometric immersion, and all the usual constructions become possible.

Equivalently, we may say that the vectors $T_q {\textbf x}(\tfrac{\partial}{\partial u^i})$, $i=1,2$, form a basis of the tangent space to the surface at each point $p \in U$, where $(u^i)$, $i=1,2$, are some (or, the standard) coordinates on $U \subseteq \mathbf{R}^2$.

Answer 2

Yuri Vyatkin already gave a very good answer. Let me just add a few things.

1. I don't particularly like the terminology of what you defined being a "regular surface". (Though this is something that you may have to live with.) The distinction I am trying to draw is the following:

   • A regular surface should be considered as a particular subset of $\mathbb{R}^3$, independent of the parametrisation.
   • A regular parametrisation of a regular surface is one that has injective tangent map.

   I draw this distinction because, consider now the case of a curve in $\mathbb{R}^2$: the curve defined by the graph $\{(x,x^2)\}$ is clearly a regular curve. It is the graph of a smooth function, and hence the parametrisation $x\mapsto (x,x^2)$ is a smooth homeomorphism with injective derivative (in this case: $\mathrm{d}f(x) = (1,2x) \neq 0$). But we can easily give an irregular parametrisation of the same curve: consider the map $t\mapsto (t^3, t^6)$. The image of this map is still the parabola, so is still the "regular curve" from before. But the parametrisation is not regular when $t = 0$.

   The same thing can be done for surfaces: given every regular surface it is possible to find a reparametrisation of it that your conditions (1) and (2) still hold, but the condition (3) fails. Therefore in terms of terminology I prefer to consider as regular surfaces the intrinsic property that the surface admits a regular parametrisation, while not requiring that the given parametrisation that I am looking at is regular.

2. To give another example: there exists curves which do not admit regular parametrisations. Consider the curve $\{(x,|x|)~|~x\in (-1,1)\}$ in $\mathbb{R}^2$. This curve has a corner at the origin and you will not consider it as a smooth curve. If you consider the map $t \mapsto \exp(-1/t^2)\cdot (\mathrm{sgn}(t),1)$ on $(-1,1)\setminus\{0\}$ extended to $0 \mapsto 0$, you easily see that the map is smooth and a homeomorphism. But the regularity condition fails.

   That is to say: if you drop the third condition, it is possible to have the image of $\mathbf{x}$ be some obviously non-smooth thing, even though it is the image of a smooth map.

3. The regularity condition allows you to establish the equivalence between the definition you listed, and the following alternative definition of a smooth surface in $\mathbb{R}^3$: "$S\subset \mathbb{R}^3$ is a regular surface if for every point $x\in S$ there exists an open neighborhood $U\subset \mathbb{R}^3$ and a smooth function $f:U\to \mathbb{R}$ such that

   • $x\in U$,
   • $U\cap S = f^{-1}(0)$,
   • $\mathrm{d}f \neq 0$ along $U\cap S$."

   The equivalence of the two definitions is an immediate consequence of the implicit function theorem.