Regular Surfaces and Parametrizations

The regular-surface / chart framework is the direct prototype for the definition of a smooth $n$-manifold used throughout modern geometry and topology, where "surface patches" are replaced by abstract coordinate charts glued by smooth transition maps — precisely the pattern proved for surfaces in the Corollary above. In general relativity, spacetime is modeled as a 4-dimensional smooth manifold covered by such charts, with the transition-map smoothness condition guaranteeing that physical laws written in local coordinates are consistent across observers. In computer graphics and computer-aided design, surface patches (NURBS, Bézier patches) are literally parametrizations $\mathbf{x}(u,v)$ in the sense defined here, and mesh-generation algorithms must respect regularity (nonvanishing normal $\mathbf{x}_u\times\mathbf{x}_v$) to avoid degenerate, folded, or self-intersecting rendered geometry. In robotics and computer vision, configuration spaces and image manifolds are frequently modeled as regular surfaces or higher-dimensional manifolds, with local charts used for optimization (gradient descent constrained to a surface, e.g., pose estimation on $SO(3)$). The graph criterion proved above underlies numerical implicit-surface and level-set methods, where a surface is represented as $f^{-1}(0)$ for a scalar field $f$, and regularity of the surface corresponds exactly to $0$ being a regular value of $f$, i.e., $\nabla f \ne 0$ on the level set — the same criterion used to build charts locally as graphs, exactly as in this article's proof technique. This entire framework is also the immediate launching point for the First Fundamental Form, since once a chart $\mathbf{x}(u,v)$ is fixed, lengths and angles on $S$ are computed by pulling back the ambient dot product to $U$ via $\mathbf{x}_u,\mathbf{x}_v$.

Ask the tutor: Can you show me step by step how to build explicit local charts for a level set $f^{-1}(0)$ using the Implicit Function Theorem, the way we built graph charts here?

Think of the plane $U$ as a flat sheet of rubber, and $\mathbf{x}$ as an instruction for bending it into space. Condition 1 says the bending is smooth — no kinks. Condition 2 says the bending does not crush any direction: at each point of the sheet, two independent tangent directions $u$ and $v$ map to two independent tangent vectors $\mathbf{x}_u, \mathbf{x}_v$ in $\mathbb{R}^3$, so their cross product is nonzero and spans a genuine tangent plane, rather than the bent sheet locally collapsing to a curve or a point. Condition 3 is the subtlest and the one most often shortchanged in a first pass. It is not enough for $\mathbf{x}$ to be an injective, smooth, regular map — you must also forbid the image from "coming back and touching itself" arbitrarily closely in a way invisible in the parameter domain. The homeomorphism condition guarantees that convergence of points on the surface corresponds exactly to convergence of parameters, which is precisely what rules out a figure-eight curve dragged into a self-intersecting sheet, or a line densely wound on a torus. The role of multiple charts is that no single parametrization can, in general, cover all of $S$ while staying a homeomorphism onto an open subset — a sphere parametrized by spherical coordinates degenerates at the poles, so we cover $S$ with an atlas of overlapping charts, exactly as a globe is drawn with several map sheets. Where two charts overlap, the change of coordinates map between them turns out to always be a diffeomorphism (proved below), which is the fact that makes it possible to define smooth functions, tangent vectors, and curvature on $S$ without reference to any particular chart.

Example 1: The unit sphere $S^2$, via six graph charts and the stereographic atlas. Let $S^2 = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 = 1\}$. We verify it is a regular surface using graph charts, which is the most robust technique in practice, guaranteed to work by the theorem above. Define

smooth on the open disk $A$ (the expression under the root is strictly positive there, and $\sqrt{\cdot}$ is smooth on $(0,\infty)$). Let

parametrize the open upper hemisphere $\{z>0\}$. By the theorem, this graph chart is automatically a valid parametrization satisfying all three conditions, with image exactly $S^2 \cap \{z > 0\}$, open in $S^2$ (it equals $S^2$ intersected with the open half-space $\{z>0\}$ of $\mathbb{R}^3$). Similarly define five more charts using $g(u,v) = -\sqrt{1-u^2-v^2}$ for the lower hemisphere $\{z<0\}$, and the analogous formulas solving for $y$ in terms of $(x,z)$ (covering $\{y>0\}$, $\{y<0\}$) and for $x$ in terms of $(y,z)$ (covering $\{x>0\}$, $\{x<0\}$). Every point of $S^2$ has at least one nonzero coordinate, hence lies in at least one of these six open hemispheres, so the six charts cover $S^2$. Thus $S^2$ is a regular surface. To see the overlap and transition maps concretely, take $\mathbf{x}_1(u,v)=(u,v,\sqrt{1-u^2-v^2})$ on the disk $A$, and the "$x>0$" chart

which parametrizes the open hemisphere $\{x>0\}$ by $(y,z)=(s,t)$. Their images overlap on

an open "octant" of the sphere. For a point on $W$, $\mathbf{x}_1^{-1}(x,y,z) = (x,y)$ and $\mathbf{x}_2^{-1}(x,y,z) = (y,z)$. The transition map $h = \mathbf{x}_1^{-1}\circ \mathbf{x}_2$ sends $(s,t) \mapsto \big(y,z\big)$ where $(x,y,z) = \mathbf{x}_2(s,t) = (\sqrt{1-s^2-t^2}, s, t)$, i.e.

Here $h$ is literally the identity on the parameter overlap $\{(s,t): s^2+t^2<1, \sqrt{1-s^2-t^2}>0\}$, manifestly a diffeomorphism, consistent with Corollary: Change of coordinates is a diffeomorphism. (In general transition maps are not this trivial; this example is unusually clean because both charts use two of the same ambient coordinates.) As a second atlas for the same sphere, minus one point, consider stereographic projection from the north pole $N=(0,0,1)$:

which is a bijection from $S^2 \setminus \{N\}$ onto $\mathbb{R}^2$, with inverse

One checks directly $x^2+y^2+z^2=1$ by summing the numerators' squares over the common denominator squared, confirming the image lies in $S^2$; it is a standard exercise in Multivariable differentiation to compute $\mathbf{x}_{N,u}\times \mathbf{x}_{N,v} \ne 0$ everywhere on $\mathbb{R}^2$, and that $\mathbf{x}_N$ is a smooth homeomorphism onto $S^2\setminus\{N\}$. Together with the analogous chart $\mathbf{x}_S$ from the south pole, these two charts already cover all of $S^2 \setminus \{N\}$ and $S^2\setminus\{S\}$ respectively, hence all of $S^2$ with just two charts — the minimum possible, since $S^2$ is compact and cannot be covered by a single chart (a single chart's image would be homeomorphic to an open planar set, but $S^2$ is compact and no compact set is homeomorphic to a nonempty open subset of $\mathbb{R}^2$).

Example 2: The torus of revolution. Fix radii $0 < r < R$ (the tube radius $r$ strictly less than the axis radius $R$, an asymmetric, non-degenerate configuration — this is what prevents self-intersection). Define

Take $U = (0,2\pi) \times (0,2\pi)$ and

Smoothness. Each component is a finite composition of sine, cosine, and polynomial functions, all $C^\infty$, so $\mathbf{x}$ is $C^\infty$ on $U$. Regularity. Compute

Their cross product has third component

A short computation of the other two components gives, after factoring,

Its length is $r(R+r\cos u)$ (since $(\cos u\cos v,\cos u\sin v,\sin u)$ is a unit vector, as $\cos^2u\cos^2v+\cos^2u\sin^2v+\sin^2u=\cos^2u+\sin^2u=1$), and since $r>0$ and $R+r\cos u \ge R - r > 0$ (using $r<R$, the asymmetric hypothesis), this length is strictly positive everywhere. Hence $\mathbf{x}_u \times \mathbf{x}_v \neq 0$ on all of $U$. Homeomorphism onto image. Injectivity: suppose $\mathbf{x}(u_1,v_1) = \mathbf{x}(u_2,v_2)$ with $u_i,v_i \in (0,2\pi)$. Writing $\rho_i = R+r\cos u_i > 0$ and $z_i = r\sin u_i$, the point's distance from the $z$-axis is $\rho_i$ and its height is $z_i$; matching $(x,y,z)$ forces $\rho_1=\rho_2$, $z_1=z_2$, hence $\cos u_1=\cos u_2$ and $\sin u_1=\sin u_2$ (from $\rho_i = R+r\cos u_i$, $z_i=r\sin u_i$, using $r\ne 0$), so $u_1=u_2 \pmod {2\pi}$; since both lie in $(0,2\pi)$, $u_1=u_2$. Then matching $x=\rho\cos v, y=\rho\sin v$ with $\rho=\rho_1=\rho_2>0$ forces $\cos v_1=\cos v_2,\sin v_1=\sin v_2$, so likewise $v_1=v_2$. Continuity of $\mathbf{x}$ is clear (smooth $\Rightarrow$ continuous). For continuity of the inverse, one recovers $u,v$ from $(x,y,z)$ by continuous formulas away from the branch ambiguities of $\arctan$: e.g. $v = \operatorname{atan2}(y,x)$ (continuous on the image, since $\rho>0$ always, so $(x,y)\ne(0,0)$), and then $u$ is recovered continuously from $z = r\sin u$ and $\rho - R = r\cos u$ via $u=\operatorname{atan2}(z/r,(\rho-R)/r)$. These are continuous on $\mathbf{x}(U)$, so $\mathbf{x}^{-1}$ is continuous, and $\mathbf{x}(U)$ (the torus minus one meridian and one longitude circle) is open in $T$. This single chart misses a "seam" (where $u$ or $v \in \{0,2\pi\}$), so it does not cover all of $T$; one uses a second chart with domain $(-\pi,\pi)\times(-\pi,\pi)$ (shifted ranges) to cover the seam and its overlap with the first chart, exactly analogous to the sphere. Together, finitely many such shifted charts show $T$ is a regular surface.