Deciding whether a point on a surface is singular

Question

Consider $S^2 \in \mathbb{R}^2$. It is well known that it is a regular surface. We can parametrize (a patch of it) by (EDIT: this seems wrong, see comment)

$$x(u,v) = \cos u \cos v$$ $$y(u,v) = \sin u \cos v$$ $$z(u,v) = \sin v$$

Here we consider $(u,v) \in (0,2\pi) \times (0,\pi) $ so we do get a nice smooth map from an open set in $\mathbb{R}^2$ to an open set in $S^2$.

The derivative of the above map is

$$ \begin{pmatrix} -\sin u \cos v & -\cos u \sin v \\ \cos u \cos v & -\sin u \sin v \\ 0 & \cos v \end{pmatrix} $$ which has rank 2 always except when $v = \pi /2$, where it has rank 1. However, these points nevertheless are regular points; it is just that the particular choice of parametrization we chose somehow breaks down at these points, and we need another parametrization to show that the surface is singular at these points.

For the cone $x^2+y^2=z^2$, however, it is well known that $(0,0,0)$ is truly a singular point, and for any parametrization we'll choose, we'll get a rank-1 jacobian there. This shows geometrically on the surface of the cone, which is why I'm interested in this phenomenon.

My question is: how to understand the phenomenon happening with the sphere, that a specific parametrization we chose caused some points to have a singular jacobian? Why does it occur, intuitively? And more importantly, is there a general method for deciding whether a certain point on a surface (given in parametric form) is truly singular, as opposed to merely "coordinate" singular (as always finding a parametrization in which the singularity disappears seems impossible)?

Answer 1

Generally speaking, the function that parameterizes a coordinate patch on a surface cannot be extended indefinitely: it might fail to be one-to-one onto an open subset if extended any further; it might become singular if extended further; the formula which defines the coordinate patch might not even be defined on a larger domain. There's no particular rhyme or reason. Functions do these things and we get used to it. One of the lessons of manifold theory is that it is unreasonable to expect that a single coordinate system to serve for the whole manifold, and for some manifolds it is provably impossible. That well known phenomenon is the reason that atlases for manifolds usually consist of more than one coordinate chart.

On $S^2$, those points where the $u,v$ coordinate system becomes singular are contained in other coordinate charts where the parameterizing function is not singular. That's why $S^2$ is a manifold, and that's why those points are not singular.

I'll just point out that is very much possible to have an open set $U \subset \mathbb R^2$, and a smooth function $f : U \to S^2$ expressed if you like as a parameterization $f(u,v) = (x(u,v),y(u,v),z(u,v))$, such that $f$ is surjective, and the Jacobian of $f$ has rank $2$ at each point in $U$. Of course, this $f$ fails to be one-to-one, as it must.

And as you say, for the cone there does not exist a coordinate chart which is nonsingular at that cone point. But what you say about a "rank-1 Jacobian" is not true, you can easily produce a parameterization with a rank 0 Jacobian at that point. What you cannot do is to produce a parameterization with a rank 2 Jacobian.

One last thing: nonsingularity can be characterized in a pretty simple manner. The implicit function theorem guarantees that at each nonsingular point there is a neighborhood of that point which is the graph of a smooth function of one of three forms: $z=f(x,y)$; $y=g(z,x)$; or $x=h(y,z)$. So, all you have to do is to figure out what $f,g,h$ are and check their smoothness at the coordinate values for your point in question. That's one way to prove that the cone point of the cone is indeed singular.

Answer 2

I think you may be confusing two concepts, and the parameterizations are making things worse. If this is not what you re looking for, please advise and I will delete my answer. I look at this problem like this: you want to find local diffeomorphisms (charts) from the surfaces to $\mathbb R^2$ because they should intuitively be $2$-dimensional manifolds. So, let's try:

take the sphere, $S^2=\{(x,y,z):x^2+y^2+z^2=1\},$ which is the zero set of $f(x,y,z)=x^2+y^2+z^2-1.$ Now, from this, we get the map $\varphi:\mathbb R^3\to \mathbb R^3:(x,y,z)\mapsto (f(x,y,z),y,z).$ Then,

$$\mathcal J(\varphi)=\begin{pmatrix} 2x &2y &2z \\ 0&1 &0 \\ 0& 0 &1 \end{pmatrix}$$

and so as long as $x\neq 0$, we get a (slice) chart in a neighborhood $U$ of $(x,y,z)$ in $\mathbb R^3:\ (U\cap S^2,\varphi)$ where $\varphi(x,y,z)=(f(x,y,z),y,z).$

Similarly, if $y\neq 0,$we get a chart by taking $\varphi(x,y,z)=(x,f(x,y,z),z)$ and if $z\neq 0$, we may take $\varphi(x,y,z)=(x,y,f(x,y,z))$ and get a chart.

It's easy to see that these three cases provide a choice of charts (an atlas) for $\textit{any}$ point on the sphere. So, if $\mathcal J(\varphi)=0$ at a given point, we can always fix it up by moving $f$ to another position.

On the other hand, in the case of the cone $C=\{(x,y,z):z=\sqrt{x^2+y^2}\},\ C$ is the zero set of $f(x,y,z)=\sqrt{x^2+y^2}-z$ so the above procedure will fail at $(0,0,0)$ because $\mathcal J(\varphi)$ is $\textit{undefined}$ at $(0,0,0)$ and there is no way to fix this problem.

In short, the difference is that in the first case, the Jacobian may be zero and we can get around that problem. In the second case, the Jacobian is not even defined at $(0,0,0)$ because the function we are using is not differentiable there, and there is no way to remedy this problem.