Singular Sets and Valid Coordinate Patches

Why Singular Points Need Special Attention

A coordinate map describes how points in a parameter domain correspond to points in physical space. Its Jacobian determinant measures local volume scaling, including whether the map locally preserves or reverses orientation. When this determinant is zero, the map collapses at least one infinitesimal direction: it cannot be treated as an ordinary smooth coordinate change there.

The issue is local. A map may be well behaved on most of its domain but fail at a few points, or along a curve or surface. Identifying that exceptional set helps determine where a map can be used as a valid coordinate patch and what additional care is needed near the excluded set.

Definition: Let $F:U\to\mathbb R^n$ be continuously differentiable on an open set $U\subseteq\mathbb R^n$. Its Jacobian matrix at $\mathbf u\in U$ is $DF(\mathbf u)$, and its Jacobian determinant is $J_F(\mathbf u)=\det DF(\mathbf u)$. The singular set of $F$ is $$ S_F=\{\mathbf u\in U:J_F(\mathbf u)=0\}. $$ A point of $S_F$ is a singular point of the map. A point outside $S_F$ is nonsingular.

A singular set can have different shapes: it may be a point, a curve, a surface, or a larger subset. This tutorial focuses on isolated singularities, meaning singular points that have a neighborhood containing no other singular points. The set can be isolated even though the map is defined and nonsingular at every nearby point other than the singular point itself.

What a Nonzero Jacobian Guarantees

The inverse function theorem gives the key local test. If $F$ is continuously differentiable near $\mathbf u_0$ and $J_F(\mathbf u_0)\ne 0$, then there are neighborhoods of $\mathbf u_0$ and $F(\mathbf u_0)$ on which $F$ has a continuously differentiable inverse. Thus, near a nonsingular point, the map really does provide local coordinates.

Where the inverse exists, the chain rule applied to $F^{-1}(F(\mathbf u))=\mathbf u$ gives $$ D(F^{-1})(F(\mathbf u))\,DF(\mathbf u)=\mathbf I. $$ Taking determinants shows that the inverse Jacobian determinant is the reciprocal of the original one: $$ \det D(F^{-1})(F(\mathbf u))=\frac{1}{J_F(\mathbf u)}. $$ This reciprocal is meaningful only when $J_F(\mathbf u)\ne 0$.

If the determinant vanishes, the inverse function theorem does not provide a local differentiable inverse there. Moreover, the reciprocal-Jacobian expression is undefined. These facts explain why a change-of-variables argument cannot simply include a singular point as if the map were an ordinary coordinate patch.

Important distinction: A vanishing Jacobian means that the usual local-inverse test fails at that point. It does not, by itself, prove that every possible integral identity involving the map is false. It does mean that the ordinary nonsingular coordinate-patch reasoning cannot be applied there without separate justification.

Worked Example: Finding the Singular Set of a Squaring Map

Consider $F:\mathbb R^2\to\mathbb R^2$ defined by $$ F(u,v)=(u^2-v^2,\,2uv). $$ Write its components as $x=u^2-v^2$ and $y=2uv$. The Jacobian matrix is $$ DF(u,v)= \begin{pmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 2u&-2v\\ 2v&2u \end{pmatrix}. $$ Therefore, $$ J_F(u,v)=(2u)(2u)-(-2v)(2v)=4u^2+4v^2=4(u^2+v^2). $$ This is zero exactly when $u^2+v^2=0$, which over the real numbers requires $u=0$ and $v=0$. Hence $S_F=\{(0,0)\}$, an isolated singularity.

At $(0,0)$, the Jacobian matrix is the zero matrix, so its rank is zero and its determinant is zero. At every other point, the determinant is positive. The inverse function theorem therefore guarantees a local differentiable inverse near every nonzero point, but not at the origin.

Check: Substituting $(u,v)=(1,0)$ gives $J_F(1,0)=4(1+0)=4$, while substituting $(0,0)$ gives $J_F(0,0)=0$, consistent with the singular set found above.

Building Valid Patches Around an Isolated Singularity

A valid coordinate patch is a region on which the map is continuously differentiable, one-to-one, and has nonzero Jacobian determinant. The inverse function theorem then ensures a differentiable inverse locally. If an isolated singularity lies in the domain, a natural first step is to remove it and work on regions that do not contain it.

Removing the singular point is not always enough to make an entire remaining domain a valid patch. A map can still send two distinct points to the same image. In that case, the domain must also be restricted to smaller regions where the map is one-to-one. This is a local issue distinct from locating the zeros of the determinant.

Worked Example: Choosing a One-to-One Patch for the Squaring Map

Use the map $F(u,v)=(u^2-v^2,2uv)$, whose only singular point is the origin. Choose the open sector $$ U=\{(u,v):0<r<2,\ -\pi/4<\theta<\pi/4\}, $$ where $u=r\cos\theta$ and $v=r\sin\theta$. In these polar descriptions of points in the parameter plane, the map becomes $$ x=r^2\cos(2\theta),\qquad y=r^2\sin(2\theta). $$ Since $-\pi/2<2\theta<\pi/2$, we have $x>0$. Also $x^2+y^2=r^4<16$. Thus the image is the open half-disk $$ V=\{(x,y):x>0,\ x^2+y^2<16\}. $$

For $(x,y)\in V$, define $\rho=\sqrt{x^2+y^2}$. Because $\rho=r^2$, the parameter radius is $r=\sqrt{\rho}=(x^2+y^2)^{1/4}$. Since $x>0$, the angle $\arctan(y/x)$ lies in $(-\pi/2,\pi/2)$; it equals $2\theta$. Therefore the inverse is $$ u=(x^2+y^2)^{1/4}\cos\left(\frac12\arctan\frac{y}{x}\right),\qquad v=(x^2+y^2)^{1/4}\sin\left(\frac12\arctan\frac{y}{x}\right). $$ This inverse is differentiable throughout $V$, and the Jacobian determinant on $U$ is $4r^2>0$. So $U$ is a valid patch.

Check: The inverse gives a point with radius $r=(x^2+y^2)^{1/4}$ and angle $\theta=\tfrac12\arctan(y/x)$. Applying $F$ produces an image with radius $r^2=\sqrt{x^2+y^2}$ and angle $2\theta=\arctan(y/x)$, so its Cartesian coordinates are exactly $(x,y)$. The formulas recover the original point on this sector.

Other angular sectors can provide patches for points around the omitted origin. Their angular widths must be chosen so that two parameter points in the same sector do not produce the same image. Patches may overlap; if a later calculation needs to use several patches together, the overlap must be handled rather than counted twice.

A Three-Dimensional Isolated Singularity

The same analysis applies in three dimensions. The determinant is now the local volume-scaling factor, and its zero set identifies points where that factor vanishes. The next example has an isolated singular point and is one-to-one on its entire punctured domain.

Worked Example: A Radial Map with an Isolated Singular Point

Define $F:\mathbb R^3\to\mathbb R^3$ by $$ F(x,y,z)=(x^2+y^2+z^2)(x,y,z). $$ Set $s=x^2+y^2+z^2$. Differentiating each component gives $$ DF(x,y,z)= \begin{pmatrix} s+2x^2&2xy&2xz\\ 2xy&s+2y^2&2yz\\ 2xz&2yz&s+2z^2 \end{pmatrix} =s\mathbf I+2\mathbf q\mathbf q^{\top}, $$ where $\mathbf q=(x,y,z)^{\top}$.

At a nonzero point, the direction $\mathbf q$ is an eigenvector of this matrix: $DF\,\mathbf q=(s+2s)\mathbf q=3s\mathbf q$. Any vector perpendicular to $\mathbf q$ is sent to $s$ times itself, because $\mathbf q^{\top}\mathbf w=0$ implies $DF\,\mathbf w=s\mathbf w$. Thus the three eigenvalues are $3s,s,s$, and $$ J_F(x,y,z)=(3s)s^2=3s^3=3(x^2+y^2+z^2)^3. $$ This determinant is positive except at the origin, where it is zero. The singular set is therefore the isolated point $\{(0,0,0)\}$.

For a nonzero vector $\mathbf q$ of length $r$, the image vector has length $r^3$ and the same direction. Consequently the map is one-to-one on the punctured ball $0<r<R$, mapping it onto $0<|\mathbf y|<R^3$. Its inverse at a nonzero image point is $$ \mathbf q=\frac{\mathbf y}{|\mathbf y|^{2/3}}. $$ Indeed, this inverse has length $|\mathbf y|^{1/3}$, and applying $F$ multiplies it by its squared length, $|\mathbf y|^{2/3}$, giving $\mathbf y$ exactly. The inverse is differentiable away from the origin, so the punctured ball is a valid patch.

Check: At $(1,0,0)$, $s=1$, so the three eigenvalues are $3,1,1$ and the determinant is $3$. The determinant formula gives $3(1)^3=3$ as well. At the origin, the derivative matrix is zero and the determinant formula gives zero.

Why the Singular Point Cannot Be Included Casually

At a singular point, the differential sends at least one nonzero direction to zero. This follows from elementary linear algebra: a square matrix with determinant zero is not invertible, so it has a nonzero vector in its null space. In local terms, an infinitesimal displacement in that direction has no first-order displacement in the image. That collapse is incompatible with a differentiable local inverse, whose derivative would have to undo it.

The reciprocal Jacobian also signals the difficulty. On a nonsingular patch, the determinant of the inverse derivative is $1/J_F$. As a singular point is approached, this reciprocal may become unbounded. A formula derived on nonsingular patches therefore cannot simply be extended across the point by substitution. Whether an integral can be evaluated by excluding the singular point and taking an appropriate limit requires its own justification.

An isolated singular point is geometrically much smaller than a curve or surface of singularities, but that observation is not a substitute for checking the hypotheses of a coordinate argument. The map must still be one-to-one on each patch, have a nonzero Jacobian there, and have a suitable inverse. When a region reaches the singular point, one must also establish that any limiting procedure is legitimate for the integral under consideration.

A Common Mistake

A frequent error is to compute a correct Jacobian and then use its reciprocal at a point where the Jacobian is zero.

Worked Example: Correcting an Invalid Reciprocal

For the squaring map, $J_F(u,v)=4(u^2+v^2)$. At $(1,0)$, the reciprocal is $1/4$, which is defined because $J_F(1,0)=4$. At the origin, however, substitution gives $$ \frac{1}{J_F(0,0)}=\frac{1}{4(0^2+0^2)}=\frac{1}{0}. $$ This is undefined, not zero and not an allowable Jacobian factor.

The correction is to exclude the origin from any ordinary nonsingular patch. Work instead on regions where $u^2+v^2>0$, and restrict further if necessary to make the map one-to-one. If a desired domain includes the origin, a separate limiting argument would be needed; the reciprocal formula alone does not provide one.

Check: At every point in the punctured plane, $4(u^2+v^2)>0$, so its reciprocal exists. At the origin, the determinant is exactly zero, confirming that the issue is the singularity, not an arithmetic simplification.

A Practical Patch Checklist

Before using a coordinate map on a region, check the following in order:

Compute the Jacobian matrix and determinant.
Solve $J_F=0$ to identify the singular set.
Restrict to open regions that avoid that set.
Check that the map is one-to-one on each chosen region.
Use the inverse function theorem at points where the determinant is nonzero.
When a region approaches an excluded singularity, justify any limiting step separately.

This checklist separates three questions that should not be confused: where the determinant vanishes, where the map has a differentiable local inverse, and where the map is one-to-one across an entire chosen patch.

Check Your Understanding

Answer each question, showing the relevant calculation or reasoning.

For $F(u,v)=(u^2-v^2,2uv)$, what is the singular set, and what is the determinant at $(1,1)$?
State the local conclusion supplied by the inverse function theorem when $J_F(\mathbf u_0)\ne 0$.
Why does $J_F(\mathbf u_0)=0$ make the reciprocal-Jacobian expression undefined at $\mathbf u_0$?
For the map $F(x,y,z)=(x^2+y^2+z^2)(x,y,z)$, where does its determinant vanish?
Why might removing singular points still fail to produce a valid patch for a map that is not one-to-one?