Regular Surface Patches and the Rank Condition

Why Check the Rank of a Parametrization?

A surface parametrization assigns a point in space to each pair of parameters. To describe a surface patch regularly, its two parameters must provide two independent directions of change at every parameter value under consideration. If one direction disappears or duplicates the other, the parametrization loses rank there.

This is a local condition on the map, not simply a visual judgment about the surface. A parametrization can fail the condition even when its image is a familiar smooth surface. Conversely, a quick check that both parameter derivatives are nonzero is not enough: two nonzero vectors can still be linearly dependent.

Definition and Rank Condition

Let \(D\) be an open subset of \(\mathbb{R}^2\), and let \(\mathbf{r}:D\to\mathbb{R}^3\) be continuously differentiable. Write the parameters as \(u\) and \(v\), and write the map in components as \(\mathbf{r}(u,v)=(x(u,v),y(u,v),z(u,v))\). Its parameter derivatives are the vectors \(\mathbf{r}_u\) and \(\mathbf{r}_v\).

Definition: The parametrization \(\mathbf{r}\) is regular at \((u_0,v_0)\in D\) if \(\mathbf{r}_u(u_0,v_0)\) and \(\mathbf{r}_v(u_0,v_0)\) are linearly independent. It is a regular parametrization on \(D\) if it is regular at every point of \(D\).

Equivalently, form the \(3\times 2\) derivative matrix whose columns are the two parameter derivatives. The rank condition says that this matrix has rank \(2\) at each parameter value. In practical terms, independence means that the only scalars \(a\) and \(b\) satisfying \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\) are \(a=b=0\).

D\mathbf{r}(u,v)= \begin{bmatrix} \vert & \vert\\ \mathbf{r}_u(u,v) & \mathbf{r}_v(u,v)\\ \vert & \vert \end{bmatrix}, \qquad \operatorname{rank}D\mathbf{r}(u,v)=2.

Regularity is separate from global one-to-one behavior. The rank condition checks the derivatives at each parameter value; by itself it does not say whether distinct parameter pairs can map to the same point. If a definition of a surface patch also requires the parametrization to be one-to-one, that requirement must be checked separately.

A Practical Test Using Minors

To check independence directly, one can solve \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\). There is also a useful equivalent test: at least one of the three \(2\times2\) minors of the derivative matrix must be nonzero.

Theorem: Two columns of a \(3\times2\) matrix are linearly independent if and only if at least one determinant formed by selecting two rows and both columns is nonzero.

To justify the test, suppose first that one such determinant is nonzero. If a linear combination of the columns is zero, then restricting to the two selected coordinates gives a \(2\times2\) homogeneous system with nonzero determinant. Its only solution has both coefficients zero, so the columns are independent.

For the converse, suppose every \(2\times2\) minor is zero. If the first column is zero, the columns are dependent. Otherwise choose a nonzero entry of the first column, in row \(i\). For each other row \(j\), the minor using rows \(i\) and \(j\) is zero. That equation says the second-column entry in row \(j\) is the same scalar multiple of the first-column entry in row \(j\) as in row \(i\). Thus the second column is a multiple of the first, and the columns are dependent. This proves the criterion.

\mathbf{r}_u= \begin{bmatrix}x_u\\y_u\\z_u\end{bmatrix}, \qquad \mathbf{r}_v= \begin{bmatrix}x_v\\y_v\\z_v\end{bmatrix}.

The three minors to check are \(x_u y_v-y_u x_v\), \(x_u z_v-z_u x_v\), and \(y_u z_v-z_u y_v\). For regularity at a parameter value, it is enough that at least one is nonzero there. Which minor works can vary from point to point.

Worked Examples

Worked Example: A Graph-Type Parametrization

Determine whether \(\mathbf{r}(u,v)=(u,v,u^2+v^2)\) is regular on \(D=\mathbb{R}^2\).

Differentiate each component with respect to each parameter:

\mathbf{r}_u=(1,0,2u), \qquad \mathbf{r}_v=(0,1,2v).

Suppose \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\). Comparing the first coordinates gives \(a=0\). Comparing the second coordinates gives \(b=0\). Therefore the two derivatives are independent for every \((u,v)\), so the parametrization is regular on all of \(\mathbb{R}^2\).

As a minor check, the minor from the first two rows is \(1\cdot1-0\cdot0=1\), which is nonzero everywhere. This verifies the same conclusion.

Worked Example: Finding a Nonzero Minor

Test \(\mathbf{r}(u,v)=(u+v,u-v,2u)\) for regularity on \(\mathbb{R}^2\).

The parameter derivatives are constant:

\mathbf{r}_u=(1,1,2), \qquad \mathbf{r}_v=(1,-1,0).

Use the first two rows of the derivative matrix. The resulting determinant is

\begin{vmatrix} 1&1\\ 1&-1 \end{vmatrix} =1(-1)-1(1)=-2.

Since \(-2\ne0\), the two columns are independent at every parameter value. Thus \(\mathbf{r}\) is regular throughout its domain.

For a direct check, if \(a(1,1,2)+b(1,-1,0)=(0,0,0)\), then the first two coordinates give \(a+b=0\) and \(a-b=0\). Adding these equations gives \(2a=0\), so \(a=0\); then \(b=0\). The direct calculation confirms the minor test.

Worked Example: Locating a Rank Failure

For \(\mathbf{r}(u,v)=(u^3,v,0)\), find the parameter values where regularity fails.

Differentiate:

\mathbf{r}_u=(3u^2,0,0), \qquad \mathbf{r}_v=(0,1,0).

If \(u\ne0\), then \(3u^2\ne0\). The first derivative has a nonzero first coordinate, while the second derivative has zero first coordinate and a nonzero second coordinate. They cannot be scalar multiples, so they are independent.

If \(u=0\), then \(\mathbf{r}_u=(0,0,0)\). A pair containing the zero vector is dependent, so regularity fails for every \((0,v)\) in the domain. Equivalently, the minor from the first two rows is

\begin{vmatrix} 3u^2&0\\ 0&1 \end{vmatrix} =3u^2,

which vanishes exactly when \(u=0\); the other two minors are also zero. The failure set is therefore the line \(u=0\) in parameter space.

This rank failure does not mean the image itself must have a defect. On \(D=(-1,1)^2\), the image is \(\{(x,y,0):-1<x<1,\,-1<y<1\}\), an open square in the plane \(z=0\). Indeed, \(x=u^3\) ranges over \((-1,1)\), and \(y=v\) ranges over \((-1,1)\). The image is a smooth planar patch, although this particular parametrization is not regular along \(u=0\).

What Rank Failure Tells You

At a point where the two derivatives are dependent, the parametrization does not provide two independent first-order changes in the surface point. One derivative may be zero, or both may be nonzero but point in dependent directions. The failure is specific to the parameter value; a parametrization can be regular on part of its domain and fail elsewhere.

It is important not to conclude more than the calculation warrants. Rank failure proves that the given map is not regular at that parameter value. It does not, by itself, prove that the image cannot be described by another regular parametrization. The example \(\mathbf{r}(u,v)=(u^3,v,0)\) shows why this distinction matters.

Common Mistake: Checking Only That Both Derivatives Are Nonzero

Consider the map \(\mathbf{r}(u,v)=(u+v,2u+2v,0)\). Its derivatives are

\mathbf{r}_u=(1,2,0), \qquad \mathbf{r}_v=(1,2,0).

Both derivatives are nonzero, but they are equal. Taking \(a=1\) and \(b=-1\) gives \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\) with coefficients not both zero. Thus the derivatives are dependent and the rank is only \(1\), not \(2\). In fact, every \(2\times2\) minor is zero.

The correction is to test independence, not merely nonvanishing. Solve the linear-combination equation or find a nonzero \(2\times2\) minor. Either test detects dependence that a check of the two vectors separately would miss.

A Reliable Workflow

Differentiate.
Compute \(\mathbf{r}_u\) and \(\mathbf{r}_v\) on the domain.

Test independence.
Either solve \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\), or compute the \(2\times2\) minors.

Locate any failures.
Set the relevant minors to zero, or identify where the linear-combination test permits a nontrivial solution.

State the conclusion precisely.
Specify whether regularity holds everywhere, on a subset, or fails at identified parameter values. Do not infer an intrinsic defect of the image from this test alone.

Why the Rank Condition Matters

The rank test is a precise way to check that the two parameters genuinely supply two independent directions of variation locally. It identifies where a parametrization loses that property and provides a foundation for studying the local geometry of parameterized surfaces. The essential distinction is between the quality of a particular parametrization and the shape of the set of points it covers: the rank condition directly assesses the former.

Check Your Understanding

For each question, justify your answer using parameter derivatives, linear independence, or a minor.

What does it mean for a parametrization \(\mathbf{r}:D\to\mathbb{R}^3\) to be regular at a parameter value?
Why does a nonzero \(2\times2\) minor prove that the two parameter derivatives are independent?
For \(\mathbf{r}(u,v)=(u,v,u^2)\), is the parametrization regular on \(\mathbb{R}^2\)? Identify a minor that answers the question.
For \(\mathbf{r}(u,v)=(u^2,v,0)\), at which parameter values does the rank condition fail?
Why does rank failure for a parametrization not necessarily imply that its image is intrinsically nonsmooth?