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Parametrized Surfaces · Tutorial 839 of 999

Reparametrizing The Same Surface Patch

Learn to compare surface patches by tracking which points of the original parameter domain—and ultimately which points of space—a new parameter range reaches.

Intermediate 9 min read

What You'll Learn

  • Define reparametrization by composing a surface map with a map between parameter domains.
  • Test whether a changed parameter range produces the same surface image.
  • Explain when covering the original parameter domain is sufficient or necessary.
  • Compare parameter ranges for a cylinder, a plane patch, and a graph over a disk.
  • Identify how repeated parameter values affect claims about equal patches.

Introduction

A surface parametrization assigns a point in space to each point in a parameter domain. The same geometric patch can often be described using different parameter coordinates or different parameter ranges. Comparing those descriptions means checking which points in space each parametrization actually reaches.

The distinction between parameter values and surface points matters. A parametrization need not assign different parameter values to different points: for instance, angle coordinates on a cylinder repeat after one full turn. Consequently, matching the parameter domains is a useful sufficient test for equal patches, but it is not always a necessary one.

Central idea: If a new parametrization is obtained by composing an existing one with a parameter map, its image is determined by the values that the parameter map reaches and by whether the original parametrization identifies distinct parameter values.

Reparametrization and the Image of a Patch

Let \(D\) be a parameter domain in \(\mathbb{R}^2\), and let \(\mathbf{X}:D\to\mathbb{R}^3\) be a surface parametrization. Its image is the set of points in space reached by the parametrization:

$$ \mathbf{X}(D)=\{\mathbf{X}(u,v):(u,v)\in D\}. $$

Suppose a second parameter domain is \(E\), and a map \(\boldsymbol{\phi}:E\to D\) sends new parameter values to old ones. Composing the maps gives a new parametrization \(\mathbf{Y}:E\to\mathbb{R}^3\):

$$ \mathbf{Y}=\mathbf{X}\circ\boldsymbol{\phi}, \qquad \mathbf{Y}(s,t)=\mathbf{X}\bigl(\boldsymbol{\phi}(s,t)\bigr). $$
Definition: A reparametrization of \(\mathbf{X}\) by \(\boldsymbol{\phi}\) is the composition \(\mathbf{Y}=\mathbf{X}\circ\boldsymbol{\phi}\), where \(\boldsymbol{\phi}:E\to D\). The two parametrizations describe the same surface image precisely when \(\mathbf{Y}(E)=\mathbf{X}(D)\).

The image of a composition is obtained by first finding the values of \(D\) reached by \(\boldsymbol{\phi}\), and then applying \(\mathbf{X}\). Thus

$$ \mathbf{Y}(E) = \mathbf{X}\bigl(\boldsymbol{\phi}(E)\bigr). $$

It follows that the exact test for equality of the images is

$$ \mathbf{X}\bigl(\boldsymbol{\phi}(E)\bigr)=\mathbf{X}(D). $$

This test concerns points in space, not just parameter values. If \(\boldsymbol{\phi}(E)=D\), then the test is automatically satisfied. But if \(\mathbf{X}\) assigns the same surface point to multiple parameter values, \(\boldsymbol{\phi}\) may omit some values of \(D\) and still reach every point in \(\mathbf{X}(D)\).

When Does the Parameter Range Have to Be Covered?

A map \(\mathbf{X}\) is injective if \(\mathbf{X}(p)=\mathbf{X}(q)\) implies \(p=q\) for all \(p,q\in D\). For an injective parametrization, omitting any parameter value omits its corresponding point in space. This gives a precise distinction between the injective and non-injective cases.

Theorem: Suppose \(\boldsymbol{\phi}:E\to D\) and \(\mathbf{Y}=\mathbf{X}\circ\boldsymbol{\phi}\). If \(\boldsymbol{\phi}(E)=D\), then \(\mathbf{Y}(E)=\mathbf{X}(D)\), whether or not \(\mathbf{X}\) is injective. If \(\mathbf{X}\) is injective, then the two images are equal only if \(\boldsymbol{\phi}(E)=D\). Without injectivity, the necessary and sufficient condition is \(\mathbf{X}(\boldsymbol{\phi}(E))=\mathbf{X}(D)\).

Proof. First, suppose \(\boldsymbol{\phi}(E)=D\). Applying \(\mathbf{X}\) to both sets gives \(\mathbf{X}(\boldsymbol{\phi}(E))=\mathbf{X}(D)\), so \(\mathbf{Y}(E)=\mathbf{X}(D)\).

Now suppose \(\mathbf{X}\) is injective and \(\mathbf{Y}(E)=\mathbf{X}(D)\). Take any \(d\in D\). Since \(\mathbf{X}(d)\in\mathbf{X}(D)=\mathbf{X}(\boldsymbol{\phi}(E))\), there is some \(q\in\boldsymbol{\phi}(E)\) such that \(\mathbf{X}(q)=\mathbf{X}(d)\). Injectivity implies \(q=d\). Therefore \(d\in\boldsymbol{\phi}(E)\). Since this holds for every \(d\in D\), \(D\subseteq\boldsymbol{\phi}(E)\). Also \(\boldsymbol{\phi}(E)\subseteq D\), because \(\boldsymbol{\phi}\) maps into \(D\). Hence \(\boldsymbol{\phi}(E)=D\). Finally, for any \(\mathbf{X}\), the identity \(\mathbf{Y}(E)=\mathbf{X}(\boldsymbol{\phi}(E))\) shows directly that the exact equality test is \(\mathbf{X}(\boldsymbol{\phi}(E))=\mathbf{X}(D)\). This proves all claims.

Worked Example: An Affine Change of Parameters

Worked Example: An Affine Change of Parameters

Consider the plane patch parametrized by

$$ \mathbf{X}(u,v)=(u,v,u+2v), \qquad D=[0,2]\times[0,1]. $$

Use the new domain \(E=[-1,1]\times[0,1]\) and the parameter map \(\boldsymbol{\phi}(s,t)=(s+1,t)\). Its first coordinate ranges from \(-1+1=0\) to \(1+1=2\), while its second coordinate ranges from \(0\) to \(1\). Thus \(\boldsymbol{\phi}(E)=D\). The reparametrization is

$$ \begin{aligned} \mathbf{Y}(s,t) &=\mathbf{X}(s+1,t)\\ &=(s+1,t,(s+1)+2t)\\ &=(s+1,t,s+1+2t). \end{aligned} $$

Every \((u,v)\in D\) is reached by \(s=u-1,\ t=v\). Indeed, \(0\leq u\leq2\) gives \(-1\leq s\leq1\), and \(0\leq v\leq1\) gives \(t\in[0,1]\). Substituting these values back gives \(\boldsymbol{\phi}(u-1,v)=(u,v)\). Therefore the parameter map reaches all of \(D\), and the two surface images are equal. The reverse substitution also checks the formula: \(\mathbf{Y}(u-1,v)=(u,v,u+2v)=\mathbf{X}(u,v)\).

Worked Example: The Cylinder and an Omitted Parameter Value

Worked Example: The Cylinder and an Omitted Parameter Value

Let

$$ \mathbf{X}(u,v)=(\cos u,\sin u,v), \qquad D=[0,2\pi]\times[0,1]. $$

Take \(E=[0,2\pi)\times[0,1]\) and let \(\boldsymbol{\phi}(s,t)=(s,t)\). The new parametrization is \(\mathbf{Y}(s,t)=(\cos s,\sin s,t)\). Here \(\boldsymbol{\phi}(E)=E\), which omits the parameter values \((2\pi,v)\) from \(D\).

Nevertheless, \(\mathbf{X}(2\pi,v)=(\cos 2\pi,\sin 2\pi,v)=(1,0,v)=\mathbf{X}(0,v)\). For any \((u,v)\in D\) with \(u<2\pi\), take \((s,t)=(u,v)\in E\); then \(\mathbf{Y}(s,t)=\mathbf{X}(u,v)\). If \(u=2\pi\), take \((s,t)=(0,v)\in E\); the same equality follows from the calculation above. Thus every point of \(\mathbf{X}(D)\) belongs to \(\mathbf{Y}(E)\). Conversely, \(E\subseteq D\), so every point of \(\mathbf{Y}(E)\) belongs to \(\mathbf{X}(D)\). Therefore \(\mathbf{Y}(E)=\mathbf{X}(D)\), even though \(\boldsymbol{\phi}(E)\ne D\).

This example shows why parameter coverage is not always necessary. The endpoints \(u=0\) and \(u=2\pi\) are different parameter values, but they give the same points on the cylinder.

Worked Example: Polar Coordinates for a Graph Patch

Worked Example: Polar Coordinates for a Graph Patch

Start with the graph patch

$$ \mathbf{X}(u,v)=(u,v,u^2+v^2), \qquad D=\{(u,v):u^2+v^2\leq1\}. $$

Use \(E=[0,1]\times[0,2\pi]\) and define \(\boldsymbol{\phi}(r,\theta)=(r\cos\theta,r\sin\theta)\). Since \(0\leq r\leq1\),

$$ (r\cos\theta)^2+(r\sin\theta)^2 =r^2(\cos^2\theta+\sin^2\theta) =r^2\leq1, $$

so \(\boldsymbol{\phi}(E)\subseteq D\). Conversely, take any \((u,v)\in D\). Set \(r=\sqrt{u^2+v^2}\), so \(0\leq r\leq1\). If \(r>0\), the point \((u/r,v/r)\) lies on the unit circle, so it equals \((\cos\theta,\sin\theta)\) for some \(\theta\in[0,2\pi]\). Then \(\boldsymbol{\phi}(r,\theta)=(u,v)\). If \(r=0\), then \((u,v)=(0,0)=\boldsymbol{\phi}(0,0)\). Hence \(\boldsymbol{\phi}(E)=D\).

Composing gives

$$ \begin{aligned} \mathbf{Y}(r,\theta) &=\mathbf{X}(r\cos\theta,r\sin\theta)\\ &=(r\cos\theta,r\sin\theta, r^2\cos^2\theta+r^2\sin^2\theta)\\ &=(r\cos\theta,r\sin\theta,r^2). \end{aligned} $$

For a direct check, the first two coordinates of \(\mathbf{Y}\) have squared distance \(r^2\cos^2\theta+r^2\sin^2\theta=r^2\), and the third coordinate is \(r^2\). Thus every point of \(\mathbf{Y}(E)\) lies on the stated graph patch. The reverse argument above represents each \((u,v)\in D\) as \(\boldsymbol{\phi}(r,\theta)\), and then \(\mathbf{Y}(r,\theta)=\mathbf{X}(u,v)\). The images are equal. The parameter map repeats the origin for every angle when \(r=0\), and repeats the angle’s endpoint values; these repetitions do not prevent it from covering the disk.

A Common Mistake: Confusing Inclusion with Equality

Showing that a new parametrization reaches only points on the original patch proves containment, not equality. For example, retain the cylinder \(\mathbf{X}(u,v)=(\cos u,\sin u,v)\) on \(D=[0,2\pi]\times[0,1]\), but restrict the new domain to \(E=[0,\pi]\times[0,1]\) and use the identity parameter map. The new image is contained in the cylinder, but it misses points on the lower half.

$$ \mathbf{X}\left(\frac{3\pi}{2},\frac12\right) = \left(\cos\frac{3\pi}{2},\sin\frac{3\pi}{2},\frac12\right) = \left(0,-1,\frac12\right). $$

For every \(s\in[0,\pi]\), \(\sin s\geq0\). Therefore the second coordinate of \(\mathbf{Y}(s,t)=(\cos s,\sin s,t)\) is never \(-1\), and \((0,-1,\frac12)\) is not in the new image. The patches are not equal. Periodicity identifies angles separated by \(2\pi\); it does not make an interval of length \(\pi\) cover all cylinder angles.

Pitfall: Do not conclude that two parametrizations give the same patch just because both lie on the same surface, or because one image is contained in the other. Check that every point of each image is reached by the other parametrization.

A Reliable Comparison Procedure

When a parametrization is given by composition, use the parameter map to organize the comparison. First identify the original domain \(D\) and the new domain \(E\). Next compute \(\boldsymbol{\phi}(E)\), taking care with open or closed endpoints and repeated parameter values. Then use \(\mathbf{X}(\boldsymbol{\phi}(E))=\mathbf{Y}(E)\) to test equality in space.

1
Write the composition.
Confirm that \(\mathbf{Y}=\mathbf{X}\circ\boldsymbol{\phi}\) and that \(\boldsymbol{\phi}\) maps \(E\) into \(D\).
2
Track the parameter range.
Determine exactly which points of \(D\) are in \(\boldsymbol{\phi}(E)\).
3
Compare spatial images.
Test whether \(\mathbf{X}(\boldsymbol{\phi}(E))=\mathbf{X}(D)\), rather than assuming that parameter-domain equality is required.
4
Use injectivity appropriately.
If \(\mathbf{X}\) is injective, equality of images forces \(\boldsymbol{\phi}(E)=D\). If it is not, check whether omitted parameters give points already reached elsewhere.

Why the Distinction Matters

In applications, a parametrization is a way to describe a geometric set, while its domain controls which part is described. A change of coordinates can make a calculation or a geometric description more convenient, but the new coordinates must still reach the intended patch. The cylinder example illustrates the endpoint issue; the disk example shows how a different coordinate system can cover a patch while repeating some parameter values.

For comparing surface images, the decisive question is always whether the same points in space are reached. The condition \(\boldsymbol{\phi}(E)=D\) is an efficient guarantee. When the original parametrization is injective, it is also required. When it is not injective, equality of the parameter ranges can be stronger than necessary, so use the exact image condition instead.

Check Your Understanding

For each question, distinguish carefully between parameter values and points in space.

  1. If \(\boldsymbol{\phi}(E)=D\), why must \(\mathbf{X}\circ\boldsymbol{\phi}\) and \(\mathbf{X}\) have the same image?
  2. What additional assumption makes \(\boldsymbol{\phi}(E)=D\) necessary for equality of the surface images?
  3. For the cylinder parametrization, which distinct angle values give the same point at a fixed height?
  4. Why does the parameter interval \([0,\pi]\) fail to cover the full cylinder, even though it covers a continuous half-turn?
  5. In the polar parametrization of the graph patch, why does the origin have multiple parameter representations?

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