# Implicitization

#### Glenn Davis

#### 2025-06-10

# Introduction

In the User Guide vignette it is shown that there is are homeomorphisms \[\begin{equation} A_n ~~ \longleftrightarrow ~~ \partial Z_n ~~ \longleftrightarrow ~~ \mathbb{S}^{2n} \end{equation}\] where \(A_n\) is the space of \(n\) or fewer pairwise disjoint arcs in the circle, and \(Z_n\) is the polar zonoid in \(\mathbb{R}^{2n+1}\). The forward direction \(\partial Z_n ~~ \to ~~ \mathbb{S}^{2n}\) is the trivial projection onto the unit sphere centered at \((0,...0,\pi)\). But the reverse direction requires intersecting a ray with \(\partial Z_n\), which is non-trivial. The function in the package that performs this intersection is `boundaryfromsphere()`. It expresses \(\partial Z_n\) as the level set of a real-valued function, and uses `uniroot()` to solve for the intersection point.

In this vignette, we outline how \(\partial Z_n\) is defined parametrically and how the real-valued function is derived. This process is called implicitization, see [1]. In the current version of the package, the implicitization has only been computed for \(n\)=1,2, and 3.

Throughout this vignette we think of \(\partial Z_n \subseteq \mathbb{R}^{2n+1}\) and identify the latter with points \((z_1,...,z_n,L)\) where \(z_i\) are complex numbers and \(L\) is real.

Assume first that \(n{=}1\) so there is a single arc in the circle, i.e. a point in \(A_1\). We parameterize the arc by its first and last endpoints \(u_1\) and \(v_1\), both on the unit circle in the complex plane, and proceeding counterclockwise from \(u_1\) to \(v_1\). Let the length of the arc be \(l_1\), so that \(v_1 = u_1 e^{i l_1}\). It is not hard to show that the map \(A_1 \to \partial Z_1\) is given by \[\begin{equation} (u_1,v_1) \mapsto ( ~(v_1{-}u_1)/i, ~ l_1 ~) ~=~ (z_1,L) \end{equation}\] where \(z_1 := (v_1{-}u_1)/i\). Similarly, when \(n{=}2\), it is easy to show that the map \(A_2 \to \partial Z_2\) is given by: \[\begin{equation} (u_1,v_1,u_2,v_2) \mapsto ( ~ \left( v_1{-}u_1 + v_2{-}u_2 \right)/i, ~ \left( v_1^2{-}u_1^2 + v_2^2{-}u_2^2 \right)/(2i), ~ l_1{+}l_2 ~) ~=~ (z_1,z_2,L) \end{equation}\] and so on for \(n \ge 3\). Since \(u_i\) and \(v_i\) are unit vectors, \(u_i \overline{u_i} = 1\) and \(v_i \overline{v_i} = 1\) where the overline denotes complex conjugate. It follows easily that: \[\begin{align} \overline{z_1} &= \big( u_2 v_2 (v_1 - u_1) + u_1 v_1(v_2 - u_2) \big) \left( -i / (u_1 u_2 v_1 v_2) \right) \\ \overline{z_2} &= \big( u_2^2 v_2^2 (v_1^2 - u_1^2) + u_1^2 v_1^2(v_2^2 - u_2^2) \big) \left( -i / (2 u_1^2 u_2^2 v_1^2 v_2^2) \right) \end{align}\]

And using Euler’s formula, one can show: \[\begin{align} \sin(L/2) &= \frac{v_1 v_2 - u_1 u_2}{2i} ~~ \frac{e ^{-iL/2}}{u_1 u_2} \\ \cos(L/2) &= \frac{v_1 v_2 + u_1 u_2}{2} ~~ \frac{e ^{-iL/2}}{u_1 u_2} \end{align}\]

The expressions for \(n{=}1\) and \(n{=}3\) are similar. These right-hand-sides are “rational enough” that algebraic geometry can be used for implicitization. We use the Rational Implicitization algorithm in Chapter 3, Section 3 of [1]. The fact that second term in each RHS is a unit complex number plays a simplifying role. The results are in the next section.

# The Equations for \(n\) = 1,2, and 3 arcs

Throughout this section we assume that \(L \in [0,2\pi]\). With lots of help from Macaulay2 [2], followed by manual rearrangement, one can show that for \(n\) = 1,2, and 3 that \[\begin{equation} (z_1,...,z_n,L) \in \partial Z_n ~~~ \text{ iff } ~~~ | \gamma_n(z_1,...,z_n,L) | = \rho_n(z_1,...,z_n,L) \end{equation}\] and \[\begin{equation} (z_1,...,z_n,L) \in Z_n ~~~ \text{ iff } ~~~ | \gamma_n(z_1,...,z_n,L) | \le \rho_n(z_1,...,z_n,L) \end{equation}\] where \(\gamma_n()\) is a complex-valued function, and \(\rho_n()\) is a real-valued function. The details of this are complicated, especially for \(n{=}3\), and are omitted. For \(n{=}4\), Macaulay2 runs out of memory and cannot finish; but I conjecture that these two functions exist for all \(n\). For small \(n\), they are given in this table:

n n n | γ n(z 1,...,z n,L)\gamma_n(z_1,...,z_n,L)γ n​(z 1​,...,z n​,L) | ρ n(z 1,...,z n,L)\rho_n(z_1,...,z_n,L)ρ n​(z 1​,...,z n​,L)
1 1 1 | z 1 z_1 z 1​ | 2 sin⁡(L/2)2\sin(L/2)2 sin(L/2)
2 2 2 | cos⁡(L/2)z 1 2 − 2 sin⁡(L/2)z 2\cos(L/2)z_1^2 ~-~ 2 \sin(L/2)z_2 cos(L/2)z 1 2​ − 2 sin(L/2)z 2​ | 4 sin⁡2(L/2) − ∣z 1∣2 4\sin^2(L/2) ~-~ |z_1|^2 4 sin 2(L/2) − ∣z 1​∣2

3 3 3

12(4 sin⁡2(L/2) − ∣z 1∣2)z 3 + (8 cos⁡2(L/2)−∣z 1∣2+4)z 1 3 − 48 sin⁡(L/2)cos⁡(L/2)z 1 z 2 + 12 z 1‾z 2 2 12 ( 4\sin^2(L/2) ~-~ |z_1|^2 ) z_3 \newline { ~+~ (8\cos^2(L/2) - |z_1|^2 + 4) z_1^3 } \newline ~-~ 48 \sin(L/2)\cos(L/2) z_1 z_2 ~+~ 12 \overline{z_1} z_2^2 12(4 sin 2(L/2) − ∣z 1​∣2)z 3​ + (8 cos 2(L/2)−∣z 1​∣2+4)z 1 3​ − 48 sin(L/2)cos(L/2)z 1​z 2​ + 12 z 1​​z 2 2​

6 sin⁡(L/2)(∣z 1∣4−8∣z 1∣2−4∣z 2∣2) + 12 cos⁡(L/2)(z 1‾2 z 2+z 1 2 z 2‾) + 96 sin⁡3(L/2){ 6\sin(L/2) (|z_1|^4 - 8|z_1|^2 - 4|z_2|^2 ) } \newline { ~+~ 12\cos(L/2)(\overline{z_1}^2z_2 + z_1^2\overline{z_2}) } \newline ~+~ 96\sin^3(L/2)6 sin(L/2)(∣z 1​∣4−8∣z 1​∣2−4∣z 2​∣2) + 12 cos(L/2)(z 1​​2 z 2​+z 1 2​z 2​​) + 96 sin 3(L/2)

The functions \(\rho_1()\) and \(\rho_2()\) are obviously real, and \(\rho_3()\) is real because \(\overline{z_1}^2z_2 + z_1^2\overline{z_2}\) is real.

The case \(n{=}1\) is in \(\mathbb{R}^3\) and easy to visualize: \[\begin{equation} (z_1,L) \in \partial Z_1 ~~~ \text{ iff } ~~~ |z_1| = 2\sin(L/2) \end{equation}\] This is a surface of revolution - a sine wave around the L axis. It is like a football with poles at (0,0) and (0,\(2\pi\)). Note that these 2 poles are exactly the points where the right side of the equality vanishes.

If we define \(f(z_1,...,z_n,L) := | \gamma_n(z_1,...,z_n,L) | - \rho_n(z_1,...,z_n,L)\) then \(f()\) vanishes on \(\partial Z_n\), is negative in the interior of \(Z_n\), and is positive outside \(Z_n\). This level-set function \(f()\) is used in `boundaryfromsphere()` to compute the intersection of a ray based at \((0,\pi)\) and \(\partial Z_n\). The gradient of \(f()\) is used in `arcsfromboundary()`; see the next section.

# Supporting Hyperplanes of \(\partial Z_n\)

For \(x = (z_1,...,z_n,L) \in \partial Z_n\) evaluating the function \(\partial Z_n \to A_n\) requires the outward normal of a supporting hyperplane to \(\partial Z_n\) at \(x\). The coordinates of this normal become the coefficients of a trigonometric polynomial, whose roots are the endpoints of arcs. If \(\partial Z_n\) is smooth at \(x\) the supporting hyperplane is unique, and is in fact the gradient of the level-set function \(f()\) at \(x\). Also in this case the trigonometric polynomial has the full set of \(2n\) roots, defining \(n\) arcs.

But unfortunately, \(\partial Z_n\) is NOT smooth at all \(x\). We saw this already with \(n{=}1\), where the surface is not smooth at the 2 poles. At the poles, there are infinitely many supporting planes, the trigonometric polynomial has 0 roots, the “south pole” maps to the empty arc, and the “north pole” to the full circle.

The function \(\partial Z_n \to A_n\) is implemented in `arcsfromboundary()`. This R function has to deal with the non-smooth case. It turns out that \(\partial Z_n\) is smooth at \(x\) iff \(\rho_n(x) > 0\). We saw this already with \(n{=}1\). So starting at \(k{=}1\) and incrementing up to \(n\), it tests \(\rho_k(z_1,...,z_k,L)\). If \(\rho_k(z_1,...,z_k,L) < \epsilon\), it returns a collection of \(k{-}1\) arcs in \(A_n\). Otherwise the surface is smooth at \(x\) and it returns \(n\) arcs. Recall that \(A_n\) is the space of \(n\) or fewer arcs. `arcsfromboundary()` first calculates the outward pointing normal at p. The the gradient of the level-set function \(f()\) is used to compute this normal; the expressions for the gradient are more complicated than the ones in the above table, and are omitted here. This coordinates of this normal are the coefficients of a trigonometric polynomial whose roots are calculated with `trigpolyroot()`. These roots are the endpoints of the arcs.

NOTE: The differences \(X_i := \partial Z_i - \partial Z_{i-1}\) almost surely form a Thom-Mather stratification of \(\partial Z_n \subset \mathbb{R}^{2n+1}\), but I have not checked this. The conditions for such a stratification are complex, and are in [3]. Each \(X_i\) is the image of exactly \(i\) arcs. I think that the above functions \(\rho_i()\), after restriction to \(X_i\), can serve as the distance functions in the stratification.

# Summary

The forward functions: \[\begin{equation} A_n ~~ \longrightarrow ~~ \partial Z_n ~~ \longrightarrow ~~ \mathbb{S}^{2n} \end{equation}\] are straightforward. In the package, they are implemented as `boundaryfromarcs()` and `spherefromboundary()` respectively.

But the inverse functions: \[\begin{equation} A_n ~~ \longleftarrow ~~ \partial Z_n ~~ \longleftarrow ~~ \mathbb{S}^{2n} \end{equation}\] use convergent numerical iteration. The first one, `boundaryfromsphere()`, uses the implicit function defined above, and this version of the package requires that \(n\) = 1, 2, or 3. The second one, `arcsfromsphere()`, solves for the roots of a trigonometric polynomial and uses `base::polyroot()`.

# References

[1]
COX, D. A., LITTLE, J. and O’SHEA, D. Ideals, varieties, and algorithms: An introduction to computational algebraic geometry and commutative algebra. B.m.: Springer New York, 2008. Undergraduate texts in mathematics. ISBN 9780387356501.
[2]
GRAYSON, Daniel R. and STILLMAN, Michael E. Macaulay2, a software system for research in algebraic geometry. B.m.: Available at https://macaulay2.com
[3]
WIKIPEDIA CONTRIBUTORS. Thom–mather stratified space — Wikipedia, the free encyclopedia [online]. 2025. Available at: https://en.wikipedia.org/w/index.php?title=Thom%E2%80%93Mather_stratified_space&oldid=1272561764. [Online; accessed 5-June-2025]

# Session Information

This document was prepared Tue Jun 10, 2025 with the following configuration:

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