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:

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

Session Information

R version 4.5.0 (2025-04-11 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=C                          
[2] LC_CTYPE=English_United States.utf8   
[3] LC_MONETARY=English_United States.utf8
[4] LC_NUMERIC=C                          
[5] LC_TIME=English_United States.utf8    

time zone: America/Los_Angeles
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] flextable_0.9.7   polarzonoid_0.1-2

loaded via a namespace (and not attached):
 [1] zip_2.3.3               cli_3.6.5               knitr_1.50             
 [4] rlang_1.1.6             xfun_0.52               textshaping_1.0.1      
 [7] gdtools_0.4.2           jsonlite_2.0.0          data.table_1.17.2      
[10] glue_1.8.0              xslt_1.5.1              openssl_2.3.2          
[13] askpass_1.2.1           V8_6.0.3                htmltools_0.5.8.1      
[16] ragg_1.4.0              sass_0.4.10             fontquiver_0.2.1       
[19] rmarkdown_2.29          grid_4.5.0              evaluate_1.0.3         
[22] jquerylib_0.1.4         fastmap_1.2.0           yaml_2.3.10            
[25] lifecycle_1.0.4         compiler_4.5.0          officer_0.6.8          
[28] fontLiberation_0.1.0    Rcpp_1.0.14             katex_1.5.0            
[31] systemfonts_1.2.3       digest_0.6.37           R6_2.6.1               
[34] curl_6.2.2              bslib_0.9.0             logger_0.4.0           
[37] uuid_1.2-1              fontBitstreamVera_0.1.1 tools_4.5.0            
[40] equatags_0.2.1          xml2_1.3.8              cachem_1.1.0