Dual billiards, Fagnano orbits and regular polygons

TL;DR

This paper investigates Fagnano orbits in dual polygonal billiards to characterize regular polygons and analyzes the iteration of the developing map.

Contribution

It introduces the concept of Fagnano orbits for dual billiards and uses them to uniquely identify regular polygons.

Findings

Fagnano orbits characterize regular polygons.

Analysis of the developing map's iteration.

New insights into dual billiard dynamics.

Abstract

We study the notion of Fagnano orbits for dual polygonal billiards. We used them to characterize regular polygons and we study the iteration of the developing map.

Figures

a). If $i$ and $n$ are relatively prime and $i\not\in\{1,n-1\}$ then $X_{i}$ is star shaped and we also call it regular (Figures 6b and c). Finally if $i$ divides $n$ and $i\neq n/2$ then $X_{i}$ is naturally a regular $n/i$-gon which, as mentioned above, we will think of as (a multiple cover of) a regular $n$-gon.

a). If $i$ and $n$ are relatively prime and $i\not\in\{1,n-1\}$ then $X_{i}$ is star shaped and we also call it regular (Figures 6b and c). Finally if $i$ divides $n$ and $i\neq n/2$ then $X_{i}$ is naturally a regular $n/i$-gon which, as mentioned above, we will think of as (a multiple cover of) a regular $n$-gon.

Equations

$$z_i=(w_i+w_{i+1})/2$$

$$(1-2x{)}^n-(-1{)}^n.$$

$$w_i=z_i-z_{i+1}+z_{i+2}-\cdots +z_{i-1},$$

$$X_i\cdot X_{n/2}=(1+q^{2i}+q^{4i}+\cdots +q^{(n-2)i})-(q^i+q^{3i}+\cdots +q^{(n-1)i})=0-0=0.$$

$$z_1-z_2+z_3-\cdots -z_n=0.$$

$$z_1+z_3+\cdots +z_{n-1}=\frac{1}{2}\left((w_1+w_2)+(w_3+w_4)+\cdots +(w_{n-1}+w_n)\right)$$

$$=\frac{1}{2}\left((w_2+w_3)+(w_4+w_5)+\cdots +(w_n+w_1)\right)=z_2+z_4+\cdots +z_n.$$

$$\mathsf{C}:=\{(w_1,\dots ,w_n)\in {\mathbb{C}}^n:w_1+\cdots +w_n=0\}.$$

$$1+q^j=q^j(1+q^{-j})=q^{j(n+1)}(1+q^{-j})=q^{j(n+1)/2}{q}^{j(n+1)/2}(1+q^{-j})$$

$$\frac{1+q^j}{2q^{j(n+1)/2}}=\frac{1+q^{-j}}{2q^{-j(n+1)/2}}.$$

$$1+q^j=q^j(1+q^{-j})=q^{j+j(2k-1)}(1+q^{-j})=q^{kj}{q}^{kj}(1+q^{-j})$$

$$\frac{1+q^j}{2q^{kj}}=\frac{1+q^{-j}}{2q^{-kj}}.$$

$$b_j=\pm q^{j(k+3/2)}{b}_{n-j}=\pm q^{j(k+3/2)}\frac{1+q^{-j}}{2}{a}_{n-j}=\pm q^{j(k+1)}\frac{q^{j/2}+q^{-j/2}}{2}{a}_{n-j}$$

$$b_{n-j}=\pm q^{-j(k+3/2)}{b}_j=\pm q^{-j(k+3/2)}\frac{1+q^j}{2}{a}_j=\pm q^{-j(k+1)}\frac{q^{j/2}+q^{-j/2}}{2}{a}_j.$$

$$(1+q^i)/(1+q^j)=q^{k(i-j)}.$$

$$\frac{a_i}{a_{n-i}}=\frac{2\ell q^{i(k+1)}}{1+q^i}=\frac{1+q^{-i}}{2\ell q^{-i(k+1)}}.$$

$$4{\ell }^2=(1+q^i)\cdot (1+q^{n-i})=2+q^i+q^{n-i}:=f(i).$$

$$\frac{b_j}{b_{n-j}}=\pm \frac{\sqrt{f(j)}{q}^{j(k+1)}}{1+q^{-j}}=\pm \sqrt{\frac{1+q^j}{1+q^{-j}}}{q}^{j(k+1)}=\pm q^{j(k+3/2)},$$

$$\mu^{n}(Q)=\Big{(}\frac{1+q^{j}}{2}\Big{)}^{n}a_{j}X_{j}+\Big{(}\frac{1+q^{n-j}}{2}\Big{)}^{n}a_{n-j}X_{n-j}.$$

$$(1+q^{n-j}{)}^n$$

$${\left(\frac{2}{1+q^j}\right)}^m{\mu }^m(Q)=a_j{X}_j+\sum _{i=j+1}^{n-j-1}{\left(\frac{1+q^i}{1+q^j}\right)}^m{a}_i{X}_i+{\left(\frac{1+q^{n-j}}{1+q^j}\right)}^m{a}_{n-j}{X}_{n-j}.$$

Full text

b) If $n\geq 3$ is even then dedal $n$-gons exist if and only if the vertices of $P$ satisfy $z_{1}-z_{2}+z_{3}-\cdots-z_{n}=0$. If this equation is satisfied then there is a unique dedal n-gon $Q_{0}$ in the space $X_{n/2}^{\perp}:=\{z\in\mathbb{C}^{n}:\ z\cdot X_{n/2}=0\}$. The set $D:=\{Q_{0}+sX_{n/2}:\ s\in\mathbb{C}\}$ consists of the dedal $n$-gons of $P$. In particular for each $i\in\{1,\dots,n\}$ every point $w\in\mathbb{C}$ is the $i$th vertex of a unique dedal $n$-gon $Q_{i}(w)$.*

We remark that the condition $z_{1}-z_{2}+z_{3}-\cdots-z_{n}=0$ means that the center of mass of the even vertices coincides with the center of mass of the odd vertices.

Proof of Proposition 1. If $n\geq 3$ is odd then the map $\mu$ is invertible with

[TABLE]

and thus the dedal polygon exists and is unique.

On the other hand, if $n\geq 3$ is even then the map $\mu$ is not invertible. The kernel is one (complex) dimensional and is generated by the vector $X_{n/2}:=(1,-1,1,-1,\dots,1,-1)$. Dedal polygons exist if and only if $z=(z_{1},z_{2},\dots,z_{n})$ is in the range of $\mu$, i.e., the space spanned by the $X_{i}$ for $i\neq n/2$. The range and kernel of $\mu$ are orthogonal since if $i\neq n/2$ then

[TABLE]

Thus $z$ is in the range of $\mu$ if and only if it satisfies $z\cdot X_{n/2}=0,$ or equivalently

[TABLE]

Alternatively, to see this note that

[TABLE]

The uniqueness of $Q_{0}$ follows since the map $\mu$ is invertible on the space $X_{n/2}^{\perp}$. The statement about the set $D$ follows immediately since $X_{n/2}$ is the kernel of $\mu$. Let $Q_{0}:=(w_{1}^{0},\dots,w_{n}^{0})$. For each $i$, any point $w\in\mathbb{C}$ can be uniquely expressed as $w_{i}^{0}+s$ for some $s\in\mathbb{C}$. $\Box$

Suppose that a polygon $Q$ without self intersection is a dedal polygon of $P$. If $Q$ is convex then it is clearly a Fagnano orbit of $P$ since by convexity the polygon $P$ must be contained in $Q$. On the other hand if $Q$ is not convex then it cannot be a Fagnano orbit since $P$ cannot be contained in $Q$. In particular Fagnano orbits always exist for triangles, but not for polygons with more sides. An example of a pentagon without a Fagnano orbit is given in Figure 5. Although not every polygon has a Fagnano orbit, it does have a periodic orbit that is nearly as simple. Namely, consider the second iteration $T^{2}$ of the dual billiard map. Connecting the consecutive points of a periodic orbit of $T^{2}$ yields a polygon. Cutler has shown that the map $T^{2}$ has a periodic orbit which lies outside any compact neighborhood of $P$ and the polygon constructed from the orbit makes a single turn about $P$ [9]. The existence of a periodic orbit for the usual billiard in an arbitrary polygon remains open.

The set of polygons with center of mass at the origin is

[TABLE]

The eigenvector $X_{0}=(1,1,\dots,1)$ represents a polygon which degenerates to a point. The set $\mathsf{C}$ is the orthogonal complement of $X_{0}$, i.e., $\mathsf{C}=\{w=(w_{1},\dots,w_{n})\in\mathbb{C}^{n}:\ w\cdot X_{0}=0\}$. We will express polygons in the eigenbasis, i.e., we write $P=\sum_{i=0}^{n-1}a_{i}X_{i}$; the coefficients $a_{i}$ are complex numbers. Since the map $\mu$ preserves the center of mass, throughout the rest of the article we assume that the center of mass is at the origin, i.e., $a_{0}=0$.

$X_{1}$ and $X_{n-1}$ are the usual regular $n$-gon in counterclockwise and clockwise orientation (Figure 6a). If $i$ and $n$ are relatively prime and $i\not\in\{1,n-1\}$ then $X_{i}$ is star shaped and we also call it regular (Figures 6b and c). Finally if $i$ divides $n$ and $i\neq n/2$ then $X_{i}$ is naturally a regular $n/i$-gon which, as mentioned above, we will think of as (a multiple cover of) a regular $n$-gon.

Two (unoriented) polygons are called similar if all corresponding angles are equal and all distances are increased (or decreased) in the same ratio. Since we study oriented ordered polygons we additionally want the marking of the vertices of $P$ and $Q$ to correspond, in which case we will call $P$ and $Q$ $\star$-similar. Thus two polygons $P=\sum_{i=1}^{n-1}b_{i}X_{i}$ and $Q=\sum_{i=1}^{n-1}a_{i}X_{i}$ are $\star$-similar if there exists a nonzero complex constant $\ell$ such that $b_{i}=\ell\,a_{i}$ for all $i$. We will also write this as $P=\ell\,Q$. Note that if $P$ and $Q$ are $\star$-similar then they are similar. On the other hand if $P$ and $Q$ are similar then $P$ is $\star$-similar to a cyclic permutation $Q^{(k)}:=(w_{k},w_{k+1},w_{k+2},\dots,w_{k-1})$ of $Q$ or a cyclic permutation $\bar{Q}^{(k)}:=(w_{k},w_{k-1},w_{k-2},\dots,w_{k+1})$ of $Q$ with the opposite orientation.

In analogy to DeTemple and Robertson’s result we have:

Note that if $n$ is odd then $Q$ is the unique dedal polygon of $P$, while if $n$ is even then $Q$ is the unique dedal polygon $Q_{0}\in X_{n/2}^{\perp}$.

Proof. Suppose $P$ is regular, i.e., there is a nonzero complex constant $\ell$ such that $P=\ell\,X_{j}$, where $j\neq n/2$ if $n$ is even. If $n$ is odd then let $Q$ be the unique dedal polygon of $P$. If $n$ is even then since $P$ is regular it satisfies (1). Thus $P$ has dedal polygons and we choose $Q=Q_{0}\in X_{n/2}^{\perp}$. In both cases let $Q=\sum a_{i}X_{i}$. Since $P=\mu(Q)=\sum\frac{1+q^{i}}{2}a_{i}X_{i}$, we have $a_{i}=0$ for $i\neq j$, i.e., $Q=a_{j}X_{j}$. Thus $Q$ is $\star$-similar to $P=\ell\,X_{j}$.

Conversely suppose that $Q=\sum a_{i}X_{i}$ is a dedal polygon of $P=\sum b_{i}X_{i}$ and that $P$ and $Q$ are $\star$-similar, i.e., there is a nonzero complex constant $\ell$ such that $a_{i}=\ell\,b_{i}$. Since $P=\mu(Q)$ we have $\frac{1+q^{i}}{2}a_{i}=b_{i}$ for $i=1,\dots,n$. Combining this with the $\star$-similarity yields $\frac{1+q^{i}}{2}=1/{\ell}$ (a constant) for each $i$ such that $a_{i}\neq 0$. It follows $q^{i}=q^{j}$ if $a_{i}\neq 0$ and $a_{j}\neq 0$. Since $q^{i}\neq q^{j}$ for $i\neq j$ it follows that only a single $a_{j}$ is nonnull. Therefore $Q=a_{j}X_{j}$ and thus $P=\ell^{-1}\,a_{j}X_{j}$, and both are regular. Note that if $n$ is even then $i\neq n/2$, since if $i=n/2$ then $(1+q^{i})/2=0$ and thus $b_{i}=a_{i}=0$. $\Box$

Of course we would like to have the analog of Theorem 2 with the usual notion of similarity. We call an $n$-gon affinely regular if there exists a $j$ ($j\neq n/2$ when $n$ is even) such that $P$ is in the subspace $A_{j}$ generated by $X_{j}$ and $X_{n-j}$. This is equivalent to $P$ being the image of a regular $n$-gon by an affine transformation, which explains the name.

b) If $n\geq 4$ is even then an $n$-gon $P$ appears in the following list of affinely regular polygons if and only if it has a dedal polygon $Q$ which is similar to $P$.*

• i)

Regular $n$-gons

• ii)

Any $P\in A_{j}$ such that there exists $k\in\{1,,\dots,n\}\setminus~{}\{n/2\}$ such that $n$ divides $j(2k-1)$

• iii)

Any $P=b_{j}X_{j}+b_{n-j}X_{n-j}\in A_{j}$ such that there exists $k\in\{1,\dots,n\}$ with ${b_{j}}/{b_{n-j}}=\pm q^{j(k+3/2)}$

All triangles are affinely regular. Berlekamp et al. noticed that every triangle is similar to its dedal triangle [2]. One would like to know if there are other special properties of the polygons in the list.

Proof. Suppose $Q^{(1)}=Q=\sum_{i=1}^{n-1}a_{i}X_{i}$. Then it is a simple exercise in linear algebra to verify that $Q^{(k)}=\sum a_{i}q^{i(k-1)}X_{i}$ and that $\bar{Q}^{(k)}=\sum a_{n-i}q^{i(k+1)}X_{i}$. Thus the similarity of $P$ to $Q$ is equivalent to the existence of a $k\in\{1,\dots,n\}$ and a nonzero complex constant $\ell$ such that $b_{i}=\ell\,a_{i}q^{i(k-1)}$ for $i=1,\dots,n-1$ or $b_{i}=\ell\,a_{n-i}q^{i(k+1)}$ for $i=1,\dots,n-1$.

The structure of the proof is as follows. We group the even and odd cases together and start by proving that the various classes of affinely regular polygons are similar to their dedal polygons. Then we turn to the converse.

Suppose that $n$ is odd and $P=b_{j}X_{j}+b_{n-j}X_{n-j}$ is affinely regular. Let $Q=Q^{(1)}=\sum a_{i}X_{i}$ be the unique dedal polygon of $P$. Since $P=\mu(Q)=\sum\frac{1+q^{i}}{2}a_{i}X_{i}$, we have $a_{i}=0$ for $i\not\in\{j,n-j\}$. We claim that there is a nonzero complex constant $\ell$ such that $P=\ell\,Q^{((n+3)/2)}$. To see this note that

[TABLE]

or

[TABLE]

Thus $P=\ell\,Q^{((n+3)/2)}$ for $\ell:=(1+q^{j})/(2q^{j(n+1)/2})$.

The case $n$ is even and $P$ of class (ii) is similar. Consider the dedal polygon $Q=Q_{0}\in X_{n/2}^{\perp}$. Again $P=\mu(Q)$ implies that $a_{i}=0$ if $i\not\in\{j,n-j\}$. We claim that there is a nonzero complex constant $\ell$ such that $P=\ell\,Q^{(k+1)}$. Since $n$ divides $j(2k-1)$ we have

[TABLE]

or

[TABLE]

Thus $P=\ell\,Q^{(k+1)}$ for $\ell:=(1+q^{j})/(2q^{kj})$.

The case $n$ even and $P$ regular has already been treated in Theorem 2; therefore it remains to treat case (iii) of $n$ even. We suppose $P$ is of this form and want to show that $P$ is similar to $\bar{Q}^{(k)}$, i.e., that there is a nonzero complex constant $\ell$ such that $b_{i}=\ell\,a_{n-i}q^{i(k+1)}$ for all $i$. As before $P=\mu(Q)$ implies that $a_{i}=0$ if $i\not\in\{j,n-j\}$. Combining the two relations, $b_{i}=\frac{1+q^{i}}{2}a_{i}$ and $b_{j}/b_{n-j}=\pm q^{j(k+3/2)}$ yields

[TABLE]

and

[TABLE]

Choosing $\ell:=\pm(q^{j/2}+q^{-j/2})/2\in\mathbb{R}$ yields $P=\ell\,\bar{Q}^{(k)}$.

We turn now to the converse. We will first prove that for $n$ even or odd if $Q=\sum a_{i}X_{i}$ is similar to $P=\sum b_{i}X_{i}$ then $P$ is affinely regular.

We first treat that case when $P=\ell\,Q^{(k+1)}$ for some $k$ and $\ell\in\mathbb{C}\setminus\{0\}$, i.e., $b_{i}=\ell\,a_{i}q^{ik}$. Since $P=\mu(Q)$ we have $\frac{1+q^{i}}{2}a_{i}=b_{i}$ for $i=1,\dots,n-1$. Note that if $n$ is even then $(1+q^{n/2})/2=0$, and thus $b_{n/2}=\frac{1+q^{n/2}}{2}a_{n/2}=0$ and $a_{n/2}=\ell^{-1}q^{-nk/2}b_{n/2}$ is zero as well. For each $i$ such that $a_{i}\neq 0$, combining the two relations between $a_{i}$ and $b_{i}$ yields $q^{-ik}\frac{1+q^{i}}{2}=\ell$. If only a single $a_{j}$ is nonnull, then $Q$ is regular, and since $P$ is similar to $Q$ it is regular as well. Now suppose that $a_{i}$ and $a_{j}$ are nonnull. Then the previous equation implies that

[TABLE]

Taking absolute values yields $|(1+q^{i})|=|(1+q^{j})|$, which implies $i=\pm j$. Thus $P$ is affinely regular.

If $n$ is even we need to conclude more. Note that $(1+q^{-j})/(1+q^{j})=q^{-j}$. Thus taking $i=-j$ in (2) implies $1=q^{j(1-2k)}$. Thus $j(2k-1)$ is a multiple of $n$, i.e., we are in case (ii) of the list.

Finally suppose that $P=\mu(Q)$ and $Q$ are similar but have the opposite orientation, i.e., $P=\ell\,\bar{Q}^{(k)}$ for some $\ell\in\mathbb{C}\setminus\{0\}$, or equivalently $b_{i}=\ell\,a_{n-i}q^{i(k+1)}$. Combing this with $\frac{1+q^{i}}{2}a_{i}=b_{i}$ yields $\frac{1+q^{i}}{2}a_{i}=\ell\,a_{n-i}q^{i(k+1)}$. If $n$ is even and $i=n/2$ then this equation implies that $a_{n/2}=0$. For all other cases it implies that $a_{i}$ and $a_{n-i}$ are simultaneous zero or nonzero. For any $i$ such that they are nonzero there are two such equations, and they imply

[TABLE]

This in turn yields

[TABLE]

It is easy to see that $f(i)=f(j)$ if and only if $j=i$ or $j=n-i$. Since the left-hand side of Equation (4) does not depend on $i$, there is exactly one pair $(a_{j},a_{n-j})$ of nonzero coefficients, i.e., $Q$ is affinely regular. Since $P$ is similar to $Q$ it is also affinely regular.

Finally suppose that $n$ is even. Equations (3) and (4) imply that for each $j$ there exists $k\in\{1,2\dots,n\}$ such that

[TABLE]

i.e., we are in case (iii) of the list. $\Box$

Proof. Suppose $Q=a_{j}X_{j}+a_{n-j}X_{n-j}$. Then

[TABLE]

But

[TABLE]

Thus $\mu^{n}(Q)=\ell\,Q$ with $\ell=(\frac{1+q^{j}}{2})^{n}$. $\Box$

In analogy to the billiard results stated in the introduction we now study the dynamics of $\mu$. On the space $\mathsf{C}$ the dynamics are not very interesting: $\mu^{m}(Q)\to(0,\dots,0)$ as $m\to\infty$ for all $Q\in\mathsf{C}$. To get a somewhat more interesting behavior notice that $\star$-similarity is an equivalence relation. Let $[Q]:=\{\ell Q:\ \ell\in\mathbb{C}\setminus\{0\}\}$ denote the equivalence class of $Q$. By identifying $\star$-similar polygons we obtain the quotient space $\hat{\mathsf{C}}:=\{[Q]:\ Q\in\mathsf{C}\}$. If $Q_{1}$ and $Q_{2}$ are $\star$-similar then so are $\mu(Q_{1})$ and $\mu(Q_{2})$, and thus the map $\mu$ defines a map $\hat{\mu}$ of $\hat{\mathsf{C}}$ to itself.

We remark that the equivalence class $[Q]$ of a polygon $Q=\sum a_{i}X_{i}$ can be represented by the vector $(a_{1},\dots,a_{n-1})\in\mathbb{C}^{n-1}\setminus\{(0,\dots,0)\}$ with the identification $(a_{1},\dots,a_{n-1})\equiv(\ell a_{1},\dots,\ell a_{n-1})$ with $\ell\in\mathbb{C}\setminus\{0\}$. Thus the quotient space $\hat{\mathsf{C}}$ is naturally identified with the complex projective space $\mathbb{CP}^{n-2}$.

Let $\mathsf{S}$ be the unit sphere in $\mathsf{C}$, i.e., the set of all $Q=\sum a_{i}X_{i}\in\mathsf{C}$ such that $\sum|a_{i}|^{2}=1$. Each equivalence class $[Q]$ intersects the unit sphere $\mathsf{S}$ in a circle, i.e., $[Q]\cap\mathsf{S}=\{\ell Q:|\ell|=1/\sum|a_{i}|^{2}\}$. We define ${{\rm{dist}}}([P],[Q])=\inf\{d(P_{0},Q_{0}):\ P_{0}\in[P]\cap\mathsf{S},Q_{0}\in[Q]\cap\mathsf{S}\}$, where $d((a_{1},\dots,a_{n-1}),(b_{1},\dots,b_{n-1}))=(\sum|a_{i}-b_{i}|^{2})^{1/2}$ is the Euclidean distance on $S$. It is a simple exercise to verify that this defines a metric on $\hat{\mathsf{C}}$.

A $\hat{\mu}$-invariant set $A\subset\hat{\mathsf{C}}$ is an exponential attractor with basin $B$ for $\hat{\mu}$ if there exists $c,\gamma>0$ such that $\rm{dist}(\hat{\mu}^{m}([Q]),A)\leq c\exp(-\gamma m)$ for all $m\geq 0$ and all $[Q]\in B$. If $B=\hat{\mathsf{C}}$ then we say that $A$ is a global exponential attractor.

For each $j\in\{1,\dots,\lceil n/2\rceil-1\}$ let $B_{j}\subset\mathsf{C}$ be the subspace generated by $\{X_{j},X_{j+1},\dots,X_{n-j}\}$. Let $B_{\lceil n/2\rceil}:=\emptyset$, and ${\rm{Aff}}:=\cup_{j=1}^{\lceil n/2\rceil-1}{A_{j}}$. For any subset $D\subset\mathsf{C}$ let $\hat{D}:=\{[P]:P\in D\}$. We will only use this notation for sets $D$ which are maximal in the sense that if $\hat{D}=\hat{E}$ then $E\subset D$.

Proof. Suppose $Q\in B_{j}\setminus B_{j+1}$. Equation (Dual Billiards, Fagnano Orbits, and Regular Polygons.) implies that the polygon $\mu^{m}(Q)=\sum_{i=1}^{n-1}(\frac{1+q^{i}}{2})^{m}a_{i}X_{i}=\sum_{i=j}^{n-j}(\frac{1+q^{i}}{2})^{m}a_{i}X_{i}$ is $\star$-similar to

[TABLE]

Since $|(1+q^{i})/(1+q^{j})|<1$ the terms in the middle sum are exponentially small. Thus $\hat{\mu}^{m}([Q])$ is exponentially close to the equivalence class $[P_{m}]\in\hat{A_{j}}$ of the polygon $P_{m}:=a_{j}X_{j}+\left[(1+q^{n-j})/(1+q^{j})\right]^{m}a_{n-j}X_{n-j}$. The fact that $\hat{\rm{Aff}}$ is a global attractor follows from this since $\hat{\mathsf{C}}$ is the union of the $\hat{B_{j}}$. Lemma 4 immediately implies that $\hat{\mu}$ is periodic on each $\hat{A_{j}}$. $\Box$

Acknowledgments.

Many thanks to Sergei Tabachnikov and the two anonymous referees for helpful remarks.