regular N-gon to be proportional to sk = Tan(kÏ/N) for 1 ⤠k < N/2. There is a long history of interest in trigonometric functions of rational multiples of Ï ...
First Families of Regular Polygons and their Mutations
G.H. Hughes In a 1978 article called “Is the Solar System Stable?” Jurgen Moser [M2] used the landmark KAM Theorem – named after A. Kalmogorov,V. Arnold and Moser – to show that there is typically a non-zero measure of initial conditions that would lead to a stable solar system – but these initial conditions are unknown and this is still an open question. Since the KAM Theorem was very sensitive to continuity, Moser suggested a ‘toy model’ based on orbits around a polygon – where continuity would fail. This is called the ‘outer-billiards’ map and in 2007 Richard Schwartz [S1] showed that if the polygon was in a certain class of ‘kites’, orbits could diverge and stability failed. The special case of a regular polygon was settled earlier in 1989 when F.Vivaldi and A. Shaidanko [VS] showed that all obits are bounded. The author met with Moser at Stanford University in that same year to discuss the ‘canonical’ structures that always arise in the regular case - and Moser suggested that a study of these structures would be an interesting exercise in ‘recreational’ mathematics. This is exactly what it became over the years- but the evolution of these canonical ‘First Families’ proved to be a difficult issue except for regular N-gons such as N = 5 and N = 8 which have ‘quadratic’ algebraic complexity (Φ(N)/2 = 2 where Φ is the Euler totient function.). In “First Families of Regular Polygons” [H4] we showed that the basic geometry of these First Families can be derived from ‘star polygons’. This makes sense because the edges of the star polygons are the extended edges of the underlying polygon –and these form natural orbits. This paper goes one step further to show how the algebraic structure of the First Families can be derived directly from the star polygons. This implies that these families are indeed a geometric realization of the basic algebraic structure of the underlying polygon - and the outer- billiards map is just one possible tool that can be used to help understand the geometric and algebraic evolution of this structure. This is a daunting prospect for polygons with are ‘cubic’ and beyond. We will look briefly at the case of N =11 which is ‘quintic’. In the words of Richard Schwartz “A case such as N = 11 may be beyond the reach of current technology.” It is interesting that the results in [H4] are based on a rather obscure result by Carl Siegel – who was both a mentor and colleague of Moser (see Chronology).We define the ‘star points’ of a regular N-gon to be proportional to sk = Tan(kπ/N) for 1 ≤ k < N/2. There is a long history of interest in trigonometric functions of rational multiples of π (trigonometric numbers) and these sk have some interesting properties that are not shared by Sin or Cos – namely the fact that the ‘primitive’ sk with gcd(k,N) = 1 are algebraically independent over the rationals . This is a nontrivial result based on a 1949 letter from Siegel to Sarvadaman Chowla [Ch]. This independence implies that the scales based on these primitive star points form a basis for N+ - the maximal real subfield of the cyclotomic field of N. This ‘scaling field’ has order Φ(N)/2 and contains all the algebraic structure of N - so calculations are efficient and exact.
‘Star’ polygons or ‘stellated’ polygons were first studied by Thomas Bradwardine (1290-1349), and later by Johannes Kepler (1571-1630). The vertices of a regular N-gon with radius r are {rCos[2πk/N], rSin[2πk/N]} for {k,1,N} A ‘star polygon’ {p,q} generalizes this by allowing N to be rational of the form p/q so the vertices are given by: {p,q} = {rCos[2πkq/p], rSin[2πkq/p]} for {k,1,p} Using the notation of H.S.Coxeter [Co] a regular heptagon can be written as {7,1} (or just {7}) and {7,3} is a ‘step-3’ heptagon formed by joining every third vertex of {7} so the exterior angles are 2π/(7/3) instead of 2π/7. By the definition above,{14,6} would be the same as {7,3}, but there are two heptagons embedded in N = 14 and a different starting vertex would yield another copy of {7,3} - so a common convention is to define {14,6} using both copies of {7,3} as shown below. This convention guarantees that all the star polygons for {N} will have N vertices. Of course this has no effect on the algebraic complexity of the star polygons – which will always match the algebraic complexity of N – namely Φ(N)/2 – where Φ is the Euler totient function. {7,1} (a.k.a. N = 7)
{7,3}
{14,6}
{10,2}
The number of ‘distinct’ star polygons for {N} is the number of integers less than N/2 – which we write as 〈N/2〉. So for a regular N-gon, the ‘maximal’ star polygon is {N, 〈N/2〉}. (Some authors would also allow the ‘boundary’ cases such as {N,0} and {N,N} – which are isolated points or ‘asterisks’.) Our default convention for the ‘parent’ N-gon will be centered at the origin with ‘base’ edge horizontal, and the matching {N,1} will be assumed equal to N. In general sN, rN and hN will denote the side, radius and height (apothem). Typically we will use hN as the lone parameter so the ‘cyclotomic’ case of rN = 1 will have hN = Cos[π/N]. Definition: The star points of a regular N-gon are the intersections of the edges of {N, 〈N/2〉} with a single extended edge of the N-gon (which will be assumed horizontal).
By convention the star points are numbered from star[1] (a vertex of N) outwards to star[〈N/2〉] – which is called GenStar[N] - so every star[k] is a vertex of {N,k} embedded in {N,〈N/2〉}.
(It would seem natural to define the star points to be on the ‘positive’ side of N, but over the years we have chosen to use a clockwise rotation around N - which makes it convenient to use negative star points. The symmetry between these choices makes it irrelevant which one is used.) Lemma: The star points of a regular N-gon with apothem hN are star[k] = -hN{sk,1} where sk = Tan(kπ/N) for 1 ≤ k < N/2 Proof: Since Tan(kπ/N) = Tan(2kπ/2N), the indices divide N into 2N segments centered at the origin with slopes given by sk. □ Example: Here N = 7 (magenta) is circumscribed about N = 14 so they have the same height. This makes it clear that the 3 star points of N = 7 are the even star points of N = 14.
Orbits of Star Points Since every star[k] point is a vertex of the {N,k}star polygon embed in {N, } – it defines a periodic orbit around N as shown below for star[4] of N = 14. All of the star points will have orbits which are ‘linked’ with the orbit shown here. These orbits extend equal distances on either side of the vertices – and this is the defining property of the ‘outer-billiards’ map – but that map is defined relative to the vertices of N – not the edges. The ‘ideal’ orbit shown here can be converted to proper vertex form by simply displacing star[4] slightly and keeping an equal distance on each side of the target vertex - but it should be clear that it will no longer be period 7 – unless it is displaced and rotated by just the right amount. The required rotation is obviously a ‘half-turn’ around N, but a slight stretching is needed to account for the duality between edges and vertices of N: OuterDual[x, N] = RotationTransform[-π/N][x*rN/hN]
The green orbit on the right above has initial point OuterDual[star[4],14]. If all the star points are transformed in this fashion, the new points will be the centers of the First Family polygons. We will derive this independently of OuterDual map or the outer- billiards map.
To define a regular N-gon in space, it is sufficient to know its height (apothem) and its center, but both of these are determined by just knowing the co-ordinates of two star points. The following lemma is almost self-evident, and it is the only algebraic tool needed for analysis. Lemma : (Two-Star Lemma): If P is a regular N-gon, any two star points are sufficient to determine the center and height. Proof. By definition, the star points lie on an extended edge of P. There is no loss of generality in assuming that this extended edge is parallel to the horizontal axis of a known coordinate system with arbitrary center.
Since all points on this extended edge will have a known second coordinate, we will just need the horizontal coordinates of the star points – which we will call p1 and p2 with p2 > p1, so d = p2-p1 will be positive. Relative to P, p1 = starP[j][[1]] = hP*Tan[jπ/N] and p2= starP[k][[1]] = hP*Tan[kπ/N]. (These indices j and k must be known.) There are only two cases to consider: (i) If p1 and p2 are on the same side of P, there is no loss of generality in assuming that it is the right-side of P because star points always exist in their symmetric form with respect to P. In this case we can assume that 1 j < k < N/2 so hP = d/(Tan[kπ/N]-Tan[jπ/N]). (ii) If p1 and p2 are on opposite sides hP = d/ (Tan[kπ/N] + Tan[jπ/N]) and it does not matter whether jk or not. Now that hP is known, the horizontal displacement of p1 and p2 relative to P are : x = hP*Tan[jπ/N] and (x +d) = hP*Tan[kPi/N] if both are on the same side or x = hP*Tan[jπ/N] and (d-x) = hP*Tan[kPi/N] if they are on opposite sides Of course only one of these is needed to define the center of P. □ Example: (The One-Elephant Case) P below shares two edges with the elephant N = 14 – which defines the coordinate system. The star points of N are: starN[k] = -hN{Tan[kπ/14],-1} where hN is arbitrary, so the horizontal coordinates of the shared star points are p1 = -hN*Tan[4π/14] and p2= -hN*Tan[π/14]
Relative to P, p1 = -starP[3][[1]] and p2 = -starP[6][[1]] so
4π π ] − Tan[ ] 14 14 hP = (p2-p1)/(Tan[6π/14]-Tan[3π/14]) = hN and the horizontal displacement 6π 3π Tan[ ] − Tan[ ] 14 14 Tan[
of p1 is x= -hP*Tan[3π/14]. But p2 must yield this same displacement (relative to N) so x must also be -hNTan[π/14]. Algebraically this says that 4π π Tan[ ] − Tan[ ] hP Tan[π /14] ≈ 0.286208264215 14 14 = = 6 3 π π hN Tan[ ] − Tan[ ] Tan[3π /14] 14 14
This horizontal displacement x is equal to –sN/2, so it is the displacement that would result if N was being constructed based on P. These two constructions must have the same displacement because they share the same p1 and p2 points and p1 = -p2 switches between P and N. In the Conformal Existence Theorem, we will show that every star[k] has a matching S[k] tile – so P here is also known as S[4]. The green orbit above is the orbit of the center of S[4] which is cS[4] = {starN[4][[1]]-sN/2, hP-hN}. Algebraically the S[k] and N are closely related because their scaling ratios will always be elements of the ‘scaling field’ defined by N = 14 (or N = 7). This is the number field generated by GenScale[7] = Tan[π/7]*Tan[π/14]. Using Mathematica: 2 AlgebraicNumberPolynomial[ToNumberField[hS[4]]/hN,GenScale[7]],x] = x − x + 1 2
So hS[4]/hN = x − x + 1 where x = GenScale[7] and this has arbitrary accuracy.
2
2
Since N= 14 and the matching N = 7 have Φ(N)/2 = 3 they are classified as ‘cubic’ polygons. This means that any generator of the scaling field (such as GenScale[7] or Cos[2π/7]) will have minimal degree 3, so the scaling polynomials will be at most quadratic. Example:(The Two-Elephant Case) The tiles below exist in the 2nd generation of N = 11. This is a ‘quintic’ N-gon so the algebra is much more complex than N = 14. Px and DS5 share a star point which is off the page at the right, but they do not share any other star points so it was a challenge to find a second defining star point of Px – even though the parameters and star points of DS5 are known. Since these two elephants are only distantly related, it is unusual for them to share a third tile – which we call Sx. This Sx tile shares extended edges with both Px and DS5.
The coordinate system here is based on N = 11 at the origin with radius 1, so hN = Cos[π/11] and this defines the vertical coordinate of the star points above . For the Two-Star Lemma all that is required are the horizontal coordinates p1 and p2:
π π π π π π π π 5π 3π 5π p1 = starDS 5[4][[1]] = −Cos[ ]Cot[ ] + 2 Sin[ ] − Sin[ ]Tan[ ]Tan[ ] − Cot[ ]Sin[ ]Tan[ ]Tan[ ]Tan[ ] 11 22 11 11 11 22 22 11 22 11 22
−5i + 8(−1)1/22 − 15(−1)3/22 − 5(−1)5/22 + 5(−1)7/22 + 5(−1)13/22 + 15(−1)15/22 − 8(−1)17/22 + 3(−1)19/22 + 3(−1) 21/22 − starPx[3][[1]] = − p2 = π 5π 4(1 + (−1) 4/11 )(−1 + (−1)5/11 )(Cos[ ] − Sin[ ]) 11 22
(Because Px does not share any scaling with the First Family for N = 11, its parameters are algebraically far more complex than DS5. The form shown here for p2 is a simplification of the trigonometric form – which would fill most of this page. Mathematica prefers to do these calculations in ‘cyclotomic’ form as shown here. Of course p2 will have vanishing complex part.) By the Two-Star Lemma: hSx = (p2-p1)/(Tan[4π/11]+Tan[3π/11]) = 2i − 11(−1)1/22 + (−1)3/22 + (−1)5/22 − 11(−1)7/22 + 12(−1)9/22 − 4(−1)13/22 + 4(−1)17/22 − 2(−1)19/22 − 12(−1) 21/22 5π π 5π 3π 2(1 + (−1) 4/11 )(2 + (−1) 2/11 − (−1)3/11 + (−1) 4/11 − (−1)5/11 + (−1)6/11 − (−1)7/11 + (−1)8/11 − (−1)9/11 )(Cot[ ] + Cot[ ])(Cos[ ] − Sin[ ]) 22 11 22 22
As expected hSx/hN ≈ .00150329 is in the scaling field and since hN = Cos[π/11] is also in the scaling field, hSx itself is in the scaling field: AlgebraicNumberPolynomial[ToNumberField[hSx,GenScale[11],x] yields hSx = −
47 543 x 5 x3 25 x 4 where x = GenScale[11] = Tan[π/11]*Tan[π/22] + + 21x 2 + − 32 16 16 32
Here are the steps to construct Sx: (i) Using p2, the displacement is x = hSx*Tan[3π/11] so MidpointSx = {-starPx[3][[1]]-x,-hN} ≈{-6.201044900,-Cos[π/11]} (ii) cSx = MidpointSx + {0,hSx} (iii) rSx = RadiusFromHeight[hSx,11] (iv) Sx = RotateVertex[cSx + {0,rSx},11,cSx] Below is a web-scan of this region. The web W is the ‘singularity set’ of the outer-billiards map , so it is obtained by mapping the extended edges of N = 11 under or -1. Note that Sx has a clone obtained by rotation about the center of Px. This web is probably multi-fractal. Click to enlarge.
.
It is our contention that all the polygonal ‘tiles’ (regular or not) which arise from the outer billiards map of N are defined by scales which lie in the scaling field of N. For N = 11, knowing the exact parameters of ‘third-generation’ tiles like Sx allow us to probe deeper into the smallscale structure of N = 11 – which is almost a total mystery. But algebraic results like this point to the fact that each new ‘generation’ will tend to involve a significant increase in the complexity of any algebraic analysis. Therefore it may be impossible at this time to probe deeper than 10 or 12 generations for N = 11. (Since the generations scale by GenScale[11] = Tan[π/11]*Tan[π/22] ≈ 0.0422171, the 25th generation would be on the order of the Plank scale of 1.6·10-35 m.) Conforming Regular Polygons In the One-Elephant case, the resulting S[4] polygon is ‘conforming’ to the edges of the star polygons of N = 14 because S[4] shares the same base edge as N and GenStar[S[4]] is the star[1] vertex of N. When N is even, the angle determined by these two edges is always the exterior angle of N, so the extended edge of S[4] matches the edge of the star polygon. (Note that if S[4] was replaced by a regular heptagon with the same center and height, the GenStar point would not change, so it would still be ‘conforming’. It is just N that must be even.)
Definition: A regular polygon P is said to be conforming relative the regular polygon N if P shares the same base edge as N and GenStar[P] = star[1] of N. P is said to be strongly conforming if it is conforming and also shares another star[k] point with N. Lemma (Conformal Replication): For N even, every N-gon has a strongly conforming ‘dual’ DN which is congruent to N. Proof: Set -starD[1] = GenStar[N]. Replicate N with using center offset -sN/2. By symmetry the corresponding tile DN must have -GenStar[DN] = star[1] so it is strongly conforming. □ It should be clear that DN will be the largest possible strongly conforming tile relative to N as shown below for N = 14. In the outer-billiards world the D’s are globally maximal.
Corollory: For a regular N-gon, these is a unique conforming regular N-gon P with hP hN. Proof: By the Conformal Replication Lemma, there is a conformal DM for any regular N-gon M that shares the base edge and star[1] of N.This DM will also be conformal relative to N because N and M have the same exterior angle. Therefore a conforming DM must exist for any hDM hN. □
Example: (N = 11) The illustration below shows some tiles which are conforming relative to D[1] – which is the second-generation D tile of N = 11. We will define the First Family of D[1] to be the strongly conforming tiles - which are shown below in cyan. The black are the actual tiles that arise in the outer-billiards map. Tiles like M[1], DS5 and M[2] (and their symmetric counterparts at GenStar[11]) are perfect matches between cyan and black, so they are the only survivors of the ‘ideal’ First Family of D[1]. This is a very minimal second-generation.
The DS5 from the Two Elephant Case is strongly conforming because it shares the star[1] and star[5] points of D[1]. Any tile that is at least conforming to D[1] will have a web evolution that is tied to D[1], so Px and Mx will have local webs which evolve from the extended edges of D[1] – just like DS5 and M[1]. This makes it possible to trace their evolution and determine their parameters. (See Appendix C of [H3] for a derivation of Mx.) Theorem (Existence of Strongly Conforming Tiles): For a regular N-gon with N even, every star[k] point defines a unique S[k] tile which is strongly conforming and has horizontal center displacement -sN/2 relative to star[k].
Proof: For p1 = star[k] and displacement x = -sN/2, the Corollory to the Conformal Replication Lemma says that there must be a conforming regular N-gon P with cP[[1]] = star[k][[1]] + x as shown above for star[5] of N = 16. We will show that in this case P must have star[k] as a star point and starP[j] = star[k] for j = N/2-k. (This must be true for k = 1 because then star[k] = star[1] and p1= p2. Also when k = N/2-1, j = 1 and P is the DN from the Replication Lemma.) Measured from p2, the displacement of cP[[1]] is - star[k][[1]] = hNTan[kπ/N]. Since p2 is -GenStar[P] = -starP[N/2-1] = hPTan[(N/2-1)π/N], hNTan[kπ/N] = hPTan[(N/2-1)π/N] so
hP Tan[kπ / N ] = hN Tan[( N / 2 − 1)π / N ] Tan[kπ / N ] Tan[π / N ] But for 1 k < N/2,= so relative to Tan [π / N ]Tan[kπ / N ] = Tan[( N / 2 − 1)π / N ] Tan[( N / 2 − k )π / N ]
p1, the index of P is j = N/2-k and hP = hN
Tan[π / N ] = Tan[π / N ]Tan[kπ / N ] Tan[( N / 2 − k )π / N ]
This defines P relative to N and S[k] = P is strongly conforming for 1< k N/2-1.
All of the S[k] formed in this fashion share two distinct star points with N, except S[1] where the local index is the same as the global index – namely N/2-1. In this case p1 = p2, but the displacement is unchanged and hP = hN
Tan[π / N ] Tan[π / N ] = = Tan[π / N ]2 Tan[( N / 2 − k )π / N ] Tan[( N / 2 − 1)π / N ]
Therefore S[1] can be constructed without the help of star[2], but we will show that the illustration above for N = 16 is canonical and S[1] always has index 2 relative to star[2]. To see this, apply the Two-Star Lemma with opposite sides and (hypothetical) local index 2. This yields
hS [1] Tan[2π / N ] − Tan[π / N ] = = Tan[π / N ]2 as above. hN Tan[( N / 2 − 1)π / N ] + Tan[2π / N ]
Therefore all of the S[k] tiles are strongly conforming and share the same star[k] offset. □
Example (N = 16): To finish the example above, hP (and hence P) is determined using either p1 or p2 (relative to P): Using p1 the displacement is x, but to find hP it is necessary to know that the local index is N/2-k. Since P is S[5], this index is 3. Therefor x = hNTan[π/16] = hPTan[3π/16] which says that hP/hN = Tan[π/16]/Tan[3π/16]. Using p2 the index is always the maximal N/2-1 and the displacement from P is -star[5][[1]] = hNTan[5π/16] = hPTan[7π/16] so hP/hN = Tan[5π/16]/Tan[7π/16] - which is the same as the ratio from p1. Note that the local indices of the S[k] must be N/2-k because by convention the star[k] are numbered from right to left but the ‘star angles’ increase from left to right as shown below.
Definition. For a regular N-gon with N even, the (ideal) First Family consists of the N/2-1 S[k] tiles and their symmetric DS[k] obtained by reflection about the center of S[N/2-2]. Example: The (ideal) First Family for N = 16
These ‘canonical’ sN/2 displacements are derived only from the geometry of the star polygons but they are exactly the displacements that would result from applying the OuterDual map to any point on the horizontal axis. Therefore it is no coincidence that this displacement is consistent with the ‘shear’ discontinuity demanded by the outer-billiards map. Actually the shear displacement of each star point is sN itself. (This shear is illustrated below in the section on Internal Mutations.) For each S[k] the center of this shear is always the center of S[k] because each shear is followed by a rotation, so the outer shell of S[k] is formed recursively by replications of this shear and rotation. This is why all the S[k] have centers which are displaced by sN/2. In the special case where star[k] is actually a vertex of S[k], the side will be sN so S[k] will have the same side as N. For N even, this occurs at GenStar[N], which is star[N/2 -1]. This yields hS[k]/hN = 1 so S[N/2-1] is a clone of N. No shear can exceed sN, so this is the maximal possible regular N-gon tile – which we call D. Lemma (Gender Change Mutation): When N is twice-odd every S[k] of the First Family with k odd can be changed from a regular N-gon to a regular N/2 –gon with the same center and height without violating any of the results of the Existence Lemma. (This case is illustrated on the left here.)
Proof: In the Existence Theorem hS [k ] Tan[π / N ] Tan[π / N ] = = ( N / 2 − k )π N hN Tan[( N / 2 − k )π / N ] / ] Tan[ 2 2 This says that if the local indices are divided by 2 (and the shared index is unchanged) any N-gon S[k] could be swapped for a N/2-gon. This reduction of indices is what would happen iff the new N/2 gon shared the same center and height as S[k] - because then it would have star[k] = star[2k] of S[k]. This would only work for S[k] with k odd because N/2-k must be even. □ The new S[N/2-2] tile will now have local index 1, so it will have the same side as N. This tile will be the surrogate N/2 ‘parent’ for the odd case. The old and new sides are related by Tan[π/N]/Tan[π/(N/2)] - which we call ScaleChange[N/N/2]. In the twice-even case, N/2-k will be even when k is even, so in theory the even S[k] could be reduced to N/2-gons - except for the fact that the shared index would now be different - as illustrated on the right above for N = 12 and N = 6. Because of the mismatch of GenStar points, N = 12 and N = 6 will have very different dynamics under the outer billiards map – while N = 14 and N = 7 will have virtually the same dynamics.
Definition: For N twice-odd, the First Family will consist of the (ideal) S[k] when k is even and the gender-change S[k] when k is odd. When N is odd the First Family is identical to the twiceodd case with the origin translated to the center of S[N/2-2] – which is now regarded as the ‘parent’ N-gon. When N is twice-even, the First Family is the same as the ideal First Family. Note: Since all of this began as a geometric fairy-tale, it was our (politically dubious) tradition to assign male and female to the even and odd cases. This really only makes sense for N odd or twice-odd , but the maximal D tiles always exist and it was natural to call them D for Dad. For the odd or twice-odd case, each D does indeed have a matching M tile which is odd but with the same side length as D. These two occur naturally in the First Families for N odd or twice-odd. The even and odd cases are fundamentally different under the outer-billiards map because is a reflection relative to a vertex c of the ‘base’ polygon, so τ(p) = 2c – p as shown on the left below. The points inside a polygonal tile of the web must all map together under , so they will all have the same period q - and it takes an even number of reflections to have a periodic orbit for the tile - so q will be even. The one possible exception is the ‘center’ of the tile – which would still have period q, but by symmetry it might have minimal period q/2. A regular M-type tile with an odd number of edges would not have the required reflective symmetry for this to occur - so every point inside would have to be even period. For D-type tiles like S[4] of N = 14, the center can be period 7 – as shown earlier- but all the other points inside S[4] must be period 14. Example: The orbits of S[1] and DS[1] in N = 5 illustrate the contrast between tiles with an even number of edges and odd number of edges. Both orbits involve 10 inversions so they are period 10. It looks like S[1] has only been inverted 5 times because S[1] can map to itself (inverted) after 5 inversions – but a pentagon cannot do this - so it takes the full 10 inversions for DS[1] to map to itself. Every point in S[1] is indeed period 10 – except the center – so these are called period-doubling orbits. D = S[2] has a similar orbit – which is step-2 instead of step-1.
Every First Family has the potential to support a 2nd generation on the edge of D as shown above for N = 5. For an odd N-gon, the scale is GenScale[N] = Tan[π/N]*Tan[π/2N]. (This is also the generator of the ‘scaling field’ for N). For N = 5, GenScale[5] ≈ .23607. Since the geometric scaling is known here, all that is needed to determine the fractal dimension of the web is the ‘temporal’ scaling – which is the growth of the periods of these tiles. The periods of the D[k] tiles are: 5,35,205,1235,7405 (using centers) and they satisfy = P(k ) (5 / 7)(8*6k −1 + (−1) k ) . The limiting temporal scaling is therefore 6 and the Hausdorff-Besicovitch fractal dimension is Ln[6]/Ln[1/GenScale[5]] ≈ 1.2411 (See [T].) All ‘quadratic’ polygons have a simple fractal web.
Internal Mutation of First Family Members The gender-change transformation could be called a ‘mutation’ of the S[k] tile, but there is another class of mutations which yield non-regular tiles. These will be called ‘internal’ and ‘external’ mutations respectively. We have seen that the internal mutations do not violate any of the results from the Two-Star Lemma or the Existence Theorem and here we will explain why these mutations arise from the outer-billiards map. Below is the First Family for N = 14, showing the gender-change mutations in the cyan odd S[k]. In each case the result is a magenta heptagon with the same center and height.
As indicated earlier, the ‘shear’ under the outer billiards map is sN – the side of N. On each iteration of the web this shear is followed by a position-dependent rotation because every edge of N is being iterated and the domains map to each other. This ‘shear and rotate’ scenario is common for any piecewise isometry acting on a polygon. Each domain is partitioned by the star polygon edges and for N even, the rotation angles are the ‘star-angles’ kφ where φ =2π/N. D is constructed by a repetition of this ‘shear and rotate’ process. Since the rotation angle is φ, D is a clone of N. For S[5] the rotation is 2φ, so repetition will form a N/2-gon with the same side length as N. But the S[k] are actually formed from two competing shears – one relative to N and the other relative to D as shown above. Sequentially these shears are φ apart because they occur on consecutive edges of N. When N is twice-odd as shown here and S[k] is odd, it takes an even number steps to go from the bottom edge to the top shear and each step is 2φ, so the top shear and bottom shear are a perfect match and either one would form S[k] independently - so the S[k] are N/2-gons. This can be regarded as a ‘mutation’ relative to the twice-even case described below. The twice-even case has the same rotation angle and the same two shears, but now it takes an odd number of steps to go from the bottom edge to the top shear, so these two are off by φ relative to each other. Therefore they will form the interlaced even and odd edges of an N-gon. This 2-step method of creation of S[N/2-2] is still not ‘normal’ compared with the simple step-1 process by which S[N/2-1] (D) is formed. This applies to all the S[k] except D. Their evolution will be multi-step and this will imply that their local First Families are also multi-step Note: In the illustration above, the origin could be shifted to N = 7 and nothing would change because for N odd the star angle is φ/2 - so N = 7 would generate the First Family of N = 14.
Evolution of First Family Tiles - Local First Families Every regular tile has a corresponding First Family so each S[k] tile of a regular N-gon has a ‘local’ First Family which is simply a scaled version of the First Family of N. This is an invitation to recursion. For ‘quadratic’ polygons like N = 5 or N = 10 , these local First Families actually exist on all scales – as long as the families evolve from D – so the ‘second generation’ would be based on DS[2] (or S[2]) acting as D[1]. We saw earlier that this evolution no longer works for N = 11 because very few First Family tiles survive the transition to the second generation – and this implies that there is no obvious path of evolution. For 4k+1 primes like N = 13, the path is smoother because the critical tiles such as M[2] and D[2] survive the transition – as shown below. (Conforming tiles like Px are actually quite common. We call them ‘volunteers’ - because they are part of the extended gene pool.)
Even in cases where it is not possible to predict the long-term evolution of S[k] tiles, it is possible to make predictions about their local First Families. Under the outer-billiards map, the S[k] tile of a regular N-gon for N even will be formed with rotation angle equal to the interior angle determined by star[k] of N - namely (N/2-k)φ, where φ = 2π/N is the external angle of N. Therefore relative to N, S[k] will be formed in a ‘step-(N/2-k)’ fashion as illustrated below for N = 22. This rotation angle is independent of whether the S[k] is mutated or not, so S[10] (D) will have a step-1 First Family (which is the actual First Family of D ), S[9] (N = 11) will have a step-2 First Family, S[8] will have a step-3 First Family and S[7] will have a step-4 family. These all must be compatible with the First Family of N= 22.
This says that the local (step-2) First Family for N = 11 consists of clones of S[2],S[4] and S[6] – as well as the existing S[8] – and of course S[10] which is N = 22. This is indeed the correct First Family for N = 11 – so we will add the 3 missing tiles and call it the Full First Family for N = 22. Adding these files is trivial because the canonical displacements of sN/2 are still valid. We could also add S[1] and S[4] of S[8] and S[2] of S[7] – knowing that these tiles will indeed exist. The case for N twice-even is very similar – with local S[k] which are odd.
As remarked in the introduction to [H2] – very little is known about how step-k webs evolve relative to the ‘normal’ step-1 webs of the outer-billiards map – but we do know something about step-2 webs from experience with the Digital Filter map – where the algorithms allow variable step size. We showed in [H3] how certain step-2 Df webs are related to outer-billiards webs and this may allow us to make some progress in difficult cases like N = 11. Example: The Full First Family for N = 26 is shown below - with the added tiles in magenta.
In the twice-odd case, using the Full First Family also yields a more complete First Family for N odd. With N =13 shown here at the origin, symmetry would allow the First Family to be reduced by half if desired. When N is twice-even, the First Family can also be extended in this fashion. This is done below for the central S[10] of N = 24 where the added Sk] are odd instead of even. External Mutations of S[k] Tiles The First Family tiles that actually occur in the outer billiards map for a regular N-gon may involve further ‘mutations’ of the S[k]. These ‘external’ mutations can only occur when gcd(k,N) >1. In these cases the period of the S[k] tile will be reduced and this may cause an incomplete local web to form. Therefore all such mutations are formed from extended edges of the ideal S[k] (in a manner similar to the ‘internal’ mutations of the twice-odd case) and these equilateral mutations do not affect the center. Example: Mutations in the First Family of N = 24. (Click for more detail)
For example the mutated S[8] has bottom edge which extends from the right-side star[3] of the underlying S[8] to the normal left-side star[1] vertex. This edge is then rotated in the web and merged with the extended top edge to form equilateral edges. The result is the weave of two regular hexagons at slight different radii. Here are the steps to construct the mutated S[8]: (i) Find the star[3] point: MidS8= {cS[8][[1]],-hN}; star[3] = MidS8+{Tan[3*Pi/24]*hS[8],0}; (ii) H1=RotateVertex[star[3], 6, cS[8]] (magenta); H2 =RotateVertex[S[8][[7]], 6, cS[8]] (blue) (iii) MuS8 = Riffle[H1,H2] (black) (This weaves them as in a card shuffle.) The secondary mutation above is known as S94 because it is actually based on S[4]. (Note that S[4] itself has a different mutation.) S[9] and the embedded S[4] share a close connection with the local S[1] – which is not shown above. This local S[1] is an S[3] relative to S[9] and S[8]
relative to the embedded S[4]. This S[1] tile occurs more frequently than any other tile because it has algebraic and geometric connections with many of the S[k]. One way to explore the connections between S[k] tiles, is to look at the canonical nesting of the S[k] as shown below.
This nesting is due to the N-D ‘duality’ discussed earlier. These D’s share star[1] of N because the S[k] are conforming. Even though the D’s are ‘virtual’ they often have First Family members which match the actual First Family of N, and it makes sense that these matches will often involve S[1]. By symmetry, a match at the foot of N usually translates to a match at S[k]. For N = 24, S[1] occurs at S[10], S[9], S[8], S[7], S[6], S[5] (the last 3 are shared), S[3], and the most important match at S[2] where S[1] is a ‘second-generation’ S[10] relative to S[2]. In every location, the dynamics are different, but the members of the local First Family of S[1] often survive to create complex inter-family relationships – such as those observed with S94. For N twice-odd the generation scaling of the First Families involves S[2] serving as the second generation D tile - and S[1] is the matching M tile. For the twice-even case, S[1] is the key to the generation evolution but there is still a close relationship between S[1] and S[2] (which will be made explicit in the next section). In truth a case like N = 24 is very similar to N = 14 because for N = 14, S[2] is indeed the key to the first family evolution, but relative to S[N/2-2] (N = 7), this is an S[1] tile because of the gender issue. Therefore the key tile for N = 14 is actually N = 7 - which determines the generation scaling. This is also true for N = 24 where S[10] determines the generation scaling and S[1] is the next generation S[10]. In general very little is known about the extended family structure of the twice-even family. Even for the 2k family it seems that the members have little in common – but that may be due to our lack of understanding. Much of the algebraic and geometric complexity for any regular N-gon can be traced to the cyclotomic field N. (See the definition in the next section.) For N = 24 this field is generated by { 2 (2Cos[2π / 8]), 3 (2Cos[2π /12]) and i} so it is closely related to 8 and 12 – which are generated by {√2, i} and {√3, i} respectively. For example there are prominent octagons in the web of N = 24 and the S94 mutation is same as the mutation of S[2] of N = 12.
These mutated tiles have very interesting dynamics in their own right and it is an open question to determine if there is any relationship between their local ‘in-situ’ dynamics and their ‘in-vitro’ dynamics when they are the ‘parent’. Of course this same question applies to any tile that arises in the outer-billiards map. In general there may be no correlation, but for tiles such as D, star point symmetry yields a simple reflective duality between D and N. With step-2 adjustments we have seen that it is possible to relate the ‘in-situ’ and ‘in-vitro’ dynamics of S[N/2-2] and it is our hope that this can be extended to more general step-k tiles
Scaling of Tiles Scaling is a very important issue in any ‘dynamical system’ because it provides insight into how the system evolves. Here we look at the scaling determined by the star polygons of any regular polygon and show how this scaling is related to the First Family scaling. Recall that the star points of a regular polygon are: star[k] = -hN{sk,1} where sk = Tan(kπ/N) for 1 ≤ k < N/2 The ‘primitive’ sk are those with gcd(k,N) = 1 and the remaining sk are ‘degenerate’. Example: For N = 14 the primitive star points are star[1], star[3] and star[5], and the magenta N = 7 shown here circumscribed about N = 14 has primitive star points star[2], star[4] and star[6].
Definition: The ‘canonical scales’ of an N-gon are scale[k] = Tan[π/N[/Tan[kπ/N] = s1/sk. The primitive scales are those where sk is primitive and the remaining scales are ‘degenerate’. GenScale[N] is defined to be scale[]. By definition star[k][[1]] = -hNTan[kπ/N], so scale[k] = star[1][[1]]/star[k][[1]]. Therefore the scales are the relative horizontal displacements of the star[k]. This makes them a natural choice for use with the outer-billiards map where the discontinuities are linear displacements. Note: For a given N-gon with central angles as shown here, it would also be possible to use Cos or Sin to define the star points and scales, but the crucial independence results of Carl Siegel only apply for Tan - and the use of rN instead hN as a reference would be awkward. By definition scale[1] is always 1 and the scales are decreasing, so GenScale[N] is the minimal scale. Each star[k] point defines a scale[k] and also an S[k] tile, and there is a simple relationship between them which will be explained below. By the independence result of Carl Siegel, the primitive star points and primitive sales are both independent over the rationals , so any degenerate scale can be written as a linear combination of the primitive scales – but this is not always easy to do. Example: For N = 9, the sk are {Tan(π/9), Tan(2π/9, Tan(3π/9), Tan(4π/9} and they are all primitive except for s3 = Tan(3π/9) = √3 and this must be a linear combination of the remaining sk. Here the solution follows from the fact that Tan(π/9) – Tan(2π/9) + Tan(4π/9) = 3√3. The matching cotangent relationship can be used to relate the scales: 3scale[3] = (scale[1] - scale[2] +
scale[4]). Here GenScale[9] = scale[4] which is primitive, but GenScale[18] = scale[8] – which is not primitive. Definition: For a regular N-gon, the cyclotomic field, N , is the algebraic number field generated by z = Cos(2π/N) + iSin(2π/N). As a vector space N is the direct sum of its real and imaginary parts - N+ and N-, Since z + z-1 = 2Cos(2π/N), N+ can be generated by Cos(2π/N). Since z has (minimal) degree Φ(N) this subfield has degree Φ(N)/2. Because of the Lemma below this is called the ‘scaling field’ of N – written [N]. (Because N is equivalent to N/2 for N twice-odd, [N] = [N/2] in that case.) Scaling Field Lemma (Corollory to Theorem 2 of H[4]). The scaling field for N has a basis consisting of the primitive scales. When N is odd or twice-even, one of the primitive scales is GenScale[N] and it has degree Φ(N)/2 so it is a generator of [N] . When N is twice-odd GenScale[N/2] is primitive and degree Φ(N)/2 so it is a generator of both [N] and [N/2]. Since the canonical scales of N are linear combinations of the primitive scales, they are all in the scaling field. Because [N] is a field ,every scale[k] has a matching cotangent scale 1/s[k] which is in [N]. These are called the ‘dual’ scales and the primitive dual scales are independent. Definition (Canonical polygons) Every regular N-gon defines a coordinate system, and any line segment or polygon P (convex or not) that exists in this co-ordinate system will be called canonical relative to N if for every side sP, the ratio sP/sN is in [N]. (When P is a regular Ngon (or N/2-gon when N is twice-odd), it will be canonical when hP/hN is in [N] and in this case every Q that is canonical with respect to P will also be canonical with respect to N because [N] is a field. Example: Any line segment defined by a linear combination of star points is canonical because if T is such a linear combination T/s1 is a linear combination of dual scales, so it is in [N]. Therefore every internal mutation or external mutation is canonical. In addition every star polygon based on N is canonical. It is our conjecture that all tiles and line segments which arise in the web W of the outer-billiards map canonical. Below are iterations 0,1,2,3,4 and 10 of the star polygon web for N = 14. For any regular polygon, this ‘inner-star’ region is invariant under W and it is a ‘template’ for the global web. When N is even, symmetry allows this template to be reduced to half of the magenta rhombus as shown on the right,and the Digital Filter map folds the global web into this rhombus.
Lemma: For a regular N-gon, the First Family tiles are canonical regular N-gons or N/2-gons. Proof: By the definition of the First Family, there is no loss of generality in assuming that N is even – because any odd N-gon can be regarded as the S[N/2-2] tile of the matching 2N-gon. There is also no loss of generality is assuming that the tile in question is an S[k] – because all of the First Family tiles are either an S[k] or a clone of an S[k. By the Existence Theorem (and the Gender- Change Lemma): for 1 k N/2-1 hS [k ] Tan[π / N ] Tan[π / N ] =( for N twice-odd and k odd) = ( N / 2 − k )π N hN Tan[( N / 2 − k )π / N ] / ] Tan[ 2 2 Since hS[k]/hN has the form s1/sj it is in the scaling field S[N] - so when S[k] is an N-gon it is canonical because hS[k]/hN = sS[k]/sN. When S[k] is an N/2-gon, hS[k]/hN is unchanged because these mutations do not affect the center or height. But in the definition of ‘canonical’ tiles, the line segments and tiles are scaled by sides, so to guarantee that the mutations are canonical N/2-gons, it is necessary to show that the ratio of the sides is in [N]. As pointed out earlier, this ratio is ScaleSwap[N,N/2] which is always in [N] when N is twice-odd. □ Since the S[k] are always canonical regular N-gons or N/2 gons, the First Families generated by the S[k] will also be canonical regular N-gons or N/2 gons. The linear or quadratic cases like N = 3,4,5,6, 8, 10, 12, 14 and 18 have webs consisting of only First Family tiles with possible external mutations, so all tiles are canonical. For all N, the initial web is partitioned by the star points of N so it is also partitioned by the S[k] and therefore the web evolves along with these S[k]. This does not imply that the web evolution matches the First Family evolution – but they are related and it makes sense that they would share the same scaling. We will define ScaleS[k] = hS[k]/hN. For N even, the relationship between ScaleS[k] and scale[k] is ScaleS[k] = scale[N/2-k]. Using GenScale[N] this ‘retrograde’ connection can be expressed as follows: for N even and 1 k N/2-1 hS [k ] Tan[π / N ] Tan[π / N ]2 GenScale[ N ] [k ] = ScaleS / N] = = Tan[π / N ]Tan[kπ= = h[ N ] Tan[( N / 2 − k )π / N ] scale[k ] Tan[π / N ] Tan[kπ / N ] This retrograde relationship is due to the symmetry of the N-even case - so the number of scales is effectively reduced by ‘half’. Because of this connection it is natural to associate S[k] with scale[k] and the S[k] can be regarded as geometric realizations of the scales. The two most important S[k] and scales for N even are: (i) hS[1]/hN = ScaleS[1] = scale[N/2-1] = GenScale[N]/scale[1] = GenScale[N] = Tan[π/N]2 (ii) hS[2]/hN = ScaleS[2] = scale[N/2-2] = GenScale[N]/scale[2] = Tan[π/N]Tan[π/(N/2)] When N is twice-even, GenScale[N] is the generator of [N] so the matching tile is S[1], but S[1] and S[2] are always related by: ScaleS [1] Tan[π / N ] GenScale[ N ] = ScaleSwap[ N , N / 2] = (The last equality = Scale[ S [2] Tan[π ( N / 2)] GenScale[ N / 2] is only valid when N is twice-odd)
This implies that either GenScale[N] or GenScale[N/2] could be used as the generator of the scaling field when N is twice-odd, but the natural choice is GenScale[N/2] = ScaleS[2] = Tan[π/N]Tan[π/(N/2)] – because in this case GenScale[N] is not primitive and not a unit. Geometrically the issue with GenScale[N] is the gender-difference between N and S[1]. The ScaleSwap formula always expresses the ratio of the scales of S[1] and S[2] so it is still valid in the twice-even case - but it is no longer the ratio of the GenScales – and this is one reason why cases like N = 12 and N = 6 are very different. Example: N = 6 has linear algebraic complexity and this means that the star polygon web does not evolve past the First Family as shown here.
hS[1]/hN= ScaleS[1] = scale[2] = GenScale[6] = Tan[π/6]2 = 1/3, but this is neither primitive nor a unit, so a better choice of scaling is hS[2]/hN= ScaleS[2] = GenScale[3] = Tan[π/6]Tan[π/3] = 1. This avoids the gender mismatch between N = 6 and S[1] and assigns the correct scaling to the web. Example: N = 12 has quadratic algebraic complexity and this is the minimal complexity for accumulation points to exist in the web because the web consists of rays or segments parallel to the sides of N and under the outer-billiards map, these segments are bounded apart by linear combinations of the vertices. When N is regular the linear space determined by the vertices has the same rank as N - namely Φ(N). Therefore the coordinate space of N-gons with linear complexity will have rank 2 and affinely rational coordinates with no limit points under iteration. The web for N =12 is fractal just as in the case of N = 5 and N = 8. The enlargement on the right shows the first few generations. Note that S[2] is mutated in the same fashion as S94 of N =24. Here ScaleS[2] = scale[4] = Tan[π/6]Tan[π/12] = 2/√3 -1 could serve a generator of the scaling field with no gender issue, but it is not a unit and not a primitive scale. (Any S[k] corresponding to a primitive scale[k] would not be mutated.). A better choice of generator is ScaleS[1] = GenScale[12] = Tan[π/12]2 = 7 − 4 3 . As the name implies, this is indeed the ‘generation’ scaling for N = 12. The limiting ‘temporal’ scaling appears to be 27 (using S[1] or S[2] periods), so the fractal dimension is -Ln[27]/Ln[GenScale[12]] ≈ 1.2513
Example: N = 18 has cubic algebraic complexity along with the matching N = 9. As with N = 14 and N = 7, there are two non-trivial primitive scales and they can interact in an unpredictable fashion so the web is a mixture of quadratic-type self-similarity (when one scale is dominant) and multi-fractal behavior when both scales interact.
hS[1]/hN = ScaleS[1] = scale[8] = GenScale[18] – but this a nonagon and a gender-mismatch with N. This could be rectified with ScaleChange[9,18] but a much better choice of generator is ScaleS[2] = scale[7] = GenScale[18]/scale[2] = GenScale[9]. This is always the correct First Family scaling in a twice-odd case like N = 18 because it is hS[2]/hN where S[2] is conjugate to DS[2] - the ‘second generation’ D tile. (As pointed out earlier, this S[2] is technically an S[1] relative to N = 9, so in a sense the S[1] scaling is the ‘right’ scaling for all regular N-gons.) The primitive scales of N = 18 are scales 1, 5 and 7. N = 9 has primitive scales 1,2 and 4 which correspond to 1,5,7 of N = 18 under the transformation scale[k] = scale[2k]*ScaleSwap[9,18]. This is the ‘gender-change’ transformation for scales – and it is also the key to the odd & twiceodd equivalence. Recently we have shown that a number of weakly conforming tiles that arise in the outerbilliards map are also canonical (see Appendix C of H[4]). The Two-Elephant case earlier contains two examples since Px and Sx are both canonical. Their scaling is complex but it still lies in the scaling field [11] because it is based on local star points – which evolve in a predictable fashion from the initial star points. The only points of discontinuity of the web are these star points - so it is sometimes possible to track the evolution of the web using these ‘hyperbolic’ star points as markers. As indicated earlier, this evolution begins with the star points of N which are a subset of the First Family star points. Since the star-point scaling predicts the ‘ideal’ evolution of these First Families, this same scaling should apply to the web in general.
Summary This paper and the previous H[4] are motivated by the desire to find a basis for the geometric scaling that occurs when the extended edges of a regular N-gon are iterated under a convex piecewise isometry such as the outer-billiards map . In the limit, this iteration defines the singularity set W (web) – which consists of all points where k is not defined for some k. The truncated extended edges of N define nested star polygons and these star polygons have a natural scaling based on the horizontal displacement of the intersection (‘star’) points. In H[4] we show that this ‘star-point’ scaling is a basis for N+ - the maximal real subfield of the cyclotomic field N. This subfield has order Φ(N)/2 and is generated by Cos[2π/N] so it defines the (real) coordinates of N. Therefore the star-point scaling can be regarded as a natural scaling of N itself and we call N+ the ‘scaling field’ of N, written [N]. We show here that this scaling is consistent with a ‘family’ of regular polygons which arise naturally from the constraints of the star polygons. This is what we call the First Family of N. Every star[k] point defines an associated scale[k] and also an S[k] tile of the First Family. This progression is shown below for N = 14:
This derivation implies that the geometric and algebraic properties of these First Families are inherent in the star polygons – and hence inherent in the underlying polygon N. Therefore it is reasonable to expect that the First Families can provide insight into the geometric scaling of W. Indeed the geometry of the S[k] is consistent with the web evolution, so these First Family tiles do arise in the web for all regular N-gons. We conjecture that all line segments or tiles P that arise in the web have sP/sN in [N]- so they are ‘canonical’ with respect to N. This is easy to see for the linear or quadratic cases with Φ(N)/2 2, but in general the webs are poorly understood. For the outer-billiards map, the progression in complexity with Φ(N)/2 appears to be the salient feature of the topological evolution of W – so all ‘quadratic’ polygons have similar web structure and the ‘cubic’ cases of N = 7 and N = 9 have similar web structure. More subtle features like the important quadratic reciprocity theorem of C.F. Gauss [G1] apparently contribute to the distinction between the evolution of First Families for 4k+1 and 4k+3 primes. (See the 4k+1 Conjecture in H[3].) Cases such as N = 11 present difficulties beyond the more predictable 4k+1 cases such as N = 13. In general it is clear that no one understands the extended algebraic or geometric structure of ‘most’ regular polygons. Standard renormalization methods have no natural extension beyond the simple quadratic cases and it appears likely that the web of a typical regular N-gon will be multi-fractal. This presents great difficulties for both geometric and ‘temporal’ scaling. But exploring the amazing geometry of a polygon like N = 11 is a worthy intellectual pursuit – and no doubt there will be more surprises like the tiny Sx island immersed in an alien sea.
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[SM] Siegel C.L, Moser, J.K. Lectures on Celestial Mechanics, Springer Verlag,1971, ISBN 3540-58656-3 [Sg] Siegel C.L. Iteration of (Complex) Analytic Functions, Ann of Math 42: 607-612 [Sg] Siegel C.L,Transcendental Numbers, Princeton University Press, 1949 - 102 pages [St1] Stillwell J., Mathematics and Its History, UTM, Springer Verlag,1989 [T] Tabachnikov S., On the dual billiard problem. Adv. Math. 115 (1995), no. 2, 221–249. MR1354670 [VS] Vivaldi F., Shaidenko A., Global stability of a class of discontinuous dual billiards. Comm. Math. Phys. 110 , 625–640. MR895220 (89c:58067) [W] Washington L.C. Introduction to Cyclotomic Fields – GTM 83 – Springer Verlag 1982 Links: (i) The author’s web site at DynamicsOfPolygons.org is devoted to the outer billiards map and related maps from the perspective of a non-professional. (ii) A Mathematica notebook called FirstFamily.nb will generate the First Family and related star polygons for any regular polygon. It is also a full-fledged outer billiards notebook which works for all regular polygons. This notebook includes the Digital Filter map (which is only applicable for N even). The default height is 1 - to make it compatible with the Digital Filter map. For investigations without the Digital Filter map, it may be preferable to use one of the notebooks described below – with the more natural convention of radius 1. Of course the scaling is independent of these choices, so it is an easy matter to mix the conventions. (iii) Outer Billiards notebooks for all convex polygons (radius 1 convention for regular cases). There are four cases: Nodd, NTwiceOdd, NTwiceEven and Nonregular. (iv) The open source PARI software at pari.math.u-bordeaux.fr has impressive facilities for computer algebra and algebraic number theory. There is an excellent introduction to Galois Theory and PARI in Fields and Galois Theory by J.S Milne – which is available at www.jmilne.org/math/
(v) We also recommend the open source (GPL- GNU) Sage software which was originally devised in 2005 by William Stein at U.C. San Diego, as a Python and C++ based library. It currently has hundreds of developers around the world and an extensive library of routines for number theory, algebra and geometry. Sage runs on an Oracle Virtual Machine which will install on almost any operating system.
