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
‘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〉}.
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, <N/2>} – 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 jk 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 displacements 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
hP = (p2-p1)/(Tan[6π/14]-Tan[3π/14]) = hN
4π
π
] − Tan[ ]
14
14 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 First Family 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 this page (see [H5]). 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
5π
π
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
+
+ 21x 2 +
−
where x = GenScale[11] = Tan[π/11]*Tan[π/22]
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: (i) If N is even and P is a regular N-gon, P is conforming relative to N if P shares the
same base edge as N and star[<N/2>] = star[N/2-1] = GenStar of P is  star[1] of N.
(ii) If N is odd and P is a regular 2N-gon then P is conforming relative to N if P shares the same
base edge as N and star[<2N/2>-1] = star[N-2] of P is  star[1] of N.
In both cases P is said to be strongly conforming if it is conforming and also shares another
star[k] point with N.
Example: (N = 11 & N = 22) Any tile conforming to N will share the star polygon edges of N –
so they are candidates for the First Family of N – and we will define the First Family of N to be
the strictly conforming tiles. The D = D[0] tile of the First Family of N = 11 is shown below, and
D[1] is the second generation D tile so it is also a 22-gon. The First Family of D[1] will include
the second generation N = 11 – which is M[1] shown below. But this cyan First Family is just a
just a scaled version of the First Family of D. There is never any guarantee that tiles of the
original First Family will survive the transition to ‘next’ generation under the outer-billiards
map, and here the survival rate is pretty minimal. The actual tiles that survive are in black and
any tile that is both cyan and black is a First Family survivor. This includes M[1], DS5, M[2]
(and their symmetric counterparts at GenStar).
This is the DS5 from the Two Elephant Case. It 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 [H5] for a derivation of Mx, Px and Sx.)
Example: (N = 13 & N = 26) On the left below both S[1] and S[2] are strictly conforming so
they are part of the First Family of N = 13. This means that they are regular 26-gons which have
star[11] = star[1] of N = 13 and they are strongly conforming because they also share star[2] of
N = 13. In the diagram on the right, M[1] is the 2nd generation N = 13. Py is a regular N-gon but
not part of the First Family of M[1]- so its parameters are unknown. The displaced S[2] is known
and it is conforming relative to Py because star[11] of S[2] is star[1] of Py (green) - but this S[2]
is not strictly conforming so it could not be used to find the center and height of Py (See [H6]
for a derivation of Py.)
Lemma (Conformal Replication): Every regular N-gon has a strongly conforming ‘dual’ DN –
which is identical to N for N even and a regular 2N-gon with same side as N for N odd.
Proof: Set -starD[1] = GenStar[N] and center offset sN/2. When N is even, the minimal star
polygon angle is φ = 2π/N. Recursively extend the current edge by sN and rotate by φ. After N-1
iterations it will form a clone of N – which we call DN. The external angle of N is also φ = 2π/N
with π complement (N-2)π/N = (2π/N)(N/2 -1). So after N/2-1 rotations DN will have star[N/2-1]
= star[<N/2>] equal to star[1] of N. Therefore DN will be strongly conforming relative to N .
When N is odd φ is 2π/2N so repeat the construction above using GenStar[N] and this will create
a regular 2N-gon with the same side as N – which we call DN. Now the exterior angle of N is 2φ
= 2π/N with π complement (N-2)π/N = (2π/2N)(N-2) so after N-2 rotations the new DN will
have star[N-2] = star[<2N/2>-1] = star[1] of N. □
Example: When N is twice-odd we will see that the First Family contains a scaled copy of N/2
and in this case they share the same D tile as shown below for N = 26 and N = 13.
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.
Corollary: For a regular N-gon, these is a unique conforming regular N-gon or 2N-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: Below are 6 conforming DM tiles for N = 14 and N = 13 – with the matching M tiles
shown in blue. For N = 13 the DM are all regular 26-gons even though the blue M’s are 13-gons.
N = 14 and N = 13 will be one of these blue tiles when DM is the D tile on the left. All the DM
tiles shown here are actually strongly conforming and the First Family Theorem below shows
that they must esist. They are called S[1]-S[6] and they are the nucleus of the First Family for N
= 14 and N = 13.
First Family Theorem: For a regular N-gon 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: The cases of N even and N odd are slightly different so we will do them separately and
then show how they are related. In both cases, the existence of a conformal S[k] is not an issue
because of the Corollary to the Conformal Replication Lemma.
(i) For N even and 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 below 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 P is conforming and star[k] = star[1] of N and p1=
p2. Also when k = N/2-1, j = 1 and P is the DN from the Replication Lemma.)
Measured from the origin, the displacement of star[k] is –hNTan[kπ/N]. Therefore measured
from p2 = star[1], the displacement of cP is also -hNTan[kπ/N]. Since P is conforming, p2 is the
right-side starP[N/2-1], so this displacement is also -hPTan[(N/2-1)π/N]. Therefore
hNTan[kPi/N] = hPTan[(N/2-1)π/N], so
hP
Tan[kπ / N ]
=
hN Tan[( N / 2 − 1)π / N ]
[ kπ / N ]
Tan
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 ]
Tan[π / N ]
p1, the index of P is j = N/2-k and hP =
.
hN 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 offset.
(ii) Suppose that N is odd. For p1 = star[k] and displacement x = -sN/2, the Corollory to the
Conformal Replication Lemma says that there must be a conforming regular 2N-gon P with
cP[[1]] = star[k][[1]] + x as shown below for N = 13 and k = 2.
The only difference between the diagram below and the N-even case is the label of p2 - which
was starP[N/2-1] and is now doubled to starP[N-2]. We will once again show that the given P
must have star[k] as a right-side star point – and now the required ‘index’ is N-2k – which is
twice the old index of N/2-k (because the minimal star angle is now 2φ instead of φ). So we will
show that starP[j] = star[k] for j = N-2k as shown here for k = 2.
The argument is the same as the N-even case - just write the displacement of cP from p2 in terms
of N and P. Relative to N, this displacement is - star[k][[1]] = hNTan[kπ/N]. Relative to P, this
displacement is -starP[N-2] = hPTan[(N-2)π/2N] so
hP
Tan[kπ / N ]
=
hN Tan[( N − 2)π / 2 N ]
Tan[kπ / N ]
Tan[π / N ]
=
Just as in the even case, for 1  k < N/2,
Tan[( N − 2)π / 2 N ] Tan[( N − 2k )π / 2 N ]
hS [k ]
Tan[π / N ]
=
Therefore
and the ‘index’ j is N-2k as desired.
hN
Tan[( N − 2k )π / 2 N ]
Once again all the S[k] will be strongly conforming except S[1] which has global index
hS [k ]
Tan[π / N ]
matching the local index. In this case
= Tan[Pi/N]2 so S[1] can be
=
hN
Tan[( N − 2)π / 2 N ]
constructed without the need of a second star point – but we will show that S[1] is always index
4 relative to star[2]. To do this use the Two-Star Lemma with opposite sides and local index 4 to
get
hS [1]
Tan[2π / N ] − Tan[π / N ]
= = Tan[π / N ]2 as above. □
hN
Tan[( N − 2)π / 2 N ] + Tan[4π / 2 N ]
Example (N = 16): To finish the N-even case of 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.
When N is even, the First Family Theorem says that all the S[k] will be regular N-gons as shown
above for N = 16. But sometimes it is possible to replace an S[k] with a N/2-gon with the same
center and height. These are called ‘gender-duals’.
Definition: For N even, a regular N-gon and N/2- gon are gender-dual if the N/2-gon is
circumscribed about the N-gon with matching ‘base’ edge, center and height. The ratio of their
sides will be Tan[π/N]/Tan[2π/N] - which we call ScaleChange(N) or SC(N) < 1/2.
Example: Below are the gender-duals for N = 12 and N = 14
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 First Family Theorem. (This case is illustrated
on the right above.)
Proof: In the First Family Theorem for N even and k odd:
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 divisible by 2 (and the shared index is unchanged) any Ngon 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. N must
be twice-odd because gender duals in the twice-even case have different GenStar points. □
Example: Below are the three gender-change mutations for N = 14. Note that the new S[k] are
still strongly conforming. By default we will make these mutations and call this the nucleus of
the First Family in the twice-odd case. This is consistent with the ‘reality’ of the outer-billiards
map and we will explain later why these mutations arise in the evolution of the ‘web’.
Definition: If N is even and P is a regular N-gon and Q = S[N/2-2] of P, then hQ/hP =
Tan[π/N]/Tan[2π/N] = SC[N] and Q and P are called an ‘M-D’ pair.
Definition (First Family) For any regular N-gon, the First Family Theorem defines the strongly
conforming S[k] tiles for 1 k  <N/2>. These tiles will be called the First Family Nucleus and
S[<N/2>] will be called DN or simply D. Based on the definition above, the S[N/2-2] tile of D
will be called M and DS[1] and DS[2] will be called M[1] and D[1]. (It is easy to see that they
are also an M-D pair –so they provide a possible path toward self-similarity.)
(i) For N twice-even, the First Family will consist of the S[k] in the First Family Nucleus
together with the (right-side) First Family Nucleus of D – called the DS[k].
(ii) For N twice-odd, the odd S[k] in the First Family Nucleus will be replaced with their N/2
counterparts, and the First Family will consist of this revised First Family Nucleus together with
the (right-side) revised First Family nucleus of D – called the DS[k]. (Here SC(14) involves a
‘gender change’ so sM *SC(14) = sD*SC(14) and the sides are equal – and ditto for (iii) below. )
(iii) For N odd, the First Family will consist of the First Family Nucleus together with the (rightside) revised First Family Nucleus of D – called the DS]k]. (This is essentially ’half’ of case ii.)
When N is even, the DS[k] are always clones of the corresponding S[k], but when N is odd, the
DS[k] have to be computed independently of the even S[k], because D is a 2N-gon. However
there is a simple relationship between the even S[k] of D & N - namely hDS[k/2] = hS[k]. This
occurs because the S[k] of the even and odd cases of the First Family Theorem are related by a
factor of 2. Therefore DS[2] in case (iii) above is a clone of S[1] of N = 7.
Of course the First Family Theorem will give the correct heights in all cases, but for N odd it will
require two steps to find a typical DS[k]. For example to find hDS[1] above for N = 7 it is first
necessary to find hD = hS[3]. By the First Family Theorem hD = hN* Tan[π/7]/Tan[π/14. Then
use the twice-odd case, and S[1] of D will be a heptagon with height hDS[1] = hD*Tan[π/14]2,
so hDS[1] = hN*Tan[π/7]Tan[π/14]. Therefore hDS[1]/hN = Tan[π/7]Tan[π/14] – which we call
GenScale[7]. This is why DS[1] is also known as M[1] – it is a second generation M. The
matching tile is DS[2] which plays the role of D[1] and together these form a 2nd generation M-D
pair. This is generic for all N.
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 there is always a matching M. In addition ,
will show that DS[1] and DS[2] are always and M-D pair – so every regular N-gon has at least
two ‘generations’ of M-D pairs.
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: For N = 5, the orbits of DS[1] and DS[2] 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 DS[2] has only been inverted 5 times because DS[2] 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 DS[2] 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 M[1] and D[1] tiles which can serve as the foundation for a 2nd
generation on the edge of D as shown above for N = 5. The generations are scaled by
hD[1]/hD =Tan[π/N]Tan[π/2N] – which is called GenScale[N]. (This is also the generator of the
‘scaling field’ for N odd). For N = 5, GenScale[5] =√5 − 2. 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. 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 First Family 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.
The outer-billiards map is discontinuous on the ‘trailing’ edge of the N-gon as the ‘target’ vertex
changes abruptly. This is called a ‘shear’ discontinuity and its magnitude is clearly sN – the side
of N. (Note that this shear is consistent with the canonical displacements of sN/2 – because the
shear operates on the center of each S[k].)
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.
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 the First Family of N scaled by hS[K]/hN. 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 shown above for N = 5.
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. (There is a close match between the
actual tile that forms in the DS5 position and the predicted. These ‘almost’ matches are not
unusual. Back in N = 11, Mx is ‘almost’ in the First Family of DS5.) Conforming tiles like Px
are similar to the Px and Mx tiles that occur for N = 11. Py is not conforming to either D[1] or
M[1] – yet this tile seems to survive in all subsequent even-generations. See [H6] for a derivation
of Px and Py.
In most cases it is not possible to predict the long-term evolution of S[k] tiles but is sometimes
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. These all must be compatible with the First Family of N= 22.
Therefore 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.
This says that the local 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 consistent with the First
Family Theorem, but of course these tiles are called S[1]-S[5] relative to N = 11. When working
inside N = 22 we will often add the 3 missing tiles and call it the Extended 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. More on this later.
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] – which 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]. When N is twice-even, it is easy to show that S[1] is a member of the First
family of every S[k] but this does not guarantee that there will be clones of S[1] in proximity to
the S[k] as we see here with S[9]. For an M-tile like S[10], the local S[2] will always be
conjugate to S[1]. 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 N even there is always
a reflective duality between D and N – with cM as the reflection point. With step-(N/2-k)
adjustments we have seen that it is possible to relate the ‘in-situ’ and ‘in-vitro’ dynamics of S[k]
and this it may be possible to extend this connection to future generations.
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[<N/2>].
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.
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.
Lemma: GenScale[N] is Tan[π/N]2 for N even and Tan[π/N]Tan[π/2N] for N odd. For all N,
scale[2] is Tan[π/N]/Tan[2π/N] and when N is even this is Tan[π/N]/Tan[π/(N/2)] = SC(N). The
symmetry of the N-even case says that scale[N/2-k] = GenScale[N]/scale[k] – so scale[N/2-2] =
GenScale[N]/scale[2] = Tan[π/N]Tan[2π/N] – which is GenScale[N/2] when N is twice odd.
By the independence result of Carl Siegel, the primitive sk and primitive scales 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 SN. (Because N is equivalent to N/2 for N
twice-odd, SN = SN/2 in that case.)
Scaling Field Lemma (Corollory to Theorem 2 of [H4]). 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 SN . When N is twice-odd
GenScale[N/2] is primitive and degree Φ(N)/2 so it is a generator of both SN and SN/2.
Since the canonical scales of N are linear combinations of the primitive scales, they are all in the
scaling field. Because SN is a field, every scale[k] has a matching ‘cotangent’ scale sk/s1 which is
in SN. 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 SN
Examples:
(i) If N is a regular N-gon and P is a regular N-gon in the coordinate system of N, then sP/sN =
hP/hN . If Q is a regular N/2-gon in this same coordinate system, then sQ/sN =
hQ*Tan[π/(N/2)]/hNTan[π/N] = (hQ/hN)/scale[2] (of N) and since scale[2] is in SN, sQ/sN is in
SN iff hQ/hN is in SN. Therefore for regular N-gons or N/2-gons, side-scaling can be replaced
with height-scaling. Since SN is a field any polygon that is canonical with respect to P or Q is
also canonical relative to N.
(ii) Any line segment defined by a linear combination of star points of N is canonical because if
T is such a linear combination T/s1 is a linear combination of dual scales, so it is in SN. Therefore
every internal mutation or external mutation is canonical. In addition every star polygon based
on N is canonical.
(iii) It is our conjecture that all tiles and line segments which arise in the web W of the outerbilliards 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.
(iv) Lemma: For a regular N-gon, the First Family tiles are canonical.
Proof: When N is even all the S[k] are either N-gons or N/2-gons with the same height as the
matching N-gon. Therefore by (i) above, these are canonical iff hS[k]/hN is in SN. By the First
hS [k ]
Tan[π / N ]
=
Family Theorem for N even,
= scale[N/2-k] which is in SN
hN
Tan[( N / 2 − k )π / N ]
hS [k ]
Tan[π / N ]
Tan[2π / 2 N ]
When N is odd
=
= scale[N-2k]/scale[2] of
=
hN
Tan[( N − 2k )π / 2 N ] Tan[( N − 2k )π / 2 N ]
2N so it is in S2N = SN. □
(v) Since the S[k] are canonical regular N-gons or N/2 gons, the First Families or subsequent
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, and 12 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. In the First Family Theorem we showed that for N even
ScaleS[k] = scale[N/2-k] and for N odd, ScaleS[k] = scale[N-2k]/scale[2] of 2N.
Recall that for N even the scales are related by scale[N/2-k] = GenScale[N]/scale[k]- so the
number of independent scales is essentially reduced by ‘half’. Therefore for N even and
1  k  N/2-1
hS [k ]
Tan[π / N ]
GenScale[ N ]
ScaleS [=
k]
=
= scale[ N / 2 − =
k]
h[ N ] Tan[( N / 2 − k )π / N ]
scale[k ]
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 the scaling field SN so the matching tile
is S[1]. For N twice-odd, GenScale[N] is still Tan[π/N]2 but it is no longer primitive and not a
unit may not generate SN. This occurs because S[1] is now a N/2-gon with a different ‘gender’
than N. The solution is to use the scaling of S[2] instead. This makes sense because S[2] of N is a
clone of S[1] of N/2 – so this is essentially S[1] scaling - just as in the twice-even case. Therefore
for N twice-odd, ScaleS[2] = hD[1]/hD =Tan[π/N]Tan[π/(N/2)] = GenScale[N/2] is the generator
of SN. (Therefore for N odd – the scaling define by D is simply GenScale[N].)
In practice the most important issue is the relationship between DS[1] and DS[2] – because that
is what determines the evolution of ‘generations’ as defined below. For N is even, this
relationship is the same as S[1] and S[2] – but for N odd, DS[1] and DS[2] will act as surrogate
S[1] and S[2].
Lemma: For all regular N-gons, S[1] and S[2] of D are an M-D pair.
Proof: (i) If N is even, D is a clone of N so DS[2] and DS[1] are an M-D pair iff S[1] and S[2]
are an M-D pair. By definition, this is true iff P = S[N/2-2] of S[2] is S[1]. By the First Family
Theorem, P will be an N-gon with hP/hS[2] = scale[(N/2-(N/2-2)] = scale[2] of S[2], so hP =
scale[2]*hS[2] = scale[2]*(GenScale[N]/scale[2])*hN = GenScale[N]*hN. The only tile with
hP/hN = GenScale[N] is S[1]. (ii) If N is odd, D is a 2N-gon so (i) applies with DS[2] and DS[1].
Note: This M-D result depends only on the existence of complementary scales, so it generalizes
as follows: For N twice-even, every S[k] has S[1] in its First Family and for N twice-odd, the
even S[k] will have S[1] in their First Family but the odd S[k] will have S[2]. When N is odd, the
twice-odd case is valid for D. See the examples below.
Example: For N = 26, the First Family of S[2] is defined to be FFS[2] =
TranslationTransform[cS[2]][hS[2]*First Family of N] where hS[2] = hN*scale[N/2-2] =
hN*scale[11] (in cyan below). To verify that S[11] of S[2] is S[1]: hS[11] = hS[2]*scale[2] =
hN*scale[2]*scale[11]. These are ‘complementary’ scales mod 13 so their product is
GenScale[26]. Therefore hS[11] = hN*GenScale[26] which is hS[1].
For N = 13, D would be a scaled copy of N = 26, so DS[1] and DS[2] would have the same
relationship as S[1] and S[2] of N = 26 - but of course S[1] and S[2] of N = 13 will no longer
have an M-D relationship. Therefore the ‘next-generation’ for N odd is always defined using D.
(See the definition below.)
Example: N = 11. For N odd we noted that all the DS[k] will have either DS[1] or DS[2] in
their First Families depending on whether k is even or odd. Here we will show that DS[3] has
S[2] in its First Family. FFDS[3] = TranslationTransform[cDS[3]][hDS[3]*FirstFamily of N]
where hDS[3] = hD*scale[8] and hD is hS[5] = hN/scale[2].Therefore hDS[3] =
hN*scale[8]/scale[2] (all scales are N = 22) This FFDS[3] is shown below in cyan and it appears
that S[4] of DS[3]is DS[2]. We will verify this below – once again using the result about
complementary scales.
scale[3]
scale[8]* scale[3]
GenScale[22]
[3]
=
=
hN =
hN=
hN *[Tan[π /11]2
hS [4] of DS[3] hDS
2
2
scale[2]
scale[2]
scale[2]
2
Since hN here is N = 11, hDS[2] = hS[1] = hN*Tan[π/11] as above.
We made use of this relationship between DS[3] and DS[2] when deriving the parameters of Px
for the Two-Elephant Case. It allowed us to use the use the local web of DS[3] instead of DS[2]
because these webs overlap. Note that star[4] of DS[3] is shared with DS[2]. See [H5].
Definition (Generations): For any regular N-gon, define D[0] = D and M[0] = DS[N/2-2]. Then
for a natural number k, M[k+1] and D[k+1] are DS[1] and DS[2] of D[k]. The pair {M[k],D[k]}
is always an M-D pair and the (ideal) kth generation of N is defined to be the (ideal) First
Family of D[k-1] so M[k+1] and D[k+1] are ‘matriarch’ and ‘patriarch’ of the kth generation.
This implies that these ideal generations will converge to star[1] of D which is also GenStar[N].
When N is even, this same convergence takes place at star[1] of N and in this case S[1] and S[2]
can be used interchangeably with DS[1] and DS[2] so the generation structure can be studied at
either star[1] of D or star[1] of N.
The tiles of this kth generation can be scaled relative to D[k] or M[k] and the results will always
Tan[π / N ]
ScaleM [k ]
Tan[π / N ]
be in the scaling field because
=
= scale[2] = SC(N).
=
ScaleD[k ] Tan[π ( N / 2)] Tan[2π / N ]
Since this is in SN = SN/2 we can use either scale as the ‘generation scale’ – but as noted earlier,
ScaleS[1] = GenScale[N] is the obvious choice for N twice-even because it is primitive and a
generator of SN. When N is twice-odd, GenScale[N] is no longer primitive or a unit so ScaleS[2]
= GenScale[N/2] is the obvious choice of generator of SN. When N is odd. the natural scaling is
through DS[1] and DS[2] – exactly as in the twice-even case – so GenScale[N] is the ‘generation
scaling’ here also.
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 self-similar like the other quadratic cases of N = 5 and N = 8. Note that
S[2] and DS[2] are mutated in the same fashion as S94 of N = 24. The enlargement on the right
shows the 2nd generation at the foot of N =12 – with S[2] as the new D and S[1] as the S[4].
The virtual First Family of S[2] is shown in cyan and indeed it shows that S[1] is a step-4 of
S[2]. Therefore S[1] and S[2] are an M-D pair. This S[1], S[2] pairing is generic for all even Ngons – but it does not always lead to the extended self-similarity as shown here.
We saw earlier that for N twice-even every S[k] or DS[k] has S[1] in its First Family – so S[1]
above is an S[3] of S[3]. This occurs also in the twice-odd case but the even and odd S[k] have
S[1] or S[2] in their families.
Since N is twice-even the natural choice of generator for the scaling field is ScaleS[1] =
GenScale[12] = Tan[π/12]2 = 7 − 4 3 . By contrast, ScaleS[2] = scale[4] = Tan[π/6]Tan[π/12] =
2/√3 -1 is not a unit and not a primitive scale. Therefore the ‘generation’ scaling for N = 12 is
GenScale[12] and the matching ‘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. (The 4k+1 Conjecture predicts
that when N is prime, the limiting temporal scaling is simply N +1, but in general we have no
way to predict the temporal scaling of a regular N-on.)
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. Note that S[3] is mutated because it has
period 6 instead of 9 – so the construction workers quit early.
hS[1]/hN = ScaleS[1] = scale[8] = GenScale[18] – but this a nonagon and a gender-mismatch
with N. This could be rectified using scale[2] of N = SC(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, 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. For N = 14 and N = 18, there are scaled copies of the
First Family which develop on the edges of S[1] of M because the geometry is compatible with
S[2], but by definition the main generation scaling is always on the edges of D (or N).
Summary
This paper and the previous [H4] 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 [H4]
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 SN.
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 SN 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 [H3].)
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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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.