The octagonal system catches all substitution tilings

Why every substitution rule in the octagonal family arises from the continued fraction system.

In When is a cut and project set substitutional? Koivusalo and Walton and I show that canonical projection tilings admitting a substitution rule are exactly those coming from a repeated quadratic element of GL4(Z)GL_4(\mathbb{Z}) that preserves the physical and internal spaces.

For our continued fraction we need those elements of GL4(Z)GL_4(\mathbb{Z}) that also commute with RR and MM.

The suitable matrices

Solving for commuting with RR and MM gives the four-parameter family X\mathcal{X}:

X(a,β,γ,d)=(aβ0−βγdγ00βaβ−γ0γd),a,β,γ,d∈Z,X(a,\beta,\gamma,d) = \begin{pmatrix} a & \beta & 0 & -\beta \\ \gamma & d & \gamma & 0 \\ 0 & \beta & a & \beta \\ -\gamma & 0 & \gamma & d \end{pmatrix}, \qquad a,\beta,\gamma,d \in \mathbb{Z},

with Q:=ad−2βγ=±1Q := ad - 2\beta\gamma = \pm 1; det⁡X=Q2=1\det X = Q^2 = 1. In Z[i]\mathbb{Z}[i]-linear coordinates in which RR gives multiplication by ii,

C(a,β,γ,d)=(a(1+i)β(1−i)γd)∈GL2(Z[i]),C(a,\beta,\gamma,d) = \begin{pmatrix} a & (1+i)\beta \\ (1-i)\gamma & d \end{pmatrix} \in GL_2(\mathbb{Z}[i]),

isomorphic to Γ0(2)±\Gamma_0(2)^{\pm}, the integer matrices with even lower-left entry, via

Γ0(2)±={g∈GL2(Z):g21≡0 mod 2},C↦(aβ2γd).\Gamma_0(2)^{\pm} = \left\{ g \in GL_2(\mathbb{Z}) : g_{21} \equiv 0 \bmod 2 \right\}, \qquad C \mapsto \begin{pmatrix} a & \beta \\ 2\gamma & d \end{pmatrix}.

The two distinguished elements are

S1=X(1,0,1,1),S2=X(1,1,1,1),S_1 = X(1,0,1,1), \qquad S_2 = X(1,1,1,1),

with S1−1=X(1,0,−1,1)S_1^{-1} = X(1,0,-1,1) and S2−1=X(−1,1,1,−1)S_2^{-1} = X(-1,1,1,-1). S1S_1 is parabolic with Q=1Q = 1; S2S_2 is a glide with Q=−1Q = -1 and eigenvalues 1±21 \pm \sqrt2. These are the two matrices of the continued fraction branches.

To have a substitution rule on tiles we need that the 2-minors (determinants of 2x2 sub-matrices) of the matrix are non-negative. The 2-minors are a2,d2,2β2,2γ2,aβ,aγ,βd,γd,βγ,a^2, d^2,2\beta^2,2\gamma^2, a\beta, a\gamma, \beta d, \gamma d, \beta \gamma, and ad−βγad-\beta\gamma.

This happens when a,d,β,γa, d, \beta, \gamma and ad−βγad - \beta\gamma are non-negative (WLOG, due to multiplying by -1 not changing lattices). As ad−βγ=βγ+Qad - \beta \gamma = \beta \gamma + Q so the only negative case is βγ=0\beta \gamma = 0 and Q=−1Q=-1, but that gives ad=−1ad=-1 but both aa and dd are non-negative. The only non-positive 2-minor cases are X(1,0,0,1)X(1,0,0,1) which is the identity, S1nS_1^n, S2S_2 and Si1n=X(1,n,0,1)Si_1^n = X(1,n,0,1) (Si1Si_1 is S1S_1 for b>ab>a in the continued fraction picture). We now want to show that S1,S2S_1, S_2 and Si1Si_1 generate this positive set X+\mathcal{X}^+.

Generating X+\mathcal{X}^+

Theorem.

S1,S2S_1, S_2 and Si1Si_1 generate X+\mathcal{X}^+.

Proof.

The three actions on the coefficients for elements of X+\mathcal{X}^+.

1. Left action of S1−1S_1^{-1}. Preserving QQ.

s1(a,β,γ,d)↦(a,β,γ−a,d−2β)s_1 \quad (a,\beta,\gamma,d) \mapsto (a, \beta, \gamma - a, d - 2\beta)

2. Left action of S2−1S_2^{-1}. Flips the sign of QQ.

s2(a,β,γ,d)↦(2γ−a,d−β,a−γ,2β−d)s_2 \quad (a,\beta,\gamma,d) \mapsto (2\gamma - a, d - \beta, a - \gamma, 2\beta - d)

3. Left action of Si1−1Si_1^{-1}. Preserving QQ.

si1(a,β,γ,d)↦(a−2γ, β−d, γ,d)si_1 \quad (a,\beta,\gamma,d) \mapsto (a - 2\gamma,\ \beta - d,\ \gamma, d)

If γ≥a\gamma \ge a and d≥2βd \ge 2\beta apply s1s_1, if a≥2γa \ge 2\gamma and β≥d\beta \ge d apply si1si_1, otherwise we are at the identity or can apply s2s_2 (by Lemma 1). Each move keeps all four entries non-negative and strictly reduces the total a+β+γ+da+\beta+\gamma+d. So the process terminates, and only at the identity.

Lemma.

If neither s1s_1 nor si1si_1 can be applied we are either at the identity or can apply s2s_2.

Proof.

If γ≥a\gamma \ge a but d<2βd < 2\beta, consider Q=ad−2βγ=a(d−2β)+2β(a−γ)Q = ad - 2 \beta\gamma = a(d-2\beta) + 2\beta(a-\gamma). Both terms must be non-positive, and Q=±1Q=\pm1. If β=0\beta=0 then aa or dd must be negative, so a=γ=1a=\gamma=1 (as a−γa-\gamma gets multiplied by 22, and aa multiplies d−2βd-2\beta) and d−2β=−1d-2\beta = -1 so the matrix is X(1,β,1,2β−1)X(1,\beta,1,2\beta-1) and s2s_2 can be applied.

If d≥2βd \ge 2\beta but γ<a\gamma < a, both terms must be non-negative and a similar argument forces the identity.

The argument for si1si_1 just swaps the roles that aa and dd play and β\beta and γ\gamma. This leaves the case where γ<a<2γ\gamma < a < 2 \gamma and β<d<2β\beta < d < 2\beta, which is exactly when s2s_2 leaves positive values.