Ammann–Beenker Dual

Visualization

@
$dim=4
$subspace=[s,1]
s=$companion([1,0,0,0])
r=$exchange()
g=s-s^3+1
h0=s^4*r*[-1,-1,-1,-1]
h1=s^5*[-1,-1,-1,-1]
h2=s^3*[0,0,-2,-2]
h3=s^5*[0,-2,0,0]
h4=s^3*[1,-1,-1,-1]
h5=s^4*[-1,-1,-1,-1]
h6=s^3*r*[-1,-1,-1,-1]
h7=[1,1,1,-1]
h8=r*s*[1,1,1,-1]
h9=s^4*[0,-2,0,0]
h10=s*r*[-1,-1,1,1]
h11=r*s*[2,0,2,0]
$root=A
A = h0^-1*g*A | h1^-1*g*A | h2^-1*g*A | h5^-1*g*B | h6^-1*g*B | h7^-1*g*B | h8^-1*g*B
B = h3^-1*g*A | h4^-1*g*A | h9^-1*g*B | h10^-1*g*B | h11^-1*g*B

Overview

The Ammann–Beenker Dual is obtained from the Ammann–Beenker tiling by swapping the roles of the two orthogonal spaces in the cut-and-project scheme (CPS).

In the algebraic framework, both tilings share the same 4D rational space $\mathbb{Z}[e^{i\pi/4}]^2$ (with $s^4 = -1$) and the same tile maps $h_0, \ldots, h_{11}$. The difference is the projection subspace:

• Ammann–Beenker: projects onto eigenplane of $e^{i\pi/4}$ ($subspace=[s,0])
• Ammann–Beenker Dual: projects onto eigenplane of $e^{3i\pi/4}$ ($subspace=[s,1])

This second projection gives the internal space of the cut-and-project scheme, whose cross-section defines the window (acceptance domain).

The dual uses inverse maps: $h_k^{-1} \cdot g$ instead of $g^{-1} \cdot h_k$. This reflects the dual GIFS structure where the expansion is applied on the right.

Dual GIFS Structure

The dual GIFS uses the inverse-map form:

The maps $h_k^{-1} g$ are contractions because $|g|^{-1} = (1+\sqrt{2})^{-1} < 1$ on the eigenplane of $e^{3i\pi/4}$ (both eigenplanes of $s$ have the same modulus structure for $g$).

References

• Ammann–Beenker tiling — Tilings Encyclopedia
• Beenker, F. P. M. (1982). Algebraic theory of non-periodic tilings of the plane by two simple building blocks. Eindhoven TH report 82-WSK04.
• Ammann–Beenker tiling — Wikipedia

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