WP-103 established that a lattice in dimension d carries a rotation of order n if and only if φ(n) ≤ d, and that in the plane this leaves n ∈ {1, 2, 3, 4, 6}. It closed by saying that no symmetry is forbidden outright — each one has a price, denominated in dimensions — and handed the four-dimensional case to WP-104, which spent it on Saturn instead.
What was never said is what the excluded orders do. They are not absent from the plane. Penrose tilings, Ammann–Beenker tilings and the twelve-fold shield tilings all live in the ordinary Euclidean plane and all have sharp diffraction. They simply have no lattice. What they have in its place is a scaling symmetry: a map that takes the tiling to itself after multiplication by a fixed real number > 1. That number is the inflation factor, and this note’s only content is the observation that it is never an arbitrary number.
An n-fold rotation of the plane acts on any point set by multiplication by ζn = e2πi/n. If a point set is to be closed under that rotation and under addition, the smallest ring it can live in is ℤ[ζn]. Distances and traces — everything a metric can see — live in the maximal real subfield
WP-103’s condition φ(n) ≤ 2 is exactly the condition that this real subfield is ℚ itself. That is the whole of the crystallographic restriction, restated: a planar lattice exists when the rotation asks for no new real numbers. COMPUTED
| n | 2 cos(2π/n) | minimal polynomial | ℚ(ζn)+ |
|---|---|---|---|
| 3 | −1 | x + 1 | ℚ |
| 4 | 0 | x | ℚ |
| 6 | 1 | x − 1 | ℚ |
| 5 | (√5 − 1)/2 | x² + x − 1 | ℚ(√5) |
| 8 | √2 | x² − 2 | ℚ(√2) |
| 10 | (1 + √5)/2 = φ | x² − x − 1 | ℚ(√5) |
| 12 | √3 | x² − 3 | ℚ(√3) |
Row n = 10 is worth stopping on. The trace of a ten-fold rotation is not merely related to the golden ratio; it is φ, on the nose. COMPUTED
The observed quasicrystal symmetries are five-, eight-, ten- and twelve-fold. This is normally reported as an empirical list. It is not a list; it is a solution set.
Both are complete over all n, not just over a search range: φ(n) grows at least like √(n/2), so no n > 32 can have φ(n) ≤ 4, and an exhaustive check to n = 2000 returns nothing further. COMPUTED The four realised orders are precisely the four integers whose totient is four — the first rung above the plane.
φ(n) is even for n > 2, so there is no φ(n) = 3 rung. The step from the plane goes 2 → 4, and it goes there for every excluded order at once. That is why the quasicrystal orders arrive as a block of four rather than one at a time.
Dirichlet’s unit theorem gives the unit group of the ring of integers of a number field K as μ(K) × ℤr with
Three cases exhaust everything this corpus touches, and they are the three rows of the following table. COMPUTED
| field | r1 | r2 | rank | unit group | what the plane does |
|---|---|---|---|---|---|
| ℚ (n = 3, 4, 6) | 1 | 0 | 0 | {±1}, order 2 | periodic — a lattice, no scaling symmetry |
| ℚ(√−3) ℤ[ω] | 0 | 1 | 0 | μ6, order 6 | closes — six units, twelve defects (Ch 21) |
| ℚ(√5), ℚ(√2), ℚ(√3) | 2 | 0 | 1 | ±εk, infinite | quasiperiodic — a fundamental unit ε to inflate by |
Rank zero means the only units are roots of unity: finitely many, all of modulus one, none of which can scale anything. Rank one means there is a unit ε > 1, and then ±εk for every integer k — an infinite multiplicative symmetry, sitting inside the ring, preserving it exactly. A tiling cannot inflate unless its field has somewhere to inflate to.
ℚ is rank zero because it has no room; ℚ(√−3) is rank zero because it is imaginary, and its two extra dimensions of unit group went into torsion instead — six sixth roots of unity, which is where Chapter 21’s twelve pentagons come from (4π ÷ 2π/6 = 12). COMPUTED The same theorem that gives the hexagonal sheet its closure gives the Penrose tiling its inflation; it just lands on the torsion side of the group rather than the free side.
Solve the Pell equation for each of the three real quadratic fields — that is, find the smallest unit > 1 — and compare against the inflation factor of the tiling with that symmetry. COMPUTED CHECKED
| n | ℚ(ζn)+ | fundamental unit ε | N(ε) | tiling | inflation λ |
|---|---|---|---|---|---|
| 5 | ℚ(√5) | (1+√5)/2 = φ | −1 | Penrose (rhombic, P3) | 1.618034 |
| 8 | ℚ(√2) | 1 + √2 | −1 | Ammann–Beenker | 2.414214 |
| 10 | ℚ(√5) | (1+√5)/2 = φ | −1 | decagonal phase | 1.618034 |
| 12 | ℚ(√3) | 2 + √3 | +1 | Stampfli / shield | 3.732051 |
For each n with φ(n) = 4, the inflation factor of the corresponding quasiperiodic tiling equals the fundamental unit of ℚ(ζn)+. The golden mean, the silver mean and 2 + √3 are not three numerological coincidences; they are the answers to three Pell equations, and the Pell equation is what Dirichlet rank 1 looks like when you write it down.
The same numbers arrive from the combinatorial side. A substitution rule has an abelianised matrix M whose Perron–Frobenius eigenvalue is the inflation factor, and in every case det M = ±1 — the inflation sits in GL(2, ℤ), multiplying the tile counts while preserving an integer volume. COMPUTED
“Unit” and “det = ±1” are the same condition seen twice: once in the ring of integers, once in the matrix. Book 6’s Aperiodic Multiplying Media already records det M = ±1 and calls it “the first sign that a conserved criticality rides along with the growth”. This note supplies the missing half of that sentence: the conserved quantity is the field norm, and λPF is a unit because a unit is precisely an element of norm ±1.
Being a unit is not enough on its own — it says the algebraic conjugate has modulus 1/λ, but a cut-and-project construction needs more: the window in the internal space must stay bounded, which requires every conjugate to lie strictly inside the unit disc. That is the Pisot condition, and for a real quadratic unit > 1 it is automatic. COMPUTED
| n | λ | conjugate λ′ | |λ′| | λ·λ′ | Pisot |
|---|---|---|---|---|---|
| 5, 10 | 1.618034 | −0.618034 | 0.618034 | −1 | yes |
| 8 | 2.414214 | −0.414214 | 0.414214 | −1 | yes |
| 12 | 3.732051 | 0.267949 | 0.267949 | +1 | yes |
The last two columns are one fact written twice: |λ·λ′| = 1 is what makes λ a unit, and it is also what forces |λ′| < 1 once λ > 1. Unit and Pisot are not two conditions on the inflation factor in the quadratic case; they are the same condition. The internal window is bounded because the external inflation is a unit — the tiling is discrete for the same reason it can scale.
This is where Chapter η’s ladder attaches. The tribonacci constant η = 1.839287… is the real root of x³ − x² − x − 1; its constant term is −1, so η is a unit, and its two complex conjugates lie inside the unit disc, so it is Pisot. COMPUTED But its minimal polynomial has degree three. η is a unit of a cubic field, and cubic fields are not maximal real subfields of planar cyclotomic ones. The recurrence ladder φ < η < … < τ = 2 climbs out of the plane at the second rung — which is exactly why Chapter η’s tribonacci chain is one-dimensional and its Penrose cousin is two.
Σ · Pentanacci. The chapter says five-fold is forbidden in lattices and that quasicrystals are the exception. Both halves now have one reason rather than two: ℚ(ζ5)+ = ℚ(√5) is not ℚ, so there is no lattice; and it has unit rank 1, so there is an inflation. The prohibition and the exception are the same theorem read in opposite directions.
Chapter 21. The closing field’s six units and the Penrose tiling’s one fundamental unit are the two halves of Dirichlet’s theorem for the two kinds of quadratic field. A shell closes when the units are torsion; a tiling inflates when they are free.
Saturn’s south pole. A decagon puts n = 10 on the table, and ℚ(ζ10)+ = ℚ(√5) with fundamental unit φ. If the decagonal structure is quasiperiodic in any sense that survives contact with a fluid, the scale ratio available to it is φ, and only φ — there is no second choice in that field. OPEN Whether that is a statement about Saturn or only about ℚ(√5) is not established here, and this note does not claim it is. What it does claim is that the number is not free: if a ratio shows up, the arithmetic says which one it must be.
Everything above is kinematics. It says which scaling factors a symmetry makes available; it says nothing about which one a physical system will take, or what it costs. WP-104 named that gap in the six-fold case — the corpus has stability results and Saturn’s question is selection — and this note does not close it in the ten-fold case either. The selection principle needs a thermodynamic argument, which is chapter work (§21.8’s heat term is where it would attach), not a working paper. OPEN
Nothing here is novel and none of it is claimed as such. Dirichlet’s unit theorem is 1846. The maximal real subfield degree is textbook cyclotomic theory. The identification of Penrose inflation with φ, Ammann–Beenker with 1 + √2 and the twelve-fold tilings with 2 + √3 is standard in the aperiodic-order literature, as is the Pisot condition on cut-and-project windows. The note exists because those facts are stated in four places in this corpus, in four different vocabularies, and never as one sentence.
It says nothing about dm³, nothing about whether Saturn’s decagon is quasiperiodic, and nothing about selection. OPEN
[1] Dirichlet, unit theorem (1846); see Neukirch, Algebraic Number Theory, I.§7.
[2] Washington, Introduction to Cyclotomic Fields, for
[ℚ(ζn)+ : ℚ] = φ(n)/2.
[3] Baake & Grimm, Aperiodic Order, Vol. 1 (CUP 2013), for the substitution matrices,
the Pisot condition on cut-and-project windows, and the inflation factors of the Penrose,
Ammann–Beenker and twelve-fold tilings.
[4] Shechtman, Blech, Gratias & Cahn, Phys. Rev. Lett. 53, 1951 (1984);
observation 8 April 1982.
[5] This corpus: book6/wp103-why-six-and-why-not-five.html;
book6/wp104-six-was-an-input.html;
book6/ch-aperiodic-multiplying-media.html;
ch-eta-dnls.html;
book4/ch21-the-closing-field.html;
chSigma-pentanacci.html.
[6] Every number in this note is regenerated by wp105-verify.py
(sympy 1.14; all checks pass).