Argon Compact Clockwork Notation · 17 Wallpaper Groups → 68 Forward Groups
Every one of the 68 forward clockwork groups (the 68 polar 3D space groups / cyclic wallpaper colourings)
specified by assigning independent time shifts τ(gᵢ) ∈ ℝ/ℤ to a minimal decoupled generating set
of each of the 17 wallpaper groups.
In The Symmetries of Things, a wallpaper group's orbifold presentation lists one generator per orbifold feature,
which couples their time shifts via linear congruence relations (such as τ(α) + τ(β) + τ(γ) ≡ 0 (mod 1) in
333, 442, 632, or τ(Q) = τ(R) in *632).
When we switch to Argon minimal generators — eliminating the redundant orbifold generator
(γ = (αβ)⁻¹, α = (βγ)⁻¹, δ = (αβγ)⁻¹, β = αZ⁻²) and replacing odd-corner mirror pairs
by rotations (α = PQ in *632, α = PQ, β = QR in *333, s = PZ in *×) —
all coupling equations disappear across all 17 wallpaper groups: the valid clockwork homomorphisms form an
unconstrained Cartesian product grid ℤd₁ × … × ℤdₖ of 104 total cells.
Why do those 104 unconstrained grid cells yield 68 distinct forward clockwork groups?
Because the size of the unconstrained homomorphism grid |Γabcw| ∈ {1, 2, 4, 6, 8, 9, 16} is an intrinsic invariant of Γ,
and two grid cells define the same spacetime group if and only if an outer crystallographic symmetry of Γ
(permuting equivalent rotation centres or perpendicular mirrors) maps one tuple to the other:
100% 1-to-1 bijection (orbit size = 1 for every cell):
every point of ℤd₁ × … × ℤdₖ is a distinct forward clockwork group.
Includes p6m (*632) on (α, R) ∈ ℤ₂ × ℤ₂ (2×2 = 4),
p6 (632) on (β, γ) ∈ ℤ₃ × ℤ₂ (3×2 = 6),
cm (*×) on (P, s) ∈ ℤ₂ × ℤ₁ (2×1 = 2),
p3m1 (*333) on (P, α, β) ∈ ℤ₂ × ℤ₁² (2×1×1 = 2),
and the 3 drift-pinned groups p1, pm, pg (1 each).
Exact 2D product grid after τ(α) ~ −τ(α):
generated by an off-mirror rotation α and a mirror P.
Because P reflects the rotation centre α into the second rotation centre α' = Pα⁻¹P in the cell with
τ(α') = −τ(α), shifting between the two centres folds ℤₙ to ℤₙ/±:
p31m (3*3) becomes {0, ±⅓} × {0, ½} (2×2 = 4) and
p4g (4*2) becomes {0, ±¼, ½} × {0, ½} (3×2 = 6).
Exact 3 × 2 = 6 product grid after swapping 2 identical involutions:
Γab ≅ ℤ₂³ has size 8 (a 2-group cannot have size 6), with 3 independent binary generators.
Two of the 3 involutions are geometrically interchangeable (α ↔ β in pmg,
P ↔ Q in cmm, P ↔ R in p4m), collapsing their
2×2 = 4 factor to {00, 0½, ½½} of size 3 and leaving an exact 3 × 2 = 6 grid!
Unconstrained homomorphism grid modded by cell permutations: relations still impose zero coupling between the Argon generators, and cell symmetries merge off-diagonal tuples into centred lattices.
p4 (442): 6 of 4×2 = 8;
pgg (22×): 4 of 2×4 = 8;
p3 (333): 4 of 3×3 = 9;
p2 (2222): 3 of 2³ = 8;
pmm (*2222): 6 D₄-necklaces of 2⁴ = 16.
2222 (p2) Has an Unconstrained 2³ = 8 Cube yet 3 Colourings (1 + 6 + 1)
16 raw → 8 valid homomorphisms → 3 distinct groups
2⁴ = 16 → 2³ = 8, zero coupling equations)
In Chaim's 4-generator orbifold presentation ⟨α, β, γ, δ | α² = β² = γ² = δ² = αβγδ = 1⟩ for the four half-turn centres at (0,0), (½,0), (0,½), (½,½), the relation αβγδ = 1 forces the linear parity constraint
τ(α) + τ(β) + τ(γ) + τ(δ) ≡ 0 (mod 1), ruling out odd numbers of flipping centres (only 8 of the 2⁴ = 16 binary combinations are valid).
Eliminating the redundant 4th half-turn δ = (αβγ)⁻¹ leaves the 3 Argon minimal generators (α, β, γ) with presentation ⟨α, β, γ | α² = β² = γ² = (αβγ)² = 1⟩.
Because 2(τ(α) + τ(β) + τ(γ)) ≡ 0 (mod 1) holds automatically for every (τ(α), τ(β), τ(γ)) ∈ {0, ½}³, all 2 × 2 × 2 = 8 cube cells are valid homomorphisms with zero coupling equations, and the 4th centre is reconstructed via τ(δ) ≡ τ(α) + τ(β) + τ(γ) (mod 1).
8 cells → 1 + 6 + 1 = 3 colourings)
Why do those 8 valid homomorphisms yield 3 colourings (P2, C2, P2₁) instead of 8?
Because the four 2-fold centres (α, β, γ, δ) in a p2 unit cell are geometrically interchangeable: shifting the origin by a half-period or changing the lattice basis permutes the 4 centres by S₄.
Once we append the induced 4th shift τ(δ), the 8 tuples partition into 3 orbits according to how many of the 4 centres in the unit cell flip (Hamming weight w ∈ {0, 2, 4}) — all C(4,2) = 6 ways of choosing which 2 of the 4 centres flip produce the same centred group C2:
(τ(α), τ(β), τ(γ)) in 2³ Cube |
Induced τ(δ) |
Full 4-Centre Pattern (α, β, γ, δ) |
Flipping Centres | Cube Orbit | Polar Space Group / Colouring |
|---|---|---|---|---|---|
(0, 0, 0) |
0 |
(0, 0, 0, 0) |
0 of 4 flip (all static) | 1/8 | P2 (2222, g5) |
1 of (α,β,γ) flips: (0,0,½), (0,½,0), (½,0,0)2 of (α,β,γ) flip: (0,½,½), (½,0,½), (½,½,0)
|
½ (×3)0 (×3)
|
(0,0,½,½), (0,½,0,½), (½,0,0,½)(0,½,½,0), (½,0,½,0), (½,½,0,0)
|
2 of 4 flip (all C(4,2) = 6 pairs) |
6/8 | C2 (c222₁2₁, g7) |
(½, ½, ½) |
½ |
(½, ½, ½, ½) |
4 of 4 flip (all flip) | 1/8 | P2₁ (2₁2₁2₁2₁, g6) |
| Wallpaper Group (17 Rows) · Argon Minimal Generators & Grid | Argon Compact Notation | Induced Shift | Grid Orbit | Polar Space Group | Clockwork & Chaim | Period & Kernel | |||||||||||||||
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I · Exact 1×1 Grid
1 × 1 = 1 cells → 1 group
Argon generators (2):
X (translation (1,0), Z1), Y (translation (0,1), Z1)Relations:
⟨X, Y | XY = YX⟩Reconstruction:
No redundant orbifold generators.Pure translations on a primitive lattice are gauge-pinned to τ(X) = τ(Y) = 0 by a Galilean frame boost. τ(X) × τ(Y)
|
X0 Y0 τ(X)=0, τ(Y)=0 |
— |
1/1 {(0, 0)} |
P1 #1 |
o
Chaim:
◦ |
N = 1
K = p1 (◦)
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IV · 2³ Cube Orbits (S₄)
2 × 2 × 2 = 8 cells → 3 groups
Argon generators (3):
α (half-turn at (0,0), Z2), β (half-turn at (½,0), Z2), γ (half-turn at (0,½), Z2)Relations:
⟨α, β, γ | α² = β² = γ² = (αβγ)² = 1⟩Reconstruction:
δ = (αβγ)⁻¹ ⇒ τ(δ) ≡ τ(α) + τ(β) + τ(γ) (mod 1)Eliminates the 4th half-turn δ = (αβγ)⁻¹ from Chaim's αβγδ = 1 (reducing 2⁴ = 16 raw tuples to an unconstrained 2³ = 8 cube where α, β, γ are independently 0 or ½). Why 3 colourings instead of 8? Because the four 2-centres (α, β, γ, δ) in the unit cell are geometrically interchangeable (S₄): including the induced τ(δ), the 8 cube cells partition into 1 + 6 + 1 = 8 by how many of the 4 centres flip (0, 2, or 4 — see worked walkthrough above). τ(α), τ(β) × τ(γ) [with induced τ(δ)]
|
α0 β0 γ0 τ(α)=0, τ(β)=0, τ(γ)=0 |
τ(δ) = 0 |
1/8 (0,0,0) ⇒ δ=0: 0 of 4 centres flip |
P2 #3 |
2222
Chaim:
2222 |
N = 1
K = p2 (2222)
|
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| α½ β½ γ½ τ(α)=½, τ(β)=½, τ(γ)=½ |
τ(δ) = ½ |
1/8 (½,½,½) ⇒ δ=½: all 4 centres flip |
P2₁ #4 |
2₁2₁2₁2₁
Chaim:
²2²2²2²2 |
N = 2
K = p1 (◦)
|
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| α0 β0 γ½ τ(α)=0, τ(β)=0, τ(γ)=½ |
τ(δ) = ½ |
6/8 3 with 1 of (α,β,γ)=½ (δ=½) + 3 with 2 of (α,β,γ)=½ (δ=0) → all C(4,2)=6 pairs of 2/4 flipping centres |
C2 #5 |
c222₁2₁
Chaim:
¹2¹2²2²2 |
N = 2
K = p2 (2222)
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I · Exact 1×1×1 Grid
1 × 1 × 1 = 1 cells → 1 group
Argon generators (3):
P (mirror x = 0, Z1), Q (mirror x = ½, Z1), α (translation (0,1), Z1)Relations:
⟨P, Q, α | P² = Q² = 1, αP = Pα, αQ = Qα⟩Reconstruction:
Matches the 3 minimal generators of **.Continuous drift along the mirrors is gauge-pinned (τ(α) = 0); flipping a mirror in ** without perpendicular pinning either rescales the cell or changes the 2D projection to pg/cm, leaving the single polar group Pm. τ(P), τ(Q) × τ(α)
|
P0 Q0 α0 τ(P)=0, τ(Q)=0, τ(α)=0 |
— |
1/1 {(0, 0, 0)} |
Pm #6 |
**
Chaim:
** |
N = 1
K = pm (**)
|
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I · Exact 1×1 Grid
1 × 1 = 1 cells → 1 group
Argon generators (2):
Y (glide along x = 0, Z1), Z (glide along x = ½, Z1)Relations:
⟨Y, Z | Y²Z² = 1⟩Reconstruction:
Matches the 2 minimal glide generators of ××.Along the parallel glides, Y² = T_(0,1) and Z² = T_(0,-1) are pure translations pinned to 0, and a frame shift along the glide axis gauges τ(Y) = τ(Z) = 0. τ(Y) × τ(Z)
|
Y0 Z0 τ(Y)=0, τ(Z)=0 |
— |
1/1 {(0, 0)} |
Pc #7 |
××
Chaim:
×× |
N = 1
K = pg (××)
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I · Exact 2×1 Grid
2 × 1 = 2 cells → 2 groups
Argon generators (2):
P (mirror y = x, Z2), s (primitive slide PZ, Z1)Relations:
⟨P, s | P² = 1, (Ps)² = sPsP⟩Reconstruction:
Z = Ps ⇒ τ(Z) ≡ τ(P) + τ(s) = τ(P) (mod 1)Replaces Chaim's coupled (P, Z) (where τ(P) = τ(Z)) with the mirror P and primitive translation s = PZ. Since τ(s) = 0 on the pinned lattice, τ(P) ∈ {0, ½} is a single unconstrained binary flip! |
P0 s0 τ(P)=0, τ(s)=0 |
τ(Z) = 0 |
1/2 (0, 0) — mirror static |
Cm #8 |
*×
Chaim:
*× |
N = 1
K = cm (*×)
|
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| P½ s0 τ(P)=½, τ(s)=0 |
τ(Z) = ½ |
1/2 (½, 0) — mirror flips |
Cc #9 |
~*×½
Chaim:
*²×² |
N = 2
K = p1 (◦)
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IV · 2⁴ Hypercube Orbits (D₄)
2 × 2 × 2 × 2 = 16 cells → 6 groups
Argon generators (4):
P (mirror x = 0, Z2), Q (mirror y = 0, Z2), R (mirror x = ½, Z2), S (mirror y = ½, Z2)Relations:
⟨P, Q, R, S | P² = Q² = R² = S² = (PQ)² = (QR)² = (RS)² = (SP)² = 1⟩Reconstruction:
All 4 mirrors are independent binary flips; centring translations are PR (along x) and QS (along y).*2222 requires 4 generators (cannot be generated by 3). Because all four corners have even order 2, there are ZERO coupling equations: all 2⁴ = 16 binary combinations (τ(P),τ(Q),τ(R),τ(S)) ∈ {0,½}⁴ are valid, and the 6 groups are the 6 D₄ necklaces of the rectangle's 4 sides! |
P0 Q0 R0 S0 τ(P)=0, τ(Q)=0, τ(R)=0, τ(S)=0 |
0/4 mirrors flip |
1/16 (0,0,0,0) — 0 sides flip |
Pmm2 #25 |
*2222
Chaim:
*2222 |
N = 1
K = pmm (*2222)
|
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| P½ Q½ R½ S½ τ(P)=½, τ(Q)=½, τ(R)=½, τ(S)=½ |
4/4 mirrors flip |
1/16 (½,½,½,½) — all 4 sides flip |
Pcc2 #27 |
*~2~2~2~2
Chaim:
*²2²2²2²2 |
N = 2
K = p2 (2222)
|
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| P0 Q½ R0 S½ τ(P)=0, τ(Q)=½, τ(R)=0, τ(S)=½ |
2/4 mirrors flip |
2/16 (0,½,0,½) ~ (½,0,½,0) — 2 opposite sides flip |
Pmc2₁ #26 |
*2₁~2₁2₁~2₁
Chaim:
*¹2²2¹2²2 |
N = 2
K = pm (**)
|
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| P0 Q0 R0 S½ τ(P)=0, τ(Q)=0, τ(R)=0, τ(S)=½ |
1/4 mirrors flip |
4/16 4 tuples with 1 side = ½ |
Amm2 #38 |
c*222₁2₁ᵃ
Chaim:
*¹2¹2¹2²2 |
N = 2
K = pmm (*2222)
|
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| P0 Q½ R½ S½ τ(P)=0, τ(Q)=½, τ(R)=½, τ(S)=½ |
3/4 mirrors flip |
4/16 4 tuples with 3 sides = ½ |
Aem2 #39 |
c*222₁2₁ᵇ
Chaim:
*¹2²2²2²2 |
N = 2
K = pmg (22*)
|
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| P0 Q0 R½ S½ τ(P)=0, τ(Q)=0, τ(R)=½, τ(S)=½ |
2/4 mirrors flip |
4/16 4 tuples with 2 adjacent sides = ½ |
Fmm2 #42 |
c*22₁22₁
Chaim:
*¹2¹2²2²2 |
N = 2
K = cmm (2*22)
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III · Factored 3×2 Grid (2³ / S₂)
2 × 2 × 2 = 8 cells → 6 groups
Argon generators (3):
α (half-turn at (0,0), Z2), β (half-turn at (0,½), Z2), P (mirror x = ¼, Z2)Relations:
⟨α, β, P | α² = β² = P² = 1, (αβ)P = P(αβ)⟩Reconstruction:
Centring shift along y is τ(αβ) ≡ τ(α) + τ(β) (mod 1).22* requires 3 generators. The commutation relation (αβ)P = P(αβ) is trivial in Γᵃᵇ, so (τ(α), τ(β), τ(P)) ∈ {0, ½}³ is an unconstrained 2³ = 8 cube. Swapping the two 2-centre classes α ↔ β merges only (0,½,τ(P)) ~ (½,0,τ(P)), factoring into an exact 3 × 2 = 6 grid! |
α0 β0 P0 τ(α)=0, τ(β)=0, τ(P)=0 |
τ(αβ) = 0 |
1/8 (0, 0, 0) |
Pma2 #28 |
22*
Chaim:
22* |
N = 1
K = pmg (22*)
|
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| α0 β0 P½ τ(α)=0, τ(β)=0, τ(P)=½ |
τ(αβ) = 0 |
1/8 (0, 0, ½) |
Pnc2 #30 |
22~*
Chaim:
¹2¹2*² |
N = 2
K = p2 (2222)
|
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| α½ β½ P½ τ(α)=½, τ(β)=½, τ(P)=½ |
τ(αβ) = 0 |
1/8 (½, ½, ½) |
Pca2₁ #29 |
2₁2₁~*
Chaim:
²2²2*² |
N = 2
K = pg (××)
|
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| α½ β½ P0 τ(α)=½, τ(β)=½, τ(P)=0 |
τ(αβ) = 0 |
1/8 (½, ½, 0) |
Pmn2₁ #31 |
2₁2₁*
Chaim:
²2²2*¹ |
N = 2
K = pm (**)
|
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| α0 β½ P0 τ(α)=0, τ(β)=½, τ(P)=0 |
τ(αβ) = ½ |
2/8 (0, ½, 0) ~ (½, 0, 0) [α ↔ β] |
Ama2 #40 |
c22₁*ᵃ
Chaim:
¹2²2*¹ |
N = 2
K = pmg (22*)
|
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| α0 β½ P½ τ(α)=0, τ(β)=½, τ(P)=½ |
τ(αβ) = ½ |
2/8 (0, ½, ½) ~ (½, 0, ½) [α ↔ β] |
Aea2 #41 |
c22₁*ᵇ
Chaim:
¹2²2*² |
N = 2
K = pgg (22×)
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IV · 2×4 Grid Orbits
2 × 4 = 8 cells → 4 groups
Argon generators (2):
α (half-turn at (0,0), Z2), Z (glide (-x+½, y+½), Z4)Relations:
⟨α, Z | α² = (αZ⁻²)² = 1⟩Reconstruction:
β = αZ⁻² ⇒ τ(β) ≡ τ(α) − 2τ(Z) (mod 1)Eliminates the second half-turn β = αZ⁻² from Chaim's α² = β² = αβZ² = 1. In Γᵃᵇ, 2τ(α) ≡ 0 and 4τ(Z) ≡ 2τ(α) ≡ 0, giving an unconstrained Z₂ × Z₄ (2 × 4 = 8) grid with τ(β) ≡ τ(α) − 2τ(Z) (mod 1). |
α0 Z0 τ(α)=0, τ(Z)=0 |
τ(β) = 0 |
1/8 (0, 0) |
Pba2 #32 |
22×
Chaim:
22× |
N = 1
K = pgg (22×)
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| α0 Z½ τ(α)=0, τ(Z)=½ |
τ(β) = 0 |
1/8 (0, ½) |
Pnn2 #34 |
22×½
Chaim:
¹2¹2ײ |
N = 2
K = p2 (2222)
|
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| α½ Z0 τ(α)=½, τ(Z)=0 |
τ(β) = ½ |
2/8 (½, 0) ~ (½, ½) [x ↔ y axis swap] |
Pna2₁ #33 |
2₁2₁×
Chaim:
²2²2×¹ |
N = 2
K = pg (××)
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| α0 Z¼ τ(α)=0, τ(Z)=¼ |
τ(β) = ½ |
4/8 {0, ½} × {¼, ¾} — all 4 quarter-glide cells |
Fdd2 #43 |
c22₁×¼
Chaim:
¹2²2×⁴ |
N = 4
K = p2 (2222)
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III · Factored 2×3 Grid (2³ / S₂)
2 × 2 × 2 = 8 cells → 6 groups
Argon generators (3):
α (off-mirror half-turn, Z2), P (mirror x = 0, Z2), Q (mirror y = 0, Z2)Relations:
⟨α, P, Q | α² = P² = Q² = (PQ)² = (QαPα⁻¹)² = 1⟩Reconstruction:
Corner 2-centre PQ has τ(PQ) ≡ τ(P) + τ(Q) (mod 1).2*22 requires 3 generators. Both (PQ)² = 1 and (QαPα⁻¹)² = 1 vanish in Γᵃᵇ, so (τ(α), τ(P), τ(Q)) ∈ {0, ½}³ is an unconstrained 2³ = 8 cube. Swapping the perpendicular mirrors P ↔ Q merges only (τ(α),0,½) ~ (τ(α),½,0), giving an exact 2 × 3 = 6 grid! |
α0 P0 Q0 τ(α)=0, τ(P)=0, τ(Q)=0 |
τ(PQ) = 0 |
1/8 (0, 0, 0) |
Cmm2 #35 |
2*22
Chaim:
2*22 |
N = 1
K = cmm (2*22)
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| α0 P½ Q½ τ(α)=0, τ(P)=½, τ(Q)=½ |
τ(PQ) = 0 |
1/8 (0, ½, ½) |
Ccc2 #37 |
2*~2~2
Chaim:
¹2*²2²2 |
N = 2
K = p2 (2222)
|
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| α½ P0 Q½ τ(α)=½, τ(P)=0, τ(Q)=½ |
τ(PQ) = ½ |
2/8 (½, 0, ½) ~ (½, ½, 0) [P ↔ Q] |
Cmc2₁ #36 |
2₁*2₁~2₁
Chaim:
²2*¹2²2 |
N = 2
K = cm (*×)
|
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| α½ P0 Q0 τ(α)=½, τ(P)=0, τ(Q)=0 |
τ(PQ) = 0 |
1/8 (½, 0, 0) |
Imm2 #44 |
c2₁*22ᵃ
Chaim:
²2*¹2¹2 |
N = 2
K = pmm (*2222)
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| α½ P½ Q½ τ(α)=½, τ(P)=½, τ(Q)=½ |
τ(PQ) = 0 |
1/8 (½, ½, ½) |
Iba2 #45 |
c2₁*22ᵇ
Chaim:
²2*²2²2 |
N = 2
K = pgg (22×)
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| α0 P0 Q½ τ(α)=0, τ(P)=0, τ(Q)=½ |
τ(PQ) = ½ |
2/8 (0, 0, ½) ~ (0, ½, 0) [P ↔ Q] |
Ima2 #46 |
c2*2₁2₁
Chaim:
¹2*¹2²2 |
N = 2
K = pmg (22*)
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IV · 4×2 Grid Orbits
4 × 2 = 8 cells → 6 groups
Argon generators (2):
α (quarter-turn at (0,0), Z4), γ (half-turn at (½,0), Z2)Relations:
⟨α, γ | α⁴ = γ² = (αγ)⁴ = 1⟩Reconstruction:
β = (αγ)⁻¹ ⇒ τ(β) ≡ −τ(α) − τ(γ) (mod 1)Eliminates the second 4-centre β = (αγ)⁻¹ from Chaim's α⁴ = β⁴ = γ² = αβγ = 1. Since 4τ(α) + 4τ(γ) ≡ 0 (mod 1) holds automatically for τ(α) ∈ Z₄, τ(γ) ∈ Z₂, (α, γ) forms an unconstrained 4 × 2 = 8 grid! Swapping the two 4-centres α ↔ β merges only the 2 centred pairs where τ(α) ≠ τ(β). |
α0 γ0 τ(α)=0, τ(γ)=0 |
τ(β) = 0 |
1/8 (0, 0) → τ(β) = 0 |
P4 #75 |
442
Chaim:
442 |
N = 1
K = p4 (442)
|
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| α½ γ0 τ(α)=½, τ(γ)=0 |
τ(β) = ½ |
1/8 (½, 0) → τ(β) = ½ |
P4₂ #77 |
4₂4₂2
Chaim:
²4²4¹2 |
N = 2
K = p2 (2222)
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| α¾ γ½ τ(α)=¾, τ(γ)=½ |
τ(β) = ¾ |
1/8 (¾, ½) → τ(β) = ¾ (left-handed 4₁) |
P4₁ #76 |
4₁4₁2₁
Chaim:
⁴4⁴4²2 |
N = 4
K = p1 (◦)
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| α¼ γ½ τ(α)=¼, τ(γ)=½ |
τ(β) = ¼ |
1/8 (¼, ½) → τ(β) = ¼ (right-handed 4₃) |
P4₃ #78 |
4₃4₃2₁
Chaim:
⁴4⁴4²2 |
N = 4
K = p1 (◦)
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| α0 γ½ τ(α)=0, τ(γ)=½ |
τ(β) = ½ |
2/8 (0, ½) ~ (½, ½) [α ↔ β swap: {0, ½}] |
I4 #79 |
c44₂2₁
Chaim:
¹4²4²2 |
N = 2
K = p4 (442)
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| α¼ γ0 τ(α)=¼, τ(γ)=0 |
τ(β) = ¾ |
2/8 (¼, 0) ~ (¾, 0) [α ↔ β swap: {¼, ¾}] |
I4₁ #80 |
c4₁4₃2
Chaim:
⁴4⁴4¹2 |
N = 4
K = p2 (2222)
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III · Factored 3×2 Grid (2³ / S₂)
2 × 2 × 2 = 8 cells → 6 groups
Argon generators (3):
P (leg mirror y = 0, Z2), Q (diagonal mirror y = x, Z2), R (leg mirror x = ½, Z2)Relations:
⟨P, Q, R | P² = Q² = R² = (PQ)⁴ = (QR)⁴ = (RP)² = 1⟩Reconstruction:
4-turns have τ(PQ) = τ(P)+τ(Q) and τ(QR) = τ(Q)+τ(R).*442 requires 3 generators. Because all three corner orders (4, 4, 2) are even, there are ZERO coupling constraints: (τ(P), τ(Q), τ(R)) ∈ {0, ½}³ is an unconstrained 2³ = 8 cube. Swapping the two 4-corner leg mirrors P ↔ R merges only (0,τ(Q),½) ~ (½,τ(Q),0), factoring into an exact 3 × 2 = 6 grid! |
P0 Q0 R0 τ(P)=0, τ(Q)=0, τ(R)=0 |
τ(PR) = 0 |
1/8 (0, 0, 0) |
P4mm #99 |
*442
Chaim:
*442 |
N = 1
K = p4m (*442)
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| P0 Q½ R0 τ(P)=0, τ(Q)=½, τ(R)=0 |
τ(PR) = 0 |
1/8 (0, ½, 0) |
P4₂mc #105 |
*4₂~4₂2
Chaim:
*¹4²4¹2 |
N = 2
K = pmm (*2222)
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| P½ Q0 R½ τ(P)=½, τ(Q)=0, τ(R)=½ |
τ(PR) = 0 |
1/8 (½, 0, ½) |
P4₂cm #101 |
*~4₂4₂~2
Chaim:
*²4¹4²2 |
N = 2
K = cmm (2*22)
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| P½ Q½ R½ τ(P)=½, τ(Q)=½, τ(R)=½ |
τ(PR) = 0 |
1/8 (½, ½, ½) |
P4cc #103 |
*~4~4~2
Chaim:
*²4²4²2 |
N = 2
K = p4 (442)
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| P0 Q0 R½ τ(P)=0, τ(Q)=0, τ(R)=½ |
τ(PR) = ½ |
2/8 (0, 0, ½) ~ (½, 0, 0) [P ↔ R] |
I4mm #107 |
c*44₂2₁ᵃ
Chaim:
*¹4¹4²2 |
N = 2
K = p4m (*442)
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| P0 Q½ R½ τ(P)=0, τ(Q)=½, τ(R)=½ |
τ(PR) = ½ |
2/8 (0, ½, ½) ~ (½, ½, 0) [P ↔ R] |
I4cm #108 |
c*44₂2₁ᵇ
Chaim:
*¹4²4²2 |
N = 2
K = p4g (4*2)
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II · Factored 3×2 Grid (Z₄/± × Z₂)
4 × 2 = 8 cells → 6 groups
Argon generators (2):
α (quarter-turn at 4-centre, Z4), P (nearest mirror, Z2)Relations:
⟨α, P | α⁴ = P² = (PαPα⁻¹)² = 1⟩Reconstruction:
Mirror-image 4-centre α' = Pα⁻¹P has τ(α') ≡ −τ(α) (mod 1).(PαPα⁻¹)² = 1 is a commutator squared, imposing ZERO constraint in Γᵃᵇ: (τ(α), τ(P)) ∈ Z₄ × Z₂ is an unconstrained 4 × 2 = 8 grid. Because the mirror P maps the 4-centre α to the second 4-centre α' = Pα⁻¹P with τ(α') = −τ(α), shifting between the two 4-centres identifies τ(α) = ¼ with ¾, yielding an exact 3 × 2 = 6 grid! |
α0 P0 τ(α)=0, τ(P)=0 |
τ(α') = 0 |
1/8 (0, 0) |
P4bm #100 |
4*2
Chaim:
4*2 |
N = 1
K = p4g (4*2)
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| α½ P½ τ(α)=½, τ(P)=½ |
τ(α') = ½ |
1/8 (½, ½) |
P4₂bc #106 |
4₂*~2
Chaim:
²4*²2 |
N = 2
K = pgg (22×)
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| α½ P0 τ(α)=½, τ(P)=0 |
τ(α') = ½ |
1/8 (½, 0) |
P4₂nm #102 |
4₂*2
Chaim:
²4*¹2 |
N = 2
K = cmm (2*22)
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| α0 P½ τ(α)=0, τ(P)=½ |
τ(α') = 0 |
1/8 (0, ½) |
P4nc #104 |
4*~2
Chaim:
¹4*²2 |
N = 2
K = p4 (442)
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| α¼ P0 τ(α)=¼, τ(P)=0 |
τ(α') = ¾ |
2/8 (¼, 0) ~ (¾, 0) [α ↔ Pα⁻¹P] |
I4₁md #109 |
c4₃*2
Chaim:
⁴4*¹2 |
N = 4
K = pmm (*2222)
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| α¼ P½ τ(α)=¼, τ(P)=½ |
τ(α') = ¾ |
2/8 (¼, ½) ~ (¾, ½) [α ↔ Pα⁻¹P] |
I4₁cd #110 |
c4₁*2
Chaim:
⁴4*²2 |
N = 4
K = pgg (22×)
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IV · 3×3 Grid Orbits (S₃)
3 × 3 = 9 cells → 4 groups
Argon generators (2):
α (third-turn at (0,0), Z3), β (third-turn at (⅓,⅔), Z3)Relations:
⟨α, β | α³ = β³ = (αβ)³ = 1⟩Reconstruction:
γ = (αβ)⁻¹ ⇒ τ(γ) ≡ −τ(α) − τ(β) (mod 1)Eliminates the third 3-centre γ = (αβ)⁻¹ from Chaim's α³ = β³ = γ³ = αβγ = 1 (which started with 3³ = 27 price lists and used the sum rule to reach 9). With 2 minimal generators (α, β), 3τ(α) + 3τ(β) ≡ 0 holds automatically, so you start directly at the unconstrained 3 × 3 = 9 grid! The 3 diagonal cells give P3, P3₁, P3₂, and all 6 off-diagonal cells (τ(α) ≠ τ(β)) give R3. |
α0 β0 τ(α)=0, τ(β)=0 |
τ(γ) = 0 |
1/9 (0, 0) → τ(γ) = 0 |
P3 #143 |
333
Chaim:
333 |
N = 1
K = p3 (333)
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| α⅔ β⅔ τ(α)=⅔, τ(β)=⅔ |
τ(γ) = ⅔ |
1/9 (⅔, ⅔) → τ(γ) = ⅔ |
P3₂ #145 |
3₂3₂3₂
Chaim:
³3³3³3 |
N = 3
K = p1 (◦)
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| α⅓ β⅓ τ(α)=⅓, τ(β)=⅓ |
τ(γ) = ⅓ |
1/9 (⅓, ⅓) → τ(γ) = ⅓ |
P3₁ #144 |
3₁3₁3₁
Chaim:
³3³3³3 |
N = 3
K = p1 (◦)
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| α⅓ β⅔ τ(α)=⅓, τ(β)=⅔ |
τ(γ) = 0 |
6/9 All 6 off-diagonal cells where τ(α) ≠ τ(β) |
R3 #146 |
r33₁3₂
Chaim:
³3³3¹3 |
N = 3
K = p3 (333)
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I · Exact 2×1×1 Grid
2 × 1 × 1 = 2 cells → 2 groups
Argon generators (3):
P (mirror, Z2), α (3-turn PQ, Z1), β (3-turn QR, Z1)Relations:
⟨P, α, β | P² = α³ = β³ = (Pα)² = (Pβ)² = (αβ)³ = 1⟩Reconstruction:
Q = Pα, R = Qβ ⇒ τ(Q) = τ(R) = τ(P)*333 requires 3 generators, and in Chaim's 3-mirror basis (P, Q, R) the odd corners (PQ)³ = (QR)³ = (RP)³ = 1 force τ(P) = τ(Q) = τ(R). Replacing Q, R by the corner rotations α = PQ, β = QR diagonalizes Γᵃᵇ ≅ Z₂: τ(α) = τ(β) = 0 are pinned by dihedral inversion, leaving τ(P) ∈ {0, ½} as a single unconstrained binary flip! |
P0 α0 β0 τ(P)=0, τ(α)=0, τ(β)=0 |
τ(Q) = τ(R) = 0 |
1/2 (0, 0, 0) — mirrors static |
P3m1 #156 |
*333
Chaim:
*333 |
N = 1
K = p3m1 (*333)
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| P½ α0 β0 τ(P)=½, τ(α)=0, τ(β)=0 |
τ(Q) = τ(R) = ½ |
1/2 (½, 0, 0) — all 3 mirrors flip |
P3c1 #158 |
*~3~3~3
Chaim:
*²3²3²3 |
N = 2
K = p3 (333)
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II · Factored 2×2 Grid (Z₃/± × Z₂)
3 × 2 = 6 cells → 4 groups
Argon generators (2):
α (off-mirror third-turn, Z3), P (nearest mirror, Z2)Relations:
⟨α, P | α³ = P² = (PαPα⁻¹)³ = 1⟩Reconstruction:
Mirror-image 3-centre α' = Pα⁻¹P has τ(α') ≡ −τ(α) (mod 1).(PαPα⁻¹)³ = 1 is a commutator cubed, imposing ZERO constraint in Γᵃᵇ: (τ(α), τ(P)) ∈ Z₃ × Z₂ is an unconstrained 3 × 2 = 6 grid. Because the mirror P reflects the off-mirror 3-centre α to the second 3-centre α' = Pα⁻¹P with τ(α') = −τ(α), shifting between the two 3-centres identifies τ(α) = ⅓ with ⅔, yielding an exact 2 × 2 = 4 grid! |
α0 P0 τ(α)=0, τ(P)=0 |
τ(α') = 0 |
1/6 (0, 0) |
P31m #157 |
3*3
Chaim:
3*3 |
N = 1
K = p31m (3*3)
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| α0 P½ τ(α)=0, τ(P)=½ |
τ(α') = 0 |
1/6 (0, ½) |
P31c #159 |
3*~3
Chaim:
¹3*²3 |
N = 2
K = p3 (333)
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| α⅔ P0 τ(α)=⅔, τ(P)=0 |
τ(α') = ⅓ |
2/6 (⅔, 0) ~ (⅓, 0) [α ↔ Pα⁻¹P] |
R3m #160 |
r3₂*3
Chaim:
³3*¹3 |
N = 3
K = p3m1 (*333)
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| α⅔ P½ τ(α)=⅔, τ(P)=½ |
τ(α') = ⅓ |
2/6 (⅔, ½) ~ (⅓, ½) [α ↔ Pα⁻¹P] |
R3c #161 |
r3₂*~3
Chaim:
³3*²3 |
N = 6
K = p3 (333)
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I · Exact 3×2 Grid
3 × 2 = 6 cells → 6 groups
Argon generators (2):
β (third-turn at (⅓,⅔), Z3), γ (half-turn at (½,½), Z2)Relations:
⟨β, γ | β³ = γ² = (βγ)⁶ = 1⟩Reconstruction:
α = (βγ)⁻¹ ⇒ τ(α) ≡ −τ(β) − τ(γ) (mod 1)Eliminates the sixth-turn α = (βγ)⁻¹ from Chaim's α⁶ = β³ = γ² = αβγ = 1. Since 6τ(β) + 6τ(γ) = 2(3τ(β)) + 3(2τ(γ)) ≡ 0 (mod 1) holds automatically for every (τ(β), τ(γ)) ∈ Z₃ × Z₂, the two generators form an EXACT unconstrained 3 × 2 = 6 grid in 1-to-1 bijection with the 6 polar P6_k groups! |
β0 γ0 τ(β)=0, τ(γ)=0 |
τ(α) = 0 |
1/6 (0, 0) → τ(α) = 0 |
P6 #168 |
632
Chaim:
632 |
N = 1
K = p6 (632)
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| β⅔ γ0 τ(β)=⅔, τ(γ)=0 |
τ(α) = ⅓ |
1/6 (⅔, 0) → τ(α) = ⅓ |
P6₂ #171 |
6₂3₂2
Chaim:
³6³3¹2 |
N = 3
K = p2 (2222)
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| β⅓ γ0 τ(β)=⅓, τ(γ)=0 |
τ(α) = ⅔ |
1/6 (⅓, 0) → τ(α) = ⅔ |
P6₄ #172 |
6₄3₁2
Chaim:
³6³3¹2 |
N = 3
K = p2 (2222)
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| β0 γ½ τ(β)=0, τ(γ)=½ |
τ(α) = ½ |
1/6 (0, ½) → τ(α) = ½ |
P6₃ #173 |
6₃32₁
Chaim:
²6¹3²2 |
N = 2
K = p3 (333)
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| β⅔ γ½ τ(β)=⅔, τ(γ)=½ |
τ(α) = ⅚ |
1/6 (⅔, ½) → τ(α) = ⅚ |
P6₅ #170 |
6₅3₂2₁
Chaim:
⁶6³3²2 |
N = 6
K = p1 (◦)
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| β⅓ γ½ τ(β)=⅓, τ(γ)=½ |
τ(α) = ⅙ |
1/6 (⅓, ½) → τ(α) = ⅙ |
P6₁ #169 |
6₁3₁2₁
Chaim:
⁶6³3²2 |
N = 6
K = p1 (◦)
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I · Exact 2×2 Grid
2 × 2 = 4 cells → 4 groups
Argon generators (2):
α (sixth-turn PQ at (0,0), Z2), R (mirror opposite 6-centre, Z2)Relations:
⟨α, R | α⁶ = R² = (αRα⁻¹R)² = (RαRα²)³ = 1⟩Reconstruction:
P = αRα⁻¹Rα, Q = α⁻¹P ⇒ τ(P) ≡ τ(α) + τ(R), τ(Q) = τ(R)Replaces Chaim's 3 coupled mirrors (P, Q, R) (where (QR)³ = 1 forces τ(Q) = τ(R)) with the sixth-turn α = PQ and mirror R. In Γᵃᵇ ≅ Z₂ × Z₂, both α and R have order 2: each generator is independently non-flipping (0) or flipping (½), forming an EXACT 2 × 2 = 4 unconstrained binary grid! |
α0 R0 τ(α)=0, τ(R)=0 |
τ(P)=0, τ(Q)=0 |
1/4 (0, 0) — neither flips |
P6mm #183 |
*632
Chaim:
*632 |
N = 1
K = p6m (*632)
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| α½ R½ τ(α)=½, τ(R)=½ |
τ(P)=0, τ(Q)=½ |
1/4 (½, ½) — both α and R flip |
P6₃cm #185 |
*6₃~3~2₁
Chaim:
*¹6²3²2 |
N = 2
K = p31m (3*3)
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| α0 R½ τ(α)=0, τ(R)=½ |
τ(P)=½, τ(Q)=½ |
1/4 (0, ½) — only R flips |
P6cc #184 |
*~6~3~2
Chaim:
*²6²3²2 |
N = 2
K = p6 (632)
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| α½ R0 τ(α)=½, τ(R)=0 |
τ(P)=½, τ(Q)=0 |
1/4 (½, 0) — only α flips |
P6₃mc #186 |
*~6₃32₁
Chaim:
*²6¹3¹2 |
N = 2
K = p3m1 (*333)
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