Symmetric Animation Analogs

1 · The smallest clockworkTurn the triangle and the clock jumps a third of a period

Three identical oscillators on a ring, each pushed away from its neighbours: that is the local rule. They settle into a wave, each one a third of a period behind the last. Turn the ring by one cell and you see the film a third of a period later, $\tau(\text{turn}) = \tfrac13$. Nothing in the equation asked for that. It is the compromise in which every bond is equally unhappy.

On a ring of $N$ cells the whole global rule is one integer, the twist $q$, with $\tau(\text{turn}) = q/N$. There are exactly $N$ such clocks, because there are exactly $N$ homomorphisms $\mathbb{Z}_N \to \mathbb{R}/\mathbb{Z}$. The centre is the one point every turn fixes, so for every $q \neq 0$ the average there must vanish (it reads about $10^{-16}$). That is the forced zero of the magic theorem’s Theorem B, the reason the film is still at α and β.

Drag the coupling through zero and the same twisted wave turns from attractor into repeller. Which twists a simulation can find is set by the local rule, not by the symmetry: with attractive coupling a twist is stable only when $\lvert q\rvert \lt N/4$, counting $q$ from $-N/2$ to $N/2$.

Theorem A (the time cost is a character of the group) and Theorem B (a point fixed by a turn that costs time must be dark), in a system with three moving parts.

2 · The film’s own symmetry, without a PDEA lattice of repelling oscillators carries g227 exactly

Now put an oscillator on every site of a triangular lattice, each pushing away from its six neighbours. Two neighbours would like to be half a period apart, but around a triangle that cannot work: the lattice is frustrated. The compromise is 120° on every bond. Three interleaved sublattices form, and the whole pattern turns steadily. (Physicists know it as the ground state of the triangular XY antiferromagnet.)

That rotating state has exactly the symmetry r33₁3₂, catalogue entry g227, the symmetry of the published film. A turn about a lattice site costs nothing (γ), about an up-triangle centre a third of a period (α), and about a down-triangle centre two thirds (β). A slide by one lattice vector costs a third or two thirds, depending on its direction: the r prefix, derived rather than assumed. Measured against the project’s own group file, all nine time costs agree to $1.7\times10^{-16}$.

Everything the magic theorem predicts can be seen here, with no differential equation in sight. The smoothed field vanishes at α and β (to $10^{-16}$), the phase winds $-1$ and $+1$ around them and $0$ around γ, and the windings in a cell sum to zero. The character law holds with the minus sign; with the plus sign it is off by exactly $\sqrt3$, the same number the magic-theorem page measured on the published film.

One sign separates this toy from the real thing. Here the clockwork is the attractor and unison is the repeller. For the Brusselator it is the other way round, which is why the tutorial has to solve for its film rather than simulate it. Flip the coupling in the demo and you watch the Brusselator’s failure mode in ten seconds.

Imposing more of the symmetry buys reliability. From 100 random starts, free relaxation reaches the perfect state 13 times (the rest freeze with walls between the two chiralities; the demo opens on one of the lucky 13, so press Scramble to meet the usual outcome). Averaging once over the nine operations of one box gives 74, and projecting onto the whole group gives 100. This lattice holds 48 such boxes, so there the nine are only 9 of its 432 operations and do not form a group. On the catalogue’s single box the nine are the whole group, and step 2’s average is an exact projector.

Step 2’s projector and Theorems A, B and C, on the project’s own symbol. The stability is inverted, so you can watch the film assemble instead of fall apart.

3 · How a global rule gets imposedFold it, project onto it, or twist time

Part of the rule fits in a single frame (the time-free symmetries $K$), and part relates different moments. Every method on this page, and every step of the tutorial, is one of three moves.

The five steps use all three: step 2 projects (and step 3 keeps what it projected), step 4 projects the kernel and twists the rest (which is why one relation is the equation and the other five follow), and step 5 checks. The delayed glue swaps step 4’s twist for a fold.

4 · TwistRun a third of a period, turn, land where you started: the tutorial’s method on 7 numbers

A ring of three Brusselator cells, with the project’s own $a = 1$ and $b = 2.3$, coupled by diffusion. Step 1 gives the job’s numbers exactly: eigenvalues $0.15 \pm 0.9887\,i$ and $T_0 = 6.3551$. Step 4 is the same equation,

in 7 unknowns instead of 2,593. Newton converges in 2 to 6 steps onto a loop whose disturbances grow ×1.47 per period. A free simulation started on that loop collapses into unison within 100 periods: the published film’s ×1.16 per period and “by period 135 none are left”, in miniature.

Two other solvers land on the same unstable loop. One makes the whole loop the unknown, as a Fourier series in time; it agrees to $8.5\times10^{-9}$, a gap that is all the shooting route’s time step, since at a tenth of the step the two agree to $8.6\times10^{-13}$ (the demo’s Fourier button does both). The other, run in the research code rather than in this demo, is the delayed glue: one cell fed its own history from $T/3$ and $2T/3$ ago. It agrees to $8\times10^{-10}$.

Step 4 entire: the kernel averaged in, one time-costing symmetry as the equation, the rest following, and the instability that defeats plain simulation reproduced.

5 · FoldDivide by the group and the global rule becomes a gluing rule on one tile

Folding the plane by p3 gives Conway’s 333 turnover: one hexagonal tile whose six edges are glued in pairs by the three 120° turns. Drag the six dots on three of the edges into any shape you like, and the group writes the other three. The global symmetry has become a local rule at the seams. This is the delayed glue with the delay set to zero. Switch on the clockwork and every tile takes the time offset its symmetry costs.

The folded domain is also the right size for the solver. Burnside’s lemma over the nine box operations counts the orbits of the 1,296 mesh nodes:

That is exactly the 146 of 1,296 nodes the delayed-glue code simulates ($1296/9 = 144$ would be wrong). The three time-free symmetries alone leave $(1296 + 3 + 3)/3 = 434$ orbits, and $2 \times 434 = 868$ is the tutorial’s “868 free numbers”. Both sizes are point counts of quotients, so for each of the other 67 entries they follow from counting fixed points, before any solver runs.

Fold: the delayed glue solves on the orbifold (146 nodes), step 4’s averaging on the quotient by the time-free symmetries (434 nodes, 868 numbers), and Burnside fixes both counts before any solver runs.

6 · ProjectChange the global rule and the same solver returns a different orbit

Three equal masses chase one another around one closed curve. Newton’s law is the local rule; a group acting on loops is the global one. Under the figure-eight group (“half a period later is a mirror image”) the action falls to the Chenciner–Montgomery figure eight, and its period matches Simó’s published value to three parts in $10^{9}$. Under “half a period later is a half turn”, the same solver, started from noise, falls to Lagrange’s rotating triangle. Same equation, same solver: the global rule picks the answer.

The rule never appears as a residual. Solved once by hand, it says which Fourier modes may appear: in $x$ only cosines with $k \equiv \pm 1 \pmod 6$, in $y$ only sines with $k \equiv \pm 2 \pmod 6$. Those surviving coefficients are the unknowns, 20 instead of 80, and every relation then holds identically. Palais’ principle of symmetric criticality is what makes this legal: a critical point of the action among the symmetric loops is a true orbit.

Project at full strength. Theorem A already says which space-time Fourier modes of a g227 film can be non-zero. Making only those the unknowns would turn all nine relations into identities, instead of averaging three and shooting one. Sized on the published film’s own spectrum (91% of its time-varying energy is in the first harmonic), that is 4,917 unknowns at eight harmonics.

7 · BilliardsA path can close in space and still not close in time

A ball in a mirrored equilateral triangle: go straight, bounce at equal angles. Unfold the triangle instead of the ray and the path becomes a straight line through a field of reflected triangles. It closes exactly when the group element it has collected carries the start point forward along that same line (a translation along it, or a glide along it), so a loop on the orbifold is the same thing as a group element.

Now hang a clock on each mirror. A reflection done twice is nothing, so a mirror can cost only 0 or half a period, and the relations of the triangle group force all three mirrors to agree. Exactly 2 of the 8 decorations are legal, and they are the catalogue’s two p3m1 entries, g230 and g231. Under the half-period one, Fagnano’s three-bounce orbit closes in space and comes back half a period late. It takes two circuits to close in spacetime, and that 2 is $m$.

Step 4’s closure condition in two dimensions (the path must land on a moved copy of where it started), and Theorem E (a mirror costs 0 or ½) as something you can watch.

8 · Conway’s LifeThe blinker is a clockwork and the pulsar is not

Life is the clockwork problem in exact integers. Measured, not remembered: the blinker’s two phases are quarter turns of each other, so a quarter turn costs half a period ($m = 2$), and its always-on centre cell is Theorem B’s forced zero, $A_1 = 0$ exactly. The pulsar, the most symmetric small oscillator anyone names, is not a clockwork at all: every symmetry it has shows in a single frame.

Life’s search programs take the global rule as two declared settings: a symmetry that merges cell variables, and a transform applied after the last generation of the period. Those are step 4’s two ingredients. A 5×5 box with two mirrors and a quarter-turn twist has 9 free cells, and its 512 seeds give the blinker in milliseconds. Imposing $D_4$ on a 13×13 box cuts 169 unknowns to 28; its 268 million seeds took 171 s here (the demo times your own machine first) and find 13 patterns, the pulsar among them.

Step 4’s interface as a declared triple: the frame symmetry $K$, the twist $g$, the clock order $m$. One solver could take that triple for all 68 catalogue entries, the way Life searches take a symmetry and a transform.

9 · Which global rules can exist9 of the 27 ways to hang offsets on 333 survive, and they are the catalogue’s p3 groups

Which global rules exist is arithmetic, settled before any solver runs. Hang a time offset on each of the three cone points of 333: 27 ways. The loop that goes once round all three is trivial on the orbifold, so its clock must read a whole number of periods, $\sum_c k_c/n_c \in \mathbb{Z}$. That is the magic theorem’s residue identity. 9 of the 27 pass, and up to relabelling they are 4: exactly the catalogue’s four p3 groups, g224 to g227.

Because $\tau$ lands in an abelian group, a clockwork structure is a character of $G$, and the count is the number of characters, $\lvert G^{\mathrm{ab}}\rvert$: 8 for 2222, 9 for 333, 8 for 442, 6 for 632. Up to relabelling those are 3, 4, 6 and 6 distinct entries, and adding the torus gives the catalogue’s 20 orientation-preserving entries, with no misses and no extras.

The general answer: the H/K theorem

Let $\Gamma$ be a finite group acting on $\mathbb{R}^n$. Some $\Gamma$-equivariant system of ODEs has a periodic solution with spatio-temporal symmetries $H$ and frame symmetries $K$ if and only if

1. $H/K$ is cyclic;
2. $K$ is an isotropy subgroup;
3. $\dim \operatorname{Fix}(K) \ge 2$, and if it equals 2 then $H = K$ or $H = N_\Gamma(K)$;
4. $H$ fixes a connected component of $\operatorname{Fix}(K) \setminus L_K$, where $L_K$ is the union of $\operatorname{Fix}(\gamma)$ over $\gamma \in H \setminus K$.

Moreover, when these hold, some such system has an asymptotically stable limit cycle with exactly those symmetries. The job as run is such a system: the Brusselator on the 36×36 mesh is an ODE in 2,592 variables, $\Gamma$ is the nine operations of the box, $\lvert K\rvert = 3$, $H/K = \mathbb{Z}_3$, and $\dim\operatorname{Fix}(K) = 868$, the tutorial’s own count. The published film is a periodic solution with exactly these symmetries, so by the “only if” the four conditions hold for r33₁3₂, and by the “moreover” some equivariant system has a stable film with this symmetry. The published film’s ×1.16 instability is a fact about the Brusselator, not about the symbol. The caveat: the theorem’s system need not be a reaction–diffusion equation. For the 22 catalogue entries with a non-zero offset and no Brusselator record, the conditions have not been checked.

Other classifications, for contrast. Planar choreographies come in two infinite families plus three exceptional groups for odd $n$ (Montaldi and Steckles); that list is infinite because rotation orders are unbounded there, while the crystallographic restriction caps them for the films. On Life’s square grid the clock can only have 1, 2 or 4 positions, because 3 does not divide 8. And all of it sits inside Xu and Wu’s 275 spacetime groups in 2+1 dimensions, whose 68 forward-in-time groups are the clockwork catalogue.

Theorem D is a character count, $\operatorname{Hom}(H_1(\mathcal{O}), \mathbb{R}/\mathbb{Z})$. The H/K theorem is the tool that could separate “the symbol forbids it” from “this solver did not find it” for the 22 catalogue entries with no Brusselator record.

10 · Side by sideEvery analog carries the group theory exactly, and the dynamics only sometimes

The last three rows were researched and checked but are not built as demos here. The seven-link kaleidocycle has no symmetry in any single frame, yet a relabelling by three tetrahedra costs exactly a seventh of the period (to $10^{-10}$). Jeandel and Rao’s 11 Wang tiles fill a 32×32 square, but close onto no torus up to 12×12.

11 · Honest limitsWhat does not transfer

• A ring has no cone points. $\mathbb{Z}_N$ moves every cell, so the ring’s “forced zero” is an average, not a point of the medium. The lattice is the first toy here with real cone points.
• Collisions are not cone points. The figure eight’s forced moment is the collinear instant at $t = \tfrac14$; collisions are the edge of its loop space, and the Brusselator has no such edge.
• Space and time trade places in the choreography and the billiard: there an instant of time is fixed by a symmetry, while in the film a point of space is.
• The glue is not always safe. On the figure eight, which conserves energy, the delayed glue amplifies a disturbance 46 to 129 times per period; on the dissipative Brusselator ring it settles onto a loop the full ring cannot hold. Nobody has yet seeded a disturbance into the g227 glue and measured it.
• Life cannot host g227. On a square grid the clock has 1, 2 or 4 positions, never 3. A third-period clock needs a triangular or hexagonal grid, which is the grid the film already uses.

12 · For the callQuestions for Tuesday

1. Chaim: Theorem D’s count is exactly the number of characters, $\lvert G^{\mathrm{ab}}\rvert$, on all four cone-point orbifolds. Is the right statement “clockwork structures on an orbifold are $H^1(\mathcal{O};\mathbb{R}/\mathbb{Z})$”, and is it already known in that form?
2. Chaim: every dynamics classification splits into symmetries that shift time and symmetries that reverse it; the catalogue of 68 is the shifting half. Is there a sensible orbifold decoration for a symmetry that costs a reversal of time, and does a magic theorem still close on it?
3. Chaim: both solvers’ sizes are point counts of quotients, by Burnside: 146 nodes for the glue, 868 numbers for step 4. Worth stating on the project’s pages, so the other 67 entries can be sized before any solver runs?
4. Scott: Life searches take the global rule as two declared inputs. Should a clockwork solver take one declared triple (K, g, m), and is that the right object for vZome or Orthohedra to draw as a decorated fundamental domain?
5. Scott: could Orthohedra carry a time offset on each gluing, so that filling space from a decorated fundamental domain gives a clockwork animation directly? On the Klein quartic the Magic Theorem’s total changes, but the residue identity is only τ applied to the orbifold relation, so it should hold unchanged.
6. Vladimir: the lattice toy turns the stability around: there the clockwork assembles itself from noise, while the Brusselator’s film has to be solved for. Is “flip one sign” the right way to show the difference between a pattern nature makes and a pattern you have to solve for? Which of these demos would you want to play with together next week?
7. Everyone: the H/K theorem promises an equivariant system with a stable r33₁3₂ film. Worth hunting for a reaction–diffusion system where the film is an attractor, which would make the solver unnecessary?
8. Everyone: two cheap experiments nobody has run: seed a disturbance into the g227 delayed glue and measure its growth; and try a space-time Fourier solve whose unknowns are only the modes Theorem A allows (4,917 at eight harmonics), with no time stepping inside the residual.