Contents
Where this sits
The coupler draws on the frame layer and on version 1’s pump, and binds the two structures. Its interface to the calculations is the 3D decoder of Explainer 9, which reads positions only.
The idea
Six calculations must not be touched, yet their paths should be free to turn. So keep two kinds of state apart: where a particle is (which the calculations read) and how it is oriented (which they never read). The coupler owns the orientations. It takes the torsion one structure builds up and hands the inverse to the other. The two memory structures are thus bound through a third system, and nothing is created or lost: every twist issued is matched by its inverse, exactly.
Frames and exact transport
Each particle’s factor carries a frame, a quaternion \(q\), measured against the canonical frame at its point (the boost \(L_x\) from the origin). A particle that has never moved has \(q = 1\). When it moves from \(x\) to \(y\), its frame is carried along the geodesic by the transvection
$$T_{xy}(v) = v + \frac{\langle v, x+y\rangle}{1+c}\,(x+y) - 2\langle v, x\rangle\, y,\qquad c = -\langle x, y\rangle = \cosh d(x,y),$$
and against the canonical frames the frame turns by the twist \(W = L_y^{-1}\,T_{xy}\,L_x\).
F1 · Transport is exact PROVEN
\(T_{xy}\) is a proper orthochronous Lorentz matrix with \(T_{xy}x = y\), fixing every vector orthogonal to both \(x\) and \(y\). For rational \(x, y\) it is rational. The twist \(W\) fixes the origin, so it is a rotation of the tangent space there, and it too is rational. Its quaternion \((1 + \operatorname{tr} W,\ W_{21}-W_{12},\ W_{02}-W_{20},\ W_{10}-W_{01})\) is rational with a positive first entry. No square root is needed anywhere.
Let \(\rho_u(v) = v - 2\frac{\langle v,u\rangle}{\langle u,u\rangle}u\), the Lorentz reflection in the hyperplane orthogonal to \(u\). With \(\langle x,x\rangle = -1\), \(\rho_x(v) = v + 2\langle v,x\rangle x\). With \(s = x + y\), \(\langle s,s\rangle = -2(1+c)\), so \(\rho_s(v) = v + \langle v,s\rangle s/(1+c)\). Expanding,
$$\rho_s\rho_x(v) = v + 2\langle v,x\rangle x + \frac{\langle v,s\rangle - 2(1+c)\langle v,x\rangle}{1+c}\,s = v + \frac{\langle v,s\rangle}{1+c}s - 2\langle v,x\rangle y = T_{xy}(v).$$
So \(T_{xy}\) is a product of two Lorentz reflections: Lorentz, with determinant 1. \(T_{xy}x = x - x - y + 2y = y\), so it keeps the upper sheet. A vector orthogonal to \(x\) and \(y\) is fixed, so \(T_{xy}\) is the pure boost along the geodesic, which is parallel transport. Every entry is a rational function of \(x\), \(y\) and \(1 + c\). \(L_x\) and \(L_y^{-1} = JL_y^{\top}J\) are rational, and \(W o = L_y^{-1}T_{xy}x = L_y^{-1}y = o\). For a rotation with angle \(\theta\), \(1 + \operatorname{tr}W = 2 + 2\cos\theta\), which is positive below \(\pi\). F2 shows \(\theta < \pi\).
A twist is an area
F2 · The twist’s angle is the area of the triangle (o, x, y) PROVEN
The twist of a move from \(x\) to \(y\) is a rotation about the normal of the plane through \(o\), \(x\) and \(y\). Its angle is that geodesic triangle’s area, \(\pi - (\alpha + \beta + \gamma)\), less than \(\pi\).
The canonical boost \(L_x\) is itself the transvection \(T_{ox}\). So \(W = T_{yo}T_{xy}T_{ox}\) is parallel transport once around the geodesic triangle \(o \to x \to y \to o\): its holonomy. The triangle lies in a totally geodesic copy of H², along which the plane’s normal is carried parallel and so fixed. In that plane, of curvature \(-1\), Gauss–Bonnet makes the holonomy a rotation by the triangle’s area, \(\pi - (\alpha+\beta+\gamma) < \pi\).
So the twist issued by a move is the area the move sweeps against the origin. A particle wandering far out sweeps large triangles and issues large twists. The simulation plots the twist each structure issues per zero point.
Spin
S · Quaternions keep the spin PROVEN
A rotation matrix cannot tell a turn of \(2\pi\) from a turn of \(4\pi\); a quaternion can. Its sign is the spin, and it returns to itself only after \(4\pi\). Each twist is lifted to the quaternion with positive first entry, its unique lift of angle below \(\pi\). Frames are then composed as quaternions, never as matrices. So the composed quaternion is the lift of the whole path, and \(q\) and \(-q\) record opposite spins of the same rotation. Exact frames are kept as coprime integers up to a positive scale, which changes neither the rotation nor the spin.
The coupler: the mirror law
Pair structure 1’s channel \(j\), particle \(p\), factor \(i\) with structure 2’s same. Structure 1 keeps its frame \(q_1\) as it is. Structure 2 keeps its frame mirrored: \(q_2\) is the inverse of its geometric frame. Whenever either structure issues a twist \(w\) at a paired factor:
$$q_1 \leftarrow w\,q_1,\qquad q_2 \leftarrow q_2\,w^{-1}.$$
K1 · The binding never changes PROVEN
For every pair, the binding \(B = q_2q_1\) stays at its starting value, the identity, through every twist either structure issues, in carry and in steer mode. In exact runs this holds exactly; in float runs, to rounding.
After a twist, \(B' = (q_2w^{-1})(wq_1) = q_2q_1 = B\) by associativity, whichever structure issued \(w\). In exact arithmetic the inverse of an integer quaternion is its conjugate up to a positive scale, so \(B'\) equals \(B\) as a normalised integer quaternion. The coupler checks this for every pair at every zero point and writes the result into its own memory word.
Why the mirror. Quaternions do not commute. Applying \(w^{-1}\) on the same side, \(q_2 \leftarrow w^{-1}q_2\), gives \(B' = w^{-1}q_2\,w\,q_1\), which is \(B\) only when \(w\) commutes with \(q_2\). Applying \(w\) itself to the partner fails too. Keeping one structure’s frames mirrored is exactly what makes one rule conserve the binding in both directions. In geometric terms, both structures’ frames receive every twist, so their relative orientation is what stays bound.
The coupler is a group action. Every update is a bijection, and every twist it issues is matched by its inverse, so it erases no information. That is the precise sense in which the offload “does not violate thermodynamics”: logical reversibility, with an exact conservation law. It is not a statement about physical energy. Tension, the boosts that move points, is not routed through the coupler: it moves positions, and the calculations see positions.
Carry and steer
K2 · Carry mode is invisible PROVEN
With the coupler in carry mode, every position, incumbent, certified cost and memory hash of all six calculations is bit for bit what it would be with the coupler off.
In carry mode the coupler reads each structure’s states after every frame and writes only its own frames and ledger. It never writes a state, a generator or an objective. The decoders and Helix B read positions only (Explainer 9). So by induction on frames, the six calculations receive the same inputs and draw the same random numbers, and produce the same outputs.
K3 · Steering keeps everything exact PROVEN
In steer mode each particle’s rotation is turned by its frames before the step: \(R \leftarrow \operatorname{blockdiag}(\operatorname{rot} f_i)\,R\) over the factors of the block. In exact runs the steered rotation is still rational and orthogonal, every state stays an exact point of the sheet, and every register’s pump closes with \(H = P^{-1}\).
\(\operatorname{rot}(f)\) of an integer quaternion is a rational orthogonal matrix, so the block-diagonal product with an exact rotation is exact and orthogonal. H1 holds for any exact rotation, so corners and states stay on the sheet. H7 is group algebra: it holds for any increments, so the pump closes. K1 does not depend on the mode at all.
What steering changes. Paths. Each structure’s polytopes now turn with frames that also carry the other structure’s twists, so the six calculations influence each other through the coupler alone. With Haar-random \(R\), \(\operatorname{rot}(f)R\) is still Haar-random for any fixed frame, by left invariance. But frames depend on the past, and the browser uses Cayley rotations, which are not Haar. So the search’s law does change. Whether that helps is for a preregistered run to decide.
Live checks and parity
| Property | Trials passed | Control rejected | What is measured |
|---|---|---|---|
| Running… | |||
The reference’s trace reference/traces/twin3d.json holds 30 exact transports, and two structures plus the coupler run for a whole exact register in carry and steer mode. It records every memory word, both pumps, every final state and every frame. The run takes about a minute.
The 3D simulation, manual
The simulation opens paused, on B1 in H³ with 30 vertices, three particles per channel and the coupler steering. Every setting except speed rebuilds the run from frame 0.
| Control | What it does |
|---|---|
| Play, +1 frame, → zero point, Reset | As in version 1. Keys: Space, . and z. |
| coupler: steer | carry | off | The switch. Steer turns each polytope by its frames; carry only transports and checks them; off runs the two structures unbound. |
| problem, size | B1, 3-colouring in H³ (30 or 60 vertices), or B2, MAX-3-SAT in H³ (30, 60 or 120 variables). Both structures solve the same problem type on their own instances: seed and seed + 1. |
| particles, ε, r, μ | Particles per channel, the softmax temperature, the step radius and the membrane’s pull. |
| exact track | B1 with 9 vertices, one particle and one factor per step: exact states, exact frames, the exact binding check and the register pump. It is slow, as BigInt arithmetic is. |
| no mirror (control) | Sends structure 2 the twist itself instead of its inverse. The binding must break, and the log says when it does. |
| Panel | What it shows |
|---|---|
| Structure 1, structure 2 | The Poincaré ball of one factor of each structure, with each channel’s particles and their frames as three short arms. The view zooms to fit the particles; a dashed circle is the ball’s edge. Drag either ball to turn both. Click a vertex in an answer panel to choose the factor shown. |
| Coupler | The twist each structure issued per zero point; the number of unbound pairs per zero point (zero while the law holds); the shared clock and register; the binding, the largest deviation, the twists issued, the total twist and both pumps. |
| The answers | Each structure’s best certified colouring or assignment, decoded by the anchor decoders and counted exactly by Helix B. |
| Three memory chains | The latest words of structure 1, structure 2 and the coupler, with both structures’ chains re-verified on every redraw. |
| Self-test, log | The decoder trace on load (28 lines); the twin trace on request (21 lines). Builds, broken bindings and closed registers. |
Experiments
1. Carry changes nothing carry off
Run both to the same frame and compare the memory words of structure 1. The hashes are identical (K2), although the coupler has transported thousands of twists.
2. Steering binds the paths steer
Same seeds, steer mode: the hashes now differ from carry’s, but every pair stays bound at every zero point (K1).
3. Remove the mirror no mirror
The coupler sends the twist itself. Within the first zero points the binding breaks on most pairs: the law needs the inverse on the mirrored side.
4. Exactly exact track
Every check is exact: the binding equals the identity as integer quaternions, and at frame 720 both structures’ registers close with H = P⁻¹ under steering (K3).
5. Twists grow with reach B2, 60 variables
Watch the twist chart: as particles move away from the origin they sweep larger triangles, so each move issues a larger twist (F2).
What is not claimed
- No claim that steering helps the search. That needs a preregistered round (phase E) with the controls already built: coupler off, carry only, no mirror.
- B3 and B4 are not in the twin yet. B3’s anchored charts change coordinates when a particle is re-anchored, and the frame would have to be carried through that change. B4’s encoder factors are Euclidean.
- Pairing needs equal shapes. Both structures must have the same channels, particles and factors.
- Physics. “Torsion” here is frame holonomy, and “thermodynamics” is logical reversibility. Neither is a physical energy statement.