Cross the wall. Mix the eyes.
One rank-one projector moves on a small chart: z = x + iy, w = x + iy², P = (1+zw)⁻¹ [[1, w], [z, zw]]. Its metric is positive on one side of y = 0 and indefinite on the other. The operator H = (2 + y/10)P + (5 + y/5)(I − P) keeps its two eigenvalues apart the whole way, so the contour count and the trace-log never flinch. Each eye below reads one quantity along the same dial.
Each point is one value of y. Blue is before the wall, red after. Metric eye against ledger eye shows the whole thesis in one curve: the horizontal coordinate changes sign while the vertical one runs straight through.
Write what you expect before you move a dial. Snapshot, move, then record what moved. Trials stay in this page until you copy them out.
What every eye sees
Click a card to solo that eye on the scope.The licence expires first. The record never notices.
A fixed six-level base D₀ = diag(2, 3, 20, −20, 35, −35) is probed along two tangent directions that tilt from Hermitian toward anti-Hermitian with a dial κ. The exact entry formula gives g(κ) = G_H − κ² G_A. The matrices G_H, G_A and the curvature coefficients here are recomputed from the gate's own seeded family, and they return the gate's wall κ* = 0.6271277514757786.
From the gate log: 19 of 20 seeded operators produced a wall in the admissible range (seed 9 did not). Every block's licence wall lies inside its metric-definite region, and the summed record stays within 1.6×10⁻¹⁴ of its reference on every row, including on the walls.
Only the round trip can carry a sign
A one-way transition strength is a square and has no sign. The sign lives in the round trip Q = M₊₋ M₋₊ / Δ². At a Hermitian point the return leg is the mirror of the outgoing leg and the gap is real, so Q = |M|²/Δ² ≥ 0. Open the system and the phases stop cancelling.
| W at the instrument floor (28 points) | ≤ 1.1×10⁻⁷ |
| |det g| on the reciprocal gate | ≤ 8.9×10⁻¹⁶ |
| Non-reciprocal control W | 0.24 – 0.98 |
| Parity suppression of the even bump | 2.4×10⁷ |
| ROT-01 slope of W against a | 1.000 |
| ROT-01 predicted sign change of det g | failed, kept |
The rotation run lifted W exactly as predicted, but det g stayed negative at every tested point, so the predicted wall never appeared. The paper keeps that failure in its ledger.
The full tangent space is split exactly in half
Put an idempotent in the form diag(I_k, 0). Every tangent is off-diagonal, D = [[0, X], [Y, 0]], and the trace form is q = Re Tr(XY). With U = (X + Y†)/2 and V = (X − Y†)/2 it becomes q = ‖U‖² − ‖V‖²: as many positive directions as negative, 2k(n − k) each. The self-adjoint tangents are the slice V = 0, which is why a Hermitian family never sees a negative direction.
Recorded, not recomputed: the self-adjoint control reads (48, 0, 49) on its 97 directions; the open horizon generator reads (96, 98, 0) on its 194 potential-plus-damping directions, and (86, 11, 0) on potentials alone. Among random planes of that 194-direction form, 24.0% of 2-planes and 3.2% of 3-planes read as definite, and none from dimension 5. The live census above draws planes from a toy neutral form so you can watch the same effect happen; its percentages belong to the toy.
A negative direction pushes the overlap above the cap
For orthogonal projectors the overlap Tr(PR) is a sum of squared cosines and cannot exceed the rank. An oblique projector projects along a slanted kernel, and its overlap can. Theorem 8 turns this into a local law: Re Tr(P(0)P(t)) = k − ½ Re Tr(D²) t² + o(t²), so a direction with g(v, v) < 0 climbs above k for every small step.
Re Tr(P(0,y) P(0,y+h)) − 1 = h²(2y + h) / [(1 − y³)(1 − (y + h)³)]
The sign of the excess is the sign of 2y + h. On the positive side the local quadratic term is negative, but a finite step can still climb above the cap: at y = −1/10 and h = 1/4 the excess is the exact rational 25000/7980973. On the wall itself the first term is cubic, h³/(1 − h³), and changes sign with h.
What the record keeps, what it forgets
Three short stories about the scalar record. A Jordan block breaks the eigenvectors but not the trace. A determinant forgets whole turns of 2πi that the trace-log keeps. And a vanishing order sets the power law you would read off as a clock.
At t = 0 the two eigenvalues meet and only one eigenvector is left. The contour around both still counts 2 and the trace-log still reads log 4. What the trace loses is the nilpotent part, which the full matrix f(J) keeps.
With UV = e^(2πi/d) VU, the relator R = UVU†V† is e^(2πi/d) times the identity. Its determinant is exactly 1 at every d, while the sum of its eigenvalue angles is one full turn. As d grows the defect collapses; the integer stays. At d = 2 the eigenvalue sits on the branch cut and the wrap is undefined.
√|λ| grows as |t|^(m/2). A transverse crossing (m = 1) gives 1/2; a touching zero (m = 2) gives 1. The gates measured 0.5154 on the crossing case and 0.9934 on the touching case. The paper keeps these as exponents only: matching a power to a horizon lapse needs a physically fixed parameter first.
Press on the operator and read the record move
The bridge to Light Keeps the Ledger. Add a source, K → K + t·bb†, and the selected trace-log changes at the rate b† P_C K⁻¹ b. For a positive self-adjoint K that is a positive spectral weight, Σ λ⁻¹ ‖P_λ b‖², and a family of such readings rebuilds the selected subspace. For an oblique K the same rate can go negative.
K = S diag(2, 5) S⁻¹ = [[2, 3s], [0, 5]] and P_C = [[1, −s], [0, 0]]. For b = (1, 1) the rate is exactly 1/2 − s/2.
At s = 3 the rate is −1 and the selected eigenvalue's derivative is −2, as in the paper. The kernel of the oblique projector tilts, so b is projected onto the negative side of the range line. Nothing thermodynamic is negative here; the positive readings of Corollary 27 need a self-adjoint K.
A screen that reads Newton up close and MOND far out
Assume a spherical screen with N_A = 4πr²c³/(Għ) slots, the Unruh relation k_BT = ħa/(2πc), and a thermal mode density ρ_p(ω) ∝ ωᵖ up to a cutoff. The energy balance becomes a·f_p(a/a_D) = GM/r². With a flat density (p = 0), f → 1 gives Newton and f ≈ (π²/6)(a/a_D) gives the deep-MOND law a ≈ √(a₀ a_N), with a₀ = 6a_D/π².
| a_N/a₀ | flat-density a/a₀ | empirical fit a_RAR/a₀ | log₁₀(a/a_RAR) dex |
|---|
The table is recomputed here and matches the paper's to every printed digit. Over an 801-point sweep the largest gap is −0.07954 dex near a_N/a₀ = 0.372, where the screen gives about 0.833 times the fit. SPARC's own residual width about its fit is 0.11 dex.
Eleven slots and one Einstein coefficient
Take the form space of a four-dimensional spacetime: scalars, one-forms and two-forms, 1 + 4 + 6 = 11 slots. Put the curvature on it and expand the trace-log: −Tr₁₁ log(I − βℝ) = 3βR + (β²/2)(R² + R_μνR^μν + R_μνρσR^μνρσ) + …. Each sector's trace is R, so the leading term is 3βR, and β = ℓ_P²/(48π) gives the Einstein–Hilbert normalization 1/(16πG).
The record is not a relative entropy, and it is not the geometry
Two limits on what the scalar record can be. Proposition 21: with H = λP + μ(I − P) the projector and the selected eigenvalue can be varied separately, so the trace-log and the metric do not determine each other. Proposition 22: a same-reference relative entropy is at a minimum when the states coincide, so its first derivative is zero; the moving wall's record increment starts with slope 1/20. No constant multiple can match them.
The ride
Claim ledger
| Result | Where | Status | Scope |
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