Split an operator, and price its weakest direction
Every real square matrix is a symmetric part plus a skew part, C = S + K. Expanding (S + K)(S − K) − (S − K)(S + K), the squares cancel and [C, Cᵀ] = −2[S, K]: the matrix is normal exactly when its two parts commute. The raw stress Φ = ½‖[C, Cᵀ]‖² = 2‖[S, K]‖² scales as the fourth power and is not a squared sine. Rigidity is a different reading: R_η = 1/(σ_min² + η), the inverse of the weakest transport singular value, regularized, and capped at 1/η.
A band's subspace moves: the metric prices it, the curvature twists it
For a projector P the tangent is off-diagonal, V = PV(I − P) + (I − P)VP, so the tangent pairing is a Gram matrix and the curvature is bounded by the metric: |Ω| ≤ 2√det g. A two-band family saturates it at every point. Here the lower band of d(k) = (sin kₓ, sin k_y, m + 2 − cos kₓ − cos k_y) is carried across the Brillouin zone. Its Chern number is the total curvature over 2π, an integer that changes only where the gap closes.
An integer that holds while the branch guard stays open
For unitaries U and V, the commutator 𝒰 = UVU†V† has determinant one, so the sum of its principal eigenphases is 2π times an integer w. While no eigenvalue reaches −1 (branch guard g_br > 0), w cannot change, |w| ≤ ⌊(d − 1)/2⌋, and the defect obeys δ + g_br² ≤ 4. Clock and shift sit exactly on that edge. Deform V and watch the eigenvalues move while the integer stays put, until one of them touches −1.
Which way, and how well: two readings of one marker
A photon scattered by a trapped atom leaves the atom kicked one way or the other. The marker states' overlap v sets the fringe visibility V = |v|; the best which-way guess sets the distinguishability D = √(1 − V²). For a pure marker D² + V² = 1 and the marker Gram has determinant D². For a warm, mixed marker the sum falls below one, which is Englert's inequality, but the Gram G_ab = Tr(ρU_a†U_b) stays positive at every rank.
An electric field tilts the ground state, and the tilt sets the response
Put a field ℰ on H(ℰ) = H₀ − ℰD. Each transition m contributes a metric piece g⁽ᵐ⁾ = |⟨m|D|0⟩|²/Δₘ², and the lossless polarizability is the same pieces with energy weights: α_s(x) = Σ 2Δₘ³/(Δₘ² − x²) · g⁽ᵐ⁾. Below the first gap the weights are positive, so the silent directions of the response and of the metric are the same. Between resonances the transitions can cancel: a tune-out. A second frequency in the same pole-free interval tells the two apart.
Peel a channel: what is removed, and what becomes silent
Lower the strengths of fixed transition channels. The metric can only fall, its silent space can only grow, and the rank drops only when the surviving channels stop spanning. Removing a redundant channel changes nothing you can see in the rank. Four optical readings, two frequencies before and two after, detect the removed coupling anyway, even across a tune-out cancellation.
The exact oscillator: stiffen the trap and the whole budget moves up
One bound charge in H = p²/2m + mΩ²x²/2 − qℰx. The field only slides the Gaussian ground state, so the fidelity, the metric g = q²/(2mħΩ³) and the polarizability α(ω) = q²/(m(Ω² − ω²)) are all exact, and derived two independent ways. The oscillator strength F = q²ħ²/2m never changes: stiffening the trap carries all of it upward. That is the paper's conserved upward redistribution, realized with the complete sum rule.
| Quantity | Ω = 1 | Ω = 2 |
|---|---|---|
| total oscillator measure F | 1/2 | 1/2 |
| electric-field metric g | 1/2 | 1/16 |
| response α(1/2) | 4/3 | 4/15 |
| secant B₀,¼ | 4/3 | 1/15 |
| squared index n² | 76/75 | 376/375 |
From a transition weight to a refraction angle
Rotate the electronic states of a two-level system against a fixed probe D = dσ_z. The energies never move, yet the field metric g = d² sin²ϑ/Δ² does, and with it the response, the permittivity R_opt = 1 + (N/ε₀)α, and the angle light bends at an interface. The optical law R₁sin²θ₁ = R₂sin²θ₂ has the same form as the rigidity law. Whether the two coefficients are the same thing is the open identification: for the natural field-Jacobian choice, no fixed calibration can do it.
| rotation ϑ | g | R_opt | angle from 45° |
|---|---|---|---|
| 0 | 0 | 1 | 45.000000° |
| π/4 | 1/8 | 7/6 | 40.893395° |
| π/2 | 1/4 | 4/3 | 37.761244° |
The boundary census is a count of couplings across the cut
Split a ground-state projector across a region A and its outside. Idempotence gives C_A − C_A² = BB†, so the eigenvalues of C_A inside the window [ε, 1 − ε] are exactly the singular values of the cross block with s² ≥ ε(1 − ε). The count lives on the boundary. Here is the paper's own 32 × 32 lattice, mass 0.3, with discs of radius 4 and 7, recomputed from the package's definitions.
Where the needles cross, the dominant mode gathers
A 28 × 28 lattice carries H = A + D₀ + εE: a Laplacian, a detuning that follows the light's intensity, and a skew flow along its gradient. As more needles cross the centre, the dominant pseudomode gathers into the 3 × 3 crossing cell, 1.15% of the area. The normal control, with the same detuning and no skew part, decides what the skew flow adds. The light is prescribed here: this is a finite model, not a solved emitter–field–medium evolution.
At an interface: a matching law, a threshold, and two ways to evolve
For the scalar action ½R|q̇|² − V, crossing a flat interface keeps the tangential momentum, so R₁sin²θ₁ = R₂sin²θ₂. A real transmitted angle needs R₁sin²θ₁ ≤ R₂: past that barrier the path turns back. The optical version has a square-root threshold, and on the Smith chart the reflection runs up the real axis to 1 and then around the rim once the wave is evanescent. Separately, a periodic skew generator keeps every norm, while a one-way chain with uniform damping rises and then decays.
The scalar ledger is even to every order; the directed response is not
Take H(t) = H₀ + tE with H₀ symmetric and E skew. Then H(−t) = H(t)ᵀ, the trace of the resolvent is the same at t and −t, and the contour ledger is even. For real positive H₀ it is exact: Z(t) − Z(0) = Σ log(1 + t²ωⱼ²). Even is not constant. And the boundary response built from the same family obeys S_b(−t) = S_b(t)ᵀ: its diagonal is even, but the difference S₁₂ − S₂₁ is odd and moves at first order.
What survives a wall, and what a boundary measures
At a collision of two modes, H(γ) = σₓ + iγσ_z, each mode's own projector blows up and its left–right overlap falls as √(1 − γ²). The pair's projector stays rank two, and the pair's trace-log stays smooth. At a boundary, eliminating the interior pulls the lost power back to the Hermitian part of the interior operator, I − S†S = 4(WV)†Herm L(WV), with no normality needed. And a thin sheet of dipoles obeys R + T + A = 1 with A ≤ ½: no reflection is not the same as complete absorption.
Needles of light: three questions, one open estimate
A Kakeya set holds a unit segment in every direction; a tube problem asks how thickened segments overlap; a wave estimate controls a coherent sum with its interference. They are related, and they are different. A light packet has a wavelength, a width and a propagation length, and staying a needle over length L needs w² ≳ λL. Wang and Zahl proved the three-dimensional Kakeya dimension; the critical flat-space wave estimate (6.2) is still open, and no finite census bound proves it.
Claim ledger
Every row is a statement of the paper with its grade. [T] has a written proof; [L] marks a finite form certified by kernel-checked Lean at a named revision; [V] is an executed computation; [C] is cited literature; [H] is an interpretation or open physical identification.
| Statement | § | Grade | Reading |
|---|