Amplification: the corner strength in the chain frame is zeta_step (1 + z')
Status: proved — a complete structured proof.
Area: Lattice, thermal, massive and gapped geometries (O7, O9, O10) · Card type: theorem · Knowledge-base id: frame-strength-amplification
Statement
For a corner (p, sigma) and two intervals I_step subset I containing p, the strengths zeta^(I') = sigma beta_{I'}(p) (beta_{I'}(x) = (x-alpha)(omega-x)/(omega-alpha), the modular-frame density of I' = (alpha,omega)) satisfy zeta^(I) >= zeta^(I_step), with equality iff I = I_step, because beta_{(alpha,omega)}(p) is increasing in omega and decreasing in alpha. If the step's kept block reaches the left end of the chain, I_step = (p_1, f) with f the far end of the moving part, then zeta^(I)/zeta^(I_step) = beta_I(p)/beta_{I_step}(p) = (a+b)(b+c)/(b(a+b+c)) = 1 + z' with a = p - p_1, b = f - p, c = p_{n+1} - f; z' = ac/(b(a+b+c)) is simultaneously the cross ratio of (p_1, p, f, p_{n+1}) and paper 1's Markov variable z of the coarse triple (kept block | A_B u A_new | rest of the chain). z' >= 0, and z' = 0 iff f = p_{n+1}, i.e. iff the step is the last one. Underlying exact fact (Lem 3.3, 3.4): the parabolic parameter is a (-1)-density, sigma' = sigma/M'(p) under a Moebius change of coordinate, so zeta = sigma beta_I(p) is Moebius invariant and is exactly the coefficient of w(y-y_p) = -4 sinh^2((y-y_p)/2) 1_{y>y_p} in the first-order field in the modular frame of I; for I = (-a,L), p = 0, sigma = s/L it is paper 1's zeta = as/(a+L).
How to verify
rigor/g1a_calculus.py section 6 (20000 exact rational cases, 0 bad); rigor/ref_p6_0_calculus.py item (d); the field identity in rigor/ref_p6_0_field_exact.py (50000 rational cases).
Evidence
- rigor/network_corner_calculus.tex — document, at
lem:ampl - rigor/g1a_calculus.py — numerics
- rigor/ref_p6_0_calculus.py — numerics
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Review
Review record: passed 2026-09-23 by REF-DEP-CFT-B1.
The referee's report and the card's event log stay in the private knowledge base; only the verdict, the date and the referee are published here.
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