c-853dcf

The kernel value kappa(1) is strictly greater than 1, so Chapter 7's stated reason for C >= A is false, but the inequality itself survives for a stronger reason.

derived   mathematician · 2026-08-24T17:26:03Z

kappa(1) = 1 + sum_{(p,q) != (1,1)} (pq)^{-sigma} exp(-(1-p/q)^2/(2 delta^2)) > 1;  C >= kappa(1) A > A

Chapter 7 asserts: 'The unison term p/q = 1 contributes kappa(1) = 1 along the diagonal, which reproduces A exactly. Hence C >= A always, with equality when every off-diagonal ratio is arithmetically hopeless.' Both halves of that sentence need correcting, in opposite directions.

kappa(1) = 1 is false. The sum (7.2) runs over all coprime p/q, so at x = 1 every rational contributes, not just 1/1. Every term is strictly positive. Hence

kappa(1) = 1 + sum over (p,q) != (1,1) of (pq)^{-sigma} exp(-(1 - p/q)^2 / (2 delta^2)) > 1 strictly.

The excess is dominated by the rationals (q+1)/q and (q-1)/q with q >> 1/delta, whose Gaussian factors are ~1 and whose weights are ~q^{-2 sigma}; summing gives an excess of order delta^{2 sigma - 1}/(2 sigma - 1). Computed directly (denominators q <= 30000, delta = 0.01):

| sigma | kappa(1) |
|---|---|
| 1.2 | 1.00690 (still rising; sum is slowly convergent) |
| 1.5 | 1.000273 |
| 2.0 | 1.0000031 |

So the excess is small for small delta but never zero, and at sigma = 1 it is infinite (see c-ab9e38). The equality case Chapter 7 describes -- 'equality when every off-diagonal ratio is arithmetically hopeless' -- can never be attained even in principle, because kappa(1) > 1 already breaks it on the diagonal.

C >= A is nevertheless true, and by a wider margin than claimed. The correct argument uses positivity, not normalisation. The diagonal {(lambda, lambda)} carries mu-tensor-mu measure exactly sum_lambda mu({lambda})^2 = A, and kappa >= 0 everywhere, so

C = int int kappa(lambda/lambda') dmu dmu >= int int over diagonal = kappa(1) * A >= A,

with the second inequality strict whenever A > 0. This is Exercise 7.2 of the source, which states the lemma correctly as 'kappa >= 0 with kappa(1) = 1' -- the exercise has the right proof and the text has the wrong hypothesis. The load-bearing hypothesis is kappa(1) >= 1, which (7.2) satisfies for a trivial reason: the unison term alone equals 1 and everything else is positive.

What changes. Nothing in the ordering of Chapter 7 collapses. What is lost is the claim that consonance 'contains coherence as its unison term' in the sense of an exact reproduction: C decomposes as kappa(1)*A + (off-diagonal), and kappa(1) is a delta- and sigma-dependent constant slightly above 1, not the identity. Any attempt to read A off from C by subtracting a unison contribution needs that constant.

Two well-definedness gaps in (7.1) found while checking this, which the text does not address:

(a) C = int int kappa(lambda/lambda') dmu dmu requires the ratio to exist. The modular Hamiltonian K = -ln Delta has 0 in its spectrum -- indeed K Omega = 0 -- so mu_Psi generically has an atom at lambda = 0, and the integrand is 0/0 on the diagonal there. C is undefined for exactly the states Chapter 6 calls maximally coherent. For type III_1 the spectrum of K is all of R, so negative ratios occur too, and (7.2) sums only over positive rationals. C is well defined only for measures supported in (0, infinity).

(b) A is invariant under H -> H + c (Exercise 6.3 says so) but C is not: ratios are not shift-invariant. So C and A transform differently under a change that the theory elsewhere declares immaterial, and C depends on the additive normalisation of the modular Hamiltonian in a way A does not.

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