The Theory of Persistence
Théorème

T4 — Spectral convergence

$\alpha_k \to 1/2$ as the sieve depth $k \to \infty$.

Statement

Consider the deep sieve at kk stages (eliminating multiples of p1,,pkp_1, \ldots, p_k). Define the mixed-transition fraction:

αk=n12(k)n12(k)+n21(k),\alpha_k = \frac{n_{12}(k)}{n_{12}(k) + n_{21}(k)},

where nij(k)n_{ij}(k) counts transitions from class ii to class jj among survivors at depth kk. Then:

limkαk=s=12.\boxed{\lim_{k \to \infty} \alpha_k = s = \frac{1}{2}.}

The proof rests on three pillars:

  1. Spectral annihilation r2(0)=0r_2(0) = 0 — the antisymmetric eigenvector vanishes at site 0 (structural result).
  2. Mertens compactness — asymptotic bounds on survivors (11/p)\prod (1 - 1/p).
  3. Gordin decomposition — the sequence {αk}\{\alpha_k\} is a quasi-martingale plus a bounded remainder.
Théorème

Plain reading. The deeper we refine the sieve (the more multiples we remove), the closer the fraction of “1 → 2” transitions vs. “2 → 1” gets to 50/50. At infinite depth, it’s exactly 50/50. That 1/21/2 is the stationary value defining ss. T4 proves that the sieve “finishes the job”: no residual bias remains.

Why it matters

T4 is the bridge between the finite sieve (at depth kk) and the ideal sieve (at infinite depth). T1, T2, T3 are exact statements about the ideal transfer matrix; T4 guarantees that the empirical matrix computed at finite depth converges to that ideal matrix.

Without T4, αk1/2>0|\alpha_k - 1/2| > 0 would persist, meaning a new constant (different from s=1/2s = 1/2) would emerge in the limit. PT would then be ill-defined. T4 closes this objection.

Proof — outline

  1. Reduce the convergence of αk\alpha_k to convergence of a Markov chain on the transition state.
  2. Decompose αk\alpha_k into quasi-martingale plus bounded remainder (Gordin).
  3. Bound the remainder via Mertens compactness.
  4. Annihilate the antisymmetric mode at site 0 by the r2(0)=0r_2(0) = 0 argument.
  5. Conclude: only the symmetric mode survives, eigenvalue λ1=1\lambda_1 = 1, uniform stationary distribution — hence α=1/2\alpha_\infty = 1/2.

Detailed proof

Step 1 — Exact recurrence

At each depth kk, the deviation D(k)=αk1/2D(k) = \alpha_k - 1/2 satisfies an exact recurrence (Recurrence Theorem, ch. 7):

D(k+1)=(pk3)D(k)+Δk,D(k + 1) = (p_k - 3) D(k) + \Delta_k,

where Δk\Delta_k is a remainder depending on correlations between primes eliminated at step k+1k+1.

This recurrence is verified for k=3,,11k = 3, \ldots, 11 by direct computation on survivors.

Step 2 — Spectral factorisation

The characteristic polynomial of the recurrence operator factors as:

f(1)=4(α1/2)2(α2+(p3)α+1).f(1) = 4 (\alpha - 1/2)^2 (\alpha^2 + (p - 3)\alpha + 1).

The first factor (α1/2)2(\alpha - 1/2)^2 has a double zero at α=1/2\alpha = 1/2, forcing the asymptotic fixed point. The second factor has complex roots for p3<2|p - 3| < 2 and real roots in (1,0)(-1, 0) for p3>2|p - 3| > 2 — in all cases, Rspec<1|R_\text{spec}| < 1.

Step 3 — Spectral bound

The global spectral bound is:

Rspec0.2871.R_\text{spec} \approx 0.287 \ll 1.

Strictly less than 1, ensuring geometric convergence of D(k)0D(k) \to 0.

Step 4 — Annihilation r2(0)=0r_2(0) = 0

To rigorously close the convergence, the remainder Δk\Delta_k must not contribute to the dominant mode λ220.36\lambda_2^2 \approx 0.36. Here the spectral annihilation intervenes:

The antisymmetric eigenvector r2r_2 associated with λ2\lambda_2 has an exactly zero value at state 0, hence:

r2,δ0=r2(0)=0.\langle r_2, \delta_0 \rangle = r_2(0) = 0.

This result is structural (cf. ch. 7, Spectral Closure Theorem) and unconditional on PNT. Consequence: only the mode λ120.012\lambda_1^2 \approx 0.012 survives in correlations starting from state 0, and convergence is exponential with strongly damped exponent.

Step 5 — Gordin decomposition

The sequence αk\alpha_k writes:

αk=Mk+Rk,\alpha_k = M_k + R_k,

with MkM_k a martingale relative to the survivor filtration, and RkR_k a remainder bounded by Mertens compactness:

RkCik(11pi)0.|R_k| \leq C \cdot \prod_{i \leq k} \left(1 - \frac{1}{p_i}\right) \to 0.

By the martingale convergence theorem and Rk0R_k \to 0, αk\alpha_k converges a.s. to a limit. Identification of that limit with 1/21/2 is forced by steps 1–4.

Consequence: closing the hierarchy problem

The result r2(0)=0r_2(0) = 0 has an unexpected consequence: it closes the PT hierarchy problem. The mode λ220.36\lambda_2^2 \approx 0.36 would have contributed a hierarchical phase shift between scales; its annihilation at site 0 ensures no new scale spontaneously emerges. This is what allows T5 to fix a unique fixed point μ=15\mu^* = 15 without hierarchical corrections.

For the complete derivation and auxiliary proofs (recurrence, factorisation, spectral bound, annihilation), see chapter 7 of the monograph.

See also