Théorème

T5 — Unique fixed point μ* = 15

The sieve admits a unique stable fixed point at $\mu^* = 3+5+7 = 15$.

Statement

The sieve self-consistency equation

\mu^* = \sum_{p\,:\,\gamma_p(\mu^*) > s} p, \qquad s = 1/2,

admits a unique stable solution, namely:

\boxed{\mu^* = 3 + 5 + 7 = 15, \qquad \text{active set} = \{3, 5, 7\}.}

The proof combines an exact rational exhaustive scan over $\mu \in [3, 100]$ with an analytic monotonicity bound for $p \geq 7$ , excluding any fixed point outside the active primorial $\{3, 5, 7\}$ .

Théorème

Plain reading. We have a self-referential equation: ” $\mu$ must equal the sum of primes active at $\mu$ ”. How many solutions are there? T5 says: exactly one. It’s $\mu = 15$ with active primes $\{3, 5, 7\}$ . No other choice of primes closes the equation. This uniqueness is what makes PT “zero-parameter”: if several solutions existed, you’d have to choose, and that choice would be a parameter.

Why it matters

T5 is the summit of the chain of theorems. All preceding ones (T0–T6, GFT, L0) are needed to formulate T5, and T5 closes the sieve: at $\mu^* = 15$ , the parity split seeded by $p = 2$ is frozen as the $q_+/q_-$ bifurcation, and the 43 physical observables are then derived.

Three regimes exist in fact:

$\mu = 10$ , set $\{2, 3, 5\}$ — binary/spin sector,
$\mu = 17$ , set $\{2, 3, 5, 7\}$ — transitional sector,
$\mu^* = 15$ , set $\{3, 5, 7\}$ — information sector, that of our physics.

It is $\mu^* = 15$ that drives PT physics, after $p = 2$ has “crystallised” into the binary infrastructure.

Proof — outline

Define $\gamma_p(\mu) = -d\ln(\sin^2\theta_p)/d\ln\mu$ (anomalous dimension).
Compute $\gamma_p$ via exact rational arithmetic at $\mu = 15$ : $\gamma_3, \gamma_5, \gamma_7 > 1/2$ , $\gamma_{11}, \gamma_{13} < 1/2$ .
Verify $3 + 5 + 7 = 15$ — consistency reached.
Analytic bound: $\gamma_p$ strictly decreasing in $p$ for $p \geq 7$ , so no prime $\geq 11$ is active at any reasonable $\mu$ .
Exhaustive scan of finite subsets of $\{3, 5, 7, 11, 13, ...\}$ to confirm uniqueness.

Detailed proof

Step 1 — Exact rational computation of $\gamma_p$ at $\mu = 15$

At $q = 13/15$ , every quantity is an exact rational number (fractions.Fraction, zero floating-point error). Values are:

$p$	$\gamma_p$ (exact fraction)	$\gamma_p$ (decimal)	$> 1/2$ ?
3	$4\,536\,129 / 5\,616\,704$	0.80761…	✓
5	$486\,792\,684\,365 / 699\,097\,512\,194$	0.69632…	✓
7	$2\,827\,519\,972\,576\,117 / 4\,748\,396\,022\,746\,468$	0.59547…	✓
11	(exact rational, omitted)	0.42573…	✗
13	(exact rational, omitted)	0.35624…	✗

Differences $\gamma_3 - \gamma_5$ , $\gamma_5 - \gamma_7$ , $\gamma_7 - \gamma_{11}$ are strictly positive (verified by exact fraction subtraction).

Step 2 — Numerical consistency

The set $\{3, 5, 7\}$ is active at $\mu = 15$ . Its sum:

3 + 5 + 7 = 15 = \mu.

The self-consistency equation is satisfied. This is a fixed point.

Step 3 — Analytic monotonicity for $p \geq 7$

$\gamma_p$ factors as $\gamma_p = F(p) \cdot G(p)$ with:

$F(p) = 4 q^{p-1}/\mu$ — exponentially decreasing,
$G(p) = (1 - \delta_p)/(\delta_p (2 - \delta_p))$ — sign analysis.

The gap fraction $\delta_p = (1 - q^p)/p$ is strictly decreasing in $p$ (real-variable analysis $x \mapsto (1 - e^{-Lx})/x$ with $L = \ln(15/13) > 0$ ).

Consequence: for every $p \geq 7$ , $\gamma_p < \gamma_7 \approx 0.595$ . Adding a prime $p \geq 11$ to the active set is therefore impossible: $\gamma_p < 1/2$ would violate activity.

Step 4 — Exhaustive scan of finite candidates

Enumerate all finite subsets $S \subset \{3, 5, 7, 11, 13, 17, 19, 23, ...\}$ , compute $\mu_S = \sum_{p \in S} p$ , and test the consistency $\gamma_p(\mu_S) > 1/2$ for $p \in S$ .

Exhaustive scan results (cf. proof_T5_fixed_point.py, 33 PASS):

$S = \{3, 5, 7\}$ : $\mu = 15$ , consistent ✓
$S = \{3, 5\}$ : $\mu = 8$ , but $\gamma_3(8), \gamma_5(8), \gamma_7(8) > 1/2$ imply 7 should be included — contradiction.
$S = \{3, 5, 7, 11\}$ : $\mu = 26$ , but $\gamma_{11}(26) < 1/2$ — inconsistent.
$S = \{2, 3, 5\}$ : $\mu = 10$ , binary/spin sector (distinct fixed point, $p = 2$ included).
$S = \{2, 3, 5, 7\}$ : $\mu = 17$ , transitional sector.

No other subset closes the equation in the information sector (without $p = 2$ ). $\mu^* = 15$ is unique for the info sector.

Step 5 — Linear stability

Stability of the fixed point is checked by computing the Jacobian:

J_{ij} = \frac{\partial \gamma_{p_j}}{\partial \mu_i} \bigg|_{\mu = \mu^*}.

The spectral radius $|\lambda_\max(J)| < 1$ confirms $\mu^* = 15$ is an attractor (cf. ch. 8, Stability Theorem).

J1–J6: six independent justifications

The monograph lists six independent justifications of $\mu^* = 15$ :

Exhaustive scan (step 4).
Linear stability (step 5).
$N_c = 3$ : unique solution of $(N_c + 1)! / (N_c + 3) = 2^{N_s - 1}$ with $N_s = 3$ .
Prime 7 = integer oracle of $e^2$ : drift $A_p = \ln p / \sqrt{p}$ is maximal at $p = 7$ .
Echo product identity: $\prod_{p \in \{3,5,7\}} q_-^p = 1/e$ exactly.
Cosmogony: the sequence $\mu = 10 \to 17 \to 15$ gives the evolutionary scenario.

For the complete derivation, see chapter 8 of the monograph.