Théorème

T2 — Spectral conservation

Exact spectral identity $|\lambda_2(T_{30})| = s^2 = 1/4$.

Statement

Let $T_{30}$ be the transfer matrix of the mod-30 sieve (the primorial $2 \cdot 3 \cdot 5$ ). Its second-largest eigenvalue $\lambda_2$ satisfies the exact algebraic identity:

\boxed{|\lambda_2(T_{30})| = s^2 = \frac{1}{4}.}

This identity comes from the Chinese Remainder Theorem (CRT) factorisation:

T_{30} \cong T_3 \otimes T_5,

and the dominant non-trivial eigenvalue $|\lambda_2(T_3)| = |\lambda_2(T_5)| = 1/2 = s$ on the antidiagonal blocks.

Théorème

Plain reading. If we look at how “information” propagates through the mod-30 sieve, we find it decreases by exactly a factor $1/4$ at each step. Not $1/3$ , not $1/5$ — exactly $1/4 = (1/2)^2$ . And that $1/2$ is the fundamental symmetry $s$ . This “spectral conservation” says PT has no hidden degree of freedom changing the propagation speed.

Why it matters

T2 is the conservation identity in the spectral sense: it expresses quantitatively how information persists across sieve steps. It is a precursor to the Gap Fundamental Theorem (GFT), which states the same conservation in entropic form: $\log_2 m = D_{KL} + H$ .

T2 also serves as a numerical touchstone: any empirical computation on $T_{30}$ must return $|\lambda_2| = 0.25$ exactly (not approximate). It is an internal consistency test of PT.

Proof — outline

Factor $T_{30}$ via CRT: $\mathbb{Z}/30 \cong \mathbb{Z}/2 \times \mathbb{Z}/3 \times \mathbb{Z}/5$ .
Reduce to the cascade sub-block: after excluding the spin/parity factor $p = 2$ , $T_{30}|_\text{dyn} = T_3 \otimes T_5$ .
Compute the eigenvalues of $T_3$ and $T_5$ on surviving classes: $|\lambda_2(\tilde{T}_3)| = |\lambda_2(\tilde{T}_5)| = 1/2 = s$ .
Multiply: $\lambda_2(\tilde{T}_3 \otimes \tilde{T}_5) = (1/2) \cdot (1/2) = 1/4$ .

Detailed proof

Step 1 — CRT factorisation

The Chinese Remainder Theorem gives the ring isomorphism:

\mathbb{Z}/30\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.

The sieve transfer matrix respects this factorisation because divisibility conditions for 2, 3, 5 are independent. Hence:

T_{30} = T_2 \otimes T_3 \otimes T_5.

Step 2 — Excluding the spin/parity factor $T_2$

By U4 (cf. T0), the factor $T_2$ carries the spin/parity infrastructure and does not enter the cascade spectrum. It fixes parity (survivors are odd), restricting the state space without adding a non-trivial transition. The relevant dynamical matrix is therefore:

T_{30}|_\text{dyn} = T_3 \otimes T_5.

Step 3 — Spectrum of $\tilde{T}_3$ (normalised)

By T1 + T3, $T_3 = \mathrm{antidiag}(1, 1)$ on $\{1, 2\} \pmod 3$ . After normalisation to a stochastic matrix (rows summing to 1), the dominant eigenvalue is $\lambda_1 = 1$ (stationary mode) and the second eigenvalue is:

|\lambda_2(\tilde{T}_3)| = 1/2 = s.

This is the eigenvalue of the antisymmetric mode on $\{1, 2\}$ .

Step 4 — Spectrum of $\tilde{T}_5$

For $T_5$ , the sieve eliminates $0 \pmod 5$ , leaving 4 surviving classes $\{1, 2, 3, 4\}$ . The transfer matrix on these classes is circulant (invariant under cyclic shift), with eigenvalues:

\sigma(\tilde{T}_5) = \{1,\, 1/2,\, 1/2,\, 0\}.

Direct computation by Fourier diagonalisation on $\mathbb{Z}/4\mathbb{Z}$ — see ch. 3 of the monograph, p. 145.

So $|\lambda_2(\tilde{T}_5)| = 1/2 = s$ (same as $T_3$ ).

Step 5 — Tensor product

For two matrices $A$ and $B$ , the spectrum of $A \otimes B$ is:

\sigma(A \otimes B) = \{\lambda_i \mu_j \mid \lambda_i \in \sigma(A),\ \mu_j \in \sigma(B)\}.

So:

\sigma(\tilde{T}_3 \otimes \tilde{T}_5) \supseteq \{1 \cdot 1, 1 \cdot 1/2, 1/2 \cdot 1, 1/2 \cdot 1/2, \ldots\} = \{1, 1/2, 1/2, 1/4, \ldots\}.

The largest is $\lambda_1 = 1$ (stationary). The second is:

|\lambda_2(\tilde{T}_{30}|_\text{dyn})| = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = s^2.

QED.

Consequences

Exact conservation. The contraction factor of the dominant non-trivial mode is exactly $s^2 = 1/4$ , not a number close to it.
No free parameter. The identity $|\lambda_2| = s^2$ is purely algebraic: it depends neither on $\mu$ , nor on $q$ , nor on any physical observable.
Numerical touchstone. Any simulation of the mod-30 sieve must reproduce $|\lambda_2| = 0.25$ to machine precision.

For the complete derivation, see chapter 3 of the monograph.