Théorème

T3 — Antidiagonal transfer

$T_3 = \mathrm{antidiag}(1,1)$ — the mod-3 matrix is purely off-diagonal.

Statement

The mod-3 sieve transfer matrix on 6-rough survivors is:

\boxed{T_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \mathrm{antidiag}(1, 1).}

This purely antidiagonal structure is a direct consequence of T1 (forbidden transitions): the diagonal coefficients being zero (T1) and each row needing to sum to 1 (stochastic matrix), the off-diagonal coefficients must equal 1.

Théorème

Plain reading. Given T1 (two consecutive 6-rough integers are never in the same mod-3 class), the “matrix” describing how to move between classes is ultra-simple: we always switch, never stay. As a 2×2 table, it’s a checkerboard: 0s on the diagonals, 1s elsewhere.

Why it matters

T3 makes explicit the matrix $T_3$ used everywhere downstream: T2 uses it for spectral conservation, T6 for holonomy, T4 for convergence. It is the simplest possible form of a non-trivial 2×2 stochastic matrix — a pure flip operator.

That $T_3$ is purely antidiagonal also has important geometric consequences: its eigenvalues are $\pm 1$ , its square is the identity $T_3^2 = I$ , and its action on the basis $\{1, 2\}$ is an involution.

Proof — outline

Start from T1: $T_3[1,1] = T_3[2,2] = 0$ .
Impose row sum = 1 (stochastic matrix).
Conclude $T_3[1,2] = T_3[2,1] = 1$ .

Detailed proof

Step 1 — Stochastic matrix

A Markov-chain transfer matrix $T$ satisfies $\sum_j T[i, j] = 1$ for each row $i$ (outgoing transition probabilities sum to 1).

For $T_3$ on $\{1, 2\}$ :

T_3[1, 1] + T_3[1, 2] = 1, \qquad T_3[2, 1] + T_3[2, 2] = 1.

Step 2 — Applying T1

T1 asserts $T_3[1, 1] = T_3[2, 2] = 0$ . Substituting:

T_3[1, 2] = 1, \qquad T_3[2, 1] = 1.

Hence:

T_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Spectrum

The characteristic polynomial is:

\det(T_3 - \lambda I) = \lambda^2 - 1.

So $\lambda \in \{+1, -1\}$ . Associated eigenvectors:

$\lambda_1 = +1$ : $v_1 = (1, 1)^T$ (stationary, uniform on $\{1, 2\}$ ),
$\lambda_2 = -1$ : $v_2 = (1, -1)^T$ (antisymmetric mode).

Involution

Since $T_3^2 = I$ , two successive applications return to the starting state. Consistent with the plain reading: “flip, flip, back”.

Articulation with T6

On the branch $q = 1 - 2/\mu$ , the generalised transfer matrix at non-trivial depth takes the form:

T_3(q) = \begin{pmatrix} 1 - \delta_3 & \delta_3 \\ \delta_3 & 1 - \delta_3 \end{pmatrix}

with $\delta_3 = (1 - q^3)/3$ . The diagonal $1 - \delta_3 = \cos\theta_3$ appears naturally, and the T6 formula $\sin^2\theta_3 = \delta_3 (2 - \delta_3)$ emerges as the squared off-diagonal transition amplitude.

In the ideal case $q \to 1$ (pure sieve limit), $\delta_3 \to 0$ and $\cos\theta_3 \to 1$ , $\sin^2\theta_3 \to 0$ . At $q = 13/15$ (fixed point $\mu^* = 15$ ), $\delta_3 = 0.1163$ and $\sin^2\theta_3 = 0.2192$ — the value entering the calculation of $\alpha_{\mathrm{EM}}$ .

T3 is therefore the limit case of T6 where $\delta = 1/2$ (stationary half-flip line), or more precisely the combinatorial envelope of which T6 is the analytically parametrised version.

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