Essay · Plain · 6 min

Where does s = 1/2 come from?

Why the only input of the model, the fundamental symmetry s = 1/2, is not a choice but a forced arithmetic consequence. A guided tour of theorem T1 (mod-3 forbidden transitions).

Go deeper: T1 , T3 , L0

The unique input

PT contains exactly one numerical input: the symmetry $s = 1/2$ . Everything else — $\mu^* = 15$ , $\alpha_{\rm EM}$ , masses, metric, the 43 observables — descends from it by deduction.

The natural question: why $1/2$ ? Why not $1/3$ , $0.47$ , or some fitted number?

PT answer: $s = 1/2$ is not chosen. It is forced by an elementary arithmetic theorem, the T1 theorem of mod-3 forbidden transitions.

The crucial observation

Take the list of “6-rough” integers — those whose only divisors are neither 2 nor 3. Modulo 30:

\{1, 7, 11, 13, 17, 19, 23, 29\}.

Each of these eight numbers has a mod-3 class: 1 or 2 (never 0, since they are not divisible by 3). Their classes alternate:

1 \to 1 \to 2 \to 1 \to 2 \to 1 \to 2 \to 2.

At first glance, almost uniform. What is remarkable is what we never see: three consecutive 6-rough integers are never all in the same mod-3 class. Not a coincidence — a theorem.

Why it is forbidden

Take three consecutive 6-rough integers $p$ , $p + g$ , $p + g + g'$ , and suppose both gaps $g$ and $g'$ are $\equiv 0 \mod 3$ . Then the three integers cover the three classes $\{0, 1, 2\}$ mod 3, and one of them is divisible by 3 — hence not 6-rough. Contradiction.

Rigorous consequence:

\boxed{P(1 \to 1 \mod 3) = P(2 \to 2 \mod 3) = 0.}

This is exact, not statistical. Not an average, not an approximation: a strict arithmetic zero.

The matrix $T_3$ and its spectrum

Place the allowed transitions in a matrix:

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

It is the unique $2 \times 2$ doubly-stochastic matrix of vanishing trace. Two eigenvalues: $+1$ (stationary vector) and $-1$ (antisymmetric mode).

The involution $T_3^2 = I$ means two successive applications return to the starting state: the class “flip, flip, back”. This pure-flip symmetry produces $s = 1/2$ .

Why $1/2$ and nothing else

The stationary distribution of $T_3$ is uniform: half the transitions go from class 1 to class 2, the other half from class 2 to class 1. This exchange symmetry measures exactly $1/2$ .

In filter language:

s = \frac{n_{12}}{n_{12} + n_{21}} \xrightarrow[k \to \infty]{} \frac{1}{2}.

Theorem T4 (spectral convergence) closes this limit. At infinite sieve depth, the fraction is exactly $1/2$ . No other number is compatible.

What this then forces

Once $s = 1/2$ is fixed, the cascade unfolds with no further choice:

$\alpha(3) = s^2 = 1/4$ (T2: spectral conservation);
$L_0$ (maximum entropy) imposes the geometric distribution with two natural branches $q_+ = 1 - 2/\mu$ and $q_- = e^{-1/\mu}$ ;
T6 (holonomy) gives $\sin^2\theta_p = \delta_p(2 - \delta_p)$ ;
T5 (fixed point) selects $\mu^* = 15$ ;
BA5 (Pontryagin) sets $\alpha_{\rm EM} = \prod \sin^2\theta_p$ .

At no step is a new parameter introduced. The rigidity is total.

In one sentence

$s = 1/2$ is not a constant of nature in the usual sense. It is the signature of the involution $\{1 \leftrightarrow 2\}$ on the mod-3 classes of 6-rough integers — an arithmetic fact. The rest of physics follows because the rest of physics needs it.