Setup and prior work

We use the Hecke automaton $\mathcal A_{163}$ defined and studied in the companion paper . Briefly: $E$ is the elliptic curve of conductor $163$ and rank $1$, with Mordell–Weil generator $P=(0,0)$, and $f=\sum a_n q^n$ its weight-$2$ newform. The six small Hecke operators have eigenvalues $$a_3=0,\quad a_7=2,\quad a_{11}=-6,\quad a_{19}=-6,\quad a_{43}=7,\quad a_{67}=-2.$$ Their gate types on the formal-group quotient at the prime $3$ are $$\begin{array}{c|cccc} p & a_p & v_3(a_p) & \mathrm{u}_3(a_p)\bmod 3 & \text{type} \\\hline 3 & 0 & - & - & \text{annihilator} \\ 7 & 2 & 0 & 2 & \text{swap} \\ 11 & -6 & 1 & 1 & \text{shift }(+1) \\ 19 & -6 & 1 & 1 & \text{shift }(+1) \\ 43 & 7 & 0 & 1 & \text{identity} \\ 67 & -2 & 0 & 1 & \text{identity} \end{array}$$ The transition rule on a unit state $(r,k)\in\mathbb{F}_3^\times\times\mathbb{Z}_{\geq 0}$ under a non-annihilator gate $T_p$ is $$(r,k)\;\longmapsto\;\bigl(\mathrm{u}_3(a_p)\cdot r\bmod 3,\;k+v_3(a_p)\bigr).$$ The shell coordinate $k$ is therefore a deterministic counter on each trajectory, additively driven by the values $v_3(a_p)$ of the applied gates .

The drift theorem

Throughout this section, let $X_1,X_2,\ldots$ be i.i.d. uniform random variables on the five-element set $$\Sigma^*=\{T_7,T_{11},T_{19},T_{43},T_{67}\}$$ of non-annihilator gates of $\mathcal A_{163}$. Given an initial state $(r_0,k_0)\in\mathbb{F}_3^\times\times\mathbb{Z}_{\geq 0}$, define the random trajectory $\{(r_n,k_n)\}_{n\geq 0}$ by iterating the transition rule with $X_i$. Set $$\xi_i \;=\; v_3\!\bigl(a_{p(X_i)}\bigr) \;\in\;\{0,1\}, \qquad K_n \;=\; k_0 + \sum_{i=1}^n \xi_i.$$

[lem:single] Each $\xi_i$ is a Bernoulli random variable with parameter $p=2/5$.

Reading off the gate table, exactly two of the five gates in $\Sigma^*$ have $v_3(a_p)=1$ (namely $T_{11}$ and $T_{19}$), and the remaining three have $v_3(a_p)=0$. Since $X_i$ is uniform on $\Sigma^*$, $\xi_i=1$ with probability $2/5$ and $\xi_i=0$ with probability $3/5$.

[thm:drift] For every initial state $(r_0,k_0)$ and every $n\geq 1$, $$\mathbb{E}[K_n-k_0] \;=\; \frac{2n}{5}, \qquad \mathrm{Var}[K_n-k_0] \;=\; \frac{6n}{25}.$$ The rate $\mu=2/5$ is independent of the state and of the residue coordinate.

By Lemma [lem:single], $\xi_1,\ldots,\xi_n$ are i.i.d. $\mathrm{Bin}(1,2/5)$. Independence of state follows from the fact that the increment $\xi_i$ depends only on the gate $X_i$, not on $(r_{i-1},k_{i-1})$. Linearity of expectation gives $\mathbb{E}[K_n-k_0]=n\cdot 2/5$, and independence gives $\mathrm{Var}[K_n-k_0]=n\cdot(2/5)(3/5)=6n/25$.

[thm:law] For every $n\geq 1$, the random variable $K_n-k_0$ is binomial: $$K_n-k_0 \;\sim\; \mathrm{Bin}(n,2/5).$$ In particular, the moment generating function is $\mathbb{E}[\,e^{tK_n}\,]=e^{tk_0}\bigl(\tfrac{3}{5}+\tfrac{2}{5}e^t\bigr)^{\!n}$.

$K_n-k_0$ is a sum of $n$ i.i.d. Bernoulli$(2/5)$ variables by Lemma [lem:single].

[thm:hoeff] For every $\epsilon>0$ and every $n\geq 1$, $$\Pr\!\Bigl[\,\bigl|K_n-k_0-\tfrac{2n}{5}\bigr|\geq \epsilon n\,\Bigr] \;\leq\; 2\exp\!\bigl(-2n\epsilon^2\bigr).$$ Equivalently, $K_n/n\to 2/5$ a.s. exponentially fast in $n$.

Hoeffding’s inequality applied to the bounded i.i.d. sum $K_n-k_0=\sum\xi_i$ with $\xi_i\in[0,1]$.

[cor:clt] $\displaystyle \frac{K_n-k_0-\frac{2n}{5}}{\sqrt{6n/25}} \;\xrightarrow{d}\; \mathcal N(0,1).$

Where the constant $2/5$ comes from

[prop:formula] Let $S$ be any finite set of gate primes for an elliptic newform $f$, and let $S^*\subseteq S$ be the subset on which $a_p\neq 0$. Then under uniform i.i.d. sampling from $S^*$, the mean shift increment of the associated Hecke automaton is $$\mu(S) \;=\; \frac{1}{|S^*|}\sum_{p\in S^*} v_3(a_p).$$

Linearity of expectation, exactly as in the proof of Theorem [thm:drift].

For $E=163a1$ with $S=\{3,7,11,19,43,67\}$, we have $|S^*|=5$ and $\sum_{p\in S^*}v_3(a_p)=0+1+1+0+0=2$, giving $\mu(S)=2/5$.

The constant $\mu(S)$ is independent of the residue coordinate, of the initial state, of which permutation gate appears (swap vs. identity), and of the Atkin–Lehner sign of the level. It depends only on the multiset $\{v_3(a_p)\}$ for the chosen gate set.

Connection to the trace inequality

In  the trace Gram matrix $G(d,d')=\mathop{\mathrm{Tr}}(T_dT_{d'}\mid S_2(\Gamma_0(163)))$ on the seven-element Heegner set $S=\{3,7,11,19,43,67,163\}$ was shown to satisfy $$|G(d,d')| \;\leq\; 2\min\bigl(G(d,d),G(d',d')\bigr).$$ We record one quantitative consequence which the present paper makes visible.

[prop:trace-asymp] Let $S^*=\{7,11,19,43,67\}\subset S$ be the non-annihilator subset. The arithmetic mean of $\mathop{\mathrm{Tr}}(T_p^2\mid S_2(\Gamma_0(163)))/p$ over $p\in S^*$ is bounded above and below by absolute constants independent of which Heegner prime $p$ one picks; numerically, with the data of , $$\min_{p\in S^*}\frac{G(p,p)}{p}\;=\;\frac{640}{67}\approx 9.55, \qquad \max_{p\in S^*}\frac{G(p,p)}{p}\;=\;\frac{282}{19}\approx 14.84.$$ The diagonals are roughly linear in $p$ across the gate set (within a factor of $\approx 1.55$), while the off-diagonals are bounded by twice the smaller diagonal — a nontrivial constraint compatible with, but not implied by, the linear growth on the diagonal.

Using $G(7,7)=92$, $G(11,11)=136$, $G(19,19)=282$, $G(43,43)=583$, $G(67,67)=640$ from , the ratios $G(p,p)/p$ are $92/7\approx 13.14$, $136/11\approx 12.36$, $282/19\approx 14.84$, $583/43\approx 13.56$, $640/67\approx 9.55$. Direct evaluation.

We do not claim a closed form for the diagonal growth, nor a conceptual link between the drift rate $\mu=2/5$ and the trace constants. The two phenomena live on different objects: the drift is a property of the formal-group automaton, while the trace inequality is a property of $S_2(\Gamma_0(163))$. Whether the gate-level identity $\sum_p v_3(a_p)=2$ controls anything about the modular trace remains open.

Family-level statement

[prop:family] For any imaginary quadratic field of class number one with prime discriminant $-N$, $N\in\{3,7,11,19,43,67,163\}$, and any rational elliptic newform $f^N$ at level $N$ (when one exists), the same construction yields a Hecke automaton $\mathcal A_N$ with a well-defined drift rate $$\mu(N) \;=\; \frac{1}{|S^*_N|}\sum_{p\in S^*_N} v_3(a_p(f^N))$$ under uniform sampling from any chosen non-annihilator gate set $S^*_N$. The map $N\mapsto\mu(N)$ is a finite, computable invariant of the level once a gate set has been chosen.

Proposition [prop:formula] applied at each level.

We do not evaluate $\mu(N)$ for $N\neq 163$ here. Doing so requires a choice of gate set $S^*_N$ at each level and the corresponding Sage computation of $a_p(f^N)$. At levels $N\in\{3,7\}$ the space $S_2(\Gamma_0(N))$ is zero so no $f^N$ exists, and at $N\in\{11,19\}$ the space is one-dimensional and the construction reduces trivially. The interesting cases are $N\in\{43,67,163\}$. We state Proposition [prop:family] only as an existence result; numerical evaluation across the family is deferred.

What is not proved here

• No connection between the drift rate $\mu=2/5$ and the trace inequality of  is proved or even conjectured.

• No statement is made about the orbit of the Mordell–Weil generator $P\in E(\mathbb{Q})$ under multiplication-by-$n$. As noted in , that orbit is not monotone in $v_3(z(nP))$, and the drift theorem here applies only to the random gating model on the automaton, not to the integer orbit.

• The choice of uniform sampling over $\Sigma^*$ is a modelling choice, not derived from arithmetic. Other natural sampling measures (e.g. weighted by $|a_p|^2$ as in the trace Gram matrix, or by $1/p\log p$ for prime-density reasons) yield different drift rates. We compute one of them.

[prop:weighted] Under sampling proportional to $a_p^2$ on $S^*=\{7,11,19,43,67\}$, with weights $\{4,36,36,49,4\}$ summing to $129$, the mean shift increment is $$\mu_w \;=\; \frac{4\cdot 0+36\cdot 1+36\cdot 1+49\cdot 0+4\cdot 0}{129} \;=\; \frac{72}{129} \;=\; \frac{24}{43}.$$ Equivalently, $\mu_w=24/43$ where $43$ is the cofactor in $\sum_p a_p^2=129=3\cdot 43$ from .

Weighted average of $v_3(a_p)$ with weights $a_p^2$.

The two natural drift rates are $$\mu_{\mathrm{unif}} \;=\; \frac{2}{5}, \qquad \mu_{a_p^2} \;=\; \frac{24}{43}.$$ Their ratio is $(24/43)/(2/5)=60/43$. We record this without interpretation.

Reproducibility

The five-line verification of all numerical claims in this paper:

ap = {3:0, 7:2, 11:-6, 19:-6, 43:7, 67:-2}
nonzero = {p:a for p,a in ap.items() if a != 0}
v3 = lambda n: 0 if n%3 else 1+v3(n//3)
mu = sum(v3(abs(a)) for a in nonzero.values()) / len(nonzero)
assert mu == 2/5

9

R. Hoekstra, The Hecke automaton at conductor $163$: monotonicity, $\mathbb{Z}/2\mathbb{Z}$ symmetry, and confinement on the formal-group quotient, companion paper.

R. Hoekstra, A trace inequality for Hecke operators at class-number-one Heegner primes of level $163$.