Words obtained from a linear flow
We apply the interchange argument to the order in which finitely many
linear functions take integer values. Their slopes will be assumed
linearly independent over ℚ; this
hypothesis is not assumed for reciprocal prime logarithms. Interchanging
two late crossings then determines the rational relations among the
corresponding subseries, without requiring recovery of every letter. The
arguments in this section remain ordinary proofs awaiting
formalisation.
Let 𝐴 be a finite alphabet. Give
𝑎 ∈𝐴 an integer radix 𝑏𝑎 ≥2 and a rational coefficient 𝑐𝑎. For 𝑤 =𝑤1𝑤2⋯, define
𝑄0=1,𝑄𝑛=𝑛∏𝑗=1𝑏𝑤𝑗,𝑉𝑐(𝑤)=∑𝑛≥1𝑐𝑤𝑛𝑄𝑛,𝑇𝑎(𝑥)=𝑐𝑎+𝑥𝑏𝑎.
Write 𝜎𝑤 =𝑤2𝑤3⋯ and 𝑋𝑤 =――――――――{𝜎𝑛𝑤:𝑛≥0} in the
one-sided product topology. A word belongs to 𝑋𝑤 precisely when each of its finite
prefixes begins some shifted word. It need not itself be a shift. The
first letter acts by the outermost map: if 𝑢 =𝑢1⋯𝑢𝑚, then 𝑉𝑐(𝑢𝑧) =𝑇𝑢1 ∘⋯ ∘𝑇𝑢𝑚(𝑉𝑐(𝑧)). Put 𝐵(𝑢) =∏𝑗≤𝑚𝑏𝑢𝑗, with 𝐵(∅) =1.
Proof. First clear the common denominator of the
coefficients, so that all 𝑐𝑎 are
integers. The recurrence 𝑉𝑐(𝜎𝑤) =𝑏𝑤1𝑉𝑐(𝑤) −𝑐𝑤1 shows that, if 𝑉𝑐(𝑤) =𝑝/𝐾, then every shifted value
belongs to 𝐾−1ℤ. For 𝜆𝑎 =𝑐𝑎/(𝑏𝑎 −1), the positive
weights
𝑏𝑤𝑛−1𝑄𝑛=1𝑄𝑛−1−1𝑄𝑛
sum to one. Hence 𝑉𝑐(𝑣) lies
between min𝑎𝜆𝑎 and max𝑎𝜆𝑎 for every word 𝑣. Sharing 𝑚 initial letters changes a value by at
most the width of this interval divided by 𝐵(𝑢). Thus 𝑉𝑐 is continuous. Since 𝐾−1ℤ is closed, the lattice condition
passes from the actual shifts to all of 𝑋𝑤.
The order of the maps matters. A direct calculation gives, for every
𝑥,
𝑇𝑎(𝑇𝑏(𝑥))−𝑇𝑏(𝑇𝑎(𝑥))=𝑐𝑎(𝑏𝑏−1)−𝑐𝑏(𝑏𝑎−1)𝑏𝑎𝑏𝑏.
The values of 𝑢𝑎𝑏𝑧 and 𝑢𝑏𝑎𝑧 therefore differ by this number
divided by 𝐵(𝑢). Their difference
belongs to 𝐾−1ℤ. If its
numerator were nonzero, its absolute value would be positive but smaller
than 1/𝐾 for a sufficiently long
𝑢, a contradiction. On a connected
interchange graph all 𝜆𝑎
agree. Conversely, for 𝑐𝑎 =𝜆(𝑏𝑎 −1) the displayed positive
weights telescope and give 𝑉𝑐(𝑣) =𝜆 for every 𝑣. Dividing by the earlier common
denominator restores the statement for rational coefficients. ◻
The common suffix is essential: distinct words can have the same
affine map. For example, the blocks (5,7,3,2) and (7,2,3,5) have the same map (51 +𝑥)/210. Choosing the second block at
block indices 1,2,4,8,…
produces a non-eventually-periodic word with rational value 51/209. Thus aperiodicity and bounded
tails do not suffice, even with integer bases; the interchange
hypothesis is essential to the argument. Nor does the lemma assert
sufficiency when its graph is disconnected.
Now let 𝑟1,…,𝑟𝑑 >0,
𝑑 ≥2, be linearly independent over
ℚ. Record the letter 𝑖 at each positive time 𝑡 for which 𝑟𝑖𝑡 +𝜃𝑖 ∈ℤ, where 𝜃𝑖 ∈[0,1). Assume that no two of
these functions take integer values simultaneously at positive time. In
chronological order, the crossings give an infinite word 𝑤 on {1,…,𝑑}.
Proof. Fix a label 𝑎.
Immediately after its sufficiently late crossings, the remaining
coordinates of the flow are
𝜃𝑖−𝑟𝑖𝑟𝑎𝜃𝑎+𝑘𝑟𝑖𝑟𝑎(mod1)(𝑖≠𝑎),
where 𝑘 runs through all sufficiently large
integers. These points are dense in the (𝑑 −1)-torus. Indeed, 1 and the 𝑟𝑖/𝑟𝑎 for 𝑖 ≠𝑎 are rationally independent. For
every nonzero integral character of the torus, its average along this
orbit is a geometric sum divided by its length and tends to zero.
Trigonometric approximation of continuous functions then shows that
every nonempty open set is visited infinitely often. This also proves
density after any finite initial segment is discarded.
Fix 𝑏 ≠𝑎 and an arbitrarily
large integer 𝐿, and work on the
section just after an 𝑎-crossing.
Set 𝑇 =𝐿/𝑟𝑎 and choose the phase of
𝑏 so that the 𝑏-coordinate crosses at 𝑇. This phase is nonzero, since 𝐿𝑟𝑏/𝑟𝑎 ∉ℤ. Choose the remaining
phases outside the countable sets that cause a crossing at zero, a third
crossing at 𝑇, or any other
positive simultaneous crossing. The resulting singular configuration has
exactly the one double crossing of 𝑎,𝑏 at 𝑇: a second meeting of these two
coordinates would make 𝑟𝑏/𝑟𝑎
rational.
Perturb the 𝑏 phase on opposite
sides. On each bounded time interval, a sufficiently small perturbation
changes only this pair’s order. Taking one-sided limits gives 𝑢𝑎𝑏𝑧 and 𝑢𝑏𝑎𝑧 with the same infinite
suffix 𝑧. No fixed nonzero
perturbation is required to preserve all later crossings: agreement on
each finite interval suffices for these limiting words. Their common
prefix has at least 𝐿 −1 letters,
from the earlier 𝑎-crossings. For
each finite prefix of either limit, a small open set of nonsingular
section phases realises that prefix. The dense actual 𝑎-section orbit meets that set, so the
prefix occurs at the beginning of an actual shifted word. Consequently
both limiting words belong to 𝑋𝑤.
Letting 𝐿 grow proves the
assertion. ◻
The following argument proves the assertion of the remark in ordinary
mathematics; formalisation and external peer review remain
outstanding.
Proof. Absolute convergence gives ∑𝑖𝑐𝑖𝑆𝑖 =𝑉𝑐(𝑤). The previous two
remarks make the interchange graph complete, so they give both
directions and the stated value. The relation space follows by moving
the rational value to the constant coordinate. An indicator vector of a
nonempty proper set has both zero and one entries and cannot be a
multiple of the strictly positive vector (𝑏𝑖 −1)𝑖. The all-ones vector is such a
multiple precisely when all the 𝑏𝑖
agree. ◻
For example, take
(𝑟1,𝑟2,𝑟3,𝑟4)=(1,√2,√3,√6),(𝑏1,𝑏2,𝑏3,𝑏4)=(2,3,5,7),𝜃𝑖=0.
The four rates
form the standard basis of the biquadratic field ℚ(√2,√3) over ℚ; the ratio of any two distinct rates
is irrational, so positive crossings never coincide. Write (𝑑1,𝑑2,𝑑3,𝑑4) =(1,2,3,6). Counting
crossings up to the 𝑛th crossing of
the 𝑖-th coordinate gives the
explicit series
𝑆𝑖=∑𝑛≥1(4∏𝑗=1𝑏⌊𝑛√𝑑𝑗/𝑑𝑖⌋𝑗)−1.(9)
The only rational relation among 1,𝑆1,𝑆2,𝑆3,𝑆4 is generated by 𝑆1 +2𝑆2 +4𝑆3 +6𝑆4 =1. In particular their
rational span has dimension four and ∑𝑖𝑆𝑖 is irrational. This does not
provide an irrationality measure or transcendence result. In the related
two-letter Hecke–Mahler setting, stronger transcendence results exist
[15]; the
example’s four independent rates do not satisfy that two-letter
hypothesis. The ordinary two-wall argument for the distinct running-LCM
sum avoids this assumption on reciprocal prime logarithms. The repeated
running-LCM question remains open.