Dual Group Structures in Diophantine Approximations

From the MLA(Mesopotamian Logarithm Algorithm) for logarithmic convergents, a similar property appears in other irrationals when analyzed in their corresponding space.

Logarithm Case Recap:

Irrational: \(\alpha = \log_b(a)\)

Convergent: \(p/q \approx \log_b(a) \Rightarrow q \times log_b(a) \approx p \Rightarrow a^q \approx b^p\)

Sequence: \(r_x = a^x \times b^{y_x}\) reduced to \([1, b)\). This is like looking at \(a^x\) "modulo \(b\)" multiplicatively. \(y_x\) tracks the 'overflow' exponent of \(b\). (This highlights the absence of a standard shorthand notation for multiplicative modulus; see link)

Sorted Sequence: Sorting \(r_x\) for \(x=1\ldots q\) gives indices \(x_k\).

Structure: \(x_k\) forms \(\mathbb{Z}/q\mathbb{Z}\) (gen \(p^{-1} \mod q\)), \(y_{x_k}\) forms \(q\) terms of \(\mathbb{Z}/p\mathbb{Z}\) (gen \(q^{-1} \mod p\)).

(Dual cyclic structure at convergents)

Let \(\alpha \in \mathbb{R}\setminus \mathbb{Q}\) with continued fraction convergent \(p/q\). Consider the rotation sequence \(r_x = \{x\alpha\}\in [0,1),\quad x=1,\dots,q\),

and let \(\sigma\) be the permutation that sorts \(r_x\) in increasing order: \(r_{\sigma(1)} < r_{\sigma(2)} < \cdots < r_{\sigma(q)}\).

Then:

(Index cycle) \(\sigma\) is an arithmetic progression modulo \(q\): \(\sigma(k) \equiv k\cdot p^{-1} \pmod q\),

where \(p^{-1}\) is the multiplicative inverse of \(p\) modulo \(q\).


(Overflow cycle / floor terms) Writing \(x\alpha = y_x + r_x\) with \(y_x=\lfloor x\alpha\rfloor\), the sequence \(y_{\sigma(k)}\) (as \(k=1,\dots,q\)) takes exactly two adjacent values that differ by 1 and forms \(q\) samples from a cycle in \(\mathbb{Z}/p\mathbb{Z}\) whose step is \(q^{-1}\pmod p\).


(Gap control) The consecutive differences \(r_{\sigma(k+1)}-r_{\sigma(k)}\) take two values (the “short” and “long” gaps) determined by \(p/q\); this is the Three Gap Theorem specialized at a convergent, where only two gaps appear across the first \(q\) points.

 
Proof sketch

Because \(p/q\) is a convergent, \(\|q\alpha-p\|\) is minimal in its range. The return map of the rotation by \(\alpha\) to the set of \(q\) points partitions the circle into two gap lengths. (TGT gives gap sizes.)


The order of the points is governed by the congruence \(x\alpha \approx x\frac{p}{q}\) modulo \(1\), so sorting by \(x\alpha\) matches sorting by \(xp/q\) modulo \(1\). The residues \(xp \bmod q\) run through \(\mathbb{Z}/q\mathbb{Z}\) in steps of \(p\), hence the sorting permutation is
\(\sigma(k)\equiv k\cdot p^{-1}\ (\bmod q)\). (This gives gap order.)


The floor/overflow terms satisfy \(y_{\sigma(k+1)}-y_{\sigma(k)} \in \{\lfloor p/q\rfloor, \lceil p/q\rceil\}\),

and, tracked modulo \(p\), they advance by \(q^{-1}\) because
\(q\alpha\approx p\) forces \(p\) steps in \(\alpha\)-space to coincide with \(q\) wraps. This yields the dual \(\mathbb{Z}/p\mathbb{Z}\) cycle.

 
(Logarithmic case via an isomorphism)

Let \(a,b>1\) and set \(\beta=\log_b{a}\). Define the multiplicative sequence \(R_x \;=\; a^x\, b^{-y_x} \in [1,b),\qquad y_x=\big\lfloor x\beta\big\rfloor\).

Then \(R_x = b^{\{x\beta\}}\). Hence ordering the \(R_x\) is the same as ordering \(\{x\beta\}\), and all claims of the Theorem transfer with \(\alpha=\beta\):

Sorting indices are \(\sigma(k)\equiv k\cdot p^{-1}\ (\bmod q)\) for any convergent \(p/q\) of \(\beta\).

The overflow exponents \(y_{\sigma(k)}\) form \(q\) samples from a \(\mathbb{Z}/p\mathbb{Z}\) cycle with step \(q^{-1}\ (\bmod p)\).

The MLA’s “stack-and-fold” is just rotation on the circle in log-coordinates, so its consecutive outputs are convergents whenever you use windows aligned with denominators \(q\).

Every Diophantine approximation problem generates a dual pair of cyclic group structures, one indexed by the convergent’s denominator, one by its numerator. a lattice of relationships between \((p,q)\) and their inverses modulo each other. It's not just about inequalities, but about explicit dynamical group actions tied to each irrational. For irrational \(\alpha\), from the overflow sequence of its natural dynamical action produces exactly the convergents of its continued fraction.

Trigonometric Case (Angle)

Irrational: We need an irrational quantity related to the angle. Let's use \(\alpha = \theta / (2\pi)\). (assuming \(\theta\) is not a rational multiple of \(2\pi\)).

Convergent: \(p/q \approx \theta / (2\pi) \Rightarrow q \times \theta / (2\pi) \approx p \Rightarrow q\theta ≈ 2\pi p\). This means \(q\) rotations by \(\theta\) is close to \(p\) full \(2\pi\) rotations.

Sequence: What's the equivalent of \(a^x \mod 1:b\)? The natural analogue for angles is \(x\theta \mod 2\pi\). Let \(r_x = (x\theta) \pmod{2\pi}\). This sequence lives in \([0, 2\pi)\).

What is \(y_x\) ? It's the number of full rotations removed: \(xθ = y_x \times 2\pi + r_x\). So, \(y_x = \lfloor x\theta / (2\pi)\rfloor\).

Sorted Sequence: Sort \(r_x\) for \(x=1\ldots q\) to get indices \(x_k\).

Structure: \(x_k\) forms \(\mathbb{Z}/q\mathbb{Z}\) (gen \(p^{-1} \mod q\)), \(y_{x_k}\) forms \(q\) terms of \(\mathbb{Z}/p\mathbb{Z}\) (gen \(q^{-1} \mod p\)).




This directly mimics the log case by replacing the multiplicative group \((\mathbb{R}^+, \cdot)\) modulo \(b\) with the additive group \(\mathbb{R} \mod 2\pi\) (the circle group \(S^1\)). The relationship \(q\theta \approx 2\pi p\) is the direct analogue of \(a^q \approx b^p\). The Three Gap Theorem describes the structure of the sorted \(r_x\) values (the points \(x\theta \mod 2\pi\) on the circle), and their ordering is intimately linked to the continued fraction convergents \(p/q\). The generators likely arise from the relationship \(q(p'/q') - p(q'/q') = \pm \)1 between consecutive convergents.


(Need to test which inverse/element works. The structure \(p_{n-1} q_n - p_n q_{n-1} = (-1)^n\) from continued fractions is key here, likely determining the specific generators.)