gpt 6 astra ultra attempt on Math-466

Pisot Substitution Conjecture

Definitions and mathematical statement

A substitution assigns to each letter in a finite alphabet a nonempty word and extends by concatenation. Its incidence matrix $M$ has entry $M_{ij}$ equal to the number of occurrences of letter $i$ in the image of letter $j$. In the irreducible unimodular Pisot setting, the characteristic polynomial is irreducible over ${\mathbb{Q}}$, $\det M=\pm1$, and its dominant eigenvalue $\beta>1$ is a Pisot number: every other algebraic conjugate has modulus less than one.

The substitution subshift consists of bi-infinite words whose finite blocks occur in iterated substituted words, with the left shift and its invariant probability measure in the primitive setting of the source. Pure discrete spectrum means that the eigenfunctions of the Koopman operator $Uf=f\circ\sigma$ span $L^2$. The conjecture asserts this property for every substitution in the specified class. A symbolic periodicity statement or a pure-point diffraction assertion without the needed dynamical equivalence is not substituted for it.

Formulation and concept references: [S1].

Short English statement

Must every irreducible unimodular Pisot substitution system have a complete basis of measurable eigenfunctions?

Research attempt

Formulation and source audit (27 September 2026).

The target is the measurable spectrum of the symbolic shift, with its invariant probability measure. It is not a claim that the sequence is periodic. An irrational rotation has infinitely many distinct orbit points and nevertheless has pure discrete spectrum. Conversely producing some eigenfunctions does not prove that they span all of $L^2$.

There is a bibliographic mismatch in the inherited source entry. Its EMS URL resolves to Berthé--Steiner--Thuswaldner, Multidimensional continued fractions and symbolic codings of toral translations, JEMS 25 $2023$, 4997--5057, rather than the title printed in [S1]. The PDF does discuss the selected Pisot conjecture, notably Section 2.2. The entry is preserved and the actual document identified in [E1]. Hollander--Solomyak [E2] prove the two-symbol case. These statements do not extend that theorem to an arbitrary alphabet. The special example below is proved directly through a rotation coding, rather than by quoting its membership in that solved class.

The attack has two levels. Pisot contraction gives a uniform discrepancy bound for letter counts. In a worked Fibonacci example, one can additionally construct an almost-everywhere one-to-one rotation coding, which gives the full spectral conclusion. The general missing step is the analogue of that one-to-one assertion; contraction alone does not prove it.

The matrix hypotheses really imply primitivity here.

The incidence matrix is a nonnegative integer matrix. If its directed graph were not strongly connected, a simultaneous permutation of rows and columns would put it in proper block triangular form. Its characteristic polynomial would then factor over $\mathbb Q$, contrary to irreducibility. Thus $M$ is irreducible as a nonnegative matrix.

If its period were $h>1$, the cyclic decomposition of that graph would give the eigenvalues $\beta\zeta$ for all $h$th roots of unity $\zeta$. For example, multiply coordinates in successive cyclic classes by successive powers of $\zeta$ to transform a Perron eigenvector into an eigenvector with eigenvalue $\beta\zeta$ or its inverse-phase version. At least one other eigenvalue would have modulus $\beta>1$. But irreducibility of the characteristic polynomial makes all its roots conjugates of $\beta$, and the Pisot hypothesis puts every other root strictly inside the unit circle. Therefore $h=1$ and $M$ is primitive. The standard Perron--Frobenius theorem consequently supplies positive left and right eigenvectors.

This argument uses the irreducible characteristic polynomial, not merely the existence of a Pisot Perron eigenvalue in an otherwise arbitrary substitution. Additional invariant blocks or additional eigenvalues are not allowed in the selected formulation.

Lemma 462.1: bounded letter discrepancy from contraction.

For a word $w$, let $\ell(w)\in\mathbb Z^d$ be its vector of letter counts, and write $|w|=\mathbf1^{\mathsf T}\ell(w)$. There is a positive vector $f$ with $\mathbf1^{\mathsf T}f=1$ and $Mf=\beta f$, and a constant $C$ depending on the substitution, such that every word in its language satisfies \begin{equation}\label{wsref-0-61-34-36-32-3a-62-61-6c-61-6e-63-65} \|\ell(w)-|w|f\|\le C. \end{equation}

Proof. Choose the positive left eigenvector $u^{\mathsf T}$ so that $u^{\mathsf T}f=1$, and define the spectral projection $Q=I-fu^{\mathsf T}$. It commutes with $M$ and annihilates the Perron direction. Because the irreducible polynomial is separable in characteristic zero, $M$ is diagonalizable over $\mathbb C$; all eigenvalues on the image of $Q$ have modulus less than one. Thus, for some $0<r<1$ and $A<\infty$, \[ \|M^jQ\|\le Ar^j\quad(j\ge0). \]

Let $p$ be a proper prefix of $\sigma^n(a)$. Inspect its cut at the outermost substitution level. It consists of complete images under $\sigma^{n-1}$ of a proper prefix of $\sigma(a)$, followed by a prefix inside one remaining substituted letter. Iterating this decomposition gives \[ \ell(p)=\sum_{j=0}^{n-1}M^j\ell(p_j), \] where each $p_j$ is a proper prefix of the image of a single letter; empty prefixes are permitted. There are finitely many possibilities for these prefix vectors. If $B$ bounds their norms, then \[ \|Q\ell(p)\|\le AB\sum_{j\ge0}r^j=AB/(1-r). \] For a full prefix $p=\sigma^n(a)$, instead use $Q\ell(p)=M^nQe_a$, which is uniformly bounded too. Every legal factor is the difference of two prefixes of some $\sigma^n(a)$, so $\|Q\ell(w)\|$ is uniformly bounded for all legal $w$.

It remains to use ordinary word length correctly. The projection $Q$ uses the left Perron vector $u$, which need not be $\mathbf1$. Writing $z=Q\ell(w)$ gives $\ell(w)=f(u^{\mathsf T}\ell(w))+z$ and $|w|=u^{\mathsf T}\ell(w)+\mathbf1^{\mathsf T}z$. Consequently \[ \ell(w)-|w|f=z-f\mathbf1^{\mathsf T}z. \] The preceding uniform bound on $z$ proves (a462:balance). $\square$

In particular two legal words of the same length have letter-count vectors differing by at most $2C$. Also $f$ is their common limiting letter-frequency vector. These are substantial combinatorial restrictions, but the proof has only controlled $d$ counting observables. It has not constructed a complete orthonormal family of eigenfunctions for all measurable observables of the subshift.

A fully worked example: Fibonacci substitution.

Take \[ \sigma(a)=ab,\qquad\sigma(b)=a, \qquad M=\begin{pmatrix}1&1\\1&0\end{pmatrix}. \] Its characteristic polynomial $z^2-z-1$ is irreducible, its determinant is $-1$, and its roots are $\varphi=(1+\sqrt5)/2$ and $-1/\varphi$. Thus it lies in the selected class. Set $\alpha=1/\varphi^2$. We will identify its symbolic system with rotation by $\alpha$ on the circle, up to sets of measure zero.

First define a one-sided word whose $b$'s occur at the zero-based positions \begin{equation}\label{wsref-0-61-34-36-32-3a-62-65-61-74-74-79} B_k=\lfloor k\varphi^2\rfloor-1,\qquad k\ge1, \end{equation} and whose other letters are $a$. Its $a$-positions are $A_k=\lfloor k\varphi\rfloor-1$. Here is a direct proof of that partition. For every positive integer $m$, the numbers in the two Beatty sequences $\lfloor k\varphi\rfloor$ and $\lfloor k\varphi^2\rfloor$ that are at most $m$ have total count \[ \left\lfloor\frac{m+1}{\varphi}\right\rfloor+ \left\lfloor\frac{m+1}{\varphi^2}\right\rfloor=m, \] since $1/\varphi+1/\varphi^2=1$ and neither summand before taking floors is integral. Subtracting the same identity at $m-1$ shows that exactly one sequence contains each positive integer $m$.

Under substitution, the $k$th $a$ creates the $k$th $b$. Before that $a$ there are $k-1$ letters $a$ and $\lfloor k\varphi\rfloor-k$ letters $b$. Their images have total length $2(k-1)+\lfloor k\varphi\rfloor-k$. The new $b$, second in the image of the $k$th $a$, is therefore at position \[ \lfloor k\varphi\rfloor+k-1 =\lfloor k\varphi^2\rfloor-1=B_k. \] The substituted word has exactly the same $b$-positions, so it is fixed by $\sigma$. It starts with $a$ and hence agrees with the increasing prefix limit of $\sigma^n(a)$: every fixed word beginning with $a$ must begin with each of those prefixes, whose lengths tend to infinity.

From the exact word to a measure-theoretic rotation.

The number of $b$'s in the first $n$ letters of (a462:beatty) is $\lfloor(n+1)\alpha\rfloor$. Thus its letter at position $n$ is $b$ exactly when \[ \lfloor(n+2)\alpha\rfloor-\lfloor(n+1)\alpha\rfloor=1. \] For rotation $T(x)=x+\alpha\pmod1$, partition the circle into $I_a=[0,1-\alpha)$ and $I_b=[1-\alpha,1)$. The displayed word is the forward itinerary of $x=\alpha$ under this partition.

For a general $x$, code all iterates $T^nx$, $n\in\mathbb Z$. The set of $x$ whose orbit meets a partition endpoint is countable and has Lebesgue measure zero. Away from that set the coding commutes with rotation and the shift. Every finite cylinder condition pulls back to an intersection of finitely many translated intervals, with finitely many boundary points. If it has nonempty interior, the dense orbit of $\alpha$ visits that interior, so its word occurs in the fixed word. Conversely a block of the fixed word is realized on a one-sided interval next to its coding point, even if that point is an endpoint. Hence it has positive Lebesgue measure and belongs to the rotation language. The subshift obtained by closing the shift orbit has exactly this language.

For completeness, the fixed-word language equals the substitution language. Every fixed-word block lies in a long enough $\sigma^n(a)$. Conversely, $b$ occurs in $\sigma(a)$, so $\sigma^n(b)$ occurs in $\sigma^{n+1}(a)$; all substituted-letter blocks therefore occur in the fixed word. No unrelated coding system has been inserted.

The coding is one-to-one away from the endpoint orbits. To see this, take distinct points $x,y$ and put $d=y-x\pmod1\ne0$. The proper interval $I_b$ and its translate $I_b-d$ have symmetric difference with nonempty interior: a proper single interval cannot be invariant under a nonzero circle translation. The dense orbit of $x$ under $T$ meets that interior, so for some $n$ exactly one of $T^nx,T^ny$ belongs to $I_b$. Their itineraries differ. Thus the coding is injective off the countable orbit of the endpoints. It is a Borel map between standard Borel spaces, so its inverse on this full-measure image is measurable.

Irrational rotation has unique invariant probability measure, Lebesgue measure: if a measure is invariant, its $k$th Fourier coefficient equals $e^{2\pi i k\alpha}$ times itself and hence is zero for $k\ne0$; density of trigonometric polynomials determines the measure. Uniform frequencies for interval cylinders follow from the geometric-sum proof of equidistribution and approximation of interval indicators by continuous functions. They identify the invariant measure of the coded subshift with the pushforward of Lebesgue measure. Thus the almost-everywhere bijection above is a measure-theoretic conjugacy for the measure in the problem statement.

The spectral conclusion in this subcase.

On the circle, the functions $e_k(x)=e^{2\pi i kx}$, $k\in\mathbb Z$, form an orthonormal basis of $L^2$ and satisfy \[ e_k(Tx)=e^{2\pi i k\alpha}e_k(x). \] Transport them through the measurable inverse coding. They remain a complete orthonormal basis of eigenfunctions on the Fibonacci subshift. This proves pure discrete spectrum in the precise symbolic sense of the catalogue. It does not merely establish pure-point diffraction or a property of a suspension flow. It also exhibits why the sequence need not be periodic: $\alpha$ is irrational.

The proof reconstructs a classical example, with no novelty claim. The general two-symbol theorem [E2] covers many additional substitutions; the elementary Beatty argument here is not asserted to cover them all.

Where the general geometric route stops.

Lemma 462.1 puts projected prefix vectors in a bounded part of the contracting space. Taking their closure suggests compact geometric pieces and a torus factor. Establishing a factor is not enough: the factor map could identify several symbolic configurations on a set of positive measure. Eigenfunctions pulled back from the factor then span only the functions constant on its fibers, rather than all of $L^2$.

The logical issue is visible without substitution terminology. The product of an irrational circle rotation with a two-sided independent binary shift has the circle rotation as a factor. That factor has a complete Fourier basis, but the product system is not pure discrete. Indeed the centered coordinate function $F(x,\omega)=\omega_0-1/2$ has $\langle U^nF,F\rangle=0$ for every nonzero integer $n$. If a nonzero function belonged to the pure-point subspace, expanding it in an orthonormal eigenbasis would make its correlations $\sum_j c_j\lambda_j^n$ with $c_j\ge0$ and $\sum c_j>0$; their Cesaro mean squared modulus has a positive contribution from at least one eigenvalue and cannot vanish. This contradicts the displayed correlations. The product is not claimed to be a Pisot substitution; it is a precise counterexample to the abstract "pure-point factor implies pure-point extension" shortcut.

In the Fibonacci proof, separation by interval translates supplied almost-everywhere injectivity. In higher-dimensional contracting spaces that role requires control of overlaps and multiplicities of geometric pieces. Neither a bounded projected orbit nor a numerical picture with small apparent overlap certifies that overlap has measure zero. An algorithm checking one substitution also does not establish successful termination with the necessary coincidence property for every substitution in the class.

Finite checks and outcome.

The companion computation compares a long exact Fibonacci fixed-word prefix with the two Beatty descriptions, checks the substitution matrix and its polynomial identity, and verifies the letter-count discrepancy on all factors of bounded length in the generated word. The Beatty floors are computed with integer square-root comparisons, not rounded floating point values near integer boundaries. Finite agreement is a diagnostic; the preceding identities prove the infinite coding.

Outcome: UNRESOLVED, with a complete classical subcase. The attempt proves bounded letter discrepancy throughout the selected class and proves the full measurable spectral assertion for the Fibonacci substitution. The unresolved step for the general class is to turn contraction and the resulting geometric factor into a sufficiently one-to-one measurable model, or otherwise establish completeness of the eigenfunctions. No general coincidence theorem is proved here.

Sources

• S1 Substitutions, Rauzy fractals and tilings. Submitted scholarly source, accessed 2026-08-31. https://ems.press/content/serial-article-files/32990?nt=1. Formulation source.
• E1 Valérie Berthé, Wolfgang Steiner, and Jörg M. Thuswaldner, Multidimensional continued fractions and symbolic codings of toral translations, Journal of the European Mathematical Society 25 $2023$, 4997--5057, Section 2.2. https://ems.press/content/serial-article-files/32990?nt=1. Actual primary document at the inherited URL, inspected 27 September 2026; the inherited bibliographic mismatch is recorded above.
• E2 Michael Hollander and Boris Solomyak, Two-symbol Pisot substitutions have pure discrete spectrum, Ergodic Theory and Dynamical Systems 23 $2003$, 533--540. https://doi.org/10.1017/S0143385702001384. Primary publisher abstract inspected 27 September 2026; the two-symbol theorem is attributed, not generalized here.