gpt 6 astra pro attempt on Math-188

Furstenberg's x2 x3 conjecture on invariant measures

Definitions and mathematical statement

On $\mathbb T={\mathbb{R}}/{\mathbb{Z}}$, let $T_2(x)=2x$ and $T_3(x)=3x$. A Borel probability measure $\mu$ is invariant if $(T_2)_*\mu=(T_3)_*\mu=\mu$. Joint ergodicity means every measurable set invariant modulo null sets under both maps has measure zero or one. The selected assertion is \[ \mu\text{ invariant and jointly ergodic} \quad\Longrightarrow\quad \mu=\text{Haar measure or }|\operatorname{supp}\mu|<\infty. \] Ergodicity for the joint action must not be silently strengthened to ergodicity for each map separately. The catalogue also mentions a coprime $(a,b)$ formulation; that broader parameter statement is recorded in its quotation, but an equivalence with the single pair $(2,3)$ is not established merely by these definitions.

Formulation and concept references: [S1].

Short English statement

Must a jointly ergodic circle measure preserved by both doubling and tripling be either uniform or concentrated on finitely many points?

Research attempt

Current frontier.

Rudolph's positive-entropy theorem uses joint ergodicity, exactly as in this entry: a measure invariant under a coprime pair and ergodic for its semigroup is Lebesgue if either generator has positive entropy [E1], abstract and Sections 1 and 3. Thus the atomless zero-entropy case is the missing one, not a case excluded by the definition. The 2024 Burton--Panangaden paper supplies equivalent complex-analytic and operator-algebraic formulations; it does not prove the classification or its separately stated periodic-approximation conjecture [S1], Section 1.1. The argument below tests a Fourier-energy approach directly on the measure, without strengthening joint ergodicity to ergodicity for each map.

Attack route.

An entropy route needs to force positive entropy in a nonatomic case where zero entropy is currently permitted. A periodic-approximation route needs both approximation and control of the limiting periodic measures. The selected route uses positive definiteness and the exact replication of Fourier coefficients under multiplication by two and three. The hoped-for input is an upper bound on low-frequency Fourier energy sharp enough to contradict that replication. We derive the required scale, prove why nonatomicity alone is far too weak, and test a simple coefficient-amplifying inequality.

Attempt: completely dispose of the atomic case.

Suppose an invariant jointly ergodic probability measure $\mu$ has an atom. Its largest atom mass $p>0$ is attained: only finitely many atoms can have mass at least half the positive supremum. Let $S$ be the nonempty finite set of atoms of mass $p$. For $T=T_2$ or $T_3$, invariance implies $\mu(\{Tx\})\ge\mu(\{x\})=p$ when $x\in S$, so $T(S)\subseteq S$. Two points of $S$ cannot have the same image, since its mass would then be at least $2p$. Thus each $T$ permutes $S$. Moreover $S\subseteq T^{-1}S$ and $\mu(T^{-1}S)=\mu(S)$, so $S$ is invariant modulo null sets. Joint ergodicity gives $\mu(S)=1$.

The measure is consequently uniform on a single joint orbit in $S$. Each point is periodic under both maps separately, so its reduced rational denominator divides some $2^r-1$ and some $3^s-1$, and is coprime to six. These conclusions were proved using joint ergodicity only. They agree with the finite-support analysis in [S1], Section 2.6. The unresolved part is therefore exactly the nonatomic case.

Attempt: a logarithmic-square Fourier obstruction.

Write \[ c(k)=\int_{\mathbb T}e^{2\pi ikx}\,d\mu(x),\qquad E_\mu(N)=\sum_{1\le |k|\le N}|c(k)|^2. \] Invariance gives the exact identities $c(2k)=c(3k)=c(k)$. If $\mu$ is not Haar, some $k_0\ne0$ has $c(k_0)\ne0$: otherwise integration agrees with Haar on trigonometric polynomials and hence, by uniform approximation, on every continuous function. Set $K=|k_0|$. Every positive frequency $2^a3^jK$ has the same absolute coefficient, so \[ E_\mu(N)\ge |c(k_0)|^2\, \#\{(a,j)\in{\mathbb{Z}}_{\ge0}^2:2^a3^j\le N/K\}. \] The frequencies are distinct by unique prime factorization. If $L=\log(N/K)$, the lattice count is \begin{equation}\label{wsref-0-65-71-31-38-33-2d-63-6f-75-6e-74} \#\{a\log2+j\log3\le L\} =\frac{L^2}{2\log2\log3}+O(L+1). \end{equation} To justify the error without equidistribution assumptions, tile the nonnegative quadrant by unit squares anchored at lattice points. The triangle of weighted radius $L$ is contained in the union of squares whose anchors lie in that triangle, and that union is contained in the triangle of weighted radius $L+\log2+\log3$. Their areas differ by $O(L+1)$, proving (eq183-count).

It follows that every non-Haar invariant measure satisfies \begin{equation}\label{wsref-0-65-71-31-38-33-2d-6c-6f-77-65-72} \liminf_{N\to\infty}\frac{E_\mu(N)}{(\log N)^2} \ge\frac{|c(k_0)|^2}{2\log2\log3}>0. \end{equation} Thus the following precise estimate would settle the remaining case: \begin{equation}\label{wsref-0-65-71-31-38-33-2d-6d-69-73-73-69-6e-67} E_\mu(N)=o((\log N)^2) \quad\text{for every nonatomic jointly ergodic invariant }\mu. \end{equation} This is a conditional criterion, not an established decay estimate. Indeed it is equivalent to the desired nonatomic conclusion by the argument just given and the fact that Haar has $E_\mu(N)=0$. The replication step itself uses no ergodicity and extends to any multiplicatively independent integer pair, with the corresponding logarithmic constant.

Attempt: what nonatomicity actually supplies.

For any probability measure, the elementary Wiener calculation gives \begin{equation}\label{wsref-0-65-71-31-38-33-2d-77-69-65-6e-65-72} \lim_{N\to\infty}\frac{1}{2N+1} \sum_{|k|\le N}|c(k)|^2 =\sum_x\mu(\{x\})^2. \end{equation} Here is the limiting justification. Expand the finite Fourier sum as the integral over $\mu\times\mu$ of $D_N(x-y)/(2N+1)$, where $D_N(t)=\sum_{k=-N}^{N}e^{2\pi ikt}$. Its absolute value is at most one; it tends to zero for $t\ne0$ in $\mathbb T$ and equals one for $t=0$. Dominated convergence gives the mass of the diagonal, which is the right side of (eq183-wiener). Hence a nonatomic measure has only $E_\mu(N)=o(N)$ by this argument. This is compatible with (eq183-lower); dividing by $N$ has discarded the required scale.

To test whether positivity or singularity alone can improve it, take the usual middle-thirds Cantor measure $\nu$: its ternary digits are independent and equally likely to be zero or two. It is nonatomic and invariant under $T_3$. Each of its $2^m$ level-$m$ intervals has length $3^{-m}$ and mass $2^{-m}$. Let $J=\lfloor3^m/4\rfloor$, for $m\ge2$, and use the nonnegative Fejér kernel \[ F_J(t)=\frac1{J+1}\left(\frac{\sin\pi(J+1)t}{\sin\pi t}\right)^2. \] For $|t|\le3^{-m}$ the sine inequalities on a fixed subinterval of $(-\pi/2,\pi/2)$ give $F_J(t)\ge c(J+1)$, with an absolute $c>0$. Pairs in the same level-$m$ interval have total $\nu\times\nu$ mass $2^{-m}$. Since $F_J\ge0$ elsewhere, \[ \sum_{|k|\le J}\left(1-\frac{|k|}{J+1}\right)|\widehat\nu(k)|^2 =\iint F_J(x-y)\,d\nu(x)d\nu(y) \ge c(J+1)2^{-m}. \] The weighted sum is at most $1+E_\nu(J)$. Therefore along these $J$, \begin{equation}\label{wsref-0-65-71-31-38-33-2d-63-61-6e-74-6f-72} E_\nu(J)\ge c'J^{1-\log2/\log3}-1. \end{equation} This is much larger than $(\log J)^2$, despite nonatomicity, positivity, and one expanding-map invariance. It is not a counterexample to the conjecture: $\nu$ is not doubling-invariant. For example $2/9$ is in its support whereas its double $4/9$ lies in the omitted middle third; continuity and the definition of support rule out $T_2$-invariance.

Attempt: try to amplify one coefficient.

For characters $X=e^{2\pi iux}$ and $Y=e^{2\pi ivx}$, Cauchy--Schwarz applied after subtracting their expectations gives \begin{equation}\label{wsref-0-65-71-31-38-33-2d-63-6f-76-61-72-69-61-6e-63-65} |c(u-v)-c(u)\overline{c(v)}| \le\sqrt{(1-|c(u)|^2)(1-|c(v)|^2)}. \end{equation} If $u,v$ belong to the same multiplicative orbit of $k_0$, then $c(u)=c(v)$, and $\operatorname{Re}c(u-v)\ge2|c(k_0)|^2-1$. This can produce nontrivial additional coefficients when $|c(k_0)|>1/\sqrt2$. But a nonzero coefficient need not be that large, and no amplification to this threshold has been proved. Moreover, taking all pairwise differences of the $O((\log N)^2)$ replicated frequencies below $N$ produces at most $O((\log N)^4)$ frequencies. Even persistent coefficients on that many frequencies do not contradict the $o(N)$ bound. An iterative argument would need both sustained amplitudes and a quantitative growth theorem for new frequencies, neither of which follows from (eq183-covariance) alone.

Stress test.

The lower bound has a measure-dependent constant, which is enough to contradict a little-oh upper bound for that same measure. No uniform lower bound over all non-Haar measures is asserted. The lattice count concerns a semigroup of nonnegative powers, not all integer frequencies, and its zero density is precisely why the Wiener estimate does not settle the problem. The Fejér calculation is exact; no finite simulation has been promoted to a classification theorem.

The same conditional calculations apply to any coprime $a,b\ge2$, replacing $2,3$ throughout. This does not prove that solving the single $(2,3)$ problem logically implies the whole parameterized family mentioned in the catalogue quotation. Both that scope distinction and the original joint rather than individual ergodicity are retained.

Outcome.

Outcome: Conditional reduction.

The attempt proves the atomic classification, the logarithmic-square Fourier obstruction, and two tests explaining why basic Fourier positivity and nonatomicity estimates do not close the gap. The criterion (eq183-missing) makes the missing quantitative input explicit; it is not a newly proved decay theorem or a claim of priority for Fourier reformulations.

What remains is an argument using both expanding maps to force that decay, or another rigidity mechanism in the nonatomic zero-entropy case. The Cantor construction violates only a weaker one-map proposal, not the selected conjecture. Neither the full $(2,3)$ classification nor the broader coprime family is resolved here.

Sources

• S1 P. Burton, J. Panangaden, 'Formulations of Furstenberg's x2 x3 conjecture in complex analysis and operator algebras', arXiv:2410.22701 $2024$. https://ar5iv.labs.arxiv.org/html/2410.22701. Catalogue formulation source.
• E1 Daniel J. Rudolph, $\times2$ and $\times3$ invariant measures and entropy, Ergodic Theory and Dynamical Systems 10$2$ $1990$, 395--406, abstract and Sections 1 and 3. https://doi.org/10.1017/S0143385700005629. Consulted 27 September 2026. The positive-entropy theorem requires joint ergodicity, not individual ergodicity of both generators.