Furstenberg’s ×2, ×3 conjecture

Mark a position by the fraction of one turn around a circle. Doubling 2/5 gives 4/5; tripling it gives 6/5, which wraps around to 1/5. Individual positions move, but a probability distribution can stay unchanged when all positions are moved this way.

Could a nonuniform distribution stay unchanged under both doubling and tripling if no single position receives any positive probability? The conjecture says only the uniform distribution, which assigns an arc its fraction of the circumference, can do so.[1]

Could doubling and tripling positions around a circle preserve a nonuniform probability distribution that puts zero probability on every individual point? Doubling $2/5$ gives $4/5$; tripling gives $1/5$ after wrapping around. The equal-weight distribution on $1/5,2/5,3/5,4/5$ is preserved by both maps because each map permutes these four positions. It has atoms and is therefore excluded.

Write the circle as $\mathbb R/\mathbb Z$: positions differing by a whole turn are identified. A Borel probability measure assigns probabilities to the usual measurable events on this circle. An atomless measure gives every singleton probability zero. For an invariant measure under a map $T$, an event $E$ satisfies $\mu(T^{-1}E)=\mu(E)$; moving a random position leaves its distribution unchanged.

For $T_2(x)=2x\pmod1$ and $T_3(x)=3x\pmod1$, must every atomless measure invariant under both equal normalized Lebesgue measure? The latter is uniform, so an arc’s probability is its fraction of the circumference. The unknown concerns probabilities, not only whether individual orbits spread densely.[1]

The known theorem covers distributions that cannot be split into invariant parts for the two maps together and that have positive entropy: orbit records generate information at a positive rate. Splitting into such parts reduces the original question to this setting, but the theorem does not cover zero entropy. What remains is the atomless zero-entropy case; having no point masses does not supply the theorem’s positive information rate.[1]

Ergodic decomposition gives an equivalent formulation requiring the measure to be ergodic for the joint doubling-and-tripling action: every event invariant under both maps has probability zero or one. Under that requirement, the known theorem forces uniformity if entropy is positive for doubling, equivalently for tripling. Entropy measures the per-step information rate in increasingly long orbit records.

The positive-entropy hypothesis cannot be dropped by citing that theorem. The atomless, joint-ergodic zero-entropy case remains unresolved; the page does not add an ergodicity assumption to its original full atomless scope.[1]

Preserving a distribution is weaker than keeping each point fixed. The two maps must agree on which distribution to preserve even though doubling and tripling stretch the circle in different ways. This makes the problem a concrete test of whether simultaneous dynamical symmetries can force a statistical description to be unique.[1]

The common invariant measure is a statistical object constrained by two different expanding maps. Their arithmetic relation matters: tripling is not an iterate of doubling, whereas preserving a distribution under doubling already preserves it under quadrupling. Merely adding a repeated version of one map would add no comparable constraint.

This is a basic example of measure rigidity: several compatible dynamics can restrict their invariant distributions much more than one does. The known positive-entropy theorem makes that connection precise in one regime, while zero entropy shows that the information-rate condition is substantive rather than a synonym for having no atoms.[1]