Mixing of all orders
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Definitions and mathematical statement
Let $X$ be a compact connected metrizable abelian group with normalized Haar measure $\mu$, and let $\alpha:\mathbb Z^d\to\operatorname{Aut}(X)$ be an action by continuous group automorphisms, where $d\ge1$. The action is mixing if \[ \mu(A\cap\alpha^{-h}B)\longrightarrow\mu(A)\mu(B) \quad\text{as }\|h\|\to\infty \] for all measurable $A,B\subset X$. It is mixing of all orders if, for every $k\ge2$ and measurable $A_1,\ldots,A_k$, \[ \mu\!\left(\bigcap_{i=1}^k\alpha^{-h_i}A_i\right) \longrightarrow\prod_{i=1}^k\mu(A_i) \quad\text{as }\min_{i\ne j}\|h_i-h_j\|\to\infty. \] The theorem that mixing implies mixing of all orders in this connected algebraic setting is established. Ward's question is to find a proof using simpler dynamical ideas that avoids the deep Diophantine estimates used in the known argument.Accordingly the desired output is a new proof method, not a counterexample to the theorem or another statement of its conclusion. Any proposed proof must retain arbitrary order $k$ and arbitrary mutually separating translation vectors; a proof for three sets or for one prescribed family of directions is only partial. Connectedness is essential to the stated setting, since zero-dimensional algebraic actions can have a finite mixing order. The abelian-group formulation fixes the standard algebraic-dynamics class to which the quoted theorem applies and makes the Haar measure and all translation limits explicit.
Short English statement
Find a dynamical proof that connected mixing algebraic actions are mixing of all orders, avoiding the known Diophantine machinery.Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 4300002 (AMR-042-0002), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 Ward - Six problems in Algebraic Dynamics (2006), Problem B. Original source indexed by AMR-042-0002. https://www.imath.kiev.ua/~skolyada/kevin.pdf.
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