A. Julien: Relationship between Complexity and Cohomology — Problem

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Let $T$ be an aperiodic repetitive tiling of $\mathbb R^d$, $d\geq1$. Aperiodic means it has no nonzero translation period; repetitive means every finite patch recurs with bounded gaps. Let $p(n)$ count radius-$n$ patches up to translation. Its tiling hull $\Omega$ is the closure of all translates of $T$ in the local tiling topology (agreement on increasingly large balls after increasingly small shifts). Cohomology here is rational Čech cohomology, and $\operatorname{rk}H^*(\Omega;\mathbb Q)$ denotes the total dimension of these rational cohomology groups. The estimate $p(n)=O(n^d)$ means $p(n)\leq Cn^d$ for all sufficiently large integers $n$.

Statement recorded in the supplied source.

Does $p(n)=O(n^d)$ imply $\operatorname{rk}\check H^*(\Omega;\mathbb Q)<\infty$ for every aperiodic repetitive $d$-dimensional tiling? [S2]

Short English statement

Does complexity of at most order volume force finite rational cohomology for a repetitive aperiodic tiling?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 6600013 (AMR-065-0013). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Faustin Adiceam (compiler), Open Problems and Conjectures related to the Theory of Mathematical Quasicrystals, arXiv:1604.06280v2 (2016), Section 2.5, Problem 2.5.1, p. 8. https://arxiv.org/pdf/1604.06280.