Nonnegative Ricci curvature, nilpotency, and Hausdorff dimension

TL;DR

This paper links the nilpotency step of a fundamental group in nonnegatively Ricci curved manifolds to the Hausdorff dimension of certain orbits in the asymptotic cone of the universal cover, revealing geometric implications of algebraic properties.

Contribution

It demonstrates that the nilpotency step of the fundamental group's nilpotent subgroup is reflected in the asymptotic geometry of the universal cover via Hausdorff dimension of isometric orbits.

Findings

The nilpotency step is at most the Hausdorff dimension of an isometric orbit.

Existence of an asymptotic cone with an $ ext{R}$-orbit reflecting the nilpotency step.

Addresses a question posed by Wei and the author.

Abstract

Let $M$ be an open (complete and non-compact) manifold with $\mathrm{Ric}\ge 0$ and escape rate not $1/2$. It is known that under these conditions, the fundamental group $\pi_1(M)$ has a finitely generated torsion-free nilpotent subgroup $\mathcal{N}$ of finite index, as long as $\pi_1(M)$ is an infinite group. We show that the nilpotency step of $\mathcal{N}$ must be reflected in the asymptotic geometry of the universal cover $\widetilde{M}$, in terms of the Hausdorff dimension of an isometric $\mathbb{R}$-orbit: there exist an asymptotic cone $(Y,y)$ of $\widetilde{M}$ and a closed $\mathbb{R}$-subgroup $L$ of the isometry group of $Y$ such that its orbit $Ly$ has Hausdorff dimension at least the nilpotency step of $\mathcal{N}$. This resolves a question raised by Wei and the author.