Nonnegative Ricci curvature, nilpotency, and Hausdorff dimension

Abstract Let M be an open (complete and non-compact) manifold with $$\textrm{Ric}\ge 0$$ Ric ≥ 0 and escape rate not 1/2. It is known that under these conditions, the fundamental group $$\pi _1(M)$$ π 1 ( M ) has a finitely generated torsion-free nilpotent subgroup $$\mathcal {N}$$ N of finite index, as long as $$\pi _1(M)$$ π 1 ( M ) is an infinite group. We show that the nilpotency step of $$\mathcal {N}$$ N must be reflected in the asymptotic geometry of the universal cover $$\widetilde{M}$$ M ~ , in terms of the Hausdorff dimension of an isometric $$\mathbb {R}$$ R -orbit: there exist an asymptotic cone ( Y , y ) of $$\widetilde{M}$$ M ~ and a closed $$\mathbb {R}$$ R -subgroup L of the isometry group of Y such that its orbit Ly has Hausdorff dimension at least the nilpotency step of $$\mathcal {N}$$ N . This resolves a question raised by Wei and the author (see Pan and Wei in Geom Funct Anal 32:676–685, 2022, Remark 1.7 and Pan in Geom Topol 28:1409–1436, 2024, Conjecture 0.2).