Let $G$ be a free Carnot group of homogeneous dimension $d$, equipped with the Carnot–Carathéodory metric. Is $(G,d)$ ever $\operatorname{CAT}(\kappa)$ for some $\kappa\in \mathbb{R}$?
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1$\begingroup$ No: a result of Pauls even asserts that any nonabelian Carnot group has no quasi-isometric embedding into any CAT(0) space. (Certainly it's easier to directly prove it's not CAT(0)). [Note: free Carnot group has no sense: you need to specify the nilpotency length $k$ and number of generators $r$. It's non-abelian iff $\min(k,r)\ge 2$.] $\endgroup$– YCorCommented Apr 24, 2022 at 15:05
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$\begingroup$ @YCor Do you have a reference to this; I'ld like to read more. $\endgroup$– Carlos_PettersonCommented Apr 24, 2022 at 18:10
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1$\begingroup$ Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. $\endgroup$– YCorCommented Apr 24, 2022 at 21:03
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