# When are Carnot groups negatively curved and homeomorphic to Euclidean space

When are Carnot groups complete and negatively curved (in the sense of $$CAT(\kappa)$$ spaces)?

• Only abelian ones. It's a theorem of Pauls that a nonabelian simply connected nilpotent groups can't even QI embed into any CAT(0) space. – YCor May 1 at 11:43
• Interesting! This i did not know. – AIM_BLB May 1 at 11:46
• Abelian Carnot groups are isometric to Euclidean spaces. So no, they're not negatively curved (in dimension $\ge 2$) although they're non-positively curved. – YCor May 1 at 11:47
• Do you have a reference to this paper, by any chance? – AIM_BLB May 1 at 12:05
• Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. However, the result that every Carnot group of dimension $\ge 2$ is not CAT($-\kappa$) for $\kappa>0$ is straightforward. Indeed, since it has a non-isometric self-homothety, it would imply that it is CAT($-\kappa'$) for every $\kappa'>0$, and hence CAT($-\infty$), which for a geodesic space means an $\mathbf{R}$-tree, which cannot have any subset homeomorphic to the plane. – YCor May 1 at 12:12

The result that every Carnot group of dimension $$\ge 2$$ is not CAT($$−\kappa$$) for any $$\kappa>0$$ is straightforward. Indeed, since it has a non-isometric self-homothety, it would imply that it is CAT($$−\kappa'$$) for every $$\kappa'>0$$, and hence CAT($$-\infty$$), which for a geodesic space means an $$\mathbf{R}$$-tree, which cannot have any subset homeomorphic to the plane.