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All Carnot groups are complete metric spaces, since they have all closed balls compact ("proper" metric space). In general, any metric space with a transitive isometry group, and having a compact subset with nonempty interior, is complete (easy exercise).

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.

Actually, a non-abelian Carnot group is not even CAT(0), and does not even have a quasi-isometric embedding into any CAT(0) orspace (or any uniformly convex Banach space). The latter fact was established in:

Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. However,

All Carnot groups are complete metric spaces, since they have all closed balls compact ("proper" metric space). In general, any metric space with a transitive isometry group, and having a compact subset with nonempty interior, is complete (easy exercise).

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.

Actually, a non-abelian Carnot group is not even CAT(0), and does not even have a quasi-isometric embedding into any CAT(0) or Banach space. The latter fact was established in:

Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. However,

All Carnot groups are complete metric spaces, since they have all closed balls compact ("proper" metric space). In general, any metric space with a transitive isometry group, and having a compact subset with nonempty interior, is complete (easy exercise).

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.

Actually, a non-abelian Carnot group is not even CAT(0), and does not even have a quasi-isometric embedding into any CAT(0) space (or any uniformly convex Banach space). The latter fact was established in:

Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. However,

Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 286

All Carnot groups are complete metric spaces, since they have all closed balls compact ("proper" metric space). In general, any metric space with a transitive isometry group, and having a compact subset with nonempty interior, is complete (easy exercise).

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.

Actually, a non-abelian Carnot group is not even CAT(0), and does not even have a quasi-isometric embedding into any CAT(0) or Banach space. The latter fact was established in:

Scott D. Pauls. The large scale geometry in nilpotent Lie groups. Commun. Anal. Geom. 9(5), 951-982, 2001. However,