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Consider a metric space $(X,d)$ and fix a base point $w$. A horofunction is a function of the form $$\beta_y(x)=d(x,y)-d(w,y).$$ The map $y\mapsto \beta_y$ is an embedding of $X$ into the space of $1$-Lipschitz map vanishing at $w$. The closure of $X$ into this space is called the horofunction compactification of $X$, which is denoted by $\overline{X}^h$ in the following. Then, $\partial^hX:=\overline{X}^h\setminus X$, the complement of $X$, is called the horofunction boundary.

Finding the homeomorphism type of the horofunction boundary is often a difficult problem. We can still say something in some examples.

The horofunction boundary for $\mathbb{R}^d$ endowed with the Euclidean distance is a sphere of dimension $d-1$. We can actually describe the topology of the compactification itself: a sequence $x_n$ converges to a point in the horofunction boundary if and only if $x_n$ goes to infinity and $\frac{x_n}{\|x_n\|}$ converges to some $\theta\in \mathbb{S}^{d-1}$.

There are several possible generalizations of this example. For instance, the horofunction compactification of a CAT(0) space coincide with the visual compactification, see Bridson and Haefliger book metric spaces of non-positive curvature for more details.

Another way of generalizing the Euclidean case is to consider a nilpotent Lie group. Let us focus on Carnot groups. Such a group is endowed with a sub-Riemannian metric, which comes from the first step of nilpotency. The path-length distance associated to this sub-Riemannian metric is called the Carnot-Caratheodory distance.

My questions are the following:

1) Is there a description of the topology of the horofunction compactification/boundary for the Carnot-Caratheodory distance on a Carnot group ? The answer is positive for the Heisenberg group (see the paper of Klein and Nicas The horofunction boundary of the Heisenberg group: the Carnot-Caratheodory metric.

2) At least, can we construct two distinct Carnot-Caratheodory distances on a Carnot group such that the horofunction compactifications/boundaries do not have the same homeomorphism type ?

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    $\begingroup$ By Klein-Nicas you quote (ams.org/journals/ecgd/2010-14-15/S1088-4173-2010-00217-1/…), the answer to 2 is negative for the $2n+1$-dimensional Heisenberg group for every $n$ (all compactifications/boundaries are homeomorphic, if I understand correctly). Most likely it extends to product of the latter with a Euclidean group, and hence settles the (very) special case when the derived subgroup has dimension $\le 1$. The first next case would be the 4-dimensional filiform group. $\endgroup$ – YCor Feb 15 at 14:20
  • $\begingroup$ Thanks. This actually answers my third question (I did not ask): what is the next example after Heisenberg ? $\endgroup$ – M. Dus Feb 15 at 15:19

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