Billera Tree Space I am studying the tree space of Billera and I do not really understand why it is an Hadamard Space. I have already read L. Billera, S. Holmes, K. Vogtmann, Geometry of the space of phylogenetic trees, Advances in Applied Math, 27 (2001) and I do not understand what Billera means with k-fold suspension in the proof of that the tree space is a Hadamard Space:
"In order to apply 2.1.19 we need to show that the link of each vertex is a flag complex. Note that we have already said that the link Ln of the origin is a simplicial complex whose k-simplices are sets of k+1 compatible splits of {0, . . . , n} that correspond to k+1 pairs of compatible thick partitions. Let v be an arbitrary vertex of the cube complex, which lies in the interior of a (unique) orthant of dimension k. This orthant corresponds to a tree with k interior edges, and thus to a set S of k pairwise compatible partitions of 0, . . . , n. If k is maximal, i.e., k = n − 2 the link of v is a triangulates sphere. 

We think about it as the k-fold suspension of the empty set.

In general the link of v is the k-fold suspension of the subcomplex of Ln, spanned by all partitions compatible with S. Since this itself is a flag complex, and since the suspensione of a flag complex is a flag. We have proved that the link of each vertex is a flag. Since the Tree space is obviously simply connected we have completed the proof."
 A: Here are some examples of the k-fold suspension of the empty set.


*

*The 1-fold suspension of the emptyset is 2 vertices. This is a triangulation of the 0-sphere.

*The 2-fold suspension of the emptyset has 4 vertices. Its maximal simplices consist of four edges arranged to form the boundary of a diamond (or square). This is a triangulation of the 1-sphere.

*The 3-fold suspension of the emptyset has 6 vertices. Its maximal simplices consist of 8 2-simplices (triangles) arranged to form the boundary of the octahedron. This is a triangulation of the 2-sphere.

*More generally, the k-fold suspension of the emptyset has 2k vertices. Its maximal simplices consist of 2k k-simplices arranged to form the boundary of the cross-polytope on 2k vertices:
https://en.wikipedia.org/wiki/Cross-polytope.
This is a triangulation of the (k-1)-sphere.
To summarize, the k-fold suspension of the emptyset is the boundary of the cross-polytope on 2k vertices (https://en.wikipedia.org/wiki/Cross-polytope), which is a triangulation of the (k-1)-sphere.
How do these objects arise in the paper by Billera, Holmes, and Vogtmann? They have subdivided an orthant of Rk into unit cubes. In the case when "k is maximal, i.e., k = n-2", vertex v is a lattice point v=(v1,v2,...,vk) in Rk with each coordinate a positive integer. The link of vertex v in this cubical complex therefore has 2k vertices, which are the 2k vectors of the following form: (v1-1,v2,...,vk), (v1+1,v2,...,vk), (v1,v2-1,...,vk), (v1,v2+1,...,vk), ..., (v1,v2,...,vk-1), (v1,v2,...,vk+1). The link ends up being the boundary of the cross-polytope on these 2k vertices, or in other words, the k-fold suspension of the emptyset.
