Billera Tree Space

Question

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."

Answer 1

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 2^k 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 R^k into unit cubes. In the case when "k is maximal, i.e., k = n-2", vertex v is a lattice point v=(v₁,v₂,...,v_k) in R^k with each coordinate a positive integer. The link of vertex v in this cubical complex therefore has 2^k vertices, which are the 2^k vectors of the following form: (v₁-1,v₂,...,v_k), (v₁+1,v₂,...,v_k), (v₁,v₂-1,...,v_k), (v₁,v₂+1,...,v_k), ..., (v₁,v₂,...,v_k-1), (v₁,v₂,...,v_k+1). The link ends up being the boundary of the cross-polytope on these 2^k vertices, or in other words, the k-fold suspension of the emptyset.