# is this sufficient to define a simplex?

I want to define a simplex based on the following properties

1. A convex polytope
2. All vertexes share an edge with all others
3. For a given vertex $v_i$ the set of all facets that the vertex belongs to is denoted by $\mathcal{F}_i$. For all vertices, the sets $\mathcal{F}_i$ are isomorphic, i.e. the sets are identical and the facets in the sets are isomorphic

For example, in the 2 dimensional regular polytopes satisfy condition 3 as each vertex belongs to two one dimensional faces that are edges and one 2 dimensional face that is the polytope

• Are we worried about edge lengths and metric properties, or just thinking about polytopes' face lattices combinatorially? I'm having trouble finding a non-simplex that satisfies (2) if we're just thinking about face lattices. If we do care about metric properties, then I'm not sure what counts as an isomorphism (of facets). Combinatorially, I'd imagine it's just a poset isomorphism. Jan 31, 2016 at 21:03
• so were not concerned with the specific metric properties, just the "lattice theoretic" perspective. I have heard that there may be cyclic polytopes that may be an example polytopes that are not simplexes but satisfy 2. see theorems 8.3 and 8.4 of math.uni-magdeburg.de/~kaibel/ALT/Downloads/cyclaut.pdf Jan 31, 2016 at 22:36
• if one can show that if a convex polytope satisfying 2. are simplexes that would be great! Jan 31, 2016 at 22:37
• @jdizzle I put a counterexample in comment to your other question which shows the set of polytopes satisfying 2 but not being simplex is not empty. Nov 6, 2016 at 14:06

No. Any cyclic polytope $C(n,d)$ where $d$ is even and greater than 3 is a counterexample.
Cyclic polytopes are convex polytopes, and are $d/2$-neighborly (your condition 2 is 2-neighborliness). They are simplicial, so all the facets are isomorphic. And for even $d$, the vertex figure at each vertex is a cyclic polytope of type $C(n-1, d-1)$, so all the vertices are surrounded by the same number of facets (and all the facets are isomorphic.)