# If $P$ is a polytope that is both simplicial and simple, then $P$ is a simplex or an $n$-gon

Question: Show that polytopes which are both simple and simplicial are either simplicies or two-dimensional $n$-gons.

• A polytope $P$ is simple if every vertex is contained in exactly $d$ facets.
• A polytope $P$ is simplicial if every face is a simplex, that is, each facet of $P$ contains exactly $d$ vertices.

Attempt: The case for $d=2$ is immediate, so assume that $d\geq 3$. It suffices to show that if $P$ is not a simplex, then $P$ is an $n$-gon (i.e. $d$=2). Since $P$ is not a simplex, it contains at least $d+2$ vertices. At this point I want to use some sort of Pigeonhole principle argument to assert that at least one of the facets of $P$ contains at least $d+1$ vertices, contradicting that $P$ is simplicial. However, I am stuck on the precise argument to be used.

Would appreciate any help!

• I am also stuck on this. By double counting we know the number of edges are equal to the number of vertices, but this has not helped me so far... Sep 17, 2018 at 20:52
• Where is this question from? As it reads, it seems to be asked as an exercise in some book. Feb 24, 2019 at 1:39

Suppose $$P \subseteq \mathbb{R}^d$$ is some $$d$$-dimensional polytope, which is both simple and simplicial. Suppose $$d \geq 3$$ and hence, $$P$$ is not a polygon. Let us look a some vertex $$v$$ of $$P$$. As $$P$$ is simple, we know that $$v$$ has exactly $$d$$ neighbors $$v_1, \ldots, v_d$$, which is equivalent to $$v$$ lying in exactly $$d$$ facets. $$P$$ is simplicial, and hence every facet is a $$d-1$$ simplex, containing exactly $$d$$ vertices. Note that the vertex-edge-graph of a simplex is the complete graph, every vertex is connected to every other. $$v$$ and every selection of $$d-1$$ of its neighbors form one of the $$d$$ many facets containing $$v$$ and hence, all vertices in the facets containing $$v$$ are connected to each other (using $$d \geq 3$$). In particular, all neighbors $$v_i$$ of $$v$$ have $$v, v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_d$$ as their $$d$$ neighbors.
But this means that there cannot be any other vertex $$w$$ and we conclude $$vertP = \{v, v_1, \ldots, v_d\}$$. Hence, $$P$$ is a simplex.