# Why can't a vertex of a $d$-dimensional polytope be in fewer than $d$ edges?

This is motivated by the definition of simple polytopes: if all vertices of a $$d$$-dimensional convex polytope $$P$$ are in exactly $$d$$ edges (i.e. $$1$$-dimensional faces of $$P$$), then $$P$$ is simple.

I struggle to show that no vertex of $$P$$ can be in fewer than $$d$$ edges. Clearly, if this would be the case for some vertex $$v$$, the (translated) cone generated by $$v$$ and its adjacent vertices would have a lower dimension than $$d$$.

So if I can only show that the cone contains $$P$$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.

Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $$P$$ being $$d$$-dimensional, I mean that its affine hull is $$d$$-dimensional, not the ambient space.

• Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using? Dec 3, 2018 at 10:38
• I edited my post. Happy to clarify further! Dec 3, 2018 at 10:44
• At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else. Dec 3, 2018 at 11:01
• If a polytope $P$ is the convex hull of finitely many points $x_1,\dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$. Dec 3, 2018 at 14:08
• @PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$. Dec 4, 2018 at 2:48

Any vertex figure of a $$d$$-polytope happens to be a $$(d-1)$$-polytope. The smallest known $$(d-1)$$-polytope is a $$(d-1)$$-simplex, which has exactly $$d$$ vertices. As those vertices in turn represent the edges of that $$d$$-polytope, which emanate from the vertex of consideration, you are done.
• Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.) Dec 4, 2018 at 2:26
• Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope. Dec 12, 2018 at 18:28