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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.

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  • $\begingroup$ 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? $\endgroup$ Dec 3, 2018 at 10:38
  • $\begingroup$ I edited my post. Happy to clarify further! $\endgroup$
    – Stefan
    Dec 3, 2018 at 10:44
  • $\begingroup$ 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. $\endgroup$ Dec 3, 2018 at 11:01
  • $\begingroup$ 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 $x_j$. $\endgroup$
    – Paul Frost
    Dec 3, 2018 at 14:08
  • $\begingroup$ @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 $P$. $\endgroup$
    – Stefan
    Dec 4, 2018 at 2:48

1 Answer 1

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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.

--- rk

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  • $\begingroup$ 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.) $\endgroup$
    – Stefan
    Dec 4, 2018 at 2:26
  • $\begingroup$ 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. $\endgroup$ Dec 12, 2018 at 18:28

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