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I have a convex polytope of arbitrary dimension. Let $\mathcal{F} (A)$ denote the set of facets that vertex $A$ belongs to. If two vertices share an edge, is it true that the disunion of $\mathcal{F} (A)$ and $\mathcal{F} (B)$ has a single element only, i.e. the share all but a single supporting facet? If so, where can I find the proof of this? Many thanks.

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  • $\begingroup$ What does "disunion" mean? $\endgroup$ Commented Nov 3, 2015 at 17:57

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Assuming that "disunion" means the symmetric difference (the set of facets containing either the vertex $A$, or the vertex $B$, but not both), then no. E.G. take a cube, and two vertices $A$ and $B$ connected by an edge. There is a face containing $A$ but not $B$, and a face containing $B$ but not $A$, so there are two facets in the symmetric difference $\mathcal{F}(A) \mathbin{\triangle}\mathcal{F}(B)$. There will always be at least two (otherwise one vertex would be in all the facets containing another vertex, which is impossible.)

Furthermore, there can be arbitrarily many facets not shared between two adjacent vertices. Consider 3-dimensional polytopes. Only two facets are on the shared edge; any other facets are not shared by the two vertices. So if the vertices are incident to more than three edges, then there are lots of non-shared facets.

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