Let $P$ be a simple convex polytope in $\mathbb{R}^d$ (that is, any vertex belongs to exactly $d+1$ facets). Given the collection of outer normals to facets of $P$, combinatorics of $P$ may be different, of course. But is this information enough to reconstruct number of faces of $P$ of all dimensions? If yes, what is specific procedure to do this? I am looking at what happened when we move facets parallel and pass through `singularity', i.e. not-simple polytope, and on first glance it looks that $f$-vector preserves, but I am always not sure with such things. And this is not good proof anyway, I would prefer more direct argument.

If the statement is however false, I wonder for which specific collections it is still true.

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    $\begingroup$ I presume the statement you wish to investigate is this: Every simple polytope with the same set of facet normals has the same $f$-vector. $\endgroup$ – Joseph O'Rourke Oct 17 '15 at 19:03
  • $\begingroup$ @Joseph Yes, exactly. Sorry if it was not clear. $\endgroup$ – Fedor Petrov Oct 17 '15 at 19:26

Consider a simplex $S$ in five dimensions, and one more face normal $\pi$ in general position with respect to $S$ (no hyperplane perpendicular to $\pi$ passes through more than one vertex of $S$). Intersect $S$ with a halfspace $H$ perpendicular to $\pi$, and translate $H$ continuously towards $S$, starting from a position where it contains all of $S$.

Then when the intersection with $H$ first cuts off one vertex of $S$, the one cut-off vertex is replaced by five new vertices (on the edges of $S$ connecting the cut-off vertex to the others) so the total number of vertices becomes $6-1+5=10$. When the intersection with $H$ cuts off two vertices of $S$, they are replaced by eight new vertices (again, one for each edge of $S$ that connects one of the two cut-off vertices to the four remaining ones), so the total number of vertices becomes $6-2+8=12$. And when the intersection with $H$ cuts off three vertices of $S$, they are replaced by nine new vertices, so the total number of vertices becomes again $6-3+9=12$. All of these are simple $5$-polytopes.

So no, the $f$-vector is not preserved. Not even the number of vertices is preserved.

  • $\begingroup$ David, did you use $5$ dimensions because it is not so clear in dimension $4$? As Fedor observed, the statement is true in $\mathbb{R}^3$, and you have given an $\mathbb{R}^5$ counterexample. That leaves $\mathbb{R}^4$...? $\endgroup$ – Joseph O'Rourke Oct 18 '15 at 21:04
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    $\begingroup$ It is also false in $\mathbb{R}^4$ but the counterexamples are a little more complicated. E.g. consider a hypercube and a face normal parallel to one of its long diagonals. Cutting off one vertex leaves $16-1+4=19$ vertices. Cutting off five vertices leaves $16-5+12=23$ vertices. $\endgroup$ – David Eppstein Oct 18 '15 at 23:03

Not an answer; just an illustration:

          $f$-vector left & right: $(V,E,F)=(6,9,5)$. Middle polyhedron is not simple.

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    $\begingroup$ Well, thank you, but warning: for dimension 3 the answer is positive just by Euler formula (and number of facets is enough, not even their directions), so it may be somehow misleading. $\endgroup$ – Fedor Petrov Oct 17 '15 at 16:07

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