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Let P be a simple polytope defined as an intersection of n halfspaces.

A facet F of P, supported by halfspace H, is removable if the intersection of the remaining (n-1) halfspaces is bounded. F is projectively removable if there exists a projective transformation π such that π(F) is removable from π(P).

It is easy to show that every facet of a simple d-polytope with at least (d+2) facets is projectively removable, since there is a projective transformation mapping (d+1) of the remaining halfspaces into a (d)-simplex.

Consider a vertex v defined by the intersection of d halfspaces and lying on a removable facet F supported by halfspace H. Suppose we translate H along its normal axis away from the center of the polytope out towards infinity. As we do so, some vertices will disappear from F as the corresponding halfspaces no longer intersect H within the polytope. At the time just before H leaves the polytope entirely there will be exactly d vertices left on F.

Define v to be a final vertex if, when H is translated out of P in this way, v is one of the d vertices remaining on F.


In a given realization of a polytope, some set of d vertices on a removable facet will be final. But if we apply an appropriate projective transformation, can any vertex v be made into a final vertex? In other words,

for any vertex v on a facet F of a simple polytope, is there a projective transformation πv such that both π(F) is removable and π(v) is final in π(P)?


Based on what I believe I understand about projective transformations, I can imagine that there is a projective transformation that shrinks F to an arbitrarily small point, and another transformation that perturbs F so that a given vertex "sticks out" enough to be a final vertex. However, I am not clear how to show from a formal definition of projective transformations that such transformations always exist or that there exists a given transformation that imposes both properties.

As a continuation of this question, let me ask: how can I gain more intuition about what properties of a polytope can be modified by projective transformation? Can facets be scaled arbitrarily and edges shifted around as I have suggested? I have taken a look at some texts suggested in Ziegler about projective geometry, but I'm interested in knowing more precisely what kind of things projective transformations can and cannot do to polytopes.

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1 Answer 1

Yes.

Let $\phi$ be a $(d-2)$-flat that lies in the supporting hyperplane $H$ but does not intersect $F$. Rotating $H$ around $\phi$ is projectively equivalent to translating $H$; just apply a projective transformation that sends $\phi$ to the hyperplane at infinity without everting $P$. Pick an arbitrary ridge $R$ that is a facet of $F$, and place $\phi$ infinitesimally close to and parallel to $R$. If we rotate $H$ outward around $\phi$, the vertices of $R$ will be the last to disappear.

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Thanks for this straightforward solution! Apologies if these two follow-up questions are naive: 1) How do we know that there exists a projective transformation sending $\phi$ to infinity that retains the property that $F$ is projectively removable? 2) Your solution finds a ridge such that some d of its vertices are final, but the original question asks about a single vertex. I presume this is a trivial difference -- ie, if we choose a point $\phi$ lying in $H$ instead of a (d-2)-flat and place it next to a desired vertex $v$ prior to applying the rotation, is it true that $v$ is final? –  Anand Kulkarni Aug 27 '10 at 15:52

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