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.



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.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – Anand Kulkarni Aug 27 '10 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.