MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let P and Q be simple polytopes such that P = Q ∩ H and let H be a halfspace with normal vector n. Let projn(e) denote the length of the projection of edge e onto vector n.

Consider the set E of edges of Q that cross through H, ie, edges with one endpoint in H and one outside H.

For any e ∈ E, does there always exist a projective transformation φ of P such that for every f ∈ E, projφ(n)(φ(e)) ≥ projφ(n)(φ(f))?

In other words, is there always a projective transformation making an arbitrary edge crossing H maximal with respect to H?

My first approach would be to choose φ to extend the endpoint of e as far as possible, but this does not appear to be sufficient to guarantee that projφ(n)(φ(e)) is maximized without adjusting other edges of Q.

share|cite|improve this question
up vote 1 down vote accepted

Let $e$ be the desired edge to be maximized. let point $a$ and $b$ be the endpoint of the edge. Let $a$ lie in $H$, $b$ outside of $H$. Let $e$ pass through the boundry of $H$, $B$ at $c$. Let $Q$ and $H$ lie in a space of dimension $n$ let this space be in a space of dimension $n+1$. Take point $a'$ which lies in this higher dimensional space and has the coordinates same as $a$ except that the coordinate in the extra dimension. Now recall that $B$ was the boundary of $H$ and has dimension one less than $n$. Add the new coordinate to the $n-1$ dimensional hyperplane that intersects $H$ in $B$ this gives $n$ dimensional space $G$. then project $Q$ onto $G$ through point $a'$. This will map point $a$ to the point at infinity. So if we take the original $n$-dimensional space and roatate it to $G$ in the $n+1$ dimensional space, with the rotation keeping the hyperplane of dimension $n-1$ fixed then the image of $a$ will go to the point at infinity as the angle approaches 90 degrees. Furthermore the the image of $a$'s projection on $n$ will go to infinity and be greater than all the other projections on $n$ of the points of $Q$. Because of this the image of $e$ will maximize all edges of the image of $Q$ that do not contain $a$. For any edge f that passes through a in Q the image of e will maximize f, in fact $ac$ will have the same normal vector and the fact that f is inside q means that $ab$ will have the greater projection.

share|cite|improve this answer
Thanks, Kristal! I understand the argument broadly, but would appreciate a slight clarification. Apologies if these questions are naive. Is G intended to be the n-dimensional space with coordinates {(x,0): x ∈ B} and a' the point (a,0), ie, does it matter whether the coordinates are identical? And what do you mean, formally, by projection through the point a - simply that a lies between Q and G? – Anand Kulkarni Mar 26 '10 at 11:03
Am I correct that the argument can be stated as follows, or did I misunderstand? For any vertex of a polytope, there exists a projective transformation mapping that vertex to infinity. Thus, there exists a projective transformation T mapping the endpoint a of e within Q to infinity. Since a is the unique vertex being mapped to infinity, T(e) has infinite length. Thus, any infinitesimal perturbation p of T will place p(T(a)) in a finite position such that the edge p(T(e)) will be larger than any other edge f when they are both projected onto any vector n (since p(T(e)) is arbitrarily large). – Anand Kulkarni Mar 26 '10 at 11:38
I followed roughly the outline you give. I tried to deal with some issues: I tried to take care that the line wasn't normal to the line it would be projected to. I also had to deal with the fact all the edges f that share point a with e will have length going to infinity so they will have to be dealt with as a separate case. – Kristal Cantwell Mar 26 '10 at 19:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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