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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Mar 26, 2010 at 11:03
  • $\begingroup$ 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). $\endgroup$ Mar 26, 2010 at 11:38
  • $\begingroup$ 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. $\endgroup$ Mar 26, 2010 at 19:45

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