Edge-maximizing projective transformation on polytopes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:16:04Z http://mathoverflow.net/feeds/question/17126 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17126/edge-maximizing-projective-transformation-on-polytopes Edge-maximizing projective transformation on polytopes Anand Kulkarni 2010-03-04T19:54:05Z 2010-03-09T19:54:07Z <p>Let P and Q be simple polytopes such that P = Q &cap; H and let H be a halfspace with normal vector n. Let proj<sub>n</sub>(e) denote the length of the projection of edge e onto vector n.</p> <p>Consider the set E of edges of Q that cross through H, ie, edges with one endpoint in H and one outside H. </p> <p>For any e &isin; E, does there always exist a projective transformation &phi; of P such that for every f &isin; E, proj<sub>&phi;(n)</sub>(&phi;(e)) ≥ proj<sub>&phi;(n)</sub>(&phi;(f))? </p> <p>In other words, is there always a projective transformation making an arbitrary edge crossing H maximal with respect to H? </p> <p>My first approach would be to choose &phi; to extend the endpoint of e as far as possible, but this does not appear to be sufficient to guarantee that proj<sub>&phi;(n)</sub>(&phi;(e)) is maximized without adjusting other edges of Q.</p> http://mathoverflow.net/questions/17126/edge-maximizing-projective-transformation-on-polytopes/17649#17649 Answer by Kristal Cantwell for Edge-maximizing projective transformation on polytopes Kristal Cantwell 2010-03-09T19:54:07Z 2010-03-09T19:54:07Z <p>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.</p>