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 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_n(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_&phi;(n)(&phi;(e)) ≥ proj_&phi;(n)(&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_&phi;(n)(&phi;(e)) is maximized without adjusting other edges of Q.</p>

http://mathoverflow.net/questions/17126/edge-maximizing-projective-transformation-on-polytopes/17649#17649 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>