Are point sets of the same order type connected by continuous (order type)-preserving motion? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:10:42Z http://mathoverflow.net/feeds/question/98283 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98283/are-point-sets-of-the-same-order-type-connected-by-continuous-order-type-preser Are point sets of the same order type connected by continuous (order type)-preserving motion? Nima Hoda 2012-05-29T14:20:55Z 2012-05-29T23:56:12Z <p>Given two general position point sets in \$\mathbb{R}^2\$ of the same size and <a href="http://www.ist.tugraz.at/staff/aichholzer/research/publications/#aak-eotsp-01" rel="nofollow">order type</a>, is it possible to continuously move the points of one set until they coincide with those of the other set in such a way that order type is preserved (equivalently, no three points become collinear) throughout the operation?</p> http://mathoverflow.net/questions/98283/are-point-sets-of-the-same-order-type-connected-by-continuous-order-type-preser/98303#98303 Answer by Joseph O'Rourke for Are point sets of the same order type connected by continuous (order type)-preserving motion? Joseph O'Rourke 2012-05-29T20:13:14Z 2012-05-29T23:56:12Z <p>I believe the answer is (in general) <b>No</b> for the following reason. A configuration of points in \$\mathbb{P}^2\$ is projectively dual to an arrangement of lines in \$\mathbb{P}^2\$. The question you ask, translated to arrangements of lines, is whether an arrangement can always be continuously moved to any isomorphic arrangement, all the while remaining isomorphic. This was asked by Ringle in 1956 (the <em><a href="http://en.wikipedia.org/wiki/Homotopy#Isotopy" rel="nofollow">isotopy</a> conjecture</em>), and answered negatively by <a href="http://en.wikipedia.org/wiki/Mnev%27s_universality_theorem" rel="nofollow">Mnev's Universality Theorem</a> in 1985: not only can the space of line arrangments isomorphic to a given arrangement be disconnected, it can have the homotopy type of any algebraic variety.</p> <p>Of course it is not difficult to change from \$\mathbb{P}^2\$ to \$\mathbb{R}^2\$. I find this explicit claim in a 1988 paper by Suvorov (<a href="http://www.springerlink.com/content/q024m7649g513x04/" rel="nofollow">Springer link</a>): <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/Suvorov13.jpg" alt="Suvorov"><br /> Here, "nondegenerated" means, I believe, what we would call today "simple": no three lines share a point. A "rigid" isotopy keeps the lines straight (in a "flexible" isotopy, the objects are pseuodolines).</p> <p>Suvorov's example has 14 lines in \$\mathbb{P}^2\$. Later (1996) Richter-Gebert found another 14-line/point example, presented in his paper "Two Interesting Oriented Matroids" (<a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.43.3848" rel="nofollow">CiteSeer link</a>): <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/JurgenR-G14.jpg" alt="Jurgen R-G Fig 1"><br /> He shows that </p> <blockquote> <p>[its] realization space [...] is an open interval from which one point has been deleted.</p> </blockquote>