Are all homogeneous metric spaces bihomogeneous? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:47:32Z http://mathoverflow.net/feeds/question/40249 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40249/are-all-homogeneous-metric-spaces-bihomogeneous Are all homogeneous metric spaces bihomogeneous? Ricky Demer 2010-09-28T01:54:25Z 2010-09-28T03:55:24Z <p>Let (X,d) be a metric space such that for all points p and q in X, there exists an isometry f such that f(p) = q. Does it follow that for all points p and q in X, there exists an isometry f such that f(p) = q and f(q) = p?</p> <p>This seems like an obvious enough question that I would be surprised if the answer isn't simply a reference, but I haven't found it mentioned anywhere.</p> http://mathoverflow.net/questions/40249/are-all-homogeneous-metric-spaces-bihomogeneous/40254#40254 Answer by Richard Borcherds for Are all homogeneous metric spaces bihomogeneous? Richard Borcherds 2010-09-28T02:42:50Z 2010-09-28T03:27:26Z <p>The vertices of a <a href="http://en.wikipedia.org/wiki/Snub_cube" rel="nofollow">snub cube</a> form a metric space with 24 points that is homogeneous but not bihomogeneous: the edges of the squares have a "direction" associated with them. </p> <p>Added later: here is an example with just 6 points: take an equilateral triangle with sides of length 1, and take the 6 points on the edges that are distance 1/4 from a vertex. </p> <p>Added later: There are no examples with less than 6 points; for example, for 5 points there are 10 edges so there are at most 2 possible lengths with 5 edges of each length, which gives essentially only 1 configuration and this is bihomogeneous. Less than 5 points is easy to do case by case. </p> http://mathoverflow.net/questions/40249/are-all-homogeneous-metric-spaces-bihomogeneous/40257#40257 Answer by Joel David Hamkins for Are all homogeneous metric spaces bihomogeneous? Joel David Hamkins 2010-09-28T03:36:05Z 2010-09-28T03:55:24Z <p>Here is a one-dimensional analogue of Richard's triangle example, obtaining a counterexample in the set of reals. Namely, replace every integer $n$ with two numbers at fixed small distance $n\pm\epsilon$. One can suitably translate and reflect to realize homogeneity, but there is no isometry swapping $\epsilon$ and $1+\epsilon$.</p>