Subject: Re: [math-fun] Fano Plane puzzle (Fred Lunnon) Message-ID: <CAN57YqvtqfqOvaYUjHraaUeXdSdch6Fks3xXAiNnA61wbdjhkw@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
If each projective line meets n+1 points, you can always embed lines as (n-1)-spheres in n-space; and there is so much freedom in choosing point coordinates that you can surely then constrain all radii to unity in principal.
If you want to reduce the number of dimensions, it becomes much harder: the Heawood embedding in 2-space was only solved in 2009 by Gerbracht.
--well as I'd explained in a previous post, any bipartite graph can be embedded as a unit distance graph in 4 dimensions so projective planes can be embedded in 4D using unit-radius spheres as the "lines." Lunnon's claim here about general n was therefore obsolete before issuance because n=4 suffices. However, the interesting question then is: for which projective planes does n=2 or n=3 suffice? One might suspect based on crude degree-of-freedom counting that only a finite number of projective planes can be embedded in 2D or 3D. However, clearly such crude arguments are not actually valid! Furthermore, my 4D solution could be considered undesirable since although all the spheres are different, the N points all happen to lie on a common circle which is the intersection of all N spheres. So then you might want to go to 5D, or whatever... in 5D the analogous solution has all points lying on a common surface-dimension-2 sphere which is the intersection of the spheres...