Correction: under D_6 the second equilateral triangle has freedom 2, so a 6-fold symmetric Fano plane has only freedom 1 . WFL On 6/30/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 6/30/15, Dan Asimov <asimov@msri.org> wrote:
I think this is just gorgeous.
Is Fred's solution the unique* set of points in R^3 having the combinatorial type of the Fano plane (with unit circles for the lines) and the symmetry group of D_3 == S_3 ???
—Dan _____________________________________ * Up to isometries of R^3, of course.
An equilateral triangle in 3-space has freedom 7 , a second one with the same axis line freedom 3 , a single point on the axis freedom 1 . Via C_3 symmetry the 7 unit radii impose only 3 constraints; so modulo isometry, a 3-fold symmetric Fano plane has freedom 11 - 3 - 6 = 2 . Since the orbits remain unchanged under D_6 , that would yield the same freedom.
Or so I reckon ...
WFL