🚀 go-pugleaf

I have made a mistake.   I said that Mike's matrices have the property
that the k'th row has the same sum as the k'th column, and that
this was a linear constraint on the matrices in the space of all
anti-symmetric matrices.  Both halves are false in spirit.  The sum
of the k'th row is the negative of the sum of the k'th column, but
this is true for all matrices.  So there is no constraint at all.

Rather, Mike's matrices span the set of all anti-symmetric matrices.
If you take our original matrix A, which I would actually prefer 
to call a here, and let tau be the transposition of 1 and 2; tau's
effect on a is two switch the first two rows and columns.  Then
a-tau(a) = 2e_12 - 2e_21, where e_ij is an elementary matrix.
The orbit of e_12-e_21 under conjugation by permutation matrices
clearly spans all antisymmetric matrices.

The rest of what I said is still true I think, except that the polytope
U is even bigger than I said it was.

				Greg