On page 436 of the remarkable book about multi dimensional determinants, by Gelfand, Kapranov and Zelevinsky, they describe how the equivalence between a deformed triangle with marked midpoints and a hexagon

is related to the description of the $3d$ associahedron (labelled as usual by chorded hexagons) as a secondary polytope. To the triangle one associates a set $A$ of six (three dimensional) coordinates, namely

$(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$

and these stand for three quadratic forms, namely

$f_1 = a_{11} X^2 + a_{12} Y^2 + a_{13} Z^2 + a_{14} XY + a_{15} XZ + a_{16} YZ$

$f_2 = a_{21} X^2 + a_{22} Y^2 + a_{23} Z^2 + a_{24} XY + a_{25} XZ + a_{26} YZ$

$f_3 = a_{31} X^2 + a_{32} Y^2 + a_{33} Z^2 + a_{34} XY + a_{35} XZ + a_{36} YZ$

in terms of a $3 \times 6$ matrix. The associahedron secondary polytope is associated to the set $A$, and there is a resultant $R(f_1,f_2,f_3)$ which is a degree $12$ polynomial in the $a_{ij}$. The $14$ allowed triangulations of the hexagon, and actual volumes associated to the polytope, correspond to the structure of $R(f_1,f_2,f_3)$. That is, if we let $[j_1 j_2 j_3]$ be the ordinary determinant of the $3 \times 3$ submatrix $a_{ij_{k}}$ then these $14$ expressions include, for example

$[145][245][256][356], [134]^{2}[246][346].$

Twistor theorists will recognise the (Grassmann) process of taking square submatrices for nice coordinate systems. Each digit $j_{k}$ stands for a point on the triangle. If we choose to label the three vertices $1,2,3$ and the midpoints $4,5,6$ then the two internal triangle configurations correspond to the polynomials

$-[145][246][356][456]$ and $[123]^{4}$

since these contain the terms $[456]$ and $[123]$. These terms look different to the other $12$ terms: the first because it is negative and the second because it only involves one triangle.

7 years ago

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