Binomial Start System

Find the binomial start systems and solutions for

(1)
\begin{align} x^2y+xy^2+1 &= 0\\ x^2y+x^2+x+1 &= 0 \end{align}

for your choice of lifts of Newton polytopes.

Solution:

I apologize for the lack of images…

Set

(2)
\begin{align} f= & x^2 y+xy^2+1\\ g= & x^2y+x^2+x+1\\ P= New(f)= & conv\{(0,0),(2,1),(1,2)\}\\ Q= New(g)= & conv\{(0,0),(1,0),(2,0),(2,1)\}\\ M=P+Q= & conv\{(0,0),(1,2),(3,3),(4,2),(4,1),(2,0)\} \end{align}

Choose lifts P' of P and Q' of Q as follows:

(3)
\begin{align} P'= & conv\{(0,0,0),(1,2,0),(2,1,0)\}\\ Q'= & conv\{(0,0,4),(1,0,3),(2,0,2),(2,1,1)\} \end{align}

Note that these lifts are both triangles. Taking the lower hull of the Minkowski sum P'+Q' gives a mixed subdivision of M=P+Q with two mixed cells M1 and M2 with areas 3 and 2 respectively:

(4)
\begin{align} M1=conv\{(0,0),(1,2),(3,3),(2,1)\}\\ M2=conv\{(2,0),(4,1),(4,2),(2,1)\} \end{align}

The parametrized system of equations corresponding to the lifts P' and Q' is:

(5)
\begin{align} x^2y+xy^2+1 & =0\\ tx^2y+t^2x^2+t^3x+t^4 & =0 \end{align}

The tropical hypersurfaces TP' and TQ' (using inward pointing normals) corresponding to the lifts P' and Q' are described as follows. TP' is centered at (0,0) and consists of the three rays k(2,-1),k(-1,2), and k(-1,-1), where k is greater than or equal to 0. TQ' is centered at (1,1) and consists of the three rays k(-1,0),k(0,1), and k(1,-2), where again k is greater than or equal to 0.

Note: The center (1,1) of TQ' can be obtained in a couple of equivalent ways. First, TQ' is the nonlinear locus of the function trop(tx^2y+t^2x^2+t^3x+t^4)=min{1+2x+y,2+2x,3+x,4} =min{1+2x+y,2+2x,4}. So the center of TQ' occurs when 1+2x+y=2+2x=4, or when x=y=1.
Alternatively, take the upward pointing normal to the lifted polytope Q' (this is just a triangle so it makes sense to talk about its upward pointing normal) and scale it so that the z-coordinate is 1. Then the x and y coordinates of the scaled normal will be the coordinates of the center of TQ'.

TP' and TQ' intersect at the points A=(-1/2,1) and B=(2,-1). This implies that we should try two substitutions for x and y. Corresponding to A take:

(6)
\begin{align} x= & c_1t^{-1/2}+HOT\\ y= & c_2t+HOT \end{align}

Corresponding to B take:

(7)
\begin{align} x= & c_1t^2 + HOT\\ y= & c_2t^{-1} + HOT \end{align}

Plugging these into (5) yields two binomial start systems in the c-variables corresponding to cancellation of lowest order terms in t. The system coming from A is:

(8)
\begin{align} c_1^2c_2+1 & =0\\ c_1^2c_2+c_1^2 &=0 \end{align}

This has the 2 solutions (1,-1) and (-1,-1) in the nonzero complex numbers. We see this corresponds to the mixed cell M2 in the mixed subdivision of P+Q given above. The system coming from B is:

(9)
\begin{align} c_1c_2^2+1 & =0\\ c_1^2c_2+1 & =0 \end{align}

Let w be a primitive 6th root of unity. This system has the 3 solutions

(10)
$$(1,-1),(-w^4,w),(-w^2,w^5)$$

So we see this corresponds to the mixed cell M1.