Unsections/Cones and “Tom vs Jerry” ambiguity

Unsections/Cones and “Tom vs Jerry” ambiguity:
why no single-valued ansatz is possible and Minkowski ambiguity is the thing to expect

[Miles Reid-like notation]
Consider two del Pezzo threefolds of degree 6.
Let Jerry be P^1 \times P^1 \times P^1
and Tom be W = P(T_{P^2}) = X_{1,1} \subset P^2 \times P^2 (hyperplane section of product of two planes in Segre embedding).
It is known that Tom and Jerry are not fibers of a flat family.

Tom has period sequence 6,
Jerry is grdb[520140 and has period sequence 21.

Their half-anticanonical section is S = S_6 (del Pezzo surface of degree 6).
So both Tom and Jerry can be degenerated to the same Gorenstein toric Fano threefold — anticanonical cone over S.

This cone has just one integral point except origin and vertices.
Let u = x + y + xy + \frac{1}{x} + \frac{1}{y} +\frac{1}{xy}
be the normalized Laurent polynomial for the honeycomb (fan polytope of S).

Note that honeycomb has two different Minkowski decompositions — as sum of three intervals and as sum of two triangles.

These decompositions correspond to two different decompositions of (u+G) into the product of Laurent polynomials [for two different values of G (G=2 and G+3)]:

u+2 = (1 + x) (1 + y) (1 + \frac{1}{xy})


u+3 = (1 + x + y) (1 + \frac{1}{x} + \frac{1}{y})

General Laurent polynomial for the cone over S
has the form

w_G = z (u + G) + \frac{1}{z}

The most interesting thing is the following:

if we choose $G=2$ then w is mirror of Jerry,
but if we choose $G=3$ then w is mirror of Tom.

Moreover applying mutation we can transform w to terminal Gorenstein polynomials:


z (u+2) + \frac{1}{z} = z (1+x)(1+y)(1+\frac{1}{xy}) + \frac{1}{z}

becomes z (1+x)(1+y) + \frac{(1+\frac{1}{xy})}{z} = z + zx + zy + zxy + \frac{1}{z} + \frac{1}{xyz}

by applying (x,y,z) \to (x,y,\frac{z}{1+\frac{1}{xy}}).

This corresponds to STD of Tom.
It looks nicer after monomial transformation (x,y,z) \to (x,y,\frac{z}{xy}):
\frac{z}{xy} + \frac{z}{x} + \frac{z}{y} + \frac{xy}{z} + \frac{1}{z}


z (u+3) + \frac{1}{z} = z (1+x+y)(1+\frac{1}{x}+\frac{1}{y}) + \frac{1}{z}

becomes z (1+x+y) + \frac{1 + \frac{1}{x} + \frac{1}{y}}{z} = z + zx + zy + \frac{1}{z} + \frac{1}{zx} + \frac{1}{zy}

by applying (x,y,z) \to (x,y,\frac{z}{1+\frac{1}{x}+\frac{1}{y}})

This is simply Laurent polynomial for the smooth model of Jerry:

x+y+z+\frac{1}{x} + \frac{1}{y} + \frac{1}{z}
after monomial transformation
(x,y,z) \to (xz,yz,z).


So Laurent phenomenon distinguishes degenerations of different varieties to the same singular and does not mix them.

One Comment

  1. Sergey says:

    Tom, Jerry and Spike.

    If in the previous scenario we look for bigger brothers of Tom and Jerry with index 1,
    then we have a pleasant surprise – not only we get 3.1 and 2.6, but also V_{12}.

    For W and \PP^1 \times \PP^1 \times \PP^1 we constructed mirrors
    as w_G = \frac{u+G}{z} + z where u = x+y+xy + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy},
    and G is either 2 or 3.

    Regularized I-series for 2.6 and 3.1 has the same coefficients a_n = b_{2n},
    so their mirrors should be related to squares of mirrors for W and \PP^1 \times \PP^1 \times \PP^1.

    Consider w_G^2 = \frac{(u+G)^2}{z^2} + 2(u+G) + \frac{1}{z^2}.
    Obviously it has the same J_w = \int \frac{\omega}{1-tw} as its isogeneous
    w'_G = \frac{(u+G)^2}{z} + 2(u+G) + \frac{1}{z} = \frac{(z+u+G)^2}{z}

    Newton polytope for this kind of Laurent polynomials have number 432464 in grdb.

    We can check that
    w'_2 = \frac{(z+u+2)^2}{z} has period sequence 11 and is mirror to 2.6,
    w'_3 = \frac{(z+u+3)^2}{z} has period sequence 22 and is mirror to 3.1.

    But we can have even more – I find it very amusing that
    w' = \frac{(z+u+2)(z+u+3)}{z} has period sequence 7 and is mirror to V_{12}.

    Since singular toric variety in question is again a cone over surface, probably it won’t be hard to prove the existence of degenerations.

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