Posts tagged ‘cluster’

Beyond Minkowski ansatz

There are two examples of correct polynomials that don’t fit into ansatz stated below.
I think I have shown these examples to Alessio in April.

More info in the notes of my talk (page 2, polynomials w_1 and w_2): pdf (or follow the link from here).

Both examples are degenerations of projective space P^3 (grdb[547386]).

Period sequence 12
First 10 period coefficients: [1, 0, 0, 0, 24, 0, 0, 0, 2520, 0]
The PF operator has N=3, r=4
This sequence has a smooth toric Fano representative
It arises from the following polytopes [(PALP id, grdb id, smoothness)]:
(0, 547386, ‘smooth’)
The PF operator for this sequence is:
256*t^4*D^3 + 1536*t^4*D^2 + 2816*t^4*D + 1536*t^4 – D^3

So we start from the familiar Laurent polynomial
w = x + y + z + \frac{1}{xyz}

I. Argument against “lattice” decomposition.

a. make monomial transformation
m_1: (x,y,z) \to (xz,yz,\frac{1}{z})

w goes to w_1 = z (x+y) + \frac{1}{z} (1+ \frac{1}{xy})

b. mutate by
f_1: (x,y,z) \to (x,y,\frac{x+y}{z})

w_1 goes to \hat{w_1} = \frac{1}{z} + z (x+y) (1 + \frac{1}{xy})

Since \hat{w_1} is derived from w by transformation from group SCr_3 it is a mirror for projective space P^3 (“weak Landau-Ginzburg model” in Przyalkowski’s notations).

Newton polygon of \hat{w_1} is fan polytope of a Gorenstein toric variety – (grdb[544357]).
This variety is anticanonical cone over smooth quadric P^1 \times P^1, i.e. a cone over section of \nu_2(P^3) (P^3 embedded into P^9 by complete linear system of quadrics),
and hence it is a geometric degeneration of P^3.

Consider the quadrangular face, corresponding to a singular point. Restriction of \hat{w_1} to this face is u = z (x+y) (1 + \frac{1}{xy}) = z ( x + y + \frac{1}{x} + \frac{1}{y} ). It is not friendly to Minkowski ansatz’s condition of lattice Minkowski decomposition (this is exactly the example of Minkowski decomposition that is not a lattice Minkowski decomposition given in the definition of the ansatz).

II. Argument against “admissible triangles” and decomposing polytopes completely.

Example w_2 from the same notes.

This one is degeneraiton to P(1,1,2,4) (grdb[547363]).

By monomial transformation

(x,y,z) \to (x,yx,z)

transform w to

w_2 = z + y (x+1) +\frac{1}{z x y^2}

then by mutation

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

transform w_2 to

\hat{w_2} = z + y + \frac{(1+x)^2}{z x y^2}

P(1,1,2,4) is embedded as a quadric in P(1,1,1,1,2) by linear system O(2), so it is a degeneration of a general quadric in this space i.e. P^3.
This variety P(1,1,2,4) is also the anticanonical cone over singular quadratic surface P(1,1,2).

Restriction to the face equivalent to this surface is equal to
\frac{(x+2+1/x)}{y} + y, so it is not admissible.

Z. How to tune the ansatz?

Universal fix:
allow change of the lattice after creating some of the good polynomials

less universal:
a. Allow non-lattice Minkowski decompositions
b. Increase the set of admissible figures

Update on June 22:
III. Examples further beyond

By combining technique from examples in this and previous post we can construct some more sophisticated
mirrors for Tom and Jerry. These mirrors should correspond to degenerations of these guys to Gorenstein cones over singular (Gorenstein or not) del Pezzo surfaces of degree 6.
I’ll write only numerical details and maybe will provide some geometry later in the comment.

Start from a honeycomb and
u = x+y+\frac{y}{x}+\frac{1}{x}+\frac{1}{y} +\frac{x}{y}

Using cluster transformations it can be transformed to mirrors constructed from Gorenstein toric degenerations of del Pezzo surface S = S_6.

first to pentagon

u_5 = y + x + \frac{1}{x} + \frac{(1+x)^2}{xy}

then to quadruple

u_4 = \frac{1}{xy} + \frac{2}{x} + \frac{2}{y} + \frac{x}{y} + \frac{y}{x} + y

then to triangle

u_3 = xy + 2x + \frac{x}{y} + \frac{3}{y} + \frac{3}{xy} + \frac{1}{x^2y}

The triangle is fan polytope of Gorenstein weighted projective plane P(1,2,3).

We can mutate it further to get non-Gorenstein weighted projective plane P(1,3,8)

u' = y + 3x + 3\frac{(x+1)^2}{y} +\frac{(x+1)^4}{y^2}

Then we choose G equal to 2 or 3
and take

w = z + \frac{u+G}{z}

This will be weak mirror for Jerry or Tom,
all underlying toric threefolds are Gorenstein.

Last two are P(1,2,3,6) (grdb[547331]) and P(1,3,8,12) (grdb[547474]).

Triangles may be Minkowski decomposable only when they are multiples of smaller triangles, which is not the case in these examples.

Altmann’s results on relations between Minkowski decompositions and deformations does not apply here since we have non-isolated singularity (it is a cone over already singular space).

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.