Fano Varieties and Extremal Laurent Polynomials

Beyond Minkowski ansatz

June 18, 2010, 1:10 pm

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
AND/OR
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).

Tags: cluster, example, geometry
Posted by: Sergey Galkin | Comment

Unsections/Cones and “Tom vs Jerry” ambiguity

June 18, 2010, 4:00 am

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)]:

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

and

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

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:

————
[tom]

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

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

————
[jerry]

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

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.

Tags: cluster, example, geometry, theory
Posted by: Sergey Galkin | 1 Comment

Should start groping for 5-dim’l cousin of V_22

June 16, 2010, 1:39 pm

$V_{22}$ can be characterized as the smooth Fano of index 1 with minimal cohomology. One way to construct its D3 is by assigning 1’s to the vertices of a terminal self-dual polytope.

Are there terminal self-dual polytopes in dim 5 that produce D5’s whose matrix entries are positive integers?

Am I right in believing that one can search for self-duals considerably more easily than for arbitrary reflexives?

Posted by: Vasily | 3 Comments

Slicing the data

May 18, 2010, 9:42 pm

Sergei: I sliced the Minkowski period data as you requested, here.

Update: here is a new version which also contains the Picard ranks.

Posted by: Tom | 8 Comments

Correction to the new ansatz

May 14, 2010, 7:41 pm

My earlier post describing the new ansatz for constructing extremal Laurent polynomials contained a mistake. This was right at the end, where I described how to go from a list of Laurent polynomials for each facet to a list of Laurent polynomials for the original polytope. I have corrected the post below. Sorry!

Posted by: Tom | Comment

D3 forms for the Minkowksi ansatz

May 12, 2010, 4:46 pm

As mentioned below, the Minkowksi ansatz gives rise to 23 D3 forms. The connection matrices and various other useful data about these D3 forms are here. This list contains 15 of the 17 minimal Fano varieties and several other currently-mysterious examples. Some of these look like they could be G-Fanos. The original data, without annotations, are here: MinkowskiD3Matrices.

Sergei: please can you take a look at this list and see whether you can fill in the gaps.

Tags: D3, data, extremal
Posted by: Tom | Comment

New data

May 12, 2010, 4:32 pm

I ran the Minkowski ansatz for all 4319 3-dimensional reflexive polytopes, calculating the first 100 power series coefficients of the principal period for each of them. The results were as follows.

3025 of the polytopes had a non-empty list of Minkowski polynomials, and 670 of them had more than one Minkowksi polynomial.

There were 165 distinct sequences of power series coefficients. I then looked for differential operators $P(t,D)$ annilihating the principal period such that $P(t,D)$ has degree $N$ in $D$ and $r$ in $t$ , where $N \leq 5$ and $(N+1)(r+1)\leq 90.$ The results were:

N=3:	23 examples in total

r=2:	3 examples
r=3:	2 examples
r=4:	18 examples

N=4:	49 examples in total

r=7:	5 examples
r=8:	3 examples
r=9:	14 examples
r=10:	27 examples

N=5:	10 examples in total

r=11:	1 example
r=12:	2 examples
r=13:	0 examples
r=14:	7 examples

So in particular we found 23 D3 forms.

Tags: data
Posted by: Tom | 3 Comments

A new ansatz for extremal Laurent polynomials

May 12, 2010, 4:14 pm

This post describes a new method for generating Laurent polynomials in 3 variables. Many of these Laurent polynomials are extremal or of low ramification, and they include the extremal Laurent polynomials mirror to 15 of the 17 minimal Fano 3-folds. We call this method the Minkowksi ansatz.

Let $P$ be a 3-dimensional reflexive polytope. We will construct a Laurent polynomial with Newton polytope equal to $P$ , or in other words we will explain how to assign a coefficient to each integer point in $P$ . This goes as follows.

Lattice Minkowski sums

We say that a polygon $Q$ is the lattice Minkowski sum of polygons $R$ and $S$ if and only if both:

$Q = R + S$ , so that $Q$ is the Minkowski sum of $R$ and $S$ as usual
the integer lattice in $Q$ is the sum of the integer lattices in $R$ and $S$ .

Note that any of the the polygons $Q, R, S$ here are allowed to be degenerate.

Examples: here are two lattice Minkowksi decompositions $P = Q+R$ of a hexagon:

A lattice Minkowksi decomposition

Another lattice Minkowski decomposition

Note that the same lattice polygon can have more than one lattice Minkowski decomposition. Note also that the first decomposition here is not a complete decomposition into lattice-Minkowksi-irreducible pieces, because the square $Q$ can be further decomposed as the sum of a vertical and a horizontal line.

This is not a lattice Minkowski decomposition

The example above is not a lattice Minkowski decomposition, because the lattice in $P$ is not the sum of the lattices in $Q$ and $R$ . In fact $P$ is lattice Minkowski irreducible.

Decompose the facets into irreducible pieces

There are 4319 3-dimensional reflexive polytopes. These polytopes contain a total of 344 distinct facets, where we regard two facets as the same if and only if they differ by a lattice-preserving automorphism. Of these facets, 79 are lattice Minkowski irreducible. These 79 facets are also the non-degenerate polygons which occur when the 344 total facets are decomposed into lattice Minkowksi irreducible pieces. Of those 79 facets, exactly 8 contain no interior lattice points. Those 8 triangles, which we call admissible triangles are all of type $A_n$ :

The eight admissible triangles

In other words, the cones over these triangles give affine toric varieties that are transverse $A_n$ singularities, for $1 \leq n \leq 8$ .

The ansatz

Given a 3-dimensional reflexive polytope $P$ , we construct a possibly-empty list of Laurent polynomials as follows. For each facet $F$ of $P$ , decompose $F$ into lattice-Minkowksi-irreducible pieces in all possible ways. Discard any such decomposition of $F$ which contains a non-degenerate polygon that is not an admissible triangle. Any remaining decomposition of $F$ will consist of line segments and admissible triangles. To this decomposition we associate a Laurent polynomial which is the product of certain basic Laurent polynomials corresponding to line segments and to admissible triangles. The basic Laurent polynomials for admissible triangles are:

The coefficients of the basic Laurent polynomials for admissible triangles.

and so on for the remaining admissible triangles. The basic Laurent polynomials for line segments are:

The coefficients of the basic Laurent polynomials for line segments

and so on for other line segments.

So now, for each facet $F$ of $P$ , we have a list $L_F$ of Laurent polynomials; this list will be empty if $F$ cannot be written as a lattice Minkowksi sum of line segments and admissible triangles. In other words for each facet $F$ we have list of ways of assigning coefficients to each integer point in $F$ . We seek a list of Laurent polynomials with Newton polytope equal to $P$ , or in other words a list of ways of assigning coefficients to each integer point in $P$ . This is produced by assigning the coefficient zero to the origin (which is the only interior point of $P$ ) and then assigning coefficients to the integer points on facets of $P$ as specified in the facet lists (but amalgamated in all possible ways, so if there are $n_F$ elements in the list for facet $F$ then the number of elements in the list for $P$ is $\prod_{\text{facets F}} n_F$ ).

Points to Note

This ansatz almost generalizes the earlier recipes given by Pryjzalkowski and Galkin, but differs a little because of the difference between Minkowski decomposition and lattice Minkowksi decomposition.
Altman has studied the deformation theory of affine toric varieties and discovered a close connection with Minkowski decompositions. Since we expect to find the local system associated to an extremal Laurent polynomial $f$ as a piece of the quantum cohomology local system associated to a smoothing of the Newton polytope of $f$ , this is encouraging. But note that Minkowski decomposition and lattice Minkowksi decomposition are not the same.
We suspect that if $P$ is a 3-dimensional reflexive polytope containing a facet with no admissible lattice Minkowski decompositions then the toric variety corresponding to $P$ does not smooth. More on this later.
This ansatz also fits well with Kouchnirenko’s criterion for a Laurent polynomial to be degenerate.

(I learned this last point from Hiroshi.)

Tags: theory
Posted by: Tom | 7 Comments

Hello World!

May 12, 2010, 1:22 pm

The blog is live. LaTeX is working:

$\dim V = 27$

All is well in the world.