Posts tagged ‘example’

## Quantum Lefshetz for non-split bundle via “abelian/non-abelian correspondence”.

Bumsig Kim explained me how their beautiful theory provides a tool for computing J-series of $V_{22}$ and many other Fano threefolds.

The computation for $V_{22}$ can be reproduced by the following pari/gp code (I omit checking that mirror map is almost trivial):

 N=9 o = O(t^(N+1)) h(n) = sum(k=1,n,1/k) hh(n) = sum(k=1,n,sum(l=k+1,n,1/k/l)) gg(x, p) = (x!*(1+h(x)*p+hh(x)*p^2+O(u^4))) simplemirrormap(F) = F * exp(-polcoeff(F,1,t)*t) reg(F) = sum(n=0,N,t^n*polcoeff(F,n,t)*n!)+O(t^(N+1)) period(F) = reg(simplemirrormap(F))

 gp > v22 = period( polcoeff( sum(a=0,N,sum(b=0,N,sum(c=0,N,o+ t^(a+b+c)* ( gg(a+b,u*(A+B)) * gg(a+c,u*(A+C)) * gg(b+c,u*(B+C)) )^3 / (gg(a,u*A) * gg(b,u*B) * gg(c,u*C) )^7 *(c-b+u*(C-B))*(c-a+u*(C-A))*(b-a+u*(B-A)) ))) + O(u^4) , 3,u) /(C-B)/(C-A)/(B-A) + o ) 

%2 = 1 + 12*t^2 + 60*t^3 + 636*t^4 + 5760*t^5 + 58620*t^6 + 604800*t^7 + 6447420*t^8 + O(t^9) 
Indeed, period sequence 17.
Note that $gg(x,p) = \frac{\Gamma(1 + x + \frac{p}{z})}{\Gamma(1 + \frac{p}{z})} + o(\frac{1}{z^3})$ is the familiar Gamma-factor with $u=\frac{1}{z}$.

——————————————————-

So, how does it works?

Consider 3-dimensional vector space $U = \CC^3$ with a fixed base, 7-dimensional vector space $V = \CC^7$, and space $M = \CC^{21}$ of 3×7 matrices $M = Hom(U,V) =Hom(\CC^3, \CC^7)$

Group $G = Aut(U) = GL(3)$ acts on M by left multiplication. It has a subgroup $T = (\CC^*)^3$ of diagonal matrices and one may restrict the action to this smaller subgroup.

Let $M_{na}$ be the subset of matrices of maximal rank and $M_{ab}$ be the subset of matrices with non-vanishing rows, $M_{na}$ is an open subset in $M_{ab}$.

Consider quotient spaces $X_{na} = M_{na} / G$ and $X_{ab} = M_{ab} / T$. Note that $X_{na} = Gr(3,V) = Gr(3,7)$ and $X_{ab} = (\PP(V))^3 = (\PP^6)^3$.

Since T is a subgroup of G, there is a natural rational map $\pi: X_{ab} -> X_{na}$: a triple of points in $\PP^6$ is sent to their linear span.

Weyl group (symmetric group $S_3$) acts on $X_{ab} = (\PP^6)^3$ and hence it acts on the cohomology $H(X_{ab}) = H((\PP^6)^3)$, so cohomology space is decomposed into representations of $S_3$.

[I’ll omit the part of the story with the partial flag space and non-holomorphic map].

1. It turns out that cohomology $H(X_{na})$ can be identified with antisymmetric part of $H(X_{ab})$ as a graded vector space (with grading shifted by 3).
Explicitly, $H(X_{ab})$ is generated by 3 pullbacks $H_1, H_2, H_3$ of hyperplane sections on $\PP^6$; cohomologies of Grassmanian are known to be quotient of symmetric polynomials. Vector space of anti-symmetric polynomials is obtained from vector space of symmetric polynomials via multiplication by anti-symmetric polynomial of the smallest degree $Formula does not parse: \Delta = \prod_{i2.Also we can compare vector bundles on [latex]X_{ab}$ and $X_{na}$ by pulling them back to M, and considering as G-linearized.
It turns out that universal bundle U over Gr(3,7) decomposes into sum of 3 line bundles on $X_{ab}: U <-> O(1,0,0) \oplus O(0,1,0) \oplus O(0,0,1)$.
So $O_{Gr(3,7)}(1) <-> O(1,1,1)$ and $U^*(1) <-> O(1,1,0) \oplus O(1,0,1) \oplus O(0,1,1)$.

3. On domain of $\pi$ one may define a relative tangent bundle $T_{\pi}$ (“traceless” part of $Hom(U,U)$). It turns out this vector bundle can be extended as a split vector bundle to whole $X_{ab}$: $T_{\pi} = O(1,-1,0) \oplus O(-1,1,0) \oplus O(1,0,-1) \oplus O(-1,0,1) \oplus O(0,1,-1) \oplus O(0,-1,1)$.
Consider “square root” of relative tangent bundle $t_{\pi} = O(0,-1,1) \oplus O(-1,0,1) \oplus O(-1,1,0)$.

4.Recall that Fano threefolds $V_{22} = V_{na}$ are sections of homogeneous vector bundle $E = 3 U^*(1)$ on $Gr(3,7)$. Comparision (2) shows these threefolds has 9-dimensional abelianizations $V_{ab}$ — complete intersections of $9$-dimensional split bundle $E_{ab} = (O \oplus O \oplus O) \otimes (O(1,1,0) \oplus O(1,0,1) \oplus O(0,1,1))$ on $X_{ab}$.

5. Abelian/non-abelian correspondence is similar for pairs $X_{ab}/X_{na}$ and $V_{ab}/V_{na}$.
J-series for Gr(3,7) can be obtained as twisted by relative tangent bundle $T_{\pi}$ I-series for $e(t_{\pi}) \cup I_{X_{ab},T_{\pi}}$ after the comparision of cohomologies described in (1).
Similarly, J-series for $V_{22}$ can be obtained via mirror map from twisted by $T_{\pi} + E_{ab}$ I-series $e(t_{\pi}) \cup I_{X_{ab},T_{\pi}+E}$ after the “pullbacked” comparision (1).

6. The sign comes from considering closely the Gamma-factor for relative tangent bundle $T_{\pi}$. Note that fibers of abelian/non-abelian correspondence are in some sense holomorphic symplectic (relative tangent bundle contains both O(D) and O(-D)), so they behave like varieties with trivial canonical class.
Consider the factor $\frac{\Gamma(1+D+d) \Gamma(1-D-d)}{\Gamma(1+D)\Gamma(1-D)}$. Since $\Gamma(1+x) \Gamma(1-x) = \frac{\pi x}{sin (\pi x)}$ and $sin (\pi (x+d)) = (-1)^d sin(\pi x)$ we have $\frac{\Gamma(1+D+d) \Gamma(1-D-d)}{\Gamma(1+D)\Gamma(1-D)} = (-1)^d \frac{d+D}{D}$.

The same method can also be applied to complete intersections of homogeneous bundles in orthogonal isotropic and symplectic isotropic Grassmanians, since these Grassmanians themselves are just sections of some homogeneous bundles (wedge or symmetric powers of universal bundle) on ordinary Grassmanians of type A. Also this can be uprgaded to treat different blowups of these varieties. In particular, in the comments to this post I compute J-series for Fano threefolds $V_5$, #2.14, #2.17, #2.20, #2.21 and #2.22.

Also Bumsig points out that one can express the ab/non-ab twist as a differential operator applied to abelian multi-parameter J-function (basically, just Vandermonde $\prod_{i>j} (\frac{d}{d q_i} - \frac{d}{d q_j})$. This interpretation is more useful for dealing with Frobenius manifolds.

References:
Gromov-Witten Invariants for Abelian and Nonabelian Quotients by Aaron Bertram, Ionut Ciocan-Fontanine, Bumsig Kim
The Abelian/Nonabelian Correspondence and Frobenius Manifolds by Ionut Ciocan-Fontanine, Bumsig Kim, Claude Sabbah
Quantum cohomology of the Grassmannian and alternate Thom-Sebastiani by Bumsig Kim, Claude Sabbah

## A riddle wrapped in a mystery inside a balls-up

Consider #2 on the Mori-Mukai list of rank-3 Fano 3-folds.  This has been giving us some difficulty, which I have now resolved.  We were making a combination of mistakes.  Mori and Mukai describe the variety $X$ as follows.

A member of $|L^{\otimes 2} \otimes_{\cO_{\PP^1 \times \PP^1}} \cO(2,3)|$ on the $\PP^2$-bundle $\PP(\cO \oplus \cO(-1,-1)^{\oplus 2})$ over $\PP^1 \times \PP^1$ such that $X \cap Y$ is irreducible, where $L$ is the tautological line bundle and $Y$ is a member of $|L|$.

Our first mistake, as Mukai-sensei pointed out in an email to Corti, was using the wrong weight convention for projective bundles.  Mori and Mukai use negative weights, so the ambient $\PP^2$-bundle $F$ is the toric variety with weight data:
$\begin{array}{cccccccc} x_0 & x_1 & y_0 & y_1 & t_0 & t_1 & t_2 & \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & \\ 0& 0 & 1 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 0 & 0& 1 &1 & 1 & \end{array}$
For later convenience we change basis, expressing $F$ as the toric variety with weight data:
$\begin{array}{cccccccc} x_0 & x_1 & y_0 & y_1 & t_0 & t_1 & t_2 & \\ 1 & 1 & 0 & 0 & -1 & 0 & 0 & A \\ 0& 0 & 1 & 1 & -1 & 0 & 0 & B \\ 0 & 0 & 0 & 0& 1 &1 & 1 & C \end{array}$
$X$ is a section of $|B+2C|$.

Our second mistake was failing to accurately account for the fact that although $X$ is Fano, the bundle $A+C$ (which restricts to $-K_X$ on $X$) is only semi-positive on $F$.  Thus we are in the situation described in this post and, in the notation defined there, we have:
$\begin{cases} F(q) = 1 \\ G(q) = 2q_3+6q_2 q_3 \\ H_1(q) = \sum_{b>0} {(-1)^{b} \over b} q_2^b = {-\log(1+q_2)} \\ H_2(q) = {-\log(1+q_2)} \\ H_3(q) = \log(1+q_2) \end{cases}$
and hence:
$\begin{cases} q_1 = {\hat{q}_1 \over 1 - \hat{q}_2} \\ q_2 = {\hat{q_2} \over 1 - \hat{q_2}} \\ q_3 = \hat{q}_3 (1-\hat{q_2}) \end{cases}$
Thus the cohomological-degree-zero part of the J-function is:
$\exp({-2}\hat{q}_3(1-\hat{q}_2)-6\hat{q}_2 \hat{q}_3) \sum_{a,b,c\geq 0} \hat{q}_1^a \hat{q}_2^b \hat{q}_3^c(1-\hat{q}_2)^{c-b-a} {1 \over z^{a+c}} {(b+2c)! \over a!a!b!b!c!c!(c-b-a)!}$
and setting $\hat{q}_1 = t$, $\hat{q}_2 = 1$, $\hat{q}_3 = t$, $z=1$ yields:
$\exp(-6t) \sum_{a,b\geq 0} t^{2a+b} {(2a+3b)! \over a!a!b!b!(a+b)!(a+b)!}$
Regularizing this gives:
$I_{reg}(t) =1+58 t^2+600 t^3+13182 t^4+247440 t^5+5212300 t^6+111835920 t^7+ \cdots$
As Galkin conjectured, this is period sequence 97.

## 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
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

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

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

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)]:

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

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.