Mori flips of type A Gavin Brown and Miles Reid September 7, 2006 Contents 1 Introduction 2 1.1 Mori flips ............................. 2 1.2 Overview ............................. 2 1.3 An example of the construction ................. 3 2 Graded rings and flip diagrams 5 2.1 The basic correspondence .................... 5 2.2 Canonical covers of Mori flips .................. 5 2.3 Two partial resolutions of a Du Val singularity ........ 5 3 Toric geometry 5 3.1 Basic decompositions of the plane ................ 6 3.1.1 A classification of some decompositions ........ 7 3.2 Affine toric Gorenstein threefolds ................ 10 3.3 Long rectangles .......................... 12 3.3.1 Long rectangles as pictures ............... 12 3.3.2 The equations ....................... 13 3.3.3 The dual continued fractions .............. 13 3.3.4 The Gorenstein projections ............... 13 3.4 The big variety .......................... 13 4 Equivariant deformation of Gorenstein varieties 13 4.1 Flips in low codimension ..................... 14 4.2 Long rectangle pairs ....................... 14 4.2.1 What is a long rectangle pair? ............. 14 4.2.2 Pairs from flips ...................... 14 4.3 Long rectangles as glued staircases ............... 15 4.4 The numerical classification of flips ............... 18 4.4.1 Pairs that flip ....................... 18 4.4.2 Proof of the numerical classification .......... 22 5 Gorenstein projection 23 5.1 Special case of the theory .................... 24 5.2 Cascading pentagrams ...................... 24 5.3 The three primary flip families ................. 26 6 Unprojection 29 6.1 Discussion of the theory ..................... 29 6.2 Models of unprojection ...................... 30 6.2.1 Easy case: unprojection type P 2 × P 2 ......... 30 6.2.2 Hard case: unprojection type P 1 × P 1 × P 1 ...... 32 1
We describe the general theory of canonical C∗ covers of flip diagrams andwork through the special case of classifying Mori flips whose elephant isan An singularity, the so-called Mori flips of type A. It is this particularcalculation that is the main substance of the paper, and in particular our in-troduction of a class of ‘nearly toric’ varieties represented by ‘long rectanglepairs’.
Our classification is not a list of all flips of type A, rather a classifica-tion of certain numerical data that can occur together with a collection ofmethods for constructing flips of each numerical type.1 These constructionsshould correspond to deformation families of flips in a straightforward way,although particular calculations appear prohibitively complicated. The nu-merical data already suggests interesting structures within these flips someof which we study. In particular, we find that codimension is not necessarilya good measure of complexity for flips. More likely is the way that they candecompose into families of distinct projection types.
1.1 Mori flips
2
1.2 Overview
3
1Perhaps this statement is precisely false.2Culture and point of view.3What’s here at the moment is generally organisational garbage waiting to be removed.
Replace with more point of view. Also mention structure of gor rings.
2
Graded rings, C∗ covers, flip diagrams
Quotients of an affine variety by a C∗.As N-graded rings correspond to projective varieties so Z-graded rings
correspond to flip diagrams.Two partial resolutions of a DuVal singularity.Canonical diagrams of 3-folds; toric geometry; hypersurfaces. Gorenstein
property; finite covers and reduction.References: Thaddeus, D&H.We keep the flipping case in mind throughout. While this restriction
is severe in the discussion of the general theory, in a number of differentcontexts it seems to determine with remarkable precision certain classes ofvariety which we can understand and classify by our methods.
Numerical flipping data
Toric Gorenstein 3-folds; long rectangles; basic equations and the full idealof equations; continued fractions; Gorenstein projection.
Pairs of long rectangles as correct deformations; skeleton of equations;codimension 3; basic boundary equations.
Numerical classification; continued fractions and staircase; number the-oretic proof.
Mori flips of type A
Gorenstein projection; cascading pentagrams.Unprojection and codimension 4; Tom and Jerry.Basic equations of a pair of long rectangles; the existence of genuine flips
for each numerical solution.Building block flips and other constructions.We sometimes abbreviate the equations by putting them in special forms.
The appearance of these special forms is not always a mystery, but in generalwe do not have good slogans for the ‘shape’ of the equations of canonicalcovers of flips. The varieties determined by the equations have Gorensteinsingularities of high embedding dimension at the origin. Apart from theBuchsbaum–Eisenbud structure theorem in codimension 3 and a numberof familiar particular examples in higher codimension, we don’t know whatkinds of resolutions to expect for such rings. We seem to encounter annumber of new syzygy structures.
Lists of results
Complete details for all flips codimenions 3–6; lists of equations; specialformats; projection and unprojection details; other favourite examples.
Comparison with [Kollar–Mori]; hyperplane sections of flips.First factorisations of flips.
1.3 An example of the construction
We run through some typical calculations.4 The constituents of this exampleare the following. First we draw a picture, called a long rectangle pair. Then
4This section might actually not be working. The idea is to get some trivia off mychest here in the guise of an example, although that’s not really happening and the textmight appear to be meaningless.
3
we write down the equations of a variety which are derived from the pair.Most of this paper is devoted to explaining the origin of these pictures andworking out how to interpret them and calculate with them. This varietywill be the canonical cover of a flip. So we finish by adding a flipping C∗
action. The construction is then complete.For the time being one can take the relationship between the pictures
and the equations for granted, treating the data of the example as a whole,although anyone versed in Newton polygons, toric geometry or finite quo-tient singularities will see what we are doing.
The name of the example, II(3, 2, 3), refers to the numerical classificationof varieties in section 4.4.1. The subscript is the codimension of the flip.
II9(3, 2, 3)
3 1 0 −12323
243
2323
243
0 −2 2 1
We imagine this picture to be of two vertical rectangles with a few markedpoints on their boundaries labelled by integers.5 The left-hand rectangle iscalled the elephant and the right-hand is the section.
There are some hidden extra labels: α and β at the top left and top rightcorner of the elephant; λ and µ at the bottom left and bottom right handcorner of the section. We give the extra label 0 to all side points in thepicture while the other corners get extra labels calculated by a continuedfraction method from α, β, λ, µ and the labels. In this case, for instance,the top left corner extra label of the section is 19λ+ 18µ. See Section 4.2.1for details of these definitions and calculations.
Together with all its labels, this long rectangle pair is intended to rep-resent equations in 13 variables, x0,. . . , x5, y0,. . . , y4, u, t. The x variablescorrespond to the marked points up the left-hand side of the rectangles(which between them are thus displaying two labels for each variable, al-though there is only a difference at the corners). The y variables correspondto those on the two right-hand sides. For each x or y variable there is abasic equation. The basic equations in the corners are the most important.At x5, the equation is
x4y4 = x35uα + t19λ+18µ.
The x4y4 monomial is the product of the variables adjacent to x5 in therectangles. The x3
5uα monomial is read from the elephant: the power of x5
appearing is the label at x5 while the power of u appearing is the extra labelat x5. The final term in t is calculated from the section: there is a factorx0
5 corresponding to the label, while the power of t appearing is again theextra label on the section at x5.
You notice that this equation is similar to that of a terminal singular-ity before taking a finite quotient, and indeed it will be the equation of asingularity on X− before the flip.
5Is this clearly a pair of rectangles?
4
Basic equations for non-corner variables are harder. For instance, at x1
the equation is in fact
x0x2 = x31 + u5α+12βtµ.
The final term cannot be read directly from the picture but must be cal-culated. In this case, our method of Gorenstein projection and pentagramsmakes the calculation tolerable.
Eventually, at least in calculable cases, the result is the ideal of equationsof a Gorenstein 4-fold in a 13-dimensional affine space, a codimension 9variety as expected. We denote this 4-fold by A. To make a flip we add aC∗ action. We do this by taking the variables as eigencoordinates and givingthem weights in such a way that all the equations are homogeneous withrespect to these weights. It is enough to choose weights for just four variablessince the homogeneity of the equations determines all other weights. Wedefine the weight of u to be 0, that of t to be 1, x0 to be −a and y0 to be −bfor positive integers a and b. These are the choices that we will often make,but beware that a and b will be either determined by other variables alreadychosen, or else satisfy some divisibility conditions. The homogeneity of thebasic equation at x0,
x1y0 = u? + x20tλ
implies that the weight of x1 is a since u? has weight 0 whatever power itis. Notice too that a is exactly λ/2: there are lots of relations between allthe labels and here we have been careless in noting them.
Similarly we find that all other weights are strictly positive. A flip arisesby taking the different C∗ quotients of A. The base of the flip is the cate-gorical quotient while the X+ in the flipping diagram is the relative Projof the negatively weighted variables. As will always be the case, this hastwo patches coming from the two negative variables x0 and y0 which coveran exceptional P(a, b) = P1 having x0, y0 as homogeneous coordinates. Forinstance, the x0 patch is the hypersurface
x1y0 = u? + tλ
(coming from the basic equation above) divided by a finite cyclic stabiliserof order a. This singularity is clearly isolated so is also terminal, as in [2]§6.
The main result of this paper is to associate a long rectangle pair to anyflip and classify those long rectangle pairs which admit a flipping C∗ action.
2 Graded rings and flip diagrams
2.1 The basic correspondence
2.2 Canonical covers of Mori flips
2.3 Two partial resolutions of a Du Val singularity
3 Toric geometry
We identify a certain class of threefolds whose equations correspond closelyto a continued fraction calculation. These will be the canonical covers ofelephants of flips with only An singularities.
5
Everything in this chapter is standard toric geometry, although we ex-press it in even more elementary terms. The point is that three dimen-sional Gorenstein cones correspond to basic subdivisions of the plane. Someproperties of these decompositions correspond to properties elephants of flipdiagrams.
3.1 Basic decompositions of the plane
Let N be the lattice Z2. A basic cone in N is the semigroup span of a pairof vectors n,n′ which form a basis of the lattice.
A basic decomposition of N , or simply decomposition, is a collection ofbasic cones whose union is N and which intersect pairwise in either theorigin or a one-dimensional ray. Note that a decomposition must compriseat least three basic cones.
The collection of primitive vectors n1, . . . , nk which span the basic conesof a decomposition are called the generators of the decomposition. Theyare always written so that any consecutive pair ni, ni+1 generate one of thebasic cones of the partition. The indices are read cyclically so that nk,n1
also bound a basic cone.Each generator nj has an integer label `j , uniquely determined by the
formulanj−1 + nj+1 = `jnj .
A decomposition is determined up to orientation and the action of SL(2,Z)by the sequence of labels. Thus we often denote a decomposition by itssequence of labels [`1, . . . , `k] and consider all representations equivalent.Notice that this notation should be read cyclically so that, for instance,[`k, `1, . . . , `k−1] denotes the same class of decompositions.
A basic decomposition can also be thought of in terms of particularfactorisations of the identity in SL(2,Z). The change of basis matrix fromthe ordered basis n1,n2 to the ordered basis n2,n3 is of the form(
0 1−1 `2
).
Changing basis right round the plane from basic cone to basic cone even-tually results in the change of basis from nk,n1 back to n1,n2. So thiscomposition of basis changes is the identity and(
0 1−1 `1
)(0 1−1 `k
)· · ·(
0 1−1 `2
)=
(1 00 1
).
Conversely, such a factorisation determines a decomposition, again up tothe action of SL(2,Z).
There are two simple methods of creating new decompositions from oldones: blowup and blowdown. If C is a basic cone of a decomposition Pgenerated by two adjacent generators n′ and n′′, the blowup σC of P at Cis the decomposition obtained when C is replaced with its subdivision intotwo basic cones by including the new generator n = n′ + n′′. Clearly thelabel at n is 1, and it is easy to see that the labels at n′ and n′′ are eachincreased by 1.
Conversely, if a generator n has label 1 then it can be removed, the twobasic cones on either side being amalgamated. The effect on the neighbour-ing labels is to lower them both by 1.
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A sequence of blowups σ1,. . . ,σm is called simple if each σi+1 is a blowupof one of the two cones introduced by the blowup σi.
We distinguish some special decompositions: [−1,−1,−1] is denoted P2
and [0, d, 0,−d] for any nonnegative integer d is denoted Fd. Of coursethese decompositions exist and are represented by the toric descriptions ofthe corresponding varieties. In addition, the decomposition [0, 0, 1, 1, 1] iscalled the standard pentagon and the decomposition [1, 1, 1, 1, 1, 1] is calledthe standard hexagon. Both can be constructed easily. These correspond toDel Pezzo surfaces of degrees 7 and 6 respectively.
3.1.1 A classification of some decompositions
[Basic strategy:1. Any ‘linear’ halfspace must contain a generator with label ≤ 1.2. Generators with label >= 1 span an angle of < π; those with label 0
span an angle of π; those with label <= −1 span an angle strictly greaterthan π.
3. A sequence of labels >= 2 span an angle < π.4. P is a subdivision of the plane on n generators. If n = 3 then P = P2.
If n = 4 then P = Fd for some d.5. Case by case study of P having a negative label, a 0 or a 1.]Let P be a decomposition and let B be a basic cone of P . The shed of B
is the triangle which is the compact hull of the origin and the basic latticepoints along the two sides of B. The angle at the origin in the shed of B isidentical to the angle at the origin in B. The shed of P is the union of thesheds of the basic cones comprising P . The roof of P is the boundary of theshed of P .
Denote by ∆(n) the union of the sheds of the two basic cones whichcontain the generator n of P . Let θ(n) be the angle in ∆(n) at the origin.We say that n subtends the angle θ(n) in P . Generally, if n1,. . . ,nr areconsecutive generators then ∆(n1, nr) denotes the union of sheds of basiccones spanned by pairs ni, ni+1 for i = 1, . . . , r−1 and θ(n1, nr) is the angleat the origin of this polygon. Of course, this notation is ambiguous: the twodirections round the origin are not distinguished. When necessary we willsay angle through n2 to fix this choice, although it is usually clear which ismeant.
A halfspace of N is the subset
{n ∈ N | f(n) > 0}
for some linear form f .
Lemma 1 Let P be a decomposition with generators n1,. . . ,nk and corre-sponding labels `1,. . . ,`k.
(a) If n is a generator of P with label ` then ` ≥ 1, respectively ≤ −1,if and only if θ(n) < π, respectively > π. Moreover ` = 0 if and only if theneighbouring generators n′, n′′ satisfy n′ = −n′′.
(b) If nr,. . . ,ns is a sequence of at least three consecutive generators of Pand the labels `r+1,. . . ,`s−1 are all at least 2 then the angle θ(nr, ns) throughnr+1 is less than π.
(c) The generators of P lying in any halfspace cannot all have label atleast 2.
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Consider a single generator n with label at least 2. The shed of P at nis convex when viewed from the origin.
The angle referred to in (a) and (b) is the angle at the origin in someA = ∆(nr, ns). As a polygon, A has a vertex at the origin and two sidescoming from that whose ends are joined by a convex boundary, again viewedfrom the origin. This cannot happen unless the angle at the origin is lessthan π.
Part (a) is now clear and part (c) follows immediately from (b).
Proposition 2 The only decomposition with three generators is P2. More-over, P2 is also the only decomposition which has two negative labels.
The only decompositions with four generators are Fd for nonnegativeintegers d.
The basic construction used for this proof and later is to construct a linethrough the origin and one of the generators. The two halfspaces on eitherside of this line must each contain contain a generator with label at most 1by part (c) of the lemma.
When P is a decomposition having only three generators, this argumentshows that each one subtends a reflex angle so must each have a negativelabel, say −a. So there are three relations of the form n′ + n′′ + an = 0 inN whence each label must be −1 and P is the decomposition P2.
If a decomposition P has two negative labels, they must be adjacentotherwise the angle sum at the origin is bigger than 2π. When P has atleast four generators, the halfspace argument shows that one of the otherlabels is a 1. Contracting the corresponding generator results in anotherdecomposition with two adjacent negative labels. Repeating the argumenteventually results in a decomposition with three generators. This must beP2 so P is some multiple blowup of P2. But since blowups increase adjacentlabels, P can have had at most one negative label in the first place.
Now suppose that P has four generators. The halfspace argument showsthat it has a generator with label less than or equal to 1. If P has a gen-erator with label 1 then contract it. The resulting decomposition has threegenerators so must be P2 and P must have been F1. If P has a generatorwith label 0 then part (a) of the lemma immediately shows that P = Fd forsome d. Finally, if some generator has negative label, the other three mustlie in a single halfspace. So one of them must have label at most one, butit cannot be negative by the previous argument. So P has a label 0 or 1 inany case.
Using the halfplane argument to find labels that are 1 and then repeat-edly contracting them we see that
Corollary 3 Let P be a decomposition with five or more generators. ThenP is a blowup of some Fd.
Proposition 4 Let P be a decomposition with five or more generators allof whose labels are strictly positive. Then P is a blowup of the standardhexagon. If the labels are only nonnegative then P is a blowup of the standardpentagon.
Suppose that P is the decomposition [`1, . . . , `k] with k ≥ 5 where all `iare strictly positive.
The halfplane argument shows that there must be at least three labelsequal to 1 in a ‘Y’ configuration. If there are two adjacent labels 1 they must
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correspond to the only two generators in some halfplane since contractingone of them leads to a 0 label.
We repeatedly blowdown 1s until two are adjacent. Nonpositive labelscan only occur when blowing down a 1 adjacent to something no bigger than1 so they won’t arise from this procedure. This results in a decompositionof the form [1, 1, `3, . . . , `k] where, by the halfplane argument again, at leastone of `4,. . . ,`k is 1.
We choose a generator with label 1 in the halfplane opposite n1,n2 andcontract without creating nonpositive labels until it is also adjacent to an-other 1. If both corresponding generators lie in the same halfspace then theresulting decomposition is of the form [1, 1, `3, 1, 1, `6] where `3 and `6 areboth strictly positive. Contracting to a four generator decomposition, someFd, shows that in fact `3 = `6 = 1 as required.
Otherwise the result is of the form [1, 1, 1, 1, `5, . . . , `k]. If `5 or `k is 1this must already be the standard hexagon. If not there must again be a 1among the remaining labels which we can repeatedly contract until we reachthe hexagon.
If P had had some 0 labels, one sees immediately that there can be eitherone 0 label or two adjacent ones. A single blowup next two or between themmakes a decomposition with all positive labels. Moreover, the argumentabove doesn’t ever contract these two or three generators so they remain inthe hexagon that results from blowing down. But now the extra generatorcan be blown down resulting in the standard pentagon. Indeed, the sequenceof blowdowns did not rely on this extra generator at all so it might as wellnot have been blown up in the first place.
Let P be a decomposition with label sequence [`1, . . . , `k]. A subsequenceof big labels, or a big subsequence, is a maximal subsequence of consecutivelabels [`r, . . . , `s] each of which is at least 2. Note that the label sequenceis read cyclically so that r could be bigger than s and maximality must beinterpreted in this sense.
By considering the effect of a blowup on the labels we see that
Corollary 5 Let P be a decomposition with five or more generators all ofwhose labels are nonnegative. If P has no more than two subsequences ofbig labels then it is a simple sequence of blowups of a cone in the standardpentagon or standard hexagon.
A basic cone is called positive if neither of the labels on its boundarygenerators is negative.
Proposition 6 Let P be a decomposition with five or more generators, oneof which has a negative label, but no more than two subsequences of biglabels. Then P is a simple sequence of blowups of a positive cone of someFd possibly with a blowup in the opposite nonpositive cone.
The possible pictures are6
−d 1 −(d− 1)| \ || −− + and | −− +
/ | ∗ / | ∗0 + 1 +
6Temporary pictures of fans.
9
where + is some positive label, ∗ indicates a simple sequence of blowups andd ≥ 1.
Suppose that P is the decomposition [`1, . . . , `k] with k ≥ 5 and that` = `1 < 0 is the minimum of the `i. We can also assume that `2 ≤ `k.
If `1 < 0 then since the angle θ(n1, nk) is less than π one of `2,. . . ,`kmust be 1. Contracting this preserves the condition that both `, `2 benegative so one can continue the argument. Eventually the result has onlyfour generators, two of which have negative labels which is impossible.
If `2 = 0 then the remaining angle requires `3 > 0. It cannot be 1for contracting it would produce the previous case of two adjacent negativelabels again which was impossible. Since P has at least 5 generators, allthe remaining labels are positive and at least one must be 1. We claim thatthere is a label 1 not adjacent to `. Indeed if the label adjacent to ` is 1,we can contract it, continuing this until the adjacent label is 0 or ≥ 2. Inthe former case there must have been a 1 before the final blowdown while inthe latter the required 1 comes from the halfplane argument. Blowing downthis internal 1 is the inductive step since this operation preserves the caseconditions.
Note that there cannot be two such internal 1s: they could not be ad-jacent for contracting one would leave 0 but the angle is too small for that;if they are not adjacent there must be at least three subsequences of biglabels.
If `2 ≥ 2 we use the halfplane through `. This puts a 1 on either side.But these must be adjacent otherwise there would be at least three bigsubsequences. But that is impossible because the angle is too small.
The remaining subcase is `2 = 1. As before, we repeatedly blowdownthis adjacent 1 resulting in a label either 0 or ≥ 2. Do the same for `k whichis positive by assumption. The resulting decomposition has a negative labelone of whose adjacent labels must be 0 by the previous case. Both of themcannot be 0 since that results in at least three big subsequences. In addition,the nonzero side must have a label 1. But now there are at least three bigsubsequences unless the blowdown to 0 was simply of the form of contractingone of two adjacent 1s. This is the second case in the statement.
3.2 Affine toric Gorenstein threefolds
Let k be any field, M be the lattice Z3 and MR be the vector space M ⊗Rnaturally containing M . A cone Σ is defined to be a lattice cone in MR whichhas a vertex at the origin and for which the corresponding toric variety XΣ
is Gorenstein. The semigroup Σ∩M has a unique minimal set of generatorscontaining at most, and typically, one not lying on the boundary of Σ. Wewrite addition in the lattice multiplicatively, writing x2 rather than x + x,for instance.
It is well-known that
Lemma 7 A lattice cone with vertex Σ is Gorenstein if and only if it hasa unique internal point u such that for any pair of consecutive boundarygenerators x, y, the triplet u,x,y form a basis for the lattice.
Suppose that the boundary generators are x1,. . . ,xm written cyclicallyround the boundary and u is the internal Gorenstein generator. The product
10
of two generators xi−1xi+1 is some expression of the form xaii uαi . We say
that ai is the label at xi and that αi is the extra label at xi. The equation
xi−1xi+1 = xaii uαi (1)
is called the basic equation at xi. The cone Σ is locally planar near xiprecisely when the extra label at xi is zero.
Using the same notation the change of basis matrix from the orderedbasis xi−1,xi,u to the ordered basis xi,xi+1,u is of the form 0 1 0
−1 ai αi0 0 1
.Again note that notation is multiplicative so that multipying the basis vectorby this matrix yields
xi+1 = x−1i−1x
aii u
αi . (2)
As for plane decompositions, changing basis right round the boundary ofthe cone, finally making the change xm,x1,u to x1,x2,u exhibits the identitychange of basis matrix as a product of matrixes of this form. In particular,the principal 2× 2 submatrix of this factorisation is of the form
∏i
(0 1−1 ai
)=
(1 00 1
).
This determines a basic decomposition of the plane in which the labels,denoted `i in the last section, are equal to the labels ai here.
It is not true that conversely a decomposition determines a Gorensteincone, but it nearly is: the extra data required is the α integers in the 3-dimensional change of basis matrixes. These cannot be chosen entirely in-dependently: any pair of them are determined by the others according tothe factorisation. Also, the assumption that the cone has a vertex and thatu is an internal point mean that labels `i which are less than or equal to 0cannot have corresponding α = 0. This can be strengthened to all `i ≤ 1if we assume minimality of the generating set, which of course we do fromnow on.
The toric variety XΣ corresponding to Σ will be three dimensional andnaturally embedded in affine space on which the generators of Σ are inde-pendent coordinate functions. The equations of this embedding will includethe basic equations at each boundary generator as well as others arisingfrom syzygies.7
The following theorem is the starting point for our constructions.
Theorem 8 Let Σ be the cone of an affine toric Gorenstein 3-fold in codi-mension at least 3 which covers a reduced flip diagram of partial resolutionsof a Du Val singularity.
Then Σ is a cone on a rectangle. Its boundary lattice generators are allcontained in two opposite sides. The labels on Σ are of the form [−d, 0, d+1, `4, . . . , `n] for some positive integers d and `i. If n = 5 then `4 = `5 = 1,otherwise exactly one of the `i is equal to 1 and the labels blowdown uniquelyas a sequence to the n = 5 values without altering d.
7Say something about C∗ actions.
11
Proof outline:1. C∗ action is hyperplane through monomial cone.2. Exceptional curves in the flip diagram are faces of the polygon; more-
over there must be no generators along those sides (by reduced hypothesis).So corresponding decomposition has no more than two subsequences of biglabels.
3. Hypothesis that both sides of elephant contain a P1 so looking atrectangles cut by C∗. In any case, since flip is reduced and labels ≤ 1cannot appear on sides the corresponding decomposition must be of thefirst kind in Proposition 6.
4. Corners must include 0, −d and 1 as labels, so labels are fixed up tosymmetries and the choice between the two possibilities for the position ofthe 1.
3.3 Long rectangles
Long rectangles, which we describe in this section, are the fundamental unitof this paper. We discuss the basic issues close to the heart of a long rect-angle: its dual pair of continued fractions, its equations, its Gorensteinprojections. These all come directly out of Theorem 8. Later we introducecoupled pairs of long rectangles in which the the resolution of similar issuesresults in the explicit construction of Mori flips.
3.3.1 Long rectangles as pictures
Suppose Y is a toric affine Gorenstein 3-fold. We invariably represent Yby the projectivisation of its monomial cone.8 According to Theorem 8 areduced Du Val cover9, cB, is an affine Gorenstein 3-fold represented by arectangle with the additional condition that two opposite sides, which wealways take to be the top and bottom, contain no generators other thanthose at the corners. So the monomial cone of cB is of the form
CONE PICTURE
Rather than drawing the rectangle we usually draw a more schematicpicture replacing the generator blobs with their corresponding labels. Theonly nonzero extra labels are those at the corners. The relations amongthem coming from the factorisation in SL(2,Z) mean that it is sufficient torecord just two of them. In the example below we indicate that the extralabels at the top two corners are α and β.
α 1 2 β33
2
0 −2
Any such picture is called a long rectangle. As a piece of data, it is com-pletely equivalent to the equations of a certain type of toric variety whichwe calculate next.
8Compare with Altmann.9Think up a better name.
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3.3.2 The equations
We illustrate the general calculation of equations of toric varieties in thisexample. Starting at the top, where the extra labels are written, we fix abasis of the lattice: x = x3, y = y2, u. The basic equations give expressionsfor the other generators in terms of these by an inductive procedure. Forexample, the labels say that the basic equation at y2 is x3y1 = y2
2uβ. So
in the chosen basis y1 = x−1y2uβ. Working down the rectangle we getexpressions for all boundary generators as follows.
x = x3
xy−1uα = x2
x2y−3u3α = x1
x5y−8u8α = x0
y2 = yy1 = x−1y2uβ
y0 = x−2y3u2β
It is easy to see that the equations are generated by those of the form ξη = . . .where ξ,η are distinct boundary generators. These equations can be readfrom the array. For instance,
x0y1 = x4y−6u8α+β = x21u
2α+β
which agrees with the label at y0 and determines the extra label there. Anon-basic equation is
x2y0 = x−1y2uα+2β = y1uα+β.
10
3.3.3 The dual continued fractions
Immediate consequence of Theorem 8; explain Riemenschneider staircases.
3.3.4 The Gorenstein projections
Unique sequence of Gorenstein projections of a long rectangle; unprojectionideal; first example of a pentagram.
3.4 The big variety
Say something about the big variety with A = uα, B = uβ.It has a monomial cone which is easy to draw. cf Altmann, although
evidently non-isolated singularities.The Gorenstein projection can be seen nicely.
4 Equivariant deformation of Gorenstein varieties
Given an elephant coming from a single long rectangle, we want to makea graded one parameter deformation so that the resulting flip diagram isa Mori flip. In this section we show that there is a second long rectangleassociated to such a deformation. We classify these long rectangle pairsand consider this statement to be the numerical solution to the problem ofconstructing a flip.
10Say something about easy tricks as for quotient singularities.
13
4.1 Flips in low codimension
COD 2: Comment on long rectangle; isolatedness leads to certain monomialsappearing; second rectangle.11
Workout of the whole C∗ procedure and our point of view using codimen-sion 3 as an example: affine model of an analytic phenomenon; modellinga representative of a flipping singularity class; more-or-less independence ofthe C∗ action; background distinction between 4-fold cover and 6-fold ‘big’variety.
Fix a favourite matrix: the usual long rectangle pair will appear fromthe matrix
y1 −L −A y0
x2 −M −xd1x1 −B
x0
.Comment on second long rectangle at A = B = 0.12
Something about PENTAGRAMS since this is an example.13 The namepentagram comes from the fact that the five equations of the codimension3 flip correspond to the five sides of a pentagram whose points are the fiveboundary generators of the rectangle.
PICTURE of a pentagram.
4.2 Long rectangle pairs
4.2.1 What is a long rectangle pair?
Definition of long rectangle pair; section/elephant plus remark about sym-metry; labels, torso, extra labels; boundary/basic and corner equations.Remark about not knowing the equations or the existence of a variety for agiven pair; reference ahead.
The codimension 3 case of the last section motivates our main definition.A long rectangle pair, or ???14, consists of a pair of long rectangles, havingthe same generators and torso labels, but distinct corner labels.
Define SKELETONS and figure out the correct corner labels from con-tinued fraction calculation. Again say something about the big variety.15
4.2.2 Pairs from flips
PROOF of the existence of the second long rectangle.16
11To do.12Check what happens with xd1 + L1. Seems fine.13To do.14Could think up a good name.15To do.16To do.
14
4.3 Long rectangles as glued staircases
Consider the codimension 9 long rectangle pair II(3, 2, 3).
3 1 0 −12323
243
2323
243
0 −2 2 1
Looking at the elephant, the labels read from the 1 round to the 0 comprisethe coefficients of a pair of dual continued fractions:
[(3), 2, 3, 2, 3] and [2, 4, 3, (−2)]
where we put brackets around the corner labels. Being dual, these arerepresented by the same Riemenschneider staircase
× ××× ××× ×
(3)
where the first continued fraction is read left to right and the second is readtop to bottom. There is a minor inconvenience with the negative coefficient.Clearly the natural continued fraction [2, 4, 4, 2] read vertically from staircase(3) is equal to the one recovered from the long rectangle. It is easy to checkthat one can read the columns of the staircase in the usual way, but translatethe final row of two ×s into a −2 coefficient. This works completely generallyand is correct for reading the rows of the staircase too: if we wanted to dothat in this case we would read [3, 2, 3, 2, 2,−1].17
Reading from the top, the labels of the two sides of the torso are the twofollowing sequences which again we think of as being continued fractions:
[2, 3, 2, 3] and [2, 4, 3].
As Riemenschneider staircases, these look like
×× ××× ×
and
+ +++ +
+
where the first is read left to right, the second is read top to bottom. Wewill consistently use × for elephant torso staircases and + for section torsostaircases.
The two torsos form part of each of the continued fractions correspondingto the big staircase (3) and one immediately sees the two smaller staircaseslying inside the big one. We draw the elephant staircase as the staircase(3) with ×s and +s exhibiting the two torsos. We use ∗ for staircase nodeswhich lie in both torsos.
+ +∗∗ ∗∗× ×
17Or use the extra blowups argument.
15
To read this as two continued fractions one can use the row–column switchingmethod for reading negative coefficients: to read left to right, the + nodesare read as a (d − 1) if they form a row of d nodes and as a (−d) if theyform a column of d nodes. These are the corner labels since they are notpart of the torso. Other nodes, either ×s or ∗s, are read positively as rowsin the usual way.
So we read the two fractions as
[(3), 2, 3, 2, 3] and [2, 4, 3, (−2)]
as required, where we also know that the initial (3) and the final (−2)correspond to corner labels.
In sum, then, we think of the continued fraction staircase correspond-ing to a single long rectangle as arising by glueing together two continuedfraction staircases which are to correspond to the two sides of the torso.Moreover, we remember how this glueing happened with the novel stair-case nodes ×, ∗, +. Since we are always dealing with long rectangles, andnot with polygons having many corners at the top and bottom, the overlapin glueing must be substantial, leaving only a single row or column to beinterpreted as a single corner label.
If the elephant is to be partnered by a section, there must be a differentway of glueing the two torso sequences together. Given that there cannotbe a great overlap at either end, one immediately sees that there is exactlyone other way of making such a glueing in this case. The proposed sectionstaircase, then, is
×∗ ∗∗∗ ∗
+
Reading this as two continued fractions produces
[2, 3, 2, 3, (2)] and [(−1), 2, 4, 3]
where the final (2) of the first fraction and the initial (−1) of the secondcorrespond to corner labels. Putting the 1 and 0 in the remaining corners inthe obvious way produces the section long rectangle of Example 2 as hopedfor.
Proposition 9 A pair of staircases, S1 and S2, determine a long rectanglepair if they can be glued together in two distinct ways so that the excesscomprises two disjoint and linear components. In this case, S1 and S2 cor-respond to the two sides of the torso of the resulting long rectangles.
Elementary arguments about glueing torso staircases together producethe following classification of long rectangle pairs in staircase notation. Inthe next section we translate these pictures into long rectangle pairs andgive a precise proof of this claim. It is striking how all the staircases listedbelow are long and thin.
Note that we haven’t yet made a decision about which long rectangle ofa pair will correspond to the elephant in the flip. Reflecting a staircase in itsantidiagonal produces another staircase but very often it is identical to theoriginal up to an obvious symmetry. For instance, reflecting a staircase infamily I below makes a staircase that isn’t listed, but whose long rectangle
16
pair is that of the original staircase but with elephant and section switchedand turned upside down. So we do not include the reflected staircase in thislist This self-adjoint type property also holds for family II below. Reflectionpairs up the families III and IV.
Family I(d, k)
In each case the diagonal dots denote k repetitions of the common row ofd− 1 nodes.
Elephant
+d· · · +
∗ d−1· · · ∗. . .
∗ d−1· · · ∗× d· · · ×
Section
+ ∗ d−1· · · ∗. . .
∗ d−1· · · ∗∗ d−1· · · ∗ ×
See how the two torso staircase are shunted up against one another.
Family II(d, e, k)
The elephant torso is of the form
× d−1· · · ×× e−1· · · ×
. . .× · · · ×
having k+ 1 rows of alternating length and ending with d− 1 or e− 1 nodesaccording to the parity of k. The section torso is similar with the roles of dand e exchanged. These pairs clearly overlap in two ways.
Family III(d, k)
In each case the diagonal dots denote k repetitions of the common row ofd− 1 nodes.
Elephant
× ∗ d−1· · · ∗∗ d−1· · · ∗
. . .
∗ d−1· · · ∗× d· · · ×
17
Section
× d· · · ×∗ d−1· · · ∗
. . .
∗ d−1· · · ∗∗ d−1· · · ∗ ×
See how the section torso staircase shifts its position by one row inside theelephant staircase.
Family IV(d, k)
INSERT TRANSPOSITION OF III.18
Exercise Confirm that the examples of long rectangle pairs seen so far arecorrectly labelled and named.
4.4 The numerical classification of flips
4.4.1 Pairs that flip
Throughout this section we omit the variables A,B,L,M corresponding tothe extra labels. The shorthand 2d is used to denote a column of d twos inthe torso of a long rectangle. Often the torso is only written on the lefthandrectangle and is taken as read on the righthand one.
Sequences of terms on either side of the torso which repeat are bracketedtogether in such a way that after multiple blowing down the entire bracketis reduced to a 1.
Family I(d, k)
The parameters range over d ≥ 1, k ≥ 0. These diagrams are nonsymmetric.Flips in this family have codimension (k + 1)(d− 1) + 4.
Case d ≥ 3
d+ 1 1 −1 02d−1{
d3
2d−3
}...
...{d
32d−3
}d+ 1 2
0 −d 1 2
The dotted lines indicate k repetitions of d on the left and of the bracketedexpression on the right.
18to do
18
Case d = 2
3 1 −1 02k 2 2k 23 k + 2 3 k + 20 −2 1 2
Case d = 1
In this case k is no longer a parameter and will be dropped from the notation.
2 1 −1 02 2 2 20 −1 1 2
Family II(d, e, k)
The parameters range over d, e ≥ 2, k ≥ 0. The cases k even and odd aretreated separately. The case d = 1 can be interpreted as a codimension 3flip.
Case k = 2h even, d, e ≥ 3
These rectangles are symmetric and the partner is the vertical flip. Flips inthis family have codimension h(d+ e− 2) + d+ 2.
d 1 0 −(e− 1)2d−2
{ed
32e−3
32d−3
...
...{ed
32e−3
32d−3
e 20 −(e− 1) d 1
The dotted lines denote h repetitions of the bracketed expression.
Case d ≥ 3, e = 2
Replace 3, 2e−3, 3 fragments in the right of the torso by 4. This is h substi-tutions. The result is the following pair.
d 1 0 −12d−2{
2d
42d−3
}...
...{2d
42d−3
}2 20 −1 d 1
19
Case d = 2, e ≥ 3
Replace 3, 2d−3, 3 fragments in the right of the previous torso by 4. This ish − 1 substitutions. All the 2s at the bottom of the right hand torso willhave vanished.
Case d = e = 2
2 1 0 −12k+1 k + 2 2k+1 k + 2
0 −1 2 1
Case k = 2h+ 1 odd, h ≥ 0, d, e ≥ 3
These rectangles are not symmetric except when d = e. The partner is thevertical flip with the roles of d and e reversed. Flips in this family havecodimension (h+ 1)(d+ e− 2) + 3.
d 1 0 −(e− 1)2d−2
{ed
32e−3
32d−3
...
...{ed
32e−3
32d−3
e 3d 2e−2
0 −(d− 1) e 1
There are h repetitions of the bracketed expression.The special cases when d or e equals 2 are handled in the same way as
the even case.
Family III(d, k)
The parameters range d ≥ 1, k ≥ 0. The rectangles are symmetric and thepartner is the vertical reflection. These flips have codimension (k + 1)(d −1) + 4.
The construction can be interpreted for d = 0 to give the codimension 3flip having a 2 in its torso.
20
d ≥ 3
1 d+ 1 0 −12d−1{
32d−3 d
}...
...{3
2d−3 d}
32d−1 d
0 −1 1 d+ 1
There are k − 1 repetitions of the bracketed expressions. When k = 0 therighthand torso is empty while the lefthand torso is sequence of d+ 1 twos.
Case d = 2
1 3 0 −12 2
k + 2 2k k + 2 2k
2 20 −1 1 3
Case d = 1
In this case k is no longer a parameter.
1 2 0 −12 22 20 −1 1 2
Family IV(d, k)
Case d ≥ 3
1 2 0 −dd+ 1 2d−2{d
32d−3
}...
...{d
32d−3
}d+ 1 2
0 −d 1 2
There are k repetitions of the bracketed expressions.
21
Case d = 2
1 2 0 −13 32k k + 2 2k k + 23 30 −1 1 2
The case d = 1 is like Family III.
4.4.2 Proof of the numerical classification
We leave the proof in the style of staircases as an exercise, preferring topresent a rather more precise proof in terms of the continued fractions them-selves below. However, the staircase argument does help us in two ways.
First, as we have seen, the classification of staircases above is very con-cise, whereas its translation into long rectangle pairs degenerates into listsof special case when the parameters d, e and k are small. Second, the setupof the staircase argument has an immediate corollary about the possiblerelative positions of 0s and 1s in long rectangle pairs which is the first stepof the argument below.
Lemma 10 There are three distinct cases for the relative position of the 1and the 0 in long rectangle pairs. Up to symmetries of the rectangles theyare
· 1 · 00 · 1 ·
· 1 0 ·0 · · 1
1 · 0 ·0 · 1 ·
where a dot denotes an unknown corner label and the torsos are omitted.These are referred to as Case 1, 2, 3 respectively.
Now we prove the classification using the elementary number theory ofcontinued fractions. We consider the three cases of Lemma 10 separately.
Case 1
Suppose given a long rectangle pair as in Case 1 of Lemma 10.
a 1 −(f − 1) 0a1...
ak−1
e
fbl−1
...b1
a1...
ak−1
e
fbl−1
...b1
0 −(e− 1) 1 b
Suppose that the continued fractions along the torso’s sides are
n
q= [a1, . . . , ak−1] and
m
p= [b1, . . . , bl−1].
22
The −(e − 1) and −(f − 1) corner labels are another easy consequence ofthe staircase picture.
The fact19 that the two continued fractions running round the long rect-angle from the label 1 are dual is expressed by
N
Q= a− 1
n/qand
N
N −Q= f − 1
m/p∗
where pp∗ ≡ 1 mod m. Since n and q are coprime we see that N = an− q.Similarly N = fm − p∗ and adding denominators shows that N = n + m.Together with the same argument on the right-hand side we obtain equations
bm− p = en− q∗ = n+m (4)
fm− p∗ = an− q = n+m (5)
where qq∗ ≡ 1 mod n. Equating (4) and (5) one sees that bm−p = fm−p∗ sob = f , p = p∗ and p2 ≡ 1 mod m. Similarly a = e, q = q∗ and q2 ≡ 1 mod n.This final fact is equivalent to [a1, . . . , ak−1] being a palindrome: a1 = ak−1,a2 = ak−2 and so on.
Now suppose that m ≥ n. By (4), b = 2 and n = m − p. In particular,n2 ≡ p2 ≡ 1 mod m. Now extending the continued fraction a little showsthat
[a, a1, . . . , ak−1] = a− 1n/q
=an− qn
=n+m
n= 1 +
m
n.
We know that n2 ≡ 1 mod m so [a − 1, a1, . . . , ak−1] is a palindrome. But[a1, . . . , ak−1] already was a palindrome whence
a1 = · · · = ak−1 = a− 1.
This determines the staircase down one side of the torso agreeing with thatof Family I(a − 1, k), albeit after a rotation. The case m < n is identical,reflecting the symmetry of the long rectangle pair.
Case 2
20
Case 3
This is similar to case 1 using the notation of this picture.
a 1 −(f − 1) 0a1
...
ak
fbl−2
...b1e
a1
...
ak
fbl−2
...b1e
−(e− 1) 0 b 1
5 Gorenstein projection
We describe how to make Gorenstein projections of pairs of long rectangles.This immediately leads to the discovery of special subpairs which form thebasis of our understanding of the equations of flips.
19make a proper reference in the text20ADD: the missing case.
23
5.1 Special case of the theory
In section 3.3.4 we noted the simple fact that eliminating a corner variablelabelled with a 1 in a single long rectangle is a projection from one Gorensteinvariety to another. Moreover, the generator of the canonical class of theprojection pulls back to a generator upstairs: in our notation there, thisis just the statement that u generates the internal ideal before and afterprojection. The same is true for a pair of long rectangles. We define anelephant projection to be the elimination of a variable from the top of a longrectangle pair whose label in the elephant rectangle is 1. Similarly a sectionprojection is the elimination of a variable whose section label is 1. If z issuch a projection variable we say that the Gorenstein projection away fromthe z–axis is a projection from z.
Lemma 11 The image of both elephant and section projections is Goren-stein. Its equations can be calculated from a modified long rectangle pair.21
Considering only left projections at first, one sees that at each stagethe result is another long rectangle pair. However, there are two distinctdifferences between a long rectangle arising in this way and those in ournumerical classification of section 4.4.1.
The first is that the extra labels on the corners are no longer simply A, Band so on, but are monomial terms in these and the corner variables. Theseintermediate long rectangles inherit a C∗ from the original one — indeed,the variables still have a Z-weighting and the equations among them are asubset of the original equations so still homogeneous — so can be used tomake flip diagrams. The effect of the more complicated corner labels is thatthese flip diagrams do not have isolated singularities.
The second is that the two sides of the long rectangle are still toric, eventhough they don’t seem to fit into the classification. Indeed, in terms of thevariables u and t the pictures are misleading since these variables are notusually even inside the projected cone. The point is, again, that the modifiedadditional labels in the corners interfere with the automatic numerics of thegenerator labels.
In practice, we treat the projected long rectangle as all others withoutbeing concerned about how it actually represents a Gorenstein variety.
5.2 Cascading pentagrams
The elephant projections start at the top of the long rectangle and worktheir way down, at each stage knocking off one of the two upper corners in away determined by a continued fraction calculation. The section projectionsare similar, but start at the bottom and work their way up.
At any stage in the sequence of elephant projections we can make oneor more section projections. In this way, we identify particular pairs oflong rectangles lying inside the starting pair. Our favourite subrectanglesare those on 5 generators, the codimension 3 images of multiple Gorensteinprojections. These are pentagrams22 so we understand their equations well.
21Insert proof which will supercede the following paragraphs of woffle. There is a logicalcorner here. We have two points of view. In the first we have a flip cover and are genuinelyprojecting something. In the second we have a long rectangle pair without knowing thatthere really is a variety which realises it. In either case we do not have the nice lookingequations and easy elimination: that’s a result of the existence of these projections andnot an ingredient.
22ref
24
Consider the long rectangle pair of type I6(3, 0):
4 1 −1 0
4222
4222
0 −3 1 2
and assume that the four corner equations are equal to their skeletons:
x1y4 = x42A+ y3LM x2y3 = y4B + L4M3
x1y0 = AB4 + x0L x0y1 = x31AB
3 + y20M
The label 1 on the elephant rectangle indicates that we can make aprojection from y4. The image is described by a long rectangle pair withlabels
3 1 · ·
422
422
0 −3 1 2
where we put a dot in place of labels that we don’t yet know and we do notcalculate the extra corner labels yet.
So now we can project from y3 since its label is 1 and then once morefrom y2. The result is the codimension 3 long rectangle pair
1 1 · ·4 40 −3 1 2
with some extra corner labels not yet calculated.In any case, this is some codimension 3 Gorenstein variety with two
known equations — the two equations at the bottom corners do not involvevariables higher up the long rectangles so remain intact throughout projec-tion. As before, these two equations are enough to fill in an antisymmetricmatrix whose Pfaffians are the equations of the image:
y1 −L −AB3 y0
x2 −y0M x31
x1 −Bx0
.
Exercise Compute the Pfaffians of this matrix and realise these equationsas those of the standard codimension 3 long rectangles with extra cornerlabels AB3, B, L, y0M . We think of these new terms as describing a blowupof the standard codimension 3 space. Compare the matrix above with thestandard codimension 3 matrix in section 4.1.
One can also compute these new corner labels directly from the originallong rectangle pair. The basic equation at x2 in the image is of the formx1y1 = . . .. An equation like this exists in the original pair of long rect-angles. Its skeleton includes the term x2AB
3 so irrespective of the variouseliminations we expect this term to survive projections. The extra cornerlabel is therefore AB3, the term appearing next to a power of x2 in the basicequation there.
25
Had we stopped projecting down the elephant rectangle sooner andstarted working up the section we would have seen other pentagrams. Infact, we would see the following four, where we display a pentagram bylisting the boundary generators appearing in it.
x2 x2 x2 x2 y4
y3 y3
x1 x1 y2 x1 y2 x1 y2
y1 y1 y1
x0 y0 y0
Reading from the right, you can see how the pentagrams fall down the longrectangles — compare with a short Jacob’s Ladder — in a way determined bythe continued fractions down the sides. We call this sequence of codimension3 Gorenstein projections the cascading pentagrams of the long rectangle pair.
Notice that whether we start calculating Pfaffians from the top or thebottom, as long as we make the calculations in sequence the equations ap-pearing in each pentagram are determined uniquely by the two equationsthat they share with the previous pentagram. In particular, the top cornerequations already determine those in the bottom corner. Moreover, if thoseat the top are equal to their skeletons, so define isolated singularities beforethe flip, then those at the bottom are also equal to their skeletons and soalso define isolated, hence terminal, singularities after the flip. Thus we passthe first sanity check on our method for constructing flips.23
Lemma 12 If the top corner equations are equal to their skeletons thenmany significant equations, including the equations after the flip, will beequal to their skeletons in some pentagram. . . 24
EXPLAIN: equations of form xiyj = . . . which eventually appear at thetop of a pentagram will be calculated by monomial Pfaffians so be 3-termpolynomials with no choices to be made.
5.3 The three primary flip families
The cascading pentagrams of the last section did not all lie in the longrectangles in the same way. Three of them had points at two xs and threeys while the bottom pentagram had three x points and two y points. Lookingat the classification of long rectangle pairs in section 4.4.1 you see that thisis typical.
However there are three families of flips which have more regular pen-tagram behaviour. Up to the symmetries in choice of notation, they arecovered by cascading pentagrams all having exactly three, two or one xpoints respectively.
These three families II(d, e, 0), III(d, 0) and II(2, 2, k) are all verticallysymmetric.25 One immediately sees the pentagrams dance up and down thelong rectangles.
23Reference to introduction comments about flips.24Point: for projection calculation we’re interested in exactly those equations which lie
at the corner of some pentagram. I’m just saying that these are clearly all equal to theirskeleton when calculated in the pentagram.
25Rethink now we have Family IV as well.
26
Running pairs: II(d, e, 0)
There are a sequence of d − 1 2s down the right-hand side of the torso ofthese rectangles so the resulting variety has codimension d+ 2.
d 1 0 −(e− 1)
e
2...2
e
2...2
0 −(e− 1) d 1
The pentagrams all have collections of points of the form x0, x1, x2, yi, yi+1:
x2 x2 x2 yd+2
yd+1
x1 x1 · · · x1
y2
y1 y1
x0 y0 x0 x0
All equations with leading term xy lie in pentagrams so are equal totheir skeletons. For appropriate i one calculates that
x2yi = yi+1B + xd−i−10 xe−1
1 LM i,
x1yi = xi2ABd−i + xd−i0 LM i
andx0yi = xi−1
2 xe−11 ABd−i + yi−1M.
So the pentagram equations are the Pfaffians of matrixes of the formyi+1 −xd−i−1
0 LM i −xi2ABd−i−1 yix2 −M −xe−1
1
x1 −Bx0
.In particular, the boundary equation in the x variables is
x0x2 = xe1 +BM.
Later we will see how to calculate the boundary equations in the y variablesusing a single unprojection. For the record, they are of the form
yi−1yi+1 = y2i + x
2(e−1)1 (x0B)d−i−1(x2M)i−1AL.
One can see that the extra term is a product of terms in pentagram equa-tions. This comes from a syzygy relating four equations.
Running triplets: III(d, 0)
There are a sequence of d + 1 2s down the right-hand side of the torso ofthese rectangles so the resulting variety has codimension d+ 3.
d+ 1 1 0 −12...2
2...2
0 −1 d+ 1 1
27
The pentagrams all have collections of points of the form x0, x1, yi−1, yi,yi+1:
x1 x1 x1 yd+2
yd+1
ydy3 · · ·
y2 y2
y1 y1
x0 y0 x0 x0
All equations with leading term xy lie in pentagrams so are equal to theirskeletons. For sensible i,
x1yi = yi+1B + xd+1−i0 LM i
andx0yi = xi−1
1 ABd+2−1 + yi−1M.
So the pentagram equations are the Pfaffians of matrixes of the formx1 −M −B x0
yi+1 −xd+1−i0 LM i−1 −yi
yi xi−11 ABd+1−i
yi−1
.This in turn produces the basic equations down the right-hand side of therectangles:
yi−1yi+1 = y2i + (x0B)d+1−i(x1M)i−1AL.
Running quads: II(2, 2, k)
There are a sequence of k + 1 2s down the right-hand side of the torso ofthese rectangles so the resulting variety has codimension k + 4.
1 2 −1 0
k + 22...2
k + 22...2
−1 0 1 2
This sequence is not quite regular. The classification of Gorenstein conesimplies in particular that the top and bottom pentagrams have three pointson one side and two on the other. However, the internal pentagrams all havecollections of points of the form x1, yi−1, yi, yi+1, yi+2:
yk+2 x2 yk+2
yk+1 yk+1
yk ykyk−1
x1 x1 y3 · · · x1 x1
y2 y2
y1 y1
x0 y0 y0
28
A typical central pentagram has equations coming from the matrixyi+2 −LM −yi+1 x1
yi+1 −Ak+1−iBk−iLiM i−1 −yiyi −AB
yi−1
.Only one boundary equation does not arise in this way, the equation at x1.
6 Unprojection
For any long rectangle pair, the cascading pentagrams compute many equa-tions of the flip but not all of them. To prove the existence of flips cor-responding to our numerical solutions of long rectangle pairs, we calculatethe equations of flips, or even just the basic ones The method we use is toproject down to small codimension, the pentagrams, and then add equationsby ‘unprojecting’ one variable at a time. Here we discuss what ‘unprojec-tion’ should mean in a general setup and then demonstrate it explicitly ina number of cases. The formal theory uses the material of [Hartshorne, III6] and [1], Appendix to §1.
6.1 Discussion of the theory
Suppose that C ⊂ Y is a divisor and that Y is embedded so that both Cand Y are projectively Gorenstein. In other words, both ωC and ωY aretrivial up to a global twist, ωC = OC(kC) and ωY = OY (kY ) say.
extended by zeros at either end. The first zero is because ωY is torsion-free,see [C3-f, appendix to §1], the last because we are hom-ing from somethingtrivial. Reading the Homs yields
0→ OY (kY − kC)→ ωY (C)(−kC)→ OC → 0.
There is a little point of care here. Twisting by IC in the Hom isn’t auto-matic since it’s not locally free. But it will work since differentials and globalsections are defined in codimension 1 on our normal, Gorenstein varieties;see [C3-f, appendix to §1].26
If on global sections the restriction to OC is surjective — when kC isnegative, for instance — then we obtain a rational form s on Y of degree−kC having exactly a single pole along C. Tracing s back through theidentifications we see it as a global section
s ∈ HomOY (IC ,OY (kY − kC)).
We use this rational function to unproject C in Y as follows.26I haven’t checked this.
29
If k[Y ] is the homogeneous coordinate ring of Y define k[X] to be thek-algebra k[Y ][s] with additional relations
s · f1 = g1, . . . , s · fn = gn
where IC = (f1, . . . , fn) and gi is the image of fi in O(kY − kC) under s.We define the unprojection of Y along C to be the corresponding X. The
ideal IC is called the unprojection ideal. This graded ring description of Xautomatically gives it a preferred embedding in a weighted projective spacetogether with a point p, the s-axis. Now by definition Y is the projection ofX from p and C is contracted rationally to p by the inverse birational map.
The main point of this construction is
Theorem 13 The unprojection X of Y along C is projectively Gorenstein.
We don’t prove this theorem, but see [Nagoya] and [Stavros] for special cases.Instead we show how to make an unprojection explicitly from a pentagramlying inside a long rectangle pair. There are three or four apparently differentcases that can happen, but they are all controlled by the syzygies of twoexamples: P2 ×P2 and P1 ×P1 ×P1.
6.2 Models of unprojection
We unproject as a method for calculating the equations of long rectanglepairs. At least to calculate the basic equations there are just two models forhow the new variable can be introduced. We do all calculations in the bigvariety where A, B, L, M are variables. This makes it easier to be moreprecise about what exactly are the generators of the unprojection ideal.
6.2.1 Easy case: unprojection type P2 ×P2
In the following example, consider the final projection to the bottom pen-tagram.
2 1 0 −23 3 3 32 2 2 20 −1 3 1
In terms of coordinate rings, this step is the elimination of y2 after y3 andx3 have already been eliminated. The equations of the bottom pentagramare the Pfaffians of the matrix
m =
x2 −M −x1 y1
x1 AB2 −x20L
x0 −A2B3
y0
.The unprojection ideal of y2 — the ideal of poles of y2 as a rational
function on the bottom pentagram — is generated by those (combinationsof) variables next to which y2 appears linearly in some equation. Explicitly,it is generated by x0, x1, y0, AB2 — just look at the skeleton equations tosee which monomials in the equations involving bottom pentagram variablescontain y2. All entries of m lie in this ideal except three of those in the firstrow (and column).
30
We write
y2 ·
x1 −AB2 −x20L
x0 −A2B3
y0
=2∧(
x2 −M −x1 y1
· · · ·
)
where the dots on the righthand side represent unknown terms for which weattempt to solve. In this expression the 4×4 matrix on the left is called theunprojection minor of m. Its entries all lie in the unprojection ideal. Thetop row on the right is the remaining row of m, the row which contains allentries of m which do not lie in the unprojection ideal.
We can indeed solve for these four entries using equations we know fromthe cascading pentagrams alone. The resulting righthand side is
2∧(x2 −M −x1 y1
x0LM x2AB −y1 −x0x1ABL
)
For example, the 1, 2 entry of the product is the equation x1y2. This appearsin the second Pfaffian so allows us to calculate the 2, 1 and 2, 2 entries on theright. Similarly we know the 2, 3 entry of the product so can also calculatethe 2, 3 entry on the right. The remaining entry is calculated as the productof x0L with the x0y2 equation. All other equations are consistent with thepentagrams and we have gained the y0y2 equation in the process.
These equations can be collected as the 4× 4 Pfaffians of a single 6× 6matrix
y2 x2 −M −x1 y1
x0LM x2AB −y1 −x0x1ABLx1 −AB2 −x2
0Lx0 −A2B3
y0
.See how we have packed the righthand 2 × 4 on top of the lefthand 4 × 4and finished off the corners with y1: add a tube of grout and we’d be tilingthe bathroom.
Exercise This example is very like the unprojection of the cone on a singleprojection of P2 ×P2. [Hint: knock a corner variable s off the 3× 3 matrixwhose minors define the image of the Segre embedding and put the resultabove the diagonal to make an antisymmetric 5×5 matrix m. The Pfaffiansof m are among the Segre equations. Now the unprojection ideal of s missesa single row and column in m. . . ]
There is a point to notice about these equations. There is no reason whythey should generate the entire ideal of equations of the double projectionaway from y3 and x3. This happens exactly when the unprojection minorgenerates the whole of the unprojection ideal. In this case it does since wecan see the four generators appearing as entries in the minor.27
Now we could be thorough and calculate all possible codimension 4 un-projections of the Pfaffians. In each case, the unprojection will be of thistype28 But instead we will continue to work with this matrix, unprojectingtwice more to find all the equations.
27cf The symmetric cod 4 Pfaffian flip occurs exactly like this.28Why are unprojections always of the same type? Is this something that we can detect
from the long rectangles? What does it say about the final variety?
31
The unprojection ideal of x3 is generated by x0, x1, y0, y1, and AB.Again it is only the top row which contains elements not lying in this idealand again there are three of them. Repeating the procedure above we findmore equations as the 4-Pfaffians of
x3 y2 x2 −M −x1 y1
? ? ? ? ?x0LM x2AB −y1 −x0x1ABL
x1 −AB2 −x20L
x0 −A2B3
y0
OLD STUFF FOR A FEW LINES....... where m2,4 = x2A(x1x3 +BLM2).The sign in the big, nonmonomial term is correct, however tempting it is tosimplify using the x1x3 equation. This calculation was not as straightforwardas the previous one. For instance, the 12.47 Pfaffian is
which is part of a syzygy in the ideal rather than a particular equation.The equations recovered at this stage are not the equations of the image
after a single projection since the x1y1 equation is not generated. But itcan be deduced easily being the 12.36 Pfaffian divided by x1 or the 12.57Pfaffian divided by M . END OF ERRORS
Finally we unproject y3. Once more the unprojection ideal contains thelower minor and we write y3 times that submatrix as
2∧(x3 y2 x2 −M −x1 y1
· · · · · ·
)
and solve for the entries of the second row. Looking at the unprojectionminor, we expect to hit all the remaining equations except ?? and possiblyalso ??.29
6.2.2 Hard case: unprojection type P1 ×P1 ×P1
To see this in action we look at a different example.
1 2 0 −23 32 3 2 33 30 −2 1 2
This is a symmetric pair of long rectangles: the elephant and the section areidentical before adding the group action.
The cascading pentagrams in this example are
x4 y2 y2
x3 x3 x3
x2 y1 x2 y1 x2 y1 x2 y1
x1 x1 x1
y0 x0 y0
29A little calculation check to do here.
32
The top pentagram corresponds to the following matrix.y1 −x2
3 −y2B x2
y2 −L2M −Ax4 −L3M2
x3
.The unprojection ideal for x1 is generated by x3, x4, y2 and L2M (since thex2y1 equation contains the term x1L
2M).The first two rows (and columns) do not lie in the unprojection ideal. . . 30
Exercise This is like the unprojection of the cone on a single projection ofP1 ×P1 ×P1.
6.2.3 Which unprojection type occurs?
One can tell in advance what type of unprojection is going to be involvedby looking at the shape of the cascading pentagrams.
Imagine the corners of a pentagram lying inside a long recangle pairof codimension at least 4. Up to symmetry, it must be one of the follow-ing forms each of which have label 1 at the top left-hand corner xi. Thecodimension condition implies that the label at the top right-hand cornergenerator yj is strictly bigger than 1, by the classification if you like.
× × × × × ×Type 1 × × Type 2 × × Type 3 ×
× × ××
We set L = M = 0 to look at the elephant — arguments with the sectionare identical. We want to see how an extra generator can be added to thebottom of a pentagram, extending it as a long rectangle. It turns out that itdoesn’t matter which on which side of the long rectangle the new generatorappears.
Lemma 14 Let P be a pentagram as above contained inside a long rectangleof codimension at least 4. The unprojection of a new variable at the bottomof P is like P1 ×P1 ×P1 if P is of Type 1 and like P2 ×P2 otherwise.
We simply calculate the 5× 5 matrix determining the pentagram equa-tions and look for elements of the unprojection ideal. Let P be of type 1and suppose that the label at yj is d + 1 and the label at xi−1, the secondleft-hand generator, is e + 1. Denote by A′ and B′ the two extra labels atthe top corners. These are monomials in the original A and B appearing atthe top of the unprojected long rectangle. Then the matrix determining Pis
xi 0 −xei−1 yjxi−1 ydjB
′ 0xi−2 A′
yj−1
.Irrespective of the side on which the unprojected variable appears, the un-projection ideal is generated by xi, xi−1, yj and some monomial in L and
30To do.
33
M . So we see that there is no 4× 4 minor of this matrix with entries in theunprojection ideal.
The argument for other pentagrams is similar. 31
6.2.4 Two tricks
32
Changing unprojection type
When an entry of a matrix of a pentagram is a square in the unprojectionideal, one of its factors can be transferred to its three partner entries. Theresulting Pfaffians generate a slightly smaller ideal than the original ones —two of them are multiplied by this factor. However, the unprojection idealnow contains more entries of the matrix so it can be unprojected as an easyPfaffian. . . 33
Glueing Pfaffians together
We can find maps to bigger Grassmannians either by forcing unprojectionof type 2P2 or by glueing Pfaffians together.
Sometimes the matrixes of adjacent pentagrams overlap. In this caseone can realise more equations as the Pfaffians of a bigger matrix.
Lemma 15 If X is a normal affine variety admitting two maps to S =ConeGr(2, 4) which are compatible with two distinct embeddings of S inT = ConeGr(2, 5) then X admits a map to T extending these two.
Choose coordinates on T so that the two embeddings of S are as twoSchubert cycles intersecting in P2. Then the compatability means that wehave most of a map from X to T given by the equations
f1 f2 f3 ?f4 = g1 f5 = g2 g3
f6 = g4 g5
g6
.This extends to a rational map X −→ T by setting
? =−f1g5 + f2g3
f4=−f1g6 + f3g3
f5=−f2g6 + f3g5
f6.
This map is not yet defined on the locus f4 = f5 = f6 = 0 in X. . . 34
This extends to the bigger case of a 6×6 matrix housing two overlapping5× 5 matrixes.35
31I should really include L, M and check that the whole thing works properly. That’sjust a matter of making sure that the two new terms contain high enough powers of Land M which will hold by the syzygy centred on the new variable in the unprojection.
32Old stuff; maybe irrelevant but often useful when strings of hard unprojections arerequired. Two or more Jerry unprojections is still something which we don’t do automat-ically.
33The point seems to be that we can ‘borrow’ syzygies from Grassmannians in morethat one way. This type of calculation typically reduces a hard multiterm syzygy to thecomputation of a couple of trivial ones.
34Not helpful as it stands, but this is often a method for avoiding syzygy computations.35Put in the appropriate cod 4 example.
34
6.2.5 The equations of codimension 4 flips
6.3 Basic equations of numerical solutions
6.3.1 The equations of the primary families
6.3.2 Proof of the existence of long rectangle deformations
Finally we see that long rectangle pairs do in fact define varieties withgood properties. In other words, they really do determine deformationsof elephants. As usual, we name the generators of the two long rectanglesx0, . . . xn, y0, . . . , ym, u, t. We denote by A the affine space on which theseare independent coordinate functions.
Theorem 16 For each long rectangle pair with generators as above thereexists an irreducible normal affine 4-fold A embedded in A so that the pen-tagram and basic equations of the long rectangle lie in its defining ideal. Thesubvariety B: (t = 0) ⊂ A is equal to the embedded elephant and A is a flatdeformation of B.
Moreover, the grading on A described by any system of weights on thelong rectangle pair determines a C∗ action on A which has the origin as afixed point at which A is Cohen-Macaulay.
This is easy. Choose four of the generators which rationally determineall the others, x0, y0, u, t say. Given the pentagram equations alone, definea generically injective graded rational map
Φ: A4 · · · → A
mapping (x0, y0, u, t) to the tuple of rational expressions of the generatorsin terms of these four. For instance, a bottom corner equation of the form
x0y1 = ya0uα + tδ
determines y1 rationally as
y1 =ya0u
α + tδ
x0.
The multiple projections show in particular that one can inductively expressthe other generators in a similar way working up the rectangle.
Now A is the closure of the image of this map. This is 4 dimensionaland irreducible. The fact that the origin is Cohen-Macaulay is free since thefunction u is a nonzero divisor whose zero locus is a Gorenstein toric variety.The pentagram equations vanish on A by definition. The basic equationsare derived from syzygies in the pentagram equations so all of these are inthe defining ideal.36
Choose any C∗ action having the coordinate functions as a system ofeigencoordinates. In other words, choose a system of weights on the longrectangle pair for which the pentagram equations are homogeneous. Theideal defining A is graded with respect to these weights. Consider a neigh-bourhood of the origin in A. The singular locus of A is invariant under theC∗ action so singularities near the origin must lie in the positive or nega-tively weighted eigenspaces. But the basic equations show that, at most,A meets these spaces along the axes corresponding to the corners of the
36Say this properly; trivially false as stated.
35
rectangles so its singular locus is 1 dimensional. Since A is Cohen-Macaulayit is also normal.
The deformation is flat since this whole argument can be applied to anyparticular value of t to show that the map to the t line is an algebraic familyof normal affine 3-folds.
6.4 Hyperplane sections of flips
These are easy to calculate. Their continued fraction, or perhaps its dual,is built from two others coming from the weights of the group action at thebottom two corners of the long rectangle pair patched together according tothe corner labels. There is a precise statement and proof.37
This is all rather irrelevant. But it does represent some comparison with[Kollar–Mori] and also show that one can hit lots of quotient singularitieswith a given big variety as one varies the C∗ action. It would also keep trackof the section when comparing flips coming from a long rectangle pair andsome elephant projection of it.
7 The equations of flips
SKETCHY WOFFLE.Long rectangle pairs are like worms. They are composed of a number of
extendable flip-like segments which are bound together with compact exoticsaddles. Moreover, small pieces of long rectangles contain information aboutthe whole thing, so can regenerate portions that are severed.
7.1 Factorisation of long rectangles
Consider example III10(4, 1).
1 5 0 −12223222
4
2223222
4
0 −1 1 5
Using × for each generator and ignoring the vertical dividing lines, thecascading pentagrams are
× × × × ××
× . . . ×× ×
× × × × × × × × × ×× ×
× . . . ××× ×37To do.
36
In this case, the dotted lines replace a single pentagram in the centre of asequence of three running triplets. In higher codimension examples of thefamily III(d, 1) this sequence is longer.
There are a number of ways of factorising this sequence into subsequenceshaving similar unprojection behaviour which come from smaller codimensionflips. In the diagram above we have indicated one factorisation with thevertical dividing lines.
Reading from the left, the first three pentagrams make up a long rectan-gle pair of type III(2, 0). However, since they have come from a projection,the extra corner labels on the rectangles will be modified. One can checkthat these three pentagrams do indeed make a III(2, 0) with extra cornerlabels A5B, A4B, L, y2
0M . The top labels come directly from the continuedfraction. The bottom labels would simply be L and M , except that the oneon the right needs to absorb excess y0 factors coming from the corner labelbeing 5 rather than 3. The three pentagrams on the right also come from aIII(2, 0).
The two central pentagrams form a sequence of running pairs. So theycan be realised as a blowup of a III(1) flip.
This kind of flip has symmetry about the central horizontal line. So wealso see immediately that it contains sequences of pentagrams correspondingto flips III(3, 1) and III(2, 1).
7.2 Patching rectangle segments together
If we try to patch segments together we should find that we get problemsif we make pieces that don’t obey the empirical rules of the generation ofbasic side equations. That is, we know a priori that the side relations notgenerated always lie on one side. And so on.
7.3 Flips with short-side torsos
The primary families all have torsos one side of which have at most onegenerator. There are three families of compound flips which also have sucha short side: I(d, 0) as in Example 1 of section 1.3, III(d, 1) and III∗(2, k).
Each of these factorises in a different way. Families I and III∗ manage totag a single pentagram onto one or both ends of a type II primary family.Type III is more interesting, being two long running triplets patched togetherover a short exotic thing. No reason to pursue this except as first smallexamples of the general case.
7.4 A bad–looking example
II6(3, 2, 1) doesn’t factorise in the way many other flips do. At first sight itappears to be a running quad together with a running triplet. However, thequad is illegal: its side labels do not appear in the genuine running quadexample because the 3 down the side is unusual. The cascading pentagramsare as follows.
× × × ×× × × × × ×× × × × × × ×× × ×
The codimension 4 Jerry example is also bad.
37
7.5 Proof that long rectangle deformations are Gorenstein
Use the chunks of equations to embed long rectangles into product of primaryor pentagram projections??
8 The geometry of flips
Want to say something about projections and factorisations.
9 Results
Recall that the x variables are indexed from zero starting in the bottom lefthand corner of the long rectangle and working up the lefthand side. The yvariables work up the righthand side, also indexed from zero.38
9.1 Flips in codimensions 3–6
Codimension 5
There are five pairs of long rectangles to consider. First, a summary of thecases and the resulting calculations. We say that a variety is a 4-Pfaffian ifits equations are the 4× 4 Pfaffians of an antisymmetric matrix whose sizedepends on the codimension. It is an almost 4-Pfaffian if all but one of itsequations are the 4× 4 Pfaffians of some matrix.
All of the solutions here have been confirmed using Macaualay2, [??],typically in about 1/50000th of the time they took to compute, excludingtyping time.
Name Brief description
I(2)
Recipe (so non-symmetric) on torso {[2, 2], [3]}.A single projection is the 4-Pfaffians of a 5× 5 matrixtogether with two 6 × 6 matrixes and can be unpro-jected in a number of different ways.
II(3, d+ 1, 0) Symmetric flip on torso {[d+ 1], [2, 2]}.This is 4-Pfaffian.
II(2, 2, 1)Symmetric flip on torso {[2, 2], [3]}.Almost 4-Pfaffian. A single projection is a 4-Pfaffiannon-flip.
III(2, 0)
Symmetric flip on torso {[2, 2, 2], []}.A single projection is a blowup of Inductive4-space.Equations are Pfaffians in a number of different ma-trixes.
III(2, 0)∗Symmetric flip on torso {[3, 3], [2]}.All equations appear as some 4-Pfaffian, although onlycodimension 3 projections are actually Pfaffian.
38Beware that the naming of the flips below is old so might be mildly incorrect. Alsopossible that elephant and section have been switched sometimes. But the results arefairly robust not.
38
I(2)
3 1 −1 0
322
322
0 −2 1 2
x1y3 = x32A+ y2LM
x0x2 = x31 + y0BM
x1y0 = AB3 + x0L
x2y2 = y3B + L3M2
y1y3 = y22 + x2
2AL2M
y0y2 = y21 + x1x2ABL
x0y1 = x21AB
2 + y20M
x2y1 = y2B + x1L2M
x2y0 = y1B + x21L
x0y3 = η2A+ y21M
x0y2 = x1ABη + y0y1M
y0y3 = y1y2 + x2ALηx1y2 = x2
2AB + y1LMx1y1 = x2AB
2 + y0LM
where η = x1x2 +BLM .
II(2, d+ 1, 0)
3 1 0 −d
d+ 122
d+ 122
0 −d 3 1
x1y3 = x32A+ LM3
x0x2 = xd+11 +BM
x1y0 = AB3 + x20L
x2y2 = y3B + xd1LM2
y1y3 = y22 + xd1x2ABLM
y0y2 = y21 + x0x
2d1 ABL
x0y1 = xd1AB2 + y0M
x2y1 = y2B + x0xd1LM
x2y0 = y1B + x20xd1L
x0y3 = xd1x22A+ y2M
x0y2 = xd1x2AB + y1M
y0y3 = y1y2 + x2d1 ALη
x1y2 = x22AB + x0LM
2
x1y1 = x2AB2 + x2
0LM
where η = x0x2 +BM .
II(2, 2, 1)
2 1 0 −122
322
3
0 −1 2 1
x2y2 = x23A+ L2M3
x1x3 = x22 +BLM2
x0x2 = x21 +AB2M
x1y0 = A2B3 + x20L
x3y1 = y2B + x2LMy0y2 = y3
1 − y1ABLM + x1x2ALx0y1 = x1AB + y0M
x3y0 = y21B + x2
1Lx2y1 = x3AB + x1LMx2y0 = y1AB
2 + x0x1L
x0x3 = x1x2 + y1BMx1y2 = x2x3A+ y1LM
2
x1y1 = x2AB + x0LMx0y2 = x2
2A+ y21M
39
In this case, the equations can be deduced from the 4× 4 Pfaffians of
y2 −LM2 −x3A y1M x2A −(y21 −ABLM)
x3 −LM −x2 y1 x1Lx2 −BM −x1 y1B
x1 −AB −x0Lx0 −AB2
y0
.
This matrix was calculated from the double projection of y0 and y2, a sym-metric operation in this case. The resulting 5× 5 matrix can be seen in thecentre of the matrix above. This was then unprojected separately to makethe first row and last column.
In this form, two equations are missed: x0y1 and x3y1. An almost 4-Pfaffian description — that is, Pfaffians hitting all but one of the equations— is achieved by the less symmetric projection of y2 and y1.
III(2, 0)∗1 2 0 −233
233
2
0 −2 1 2
x2y2 = x3A+ L3M2
x1x3 = x32 + y2BLM
x0x2 = x31 + y0ABM
x1y0 = A3B2 + x0L
x3y1 = y22B + x2
2L2M
y0y2 = y21 + x1x2AL
x0y1 = x21A
2B + y20M
x3y0 = y1y2B + x2Lηx2y1 = y2AB + x1L
2Mx2y0 = y1AB + x2
1L
x0x3 = η2 + y21BM
x1y2 = x22A+ y1LM
x1y1 = x2A2B + y0LM
x0y2 = x1Aη + y0y1M
where η = x1x2 +ABLM .
III(2, 0)
1 3 0 −1222
222
0 −1 1 3
x3y1 = x4A+ L3Mx2x4 = x2
3 + y21BL
2Mx1x3 = x2
2 + y0y1ABLMx0x2 = x2
1 + y20A
2BMx1y0 = A3B + x0L
x4y0 = y31B + x3L
x0y1 = x1A+ y30M
x1x4 = x2x3 + y1BLMηx0x4 = x2
2 + η2BMx0x3 = x1x2 + y0ABMη
x3y0 = y21AB + x2L
x2y1 = x3A+ y0L2M
x2y0 = y1A2B + x1L
x1y1 = x2A+ y20LM
40
where η = y0y1+AL. The left-hand side equations are among the 4-Pfaffiansof the anti-symmetric matrix
y0AB ηB x0 x1 x2
y1BL x1 x2 x3
x2 x3 x4
y0AM ηMy1LM
.
Codimension 6
IV(3, 0)
1 2 0 −344
22
44
22
0 −3 1 2
x2y3 = x3A+ L4M3
x1x3 = x42 + y3BLM
x0x2 = x41 + y0ABM
x1y0 = A4B3 + x0L
x3y2 = y23B + x3
2L3M2
y1y3 = y22 + x1x
22AL
2My0y2 = y2
1 + x21x2A
2BLx0y1 = x3
1A3B2 + y2
0M
x0x3 = η3 − x1x2ABLMη + y1y2BMx3y1 = y2y3B + x2L
2Mηx3y0 = y1y3 + x2
1x32L
x2y2 = y3AB + x1L3M2
x2y1 = y2AB + x1L3M2
x2y0 = y1AB + x31L
y0y3 = y1y2 + x1x2ALηx1y3 = x3
2A+ y2LMx1y2 = x2
2A2B + y1LM
x1y1 = x2A3B2 + y0LM
x0y3 = x1Aη2 + y2
1Mx0y2 = x2
1A2Bη + y0y1M
where η = x1x2 +ABLM .
I(2, 1)
2 1 −2 032
32
32
32
0 −1 1 3
39
x2y3 = x23A+ y2
2L2M
x1x3 = x32 + y2
1BLMx0x2 = x2
1 + y20AB
2Mx1y0 = A3B5 + x0L
x3y2 = y3B + L5M2
y1y3 = y32 + x3AL
3My0y2 = y2
1 + x2ABLx0y1 = x1A
2B3 + y30M
x2y2 = x3AB + y1L2M
x2y1 = y2AB2 + y0L
2Mx2y0 = y1AB
2 + x1L
x3y1 = y22B + x2L
3Mx1y2 = x2
2AB + y0y1LMx1y1 = x2A
2B3 + y20LM
x0y2 = x1x2AB + y20y1M + y0A
2B3LMx0x3 = x1x
22 + y2
0y2BM − y1A2B4LM
x0y3 = x1x2x3A+ y20y
22M + x1y2ABL
2M + (y1y2 +ABL3M)A2B3LMx1y3 = x2
2x3A+ y21y2LM + x2y2ABL
2M + y1ABL4M2
y0y3 = y1y22 + x2x3AL+ y2ABL
3Mx3y0 = y1y2B + x2
2L.
39This example is interesting in that it can be calculated without resort to a singlesyzygy once one allows “3-vector” Jerry unprojections.
For the time-being, an arbitrary list of questions together with junk.
• Other elephant types. These are types Dn and E6, each of which is asingle unprojection away from an An case. But how does this work atthe level of covers?
• Haven’t yet understood how equivariant birational maps between thecovers or the big varieties translate into maps between the flip diagramsor parts of them. At first sight it does look like a birational map ofcovers An → Am should factorise a flip as
X−n → X−m → (Am flip) → X+m ← X+
n
although I don’t understand the maps even in easy cases. However,I do know that if something like this works then in the easy casesone can arrange for the central map to be a genuine flip rather thananything coming from an unusual C∗ action on a long rectangle pair.
• Favourite examples: what can we say when two flips have a commonelephant? Is there a Tom & Jerry phenomenon here?
• Are there general formats extending low cod flips (like the equationsof Grassmannians) or anything else which will help to calculate thefull ideal of equations?
• Are these flips finite quotients of flops as for easy toric examples?
• Does t = a give a semisimple structure to these flips?
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• Are there rules for glueing long rectangles together in the spirit of thedecompositions into basic families?
• What is the change in K3?
• Can we make deformations of these flips? Into toric flips, for instance.
• What is the Poincare polynomial of the graded ring?
• Compare the number theory of the continued fractions with the geom-etry of the elephant and section. Compare the numerical classificationwith Mori&Kollar.
• Each pair of cascading pentagrams determines an automorphism ofthe big codimension 3 space. What is it?
• Compare the equations down the sides of long rectangles with thosecoming from the deformation of the corresponding quotient singularity.
• Find model unprojections from cod 4 to cod 5. There are lots of cod 5examples as hypersurfaces in scrolls, for instance — can you see howto make a natural Gorenstein projection?
• Can the “transpose” relationship between various pairs of long rect-angles be seen in their equations? How is one set of equations relatedto the other?
• Symmetries in flips: elephant vs section; two C∗ actions; simple per-mutation actions on variables coming from symmetries of the longrectangle pairs. Do these mean anything?
• Say something about long triangles and pulling out corners: P(1, 2, 3).
