# The geography of negative curves

Javier González Anaya, José Luis González and Kalle Karu  J. González-Anaya,
Department of Mathematics, University of California, Riverside, Riverside, CA
92521, United States.
J.L. González, Department of Mathematics, University of California, Riverside,
Riverside, CA 92521, United States.
K. Karu, Department of Mathematics, University of British Columbia, Vancouver,
BC V6T1Z2, Canada<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study the Mori Dream Space (MDS) property for blowups of weighted
projective planes at a general point and, more generally, blowups of toric
surfaces defined by a rational plane triangle. The birational geometry of
these varieties is largely governed by the existence of a negative curve in
them, different from the exceptional curve of the blowup.

We consider a parameter space of all rational triangles, and within this space
we study how the negative curves and the MDS property vary. One goal of the
article is to catalogue all known negative curves and show their location in
the parameter space. In addition to the previously known examples we construct
two new families of negative curves. One of them is, to our knowledge, the
first infinite family of special negative curves.

The second goal of the article is to show that the knowledge of negative
curves in the parameter space often determines the MDS property. We show that
in many cases this is the only underlying mechanism responsible for the MDS
property.

The second author was supported by a grant from the Simons Foundation (Award
Number 710443) and by the UCR Academic Senate. The third author was supported
by a NSERC Discovery grant.

##  1\. Introduction

We work over an algebraically closed field $k$ of characteristic zero.

Let us consider the blowup of a weighted projective plane $\mathbb{P}(a,b,c)$
at a general point, and more generally the blowup of a projective toric
surface $X_{\Delta}$ with Picard number $1$, again at a general point. The
main problems we are studying are: (i) find the Mori cone of curves of this
variety, and (ii) determine if the variety is a Mori Dream Space (MDS). While
previous work has concentrated on constructing examples and non-examples of
MDS, in this article we start a more systematic study of families of such
varieties.

Let us describe the setup. Let $\Delta$ be a rational plane triangle, defining
a projective toric surface $X_{\Delta}$ with Picard number $1$, and let $X$ be
the blowup of this surface at a general point $e$,

$X=\operatorname{Bl}_{e}X_{\Delta}.$

A negative curve $C$ in $X$ is an irreducible curve of non-positive self-
intersection, $C\cdot C\leq 0$, different from the exceptional curve $E$ of
the blowup. The classes of $C$ and $E$ generate the Mori cone of curves of
$X$. Hence, finding such a negative curve $C$ determines the Mori cone as well
as the dual nef cone. Moreover, if $C$ has strictly negative self-
intersection, $C\cdot C<0$, then this $C$ is unique in $X$.

Once a negative curve $C$ in $X$ is known, there is one more step to deciding
if $X$ is a MDS. By a result of Cutkosky [2], $X$ is a MDS if and only if it
contains another curve $D$ such that $C\cap D=\emptyset$. The class of the
curve $D$ and the hyperplane class $H$ in $X_{\Delta}$ generate the nef cone
of $X$.

Our aim in this article is to study all triangles $\Delta$ and the pairs of
non-intersecting curves $C$ and $D$ in them. Our main object of interest is
not the negative curve $C\subset X$, but the restriction $C^{\circ}$ of this
curve to the torus $T\subseteq X_{\Delta}$. Different varieties $X$ may
contain negative curves with the same restrictions to the torus. More
precisely, a negative curve $C$ in $X$ is defined by an irreducible Laurent
polynomial $\xi(x,y)$ vanishing to order $m$ at $e$ and with support in a
triangle $\Delta$ with area $\leq\frac{m^{2}}{2}$. For each such $\xi$ we may
deform the triangle $\Delta$ and hence possibly get many other triangles for
which $\xi$ defines a negative curve. All these negative curves are obtained
from the same curve $C^{\circ}$ using the operations of closure and strict
transform.

We consider the space of all rational triangles $\Delta$, and in this space we
consider the subset of triangles that support a negative curve defined by some
fixed polynomial $\xi$. Two triangles that differ by scaling and translation
define the same toric variety and hence should be considered the same; we call
such triangles parallel. A natural choice of parameters for this space are the
three slopes of $\Delta$. Without loss of generality we will restrict our
attention to a slice of this parameter space and assume that all our triangles
$\Delta$ have a horizontal base, left slope $s$, right slope $t$ and $0<s<t$.
Figure 1 shows the $st$-plane with many diamond shaped regions. Each such
region corresponds to a polynomial $\xi$ and consists of all triangles
$\Delta_{s,t}$ for which $\xi$ defines a negative curve.

Figure 1. Negative curves in the parameter space of triangles.

The rest of the article is about Figure 1. It turns out that the location of
the diamonds on the map explains many results and open problems in the field.
We will briefly list some of these, with more details in subsequent sections.

###  1.1. Infinite families of negative curves

In our previous work [4, 3, 5], we found infinite families of polynomials
$\xi$ that define negative curves in some triangles. These families are
unified in [5] as follows. For every integer $K\geq 4$, there are two infinite
families of polynomials $\xi$ described by pairs of integers $(M_{n},N_{n})$,
$n=1$,$2$,…. The first family of polynomials is supported in the integral
triangles $IT_{K}(M_{n},N_{n})$, the second family in rational triangles
$RT_{K}(M_{n},N_{n})$. These negative curves can all be found on the map. The
diamonds corresponding to $(K,M_{n},N_{n})$ have centres with coordinate
$t=K$. Along each line $t=K$ the diamonds alternate: $IT_{K}(M_{1},N_{1})$,
$RT_{K}(M_{1},N_{1})$, $IT_{K}(M_{2},N_{2})$, $RT_{K}(M_{2},N_{2})$, $\ldots$,
$IT_{K}(N_{2},M_{2})$, $RT_{K}(N_{2},M_{2})$, $IT_{K}(N_{1},M_{1})$,
$RT_{K}(N_{1},M_{1})$.

We will describe below a new infinite family of negative curves. For each
$K\geq 4$ there is a triangle with slopes

$s=\frac{2K-3}{2K-5},\quad t=K-\frac{1}{2}$

that supports a negative curve with vanishing order

$m=\left\lceil K^{2}-\frac{7}{2}K+2\right\rceil.$

We summarize in Table 1 all known infinite families of negative curves. The
family of special negative curves is described in the next subsection.

Name | Order of vanishing $m$ | $s$ | $t$
---|---|---|---
$IT_{K}(M,N)$ | $M+N$ | $\frac{NK}{M+N}$ | $K$
$RT_{K}(M,N)$ | $M$ | $\frac{M+N}{M}$ | $K$
New non-special | $\lceil K^{2}-\frac{7}{2}K+2\rceil$ | $\frac{2K-3}{2K-5}$ | $K-\frac{1}{2}$
New special | $(K-3)(K-1)^{2}$ | $\frac{K(K-2)}{(K-1)(K-2)-1}$ | $K-\frac{1}{(K-2)^{2}}$

Table 1. Infinite families of negative curves.

There are a few other negative curves that can be found using computer search,
but we have not been able to fit them into infinite families. (See Table 2 on
page 2 below for a list of these curves.) In general, for each vanishing order
$m$ there is a finite number of isomorphism classes of negative curves. For
$m=1,2$ there is a unique curve, for $m=3$ there are two curves, for $m=4,5,6$
there are $4$ curves. All these curves for $m\leq 6$ can be found on the map.
Computer experiments suggest that as $m$ gets larger, it becomes more
difficult to find a negative curve that is not in one of the infinite
families. Among the negative curves that we have found which do not yet fit
into an infinite family, the largest value of $m$ that occurs is $m=99$.

###  1.2. Special negative curves.

To find a negative curve $C$, we need an irreducible polynomial vanishing to
order at least $m$ at $e$ and supported in a triangle $\Delta$ with area
$\leq\frac{m^{2}}{2}$. The vanishing condition gives ${m+1\choose 2}$ linear
equations in the coefficients of $\xi$. Thus, if $\Delta$ contains at least
${m+1\choose 2}+1$ lattice points, then a polynomial vanishing to order at
least $m$ can always be found. This polynomial may be reducible. If it is
irreducible we call the polynomial $\xi$ (or the negative curve $C$ it
defines) special if the triangle $\Delta$ contains fewer than ${m+1\choose
2}+1$ lattice points. In that case we call the difference between ${m+1\choose
2}+1$ and the number of lattice points in $\Delta$ the deficiency of $\xi$.

Two special negative curves were found by Kurano and Matsuoka [8]. These are
the unique negative curves in the blowups of $\mathbb{P}(8,15,43)$ and
$\mathbb{P}(5,33,49)$. They have vanishing orders $m=9$ and $m=18$,
respectively, and deficiency $1$. These two weighted projective planes
correspond to triangles with slopes $(s=1.6,t=3.75)$ and
$(s=1.6\overline{6},t=3.3)$ in our map.

We generalize the first example of Kurano and Matsuoka to an infinite family
of special negative curves, one for each $K\geq 4$. The minimal triangles for
these curves have slopes

$s=\frac{K(K-2)}{(K-1)(K-2)-1},\quad t=K-\frac{1}{(K-2)^{2}},$

vanish to order $m=(K-3)(K-1)^{2}$ and have deficiency $\frac{(K-2)(K-3)}{2}$.

We do not know if the $m=18$ example of Kurano and Matsuoka can be extended to
an infinite family.

The specialty of a curve is related to an interpolation problem. A curve
defined by $\xi$ is special iff there is a degree $\leq m-1$ bivariate
polynomial that vanishes at all lattice points of the triangle $\Delta$. In
general, the deficiency is the dimension of the space of polynomials of degree
$\leq m-1$ that vanish at all lattice points of $\Delta$. In the infinite
family of special curves mentioned above, the deficiency grows quadratically
with $K$. Hence, the triangles of these curves have more and more relations
among their lattice points.

Kurano [7] studied the genus of negative curves, noting in particular that all
known negative curves are rational. Similarly, all negative curves that we
have listed are rational, except possibly the special curves. Consider a
negative curve defined by $\xi$, and let $Y$ be the blowup of the toric
variety defined by the Newton polygon of $\xi$. Here we assume $m>1$ to avoid
the case where the Newton polygon is a line segment. Then, $\xi$ defines a
negative curve $C$ that lies in the smooth locus of $Y$. One can compute that
$K_{Y}\cdot C<0$ for all known negative curves except the special ones. This
implies that such curves are rational and smooth in $Y$. For the special
family we find that $K_{Y}\cdot C=(K-1)(K-4)\geq 0$, and the arithmetic genus
of the curve in $Y$ is $\frac{(K-1)(K-4)}{2}$. We do not know the geometric
genus of these curves when $K\geq 5$.

###  1.3. No negative curves.

It is possible that for some triangle $\Delta$ there is no negative curve at
all. In that case the boundary of the Mori cone lies on a ray with self-
intersection number zero. Kurano and Matsuoka [8] computed that the blowup of
$\mathbb{P}(9,10,13)$ does not have a negative curve for $m\leq 24$. One can
also see, by computation or from the Pick’s theorem, that as $m$ grows larger,
a negative curve with vanishing order $m$ must have higher and higher
deficiency. This strongly suggests that this blowup has no negative curve, but
we do not have a proof of it. Another interesting fact is that if there is no
negative curve in $X$, then the Mori cone of $X$ must have an edge with
irrational slope.

We conjecture that most of the $st$-plane that is not already covered by
diamonds contains no negative curve. Nevertheless, we do not have a proof of
this for any triangle.

###  1.4. Mapping $\mathbb{P}(a,b,c)$.

There is a large body of work in commutative algebra and algebraic geometry on
the MDS property of blowups of $\mathbb{P}(a,b,c)$. For example, Huneke [6],
Cutkosky [2] and Srinivasan [10] have shown that these blowups are always MDS
when one of the weights equals $1,2,3,4$ or $6$. In particular, such blowups
contain a negative curve.

In a recent work, McKinnon, Razafy, Satriano and Sun [9] compute bounds for
the nef cone of $\operatorname{Bl}_{e}\mathbb{P}(a,b,c)$ for small values of
$a$.

It is therefore interesting to see where the weighted projective planes
$\mathbb{P}(a,b,c)$ lie on the map and if we can find negative curves in them.

We will show that if $a$ is fixed, then all $\mathbb{P}(a,b,c)$ correspond to
points on the $st$-plane with either $s$ or $t$ of the form $\frac{a}{d}$,
where $1\leq d\leq a-1$ and $\gcd(a,d)=1$. There are further symmetries that
can be used to reduce the number points on the $st$-plane that we need to
consider. For example, the $a=4$ triangles lie on the horizontal line $t=4$
and on the vertical lines $s=\frac{4}{3},\frac{4}{1}$. The ones on the
horizontal line $t=4$ contain a negative curve from one of the infinite
families $IT_{4}(M_{n},N_{n})$, $RT_{4}(M_{n},N_{n})$. The ones lying on the
vertical lines can only contain the negative curve with $m=1$ defined by a
binomial. One can similarly find negative curves in all cases when $a<4$ or
$a=6$, see Example 3.2. Moreover, once the negative curves have been
identified, the MDS property for these spaces follows from Proposition 6.1.

It is not known if all blowups of $\mathbb{P}(a,b,c)$ with $a=5$ are MDS or
even if they contain a negative curve. In our picture, all triangles
corresponding to $a=5$ are equivalent to the ones on the horizontal line $t=5$
or on the vertical line $s=5/3$. The ones on the line $t=5$ contain a negative
curve from the infinite families $IT_{5}(M_{n},N_{n})$ and
$RT_{5}(M_{n},N_{n})$, and they are all MDS. The vertical line $s=5/3$ is the
interesting one. See Figure 4 on page 4 below for a closer look at this
vertical line. The line contains the curve in $RT_{4}(3,2)$ and the special
curve with $m=18$ of Kurano and Matsuoka. Along this line there is a sequence
of diamonds with some gaps between them. We suspect that most triangles in
these gaps do not contain negative curves. Srinivasan [10] studied the case
$a=5$ and gave bounds on $b/c$ for the blowup to be a MDS. These bounds can be
related to the location of the triangle on the line $s=5/3$.

###  1.5. The MDS property

We fix one diamond corresponding to a negative curve $C^{\circ}\subset T$ and
ask which triangles in it have the MDS property. One expects that the MDS
region in each diamond is divided into pieces corresponding to different
curves $D$.

It turns out that the MDS property in a diamond is often determined by
neighbouring diamonds. Suppose a triangle $\Delta$ lies in the intersection of
two diamonds. This $X$ contains two irreducible curves with zero self-
intersection: the negatives curve of each diamond. Hence, $X$ is a MDS. The
MDS property is sometimes preserved as we move from the intersection point
into the interior of one of the diamonds. The reason is that the two curves
that were disjoint in $X$ sometimes remain disjoint; one of them becomes the
negative curve $C$, the other one the curve $D$.

In our previous work [4, 3, 5] we have determined the MDS property in some
cases for the negative curves in $IT_{K}(M,N)$ and $RT_{K}(M,N)$. It turns out
that in all these cases the MDS property can be explained by the intersection
of diamonds. For example, as we move along the horizontal line $t=K$, the
diamonds for $IT_{K}(M,N)$ and $RT_{K}(M,N)$ are stringed together. As we move
from one diamond to the next, the $C$ and $D$ curves switch their roles. For
triangles in the diamond of $IT_{K}(M_{n},N_{n})$, $n>1$, we can determine
completely which ones are MDS and which ones are not. For the remaining
triangles in the two infinite families we can do this only for the upper half
of the diamond.

The intersection of diamonds can be used to prove the MDS property in many
other cases. This includes the minimal triangles supporting the special
curves, and the new family of non-special curves for $K$ even. In addition to
these examples, we will prove that all triangles on the horizontal line
$t=K\in{\mathbb{Z}}$ have the MDS property. In the region where the negative
curve is defined by $\xi=1-y$, it is interesting that points on the line $t=K$
with different $s$ value have a different $D$ curve (even when restricted to
the torus). Indeed, the Newton polygons of the polynomials defining the curves
$D$ have left edge with slope $s$. This shows somewhat unexpectedly that the
interval is not divided into discrete subintervals corresponding to different
$D$. Instead, every rational point corresponds to a different polynomial
defining the curve $D$.

We refer again to Figure 4 on page 4 below for some of the intricacies
involved in determining the MDS regions in each diamond.

###  1.6. Positive characteristic.

We work over a field $k$ of characteristic $0$. However, one can draw a
similar picture of diamonds when working over any field. In positive
characteristic the existence of a negative curve implies that the blowup is a
MDS [2]. Negative curves in characteristic $0$ can all be defined by integer
coefficients. It follows from the proofs given below that the four infinite
families of curves stay irreducible in any characteristic. Hence, these same
diamonds appear when we work in positive characteristic. Some negative curves
in characteristic $0$ are reducible in characteristic $p$. For example, the
curve with $m=16$ in Table 2 on page 2 decomposes as a union of an irreducible
negative curve with $m=15$ and a curve with $m=1$ in characteristic $2$. In
positive characteristic there are in general more negative curves. For
example, the blowup of $\mathbb{P}(9,10,13)$, which is not known to contain a
negative curve in characteristic $0$, does contain negative curves in
characteristics $2$ and $3$.

##  2\. The Parameter space of triangles

As explained in the introduction, we consider rational triangles $\Delta$ in
$\mathbb{R}^{2}$ with one edge horizontal and the other two edges with slopes
$s$ and $t$. In this section we study the $st$-plane that parametrizes these
triangles. We call two triangles isomorphic if they differ by a translation,
scaling, and action of $\operatorname{GL}_{2}({\mathbb{Z}})$ on the plane
$\mathbb{R}^{2}$. Two triangles are isomorphic if and only if the
corresponding toric varieties are isomorphic.

For a more complete picture, one should consider the slopes of all three edges
of $\Delta$. For each triangle we can choose coordinates so that one of its
edges is horizontal. This implies that every triangle is isomorphic to one
with slopes $\\{0,s,t\\}$, corresponding to the point $(s,t)$ or $(t,s)$ in
our parameter space. However, we should warn that considering the $st$-plane
only has several shortcomings. The most serious of them being that making an
edge horizontal is not a continuous operation: two triangles may have slopes
very close to each other, but the corresponding points on the $st$-plane can
be very far apart. Nevertheless, we consider the $st$-plane as a good way to
visualize the parameter space.

###  2.1. The fundamental domain

We consider triangles with one edge horizontal, and isomorphisms between these
triangles that map the horizontal edge to another horizontal edge. Let us
first find a fundamental domain on the $st$-plane containing one point for
each isomorphism class.

We will make one more restriction by ignoring the triangles where $s$ and $t$
have different signs, or where one of them is infinite. Such triangles are
very simple for the problem we consider. In these cases the variety
$X=\operatorname{Bl}_{e}X_{\Delta}$ contains two non-intersecting curves,
defined by the vanishing of $1-y$ and $1-x$. Hence, $X$ is always a MDS with
the negative curve defined by either $1-y$ or $1-x$.

######  Lemma 2.1.

A fundamental domain in the coordinates $u=1/s,v=1/t$ can be chosen as the
triangle in the $uv$-plane with vertices

$(0,0),(1,0),(1/2,1/2).$

###### Proof.

Since we consider $u$ and $v$ as an unordered pair, we may order them so that
$u>v$. The operations of translation and scaling do not change the slopes. The
subgroup of $GL_{2}({\mathbb{Z}})$ that preserves the zero slope is generated
by the shear transformation

$(u,v)\mapsto(u+1,v+1),$

and multiplication with $-1$, which after reordering so that $u>v$ can be
written as

$(u,v)\mapsto(-v,-u).$

This last operation is the reflection of the plane across the line $u=-v$.

A fundamental domain for this action is the strip

$0\leq u+v\leq 1.$

Restricting to the half-plane $u>v$ and removing points of the form $u>0,v<0$
gives the triangle as in the statement of the lemma. ∎

Figure 2 shows the fundamental domain in the coordinates $u,v$. In the
coordinates $s,t$ the fundamental domain is defined by

$t\geq s\geq 1,\quad t\geq 1+\frac{1}{s-1}.$

For a point in this domain, we may assume that the base of the triangle is
horizontal, $s$ is the slope of the left edge and $t$ the slope of the right
edge.

Figure 2. Negative curves in the parameter space of triangles.

Any triangle corresponds to a unique point $(s,t)$ in the fundamental domain
above once we choose which edge we make horizontal. Since there are three
choices for the horizontal edge, there are up to three points that correspond
to the same triangle. We will consider some examples of this.

######  Example 2.2.

Consider the set of points $(s,t)$ where $t=K$ is an integer. We can make the
edge with slope $t$ horizontal by applying a shear transformation in the
vertical direction. The resulting triangle has coordinates $(t-s,t)$. Thus,
the triangles with $K/2\leq s<K$ are isomorphic to triangles with $0<s\leq
K/2$. Moreover, if $s\leq 1$ then we can transform the triangle to one that
has $s$ and $t$ with different signs or one of them infinite. Hence, when
$t=K$ it is enough to consider points with $1<s<K/2$. This isomorphism
explains the symmetry in Figure 1 on the horizontal line $t=K$.

If $s=L$ is an integer, we can similarly make the edge with slope $s$
horizontal. The resulting triangle has slopes $\\{s-t,s\\}$, where $s>0$ and
$s-t<0$, so we may ignore these triangles.

The triangle with slopes $s=2.25,t=3.5$ has the property that no matter which
edge we make horizontal, we get the same point $(s,t)$. This triangle appears
in the first entry of Table 2 below as the minimal triangle supporting a
negative curve.

##  3\. Weighted projective planes $\mathbb{P}(a,b,c)$.

Weighted projective planes $\mathbb{P}(a,b,c)$ are special cases of the toric
varieties $X_{\Delta}$. We consider here which points $(s,t)$ correspond to
weighted projective planes and how to determine $a,b,c$ from $s,t$.

For a rational triangle $\Delta$, let $\vec{n}_{i}$, $i=1,2,3,$ be its outer
normal vectors, which we choose to be integral and primitive. Recall that
$X_{\Delta}$ is isomorphic to $\mathbb{P}(a,b,c)$ if the normal vectors
generate the lattice ${\mathbb{Z}}^{2}$, in which case $a,b,c$ are given by a
relation

$a\vec{n}_{1}+b\vec{n}_{2}+c\vec{n}_{3}=0.$

The positive integral vector $(a,b,c)$ is usually also chosen primitive.

######  Lemma 3.1.

Let the triangle $\Delta$ have rational slopes $0<s<t$. Write $s=a/p$, $t=b/q$
in lowest terms. Then $X_{\Delta}$ is isomorphic to a weighted projective
plane if and only if $a$ and $b$ are relatively prime. In this case
$X_{\Delta}\cong\mathbb{P}(a,b,c)$, where $c=bp-aq$.

###### Proof.

The normal vectors are $(b,-q)$, $(-a,p)$, $(0,-1)$. These generate the
lattice ${\mathbb{Z}}^{2}$ if and only if $a$ and $b$ generate the lattice
${\mathbb{Z}}$. The three vectors satisfy the relation with coefficients
$a,b,c=bp-aq$. ∎

The previous lemma tells us exactly which points $s,t$ correspond to weighted
projective planes. Conversely, given a weighted projective plane
$\mathbb{P}(a,b,c)$, it corresponds to a triangle $\Delta$. We can make the
side corresponding to $c$ horizontal. Then the other sides must have slopes
$a/p$ and $b/q$ for some $p$ and $q$. Consider now a fixed $a$ and varying $b$
and $c$. All such $\mathbb{P}(a,b,c)$ must have either $s$ or $t$ of the form
$a/q$ for some $q$ that is relatively prime to $a$. If the point $(s,t)$ lies
in the fundamental domain, then $1\leq q<a$.

It may happen that the point $(s,t)$ corresponding to some $\mathbb{P}(a,b,c)$
has $s$ and $t$ with different signs and hence cannot be moved to the
fundamental domain. This case is well-known and often discarded in the
commutative algebra literature. It means that the point $e=(1,1)$ that is
blown up is a complete intersection in $\mathbb{P}(a,b,c)$, with homogeneous
ideal generated by two elements, $1-x$ and $1-y$ in our notation.

######  Example 3.2.

Consider the weighted projective planes $\mathbb{P}(a,b,c)$ where $a=2,3,4,6$.
These correspond to points where either $s$ or $t$ is equal to $a$ or
$a/(a-1)$. The cases where $s$ is an integer are isomorphic to triangles where
$s$ and $t$ have different signs and are hence not interesting. When $t$ is an
integer, we may further restrict $s\leq t/2$. Thus, all interesting examples
of such $\mathbb{P}(a,b,c)$ have coordinates $t=a,1<s\leq t/2$ or
$s=a/(a-1),t>a$. In Figure 1 these points lie on the horizontal line $t=a$
covered by the diamonds of $IT_{a}(M,N),RT_{a}(M,N)$ and on the vertical line
$s=a/(a-1)$ covered by the diamond of $IT_{a}(a-2,1)$ and the diamond of
$\xi=1-y$.

######  Example 3.3.

Consider the weighted projective planes ${\mathbb{P}}(a,b,c)$ with $a=5$.
These correspond to points $(s,t)$ where either $s$ or $t$ has the form $5/q$,
where $q=1,2,3,4$. We proceed to explain how these cases are isomorphic to
ones with $s=5/3,t>2$ and $t=5,1<s<5/2$.

First consider points $(s,t)$ where $s=5/2$. Choosing coordinates
appropriately, we can make the edge with slope $s$ horizontal. The other two
edges will then have slopes $5/3$ and $2+1/(t-3)$. This shows that triangles
on the vertical line $s=5/2$ are isomorphic to triangles on the vertical line
$s=5/3$. The interesting line segments here are $s=5/2,t>3$ and $s=5/3,t>2$.
These are mapped to each other bijectively. The line segments $s=5/2,5/2<t<3$
and $s=5/3,5/3<t<2$ are mapped to points where $s$ and $t$ have different
signs and hence can be ignored.

Now consider the case $t=5/2$. This lies in the fundamental domain if
$5/3<s<5/2$. By the group action these are isomorphic to $s=5/3$, $5/3<t<5/2$,
which we considered before.

When $s=5$, we can make this side horizontal and get the two slopes with
different signs. These can be ignored.

When $s=5/4$ and we make this edge horizontal, the remaining two slopes are
$t=5,s=4-1/(t-4)$.

In summary, all interesting cases of $(s,t)$ in the fundamental domain where
either $s$ or $t$ is equal to $5/q$ for some $q=1,2,3,4$ are isomorphic to
triangles with coordinates $s=5/3,t>2$ and $t=5,1<s<5/2$. In Figure 1 the
horizontal line $t=5$ is covered by diamonds of $IT_{5}(M,N),RT_{5}(M,N)$. The
vertical line $s=5/3$ is magnified in Figure 4 below. The vertical line
contains the vertical axis of $RT_{4}(3,2)$ and several other diamonds, with
gaps between them.

It is interesting that the values of $b/c$ not covered by Srinivasan’s bounds
in [10] exactly lie on the vertical line $s=5/3$. Moreover, the slope $t$ of
these triangles is a function of $b/c$ and the bounds given by her precisely
correspond to the bottom vertex of the diamond of $RT_{4}(3,2)$ and the top
boundary point of the diamond of $IT_{3}(1,1)$.

##  4\. Infinite families of negative curves

In [5] we described two infinite families of negative curves, each family
depending on two parameters. We start by recalling these families. We then
construct a new infinite family of non-special negative curves.

###  4.1. The families $IT_{K}(M,N)$ and $RT_{K}(M,N)$.

Consider positive integers $K$, $M$ and $N$ satisfying the equation

(4.1) $(M+N)^{2}=KMN+1.$

For $K=3$ there is a unique pair $(M,N)=(1,1)$, and for $K\geq 4$ there is an
infinite number of pairs $(M,N)$ satisfying the equation. We describe these
pairs below. Since $K$ is determined by $M$ and $N$ if $M,N>0$, we will often
omit it from the notation.

For each $K,M,N$ satisfying the Equation 4.1 there are two negative curves.
The first one is supported in the integral triangle $IT_{K}(M,N)$ with
vertices

$(0,0),(M,0),(M+N,KN).$

This triangle has coordinates $s=KN/(M+N),t=K$, and the negative curve has
vanishing order $m=M+N$. The second curve is supported in the rational
triangle $RT_{K}(M,N)$ with vertices

$(0,0),\left(M-\frac{M+N}{K},0\right),(M,M+N).$

The triangle has coordinates $s=(M+N)/M,t=K$, and the negative curve has
vanishing order $m=M$.

On the $st$-plane the triangles $IT_{K}(M,N)$ and $RT_{K}(M,N)$ lie on the
horizontal line with $t=K$. The curves with $M>N$ are isomorphic to those with
$M<N$. Nevertheless, both diamonds appear on the $st$-plane because points in
one diamond may not be isomorphic to points in the other.

For small values of $M$ and $N$ there are isomorphisms between pairs of these
curves. For example, the curve in $IT_{K-1}(K-3,1)$ is isomorphic to the curve
in $RT_{K}(K-2,1)$. Similarly, all curves with $m=1$ are isomorphic. This
includes $IT_{K}(0,1)$ and $RT_{K}(1,K)$ for all $K$. Other than these
coincidences, the negative curves are pairwise non-isomorphic.

We will recall how to find all solutions $K,M,N$ to Equation 4.1, and how to
construct the equations of the negative curves recursively. These equations
will be used below.

Fix $K\geq 3$ and consider the sequence of integers
$F_{0}=0,F_{1}=1,F_{2}=K-2,\ldots$ defined by the recurrence

$F_{n+2}=(K-2)F_{n+1}-F_{n}.$

Then the non-negative integral solutions $M>N$ of the equation above are

$(M_{n},N_{n})=(F_{n+1},F_{n}),\quad n=0,1,\ldots.$

Since the equation is symmetric, $(M=N_{n},N=M_{n})$ is also a solution.
However, we have isomorphisms of triangles and hence isomorphisms of negative
curves

$IT(N_{n},M_{n})\cong IT(M_{n},N_{n}),\quad RT(N_{n},M_{n})\cong
RT(M_{n-1},N_{n-1}).$

We let $\xi_{n}^{int}$ and $\xi_{n}^{rat}$ be the polynomials defining the
negative curves for $IT(M_{n},N_{n})$ and $RT(M_{n},N_{n})$, respectively,
normalized so that their constant terms are $1$. We will sometimes use the
notation $\xi_{n,K}^{int}$ or $\xi_{n,K}^{rat}$ to make the value of $K$
explicit.

For every $n>0$ these polynomials satisfy the recurrence relations

(4.2)
$\displaystyle\begin{split}\xi^{int}_{n}&=\xi^{rat}_{n}\xi^{rat}_{n-1}-\varepsilon^{rat}_{n}x^{M_{n}}(y-1)^{M_{n}+N_{n}},\\\
\big{(}\xi^{rat}_{n-1}\big{)}^{K}&=\xi^{int}_{n}\xi^{int}_{n-1}-\varepsilon^{int}_{n}x^{M_{n}+N_{n}}(y-1)^{KN_{n}}.\end{split}$

Here $\varepsilon^{rat}_{n}$ and $\varepsilon^{int}_{n}$ are the coefficients
of the highest degree terms of the polynomials. For $K$ even they are

$\varepsilon^{int}_{n}=-1,\quad\varepsilon^{rat}_{n}=(-1)^{n+1}.$

For $K$ odd,

$\varepsilon^{int}_{n}=\begin{cases}\hphantom{-}1&\text{if $n\equiv 1\mod
3$,}\\\
-1&\text{otherwise,}\end{cases}\qquad\varepsilon^{rat}_{n}=\begin{cases}\hphantom{-}1&\text{if
$n\equiv 2\mod 3$,}\\\ -1&\text{otherwise.}\end{cases}$

From these equations and the initial conditions $\xi^{int}_{0}=1-x$ and
$\xi^{rat}_{0}=1-xy$ one can compute all these polynomials recursively.

###  4.2. A new family of negative curves.

We describe here a new infinite family of negative curves, complementing the
two-parameter families described above.

Let us denote the parameters

$a=2K-3,\quad b=2K-1,\quad c=4K^{2}-16K+11,\quad d=2K-5,\quad m=\left\lceil
K^{2}-\frac{7}{2}K+2\right\rceil.$

######  Theorem 4.1.

For each $K\geq 4$ there is a negative curve in the blowup of
$\mathbb{P}(a,b,c)$, with $m$ the order of vanishing at $e$. The triangle
$\Delta_{K}$ corresponding to the weighted projective plane has horizontal
base, and left and right slopes

$s=\frac{a}{d}=1+\frac{2}{2K-5},\quad t=\frac{b}{2}=K-\frac{1}{2}.$

For even $K$, the triangle has vertices

$\left(-\frac{1}{a},0\right),\quad(m-K+2,0),\quad\left(m,\frac{(K-2)b}{2}\right).$

The triangle has two integral vertices and the lower left vertex rational.

For odd $K$, the triangle has vertices

$(0,0),\quad(m-K+2,0),\quad\left(\frac{K-1}{2}+\frac{1}{c}\right)(d,a).$

This triangle has integral base and rational top vertex.

The rest of this subsection consists of the proof of the theorem.

In both even and odd cases it is easy to check that the triangle contains
exactly ${{m+1}\choose{2}}+1$ lattice points. To see this, one considers the
convex hull of lattice points in the triangle and computes its area
$(m^{2}-1)/2$ and lattice perimeter $m+1$. Then Pick’s theorem gives the
number of lattice points. From the number of lattice points and the area of
the triangle being $<m^{2}/2$, we know that the triangle supports a negative
curve. The only problem is to show that this curve is irreducible. We will do
this below for even and odd cases separately. Note also that these negative
curves are not isomorphic to the negative curves in the previous infinite
families. The triangles $RT(M,N)$ have width exactly $m$. For these triangles
it is not possible to choose coordinates in which their width is $m$.

A special feature of this family of triangles, as well as the triangles
$IT(M,N)$ and $RT(M,N)$ above, is that the convex hull of the lattice points
in the triangle has width exactly $m$. In fact, to get a nontrivial negative
curve, the width has to be at least $m$.

######  Lemma 4.2.

Let $f(x,y)\in k[x,y]$ be a polynomial that vanishes to order $m$ at
$e=(1,1)$. Assume that the polynomial $f$ has degree less than $m$ in $x$.
Then $1-y$ divides $f(x,y)$.

###### Proof.

The polynomial $f(x,1)$ must have multiplicity at least $m$ at $x=1$. Since
the degree of this polynomial is less than $m$, it must vanish identically. ∎

Let us denote by $f(x,y)$ a nonzero polynomial supported in the triangle
$\Delta=\Delta_{K}$ that vanishes to order at least $m$ at $e$. We will prove
that this $f$ is irreducible. Let us start with the case of even $K$.

######  Lemma 4.3.

Assume that $K$ is even. Then $f(x,y)$ is irreducible.

###### Proof.

We will use the following simple fact. Suppose we have a possibly reducible
curve $C$ in $X$ with negative self-intersection, and an irreducible curve $D$
such that $C\cdot D=0$. Then $C$ lies on the boundary of the Mori cone of
curves and hence is an integer multiple of an irreducible negative curve.
Indeed, the only effective curve classes in $X$ that multiply to zero must lie
on the boundaries of the Mori cone and the nef cone.

To find this curve $D$, we use the observation that on the $st$-plane the
diamond of $f(x,y)$ meets the diamond of the negative curve defined by
$\xi_{1}^{int}$ supported in $IT_{K-1}(K-3,1)$. Let us check that the curve
$D\subset X$ defined by $\xi_{1}^{int}$ is orthogonal to $C$. We denote by
$b_{C},h_{C}$ the base length and the height of the triangle $\Delta$.
Similarly, let $b_{D},h_{D}$ be the base and height of the smallest triangle
parallel to $\Delta$ that supports $\xi_{1}^{int}$. Then the classes of $C$
and $D$ are

$\displaystyle[C]$ $\displaystyle=H-mE,$ $\displaystyle[D]$
$\displaystyle=\frac{b_{D}}{b_{C}}H-(K-2)E.$

Here $H$ is the class corresponding to the triangle $\Delta$, and $K-2$ is the
order to which $\xi_{1}$ vanishes at $e$. Now using that $H\cdot
H=b_{C}h_{C}$, we get

$C\cdot D=b_{D}h_{C}-(K-2)m=b_{D}\frac{(K-2)(2K-1)}{2}-(K-2)m.$

We need to find $b_{D}$ and check that this intersection number is zero. The
Newton polygon of $\xi_{1}^{int}$ is the triangle with vertices
$(0,0),(K-3,0),(K-2,K-1)$, with the slope of the right edge $K-1$. To make the
sides parallel to the sides of $\Delta$, we increase the left slope (which
does not affect the base), and also increase the right slope to
$K-\frac{1}{2}$. The new triangle will have base

$b_{D}=(K-2)-\frac{2(K-1)}{2K-1}.$

Now one easily checks that $C\cdot D=0$.

This leaves us with the possibility that $C$ is not irreducible, but it is a
multiple of an irreducible curve. Let us rule this out. First notice that $C$
and $D$ must be disjoint. Indeed, if not, then $D$ would contain as a
component the irreducible negative curve. But $D$ is irreducible and has
positive self-intersection. Now since $D$ passes through the torus fixed
points corresponding to the right and top vertices of the triangle $\Delta$,
it follows that $C$ must not pass through these points, hence these lattice
points are in the support of $f(x,y)$. From the fact that $1-y$ does not
divide $f$ we get that the extreme left lattice point $(0,0)$ must also be in
the support of $f$. In summary, the Newton polygon of $f$ has horizontal base
of length $m-K+2$ and right edge of lattice length $\frac{K-2}{2}$. If
$f=g^{p}$ for some integer $p>1$ then $p$ must divide the lengths of these
sides. However, one easily checks that the lengths are relatively prime. ∎

Let us now turn to the case of odd $K$. We would like to prove as before that
the Newton polygon of $f$ contains the two integral vertices and the highest
lattice point $\frac{K-1}{2}(d,a)$. We can only prove this for the top lattice
point and the left vertex by showing that $1-y$ does not divide $f$. This will
be enough to prove that $f$ is irreducible.

######  Lemma 4.4.

Let $K$ be odd. Then $1-y$ does not divide $f$.

###### Proof.

We start by showing that if $1-y$ divides $f$ then also
$(1-y)^{d}=(1-y)^{2K-5}$ divides $f$. If $1-y$ divides $f$ then the lattice
points in the Newton polygon of $f$ lie in fewer than $m$ columns. Writing
$f=(1-y)g$, the lattice points in the support of $g$ are obtained by starting
with the lattice points in the support of $f$ and removing one point from each
column. If the number of columns decreases, we can repeat this step.

The triangle $\Delta$ has left slope

$\frac{2K-3}{2K-5}=1+\frac{1}{K-\frac{5}{2}}.$

Counting the lattice points in $\Delta$ in the columns $x=0,1,2,\ldots$, we
get the numbers:

$1,2,3,\ldots,K-2,K,K+1,\ldots,2K-4,2K-2,\ldots.$

Counting similarly columns from the right we get

$1,K-1,\ldots.$

Assuming that one of the extreme left or right columns with $1$ lattice point
is not in the support of $f$, when we order the remaining columns in $\Delta$
by the number of lattice points, we get $1,2,\ldots,2K-4,\ldots$. Every time
we remove one point from each column, we decrease the number of columns. We
can do at least $2K-4$ such steps, which is more than $2K-5$ that we need.
(One can in fact prove that the columns of lattice points in $\Delta$ have
size exactly $1,1,2,3,\ldots,m$. This can be used to prove the claim of the
lemma. We will instead use a different argument that only needs the $2K-5$
shortest columns.)

Let $C_{1}$ be the irreducible curve in $X$ defined by $1-y$. Note that $C$ is
irreducible with positive self-intersection, which implies that it lies in the
nef cone. Now write $C=(2K-5)C_{1}+C_{2}$ for an effective curve $C_{2}$.
Then, since $C_{1}$ lies in the nef cone, $C_{1}\cdot C_{2}\geq 0$. We will
check that this inequality is not satisfied, giving a contradiction to the
assumption that $1-y$ divides $f$. This is a straight-forward computation.

Let $H$ be the curve class in $X_{\Delta}$ corresponding to the triangle
parallel to $\Delta$ and the left side integral of lattice length $1$. The
triangle has base of length

$B_{H}=\frac{c}{b}$

and vertices

$(0,0),\quad(B_{H},0),\quad(d,a).$

We can compute the self-intersection number

$H^{2}=B_{H}a=\frac{ac}{b}.$

The class of $C_{1}$ can be written in terms of the class of $H$ as:

$\frac{bd}{ac}H-E.$

The class of $C_{2}$ is

$[C]-d[C_{1}]=\left(\frac{K-1}{2}+\frac{1}{c}-\frac{bd^{2}}{ac}\right)H-(m-d)E.$

Now the inequality $C_{1}C_{2}\geq 0$ is equivalent to

$\frac{K-1}{2}+\frac{1}{c}-\frac{bd^{2}}{ac}\geq\frac{m-d}{d}=\frac{K-3}{2}$

This can be simplified to the inequality

$bd^{2}-a\leq ac.$

Writing it out in terms of $K$ we get

$(2K-1)(2K-5)^{2}-(2K-3)\leq(2K-3)(4K^{2}-16K+1),$

Which simplifies to

$-22K+25\geq 0.$

Clearly the inequality is false when $K\geq 5$. ∎

######  Lemma 4.5.

Let $K$ be odd. Then $f$ is irreducible.

###### Proof.

Suppose that $f$ is reducible, $f=f_{1}f_{2}$, corresponding to
$C=C_{1}+C_{2}$. The Newton polygon of $f$ contains the left vertex $(0,0)$ of
$\Delta$ and its highest lattice point $\frac{K-1}{2}(d,a)$, which also lies
on the left edge. This integral edge of the Newton polygon of $f$ must be the
sum of the corresponding integral edges of the Newton polygons of $f_{1}$ and
$f_{2}$. It follows that the Newton polygons of $f_{1}$ and $f_{2}$ must have
lower left vertex $(0,0)$, the highest lattice point on the left edge
$w_{i}(d,a)$ for integers $w_{1},w_{2}$ such that $w_{1}+w_{2}=\frac{K-1}{2}$.

Let $H$ be the curve class in $X_{\Delta}$ as in the previous proof,
corresponding to a triangle with an integral left edge of lattice length $1$.
Then the class of $C$ is

$[C]=\left(\frac{K-1}{2}+\frac{1}{c}\right)H-mE$

Since $1-y$ does not divide $f$, the Newton polygon of $f$ has width exactly
$m=d\frac{K-1}{2}$. Then it must also be true that the width of the Newton
polygon of $f_{i}$ is equal to its vanishing order. Hence, we may write the
classes

$[C_{1}]=\left(w_{1}+\frac{\varepsilon_{1}}{c}\right)H-dw_{1}E,$
$[C_{2}]=\left(w_{2}+\frac{\varepsilon_{2}}{c}\right)H-dw_{2}E,$

where $w_{1},w_{2}\geq 1$ are integers, $w_{1}+w_{2}=\frac{K-1}{2}$, and
$\varepsilon_{1},\varepsilon_{2}\geq 0$ are rational numbers such that
$\varepsilon_{1}+\varepsilon_{2}=1$.

Another restriction we need to have is that the triangle corresponding to the
class $\left(w_{i}+\frac{\varepsilon_{i}}{c}\right)H$ must be the smallest
triangle parallel to $\Delta$ that contains the Newton polygon of $f_{i}$.
This implies that the triangle must contain at least one lattice point on the
right edge, and that means that $\varepsilon_{i}$ can only be $0$ or $1$. Let
us assume that $\varepsilon_{1}=0$ and $\varepsilon_{2}=1$. Then $C_{1}$ has
negative self-intersection and $C_{2}$ positive self-intersection.

In the proof for the even $K$ we used the fact that the irreducible curve $D$
defined by $\xi_{1}^{int}$ has zero intersection with the negative curve. This
is no longer true in the case of odd $K$. However, $D$ is an irreducible curve
in $X$, and since it has positive self-intersection, it must lie in the nef
cone. We get a contradiction to reducibility of $f$ by showing that $D\cdot
C_{1}<0$.

We compute the class of $D$ as in the case of even $K$. Let $B_{D}$, $B_{H}$
be the lengths of the bases of the triangles corresponding to the classes
$D,H$. Then

$[D]=\frac{B_{D}}{B_{H}}H-(K-2)E.$

Here as before

$B_{D}=(K-2)-\frac{2(K-1)}{2K-1}.$

Now using that $H^{2}=B_{H}a$, we compute

$C_{1}\cdot
D=\frac{B_{D}}{B_{H}}w_{1}H^{2}-dw_{1}(K-2)=B_{D}w_{1}a-dw_{1}(K-2).$

The claim that this product is negative is equivalent to

$B_{D}a<d(K-2),$

which one checks easily. ∎

###  4.3. Additional negative curves

We list here negative curves that do not fit into the infinite families
described above. These examples were found with the help of a computer. We
only list non-special negative curves. The section below treats the special
negative curves.

The finite set of negative curves is listed in Table 2. The columns of the
table contain the vanishing order $m$, the slopes $(s,t)$ of the triangle, and
the $\mathbb{P}(a,b,c)$ whose blowup contains the negative curve. For some
triangles it is more natural to give the triangle rather than the
$\mathbb{P}(a,b,c)$. For example, the negative curve with vanishing order
$m=8$ is supported in a triangle that is three times the integral triangle
with vertices $(0,0),(1,0),(3,7)$. This curve also appears as a negative curve
in the blowup of $\mathbb{P}(7,65,66)$, defined by a homogeneous polynomial of
degree $1386$. The curve with $m=5$ is supported in a triangle with slopes
$s=13/4,t=13/3$. This triangle does not correspond to a weighted projective
plane. The toric variety is the quotient of $\mathbb{P}^{2}$ by the cyclic
group of order $13$. Deforming the right side slightly will give us a weighted
projective plane $\mathbb{P}(13,186,185)$. The negative curve is defined by a
polynomial of degree $3344$ in the homogeneous coordinate ring.

$m$ | $(s,t)$ | Triangle | WPP | degree
---|---|---|---|---
$4$ | $(2.25,3.5)$ |  | $\mathbb{P}(7,9,10)$ | $100$
$5$ | $(3.25,4.3\overline{3})$ | $\left[\left(-\frac{4}{13},0\right),\left(\frac{14}{13},0\right),\left(\frac{68}{13},18\right)\right]$ | $\mathbb{P}(13,186,185)$ | $3344$
$6$ | $(1.8,3.5)$ |  | $\mathbb{P}(7,9,17)$ | $196$
$6$ | $(2.2,3.5)$ |  | $\mathbb{P}(7,11,13)$ | $189$
$7$ | $(1.6\overline{6},3.6\overline{6})$ |  | $\mathbb{P}(5,11,18)$ | $220$
$8$ | $(2.3\overline{3},3.5)$ | $3*[(0,0),(1,0),(3,7)]$ | $\mathbb{P}(7,65,66)$ | $1386$
$9$ | $(1.429,4.5)$ |  | $\mathbb{P}(9,10,43)$ | $559$
$11$ | $(1.6\overline{6},3.3\overline{3})$ | $2*[(0,0),(3,0),(6,10)]$ | $\mathbb{P}(5,47,71)$ | $1420$
$12$ | $(1.75,3.3\overline{3})$ |  | $\mathbb{P}(7,10,19)$ | $437$
$15$ | $(1.75,3.5)$ | $4*[(0,0),(2,0),(4,7)]$ | $\mathbb{P}(7,114,227)$ | $6384$
$16$ | $(1.6\overline{6},3.714)$ |  | $\mathbb{P}(5,26,43)$ | $1196$
$21$ | $(1.3,5.5)$ |  | $\mathbb{P}(11,13,84)$ | $2301$
$22$ | $(1.286,5.6\overline{6})$ |  | $\mathbb{P}(9,17,92)$ | $2610$
$26$ | $(1.6\overline{6},3.75)$ | $3*[(0,0),(5,0),(9,15)]$ | $\mathbb{P}(5,139,232)$ | $10440$
$31$ | $(1.4,4.75)$ |  | $\mathbb{P}(7,19,67)$ | $2926$
$99$ | $(1.4,4.6\overline{6})$ | $10*[(0,0),(7,0),(10,14)]$ |  |

Table 2. Additional negative curves.

One can list all negative curves up to isomorphism for small values of $m$.
This can be done by classifying all integral polygons containing a small
number of lattice points (see, for example, [1]), and checking which one of
them supports a negative curve. We list here all negative curves up to $m=6$.

For $m=1$ there is a unique negative curve defined by the polynomial $1-y$.
For $m=2$ there is also a unique curve with coordinates $s=1.5,t=3$ on the
map. This is the curve that is supported in $IT_{3}(1,1)$, $RT_{4}(2,1)$,
$RT_{4}(2,3)$.

For $m=3$ there are two negative curves. The first one has coordinates
$(1.3\overline{3},4)$ and is supported in $IT_{4}(2,1)$, $RT_{5}(3,1)$. The
other one has coordinates $(1.6\overline{6},4)$ and is supported in
$RT_{4}(3,2)$.

For $m=4$ there are four non-isomorphic curves. These are $(1.25,5)$ in
$IT_{5}(3,1)$, $(1.75,4)$ in $RT_{4}(4,3)$, the curve in the table above, and
the first curve for $K=4$ in the new family described in the previous section.

For $m=5$ there are four curves, $(1.2,6)$ in $IT_{6}(4,1)$, $(1.6,4)$ in
$IT_{4}(3,2)$, $(1.8,4)$ in $RT_{4}(5,4)$, and the curve in the table.

For $m=6$ there are again four curves, in $IT_{7}(5,1)$, $RT_{4}(6,5)$, and
the two curves in the table.

For $m=7$ we have found four curves, in $IT_{8}(6,1)$, $RT_{4}(7,6)$,
$IT_{4}(4,3)$, and the one in the table. There may be more curves with $m=7$.

##  5\. Special negative curves

Consider a triangle $\Delta$ supporting a negative curve with equation $\xi$
vanishing to order $m$ at $e$. Recall that we call $\xi$ (or the negative
curve $C$ it defines) special if $\Delta$ contains less than ${m+1\choose
2}+1$ lattice points. Its deficiency is the difference between ${m+1\choose
2}+1$ and the number of lattice points in $\Delta$.

Kurano and Matsuoka first discovered two examples of special negative curves
with deficiency $1$ by numerical methods. These live in the blowups of
$\mathbb{P}(8,15,43)$ and $\mathbb{P}(5,33,49)$, vanish to orders $m=9$ and
$m=18$ and lie in degrees $645$ and $1617$, respectively.

In this section we present the construction of an infinite family of special
negative curves. This family generalizes the $m=9$ curve of Kurano-Matsuoka
and has the notable property that the deficiency of its members becomes
arbitrarily large.

######  Theorem 5.1.

Let $K\geq 4$ be an integer and define $N=K-2$ and $M=N^{2}-1$. Consider the
weights

$(a,b,c)=\left(KN,KN^{2}-1,KMN^{2}-(M+N)\right).$

Then, $X=\operatorname{Bl}_{e}\mathbb{P}(a,b,c)$ is a MDS.

The negative curve $C$ in $X$ is supported in the triangle $\Delta$ with top
vertex $(N(M+N)-1,KN^{2}-1)$, bottom RHS vertex $(MN-1,0)$ and LHS slope
$\frac{KN}{M+N}$. It’s bottom LHS vertex is $\left(-\frac{K-1}{KN},0\right)$.
Furthermore, this curve is special, vanishes to order $m=N(M+N)-1$ at $e$ and
has deficiency $\frac{(K-2)(K-3)}{2}$.

For $K=4$ one recovers Kurano and Matsuoka’s $(a,b,c)=(8,15,43)$ example. The
first new curve in the family appears as soon as $K>4$. For $K=5$ we obtain a
negative curve with $m=32$ and deficiency $3$, living in
$\operatorname{Bl}_{e}\mathbb{P}(15,44,349)$. For $K=6,7$ the weights are
$(a,b,c)=(24,95,1421),(35,174,4171)$ with negative curves having $m=75,144$
and deficiencies $6$ and $10$, respectively.

######  Remark 5.2.

Unlike the construction of the infinite family presented above, this one is
constructed by explicitly describing its defining equations. In the notation
of Subsection 4.1, the values of $M$ and $N$ in the theorem correspond to
$M=M_{2}$ and $N=N_{2}$ with RHS slope $K$.

Below we will use the polynomials $\xi_{2,K}^{int}$ and $\xi^{int}_{1,K-1}$
for the proof; here the first subindex corresponds to $n$ as in Subsection
4.1, while the second one is the RHS slope of the corresponding triangle. More
concretely, the first one of these polynomials is supported on the triangle
$IT_{K}(M_{2},N_{2})=IT_{K}((K-2)^{2}-1,K-2)$, having RHS slope $K$. On the
other hand, $\xi_{1,K-1}^{int}$ is supported on the triangle
$IT_{K-1}(M_{1},N_{1})=IT_{K-1}(N_{2}-1,1)$ and has RHS slope $K-1$.

The polynomials $\Upsilon_{K}$ giving the negative curves in the theorem are
defined via the equation

(5.1) $\displaystyle
x(1-y)\Upsilon_{K}=\left(\xi_{2,K}^{int}\right)^{N}-\left(\xi_{1,K-1}^{int}\right)^{M+N}.$

The polynomials $\xi_{2,K}^{int}$ and $\xi_{1,K-1}^{int}$ vanish to order
$M+N$ and $N$ at $e$, respectively. It follows that, if it exists,
$\Upsilon_{K}$ vanishes to order $m=N(M+N)-1$ at $e$.

To simplify the notation define the auxiliary polynomials

$\Upsilon^{\prime}_{K}:=\left(\xi_{2,K}^{int}\right)^{N}-\left(\xi_{1,K-1}^{int}\right)^{M+N}.$

The outline of this section is as follows. We begin by showing that $x(1-y)$
divides $\Upsilon_{K}^{\prime}$, so that the polynomials $\Upsilon_{K}$ are
well defined. The irreducibility of $\Upsilon_{K}$ is a direct consequence of
this. The remainder of the section is devoted to computing the deficiency of
these curves.

Since $\Upsilon_{K}^{\prime}=x\Upsilon_{K}-xy\Upsilon_{K}$, the Newton polygon
of $xy\Upsilon_{K}$ can be obtained from that one of $\Upsilon_{K}^{\prime}$
by removing the lowest lattice point on each column of its support. We use
this in the proof of Proposition 5.6 to compute the Newton polygon of
$\Upsilon_{K}$. Having described the Newton polygon of $\Upsilon_{K}$,
computing its deficiency is a direct application of Pick’s theorem.

######  Lemma 5.3.

The polynomial $\Upsilon_{K}^{\prime}$ is divisible by $x$ and $1-y$ for every
$K\geq 4$.

###### Proof.

By definition, every $\xi^{int}$ is normalized to have constant term $1$, so
clearly $x$ divides $\Upsilon_{K}^{\prime}$. To see that $1-y$ divides
$\Upsilon^{\prime}_{K}$ note that $\xi_{1,K-1}^{int}|_{y=1}=(1-x)^{N}$ and
$\xi_{2,K}^{int}|_{y=1}=(1-x)^{M+N}$. Indeed, the monomials $x^{N}y^{N+1}$ and
$x^{M+N}y^{KN}$ are part of the support of $\xi_{1,K-1}^{int}$ and
$\xi_{2,K}^{int}$, respectively, with coefficients $\pm 1$. Then, since
$\xi_{2,K}^{int}\in(1-x,1-y)^{M+N}$ and $\xi_{1,K-1}^{int}\in(1-x,1-y)^{N}$,
the result follows. ∎

######  Proposition 5.4.

The polynomial $\Upsilon_{K}$ is irreducible.

###### Proof.

Let $\Delta OA^{\prime}B^{\prime}$ denote the triangle $N\cdot IT_{K}(M,N)$
and $\Delta OP^{\prime}Q^{\prime}$ the triangle $(M+N)\cdot IT_{K-1}(N-1,1)$.
Here $O$ is the origin and vertices are written in counterclockwise order; see
Figure 3.

The coefficients of $\Upsilon_{K}^{\prime}$ at $A^{\prime}$ and $B^{\prime}$
come uniquely from those of $\left(\xi_{2,K}^{int}\right)^{N}$. Indeed, LHS
slope of $IT_{K}(M,N)$ is larger than that one of $IT_{K-1}(N-1,1)$, and the
base of $N\cdot IT_{K}(M,N)$ is larger than that one of $(M+N)\cdot
IT_{K-1}(N-1,1)$. By [5, Paragraph 6.2], the coefficients of $\xi_{2,K}^{int}$
at all of its vertices are $\pm 1$, so the coefficients of
$\Upsilon_{K}^{\prime}$ at $A^{\prime}$ and $B^{\prime}$ are also $\pm 1$.

Notice that $B^{\prime}=Q^{\prime}+(0,1)$, so that the coefficient at
$Q^{\prime}$ is the negative of that one at $B^{\prime}$, this is, $\pm 1$.
Furthermore, by the same token the edge $\overline{A^{\prime}Q^{\prime}}$ must
be an edge of the support of the $x\Upsilon_{K}$ term of
$\Upsilon_{K}^{\prime}$. Then, by a standard argument using Minkowski sums,
$\Upsilon_{K}$ will be irreducible if we can prove that this edge has lattice
length $1$, see for example [5, Lemma 2.3].

The fact that $\overline{A^{\prime}Q^{\prime}}$ has length $1$ is immediate
once we note that $\overline{A^{\prime}B^{\prime}}$ has integral slope $K$ and
$B^{\prime}=Q^{\prime}+(0,1)$. ∎

The remainder of this section will be concerned with computing the deficiency
of $\Upsilon_{K}$. For this we will describe the Newton polygon of
$\Upsilon_{K}^{\prime}$ and then use Pick’s theorem to obtain the result.

######  Lemma 5.5.

After removing the origin, the convex hull of the remaining lattice points in
the triangles $N\cdot IT_{K}(M,N)$ and $(M+N)\cdot IT_{K-1}(N-1,1)$ has
vertices $A^{\prime},Q^{\prime},B^{\prime},C^{\prime},D^{\prime},(1,1)$ and
$(1,0)$, where

$\displaystyle A^{\prime}$ $\displaystyle=\text{bottom RHS vertex of }N\cdot
IT_{K}(M,N)=(MN,0),$ $\displaystyle Q^{\prime}$ $\displaystyle=\text{top
vertex of }(M+N)\cdot IT_{K-1}(N-1,1)=(N(M+N),(N+1)(M+N)),$ $\displaystyle
B^{\prime}$ $\displaystyle=\text{top vertex of }N\cdot
IT_{K}(M,N)=(N(M+N),KN^{2}),$ $\displaystyle C^{\prime}$
$\displaystyle=\text{top vertex of }IT_{K}(M,N)=(M+N,KN)\text{ and}$
$\displaystyle D^{\prime}$ $\displaystyle=\text{top vertex of
}IT_{K-1}(N-1,1)=(N,N+1).$

See Figure 3. In particular, the polynomial $\Upsilon_{K}^{\prime}$ is
supported in this polygon.

$O$$P^{\prime}$$Q^{\prime}=(M+N)\cdot
D^{\prime}$$A^{\prime}=N\cdot(M,0)$$B^{\prime}=N\cdot
C^{\prime}$$C^{\prime}=(M+N,KN)$$D^{\prime}=(N,N+1)$$(1,0)$$(1,1)$ Figure 3.
Newton polygons of $\left(\xi_{1,K-1}^{int}\right)^{M+N}$ and
$\left(\xi_{2,K}^{int}\right)^{N}$ on top of each other.

###### Proof.

It follows from the discussion in the proof of Proposition 5.4 that the
vertices $A^{\prime},B^{\prime}$ and $Q^{\prime}$ all are all in the support
of $\Upsilon_{K}^{\prime}$ with coefficients $\pm 1$.

Since the LHS slope of $IT_{K}(M,N)$ is greater than that one of
$IT_{K-1}(N-1,1)$, the point $C^{\prime}$ is also a vertex of the support of
$\Upsilon_{K^{\prime}}$ (one can check that it has coefficient $\pm N$ because
the coefficient of $\xi_{2,K}^{int}$ at its top vertex is $\pm 1$). It only
remains to show that the missing vertices are $D^{\prime},(1,1)$ and $(1,0)$.

To see this we show that the (non-convex) polygon with vertices
$O,D^{\prime},(1,1)$ and $(1,0)$ has no lattice points in its interior.
Indeed, a direct computation shows that the vectors $(1,1)$ and $D^{\prime}$
form a basis of the lattice, just as $D^{\prime}$ and $C^{\prime}$ do.∎

With the previous result we can now determine the Newton polygon of
$\Upsilon_{K}$.

######  Proposition 5.6.

The Newton polygon of $\Upsilon_{K}$ has vertices

$\displaystyle O$ $\displaystyle=(0,0),$ $\displaystyle A$
$\displaystyle=(MN-1,0),$ $\displaystyle B$
$\displaystyle=(N(M+N)-1,KN^{2}-1),$ $\displaystyle C$
$\displaystyle=(M+N-1,KN-1)\text{ and}$ $\displaystyle D$
$\displaystyle=(N-1,N).$

###### Proof.

Since $\Upsilon^{\prime}_{K}=x(1-y)\Upsilon_{K}$, the polynomial
$\Upsilon_{K}$ is supported in the polygon obtained by removing the bottom
layer of lattice points in the polygon from Lemma 5.5 and then translating the
result by $(-1,-1)$. We show that the resulting polygon has vertices $O,A,B,C$
and $D$ and that the coefficients of $\Upsilon_{K}$ at these points are
nonzero.

First note that every lattice point along the bottom edge of the polygon
supporting $\Upsilon_{K}^{\prime}$ is removed by this process. All the other
points that get removed are $Q^{\prime}$ and all lattice points along the edge
$\overline{A^{\prime}B^{\prime}}$ except for $B^{\prime}$. Indeed,
$\overline{A^{\prime}B^{\prime}}$ is the RHS edge of $N\cdot IT_{K}(M,N)$ and
has integral slope, so the only possibility is for the points along it or
below it to get removed by this process. However, since
$B^{\prime}=Q^{\prime}+(0,1)$, it’s clear that there are no lattice point
points in the support of $\Upsilon_{K}^{\prime}$ lying below those along the
edge.

It follows that the Newton polygon of $xy\Upsilon_{K}$ is contained in the
convex hull of $(1,1)$, $A^{\prime}+(0,1)$, $B^{\prime}$, $C^{\prime}$ and
$D^{\prime}$. Translating by $(-1,-1)$ yields the polygon with vertices
$O,A,B,C$ and $D$.

Now we show that $\Upsilon_{K}$ has nonzero coefficients at these points.

From the proof of Lemma 5.5 we know that the coefficients of $\Upsilon_{K}$ at
$A$ and $B$ are $\pm 1$. Similarly, the coefficient of $\Upsilon_{K}^{\prime}$
at $C^{\prime}$ is $\pm N$ because the coefficient of $\xi_{2,K}^{int}$ at its
top vertex is $\pm 1$, so the same is true for the coefficient of
$\Upsilon_{K}$ at $C$.

The constant term of $\Upsilon_{K}$ is equal to $\pm 1$. It’s nonzero because
otherwise the polynomial would be divisible by $1-y$ by Lemma 4.2. This is
impossible because $B$ is in its support with coefficient $\pm 1$ and no other
point along the same column is. Furthermore, the coefficient is equal to $\pm
1$ because the previous argument as well as the proof of Lemma 4.2 are
independent of the characteristic of the field.

To finish the proof we show that the coefficient of $\Upsilon_{K}$ at $D$ is
nonzero or, equivalently, that the same is true for the coefficient of
$\Upsilon_{K}^{\prime}$ at $D^{\prime}$. The contributions to this coefficient
come from both $\left(\xi_{2,K}^{int}\right)^{N}$ and
$\left(\xi_{1,K-1}^{int}\right)^{M+N}$. On the one hand, $D^{\prime}$ is the
top vertex of $IT_{K-1}(N-1,1)$, so $\left(\xi_{1,K-1}^{int}\right)^{M+N}$
contributes by $\pm(M+N)$. On the other hand, we will show that
$\left(\xi_{2,K}^{int}\right)^{N}$ has coefficient $rN$ at this vertex for
some $r\in{\mathbb{Z}}$. This would imply the claim because the coefficient of
$\Upsilon_{K}$ at $D$ would be, up to a sign,

$(M+N)\pm rN=M+(1\pm r)N\neq 0.$

This can’t be zero because $M$ and $N$ are relatively prime.

To compute the contribution of $\left(\xi_{2,K}^{int}\right)^{N}$ to this
coefficient note that the segment $\overline{OD^{\prime}}$ has length $1$, and
the only other lattice point in $IT_{K}(M,N)$ having slope greater than or
equal to that one of $D^{\prime}$ is its top vertex. However, the top vertex
is further away from the origin than $D^{\prime}$, so the coefficient of
$\left(\xi_{2,K}^{int}\right)^{N}$ at $D^{\prime}$ is the same as that one of
$(1+r\chi^{D^{\prime}})^{N}$, where $r$ is the coefficient of
$\xi_{2,K}^{int}$ at $D^{\prime}$. Since the polynomials $\xi_{IT}$ are
defined over ${\mathbb{Z}}$, this concludes the proof. ∎

Computing the deficiency of $\Upsilon_{K}$ is now a straightforward but
lengthy computation, which is left to the reader. We present the following
lemma as a guide.

######  Lemma 5.7.

The Newton polygon of $\Upsilon_{K}$ has:

  1. (i)

Lattice perimeter $N(M+1)+1$, and

  2. (ii)

$\operatorname{Area}=\frac{1}{2}\left(KMN^{3}-KN^{2}-MN+M+N\right)$.

By Pick’s theorem, its number of lattice points is:

$\frac{1}{2}\left(KMN^{3}-KN^{2}+2N+M+3\right).$

As a consequence, the deficiency of $\Upsilon_{K}$ is $\frac{(K-2)(K-3)}{2}$.

###### Proof.

We show only (i). Let $A,B,C$ and $D$ be as in Proposition 5.6.

  1. (i)

From the proof of Lemma 5.5 we know that the segments $\overline{OD}$ and
$\overline{CD}$ have lattice length $1$ and $\overline{OA}$ has lattice length
$MN-1$. Similarly, we showed above that the edge $\overline{AB}$ has lattice
length $1$. Finally,

$B=C+(N-1)(M+N,KN),$

where $M+N$ and $KN$ are relatively prime. Therefore, $\overline{BC}$ has
length $N-1$. Adding up these lengths yields a perimeter of $N(M+1)+1$.

∎

To conclude this section we piece together all the previous result to prove
Theorem 5.1.

###### Proof of Theorem 5.1..

By equation 5.1 we know that $\Upsilon_{K}$ has $m=N(M+N)-1$. The previous
lemma shows that this curve has the desired deficiency. It only remains to
show that the polynomial is supported in the triangle described in the
statement. The vertices $A$ and $B$ from Proposition 5.6 are precisely the
ones described in the statement of the theorem, and a direct computation shows
that the edge $\overline{BC}$ has slope $\frac{KN}{M+N}$.

The self-intersection of the negative curve $C$ defined by $\Upsilon_{K}$ is:

$C\cdot
C=\left(MN-1+\frac{K-1}{KN}\right)(KN^{2}-1)-(N(M+N)-1)^{2}=-\frac{K-1}{KN}.$

∎

##  6\. The Mori Dream Space property

In this section we consider the Mori Dream Space property of the blowups $X$.
The main observation is that this property is often determined by the location
of the diamonds of the negative curves on the $st$-plane.

By a result of Cutkosky, $X$ is a MDS if and only if it has a negative curve
$C$ and an irreducible curve $D$ disjoint from it. Consider a blowup $X$ that
corresponds to a point on the $st$-plane where two diamonds meet. The two
negative curves are disjoint in $X$, hence $X$ is a MDS. If we now move to the
interior of one diamond, then under some conditions the two curves stay
disjoint and the variety $X$ stays a MDS. In fact, we will see that this idea
explains almost all known cases of the MDS property for $X$. We start with a
proposition that is the basis for this argument.

We say that a triangle $\Delta$ supports a negative curve $C$ if the Newton
polygon of the equation of $C$ lies in $\Delta$, and $\Delta$ is the smallest
such triangle parallel to $\Delta$.

######  Proposition 6.1.

Let $\Delta_{1}\subseteq\Delta_{2}$ be two triangles that support the same
negative curve $C$. If $X_{2}=\operatorname{Bl}_{e}X_{\Delta_{2}}$ is a MDS
then so is $X_{1}=\operatorname{Bl}_{e}X_{\Delta_{1}}$.

###### Proof.

Consider a variety $Y$ with proper birational morphisms $\pi_{i}:Y\to X_{i}$,
$i=1,2$. Here $Y=\operatorname{Bl}_{e}Z$, where $Z$ is the toric surface whose
fan is the coarsest refinement of the normal fans of $\Delta_{1},\Delta_{2}$.
We consider curves in $Y$ and $X_{i}$ as either $\mathbb{Q}$-Cartier divisors
or as Chow classes. We then have the pullback $\pi_{i}^{*}$ of divisors and
the proper pushforward $(\pi_{i})_{*}$ of the Chow classes.

Let $C_{i}\subset X_{i}$ be the negative curves. The assumption that
$\Delta_{1}\subseteq\Delta_{2}$ means that $\pi_{1}^{*}(C_{1})$ is an
irreducible curve in $Y$. Indeed, the smallest polygon that supports the
equation of $C_{1}$ and has sides parallel to the sides of $\Delta_{1}$ and
$\Delta_{2}$ is the triangle $\Delta_{1}$. Then
$C_{2}=(\pi_{2})_{*}\pi_{1}^{*}(C_{1})$.

Let $D_{2}$ be an irreducible curve in $X_{2}$ orthogonal to $C_{2}$, and let
$D_{1}=(\pi_{1})_{*}\pi_{2}^{*}(D_{2})$. This $D_{1}$ is an effective curve in
$X_{1}$, possibly reducible. It is orthogonal to $C_{1}$:

$C_{1}\cdot
D_{1}=C_{1}\cdot(\pi_{1})_{*}\pi_{2}^{*}(D_{2})=\pi_{1}^{*}(C_{1})\cdot\pi_{2}^{*}(D_{2})=(\pi_{2})_{*}\pi_{1}^{*}(C_{1})\cdot
D_{2}=C_{2}\cdot D_{2}=0.$

The curve $D_{1}$ consists of the strict transform of the curve $D_{2}$ plus a
linear combination of boundary divisors. There is no component $C_{1}$ in it,
hence $D_{1}$ and $C_{1}$ are disjoint. This implies that $D_{1}$ is in fact
irreducible, equal to the strict transform of $D_{2}$. ∎

The proposition shows that if the triangles supporting the negative curve $C$
satisfy the containment $\Delta_{1}\subseteq\Delta_{2}$, then the equation
defining the curve $D$ in $X_{2}$ also defines the curve $D$ in $X_{1}$. This
implies that the triangles supporting the equation of $D$ satisfy the reverse
containment $\Delta^{\prime}_{2}\subseteq\Delta^{\prime}_{1}$.

Next we explain how to check if $\Delta_{1}\subseteq\Delta_{2}$. Let
$(s_{i},t_{i})$ be the slopes of $\Delta_{i}$. We start with the cases where
either $s_{1}=s_{2}$ or $t_{1}=t_{2}$. These are determined as follows:

  1. (1)

Assume that $\Delta_{1}$ has the lower left vertex integral. If
$s_{2}>s_{1},t_{2}=t_{1},$ then $\Delta_{1}\subseteq\Delta_{2}$. Indeed, we
pivot the left edge of $\Delta_{1}$ around its lower left vertex, making the
triangle larger.

  2. (2)

Assume that $\Delta_{1}$ has the lower right vertex integral. If
$s_{2}=s_{1},t_{2}<t_{1},$ then $\Delta_{1}\subseteq\Delta_{2}$.

  3. (3)

Assume that $\Delta_{1}$ has the top vertex integral. If
$s_{2}=s_{1},t_{2}>t_{1},$ or $s_{2}<s_{1},t_{2}=t_{1},$ then
$\Delta_{1}\subseteq\Delta_{2}$.

All other cases do not yield $\Delta_{1}\subseteq\Delta_{2}$ when varying
$s_{1}$ or $t_{1}$. We can combine these cases and vary both $s_{1}$ and
$t_{1}$:

  1. (1)

Assume that $\Delta_{1}$ has the top vertex integral. If $s_{2}\leq
s_{1},t_{2}\geq t_{1},$ then $\Delta_{1}\subseteq\Delta_{2}$. We first change
$s_{1}$ to $s_{2}$ by pivoting around the top vertex. This keeps the top
vertex integral, so we can change $t_{1}$ to $t_{2}$. Similarly, if
$\Delta_{1}$ has its lower left and right vertices integral and $s_{2}\geq
s_{1},t_{2}\leq t_{1},$ then $\Delta_{1}\subseteq\Delta_{2}$.

  2. (2)

Consider the case $s_{2}>s_{1},t_{2}>t_{1}$. Even if all vertices of
$\Delta_{1}$ are integral, the condition $\Delta_{1}\subseteq\Delta_{2}$ may
not be satisfied. Indeed, if we first change $s_{1}$ to $s_{2}$, then the top
vertex may become rational, and we are not allowed to change $t_{1}$ to
$t_{2}$.

We now consider various diamonds of negative curves and show that the previous
proposition explains most known examples where the MDS property holds, at
least for the case $m>1$.

  1. (1)

Consider the triangle $IT(M,N)$. It supports a negative curve with Newton
polygon equal to the triangle itself. Let $(s_{0},t_{0})$ be its coordinates.
This is the point in the centre of the diamond. Consider the line segment from
$(s_{0},t_{0})$ to $(s_{0},t_{1})$, where the second point lies at the top
vertex of the $IT(M,N)$ diamond. The point $(s_{0},t_{1})$ is the intersection
point of the $IT(M,N)$ diamond and the region of the negative curve defined by
$1-y$. As we move up from $(s_{0},t_{0})$ towards $(s_{0},t_{1})$, we increase
the triangle supporting the negative curve $C$. Since the variety
corresponding to $(s_{0},t_{1})$ is a MDS, so is every other variety $X$
corresponding to a point on the line segment.

The left and right vertices of the $IT(M_{n},N_{n})$ diamond meet diamonds of
$RT(M_{n-1},N_{n-1})$ and $RT(M_{n},N_{n})$. By a similar argument, the whole
horizontal line in the $IT(M,N)$ diamond through $(s_{0},t_{0})$ has the MDS
property. The lower vertex of the $IT(M,N)$ diamond meets another diamond only
for $(M,N)=(M_{n},N_{n})$, $n=1,2$ and for $(M,N)=(N_{1},M_{1})$. For
instance, the diamond of $IT_{K}(M_{2},N_{2})=IT_{K}((K-2)^{2}-1,K-2)$ meets
the diamond of the special negative curve in the infinite family. In these
cases the lower half of the vertical axis in the $IT(M,N)$ diamond has the MDS
property.

  2. (2)

For triangles $RT(M,N)$, the same argument as in the integral case shows that
points on the horizontal axis and the upper half of the vertical axis have the
$MDS$ property. When we move down from $(s_{0},t_{0})$, then the triangle
supporting $C$ does not get bigger, hence we cannot apply the proposition to
this case. Except for the small values $(M_{1},N_{1})=(K-2,1)$ and
$(N_{1},M_{1})$, the lower vertex of the $RT(M,N)$ diamond is irrational,
hence one does not expect it to meet another diamond.

  3. (3)

Consider the diamond of the negative curve that is supported in $RT_{4}(2,1)$
and $IT_{3}(1,1)$ (see Figure 4). This diamond meets other diamonds not only
at its four vertices but also at other points on its boundary. Let
$(s_{0},t_{0})=(1.5,3)$ be the centre of the diamond and $(s_{1},t_{1})$ a
point on the boundary that also lies in another diamond. If the point
$(s_{1},t_{1})$ lies either on the north-west or south-east boundary, then all
points in the rectangle $[s_{0},s_{1}]\times[t_{0},t_{1}]$ have the MDS
property. Indeed, for all these points the triangle supporting the negative
curve $C$ lies in the triangle at $(s_{1},t_{1})$. If the $(s_{1},t_{1})$ lies
on the north-east boundary, then we can only conclude the MDS property for the
horizontal line $s_{0}\leq s\leq s_{1},t=t_{1}$. Along this line we have
containment of triangles. Similarly, when $(s_{1},t_{1})$ lies on the south-
west boundary, we get the MDS property on the vertical line $s=s_{1},t_{1}\leq
t\leq t_{0}$.

The triangle $RT_{4}(2,1)$ contains the same negative curve, but has
coordinates $(1.5,4)$. The general proof for triangles $RT(M,N)$ in this case
shows that points above the line $t=4$ are non-MDS, unless they lie on the
vertical axis $s=1.5$.

There is an obvious symmetry in the diamond. This stems from the fact that
only the north-east half of the diamond lies in the fundamental domain. There
is an isomorphism of triangles that maps the diamond to itself. It is given by
first transforming the triangle by the shear transformation
$(x,y)\mapsto(x-y,y)$ and then reflecting in the $y$ axis.

A similar argument applies to the negative curves supported in $RT(K-2,1)$ and
$IT(K-3,1)$ for $K>4$.

Figure 4. The big diamond of the negative curve in $IT_{3}(1,1)$ and
$RT_{4}(2,1)$ with $m=2$. Figure is not to scale.

  4. (4)

Consider the new family of non-special curves constructed in Section 4.2. For
even $K$, the diamonds of these curves meet the diamonds of
$RT_{K}(M_{1},N_{1})$. Hence, the left half of their horizontal axis has the
MDS property. In particular, the centres of the diamonds, corresponding to the
minimal triangles that support the negative curve, have the MDS property.

  5. (5)

The diamonds of the infinite family of special curves meets the diamond of
$IT_{K}(M_{2},N_{2})$ at its highest point and the diamond of
$RT_{K}(M_{1},N_{1})$ at its left vertex. This means that the top half of the
vertical axis and the left half of the horizontal axis have the MDS property.
In particular, the centre points $(s_{0},t_{0})$ have the MDS property.

The $m=18$ special curve from [8] is similar to the other special curves. Its
diamond meets the $RT(2,1)$ diamond at its left vertex and the $m=11$ diamond
at the top vertex. In particular, the blowups of the two examples in [8],
$\mathbb{P}(5,33,49)$ and $\mathbb{P}(8,15,43)$ are both MDS.

  6. (6)

In the column $s=5/3$ there is a continuous string of diamonds corresponding
to curves with $m=7,4,11,18$. We get the MDS property in the line segment from
the centre of the $m=7$ diamond to the centre of the $m=18$ diamond. This
gives a range of $b,c$ such that $\operatorname{Bl}_{e}\mathbb{P}(5,b,c)$ is a
MDS.

  7. (7)

Consider the region where the negative curve is defined by the vanishing of
$1-y$. Here if we have a point $(s,t)$ with the MDS property, then any point
above it, $(s,t^{\prime})$ with $t^{\prime}>t$, also has this property. We
will see in the following lemma that all points that lie on the horizontal
lines having $t=K$ an integer are MDS. Other than these examples, we also get
vertical lines above the centres of $IT(M,N)$ and $RT(M,N)$.

######  Lemma 6.2.

Let $\Delta$ support the negative curve $C$ defined by the polynomial $1-y$.
If the right slope $t$ of $\Delta$ is an integer, then
$X=\operatorname{Bl}_{e}X_{\Delta}$ is a MDS.

###### Proof.

The class of $D$ that is orthogonal to $C$ must have the form

$[D]=H^{\prime}-nE,$

where $H^{\prime}$ corresponds to a triangle $\Delta^{\prime}$ parallel to
$\Delta$ and with width equal to $n$. We take for $\Delta^{\prime}$ the
smallest such triangle that has the left edge with integral vertices and claim
that this triangle supports the required curve $D$.

To prove the claim, we use the same argument as in [5, Section 2]. Notice that
$\Delta^{\prime}$ contains $n+1$ columns of lattice points. If the count of
lattice points in columns is $1,1,2,3,4,\ldots,n$, then $\Delta^{\prime}$
supports a unique polynomial with vanishing order $n$ at $e$, and whose
coefficients at the left and top lattice points are nonzero. This polynomial
then defines the curve $D$ that is disjoint from $C$. We will show that this
count of lattice points in columns holds after we remove some lattice points
in $\Delta^{\prime}$. The remaining points then support the polynomial
defining $D$.

The vertical line through the lower right vertex divides $\Delta^{\prime}$
into two sub-triangles. We transform $\Delta^{\prime}$ by applying a shear
transformation to the right sub-triangle, making the right edge with slope $t$
horizontal. This operation does not change the number of lattice points in
columns. The result of this shear is a new triangle with horizontal base
$[0,n]$, and height $\geq n$. Indeed, if $C^{2}\leq 0$ then the area of
$\Delta^{\prime}$ is $\geq\frac{n^{2}}{2}$. Choose a sub-triangle
$\Delta^{\prime\prime}$ in the sheared triangle with the same base $[0,n]$ and
height equal to $n$. We consider all lattice points in $\Delta^{\prime\prime}$
except those on the left edge other than the left vertex. These lattice points
have as their row count $n+1,n-1,n-2,\ldots,3,2,1$, and hence they have the
correct count in columns. ∎

Notice that in the case where the curve $C$ is defined by $1-y$, the curve $C$
passes through the torus-fixed points corresponding to left and top vertices
of the triangle. The curve $D$ must then have these vertices in its Newton
polygon. This means that if the two triangles have different slopes $s$ of the
left edge then the curves $D$ must be defined by different polynomials. Hence,
for every rational point in the line $t=K$ we get a different polynomial
defining the curve $D$.

The case where the curve $C$ is defined by $1-y$ is also interesting for the
following reason. In all other examples listed above, the curve $D$ is defined
by a polynomial that also defines a negative curve in a neighbouring diamond.
This is not necessarily true in the $1-y$ case.

###  6.1. The non-MDS property

Proving that a variety $X$ is not a MDS is more difficult than proving that it
is a MDS. We expect that all $X$ are non-MDS unless there is a clear reason
for them to be a MDS, like the ones listed above. We prove this for the
diamonds of $IT(M,N)$ and the upper half of the diamonds of $RT(M,N)$, except
for some small values of $M$ and $N$.

######  Lemma 6.3.

The following points in the diamonds of $IT_{K}(M,N)$ and $RT_{K}(M,N)$ have
the non-MDS property:

  1. (1)

When $N>K-2$, all points in the diamond of $IT_{K}(M,N)$ except those on the
horizontal axis and the upper half of the vertical axis.

  2. (2)

When $N=K-2$ for $IT_{K}(M,N)$ and $M+N>1$ for $RT_{K}(M,N)$, all points in
the upper part of the diamond, $t>K$, except those on the vertical axis.

###### Proof.

These results were proved in [5, Lemma 5.5] for the upper left quadrant of the
diamonds. The same proof works also for the upper right quadrant. We will
recall the main steps of the proof and then use a similar argument for the
lower half of the diamond in the integral case.

Let $(s_{0},t_{0}=K)$ be the coordinates of the triangle $IT_{K}(M,N)$ or
$RT_{K}(M,N)$, and let $(s_{1},t_{1}>K)$ be a point in the upper half of the
diamond. We let the corresponding triangles that support the same negative
curve $C$ be $\Delta_{0}$ an $\Delta_{1}$. Then
$\Delta_{0}\subseteq\Delta_{1}$. The proof of Proposition 6.1 shows that if a
polynomial $\zeta$ defines the curve $D$ for $\Delta_{1}$, then the same
polynomial also defines the curve $D$ for $\Delta_{0}$. However, we already
know several curves $D$ for $\Delta_{0}$, for example the curve defined by
$1-y$. For suitable integers $q,r$, the polynomials $\zeta$ and
$x^{q}(1-y)^{r}$ lie in the same class and hence are sections of the same line
bundle. Since these sections do not vanish when restricted to the curve
$C\subseteq\operatorname{Bl}_{e}X_{\Delta_{0}}$, one is a constant multiple of
the other when restricted to $C$. Let $\xi$ be the polynomial defining $C$.
Then we can write

$\zeta=cx^{q}(1-y)^{r}+\xi g$

for a suitable constant $c$ and polynomial $g$. The contradiction to the
existence of $\zeta$ comes from checking the class of the curve $\overline{D}$
defined by $g$. Assuming that $t_{1}>t_{0}$, this curve $\overline{D}$
intersects $C$ negatively in $\operatorname{Bl}_{e}X_{\Delta_{0}}$ by the
argument in [5, Lemma 5.3 and Lemma 5.5], which is stated for the upper left
quadrant of the diamond but runs word-by-word also for the upper right
quadrant. Hence, $D$ must have $C$ as a component, which means that $\xi$
divides $g$. However, this can be ruled out by restricting the polynomials to
the left edge of the triangle $\Delta_{0}$, and showing that the restriction
of $\zeta-cx^{q}(1-y)^{r}$ is not divisible by the restriction of $\xi^{2}$.
Indeed, assuming that $s_{1}\neq s_{0}$, $\zeta-cx^{q}(1-y)^{r}$ is supported
at two lattice points on the left edge. The multiple roots of a single
variable polynomial that contains exactly two nonzero terms must all be zero,
but the roots of $\xi$ are not all zero because $\xi$ also has two nonzero
terms on the left edge.

We will now use a similar argument for the lower half of the $IT(M,N)$
diamonds. We let $\Delta_{0},\Delta_{1}$ be as before, with $(s_{1},t_{1}<K)$
now in the lower half of the diamond. Then in the integral case
$\Delta_{0}\subseteq\Delta_{1}$. (This is no longer true for rational
triangles and that is why we cannot say anything about the lower half of these
diamonds.) Assuming that there is a curve $D$ for $\Delta_{1}$ defined by a
polynomial $\zeta$, we get two such curves $D$ for the triangle $\Delta_{0}$,
defined by $\zeta$ and $1-y$. As before, these must satisfy an equation

$\zeta=cx^{q}(1-y)^{r}+\xi g$

for a constant $c$ and a polynomial $g$. We can assume that $\zeta$ has
rational coefficients, since defining a (possibly reducible) $D$ curve is
characterized by the coefficients of $\zeta$ satisfying a system of
homogeneous linear equations with rational coefficients, so if there is a
nonzero solution over $k$ there is one over $\mathbb{Q}$. Since the monomial
$x^{q}$ has a zero coefficient in $\xi g$, we deduce that in this case
$c\in\mathbb{Q}$. We get a contradiction by showing that the difference
$\zeta-cx^{q}(1-y)^{r}$ is not divisible by $\xi$. This can be detected
already when we restrict the polynomials to the right edge of the triangle
$\Delta_{0}$. These restrictions lie in the ring $\mathbb{Q}[z]$, where we let
the lower right vertex correspond to $1$, the next lattice point to $z$, and
so on. The restriction of $\zeta-cx^{q}(1-y)^{r}$ has the form $dz^{p}-c$ for
some integer $p>0$ and some $d\in\mathbb{Q}$. Notice that all roots of this
polynomial have the same absolute value. Now it suffices to show that $\xi$
has roots with different absolute values.

To help the flow of this argument we postpone some computational details to
Lemma 6.4 and Lemma 6.5 below. Let $(M_{n},N_{n})=(F_{n+1},F_{n})$ be as
defined in Subsection 4.1. We will show in the proof of Lemma 6.5 that the
polynomial $\xi=\xi^{int}_{M,N}$, when restricted to the right edge of the
triangle, has the form

$\pm 1+a_{n}z+\ldots+b_{n}z^{N-1}\pm z^{N}.$

By Lemma 6.4, if all (complex) roots of $\xi$ have the same absolute value,
then $|a_{n}|=|b_{n}|$. The coefficient $b_{n}$ was computed in [5, Lemma 6.3]
and we will bound the coefficient $a_{n}$ in Lemma 6.5. Then, these two
coefficients are as follows.

First assume that $M>N$, so that $(M,N)=(M_{n},N_{n})=(F_{n+1},F_{n})$. Then,

$|b_{n}|=F_{n-1}+F_{n-2},\quad|a_{2n}|=|a_{2n-1}|\leq n.$

For $K\geq 4$ and $n\geq 3$ this implies $|b_{n}|>|a_{n}|$.

The case $M<N$ corresponds to $(M,N)=(N_{n},M_{n})=(F_{n},F_{n+1})$. In this
case

$|b_{n}|=F_{n}+F_{n-1},\quad|a_{2n+1}|=|a_{2n}|\leq n+1.$

Now, for $K\geq 4$ and $n\geq 2$ we have $|b_{n}|>|a_{n}|$. ∎

To summarize, the only cases where we don’t know about the MDS property for
$IT_{K}(M,N)$ and $RT_{K}(M,N)$ are (assume $m>1$):

  1. (1)

The lower half, $t<K$, of the diamonds of $RT_{K}(M,N)$.

  2. (2)

The lower half of the diamonds of the first integral triangles
$IT_{K}(M_{2},N_{2})=IT_{K}((K-2)^{2}-1,K-2)$ and
$IT_{K}(N_{1},M_{1})=IT_{K}(K-2,1)$. For example, $IT_{4}(3,2)$,
$IT_{4}(1,2)$. We know that at least the vertical axis in these diamonds
consists of MDS points.

  3. (3)

The diamonds of the first rational triangles
$RT_{K}(M_{1},N_{1})=RT_{K}(K-2,1)$. (These are the same as the diamonds
$IT_{K-1}(M_{1},N_{1})$). For example, the diamond of $RT_{4}(2,1)$ and
$IT_{3}(1,1)$ that is shown in Figure 4.

######  Lemma 6.4.

Consider a real polynomial of the form

$\xi=\pm 1+az+\ldots+bz^{N-1}\pm z^{N}.$

If all roots of the polynomial have the same absolute value then $|a|=|b|$.

###### Proof.

Since the highest and lowest degree terms have coefficients $\pm 1$, if all
roots have the same absolute value, this absolute value must be $1$. Let
$\alpha_{i}$ be the (complex) roots of the polynomial. Then

$|a|=\left|\sum\frac{1}{\alpha_{i}}\right|=\left|\sum\overline{\alpha_{i}}\right|=\left|\sum\alpha_{i}\right|=|b|.$

∎

Let us now bound the coefficient $a$ in $\xi^{int}$. Fix $K\geq 4$ and
consider $M>N$, $(M,N)=(M_{n},N_{n})$. We are interested in the polynomial
$\xi_{n}^{int}$ restricted to the right edge of the triangle. The isomorphism
of triangles $IT(M,N)\cong IT(N,M)$ in [5, Lemma 3.3] exchanges the lower
edges. Hence, $\xi$ restricted to the right edge in the case $M<N$ is equal to
$\xi$ restricted to the bottom edge in the case $M>N$. To cover both cases, it
suffices to study the coefficients of $\xi_{n}^{int}$ restricted to the lower
two edges.

Denote the terms in $\xi_{n}^{int}$ supported on the lower two edges of the
triangle as

$\xi_{n}^{int}=1+\ldots+a^{\prime}_{n}x^{M-1}+\delta_{n}x^{M}+a_{n}x^{M+1}y^{K}+\cdots\pm
x^{M+N}y^{NK}.$

If we restrict this polynomial to the right edge we get
$\delta_{n}+a_{n}z+\ldots\pm z^{N}$, where we set $z^{j}=x^{M+j}y^{jK}$. If we
restrict the polynomial to the base of the triangle we get
$z^{M}+\cdots+a^{\prime}_{n}z+\delta_{n}$, where we set $z^{j}=x^{M-j}$. The
coefficients $a_{n}$ and $a_{n}^{\prime}$ are the ones we need to bound.

######  Lemma 6.5.

With notation as above,

$|a_{2n}|=|a_{2n-1}|\leq n,\quad|a^{\prime}_{2n+1}|=|a^{\prime}_{2n}|\leq
n+1.$

###### Proof.

We will use the recurrence relations (4.2). Let us start by showing that the
coefficient of $x^{M}$ in $\xi_{n}^{int}$ (corresponding to the right vertex)
is $\delta_{n}=\pm 1$. In the first relation

$\xi^{int}_{n}=\xi^{rat}_{n}\xi^{rat}_{n-1}-\varepsilon^{rat}_{n}x^{M_{n}}(y-1)^{M_{n}+N_{n}},$

the terms $\xi^{rat}_{n}\xi^{rat}_{n-1}$ do not involve the right vertex of
the triangle. Hence, the coefficient of $x^{M}$ is

$-\varepsilon^{rat}_{n}(-1)^{M_{n}+N_{n}}=\pm 1.$

Let us now consider restriction of the polynomials to the right edge of the
triangle. These restrictions will be in the ring $k[z]$, where the constant
$1$ corresponds to the right vertex and $z$ corresponds to the next lattice
point $(M+1,K)$ on the edge. Then we may write the restriction of
$\xi^{int}_{n}$ as

$\pm 1+a_{n}z+\cdots\pm z^{N}.$

We will also need the restrictions of $\xi^{rat}_{n}$ to the right edge of
their triangles. Note that these triangles have the right vertex rational,
hence there is no constant term in the polynomials. By [5, Lemma 3.1], for $n$
even the first lattice point has $y$-coordinate $1$, and for $n$ odd the first
lattice point has $y$-coordinate $K-1$. We will write the restriction of
$\xi^{rat}_{n}$ as $c_{n}z^{\frac{1}{K}}+\cdots$ in the case of even $n$ and
$c_{n}z^{\frac{K-1}{K}}+\cdots$ in the case of odd $n$.

Considering the coefficients of $z$ in the two recurrence relations in (4.2),
we get

$\displaystyle\delta_{n}a_{n-1}+\delta_{n-1}a_{n}$
$\displaystyle=\begin{cases}c_{n-1}^{K}&\text{if $n$ is odd,}\\\ 0&\text{if
$n$ is even,}\end{cases}$ $\displaystyle c_{n}c_{n-1}$ $\displaystyle=a_{n}.$

The two cases in the first equation come from the fact that
$(\xi^{rat}_{n-1})^{K}$ involves the monomial $z$ only in the case where $n$
is odd. The initial values are $a_{0}=0$, $c_{0}=-1$.

The first equation implies that $a_{2n}=\pm a_{2n-1}$. From the second
equation we can expand

$c_{n}=\frac{a_{n}}{c_{n-1}}=\frac{a_{n}c_{n-2}}{a_{n-1}}=\cdots.$

Using the equality $a_{2n}=\pm a_{2n-1}$ from the first equation and the
initial value $c_{0}=-1$, we see that

$|c_{2n}|=\left|\frac{a_{2n}}{a_{2n-1}}\right|\cdots\left|\frac{a_{2}}{a_{1}}\right|=1\quad\text{and}\quad|c_{2n+1}|=|a_{2n+1}|.$

The inequality from the first relation

$|a_{2n+1}|=|\pm a_{2n}\pm c_{2n}^{K}|=|\pm a_{2n}\pm 1|\leq|a_{2n}|+1$

now proves the claim for $a_{n}$.

The proof for $a^{\prime}_{n}$ is very similar. We now restrict the
polynomials to the base of the triangles. These restrictions lie in $k[z]$
where we denote by $1$ the monomial corresponding to the right vertex and by
$z$ the monomial corresponding to the next lattice point. Then the restriction
of $\xi^{int}_{n}$ is

$\pm 1+a^{\prime}_{n}z+\cdots+z^{M}.$

The restriction of $\xi^{rat}_{n}$ is $c^{\prime}_{n}z^{\frac{K-1}{K}}+\cdots$
in the case of even $n$ and $c^{\prime}_{n}z^{\frac{1}{K}}+\cdots$ in the case
of odd $n$. (This is because in the even case the first lattice point on the
bottom edge is distance $(K-1)/K$ from the right vertex in the even case, and
distance $1/K$ in the odd case.)

We again consider the coefficient of $z$ in the recurrence relations (4.2).
These give

$\displaystyle\delta_{n}a^{\prime}_{n-1}+\delta_{n-1}a^{\prime}_{n}$
$\displaystyle=\begin{cases}(c^{\prime}_{n-1})^{K}&\text{if $n$ is even,}\\\
0&\text{if $n$ is odd,}\end{cases}$ $\displaystyle
c^{\prime}_{n}c^{\prime}_{n-1}$ $\displaystyle=a^{\prime}_{n}.$

The initial conditions are now $a^{\prime}_{0}=-1$, $c^{\prime}_{0}=1$. As
before, we compute

$a^{\prime}_{2n+1}=\pm a^{\prime}_{2n},\quad c^{\prime}_{2n+1}=\pm 1,\quad
c^{\prime}_{2n}=\pm
a^{\prime}_{2n},\quad|a^{\prime}_{2n}|\leq|a^{\prime}_{2n-1}|+1.$

This proves the claim. ∎

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