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2000:\ Geometry of Toric Varieties} \strut}

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\medskip
\centerline{\bigbf Lecture 7}
\medskip
\centerline{\bigbf Toric Surfaces}
\medskip
\centerline{\bigrm David A. Cox}
\bigskip
\medskip

This lecture will discuss toric surfaces and is based on portions of
Sections 2.2, 2.5 and 2.6 of [2].  We will also discuss lattice
polygons and the number 12, following [6].
\bigskip
\medskip

\noindent {\bigbf \S1. Singularities of Toric Surfaces}
\bigskip

At the end of Lecture 5, we noted that a toric surface has at most
finitely many singular points. These points correspond to the
2-dimesional cones of the fan and are at worst finite Abelian quotient
singularities.  We now study these singularities in more detail.
\bigskip
\smallskip

\noindent \ul{C\sc{yclic} Q\sc{uotient} S\sc{ingularities}}.  Let
$\sigma \subset N_\R \simeq \R^2$ be a 2-dimensional strongly convex
rational polyhedral cone in the plane, and let $v_1,v_2 \in N$ be its
minimal generators.  Since $v_2$ is primitive, we can choose a basis
$e_1,e_2$ of $N$ such that $v_2 = e_2$.  Then $v_1 = me_1 + le_2$ for
some $m,l \in \Z$.  Note that $m,l$ are relatively prime since $v_1$
is also primitive.
\bigskip

\ex{1.1} Explain why $m \ne 0$.  Also explain why we may assume $m >
0$, and in this case, show that if $N' \subset N$ is the sublattice
generated by $v_1,v_2$, then $N/N' \simeq \Z/m\Z$. 
\bigskip

By Lecture 5, $U_\sigma \simeq \C^2/G$, where $G = N/N'$.
Exercise~1.1 shows that $G$ is cyclic, so that $U_\sigma$ has at worst
cyclic quotient singularities.  Hence this is true for all toric
surfaces.
\bigskip
\smallskip

\noindent \ul{N\sc{ormal} F\sc{orm of} T\sc{wo-}D\sc{imensional}
C\sc{ones in the} P\sc{lane}}.  We next find a normal form for the
cone $\sigma$.  So far, we have a basis $e_1,e_2$ of $N$ such that the
minimal generators of $\sigma$ are $v_1 = me_1 + le_2$ and $v_2 =
e_2$, where $m > 0$ is the index of $N'$ in $N$ and $\gcd(m,l) = 1$.

We get a unique choice of $l$ as follows.

\bigskip

\ex{1.2} As above, suppose that $v_1 = me_1 + le_2$, $v_2 = e_2$,
where $m > 0$ and $\gcd(m,l) = 1$.
\item{a.} Let $r \in \Z$.  Show that $e_1' = e_1 + re_2$ and $e_2$ is
a basis of $N$. 
\item{b.} Relative to the basis of part a, show that $v_1 = me_1' +
(l-rm)e_2$.
\item{c.} Explain why there is a unique choice of $r$ so that $v_1 =
me_1' - ke_2$, where $0 \le k < m$ and $\gcd(m,k) = 1$.
\bigskip

{}From here on, we will assume that a basis $e_1,e_2$ of $N$ has been
chosen so that the minimal generators of $\sigma$ are $v_1 = me_1 -
ke_2$, $v_2 = e_2$, where $0 \le k < m$ and $\gcd(m,k) = 1$.
We call $m$, $k$ the {\it parameters\/} of $\sigma$.
\bigskip

\ex{1.3} Let $\sigma$ be a cone with parameters $m$, $k$.  Prove that
$\sigma^\vee$ has parameters $m$, $m-k$.
\bigskip

We can also show that the basis $e_1,e_2$ is uniquely determined.
\bigskip

\ex{1.4} Let $\tilde e_1,\tilde e_2$ be another basis of $N$ such that
$v_1 = m\tilde e_1 - k\tilde e_2$ and $v_2 = \tilde e_2$, where $0 \le
k < m$ and $\gcd(m,k) = 1$.  Prove that $\tilde e_i = e_i$ for $i =
1,2$.
\bigskip

We call $e_1,e_2$ the {\it normalized basis\/} of $N$ relative to
$\sigma$ when the above conditions are satisfied.   For the rest of
this section, fix a normalized basis $e_1,e_2$ and let $e_1^*,e_2^*$
be the dual basis of $M$.

We next describe how the parameters $m$, $k$ determine the action of
$\Z/m\Z \simeq N/N'$ on $\C^2$.
\bigskip

\ex{1.5} Let $M'$ be the dual of $N'$ and let
$u_1,u_2 \in M'$ be dual to $v_1,v_2 \in N'$. 
\item{a.} Prove that 
$$
\eqalign{\langle u_1,e_1\rangle &= 1/m\cr
	 \langle u_2,e_1\rangle &= k/m\cr
	 \langle u_1,e_2\rangle &= 0\cr
	 \langle u_2,e_2\rangle &= 1.\cr}
$$
\item{b.} Let $\mu_m = \{\zeta \in \C \mid \zeta^m = 1\}$ be the group
of $m$th roots of unity.  Show that $\mu_m \simeq N/N'$ via $e^{2\pi i
j/m} \mapsto je_1+N'$.
\item{c.} By Lecture 5, $N/N'$ acts on $\C^2$ via $(v+N')\cdot
\chi^{u'} = e^{2\pi i\langle u',v\rangle} \chi^{u'}$ for $u' \in
\sigma^\vee\cap M'$ and $v+N' \in N/N'$.  Using part b and the
coordinates $x = \chi^{u_1},y = \chi^{u_2}$ on $\C^2$, show that the
action of $\zeta \in \mu_m \simeq N/N'$ is given by
$$
\zeta\cdot (x,y) = (\zeta x,\zeta^{k} y).
$$
\medskip

\ex{1.6} Let $v_1 = me_1 - ke_2$ and $v_2 = e_2$ be minimal generators
of $\sigma$ as above. 
\item{a.} Let $u_1,u_2 \in M'$ be as in Exercise 1.5.  Show that $u_1
= (1/m) e_1^*$ and $u_2 = (k/m) e_1^* + e_2^*$.
\item{b.} Conclude that $\sigma^\vee \subset M_\R$ is the cone
generated by $e_1^*$ and $ke_1^*+me_2^*$.   
\bigskip

When $m = 3$ and $k = 2$, Exercise 1.6 gives the follow picture of
$\sigma$ and its dual $\sigma^\vee$:
\vglue1.9truein
\special{psfile=lect7fig1.ps}

\bigskip
\noindent In these pictures, we are using the bases $e_1,e_2$ of $N$
(on the left) and $e_1^*,e_2^*$ of $M$ (on the right).  Lattice points
are indicated by dots.

Note that $\sigma$ is regular if and only if $m = 1$, so that
$U_\sigma$ is smooth in this case.  When $m > 1$, there are two cases
when the singularity of $U_\sigma$ is especially interesting and
well-known.
\bigskip
\smallskip

\noindent \ul{T\sc{he} C\sc{one over a} R\sc{ational} N\sc{ormal}
C\sc{urve}}.  Let $\sigma$ be a cone with parameters $m > 1$ and $k =
1$ (so $k$ is as small as possible).  In this case, the action of
$\zeta \in \mu_m$ on $\C^2$ is given by
$$
\zeta \cdot (x,y) = (\zeta x,\zeta y),
$$
where we are using the notation of Exercise 1.5.  
\bigskip

\ex{1.7} For the above $\mu_m$-action, show that $\C[x,y]^{\mu_m} =
\C[x^m,x^{m-1}y,\ldots,xy^{m-1},y^m]$.
\bigskip

This exercise shows that when $k = 1$, the coordinate ring of
$U_\sigma$ is
$$
\C[U_\sigma] = \C[x^m,x^{m-1}y,\ldots,xy^{m-1},y^m] \subset
\C[x,y].\leqno(1.1)
$$
We can describe this geometrically as follows.  The function $\P^1 \to
\P^m$ defined by
$$
(x,y) \mapsto (x^m,x^{m-1}y,\ldots,xy^{m-1},y^m)\leqno(1.2)
$$
parametrizes the {\it rational normal curve of degree\/} $m$ in
$\P^m$.  If we think of (1.2) as a map $\C^2 \to \C^{m+1}$, we get a
surface called the {\it cone over the rational normal curve of
degree\/} $m$.  More precisely, Lecture~3 implies that the image of
$\P^1 \to \P^m$ is a projective variety $C \subset \P^m$.  Its ideal
${\bf I}(C) \subset \C[y_0,\ldots,y_m]$ is homogeneous, and the
variety of ${\bf I}(C)$ in $\C^{m+1}$ is the {\it cone\/} over $C$.
\bigskip

\ex{1.8} Let $S \subset \C^{m+1}$ be the image of the map (1.2).
\item{a.} Prove that $S$ is the affine variety defined by ${\bf
I}(C)$.  (This is less trivial than it sounds since in general the
image of a polynomial map is not a variety.  The key point is that
image in $\P^m$ {\it is\/} a variety.)
\item{b.} Prove that $\C[S] \simeq
\C[x^m,x^{m-1}y,\ldots,xy^{m-1},y^m] \subset \C[x,y]$.  
\item{c.} Use (1.1) to conclude that $U_\sigma$ is the cone over the
rational normal curve of degree $m$.
\bigskip

Finally, let us relate our discussion to the semigroup algebra
$\C[\sigma^\vee\cap M]$ of $\sigma$.  When $m = 4$ and $k = 1$,
$\sigma$ and $\sigma^\vee$ have the following pictures:
\vglue1.9truein
\special{psfile=lect7fig2.ps}

\bigskip
Here, the crucial observation is that the semigroup $\sigma^\vee\cap
M$ is generated by $e_1^* + ie_2^*$ for $0 \le i \le 4$ (these are
indicated by white dots in the above picture).  This generalizes as
follows.
\bigskip

\ex{1.9} Let $\sigma$ be a cone with  $m > 1$ and $k = 1$, and let
$U = \chi^{e_1^*}$ and $V = \chi^{e_2^*}$.  
\item{a.} Prove that the semigroup $\sigma^\vee\cap M$ is generated by
$e_1^* + ie_2^*$ for $0 \le i \le m$. 
\item{b.} Conclude that $\C[\sigma^\vee\cap M] =
\C[U,UV,\ldots,UV^m]$.
\item{c.} Our earlier discussion used $x = \chi^{u_1}$ and $y =
\chi^{u_2}$, where $u_1,u_2$ are dual to $v_1,v_2$ as in
Exercises~1.5 and 1.6.  Use Exercise 1.6 to show that $U = x^m$ and $V
= y/x$.
\item{d.} Show that $\C[U,UV,\ldots,UV^m] =
\C[x^m,x^{m-1}y,\ldots,xy^{m-1},y^m]$ under the identifications of
part c.  Thus the toric description of $\C[U_\sigma]$ is consistent
with (1.1).

\smallskip
\bigskip
\noindent \ul{R\sc{ational} D\sc{ouble} P\sc{oints of} T\sc{ype}
$A_k$}.  Now let $\sigma$ be a cone whose parameters $m$, $k$ satisfy
$k = m-1$ (so $k$ is as large as possible).  For reasons which will
soon become apparent, we will write $k+1$ instead of $m$.  Since $k
\equiv -1 \bmod k+1$, the action of $\zeta \in \mu_{k+1}$ on $\C^2$ is 
given by 
$$
\zeta \cdot (x,y) = (\zeta x,\zeta^{-1} y),
$$
where $x = \chi^{u_1}$ and $y = \chi^{u_2}$ as in Exercise 1.5.
\bigskip

\ex{1.10} Show that $\C[x,y]^{\mu_{k+1}} = \C[x^{k+1},y^{k+1},xy]$.
Then show that 
$$
\C[x^{k+1},y^{k+1},xy] \simeq \C[X,Y,Z]/\langle Y^{k+1}-XZ\rangle
$$
via the map defined by $X \mapsto x^{k+1}$, $Y \mapsto xy$ and $Z
\mapsto y^{k+1}$.
\bigskip

\goodbreak

The singularity of ${\bf V}(X^{k+1}+Y^2+Z^2) \subset \C^3$ at $0 \in
\C^3$ is a {\it rational double point of type\/} $A_k$.
\bigskip

\ex{1.11} Construct a coordinate change of $\C^3$ which takes
${\bf V}(X^{k+1}+Y^2+Z^2)$ into ${\bf V}(Y^{k+1}+XZ)$.  This shows
that when $m = k+1$, the singular point of $U_\sigma$ is a rational
double point of type $A_k$.
\bigskip

In general, a {\it rational double point\/} is the singularity of
$\C^2/G$ at the origin, where $G$ is a finite subgroup of ${\rm
SU}(2,\C)$.  These subgroups are well-known and are related to the
Dynkin diagrams $A_k$, $D_k$, $E_6$, $E_7$ and $E_8$.  The groups
corresponding to $D_k$, $E_6$, $E_7$ and $E_8$ are non-Abelian and
hence do not arise in the toric context.  This is why rational double
points on toric surfaces are all of type $A_k$.  See [1] for an
excellent introduction to rational double points.

Finally, let us relate Exercise~1.10 to the semigroup algebra
$\C[\sigma^\vee\cap M]$.
\bigskip

\ex{1.12} Let $\sigma$ be a cone with  $m = k + 1$, and let
$U = \chi^{e_1^*}$ and $V = \chi^{e_2^*}$.  
\item{a.} Prove that the semigroup $\sigma^\vee\cap M$ is generated by
$e_1^*, e_1^* + e_2^*$, and $ke_1^* + (k+1)e_2^*$.
\item{b.} Conclude that $\C[\sigma^\vee\cap M] =
\C[U,UV,U^kV^{k+1}]$.
\item{c.} Let $x = \chi^{u_1}$ and $y = \chi^{u_2}$ as in
Exercises~1.5 and 1.6.  Use Exercise 1.6 to show that $U = x^{k+1}$
and $V = y/x^k$.
\item{d.} Show that $\C[U,UV,U^kV^{k+1}] = \C[x^{k+1},xy,y^{k+1}]$
under the identifications of part c.  Thus the toric description of
$\C[U_\sigma]$ coincides with what we found in Exercise 1.10.
\bigskip

\ex{1.13} The special cases $k = 1$ and $m = k+1$ coincide when $m =
2$ and $k = 1$.  Show that this is the affine toric variety ${\bf
V}(xy-z^2) \subset \C^3$ introduced in Lecture~2.
\bigskip
\medskip
\goodbreak

\noindent {\bigbf \S2. Resolution of Toric Surface Singularities}
\bigskip

The singularities of the affine toric surface $U_\sigma$ are easy to
resolve.  Here, $\sigma \subset N_\R \simeq \R^2$ is a 2-dimensional
strongly convex rational polyhedral cone with parameters $m$, $k$
defined in \S1.  We will assume $m > 1$, and we begin with the two
special cases considered in \S1.
\bigskip
\smallskip

\noindent \ul{R\sc{esolving the} C\sc{one over a} R\sc{ational}
N\sc{ormal} C\sc{urve}}.  If $\sigma$ be a cone with $k = 1$, then
$U_\sigma$ is the cone over the rational normal curve of degree $m$.
We will resolve the singularity of $U_\sigma$ by subdividing $\sigma$
into cones $\sigma_1$, $\sigma_2$ along the edge $\rho = {\rm
Cone}(e_1)$.  When $m = 4$ and $k = 1$, this gives the following
picture.
\vglue1.8truein
\special{psfile=lect7fig3.ps}
 
\bigskip
\noindent Let $\Delta$ consist of $\sigma_1$, $\sigma_2$ and their
faces.  Then one easily sees that $\Delta$ is a regular fan.  This is
obvious for $\sigma_1$ and follows from $k = 1$ for $\sigma_2$.  Thus
$X = X(\Delta)$ is a smooth toric surface.  To see what $X$ looks
like, consider the following:
\item{$\bullet$} ${\rm Star}(\rho) = \{\rho,\sigma_1,\sigma_2\}$, and
the corresponding orbit closure $E = V(\rho)$ is isomorphic to $\P^1$.
\item{$\bullet$} $X - E$ is the toric variety of $\Delta-{\rm
Star}(\rho)$.
\item{$\bullet$} But $\Delta-{\rm Star}(\rho)$ is the fan consisting
of the proper faces of $\sigma$.  The corresponding toric variety is
$U_\sigma - \{x_\sigma\}$, where $x_\sigma$ is the singular point of
$U_\sigma$.
\vskip0pt

\goodbreak

\noindent This tells us that $X - E = U_\sigma - \{x_\sigma\}$, so
that we can regard $X$ as a copy of $U_\sigma$ where the singular
point $x_\sigma$ replaced by the {\it exceptional curve\/} $E \simeq
\P^1$.

A better way to understand $X$ is to realize that each cone of
$\Delta$ lies in $\sigma$, which means that we get a morphism of toric
varieties $\pi : X \to U_\sigma$.  The above discussion shows that
$\pi$ induces an isomorphism $X - E \simeq U_\sigma - \{x_\sigma\}$
and maps $E$ to the singular point $x_\sigma$.  Note also that $\pi$
is proper by Lecture 6.  We say that $\pi : X \to U_\sigma$ is a {\it
resolution of singularities\/} of $U_\sigma$.

A final comment is that the curve $E \subset X$ has a {\it
self-intersection\/} $(E\cdot E)$.  We will discuss self-intersections
in \S3.  We will also see that the identity
$$
e_2 + (me_1-e_2) = me_1\leqno(2.1)
$$
connecting the minimal generators of $\sigma_1$ and $\sigma_2$ implies
that $(E\cdot E) = -m$.  This is part of how the resolution determines
the singularity it comes from.

\smallskip
\bigskip
\noindent \ul{R\sc{esolving} R\sc{ational} D\sc{ouble} P\sc{oints of}
T\sc{ype} $A_k$}.  We next suppose that $\sigma$ is a cone with $m =
k+1$, so that $U_\sigma$ has a rational double point of type $A_k$.  In
this case, we get the nicest resolution by picking a new basis for
$N$.  
\bigskip

\ex{2.1} Let $\sigma$ be a cone with $m=k+1$ and let $e_1,e_2$ be the
normalized basis of $N$ relative to $\sigma$.  Set $\tilde e_1 = e_2$
and $\tilde e_2 = e_1 - e_2$.  Show that the minimal generators of
$\sigma$ are given by
$$
v_1 = \tilde e_1 + (k+1) \tilde e_2,\quad v_2 = \tilde e_1.
$$
\medskip

We will resolve the singularity of $U_\sigma$ by subdividing
$\sigma$ into $k+1$ cones $\sigma_1,\ldots,\sigma_{k+1}$ by introducing
the edges $\rho_j = {\rm Cone}(\tilde e_1 + j \tilde e_2)$ for $j =
1,\ldots, k$.  When $m = 4$ and $k = 3$, we get the following picture
(relative to the basis $\tilde e_1,\tilde e_2$):
\vglue1.9truein
\special{psfile=lect7fig4.ps}
 
\bigskip
\noindent Then let $\Delta$ be the fan consisting of the $\sigma_i$
and their faces.  In the following exercise, you will show how $X =
X(\Delta)$ resolves a rational double point of type $A_k$.
\bigskip

\ex{2.2} Let $\sigma_i$, $\rho_i$ and $\Delta$ be as above.
\item{a.} Show that $\Delta$ is a regular fan, so that $X$ is a smooth
toric surface.
\item{b.} Show that $E_i = V(\rho_i) \simeq \P^1$.  Let $E = E_1 \cup
\cdots \cup E_k$.
\item{c.} Show that $X - E$ is the toric variety coming from the fan 
$\Delta - ({\rm Star}(\rho_1) \cup \cdots \cup {\rm Star}(\rho_k))$.
\item{d.} Show that the fan of part d is the fan consisting of all
proper faces of $\sigma$.  Conclude that $X-E$ is isomorphic to
$U_\sigma - \{x_\sigma\}$, where $x_\sigma$ is the singular point of
$U_\sigma$.
\item{e.} Show that there is a proper toric morphism $\pi: X \to
U_\sigma$ which maps $E$ to $x_\sigma$ and induces an isomorphism $X-E
\simeq U_\sigma - \{x_\sigma\}$.
\bigskip

\ex{2.3} Using the notation of Exercise 2.2, let $E_i = V(\rho_i)
\simeq \P^1$ and $E = E_1 \cup \cdots \cup E_k$.  Draw the graph with
vertices $\{E_1,\ldots,E_k\}$, where $E_i$ and $E_j$ are linked by an
edge if and only if $E_i\cap E_j \ne \emptyset$.  The graph you get
should be the Dynkin diagram for $A_k$.
\bigskip

Note also that the minimal generators of $\sigma_j$ are $\tilde e_1 +
(j-1)\tilde e_2$ and $\tilde e_1 + j\tilde e_2$.  It follows that the
generators of $\sigma_j$ and $\sigma_{j+1}$ satisfy the relation
$$
(\tilde e_1 + (j-1)\tilde e_2) + (\tilde e_1 + (j+1)\tilde e_2) =
2(\tilde e_1 + j\tilde e_2).\leqno(2.2)
$$
As we will see in \S3, this implies that $E_j$ has self-intersection
$(E_j\cdot E_j) = -2$.  Rational double points of type $A_k$ are
characterized by having a resolution whose exceptional components have
graph $A_k$ and self-intersections all equal to $-2$.

A final observation concerns the relation between regular subdivisions
and irreducible elements.  If you compare the illustrations on pages 3
and 5, especially the pictures on the right hand side, you will see
that when $m = k+1$, the edges of the regular subdivision of $\sigma$
are generated by the irreducible elements of $\sigma\cap N-\{0\}$.
Here is another case when this is true.
\bigskip

\ex{2.4} Let $\sigma$ satisfy $k = 1$.  Prove that the edges of the
regular subdivision of $\sigma$ are generated by the irreducible
elements of $\sigma\cap N-\{0\}$.  Hint: Look at the picture on page
4.  Then put the cone $\sigma^\vee$ of Exercise 1.12 into normal form.
\bigskip

In the next section, we will see that the relation between regular
subdivisions and irreducible elements will give an elegant description
of the resolution of a toric surface.

\smallskip
\bigskip
\noindent \ul{R\sc{esolving} G\sc{eneral} T\sc{oric} S\sc{urfaces}}.
We can now prove the existence of a resolution of singularities $\pi :
X' \to X$ for an arbitrary toric surface $X$.  Suppose $X =
X(\Delta)$, where $\Delta$ is a fan in $N_\R \simeq \R^2$.  Let
$\sigma_1,\ldots,\sigma_m$ be the cones of $\Delta(2)$.  We can assume
$m > 0$ since otherwise $X$ is already smooth.  For each $i$, let $N_i
\subset N$ be the sublattice generated by the minimal generators of
$\sigma_i$.  We will use
$$
s(\Delta) = \sum_{i=1}^m ([N:N_i] - 1)
$$
as a measure of how singular $X$ is.  This is justified because
$[N:N_i] \ge 1$, with equality if and only if $\sigma_i$ is regular.
It follows that $s(\Delta) \ge 0$, with equality if and only if
$\Delta$ is regular.  We call $[N:N_i]$ the {\it multiplicity\/} of
$\sigma_i$. 

We claim that if $s(\Delta) > 0$, then $\Delta$ has a refinement
$\Delta_0$, obtained by subdividing one of the cones of $\Delta(2)$,
such that $s(\Delta_0) < s(\Delta)$.  To construct $\Delta'$, pick a
cone $\sigma_i$ with $m = [N:N_i] > 1$ ($\sigma_i$ exists because
$s(\Delta) > 0$).  Then let $e_1,e_2$ be a normalized basis of $N$
relative to $\sigma_i$.  Thus the minimal generators of $\sigma_i$ are
$v_1 = me_1-ke_2$, $v_2 = e_2$, where $0 \le k < m$ and $\gcd(m,k) =
1$.  Now, as in the case $k = 1$, subdivide $\sigma_i$ by inserting
the edge $\rho = {\rm Cone}(e_1)$.  This gives cones $\sigma_i' = {\rm
Cone}(e_1,v_1)$ and $\sigma_i'' = {\rm Cone}(e_1,v_2)$.  Since $v_2 =
e_2$, we see that $\sigma_i''$ is regular.  As for $\sigma_i'$, we can
analyze it as follows.
\bigskip

\ex{2.5} Let $\sigma_i' = {\rm Cone}(e_1,v_1) = {\rm
Cone}(e_1,me_1-ke_2)$ be as above, and write $m = a_1 k - k_1$ where
$0 \le k_1 < k$.  Prove that the parameters of $\sigma_i'$ are
$k$, $k_1$.
\bigskip

Now let $\Delta_0$ the the fan obtained from $\Delta$ by replacing
$\sigma_i$ with $\sigma_i'$, $\sigma_i''$, and $\rho$.  We claim that
$$
s(\Delta_0) < s(\Delta).
$$
This follows because $\sigma_i$ has multiplicity $m$ while $\sigma_i'$
and $\sigma_i''$ have multiplicities $k < m$ (by Execise~2.5) and $1$
respectively.

If we repeat this process as many times as necessary, we eventually
obtain a subdivison $\Delta'$ of $\Delta$ such that $\Delta'$ is
regular.  This gives a smooth toric variety $X' = X(\Delta')$ and a
toric morphism $\pi : X' \to X$ with the following properties:
\item{a.} $\pi$ is proper since $\Delta$ and $\Delta'$ have the same
support. 
\item{b.} Let $x_1,\ldots,x_s$ be the singular points of $X$ and set
$E = \pi^{-1}(\{x_1,\ldots,x_s\}) \subset X'$.  Then:
\itemitem{$\bullet$} $E$ is a union of 1-dimensional orbit closures in
$X'$, each isomorphic to $\P^1$. 
\itemitem{$\bullet$} $\pi$ induces an isomorphism $X' - E \simeq X -
\{x_1,\ldots,x_s\}$. 
\vskip0pt
\noindent Thus $\pi : X' \to X$ is a resolution of singularities of
$X$. 

\smallskip
\bigskip
\noindent \ul{H\sc{irzebruch-}J\sc{ung} C\sc{ontinued} F\sc{ractions}}.
When $\sigma \subset N_\R \simeq \R^2$ is a 2-dimensional rational
polyhedral cone with parameters $m$, $k$, we can describe the
resolution of $U_\sigma$ using the {\it Hirzebruch-Jung\/} continued
fraction 
$$
{m \over k} = a_1 - {1 \over \displaystyle{\ a_2 - \displaystyle{\strut
1 \over \ \cdots - \displaystyle{\strut  1 \over a_r}}}},\leqno(2.3)
$$
where the $a_i \ge 2$ are integers.  This is explained in the
following exercises.
\bigskip

\ex{2.6} Given relatively prime integers $0 \le k < m$, the {\it
modified Euclidean algorithm\/} considers successive divisions 
$$
\eqalign{ m &= a_1 k - k_1,\quad\ 0 \le k_1 < m\cr
	  k &= a_2 k_1 - k_2,\quad 0 \le k_2 < k_1\cr
	    &\ \,\vdots\phantom{a_2 k_1 - k_2,\quad 0 \le k_2}\vdots\cr}
$$
\item{a.} Show that $a_i \ge 2$ for all $i \ge 1$.
\item{b.} Show that eventually $k_{r-1} = 1$ and $k_r = 0$.  Thus the
algorithm terminates.
\item{c.} Let $r$ be as in part b.  Show that $r$ and $a_1,\ldots,a_r$
have the same meaning as in (2.3).  Hint: $m = a_1 k - k_1$ implies
$$
{m\over k} = a_1 - {k_1\over k} = a_1 - {1 \over k/k_1}.
$$
\bigskip

\ex{2.7} Given (2.3), define $p_i,q_i$ for $i \ge 0$ by setting
$$ 
p_0 = 1,\ \ q_0 = 0;\ \ p_1 = a_1,\ \ q_1 = 1
$$
and, for $i \ge 1$,
$$
\eqalign{p_{i+1} &= a_{i+1} p_i - p_{i-1}\cr
	q_{i+1} &= a_{i+1} q_i - q_{i-1}.\cr}
$$
\item{a.} Show that $p_{i+1} q_i - q_{i+1} p_i = -1$ for all $i \ge
0$. 
\item{b.} Show that $p_{i+1} > p_i$ and $q_{i+1} > q_i$ for all $i \ge
0$. 
\item{c.} Show that $\gcd(p_i,q_i) = 1$ for all $i \ge 0$. 
\item{d.} Assume that for some $1 \le i < r$, we have 
$$ 
{p_i  - p_{i-1} x \over q_i  - q_{i-1}x} = a_1 - {1 \over
\displaystyle{\ a_2 - \displaystyle{\strut 
1 \over \ \cdots - \displaystyle{\strut  1 \over a_i - x}}}}. 
$$
Prove that the same identity holds for $i+1$.
Hint: Let $x = 1/(a_{i+1} - y)$.
\bigskip

Setting $x = 0$ in part d of Exercise 2.7, we obtain
$$
{p_i \over q_i} = a_1 - {1 \over \displaystyle{\ a_2 -
\displaystyle{\strut 1 \over \ \cdots - \displaystyle{\strut  1 \over
a_i}}}}.
$$
Note that $p_i/q_i$ is in lowest terms by part c of Exercise 2.7.  We
call $p_i/q_i$ the $i$th {\it partial quotient\/} of $m/k$.  Note also
that $p_r/q_r = m/k$.
\bigskip

\ex{2.8} Now let $\sigma$ be a cone with parameters $m$, $k$ and let
$e_1,e_2$ be a normalized basis for $N$ relative to $\sigma$.  Thus
$$
\sigma = {\rm Cone}(me_1-ke_2,e_2).
$$
Let $\tilde v_i = p_{i-1} e_1 - q_{i-1} e_2$ for $1 \le i \le r+1$,
where $p_i,q_i$ are from Exercise 2.7.  Thus $\tilde v_1 = e_1$ and
$\tilde v_{r+1} = me_1-ke_2$.  Also let $\tilde v_0 = e_2$.  Then
consider the cones
$$
\sigma_i = {\rm Cone}(\tilde v_{i},\tilde v_{i+1}),\quad i =
	0,\ldots,r.
$$
The goal is to show that $\sigma_0,\ldots,\sigma_r$ and their faces
form a regular fan $\Delta'$ whose support is $\sigma$.
\item{a.} Prove that $\sigma_i$ is regular.  Hint: Part a of Exercise
2.7.
\item{b.} Prove that $\sigma_{i-1}\cap\sigma_{i} = {\rm Cone}(\tilde
v_i)$.  Hint: Compare slopes using part a of Exercise~2.7.
\item{c.} Show that $\sigma = \sigma_0\cup\cdots\cup\sigma_r$, and
conclude that $\Delta'$ has the desired properties.
\bigskip

If we let  $X' = X(\Delta')$, where $\Delta'$ is the fan of Exercise
2.8, then our previous arguments show that we get a toric morphism
$$
\pi : X' \to U_\sigma\leqno(2.4)
$$ 
which is a resolution of singularities.  
We get an intrinsic version of this resolution as follows.
\bigskip

\ex{2.9} Let $\sigma$ be a cone with parameters  $m$, $k$ and let
$\tilde v_i$, $0 \le i \le r+1$ be as in Exercise~2.8.  Then consider
$S = \{\tilde v_0,\ldots,\tilde v_{r+1}\} \subset \sigma\cap
N-\{0\}$. 
\item{a.} Show that $S$ generates $\sigma\cap N$ as a
semigroup. Hint:~Use Exercise 2.8.
\item{b.} If $\tilde v_i \in S$ is not irreducible, show that it can
be expressed as a non-negative integer linear combination of the
elements of $S - \{\tilde v_i\}$.
\item{c.} Show that every element of $S$ is irreducible.  Hint:~Use
part b together with part b of Exercise 2.7. 
\item{d.} Conclude that $S$ is the set of irreducible elements of
$\sigma\cap N-\{0\}$.
\bigskip

Comparing the two previous exercises, we see that the minimal
generators of the cones of Exercise~2.8 are the irreducible elements
of $\sigma\cap N-\{0\}$.  Thus, from an intrinsic point of view, the
toric morphism (2.4) is obtained by subdividing the cone $\sigma$ into
subcones whose edges are the irreducible elements of $\sigma\cap N -
\{0\}$.

We can also think of irreducible elements geometrically as follows.
\bigskip

\ex{2.10} Let $\sigma$ be as above.  Show that the irreducible
elements of $\sigma\cap N -\{0\}$ are the vertices of the convex hull
${\rm Conv}(\sigma\cap N -\{0\})$.
\bigskip

The $\tilde v_i \in \sigma\cap N - \{0\}$ from Exercise 2.8 have one
further interesting property.
\bigskip

\ex{2.11} Let $\tilde v_{0},\ldots,\tilde v_{r+1}$ be as in Exercise
2.8.  Use Exercise 2.7 to prove that
$$
a_i \tilde v_i = \tilde v_{i+1} + \tilde v_{i-1},\quad 1 \le i \le
r.\leqno(2.5)
$$
\medskip

This has the following consequence for the resolution (2.4).  The
edges of the fan $\Delta'$ are $\rho_i = {\rm Cone}(\tilde v_i)$.  For
$1 \le i \le r$, $\rho_i$ is in the relative interior of $\sigma$,
which means that the corresponding orbit closure $E_i = V(\rho_i)
\subset X'$ is isomorphic to $\P^1$.  In \S3, we will see that (2.5)
then implies that $E_i$ has self-intersection $(E_i\cdot E_i) = -a_i$
for $1 \le i \le r$.  Thus the continued fraction (2.3) determines
everything about the exceptional divisor $E = E_1 \cup \cdots \cup
E_r$ of $\pi : X' \to U_\sigma$.

Finally, we observe that the above description of irreducible
elements can be used to give explicit generators for the semigroup
algebra $\C[\sigma^\vee\cap M]$.
\bigskip

\ex{2.12} Let $\sigma \subset N_\R \simeq \R^2$ be a cone with
parameters $m$, $k$, and let $e_1,e_2$ be a normalized basis of $N$
relative to $\sigma$.  We saw in Exercise 1.3 that $\sigma^\vee
\subset M_\R$ has parameters $m$, $m-k$.  Let $U = \chi^{e_1^*}$ and 
$V = \chi^{e_2^*}$, and let  the partial quotients of
$m/(m-k)$ be $P_1/S_1,\ldots,P_s/Q_s$.
\item{a.} Show that $\C[\sigma^\vee\cap M] =
\C[U,UV,U^{P_1-Q_1}V^{P_1},\ldots, U^{P_s-Q_s}V^{P_s}]$. 
\item{b.} Use (3.2) from Lecture 5 to show that the dimension of the
Zariski tangent space at the singular point of $U_\sigma$ is $s+2$.
\item{c.} When $k = 1$, show that $s = m-1$ and $P_i = Q_i+1 = i+1$
for $1 \le i \le s$.  Conclude that the description of part a agrees
with Exercise~1.9
\item{d.} When $m = k+1$, show that $s = 1$, $P_1 = k+1$ and $Q_1 =
1$.  Conclude that the description of part a agrees with
Exercise~1.12.
\vskip0pt
\noindent There is a nice relation between the continued fraction
expansions of $m/k$ and $m/(m-k)$.  This is discussed in [5, p.\ 29].
\bigskip
\smallskip

\noindent \ul{R\sc{eversing} O\sc{rientation}}.  The attentive reader
may have noticed that the parameters $m$, $k$ of a cone $\sigma$
depend on the order in which we list the minimal generators $v_1$,
$v_2$.  In other words, $m$ and $k$ depend on the {\it orientation\/}
of $\sigma$.  To see what happens when we reverse the orientation,
assume $m > 1$ and let $0 \le k' < m$ be the multiplicative inverse of
$k$ modulo $m$, i.e., $kk' \equiv 1 \bmod m$.  We know that $k'$
exists since $\gcd(m,k) = 1$.
\bigskip

\ex{2.13} Let $\sigma$ be a cone with minimal generators $v_1 =
me_1-ke_2$, $v_2 = e_2$, where $0 \le k < m$ and $\gcd(m,k) = 1$.  Let
$k'$ be as above, and pick $m' \in \Z$ so that $mm'+kk' = 1$.  Set
$$
\eqalign{\tilde e_1 &= k'e_1 + m'e_2\cr
	\tilde e_2 &= me_1 - ke_2.\cr}
$$
\item{a.} Show that $\tilde e_1,\tilde e_2$ form a basis of $N$.
\item{b.} Show that $v_2 = m\tilde e_1 -k'\tilde e_2$ and $v_1 =
\tilde e_2$.
\bigskip

Exercise 2.13 shows that if $\sigma$ has parameters $m,k$ for
$v_1,v_2$, then its parameters for $v_2,v_1$ are $m,k'$.  Thus
reversing the orientation of $\sigma$ preserves $m$ and interchanges
$k$ and $k'$.  

This has a nice consequence for the continued fractions of $m/k$ and
$m/k'$.  Write (2.3) as
$$
{m\over k} = [a_1,\ldots,a_r].
$$
If $\sigma$ is an oriented cone with parameters $m$, $k$, then (2.5)
tells us that the $a_i$ are determined by the irreducible elements of
$\sigma\cap N -\{0\}$.  When we reverse the orientation, we change $k$
to $k'$ and traverse the irreducible elements in reverse order.  It
follows that
$$
{m\over k'} = [a_r,\ldots,a_1].
$$

Finally, suppose that $\sigma_1$ and $\sigma_2$ are oriented cones
with parameters $m_i$, $k_i$, $i = 1,2$.  Then the affine toric
varieties $U_{\sigma_1}$ and $U_{\sigma_2}$ are isomorphic as
varieties $\Leftrightarrow$ if $m_1 = m_2$ and either $k_1 = k_2$ or
$k_1k_2 \equiv \bmod\ m_1$.  The $\Leftarrow$ direction follows from
the above discussion, while the other direction is more difficult
since we are not assuming that the isomorphism $U_{\sigma_1} \simeq
U_{\sigma_2}$ is toric.  Here, one needs the fact that such an
isomorphism induces an isomorphism of minimal resolutions constructed
in (2.4).  Then the desired result follows by studying how the
isomorphism acts on the exceptional divisor.  Details of this
argument, together with other facts about Hirzebruch-Jung continued
fractions and the singularities of $U_\sigma$, can be found in [5,
1.6] and in the references given on page 134 of [2].
\bigskip
\medskip
\goodbreak

\noindent {\bigbf \S3. Classification of Smooth Complete Toric
Surfaces} 
\bigskip
\smallskip

\noindent \ul{H\sc{irzebruch} S\sc{urfaces}}. We begin our discussion
of smooth complete toric surfaces by describing the {\it Hirzebruch
surface\/} $\F_a$, which is the toric variey of the complete fan whose
edges are generated by $e_1$, $\pm e_2$ and $-e_1+ae_2$ for $a > 0$.
The fan for $\F_a$ is pictured as follows:
\vglue2.3truein
\special{psfile=lect7fig5.ps}
 
It is an easy exercise to check that this fan is regular, so that
$\F_a$ is a smooth complete toric surface.  We will soon see that the
$\F_a$, together with $\P^2$ and $\P^1\times\P^1$, are the basic
building blocks of all smooth complete toric surfaces.
\bigskip
\smallskip

\noindent \ul{T\sc{he} C\sc{lassification} T\sc{heorem}}.
Let $X = X(\Delta)$ be a smooth complete toric surface.  If $\sigma$
is a cone in $\Delta(2)$ with minimal generators $v_1,v_2$, then the
fan $\Delta'$ obtained by subdividing $\sigma$ along the edge
generated by $v_1+v_2$ is again regular.  We say that $\Delta'$ is
obtained from $\Delta$ by {\it stellar subdivision\/}.  In Lecture~6,
we saw that $X' = X(\Delta')$ is the blow-up of $X$ at the point
$x_\sigma$ corresponding to $\sigma$.  Note that $X'$ is again a
complete smooth toric surface.
\bigskip

\ex{3.1} Show that $\F_{1}$ is isomorphic to the blow-up of $\P^2$ at
a point. 
\bigskip

The classification theorem for smooth complete toric surfaces goes as
follows.

\proclaim Theorem 3.1. All smooth complete toric surfaces arise from
$\P^2$, $\P^1\times\P^1$, and the Hirzebruch surfaces $\F_a$ for $a
\ge 2$ by successive blow-ups corresponding to stellar subdivisions.

Before give the proof, we need to do some preliminary work.  Suppose
that $X = X(\Delta)$ for some complete regular fan $\Delta$, and let
the minimal generators of the cones of $\Delta$ be $v_0,\ldots,v_d =
v_0$, listed in clockwise from $v_0$.  Then the 2-dimensional cones of
$\Delta$ are generated by successive pairs.  We will write the indices
modulo $d$.
\bigskip

\ex{3.2} Prove that there are integers $a_i$ such that 
$$
v_{i-1} + v_{i+1} = a_i v_i\leqno(3.1)
$$
for all $i$.  Hint: $v_{i-1}, v_i$ and $v_i,v_{i-1}$ are bases of $N$.
Also remember that $v_{i-1}$ and $v_{i+1}$ lie on opposite sides of
the line generated by $v_i$.
\bigskip

The following lemma will play a key role in the proof of Theorem 3.1.

\proclaim Lemma 3.2. Let $\Delta'$ be a regular subdivision of a
regular cone $\sigma$.  The $\Delta'$ is obtained from $\sigma$ by a
sequence of stellar subdivisions.

\prf
Let the cones of $\Delta'$ have minimal generators $v_1 = \tilde v_0,
\tilde v_1,\ldots,\tilde v_{r},\tilde v_{r+1} = v_2$, listed
clockwise from $v_1$.  We proceed by induction on $r$, the case $r =
0$ being obvious.

If for some $1 \le i \le r$ we have $\tilde v_{i-1} + \tilde v_{i+1} =
\tilde v_i$, then $\Delta'$ comes from a stellar subdivision of a
regular fan $\Delta''$ which also subdivides $\sigma$.  Then we are
done by our inductive hypothesis.

On the other hand, if $\tilde v_{i-1} + \tilde v_{i+1} \ne \tilde v_i$
for all $i$, then (3.1) implies $\tilde v_{i-1} + \tilde v_{i+1} = a_i
\tilde v_i$ for some $a_i \ne 1$.  We also know that $a_i > 0$ since
$\sigma$ is strongly convex, and it follows that $a_i \ge 2$ for $1
\le i \le r$.  Now consider the Hirzebruch-Jung continued fraction
$$
{m\over k} = [a_1,\ldots,a_r].
$$
The results of \S2, especially Exercise 2.7, imply that $\sigma$ has
parameters $m$, $k$ (see Exercise~3.4 below).  Since $0 < k < m$, we
have $m \ge 2$.  This is impossible since $\sigma$ is regular.\qed
\bigskip

\ex{3.3} In the situation of Lemma 3.2, assume $a_i \ge 2$ and let
$p_i/q_i$ be the partial quotients of $m/k = [a_1,\ldots,a_r]$.
\item{a.} Prove that $\tilde v_i = p_{i-1} \tilde v_1 - q_{i-1} \tilde
v_0$ for $i = 1,\ldots,r+1$.
\item{b.} Conclude that $v_2 = m\tilde v_1 - k\tilde v_0$ and that
$\sigma$ has parameters $m$, $k$.  
\bigskip

\noindent {\bf Proof of Theorem 3.1.}  Let $\Delta$ be a complete
regular fan with minimal edge generators $v_0,\ldots,v_d = v_0$ as
above.  We also know that $v_{i-1} + v_{i+1} = a_i v_i$ for some $a_i
\in \Z$.  If some $a_i = 1$, the $\Delta$ is the stellar subdivision
of a fan with fewer cones.  By induction, we may assume that
$$
v_{i-1} + v_{i+1} = a_i v_i,\quad a_i \ne 1\ \hbox{for all}\
i.\leqno(3.2) 
$$

First suppose that $v_i = -v_j$ for some $i,j$.  Relabeling, we may
assume $v_1 = -v_j$.  Using the basis $v_1,v_2$, we get the following
picture:
\vglue2.3truein
\special{psfile=lect7fig6.ps}

\noindent Since $v_2,v_j$ generate a regular cone, Lemma 3.2 and (3.2)
imply that $v_j = v_3 = -v_1$.  Also note that $v_0 + v_2 = a_1v_1$
and $a_1 \ne 1$.  Then $v_0 = -v_2 + a_1v_1$.  The cones generated by
$v_0,v_1$ and $v_0,v_3 = -v_1$ are regular.  Then Lemma 3.2 and (3.2)
show that $v_0,v_1,v_2,v_3$ are the only minimal edge generators of
$\Delta$.  From here, one easily sees that $X = X(\Delta)$ is either
$\P^1\times \P^1$ (if $a_1 = 0$) or $\F_{a}$ (if $a = |a_1| \ne 0$).

To prove the theorem in general, we will use the basis $v_1,v_2$ of
$N$.  By the previous case, we can assume $v_3 \ne -v_1$.  Then,
relative to our chosen basis, $v_3$ must lie in the interior of the
third or fourth quadrant.  First suppose it lies in the third
quadrant, as in the following picture:
\vglue2.3truein
\special{psfile=lect7fig7.ps}

\noindent We know that $v_1 + v_3 = a_2 v_2$, and $a_2 < 0$ since
$v_3$ is in the third quadrant.  Let $\sigma$ be the oriented cone
generated by $v_3,v_1$.  If $a_2 = -1$, then $\sigma$ is regular, so
that $v_3 = v_0$ by Lemma 3.2 and (3.2).  This implies $v_0 =
-v_1-v_2$, and $X = \P^2$ follows.  On the other hand, suppose that
$a_2 < -1$.  Here, we put $m = -a_2$ and let $e_1 = -v_2$, $e_2 =
v_1$.  Then 
$$
v_3 = m e_1 - e_2,\quad v_1 = e_2.
$$
This shows that $\sigma$ has parameters $m$ and $k = 1$.  Now observe
that $v_3,v_4,\ldots,v_d = v_0,v_1$ lie in $\sigma$ and give a regular
subdivision.  If we apply the argument of Lemma 3.2 and Exercise 3.3
to this situation, we see that $m/1 = [a_d,\ldots,a_4]$ (can you
explain why the $a_i$'s are ordered this way?).  But $m/1 = m = [m]$,
which (by uniqueness of continued fraction expansions) implies that
$d = 4$ and $a_4 = -a_2$.  Then the indices work modulo 4, so that
$$
-a_2 v_0 = a_4 v_0 = a_4 v_4 = v_3 + v_5 = v_3 + v_1 = a_2 v_2.
$$
Hence $v_0 = -v_2$, which puts us in the previous case. 

The final case to consider is when $v_3$ lies in the fourth quadrant.
Let $v_j$, $j \ge 3$, be the last cone in the fourth quadrant, as
measured clockwise from $v_2$.  We may assume $v_j \ne -v_1$.  Then
$v_j = av_2 - bv_1$ for $a,b > 0$.  Our choice of $v_j$ implies that
$v_{j+1} = -cv_2 + dv_1$ for $c > 0$.  If $d = 0$, then $v_{j+1} =
-v_2$ (remember $v_{j+1}$ is primitive), which puts us in a case
already considered.

We will complete the proof by showing that $d \ne 0$ leads to a
contradiction.  The key point is that $v_j,v_{j+1}$ span a smooth
cone, so that
$$
\det\pmatrix{\phantom{-} a & -b\cr -c & \phantom{-}d\cr} = ad - bc =
\pm 1.\leqno(3.3)
$$
We know that $a,b,c,d \in \Z$ and $a,b,c > 0$.  If $d < 0$, then
$ad-bc < -1$, contradicting (3.3).  If $d > 0$, then $v_j$ (resp.\
$v_{j+1}$) lies in the fourth (resp.\ second) quadrant.  Since the
cone they span goes clockwise from $v_j$ to $v_{j+1}$, it follows that
$v_j + v_{j+1}$ lies in the third quadrant.  Then
$$
v_j + v_{j+1} = (av_2 - bv_1) + (-cv_2 + dv_1) = (a-c)v_2 + (d-b)v_1
$$
shows that $a > c > 0$ and $d > b > 0$.  This implies $ad-bc > 1$,
again contradicting (3.3).\qed
\bigskip

\ex{3.4} This exercise will show that $a_1,\ldots,a_d$ from (3.1)
satisfy $\sum_{i=1}^d a_i = 3d-12$. 
\item{a.} Verify the identity for $\P^2$, $\P^1\times\P^1$ and the
Hirzebruch surfaces. 
\item{b.} Prove that if the identity holds for $\Delta$, then it holds
for any stellar refinement of $\Delta$. 
\item{c.} Prove that the identity holds for all smooth complete toric
surfaces.  
\vskip0pt
\noindent The identity $\sum_{i=1}^d a_i = 3d-12$ will have an nice
application in \S4.

\smallskip
\bigskip
\noindent \ul{I\sc{ntersection} N\sc{umbers}}. We begin with a brief
discussion of the intersection number $(C \cdot D)$ of two irreducible
curves on a smooth variety $S$ of dimension 2.  If $C \ne D$, the
intersection $C\cap D$ is a finite set of points, and we define
$$
(C\cdot D) = \sum_{p \in C\cap D} I_p(C,D),\leqno(3.4)
$$
where $I_p(C,D)$ is the {\it local intersection multiplicity\/}
defined by 
$$
I_p(C,D) = \dim_\C \cO_{S,p}/\langle f_p,g_p\rangle.
$$
Here, $f_p = 0$ (resp.\ $g_p = 0$) is a local equation for $C$ (resp.\
$D$) in a Zariski neighbhorhood of $p$.
\bigskip

\ex{3.5} Let $X = X(\Delta)$ be a smooth complete toric surface.  Let
$\sigma \in \Delta(2)$ and let $\rho_i$, $i = 1,2$ be the edges of
$\sigma$.  Then let $D_i = V(\rho_i) \simeq \P^1$ be the corresponding
orbit closures in $X$.  Prove that $(D_1 \cdot D_2) = 1$.  Hint: Work
in $U_\sigma \simeq \C^2$.
\bigskip

We next discuss the self-intersection $(C\cdot C)$.  We can't define
this using (3.4) since $C\cap C$ is infinite.  This is a common
problem in intersection theory---what do we do when an intersection
has the wrong dimension?  The standard method is to take a Chern class
of an appropriate {\it excess intersection bundle\/}.  For $(C\cdot
C)$, the excess intersection bundle is the {\it normal bundle\/} of $C
\subset S$, which is defined as follows.  Using the structure sheaf
$\cO_C$ on $C$ and the inclusion $i : C \subset S$, we get the sheaf
$i_*\cO_C$ on $S$ defined by
$$
i_*\cO_C(U) = \cO_C(U\cap C).
$$
Then restriction of functions gives an exact sequence of sheaves
$$
0 \to \cI_C \to \cO_S \to i_*\cO_C \to 0.
$$
We call $\cI_C$ the {\it ideal sheaf\/} of $C$ in $S$.  In Lecture~9,
we will see that $\cI_C = \cO_S(-C)$ since $C$ is a Cartier divisor on
$S$.  The {\it conormal sheaf\/} $\cI_C/\cI_C^2$ has the natural
structure of a sheaf of $\cO_C$-modules, and we define the {\it normal
sheaf\/} of $C$ in $S$ to be the dual sheaf
$$
{\cal N}_{S/C} = \underline{Hom}_{\cO_C}(\cI_C/\cI_C^2,\cO_C).
$$
In general, this sheaf can be badly behaved, but here, ${\cal
N}_{S/C}$ is an {\it inverible sheaf\/} (also called a {\it line
bundle\/}).  Since this is a line bundle on a curve, the Chern class
$c_1({\cal N}_{S/C})$ is an integer called the {\it degree\/} of
${\cal N}_{S/C}$, written ${\rm deg}({\cal N}_{S/C})$.  (Some of these
terms will be defined in Lecture~9).  

Then the {\it self-intersection\/} $(C\cdot C)$ is defined by
$$
(C\cdot C) = {\rm deg}({\cal N}_{S/C}).\leqno(3.5)
$$
We now apply this theory to smooth toric surfaces.
\bigskip

\smallskip
\noindent \ul{S\sc{elf-}I\sc{ntersections on} S\sc{mooth} T\sc{oric}
S\sc{urfaces}}.  Let $X = X(\Delta)$ be a smooth toric surface, and
let $\sigma_1,\sigma_2 \in \Delta(2)$ be adjacent cones.  If the
minimal generators of $\sigma_1$ (resp.\ $\sigma_2$) are $v_1,v_2$
(resp.\ $v_2,v_3$), then Exercise~3.2 implies
$$
v_1 + v_3 = av_2
$$
for $a \in \Z$.  Consider the edge $\rho = \sigma_1\cap\sigma_2 = {\rm
Cone}(v_2)$.  In Lecture~9, we will prove that the normal bundle of
the orbit closure $V(\rho) \simeq \P^1$ in $X$ is $\cO_{\P^1}(-a)$.
We will also see that $\cO_{\P^1}(-a)$ has degree $-a$.  Then (3.5)
will imply the following result.

\proclaim Proposition 3.3.  $D = V(\rho)$ is a curve on $X$ with
self-intersection $(D\cdot D) = -a$.   

This proposition has some nice consequences.  In the situation of the
classification theorem, we have $X = X(\Delta)$ with minimal generators
$v_i$ satisfying
$$
v_{i-1} + v_{i+1} = a_iv_i
$$
by Exercise 3.2.  The edge $\rho_i = {\rm Cone}(v_i)$ gives the orbit
closure $D_i = V(\rho_i) \simeq \P^1$ on $X$, and then Proposition~3.3
implies that $(D_i\cdot D_i) = -a_i$ for all $i$.  Furthermore,
Proposition~3.3 justifies the claims made about self-intersections in
the discussions following (2.1), (2.2), and (2.5).
\bigskip
\medskip
\goodbreak

\noindent {\bigbf \S4. Lattice Polygons and the Number 12}
\bigskip

Our discussion will be based on the paper [6].  Many proofs will be
omitted.  One of our goals is to illustrate some of the many
interesting things connected with toric varieties. 
\bigskip
\smallskip

\noindent \ul{R\sc{eflexive} P\sc{olygons}}.  Let $P \subset N_\R
\simeq \R^2$ be a 2-dimensional polytope which is the convex hull of a
set of points in $N$.  We say that $N$ is a {\it lattice polygon}.  

\proclaim Definition 4.1. A lattice polygon $P \subset N_\R \simeq
\R^2$ is {\bf reflexive\/} if there is exactly one lattice point in
the interior of $P$.

If $P$ is a reflexive polygon and $\gamma \in {\rm Aut}_\Z(N) \simeq
{\rm GL}(2,\Z)$, then one easily sees that $\gamma(P)$ is also
reflexive.  We say that $P$ and $\gamma(P)$ are {\it equivalent\/}.
Up to equivalence (and allowing translation by elements of $N$), there
are precisely 16 reflexive polygons in the plane:
\vglue3.55truein
\special{psfile=lect7fig8.ps}

If $P \subset N_\R \subset \R^2$ is a polygon with the origin in
its interior, then its {\it dual\/} or {\it polar\/} is
$$
P^\circ = \{u \in M_\R \mid \langle u,v\rangle \ge -1\ \hbox{for all}\
v \in P\}.
$$
One can show that $0$ is an interior point of $P^\circ$, and one has
the duality $P = (P^\circ)^\circ$ for polygons in the plane containing
the origin in their interior.

In general, the dual of a lattice polygon need not be a lattice
polygon.  
\bigskip

\ex{4.1} Let $P \subset N_\R \simeq \R^2$ be the lattice polygon with
vertices $\pm2e_1,\pm2e_2$, where $e_1, e_2$ form a basis of $N$.
Prove that $P^\circ$ has vertices ${1\over2}(\pm e_1^*\pm e_2^*)$ in
$M_\R$.
\bigskip
 
We can characterize reflexive polygons in the plane as follows.

\proclaim Proposition 4.2.  Let $P$ be a lattice polygon $P$
containing 0 in its interior.  The $P$ is reflexive if and only if
$P^\circ$ is reflexive.

\medskip
\ex{4.2} In the illustration on the previous, find the dual of each
reflexive polygon.  (When computing the dual, assume that the origin
is the unique interior lattice point.)  Note that some of the polygons
are self-dual.
\bigskip

We should also mention that in higher dimensions, the definition of
reflexive requires more than just a unique interior lattice point 
(as in Definition~4.1).  This will be covered in the lectures on
mirror symmetry given next week.
\bigskip
\smallskip

\noindent \ul{L\sc{attice} P\sc{oints of} R\sc{eflexive}
P\sc{olygons}}.  Given a reflexive polygon $P$, we let $\ell(P)$
denote the number of lattice points on the boundary of $P$.  

\proclaim Theorem 4.3. Let $P$ be a reflexive polygon containing the
origin.  Then 
$$
\ell(P) + \ell(P^\circ) = 12.
$$

The paper [6] gives four proofs of this result.  The simplest uses the
illustration on the previous page together with Exercise~4.2.  There
is also a proof using modular forms.  For us, the proof of interest
uses toric varieties and the Noether formula.  In fact, we will give
two ``toric'' proofs, only one of which uses the Noether formula.

Given a reflexive polygon $P$ containing the origin, we get a fan
$\Delta_P$ by using the lattice points on the boundary of $P$ as cone
generators.  Here is an example of what this looks like: 
\vglue2.3truein
\special{psfile=lect7fig9.ps}

\noindent Note that $\Delta_P$ has $\ell(P)$ edges and $\ell(P)$
2-dimensional cones.  This fan is obviously complete, and the
following exercise shows that $\Delta_P$ is also regular.
\bigskip

\ex{4.3} Let $v_1,v_2 \in N \simeq \R^2$ be the minimal generators of
a 2-dimensional strongly convex rational polyhedral cone $\sigma$.
Prove that $\sigma$ is regular if and only if ${\rm
Conv}(0,v_1,v_2)\cap N = \{0,v_1,v_2\}$.  Hint:~Let $e_1,e_2$ be a
normalized basis for $\sigma$ and let $m$, $k$ be the parameters of
$\sigma$.  Show that $m > 1$ implies $e_1 \in {\rm Conv}(0,v_1,v_2)$.
\bigskip

In this construction, it is important to use {\it all\/} lattice
points on the boundary.  For example, if we only used the vertices of
$P$ in the above example, the resulting fan would not be regular.  

For later purposes, we will also need to discuss the {\it area\/} of a
polygon in $N_\R \simeq \R^2$.  Let $e_1,e_2$ be a basis of $N$.  Then
we will measure area so that the square with vertices
$0,e_1,e_2,e_1+e_2$ has area 1.  (Since elements of ${\rm GL}(2,\Z)$
have determinant $\pm1$, this definition is independent of the chosen
basis.)  We can then relate area to regularity as follows.
\bigskip

\ex{4.4} Let $\sigma$ and $v_1,v_2$ be as in Exercise 4.3.  Prove that
$\sigma$ is regular if and only if ${\rm Conv}(0,v_1,v_2)$ has area
${1\over 2}$.
\bigskip

\ex{4.5} Let $P$ be a reflexive polygon.  Prove that ${\rm Area}(P) =
{1\over 2}\ell(P)$.
\bigskip
\smallskip

\noindent \ul{S\sc{elf-}I\sc{ntersections on} S\sc{mooth} T\sc{oric}
S\sc{urfaces}}.  Before we present the toric proofs of Theorem~4.3, we
first discuss another way to compute self-intersections.  Suppose that
$X = X(\Delta)$ be a smooth complete toric surface, and let
$\rho_1,\ldots,\rho_d$ denote its edges, and let $v_i$ be the minimal
generator of $\rho_i$.  This gives orbit closures $D_i = V(\rho_i)
\simeq \P^1$ in $X$.

Now consider a formal sum $D = \sum_{i=1}^d b_i D_i$, where $b_i \in
\Z$.  In the terminology of Lecture~9, $D$ is a {\it $T$-invariant
divisor\/} on $X$.  From $D$, we get the polygon $Q \subset M_\R$
defined by
$$
Q = \{u \in M \mid \langle u,v_i\rangle \ge -b_i\ \hbox{for all}\
i\}.\leqno(4.1) 
$$
Then later in the week, you will see that if $b_i \ge 0$ for all $i$,
then 
$$
(D\cdot D) = 2\, {\rm Area}(Q).\leqno(4.2)
$$
(See also page 111 of [2].)  This formula can be used to prove Theorem
4.3 as follows.
\medskip

\noindent {\bf First Proof of Theorem 4.3.} Given a reflexive polytope
$P$ containing the origin, we get the complete regular fan $\Delta_P$
described above.  Then $X = X(\Delta_P)$ is a smooth complete toric
surface.  Now consider the divisor $-K = \sum_{i=1} D_i$ (the reason
for calling this $-K$ will become clear in the second proof).  For
this divisor, the polytope (4.1) is
$$
Q = \{u \in M \mid \langle u,v_i\rangle \ge -1\ \hbox{for all}\
i\}.
$$
Since $P$ is the convex hull of the $v_i$, it follows that
$$
Q = \{u \in M \mid \langle u,v\rangle \ge -1\ \hbox{for all}\
v \in P\} = P^\circ.
$$
Hence the polytope associated to the divisor $-K$ is the dual reflexive
polygon $P^\circ$.  This implies
$$
(-K\cdot -K) = 2\, {\rm Area}(P^\circ) = \ell(P^\circ),\leqno(4.3)
$$
where the first equality is (4.2) and the second is by Exercise 4.5
(since $P^\circ$ is reflexive by Proposition~4.2).  However, we can
also compute $(-K\cdot -K)$ as follows:
$$
(-K\cdot -K) = \Big(\sum_{i=1}^d D_i \cdot \sum_{i=1}^d D_i\Big) =
	\sum_{i=1}^d (D_i\cdot D_i) + 2\sum_{i<j} (D_i\cdot D_j).
$$
We saw at the end of \S3 that $(D_i\cdot D_i) = -a_i$, and we also
know that for $i < j$,
$$
(D_i\cdot D_j) = \cases{ 1 & if $v_i$ and $v_j$ are adjacent\cr 0 &
otherwise.\cr} 
$$
Thus the above computation simplifies to
$$
(-K\cdot -K) = -\sum_{i=1}^d a_i + 2d = -(3d-12) + 2d = 12 - d, 
$$
where the second equality uses Exercise 3.5.  Here, $d$ is the number
of edges in $\Delta_P$, which is $\ell(P)$ by the construction of
$\Delta_P$.  Comparing this with (4.3), we obtain
$$
12 - \ell(P) = (-K\cdot -K) = \ell(P^\circ).
$$
This completes the first proof.\qed

\goodbreak

\smallskip
\bigskip
\noindent \ul{N\sc{oether's} F\sc{ormula}}. Before we can give the
second proof, we need to explain Noether's formula.  Let $X$ be a
smooth complete surface.  Then Noether's formula states that
$$
12\chi(\cO_X) = (K\cdot K) + \chi(X).\leqno(4.4)
$$
Here are what the components of this formula mean:
\item{$\bullet$} $\chi(\cO_X) = \dim H^0(X,\cO_X) - \dim H^1(X,\cO_X)
+ \dim H^2(X,\cO_X)$.  This is {\it holomorphic Euler
characteristic\/}.  Sheaf cohomology will be discussed later in the
week.
\item{$\bullet$} $\chi(X) = \sum_{q=0}^4 (-1)^qb_q(X)$, where the {\it
$q$th Betti number\/} $b_q(X)$ is the rank of the singular homology
group $H_q(X)$.  This is the usual {\it Euler characteristic\/} from
algebraic topology.
\item{$\bullet$} $K$ is a {\it canonical divisor\/}, which means that
the sheaf $\cO_X(K)$ is isomorphic to the sheaf $\Omega_X^2$ of
algebraic 2-forms on $X$.  Lecture~9 will explain how any Cartier
divisor $D$ gives an invertible sheaf $\cO_X(D)$.
\vskip0pt
\noindent The Noether formula is discussed in detail in [3, Chapter 4]
and [4, A.4].  We should also mention that Noether's formula can be
stated using the tangent bundle ${\cal T}_X$ of $X$.  Namely, one can
show that $(K\cdot K) = \int_X c_1({\cal T}_X)\cup c_1({\cal T}_X)$ and
$\chi(\cO_X) = \int_X c_2({\cal T}_X)$, where $c_1({\cal T}_X)$ and
$c_2({\cal T}_X)$ are the {\it Chern classes\/} of the vector bundle
${\cal T}_X$.

Now let $X = X(\Delta)$ be a smooth complete toric surface.  We will
see in later lectures that the following holds:
\item{$\bullet$} $H^0(X,\cO_X) = \C$ and $H^i(X,\cO_X) = 0$ for $i \ge 
1$.  Thus $\chi(\cO_X) = 1$.
\item{$\bullet$} The Euler characteristic of $X$ is $d$, the number of
cones in $\Delta(2)$.
\item{$\bullet$} The canonical divisor of $X$ is $K = -\sum_{i=1}^d
D_i$. 
\vskip0pt
\noindent Thus, Noether's formula (4.4) for a smooth complete toric
surface can be written
$$
12 = (-K\cdot -K) + d,\leqno(4.5)
$$
where $d$ is the cardinality of $\Delta(2)$ and $-K = \sum_{i=1}^d
D_i$. 
\medskip 

\noindent{\bf Second Proof of Theorem 4.3.}  Let $X = X(\Delta_P)$ be
the smooth complete toric surface coming from the reflexive polytope
$P$.  Then $(-K\cdot -K) = \ell(P^\circ)$ by (4.3), and $d = \ell(P)$
follows from the construction of $\Delta_P$.  Then Theorem~4.3 is an
immediate consequence of (4.5).\qed
\bigskip

\ex{4.5} In the first proof of Theorem 4.3, we showed that $(-K\cdot
-K) = 2d -\sum_{i=1}^d a_i$.  Use this to prove that (4.4) is
equivalent to the identity $\sum_{i=1}^d a_i = 3d-12$ of
Exercise~3.5. 
\bigskip

In one sense, it is overkill to use Noether's formula to prove
Theorem~4.3.  Our reasons for giving the proof were to introduce some
ideas you will see later in the week and to illustrate how fundamental
invariants of toric varieties (such as $(K\cdot K)$ and $\chi(X))$ can
be computed from combinatorial invariants of the fan.  We also wanted
to give a simple example of the deep link between polytopes and toric
varieties.

\bigskip
\noindent{\bigbf References}
\bigskip

\parindent=12pt

\ref 1.~A.\ Durfee, {\it Fifteen characterizations of rational double
points and simple critical points\/}, Enseign.\ Math.\ {\bf 25}
(1979), 131--163.
\smallskip

\ref 2.~W.\ Fulton, {\sl Introduction to Toric Varieties\/}, Princeton
University Press, Princeton, NJ, 1993.
\smallskip

\ref 3.~P.\ Griffiths and J.\ Harris, {\sl Principles of Algebraic
Geometry\/}, Wiley, New York, 1978
\smallskip

\ref 4.~R.\ Hartshorne, {\sl Algebraic Geometry\/}, Springer-Verlag,
New York-Berlin-Heidelberg, 1977. 
\smallskip

\ref 5.~T.\ Oda, {\sl Convex Bodies and Algebraic Geometry: An
Introduction to Toric Varieties\/}, Sprin\-ger-Verlag, New
York-Berlin-Heidelberg, 1988. 
\smallskip

\ref 6.~B.\ Poonen and F.\ Rodriguez-Villegas, {\it Lattice polygons
and the number 12\/}, Amer.\ Math.\ Monthly {\bf 107} (2000),
238--250.
\smallskip

\end 