Another interesting topic is generalizations of ${\mathbb{C}}/L$ to higher
dimensions. For example, we may want to consider the complex manifolds
${\mathbb{C}}^{2}/L$ where $L\cong{\mathbb{Z}}^{4}$ is a discrete subgroup.
Remarkably, for a generic choice of $L$ the quotient ${\mathbb{C}}^{2}/L$ is
not an algebraic variety! For some, well-understood, conditions on $L$ the
quotients are algebraic, and are referred to as abelian surfaces.

We would be amiss not to mention the Frey elliptic curves which play a crucial
role in the proof of Fermat’s Last Theorem. For a putative integer solution to

$a^{n}+b^{n}=c^{n},$

Hellegouarch [21] and then Frey [16, 17] considered the elliptic curve

$y^{2}z=(x-a^{n}z)x(x+b^{n}z)$

and proved that it would have some weird properties. This put FLT squarely
into a well-studied area of elliptic curves over ${\mathbb{Q}}$ and eventually
lead to Wiles’ proof.13.113.113.1 I was still in high school when Frey’s paper
discussing the connection between FLT and modularity conjecture appeared, but
a mathematician I knew correctly predicted that because of it FLT will now be
proved sooner rather than later.

Exercise 1. Prove (13.2).

Exercise 2. Prove that $\theta(z)$ has zeros of first order at
$L={\mathbb{Z}}+{\mathbb{Z}}\tau$ by integrating
$\frac{\theta^{\prime}(z)}{\theta(z)}$ along the boundary of a fundamental
parallelogram of $L$.

Exercise 3. Prove that the function

$-{\mathrm{i}}\sum_{n\in{\mathbb{Z}}}(-1)^{n}{\mathrm{e}}^{\pi{\mathrm{i}}((n+\frac{1}{2})^{2}\tau+(2n+1)z)}$

is well-defined and holomorphic, and satisfies the transformation properties
(13.2).

##  14 Genus of complex algebraic curves and Riemann-Hurwitz formula.

Let $X$ be a compact connected Riemann surface. Since it is an oriented real
surface, it is homeomorphic to a sphere with $g$ handles.

For $g=0$, the only14.114.114.1 Uniqueness is not obvious. such surface is
${\mathbb{C}}{\mathbb{P}}^{1}$. For $g=1$ these are elliptic curves. They form
a one-parameter family (basically $\tau$ in
$L={\mathbb{Z}}+{\mathbb{Z}}\tau$). A general picture for $g\geq 2$ is
something like:

$g=3$

A nontrivial theorem implies that $X$ has a unique structure of a complex
algebraic curve. The difficult part is to show that there is a meromorphic
function $X\to{\mathbb{C}}{\mathbb{P}}^{1}$.

Let us discuss nonconstant holomorphic maps between Riemann surfaces. Suppose
we have one such map $f:X\to Y$. For a point $p\in X$ consider $q=f(p)$. If we
consider local charts at $p$ and $q$ with coordinates $z$ and $t$ respectively
(so that $z=0$ corresponds to $p$ and $t=0$ corresponds to $q$) then the map
is given by

$t=f(z)=c_{1}z+c_{2}z^{2}+\ldots$

and we will call the smallest $k$ such that $c_{k}\neq 0$ the multiplicity of
$f$ at $p$. One can think of the multiplicity of $f$ at $p$ as the number of
solutions to $f(z)=\varepsilon$ in the neighborhood of $z=0$ for small
$\varepsilon$, see Exercise 1. We also note that there are only finitely many
points in $X$ where the multiplicity of $f$ is larger than one. Indeed, by
compactness of $X$, we would otherwise have an accumulation point which would
lead to an accumulation point of the zeros of the derivative, thus making $f$
constant by analytic continuation.

######  Proposition 14.1.

For $q\in Y$ the number of preimage points $p\in X$, counted with
multiplicities

$\sum_{p,f(p)=q}m_{p}$ (14.1)

is finite and is independent of $q$. We call it the degree of the map $f$.

###### Proof.

We will sketch the proof and the reader is welcome to fill in the details.

Finiteness is obvious due to compactness of $X$, since an accumulation point
would imply that $f$ is locally (and hence globally) constant. Now suppose
that $p_{1},\ldots,p_{k}$ map to $q\in Y$ with multiplicities
$m_{1},\ldots,m_{k}$. Let $U$ be a small neighborhood of $q$ and consider
connected components of the open set $f^{-1}(U\setminus\\{q\\})$. By
compactness of $X$, every such connected component must have one of $x_{i}$ in
its closure, since we can find an accumulation point of a sequence of points
in $X$ whose values approach $q$. We can pick a small enough neighborhood $U$
of $q$ so that preimages of $U$ near $x_{i}$ are close enough to $x_{i}$ (in
some metric) to be disjoint. This means that that for $q$ in this $U$ all of
the preimages are close to $x_{i}$ and by Exercise 1, the total number in
(14.1) is locally constant and hence constant. ∎

######  Remark 14.2.

The invariant $\deg f$ has an algebraic meaning. The fields of meromorphic
functions on $X$ and $Y$ are transcendence degree one field extensions of
${\mathbb{C}}$ and the pullback under $f$ induces a field extension

${\rm Mer}(Y)\subseteq{\rm Mer}(X).$

Then this extension is finite and $\deg f$ is its degree, i.e. $\dim_{{\rm
Mer}(Y)}{\rm Mer(X)}$. We do not prove this claim, but see Exercise 2.

Proposition 14.1 has an important corollary.

######  Corollary 14.3.

For any nonzero meromorphic function $f$ on $X$ with zeros/poles at points
$x_{i}\in X$ of multiplicities $m_{i}$, there holds $\sum_{i}m_{i}=0$.

###### Proof.

If $f$ is a nonzero constant, then the sum is empty. Otherwise, we can think
of $f$ as a nonconstant holomorphic map $X\to{\mathbb{C}}{\mathbb{P}}^{1}$.
The sum of positive $m_{i}$ is precisely the degree of $f$ computed via
preimages of $0\in{\mathbb{C}}{\mathbb{P}}^{1}$ and the opposite of the sum of
negative $m_{i}$ is the degree of $f$ computed via preimages of
$\infty\in{\mathbb{C}}{\mathbb{P}}^{1}$. ∎

Another proof of Corollary 14.3 can be obtained by integrating

$\frac{1}{2\pi{\mathrm{i}}}\int_{\gamma}d\log
f=\frac{1}{2\pi{\mathrm{i}}}\int_{\gamma}\frac{df}{f}$

over a contour $\gamma$ which is the boundary of a region $U$ that contains
all zeros and poles of $f$. Now observe that $\gamma$ is also the boundary
(with the wrong sign) of the complement of $U$ which now has no zero or poles
of $f$ and thus the integral is zero.

Corollary 14.3 implies that the map ${\rm WeilDiv}(X)\to{\mathbb{Z}}$ which
sends

$\sum_{k}m_{k}[p_{k}]\mapsto\sum_{k}m_{k}$

induces a map ${\rm Pic}(X)\to{\mathbb{Z}}$, which is understandably called
the degree map. Isn’t it nice how there are terms “the degree of a map” and
“the degree map” which are quite different but sufficiently related to cause
confusion?

######  Remark 14.4.

The kernel of the degree map is called ${\rm Pic}^{0}(X)$. For $g=0$ we have
${\rm Pic}^{0}({\mathbb{C}}{\mathbb{P}}^{1})\cong\\{0\\}$. For $g=1$ we saw
that

${\rm Pic}^{0}({\mathbb{C}}/L)\cong{\mathbb{C}}/L.$

More generally, for any compact Riemann surface $X$ of genus $g$ the group
${\rm Pic}^{0}(X)$ has a natural structure of an algebraic variety of
dimension $g$. In fact it is isomorphic to ${\mathbb{C}}^{g}/L$ for some
lattice $L\cong{\mathbb{Z}}^{2g}$. We do not prove these statements.

We will now talk about a wonderful result known as the Riemann-Hurwitz
formula.

######  Theorem 14.5.

Let $f:X\to Y$ be a nonconstant holomorphic map of compact connected Riemann
surfaces. Then

$2g_{X}-2=(\deg f)(2g_{Y}-2)+\sum_{p\in X}(m_{p}-1)$

where $m_{p}$ is the multiplicity of $f$ at $p$.

###### Proof.

First of all, $\sum_{p}$ is really just a finite sum since we only need to
worry about the points where $m_{p}>1$. We call points in $X$ with $m_{p}>1$
the ramification points of $f$. We pick a triangulation $T_{Y}$ of $Y$ in such
a way that all images of ramification points of $X$ are among the vertices of
$T_{Y}$. Then we compare the triangulation $T_{Y}$ with its preimage $T_{X}$
on $X$. Note that each triangle $\Delta$ in $T_{Y}$ gives rise to $\deg f$
triangles in $T_{X}$ because the map $f^{-1}(\Delta)\to\Delta$ is unramified,
and the same is true for the edges. For vertices, a vertex $q\in Y$ of $T_{Y}$
gives rise to

$\sum_{p,f(p)=q}1=\sum_{p,f(p)=q}m_{p}-\sum_{p,f(p)=q}(m_{p}-1)=\deg
f-\sum_{p,f(p)=q}(m_{p}-1)$

vertices in $T_{X}$. We know that the Euler characteristics of $X$ (resp. $Y$)
is the number of triangles minus the number of edges plus the number of
vertices of $T_{X}$ (resp. $T_{Y}$). It remains to combine the observation
about the numbers of triangles, edges and faces of $T_{X}$ and $T_{Y}$ with
well-known formulas for the Euler characteristics
$\chi(X)=2-2g_{X},~{}\chi(Y)=2-2g_{Y}.$ ∎

We will illustrate the Riemann-Hurwitz formula with a couple of examples.

  * •

Let $X$ be an elliptic curve and $X\to{\mathbb{C}}{\mathbb{P}}^{1}$ be the
degree two map ramified at $4$ points (see Section 10). The Riemann-Hurwitz
formula reads

$2\cdot 1-2=2(2\cdot 0-2)+4\times(2-1).$

  * •

Let $f:{\mathbb{C}}{\mathbb{P}}^{1}\to{\mathbb{C}}{\mathbb{P}}^{1}$ be the map
$(x_{0}:x_{1})\mapsto(x_{0}^{n}:x_{1}^{n})$ which is a compactification of the
map $z\mapsto z^{n}$. Then the degree of $f$ is $n$ and there are two
ramification points $(0:1)$ and $(1:0)$ with $m_{p}=n$ each. The Riemann-
Hurwitz formula reads

$2\cdot 0-2=n(2\cdot 0-2)+(n-1)+(n-1).$

For a dimension one complex manifold $X$, the holomorphic cotangent bundle is
the canonical bundle of $X$. If $X$ is compact, it makes sense to ask about
the degree of the corresponding Weil divisor class $K_{X}$, and Riemann-
Hurwitz formula allows us to compute it.

First, we note that in the case of $X={\mathbb{C}}{\mathbb{P}}^{1}$ the degree
of $K_{X_{{\mathbb{C}}{\mathbb{P}}^{1}}}$ is $(-2)$. Indeed, if we look at the
differential form $dz$ which has no zeros/poles in ${\mathbb{C}}$, it has a
second order pole at infinity, because
$dz=-\left(\frac{1}{z}\right)^{-2}\,d\left(\frac{1}{z}\right)$ near $\infty$.

######  Proposition 14.6.

For a smooth compact Riemann surface $X$ we have $\deg K_{X}=2g_{X}-2$.

###### Proof.

For any nonconstant map $f:X\to Y$ we have

$\deg K_{X}=(\deg f)\deg K_{Y}+\sum_{p\in X}(m_{p}-1).$ (14.2)

Indeed, if we take a meromorphic form on $Y$ and compare its zeros and poles
with zeros and poles of its preimage, we will get additional $(m_{p}-1)$ for
every ramification point, because $dz^{k}=kz^{k-1}\,dz$.

Because every $X$ has a holomorphic map $f:X\to{\mathbb{C}}{\mathbb{P}}^{1}$,
we combine Riemann-Hurwitz formula for $f$ with (14.2) to get the desired
result. ∎

Let us now talk about automorphisms of compact Riemann surfaces.

For $g(X)=0$, we have ${\rm Aut}({\mathbb{C}}{\mathbb{P}}^{1})={\rm
PGL}(2,{\mathbb{C}})$, given by the Möbius transformations
$z\mapsto\frac{az+b}{cz+d}$.

For $g(X)=1$, the automorphism group of $X={\mathbb{C}}/L$ is a semidirect
product of ${\mathbb{C}}/L$ with a finite group, which could be
${\mathbb{Z}}/2{\mathbb{Z}}$, ${\mathbb{Z}}/4{\mathbb{Z}}$ or
${\mathbb{Z}}/6{\mathbb{Z}}$. The normal subgroup ${\mathbb{C}}/L$ is obtained
by parallel transports $\varphi_{w}:z\mapsto z+w$, and the finite group comes
from scalar multiplications that preserve the lattice $L$. It is usually just
$\\{\pm 1\\}$, but it is $\\{1,{\mathrm{i}},-1,{\mathrm{i}}\\}$ if
$L={\mathbb{Z}}+{\mathbb{Z}}{\mathrm{i}}$ and it is the group of sixth roots
of $1$ for the hexagonal lattice
$L={\mathbb{Z}}+{\mathbb{Z}}(\frac{1}{2}+\frac{\sqrt{3}}{2}{\mathrm{i}})$. We
do not prove this, but it is not too hard to derive from what we know about
elliptic curves.

For $g(X)\geq 2$, the automorphism group of $X$ is always finite. One way to
prove it (we omit details) is to argue that $X$ can be embedded into a
projective space in a way that automorphisms of $X$ extend to the ambient
space. Then the automorphism group of $X$ will be an algebraic subgroup of
some ${\rm PGL}(n,{\mathbb{C}})$, so it will have a finite number of connected
components. Finally, the identity component must be just a point, otherwise
$X$ would admit a nonzero holomorphic vector field, which would contradict
$\deg K_{X}>0$.

There is an even stronger statement with a surprisingly simple proof.

######  Theorem 14.7.

Let $X$ be a compact connected Riemann surface of genus $g\geq 2$. Then

$|{\rm Aut}(X)|\leq 84(g-1).$

###### Proof.

Let $G$ be the automorphism group of $X$. One can show14.214.214.2 We will
indeed show this later in Corollary 16.3, although a motivated reader can try
to do it now. that the set of $G$-orbits of $X$ can be given the structure of
a Riemann surface so that $X\to Y=X/G$ is a degree $|G|$ holomorphic map. An
orbit of $G$ with a stabilizer of size $n$ contributes $|G|/n$ points of
multiplicity $n$ to the set of ramification points of $X\to Y$. Therefore, the
Riemann-Hurwitz formula gives

$2g-2=|G|(2g_{Y}-2)+|G|\frac{n_{1}-1}{n_{1}}+|G|\frac{n_{2}-1}{n_{2}}+\ldots+|G|\frac{n_{k}-1}{n_{k}}$

for some integers $n_{1},\ldots,n_{k}$ that encode stabilizer sizes of the
action of $G$.

If $g_{Y}\geq 2$, then we have $2g-2\geq|G|\cdot 2$, and the claim follows. If
$g_{Y}=1$, then we have

$2g-2=|G|\sum_{i=1}^{k}(1-\frac{1}{n_{i}}).$

Since the left-hand side is positive, the sum on the right is nonempty. Each
of the terms is at least $|G|\frac{1}{2}$, so $|G|\leq 4(g-1)$.

If $g_{Y}=0$, we see that

$\frac{2g-2}{|G|}=-2+\sum_{i=1}^{k}(1-\frac{1}{n_{i}}).$

So it remains to observe that the smallest possible positive value of the
right hand side is $\frac{1}{42}$, achieved by

$-2+\frac{1}{2}+\frac{2}{3}+\frac{6}{7},$

see Exercise 3. ∎

Exercise 1. Prove that for a nonconstant holomorphic function $f$ with
$f(0)=0$ the number of solutions to $f(0)=\varepsilon$ is equal, for all small
nonzero $\varepsilon$, to the smallest $k$ such that the $k$-th derivative of
$f$ at $z=0$ is nonzero. _Hint:_ Since $f^{\prime}(z)$ is holomorphic, there
is a neighborhood where $f^{\prime}(z)$ has no other zeros than perhaps $z=0$.
Thus any zeros of $f(z)-\varepsilon$ will be simple. Then you can count the
their number in a circle $|z|<\alpha$ by

$\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=\alpha}\frac{f^{\prime}(z)}{f(z)-\varepsilon}\,dz.$

It is continuous in $\varepsilon$ for $\varepsilon$ close to $0$ and is equal
to
$\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=\alpha}\frac{f^{\prime}(z)}{f(z)}\,dz.$

Exercise 2. Check the claim of Remark 14.2 for the map
$f:{\mathbb{C}}{\mathbb{P}}^{1}\to{\mathbb{C}}{\mathbb{P}}^{1}$ that sends
$(x_{0}:x_{1})$ to $(x_{0}^{n}:x_{1}^{n})$ for some positive integer $n$.

Exercise 3. Prove the last claim of the section. _Hint:_ First argue that for
$k\geq 5$ the right hand side is at least $\frac{1}{2}$. Then for $k=4$ the
right hand is positive, so at least one of $n_{i}$ is $3$ or more, which gives
at least $-2+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{2}{3}=\frac{1}{6}$. If
$k\leq 2$, the right hand side is negative, so we must have $k=3$, etc.

##  15 Blowup of a point in ${\mathbb{C}}^{2}$. Birational equivalence of
algebraic varieties.

We are now switching gears to talk about phenomena that appear in complex
dimension two and higher. We start with a wonderful and initially puzzling
construction of the blowup of a point in ${\mathbb{C}}^{2}$.

Consider the subset
$S\subset{\mathbb{C}}^{2}\times{\mathbb{C}}{\mathbb{P}}^{1}$ defined by

$S=\\{(x,y)\times(u:v)\in{\mathbb{C}}^{2}\times{\mathbb{C}}{\mathbb{P}}^{1},{\rm~{}such~{}that~{}}xv-
yu=0\\}.$

######  Proposition 15.1.

The set $S$ is a smooth complex algebraic surface.

###### Proof.

Recall that ${\mathbb{C}}{\mathbb{P}}^{1}$ is covered by two copies of
${\mathbb{C}}^{1}$, with points $(\frac{u}{v}:1)$ and $(1:\frac{v}{u})$. This
gives a covering of ${\mathbb{C}}^{2}\times{\mathbb{C}}{\mathbb{P}}^{1}$ by
two copies of ${\mathbb{C}}^{3}$. In the first open set, when using
$z=\frac{u}{v}$ as the third coordinate, we get

$S\cap{\mathbb{C}}^{3}=\\{(x,y,z)\in{\mathbb{C}}^{3},{\rm~{}such~{}that~{}}x-yz=0\\}$

which is isomorphic to ${\mathbb{C}}^{2}$ (we can solve for $x$). Similarly,
in the second open set we get

$x\left(\frac{v}{u}\right)-y$

and we can solve for $y$. It is also clearly algebraic, for example because it
is given by a polynomial equation in
${\mathbb{C}}^{2}\times{\mathbb{C}}{\mathbb{P}}^{2}$. ∎

Whenever we have a subset in a product, it is natural to consider the two
projections of it. We will first look at the projection

$S\to{\mathbb{C}}{\mathbb{P}}^{1}$

that forgets $(x,y)$. The fiber of $S$ over $(u:v)$ is the set of all $(x,y)$
such that $xv=yu$. Since $(u,v)\neq 0$, this is equivalent to
$(x,y)=\lambda(u,v)$ for some $\lambda\in{\mathbb{C}}$. We can see from this
that $S$ is the line bundle $\mathcal{O}(-1)$ over
${\mathbb{C}}{\mathbb{P}}^{1}$.

The other projection

$\pi:S\to{\mathbb{C}}^{2}$

is even more interesting. We get

$\pi^{-1}(a,b)=\left\\{\begin{array}[]{ll}\\{(a,b)\times(a:b)\\},&{\rm
if~{}}(a,b)\neq(0,0);\\\ \\{(0,0)\\}\times{\mathbb{C}}{\mathbb{P}}^{1},&{\rm
if~{}}(a,b)=(0,0).\end{array}\right.$

So we have drastically different fibers – a single point everywhere except
$(0,0)$ and a whole ${\mathbb{C}}{\mathbb{P}}^{1}$ over $(0,0)$.

How should we visualize the blowup surface $S$? It is important to realize
that the naive picture on the left in the figure below

$S$$\hskip-6.5pt\cdot(0,0)$${\mathbb{C}}^{2}$$S$$\hskip-6.0pt\cdot(0,0)$${\mathbb{C}}^{2}$

is wrong, because $S$ would not be smooth. The correct way of thinking about
it is the picture on the right, where I also indicate that approaching the
origin along different directions gives different limiting points on $S$.

Terminology: The map $S\to{\mathbb{C}}^{2}$ is called the blowdown map or the
blowdown morphism, and $S$ is called the blowup of ${\mathbb{C}}^{2}$ at
$(0,0)$ because that point is “blown up” to a
${\mathbb{C}}{\mathbb{P}}^{1}$.15.115.115.1 Algebraic geometers waiting for
their flight in the airport should refrain from discussing blowing up the
plane at any number of points.

The construction generalizes, of course. First of all, we can blow up a point
in ${\mathbb{C}}^{n}$. Consider
$S\subset{\mathbb{C}}^{n}\times{\mathbb{C}}{\mathbb{P}}^{n-1}$ given by

$S=\\{(x_{1},\ldots,x_{n})\times(y_{1}:\ldots:y_{n}),{\rm~{}such~{}that~{}rk}\hskip-2.0pt\left(\begin{array}[]{ccc}x_{1}&\cdots&x_{n}\\\
y_{1}&\cdots&y_{n}\end{array}\right)=1\\}.$

Since not all $y_{i}$ are zero, the rank of the matrix is at least one, so $S$
is cut out by all $2\times 2$ minors of the matrix, i.e.

$x_{i}y_{j}-x_{j}y_{i}=0,~{}1\leq i<j\leq n.$

We claim that $S$ is smooth. Indeed, suppose we are working in the chart of
${\mathbb{C}}{\mathbb{P}}^{n-1}$ where $y_{n}\neq 0$. Then the rank condition
is equivalent to

$(x_{1},\ldots,x_{n})=\lambda(\frac{y_{1}}{y_{n}},\ldots,\frac{y_{n-1}}{y_{n}},1).$

This means that $\lambda=x_{n}$ and then in this chart $S$ is isomorphic to
${\mathbb{C}}^{n}$ with coordinates

$x_{n},\frac{y_{1}}{y_{n}},\ldots,\frac{y_{n-1}}{y_{n}}.$

Looking at the fibers of $\pi:S\to{\mathbb{C}}^{n}$, we see that

$\pi^{-1}(x_{1},\ldots,x_{n})=\left\\{\begin{array}[]{ll}\\{(x_{1},\ldots,x_{n})\times(x_{1}:\ldots:x_{n})\\},&{\rm
if~{}}(x_{1},\ldots,x_{n})\neq(0,\ldots,0),\\\
\\{(0,\ldots,0)\\}\times{\mathbb{C}}{\mathbb{P}}^{n-1},&{\rm
if~{}}(x_{1},\ldots,x_{n})=(0,\ldots,0),\end{array}\right.$

so the point $(0,\ldots,0)\in{\mathbb{C}}^{n}$ is blown up to
${\mathbb{C}}{\mathbb{P}}^{n-1}$.

######  Remark 15.2.

In general, it is possible to blow up a smooth subvariety $Z$ in a smooth
variety $X$ to get a blowdown morphism

$\pi:{\rm Bl}(Z\subset X)\to X$

with the following properties.

  * •

${\rm Bl}(Z\subset X)$ is a smooth algebraic variety.

  * •

Fibers of $\pi$ over points in $X\setminus Z$ are points, i.e. $\pi$ induces
an isomorphism $\pi^{-1}(X\setminus Z)\to X\setminus Z$.

  * •

The morphism $\pi^{-1}Z\to Z$ can be naturally identified with the map
${\mathbb{P}}N(Z\subset X)\to Z$. Here $N(Z\subset X)$ is the normal
bundle15.215.215.2 A fiber of this bundle at a point $z\in Z$ is the quotient
of the tangent space to $X$ at $z$ by the tangent space to $Z$ at $z$. to $Z$
in $X$ and ${\mathbb{P}}W$ means the space of lines in the fibers of a vector
bundle $W$.

In other words, a subvariety $Z$ gets blown up to the projectivization of its
normal bundle.

Even more generally, one can blow up arbitrary (i.e. singular) subvarieties of
arbitrary varieties, but then there is no guarantee that the result is smooth.
In fact, it usually isn’t.

The blowup construction inspires the concept of birational equivalence of
algebraic varieties.

######  Definition 15.3.

We call complex algebraic varieties $X$ and $Y$ birationally equivalent, if
their fields of rational functions are isomorphic as field extensions of
${\mathbb{C}}$.

######  Remark 15.4.

An equivalent, more geometric, definition is that $X$ and $Y$ are birational
if and only if there exist Zariski open subsets $U\subseteq X$ and $V\subseteq
Y$ with $U$ isomorphic to $V$ as a complex algebraic variety. We also call a
morphism of algebraic varieties $X\to Y$ a birational morphism if it induces
an isomorphism of Zariski open subsets.

The second definition makes it clear that blowups (and blowdowns) do not
change the birational equivalence class. Study of birational equivalence is
known as birational geometry and is a big active area of Algebraic Geometry,
with multiple Fields medals awarded for it (Hironaka, Mori, Birkar). A major
part of birational geometry is the so-called Minimal Model Program which aims
to find a “nice” representative in each birational class. Some singular
varieties naturally occur in the Minimal Model Program.

I will state some accessible results of birational geometry, without proof.

  * •

For every, possibly singular, complex algebraic variety $X$ there exists a
smooth complex algebraic variety $Y$ with a birational morphism $Y\to X$
(Hironaka). This also holds over other fields of characteristic zero, but is a
very famous open problem in positive characteristics!

  * •

$\dim=1$. Two smooth complex projective algebraic varieties $X$ and $Y$ of
dimension one are birational to each other if and only if they are isomorphic.

  * •

$\dim=2$. Two smooth complex projective algebraic varieties $X$ and $Y$ of
dimension two are birational to each other if and only if they are connected
by a sequence of blowups of points followed by a sequence of blowdowns, as
indicated below.

$X\leftarrow\cdots\leftarrow Z\rightarrow\cdots\rightarrow Y$ (15.1)

  * •

$\dim={\rm any}$. (Weak Factorization Theorem, by Abramovich, Karu, Matsuki,
Włodarczyk, see [1]) Two smooth complex projective algebraic varieties $X$ and
$Y$ of any dimension are birational to each other if and only if they are
connected by a sequence of blowups and blowdowns of smooth subvarieties as
indicated below

$\displaystyle X=X_{0}\dashleftarrow\hskip-5.0pt\dashrightarrow
X_{1}\dashleftarrow\hskip-5.0pt\dashrightarrow\cdots\dashleftarrow\hskip-5.0pt\dashrightarrow
X_{n-1}\dashleftarrow\hskip-5.0pt\dashrightarrow X_{n}=Y$

where each $\dashleftarrow\hskip-5.0pt\dashrightarrow$ is either a blowup or a
blowdown morphism.15.315.315.3 The precise statement is stronger but a bit
more technical. It is an open problem, even in dimension $3$, whether the weak
factorization statement can always be upgraded to strong factorization, as in
(15.1).

When one has a birational morphism $\pi:X\to Y$, one can talk about the
exceptional locus ${\rm Exc}$ of $\pi$, which are the points $x\in X$ where
$\pi$ is not an isomorphism in any neighborhood of $x$. One is also interested
in the image $\pi({\rm Exc})$ in $Y$. If $Z$ is any closed subset of $Y$ which
is not contained in $\pi({\rm Exc})$ then the proper preimage of $Z$ is
defined as the closure

$\overline{\pi^{-1}\Big{(}C\setminus(C\cap\pi({\rm Exc}))\Big{)}}.$ (15.2)

This construction could be used to desingularize algebraic varieties by
looking at their proper preimages in carefully chosen blowups of some ambient
varieties.

Exercise 1. Check that the blowup $S$ of ${\mathbb{C}}^{n}$ at the origin can
be identified with the line bundle
$\mathcal{O}(-1)\to{\mathbb{C}}{\mathbb{P}}^{n-1}$.

Exercise 2. Let $W$ be the subset of
${\mathbb{C}}{\mathbb{P}}^{2}\times{\mathbb{C}}{\mathbb{P}}^{2}$ given by the
conditions that the points $(x_{0}:x_{1}:x_{2})\times(y_{0}:y_{1}:y_{2})$
satisfy

$x_{0}y_{0}=x_{1}y_{1}=x_{2}y_{2}.$

Prove that $W$ is a smooth surface. Verify that the projection maps
$W\to{\mathbb{C}}{\mathbb{P}}^{2}$ are birational and find the image of the
exceptional locus (i.e. points in ${\mathbb{C}}{\mathbb{P}}^{2}$ with more
than one preimage point). _Remark:_ $W$ is the closure of the graph of the
Cremona transformation
$(x_{0}:x_{1}:x_{2})\mapsto(\frac{1}{x_{0}}:\frac{1}{x_{1}}:\frac{1}{x_{2}})$.

Exercise 3. Let $x^{2}-y^{3}=0$ be a singular curve on ${\mathbb{C}}^{2}$.
Compute its proper preimage (15.2) on the blowup $S$ of ${\mathbb{C}}^{2}$ at
$(0,0)$ and verify that it is a smooth curve. _Hint:_ Work in the coordinate
charts on $S$.

##  16 Finite group actions on affine algebraic varieties and complex
manifolds. Subgroups of $SL_{2}({\mathbb{C}})$.

Let $X\subseteq{\mathbb{C}}^{n}$ be an affine complex algebraic variety.
Suppose that a finite group $G$ acts on $X$, i.e.

$G\subseteq{\rm Aut}_{{\mathbb{C}}-{\rm
alg}}({\mathbb{C}}[x_{1},\ldots,x_{n}]/I)$

where $I$ is the defining ideal of $X$. What can we say about the set of
$G$-orbits $X/G$? The following beautiful result goes back to at least
Hilbert.

######  Theorem 16.1.

The set $X/G$ has a natural structure of an affine algebraic variety. More
precisely, the subring of $G$-invariant elements

$({\mathbb{C}}[x_{1},\ldots,x_{n}]/I)^{G}$

is a finitely generated ${\mathbb{C}}$-algebra, and its maximal ideals are in
a natural bijection with $G$-orbits on $X$.

###### Proof.

Consider the elements $y_{ik}=g_{i}(x_{k})$ for all $g_{i}\in G$ and all
$1\leq k\leq n$. We can use this set of generators to write $X$ as an
algebraic subvariety of ${\mathbb{C}}^{n|G|}$, so that the action of $G$ comes
from a permutation action on the coordinates of ${\mathbb{C}}^{n}$. Then we
have

$({\mathbb{C}}[x_{1},\ldots,x_{n}]/I)^{G}\cong({\mathbb{C}}[y_{1},\ldots,y_{l}]/J)^{G}\cong{\mathbb{C}}[y_{1},\ldots,y_{l}]^{G}/J^{G}.$

In the second isomorphism we used the fact that every element $r$ of the
invariant ring $({\mathbb{C}}[y_{1},\ldots,y_{l}]/J)^{G}$ can be lifted to an
element $\hat{r}$ of ${\mathbb{C}}[y_{1},\ldots,y_{l}]$ and then

$\frac{1}{|G|}\sum_{g\in G}g(\hat{r})=r\,{\rm mod}\,J.$

Thus, to prove that $({\mathbb{C}}[x_{1},\ldots,x_{n}]/I)^{G}$ is a finitely
generated ${\mathbb{C}}$-algebra, it suffices to prove it for
${\mathbb{C}}[y_{1},\ldots,y_{l}]^{G}$, with the permutation action of $G$. We
actually don’t care too much that the action is a permutation action, but we
will use a weaker statement that the action of $G$ on
$A={\mathbb{C}}[y_{1},\ldots,y_{l}]$ preserves the natural grading on $A$ by
the total degree in $y_{i}$.

Consider the ideal $AA_{+}^{G}\subset A$ which is generated by all homogeneous
elements of $A^{G}$ of positive degree. Since $A$ is Noetherian, we can find a
finite set of generators $f_{1},\ldots,f_{k}$ of $AA_{+}^{G}$, and we can even
assume that all $f_{i}$ are $G$-invariant homogeneous elements of $A$. We will
now prove that $f_{i}$ generate the ring $A^{G}$ as a ${\mathbb{C}}$-algebra,
by induction on degree.

In degree zero, we have $A_{0}=A^{G}_{0}={\mathbb{C}}$, and there is nothing
to prove. Suppose we have proved that every homogeneous element of $A^{G}$ of
degree at most $d$ can be written as a polynomial in $f_{1},\ldots,f_{k}$ with
constant coefficients. Then for a homogeneous element $f\in A^{G}$ of degree
$d+1$ we use $f\in AA_{+}^{G}$ to write

$f=\sum_{i=1}^{k}a_{i}f_{i}$

where $a_{i}$ are some elements of $A$ of degree $d+1-\deg f_{i}$. We will now
use the averaging trick

$\displaystyle f=\frac{1}{|G|}\sum_{g\in
G}g(f)=\frac{1}{|G|}\sum_{i=1}^{k}\sum_{g\in
G}g(a_{i}f_{i})=\frac{1}{|G|}\sum_{i=1}^{k}\sum_{g\in G}g(a_{i})f_{i}$
$\displaystyle=\sum_{i=1}^{k}f_{i}\left(\frac{1}{|G|}\sum_{g\in
G}g(a_{i})\right)=\sum_{i=1}^{k}f_{i}h_{i}$

where $h_{i}$ are $G$-invariant elements of degree $d+1-\deg f_{i}$. By
induction assumption, $h_{i}$ can be written as polynomials in
$f_{1},\ldots,f_{k}$, therefore so can be $f$.

Now that we have established that $({\mathbb{C}}[x_{1},\ldots,x_{n}]/I)^{G}$
is finitely generated, let us prove the bijection between its maximal ideals
and the set of orbits of $G$-action on the set of maximal ideals of
${\mathbb{C}}[x_{1},\ldots,x_{n}]/I$.16.116.116.1 Recall that points in $X$
correspond to maximal ideals in $R={\mathbb{C}}[x_{1},\ldots,x_{n}]/I$ by
Hilbert’s Nullestellensatz. In one direction, to any maximal ideal $m\subset
R$ we associate the ideal $m^{G}=m\cap R^{G}\subset R^{G}$. The ideal $m^{G}$
is maximal because for any $x\in R^{G}\setminus m^{G}$ there exists $y\in R$
such that $xy=1\,{\rm mod}\,m$. Then the element

$\prod_{g\in G}(1-g(xy))=\prod_{g\in G}(1-xg(y))=1-x\hat{y}$

(with $G$-invariant $\hat{y}$) lies in $m^{G}$, which shows that $x\,{\rm
mod}\,m^{G}$ is invertible. It is clear that different $m$ in the same
$G$-orbit give the same $m^{G}$.

In the other direction, for a maximum ideal $I\subset R^{G}$ consider the
ideal $RI$ that it generates in $R$. If $RI$ contains $1$ then there holds

$1=\sum_{i}r_{i}x_{k}$

with $x_{k}\in I$. Averaging gives
$1=\sum_{i}\left(\frac{1}{|G|}r_{i}\right)x_{k}\in I,$ contradiction.
Therefore, $RI$ is contained in some maximal ideal $m$. It is clear that
$m^{G}$ contains $I$ and is therefore equal to it, since $I$ is maximal.

It now suffices to show that if $m_{1}$ and $m_{2}$ satisfy
$m_{1}^{G}=m_{2}^{G}$, then $m_{1}$ and $m_{2}$ are in the same $G$-orbit.
Consider $I_{1}=\bigcap_{g\in G}g(m_{1})$ and $I_{2}=\bigcap_{g\in
G}g(m_{2})$. If $I_{1}+I_{2}$ were contained in a maximum ideal $m$, we would
have

$m\supseteq\bigcap_{g\in G}g(m_{1}),~{}m\supseteq\bigcap_{g\in G}g(m_{2}).$

If $m$ is not equal to any $g(m_{1})$, then there are elements $x_{g}\in
g(m_{1}),x_{g}\not\in m$, and their product is in $\bigcap_{g\in G}g(m_{1})$
but not in $m$. Similarly, $m$ must be equal to one of $g(m_{2})$, and this is
impossible, since $m_{1}$ and $m_{2}$ are in different $g$-orbits. Therefore,
$I_{1}+I_{2}=R$ and16.216.216.2 You have been warned.

$1=a_{1}+a_{2},~{}a_{i}\in I_{i}.$

Moreover, since $I_{i}$ are $G$-invariant, when we average over $G$, we may
assume that $a_{i}$ are $G$-invariant. However, this means that $a_{i}\in
m_{i}^{G}$, which contradicts $1\not\in m_{1}^{G}=m_{2}^{G}$. ∎

######  Remark 16.2.

The invariant rings are also finitely generated when the group $G$ is a linear
reductive group. However, they don’t have to be finitely generated in the non-
reductive case, see a famous counterexample to the Hilbert’s fourteenth
problem given by Nagata [25].

We will now discuss finite group actions on smooth complex manifolds. Suppose
a finite group $G$ fixes a point $p\in X$. Then locally we have a surjective
map of $G$-representations

${\rm Hol}^{0}(p)\stackrel{{\scriptstyle
d}}{{\to}}\left(TX\right)^{\vee}_{p}\to 0$

where ${\rm Hol}^{0}(p)$ denotes the infinite-dimensional space of holomorphic
functions near $p$ which are zero at $p$. This surjection has a $G$-invariant
splitting. To see this, a person nervous around infinite-dimensional vector
spaces, can lift a basis of $\left(TX\right)^{\vee}_{p}$, look at the span of
the $G$-translates of it and thus reduce the problem to that of surjective
finite-dimensional representations of $G$. The $G$-invariant splitting then
endows $X$ near $p$ with analytic coordinates on which $G$ acts linearly.

######  Corollary 16.3.

In dimension one, all actions that fix a point $p$ locally analytically look
like $z\mapsto\alpha z,~{}\alpha^{n}=1$, and the quotient is given by
$z\mapsto z^{n}=w$.

###### Proof.

Locally near $p$, we have an action of ${\rm
GL}(1,{\mathbb{C}})={\mathbb{C}}^{*}$, and all finite subgroups of
${\mathbb{C}}^{*}$ are given by groups of $n$-th roots of $1$. ∎

######  Corollary 16.4.

For any finite group $G$ of automorphisms of a complex manifold $X$, the fixed
locus $X^{G}$ is a disjoint union of complex submanifolds of $X$.

###### Proof.

Since the action is locally linear, it suffices to look at $G\subseteq{\rm
GL}(n,{\mathbb{C}})$ acting on $X={\mathbb{C}}^{n}$. Then $X^{G}$ is the
linear subspace $({\mathbb{C}}^{n})^{G}$. ∎

We will now focus on dimension two, and more precisely on the action of finite
subgroups $G\subset{\rm SL}(2,{\mathbb{C}})$ on ${\mathbb{C}}^{2}$. (We could
have considered ${\rm GL}(2,{\mathbb{C}})$ but it is more complicated.)

We first remark that any such $G$ can be conjugated to lie inside the unitary
group ${\rm SU}(2)$. Indeed, $G\subset SU(2)$ iff $G$ fixes the standard
Hermitean on ${\mathbb{C}}^{2}$. We can take a positive definite Hermitean
form on ${\mathbb{C}}^{2}$ and average it over $G$ to prove that $G$ preserves
some positive definite Hermitean form. Since all such forms can be written as
the standard form in some basis, this change of basis makes a conjugate of $G$
unitary.16.316.316.3 This is a particular case of the Weyl’s unitarian trick.
We put a matrix version of this statement in Exercise 2.

Recall that

${\rm SU}(2)=\left\\{\left(\begin{array}[]{rr}a&-\bar{b}\\\
b&\bar{a}\end{array}\right),~{}|a|^{2}+|b^{2}|=1\right\\}$

is a real $3$-sphere. We can also think of it as the length one quaternions
$\gamma\in{\mathbb{H}}$ acting on ${\mathbb{H}}\cong{\mathbb{C}}^{2}$ by left
multiplication $\gamma:w\mapsto\gamma w$.

Now consider the conjugation action

$w\mapsto\gamma w\gamma^{-1}$

of ${\rm SU}(2)$ on ${\mathbb{H}}$. It preserves the trace zero part of
${\mathbb{H}}$ (see Exercise 3) and the norm

$|w|^{2}=w\bar{w}$

on it. This gives a group homomorphism ${\rm SU(2)}\to{\rm
SO}(3,{\mathbb{R}})$, whose kernel is $\\{\pm 1\\}$. Finite subgroups of ${\rm
SO}(3,{\mathbb{R}})$ are classically known. They include two infinite series
(rotation symmetries of a regular pyramid and a regular bipyramid) as well as
rotational symmetries of the regular tetrahedron, cube/octahedron, and
icosahedron/dodecahedron. This allows one to classify all finite subgroups
$G\subset{\rm SL}(2,{\mathbb{C}})$.

In the future sections we will study in detail the singular quotients
${\mathbb{C}}^{2}/G$ for these groups $G$ and their resolutions of
singularities.

Exercise 1 (Molien series.) Let $V$ be a complex finite-dimensional
representation of a finite group $G$. It induces a grading-preserving action
of $G$ on ${\rm Sym}^{*}(V)\cong{\mathbb{C}}[x_{1},\ldots,x_{n}]$. Prove that
the Hilbert series of the ring of invariants ${\rm Sym}^{*}(V)^{G}$ can be
computed by

$\sum_{d=0}^{\infty}\dim_{\mathbb{C}}\left({\rm
Sym}^{d}(V)\right)^{G}\,t^{d}=\frac{1}{|G|}\sum_{g\in
G}\prod_{i=1}^{n}\frac{1}{1-\lambda_{g,i}t}=\frac{1}{|G|}\sum_{g\in
G}\det(1-tg)^{-1}$

where $\lambda_{g,1},\ldots,\lambda_{g,n}$ are eigenvalues of $g$ on $V$.
_Hint:_ For a finite-dimensional $G$-representation $W$ we have
$\dim_{\mathbb{C}}W^{G}=\frac{1}{|G|}\sum_{g\in G}{\rm Tr}(g,W)$. Then use the
result that all finite order elements in ${\rm GL}(V)$ are diagonilizable.

Exercise 2. Let $M$ be a Hermitean form on ${\mathbb{C}}^{n}$ given by
$\langle v,w\rangle=\bar{w}^{T}Mv$. Prove that an endomorphism

$g:v\to Av$

satisfies $\langle Av,Aw\rangle=\langle v,w\rangle$ for all $v$ and $w$ if and
only if $\bar{A}^{T}MA=M$. Prove that if for some invertible matrix $S$ we
have $M=\bar{S}^{T}S$, then $SAS^{-1}$ is a unitary matrix.

Exercise 3. Prove that the conjugation action of quaternions preserves the
subspace ${\mathbb{R}}{\mathrm{i}}+{\mathbb{R}}{\rm j}+{\mathbb{R}}{\rm k}$
and the norm on it.

##  17 Resolutions of $A_{n}$ and $D_{4}$ singularities.

Let us now look at some specific examples of finite subgroups $G$ of ${\rm
SL}(2,{\mathbb{C}})$ and the corresponding quotients ${\mathbb{C}}^{2}/G$.

The first series of examples comes from the action of the cyclic group of
order $n\geq 2$ which necessarily17.117.117.1 Exercise 1. looks like

$G=<g>,~{}g(x,y)=(\xi x,\xi^{-1}y)$ (17.1)

where $\xi={\mathrm{e}}^{2\pi{\mathrm{i}}/n}$. As a consequence of the
definition, the action17.217.217.2 There is an evergreen issue of the action
on points versus the action on functions. We consider the action on the ring
${\mathbb{C}}[x,y]$. of the generator $g$ on the monomial $x^{a}y^{b}$ is

$g(x^{a}y^{b})=\xi^{a-b}x^{a}y^{b}$

which implies that the invariant ring ${\mathbb{C}}[x,y]^{G}$ is the linear
span of monomials $x^{a}y^{b}$ with the property $a=b\,{\rm mod}\,n$. Some
prominent examples of such monomials are $u=x^{n}$, $v=y^{n}$ and $w=xy$. We
claim that every $G$-invariant monomial $x^{a}y^{b}$ can be written as a
product of these three. Indeed, we can use $u$ and $v$ to reduce to the case
$a,b<n$, and then we must have $a=b$. It is also clear that $u,v,w$ satisfy

$uv-w^{n}=0,$ (17.2)

so the ring of invariants ${\mathbb{C}}[x,y]^{G}$ is isomorphic to
${\mathbb{C}}[u,v,w]/(uv-w^{n})$, and the quotient ${\mathbb{C}}^{2}/G$ is a
surface in ${\mathbb{C}}^{3}$ given by the equation (17.2). By looking at
partial derivatives, we see that the only singular point of this surface
occurs at the origin $(0,0,0)$. It is known as the $A_{n-1}$ singularity, and
we will soon see why.

We will first look in detail at the $A_{1}$ singularity given by

$uv-w^{2}=0$ (17.3)

This is the simplest hypersurface singularity in complex dimension two, in the
following sense. If we have a function $f(u,v,w)$ with $f(0,0,0)=0$, we can
consider its Taylor series

$f(u,v,w)=0+{\rm~{}linear~{}}+{\rm~{}quadratic~{}}+{\rm higher~{}order.}$

If the linear term is nonzero, then $f(u,v,w)=0$ is smooth at $(0,0,0)$. If
the the linear term is zero, then we can ask about the quadratic term. If the
rank of the corresponding form is the maximum possible $3$, then we get an
$A_{1}$ singularity.

We will now resolve the surface (17.3) by blowing up the ambient
${\mathbb{C}}^{3}$ at $(0,0,0)$ and looking at the proper preimage of the
surface $uv=w^{2}$. Here is the intuitive picture of what will happen, where
we indicate in blue the preimage of the singular point.

$\bullet$$\\{uv=w^{2}\\}\subset{\mathbb{C}}^{3}$$\tilde{S}\subset\tilde{\mathbb{C}}^{3}$

Let us see this with all the gory details, in the coordinate charts. Recall
that the blowup $\tilde{\mathbb{C}}^{3}$ is a submanifold of
${\mathbb{C}}^{3}\times{\mathbb{C}}{\mathbb{P}}^{2}$, given by
$\\{(u,v,w),(u_{1}:v_{1}:w_{1})\\}$ such that there exists $\lambda$ with
$(u,v,w)=\lambda(u_{1},v_{1},w_{1})$. As a result, $\tilde{\mathbb{C}}^{3}$ is
covered by three open charts, each one isomorphic to ${\mathbb{C}}^{3}$.

In the first open chart $w_{1}\neq 0$, we have ${\mathbb{C}}^{3}$ with
coordinates $\frac{u}{w},\frac{v}{w},w$ and the map to the original
${\mathbb{C}}^{3}$ is given by

$(u,v,w)=(w\left(\frac{u}{w}\right),w\left(\frac{v}{w}\right),w).$ (17.4)

######  Remark 17.1.

You may be concerned with such a bizarre notation for the coordinates.
Shouldn’t they be variables? Yes, of course. Let’s say that we have
${\mathbb{C}}^{3}$ with coordinates $t_{1},t_{2},t_{3}$ which maps to the
original ${\mathbb{C}}^{3}$ by
$(t_{1},t_{2},t_{3})\mapsto(u,v,w)=(t_{1}t_{3},t_{2}t_{3},t_{3})$. But then we
think about what the coordinate functions $t_{i}$ look like as rational
functions in $u,v,w$ and we recover (17.4). At which point we forget about
$t_{i}$-s and accept a somewhat weird notation. It does have advantages of
being easily able to write the transition functions, although one has to be
careful figuring out which subsets in the three open charts are glued
together. Fortunately, we will not worry too much about it.

What is the preimage of $S=\\{uv-w^{2}=0\\}$ in the chart with coordinates
$\frac{u}{w},\frac{v}{w},w$? We rewrite the equation of $S$ as

$0=\left(\frac{u}{w}\right)w\left(\frac{v}{w}\right)w-w^{2}=w^{2}\left(\left(\frac{u}{w}\right)\left(\frac{v}{w}\right)-1\right)$

and see that the preimage has two components: $w=0$ and
$\left(\frac{u}{w}\right)\left(\frac{v}{w}\right)-1=0$. The first component is
the preimage of $(u,v,w)=0$, a.k.a. the exceptional divisor, and the second
component is the proper preimage of $S$. Or, more precisely, these are the
intersections of the exceptional divisor and the proper preimage of $S$ with
this coordinate chart. Observe now that the proper preimage of $S$ is smooth
in this chart.

Let us now look at the coordinate chart with coordinates
$u,\frac{v}{u},\frac{w}{u}$. We similarly get

$0=u^{2}\left(\frac{v}{u}\right)-u^{2}\left(\frac{w}{u}\right)^{2}=u^{2}\left(\left(\frac{v}{u}\right)-\left(\frac{w}{u}\right)^{2}\right)$

and again we see that the proper preimage
$0=\left(\frac{v}{u}\right)-\left(\frac{w}{u}\right)^{2}$ is smooth. Obviously
the $v_{1}\neq 0$ chart is handled similarly.

We thus see that the proper preimage $\tilde{S}$ of $S$ is smooth. It is also
instructive to look at the fibers of $\tilde{S}\to S$. Outside of $(0,0,0)$,
the fibers are of course points. The preimage of $(0,0,0)$ is a subvariety of
the exceptional divisor, given by
$(u_{1}:v_{1}:w_{1})\in{\mathbb{C}}{\mathbb{P}}^{2}$ with
$u_{1}v_{1}-w_{1}^{2}=0$, so it is a ${\mathbb{C}}{\mathbb{P}}^{1}$. For
example, in the coordinate chart $w_{1}\neq 0$, we have
$\frac{u}{w}=\frac{u_{1}}{w_{1}},~{}\frac{v}{w}=\frac{v_{1}}{w_{1}}$ so
$0=\left(\frac{u}{w}\right)\left(\frac{v}{w}\right)-1$ matches
$u_{1}v_{1}-w_{1}^{2}=0$ in the appropriate chart on
${\mathbb{C}}{\mathbb{P}}^{2}$.

Now let us look at the $A_{n-1}$ case, i.e.

$S=\\{uv-w^{n}=0\\}.$

As before, we blow up ${\mathbb{C}}^{3}$ and see what happens on the three
coordinate charts, two of which are similar due to $u\leftrightarrow v$
symmetry. On the chart with coordinates $\frac{u}{w},\frac{v}{w},w$ we get

$uv-w^{n}=w^{2}\left(\left(\frac{u}{w}\right)\left(\frac{v}{w}\right)-w^{n-2}\right)$

so we may still have a singular point at the origin, but the value of $n$ has
decreased by $2$. Also note that the intersection of the proper preimage of
$S$ with the exceptional divisor is given in this chart by

$w=0,~{}\left(\frac{u}{w}\right)\left(\frac{v}{w}\right)-w^{n-2}=0,$ (17.5)

so we get two lines $w=\frac{u}{w}=0$ and $w=\frac{v}{w}=0$. In the other two
charts the proper preimage is smooth (Exercise 2). They also compactify the
lines (17.5) from ${\mathbb{C}}$ to ${\mathbb{C}}{\mathbb{P}}^{1}$. In short,
we get a map $\tilde{S}\to S$ with the preimage of $(0,0,0)$ being a union of
two ${\mathbb{C}}{\mathbb{P}}^{1}$-s, which intersect as coordinate lines at
an $A_{n-3}$ singularity, where $A_{0}$ is by definition smooth.

$\bullet$$~{}~{}A_{n-1}$$S$$\bullet$$~{}~{}A_{n-3}$$\tilde{S}$

We repeat the process. Each time, the power $n$ decreases by $2$ and two new
${\mathbb{C}}{\mathbb{P}}^{1}$-s spring out. A paranoid reader can verify that
under the blowup the previous ${\mathbb{C}}{\mathbb{P}}^{1}$-s separate and
two new lines intersect them transversely. The last step is either unnecessary
(if $n$ is odd, and we reduce to $uv=w$) or we only have one
${\mathbb{C}}{\mathbb{P}}^{1}$ (if $n$ is even and we reduce to
$A_{1}$-singularity). In both cases we end up with a resolution of
singularities $\tilde{S}\to S$ such that the preimage of the singular point is
a chain of $(n-1)$ transversal ${\mathbb{C}}{\mathbb{P}}^{1}$-s.

$\cdots$${\mathbb{C}}{\mathbb{P}}^{1}$-s as
lines$\cdots$${\mathbb{C}}{\mathbb{P}}^{1}$-s as spheres

Neither of the above pictures is entirely correct, since the Riemann spheres
intersect transversely. At any rate, we will consider the dual graph that
encodes the components of the exceptional locus of $\tilde{S}\to S$ as
vertices and their intersections as edges. It is

$\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\cdots-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ$

with $n-1$ vertices, which is the $A_{n-1}$ Dynkin diagram.17.317.317.3 There
is, of course, more to this notation. Appropriately understood intersection
pairing matrix of the components is up to the sign the Cartan matrix of
$A_{n-1}$. We will not try to define it now, but will comment on it in Section
19.

Now let us look at the next case. Here we have a subgroup $G$ of ${\rm
SL}(2,{\mathbb{C}})$ generated by

$(x,y)\mapsto(\xi x,\xi^{-1}y),~{}(x,y)\mapsto(-y,x).$ (17.6)

If we apply the second operation twice, we get $(x,y)\mapsto(-x,-y)$, so we
might as well assume that $\xi$ is the primitive $(2n)$-th root of $1$.

Let us first compute the ring of invariants. The cyclic subgroup is normal, so
we will first take the quotient by that to get $u=x^{2n},v=y^{2n},w=xy$ with
the ring

${\mathbb{C}}[u,v,w]/(uv-w^{2n}).$

The symmetry $(x,y)\mapsto(-y,x)$ now becomes the involution
$(u,v,w)\mapsto(v,u,-w)$. We can take the invariants first and then the
quotient. The action of this involution on ${\mathbb{C}}^{3}$ has positive
eigenspace $u+v$ and the $A_{1}$-type action on $u-v$ and $w$, so the ring of
invariants is generated by

$t_{1}=u+v,~{}t_{2}=(u-v)^{2},~{}t_{3}=w^{2},~{}t_{4}=(u-v)w$

with relations $t_{2}t_{3}=t_{4}^{2}$. We then have

$\displaystyle{\mathbb{C}}[x,y]^{G}\cong{\mathbb{C}}[t_{1},\ldots,t_{4}]/(t_{2}t_{3}-t_{4}^{2},\frac{1}{4}(t_{1}^{2}-t_{2})-t_{3}^{n})$
$\displaystyle\cong{\mathbb{C}}[t_{1},t_{3},t_{4}]/((t_{1}^{2}-4t_{3}^{n})t_{3}-t_{4}^{2}).$

We rescale and rename the variables to get this into the standard form of

$\\{u^{2}+wv^{2}+w^{n+1}=0\\}\subset{\mathbb{C}}^{3}$

called the $D_{n+2}$ singularity. Note that the extra $w$ in the second term
makes it different from the $A_{n}$ singularities.

Looking at the group generators or the equation of the quotient, we see that
$n=1$ case simply gives $A_{3}$ singularity, see Exercise 3. We will now
consider the simplest new case $n=2$, i.e. the $D_{4}$ singularity

$u^{2}+wv^{2}+w^{3}=0.$

It is easy to see that the only singular point occurs at $(0,0,0)$. This is
also apparent from the quotient description, since nonidentity diagonalizable
elements of ${\rm SL}(2,{\mathbb{C}})$ can not have eigenvalue $1$ and thus
act freely away from the origin. Now we blow up ${\mathbb{C}}^{3}$ at
$(0,0,0)$ and see what happens.

In the open chart with coordinates
$(z_{1},z_{2},z_{3})=(\frac{u}{v},v,\frac{w}{v})$ we get17.417.417.4 So having
raised a stink about using letters for the variables, we are now doing it.
Enjoy!

$0=u^{2}+wv^{2}+w^{3}=z_{1}^{2}z_{2}^{2}+z_{2}^{3}z_{3}+z_{2}^{3}z_{3}^{3}=z_{2}^{2}(z_{1}^{2}+z_{2}z_{3}+z_{2}z_{3}^{3})$

and the proper preimage is given by $0=z_{1}^{2}+z_{2}z_{3}+z_{2}z_{3}^{3}$.
Vanishing of partial derivatives gives

$z_{1}=0,z_{3}+z_{3}^{3}=0,z_{2}(1+3z_{3}^{2})=0,$

which gives three singular points
$(0,0,0),(0,{\mathrm{i}},0),(0,-{\mathrm{i}},0)$. Local analysis of these
points show that all three are of the type $A_{1}$. We skip the calculations
for the two other charts, but we claim that we do not see any new singular
points. So the $D_{4}$ singularity blows up to a
${\mathbb{C}}{\mathbb{P}}^{1}$ with three type $A_{1}$ singular points on it.

$\bullet$$~{}~{}D_{4}$$S$$\bullet$$~{}~{}A_{1}$$\bullet$$~{}~{}A_{1}$$\bullet$$~{}~{}A_{1}$$\tilde{S}$

When we further blow up these three points, we obtain a resolution of
singularities $\tilde{S}\to S$ with four ${\mathbb{C}\mathbb{P}}^{1}$-s in it
which intersect transversally in the $D_{4}$ Dynkin diagram (one central
Riemann sphere intersects three others).

$\displaystyle\begin{array}[]{c}\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ\\\
|\\\ \circ\end{array}$

Exercise 1. Prove that every cyclic subgroup of ${\rm SL}(2,{\mathbb{C}})$ can
be conjugated to the one in (17.1).

Exercise 2. Verify that the proper preimage of $uv-w^{n}$ in the chart with
coordinates $u,\frac{v}{u},\frac{w}{u}$ is smooth.

Exercise 3. Check that the $D_{3}$ case is the same as the $A_{3}$ case in two
ways, by looking at the group action, and by making a change of variables in
$u^{2}+wv^{2}+w^{2}=0$.

##  18 More resolutions of $ADE$ singularities.

Let us now discuss the general $D_{n+2}$ case

$S=\\{u^{2}+wv^{2}+w^{n+1}=0\\}\subset{\mathbb{C}}^{3}$

for $n>2$.

As always, we blow up the origin in ${\mathbb{C}}^{3}$. On the open chart with
coordinates $\frac{u}{v},v,\frac{w}{v}$ we get

$0=v^{2}\left(\left(\frac{u}{v}\right)^{2}+v\left(\frac{w}{v}\right)+\left(\frac{w}{v}\right)^{n+1}v^{n-1}\right)$

with the second factor giving the proper preimage of $S$. Let us find singular
points of $x^{2}+yz+z^{n+1}y^{n-1}=0$. We have to solve the
equations18.118.118.1 Yes, algebraic geometry is about solving polynomial
equations.

$2x=0,~{}z+(n-1)z^{n+1}y^{n-2}=0,~{}y+(n+1)z^{n}y^{n-1}=0,~{}x^{2}+yz+z^{n+1}y^{n-1}=0.$

We can multiply the second equation by $(n+1)y$ and the third one by $(n-1)z$,
subtract them and get $yz=0$. Since $n>2$, we use the second and third
equations to deduce that both $y$ and $z$ are $0$. So the only singular point
in this chart is $(0,0,0)$, which corresponds to $(0:1:0)$ on the exceptional
${\mathbb{C}}{\mathbb{P}}^{2}$ of $\tilde{\mathbb{C}}^{3}\to{\mathbb{C}}^{3}$
(we are in the chart $v_{1}\neq 0$). For $n>2$, we see that the singularity at
this point is of type $A_{1}$, because we can ignore higher degree terms. The
intersection with the exceptional locus $v=0$ is given by $\frac{u}{v}=0$,
i.e. we have the line $(0:*:*)$ on the exceptional
${\mathbb{C}}{\mathbb{P}}^{2}$ of $\tilde{\mathbb{C}}^{3}\to{\mathbb{C}}^{3}$.

Let us look at the $u,\frac{v}{u},\frac{w}{u}$ chart. There, the proper
preimage is given by

$0=1+\left(\frac{w}{u}\right)\left(\frac{v}{u}\right)u+\left(\frac{w}{u}\right)^{n+1}u^{n-1}$

and the exceptional locus is $u=0$. Therefore, this chart of $S$ does not
intersect the exceptional locus and is automatically smooth (one can also
check it directly).

The last chart has coordinates $\frac{u}{w},\frac{v}{w},w$. There we get

$0=\left(\frac{u}{w}\right)^{2}w^{2}+\left(\frac{v}{w}\right)^{2}w^{3}+w^{n+1}=z^{2}\left(\left(\frac{u}{w}\right)^{2}+w\left(\frac{v}{w}\right)^{2}+w^{n-1}\right).$

Clearly, we have a $D_{n-2}$ singularity at the origin, which corresponds to
the point $(0:0:1)$.

Combining the information from the three charts, we see that a single blowup
gives us a map from the proper preimage $\tilde{S}$ to $S$ so that the
exceptional locus is ${\mathbb{C}}{\mathbb{P}}^{1}$ with two singular points
on it. One of these points of type $A_{1}$ and the other is of type $D_{n-2}$.
By resolving the $A_{1}$ singularity, we replace $D_{n}$ by two transversal
${\mathbb{C}}{\mathbb{P}}^{1}$-s with a $D_{n-2}$ point on one of them.

$\bullet$$~{}~{}D_{n}$$\bullet$$~{}~{}A_{1}$$\bullet$$~{}~{}D_{n-2}$$\bullet$$~{}~{}D_{n-2}$

When we repeat the process, skipping details, we eventually get to $D_{4}$ or
$D_{3}=A_{3}$ and the resulting dual graph is given by the $D_{n+2}$ Dynkin
diagram.

$\displaystyle\begin{array}[]{c}\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\cdots-\hskip-5.0pt-\circ\\\
|\hskip 50.0pt{}\\\ \circ\hskip 50.0pt{}\\\ \end{array}$

As you may deduce from the title of the section, there are three more cases to
consider, which correspond to the remaining three Dynkin diagrams $E_{6}$,
$E_{7}$ and $E_{8}$, so that cumulatively these singularities are referred to
as $ADE$ surface singularities.18.218.218.2 They are also called du Val
singularities, canonical singularities, simple surface singularities, Kleinian
singularities, and rational double points. This should tell you how ubiquitous
these singularities are. The corresponding groups are
${\mathbb{Z}}/2{\mathbb{Z}}$ central extensions of the groups of rotational
symmetries of the regular tetrahedron, cube and dodecahedron, respectively.

We will not write down the group generators, but will list the equations
below.

$\begin{array}[]{ll}E_{6}:&u^{2}+v^{3}+w^{4}=0\\\
E_{7}:&u^{2}+v^{3}+vw^{3}=0\\\ E_{8}:&u^{2}+v^{3}+w^{5}=0\end{array}$

There is another way of looking at the resolutions of $D_{n+2}$ singularities.
The group $G$ given by (17.6) has a cyclic subgroup
$G_{1}={\mathbb{Z}}/2n{\mathbb{Z}}$. We can first take the quotient of
${\mathbb{C}}^{2}$ by $G_{1}$, then resolve the resulting $A_{2n-1}$
singularity to get a smooth surface $S_{1}$, then take the quotient by
$G/G_{1}\cong{\mathbb{Z}}/2{\mathbb{Z}}$ to get a singular surface $S_{2}$ and
then further resolve to get a resolution $S_{3}$ of ${\mathbb{C}}^{2}/G$, see
the commutative diagram below.

$\begin{array}[]{c}{\mathbb{C}}^{2}\hskip 50.0ptS_{1}\hskip 50.0ptS_{3}\hskip
5.0pt\\\ \searrow\hskip 20.0pt\swarrow\hskip 20.0pt\searrow\hskip
20.0pt\swarrow\\\ {\mathbb{C}}^{2}/G_{1}\hskip 40.0ptS_{2}\hskip 10.0pt\\\
\searrow\hskip 30.0pt\swarrow\\\ {\mathbb{C}}^{2}/G\end{array}$

Specifically, we have the $A_{2n-1}$ singularity given by

$uv-w^{2n}=0$

and the action of $G/G_{1}\cong{\mathbb{Z}}/2{\mathbb{Z}}$ which switches $u$
and $v$. The resolution $S_{1}$ has a chain of $(2n-1)$ Riemann spheres

$\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\cdots-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ$

which intersect transversely, and the involution of $G/G_{1}$ switches the
opposite spheres. Because we have an odd number of spheres, this involution
will preserve the middle ${\mathbb{C}}{\mathbb{P}}^{1}$. It acts by switching
two intersection points with two adjacent ${\mathbb{C}}{\mathbb{P}}^{1}$-s. We
can call these points $0$ and $\infty$ and observe that every such involution
can be written in the form $z\mapsto\frac{1}{z}$, see Exercise 1. Then we have
two fixed points $z=\pm 1$. Since these are isolated points, the eigenvalues
of the action on the tangent space must be $(-1,-1)$, so the quotient will
have an $A_{1}$ singularity there. Therefore, the image of the chain of
$(2n-1)$ lines on $S_{2}$ will be $n$ lines, with the first line having two
additional $A_{1}$ singularities on it. Blowing up these $A_{1}$
singularities, we get $S_{3}$ with the $(n+2)$ lines in $D_{n+2}$
configuration.

We can similarly get the $E_{6}$ singularity as the quotient of the $D_{4}$
singularity by a ${\mathbb{Z}}/3{\mathbb{Z}}$ action. Then if you resolve the
$D_{4}$ singularity to get a surface with a $D_{4}$ configurations of
projective lines, the group can be seen rotating the three “outside” lines
while fixing the middle one. By Exercise 2, any order three automorphism of
${\mathbb{C}}{\mathbb{P}}^{1}$ can be written in the form
$z\mapsto{\mathrm{e}}^{2\pi{\mathrm{i}}/3}z$, so it will have two fixed points
$0$ and $\infty$. It is then possible to compute that the quotient will be the
$A_{2}$ singularity, so we get $S_{1}\to S$ with $E_{6}$ singularity replaced
by two lines with two $A_{2}$ singular points on one of them.

$/({\mathbb{Z}}/3{\mathbb{Z}})$$\bullet$$A_{2}$$\bullet$$A_{2}$

When we resolve them, we get the $E_{6}$ diagram below, with the colors
matching the colors of the above picture.

$\begin{array}[]{c}\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\circ}-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ\\\
|\\\
{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\circ}\end{array}$

Similarly, the $E_{7}$ singularity is a ${\mathbb{Z}}/2{\mathbb{Z}}$ quotient
of the $E_{6}$ singularity.18.318.318.3 The set of eight vertices of the cube
naturally splits into two sets of four vertices of a regular tetrahedron, and
the group of rotational symmetries of the tetrahedron is an index two subgroup
of that of the cube. After resolving the singularity, the action on the
$E_{7}$ configuration of lines is by switching the opposing ends. So we get
four lines, $C_{1}$, $C_{2}$, $C_{3}$, $C_{4}$, connected in a chain, and
there are three $A_{1}$ singularities: one one $C_{1}$, one on the
intersection of $C_{1}$ and $C_{2}$ and one elsewhere on $C_{2}$.

$/({\mathbb{Z}}/2{\mathbb{Z}})$$\bullet$$A_{1}$$\bullet$$A_{1}$$\bullet$$A_{1}$

When we resolve these $A_{1}$ singularities, we get the $E_{7}$ diagram

$\begin{array}[]{c}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\circ}-\hskip-5.0pt-\,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\circ}-\hskip-5.0pt-\,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\circ}-\hskip-5.0pt-\circ-\hskip-5.0pt-{\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color<EMAIL_ADDRESS>2.0pt\circ\\\ |\hskip 21.0pt{}\\\ \circ\hskip 21.0pt{}\end{array}$

with the matching colors.

Unfortunately, the group of symmetries of the regular dodecahedron is not
solvable, so one can not obtain the $E_{8}$ resolution by this process. We can
however blow up the origin in ${\mathbb{C}}^{3}$ and look at the proper
preimage of the surface $u^{2}+v^{3}+w^{5}=0$ in it. We get in the
$\frac{u}{w},\frac{v}{w},w$ coordinate chart

$0=w^{2}\left(\left(\frac{u}{w}\right)^{2}+\left(\frac{v}{w}\right)^{3}w+w^{3}\right)$

which is the $E_{7}$ singularity. We see in Exercise 3 that the proper
preimage has no singularities in other coordinate charts. The intersection
with the special fiber is a line $\frac{u}{w}=u=0$, and one can see, with some
effort, that after resolving that $E_{7}$ singularity we get the $E_{8}$
diagram.

$\begin{array}[]{c}\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\circ-\hskip-5.0pt-\hskip
2.0pt\circ\\\ |\hskip 48.0pt{}\\\ \circ\hskip 48.0pt{}\end{array}$

Exercise 1. Prove that every involution of ${\mathbb{C}}{\mathbb{P}}^{1}$ that
sends $0$ to $\infty$ can be written in the form $z\mapsto\frac{c}{z}$ for
some nonzero constant $c$. Then argue that by scaling $z$ we can rewrite it as
$z\mapsto\frac{1}{z}$.

Exercise 2. Prove that every order three automorphism of
${\mathbb{C}}{\mathbb{P}}^{1}$ can be written as
$z\mapsto{\mathrm{e}}^{2\pi{\mathrm{i}}/3}z$ in some coordinates. _Hint:_
Consider the corresponding element in ${\rm PGL}(2,{\mathbb{C}})$ and its
eigenvalues (up to common scaling).

Exercise 3. Check that the proper preimage of $u^{2}+v^{3}+w^{5}=0$ has no
singular points in the coordinate charts with coordinates
$u,\frac{v}{u},\frac{w}{u}$ and $\frac{u}{v},v,\frac{w}{v}$.

##  19 Overview of intersection theory. Surfaces.

In this section we will talk about intersection theory. Even for these notes,
this section is unusually hand-wavy, with many of the claims partially or
totally unsubstantiated. Sorry, not sorry!

Let $X$ be a smooth complex algebraic variety or a complex manifold. We can
view it as an orientable real manifold and associate to it two collections of
abelian groups, namely its integral simplicial homology groups
$H_{i}(X,{\mathbb{Z}})$ and cohomology groups $H^{i}(X,{\mathbb{Z}})$ for
$0\leq i\leq\dim_{\mathbb{R}}X=2\dim_{\mathbb{C}}X$. All of these are finitely
generated for algebraic varieties $X$. Moreover, $H^{*}(X,{\mathbb{Z}})$ is an
associative graded, super-commutative19.119.119.1 Super-commutativity means
that for $a\in H^{i}(X,{\mathbb{Z}})$ and $b\in H^{j}(X,{\mathbb{Z}})$ we have
$\alpha\cup\beta=(-1)^{ij}\beta\cup\alpha\in H^{i+j}(X,{\mathbb{Z}})$. ring
with identity with the multiplication operation cup-product $\cup$. The
homology $H_{*}(X,{\mathbb{Z}})$ is a graded module over
$H^{*}(X,{\mathbb{Z}})$. The groups $H^{*}$ and $H_{*}$ are also related by
the Universal Coefficient Theorem, but we will not use it in these notes.

The nicest case occurs when $X$ is a smooth and compact (for example,
projective) algebraic variety. In this case we have the Poincaré duality

$H^{i}(X,{\mathbb{Z}})\stackrel{{\scriptstyle\rm
natural}}{{\cong}}H_{2\dim_{\mathbb{C}}X-i}(X,{\mathbb{Z}}).$

Under Poincaré duality, $1\in H^{0}(X,{\mathbb{Z}})$ corresponds to the class
of $[X]$ in $H_{2\dim_{\mathbb{C}}X}(X,{\mathbb{Z}})$, but the more
interesting statement is that the class of the point in
$H_{0}(X,{\mathbb{Z}})$ corresponds to a special element ${\rm P.D.(point)}$
in $H^{\rm top}(X,{\mathbb{Z}})=H^{2\dim_{\mathbb{C}}X}(X,{\mathbb{Z}})$. In
fact, we have

$H^{\rm top}(X,{\mathbb{Z}})={\mathbb{Z}}\,{\rm P.D.(point)},$

and this gives us a concept of the intersection number.

######  Definition 19.1.

Let $Y$ and $Z$ be algebraic subvarieties of $X$, of complementary dimension,
i.e.

$\dim_{\mathbb{C}}Y+\dim_{\mathbb{C}}Z=\dim_{\mathbb{C}}X.$

Consider the product of Poincaré duals

${\rm P.D}([Y])\cup{\rm P.D}([Z])=\alpha\,{\rm P.D.(point)}\in H^{\rm
top}(X,{\mathbb{Z}}).$

The integer $\alpha$ is called the intersection number of $Y$ and $Z$ and will
be denoted by $Y\cdot Z$.

######  Remark 19.2.

If $\,Y$ and $Z$ intersect transversely at $k$ points, then $Y\cdot Z=k$ (in
particular, $Y\cdot Z=0$ if $Y\cap Z=\emptyset$). More generally, if $Y\cap Z$
is a finite set, then $Y\cdot Z$ is a sum of local contributions, each of
which is positive. However, the definition makes sense even if $Y\cap Z$ is
positive-dimensional, or even $Y\subseteq Z$! In these cases, we just know
that $Y\cdot Z$ is integer but can make no claims about the sign of it.
Philosophically, we can deform $Y$ as a real cycle to some $Y^{\prime}$ with
the same homology classes $[Y^{\prime}]=[Y]$ so that $Y^{\prime}$ intersects
$Z$ transversely, and we add up $(\pm 1)$ at each intersection point,
depending on the orientations.

The simplest nontrivial example is provided by
$X={\mathbb{C}}{\mathbb{P}}^{2}$. If $Y$ and $Z$ are complex curves in $X$, we
have $\dim_{C}Y+\dim_{\mathbb{C}}Z=\dim_{\mathbb{C}}X$, so their intersection
product makes sense. The cohomology groups of $X$ are given by

$H^{1}(X,{\mathbb{Z}})=H^{3}(X,{\mathbb{Z}})=0,~{}H^{0}(X,{\mathbb{Z}})\cong
H^{2}(X,{\mathbb{Z}})\cong H^{4}(X,{\mathbb{Z}})\cong{\mathbb{Z}}$

with the generators of the latter three groups given by $1$, the Poincaré dual
of the class $[l]$ of a line, and the Poincaré dual of the class of a point,
and we have ${\rm P.D.}[l]\cup{\rm P.D.}[l]={\rm P.D.(point)}$19.219.219.2
This is a very fancy way of saying that two lines in
${\mathbb{C}}{\mathbb{P}}^{2}$ intersect transversely at a point. The class of
the curve $Y$ is $(\deg Y)[l]$, and therefore $Y\cdot Z=(\deg Y)(\deg Z)$.
This is essentially the Bezout’s Theorem 4.3 that we discussed rigorously in
Sections 4 and 5, although to make the connection precise we would need to
talk about the intersection multiplicities in the non-transversal case.

Now let us consider the blowup of ${\mathbb{C}}{\mathbb{P}}^{2}$ at one point.
We get $X\to{\mathbb{C}}{\mathbb{P}}^{2}$ with $E$ the exceptional divisor on
$X$.

$X$$\hskip-6.0pt\cdot$${\mathbb{C}}{\mathbb{P}}^{2}$$E$$L^{\prime}$$L$

We have the following curves of interest on $X$: $E$, $L$, $L^{\prime}$. The
curve $E$ was already discussed, the curve $L$ is the preimage of a line in
${\mathbb{C}}{\mathbb{P}}^{2}$ that does not pass through the point we blew up
and the curve $L^{\prime}$ is the proper preimage of line that does pass
through a point of blowup.

The key observation is that $L\sim L^{\prime}+E$, in the sense that
$[L]=[L^{\prime}]+[E]$ in $H_{2}(X,{\mathbb{Z}})$. Indeed, more generally
there is a map of Weil divisors to $H_{2}(X,{\mathbb{Z}})$ which passes
through the class group. Another way of saying the same thing is that a
divisor of a rational function is always homologically trivial. And there is a
rational function (the ratio of the equations of the equations of the two
lines) which has simple zeros at $E$ and $L^{\prime}$ and poles at
$L$.19.319.319.3 In this particular case can think of the family of smooth
preimages of lines in ${\mathbb{C}}{\mathbb{P}}^{2}$ degenerating into a
singular curve $E\cup L^{\prime}$. We also have

$L^{2}=1,~{}L\cdot E=0,L^{\prime}\cdot E=1,L\cdot L^{\prime}=1$ (19.1)

based on the transversal intersections of the corresponding curves. This
implies that

$0=L\cdot E=(L^{\prime}+E)\cdot E=L^{\prime}\cdot E+E\cdot E=1+E^{2},$

so we get

$E^{2}=-1.$

What? This certainly looked weird to me the first time I saw it. How can a
curve intersect itself negatively?

Actually, there is no problem. This curve $E$ can not be deformed
holomorphically, but you can deform it as a real $2$-cycle on $X$ to some
$\tilde{E}$ which intersects $E$ at one point transversely with the “wrong”
orientation. In general, for a smooth curve $C$ on a smooth projective surface
$X$ the self-intersection $C^{2}$ is equal to the degree of the normal bundle
to $C$ in $X$, where by the degree of a line bundle I mean the degree of the
corresponding Weil divisor class.

In a footnote to Section 17 we indicated that the trees of lines in the
resolutions of ADE singularities have an intersection matrix which is negative
of the Cartan matrix of the corresponding Dynkin diagram. Since these diagrams
are simply-laced, this simply means checking that for each of these lines $E$
there holds $E^{2}=-2$. We will not do this in full generality but will
address the case of the resolution of $A_{1}$ singularity. In this case, we
have a surface $S$ in ${\mathbb{C}}^{3}$ given by $uv-w^{2}=0$, and we blow up
${\mathbb{C}}^{3}$ at the origin. The proper preimage $X$ of $S$ is then
smooth and the map

$\pi:X\to S$

contracts a single line $E$ to the origin.

######  Remark 19.3.

If you object to doing intersection theory on a non-compact surface $X$, you
are not alone. To avoid this issue, we will compactify
${\mathbb{C}}^{3}\subset{\mathbb{C}}{\mathbb{P}}^{3}$ and consider the closure
of $S$ in it, given by

$uv-w^{2}=0$

in homogeneous coordinates $(t:u:v:w)$ of ${\mathbb{C}}^{3}$. We end up adding
a smooth conic in the plane $t=0$, and it is easy to see that there is still
only one singular point and the blowup of ${\mathbb{C}}{\mathbb{P}}^{3}$ in it
still induces a resolution of singularities. We will still use $S$ and $X$ as
our notations. Alternatively, one can argue that all we are trying to do is to
understand the degree of the normal bundle of $E$ in $X$, and this does not
require any compactification.

Consider two lines $C_{1}$ and $C_{2}$ on $S$ given by $u=w=0$ and $v=w=0$
respectively. Their proper preimages on $X$ intersect $E$ transversely at one
point each.

$\bullet$$~{}~{}A_{1}$$S$$X$$C_{1}$$~{}~{}C_{2}$$~{}~{}D$$E$$C_{2}^{\prime}$$~{}~{}D$$C_{1}^{\prime}$

We also see that the rational function $\frac{w}{t}$ has divisor

$[C_{1}^{\prime}]+[E]+[C_{2}^{\prime}]-[D]$

where $D$ is the preimage of the conic at infinity $\\{uv-w^{2}=t=0\\}$.
Therefore, we have $[E]=[D]-[C_{1}^{\prime}]-[C_{2}^{\prime}]$ in
$H_{2}(X,{\mathbb{Z}})$ and

$E^{2}=D\cdot E-C_{1}\cdot E-C_{2}\cdot E=0-1-1=-2$

since $D\cap E=\emptyset$ and $C_{i}$ intersect $E$ transversely at a single
point.

######  Remark 19.4.

It is possible to do intersection theory purely algebraically, over any field
$k$, even if $k$ is not algebraically closed or has positive characteristics.
One needs to replace the (even) homology groups $H_{*}$ with the so-called
Chow groups $A_{*}(X)$. As a matter of fact, $A_{*}(X)$ provide more refined
invariants, compared to $H_{*}$. A very good, but also quite advanced
reference for it is the Fulton’s book [14].

Exercise 1. Use (19.1) to check that on the blowup $X$ of a point in
${\mathbb{C}}{\mathbb{P}}^{2}$ there holds $(L^{\prime})^{2}=0$. What is the
geometric meaning of it?

Exercise 2. Let $C$ be a smooth curve on a smooth projective surface $S$ and
let $X\to S$ be the blowup of $S$ at a point $p\in S$. Denote by $C^{\prime}$
the proper preimage of $C$ in $X$. Prove that $(C^{\prime})^{2}=C^{2}-1$. This
generalizes the example of Exercise 1.

Exercise 3. Let $X$ be a blowup of ${\mathbb{C}}{\mathbb{P}}^{2}$ in $5$
points $p_{1},\ldots,p_{5}$ in general position. Prove that the proper
preimage of any line through two of these points has self-intersection $(-1)$.
Prove that the proper preimage of the unique conic that passes through all
five points also has self-intersection $(-1)$.

##  20 Rational surfaces. Del Pezzo surfaces.

We will now discuss arguably the simplest class of complex algebraic surfaces,
namely the rational surfaces.

######  Definition 20.1.

A complex algebraic surface $X$ is called rational if and only if it is
birational to ${\mathbb{C}}{\mathbb{P}}^{2}$ or, equivalently, the field of
rational functions on $X$ is isomorphic to the field of rational functions in
two variables.

This definition makes sense in arbitrary dimension, to give us a notion of a
rational algebraic variety. In some sense, rational varieties are what we
would wish all varieties to be: their points, at least most of them, are
precisely encoded by free variables. In higher dimension, it is a highly
nontrivial problem to determine if a variety is rational, but in dimension two
it is usually quite manageable due to the following theorem of Castelnuovo,
which we will not attempt to prove.

######  Theorem 20.2.

Let $X$ be a smooth projective complex algebraic surface. Then $X$ is rational
if and only if the square of the canonical line bundle $K_{X}^{\otimes 2}\to
X$ has no nonzero holomorphic sections.

As was mentioned before, every two projective algebraic surfaces which are
birational to each other are obtained by successive blowups of points and
blowdowns. Thus, it is natural to define minimal rational surfaces as smooth
projective rational surfaces which are not isomorphic to a blowup of any other
surface. Since blowdown decreases the rank of the second cohomology group, one
can not do them indefinitely, so every rational surface is obtained by a
repeated blowup of a minimal one.

Prominent examples of minimal rational surfaces are Hirzebruch surfaces.

######  Definition 20.3.

Let $n$ be a nonnegative integer and let $W=\mathcal{O}\oplus\mathcal{O}(-n)$
be a rank two vector bundle over ${\mathbb{C}}{\mathbb{P}}^{1}$.20.120.120.1 A
theorem due to Grothendieck states that any vector bundle over
${\mathbb{C}}{\mathbb{P}}^{1}$ is a direct sum of line bundles. This is a very
peculiar property of ${\mathbb{C}}{\mathbb{P}}^{1}$. We define the Hirzebruch
surface

$H_{n}:={\mathbb{P}}W$

as the set of lines through the origin in the fibers of $W$.

Each of the two summands $\mathcal{O}$ and $\mathcal{O}(-n)$ of $W$ gives a
line in the fiber of $W$ and thus a point in the fiber of ${\mathbb{P}}W$. As
fibers vary, we get two smooth curves in $H_{n}$ which are sections of
$H_{n}\to{\mathbb{C}}{\mathbb{P}}^{1}$ in the usual sense of the word. We call
these curves $S_{0}$ and $S_{-n}$. In addition, we will be interested in the
fiber $F_{0}$ and $F_{\infty}$ over the two points of
${\mathbb{C}}{\mathbb{P}}^{1}$.

$F_{0}$$F_{\infty}$$S_{0}$$S_{-n}$$\bullet$$0$$\bullet$$\infty$

It is easy to see (Exercise 2) that

$H_{n}\setminus\left(S_{0}\cup F_{0}\right)\cong{\mathbb{C}}^{2}.$ (20.1)

One can show that every Weil divisor on ${\mathbb{C}}^{2}$ can be given as a
divisor of a rational function. Therefore, the class group of $H_{n}$ is
generated by $[S_{0}]$ and $[F_{0}]$. Let us find their intersection numbers.

######  Proposition 20.4.

The intersection pairings of $S_{0}$ and $F_{0}$ on $H_{n}$ are given by

$F_{0}^{2}=0,~{}F_{0}S_{0}=1,~{}S_{0}^{2}=-n.$

###### Proof.

The rational function $\frac{x_{0}}{x_{1}}$ on ${\mathbb{C}}{\mathbb{P}}^{1}$
gives a rational function on $H_{n}$ with divisor $[F_{0}]-[F_{\infty}]$. We
thus get

$F_{0}^{2}=F_{0}F_{\infty}=0,$

since $F_{0}\cap F_{\infty}=\emptyset$. The curves $F_{0}$ and $S_{0}$
intersect transversally at a single point, thus $F_{0}S_{0}=1$.

The computation for $S_{0}^{2}$ is a bit more subtle. We claim that there is a
rational function on $H_{n}$ which has divisor

$[S_{0}]-[S_{-n}]+n[F_{0}].$ (20.2)

This would imply that $[S_{0}]=[S_{-n}]-n[F_{0}]$ in ${\rm Cl}(H_{n})$, so

$S_{0}^{2}=S_{0}S_{-n}-nS_{0}F_{0}=0-n=-n.$

To construct such rational function, we write
$\mathcal{O}(-n)\to{\mathbb{C}}{\mathbb{P}}^{1}$ in coordinate charts as being
glued from two copies of ${\mathbb{C}}\times{\mathbb{C}}\to{\mathbb{C}}$ over
the two charts ${\mathbb{C}}$ on ${\mathbb{C}}{\mathbb{P}}^{1}$ via

$(t_{0},(1:\frac{x_{1}}{x_{0}}))\sim((\frac{x_{0}}{x_{1}})^{n}t_{1},(\frac{x_{0}}{x_{1}}:1)).$

Then $W\to{\mathbb{C}}$ is glued from two copies of
${\mathbb{C}}^{2}\times{\mathbb{C}}\to{\mathbb{C}}$ via

$(s_{0},t_{0},(1:\frac{x_{1}}{x_{0}}))\sim(s_{1},(\frac{x_{0}}{x_{1}})^{n}t_{1},(\frac{x_{0}}{x_{1}}:1)).$

Then $H_{n}={\mathbb{P}}W$ is glued from two copies of
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}\to{\mathbb{C}}$ via

$((s_{0}:t_{0}),(1:\frac{x_{1}}{x_{0}}))\sim((s_{1}:(\frac{x_{0}}{x_{1}})^{n}t_{1}),(\frac{x_{0}}{x_{1}}:1)).$

The curve $S_{0}$ is defined by $t_{0}=0$ in the first chart and $t_{1}=0$ in
the second chart. Similarly, $S_{-n}$ is $s_{0}=0$ in the first chart and
$s_{1}=0$ in the second chart. The curve $F_{0}$ is the complement of the
first chart and is defined by $\frac{x_{0}}{x_{1}}=0$ in the second chart.

The rational function

$\frac{t_{0}}{s_{0}}=\frac{t_{1}}{s_{1}}(\frac{x_{0}}{x_{1}})^{n}$

has divisor $[S_{0}]-[S_{-n}]$ in the first chart and
$[S_{0}]-[S_{-n}]+n[F_{0}]$ in the second. ∎

The following result, which we will not prove, describes all minimal rational
surfaces.

######  Theorem 20.5.

Minimal rational surfaces are either ${\mathbb{C}}{\mathbb{P}}^{2}$ or
Hirzebruch surfaces $H_{n}$ for $n=0$ or $n\geq 2$.

######  Remark 20.6.

The Hirzebruch surface $H_{0}$ is simply
${\mathbb{C}\mathbb{P}}^{1}\times{\mathbb{C}}{\mathbb{P}}^{1}$. The reason
that the Hirzebruch surface $H_{1}$ is not included in the list is that it is
actually isomorphic to the blowup of ${\mathbb{C}}{\mathbb{P}}^{2}$ at one
point. The Hirzebruch surface $H_{2}$ is the resolution of the $A_{1}$
singularity of the singular quadric $x_{0}x_{1}=x_{2}^{2}$ in
${\mathbb{C}}{\mathbb{P}}^{3}$.

Some of the most fascinating examples of rational surfaces are the so-called
del Pezzo surfaces. Recall (see Definition 7.1) that a line bundle $L\to X$ is
called ample if global sections of some positive tensor power of it define an
embedding of $X$ into a projective space.

######  Definition 20.7.

A smooth projective surface $X$ is called a del Pezzo surface, if the
anticanonical line bundle $K_{X}^{\vee}\cong\wedge^{2}TX\to X$ is ample.

The simplest example of a del Pezzo surface is ${\mathbb{C}}{\mathbb{P}}^{2}$.
The anticanonical line bundle on ${\mathbb{C}}{\mathbb{P}}^{2}$ is
$\mathcal{O}(3)$, its sections are cubic polynomials in the homogeneous
coordinates, and they define an embedding of ${\mathbb{C}}{\mathbb{P}}^{2}$
into ${\mathbb{C}}{\mathbb{P}}^{9}$, a particular case of the Veronese
embedding. So the anticanonical line bundle is very ample. Similarly, the
anticanonical line bundle on
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}{\mathbb{P}}^{1}$ is very
ample.

It can be shown that all del Pezzo surfaces are either
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}{\mathbb{P}}^{1}$ or blowups of
at most $8$ distinct points $p_{1},\ldots,p_{k}$ on
${\mathbb{C}}{\mathbb{P}}^{2}$, provided that these points are in a
sufficiently general position (for example, three points can not be
collinear). An interesting feature of a del Pezzo surface is its set of
$(-1)$-curves, which are the smooth curves of genus zero that can be blown
down to get another del Pezzo surface. We will comment on these below.

  * •

$k=1.$ Blowing up one point on ${\mathbb{C}}{\mathbb{P}}^{2}$, we get the
Hirzebruch surface $H_{1}$. It has a unique $(-1)$-curve, namely the
exceptional curve of the blowdown morphism.

  * •

$k=2.$ When we blow up two points $p_{1}$ and $p_{2}$ on
${\mathbb{C}}{\mathbb{P}}^{2}$, we get three $(-1)$-curves: two exceptional
curves $E_{1}$ and $E_{2}$ that blow down to $p_{1}$ and $p_{2}$ respectively,
and the proper preimage $l$ of the line
$\overline{p_{1}p_{2}}\subset{\mathbb{C}}{\mathbb{P}}^{2}$. We can blow down
$l$, and the resulting surface turns out to be isomorphic to
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}{\mathbb{P}}^{1}$. The reader
can try to prove it now, or wait until we talk about toric varieties in
Section 29.

  * •

$k=3.$ Since points can not be collinear, we might as well choose to blow up
$(1:0:0)$, $(0:1:0)$ and $(0:0:1)$. There are now six $(-1)$-curves in $X$,
connected to each other like vertices in a hexagon. They are three exceptional
curves and three proper preimages of the coordinate lines. In fact, we have
seen this surface in Exercise 2 of Section 15.

  * •

$k=4.$ There is only one such surface, since every four points on
${\mathbb{C}}{\mathbb{P}}^{2}$, no three of which are collinear, can be moved
to $(1:0:0)$, $(0:1:0)$, $(0:0:1)$ and $(1:1:1)$ by the action of ${\rm
PGL}(3,{\mathbb{C}})$. This surfaces has ten $(-1)$-curves: $4$ exceptional
curves and $6$ proper preimages of the lines through two out of $4$ points.

  * •

$k=5.$ There is now a two-dimensional family of surfaces, which we looked at
in Exercise 3 of Section 19. There are $16$ $(-1)$-curves: $5$ exceptional
curves of the blowup, $10$ proper preimages of lines and one proper preimage
of the conic through $p_{1},\ldots,p_{5}$. Sections of the anticanonical line
bundle on $X$ can be identified with cubic polynomials in the homogeneous
coordinates of ${\mathbb{C}}{\mathbb{P}}^{2}$ which are zero on
$p_{1},\ldots,p_{5}$. This is a five-dimensional space, and it gives an
embedding of $X$ into ${\mathbb{C}}{\mathbb{P}}^{4}$, so that the ideal of $X$
is generated by two degree two equations.

  * •

$k=6.$ This case is a bit more famous than the others. There are $27$
$(-1)$-curves of $X$, which come from $6$ exceptional curves, $15$ proper
preimages of lines and $6$ proper preimages of conics through $5$ out of six
points we blow up. All such surfaces are embedded into
${\mathbb{C}}{\mathbb{P}}^{3}$ by the sections of $K_{X}^{\vee}$ and the image
is given by one cubic equation. The $(-1)$-curves become lines in
${\mathbb{C}}{\mathbb{P}}^{3}$ in this embedding. Conversely, every smooth
cubic surface in ${\mathbb{C}}{\mathbb{P}}^{3}$ is del Pezzo, has $27$ lines
on it, and any collection of pairwise disjoint lines on it can be contracted
to get ${\mathbb{C}}{\mathbb{P}}^{2}$. We do not prove these classical
results, but we will see how the expected number of lines can be deduced using
calculations in the cohomology ring of the Grassmannian of lines in Exercise 3
of Section 26.

  * •

$k=7.$ In this case, the surface $X$ always has an involution, so that the
quotient is ${\mathbb{C}}{\mathbb{P}}^{2}$ and the map
$X\to{\mathbb{C}}{\mathbb{P}}^{2}$ is ramified over a smooth curve of degree
$4$. There are $56$ $(-1)$-curves on $X$. In addition to the $7$ exceptional
curves, proper preimages of $21$ lines through two of the blowup points and
$21$ conics through five of the points, there are also proper preimages of $7$
cubic curves in ${\mathbb{C}}{\mathbb{P}}^{2}$ which pass through $6$ out of
seven points of the blowup and are singular at the $7$-th point. The
anticanonical line bundle is not very ample, i.e. sections of $K_{X}^{\vee}$
do not define an embedding.

  * •

$k=8.$ There are $240$ $(-1)$-curves. Specifically, there are $8$ exceptional
curves, $28$ and $56$ proper preimages of lines and conics respectively and
$64$ proper preimages of cubic curves which are singular at one of the points
and pass through six points of the blowup. Furthermore, there are proper
preimages of $56$ degree four curves which are singular at three points and
pass through the remaining five points, and $28$ degree five curves which are
singular at six points and pass through the remaining two. The anticanonical
line bundle is not very ample.

######  Remark 20.8.

Another interesting family of surfaces is obtained by blowing up $9$ points on
${\mathbb{C}}{\mathbb{P}}^{2}$ which are the intersection points of two smooth
cubic curves $f_{1}=0$ and $f_{2}=0$. The resulting surface is not del Pezzo,
but its anticanonical line bundle defines a map to
${\mathbb{C}}{\mathbb{P}}^{1}$. The generic fiber of this map is an elliptic
curve, and these surfaces are called rational elliptic surfaces.

Exercise 1. Prove that for any vector bundle $W\to X$ and a line bundle $L\to
X$ we have

${\mathbb{P}}(W\otimes L)\cong{\mathbb{P}}W.$

Conclude that the projectivization of
$\mathcal{O}(a)\oplus\mathcal{O}(b)\to{\mathbb{C}}{\mathbb{P}}^{1}$ is
isomorphic to the Hirzebruch surface $H_{|a-b|}$.

Exercise 2. Show that the complement $H\setminus F_{0}$ is isomorphic to
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}$. Use it to prove that (20.1).

Exercise 3. A $(-1)$-curve in a smooth surface $X$ is, by definition,
isomorphic to ${\mathbb{C}}{\mathbb{P}}^{1}$ and satisfies $E^{2}=-1$. Use
adjunction formula to prove that it satisfies $K\cdot E=-1$ where $K$ is the
Weil divisor that corresponds to the canonical line bundle $K_{X}\to X$.

##  21 Kodaira dimension and classification of algebraic surfaces.

We will now discuss a very important invariant of algebraic varieties known as
the Kodaira dimension.

Let $X$ and $Y$ be compact smooth $n$-dimensional algebraic varieties which
are birational to each other, i.e. they contain isomorphic Zariski closed
subsets $X\supseteq U_{X}\cong U_{Y}\subseteq Y$. Because $Y$ is compact,
there exists an open subset $V\subseteq X$ that contains $U_{X}$ such that

  * •

${\rm codim}_{\mathbb{C}}(X\setminus V)\geq 2$.

  * •

There is a birational holomorphic map $\mu:V\to Y$.

This is often phrased as “birational equivalences are isomorphisms away from a
set of codimension one and are morphisms away from a set of codimension two”.
We will not prove this statement,21.121.121.1 It is related to the valuative
criterion of properness, [20]. but will illustrate it by an example.

The algebraic surface ${\mathbb{C}}{\mathbb{P}}^{2}$ and its blowup at a point
$p$ are birational to each other, because they contain isomorphic Zariski open
subsets, namely the complement of the point and its preimage. If we take $Y$
to be the blowup of ${\mathbb{C}}{\mathbb{P}}^{2}$ and $X$ to be
${\mathbb{C}}{\mathbb{P}}^{2}$, then
$V={\mathbb{C}}{\mathbb{P}}^{2}\setminus\\{p\\}$ maps to $Y$. On the other
hand, if we take $X$ to be the blowup of ${\mathbb{C}}{\mathbb{P}}^{2}$ and
$Y$ to be ${\mathbb{C}}{\mathbb{P}}^{2}$, then we can take $V$ to be all of
$X$, although the holomorphic map $V\to Y$ is no longer an inclusion.

Now let $w\in\Gamma(Y,\Lambda^{n}TY^{\vee})$ be a global holomorphic $n$-form
on $Y$. When we pull it back to $V$ we get
$\mu^{*}w\in\Gamma(V,\Lambda^{n}TX^{\vee})$, a holomorphic $n$-form on $V$. We
now invoke a statement knowns as the Hartogs’ Lemma: a holomorphic section of
a line bundle defined outside of the locus of complex codimension $2$ or
higher can be uniquely extended to a holomorphic section.21.221.221.2 It is
often stated for functions rather than sections of line bundles, but there is
no difference locally. Thus, our assumption of $V$ implies that $\mu^{*}w$ can
be extended to an element of $\Gamma(X,\Lambda^{n}TX^{\vee})$. This gives us a
pullback-then-extension linear map

$\Gamma(Y,\Lambda^{n}TY^{\vee})\to\Gamma(X,\Lambda^{n}TX^{\vee}).$

Naturally, we also have the pullback-then-extension map in the other direction

$\Gamma(X,\Lambda^{n}TX^{\vee})\to\Gamma(Y,\Lambda^{n}TY^{\vee})$

and these two maps are inverses of each other, because the compositions (in
the appropriate order) preserve the forms on $U_{X}$ and $U_{Y}$.

Recall that the line bundle $\Lambda^{n}TX^{\vee}$ is called the canonical
line bundle of $X$ and is denoted by $K_{X}$. A similar argument works for any
positive tensor power21.321.321.3 Just because we can’t integrate expressions
like $g(z)\,(dz_{1}\wedge\cdots\wedge dz_{n})^{\otimes m}$, it doesn’t mean
that they don’t make sense. They can also be pulled back for $m\geq 0$. of
$K_{X}$ so we see that

$\Gamma(X,K_{X}^{\otimes m})$

is an invariant of birational equivalences between smooth compact algebraic
manifolds. Moreover, the graded ring

$R=\bigoplus_{m\geq 0}\Gamma(X,K_{X}^{\otimes m})$

is a birational invariant. One of the great achievements of birational
geometry is the proof in 2010, by Birkar, Cascini, Hacon and McKernan [5, 6]
that $R$ is finitely generated (over its degree zero part which is always
${\mathbb{C}}$). I believe this is still an open problem over the fields of
positive characteristics, and I am not willing to venture any guesses if it is
true there or not.

######  Warning 21.1.

For $m<0$, the vector space

$\Gamma(X,K_{X}^{\otimes m})$

is not a birational invariant. The proof that works for $m>0$ is not
applicable, because vector fields do not pull back. For instance, for $m=-1$,

$\dim_{\mathbb{C}}\Gamma({\mathbb{C}}{\mathbb{P}}^{2},K_{{\mathbb{C}}{\mathbb{P}}^{2}}^{\vee})=\dim_{\mathbb{C}}\Gamma({\mathbb{C}}{\mathbb{P}}^{2},\mathcal{O}(3))=10$

but for the blowup $X$ of ${\mathbb{C}}{\mathbb{P}}^{2}$ at a point $p$ we get
$\dim_{\mathbb{C}}\Gamma(X,K_{X}^{\vee})=9$ as these correspond to homogeneous
cubic polynomials in three variables that are zero at $p$.

Kodaira dimension concerns the “size” of $R$. More precisely, $R$ is always an
integral domain, and its quotient field is a field extension of ${\mathbb{C}}$
of finite transcendence degree.

######  Definition 21.2.

We define Kodaira dimension of $X$ by

$\kappa(X)=\left\\{\begin{array}[]{ll}{\rm tr.deg.}(Q.F.(R))-1,&{\rm
if~{}}R_{>0}\neq 0.\\\ -\infty.&{\rm if~{}}R_{>0}=0.\end{array}\right.$

######  Remark 21.3.

Kodaira dimension of $X$ is a birational invariant which takes values in

$\\{-\infty,0,1,\ldots,\dim_{\mathbb{C}}X\\}.$

After the dimension, it is arguably the most important invariant of an
algebraic variety. Occasionally, $\kappa(X)=-\infty$ is denoted by
$\kappa(X)=-1$, but the $-\infty$ notation is preferable, because it satisfies

$\kappa(X\times Y)=\kappa(X)+\kappa(Y)$

in the appropriate sense.

In $\dim_{\mathbb{C}}X=1$ case, birational equivalence of smooth compact
Riemann surfaces implies isomorphism, and Kodaira dimension separates all
Riemann surfaces into three classes that we have already seen in Section 14.

  * •

$\kappa(X)=-\infty.$ This is the case of genus $g=0$, and
$X\cong{\mathbb{C}}{\mathbb{P}}^{1}$ is the only example. The universal cover
of $X$ is ${\mathbb{C}}{\mathbb{P}}^{1}$. The automorphism group is ${\rm
PGL}(2,{\mathbb{C}})$.

  * •

$\kappa(X)=0.$ We have $g=1$, $X$ is an elliptic curve, the universal cover is
${\mathbb{C}}$. The automorphism group is one-dimensional.

  * •

$\kappa(X)=1.$ Here $g\geq 2$, the universal cover is the unit disc
(equivalently, upper half plane). The automorphism group is finite.

We will now talk about the next case, $\dim_{\mathbb{C}}X=2$. The situation is
a lot more complicated. We will make no attempts at proving anything.

First let’s talk about $\kappa(X)=-\infty.$ Examples include all of the
rational surfaces, such as ${\mathbb{C}}{\mathbb{P}}^{2}$,
${\mathbb{C}}{\mathbb{P}}^{1}\times{\mathbb{C}}{\mathbb{P}}^{1}$, Hirzebruch
and del Pezzo surfaces. In addition, this class contains products
${\mathbb{C}}{\mathbb{P}}^{1}\times Y$ for a curve $Y$, and more generally
ruled surfaces ($X\to Y$ with generic fiber isomorphic to
${\mathbb{C}}{\mathbb{P}}^{1}$).

The next case of $\kappa(X)=0$ is arguably the most fascinating. In this case
either the canonical line bundle $K_{X}$ or some positive tensor power of it
is trivial. Examples include abelian surfaces, so-called hyperelliptic
surfaces, K3 surfaces and Enriques surfaces.

Abelian surfaces are topologically equivalent to $(S^{1})^{4}$ and are
isomorphic to ${\mathbb{C}}^{2}/L$ for certain lattices
$L\cong{\mathbb{Z}}^{4}$. They form countably many three-dimensional families
inside of a four-dimensional family of complex manifolds (see Exercise 1),
most of which are not algebraic.

$\dim 4$$\cdots$$\dim 3$$\begin{array}[]{c}{\rm
You~{}try~{}drawing~{}infinitely~{}many}\\\ {\rm three-
dimensional~{}loci~{}inside}\\\ {\rm a~{}four-
dimensional~{}space!}\end{array}$

Hyperelliptic (a.k.a. bielliptic) surfaces are quotients $(E\times F)/G$ of
certain products of elliptic curves by certain finite group actions. There are
seven families of such surfaces.

K3 surfaces have been studied by many authors, but they were named as such in
1958 by Weil, to honor Kummer, Kähler and Kodaira, as well as the K2 mountain
in the Himalayas. Similarly to the abelian surfaces, they form countably many
$19$-dimensional families inside a $20$-dimensional family of complex
surfaces. In what follows, I will describe some of these families and give a
(non-rigorous) count of the parameters.

  * •

Consider the double cover $X\to{\mathbb{C}}{\mathbb{P}}^{2}$ ramified over a
smooth degree $6$ curve $f(x_{0},x_{1},x_{2})=0$ which roughly means looking
at $t^{2}=f(x_{0},x_{1},x_{2})$ in some appropriate space, see Exercise
2.21.421.421.4 We do not explain why $K_{X}\to X$ is trivial in this case. The
parameter count goes as follows. We have a $28$-dimensional space of
homogeneous polynomials of degree $6$, which gives $27$ parameters up to
scaling. Moreover, we should subtract the dimension of the automorphism group
of ${\mathbb{C}}{\mathbb{P}}^{2}$, which is $\dim{\rm PGL}(3,{\mathbb{C}})=8$.
Thus, we get $27-8=19$.

  * •

The next case is that of degree $4$ hypersurfaces $X$ in
${\mathbb{C}}{\mathbb{P}}^{3}$ that one can view as a direct generalization of
cubic curves in ${\mathbb{C}}{\mathbb{P}}^{2}$. Note that by the adjunction
formula of Proposition 9.7 we have

$K_{X}\cong\mu^{*}(K_{{\mathbb{C}}{\mathbb{P}}^{3}}\otimes\mathcal{O}(X))\cong\mu^{*}(\mathcal{O}(-4)\otimes\mathcal{O}(4))=\mu^{*}\mathcal{O}$

is trivial. The space of polynomials of degree $4$ is of dimension $35$ and we
get as before

$35-1-\dim{\rm PGL}(4,{\mathbb{C}})=35-1-15=19.$

  * •

Now let us consider the surfaces $X$ in ${\mathbb{C}}{\mathbb{P}}^{4}$ whose
homogeneous ideal is generated by one quadratic and one cubic equation. We can
think of $X$ as an intersection of a quadric $Q$ and a cubic $C$ in
${\mathbb{C}}{\mathbb{P}}^{4}$. The canonical class of $X$ is trivial by a
repeated application of the adjunction formula, first going from
${\mathbb{C}}{\mathbb{P}}^{4}$ to $C$ and then going from $C$ to $X$. The
space of quadratic polynomials in $5$ variables is of dimension $15$, so we
get a dimension $14$ parameter space of $Q$. The space of equations of $C$ is
of dimension $35$, but we need to not only account for the scaling (subtract
$1$) but also for adding any products of the equation of $Q$ and a degree one
polynomial. And let’s not forget about the automorphisms of
${\mathbb{C}}{\mathbb{P}}^{4}$. All in all, we get

$14+(35-1-5)-\dim{\rm PGL}(5,{\mathbb{C}})=14+29-24=19.$

######  Remark 21.4.

We do list one more family in Exercise 3 and there are, as I mentioned
earlier, infinitely many families, but the description gets more complicated
further on. One really needs Hodge theory for the proper treatment of K3
surfaces. What is remarkable, and not at all obvious, is that all of the above
are diffeomorphic to each other as real manifolds!

To finish the $\kappa(X)=0$ case, Enriques surfaces are quotients of certain
special K3 surfaces by ${\mathbb{Z}}/2{\mathbb{Z}}$ that acts without fixed
points. They form a single $10$-parameter family.

In the case of $\kappa(X)=1$, the surfaces in question have a structure of
elliptic fibration, i.e. there is a map $X\to Y$, where $Y$ is a curve, and
most fibers are smooth curves of genus $1$. There is a detailed classification
of what kind of special fibers could occur, due to Kodaira.

The case of $\kappa(X)=2$ is that of surfaces of general type. It includes in
particular products $C_{1}\times C_{2}$ where both $C_{i}$ are curves of genus
at least two. Other simple examples are smooth surfaces in
${\mathbb{C}}{\mathbb{P}}^{3}$ of degree at least five. Overall, we are
nowhere near classifying surfaces of general type, although there are multiple
very interesting constructions of some of them with small Betti numbers. My
hobby over the last eight years or so has been trying to construct explicit
equations of a very special type of such surfaces, known as fake projective
planes.21.521.521.5 They have been classified as $100$ group quotients of a
two-dimensional ball by [8], following [26], but this does not provide
explicit equations. For a review of related surfaces, I highly recommend [4].

Exercise 1. Count the parameters of rank $4$ lattices in ${\mathbb{C}}^{2}$,
up to linear transformations of ${\mathbb{C}}^{2}$ to show that they form a
four-dimensional family.

Exercise 2*. The weighted projective space $W{\mathbb{P}}(1,1,1,3)$ is defined
as the quotient of ${\mathbb{C}}^{4}\setminus\\{(0,0,0,0\\}$ with coordinates
$(x_{0},x_{1},x_{2},t)$ by the action of ${\mathbb{C}}^{*}$ given by
$(x_{0},x_{1},x_{2},t)\mapsto(\lambda x_{0},\lambda x_{1},\lambda
x_{2},\lambda^{3}t)$. As a set, $W{\mathbb{P}}(1,1,1,3)$ can be covered by the
charts $U_{i}=\\{x_{i}\neq 0\\}$ and $U_{t}=\\{t\neq 0\\}$ which gives it a
structure of a (singular) algebraic variety. Prove that $U_{0,1,2}$ are in
natural bijection to ${\mathbb{C}}^{3}$ and $U_{t}$ is in bijection to the
quotient of ${\mathbb{C}}^{3}$ by the group ${\mathbb{Z}}/3{\mathbb{Z}}$ that
acts by scaling coordinates by the third roots of $1$. Prove that the surface

$t^{2}=x_{0}^{6}+x_{1}^{6}+x_{2}^{6}$

is smooth and maps $2:1$ to ${\mathbb{C}}{\mathbb{P}}^{2}$, with ramification
divisor given by

$0=x_{0}^{6}+x_{1}^{6}+x_{2}^{6}.$

Exercise 3. One more accessible example of K3 surfaces is that of surfaces $X$
in ${\mathbb{C}}{\mathbb{P}}^{5}$ cut out by three quadratic equations. Do the
rough parameter count, taking into account that only the linear span of the
three equations matters. _Hint:_ Count the ordered bases of the space of
generators, which overcounts the parameter space by $\dim{\rm
GL}(3,{\mathbb{C}})=9$. If you know the formula for the dimensions of
Grassmannians, that could be used as well.

##  22 Chern classes of vector bundles. Chern character. Euler sequence.

Let $X$ be a smooth complex manifold and $W\to X$ be a rank $r$ holomorphic
vector bundle. Then it is possible, although not easy, to construct the total
Chern class of $W$

$c(W)=1+c_{1}(W)+c_{2}(W)+\ldots+c_{r}(W)\in H^{\rm
even}(X,{\mathbb{Z}})=\bigoplus_{i}H^{2i}(X,{\mathbb{Z}})$

where $1\in H^{0}(X,{\mathbb{Z}})$ and $c_{i}(W)$ is an element of
$H^{i}(X,{\mathbb{Z}})$. The individual terms $c_{i}(W)$ are called the $i$-th
Chern classes of $W$.

The case of line bundles $L\to X$ is of particular interest. Here the only
nontrivial Chern class is $c_{1}(L)$. If $L$ corresponds to the Weil divisor
$D=\sum_{i}a_{i}D_{i}$, the class $c_{1}(L)$ turns out to be the Poincaré dual
of the homology element $[D]$ in $H_{2\dim_{\mathbb{C}}X-2}(X,{\mathbb{Z}})$.
So in this sense we can think of Chern classes as some kind of higher rank
invariants that generalize the correspondence between line bundles and Weil
divisor classes.

Chern classes carry a number of properties, which we list below, obviously
without proof.

  * •

$c_{k}(W)=0$ for $k>\operatorname{rk}W$.

  * •

(Functoriality) For a morphism $f:X\to Y$ and vector bundle $W\to Y$ there
holds

$c_{k}(f^{*}W)=f^{*}(c_{k}(W))$

for all $k$. On the left hand side we have the Chern class of the pullback of
the vector bundle, and on the right we have the cohomology pullback
$H^{2k}(Y,{\mathbb{Z}})\to H^{2k}(X,{\mathbb{Z}})$ of the Chern class.

  * •

For a short exact sequence of vector bundles22.122.122.1 This means we have
maps of vector bundles that induce short exact sequences in each fiber.

$0\to W_{1}\to W_{2}\to W_{3}\to 0$

on $X$ there holds

$c(W_{2})=c(W_{1})c(W_{3}).$

As the corollary of the last statement, if a vector bundles $W$ is a direct
sum of line bundles

$W\cong L_{1}\oplus L_{2}\oplus\ldots\oplus L_{r},$

then

$c(W)=(1+c_{1}(L_{1}))(1+c_{1}(L_{2}))\cdot\ldots\cdot(1+c_{1}(L_{r})).$

Now I want to explain how one can work with Chern classes. A typical problem
might be the following. Suppose we have a rank $2$ vector bundle $W$ with the
total Chern class $c(W)=1+c_{1}+c_{2}$, and we want to compute the Chern class
of the third symmetric power ${\rm Sym}^{3}(W)$.

Because ${\rm Sym}^{3}({\mathbb{C}}^{2})\cong{\mathbb{C}}^{4}$, the vector
bundle ${\rm Sym}^{3}(W)$ has rank $4$. Suppose for a moment that $W$ is a
direct sum of line bundles

$W\cong L_{1}\oplus L_{2},$

in which case

${\rm Sym}^{3}(W)\cong L_{1}^{\otimes 3}\oplus(L_{1}^{\otimes 2}\otimes
L_{2})\oplus(L_{1}\otimes L_{2}^{\otimes 2})\oplus L_{2}^{\otimes 3}.$

If we denote $c_{1}(L_{1})=x_{1},~{}c_{2}(L_{2})=x_{2}$, then we have
$c_{1}=x_{1}+x_{2}\in H^{2}(X,{\mathbb{Z}})$ and $c_{2}=x_{1}x_{2}\in
H^{4}(X,{\mathbb{Z}})$. Since tensor product of line bundles corresponds to
the addition of the corresponding Weil divisor classes, we get

$c({\rm Sym}^{3}(W))=(1+3x_{1})(1+2x_{1}+x_{2})(1+x_{1}+2x_{2})(1+3x_{2}).$

On the right, we have a symmetric function in $x_{1},x_{2}$, which we can
write in terms of its elementary symmetric polynomials $c_{1}$ and $c_{2}$ as

$c({\rm
Sym}^{3}(W))=1+6c_{1}+(11c_{1}^{2}+10c_{2})+(6c_{1}^{3}+30c_{1}c_{2})+(18c_{1}^{2}c_{2}+9c_{2})$
(22.1)

The black magic known as the splitting principle then says that $c({\rm
Sym}^{3}(W))$ is given by the formula (22.1) even if $W$ is not isomorphic to
a direct sum of line bundles! This allows us to compute Chern classes of
vector bundles that are constructed from given vector bundles.

An important terminology convention is to write

$c(W)=\prod_{i=1}^{\operatorname{rk}W}(1+x_{i}(W))$

and call $x_{i}(W)$ the Chern roots of $W$. If $W$ is a direct sum of line
bundles, then its Chern roots are just their first Chern classes. More
generally, the Chern roots themselves are not assigned a precise meaning, but
their elementary symmetric functions are the Chern classes of $W$. As a
consequence, a degree $k$ symmetric polynomial in the Chern roots of $W$ is a
polynomial in the Chern classes of $W$, and is an element of
$H^{2k}(X,{\mathbb{Z}})$. A particular case is that of the tangent bundle
$TX\to X$, where we write

$c(TX)=\prod_{i=1}^{\dim_{\mathbb{C}}X}(1+x_{i})$

and call $x_{i}$ the Chern roots of $X$.22.222.222.2 I have also seen these
more properly referred to as the Chern roots of the tangent bundle to $X$.
Perhaps this will become the common convention in the future.

Let us now do an important example. We would like to compute the total Chern
class of the tangent bundle to ${\mathbb{C}}{\mathbb{P}}^{n}$.

######  Proposition 22.1.

There is a natural sequence of vector bundles on
${\mathbb{C}}{\mathbb{P}}^{n}$

$0\to\mathcal{O}\to\mathcal{O}(1)^{\oplus(n+1)}\to
T{\mathbb{C}}{\mathbb{P}}^{n}\to 0$

usually called the Euler sequence.

###### Proof.

I will only sketch the proof. The key to this is the presentation

${\mathbb{C}}{\mathbb{P}}^{n}=({\mathbb{C}}^{n+1}\setminus\\{{\mathbf{0}}\\})/{\mathbb{C}}^{*}.$

We can think of vector fields on ${\mathbb{C}}{\mathbb{P}}^{n}$ as
${\mathbb{C}}^{*}$-equivariant vector fields on
${\mathbb{C}}^{n+1}\setminus\\{{\mathbf{0}}\\}$ of “weight $1$”, up to the
vector field $\sum_{i=0}^{n}x_{i}\frac{\partial}{\partial x_{i}}$ along the
fibers of the quotient map. ∎

As a consequence of the Euler sequence, we get

$c(T{\mathbb{C}}{\mathbb{P}}^{n})=\frac{c(\mathcal{O}(1))^{n+1}}{c(\mathcal{O})}=\frac{(1+H)^{n+1}}{1+0}=(1+H)^{n+1}$

where $H=c_{1}(\mathcal{O}(1))$ is the standard generator of
$H^{2}({\mathbb{C}}{\mathbb{P}}^{n},{\mathbb{Z}})$. Note, however, that it
would be wrong to claim that $H,\ldots,H$ ($n+1$ times) are Chern roots of
$T{\mathbb{C}}{\mathbb{P}}^{n}$, because that would give the wrong rank.

We will now discuss a very important way of packaging the Chern class
information, known as the Chern character.

######  Definition 22.2.

Let $W$ be a rank $r$ vector bundle on $X$ with

$c(W)=\prod_{i=1}^{r}(1+y_{i})$

where $y_{i}$ are the Chern roots. Then we define the Chern character of $W$
by

$ch(W):={\mathrm{e}}^{y_{1}}+{\mathrm{e}}^{y_{2}}+\cdots+{\mathrm{e}}^{y_{r}}=r+(y_{1}+\ldots
y_{r})+(\frac{1}{2}y_{1}^{2}+\ldots+\frac{1}{2}y_{r}^{2})+\cdots$

in $H^{\rm even}(X,{\mathbb{Q}})$. We have to use the coefficient ring
${\mathbb{Q}}$ because of the denominators in the exponential. Note that the
formal power series of ${\mathrm{e}}^{y}$ can be truncated to a polynomial,
since $H^{>2\dim_{\mathbb{C}}X}(X,{\mathbb{Q}})=0$.

######  Remark 22.3.

People these days plug everything into the exponential function! My personal
favorite is ${\mathrm{e}}^{X}$, which is the disjoint union of properly
understood quotients of the products of $n$ copies of a manifold $X$ by the
symmetric group of permutations.

######  Remark 22.4.

The standard notations $ch(W)$ and $c(W)$ are close enough to be somewhat
confusing at first reading, even though total Chern class and Chern character
have very different properties.

The following properties of the Chern character can be deduced from those of
the total Chern class and the splitting principle.

  * •

The $H^{0}(X,{\mathbb{Q}})$ part of $ch(W)$ is always the rank of $W$.

  * •

For a short exact sequence

$0\to W_{1}\to W_{2}\to W_{3}\to 0$

there holds $ch(W_{2})=ch(W_{1})+ch(W_{3})$.

  * •

For $f:X\to Y$ and vector bundles $W$ on $Y$ there holds
$ch(f^{*}(W))=f^{*}(ch(W))$.

  * •

$ch(W_{1}\otimes W_{2})=ch(W_{1})ch(W_{2})$.

The last property is perhaps the most useful, it is much simpler than the
corresponding statement for the total Chern classes.

As a corollary, we can use the Euler sequence on
${\mathbb{C}}{\mathbb{P}}^{n}$ to get

$ch(T{\mathbb{C}}{\mathbb{P}}^{n})=(n+1){\mathrm{e}}^{H}-1.$

There is also some strange but important class called the Todd class of $X$.
It is defined as follows.

######  Definition 22.5.

Let $x_{i}$ be the Chern roots of $TX$, i.e.

$c(TX)=1+c_{1}+c_{2}+\ldots=\prod_{i}(1+x_{i}).$

Then

${\rm
Td}(X)=\prod_{i=1}^{\dim_{\mathbb{C}}X}\frac{x_{i}}{1-{\mathrm{e}}^{-x_{i}}}$

where we expand $\frac{x}{1-{\mathrm{e}}^{-x}}$ as a formal power series in
$x$. It can be explicitly written as

${\rm
Td}(X)=1+\frac{c_{1}}{2}+\frac{c_{1}^{2}+c_{2}}{12}+\frac{c_{1}c_{2}}{24}+\cdots,$
(22.2)

see Exercise 3.

Exercise 1. Verify (22.1)

Exercise 2. Prove that the first Chern class $c_{1}(W)$ of a vector bundle $W$
is the first Chern class of its top exterior power
$\Lambda^{\operatorname{rk}W}W$.

Exercise 3. Prove that the coefficients of the Todd class of $X$ can be
written as polynomials in the Chern classes of $TX$, without using the
dimension of $X$, and verify (22.2). _Hint:_ By looking at

$\ln c(X)=\sum_{i}\ln(1+x_{i}),$

we can write the sums $\sum_{i=1}^{\dim_{\mathbb{C}}X}x_{i}^{k}$ as
polynomials in $c_{j}$. Then consider $\ln{\rm Td}(X)$.

##  23 Cohomology of vector bundles and Hirzebruch-Riemann-Roch formula.
Example: $\chi({\mathbb{C}}{\mathbb{P}}^{n},\mathcal{O}(k))$.

Recall that for a vector bundle $\pi:W\to X$ on a smooth complex algebraic
variety, sections $s$ of it are maps $s:X\to W$ such that $\pi\circ s={\rm
id}_{X}$. Sections of $W$ have a natural structure of a complex vector space,
which we (for now) denote by $\Gamma(X,W)$.

Suppose that we have a short exact sequence

$0\to W_{1}\to W_{2}\to W_{3}\to 0$

of vectors bundles on $X$, which means that we have this short exact sequence
of vector spaces in each fiber. Any section of $W_{1}$ induces a section of
$W_{2}$ and a section of $W_{2}$ induces one of $W_{3}$ and we have the
following result.

######  Proposition 23.1.

There is a left exact sequence

$0\to\Gamma(X,W_{1})\to\Gamma(X,W_{2})\to\Gamma(X,W_{3})$

###### Proof.

The fact that $\Gamma(X,W_{1})\to\Gamma(X,W_{2})$ is injective is obvious,
since a zero in a fiber of $W_{2}$ is also a zero in a fiber of $W_{1}$. For
the exactness in the middle, suppose that $s:X\to W$ is a section of $W_{2}\to
X$ which maps to the zero section of $W_{3}\to X$. This means that for every
$x\in X$ the image of $s(x)$ in the fiber of $W_{3}\to X$ is zero. But this
means that $s(x)$ actually lies in the fiber of $W_{1}$ inside that of
$W_{2}$, and $s$ is an image of a section $s_{1}$ of $W_{1}\to X$. ∎

As you may have guessed, there is generally no right exactness. In other
words, a surjective map of vector bundles

$W_{2}\to W_{3}\to 0$

may or may not induce a surjective map of the corresponding spaces of
sections. The issue here is that given a section $s:X\to W_{3}$, we may always
pick a lift to $W_{2}$ at each fiber, but we may not be able to do it in a
consistent (say, holomorphic) fashion over all of $X$. We provide a specific
example of non-surjectivity below.

Let $X={\mathbb{C}}{\mathbb{P}}^{n}$. Consider the Euler sequence

$0\to\mathcal{O}\to\mathcal{O}(1)^{\oplus n+1}\to TX\to 0.$

We can dualize it to get

$0\to(TX)^{\vee}\to\mathcal{O}(-1)^{\oplus n+1}\to\mathcal{O}\to 0.$

The line bundle $\mathcal{O}$ is the trivial bundle ${\mathbb{C}}\times X\to
X$, whose global sections are the holomorphic functions on $X$, which are
constants. So $\Gamma(X,\mathcal{O})\cong{\mathbb{C}}$. We claim that the line
bundle $\mathcal{O}(-1)\to{\mathbb{C}}{\mathbb{P}}^{n}$ has only zero
holomorphic sections. One way of seeing it is that the coordinates
$x_{0},\ldots,x_{n}$ on ${\mathbb{C}}{\mathbb{P}}^{n}$ are sections of
$\mathcal{O}(1)$, so for any section
$s\in\Gamma({\mathbb{C}}{\mathbb{P}}^{n},\mathcal{O}(-1))$ the tensor product
$s\otimes x_{0}$ is a section of $\mathcal{O}$ and is therefore a constant.
Since it is zero at $x_{0}=0$, we see that it must be a zero constant. This is
implies that $s$ is zero for $x_{0}\neq 0$ and is thus zero everywhere by
continuity.

It turns out that there is a way of working with the lack of exactness on the
right, other than just giving up. Specifically, for a short exact sequence

$0\to W_{1}\to W_{2}\to W_{3}\to 0$

of vector bundles on $X$ there exists a long exact sequence

$\begin{array}[]{l}0\to\Gamma(X,W_{1})\to\Gamma(X,W_{2})\to\Gamma(X,W_{3})\to\\\\[5.0pt]
\to H^{1}(X,W_{1})\to H^{1}(X,W_{2})\to H^{1}(X,W_{3})\to\\\\[5.0pt] \to
H^{2}(X,W_{1})\to\cdots\end{array}$ (23.1)

where $H^{i}(X,W)$ are certain cohomology vector spaces which depend
functorially on $W$.

######  Remark 23.2.

From now on, we will use $H^{0}(X,W)=\Gamma(X,W)$ to make the long exact
sequence (23.1) look more uniform.

While we don’t have the tools needed to define $H^{i}(X,W)$ (the hardest part
are the so-called connecting homomorphisms $H^{i}(W_{3})\to H^{i+1}(W_{1})$)
we will list some of their properties.

  * •

$H^{i}(X,W)=0$, unless $0\leq i\leq\dim_{\mathbb{C}}X$.

  * •

For $X$ compact, $H^{i}(X,W)$ are finite-dimensional.

  * •

(Serre Duality) For a smooth projective variety $X$, for all vector bundles
$W$ and all $i$ we have a natural duality

$H^{i}(X,W)\cong H^{\dim_{\mathbb{C}}X-i}(X,W^{\vee}\otimes K_{X})^{\vee}$

where $K_{X}$ is the canonical line bundle on $X$.

  * •

(Kodaira vanishing theorem) Let $L\to X$ be an ample line bundle on a smooth
projective complex variety $X$, see Definition 7.1. Then $H^{i}(X,L\otimes
K_{X})=0$ for all $i>0$.

######  Definition 23.3.

For any vector bundle $W$ on a smooth projective variety $X$ we define the
Euler characteristics of $W$ by

$\chi(X,W):=\sum_{i=0}^{\dim
X}\dim_{\mathbb{C}}H^{i}(X,W)=\dim_{\mathbb{C}}H^{0}(X,W)-\dim_{\mathbb{C}}H^{1}(X,W)+\cdots.$

The main property of the Euler characteristics of a vector bundle is that it
is additive on short exact sequences. More precisely, for any

$0\to W_{1}\to W_{2}\to W_{3}\to 0$

there holds $\chi(W_{2})=\chi(W_{1})+\chi(W_{3})$, see Exercise 1. This is why
it is often much easier to compute $\chi(X,W)$ than the more geometrically
meaningful $\dim_{\mathbb{C}}H^{0}(X,W)$. Fortunately, in some cases of
interest Kodaira vanishing or other similar theorems allow one to conclude
that $\chi(X,W)=\dim_{\mathbb{C}}H^{0}(X,W)$.

Additivity of $\chi$ on short exact sequences should remind you of the
additivity of the Chern character. Not coincidentally, the famous Hirzebruch-
Riemann-Roch formula allows one to compute $\chi(X,W)$ if one has a good
understanding of $ch(W)$, ${\rm Td}(X)$ and the product in $H^{\rm
even}(X,{\mathbb{Q}})$.

######  Theorem 23.4.

Let $X$ be a smooth projective complex algebraic variety. We define
$\int_{X}:H^{\rm even}(X,{\mathbb{Q}})\to{\mathbb{Q}}$ as reading off the
coefficient at the Poincaré dual of the point in the top cohomology
$H^{2\dim_{\mathbb{C}}X}(X,{\mathbb{Q}})$. Then for any vector bundle $W$ on
$X$ we have

$\chi(X,W)=\int_{X}ch(W)\,{\rm Td}(X).$

We have no intention of proving Theorem 23.4 but we will look at some of its
corollaries.

  * •

$\operatorname{rk}=1$, $\dim_{\mathbb{C}}X=1$. We will denote the line bundle
by $L$. Since $H^{>2}(X,{\mathbb{Q}})=0$, we get

$ch(W)={\mathrm{e}}^{c_{1}(L)}=1+c_{1}(L)=1+(\deg L){\rm P.D.(point)}.$

We also have ${\rm Td}(X)=1+\frac{1}{2}c_{1}(TX)=1+\frac{1}{2}(2-2g){\rm
P.D.(point)}$ where $g$ is the genus of $X$. Therefore, we have

$\chi(X,L)=\int_{X}(1+(\deg L){\rm P.D.(point)}+(1-g){\rm P.D.(point)})=\deg
L+1-g.$

Here we used Proposition 14.6 that the degree of the canonical line bundle is
$2g-2$, and the fact that $TX$ is the dual of the canonical line bundle.

  * •

$\operatorname{rk}=1$, $\dim_{\mathbb{C}}X=2$. We have

$ch(L)=1+c_{1}(L)+\frac{1}{2}c_{1}(L)^{2},~{}{\rm
Td}(X)=1+\frac{1}{2}c_{1}+\frac{c_{1}^{2}+c_{2}}{12}$

where $c_{i}$ denotes $c_{i}(TX)$. This gives

$\chi(X,L)=\frac{1}{2}c_{1}(L)^{2}+\frac{1}{2}c_{1}(L)c_{1}+\frac{1}{2}c_{1}+\frac{c_{1}^{2}+c_{2}}{12}.$

In the particular case $L=\mathcal{O}$ we get the Noether’s
formula23.123.123.1 It is worth mentioning that $c_{1}^{2}$ is the self-
intersection of the canonical divisor and $c_{2}$ is the topological Euler
characteristics of $X$, i.e. the alternating sum of its Betti numbers.

$\chi(\mathcal{O})=\frac{c_{1}^{2}+c_{2}}{12}.$

Then for general $L$ that corresponds to a Weil divisor $D$ we get

$\chi(X,L)=\frac{1}{2}D(D-K)+\chi(\mathcal{O}).$

  * •

$X={\mathbb{C}}{\mathbb{P}}^{n}$ and $L=\mathcal{O}(k)$. The cohomology ring
of ${\mathbb{C}}{\mathbb{P}}^{n}$ is generated in degree two by the Poincaré
dual $H$ of a hyperplane, with $H^{n}={\rm P.D.(point)}$ being the generator
of the top cohomology.

$H^{*}({\mathbb{C}}{\mathbb{P}}^{n},{\mathbb{Z}})=H^{\rm
even}({\mathbb{C}}{\mathbb{P}}^{n},{\mathbb{Z}})={\mathbb{C}}\oplus{\mathbb{C}}H\oplus\cdots\oplus{\mathbb{C}}H^{n}.$

We have $ch(L)={\mathrm{e}}^{kH}$, but the computation for the Todd class is a
bit more complicated. We have

$c(TX)=(1+H)^{n+1},$

but these are not the Chern roots. Nevertheless23.223.223.2 One can define
Todd classes of vector bundles in a way that’s multiplicative for short exact
sequences and then use the Euler sequence.,

${\rm Td}(X)=\left(\frac{H}{1-{\mathrm{e}}^{-H}}\right)^{n+1}$

so Hirzebruch-Riemann-Roch formula says

$\displaystyle\chi(X,\mathcal{O}(k))=\int_{X}\frac{{\mathrm{e}}^{kH}H^{n+1}}{(1-{\mathrm{e}}^{-H})^{n+1}}={\rm
coeff.~{}at~{}}H^{n}\left(\frac{{\mathrm{e}}^{kH}H^{n+1}}{(1-{\mathrm{e}}^{-H})^{n+1}}\right)$
$\displaystyle={\rm
Res}_{H=0}\frac{{\mathrm{e}}^{kH}}{(1-{\mathrm{e}}^{-H})^{n+1}}=\frac{1}{2\pi{\mathrm{i}}}\oint_{|H|=\varepsilon}\frac{{\mathrm{e}}^{kH}}{(1-{\mathrm{e}}^{-H})^{n+1}}\,dH$
$\displaystyle\stackrel{{\scriptstyle
t=1-{\mathrm{e}}^{-H}}}{{=}}\frac{1}{2\pi{\mathrm{i}}}\oint_{|t|=\varepsilon}\frac{(1-t)^{-k-1}}{t^{n+1}}\,dt=\frac{(k+1)(k+2)\cdot\ldots\cdot(k+n)}{n!}$

where we treat $H$ as a complex variable in the second line. In the last
identity we used that it is the coefficient at $t^{n}$ of the Maclaurin series
of $(1-t)^{-k-1}$. We will see the more elementary reason behind this formula
in Exercise 3.

Exercise 1. Prove that Euler characteristics is additive on short exact
sequences. _Hint:_ Use (23.1) and the general statement about alternating sum
of dimensions of vector spaces in a long exact sequence.

Exercise 2. Use Kodaira vanishing theorem to show that for all $k>-n-1$ we
have $H^{>0}({\mathbb{C}}{\mathbb{P}}^{n},\mathcal{O}(k))=0$. _Hint:_ Use
Exercise 3 of Section 9.

Exercise 3. One can show (see Exercise 2 of Section 7) that for $k\geq 0$ the
global sections of $\mathcal{O}(k)$ on ${\mathbb{C}}{\mathbb{P}}^{n}$ are
homogeneous polynomials of degree $k$ in the homogeneous coordinates on
${\mathbb{C}}{\mathbb{P}}^{n}$. Together with Exercise 2, show that this is
consistent with the above calculation of
$\chi({\mathbb{C}}{\mathbb{P}}^{n},\mathcal{O}(k))$.

##  24 Introduction to Grassmannians.

In the next two sections we will talk about the Grassmannian varieties.

######  Definition 24.1.

Let $V$ be a complex vector space of dimension $n$. For each
$k\in\\{0,\ldots,n\\}$, the Grassmannian ${\rm Gr}(k,V)$ is defined as the set
of all dimension $k$ vector subspaces of $V$. We also use ${\rm Gr}(k,n)$
notation for it if we don’t particularly care about the nature of $V$.

Clearly, for $k=0$ or $k=n$ the set ${\rm Gr}(k,V)$ consists of one element.
For $k=1$ we have ${\rm
Gr}(k,V)={\mathbb{P}}V\cong{\mathbb{C}}{\mathbb{P}}^{n-1}$. For $k=n-1$ we
have ${\rm Gr}(k,V)={\mathbb{P}}V^{\vee}\cong{\mathbb{C}}{\mathbb{P}}^{n-1}$.
This makes ${\rm Gr}(2,4)$ the first interesting case.

It is reasonable to expect that ${\rm Gr}(k,n)$ has a natural structure of an
algebraic variety and a complex manifold. Indeed, this is the case. As with
projective spaces, we can define it by using coordinate charts. We will think
of $V$ as the space of length $n$ row vectors. Then a dimension $k$ subspace
$W$ of $V$, together with a choice of a basis can be regarded as a $k\times n$
matrix $M$ of rank $k$. Different choices of said basis of $W$ are related by
invertible row transformations of $M$. Thus we get, as sets,

${\rm Gr}(k,n)=\\{{\rm max~{}rank~{}}k\times
n{\rm~{}complex~{}matrices}\\}/{\rm GL}(k,{\mathbb{C}})$

for the left multiplication action of ${\rm GL}(k,{\mathbb{C}})$. Note, that
in the $k=1$ case, this perfectly reproduces the definition of
${\mathbb{C}}{\mathbb{P}}^{n-1}$.

Recall that $M$ has rank $k$ if and only if it has a nonzero minor. This
inspires us to consider, for each set of indices

$1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n,$

the subset $U_{i_{1},i_{2},\ldots,i_{k}}$ of ${\rm Gr}(k,n)$ such that

$\det M_{i_{1},\ldots,i_{k}}\neq 0$

where the determinant is that of the square submatrix of $M$ made from columns
$i_{1},i_{2},\ldots,i_{k}$. Note that this property is preserved under
invertible linear row transformations.

Let us look in detail at the particular case of $U_{1,2,\ldots,k}$. This means
that the first minor of $M$ is invertible, so Gaussian elimination gives

$M\sim\left(\begin{array}[]{ccccccc}1&0&\ldots&0&*&\ldots&*\\\
0&1&\ldots&0&*&\ldots&*\\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
0&0&\ldots&1&*&\ldots&*\end{array}\right)$

with the unique reduced row echelon form. This is just a fancy way of saying
that we multiply $M$ by $M_{1,2,\ldots,k}^{-1}$ to get the first submatrix to
be identity. The entries of the remaining $k\times(n-k)$ matrix can be
arbitrary complex numbers, so we can identify $U_{1,\ldots,k}$ with
${\mathbb{C}}^{k(n-k)}$. For arbitrary $i_{1}<\cdots<i_{k}$, the chart
$U_{i_{1},\ldots,i_{k}}$ can be identified with ${\mathbb{C}}^{k(n-k)}$ by
looking at the entries of $M_{i_{1},i_{2},\ldots,i_{k}}^{-1}M$ outside of the
$i_{j}$-th columns.

On the intersection of $U_{i_{1},\ldots,i_{k}}$ and $U_{j_{1},\ldots,j_{k}}$,
the transition functions are clearly24.124.124.1 To make it even more clear,
we ask the reader to find this change of variables explicitly in Exercise 1.
rational and holomorphic, which gives ${\rm Gr}(k,n)$, and more generally
${\rm Gr}(k,V)$ the structure of a smooth complex algebraic variety.

######  Remark 24.2.

Compactness of ${\rm Gr}(k,n)$ is also fairly easy and is proved similarly to
that of ${\mathbb{C}}{\mathbb{P}}^{n}$. Every $k$-dimensional subspace $W$ of
${\mathbb{C}}^{n}$ has an orthonormal basis with respect to the standard
Hermitean form on ${\mathbb{C}}^{n}$. The space of such bases is a closed
subset of $(S^{2n-1})^{k}$ (given by the orthogonality conditions) and is
therefore compact. It maps continuously and surjectively onto ${\rm Gr}(k,n)$,
which proves compactness.

We will now embed ${\rm Gr}(k,V)$ into some projective space. The idea is
rather elegant: given a dimension $k$ subspace $W\subseteq V$ we get

$\Lambda^{k}W\subseteq\Lambda^{k}V.$

Since $\dim_{\mathbb{C}}\Lambda^{k}W=1$, we naturally associate to each $W$ a
point in the projective space ${\mathbb{P}}\Lambda^{k}V$ (of dimension
$\frac{n!}{k!(n-k)!}-1$). So we get a map

$\mu:{\rm Gr}(k,V)\to{\mathbb{P}}\Lambda^{k}V.$

######  Remark 24.3.

It is not a particularly mysterious map. If we think of ${\rm Gr}(k,n)$ as the
quotient as before, then the map is induced by

$M\mapsto\left(\det M_{i_{1},\ldots,i_{k}}\right)_{1\leq
i_{1}<\cdots<i_{k}\leq n}.$

Indeed, left multiplication by a matrix $A$ scales all of the above
determinants by $\det A$, so the map is well-defined. It is also clear from
this description that $\mu$ is holomorphic (and algebraic).

######  Remark 24.4.

It can be seen from the same construction that the map is an embedding. If two
points go to the same point in the projective space, then they must have the
same nonzero minor, which we can for simplicity assume to be $M_{1,\ldots,k}$.
We can assume that $M$ is in the reduced form and observe that determinants of
minors that use $(k-1)$ of the first $k$ columns pick up precisely the
coordinates of the reduced form of $M$. It is also easy to see that tangent
vectors do not collapse under $\mu$.

Another, more abstract, way of thinking about injectivity is that

$v\wedge(\Lambda^{k}W)=0\in\Lambda^{k+1}V\iff v\in W,$

so $\Lambda^{k}W$ determines $W$ uniquely.

This embedding $\mu$ is called the Plücker embedding of the Grassmannian ${\rm
Gr}(k,V)$ and the homogeneous coordinates on ${\mathbb{P}}\Lambda^{k}V$ are
called the Plücker coordinates. I now want to discuss the equations that cut
out ${\rm Gr}(k,V)$ in terms of Plücker coordinates. Hmm, what should we call
them?

But, first, let us talk about the exterior algebra $\Lambda^{*}V$. If
$e_{1},\ldots,e_{n}$ form a basis of $V$, then $\Lambda^{*}V$ is spanned by
$e_{i_{1}}\wedge\cdots\wedge e_{i_{m}}$ with the associativity rules, and
$e_{i}\wedge e_{j}=-e_{j}\wedge e_{i}$. In fact, up to one measly sign, the
rules are the same as those for the polynomials in $e_{i}$, and the exterior
algebra can be viewed as a ring of polynomials in odd (in physics terminology
“fermionic”) variables. Just as one has partial differentiation operators
acting on the polynomial ring, there are odd analogs of them here, known as
contractions. Specifically, for $\alpha\in V^{\vee}$ we can define linear maps
${\rm contr}_{\alpha}:\Lambda^{m}V\to\Lambda^{m-1}V$ with the property that
for any $v_{1},\ldots,v_{m}\in V$ there holds

$\begin{array}[]{l}{\rm contr}_{\alpha}(v_{1}\wedge\cdots\wedge
v_{m})=\alpha(v_{1})v_{2}\wedge\cdots\wedge v_{m}\\\ \hskip
20.0pt-\alpha(v_{2})v_{1}\wedge v_{3}\wedge\cdots\wedge
v_{m}+\ldots+(-1)^{m-1}\alpha(v_{m})v_{1}\wedge\cdots\wedge
v_{m-1}.\end{array}$ (24.1)

Suppose we have a point in ${\rm Gr}(k,V)\subseteq{\mathbb{P}}\Lambda^{k}V$
or, equivalently, a decomposable element $w=v_{1}\wedge\cdots\wedge
v_{k}\subseteq\Lambda^{k}V$. Then by (24.1), for arbitrary
$\alpha_{1},\ldots,\alpha_{k-1}\in V^{\vee}$ we see that

${\rm contr}_{\alpha_{k-1}}\circ\cdots\circ{\rm contr}_{\alpha_{1}}(w)$

is a linear combination of $v_{1},\ldots,v_{k}$, so we have

$\left({\rm contr}_{\alpha_{k-1}}\circ\cdots\circ{\rm
contr}_{\alpha_{1}}(w)\right)\wedge w=0\in\Lambda^{k+1}V.$

We can further restate it by saying that if $w$ is decomposable, then for all
$\alpha_{1},\ldots,\alpha_{2k}\in V^{\vee}$ there holds the following Plücker
relation.

${\rm contr}_{\alpha_{2k}}\circ\cdots\circ{\rm
contr}_{\alpha_{k}}\left(\left({\rm contr}_{\alpha_{k-1}}\circ\cdots\circ{\rm
contr}_{\alpha_{1}}(w)\right)\wedge w\right)=0$ (24.2)

We comment that it suffices to pick $\alpha_{i}$ in (24.2) among the basis
elements of $V^{\vee}$, so we have a finite and easily computable set of
quadratic equations in coefficients of $w$. It is true, though definitely not
obvious, that Plücker relations (24.2) cut out exactly ${\rm Gr}(k,V)$. As a
matter of fact, they generate the homogeneous ideal of ${\rm Gr}(k,V)$ in
${\mathbb{P}}\Lambda^{k}V$.

Let us compute the smallest interesting example, that of ${\rm Gr}(2,4)$. We
have a vector space $V$ with the basis $e_{1},\ldots,e_{4}$ and we want to
characterize the decomposable elements $w\in\Lambda^{2}V$. Every
$w\in\Lambda^{2}V$ can be written as

$w=x_{12}\,e_{1}\wedge e_{2}+x_{13}\,e_{1}\wedge e_{3}+x_{14}\,e_{1}\wedge
e_{4}+x_{23}\,e_{2}\wedge e_{3}+x_{24}\,e_{2}\wedge e_{4}+x_{34}\,e_{3}\wedge
e_{4},$

and we think of $x_{12},\ldots,x_{34}$ as the homogeneous coordinates of
${\mathbb{C}}{\mathbb{P}}^{5}$. Since the dimension of ${\rm Gr}(2,4)$ is $4$,
its image under the Plücker embedding is a hypersurface in
${\mathbb{C}}{\mathbb{P}}^{5}$, so we expect a single quadratic equation. More
precisely, if we take $\alpha_{1}$ to be the dual basis vector $e_{1}^{\vee}$,
then we have

${\rm contr}_{\alpha_{1}}w=x_{12}\,e_{2}+x_{13}\,e_{3}+x_{14}\,e_{4}.$

We wedge it with $w$ to get

$\displaystyle({\rm contr}_{\alpha_{1}}w)\wedge
w=(x_{12}e_{2}+x_{13}e_{3}+x_{14}e_{4})\wedge(x_{12}e_{1}\wedge
e_{2}+\cdots+x_{34}e_{3}\wedge e_{4})$
$\displaystyle=x_{12}x_{13}\,e_{2}\wedge e_{1}\wedge
e_{3}+x_{12}x_{14}\,e_{2}\wedge e_{1}\wedge e_{4}+x_{12}x_{34}\,e_{2}\wedge
e_{3}\wedge e_{4}$ $\displaystyle+x_{13}x_{12}\,e_{3}\wedge e_{1}\wedge
e_{2}+x_{13}x_{14}\,e_{3}\wedge e_{1}\wedge e_{4}+x_{13}x_{24}\,e_{3}\wedge
e_{2}\wedge e_{4}$ $\displaystyle+x_{14}x_{12}\,e_{4}\wedge e_{1}\wedge
e_{2}+x_{14}x_{13}\,e_{4}\wedge e_{1}\wedge e_{3}+x_{14}x_{23}\,e_{4}\wedge
e_{2}\wedge e_{3}$
$\displaystyle=(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})e_{2}\wedge e_{3}\wedge
e_{4}.$

Therefore, the equation of ${\rm
Gr}(2,4)\subseteq{\mathbb{C}}{\mathbb{P}}^{5}$ is

$0=x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23},$

and ${\rm Gr}(2,4)$ is a smooth (rank $6$) quadric hypersurface in
${\mathbb{C}}{\mathbb{P}}^{5}$.

Exercise 1. Find explicitly the change of variables from $U_{1,3}$ to
$U_{3,4}$ in ${\rm Gr}(2,4)$.

Exercise 2. Verify that in the special case of ${\rm Gr}(2,4)$ one can write
the Plücker relation as $w\wedge w=0$.

Exercise 3. Give an example of an element $w$ in $\Lambda^{3}{\mathbb{C}}^{6}$
with $w\wedge w=0$ which can not be written as $w=v_{1}\wedge v_{2}\wedge
v_{3}$.

##  25 Grassmannians continued. Number of lines that intersect four given
lines in ${\mathbb{C}}{\mathbb{P}}^{3}$.

In this section we continue our discussion of the complex Grassmannians ${\rm
Gr}(k,n)$.

The singular cohomology ring $H^{*}({\rm Gr}(k,n),{\mathbb{Z}})$ of ${\rm
Gr}(k,n)$ is well-studied but is intricate. Cohomology only occurs in even
degrees and has a free generator set given by the Poincaré duals of the so-
called Schubert cycles which are described as follows.

Let us think about points of ${\rm Gr}(k,n)$ as full rank $k\times n$ matrices
up to row transformations and recall that every such matrix has a unique
reduced row echelon form. Such reduced row echelon form is characterized by
the location of the pivot columns, and the Schubert cycle is defined as the
closure of the locus of $M$ whose reduced row echelon form has given pivot
columns. Instead of setting up general notation, we will illustrate this in
the particular case of ${\rm Gr}(2,4)$. We can think of the points in ${\rm
Gr}(2,4)$ as parameterizing lines $l\cong{\mathbb{C}}{\mathbb{P}}^{1}$ in the
three-dimensional projective space ${\mathbb{C}}{\mathbb{P}}^{3}$, instead of
${\mathbb{C}}^{2}\subset{\mathbb{C}}^{4}$.25.125.125.1 This leads to a
competing convention where what we call ${\rm Gr}(2,4)$ is called ${\rm
Gr}(1,3)$, but I believe that ours is more widely used. There are six possible
choices of pivot columns, and we list the Schubert cycles below, together with
their geometric descriptions.

$\displaystyle\overline{\left(\begin{array}[]{cccc}1&0&*&*\\\
0&1&*&*\end{array}\right)},\hskip
50.0pt\overline{\left(\begin{array}[]{cccc}1&*&0&*\\\
0&0&1&*\end{array}\right)},\hskip
50.0pt\overline{\left(\begin{array}[]{cccc}1&*&*&0\\\
0&0&0&1\end{array}\right)},$ (25.7) $\displaystyle{\rm~{}all~{}of~{}}{\rm
Gr}(2,4)\hskip 50.0ptl\cap\\{(0:0:*:*)\\}\neq\emptyset\hskip
45.0pt(0:0:0:1)\in l$
$\displaystyle\overline{\left(\begin{array}[]{cccc}0&1&0&*\\\
0&0&1&*\end{array}\right)},\hskip
50.0pt\overline{\left(\begin{array}[]{cccc}0&1&*&0\\\
0&0&0&1\end{array}\right)},\hskip
50.0pt\overline{\left(\begin{array}[]{cccc}0&0&1&0\\\
0&0&0&1\end{array}\right)}.$ (25.14) $\displaystyle
l\subset\\{(0:*:*:*)\\}\hskip 10.0pt(0:0:0:1)\in l\subset\\{(0:*:*:*)\\}\hskip
10.0ptl=(0:0:*:*)$ (25.15)

######  Remark 25.1.

There is an equivalent description of Schubert cycles in terms of dimensions
of the intersection of the subspace $W$ with elements of a complete flag on
$V$, see [19].

There is a very nice alternative description of the cohomology ring of ${\rm
Gr}(k,n)$ in terms of the Chern classes of certain vector bundles.
Specifically, we have a natural exact sequence of vector bundles on ${\rm
Gr}(k,n)$

$0\to S\to\mathcal{O}^{\oplus n}\to Q\to 0.$ (25.16)

All fibers of $\mathcal{O}^{\oplus n}$ are naturally identified with
${\mathbb{C}}^{n}$, the fiber of $S\to{\rm Gr}(k,n)$ over the point that
corresponds to $W\subset{\mathbb{C}}^{n}$ is $W$, and the fiber of $Q$ over
this point is ${\mathbb{C}}^{n}/W$. We call $S$ and $Q$ the tautological
subbundle and the tautological quotient bundle respectively.

From the properties of Chern classes, we have

$1=c(S)c(Q)=(1+s_{1}+s_{2}+\cdots+s_{k})(1+q_{1}+q_{2}+\cdots+q_{n-k}).$

This gives relations on $s_{i}$ and $q_{i}$ and there is a beautiful result
that states that $H^{*}({\rm Gr}(k,n),{\mathbb{Z}})$ is generated by $s_{i}$
and $q_{j}$ subject to these relations.

We will illustrate this in the case of ${\rm Gr}(2,4)$. We have
$(1+s_{1}+s_{2})(1+q_{1}+q_{2})=1$. Therefore,

$s_{1}+q_{1}=0,~{}s_{1}q_{1}+s_{2}+q_{2}=0,~{}s_{1}q_{2}+s_{2}q_{1}=0,s_{2}q_{2}=0.$
(25.17)

The quotient of ${\mathbb{Z}}[s_{1},s_{2},q_{1},q_{2}]$ by the relations
(25.17) is isomorphic to

${\mathbb{Z}}[s_{1},s_{2}]/(s_{1}^{3}-2s_{1}s_{2},s_{1}^{2}s_{2}-s_{2}^{2}),$

see Exercise 1. This implies $s_{1}s_{2}^{2}=s_{1}^{3}s_{2}=2s_{1}s_{2}^{2}$,
so $s_{1}s_{2}^{2}=s_{1}^{3}s_{2}=0$. From here we get that
$s_{1}s_{2}^{2}=s_{2}^{3}=0$ and $s_{1}^{5}=0$. As the result, we get

$\displaystyle H^{0}({\rm Gr}(2,4),{\mathbb{Z}})={\mathbb{Z}},~{}H^{2}({\rm
Gr}(2,4),{\mathbb{Z}})={\mathbb{Z}}s_{1},~{}H^{4}({\rm
Gr}(2,4),{\mathbb{Z}})={\mathbb{Z}}s_{1}^{2}\oplus{\mathbb{Z}}s_{2},$
$\displaystyle H^{6}({\rm
Gr}(2,4),{\mathbb{Z}})={\mathbb{Z}}s_{1}s_{2},~{}H^{8}({\rm
Gr}(2,4),{\mathbb{Z}})={\mathbb{Z}}s_{2}^{2}.\hskip 80.0pt$

The geometric meaning of the $s_{i}$ classes is the following.

  * •

The $q_{1}=-s_{1}$ class is the Poincaré dual of the Schubert cycle of lines
in ${\mathbb{C}}{\mathbb{P}}^{3}$ that intersect a given line (the second
entry in (25.7)). To see that, we dualize the short exact sequence of vector
bundles (25.16) to get

$0\to Q^{\vee}\to(\mathcal{O}^{\oplus 4})^{\vee}\to S^{\vee}\to 0.$

Linear functions on ${\mathbb{C}}^{4}$ give sections of the bundle $S^{\vee}$
by looking at their restrictions on the fibers of $S$. Then the Plücker
coordinates give sections of $\Lambda^{2}S^{\vee}$, and one can see that this
Schubert cycle is given by $x_{12}=0$. The Poincaré dual of the divisor that
corresponds to a section of $\Lambda^{2}Q$ is therefore
$c_{1}(S^{\vee})=-c_{1}(S)=-s_{1}$.

  * •

The $s_{2}$ class is the Poincaré dual of the Schubert class of lines inside a
given plane in ${\mathbb{C}}{\mathbb{P}}^{3}$ (the fourth entry). To explain
this, we invoke the general claim the top Chern class of a vector bundle is
equal, under some transversality conditions, to the Poincare dual of the zero
locus of a section. In our case, with a section given by a coordinate on
${\mathbb{C}}^{4}$, and the vanishing locus is that of lines $l$ that lie in
the corresponding plane.

  * •

The $(-s_{1})s_{2}$ class is the Poincaré dual of the Schubert class of lines