MODULI SPACES AND INVARIANT THEORY §0. Syllabus 2 §1

Transcription

MODULI SPACES AND INVARIANT THEORY
JENIA TEVELEV
C ONTENTS
§0. Syllabus
§1. Geometry of lines
§1.1. Grassmannian as a complex manifold.
§1.2. Moduli space or a parameter space?
§1.3. Stiefel coordinates.
§1.4. Plücker coordinates.
§1.5. Grassmannian as a projective variety
§1.6. Second fundamental theorem – relations
§1.7. Hilbert polynomial
§1.8. Enumerative geometry
§1.9. Homework
§2. Algebraic curves and Riemann surfaces
§2.1. Genus
§2.2. Divisors on curves
§2.3. Riemann–Hurwitz formula
§2.4. Riemann–Roch formula and linear systems
§3. Elliptic curves: j-invariant
§3.1. J-invariant
§3.2. Monstrous Moonshine
§3.3. Families of elliptic curves: coarse and fine moduli spaces
§3.4. Homework
§4. Icosahedron, E8 , and quotient singularities
§4.1. Symmetric polynomials, one-dimensional actions
§4.2. A1 -singularity
§4.3. Chevalley–Shephard–Todd theorem
§4.4. Finite generation
§4.5. Basic properties of quotients
§4.6. Quotient singularity 1r (1, a) and continued fractions.
§4.7. Zariski tangent space
§4.8. E8 -singularity
§4.9. Resolution of singularities: cylindrical resolution
§4.10. Resolution of cyclic quotient singularities
§4.11. Homework
§5. Weighted projective spaces
§5.1. First examples
§5.2. Proj (projective spectrum)
§5.3. Abstract algebraic varieties
§5.4. Separatedness
1
2
3
3
4
5
6
8
11
12
14
17
20
22
23
24
25
27
29
31
31
36
41
41
42
42
43
44
47
50
51
54
55
58
62
62
64
65
66
2
JENIA TEVELEV
§5.5. Veronese embedding
§6. Genus 2 curves.
§6.1. Genus 2 curves: analysis of the canonical ring
§6.2. Graded algebra of an ample divisor
§6.3. GIT: Proj quotient
§6.4. Classical invariant theory of a binary sextic
§6.5. Homework
§7. GIT quotients and stability.
§7.1. Algebraic representations of reductive groups
§7.2. Finite Generation Theorem via the unitary trick
§7.3. Surjectivity of the quotient map
§7.4. Separation of orbits
§7.5. Unstable locus: Hilbert–Mumford criterion
§7.6. Hypersurfaces
§7.7. Homework
§8. Jacobians and periods
§8.1. Albanese torus
§8.2. Jacobian
§8.3. Abelian varieties
§8.4. Abel’s Theorem
§8.5. Differentials of the third kind
§8.6. Summation maps
§8.7. Theta-divisor
§8.8. Homework
§9. Torelli theorem
References
66
69
69
71
72
73
74
77
77
80
81
83
85
88
92
95
95
96
98
99
99
101
102
102
104
104
§0. Syllabus
A moduli space appears each time we want to parametrize all geometric objects of some sort. For example, elliptic curves are classified by the
so-called J-invariant, so the moduli space of elliptic curves is a line (with
coordinate J). The Jacobian of a Riemann surface is another example of a
moduli space: it classifies line bundles on a Riemann surface.
The study of moduli spaces is a very old subject and many methods were
developed to understand them: classical invariant theory, geometric invariant theory, period domains and variation of Hodge structures, etc. Every
generation of algebraic geometers contributes something new to the language of Algebraic Geometry and thus to moduli theory. Because of this,
reading modern literature on moduli theory requires deep understanding
of quite advanced topics in Algebraic Geometry (stacks, derived categories,
Mori theory, intersection theory, etc.) The standard approach to studying
moduli spaces is to master these foundations of Algebraic Geometry in
depth first. However, we believe that a lot can be learned about moduli
spaces by using minimal machinery, and in fact it is much easier to study
deep tools after these beautiful examples are digested. So our approach will
3
be to introduce classical and beautiful examples of moduli spaces (approximately one a week) and to develop techniques necessary to understand
them. Along the way we will learn a lot of Algebraic Geometry.
The recommended textbook for this class is An introduction to Invariants
and Moduli by Shigeru Mukai. We won’t follow it closely, but we will borrow heavily from it. Mukai’s contributions to moduli theory are monumental, and his highly original textbook will provide a lot of insight for a
student willing to read it with a pencil in hand. A great advantage of this
book is that fundamentals of Algebraic Geometry are introduced along the
way in a very unorthodox and intuitive way. However, the reader should
be prepared to work hard: some proofs are only sketched.
The prerequisites for this class are graduate algebra and complex analysis, basic theory of manifolds, and a class on Riemann surfaces.
Grading will be based on 6 biweekly homework sets. My philosophy
about grading the graduate topics class is that students who are willing to
work hard on homework sets should be rewarded with an A. I will try a
new system this semester with the following rules. Each homework will
have a two-week deadline and individual problems in the homework will
be worth some points (with a total of about 20-25 points for each homework). The passing standard (for an A) will be 90 points at the end of the
semester. So it will be necessary to solve about 60-75% of each homework
(or solve more in the beginning to get an A early). Homework problems
can be presented in two ways. An ideal method is to come to my office
and explain your solution at the blackboard (either during office hours or
at any other time assuming I am in the office and not very busy). If you
can show me a correct solution at the blackboard, you won’t have to turn
in this problem in the written form. This will save both me and you a lot
of time. If your solution does not work (or if you are out of ideas), I will
give you a lot of hints. Problems not discussed in my office will have to be
written down and turned in at the end of the two-week period.
§1. Geometry of lines
Let’s start with a familiar example of a moduli space. Recall that the
Grassmannian G(r, n) parametrizes r-dimensional linear subspaces of Cn .
For example,
G(1, n) = Pn−1
is a projective space1. Let’s try to understand the next case, G(2, n). The
projectivization of a 2-dimensional subspace U ⊂ Cn is a line l ⊂ Pn−1 , so
in essence G(2, n) is a moduli space of lines in the projective space. Most of
our discussion remains valid for general G(r, n), but the case of G(2, n) is
notationally easier.
§1.1. Grassmannian as a complex manifold. Thinking about G(2, n) as
just a set is boring: we need to introduce some geometry on it. We care
about two flavors of geometry, Analytic Geometry and Algebraic Geometry.
1I will often italicize various words in these lecture notes. If you see something italicized,
make a pause and ask yourself: do I know what this means? I hope that this will help you
to learn the vocabulary faster.
4
JENIA TEVELEV
We won’t need much of either in the beginning and will develop the latter
substantially as we go along. A basic object of analytic geometry is a complex manifold, i.e. a Hausdorff topological space X covered by charts Xi homeomorphic to open subsets of Cn . Coordinate functions on Cn are called local
coordinates in the chart. On the overlaps Xi ∩ Xj we thus have two competing systems of coordinates, and the main requirement is that transition
functions between these coordinate systems are holomorphic. Maps between
complex manifolds are presumed to be holomorphic maps, i.e. maps that are
holomorphic in charts.
Let’s see how this is done for the Grassmannian. Any 2-dimensional
subspace U ⊂ Cn is a row space of a 2 × n matrix A of rank 2. Let Aij
denote the 2 × 2 submatrix of A with columns i and j and let pij = det Aij
be the corresponding minor. Since rank A = 2, we can find some i < j such
that pij 6= 0 (why?) . Then (Aij )−1 A has a form
... ∗ 1 ∗ ... ∗ 0 ∗ ...
(1.1.1)
... ∗ 0 ∗ ... ∗ 1 ∗ ...
1
0
where the i-th column is
, the j-th column is
, and the remaining
0
1
n − 2 columns are arbitrary. Notice that multiplying A by an invertible
2 × 2 matrix on the left does
not change the row space.
n
We cover G(2, n) by 2 charts
Xij = {U ∈ G(2, n) | U is represented by a matrix A like (1.1.1)}.
Geometrically, this chart parametrizes 2-dimensional subspaces that surject onto the coordinate subspace hei , ej i under projection along the complementary coordinate subspace (why?) . Each subspace from Xij is a row
space of a unique matrix (1.1.1), in particular Xij can be identified with
C2(n−2) .
To show that G(2, n) is a complex manifold we have to check that the
transition functions between charts Xij and Xi0 j 0 are holomorphic. Any
2×n matrix that represents a subspace U ∈ Xij ∩Xi0 j 0 has pij 6= 0 and pi0 j 0 6=
0. In the chart Xij , the subspace is represented by a matrix A as in (1.1.1).
In the chart Xi0 j 0 , the subspace is represented by a matrix (Ai0 j 0 )−1 A. The
matrix entries of it depend holomorphically (in fact just rationally) on the
matrix entries of A, thus the transition functions are indeed holomorphic.
In this example (and in general) the structure of a topological space on
X is introduced simultaneously with constructing charts: the subset is declared open iff its intersection with each chart is open. It is easy to check
(why?) that G(2, n) is indeed Hausdorff.
§1.2. Moduli space or a parameter space? All lines in Pn−1 are isomorphic
to P1 and to each other. So G(2, n) is not really a moduli space, but rather a
parameter space: it classifies not geometric objects up to isomorphism but
rather it classifies geometric sub-objects (lines) in a fixed geometric object (projective space). This distinction is mostly philosophical (depends
on how do you decide when two objects are equivalent). Later we will
study Chow varieties and Hilbert schemes: those parametrize all geometric
sub-objects (technically called algebraic subvarieties or subschemes) of Pn .
5
As a rule, parameter spaces are easier to construct than moduli spaces.
To construct an honest moduli space M of geometric objects X, one can
• embed these objects in Pn−1 for some n;
• construct a “parameter space” H for the embedded objects;
• divide H by the equivalence relation (two embedded objects are
equivalent if they are abstractly isomorphic) to get M.
In many cases of interest objects are abstractly isomorphic if and only if
they are projectively equivalent in Pn−1 (i.e. differ by an element of GLn ).
So in effect we will have to construct an orbit space
M = H/ GLn
and a quotient map
H→M
that sends each point to its orbit. For example, G(2, n)/ GLn is a point – all
lines of Pn−1 are abstractly (and projectively!) isomorphic.
Often a more general procedure is necessary:
• first construct a parameter space H of pairs (X, v), where v is some
sort of an extra data on X (often called “marking”). For example, v
can be an embedding X ,→ Pn but often it’s something else.
• then construct a “forgetful” map H → M by “forgetting” marking v. Usually this map is a quotient map for the group action.
In any case, the basic principle is
1.2.1. P RINCIPLE . A good model for a moduli space is provided by an orbit
space for a group action.
So we will have to understand how to construct quotients by group actions. Those techniques are provided by invariant theory – the second component from the title of this course.
1.2.2. R EMARK . For now, the term “quotient map” will have a very crude
set-theoretic meaning: we just require that the fibers are exactly the orbits
for the group action. We are not going to worry (as we should) about the
relationship between geometries on the source and on the target of the quotient map. Later on, when we have more examples to play with, we will
give a much more refined definition of the “quotient map”. Likewise for
now an “orbit space” has simply a set-theoretic meaning: the set of orbits.
§1.3. Stiefel coordinates. To illustrate these ideas, let’s construct the Grassmannian itself as a quotient! We can mark a subspace by its basis: consider
triples (U, v1 , v2 ), where U ⊂ Cn is a subspace with basis {v1 , v2 }. Writing
down v1 , v2 in terms of the standard basis e1 , . . . , en of Cn , we see that these
triples are parametrized by an open subset
Mat02,n ⊂ Mat2,n
of matrices of rank 2 (with rows v1 and v2 ) . This is a very simple space – in
this business an open subset of an affine space is the easiest space you can
possibly hope for!
6
JENIA TEVELEV
Matrix coordinates on Mat02,n in times long gone were known as Stiefel
coordinates on the Grassmannian. The “forgetful” map
Mat02,n → G(2, n)
(1.3.1)
has the following meaning: rank 2 matrices X and X 0 have the same row
space if and only if X = gX 0 for some matrix g ∈ GL2 . This (1.3.1) is a
quotient map for the action of GL2 on Mat02,n by left multiplication.
§1.4. Plücker coordinates. Since we already have a good grasp of the Grassmannian, we don’t really need invariant theory to construct the map (1.3.1).
Nevertheless, let me use this example to explain how to use invariants to
construct quotient maps (and thus moduli spaces). In fact, this will tell us
something new about the Grassmannian.
1.4.1. D EFINITION . We start with a very general situation: let G be a group
acting on a set X. A function f : X → C is called an invariant function if it
is constant along G-orbits, i.e. if
f (gx) = f (x)
for any x ∈ X, g ∈ G.
Invariant functions f1 , . . . , fr form a complete system of invariants if they separate orbits. This means that for any two orbits O1 and O2 , there exists at
least one function fi such that fi |O1 6= fi |O2 .
In this case then the map
F : X → Cr ,
F (x) = (f1 (x), . . . , fr (x))
is obviously a quotient map (onto its image): its fibers are exactly the orbits!
Often we want to have a quotient map with target Pr rather than Cr .
Thus we need the following generalization:
1.4.2. D EFINITION . Fix a homomorphism
χ : G → C∗ .
A function f : X → C is called a semi-invariant of weight χ if
f (gx) = χ(g)f (x)
for any x ∈ X, g ∈ G
(notice that an invariant function is a special case of a semi-invariant of
weight χ = 1). Suppose f0 , . . . , fr are semi-invariants of the same weight χ.
We will call them a complete system of semi-invariants of weight χ if
• for any x ∈ X, there exists a function fi such that fi (x) 6= 0;
• for any two points x, x0 ∈ X not in the same orbit, we have
[f0 (x) : . . . : fr (x)] 6= [f0 (x0 ) : . . . : fr (x0 )].
The first condition means that we have a map
F : X → Pr ,
F (x) = [f0 (x) : . . . : fr (x)],
which is clearly constant along G-orbits:
[f0 (gx) : . . . : fr (gx)] = [χ(g)f0 (x) : . . . : χ(g)fr (x)] = [f0 (x) : . . . : fr (x)].
The second condition means that F is a quotient map onto its image.
7
1.4.3. E XAMPLE . G = GL2 acts by left multiplication on Mat02,n . Consider
the 2×2 minors pij as functions on Mat02,n . It is convenient to set pji := −pij
for j > i. Consider the homomorphism
det : GL2 → C∗ .
1.4.4. P ROPOSITION . The minors pij form a complete system of semi-invariants
on Mat02,n of weight det.
Proof. We have
pij (gA) = det(g)pij (A) for any g ∈ GL2 , A ∈ Mat02,n .
It follows that pij ’s are semi-invariants of weight det. For any A ∈ Mat02,n ,
at least one of the pij ’s does not vanish. So we have a map
n
F : Mat02,n → P( 2 )−1
given by the minors pij .
Now take A, A0 ∈ Mat02,n such that F (A) = F (A0 ). We have to show
that A and A0 are in the same G-orbit. Suppose pij (A) 6= 0, then certainly
pij (A0 ) 6= 0. By twisting A and A0 by some elements of GL2 , we can assume
that both A and A0 have a form (1.1.1). In particular,
pij (A) = pij (A0 ) = 1.
Since F (A) = F (A0 ), we now have
pi0 j 0 (A) = pi0 j 0 (A0 )
for any i0 , j 0 .
Now it’s really easy to see that A = A0 : a key point is a trivial observation that an element in the first (resp. second) row and the k-th column of
A (1.1.1) can be computed as pkj (resp. pik ). Thus A = A0 .
Since we already know that
Mat02,n /GL2 = G(2, n),
this gives an inclusion
n
i : G(2, n) ,→ P( 2 )−1
called the Plücker embedding. The minors pij are in this context called
Plücker coordinates on G(2, n).
1.4.5. R EMARK . A little warning: for now we have only proved that i is
the inclusion of sets. It is also clear that i is a holomorphic map of manifolds: in each chart it is given by 2 × 2 minors of a matrix representing a
2-dimensional subspace in this chart, these minors are obviously holomorphic. To show that this inclusion is an embedding of complex manifolds, a
little extra work is required, see below.
What is a rationale for considering minors pij and not something else as
a complete system of semi-invariants? Well, let’s consider all possible semiinvariants on Mat02,n which are polynomials in 2n matrix entries. In fact, by
continuity, this is the same thing as polynomial semi-invariants on Mat2,n .
Let
O(Mat2,n ) = C[a1i , a2i ]1≤i≤n
8
JENIA TEVELEV
denote the algebra of polynomial functions on Mat2,n . It is easy to see
(why?) that the only holomorphic homomorphisms GL2 (C) → C∗ are powers of the determinant. Let
2
Ri = O(Mat2,n )GL
deti
i
be a subset of polynomial
semi-invariants of weight det
that the
. Notice
t 0
t 0
= t2i . It folscalar matrix
acts on Ri by multiplying it on deti
0 t
0 t
lows that all polynomials in Ri have degree 2i, in particular
Ri = 0 for i < 0,
R0 = C.
We assemble all semi-invariants in one package (algebra of semi-invariants):
M
Ri ⊂ O(Mat2,n ).
R=
i≥0
Since the product of semi-invariants of weights χ and χ0 is a semi-invariant
of weight χ · χ0 , R is a graded subalgebra of O(Mat2,n ).
The following theorem was classically known as the First Fundamental
Theorem of invariant theory.
1.4.6. T HEOREM . The algebra R is generated by the minors pij for 1 ≤ i < j ≤ n.
Thus considering only pij ’s makes sense: all semi-invariants can not separate orbits any more effectively than the generators. We are not going to
use this theorem and the proof. But this raises some general questions:
• is the algebra of polynomial invariants (or semi-invariants) always
finitely generated?
• do these basic invariants separate orbits?
• how to compute these basic invariants?
We will see that the answer to the first question is positive under very general assumptions (Hilbert’s finite generation theorem). The answer to the second question is “not quite” but a detailed analysis of what’s going on is
available (Hilbert–Mumford’s stability and the numerical criterion for it). As
far as the last question is concerned, the generators of the algebra of invariants can be computed explicitly only in a handful of cases – from this
perspective we are lucky that we have Plücker generators.
§1.5. Grassmannian as a projective variety. What is the image of a Plücker
n
embedding G(2, n) ,→ P( 2 )−1 ? Let U be a row space of a matrix
a11 . . . a1n
a21 . . . a2n
and consider a bivector
b = (a11 e1 + . . . + a1n en ) ∧ (a21 e1 + . . . + a2n en ) =
X
pij ei ∧ ej .
i<j
n
Thus if we identify P( 2 )−1 with the projectivization of Λ2 Cn , the map i simply sends a subspace U generated by vectors u, u0 ∈ Cn to u ∧ u0 . Therefore,
the image of i is a subset of decomposable bivectors. Let’s show that this
subset is a projective algebraic variety
9
1.5.1. D EFINITION . Let f1 , . . . , fr ∈ C[x0 , . . . , xn ] be homogeneous polynomials. The vanishing set
X = V (f1 , . . . , fr ) = {x ∈ Pn | f1 (x) = . . . = fr (x) = 0}
is called a projective algebraic variety. For each chart Ui ⊂ Pn (points where
xi 6= 0), X ∩ Ui ⊂ Ui ' An is an affine algebraic variety given by vanishing of
f1 , . . . , fr dehomogenized with respect to xi .
What kind of polynomials vanish along the image of i? Notice that we
have
X
0=b∧b=
(pij pkl − pik pjl + pil pjk )ei ∧ ej ∧ ek ∧ el .
i<j<k<l
Thus, quadratic polynomals
xij xkl − xik xjl + xil xjk
vanish along the image of i. These polynomials are called Plücker relations
1.5.2. P ROPOSITION . i(G(2, n)) is a projective variety given by vanishing of
Plücker relations. The map i is an immersion of complex manifolds.
In particular, we can use this fact to redefine G(2, n) purely algebraically
as a projective algebraic variety in the Plücker projective space defined by
Plücker relations.
Proof. We already know that Plücker relations vanish along the image of i,
so we just have to work out the vanishing set of Plücker relations
n
X = V (xij xkl − xik xjl + xil xjk ) ⊂ P( 2 )−1 .
We can do it in charts. For simplicity, let’s only consider the chart U12 ,
where we have x12 = 1. What are the equations of X ∩ U12 ? Some of them
are
xkl = x1k x2l − x1l x2k , 2 < k < l ≤ n,
(1.5.3)
i.e. any xkl whatsoever is just a minor of the matrix
1 0 −x23 −x24 . . . −x2n
0 1 x13
x14 . . . x1n
It follows that this point of the Plücker vector space is a row space of a
matrix above, and all other Plücker relations in this chart are just formal
consequences of (1.5.3), i.e. X = G(2, n) set-theoretically.
But of course more is true: X12 = X ∩ U12 is defined by equations (1.5.3),
n
which can be interpreted as follows: X12 is embedded in A( 2 )−1 as a graph
of the map
n
n−2
A2(n−2) → A( 2 )−1−2(n−2) = A( 2 ) ,
1 0 a13 a14 . . . a1n
A=
7→ {pkl (A)}3≤k<l≤n .
0 1 a23 a24 . . . a2n
In particular,
X ∩ U12 ' A2(n−2)
and the transition functions between various affine charts of X are exactly
the same as transition functions between charts of G(2, n) (why?) . It follows
that X is a complex manifold isomorphic to G(2, n) via the map i.
10
JENIA TEVELEV
Notice that of course not any projective (or affine) algebraic variety X is
a complex manifold: they are often singular. How can we check if X is a
smooth manifold? For simplicity (by considering charts), we can assume
that X = {f1 = . . . = fr = 0} ⊂ An is an affine variety. Let p ∈ X. Suppose
that we can choose l equations (after reordering, let’s assume that the first
l equations work) such that
• in some complex neighborhood of p, X = {f1 = . . . = fl = 0},
• The rank of the Jacobian matrix
 ∂f1
∂f1 
∂x1 . . . ∂xn

.. 
..
J =  ...
.
. 
∂fl
∂x1
...
∂fl
∂xn
is equal to l.
Then, by the Implicit Function Theorem, X is (locally near p) a complex manifold of dimension n − l with tangent space Ker J. Notice that different
points p ∈ X may (and usually will) require different subcollections of l
equations.
Here is the Algebraic Geometry’s approach. The same variety X can
be defined by different sets of equations, and sometimes they will fail to
detect smoothness of X. For example, if we define the y-axis of A2 by the
equation x2 = 0 (rather than simply x = 0), the Jacobian matrix will have
rank 0 (rather than 1), which reflects the fact that a double line should be
thought as singular at all points.
By Hilbert’s Nullstellensatz,
the ideal of all polynomials that vanish along
√
√
X is the radical I. Let’s suppose that I is already a radical ideal, i.e. I = I.
If I is not a prime ideal then X is reducible, i.e. is a union of two algebraic
varieties (if f g ∈ I but f, g 6∈ I then V (I) = V (I, f ) ∪ V (I, g)). If we want
X to be a smooth manifold then these components of X better be smooth
individually (and don’t intersect). So let’s suppose that I is prime. In this
case X is called an irreducible affine variety.
Let f1 , . . . , fr be generators of I (recall that I has finitely many generators
by Hilbert’s basis theorem) and consider the Jacobian matrix J. For any point
p ∈ X, the kernel of J is called a tangent space Tp X (it is easy to see (why?)
that it does not depend on the choice of generators). Let s be the maximal
possible dimension of Tp X. A point p is called non-singular if dim Tp X = s,
otherwise it is called singular. The set of non-singular (resp. singular) points
is called a smooth locus Xsm (resp. singular locus Xsing ).
If X is irreducible then the coordinate algebra
O(X) = C[x1 , . . . , xn ]/I
is a domain and its quotient field is called the field of rational functions on X,
denoted by C(X). The dimension of X is defined as follows:
dim X = tr.deg.C C(X)
(the transcendence degree). For a proof of the following, see [X, II.1.1].
1.5.4. T HEOREM .
• s = dim X.
11
• Xsm ⊂ X is Zariski-open.
If X is a non-singular algebraic variety then X is also a complex manifold
of the same dimension. To distinguish X the variety from X the complex
manifold, the latter is denoted by X an .
§1.6. Second fundamental theorem – relations. We proved that G(2, n) is
defined by Plücker relations in the Plücker projective space.
We can ask
for more: is it possible to describe all polynomials in n2 variables xij that
vanish along G(2, n)? Algebraically, we consider the homomorphism of
polynomial algebras
ψ : C[xij ]1≤i<j≤n → C[a1i , a2i ]1≤i≤n ,
xij 7→ pij (A)
and we ask: what is the kernel of ψ? (Notice that the image of ψ is equal
to the algebra of GL2 -semi-invariants by the First fundamental theorem of
invariant theory, but we are not going to use this). Let I = Ker ψ. A good
way of thinking about I is that its elements are relations between 2 × 2
minors of a general 2 × n matrix.
1.6.1. T HEOREM (Second fundamental theorem of invariant theory). The kernel of ψ is generated (as an ideal) by Plücker relations
xij xkl − xik xjl + xil xjk
for all fourtuples i < j < k < l.
Proof. The proof consists of several steps.
Step 1. Plücker relations are in I. We already know this, see Proposition 1.5.2. Let I 0 ⊂ I be an ideal generated by the Plücker relations. The
goal is to show that I = I 0 .
Step 2. This is called the straightening law – it was introduced by Alfred Young (who has Young diagrams named after him). We encode each
monomial xi1 j1 . . . xik jk in a Young tableaux
i1 i2 . . . ik
(1.6.2)
j1 j2 . . . jk
The tableaux is called standard if it has increasing rows:
i1 ≤ i2 ≤ . . . ≤ ik
and j1 ≤ j2 ≤ . . . ≤ jk .
In this case we also call the corresponding monomial a standard monomial.
We claim that any monomial in xij ’s is equivalent modulo I 0 (i.e. modulo
Plücker relations) to a linear combination of standard monomials. Indeed,
suppose that x = xi1 j1 . . . xik jk is not standard. By reordering the variables
in x, we can assume that
i1 ≤ i2 ≤ . . . ≤ ik
and if il = il+1 for some l then jl ≤ jl+1 . Let l be the largest index such
that jl > jl+1 . We argue by induction on l that x is a linear combination of
standard monomials. We have
il < il+1 < jl+1 < jl .
Consider the Plücker relation
xil jl xil+1 jl+1 = −xil il+1 xjl+1 jl + xil jl+1 xil+1 jl
mod I 0
12
JENIA TEVELEV
Since the tableaux
il
il+1
jl+1
jl
and
il
jl+1
il+1
jl
are standard, once we substitute −xil il+1 xjl+1 jl +xil jl+1 xil+1 jl for xil jl xil+1 jl+1
in x we will get a linear combination of two monomials which can be
both written as linear combinations of standard monomials by inductive
assumption.
Step 3. Finally, we claim that standard monomials are linearly independent modulo I, i.e.
{ψ(x) | x is a standard monomial}
is a linearly independent subset of C[a1i , a2i ]1≤i≤n . A cool idea is to order
the variables as follows:
a11 < a12 < . . . < a1n < a21 < a22 < . . . < a2n
and to consider the corresponding lexicographic ordering of monomials in
C[a1i , a2i ]1≤i≤n . For any polynomial f , let init(f ) denote the initial monomial
of f (i.e. the smallest monomial for lexicographic ordering). Notice that
init(f ) is multiplicative:
init(f g) = init(f ) init(g)
(1.6.3)
for any (non-zero) polynomials. We have
init pij = a1i a2j ,
and therefore
init(ψ(x)) = init(pi1 j1 . . . pik jk ) = a1i1 a1i2 . . . a1ik a2j1 a2j2 . . . a2jk .
Notice that a standard monomial x is completely determined by init(ψ(x)).
However, if the set of polynomials {ψ(x)} is linearly dependent, then at
least some of the initial monomials (namely, the smallest initial monomials)
should cancel each other.
A lot of calculations in Algebraic Geometry can be reduced to manipulations with polynomials just like in the proof above. It is important to
master these (quintessential algebraic!) techniques.
§1.7. Hilbert polynomial. It is rarely the case that equations of the moduli
space are known so explicitly as in the case of the Grassmannian. But some
numerical information about these equations is often available, as we now
explain.
We start with a general situation: let X ⊂ Pn be a projective variety and
let I ⊂ C[x0 , . . . , xn ] be a homogeneous ideal of polynomials that vanish on
X. The algebra R = C[x0 , . . . , xn ]/I is known as a homogeneous coordinate
algebra of X. Note that R is graded (by degrees of polynomials):
M
R=
Rj , R0 = C,
j≥0
and R is generated by R1 as an algebra. The function
h(k) = dim Rk
13
is called the the Hilbert function of X. Notice that knowing h(k) is equivalent
to knowing dim Ik for any k:
n+k
h(k) + dim Ik =
(why?)
k
We have the following fundamental theorem:
1.7.1. T HEOREM . There exists a polynomial H(t) (called Hilbert polynomial) with
h(k) = H(k) for k 0.
This polynomial has degree r = dim X and has a form
d r
t + (lower terms),
r!
where d is the degree of X, i.e. the number of points in the intersection of X with a
general projective subspace of codimension r (a subspace is general if it intersects
G(2, n) transversally in all intersection points).
We will prove this theorem later along with other important properties
of the Hilbert function. But now let’s use it!
1.7.2. P ROPOSITION . Let n ≥ 3. The Hilbert function of G(2, n) in the Plücker
embedding is
n+k−1 2
n+k
n+k−2
h(k) =
−
.
(1.7.3)
k+1
k−1
k
The degree of G(2, n) in the Plücker embedding is the Catalan number
2n − 4
1
= 1, 2, 5, 14, 42, 132, . . .
n−1 n−2
Proof. During the proof of the Second Fundamental Theorem 1.6.1 we have
established that h(k) is equal to the number of standard monomials of degree k, i.e. to the number of standard tableaux with k columns. Let Nl be
the number of non-decreasing sequences 1 ≤ i1 ≤ . . . ≤ il ≤ n. Then we
have
n+l−1
Nl =
:
l
this is just the number of ways to choose l objects from {1, . . . , n} with
repetitions (so it is for example equal to the dimension of the space of polynomials in n variables of degree l). The number of tableaux
i1 i2 . . . ik
, 1 ≤ i1 ≤ . . . ≤ ik ≤ n, 1 ≤ j1 ≤ . . . ≤ jk ≤ n
j1 j2 . . . jk
(but without the condition that il < jl for any l) is clearly equal to
n+k−1 2
2
Nk =
.
k
Now we have to subtract the number of non-standard tableaux. We claim
that there is a bijection between the set of nonstandard tableaux and the set
of pairs (A, B), where A is a non-decreasing sequence of length k + 1 and
B is a non-decreasing sequence of length k − 1. This will prove (1.7.3).
14
JENIA TEVELEV
Suppose that l is the number of the first column where il ≥ jl . Then we
can produce two sequences:
j1 ≤ . . . ≤ jl ≤ il ≤ . . . ≤ ik
of length k + 1 and
i1 ≤ . . . ≤ il−1 ≤ jl+1 ≤ jk
of length k − 1. In an opposite direction, suppose we are given sequences
i1 ≤ . . . ≤ ik+1
and j1 ≤ . . . ≤ jk−1 .
Let l be the minimal index such that il ≤ jl and take a tableaux
j1 . . . jl−1 il+1 il+2 . . . ik
i1 . . . il−1 il
jl . . . jk−1
If il > jl for any l ≤ k − 1, then take the tableaux
j1 . . . jk−1 ik
.
i1 . . . ik−1 ik+1
After some manipulations with binomial coefficients, (1.7.3) can be rewritten as
1
(k + n − 1)(k + n − 2)2 . . . (k + 2)2 (k + 1).
(n − 1)!(n − 2)!
1
This is a polynomial in k of degree 2n−4 with a leading coefficient (n−1)!(n−2)!
.
Since the degree of G(2, n) is equal to (2n−4)! multiplied by the leading coefficient of h(k), we see that this degree is indeed the Catalan number. §1.8. Enumerative geometry. Why do we need moduli spaces? One reason is that their geometry reflects delicate properties of parametrized geometric objects. As an example, let’s try to relate the degree of the Grassmannian (i.e. the Catalan number) to geometry of lines.
1.8.1. T HEOREM . The number of lines in Pn−1 that intersect 2n − 4 general codimension 2 subspaces is equal to the Catalan number
2n − 4
1
n−1 n−2
For example, there is only one line in P2 passing through 2 general points,
2 lines in P3 intersecting 4 general lines, 5 lines in P4 intersecting 6 general
planes, and so on. This is a typical problem from the classical branch of
Algebraic Geometry called enumerative geometry, which was described by
H. Schubert (around 1870s) as a field concerned with questions like: How
many geometric figures of some type satisfy certain given conditions? If
these figures are lines (or projective subspaces), the enumerative geometry
is nowdays known as Schubert calculus, and is more or less understood. Recently enumerative geometry saw a renaissance (Gromov–Witten invariants,
etc.) due to advances in moduli theory.
Proof. The degree is equal to the number of points in
G(2, n) ∩ L,
15
n
where L ⊂ P( 2 )−1 is a general subspace of codimension 2n − 4, i.e. a subspace that intersects G(2, n) transversally in all intersection points. In other
words, it is the number of points in the intersection
G(2, n) ∩ H1 ∩ . . . ∩ H2n−4 ,
where Hi ’s are hyperplanes, as long as this intersection is transversal.
Notice that the set of lines intersecting a fixed codimension 2 subspace
can be described as the intersection with a hyperplane
D = G(2, n) ∩ H.
For example, lines intersecting W = he3 , . . . , en i are exactly the lines that
do not surject onto he1 , e2 i when projected along W . This is equivalent to
vanishing of the Plücker coordinate p12 . So D is exactly the complement of
the chart U12 ! It is called a special Schubert variety.
We can describe D explicitly by writing down the minor p12 = 0 in other
charts Xij : in most charts (when i > 2) D is a quadric of rank 4, in particular it is a singular hypersurface. This looks a bit worrisome for us because it implies that H is not everywhere transversal to G(2, n), (because
at transversal intersection points the intersection is non-singular). In particular, H is not really a general hyperplane (general hyperplanes intersect
G(2, n) everywhere transversally by Bertini’s Theorem). However, at any
point p of a smooth locus D0 ⊂ D the hyperplane H is transversal to G(2, n)
and the intersection of tangent spaces
Tp D ∩ Tp G(2, n) = Tp D.
Notice that any codimension 2 subspace in Pn−1 is GLn -equivalent to W .
So the claim that we have to check is that if g1 , . . . , g2n−4 ∈ G = GLn are
sufficiently general group elements then any point in
g1 D ∩ . . . ∩ g2n−4 D
is
• away from the singular locus of each gi D;
• a transversal intersection point of gi D’s.
Quite remarkably, the proof relies only on the fact that GLn acts transitively on G(2, n) and on nothing else. It is known as the Kleiman–Bertini
transversality argument. But first we have to explain a very powerful technique in algebraic geometry called dimension count.
1.8.2. Firstly, let’s discuss regular morphisms and rational maps between algebraic varieties. This is very straightforward but there are several delicate
points. We start with an affine case: suppose we have affine varieties
X ⊂ An
and Y ⊂ Am .
A regular function on X is a restriction of a polynomial function. Regular
functions form a coordinate algebra
O(X) = C[x1 , . . . , xn ]/I(X).
A morphism (or a regular morphism)
f : X → Am
16
JENIA TEVELEV
is a map given by m regular functions f1 , . . . , fm ∈ O(X). If the image
lies in Y ⊂ Am then we have a morphism X → Y . It defines a pull-back
homomorphism of coordinate algebras f ∗ : O(Y ) → O(X) and is completely
determined by it. An isomorphism of algebraic varieties is a morphism that has
an inverse. This happens if and only if the pullback f ∗ is an isomorphism
of algebras.
Now suppose, in addition, that X is irreducible. Suppose f ∈ C(X) is
a rational function. Recall that this means that f is a ratio of two regular
functions, and so we can write f = p/q, where p and q are some polynomials. One has to be a bit careful: the presentation f = p/q is not unique
in two different ways. A minor issue is that of course polynomial p and q
are defined only up to the ideal I(X), so one should think about them as
elements of O(X). A more serious issue is that O(X) is rarely a UFD, and
as a result there is no “canonical” way to write down a fraction. Let x ∈ X
be a point. We say that f is defined at x if it can be written as a ratio of two
regular functions such that q(x) 6= 0. The set of points where f is defined is
obviously a Zariski open subset of X.
Suppose we have m rational functions f1 , . . . , fm such that if all of them
are defined at x ∈ X then (f1 (x), . . . , fm (x)) ∈ Y . In this case we say that
we have a rational map
f : X 99K Y
(so a rational map is not everywhere defined). Here is a good exercise on
comparing definitions of regular and rational maps:
1.8.3. L EMMA . If a rational map f : X 99K Y is everywhere defined then in fact
f is a regular morphism.
Proof. It is enough to check that an everywhere defined rational function
is regular. For any point x ∈ X, choose a presentation f = px /qx with
qx (x) 6= 0. Thus X is covered by principal Zariski open sets
D(qx ) := {a ∈ X | qx 6= 0}.
Consider the ideal I = (qx ) ⊂ O(X) generated by all denominators and
choose its finite basis q1 , . . . , qr out of them. These functions don’t have any
common zeros on X, and therefore I = O(X) by Hilbert’s Nullstellensatz.
Thus we can write
1 = a1 q1 + . . . + ar qr .
It follows that
f = a1 f q1 + . . . + ar f qr = a1 p1 + . . . + ar pr
is a regular function.
Finally, let’s discuss rational maps of projective varieties. Even more
generally, a (Zariski) open subset U of a projective variety X is called a
quasi-projective variety. All algebraic varieties appearing in this course will
be quasi-projective (of course we have not really defined anything more
general at this point). If x ∈ U ⊂ X ⊂ Pn and f = P/Q is a rational
function in x0 , . . . , xn such that deg P = deg Q and Q(x) 6= 0 then f defines
a rational function on U regular at x. (We can also just work in the affine
chart An then there is no assumptions on degrees). A rational function is
17
regular on U if it is regular at any point of U . A rational map U 99K An is a
mapping with rational components.
Finally, suppose that U ⊂ Pn and V ⊂ Pm are quasi-projective varieties.
A regular morphism f : U → V is a map such that for any point x ∈ U
and y = f (x) there exists a Zariski neighborhood U 0 of x and an affine chart
y ∈ Am such that the induced map U 0 → Am is regular.
Here is an important theorem:
1.8.4. T HEOREM ([X, 1.6.3]). Let f : X → Y be a regular map between irreducible quasiprojective varieties. Suppose that f is surjective, dim X = n, and
dim Y = m. Then m ≤ n and
• dim F ≥ n − m for any irreducible component F of any fibre f −1 (y),
y ∈Y.
• there exists a non-empty Zariski-open subset U ⊂ Y such that dim f −1 (y) =
n − m for any y ∈ U .
In other words, dim f −1 (y) is an upper-semicontinuous function on Y .
Sketch of the Kleiman–Bertini transversality argument. The first part is a dimension count: denoting by Z the singular locus of D, consider a subset
W = {g1 z = g2 d2 = . . . = g2n−4 d2n−4 } ⊂ (G × Z) × (G × D) × . . . × (G × D)
Clearly, g1 and z can be arbirary, as are di for i ≥ 2, and then for each gi
with i ≥ 2 we have dim G(2, n) = 2n − 4 independent conditions. So
dim W = (2n−4) dim G+dim Z+(2n−5)(2n−5)−(2n−5)(2n−4) < (2n−4) dim G.
It follows that the projection
W → G × ... × G
(2n − 4 times)
(1.8.5)
has empty general fibers, i.e. the general translates of D intersect away from
their singular points.
So we can throw away the singular locus and assume that D is nonsingular (but not compact now). The rest of the argument repeats itself but
now we have to analyze tangent spaces a little bit, which is a part of the
argument that we will skip. Consider a subset
W = {g1 d1 = g2 d2 = . . . = g2n−4 d2n−4 } ⊂ (G×D)×(G×D)×. . .×(G×D).
Now dim W = (2n − 4) dim G. A little reflection shows that W is also nonsingular. So generic fibers of the projection (1.8.5) are either empty or a
bunch of non-critical points. A little local calculation (that we skip) shows
that this is equivalent to the transversality of the corresponding translates
g1 D, . . . , g2n−4 D.
§1.9. Homework. Problem 1. (a) Let L1 , L2 , L3 ⊂ P3 be three general lines.
Show that there exists a unique quadric surface S ⊂ P3 containing them all
and that lines that intersect L1 , L2 , L3 are exactly the lines from the ruling
of S. Try to be as specific as possible about the meaning of the word “general”. (b) Use the previous part to give an alternative proof of the fact that
4 general lines in P3 have exactly two common transversals (2 points).
18
JENIA TEVELEV
n
Problem 2. Show that the Plücker vector space C( 2 ) can be identified
with the space of skew-symmetric n × n matrices and G(2, n) with the projectivization of the set of skew-symmetric matrices of rank 2. Show that the
Plücker relations in this language are 4 × 4 Pfaffians (1 point).
Problem 3. For any line L ⊂ P3 , let [L] ∈ C6 be the corresponding Plücker
vector. The Grassmannian G(2, 4) ⊂ P5 is a quadric, and therefore can be
described as the vanishing set of a quadratic form Q, which in turn has an
associated inner product such that Q(v) = v · v. Describe this inner product
and show that [L1 ]·[L2 ] = 0 if and only if lines L1 and L2 intersect (1 point).
Problem 4. 4 In the notation of the previous problem, show that five lines
L1 , . . . , L5 have a common transversal if and only if


0
[L1 ] · [L2 ] [L1 ] · [L3 ] [L1 ] · [L4 ] [L1 ] · [L5 ]
[L2 ] · [L1 ]
0
[L2 ] · [L3 ] [L2 ] · [L4 ] [L2 ] · [L5 ]


0
[L3 ] · [L4 ] [L3 ] · [L5 ]
det 
[L3 ] · [L1 ] [L3 ] · [L2 ]
=0
[L4 ] · [L1 ] [L4 ] · [L2 ] [L4 ] · [L3 ]
0
[L4 ] · [L5 ]
[L5 ] · [L1 ] [L5 ] · [L2 ] [L5 ] · [L3 ] [L5 ] · [L4 ]
0
(4 points)
Problem 5. Prove (1.6.3). (1 point)
Problem 6. Let X ⊂ Pn be an irreducible hypersurface of degree d (i.e.
a vanishing set of an irreducible homogeneous polynomial of degree d).
Compute its Hilbert polynomial (1 point).
Problem 7. Let X ⊂ Pn be a hypersurface and let FX ⊂ G(2, n) be the
subset of lines contained in X. Show that FX is a projective algebraic variety (2 points).
Problem 8. (a) For any point p ∈ P3 (resp. plane H ⊂ P3 ) let Lp ⊂ G(2, 4)
(resp. LH ⊂ G(2, 4)) be a subset of lines containing p (resp. contained in H).
Show that each Lp and LH is isomorphic to P2 in the Plücker embedding
of G(2, 4). (b) Show that any P2 ⊂ G(2, 4) has a form Lp or LH for some p
or H. (3 points)
Problem 9. Consider the d-th Veronese map
P1 → Pd ,
[x : y] 7→ [xd : xd−1 y : . . . : y d ],
and its image, the rational normal curve. (a) Show that this map is an embedding of complex manifolds. (b) Show that the ideal of the rational normal curve is generated by 2 × 2 minors of the matrix
z0 z1 . . . zn−1
det
,
z 1 z 2 . . . zn
and compute its Hilbert polynomial. (2 points)
Problem 10. Consider the Segre map
2 −1
Pn−1 × Pn−1 → Pn
= P(Matnn ),
([x1 : . . . : xn ], [y1 : . . . : yn ]) 7→ [x1 y1 : . . . : xi yj : . . . : xn yn ]
(a) Show that this map is an embedding of complex manifolds. (b) Show
that the ideal of the Segre variety in P(Matnn ) is generated by 2 × 2 minors
aij akl − ail akj (c) Compute the Hilbert polynomial of the Segre variety. (d)
Compute the degree of the Segre variety. (3 points)
19
Problem 11. In the notation of the previous problem, give a geometric interpretation of the degree of the Segre variety in the spirit of Theorem 1.8.1.
What is the analogue of a special Schubert variety? (1 point)
Problem 12. Consider 4 lines L1 , L2 , L3 , L4 ⊂ P3 . Suppose no three of
them lie on a plane. Show that if 5 pairs of lines Li , Lj intersect then the
sixth pair of lines intersects as well. (1 point)
Problem 13. Check a “little local calculation” at the end of the proof of
Theorem 1.8.1. (3 points)
Problem 14. Let I ⊂ R = C[x0 , . . . , xn ] be a homogeneous ideal and let
V (I) ⊂ Pn be the corresponding projective variety. (a) Show that V (I) is
empty if and only if there exists s > 0 such that I contains all monomials
of degree s. (b) Show that there exists an inclusion-reversing bijection between projective subvarieties of Pn and radical homogeneous ideals of R
different from R+ := (x0 , . . . , xn ). (2 points)
Problem 15. An alternative way of thinking about a 2 × n matrix
x11 . . . x1n
X=
x21 . . . x2n
is that it gives n points p1 , . . . , pn in P1 (with homogeneous coordinates
[x11 : x21 ], . . ., [x1n : x2n ]), at least as soon as X has no zero columns.
Suppose n = 4 and consider the rational normal curve (twisted cubic)
f : P1 ,→ P3 . (a) Show that points f (p1 ), . . . , f (p4 ) lie on a plane if and
only
 3

x11 x211 x21 x11 x221 x321
x3 x2 x22 x12 x2 x3 
12
12
22
22 
F (X) = det 
x313 x213 x23 x13 x223 x323  = 0.
x314 x214 x24 x14 x224 x324
(b) Show using the first fundamental theorem of invariant theory that F (X)
is a polynomial in 2 × 2 minors of the matrix X. (c) Do the same thing
without using the first fundamental theorem (3 points).
20
JENIA TEVELEV
§2. Algebraic curves and Riemann surfaces
After the projective line P1 , the easiest algebraic curve to understand is
an elliptic curve (Riemann surface of genus 1). Let
M1 = {isom. classes of elliptic curves}.
We are going to assign to each elliptic curve a number, called its j-invariant
and prove that
M1 = A1j .
So as a space M1 ' A1 is not very interesting. However, understanding A1
as a moduli space of elliptic curves leads to some breath-taking mathematics. More generally, we introduce
Mg = {isom. classes of smooth projective curves of genus g}
and
Mg,n = {isom. classes of curves C of genus g with points p1 , . . . , pn ∈ C}.
We will return to these moduli spaces later in the course. But first let us
recall some basic facts about algebraic curves = compact Riemann surfaces.
We refer to [G] and [Mi] for a rigorous and detailed exposition.
The theory of algebraic curves has roots in analysis of Abelian integrals.
An easiest example is the elliptic integral: in 1655 Wallis began to study the
arc length of an ellipse (X/a)2 + (Y /b)2 = 1. The equation for the ellipse
can be solved for Y :
p
Y = (b/a) (a2 − X 2 ),
and this can easily be differentiated to find
−bX
Y0 = √
.
a a2 − X 2
Rp
1 + (Y 0 )2 dX for the arc length.
This is squared and put into the integral
Now the substitution x = X/a results in
Z r
1 − e2 x2
s=a
dx,
1 − x2
p
between the limits 0 and X/a, where e = 1 − (b/a)2 is the eccentricity.
This is the result for the arc length from X = 0 to X/a in the first quadrant,
beginning at the point (0, b) on the Y -axis. Notice that we can rewrite this
integral as
Z
Z
a − ae2 x2
p
dx = P (x, y) dx,
(1 − e2 x2 )(1 − x2 )
where P (x, y) is a rational function and y is a solution of the equation
y 2 = (1 − e2 x2 )(1 − x2 ).
This equation defines an elliptic curve! y is an example of an algebraic function. Namely, an algebraic function y = y(x) is a solution of the equation
y n + a1 (x)y n−1 + . . . + an (x) = 0,
(2.0.1)
where ai (x) ∈ C(x) are rational functions (ratios of polynomials). Without
loss of generality, we can assume that this equation is irreducible over C(x).
21
p
√
For example, we can get nested radicals y(x) = 3 x3 − 7x x, although
after Abel and Galois we know that not any algebraic function is a nested
radical (for n ≥ 5)! An Abelian integral is the integral of the form
Z
P (x, y) dx
where P (x, y) is some rational function. All rational functions P (x, y) form
a field K, which is finitely generated and of transcendence degree 1 over
C (because x and y are algebraically dependent). And vice versa, given a
field K such that tr.deg.C K = 1, we can let x be an element transcendent
over C. Then K/C(x) is a finitely generated, algebraic (hence finite), and separable (because we are in characteristic 0) field extension. By a theorem on the
primitive element, we have K = C(x, y), where y is a root of an irreducible
polynomial (2.0.1). Notice that of course there are infinitely many choices
for x and y, thus the equation (2.0.1) is not determined by the field extension. It is not important from the perspective of computing integrals either
(we can always do u-substitutions).
So on a purely algebraic level we can study
isomorphism classes of f.g. field extensions K/C with tr.deg.C K = 1.
Clearing denominators in (2.0.1) gives an irreducible affine plane curve
C = {f (x, y) = 0} ⊂ A2
and its projective completion, an irreducible plane curve in P2 . Recall that the
word curve here means “of dimension 1”, and dimension of an irreducible
affine or projective variety is by definition the transcendence degree of the
field of rational functions C(C). So we can restate our moduli problem as
understanding
birational equivalence classes of irreducible plane curves.
Here we use the following definition
2.0.2. D EFINITION . Irreducible (affine or projective) algebraic varieties X
and Y are called birationally equivalent if their fields of rational functions
C(X) and C(Y ) are isomorphic.
More generally, we can consider an arbitrary irreducible affine or projective curve C ⊂ An or C ⊂ Pn :
birational equivalence classes of irreducible algebraic curves.
This gives the same class of fields, so we are not gaining any new objects.
Geometrically, for any such curve a general linear projection
Pn 99K P2
is birational onto its image.
Let us remind some basic facts related to regular maps (morphisms) and
rational maps (see lectures on the Grassmannian 1.8.2 for definitions):
2.0.3. T HEOREM ([X, 2.3.3]). If C is a smooth curve and f : C → Pn is a
rational map then f is regular. More generally, if X is a smooth algebraic variety and f : X → Pn is a rational map then the indeterminancy locus of f has
codimension 2.
22
JENIA TEVELEV
2.0.4. T HEOREM ([X, 1.5.2]). If X is a projective variety and f : X → Pn is a
regular morphism then f (X) is closed (i.e. also a projective variety).
2.0.5. T HEOREM . For any algebraic curve C, there exists a smooth projective
curve C 0 birational to C.
Taken together, these facts imply that our moduli problem can be rephrased
as the study of
isom. classes of smooth projective algebraic curves.
2.0.6. R EMARK . Theorem 2.0.5 is proved in [G] by
• take a plane model C ⊂ P2 (by projecting Pn 99K P2 ).
• compute the normalization C 0 → C.
Construction of the normalization in [G] is transcendental: one first constructs C 0 as a compact Riemann surface and then invokes a general fact
(see below) that it is in fact a projective algebraic curve. Notice however
that there exist purely algebraic approaches to desingularization by either
(a) algebraic normalization (integral closure in the field of fractions) [X,
2.5.3] or (b) blow-ups [X, 4.4.1] .
The analytic approach is to consider Riemann surfaces instead of algebraic curves. It turns out that this gives the same moduli problem:
biholomorphic isom. classes of compact Riemann surfaces.
It is easy to show that a smooth algebraic curve is a compact Riemann surface. It is harder but not too hard to show that a holomorphic map between
two smooth algebraic curves is in fact a regular morphism, for example ant
meromorphic function is in fact a rational function. But a really difficult
part of the theory is to show that any compact Riemann surface is an algebraic curve. It is hard to construct a single meromorphic function, but
once this is done the rest is easy. This is done by analysis: to construct
a harmonic function on a Riemann surface one (following Klein and Riemann): “This is easily done by covering the Riemann surface with tin foil...
Suppose the poles of a galvanic battery of a given voltage are placed at the
points A1 and A2 . A current arises whose potential u is single-valued, continuous, and satisfies the equation ∆u = 0 across the entire surface, except
for the points A1 and A2 , which are discontinuity points of the function."
A modern treatment can be found in [GH], where a much more general
Kodaira embedding theorem is discussed.
§2.1. Genus. The genus g of a smooth projective algebraic curve can be
computed as follows:
• topologically: the number of handles.
• analytically: the dimension of the space of holomorphic differentials.
• algebraically: the dimension of the space of rational differentials without poles ω = a dx, where a, x are rational functions on C.
One also has the following genus formula:
2g − 2 = (number of zeros) − (number of poles)
of any meromorphic (=rational) differential ω.
(2.1.1)
23
For example, a form ω = dx on P1 at the chart x = 1/y at infinity is
dx = d(1/y) = −(1/y 2 )dy.
So it has no zeros and a pole of order 2 at infinity, which agrees with (2.1.1).
A smooth plane curve C ⊂ P2 of degree d has genus
g=
(d − 1)(d − 2)
2
(2.1.2)
(more generally, if C has only nodal singularities then g = (d−1)(d−2)
− δ,
2
where δ is the number of nodes). There is a nice choice of a holomorphic
form on C: suppose C ∩ A2x,y is given by the equation f (x, y) = 0. Differentiating this equation shows that
dx
dy
=−
fy
fx
along C, where the first (resp. second) expression is valid at points where x
(resp. y) is a holomorphic coordinate. This gives a non-vanishing holomorphic form ω on C ∩A2 . A simple calculation shows that ω has zeros at points
at infinity each of multiplicity d − 3. Combined with (2.1.1), this gives
2g − 2 = d(d − 3),
which is equivalent to (2.1.2).
§2.2. Divisors on curves. A divisor D is just a linear combination
points Pi ∈ C with integer multiplicities. Its degree is defined as
X
deg D =
ai .
P
ai Pi of
If f is a rational (=meromorphic) function on C, we can define its divisor
X
(f ) =
ordP (f )P,
P ∈C
where ordP (f ) is the order of zeros (or poles) of f at P . Analytically, if z is
a holomorphic coordinate on C centered at P then near P
f (z) = z n g(z),
where g(z) is holomorphic and does not vanish at p. Then ordP (f ) = n.
Algebraically, instead of choosing a holomorphic coordinate, we choose a
local parameter, i.e. a rational function z regular at P , z(P ) = 0, and such
that any rational function f on C can be written (uniquely) as
f = z n g,
where g is regular at P and does not vanish there (see [X, 1.1.5])2. For example, we can choose an affine chart where the tangent space TP C surjects
onto one of the coordinate axes. The corresponding coordinate is then a local parameter at P (of course this is also exactly how one usually introduces
a local holomorphic coordinate on a Riemann surface).
2This is an instance of a very general strategy in Algebraic Geometry: if there is some
useful analytic concept (e.g. a holomorphic coordinate) that does not exist algebraically,
one should look for properties (e.g. a factorization f = z n g above) that we want from
this concept. Often it is possible to find a purely algebraic object (e.g. a local parameter)
satisfying the same properties.
24
JENIA TEVELEV
We can define the divisor of a meromorphic form ω in a similar way:
X
K = (ω) =
ordP (ω)P,
P ∈C
where if z is a holomorphic coordinate (or a local parameter) at P then
we can write ω = f dz and ordP (ω) = ordP (f ). This divisor is called the
canonical divisor. So we can rewrite (2.1.1) as
deg K = 2g − 2.
§2.3. Riemann–Hurwitz formula. Suppose f : C → D is a non-constant
map of smooth projective algebraic curves. Its degree deg f can be interpreted as follows:
• topologically: number of points in the preimage of a general point.
• algebraically: degree of the induced field extension C(C)/C(D),
where C(D) is embedded in C(C) by pull-back of functions f ∗ .
It is easy to define a refined version with multiplicities: suppose P ∈ D and
let f −1 (P ) = {Q1 , . . . , Qr }. If z is a local parameter at P then
deg f =
r
X
ordQi f ∗ (z)
i=1
does not depend on P . In particular, a rational function f on C can be
thought of as a map C → P1 , Its degree is equal to the number of zeros
(resp. to the number of poles) of f , and in particular
deg(f ) = 0
for any f ∈ k(C).
A point P is called a branch point if ordQ f ∗ (z) > 1 for some Q ∈ f −1 (P ),
where z is a local parameter at P . In this case Q is called a ramification point
and eQ = ordQ f ∗ (z) is called a ramification index. So if t is a local parameter
at Q then f ∗ (z) = teQ g, where g is regular at Q and g(Q) 6= 0. Analytically,
one can compute a branch of the eQ -th root of g and multiply t by it: this
is often phrased by saying that a holomorphic map of Riemann surfaces
locally in coordinates has a form t 7→ z = te , e ≥ 1.
If ω is a meromorphic form on D without zeros or poles at branch points
then each zero or pole of ω contributes to deg f zeros or poles of f ∗ ω. In
addition, the formula
f ∗ (dz) = d(teQ g) = eQ teQ −1 g dt + teQ dg
shows that each ramification point will also be a zero of f ∗ ω of order eQ −1.
This gives a Riemann–Hurwitz formula
2.3.1. T HEOREM (Riemann–Hurwitz).
X
KC = f ∗ KD +
(eQ − 1)[Q].
Q∈C
and comparing the degrees and using (2.1.1),
2g(C) − 2 = deg f [2g(D) − 2] +
X
Q∈C
(eQ − 1).
25
§2.4. Riemann–Roch formula and linear systems. Finally, we have the
most important
2.4.1. T HEOREM (Riemann–Roch). For any divisor D on C, we have
l(D) − i(D) = 1 − g + deg D,
where
l(D) = dim L(D),
where L(D) = {f ∈ C(C) | (f ) + D ≥ 0}
and
i(D) = dim K 1 (D),
where K 1 (D) = {meromorphic forms ω | (ω) ≥ D}.
Let’s look at some examples. If D = 0 then i(D) = g: indeed K 1 (0)
is the space of holomorphic differentials and one of the characterizations
of the genus is that it is the dimension of the space of holomorphic differentials. On the other hand, l(D) = 1 as the only rational functions regular everywhere are constants. Analytically, this is Liouville’s Theorem for
Riemann surfaces (see also the maximum principle for harmonic functions).
Algebraically, this is
2.4.2. T HEOREM . If X is an irreducible projective variety then the only functions
regular on X are constants.
Proof. A regular function is also a regular morphism X → A1 . Composing
it with the inclusion A1 ,→ P1 gives a regular morphism f : X → P1 such
that f (X) ⊂ A1 . But by Theorem 2.0.4, f (X) must be closed in P1 , thus
f (X) must be a point.
One way or another, if D = 0 then we get a triviality 1 − g = 1 − g. If, on
the other hand, D = K then the RR gives (2.1.1).
2.4.3. E XAMPLE . Suppose g(C) = 0. Let D = P be a point. Then RR gives
l(P ) = i(P ) + 2 ≥ 2.
It follows that L(D) contains a non-constant function f with a unique pole
at P . It gives an isomorphism f : C ' P1 .
The last example shows the most common way of using Riemann–Roch.
We define a linear system of divisors
|D| = {(f ) + D | f ∈ L(D)}.
A standard terminology here is that divisors D and D0 are called linearly
equivalent if
D − D0 = (f ) for some f ∈ k(C).
A divisor D0 is called effective if D0 ≥ 0, (i.e. all coefficients of D0 are positive). So a linear system |D| consists of all effective divisors linearly equivalent to D. Choosing a basis f0 , . . . , fr of L(D) gives a map
φD : C → Pr ,
φD (x) = [f0 , . . . , fr ].
Since C is a smooth curve, this map is regular. More generally, we can
choose a basis of a linear subspace in LD and define a similar map. It called
a map given by an incomplete linear system. In fact any map φ : C → Pr is
given by an incomplete linear system as soon as C is non-degenerate, i.e. if
26
JENIA TEVELEV
φ(C) is not contained in a projective subspace of Pr . It can be obtained as
follows: Any map φ is obtained by choosing rational functions
f0 , . . . , fr ∈ k(C).
Consider their divisors (f0 ), . . . , (fr ) and let D be their common denominator. Then, clearly,
f0 , . . . , fr ∈ L(D).
Moreover, in this case divisors (f0 ) + D, . . . , (fr ) + D have very simple
meaning: they are just pull-backs of coordinate hyperplanes in Pr . Indeed,
suppose h is a local parameter at a point P ∈ C and suppose that P contributes nP to D. Then φ (in the neighborhood of P ) can be written as
[f0 hn : . . . fr hn ],
where at least one of the coordinates does not vanish (otherwise we can
subtract P from D, so D is not the common denominator). So pull-back
of coordinate hyperplanes are (locally near P ) given by divisors (f0 ) +
D, . . . , (fr ) + D.
If we start with any divisor D, a little complication can happen: a fixed
part (or base points) of D is a maximal effective divisor E such that D0 −E ≥ 0
for any D0 ∈ |D|. Those start to appear more often in large genus, but if they
do then |D| = |D − E|. In fact, this is an if and only if condition:
2.4.4. P ROPOSITION . D has no base points if and only if, for any point P ∈ C,
l(D − P ) = l(D) − 1.
The last question we wish to address is when φD gives an embedding
C ⊂ Pr . If this happens then we call D a very ample divisor. One has the
following very useful criterion:
2.4.5. T HEOREM . D is very ample if and only if
• φD separates points: l(D − P − Q) = l(D) − 2 for any points P, Q ∈ C.
• φD separates tangents: l(D − 2P ) = l(D) − 2 for any point P ∈ C.
This pretty much summarizes the course on Riemann surfaces!
27
§3. Elliptic curves: j-invariant
Let us recall the following basic theorem.
3.0.6. T HEOREM . The following are equivalent:
(1) C ∩ A2x,y is given by a Weierstrass equation y 2 = 4x3 − g2 x − g3 , where
∆ = g23 − 27g32 6= 0.
(2)
(3)
(4)
(5)
(6)
C
C
C
C
C
is isomorphic to a smooth cubic curve in P2 .
is isomorphic to a 2 : 1 cover of P1 ramified at 4 points.
is isomorphic to a complex torus C/Λ, where Λ ' Z ⊕ Zτ , Im τ > 0.
is a projective algebraic curve of genus 1.
is a compact Riemann surface of genus 1.
Proof. Simple implications:
(1) ⇒ (2) (just have to check that C is smooth),
(2) ⇒ (3) (project C ⊂ P2 99K P1 from any point p ∈ C).
(3) ⇒ (5) (Riemann–Hurwitz).
(5) ⇒ (6) (induced complex structure),
(4) ⇒ (6) (C is topologically a torus and has a complex structure induced
from a translation-invariant complex structure on C),
(2) ⇒ (1) (find a flex point (by intersecting C with a Hessian curve),
move it to [0 : 1 : 0] and make the line at infinity z = 0 the flex line).
Logically unnecessary but fun:
(2) ⇒ (5) (genus of plane curve formula),
(1) ⇒ (3) (project A2x,y → A1x , the last ramification point is at ∞),
Now the Riemann–Roch analysis.
Let C be an algebraic curve of genus 1. Then L(K) is one-dimensional.
Let ω be a generator. Since deg K = 0, ω has no zeros. It follows by RR that
l(D) = deg D
for
deg D > 0.
It follows that ψD has no base points for deg D > 1 and is very ample
for deg D > 2. Fix a point P ∈ C. Since 3P is very ample, we have an
embedding
ψ3P : C → P2 ,
where the image is a curve of degree 3, moreover, a point P is a flex point.
This shows that (6) ⇒ (2).
It is logically unnecessary but still fun: since 2P has no base-points, we
have a 2 : 1 map
ψ2P : C → P1 ,
with P as one of the ramification points, which shows directly that (6) ⇒
(3). Let ℘ ∈ L(2P ) be a meromorphic function with pole of order 2 at P . If
C is obtained as C/Λ (and P is the image of the origin), one can pull-back
℘ to a doubly-periodic (i.e. Λ-invariant) meromorphic function on C with
poles only of order 2 and only at lattice points. Moreover, this function is
unique (up-to rescaling and adding a constant). It is classically known as
the Weierstrass ℘-function
X 1
1
1
−
.
℘(z) = 2 +
z
(z − γ)2 γ 2
γ∈Λ, γ6=0
28
JENIA TEVELEV
Notice that ℘0 (z) has poles of order 3 at lattice points, and therefore
{1, ℘(z), ℘0 (z)}
is a basis of L(3P ). It follows that the embedding C ⊂ P2 as a cubic curve
is given (when pull-backed to C) by map
z 7→ [℘(z) : ℘0 (z) : 1],
C → C2 ,
and in particular ℘0 and ℘ satisfy a cubic relation. It is easy to check that
this relation has a Weierstrass form
(℘0 )2 = 4℘3 − g2 ℘ − g3 .
This gives another proof that (4) ⇒ (1).
The only serious implication left is to show that (6) ⇒ (4). There are
several ways of thinking and generalizing this result, and we will discuss
some of these results later in this course. For example, we can argue as
follows: we fix a point P ∈ C and consider a multi-valued holomorphic
map
Z
z
π : C → C,
z 7→
ω.
P
If the curve is given as a cubic in the Weierstrass normal form then those
are elliptic integrals
Z
dx
p
4x3 − g2 x − g3
We take the first homology group H1 (C) = Zα + Zβ and define periods
Z
Z
ω,
ω ∈ C.
α
β
The periods generate a subgroup Λ ⊂ C. If the periods are not linearly independent over R then (after multiplying ω by a constant), we can assume
that Λ ⊂ R. Then Im π is a single-valued harmonic function, which must be
constant by the maximum principle. This is a contradiction: π is clearly a
local isomorphism near P . So Λ is a lattice and π induces a holomorphic
map
f : C → C/Λ.
As we have already noticed, this map has no ramification (which also follows from Riemann–Hurwitz), thus from the theory of covering spaces f
corresponds to a subgroup of π1 (C/Λ) = Λ. Thus f must have the form
' C/Λ0 → C/Λ,
homeo
where Λ0 ⊂ Λ is a sublattice. Notice that the integration map is well-defined
on the universal cover of C, i.e. on C and gives the map F : C → C, which
should be just the identity map. But then F (Λ0 ) = Λ0 , i.e. periods belong
to Λ0 . Thus Λ = Λ0 .
C
3.0.7. R EMARK . An important generalization of the last step of the proof
is a beautiful Klein–Poincare Uniformization Theorem: a universal cover of a
compact Riemann surface is either
• P1 if g = 0, or
• C if g = 1, or
29
• H (upper half-plane) if g ≥ 2. In other words, any algebraic curve
of genus ≥ 2 is isomorphic to a quotient H/Γ, where
Γ ⊂ Aut(H) = PGL2 (R)
is a discrete subgroup acting freely on H.
§3.1. J-invariant. Now we would like to classify elliptic curves up to isomorphism, i.e. to describe M1 . As we will see many times in this course,
automorphisms of parametrized objects can cause problems. An elliptic
curve has a lot of automorphisms: since C ' C/Λ, it is in fact a group
itself! If thinking about an elliptic curve as a complex torus is too transcendental for you, observe also that if P, Q ∈ C then by our discussion of the
Riemann–Roch above, we have l(P +Q) = 2 and so we have a double cover
φ|P +Q| : C → P1
with P + Q as one of the fibers. Any double cover has an involution permuting the two branches, which shows that any two points P, Q ∈ C can
be permuted by an involution, and in particular that Aut C acts transitively
on C. In a cubic curve realization, the group structure on C is a famous
“three points on a line” group structure, but let’s postpone this discussion
until the lectures on Jacobians.
In any case, we can eliminate many automorphisms (namely translations) by fixing a point:
M1 = M1,1 .
So our final definition of an elliptic curve is: a pair (C, P ), where C is an
algebraic curve of genus 1 and P ∈ C. It is very convenient to choose P to
be the unity of the group structure if one cares about it.
Notice that even a pointed curve (C, P ) still has at least one automorphism, namely the involution given by permuting the two branches of φ2P .
In the C/Λ model this is the involution z 7→ −z (if P is chosen to be 0): this
reflects the fact that the Weierstrass ℘-function is even.
Now let’s work out when two elliptic curves are isomorphic and when
Aut(C, P ) is larger than Z2 .
3.1.1. T HEOREM .
(1) Curves given by Weierstrass equations y 2 = 4x3 −
g2 x − g3 and y 2 = 4x3 − g20 x − g30 are isomorphic if and only if there
exists t ∈ C∗ such that g20 = t2 g2 and g30 = t3 g3 . There are only two
curves with special automorphisms: the curve y 2 = x3 + 1 gives Z6 and
the curve y 2 = x3 + x gives Z4 (draw the family of cuspidal curves in the
g2 g3 -plane).
(2) Two smooth cubic curves C and C 0 are isomorphic if and only if they are
projectively equivalent.
(3) Let C (resp. C 0 ) be a double cover of P1 with a branch locus p1 , . . . , p4
(resp. p01 , . . . , p04 ). Then C ' C 0 if and only if there exists g ∈ PGL2 such
that p0i = g(pi ) for any i. In particular, we can always assume that branch
points are 0, 1, ∞, λ. There are two cases with non-trivial automorphisms,
2πi
λ = −1 (Aut C = Z4 ) and λ = ω = e 3 (Aut C = Z6 ). Modulo Z2 ,
these groups are automorphism groups of the corresponding fourtuples.
30
JENIA TEVELEV
(4) C/Λ ' C/Λ0 if and only if Λ = αΛ0 for some α ∈ C∗ . If Λ = Z ⊕ Zτ and
Λ0 = Z ⊕ Zτ 0 with Im τ, Im τ 0 > 0 then this is equivalent to
aτ + b
a b
0
τ =
,
∈ PSL2 (Z)
(3.1.2)
c d
cτ + d
There are two elliptic curves (draw the square and the hexagonal lattice)
with automorphism groups Z4 and Z6 , respectively.
Proof. (2) Suppose plane cubic realizations of C and C 0 are given by linear
systems 3P and 3P 0 , respectively. We can assume that an isomorphism of
C and C 0 takes P to P 0 . Then the linear system 3P is a pull-back of a linear
system 3P 0 , i.e. C and C 0 are projectively equivalent.
A similar argument proves (3). Notice that in this case Aut(C, P ) modulo
the hyperelliptic involution acts on P1 by permuting branch points. In fact,
λ is simply the cross-ratio:
λ=
(p4 − p1 ) (p2 − p3 )
,
(p2 − p1 ) (p4 − p3 )
but branch points are not ordered, so we have an action of S4 on possible
cross-ratios. However, it is easy to see that the Klein’s four-group V does
not change the cross-ratio. The quotient S4 /V ' S3 acts non-trivially:
λ 7→ {λ, 1 − λ, 1/λ, (λ − 1)/λ, λ/(λ − 1), 1/(1 − λ)}
(3.1.3)
Special values of λ correspond to cases when some of the numbers in this
list are equal. For example, λ = 1/λ implies λ = −1 and the list of possible
cross-ratios boils down to −1, 2, 1/2 and λ = 1/(1 − λ) implies λ = −ω, in
which case the only possible cross-ratios are −ω and −1/ω.
(4) Consider an isomorphism f : C/Λ0 → C/Λ. Composing it with translation automorphisms on the source and on the target, we can assume that
f (0+Λ0 ) = 0+Λ. Then f induces a holomorphic map C → C/Λ with kernel
Λ0 , and its lift to the universal cover gives an isomorphism F : C → C such
that F (Λ0 ) = Λ. But it is proved in complex analysis that all automorphisms
of C preserving the origin are maps z 7→ αz for α ∈ C∗ . So we have
Z + Zτ = α(Z + Zτ 0 ),
which gives
ατ 0 = a + bτ,
α = c + dτ,
which gives (3.1.2).
So finally, we can introduce the j-invariant:
j = 1728
(λ2 − λ + 1)3
g23
= 256 2
.
∆
λ (λ − 1)2
2
(3.1.4)
3
−λ+1)
It is easy to see that the expression 256 (λλ2 (λ−1)
2 does not change under the
transformations (3.1.3). For a fixed j0 , the polynomial
256(λ2 − λ + 1)3 − jλ2 (λ − 1)2
has six roots related by the transformations (3.1.3). So the j-invariant uniquely
determines an isomorphism class of an elliptic curve.
The special values of the j-invariant are j = 0 (Z6 ) and j = 1728 (Z4 ).
31
§3.2. Monstrous Moonshine. The most interesting question here is how
to compute the j-invariant in terms of the lattice parameter τ . Notice that
j(τ ) is invariant under the action of PSL2 (Z) on H. This group is called the
modular group. It is generated by two transformations,
S : z 7→ −1/z
and T : z 7→ z + 1
(pull notes from Adam’s TWIGS talk). It has a fundamental domain (draw
the modular figure, two special points). The j-invariant maps the fundamental domain to the plane A1 (draw how).
Since the j-invariant is invariant under z 7→ z + 1, it can be expanded in
a variable q = e2πiτ :
j = q −1 + 744 + 196884q + 21493760q 2 + . . .
What is the meaning of these coefficients? According to the classification
of finite simple groups, there are several infinite families of them (like an
alternating group An ) and a few sporadic groups. The largest sporadic
group is the monster group F 1 that has about 1054 elements. Its existence
was predicted by Robert Griess and Bernd Fischer in 1973 and it was eventually constructed by Griess in 1980 as the automorphism group of the
Griess (commutative, non-associative) algebra whose dimension is 196884:
so 196884 is to F 1 as n is to Sn . The dimension of the Griess algebra is one of
the coefficients of j(q)! In fact all coefficients in this q-expansion are related
to representations of the Monster. This is a Monstrous Moonshine Conjecture
of McKay, Conway, and Norton proved in 1992 by Borcherds (who won the
Fields medal for this work).
§3.3. Families of elliptic curves: coarse and fine moduli spaces. So far we
were mostly concerned with moduli spaces as sets that parametrize isomorphism classes of geometric objects. The geometric structure on the moduli
space came almost as an afterthought, even though it is this structure of
course that is responsible for all applications. The most naive idea is that
two points in the moduli space are close to each other if the objects that they
represent are small deformations of each other. There exists an extremely
simple and versatile language (developed by Grothendieck, Mumford, etc.)
for making this rigorous. The key words are family of objects, coarse moduli
space, pull-back, and fine moduli space.
What is a family, for example what is a family of elliptic curves? One
should think about it as a sort of fibration with fibers given by elliptic
curves (appropriately called an elliptic fibration).
More generally, a family of objects is a regular map f : X → Y , where
the fibers are geometric objects we care about. In practice, considering all
maps does not work, and one has to impose some conditions on f . These
required conditions in fact often depend on the moduli problem being stidied, so in the interest of drama let’s call it “Property X” for now. Let M be
the moduli “set” of isomorphism classes of these objects. We have a map
Y → M which sends y ∈ Y to the isomorphism class of the fiber f −1 (y).
Geometric structure we are imposing on M should be compatible with this
map Y → M: basically we should just ask that this map Y → M is a
regular map.
32
JENIA TEVELEV
This is a basic idea, but there is a minor complication: with this definition
the moduli space (even when it exists) is almost never going to be unique.
For example, let’s suppose that A1 is a moduli space for some problem. Let
C = {y 2 = x3 } ⊂ A2 be a cuspidal curve with the normalization map
ν : A1 → C,
t 7→ (t3 , t2 ).
Notice that ν is a bijection on points but not an isomorphism. Any family of
objects over Y will give us a regular map Y → A1 which when composed
with ν will give a regular map Y → C. To guarantee uniqueness of the
moduli space, we add an extra condition (3) to the following definition:
3.3.1. D EFINITION . We say that the algebraic variety M is a coarse moduli
space for the moduli problem if
(1) Points of M correspond to iso classes of objects in question.
(2) Any family X → Y (i.e. a regular map satisfying property X) induces a regular map Y → M .
(3) For any other algebraic variety M 0 satisfying (1) and (2), an obvious
map M → M 0 is regular.
3.3.2. R EMARK . It is rare that the moduli problem studies geometric objects without any “decorations”. For example, an elliptic curve is not just
a genus 1 curve C but also a point P ∈ C. This extra data should be built
into the definition of the family. For example, we can say that an elliptic fibration is a morphism f : X → Y (satisfying property X) plus a morphism
σ : Y → X such that f ◦ σ = IdY . A morphism like this is called a section.
3.3.3. R EMARK . It practice, it is often necessary to enlarge the category of
algebraic varieties to the category of algebraic schemes. For example, in
number theory one can look at an elliptic curve defined by equations with
integral coefficients or with coefficients in some ring of algebraic integers.
Then it is interesting to work out “reductions” of this elliptic curve modulo
various primes. Geometrically, all primes in the ring of algebraic integers R
form an algebraic scheme, called Spec R and one thinks about reductions of
an elliptic curve modulo various primes as fibers of the family E → Spec R
where E is again a scheme called an integral model of the original complex
elliptic curve. This is an arithmetic analogue of a geometric situation when
we have an elliptic fibration over an algebraic curve. Some fibers won’t be
smooth, this happens at so called primes of bad reduction.
Families can be pulled-back: if we have a morphism f : X → Y (satisfying property X) and an arbitrary morphism g : Z → Y then we define a
pull-back (or a fibered product)
X ×Y Z = {(x, z) ∈ X × Z | f (x) = g(z)} ⊂ X × Z.
We have a morphism X ×Y Z → Z induced by the second projection. Since
we only care about maps satisfying property X, we have to make sure that
Property X is stable under pull-back, i.e. if f has property X then the induced
map X ×Y Z → Z also has it (notice however that the morphism g : Z → X
can be arbitrary). If we include some decorations in the family, we have to
modify the notion of the pull-back to include decorations. For example, a
section σ : Y → X will induce a section Z → X ×Y Z, namely (σ ◦ g, IdZ ).
33
The basic point is that X → Y and X ×Y Z → Z have the same fibers,
and the map X ×Y Z → M to the moduli space factors through Y → M.
This raises a tantalizing possibility that
3.3.4. D EFINITION . M is a fine moduli space if there exists a universal family
U → M, i.e. a regular map (with property X) such that any other family
X → Y is isomorphic to a pull-back along a unique regular map Y → M.
Let’s look at various examples.
3.3.5. E XAMPLE . We know that P1 is the only genus 0 curve (up to isomorphisms). So the moduli problem of “families of genus 0 curves” has an
obvious course moduli space: a point. However, it is not a fine moduli
space. If it were, then all families X → Y (satisfying property X) with
fibers isomorphic to P1 would appear as
Y ×pt P1 = Y × P1 .
However, there exist extremely simple P1 -fibrations not isomorphic to the
product. For example, consider the Hirzebruch surface F1 . It is obtained by
resolving indeterminacy locus of a “projection from a point” rational map
f : P2 99K P1 by blowing up this point. In coordinates, the rational map is
[x : y : z] 7→ [x : y],
which is undefined at [0 : 0 : 1]. Its blow-up F1 is a surface in P2 × P1[s:t]
given by an equation
xt = ys.
The resolution of f is obtained by just restricting the second projection P2 ×
P1 → P1 to F1 . The first projection identifies F1 with P2 everywhere outside
of the point [0 : 0 : 1], the preimage of this point is a copy of P1 called the
exceptional divisor. All fibers of the resolved map F1 → P1 are isomorphic
to P1 but F1 6' P1 × P1 . For example, E has self-intersection −1 on F1 but
there are no (−1)-curves in P1 × P1 .
3.3.6. E XAMPLE . How about the Grassmannian? We claim that G(2, n) is
a fine moduli space for 2-dimensional subspaces of An . What is a family here? A family over an algebraic variety X should be a varying 2dimensional subspace of An . In other words, a family over X is just a
2-dimensional vector sub-bundle E of the trivial vector bundle X × An .
What is an r-dimensional vector bundle over an algebraic variety X? It is
an algebraic variety E, a morphism π : E → X, and a trivializing covering
X = ∪Uα , which means that we have isomorphisms
ψα : π −1 (Uα ) → Uα × Ar ,
p2 ◦ ψi = π
that are given by linear maps on the overlaps, i.e. over Uα ∩ Uβ the induced
map
p2
Uα ∪ Uβ × Ar → Uα ∪ Uβ × Ar → Ar
takes
(x, v) 7→ A(x)v,
where A(x) is an invertible matrix with entries in O(Ui ∩ Uj ). A map of
vector bundles E1 → E2 is map of underlying varieties that is given by
34
JENIA TEVELEV
linear transformations in some trivializing charts (with coefficients of these
linear transformations being regular functions on charts).
So let’s fix a 2-dimensional vector bundle E over X and let’s assume that
it is a sub-bundle of a trivial bundle X ×Ar . What is the corresponding map
to the Grassmannian? Choose a trivializing affine covering {Uα } of X such
that E|Uα ' Uα × A2 . Choose a basis u, v in A2 . An embedding E ⊂ X × An
in the chart gives an embedding Uα × A2 ⊂ Uα × An . Composing it with
projection to An gives maps
X
X
(x, u) 7→
a1i (x)ei , (x, v) 7→
a2i (x)ei ,
i
i
where a1i (x), a2i (x) ∈ O(Uα ). Then
X
u∧v =
pij (x)ei ∧ ej ,
i<j
which after the projectivization gives a morphism
Uα → P(Λ2 A2 ).
Clearly the function pij (x) satisfy the Plücker relations, thus this regular
map factors through the map
Uα → G(2, n).
This map does not depend on the choice of the basis {u, v} and thus these
maps glue on overlaps Uα ∩ Uβ to give a regular map X → G(2, n). What
is the universal family here? Quite appropriately, it is called the universal bundle over G(2, n) and is defined as follows: the fiber over a point in
G(2, n) that corresponds to a subspace U ⊂ An , is U itself. More precisely,
the universal sub-bundle is
E = {([U ], v) | v ∈ U } ⊂ G(2, n) × An .
It is trivialized in standard affine charts of the Grassmannian: for example
let’s consider the chart U12 defined by p12 6= 0. For any point of this chart,
rows of the matrix
1 0 a13 a14 . . . a1n
A=
0 1 a23 a24 . . . a2n
give a basis of the corresponding subspace in An . Choosing a basis and
trivializing if clearly the same thing.
Returning to the j-invariant, we have to define a family of elliptic curves.
It is clear that it should be a regular map of varieties π : X → Y and a section σ : Y → X such that each fiber (π −1 (y), σ(y)) is an elliptic curve.
Now it’s time to discuss a mysterious “property X”. Recall that in the definition of the vector bundle above we assumed not only that each fiber of
the map is a vector space but also that a vector bundle has “local models”,
i.e. it becomes trivial in sufficiently small neighborhoods. So in particular a vector bundle is a locally trivial fibration, which is certainly something
reasonable for a family of varieties. However, we can not require any interesting map to be a locally trivial fibration in Zariski topology, because
a locally trivial fibration has isomorphic fibers, and this is certainly something we want to avoid in our discussion of moduli! (The corresponding
35
notion in Algebraic Geometry is called an isotrivial fibration). So we have to
come up with some property of a regular map π to ensure that it behaves
like a locally trivial fibration without actually being one. In the analytic
category, i.e. when X and Y are complex manifolds, the right concept is the
notion of a proper submersion, i.e. a surjective holomorphic map with compact fibers and everywhere surjective differential. There is a well-known
theorem of Ehresmann that, even though a submersion is almost never a locally trivial fibration in the analytic category (i.e. fibers are not isomorphic
as complex analytic manifolds), it is nevertheless a locally trivial fibration
in the category of real manifolds, and so for example the fibers are diffeomorphic (assuming the base is connected). So if we only want to define
“property X” when the base is a smooth projective variety, we can ask that
π is a submersion when considered as a map of complex manifolds. It is
proved in any course on manifolds that the property of being a submersion
is preserved by pull-backs. It turns out that Algebraic Geometry allows
one to extend this notion even to the case when the base is not smooth. It
is called a smooth morphism. We will not define it here, see [Ha, 3.10]. In the
case of elliptic fibrations, the property of being smooth is equivalent to the
following extremely useful “local model” result:
3.3.7. T HEOREM ([MS, page 203]). π : X → Y is an elliptic fibration with a
section σ : Y → X if and only if the following condition is satisfied. Every point
y ∈ Y has an affine neighborhood such that π −1 (U ) is isomorphic to a subvariety
of U × P2 given by the Weierstrass normal form
y 2 z = 4x3 − g2 xz 2 − g3 z 3 ,
where g2 , g3 ∈ O(U ) are regular function on U such that ∆ = g23 − 27g32 does not
vanish on U . Moreover, g2 , g3 ∈ O(U ) are defined uniquely up to transformations
g2 7→ t4 g2 ,
g3 7→ t6 g3
(3.3.8)
for some invertible function t ∈ O∗ (U ).
3.3.9. R EMARK . Dependence on t comes from the following basic observation: multiplying y by t3 and x by t2 will induce (3.3.8). Notice that if t
is a constant function then we can take its square root and multiply by it
instead. But if t is a regular function, its square root is rarely regular, and
so can not be used.
3.3.10. R EMARK . This idea of presenting a fibration by using some normal form for equations of an algebraic variety where the coefficients are
allowed to vary is very common.
Now we can show that
3.3.11. T HEOREM . The j-line is a course moduli space for elliptic curves.
Proof. The proof resembles the corresponding argument for the Grassmannian. Assuming we have an elliptic fibration X → Y , we have to construct
a morphism Y → A1j , i.e. we have to show that j is a regular function on
Y . This is a local statement that we can check on charts of Y , thus by Theorem 3.3.7 we can assume that the fibration is in the Weierstrass normal
form. But then we can just define j by the usual formula (3.1.4): since g2
and g3 are regular functions on the charts, j is regular as well.
36
JENIA TEVELEV
Interestingly, this also shows that A1j is not a fine moduli space. Indeed,
if A1 carries a universal family then Theorem 3.3.7 would be applicable to
A1j as well. This would imply that locally at any point P ∈ A1 we have
j = 1728
g23 (j)
g23 (j) − 27g32 (j)
for some rational functions g2 and g3 . But j has zero of order 1 at 0 where
as the order of zeros of the RHS at 0 is divisible by 3. Likewise, we have
j − 1728 = 1728
27g32 (j)
.
− 27g32 (j)
g23 (j)
j −1728 has zero of order 1 at 1728 but the RHS has even order of vanishing!
We see that special elliptic curves with automorphisms prevent the j-line
from being a fine moduli space.
§3.4. Homework.
Problem 1. Let M be the set of isomorphism (=conjugacy) classes of invertible complex 2 × 2 matrices. (a) Describe M as a set. (b) Let’s define the
following moduli problem: a family over a variety X is a 2 × 2 matrix A(x)
with coefficients in O(X) such that det A(x) ∈ O∗ (X), i.e. A(x) is invertible
for any x ∈ X. Explain how the pull-back of families should be defined.
(c) Show that there is no structure of an algebraic variety on M that makes
it into a coarse moduli space (2 points).
Problem 2. Compute j-invariants of elliptic curves (1 point):
(a) y 2 + y = x3 + x;
(b) y 2 = x4 + ax3 + bx2 + cx.
Problem 3. Prove Theorem 3.1.1, (1) (2 points).
Problem 4. Show that any elliptic curve is isomorphic to a curve of the
form y 2 = (1 − x2 )(1 − e2 x2 ) (1 point).
Problem 5. Show that the two formulas in (3.1.4) agree (1 point).
Problem 6. (a) Compute the j-invariant of an elliptic curve
y 2 + xy = x3 −
36
1
x−
,
q − 1728
q − 1728
where q is some parameter. (b) Consider families of elliptic curves (defined
as in Theorem 3.3.7) but with an extra condition that no fiber has a special
automorphism group, i.e. assume that j 6= 0, 1728. Show that A1j \ {0, 1728}
carries a family of elliptic curves with j-invariant j. Is your family universal? (2 points).
2
3
Problem 7. The formula j = 256 (λλ2 −λ+1)
gives a 6 : 1 cover P1λ → P1j .
2
(λ−1)
Thinking about P1 as a Riemann sphere, let’s color P1j in two colors: color
the upper half-plane H white and the lower half-plane −H black. Draw the
pull-back of this coloring to P1λ (2 points).
Problem 8. Let f be a rational function on an algebraic curve C such
that all zeros of f have multiplicity divisible by 3 and all zeros of f − 1728
have multiplicities divisible by 2. Show that C \ {f = ∞} carries an elliptic
fibration (defined as in Theorem 3.3.7) with j-invariant f . (2 points)
37
Problem 9. Using a birational isomorphism between P1 and the circle
{x2 + y 2 = 1} ⊂ A2 given by stereographic projection from (0, 1), describe
an algorithm for computing integrals of the form
Z
p
P (x, 1 − x2 ) dx
where P (x, y) is an arbitrary rational function (2 points).
Problem 10. Let (C, P ) be an elliptic curve. (a) By considering a linear
system φ|4P | , show that C embeds in P3 as a curve of degree 4. (b) Show
that quadrics in P3 containing C form a pencil P1 with 4 singular fibers.
(c) These four singular fibers define 4 points in P1 . Relate their cross-ratio
to the j-invariant of C (4 points).
Problem 11. Let X be an affine variety and let f ∈ O(X). A subset
D(f ) = {x ∈ X | f (x) 6= 0}
is called a principal open set. (a) Show that principal open sets form a
basis of Zariski topology. (b) Show that any principal open set is itself an
affine variety with a coordinate algebra O(X)[1/f ]. (c) Show that any affine
(resp. projective) variety is quasi-compact, i.e. any open cover has a finite
subcover (2 points).
Problem 12. Solve a cross-word puzzle (1 point)
Problem 13. (a) Show that any affine (resp. projective) variety X is a
union of finitely many irreducible projective varieties X1 , . . . , Xn such that
Xi 6⊂ Xj for i 6= j (called irreducible components of X). (b) Show that
irreducible components are defined uniquely (2 points).
Problem 14. Let X ⊂ Pn be an irreducible projective variety. Show that
any morphism X → Pm is given by m + 1 homogeneous polynomials
F0 , . . . , Fm in n + 1 variables of the same degree such that, for any point
x ∈ X, at least one of the polynomials Fi does not vanish (1 point).
Problem 15. (a) Let X ⊂ An and Y ⊂ Am be affine varieties with coordinate algebras O(X) and O(Y ). Suppose these algebras are isomorphic.
Show that varieties X and Y are isomorphic. (b) Let X and Y be irreducible
quasi-projective varieties with fields of rational functions C(X) and C(Y ).
Show that these fields are isomorphic (i.e. X and Y are birational) if and
only if there exist non-empty open subsets U ⊂ X and V ⊂ Y such that U
is isomorphic to V (2 points).
Problem 16. Let (C, P ) be an elliptic curve. Let Γ ⊂ C be the ramification
locus of φ|2P | . (a) Show that Γ ' Z2 × Z2 is precisely the 2-torsion subgroup
in the group structure on C. (b) A level 2 structure on (C, P ) is a choice
of an ordered basis {Q1 , Q2 } ∈ Γ (considered as a Z2 -vector space). Based
on Theorem 3.3.7, describe families of elliptic curves with level 2 structure.
Show that P1λ \{0, 1, ∞} carries a family of elliptic curves with with a level 2
structure such that any curve with a level 2 structure appears (uniquely) as
one of the fibers. Is your family universal? (2 points).
Problem 17. Consider the family y 2 = x3 + t of elliptic curves over
1
A \ {0}. Show that all fibers of this family have the same j-invariant but
nevertheless this family is not trivial over A1 \ {0} (2 points).
38
JENIA TEVELEV
$
%
!
"
#
*
'
)
&
$(
$$
$%
$!
$"
$#
$*
$'
$)
$&
%(
%$
%%
%!
%"
%#
$
"
)
&
$$
$!
$"
$#
$*
$&
%%
%!
%"
%#
+,-.//
+012345-0/6/748079570,.--4/:.3;/07.050
:-.:4-0/<=/470.>050=5/2/0230?@AB02/0
,5114;0CCC0
D94.-480.>0E4-./0
+0>-243;160F2537092;23F023/2;407940
GH><3,72.30
+1104>>4,72I40;2I2/.-/012345-160
4J<2I5143707.050F2I430.340
+3051F4=-52,0,<-I40.I4-0K02/051/.0
L3.M305/050CCC0
N40,.234;05074-80O,.5-/408.;<120
/:5,4O0
+0,-.//H-572.02/051/.0L3.M305/0530CCC0
-572.0
P.-;0</4;0=60Q2485330>.-0503<8=4-0
.>0:5-58474-/0
+30577-2=<740.>050;2I2/.-0,.37-2=<723F0
7.07940Q248533HQ.,90>.-8<150
RS58:140.>050>23408.;<120/:5,40
A2843/2.30.>07940/:5,40.>0
9.1.8.-:92,0;2>>4-437251/0.307940
51F4=-52,0,<-I40
T.847923F06.<0,530;.0M2790>582124/0
+3051F4=-52,0I5-247602/0,5114;0CCC02>02702/0
3.7050<32.30.>07M.0:-.:4-0,1./4;0
/<=/47/0
+03.-8510>.-80.>050,<=2,0,<-I402/0
3584;05>74-09280
%
!
#
*
'
$(
$%
$'
$)
%(
%$
A.M3
D6:40.>053051F4=-52,0,<-I408./70
-4I4-4;0=603<8=4-0794.-2/7/0
A2I2/.-0.>05084-.8.-:92,0>.-80
+3051F4=-52,0F4.8474-U/05351.F<40.>050
9.1.8.-:92,0,..-;235740
N4085;4028:.-75370,.37-2=<72.3/07.07940
/7<;60.>02374F-51/0.>051F4=-52,0><3,72.3/0
+0>53,60M.-;0</4;0=6051F4=-52,0
F4.8474-/023/745;0.>0O,.8:5,7O0
N.1.8.-:92,085:0M2790/<-G4,72I40
;2>>4-4372510
D940=4/70L23;0.>0508.;<120/:5,40
V85F40.>050-582>2,572.30:.2370
D940/4,.3;0=4/70L23;0.>0508.;<120/:5,40
+0;2I2/.-0M27903.3H34F572I40
8<172:12,2724/02/0,5114;0CCC0
+304I4-6M94-40;4>234;0-572.351085:02/0
,5114;0CCC0
Problem 18. Consider the family of cubic curves
Ca = {x3 + y 3 + z 3 + axyz = 0} ⊂ P2
parametrized by a ∈ A1 . (a) Find all a such that Ca is smooth and find its
flex points. (b) Compute j as a function on a and find all a such that Ca has
a special automorphism group (2 points).
39
Problem 19. Let (C, P ) be an elliptic curve equipped with a map C → C
of degree 2. By analyzing the branch locus φ2P , show that the j-invariant
of C has only 3 possible values and find these values (3 points).
40
JENIA TEVELEV
41
§4. Icosahedron, E8 , and quotient singularities
It is time to study invariant theory and orbit spaces more systematically.
We will start with a finite group G acting linearly3 on a vector space V and
discuss the orbit space V /G and the quotient morphism V → V /G. There
are several reasons to do this:
• The space V /G is usually singular, and singularities of this form (the
so-called quotient or orbifold singularities) form perhaps the most
common and useful class of singularities.
• Moduli spaces are often constructed as quotients X/G. Here X is
usually not a vector space and G is usually not a finite group. However, many results can be generalized and hold in this more general
situation (albeit with some complications).
• In good situations, a moduli space near a point p often “looks like”
V /G, where V is a vector space (a so-called versal deformation space
of the geometric object that corresponds to p) and G is an automorphism group of this object, which is usually finite.
To give us a concrete goal, we will try to understand geometry of the
most enigmatic du Val singularity related to E8 and the icosahedron. It is
defined as follows. According to Plato, finite subgroups in SO3 correspond
to platonic solids. For example, A5 embeds in SO3 as a group of rotations
of the icosahedron. In particular, A5 acts on the circumscribed sphere of the
icosahedron. This action is obviously conformal (preserves oriented angles),
and so if we think about S 2 as a Riemann sphere P1 (for example by using
the stereographic projection), we get an embedding A5 ⊂ PSL2 (since it is
proved in complex analysis that conformal maps are holomorphic). The
preimage of A5 in SL2 is called the binary icosahedral group Γ. What is the
orbit space C2 /Γ? Let’s start with more straightforward examples.
§4.1. Symmetric polynomials, one-dimensional actions. Let G = Sn be a
symmetric group acting on Cn by permuting the coordinates. Recall that
our old recipe for computing the orbit space calls for studying the ring of
invariants
C[x1 , . . . , xn ]Sn .
These invariant polynomials are called symmetric polynomials. By the classical Theorem on Symmetric Polynomials, they are generated by elementary
symmetric polynomials
σ1 = x1 + . . . + xn ,
...
X
σk =
x i1 . . . xik ,
i1 <...<ik
...
σn = x1 . . . xn .
Thus the candidate for the quotient map is
π : An → An ,
(x1 , . . . , xn ) 7→ (σ1 , . . . , σn ).
3Recall that a linear action is given by a homomorphism G → GL(V ). In this case we
also say that V is a representation of G.
42
JENIA TEVELEV
Notice the following interesting feature:
4.1.1. P ROPOSITION . The map π is surjective and its fibers are the Sn -orbits.
Proof. By the Vieta formulas, we can recover x1 , . . . , xn from σ1 , . . . , σn as
roots of the polynomial T n − σ1 T n−1 + . . . + (−1)n σn = 0.
If G acts linearly on C then essentially we have a character, i.e. a homomorphism G → C∗ . Its image is a subgroup µd of d-th roots of unity. For
the purpose of computing invariants we can always assume that the action
is faithful, so let’s just assume that G = µd acts on C by multiplication. A
non-zero orbit has d elements x, ζx, . . . , ζ d−1 x, where ζ ∈ µd is a primitive
root. There is an obvious invariant, namely xd , and it is clear that
C[x]µd = C[xd ].
It is also clear that xd separates orbits and that the quotient morphism in
this case is just
π : A1 → A1 , x 7→ xd .
§4.2. A1 -singularity. Let Z2 act on A2 by (x, y) 7→ (−x, −y). Invariant
polynomials are just polynomials of even degree, and so
C[x, y]Z2 = C[x2 , y 2 , xy]
and the quotient morphism is
π : A2 → A3 ,
(x, y) 7→ (x2 , y 2 , xy).
It is clear that invariants separate orbits. It is also clear that the quotient
map is surjective onto the quadratic cone
(uv = w2 ) ⊂ A3 .
So the quotient A2 /Z2 is a quadratic cone. It has a very basic singularity
called “A1 -singularity” (the easiest du Val singularity).
§4.3. Chevalley–Shephard–Todd theorem. Based on the examples above,
one can ask: when is the algebra of invariants a polynomial algebra? It is
true for the natural actions of Sn on An and µd on A1 , but fails for the action
of Z2 on A2 by ±1. What is so special about the first two cases? The answer
turns out to be very pretty, but a bit hard to prove:
4.3.1. T HEOREM . Let G be a finite group acting linearly and faithfully on Cn .
Then C[x1 , . . . , xn ]G is isomorphic to a polynomial algebra if and only if G is
generated by pseudo-reflections, i.e. by elements g ∈ G such that the subspace
{v ∈ Cn | gv = v}
has codimension 1.
In other words, g is a pseudo-reflection if and only if its matrix is some
basis is equal to diag[ζ, 1, 1, . . . , 1], where ζ is a root of unity. If ζ = −1 then
g is called a reflection. For example, if Sn acts on Cn then any transposition
(ij) acts as a reflection with mirror xi = xj . Further examples of groups
generated by reflections are Weyl groups of root systems. On the other hand,
the action of µd on C is generated by a pseudo-reflection z 7→ ζz. This is
not a reflection (for d > 2) but the algebra of invariants is still polynomial.
43
The action of Z2 on C2 by ±1 is not a pseudoreflection (a fixed subspace has
codimension 2).
§4.4. Finite generation.
4.4.1. T HEOREM . Let G be a finite group acting linearly on a vector space V .
The algebra of invariants O(V )G is finitely generated.
We will split the proof into two Lemmas. The second one will be later
reused to proof finite generation for various important infinite groups.
4.4.2. D EFINITION . A linear map
R : O(V ) → O(V )G
is called a Reynolds operator if
• R(1) = 1;
• R(f g) = f R(g) for any f ∈ O(V )G and g ∈ O(V ).
In particular, the Reynolds operator is a projector onto O(V )G :
R(f ) = R(f · 1) = f R(1) = f
for any f ∈ O(V )G .
4.4.3. L EMMA . The Reynolds operator exists for any linear action of a finite group.
Proof. Since G acts on V , it also acts on the polynomial algebra O(V ). It’s
fun to check that the action has to be defined as follows: if p ∈ O(V ) then
(g · p)(x) = p(g −1 x).
This is how the action on functions is defined: if you try g instead of g −1 ,
the group action axiom will be violated (why?) . We define the Reynolds
operator R as an averaging operator:
1 X
g · p.
R(p) =
|G|
g∈G
It is clear that both axioms of the Reynolds operator are satisfied. This
works over any field as soon as its characteristic does not divide |G|.
4.4.4. L EMMA . If G is any group acting linearly on a vector space V and possessing a Reynolds operator, then O(V )G is finitely generated.
It is obvious that Theorem 4.4.1 follows from these two lemmas.
Proof of Lemma 4.4.4. This is an ingenious argument that belongs to Hilbert.
First of all, the action of G on polynomials preserves their degrees. So
O(V )G is a graded subalgebra of O(V ). Let I ⊂ O(V ) be the ideal generated
by homogeneous invariant polynomials f ∈ O(V )G of positive degree. By
Hilbert’s basis theorem (proved in the same paper as the argument we are
discussing), I is in fact generated by finitely many homogeneous invariant
polynomials f1 , . . . , fr of positive degree. We claim that the same polynomials generate O(V )G as an algebra, i.e. any f ∈ O(V )G is a polynomial in
f1 , . . . , fr . Without loss of generality, we can assume that f is homogeneous
and argue by induction on its degree. We have
f=
r
X
i=1
ai fi ,
44
JENIA TEVELEV
where ai ∈ O(V ). Now apply the Reynolds operator:
f = R(f ) =
r
X
R(ai )fi .
i=1
Each R(ai ) is an invariant polynomial, and if we let bi be its homogeneous
part of degree deg f − deg fi , then we still have
f=
r
X
bi fi .
i=1
By inductive assumption, each bi is a polynomial in f1 , . . . , fr . This shows
the claim.
§4.5. Basic properties of quotients.
4.5.1. D EFINITION . The finite generation Theorem allows us to construct the
quotient variety and the quotient map. Namely, V /G is defined as an affine
variety such that
O(V /G) = O(V )G
and the quotient morphism π : V → V /G is a morphism with the pullback of regular functions given by the inclusion π ∗ : O(V )G ⊂ O(V ). More
concretely, we choose a system of generators f1 , . . . , fr of O(V )G and write
k[x1 , . . . , xr ]/I ' O(V )G ,
xi 7→ fi .
We define V /G as an affine subvariety in Ar given by the ideal I and let
π : V → V /G ,→ Ar ,
v 7→ f1 (v), . . . , fr (v).
A different system of generators gives an isomorphic affine variety.
4.5.2. To show that this definition is reasonable, let’s check two things:
• Fibers of π are exactly the orbits, i.e. any two orbits are separated by
polynomial invariants, and
• All points of V /G correspond to orbits, i.e. π is surjective.
4.5.3 (S EPARATION OF ORBITS). To show the first property, we are going to
use significantly that the group is finite. (In fact, we will see later on that
separation of orbits fails for infinite groups such as SLn .) Take two orbits,
S1 , S2 ⊂ V . Since they are finite, it is easy to see (why?) that there exists a
polynomial f ∈ O(V ) such that f |S1 = 0 and f |S2 = 1. Then the average
1 X
g·f
F = R(f ) =
|G|
g∈G
is an invariant polynomial but we still have F |S1 = 0 and F |S2 = 1.
Now surjectivity:
4.5.4. T HEOREM . Let G be a group acting linearly on a vector space V and possessing a Reynolds operator. Then the quotient map π : V → V /G is surjective.
45
4.5.5. To prove this in style, let’s introduce another more intrinsic point of
view on affine varieties based on Hilbert’s Nullstellensatz. Let X be an affine
variety. By the aforementioned theorem, we have equality of sets
X = MaxSpec O(X),
x ↔ {f ∈ O(X) | f (x) = 0}.
The Zariski topology can also be completely recovered here:
V (I) ↔ {m ∈ MaxSpec O(X) | I ⊂ m}.
Principal open sets? No problem:
D(f ) ↔ {m ∈ MaxSpec O(X) | f 6∈ m}.
Functions? This is a fun part: suppose f ∈ O(X) and let x ∈ X. What is
f (x) in terms of the maximal ideal m ⊂ O(X) that corresponds to x? We
have an isomorphism O(X)/m ' C and we simply have
f (x) = f + m ∈ O(X)/m ' C.
How to see a regular map π : X → Y in terms of maximal ideals? Suppose
π(x) = y. We have a pull-back homomorphism on functions
π ∗ : O(Y ) → O(X).
Now notice that a function f ∈ O(Y ) vanishes at y iff its pull-back π ∗ (f )
vanishes at x. In other words, in the language of maximal ideals we have
π : MaxSpec O(X) → MaxSpec O(Y ),
π(m) = (π ∗ )−1 (m).
For instance, how would we check algebraically that a given regular map
of affine varieties π : X → Y is surjective?
4.5.6. L EMMA . A regular map π : X → Y of affine varieties is surjective if and
only if O(X)π ∗ (n) 6= O(X) for any maximal ideal n ⊂ O(Y ).
Proof. For any point y ∈ Y (i.e. a maximal ideal n ⊂ O(Y )) we have to
show existence of a point x ∈ X (i.e. a maximal ideal m ⊂ O(X)) such
that f (x) = y (i.e. (π ∗ )−1 (m) = n). So we have to show that there exists a
maximal ideal m ⊂ O(X) that contains π ∗ (n). The image of an ideal under
homomorphism is not necessarily an ideal, so the actual condition is that
the ideal O(X)π ∗ (n) is a proper ideal.
Of course this machinery will be successful only if a regular map π is
from the beginning defined not geometrically but algebraically, in terms of
the pull-back of functions. But this is exactly the case for the quotient map!
Proof of Theorem 4.5.4. Let n ⊂ O(V )G be a maximal ideal. We have to show
that
O(V )n 6= O(V )
(recall that a pull-back of functions for the quotient map π : V → V /G is
just the inclusion O(V )G ⊂ O(V )). Arguing by contradiction, suppose that
O(V )n = O(V ). Then we have
X
ai fi = 1,
where ai ∈ O(V ) and fi ∈ n. Applying the Reynolds operator, we see that
X
bi fi = 1,
46
JENIA TEVELEV
where bi ∈ O(V )G . But n is a proper ideal of O(V )G , contradiction.
This argument only uses the existence of a Reynolds operator, and so has
a wider range of applications then the case of finite groups. But for finite
groups we can do a little bit better:
4.5.7. L EMMA . O(V ) is integral over O(V )G .
Proof. Indeed, any element f ∈ O(V ) is a root of the monic polynomial
Y
(T − g · f ).
g∈G
Coefficients of this polynomial are in O(V )G (by Vieta formulas).
4.5.8. D EFINITION . Let π : X → Y be a regular map of affine varieties such
that the pull-back map is an inclusion π ∗ : O(Y ) ⊂ O(X). Geometrically
this means that π is dominant, i.e. π(X) is dense in Y (otherwise we would
find a non-trivial function in O(Y ) that vanishes on π(X) – this function
will be in Ker π ∗ ). We say that π is finite if O(X) is integral over O(Y ).
4.5.9. R EMARK . The reason finite maps are called finite is because they have
finite fibers: suppose we fix a point y ∈ Y ⊂ An . Suppose that π(x) = y for
some x ∈ X ⊂ Am . Then any linear function l on Am is a regular function
on X, and therefore satisfies a monic equation
ls + a1 ls−1 + . . . + as = 0,
where ai ∈ O(Y ). So we have
l(x)s + a1 (y)l(x)s−1 + . . . + as (y) = 0,
which implies that l(x) can have at most s different values, and so π −1 (y)
contains at most s points.
4.5.10. R EMARK . Of course not any map with finite fibers is finite: the typical counterexample is an inclusion
A1 \ {0} ⊂ A1 .
The pull-back homomorphism is an embedding C[x] ⊂ C[x, x1 ]. Notice that
1/x is not integral over C[x] (why?) . To resolve this confusion, regular maps
with finite fibers are called quasi-finite.
We have the following
4.5.11. T HEOREM . A finite map is surjective.
Algebraically, by Lemma 4.5.6, this follows from the following fact:
4.5.12. P ROPOSITION . Let A ⊂ B be rings. Suppose B is integral and finitely
generated over A. If M ⊂ A is a proper ideal then BM ⊂ B is a proper ideal.
Proof. Arguing by contradiction, suppose BM = B. Since B is finitely generated as an A-algebra and integral over it, B is a finite A-module (why?) .
Let b1 , . . . , bs be generators. Then we have a system of equation
X
mij bj = bi for some mij ∈ M for any i,
j
or
X
(δij − mij )bj
47
for any i.
j
Multiplying by the adjoint matrix (i.e. by Cramer’s rule), we have
Dbi = 0 for any i, where
D = det(δij − mij ).
But then D · 1 = 0. Since D = 1 − m for some m ∈ M , this implies that
1 ∈ M , contradiction.
§4.6. Quotient singularity 1r (1, a) and continued fractions. How to compute the algebra of invariants? In general it can be quite complicated but
things become much easier if the group is Abelian. Let’s focus on the most
useful example of a cyclic quotient singularity 1r (1, a). It is defined as follows:
consider the action of µr on C2 , where the primitive generator ζ ∈ µr acts
via the matrix
ζ 0
0 ζa
The cyclic quotient singularity is defined as the quotient
C2 /µr = MaxSpec C[x, y]µr .
How to compute this algebra of invariants? Notice that the group acts on
monomials diagonally as follows:
ζ · xi y j = ζ −i−ja xi y j .
So a monomial xi y j is contained in C[x, y]µr if and only if
i + ja ≡ 0
mod r.
There are two cases when the answer is immediate:
4.6.1. E XAMPLE . Consider 1r (1, r − 1). Notice that this is the only case when
µr ⊂ SL2 . The condition on invariant monomials is that
i≡j
mod r
(draw). We have
C[x, y]µr = C[xr , xy, y r ] = C[U, V, W ]/(V r − U W ).
So we see that the singularity 1r (1, r − 1) is a hypersurface in A3 given by
the equation V r = U W . It is called an Ar−1 -singularity.
4.6.2. E XAMPLE . Consider 1r (1, 1). The condition on invariant monomials is
i+j ≡0
mod r
(draw). We have
C[x, y]µr = C[xr , xr−1 y, xr−2 y 2 , . . . , y r ].
The quotient morphism in this case is
A2 → Ar+1 ,
(x, y) 7→ (xr , xr−1 y, xr−2 y 2 , . . . , y r ).
The singularity 1r (1, 1) with a cone over a rational normal curve
[xr : xr−1 y : xr−2 y 2 : . . . : y r ] ⊂ Pr−1 .
4.6.3. E XAMPLE . Here is a more random example:
48
JENIA TEVELEV
F IGURE 1
To describe a general cyclic quotient singularity 1r (1, a) we need an amusing concept called Hirzebruch–Young continued fractions. It looks just like
an ordinary continued fraction but with minuses instead of pluses. More
precisely, we have
4.6.4. D EFINITION . Let r > b > 0 be coprime integers. The following expression is called the Hirzebruch–Jung continued fraction:
r/b = a1 −
1
a2 −
1
a3 −...
= [a1 , a2 , ..., ak ].
For example,
5/1 = [5],
5/4 = [2, 2, 2, 2],
5/2 = [3, 2].
Here’s the result:
4.6.5. T HEOREM . Suppose µr acts on A2 with weights 1 and a, where a and r
r
are coprime. Let r−a
= [a1 , a2 , ..., ak ] be the Hirzebruch–Jung continued fraction
4
expansion . Then C[x, y]µr is generated by
f0 = xr , f1 = xr−a y, f2 , . . . , fk , fk+1 = y r ,
where the monomials fi are uniquely determined by the following equations:
−1
fi+1 = fiai fi−1
for i = 1, . . . , k.
(4.6.6)
4.6.7. R EMARK . We see that the codimension of A2 /Γ in the ambient affine
space Ak+2 is equal to the length of the Hirzebruch–Young continued fraction. This is a good measure of the complexity of the singularity. From
4Notice that we are expanding r/(r − a) and not r/a!
49
this perspective, 1r (1, 1) (the cone over a rational normal curve) is the most
complicated singularity: the Hirzebruch–Young continued fraction
r/(r − 1) = [2, 2, 2, . . . , 2] (r − 1 times)
uses the smallest possible denominators. It is analogous to the “standard”
continued fraction of the ratio of two consecutive Fibonacci numbers, which
uses only 1’s as denominators.
Proof of Theorem 4.6.5. This is completely combinatorial: invariant monomials in C[x, y]µr correspond to the intersection of the first quadrant
{(i, j) | i, j ≥ 0} ⊂ Z2
with the lattice
L = {(i, j) | i + aj ≡ 0
mod r} ⊂ Z2 .
This intersection is a semigroup and we have to find
The lat its
generators.
0
r
tice L contains the sublattice rZ2 (generated by
and
) and modr
0
r−a
(draw the “torus” Zr × Zr ).
ulo this sublattice L is generated by
1
r
0
So L is generated by
,
, and by the monomials inside the square
0
r
{(i, j) | 0 < i, j < r}, which are precisely the monomials
((r − a)j
mod r,
j),
j = 1, . . . , r − 1.
Of course many of these monomialsare unnecessary.
The first monomial in
r−a
the square that we actually need is
. Now start taking multiples of
1
r−a
. The next generator will occur when (r − a)j goes over r, i.e. when
1
j=d
r
e = a1
r−a
(in the Hirzebruch–Young continued fraction expansion for
a)a1 mod r = (r − a)a1 − r, the next generator is
(r − a)a1 − r
.
a1
r
r−a ).
Since (r−
Notice that so far this confirms our formula (4.10.1). We are interested in
the remaining generators of Linside the r × r square. Notice that they all
r−a
lie above the line spanned by
. So we can restate our problem: find
1
generators of the semigroup obtained by intersecting
L with
points lying
r−a
in the first quadrant and above the line spanned by
.
1
Next we notice that
r r−a
r − a (r − a)a1 − r
det
= det
= r.
0
1
1
a1
50
JENIA TEVELEV
0
(r − a)a1 − r
r−a
It follows that lattice L is also spanned by
,
, and
.
r
a1
1
We are interested in generators of the semigroup
intersecting
obtained
by 0
r−a
this lattice with the “angle” spanned by vectors
and
.
r
1
Consider the linear transformation ψ : R2 → R2 such that
1
0
0
1
, ψ
= r−a .
ψ
=
1
0
− 1r
r
Then we compute
0
0
r−a
r−a
ψ
=
, ψ
=
,
r
r−a
1
0
(r − a)a1 − r
(r − a)a1 − r
ψ
=
.
a1
1
So we get the same situation as before with a smaller lattice. Notice that if
r
1
= a1 −
r−a
q
then
r−a
,
(r − a)a1 − r
so we will recover all denominators in the Hirzebruch–Jung continued fraction as we proceed inductively.
q=
§4.7. Zariski tangent space. How do we know that n1 (1, a) is a singularity?
Recall that our current definition of the tangent space to an algebraic variety is as follows. Suppose X = V (I) ⊂ Ak is an affine variety and choose
generators
n ho1 , . . . , hs of I. Let p ∈ X. Then Tp X is a kernel of the Jacobian
matrix
∂hi
∂xj
, i.e. a linear subspace of vectors (a1 , . . . , ak ) such that
X ∂hi
aj = 0 for any i.
∂xj
j
This definition is very convenient if we know the generators hi of the ideal,
but what if we don’t? For example, in our case we have generators f1 , . . . , fk
of the algebra C[x, y]µr , which allows us to write
C[x, y]µr = C[x1 , . . . , xk ]/I.
But to compute generators of I, we would have to describe all relations between f1 , . . . , fk . This is possible but it would be nice to have a description
of the tangent space entirely in terms of the algebra of functions. It turns
out that this description is available and incredibly simple:
4.7.1. L EMMA . Let p ∈ X be a point of an affine algebraic variety that corresponds to the maximal ideal m ⊂ O(X). Then the cotangent vector space Tp∗ X is
canonically identified with the Zariski cotangent space m/m2 .
4.7.2. E XAMPLE . In the setup of Theorem 4.6.5, take the ideal m ⊂ C[x, y]µr
generated by f0 , . . . , fn+1 . Since C[x, y]µr /m ' C, this ideal is maximal and
hence gives a point p ∈ X = MaxSpec C[x, y]µr . Notice that the ideal m2 is
generated by pairwise products of generators fi fj . From our description of
51
generators f0 , . . . , fn+1 , it is clear that they are minimal, i.e. none of them
can be written as a non-trivial product of invariant monomials. Hence
Tp∗ X = m/m2 = hf0 + m2 , . . . , fn+1 + m2 i = Cn+1 .
Since dim X = 2, it follows that X is always singular at p (with the exception of the trivial case r = 1).
Proof of Lemma. We can realize X = V (I) ⊂ Ak . By shifting coordinates,
for simplicity let’s also assume that p ∈ X is at the origin. Then Tp∗ X is a
quotient vector space:
Tp∗ X = Ckx1 ,...,xk /hdh1 (0), . . . , dhk (0)i,
where
dhi (0) =
X ∂hi
(0) xj
∂xj
j
P
is a differential (or linearization of hi ). Notice that if h =
gi hi is some
other function in the ideal I then we have
X
X
dh(0) =
gi (0)dhi (0)+hi (0)dgi (0) =
gi (0)dhi (0) ∈ hdh1 (0), . . . , dhk (0)i
i
i
So we can also write
Tp∗ X = Ckx1 ,...,xk /hdh(0)ih∈I .
Now let’s untangle the other side of our formula. If M = (x1 , . . . , xk ) is the
maximal ideal of the origin, then we have
X
m/m2 = M/(I + M 2 ) = (x1 , . . . , xk )/(I +
xi xj ) = Ckx1 ,...,xk /hdh(0)ih∈I .
So the formula is proved.
As an extra bonus, this also shows that our definition of the tangent
space is intrinsic to the variety X, i.e. does not depend on the choice of
embedding in Ak .
§4.8. E8 -singularity. Let’s return to computation of A2 /Γ, where Γ ⊂ SL2
is the binary icosahedral group. Recall that Γ is the preimage of A5 ⊂ PSL2
thought of as a group of conformal transformations of P1 ' S 2 preserving
the inscribed icosahedron5.
We have to compute C[x, y]Γ . There is a miraculously simple way to
write down some invariants: A5 has three special orbits on S2 : 20 vertices of
the icosahedron, 12 midpoints of faces (vertices of the dual dodecahedron),
and 30 midpoints of the edges. Let f12 , f20 , and f30 be polynomials in x, y
that factor into linear forms that correspond to these special points.
We claim that these polynomials are invariant. Since Γ permutes their
roots, they are clearly semi-invariant, i.e. any γ ∈ Γ can only multiply them
by a scalar. Since they all have even degree, the element −1 ∈ Γ does not
change these polynomials. But Γ/{±1} ' A5 is a simple group, hence has
no characters at all, hence the claim.
5It is well-known that the image of this embedding SO (R) ⊂ PSL (C) is PSU (C). This
3
2
2
is a geometric interpretation of a famous 2 : 1 homomorphism SU2 (C) → SO3 (R). We are
not going to use this.
52
JENIA TEVELEV
4.8.1. T HEOREM . C[x, y]Γ = C[f12 , f20 , f30 ] ' C[U, V, W ]/(U 5 + V 3 + W 2 ).
Proof. Let’s try to prove this using as few explicit calculations as possible.
The key is to analyze a chain of algebras
C[x, y] ⊃ C[x, y]Γ ⊃ C[f12 , f20 , f30 ] ⊃ C[f12 , f20 ].
We claim that C[f12 , f20 ] ⊂ C[x, y] (and hence all other inclusions in the
chain) is an integral extension. In other words, we claim that a regular map
A2 → A2 ,
(x, y) 7→ (f12 , f20 )
(4.8.2)
is finite. We are only going to use the fact that f12 and f20 have no common
zeros in P1 . By Nullstellensatz, this implies that
p
(f12 , f20 ) = (x, y),
i.e. xn , y n ∈ (f12 , f20 ) for some large n. This implies that C[x, y] is finitely
generated as an C[f12 , f20 ]-module by monomials xi y j with i, j < n. We
have to check that any monomial xi y j can be written as a linear combination of monomials with i, j < n with coefficients in C[f12 , f20 ]. But using the
fact that xn , y n ∈ (f12 , f20 ), we can repeatedly rewrite xi y j as a linear combination of smaller and smaller monomials with coefficients in C[f12 , f20 ]
until all the monomials have degrees less than n. It follows that C[x, y] s
integral over C[f12 , f20 ].
Now let’s consider the corresponding chain of fraction fields
C(x, y) ⊃ Quot C[x, y]Γ ⊃ C(f12 , f20 , f30 ) ⊃ C(f12 , f20 ).
(You will show in the Exercises that Quot C[x, y]Γ = C(x, y)Γ but we are not
going to use this.)
Here are some basic definitions, and a fact.
4.8.3. D EFINITION . Let f : X → Y be a dominant map of irreducible affine
varieties of the same dimension. It induces an embedding of fields
f ∗ : C(Y ) ⊂ C(X).
Since the dimension is equal to the transcendence degree of the field of
functions, this embedding is algebraic, hence finite (because C(X) is finitely
generated). We define the degree of f as follows:
deg f = [C(X) : C(Y )].
Now suppose, in addition, that f is a finite map. Then the previous
definition applies, but we also know that f has finite fibers. We want to
compare deg f with the number of points in each fiber. Here we face one
subtlety: if C ⊂ A2 is a nodal cubic then its normalization A1 → C has
degree 1, but the fiber over a node has two points in it. To avoid this sort of
situation, here is another definition:
4.8.4. D EFINITION . An irreducible affine variety X is called normal if O(X)
is integrally closed in C(X).
Now the fact:
53
4.8.5. T HEOREM . Let f : X → Y be a finite map of irreducible affine varieties.
Suppose that Y is normal. Then any fiber f −1 (y) has at most deg f points. Let
U = {y ∈ Y
| f −1 (y) has exactly deg f points }.
Then U is open and non-empty.
Let’s postpone the proof and see how we can use it. First of all, any UFD
is integrally closed, hence C[x, y] and C[f12 , f20 ] are integrally closed.
Secondly, C[x, y]Γ is integrally closed. Indeed, if f ∈ Quot C[x, y]Γ is
integral over C[x, y]Γ then it is also integral over C[x, y], but the latter is
integrally closed, hence f is actually a polynomial, and so f ∈ C[x, y]Γ .
It follows that [C(x, y) : Quot C[x, y]Γ ] = 120 ( of course if we know that
the second field is C(x, y)Γ , this formula also follows from Galois theory).
The fibers of the map (4.8.2) are level curves of f12 and f20 , and therefore
contain at most 240 points by Bezout theorem. One can show geometrically
that general fibers contain exactly 240 points or argue as follows: if this
is not the case then Quot C[x, y]Γ = C(f12 , f20 , f30 ) = C(f12 , f20 ) and in
particular f30 ∈ C(f12 , f20 ). But f30 is integral over C[f12 , f20 ], and the latter
is integrally closed, so f30 ∈ C[f12 , f20 ]. But this can’t be the case because of
the degrees! So in fact we have
Quot C[x, y]Γ = C(f12 , f20 , f30 ) and [C(f12 , f20 , f30 ) : C(f12 , f20 )] = 2.
The latter formula implies that the minimal polynomial of f30 over C(f12 , f20 )
has degree 2. The second root of this polynomial satisfies the same integral
dependence as f20 , and therefore all coefficients of the minimal polynomial
are integral over C[f12 , f20 ], by Vieta formulas. But this ring is integrally
closed, and therefore all coefficients of the minimal polynomial are in fact
in C[f12 , f20 ]. So we have an integral dependence equation of the form
2 + af + b = 0, where a, b ∈ C[f , f ]. Looking at the degrees, there is
f30
30
12 20
only one way to accomplish this (modulo multiplying f12 , f20 , and f30 by
scalars), namely
5
3
2
f12
+ f20
+ f30
= 0.
It remains to prove that C[x, y]Γ = C[f12 , f20 , f30 ]. Since they have the
same quotient field, it is enough to show that the latter algebra is integrally
closed, and this follows from the following extremely useful theorem that
we are not going to prove, see [M3, page 198].
4.8.6. T HEOREM . Let X ⊂ An be an irreducible affine hypersurface such that its
singular locus has codimension at least 2. Then X is normal.
For example, a surface S ⊂ A3 with isolated singularities is normal. It
is important that S is a surface in A3 , it is easy to construct examples of
non-normal surfaces with isolated singularities in A4 .
Proof of Theorem 4.8.5. Let y ∈ Y and choose a function a ∈ O(X) that takes
different values on points in f −1 (y). The minimal polynomial F (T ) of a
over C(Y ) has degree at most deg f . Since Y is normal, all coefficients of
the minimal polynomial are in fact in O(Y ). Arguing as in Remark 4.5.9,
we see that f −1 (y) has at most n points. Since we are in characteristic 0,
the extension C(X)/C(Y ) is separable, and hence has a primitive element.
54
JENIA TEVELEV
Let a ∈ O(X) be an element such that its minimal polynomial (=integral
dependence polynomial) has degree n:
F (T ) = T n + b1 T n−1 + . . . + bn ,
bi ∈ O(Y ).
Let D ∈ O(Y ) be the discriminant of F (T ) and let U = {y ∈ Y | D 6= 0
be the corresponding principal open set. We claim that f has exactly n
different fibers over any point of U . Indeed, the inclusion O(Y )[a] ⊂ O(X)
is integral, hence induces a finite map, hence induces a surjective map. But
over a point y ∈ Y , the fiber of
MaxSpec O(Y )[a] = {(y, t) ∈ Y × A1 | tn + b1 (y)tn−1 + . . . + bn (y) = 0}
is just given by the roots of the minimal polynomial, and hence consists of
n points. Thus the fiber f −1 (y) also has n points.
§4.9. Resolution of singularities: cylindrical resolution. No discussion
of 1r (1, a) would be complete without describing its resolution of singularities. We are looking for a smooth surface S and a birational surjective map
π : S → A2 /µr . To avoid fake resolutions which essentially just remove
singular points, we have to assume that π is proper, which in analytic geometry means “has compact fibers”. In algebraic geometry the definition
is slightly more technical, and we are going to skip it for now.
As a warm-up, let’s look at 21 (1, 1), i.e. let’s resolve singularities of the
quadratic cone
Z = (V 2 − U W ) ⊂ A3 .
Notice that the locus in P2 given by the equation above is just a smooth
conic. Our quadratic cone is a cone over this conic: over each point p of
the conic we have the corresponding ruling Lp , which is a line in A3 . All
these lines of course pass through the origin, which is exactly the point
that creates singularity. The idea of the “cylinder” resolution is simply to
take the disjoint union of lines Lp (draw the picture). More precisely, the
resolution will be a line bundle S → C with the map S → Z ⊂ A3 defined
as follows.
We cover the conic C by two copies of A1 : the piece C0 = (W 6= 0) is
parametrised by [x2 : x : 1] and the piece C1 = (U 6= 0) is parametrised by
[1 : y : y 2 ]. On the overlap we have x = y −1 (this of course agrees with the
usual identification of a conic with P1 ). We take trivial line bundles C0 × C
(with coordinate w) and C1 × C (with coordinate u). The map C0 × C → A3
is defined by (x, w) 7→ (x2 w, xw, w). The map C1 × C → A3 is defined
by (y, u) 7→ (u, yu, y 2 u). Finally, we want to define a line bundle E on C
trivialized as above and a morphism E → A3 that on the cover restricts to
morphisms above. There is only one way to do this: on the overlap C1 ∩ C2 ,
we have x, y 6= 0, and the transition function should be
(x, w) 7→ (y, u) = (x−1 , x2 w).
Notice that the preimage of the singular point is nothing but the zero section of this line bundle, which is obviously isomorphic to P1 . One can show
that its self-intersection number is −2.
55
§4.10. Resolution of cyclic quotient singularities. Next we are going to
resolve a cyclic quotient singularity 1r (1, a). We will first discuss how the
resolution should look like and then state a theorem.
In the quadratic cone example above, the surface S was a line bundle.
This turns out to be a bit of a fluke: the fact of importance is that S is
covered by two charts C0 × C and C1 × C both isomorphic to A2 . This is
how we are going to construct our resolution in general: S will be the union
S = U1 ∪ . . . ∪ Ur
of r charts each isomorphic to A2 . We will construct regular maps
ψi : Ui → Z
for each i first. Then we will describe how these charts should be glued to
give a regular map S → Z: this will be our resolution. The surface S we
are going to construct will be neither affine nor projective. In fact it will be
quasi-projective but we won’t prove this as no realization of S as a subset
of the projective space is important enough to justify our attention. We will
describe a general definition of an algebraic variety obtained by gluing after
we finish the discussion of our example: it will provide a useful illustration.
So how to construct maps these maps ψi : Ui → Z? Recall that we want
each chart Ui to be isomorphic to A2 , so let’s process one chart at a time and
denote coordinates in the chart by ζi , ηi . Algebraically, we have to define
the pullback homomorphisms
ψi∗ : C[x, y]µr → C[ζi , ηi ].
A simple way to do this would be to send “monomials to monomials”: for
any invariant monomial xα y β ∈ C[x, y]µr , we want ψi (xα y β ) to be equal to
0
0
some monomial ζiα ηiβ ∈ C[ζi , ηi ]. The condition of being a homomorphism
then simply means that the map
(α, β) 7→ (α0 , β 0 )
is a linear map. Recall that the semigroup of invariant monomials can be
described as
Λ = M ∩ (Z≥0 )2 ,
where M = {(α, β),
α + aβ ≡ 0
mod (r)} ⊂ Z2
is a lattice. So we have to define a linear map of semigroups
Ai : Λ → (Z≥0 )2 .
We want each ψi to be dominant, so we want each pull-back homomorphism ψi∗ to be injective, so we want our linear map Ai to be injective. In
fact, we want each ψi to be birational, hence we will impose a condition
that Ai is a restriction of an isomorphism of lattices
Ai : M ' Z2 ,
which we will denote by the same letter.
The trick is to construct a dual linear isomorphism A∗ instead
A∗i : Z2 ' N,
56
JENIA TEVELEV
where N is the lattice dual to M . We can then restrict A∗i and get a map of
semigroups
A∗i : (Z≥0 )2 → Λ∗ ,
where Λ∗ is the dual semigroup. All these dual objects are defined as follows. We will identifying R2 and (R2 )∗ by means of the standard inner
product. Then
N = {(a, b) ∈ R2 | aα + bβ ∈ Z for any (α, β) ∈ M }.
Since M is generated by (r, 0), (0, r), and (r − a, 1), this gives
1
N = {(i, j) ∈ Z2 | i + (r − a)j ≡ 0 mod r} ⊂ R2
r
and
Λ∗ = N ∩ (Z≥0 )2 .
1
For example, for 12
(1, 5), we will get the following semigroup Λ∗ :
N looks just like M for the dual cyclic quotient singularity 1r (1, r − a)
(after rescaling). In particular, Theorem 4.6.5 shows that Λ∗ is generated by
a 1
,
, e2 , . . . , ek , ek+1 = (0, 1),
e0 = (1, 0), e1 =
r r
where vectors ei are uniquely determined by the following equations:
ei+1 = bi ei − ei−1
(4.10.1)
for i = 1, . . . , k,
6
r
a
where = [b1 , b2 , ..., bk ] is the Hirzebruch–Jung continued fraction .
It is clear from this inductive description that the lattice N is generated
by ei , ei+1 for any i. So finally we can define our maps: the map ψi : Ui →
C2 /µr is defined by its pull-back ψ ∗ : C[x, y]µr → C[ζi , ηi ], which is defined
by the linear map of semigroups Ai : Z2≥0 → Λ∗ , which sends basis vectors
of Z2≥0 to ei and ei+1 .
This looks scary but in fact it’s just linear algebra. For example, let’s
work out the gluing. The rational map
−1
ψi+1
◦ ψi :
6Notice that here we are expanding r/a.
Ui 99K Ui+1
57
should be the identity on the overlap. This is a “monomial” map, and chasing the definition above, we see that the linear map of lattices dual to lattices of monomials is given by a matrix
bi 1
.
−1 0
It follows that the linear map of lattices of monomials is given by a transposed matrix
bi −1
, i.e. ζi+1 = ζibi ηi , ηi+1 = ζi−1 .
1 0
So the gluing is just defined by these formulas above and goes like this:
Ui \ {ζi = 0} ' Ui+1 \ {ηi+1 = 0}.
Now we can visualize the resolution as follows:
We would like to have a framework where this sort of gluing is allowed.
The smallest “universe” is to consider “toric varieties”: those are glued
from pieces isomorphic to An (or more generally by spectra of subrings
in C[x1 , . . . , xn ] generated by some monomials) where the gluing is also
monomial, see [Fu] for the introduction. This is a useful but extremely
restrictive class of varieties to work with. Practically no varieties of importance are toric. However, many interesting singular varieties “analytically”
or “formally analytically” look like a singular toric variety in the analytic
(but not Zariski!) neighborhood of a singular point. For example, the most
common singularity of a surface is a cyclic quotient 1r (1, a). Another reason
to study toric varieties is that many important varieties are naturally embedded in a toric variety, for example any plane curve is embedded in P2 ,
which is of course toric. Practically all known example of Calabi–Yau threefolds (which are important in physics) are constructed as hypersurfaces in
toric varieties. We won’t develop a general theory of toric varieties: it is
very close in spirit to the Hirzebruch–Jung resolution considered above
(which was one of the inspirations for toric geometry) and if you understand this resolution you will understand the general construction without any difficulty. In the next chapter we will study another example of
58
JENIA TEVELEV
a toric variety, namely the weighted projective space. This will help us
to understand a very general construction, called Proj, which can be used
to define varieties “by gluing”. After that it will be easy to introduce the
most general class of “algebraic varieties”, which are not necessarily quasiprojective. This class will include all previously worked out examples.
§4.11. Homework.
Problem 1. For the cyclic quotient singularity 17 (1, 3), compute generators of C[x, y]µ7 , realize A2 /µ7 as an affine subvariety of A? , and compute
the ideal of this subvariety. (2 points)
Problem 2. For the cyclic quotient singularity 51 (1, 4), show how to write
down explicitly 5 affine charts Yi ' A2 , regular morphisms Yi → A2 /µ5 ,
and gluing maps between affine charts such that Y = ∪Yi is a resolution of
singularities of A2 /µ5 ⊂ A3 . (2 points)
Problem 3. Let G be a finite group acting on an affine variety X by automorphisms. (a) Show that there exists a closed embedding X ⊂ Ar such
that G acts linearly on Ar inducing the original action on X. (b) Show that
the restriction homomorphism O(V )G → O(X)G is surjective. (c) Show
that O(X)G is finitely generated (2 points).
Problem 4. Let G be a group acting linearly on a vector space V . Let
L ⊂ V be a linear subspace. Let
Z = {g ∈ G | g|L = Id|L },
N = {g ∈ G | g(L) ⊂ L},
and W = N/Z.
(a) Show that there exists a natural homomorphism π : O(V )G → O(L)W .
(b) Suppose there G · L = V . Show that π is injective. (2 points).
Problem 5. Let G = SOn (C) be an orthogonal group preserving a quadratic form f = x21 +. . .+x2n . Show that C[x1 , . . . , xn ]G = C[f ]. (Hint: apply
the previous problem to L = Ce1 ). (1 point).
Problem 6. Let G = GLn be a general linear group acting on Matn by
conjugation. (a) Let L ⊂ Matn be the space of diagonal matrices. Show that
G · L = Matn . (b) Show that O(Matn )G is generated by coefficients of the
characteristic polynomial (2 points).
Problem 7. (a) In the notation of the previous problem, describe all fibers
of the quotient morphism π : Matn → Matn /G. (b) Show that not all
orbits are separated by invariants and find all orbits in the fiber π −1 (0).
(c) Describe all fibers of π that contain only one orbit (3 points).
Problem 8. (a) Let G be a finite group acting linearly on a vector space V .
Show that C(V )G (the field of invariant rational functions) is equal to the
quotient field of O(V )G . (b) Show that (a) can fail for an infinite group.
(c) Show that if G is any group acting linearly on a vector space V then
any invariant rational function f ∈ C(V )G can be written as a ratio of two
semi-invariant functions of the same weight. (3 points).
Problem 9. Consider the standard linear action of the dihedral group
Dn in R2 (by rotating the regular n-gon) and tensor it with C. Compute
generators of the algebra of invariants C[x, y]Dn (2 points).
Problem 10. Let R be an integral finitely generated graded algebra. Show
that MaxSpec R is smooth if and only if R is isomorphic to a polynomial
algebra. (1 point)
59
Problem 11. Show that a finite map between affine varieties takes closed
sets to closed sets. (1 point).
Problem 12. Let f : X → Y be a regular map of affine varieties such
that every point y ∈ Y has an affine neighborhood U ⊂Y such that f −1 (U )
is affine and the restriction f : f −1 (U ) → U is affine. Show that f itself is
finite. (2 points)
P
Problem 13. Let R =
k≥0 Rk be a finitely generated graded algebra.
Recall that a function
h(k) = dimC Rk
is called the Hilbert function of R and its generating function
X
P (t) =
h(k)tk
1
is called the Poincare series. (a) Show that P (t) = (1−t)
n for R = C[x1 , . . . , xn ].
S
n
(b) Let R = C[x1 , . . . , xn ] . Show that P (t) = (1−t)(1−t12 )...(1−tn ) . (1 point)
Problem 14. (a) Suppose that R is a graded algebra generated by homogeneous generators r1 , . . . , rs of degrees k1 , . . . , ks . Show that the Poincare
series P (t) is a rational function of the form
f (t)
(1 − tk1 ) . . . (1 − tks ),
where f (t) is a polynomial with integer coefficients. (b) Compute Poincare
series of the algebra C[x, y]µr , where µr acts as 1r (1, r − 1). (c) Compute
Poincare series of C[x, y]Γ , where Γ is a binary icosahedral group. (3 points).
Problem 15. Suppose that R is a graded algebra generated by finitely
many generators all in degree 1. Show that there exists a Hilbert polynomial
H(t) such that h(k) = H(k) for k 0. (2 points)
Problem 16. (a) Let F : An → An be a morphism given by homogeneous
polynomials f1 , . . . , fn such that V (f1 , . . . , fn ) = {0}. Show that F is finite.
(b) Give example of a dominant morphism A2 → A2 which is not finite
(2 points).
Problem 17. Let G be a group acting by automorphisms on a normal
affine variety X. Show that the algebra of invariants O(X)G is integrally
closed. (1 point)
Problem 18. Let A be an integrally closed domain with field of fractions
K and let A ⊂ B be an integral extension of domains. Let b ∈ B and let
f ∈ K[x] be its minimal polynomial. Show that in fact f ∈ A[x]. (1 point)
Problem 19. Consider the action of SL2 on homogeneous polynomials in
x and y of degree 6 written as follows:
ζ0 x6 + 6ζ1 x5 y + 15ζ2 x4 y 2 + 20ζ3 x3 y 3 + 15ζ4 x2 y 4 + 6ζ5 xy 5 + ζ6 y 6 .
Show that the function

ζ0
ζ1
det 
ζ2
ζ3
ζ1
ζ2
ζ3
ζ4
ζ2
ζ3
ζ4
ζ5

ζ3
ζ4 

ζ5 
ζ6
belongs to the algebra of invariants C[ζ0 , ζ1 , . . . , ζ6 ]SL2 . (2 points)
60
JENIA TEVELEV
Problem 20. Consider the action of An on An by permutations of coordinates. Show that C[x1 , . . . , xn ]An is generated by Q
elementary symmetric
polynomials σ1 , . . . , σn and the discriminant D =
(xi − xj ) (2 points).
1≤i<≤j
61
62
JENIA TEVELEV
§5. Weighted projective spaces
§5.1. First examples. Let’s move from quotient spaces by finite groups to
the actions of the easiest infinite group: C∗ . We will work out the following
extremely useful example: fix positive integers a0 , . . . , an (called weights)
and consider the action of C∗ on An+1 defined as follows:
λ · (x0 , . . . , xn ) = (ta0 x0 , . . . , tan xn ) for any t ∈ C∗ .
The quotient (which we are going to construct) is called the weighted projective space. Notation:
P(a0 , . . . , an ).
For example, we have
P(1, . . . , 1) = Pn .
5.1.1. E XAMPLE . We have met P(4, 6) before. Recall that any elliptic curve
has a Weierstrass equation y 2 = 4x3 − g2 x − g3 , ∆ = g23 − 27g32 6= 0 and
this is an extremely useful fact for studying elliptic fibrations (and elliptic
curves defined over rings of algebraic integers). Coefficients g2 and g3 are
defined not uniquely but only up to admissible transformations
g2 7→ t4 g2 ,
g3 7→ t6 g3 .
So the moduli space of elliptic curves is P(4, 6) with a point “at infinity”
removed (which corresponds to the C∗ -orbit {∆ = 0}). So it should come at
no surprise that
P(4, 6)[g2 :g3 ] ' P1[j:1] ,
where j is given by the usual formula (3.1.4). Notice however that thinking
about P(4, 6) has a lot of advantages: it encompasses the idea of Weierstrass families better and it emphacizes the role of special elliptic curves
with many automorphisms. In general, we will see that weighted projective spaces are different from usual ones: they have singularities.
5.1.2. E XAMPLE . Let’s construct P(1, 1, 2) by hand. Take the map
π : A3x,y,z \ {0} → P3[A:B:C:D] ,
(x, y, z) 7→ [x2 : xy : y 2 : z].
It is easy to see that it separates orbits, i.e. π(x, y, z) = (x0 , y 0 , z 0 ) if and only
if there exists t ∈ C∗ such that
x0 = tx, y 0 = ty, z 0 = t2 z.
The image is a quadratic cone AB = C 2 in P3 .
Now let’s discuss the general construction of the weighted projective
space. Remember the drill: we have to find all semi-invariants. Here this
is exceptionally easy: any monomial xi00 . . . xinn is a semi-invariant for the
C∗ -action of weight w = i0 a0 + . . . + in an (i.e. t ∈ C∗ acts by multiplying this monomial by tw ). So the algebra of semi-invariants is just the full
polynomial algebra
C[x0 , . . . , xn ].
However, we have to introduce a different grading on this algebra, where
each variable xi has degree ai . Here are some basic observations:
63
• There are no non-constant invariants. So we can not produce a quotient by our method of taking MaxSpec of the algebra of invariants
(by taking the image of the map to Ar given by r basic invariants).
Here is an “explanation”: all orbits contain zero in their closure.
So any invariant polynomial is just a constant equal to the value of
this polynomial in 0. This is the reason we have to remove zero,
just like in the Pn case. Notice that An+1 \ {0} is not an affine variety anymore. The procedure of taking MaxSpec won’t work after
removing the origin.
• The algebra of semi-invariants is generated by variables, which have
different degrees. So the situation is different from our experience
of writing the Grassmannian G(2, n) as a quotient Mat(2, n)/GL2 ,
where basic semi-invariants (2 × 2 minors) all had the same degree.
So we need a new approach. The idea is simple: An+1 \ {0} is covered by
principal open sets D(xi ). We will take take their quotients by C∗ first and
then glue them, just like in the definition of the usual projective space.
In the case of Pn we don’t even notice the C∗ action because we kill it by
setting xi = 1. So we quite naturally identify D(xi )/C∗ ' An . Let’s denote
the corresponding chart Dxi ⊂ Pn to distinguish it from D(xi ) ⊂ An+1 .
What will happen in a more general case? Setting xi = 1 does not quite
eliminate t: it just implies that tai = 1. This is still an achievement: it shows
that the action of C∗ on D(xi ) ⊂ An+1 is reduced to the action of µai on An .
This is a familiar ground: the quotient will be
Dxi = D(xi )/C∗ ' An /µai = MaxSpec C[x0 , . . . , x̂i , . . . , xn ]µai ,
where µai acts with weights a0 , . . . âi , . . . , an . So for example, a projective
quadratic cone P(1, 1, 2) is covered by three charts: two copies of A2 and
one copy of 21 (1, 1), which is isomorphic to an affine quadratic cone.
Here is another way of thinking about this. Notice that
1
O(D(xi )) = C x0 , . . . , xn ;
xi
and that C∗ now acts on the affine variety D(xi ). We can use our old recipe
for computing the quotient: take the algebra of invariants and compute its
spectrum. So we set
p
C∗
O(Dxi ) = O(D(xi )) =
| p ∈ C[x0 , . . . , xn ], deg p = kai
xki
(here and after the degree deg is our funny weighted degree). There are
two cases: if ai = 1 then we just have
x0
xn
O(Dxi ) = C a0 , . . . , an ' C[y1 , . . . , yn ].
xi
xi
The chart is an affine space, just like for the standard Pn . To figure out the
general case, for simplicity let’s restrict to the weighted projective plane
P(a0 , a1 , a2 ). What will be the first chart? Consider the cyclic field extension
C(x0 , x1 , x2 ) ⊂ C(z0 , x1 , x2 ),
64
JENIA TEVELEV
where x0 = z0a0 . Then we have
(
p
| p ∈ C[x0 , x1 , x2 ],
O(Dxi ) =
a0 k
z0

X

i,j
aij
x1
z0a1
i x2
z0a2
)
deg p = ka0
j
|
a1 i + a2 j ≡ 0
mod a0



=
⊂C
x1 x2
.
,
z0a1 z0a2
So we get a subalgebra in C[y1 , y2 ] spanned by monomials y1i y2j such that
a1 i + a2 j ≡ 0 mod a0 . This is our old friend, the cyclic quotient a10 (a1 , a2 ).
§5.2. Proj (projective spectrum). Let’s generalize this even further. Let
R be any finitely generated graded integral domain such that R0 = C.
We can write R as a quotient of C[x0 , . . . , xn ] (with grading given by degrees a0 , . . . , an of homogeneous generators of R) by a homogeneous (in
this grading!) prime ideal. Functions in this ideal are constant along C∗
orbits in An+1 . As a set, we simply define
Proj R ⊂ P(a0 , . . . , an )
as a set of C∗ -orbits where all functions in the ideal vanish.
Rational functions on Proj R are defined as ratios of polynomials of the
same (weighted) degree, i.e.
C(Proj R) = (Quot R)0 ,
where the subscript means that we are only taking fractions of degree 0.
We call a function regular at some point if it has a presentation as a fraction with a denominator non-vanishing at this point. It is clear that Proj R
is covered by affine charts Df for each homogeneous element f ∈ R of
positive degree, where
O(Df ) = R[1/f ]0 .
What is the gluing? Given Df and Dg , notice that
Df ∩ Dg = Df g ,
is a principal open subset in both Df (where it is a complement of a vanishdeg g
deg f
ing set of a regular function fg deg g ) and Dg (where we use fgdeg f ). Formally
speaking, we have to check that in C(Proj R) we have
deg g f
R[1/f g]0 = R[1/f ]0 deg f .
(5.2.1)
g
This kind of formulas are proved by tinkering with fractions with a sole
purpose to balance degrees of the numerator and the denominator. We
leave it as an exercise.
65
§5.3. Abstract algebraic varieties. To continue this discussion, we have to
ask ourselves: what is it that we are trying to prove? We will later see that
Proj R is in fact a projective variety, but at this point it would be useful to
give a definition of an abstract algebraic variety.
5.3.1. D EFINITION . For simplicity, we will only define an irreducible algebraic
variety X. We need
• A finitely generated field extension K of C. This will be a field of
rational functions on X.
• Topology on X.
• For each open subset U ⊂ X we need a subalgebra OX (U ) ⊂ K. It
should satisfy the condition
!
\
[
OX
OX (Ui ).
Ui =
i∈I
i∈I
OX is called the structure sheaf.
• Finally, X should admit a finite cover {Ui } such that each Ui (with
an induced topology) is an irreducible affine variety (with Zariski
topology) with function field K and for each open subset V ⊂ Ui ,
OX (V ) ⊂ K is the algebra of rational functions regular on V .
In practice, algebraic varieties are constructed by gluing affine varieties.
Suppose A and B are irreducible affine varieties with the same function
field K. Suppose, in addition, that there exists another affine variety C and
open immersions
iA : C ,→ A
and iB : C ,→ B.
Then we define the topological space X = A ∪C B by identifying points
iA (x) with iB (x) for any x ∈ C and by declaring a subset U ⊂ X open if
U ∩ A and U ∩ B is open. Finally, we set
OX (U ) = OA (U ∩ A) ∩ OB (U ∩ B)
It is easy to generalize this to several affine charts: we need irreducible
affine varieties
U0 , . . . , U r ,
with the same function field. For each pair Ui , Uj we have affine open subsets
Uij ⊂ Ui , Uji ⊂ Uj
and an isomorphism
φij : Uij → Uji .
This isomorphism should satisfy (draw pictures)
• φij = φ−1
ji ,
• φij (Uij ∩ Uik ) = Uji ∩ Ujk , and
• φik = φjk ◦ φij on Uij ∩ Uik .
5.3.2. L EMMA . Proj R is an algebraic variety.
66
JENIA TEVELEV
Proof. We have K = (Quot R)0 . For any homogeneous f ∈ R we have an
affine variety
Df = MaxSpec R[1/f ]0 .
To get a finite atlas, take only homogeneous generators of R. To see the gluing condition, notice that Df g is a principal open subset in both Df and Dg .
The compatibility conditions on triple overlaps are of set-theoretic nature,
and are clearly satisfied.
§5.4. Separatedness. There is one annoying phenomenon that we can discuss now and then safely ignore later on. One can take two copies of A1
and glue them along D(x) = A1 \ {0}. This produces a famous “line with
two origins” (draw). What’s happening here is that diagonally embedded
D(x) is not closed in the product of charts (draw), compare with how P1 is
glued (draw). So we give
5.4.1. D EFINITION . An algebraic variety is called separated if it has an affine
atlas such that for any pair A, B of charts with C = A ∩ B, the diagonal
inclusion of C in A × B is a closed subset of the product.
How to check this in practice?
5.4.2. L EMMA . Suppose any two affine charts A and B with C = A ∩ B have the
following property: there exists f ∈ O(A) and g ∈ O(B) such that
O(C) = O(A)f = O(B)g ⊂ K.
Then X is separated iff for any A and B, we have
O(C)
is generated by
O(A) and
O(B).
In particular, Proj R is separated.
Proof. We have O(A × B) = O(A) ⊗k O(B) (why?) , and the diagonal map
∆ : C → A × B is given by a homomorphism
f g
∆∗ : O(A) ⊗k O(B) → O(C), f ⊗ g 7→ · .
1 1
The closure (∆(C)) of the diagonal is defined by the kernel of ∆∗ . In particular, its algebra of functions is O(A)O(B) ⊂ O(C). So X is separated iff
this inclusion is an equality for any pair of charts.
The last remark follows from the formula
(Rf g )0 = (Rf )0 (Rg )0 ,
which we leave as an exercise.
(5.4.3)
§5.5. Veronese embedding. We now have two models for P(1, 1, 2): as a
weighted projective plane defined by charts and as a quadratic cone in P3 .
What is the relationship between these models? We are going to show that
in fact any Proj R is a projective variety.
P
5.5.1. D EFINITION . If R is a graded ring then its subring R(d) = d|n Rn is
called a d-th Veronese subring.
For example, for P(1, 1, 2) the second Veronese subring is generated by
and z, subject to a single quadratic relation. So Proj R(2) is a
quadratic cone in P3 in this case. The basic fact is:
x2 , xy, y 2 ,
67
5.5.2. P ROPOSITION . Proj R = Proj R(d) for any d.
Proof. First of all, we have (Quot R)0 = (Quot R(d) )0 . Indeed, any fraction
a/b ∈ (Quot R)0 can be written as abd−1 /bd ∈ (Quot R(d) )0 .
Let f1 , . . . , fr be homogeneous generators of R, so that Proj R is covered by charts Dfi . Then f1d , . . . , frd ∈ R(d) are not necessarily generators,
however Proj R(d) is still covered by charts Df d . Indeed, if all fid vanish at
i
some point p ∈ Proj R(d) then also any function in the ideal generated by
them (and hence any function in its radical) vanishes at p. But any generator g of R(d) can be expressed as a polynomial in f1 , . . . , fr , and therefore
a sufficiently high power of g belongs to the ideal (in R(d) ) generated by
f1d , . . . , frd . So we have
Proj R =
r
[
D fi
and
Proj R(d) =
i=1
r
[
Df d
i
i=1
The basic local calculation we need is that charts Dfi of Proj R and Df d
i
of Proj R(d) can be identified, i.e. that
R(d) [1/f d ](0) ' R[1/f ](0)
for any homogeneous element f of R. But indeed,
f dj−i g
g
=
fi
f dj
as soon as dj > i. So Proj R and Proj Rd have the same charts glued in the
same way.
Now another basic algebraic fact is:
5.5.3. L EMMA . For a sufficiently large d, R(d) is generated by Rd .
Proof. Let a1 , . . . , ar be degrees of homogeneous generators f1 , . . . , fr of R.
Let a = l. c. m.(a1 , . . . , ar ) and let d = ra. We claim that this d works. For
each i, let a = ai bi : then
deg fibi = a.
Now take any element f ∈ Rkd . We claim that it can be written as a polynomial in elements of Rd . It suffices to consider a monomial f = f1n1 . . . frnr .
For inductive purposes, notice that deg f = kd = (kr)a. If ni < bi for each i
then
deg f < ra = d,
a contradiction. So we can write f = fibi g, where deg g = deg f − a. Continuing inductively, we will write
bi
b
f = [fi1 1 . . . fisis ]g,
where deg g = d and degree of the first term is a multiple of d. Since
deg fibi = a for each i, we can group elements of the first term into groups of
r powers each of degree d. This shows that f can be written as a polynomial
in elements of Rd .
68
JENIA TEVELEV
By the lemma we can realize Proj R as a subvariety in PN for a sufficiently large N . Indeed, Proj R ' Proj R(d) and
R(d) = C[y0 , . . . , yN ]/I,
where I is a homogeneous ideal (in the usual sense). So
Proj R(d) = V (I) ⊂ PN .
5.5.4. C OROLLARY. Proj R is a projective variety.
69
§6. Genus 2 curves.
We are going to spend a considerable amount of time studying the moduli space M2 of algebraic curves of genus 2. Incidentally, this will also give
us the moduli space A2 of principally polarized Abelian surfaces: those are algebraic surfaces isomorphic to C2 /Λ, where Λ ' Z4 is a lattice. So Abelian
surfaces are naturally Abelian groups just like elliptic curves. We will see
that M2 embeds in A2 as an open subset (via the Jacobian construction) and
the complement A2 \ M2 parametrizes split Abelian surfaces of the form
E1 × E2 , where E1 and E2 are elliptic curves. The map Mg ,→ Ag can be
constructed in any genus (its injectivity is called the Torelli theorem) but the
dimensions are vastly different:
g(g + 1)
.
2
The characterization of Mg as a sublocus of Ag is called the Shottky problem.
dim Mg = 3g − 3
and
dim Ag =
§6.1. Genus 2 curves: analysis of the canonical ring. Let’s start with a
basic Riemann–Roch analysis of a genus 2 curve C. We fix a canonical
divisor K. We have
deg K = 2 × g − 2 = 2 and l(K) = g = 2.
So we can assume that
K≥0
is an effective divisor. by Riemann–Roch, for any point P ∈ C,
l(K − P ) − l(K − (K − P )) = 1 − 2 + deg(K − P ) = 0.
Since l(P ) = 1 (otherwise C is isomorphic to P1 ), we have l(K − P ) = 1.
So |K| has no fixed part, and therefore gives a degree 2 map
φ|K| : C → P1 .
By Riemann–Hurwitz, it has 6 ramification points called Weierstrass points.
We also see that C admits an involution permuting two branches of φ|2K| .
It is called the hyperelliptic involution.
Now consider |3K|. By Riemann–Roch, we have l(3K) = 5 and l(3K −
P − Q) = 3 for any points P, Q ∈ C. It follows that |3K| is very ample and
gives an embedding
C ,→ P4 .
To get a bit more, we observe that most of geometry of C is nicely encoded in the canonical ring
R(K) =
∞
M
L(nK).
n=0
We can give a more general definition:
6.1.1. D EFINITION . Let D ≥ 0 be an effective divisor on a curve C. Its
graded algebra is defined as follows:
R(D) =
∞
M
n=0
L(nD).
70
JENIA TEVELEV
This is a graded algebra: notice that if f ∈ L(aD) and g ∈ L(bD) then
(f g) + (a + b)D = (f ) + aD + (g) + bD ≥ 0,
so f g ∈ L(a + b)D.
6.1.2. R EMARK . We have only defined divisors on curves in this class, but
in principle it is no harder to defined a graded algebra of any divisor on an
algebraic variety of any dimension. The canonical ring R(K) of a smooth
variety of dimension n was a subject of a really exciting research in the last
30 years which culminated in the proof of a very important theorem of Siu
and Birkar–Cascini–Hacon–McKernan: R(K) is a finitely generated algebra. This does not sound like much, but it allows us to define Proj R(K),
the so-called canonical model of X. It is easy to see that it depends only on
the field of rational functions C(X). In the curve case, C is uniquely determined by its field of functions, by in dimension > 1 it is easy to modify a
variety without changing its field of rational functions (e.g. by blow-ups).
So it is very handy to have this canonical model of the field of rational functions. There exists a sophisticated algorithm, called the Minimal Model
Program, which (still conjecturally) allows one to construct the canonical
model by performing a sequence of basic “surgeries” on X called divisorial contractions and flips.
We can compute the Hilbert function of R(K) by Riemann–Roch:


1





2
hn (R(K)) = l(nK) = 3



5



2n − 1
if
if
if
if
if
n=0
n=1
n=2
n=3
n ≥ 2.
Let’s work out the generators. L(0) = C is generated by 1. This is a unity
in R(K). Let x1 , x2 be generators of L(K). One delicate point here is that
we can (and will) take x1 to be 1 ∈ C(C), but it should not be confused with
a previous 1 because it lives in a different degree in R(K)! In other words,
R(K) contains a graded polynomial subalgebra C[x1 ], where any power xn1
is equal to 1 as a rational function on C.
Any other element of first degree has pole of order 2 at K (because if it
has a pole of order 1, it would give an isomorphism C ' P1 .
A subalgebra S = C[x1 , x2 ] of R is also a polynomial subalgebra: if we
have some homogeneous relation f (x1 , x2 ) of degree d then we have
d
Y
f (x1 , x2 ) =
(αi x1 + βi x2 ) = 0 in C(C),
i=1
which implies that αi x1 + βi x2 = 0 for some i, i.e. that x1 and x2 are not
linearly independent, contradiction.
71
The Hilbert function of S is


1





2
hn (S) = 3



4



n
if
if
if
if
if
n=0
n=1
n=2
n=3
n ≥ 2.
So the next generator we need for R(K) is a generator y in degree 3.
What happens in degree 4? We need 7 elements and we have 7 elements
x41 , x31 x2 , x21 x32 , x1 x32 , x42 ,
yx1 , yx2 .
We claim that they are indeed linearly independent, and in fact we claim:
6.1.3. L EMMA . There is no linear relation in C(C) of the form
yfk (x1 , x2 ) = fk+3 (x1 , x2 ),
where the lower index is the degree. In particular, R(K) is generated by x1 , x2 , y.
Proof. Suppose the linear relation of the form above exists. Then y, as a
rational function on C, is a rational function f (x1 , x2 ). One can show that
this is impossible either by an elementary analysis of possible positions of
roots of y and this rational function f (x1 , x2 ) or by simply invoking the fact
that as we already know 3K is very ample, and in particular functions in
|3K| separate points of C. But if y is a rational function in x1 and x2 then y
takes the same values on two points from each fiber of φ|2K| .
It follows that
6.1.4. L EMMA . R(K) is isomorphic to a polynomial algebra in x1 , x2 , y modulo a
relation
y 2 = f6 (x1 , x2 ),
where f6 is a polynomial of degree 6.
Proof. We already know that R(K) is generated by x1 , x2 , y, and that y 6∈
C(x1 , x2 ). It follows that y 2 , yC[x1 , x2 ]3 , and C[x1 , x2 ]6 are linearly dependent in R(K)6 and this gives the only relation in R(K):
y 2 = yf3 (x1 , x2 ) + f6 (x1 , x2 ).
We can make a change of variables y 0 = y − 21 f3 to complete the square,
which brings the relation in the required form.
§6.2. Graded algebra of an ample divisor. Now let’s interpret these algebraic results geometrically. The basic fact is:
6.2.1. L EMMA . If D is an ample divisor on a curve C then Proj R(D) = C.
Proof. If D is very ample and R(D) is generated by R(D)1 then R(D) is
isomorphic to a a polynomial algebra in x0 , . . . , xN ∈ L(D) modulo the
relations that they satisfy, i.e. R(D) = C[x0 , . . . , xN ]/I, where I is a homogeneous ideal of C ⊂ PN . So in this case clearly Proj R(D) = C. In general,
if D is ample then kD is very ample for some k > 0. Also, we know by
Lemma 5.5.3 that the Veronese subalgebra R(lD) = R(D)(l) is generated
72
JENIA TEVELEV
by its first graded piece for some l > 0. So klD is a very ample divisor
and R(klD) = R(kl) is generated by its first graded piece. Then we have
Proj R(D) = Proj R(klD) = C. We are not using here that C is a curve, so
if you know your divisors in higher dimension, everything works just as
nicely.
As a corollary, we have
6.2.2. C OROLLARY. Let C be a genus 2 curve. Then R(K) induces an embedding
C ⊂ P(1, 1, 3)
and the image is defined by an equation
y 2 = f6 (x1 , x2 ).
(6.2.3)
The embedding misses a singularity of P(1, 1, 3) (where x1 = x2 = 0, y = 1).
In the remaining two charts of P(1, 1, 3), the curve is given by equations
y 2 = f6 (1, x2 ) and y 2 = f6 (x1 , 1).
The projection onto P1[x1 :x2 ] is a bicanonical map φ|2K| and roots of f6 are branch
points of this 2 : 1 cover. In particular, f6 has no multiple roots and any equation
of the form (6.2.3) defines a genus 2 curve.
The tricanonical embedding C ⊂ P4 factors through the Veronese embedding
P(1, 1, 3) ,→ P4 ,
(x1 , x2 , x3 , y) 7→ [x31 : x21 x2 : x1 x22 : x32 : y],
where the image is a projectivized cone over a rational normal curve.
This sets up a bijection between curves of genus 2 and unordered 6tuples of distinct points p1 , . . . , p6 ∈ P1 modulo PGL2 . We are going to
use this to construct M2 . The classical way of thinking about 6 unordered
points in P1 is to identify them with roots of a binary form f6 (x1 , x2 ) of
degree 6. Let V6 be a vector space of all such forms and let D ⊂ P(V6 )
be the discriminant hypersurface (which parameterizes binary sextics with
multiple roots). Thus we have (set-theoretically):
M2 = (P(V6 ) \ D)/ PGL2 .
§6.3. GIT: Proj quotient. We will construct the quotient P(V6 )/ PGL2 and
then through away the image of D from it to get M2 . So far we were only
taking quotients of affine varieties by the action of the group. How about
quotients of projective varieties?
6.3.1. E XAMPLE . Here is a preview: what is the quotient of the standard P2
by the action of the symmetric group S3 that acts by permuting the coordinates x1 , x2 , x3 ? We can realize P2 as the quotient of A3 by the action of C∗ ,
which commutes with the action of S3 . So we can take the quotient by the
action of S3 first, which gives A3 with coordinates given by the elementary
symmetric functions. Now we can quotient out by the action of C∗ but now
notice that it has weights 1, 2, 3! So the quotient morphism is
π : P2 → P(1, 2, 3),
[x1 : x2 : x3 ] 7→ [x1 + x2 + x3 : x1 x2 + x2 x3 + x1 x3 : x1 x2 x3 ].
73
More systematically, the procedure is as follows. Suppose a group G acts
on a projective variety X. Suppose we can write X = Proj R, where R is
some finitely generated graded algebra. This is called a choice of polarization.
Suppose we can find an action of G on R that induces an action of G on X.
This is called a choice of linearization. Then we can form a GIT quotient
X//G = Proj RG .
In the example above, P2 = Proj C[x1 , x2 , x3 ], and
P2 //S3 = Proj C[x1 , x2 , x3 ]S3 =
= Proj C[x1 + x2 + x3 , x1 x2 + x2 x3 + x1 x3 , x1 x2 x3 ] = P(1, 2, 3).
We will use this construction to describe M2 .
§6.4. Classical invariant theory of a binary sextic. We have to describe the
algebra R = O(V6 )SL2 of SL2 -invariant polynomial functions for the linear
action of SL2 on V6 . The classical convention for normalizing the coefficients of a binary form is to divide coefficients by the binomial coefficients:
f6 = ax6 + 6bx5 y + 15cx4 y 2 + 20dx3 y 3 + 15ex2 y 4 + 6f xy 5 + gy 6 .
Explicit generators for R were written down in the 19-th century by Clebsch, Cayley, and Salmon. We are not going to prove that they indeed generate the algebra of invariants but let’s discuss them to see how beautiful
the answer is. Let p1 , . . . , p6 denote the roots of the dehomogenized form
f6 (x, 1) and write (ij) as a shorthand for pi −pj . Then we have the following
generators (draw some graphs):
X
I2 = a2
(12)2 (34)2 (56)2
fifteen
X
4
I4 = a
(12)2 (23)2 (31)2 (45)2 (56)2 (64)2
ten
X
6
I6 = a
(12)2 (23)2 (31)2 (45)2 (56)2 (64)2 (14)2 (25)2 (36)2
sixty
Y
D = I10 = a10 (ij)2
i<j
I15 = a15
X
((14)(36)(52) − (16)(32)(54)).
fifteen
Here the summations are chosen to make the expressions S6 -invariant. In
particular, they can all be expressed as polynomials in C[a, b, c, d, e, f, g], for
example
I2 = −240(ag − 6bf + 15ce − 10d2 ).
(6.4.1)
Here is the main theorem:
6.4.2. T HEOREM . The algebra R = O(V6 )SL2 is generated by invariants I2 , I4 , I6 ,
I10 , and I15 . The subscript is the degree. Here D = I10 is the discriminant which
vanishes iff the binary form has a multiple root. The unique irreducible relation
among the invariants is
2
I15
= G(I2 , I4 , I6 , I10 ).
Now we use our strategy to construct M2 :
74
JENIA TEVELEV
• Compute V6 // SL2 = MaxSpec R first. By 19-th century, this is
2
C[I2 , I4 , I6 , I10 , I15 ]/(I15
= G(I2 , I4 , I6 , I10 )).
• Now quotient the result by C∗ , i.e. compute Proj R. Here we have
a magical simplification: Proj R = Proj R(2) but the latter is gen2 . Since I 2 is a polynomial in other
erated by I2 , I4 , I6 , I10 , and I15
15
invariants, in fact we have
Proj R(2) = Proj C[I2 , I4 , I6 , I10 ] = P(2, 4, 6, 10) = P(1, 2, 3, 5).
• To get M2 , remove a hypersurface D = 0, i.e. take the chart DI10 of
P(1, 2, 3, 5). This finally gives
M2 = A3 /µ5 ,
where µ5 acts with weights 1, 2, 3.
• One can show that C[A, B, C]µ5 has 8 generators. So as an affine
variety, we have
M2 = (P(V6 ) \ D)/ PGL2 ,→ A8 ,
5 3
I2 I2 I4 I2 I42 I45 I22 I6 I2 I63 I65 I4 I6
2
{y = f (x)} 7→
,
,
, 2 ,
, 2 , 3 ,
.
I10 I10 I10 I10
I10 I10
I10 I10
This of course leaves more questions then gives answers:
(1) How do we know that points of M2 correspond to isomorphism
classes of genus 2 curves? In other words, why is it true that our
quotient morphism
P(V6 ) \ D → A3 /µ5
(2)
(3)
(4)
(5)
(6)
is surjective and separates PGL2 -orbits? It is of course very easy
to give examples of quotients by infinite group actions that do not
separate orbits.
Can one prove the finite generation of the algebra of invariants and
separation of orbits by the quotient morphism without actually computing the algebra of invariants?
Is M2 a coarse moduli space (and what is a family of genus 2 curves)?
Our explicit description of M2 as A3 /µ5 shows that it is singular.
Which genus 2 curves contribute to singularities?
Our construction gives not only M2 but also its compactification by
Proj R. Can we describe the boundary Proj R \ M2 ?
Are there other approaches to the construction of M2 ?
§6.5. Homework.
Problem 1. Let C ⊂ Pd be a rational normal curve of degree d, let Ĉ ⊂
be the affine cone over it, and let C̄ ⊂ Pd+1 be its projective closure.
Show that C̄ is isomorphic to P(1, 1, d) (1 point).L
Problem 2. Let P ∈ P1 be a point. Let R =
L(kP ) be the associated
Ad+1
k≥0
ring. Describe the projective embedding of P1 given by the d-th Veronese
subalgebra of R (1 point)
Problem 3. Show that I2 (see (6.4.1)) is indeed an SL2 -invariant polynomial. (2 points)
75
Problem 4. A weighted projective space P(a0 , . . . , an ) is well-formed if no
n of the weights a0 , . . . , an have a common factor. For example, P(1, 1, 3)
is well-formed but P(2, 2, 3) is not. Consider the polynomial ring R =
C[x0 , . . . , xn ], where xi has weight ai . (a) Suppose that d = gcd(a0 , . . . , an ).
Show that R(d) = R and that P(a0 , . . . , an ) ' P(a0 /d, . . . , an /d). (b) Suppose
that d = gcd(a1 , . . . , an ) and that (a0 , d) = 1. Compute R(d) and show that
P(a0 , . . . , an ) ' P(a0 , a1 /d . . . , an /d). Conclude that any weighted projective space is isomorphic to a well-formed one (2 points).
Problem 5. Compute Proj C[x, y, z]/(x5 + y 3 + z 2 ). Here x has weight 12,
y has weight 20, and z has weight 30 (1 point).
Problem 6. Using the fact that M2 = A3 /µ5 , where µ5 acts with weights
1, 2, 3, construct M2 as an affine subvariety of A8 (1 point).
Problem 7. Let V4 be the space of degree 4 binary forms. Show that
O(V4 )SL2 is a polynomial algebra generated by invariants of degrees 2 and 3
(hint: use Problem 4 from the previous homework). (3 points).
Problem 8. (a) Prove (5.2.1). (b) Prove (5.4.3) (1 point).
Problem 9. Let P ∈ E be a point on an elliptic curve. (a) Compute
Proj R(P ) and the embedding of E in it. (b) Compute Proj R(2P ) and the
embedding of E in it. (2 points)
Problem 10. Let P ∈ E be a point on an elliptic curve. Show that φ|4P |
embeds E in P3 as a complete intersection of two quadrics (i.e. the homogeneous ideal of E in this embedding is generated by two quadrics) (2 points).
Problem 11. Show that any genus 2 curve C can be obtained as follows.
Start with a line l ⊂ P3 . Then one can find a quadric surface Q and a cubic
surface S containing l such that Q ∩ S = l ∪ C (2 points).
Problem 12. Assuming that M2 = A3 /µ5 set-theoretically, define families
of curves of genus 2 (analogously to families of elliptic curves), and show
that M2 is a coarse moduli space (2 points).
Problem 13. Assuming the previous problem, show that M2 is not a fine
moduli space (2 points).
Problem 14. Show that An \{0} is not an affine variety for n > 1 (1 point).
Problem 15. Suppose X and Y are separated algebraic varieties. Explain
how to define X × Y as an algebraic variety and show that it is separated
(2 points).
Problem 16. (a) Show that an algebraic variety X is separated if and only
if the diagonal X is closed in X × X. (b) Show that a topological space X is
Hausdorff if and only if the diagonal X is closed in X × X equipped with
a product topology. (c) Explain how (a) and (b) can be both true but A1 is
both separated and not Hausdorff (2 points).
Problem 17. Use affine charts to show that G(2, n) is an algebraic variety
without using the Plücker embedding (1 point).
Problem 18. Consider rays R1 , . . . , Rk ⊂ R2 emanating from the origin,
having rational slopes, going in the counter-clockwise direction, and spanning the angle 2π once. Suppose that each angle Ri Ri+1 (and Rk R1 ) is less
than π. This is called a (two-dimensional) fan. The angles Ri Ri+1 (and
Rk R1 ) are called (top-dimensional) cones of the fan. Rays themselves are
also (one-dimenesional) cones. The origin is a zero-dimensional cone. Now
for each cone σ of the fan, consider the semigroup Λ = σ ∩ Z2 and the dual
76
JENIA TEVELEV
semigroup
Λ⊥ = {(u, v) ∈ Z2
| ui + vj ≥ 0
for any (i, j) ∈ Λ} ⊂ Z2 .
Let K be the field C(x, y). We can think about an element (i, j) ∈ Z2
as a Laurent monomial xi y j . This gives us algebras C[σ] ⊂ K spanned
by monomials in Λ⊥ . (a) Show that for each inclusion of cones τ ⊂ σ,
MaxSpec C[τ ] is a principal open subset in MaxSpec C[σ]. (b) Show that
one can glue all MaxSpec C[σ] together. This is called a toric surface. (c)
Show that weighted projective planes are toric surfaces (3 points).
Problem 19. An algebraic curve is called bielliptic if it admits a 2 : 1
morphism C → E onto an elliptic curve; the covering transformation is
called a bielliptic involution. Let C be a genus 2 curve. (a) Show that if
C is bielliptic then its bielliptic involution commutes with its hyperelliptic
involution. (b) Show that C is bielliptic if and only if the branch locus
p1 , . . . , p6 ∈ P1 of its bi-canonical map has the following property: there
exists a 2 : 1 morphism f : P1 → P1 such that f (p1 ) = f (p2 ), f (p3 ) = f (p4 ),
and f (p5 ) = f (p6 ). (c) Show that (b) is equivalent to the following: if we
realize P1 as a conic in P1 then lines p1 p2 , p3 p4 , and p5 p6 all pass through a
point (3 points).
77
§7. GIT quotients and stability.
Let’s summarize where we stand. We want to construct M2 as an orbit
space for
SL2 acting on P(V6 ) \ D.
We use our standard approach using invariants. The classical invariant
theory tells us that O(V6 )SL2 is generated by I2 , I4 , I6 , I10 = D, and I15 with
2 = g(I , I .I , I ).
a single quadratic relation I15
2 4 6 10
So our natural candidate for the quotient is Proj O(V6 )SL2 , and the quotient map is
f 7→ [I2 (f ) : . . . : I15 (f )] ∈ P(2, 4, 6, 10, 15).
Here we got lucky: since Proj R = Proj R(2) , we can also write the quotient
map as
f 7→ [I2 (f ) : . . . : I10 (f )] ∈ P(2, 4, 6, 10) = P(1, 2, 3, 5).
Since there are no relations between I2 , . . . , I10 we actually expect the quotient to be P(1, 2, 3, 5).
If we throw away the vanishing locus of the discriminant, we get the
affine chart
{D 6= 0} ⊂ P(1, 2, 3, 5).
So our hope is that
M2 = A3 /µ5 ,
where µ5 acts with weights 1, 2, 3. We’ve seen that if we want to embed this
cyclic quotient singularity in the affine space, we need at least A8 .
Of course this construction alone does not guarantee that each point of
A3 /µ5 corresponds to a genus 2 curve and that different points correspond
to different curves: this is something we are trying to work out in general.
§7.1. Algebraic representations of reductive groups. Consider any representation
G → GL(V ).
We want to define the quotient of P(V ) by G, or even more generally a
quotient of any projective variety by some action of G.
7.1.1. D EFINITION . Suppose that G is a group and an affine algebraic variety
such that the multiplication map G × G → G and the inverse map G → G
are regular. Then G is called a linear algebraic group
Examples:
•
•
•
•
•
•
GLn = D(det) ⊂ Matn ,
SLn = V (det −1) ⊂ Matn ,
“the maximal torus” (diagonal matrices in GLn ),
“the Borel subgroup” (upper-triangular matrices in GLn ),
SOn , Spn .
finite groups.
78
JENIA TEVELEV
7.1.2. R EMARK . The terminology “linear algebraic group” can be explained
by a theorem of Chevalley: any linear algebraic group is isomorphic to a
(Zariski closed) subgroup of GLn for some n. And vice versa, it is clear
that any such subgroup is linear algebraic. If we remove “affine” from
the definition of an algebraic group, then there are other possibilities, for
example an elliptic curve (or an Abelian surface or more generally an Abelian
variety) is a projective algebraic group.
Non-examples:
• SL2 (Z) and other non-finite discrete groups.
• SUn ⊂ SLn (C) and other non-finite compact linear Lie groups.
In fact, we can show that
7.1.3. L EMMA . SUn is Zariski dense in SLn (C).
Proof. Indeed, let f be any regular function on SLn (C) that vanishes on SUn .
We have to show that it vanishes on SLn (C). It is equally easy to show
this for any function holomorphic in the neighborhood of Id ∈ SLn (C).
Consider the exponential map
A2
+ ...
2
This map is biholomorphic in the neighborhood of the origin (the inverse
map is given by log) and (locally) identifies SLn (C) with a complex vector
subspace sln of complex matrices with trace 0 and SUn with a real subspace sun of skew-Hermitian matrices (i.e. matrices such that A + Āt = 0).
So g = f (exp(A)) is a function holomorphic near the origin which vanishes
on sun . But since sun + isun = sln , this function vanishes on sln as well:
indeed, the kernel of its differential at any point of sun contains sun , and
therefore contains sln (being a complex subspace). So all partial derivatives
of g vanish along sun , and continuing by induction all higher-order partial derivatives of g vanish along sun . So g is identically zero by Taylor’s
formula.
exp : Matn (C) → GLn (C),
A 7→ exp(A) = Id + A +
7.1.4. D EFINITION . A finite-dimensional representation of a linear algebraic
group is called algebraic (or rational) if the corresponding homomorphism
G → GL(V ) is a regular morphism. In other words, an algebraic representation is given by a homomoprhism


a11 (g) . . . a1n (g)

..  ,
..
G → GL(V ),
g 7→  ...
.
. 
an1 (g) . . . ann (g)
where aij ∈ O(G).
We can generalize this definition to non-linear actions of G on any algebraic variety X: the action is called algebraic if the “action” map
G×X →X
is a regular map. Why is this definition the same as above for the linear
action? (explain). Finally, we need a notion of a linearly reductive group.
79
7.1.5. T HEOREM . An algebraic group is called linearly reductive if it satisfies
any of the following equivalent conditions:
(1) Any finite-dimensional algebraic representation V of G is completely reducible, i.e. is a direct sum of irreducible representations.
(2) For any finite-dimensional algebraic representation V of G, there exists
a G-equivariant projector πV : V → V G (which is then automatically
unique).
(3) For any surjective linear map A : V → W of algebraic G-representations,
the induced map V G → W G is also surjective.
Proof. (1) ⇒ (2). Decompose V = V1 ⊕ . . . ⊕ Vk . Suppose G acts trivially
on the first r sub-representations and only on them. Let U ⊂ V be an
irreducible subrepresentation. By Schur’s lemma, its projection on any Vi
is either an isomorphism or a zero map. It follows that if U is trivial, it is
contained in V1 ⊕ . . . ⊕ Vr and if it is not trivial, it is contained in Vr+1 ⊕
. . . ⊕ Vk . So in fact we have a unique decomposition
V = V G ⊕ V0 ,
where V0 is the sum of all non-trivial irreducible subWe will representations. The projector V → VG is the projector along V0 .
(2) ⇒ (3). Suppose that the induced map V G → W G is not onto. Choose
w ∈ W G not in the image of V G and choose any projector W G → hwi that
annihilates the image of V G . Then the composition
A
π
W
V −→ W −→
W G → hwi
is a surjective G-invariant linear map f : V → C that annihilates V G . After
dualizing, we have a G-invariant vector f ∈ V ∗ which is annihilated by all
G-invariant linear functions on V ∗ . However, this is nonsense: we can easily construct a G-invariant linear function on V ∗ which does not annihilate
f by composing a G-invariant projector V ∗ → (V ∗ )G (which exists by (2))
with any projector (V ∗ )G → hf i.
(3) ⇒ (1). It is enough to show that any sub-representation W ⊂ V
has an invariant complement. Here we get sneaky and apply (2) to the
restriction map of G-representations
Hom(V, W ) → Hom(W, W ).
The G-invariant lift of Id ∈ Hom(W, W ) gives a G-invariant projector V →
W and its kernel is a G-invariant complement of W .
Next we study finite generation of the algebra of invariants. Suppose we
have a finite-dimensional representation
G → GL(V ).
We are looking for criteria that imply that O(V )G is a finitely generated
algebra. In fact, we already know (Lemma 4.4.4) that it is enough to show
existence of a Reynolds operator O(V ) → O(V )G .
7.1.6. L EMMA . The Reynolds operator exists for any algebraic finite-dimensional
representation G → GL(V ) of a linearly reductive group. In particular, O(V )G
in this case is finitely generated.
80
JENIA TEVELEV
Proof. O(V ) is an algebra graded by degree and each graded piece O(V )n
has an induced representation of G. By linear reductivity, there exists a
unique G-invariant linear projector Rn : O(V )n → O(V )G
n for each n. We
claim that this gives a Reynolds operator R. The only thing to check is that
R(f g) = f R(g)
for any f ∈ O(V )G . Without loss of generality we can assume that f ∈
O(V )G
n and g ∈ O(V )m . Then f O(V )m is a G-sub-representation in O(V )m+n ,
so its G-invariant projector f R(g) should agree with a G-invariant projector
R(f g) on O(V )m+n .
§7.2. Finite Generation Theorem via the unitary trick.
7.2.1. L EMMA . Any algebraic representation of SLn is completely reducible, i.e. SLn
is a linearly reductive group.
Proof. We will use a unitary trick introduced by Weyl (and Hurwitz). An
algebraic representation of SLn induces a continuous representation of SUn ,
and any sub-representation of SUn is in fact a sub-representation for SLn by
Lemma 7.1.3 (explain).
So it is enough to show that any continuous complex representation
SUn → GL(V ) is completely reducible. There are two ways to prove this.
One is to use the basic lemma above and to construct an equivariant projector V → V G for any finite-dimensional continuous representation. Just
like in the case of finite groups, one can take any projector p : V → V G and
then take it average
R
p(gv) dµ
.
π(v) = SURn
SUn dµ
Here µ should be an equivariant measure on SUn (a so-called Haar measure),
and then of course we would have to prove its existence. We will follow a
more naive approach.
We claim that V has an SUn -invariant positive-definite Hermitian form.
If this is true then for any complex subrepresentation U ⊂ V , the orthogonal complement U ⊥ is also SUn -invariant, and we will keep breaking V
into pieces until each piece is irreducible. To show the claim, consider the
induced action of SUn on all Hermitian forms (·, ·) on V by change of variables. Let S be the set of positive-definite Hermitian forms. The action of
SUn preserves the set S, which is convex because any positive linear combination of positive-definite Hermitian forms is positive-definite. So it is
enough to prove the following Lemma.
7.2.2. L EMMA . Let S ⊂ Rn is a convex set preserved by a compact subgroup K
of the group of affine transformations (i.e. compositions of linear transformations
and translations) of Rn . Then K has a fixed point on S.
Proof. Without loss of generality we can assume that S is convex and compact. Indeed, since K is compact, any K-orbit in S is compact as well (being the image of K under a continuous map). The convex hull S 0 of this
K-orbit is therefore a compact, convex, and G-invariant subset of S. A Kfixed point in S 0 will of course also be a K-fixed point in S.
81
If the minimal affine subspace containing S is not the whole of Rn (draw
the picture), then take this linear span Rk instead of Rn (since K preserves S,
it also preserves its affine span). K clearly has an induced action there by
affine transformations).
Now let p be the center of mass of S with coordinates
R
xi dV
pi = RS
S dV
(here dV is the standard measure on Rk ). Since S is convex, the Riemann
sum definition of the integral shows that p ∈ S and that p is preserved by
any affine transformation of Rn that preserves S. So, p is fixed by K.
7.2.3. C OROLLARY. For any finite-dimensional algebraic representation V of SLn ,
the algebra of invariants O(V )SLn is finitely generated.
§7.3. Surjectivity of the quotient map. Let’s consider any algebraic finitedimensional representation of a linearly reductive group G → GL(V ). The
existence of the Reynolds operator alone implies surjectivity of the quotient
map
V → MaxSpec O(V )G ,
see Theorem 4.5.4. However, this is not the quotient that we want!
7.3.1. D EFINITION . Let’s fix the following terminology. If G acts on the
affine variety X then we call X//G = MaxSpec O(X)G the affine quotient.
If G acts on the projective variety X = Proj R and this action lifts to the
action of G on R then we will call X/G = Proj RG the GIT quotient. In most
of our examples X = P(V ) and the action of G comes from the linear action
of G on V . Often we will just restrict to this case.
These constructions are related. For example, the principal open subset
of P(V6 )/ SL2 (isomorphic to M2 ) is the affine quotient U// SL2 , where
U = {D 6= 0} ⊂ P(V6 )
(D = I10 is the discriminant).
Let’s generalize. Take any algebraic representation G → GL(V ) and the
induced action of G on P(V ) (even more generally, we can take any algebraic action of G on a graded algebra R without zero-divisors and the
induced action of G on Proj R). Our quotient is given by Proj O(V )G (even
more generally, we take Proj RG ). On charts this quotient looks as follows.
Let f1 , . . . , fn ∈ O(V )G be homogeneous generators (or in fact any homogeneous elements that generate the irrelevant ideal after taking the radical).
For example, for M2 we can take I2 , I4 , I6 , D = I10 . We have quotient maps
D(fi ) = MaxSpec(O(V )fi )0 → D0 (fi ) = MaxSpec(O(V )G
fi )0
induced by the inclusion O(V )G ,→ O(V ).
Notice that many elements of P(V ) go missing in the quotient!
7.3.2. D EFINITION . A point x ∈ P(V ) (more generally, x ∈ Proj R) is called
unstable if all G-invariant polynomials (more generally, functions in RG ) of
positive degree vanish on x. Let N ⊂ P(V ) be the locus of unstable points
(also known as the null-cone). Caution: points that are not unstable are
called semistable.
82
JENIA TEVELEV
We see that the GIT quotient is actually the map
P(V ) \ N → Proj O(V )G
(more generally, (Proj R) \ N → Proj RG ) obtained by gluing quotients on
charts. We will return to the question of describing the unstable locus later.
But now, let’s finish our discussion of surjectivity of the GIT quotient.
On charts, we have affine quotients. Let’s study them first. We start with a
very important observation
7.3.3. T HEOREM . Consider the regular action of a linear algebraic group G on an
affine variety X. Then
(1) Any function f ∈ O(X) is contained in a finite-dimensional sub-representation
U ⊂ O(X).
(2) There exists an algebraic representation of G on a vector space V and a
G-invariant closed embedding X ,→ V .
Proof. The action G × X → X induces a homomorphism (called the coaction)
φ : O(X) → O(G × X) = O(G) ⊗C O(X),
k
X
f 7→
α ⊗ fi .
i=1
Quite literally, this means that
f (gx) =
k
X
α(g)fi (x) for g ∈ G, x ∈ X.
i=1
For example, f is in the linear span of fi ’s (take g = 1). And of course any
f (gx) with fixed g is in the linear span of fi ’s. But this implies that the linear
span U ⊂ O(X) of all functions f (gx) for g ∈ G is finite-dimensional. It is
clear that U is G-invariant.
For part (b), we first use (a) to show that O(X) contains a finite-dimensional
G-invariant subspace V ∗ that contains generators of O(X). Realizing O(X)
as the quotient algebra of the polynomial algebra O(V ) gives a required
embedding X ,→ V .
Now we can finally work out surjectivity of the quotient map.
7.3.4. T HEOREM . Consider the regular action of a linearly reductive group G on
an affine variety X. Then
(1) There exists a Reynolds operator O(X) → O(X)G .
(2) O(X)G is finitely generated, let X//G be the affine quotient.
(3) The quotient map X → X//G is surjective
Proof. (1) O(X) is a sum of finite-dimensional representations of G. So the
Reynolds operator is defined just like in the case of O(V ), the only difference is that there is no grading to naturally break the algebra into finitedimensional pieces. But since the projector U → U G is unique, we will
have a well-defined projector O(X) → O(X)G .
(2) and (3) now follow from the existence of the Reynolds operator just
like in the case of the group acting on a vector space (the linearity was not
used anywhere).
83
7.3.5. C OROLLARY. Any point of A3 /µ5 represents a genus 2 curve.
§7.4. Separation of orbits. We look at the affine case first.
7.4.1. T HEOREM . Let G be a linearly reductive group acting regularly on the affine
variety X. Let O, O0 be two G-orbits. Then TFAE
(1) The closures of O and O0 have a common point.
(2) There exists a sequence of orbits O = O1 , . . . , On = O0 such that the
closures of Oi and Oi+1 have a common point for any i.
(3) O and O0 are not separated by G-invariants in O(X)G .
In particular, every fiber of π : X → X//G contains exactly one closed orbit.
Proof. (1)⇒(2)⇒(3) are clear (since any G-invariant regular function takes
the same value on any orbit and its closure). We have to show (3)⇒(1).
Let I, I 0 ⊂ O(X) be ideals of Ō, Ō0 . They are clearly G-invariant subspaces. Suppose that Ō ∩ Ō0 = ∅. By Nullstellensatz, this implies that
I + I 0 = O(X),
i.e. we can write 1 = f +g, where f ∈ I and g ∈ I 0 . Now apply the Reynolds
operator: R(f )+R(g) = 1. We claim that R(f ) ∈ I (and similarly R(g) ∈ I 0 ).
This implies that R(f ) is an invariant function which is equal to 0 on O and
1 on O0 , i.e. these orbits are separated by invariants. But the claim is clear:
recall that the Reynolds operator O(V ) → O(V )G is obtained by simply
gluing all projectors U → U G for all finite-dimensional subrepresentations
U ⊂ O(V ). In particular, R preserves any G-invariant subspace of O(V ),
for example I and I 0 .
Now we can finally describe M2 :
7.4.2. T HEOREM . There are natural bijections (described previously) between
(1) isomorphism classes of genus 2 curves;
(2) SL2 orbits in P(V6 ) \ D;
(3) points in A3 /µ5 acting with weights 1, 2, 3.
Proof. The only thing left to check is that all SL2 orbits in P(V6 ) \ D are
closed. But this is easy: for any orbit O and any orbit O0 6= O in its closure, dim O0 < dim O. However, all SL2 orbits in P(V6 ) \ D have the same
dimension 3, because the stabilizer can be identified with a group of projective transformations of P1 permuting roots of the binary sextic, which is a
finite group if all roots are distinct (or even if there are at least three distinct
roots).
This gives a pretty decent picture of the quotient P(V6 )/ SL2 , at least in
the chart D 6= 0, which is the chart we mostly care about. To see what’s
going on in other charts, let’s experiment with generators I2 , I4 , I6 , I10 (defined in §6.4). Simple combinatorics shows that (do it):
• if f ∈ V6 has a root of multiplicity 4 then f is unstable.
• if f ∈ V6 has a root of multiplicity 3 then all basic invariants vanish
except (potentially) I2 .
So we should expect the following theorem:
84
JENIA TEVELEV
7.4.3. T HEOREM . Points of P(V6 )/ SL2 = P(1, 2, 3, 5) correspond bijectively to
GL2 -orbits of degree 6 polynomials with at most a double root (there can be several
of them) plus an extra point [1 : 0 : 0 : 0], which has the following description.
All polynomials with a triple root (but no fourtuple root) map to this point in the
quotient. The corresponding orbits form a one-parameter family (draw it), with a
closed orbit that corresponds to the polynomial x3 y 3 .
To prove this theorem, it is enough to check the following facts:
(1) Any unstable form f has a fourtuple root (or worse). In other words,
semistable forms are the forms that have at most triple roots.
(2) A semistable form f ∈ P(V6 ) has a finite stabilizer unless f = x3 y 3
(this is clear: this is the only semistable form with two roots).
(3) Any semistable form f without triple roots has a closed orbit in
the semistable locus in P(V6 ), and hence in any principal open subset
D+ (I) it belongs to, where I is one of the basic invariants. Notice
that we do not expect f to have a closed orbit in the whole P(V6 ), in
fact one can show that there is only one closed orbit there, namely
the orbit of x6 .
(4) If f has a triple root then it has the orbit of x3 y 3 in its closure. Indeed, suppose f = x3 g, where g = y 3 + ay 2 x + byx2 + cx3 is a cubic
form (it has to start with
y 3 ,otherwise f has a fourtuple root). Let’s
t 0
act on f by a matrix
. We get x3 y 3 + at2 y 2 x + bt4 yx5 + ct6 x6 .
0 t−1
So as t → 0, we get x3 y 3 in the limit.
To check (1) and (3), one can use an exceptionally useful numerical criterion also known as Hilbert–Mumford criterion, which we will discuss in the
next section. First some definitions to wrap up our discussion:
7.4.4. D EFINITION . Suppose a linearly reductive group G acts regularly on
an affine variety X. A point x ∈ X is called stable if two conditions are
satisfied:
• The G-orbit of x is closed.
• The stabilizer of x is finite.
For example, any point in P(V6 ) \ D is stable. On the other hand, x3 y 3 is
an example of a semistable but not stable point in P(V6 ) \ {I2 = 0}.
7.4.5. R EMARK . The concept of stability is very general and applies in many
circumstances. In moduli questions, we call objects stable if their isomorphism class gives a point in the moduli space. Unstable objects have to be
discarded in the moduli space. Semistable but not stable objects survive
in the moduli space but a single point in the moduli space can correspond
to several isomorphism classes. The concept of stability is not intrinsic to
objects but depends on how we construct the moduli space, i.e. which objects we want to consider as isomorphic and how we define families of
objects. One can change “stability conditions” and get a different moduli
space (variation of moduli spaces). If we construct a moduli space as an orbit space for some group action, then stability can be defined as Mumford’s
GIT-stability discussed above.
85
§7.5. Unstable locus: Hilbert–Mumford criterion. Let’s focus on linear
actions on linearly reductive groups.
7.5.1. D EFINITION . The group T = (C∗ )n is called an algebraic torus.
7.5.2. T HEOREM . An algebraic torus T is a linearly reductive group. In fact, any
algebraic representation of T is diagonalizable and is isomorphic to a direct sum of
one dimensional irreducible representations Vχ , where χ : T → GL1 (C) = C∗ is
an algebraic character. Any character has a form
(z1 , . . . , zn ) → z1m1 . . . znmn
for some vector m = (m1 , . . . , mn ) ∈ Zn .
Proof. One can prove this just like for SLn . The analogue of SUn (the maximal compact subgroup) will be the real torus (S 1 )n ⊂ (C∗ )n , where
S 1 = {z, |z| = 1} ⊂ C∗ .
Just for fun, let’s give a different proof, although close in spirit. We have
O(T ) = C[z1±1 , . . . , zn±1 ]
is the algebra of Laurent polynomials. Let µ ⊂ C∗ be the subgroup of
all roots of unity. Being infinite, it is Zariski dense in C∗ . And in fact,
µn ⊂ (C∗ )n (the subgroup of all torsion elements) is also Zariski dense.
Indeed, if
X
f (z1 , . . . , zn ) =
gi (z1 , . . . , zn−1 )zni
i
µn
vanishes on
then all coefficients gi must vanish on µn−1 (why?) , hence
they are identically zero by inductive assumption. Now any representation
T → GL(V ) restricts to a representation µn → GL(V ). The image consists of commuting matrices of finite order, hence can be simultaneously
diagonalizable. But then the image of T is diagonalizable as well. As for
the description of one-dimensional representations, notice that a character T → C∗ is a non-vanishing regular function on T . We can write it as
a Laurent monomial multiplied by a polynomial f (z1 , . . . , zn ) which does
not vanish in (C∗ )n . Therefore, its vanishing locus in An is a union of coordinate hyperplanes. By factoriality of the ring of polynomials (and Nullstellensatz), it follows that f is a monomial multiplied by a constant. Since
f (1, . . . , 1) = 1, this constant is equal to 1.
7.5.3. D EFINITION . Let G be a connected linearly reductive group. An algebraic subgroup T ⊂ G is called a maximal torus if T ' (C∗ )n and T is
maximal by inclusion among this kind of subgroups.
7.5.4. T HEOREM . All maximal tori in G are conjugate.
We won’t need this theorem, so we won’t prove it. But notice that this
is clear if G = SLn . Indeed, any algebraic torus T ⊂ G will be diagonalizable in some basis of Cn , which means that (after conjugation by
a change of basis matrix), T is contained in a subgroup diag(z1 , . . . , zn ),
where z1 . . . zn = 1. So this subgroup is the maximal torus and any other
maximal torus is conjugate to it.
The Hilbert–Mumford criterion consists of two parts: reduction from G
to T and analysis of stability for torus actions. First the reduction part.
86
JENIA TEVELEV
7.5.5. T HEOREM . Consider any finite-dimensional representation G → GL(V ) of
a linearly reductive group. Let T ⊂ G be a maximal torus, and let v ∈ V . TFAE:
(1) v is unstable, i.e. any homogeneous polynomial f ∈ O(V )G of positive
degree vanishes on v.
(2) The G-orbit Gv contains 0 in its closure.
(3) There exists u ∈ Gv such that the T -orbit T u contains 0 in its closure.
Proof. It is clear that (3)⇒(2)⇒(1). Theorem 7.4.1 shows that (1)⇒(2).
We will do a difficult implication (2)⇒(3) only for G = SLn . Recall that
any matrix A ∈ SLn has a polar decomposition
A = U P,
where U ∈ K = SUn and P is positive-definite Hermitian matrix. By spectral theorem, P has an orthonormal basis of eigenvectors, so we can write
P = U 0 D(U 0 )−1 ,
where U 0 ∈ K and D is a diagonal matrix. So combining these fact, we
have a useful decomposition7
G = KT K.
By hypothesis, 0 ∈ Gv (Zariski closure). We show in the exercises that in
fact we also have 0 ∈ Gv = KT Kv (the closure in real topology). Since K
is compact, this implies that
0 ∈ T Kv
(the closure in real topology). Consider the quotient map
πT : V → V //T
and let
O = πT (0).
πT is continuous in real topology (since polynomials are continuous functions), so we have
O ∈ πT (T Kv)
⇒
O ∈ πT (T Kv) = πT (Kv).
But by compactness,
πT (Kv) = πT (Kv),
and therefore there exists g ∈ K such that πT (gv) = O, i.e. 0 ∈ T u for
u = gv.
Now we have to investigate the unstable locus for the torus action. Consider any finite-dimensional representation T = (C∗ )n → GL(V ). Let
V = ⊕m∈Zn Vm be the decomposition of V into T -eigenspaces.
P
7.5.6. D EFINITION . For any u ∈ V , let u =
um be the decomposition of u
into T -eigenvectors. Then
N P (u) = Convex Hull{m ∈ Zn | um 6= 0}
is called the Newton polytope of u.
7An analogous decomposition holds in any connected complex linearly reductive group
with K its maximal connected compact subgroup.
87
Before we state a general theorem, let’s look at some examples of Newton
polytopes and its relation to stability.
7.5.7. E XAMPLE . Finish the description of unstable elements in V6 .
7.5.8. E XAMPLE . Consider the action of SL3 on degree 2 polynomials in
three variables. Describe unstable Newton polygons, show that they correspond to singular conics. This relates nicely to the fact that O(V2 )SL3 is
generated by a single invariant (discriminant).
7.5.9. T HEOREM . Consider any finite-dimensional algebraic representation T =
(C∗ )n → GL(V ). Let v ∈ V . TFAE:
88
JENIA TEVELEV
(1) v is unstable.
(2) 0 ∈
6 N P (v).
Proof. The game here is based on the fact that there are two ways to describe convexity: using positive linear combinations or using supporting
hyperplanes. More precisely, we have the following well-known Lemma,
which goes by names of Farkas’ Lemma, Gordan Theorem, etc.
7.5.10. L EMMA . Let S ⊂ Rn be a convex hull of finitely many lattice points
v1 , . . . , vk ∈ Zn . Then
• If 0 6∈ S if and only if there exists a vector u ∈ Zn such that u · vi > 0 for
any i.
• If 0 ∈ S if and only if there exist rational numbers α1 , . . . , αk ≥ 0 such
that
X
X
0=
αi vi and
αi = 1.
Now we can prove the Theorem. If 0 6∈ N P (v) then by Lemma we can
choose a vector u = (u1 , . . . , un ) ∈ Zn such that u·vi > 0 for any i. Consider
a subgroup χ(t) = (tu1 , . . . , tun ) ⊂ T . Then we have
X
X
χ(t) · v =
χ(t) · vm =
tm·u vm .
m∈Zn
m
We see that
lim χ(t) · v = 0.
t→∞
On the other hand, let’s suppose that 0 ∈ N P (v). By lemma, we can choose
rational numbers αm ≥ 0 indexed
P by m such that vm 6= 0, and not all of
them equal to 0, such that 0 =
αm vm . By rescaling, we can assume that
all these numbers are integers. Choose linear functions fm on V for any m
such that vm 6= 0. We can assume that fm (vm ) = 1. But now consider the
function
Y
αm
I=
fm
.
We have I(v) = 1 and I is T -invariant. Therefore, 0 is not in the closure of
T v.
§7.6. Hypersurfaces. We will discuss some examples when stability is easy
to verify. Let G be a reductive group acting on the affine variety X with the
quotient
π : X → X//G = MaxSpec O(X)G .
Let
Xs ⊂ X
be the set of stable points and let
Z⊂X
be the subset of points such that Gx is not finite.
7.6.1. T HEOREM . Z is closed, X s is open, Xs is the complement in X of π −1 (π(Z)).
The quotient π induces a 1 − 1 bijection between G-orbits in X s points in π(X s .
89
Proof. Consider the map
G × X → X × X, (g, x) 7→ (gx, x).
Let Z̃ be the preimage of the diagonal it is closed. But then Z is the locus
of points in X where the fibers of π2 |Z have positive dimension. Thus Z is
closed, by semi-continuity of dimension of fibers.
Next we claim that π(Z) is closed. Here we are only going to use the fact
that Z is closed, thus Z = V (I) for some I ⊂ O(X). Then
π(Z) = V (π ∗ (I)) = I ∩ O(X)G = I G .
But the exact sequence
0 → I → O(X) → O(X)/I →
stays exact after taking G-invariants (this is one of the equivalent definitions of reductivity), so we have
O(Z)G = O(X)G /I G .
Thus the map π|Z : Z → π(Z) can be identified with the quotient map
Z → Z//G,
which as we know is surjective.
Now suppose x ∈ π −1 (π(Z)). Then the fiber of π through x contains a
closed orbit with a positive-dimensional stabilizer. Thus either Gx is not
closed or Gx is positive-dimensional. In any case x is not stable.
If x 6∈ π −1 (π(Z)) then Gx is finite. If Gx is not closed then a closed
orbit in the closure of Gx also does not belong to π −1 (π(Z)), which is a
contradiction. So in fact x is stable.
We are going to prove a classical theorem of Matsumura, Monsky, and
Mumford (MMM) concerning stability of smooth hypersurfaces. On the
group-theoretic side, we consider the representation of SLn+1 in the vector
space
Vn,d = Symd (Cn+1 )∗
which parametrizes polynomials of degree d in n + 1 variables. Let
Un,d ⊂ Vn,d
be the locus such that the corresponding hypersurface in Pn is non-singular,
and let Dn,d be the complement, the discriminant set.
7.6.2. T HEOREM . Dn,d is an irreducible hypersurface of degree (n + 1)(d − 1)n .
Its defining equation Dn,d (called the discriminant) belongs to O(Vn,d )SLn+1 .
Proof. The proof is by dimension count. Consider the incidence subset Z ⊂
P(V n,d )×Pn of pairs (F, z) such that z ∈ Sing(F = 0). This is a closed subset
(defined by vanishing of partials F0 , . . . , Fn ). All fibers of its projection onto
Pn are irreducible (in fact they are projective spaces) and have dimension
dim P(Vn,d ) − n − 1 (why?) . Therefore Z is irreducible (in fact smooth) and
dim Z = dim P(Vn,d ) − 1 by the Theorem on dimension of fibers. Notice
that the projectivization of Dn,d is the image of Z. Thus Dn,d is irreducible
and to count its dimension it suffices to show that a general hypersurface
90
JENIA TEVELEV
singular at z ∈ Pn is singular only there: this would imply in fact that the
first projection
Z → Dn,d
is birational (this is a very useful resolution of singularities of the discriminant locus). But this is easy: just take a smooth hypersurface S ⊂ Pn−1 (for
example xd1 + . . . + xdn ). The cone over it has only one singular point (the
vertex).
We won’t need the degree, but here is a quick calculation in case you
are wondering. Take a general pencil aF + bG of degree d hypersurfaces.
We have to count the number of singular hypersurfaces in this pencil. A
general singular hypersurface has a unique singularity, so we might just as
well count the total number of singular points of hypersurfaces in the pencil. Those points are intersection points of n + 1 “partial derivatives hypersurfaces” aFi + bGi = 0, i = 0, . . . , n + 1 that intersect transversally (why?) .
Quite generally, if X is a smooth projective algebraic variety of dimension
n then it has intersection theory. Its easiest incarnation is that for any n divisors (integral combinations of irreducible hypersurfaces) D1 , . . . , Dn one
can compute the intersection number
D1 · . . . · Dn
which has two basic properties:
• If the hypersurfaces intersect transversally then this is the number
of intersection points.
• If Di ∼
= Di0 are linearly equivalent divisors (i.e. there exists f ∈ k(X)
such that (f ) = Di − Di0 ) then D1 · . . . · Dn = D10 · . . . Dn0 .
Using this in our case, each partial derivatives hypersurface is linearly
equivalent to a hypersurface
L + (d − 1)H = {az0d−1 = 0}.
So the intersection number is
(L + (d − 1)H)n+1 = (n + 1)(d − 1)n
by the binomial formula.
Finally, the discriminant hypersurface is obviously SLn+1 -invariant, and
so the action of SLn+1 can only multiply it by a character. But SLn+1 has no
non-trivial characters.
So we can mimic the construction of M2 and consider
• The GIT quotient P(Vn,d )/ SLn+1 which compactifies its principal
open subset
• (P(Vn,d ) \ Dn,d )// SLn+1 , the moduli space of non-singular hypersurfaces.
In fact, in direct analogy with the case of V6 , we claim that
7.6.3. T HEOREM (Matsumura–Monsky–Mumford). Let d ≥ 3. Then all smooth
hypersurfaces are stable, i.e. the principal open subset above parametrizes smooth
hypersurfaces in Pn modulo projective transformations.
91
Proof. The first step is to reduce the question to a commutative algebra
problem. It suffices to prove that any point in P(Vn,d ) \ Dn,d has a finite
stabilizer in SLn+1 , or, which is the same, that any point F in Vn,d \ Dn,d has
a finite stabilizer in GLn+1 (why?) . If the stabilizer is infinite then the orbit
map
X
X
GLn+1 → Vn,d , A = {aij } 7→ AF = F (
a0i xi , . . . ,
ani xi )
has positive dimensional fibers. This implies that its differential is not injective. Computing it by the chain rule, we find linear forms
X
X
l0 =
a0i xi , . . . , ln =
ani xi
such that
l0 F0 + . . . + ln Fn
is a zero polynomial. Without loss of generality we can assume that l0 6= 0.
Let I ⊂ C[x0 , . . . , xn ] be the ideal generated by F1 , . . . , Fn . Then we have
p
(F0 , I) = (x0 , . . . , xn )
(because F = 0 is non-singular),
L0 6∈ I
(by degree reasons, this is where we use d ≥ 3), and
L0 F0 ∈ I.
We claim that this is impossible by some very cool commutative algebra,
which we are going to remind. We won’t give a self-contained proof, but
everything can be found in [AM] (or even better, in [Ma]).
7.6.4. D EFINITION . An ideal Q ⊂ R is called primary if
xy ∈ Q
⇒
x ∈ Q or y n ∈ Q for some n.
It √
is easy to see (why?) that the radical of a primary ideal is a prime ideal.
If Q = P then we say that Q is P -primary.
7.6.5. T HEOREM . Any ideal I in a Noetherian ring R has a primary decomposition
I = Q1 ∩ . . . ∩ Qr ,
where Qi ’s are primary, and no Qi contains the intersection of the remaining ones
(this would would make it redundant).
Intersection of P -primary ideals is P -primary, so we can and will assume that
the radicals Pi of Qi ’s are different.
The prime ideals Pi ’s are called associated primes of I. One has
• A prime ideal P ⊂ R is associated to I if and only if P = (I : x) for some
x ∈ R.
• The union of associated primes is given by elements which are zero-divisors
modulo I:
P1 ∪ . . . ∪ Pr = {x ∈ R | (I : x) 6= I}.
(7.6.6)
92
JENIA TEVELEV
What is the geometry behind this? We have
p
√
I = P1 ∩ . . . ∩ Pr , where Pi = Qi .
√
This decomposition corresponds to breaking V (I) = V ( I) into irreducible
components with one annoying caveat: it could happen that some Pi ’s are
actually redundant, i.e. Pi ⊃ Pj for some j 6= i. If this happens then we
call Pi an embedded prime. Otherwise we call Pi a minimal (or isolated prime).
Here is an example of a primary decomposition when this happens:
(x2 , xy) = (x) ∩ (x, y)2 .
Here (x) is a minimal prime and (x, y) is an embedded prime (draw the picture). It’s nice to explore situations when embedded primes don’t appear.
We have the following fundamental
7.6.7. T HEOREM (Macaulay’s unmixedness theorem). Let
I = (f1 , . . . , fr ) ⊂ k[x0 , . . . , xn ]
be an ideal such that codim V (I) = r (in general codim V (I) ≤ r). Then I is
unmixed, i.e. there are no embedded primes.
Noetherian rings that satisfy the conclusion of this theorem are called
Cohen–Macaulay rings. (For the proof that this condition is equivalent to
the standard definition of Cohen-Macaulay rings via depth and height of
localizations, see [Ma]). So k[x0 , . . . , xn ] is a Cohen–Macaulay ring. In fact
if X is a smooth affine variety then O(X) is Cohen–Macaulay.
Final step in the proof of Matsumura–Monsky. Clearly V (I) = V (F1 , . . . , Fr )
has dimension 1 (if it has a component of dimension greater than 1 then
V (F0 , . . . , Fr ) has a positive dimension but it is a point). By unmixedness,
I has no embedded primes, i.e. all associated primes of I have height r.
However, since F0 L0 ∈ I, F0 belongs to one of the associated primes of I
by (7.6.5). This implies that the ideal (F0 , . . . , Fr ) is contained in the ideal
of height r, which contradicts the fact that V (F0 , . . . , Fr ) is a point.
This shows how useful commutative algebra is even when the basic object is just a non-singular hypersurface. I will put more exercises on unmixedness in the homework. Commutative algebra serves as a technical
foundation of modern algebraic geometry through the theory of algebraic
schemes.
§7.7. Homework.
In problems 1–7, describe the unstable locus for given representations.
Problem 1. The action of SLn on Matn,m by left multiplication (1 point).
Problem 2. Show that a degree 3 form in 3 variables is semistable for the
action of SL3 if and only if the corresponding cubic curve is either smooth
or has a node (1 point).
Problem 3. The action of GLn on Matn,n by conjugation (1 point).
Problem 4. The action of SLn on quadratic forms in n variables (1 point).
action of SL3 if and only if the corresponding cubic curve is either smooth
or has a node (1 point).
93
Problem 6. Show that a degree 4 form in 3 variables is semistable for
the action of SL3 if and only if the corresponding quartic curve in P2 has
no triple points8 and is not the union of the plane cubic and an inflectional
tangent line (2 points).
action of SL4 if and only if all points of the corresponding cubic surface in
P3 are either smooth, or ordinary double points, or double points p such
that (after a linear change of variables) the quadratic part of F (x, y, z, 1) is
xy and the line x = y = 0 is not contained in S (3 points)9.
Problem P
8. Let G be a finite group with a representation ρ : G → GL(V ).
k
Let P (z) =
dim O(V )G
k z be the Poincare series of the algebra of invariants. Show that (2 points)
1 X
1
.
P (z) =
G
det(Id − zρ(g))
g∈G
Cn
Problem 9. Let X ⊂
be an irreducible affine subset and let U ⊂ X be
a Zariski-open subset. Show that U is dense in X in topology induced from
the standard Euclidean topology on Cn (1 point).
Problem 10. (a) Let G be a finite group with a representation G → GL(V ).
Show that O(V )G is generated by elements of the form
X
(g · f )d ,
g∈G
where f is a linear form. (b) Show that O(V )G is generated by polynomials
of degree at most |G| (2 points).
Problem 11. Find a genus 2 curve C such that Aut C contains Z5 and
confirm (or disprove) my suspicion that this curve gives a unique singular
point of M2 (2 points).
Problem 12. In the proof of Theorem 7.3.3, it was left unchecked that the
representation of G in V is algebraic. Show this (1 point).
Problem 13. Let X and Y be irreducible affine varieties and let f : X →
Y be a morphism. (a) Show that f factors as X → Y × Ar → Y , where the
first map is generically finite and the second map is a projection. (b) Show
that if f is generically finite then Y contains a principal open subset D such
that f −1 (D) → D is finite. (c) Show that the image of f contains an open
subset. (2 points)
Problem 14. Let G be a linear algebraic group acting regularly on an
algebraic variety X. Using the previous problem, show that any G-orbit is
open its closure (1 point).
Problem 15. Let H ⊂ G be a linearly reductive subgroup of a linear
algebraic group. Show that the set of cosets G/H has a natural structure of
an affine algebraic variety. On the other hand, show that the set of cosets
8A hypersurface F (x , . . . , x ) ⊂ Pn has a point of multiplicity d at p ∈ Pn if the folllow0
n
ing holds. Change coordinates so that p = [0 : . . . : 0 : 1]. Then F (x0 , . . . , xn−1 , 1) should
have no terms of degree less than d. A point of multiplicity 2 (resp. 3) is called a double
(resp. triple) point. A point p is called an ordinary double point (or a node) if p is a double
point and the quadratic part of F (x0 , . . . , xn−1 , 1) is a non-degenerate quadratic form.
9It is not hard to show that these last singularities are in fact A singularities
2
94
JENIA TEVELEV
GLn /B, where B is a subgroup of upper-triangular matrices, has a natural
structure of a projective variety (2 points).
Problem 16. Let G → GL(V ) be a representation of a linearly reductive
group and let π : V → V //G be the quotient. Show that the following
properties are equivalent (2 points):
• V //G is non-singular at π(0).
• V //G is non-singular.
• O(V )G is a polynomial algebra.
95
§8. Jacobians and periods
So far we have focussed on constructing moduli spaces using GIT, but
there exists a completely different approach using variations of Hodge structures. I will try to explain the most classical aspect of this theory, namely
the map
Mg → Ag , C 7→ Jac C.
Injectivity of this map is the classical Torelli theorem.
§8.1. Albanese torus. Let X be a smooth projective variety. We are going to integrate in this section, so we will mostly think of X as a complex
manifold. Recall that we have the first homology group
H1 (X, Z).
We think about it in the most naive way, as a group generated by smooth
oriented loops γ : S 1 ,→ X modulo relations γ1 + . . . + γr = 0 if loops
γ1 , . . . , γr bound a smooth oriented surface in X (with an induced orientation on loops). We then have a first cohomology group
H 1 (X, C) = Hom(H1 (X, Z), C).
This group can also be computed using de Rham cohomology
P
{complex-valued 1-forms ω =
fi dxi such that dω = 0}
1
.
HdR (X, C) =
{exact forms ω = df }
Pairing between loops and 1-forms is given by integration
Z
ω,
γ
which is well-defined by Green’s theorem. The fact that X is a smooth projective variety has important consequences for the structure of cohomology,
most notably one has Hodge decomposition, which in degree one reads
1
HdR
(X, C) = H 1,0 ⊕ H 0,1 ,
where H 1,0 = H 0 (X, Ω1 ) is the (finite-dimensional!) vector space of holomorphic 1-forms, and H 0,1 = H 1,0 is the space of anti-holomorphic 1 forms.
Integration gives pairing between H1 (X, Z) (modulo torsion) and H 0 (X, Ω1 ),
and we claim
R that this pairing is non-degenerate. Indeed, if this is not the
case then γ ω = 0 for some fixed non-trivial cohomology class γ ∈ H1
(modulo torsion)
R and for any holomorphic 1-form ω. But then of course
we also have γ ω̄ = 0, which contradicts the fact that pairing between
H1 (X, Z) (modulo torsion) and H 1 (X, C) is non-degenerate.
It follows that we have a complex torus
Alb(X) =
H 0 (X, Ω1 )∗
= V /Λ = Cq /Z2q
H1 (X, Z)/Torsion
called the Albanese torus of X. Λ is called the period lattice and
q = dim H 0 (X, Ω1 )
96
JENIA TEVELEV
is called the irregularity of X. If we fix a point p0 ∈ X, then we have a
holomorphic Abel–Jacobi map
Z p
•
µ : X → Alb(X), p 7→
p0
The dependence on the path of integrationP
is killed by taking the quotient by periods. Moreover, for any 0-cycle
aP
i pi (a formal combination
of P
points with integer P
multiplicities)
such
that
ai = 0, we can define
P
µ( ai pi ) by breaking ai pi = (qi − ri ) and defining
X
X Z qi
µ(
ai pi ) =
•.
ri
Again, any ambiguity in paths of integration and breaking the sum into
differences disappears after we take the quotient by periods.
When dim X > 1, we often have q = 0 (for example if π1 (X) = 0 or at
least H1 (X, C) = 0), but for curves q = g, the genus, and some of the most
beautiful geometry of algebraic curves is revealed by the Abel–Jacobi map.
§8.2. Jacobian. Let C be a compact Riemann surface (= an algebraic curve).
The Albanese torus in this case is known as the Jacobian
Jac(C) =
H 0 (C, K)∗
= V /Λ = Cg /Z2g
H1 (C, Z)
The first homology lattice H1 (C, Z) has a non-degenerate skew-symmetric
intersection pairing γ · γ 0 , which can be computed by first deforming loops
γ and γ 0 a little bit to make all intersections transversal and then computing the number of intersection points, where each point comes with + or −
depending on orientation of γ and γ 0 at this point. In the standard basis of
α and β cycles (draw), the intersection pairing has a matrix
0 −I
.
I 0
1 (C, C) also has a non-degenerate skew-symmetric pairing given by
HdR
Z
ω ∧ ω0.
C
We can transfer this pairing to the dual vector space H1 (C, C) and then
restrict to H1 (C, Z). It should come at no surprise that this restriction agrees
with the intersection pairing defined above. To see this concretely, let’s
work in the standard basis
δ1 , . . . , δ2g = α1 , . . . , αg , β1 , . . . , βg
of α and β cycles. We work in the model where the Riemann surface is
obtained by gluing the 4g gon ∆ with sides given by
α1 , β1 , α1−1 , β1−1 , α2 , . . .
Rp
Fix a point p0 in the interior of ∆ and define a function π(p) = p0 ω
(integral along the straight segment). Since ω is closed, the Green’s formula
97
shows that for any point p ∈ αi , and the corresponding point q ∈ αi−1 , we
have
Z
π(q) − π(p) =
ω.
βi
For any point p ∈ βi and the corresponding point q ∈ βi−1 , we have
Z
Z
ω=−
ω.
π(q) − π(p) =
α−1
i
αi
Then we have
Z
Z
Z
0
0
d(πω 0 )
(because ω 0 is closed)
dπ ∧ ω =
ω∧ω =
∆
∆
C
Z
=
πω 0
(by Green’s formula)
∂∆
XZ
XZ
0
πω 0 =
πω +
=
βi ∪βi−1
αi ∪α−1
i
=−
XZ
Z
βi
XZ
ω0 +
ω
αi
Z
ω
αi
ω0,
βi
which is exactly the pairing dual to the intersection pairing10.
Specializing to holomorphic 1-forms gives Riemann bilinear relations
8.2.1. P ROPOSITION . Let ω and ω 0 be holomorphic 1-forms. Then
Z
Z
Z
Z
X Z
0
0
ω ∧ ω 0 = 0,
ω =
ω
ω −
ω
and
X Z
Z
Z
ω̄ −
ω
αi
0
βi
C
αi
βi
βi
αi
Z
ω
βi
ω̄
αi
0
Z
=
ω ∧ ω̄ 0 .
C
We define a Hermitian form H on H 0 (C, K) by formula
Z
i
ω ∧ ω̄ 0
C
(notice an annoying i in front) and we transfer it to the Hermitian form
on V := H 0 (C, K)∗ , which we will also denote by H. The imaginary part
Im H is then a real-valued skew-symmetric form on H 0 (C, K) (and on V ).
We can view V and H 0 (C, K) as dual real vector spaces using the pairing
Re v(ω). Simple manipulations of Riemann bilinear relations give
Z
Z
Z
Z
X Z
0
0
Re
ω
Re
ω̄ − Re
ω
Re
ω̄ = Im i
ω ∧ ω̄ 0 ,
αi
βi
βi
αi
C
i.e. we have
8.2.2. C OROLLARY. The restriction of Im H on Λ := H1 (C, Z) is the standard
intersection pairing.
10Note that a general rule for computing dual pairing in coordinates is the following:
if B is a non-degenerate bilinear form on V , choose bases {ei } andP{ẽi } of V such that
B(ei , ẽj ) = δij . Then the dual pairing on V ∗ is given by B ∗ (f, f 0 ) =
f (ei )f 0 (ẽi ). It does
not depend on the choice of bases. In our example, the first basis of H1 (C, Z) is given by cycles α1 , . . . , αg , β1 , . . . , βg , and the second basis is then given by β1 , . . . , βg , −α1 , . . . , −αg .
98
JENIA TEVELEV
The classical way to encode Riemann’s bilinear identities is to choose a
basis ω1 , . . . , ωg of H 0 (C, K) and consider the period matrix
R
R
R
R

α 1 ω1 . . .
α g ω1
β1 ω1 . . .
βg ω1

.
..
. 
..
..
Ω =  ...
. R ..
. R .. 
R
R .
α 1 ωg . . .
α g ωg
β1 ωg . . .
βg ωg
Since H is positive-definite, the first minor g × g of this matrix is nondegenerate, and so in fact there exists a unique basis {wi } such that
Ω = [Id | Z] ,
where Z is a g × g matrix. Riemann’s bilinear identities then imply that
Z = Zt
and Im Z is positive-definite.
8.2.3. D EFINITION . The Siegel upper-half space Sg is the space of symmetric
g × g complex matrices Z such that Im Z is positive-definite.
To summarize our discussion above, we have the following
8.2.4. C OROLLARY. Let C be a genus g Riemann surface. Let Jac(C) R= V /Λ be
its Jacobian. Then V = H 0 (C, K)∗ carries a Hermitian form H = i C ω ∧ ω̄ 0 ,
and Im H restricts to the intersection pairing on Λ = H1 (C, Z). Any choice of
symplectic basis {δi } = {αi } ∪ {βi } in Λ determines a unique matrix in Sg .
Different choices of a symplectic basis are related by the action of the
symplectic group Sp(2g, Z). So we have a map
Mg → Ag := Sg / Sp(2g, Z).
It turns out that Ag is itself a moduli space.
§8.3. Abelian varieties.
8.3.1. D EFINITION . A complex torus V /Λ is called an Abelian variety if carries a structure of a projective algebraic variety, i.e. there exists a holomorphic embedding V /Λ ,→ PN .
One has the following theorem of Lefschetz:
8.3.2. T HEOREM . A complex torus is projective if and only if there exists a Hermitian form H on V (called polarization) such that Im H restricts to an integral
skew-symmetric form on Λ.
It is easy to classify integral skew-symmetric forms Q on Z2g :
8.3.3. L EMMA . There exist uniquely defined positive integers δ1 |δ2 | . . . |δg such
that the matrix of Q in some Z-basis is


−δ1


−δ2


..


.




−δg 

.


δ1


δ2




.


..
δg
99
Proof. For each λ ∈ Λ = Z2g , let dλ be the positive generator of the principal ideal {Q(λ, •)} ⊂ Z. Let δ1 = min(dλ ), take λ1 , λg+1 ∈ Λ such that
Q(λ, λg+1 ) = δ1 . Those are the first two vectors in the basis. For any λ ∈ Λ,
we know that δ1 divides Q(λ, λ1 ) and Q(λ, λg+1 ), and therefore
λ+
Q(λ, λg+1 )
Q(λ, λ1 )
λg+1 +
λ1 ∈ hλ1 , λg+1 i⊥
Z.
δ1
δ1
Now we proceed by induction by constructing a basis in hλ1 , λg+1 i⊥
Z.
8.3.4. D EFINITION . A polarization H is called principal if we have
δ1 = . . . = δg = 1
in the canonical form above. An Abelian variety V /Λ endowed with a principal polarization is called a principally polarized Abelian variety.
So we have
8.3.5. C OROLLARY. Ag parametrizes principally polarized Abelian varieties.
In fact Ag has a natural structure of an algebraic variety. One can define
families of Abelian varieties in such a way that Ag is a coarse moduli space.
§8.4. Abel’s Theorem. Returning to the Abel–Jacobi map, we have the following fundamental
8.4.1. T HEOREM (Abel’s theorem). The Abel–Jacobi map AJ : Div0 (C) →
Jac(C) induces a bijection
µ : Pic0 (C) ' Jac(C).
The proof consists of three steps:
(1) AJ(f ) = 0 for any rational function f ∈ k(C), hence AJ induces µ.
(2) µ is injective.
(3) µ is surjective.
For the first step, we consider a holomorphic map P1[λ:µ] given by
[λ, µ] 7→ AJ(λf + µ).
It suffices to show that this map is constant. We claim that any holomorphic
map r : P1 → V /Λ is constant. It suffices to show that dr = 0 at any
point. But the cotangent space to V /Λ at any point is generated by global
holomorphic forms dz1 , . . . , dzg (where z1 , . . . , zg are coordinates in V . A
pull-back of any of them to P1 is a global 1-form on P1 , but KP1 = −2[∞],
hence the only global holomorphic form is zero. Thus dr∗ (dzi ) = 0 for
any i, i.e. dr = 0.
§8.5. DifferentialsP
of the third kind. To show injectivity of µ, we have to
check that if D =
ai pi ∈ Div0 and µ(D) = 0 then D = (f ). If f exists
then
1 df
1
d log(f ) =
ν=
2πi
2πi f
has only simple poles, these poles are at pi ’s and Respi ν = ai (why?) . Moreover, since branches of log differ by integer multiples of 2πi, any period
Z
ν∈Z
γ
100
JENIA TEVELEV
for any closed loop γ. And it is easy to see that if ν with these properties
exists then we can define
Z p
ν).
f (p) = exp(2πi
p0
This will be a single-valued meromorphic (hence rational) function with
(f ) = D. So let’s construct ν. Holomorphic 1-forms on C with simple
poles are classically known as differentials of the third kind. They belong to
the linear system H 0 (C, K + p1 + . . . + pr ). Notice that we have an exact
sequence
ψ
0 → H 0 (C, K) → H 0 (C, K + p1 + . . . + pr ) −→ Cr ,
where ψ is given by taking residues. by Riemann–Roch, dimensions of the
linear systems are g and g + r − 1. ByPa theorem on the sum of residues, the
image of ψ lands in the hyperplane ai = 0. It follows that ψ is surjective
onto this hyperplane, i.e. we can find a differential η of the third kind with
any prescribed residues (as long as they add up to zero). The game now is
to make periods of η integral by adding to η a holomorphic form (which of
course would not change the residues). Since the first g × g minor of the
period matrix is non-degenerate, we can arrange that A-periods of η are
equal to 0.
Now, for any holomorphic 1-form ω, arguing as in the proof of Prop. 8.2.1,
we have the following identity:
X
Z
Z
Z
Z pi
r
r
X Z
X
ω
η−
ω
η =
ai π(pi ) =
ω.
ai
αi
βi
βi
αi
i=1
i=1
p0
Indeed, we can remove small disks around
R each pi to make η holomorphic
in their complement, and then compute ω ∧ η by Green’s theorem as in
Prop. 8.2.1. This gives
Z
Z pi
r
XZ
X
ω
η=
ω.
ai
βi
αi
p0
i=1
R
P
Since µ(D) = 0, we can write the RHS as γ ω, where γ =
mi δi is an
integral linear combination of periods. Applying this to the normalized
basis of holomorphic 1-forms gives
Z
Z
η = ωi
βi
Now let
0
η =η−
γ
g
X
mg+k ωk .
k=1
Then we have
Z
η 0 = −mg+i
αi
and
Z
βi
η0 =
Z
ωi −
γ
X
Z
mg+k
ωk =
βi
X
Z
mk
ωi +
X
αi
Z
ωi −
mg+k
X
101
Z
mg+k
βk
ωk = mi
βk
§8.6. Summation maps. To show surjectivity, we are going to look at the
summation maps
C d → Picd → Jac(C),
(p1 , . . . , pd ) 7→ µ(p1 + . . . + pd − dp0 ),
where p0 ∈ C is a fixed point. It is more natural to define
Symd C = C d /Sd ,
and think about summation maps as maps
φd : Symd C → Jac C.
It is not hard to endow Symd C with a structure of a complex manifold
in such a way that φd is a holomorphic map11. We endow Symd C with a
quotient topology for the map π : C d → Symd C, and then define complex charts as follows: at a point (p1 , . . . , pd ), choose disjoint holomorphic neighborhoods Ui ’s of pi ’s (if pi = pj then choose the same neighborhood Ui = Uj ). Let zi ’s be local coordinates. Then local coordinates on
π(U1 × . . . × Ud ) can be computed a follows: for each group of equal points
pi , i ∈ I, use elementary symmetric functions in zi . i ∈ I instead of zi ’s
themselves.
The main point is absolutely obvious
8.6.1. L EMMA . For D ∈ Picd , µ−1 (D) = |D|. Fibers of µ are projective spaces.
To show that µ is surjective it suffices to show that φd is surjective. Since
this is a proper map of complex manifolds of the same dimension, it suffices
to check that a general fiber is a point. In view of the previous Lemma this
boils down to showing that if (p1 , . . . , pg ) ∈ Symg is sufficiently general
then H 0 (C, p1 + . . . + pg ) = 1. For inductive purposes, lets show that
8.6.2. L EMMA . For any k ≤ g, and sufficiently general points p1 , . . . , pk ∈ C,
we have H 0 (C, p1 + . . . + pk ) = 1.
Proof. By Riemann-Roch, we can show instead that
H 0 (C, K − p1 − . . . − pk ) = g − k
for k ≤ g and for a sufficiently general choice of points. Choose an effective
canonical divisor K and choose points pi away from it. Then we have an
exact sequence
0 → L(K − p1 − . . . − pk ) → L(K − p1 − . . . − pk−1 ) → C,
where the last map is the evaluation map at the point pk . It follows that
either |K − p1 − . . . − pk | = |K − p1 − . . . − pk−1 | or dimensions of these
two projective spaces differ by 1, the latter happens if one of the functions
in L(K − p1 − . . . − pk−1 ) does not vanish at pk . So just choose pk to be a
point where one of these functions does not vanish.
8.6.3. C OROLLARY. We can identify Pic0 and Jac by means of µ.
11It is also not hard to show that Symd C is a projective algebraic variety. Since Jac C is
projective by Lefschetz theorem, it follows (by GAGA) that φd is actually a regular map.
102
JENIA TEVELEV
§8.7. Theta-divisor.
8.7.1. C OROLLARY. The image of φd−1 i a hypersurface Θ in Jac C.
8.7.2. D EFINITION . Θ is called the theta-divisor.
8.7.3. E XAMPLE . If g = 1, not much is going on: C = Jac C. If g = 2, we
have φ1 : C ,→ Jac C: the curve itself is a theta-divisor! The map φ2 is a
bit more interesting: if h0 (C, p + q) > 1 then p + q ∈ |K| by Riemann–Roch.
In other words, p and q are permuted by the hyperelliptic involution and
these pairs (p, q) are parametrized by P1 as fibers of the 2 : 1 map φ|K| :
1
C → P1 . So φ2 is an isomorphism outside of K ∈ Pic2 , but φ−1
2 (K) ' P .
2
Since both Sym C and Jac C are smooth surfaces, this implies that φ2 is a
blow-up of the point.
8.7.4. E XAMPLE . In genus 3, something even more interesting happens. Notice that φ2 fails to be an isomorphism only if C carries a pencil of degree
2, i.e. if C is hyperelliptic. In this case φ2 again contracts a curve E ' P1 ,
but this time it is not a blow-up of a smooth point. To see this, I am going to use adjunction formula. Let Ẽ ⊂ C × C be the preimage. Then Ẽ
parametrizes points (p, q) in the hyperelliptic involution, i.e. Ẽ ' C but not
a diagonally embedded one. We can write a holomorphic 2-form on C × C
as a wedge product pr∗1 (ω) ∧ pr∗2 (ω), where ω is a holomorphic 1-form on
C. Since deg KC = 2, the canonical divisor K on C × C can be chosen as a
union of 4 vertical and 4 horizontal rulings. This K · Ẽ = 8, but
(K + Ẽ) · Ẽ = 2g(Ẽ) − 2 = 4
by adjunction, which implies that Ẽ · Ẽ = −4. Under the 2 : 1 map C × C →
Sym2 C, Ẽ 2 : 1 covers our E ' P1 . So E 2 = −2. This implies that the
image of φ2 has a simple quadratic singularity at φ2 (E). So the Abel-Jacobi
map will distinguish between hyperelliptic and non-hyperelliptic genus 3
curves by appearance of a singular point in the theta-divisor.
§8.8. Homework.
Problem 1. Generalizing the action of SL(2, Z) on the upper-half plane,
give formulas for the action of Sp(2g, Z) on Sg (1 point).
Problem 2. In the proof of Lemma 8.3.3, show that indeed we have
δ1 |δ2 | . . . |δg
(1 point).
Problem 3. Show that (A2 )n //Sn is singular (with respect to the action
permuting factors) at the point that corresponds to the orbit (0, . . . , 0) (2 points).
Problem 4. Show that Symd P1 = Pd (1 point).
Problem 5. Let C be an algebraic curve. Define Symd C as an algebraic
variety (1 point).
Problem 6. Show that if φ1 (C) ⊂ Jac C is symmetric (i.e. φ1 (C) = −φ1 (C))
then C is hyperelliptic. Is the converse true? (1 point).
Problem 7. Show that either the canonical map φ|K| is an embedding or
C is hyperelliptic. (1 point).
Problem 8. Let C be a non-hyperellptic curve. C is called trigonal if it
admits a 3 : 1 map C → P1 . (a) Show that C is trigonal if and only if
103
its canonical embedding φ|K| has a trisecant, i.e. a line intersecting it in (at
least) three points. (b) Show that if C is trigonal then its canonical embedding is not cut out by quadrics12. (2 points).
Problem 9. Show that the secant lines of a rational normal curve in Pn
are are parametrized by the surface in the Grassmannian G(2, n + 1) and
that this surface is isomorphic to P2 (2 points).
Problem 10. Consider two conics C1 , C2 ⊂ P2 which intersect at 4 distinct points. Let E ⊂ C1 × C2 be a curve that parametrizes pairs (x, y) such
that the line Lxy connecting x and y is tangent to C2 at y. (a) Show that E is
an elliptic curve. (b) Consider the map t : E → E defined as follows: send
(x, y) to (x0 , y 0 ), where x0 is the second point of intersection of Lxy with C1
and Lx0 y0 is the second tangent line to C2 through x0 . Show that t is a translation map (with respect to the group law on the elliptic curve). (c) Show
that if there exists a 7-gon inscribed in C1 and circumscribed around C2
then there exist infinitely many such 7-gons, more precisely there is one
through each point of C1 (3 points).
Problem 11. Let C be a hyperelliptic curve and let R = {p0 , . . . , p2g+1 }
be the branch points of the 2 : 1 map C → P1 . We choose p0 as the base
point for summation maps φd : Symd → Jac. For any subset S ⊂ R, let
α(S) = φ|S| (S). (a) Show that αS ∈ Jac[2] (the 2-torsion part). (b) Show
that αS = αS c . (c) Show that α gives a bijection between subsets of Bg of
even cardinality defined upto S ↔ S c and points of Jac[2] (3 points).
Problem 12. A divisor D on C is called a theta-characteristic if 2D ∼
K. A theta-characteristic is called vanishing if h0 (D) is even and positive.
Show that a curve of genus 2 has no vanishing theta characteristics but a
curve of genus 3 has a vanishing theta characteristic if and only if it is a
hyperelliptic curve (1 point).
Problem 13. Show that a nonsingular plane curve of degree 5 does not
have a vanishing theta characteristic (3 points).
Problem 14. Let E = {y 2 = 4x3 − g2 x − g3 be an elliptic curve with real
coefficients g2 , g3 . Compute periods to show that E ' C/Λ, where either
Λ = Z + τ iZ or Λ = Z + τ (1 + i)Z (with real τ ) depending on the number
of real roots of the equation 4x3 − g2 x − g3 = 0 (3 points).
Problem 15. Consider a (non-compact!) curve C = P1 \ {p1 , . . . , pr }.
Since P1 has no holomorphic 1-forms, lets consider instead differentials of
the third kind and define
V := H 0 (P1 , K + p1 + . . . + pr )∗ .
Show that Λ := H1 (C, Z) = Zr−1 , define periods, integration pairing, and
the “Jacobian” V /Λ. Show that V /Λ ' (C∗ )r−1 and that C embeds in V /Λ '
(C∗ )r−1 by the Abel-Jacobi map (2 points).
Problem 16. Let C be an algebraic curve with a fixed point p0 and consider the Abel–Jacobi map φ = φ1 : C → Jac. For any point p ∈ C, we have
a subspace dφ(Tp C) ⊂ Tφ(p) Jac. By applying a translation by φ(p), we can
identify Tφ(p) Jac with T0 Jac ' Cg . Combining these maps together gives
a map C → Pg−1 , p 7→ dφ(Tp C). Show that this map is nothing but the
canonical map φ|K| (2 points).
12This is practically if and only if statement by Petri’s theorem.
104
JENIA TEVELEV
Problem 17. Let F , G be homogeneous polynomials in C[x, y, z]. Suppose that curves F = 0 and G = 0 intersect transversally at the set of points
Γ. (a) Show that associated primes of (F, G) are the homogeneous ideals
I(pi ) of points pi ∈ Γ. (b) Show that every primary ideal of (F, G) is radical
by computing its localizations at pi ’s (c) Conclude that I(Γ) = (F, G), i.e.
any homogeneous polynomial that vanishes at Γ is a linear combination
AF + BG (2 points).
§9. Torelli theorem
Will type when have some time.
R EFERENCES
[AM]
[Do]
[F]
[Fu]
[G]
[GH]
[HM]
[Ha]
[I]
[Ma]
[Mi]
[Mu]
[M1]
[M2]
[M3]
[MS]
[PV]
[R]
[R1]
[S]
[X]
[Sh]
[St]
Atiyah–Macdonald
I. Dolgachev, Topics in Classical Algebraic Geometry
R. Friedman, Algebraic surfaces and holomorphic vector bundles
W. Fulton, Introduction to Toric Varieties
P. Griffiths, Introduction to Algebraic Curves
P. Griffiths, J. Harris, Principles of Algebraic Geometry
J. Harris, I. Morrison, Moduli of Curves
R. Hartshorne, Algebraic Geometry
J. Igusa, On Siegel modular forms of genus two, Amer. J. Math. 84 (1962) 175Ð200
Matsumura
R. Miranda, Algebraic Curves and Riemann Surfaces
S. Mukai, An introduction to Invariants and Moduli
D. Mumford, Curves and their Jacobians
D. Mumford, Geometric Invariant Theory
D. Mumford, Red Book of Varieties and Schemes
D. Mumford, K. Suominen, Introduction to the theory of moduli
V. Popov, E. Vinberg, Invariant Theory
M. Reid, Surface cyclic quotient singularities and and Hirzebruch–Jung resolutions
M. Reid, Graded rings and varieties in weighted projective spaces
G. Salmon, Lessons introductory to the modern higher algebra, 2nd edition, Hodges
Smith, Dublin, 1866
I. Shafarevich, Basic Algebraic Geometry
V. Shokurov, Riemann surfaces and algebraic curves
B. Sturmfels, Algorithms in Invariant Theory

MODULI SPACES AND INVARIANT THEORY §0. Syllabus 2 §1

Transcription

Similar documents

Managers` Towering Personality with 14Qs

View a sample Math in Focus lesson

Toy Trains - Inside Mathematics

on the development of early algebraic thinking

8.4 Dividing Integers

History of Solving Polynomial Equations