Geometric Invariant TheoryMath 533 - Spring 2011

Zinovy Reichestein

Lecture notes by Pooya Ronagh

Department of mathematics, University of British ColumbiaRoom 121 - 1984 Mathematics Road, BC, Canada V6T 1Z2

E-mail address: pooya@math.ubc.ca

Preface 3

Chapter 1. Moduli spaces 41. An example 4

Chapter 2. Crash course on linear algebraic groups 81. Diagonalizable groups and characters 92. Reductive groups 123. Borel subgroups 13

Chapter 3. Action of algebraic groups 151. Relation of coarse moduli spaces and categorical quotients 15

Chapter 4. Categorical quotient of action of algebraic groupson affine spaces 18

1. Hilbert’s 14-th problem 192. Reductive and linearly reductive groups 203. Stability 25

Chapter 5. Hilbert-Mumford Criterion 29

Chapter 6. Categorical quotients for projective varieties 37

Chapter 7. Application - Construction of the moduli of elliptic curves 421. Review on curves 422. Elliptic curves as ramified coverings of the projective line 443. The moduli problem and its solution 46

Chapter 8. Construction of Mg 491. Hilbert polynomial 492. The Hilbert scheme 493. Our moduli problem 49

Appendix A. Toric varieties as GIT quotients 511. Review of toric varieties 512. GIT quotients 52

Appendix B. Moduli of representations of a quiver 54

Appendix C. Comments on homework problems 55

Appendix. Bibliography 56

Contents

Preface

The following is the lecture notes I have written from a course Professor Zinovy Reichsteintaught in University of British Columbia in Winter 2011. I am thankful to him on corre-spondences in which he gave me the references for the material presented below which Iwill cite throughout and in this preface. I am also thankful to him for reading these lecturenotes, providing me with comments and corrections.

What follows heavily and speacially at the beginning of the term relies on the textbookby Newstead [4]. A number of other sources, including papers and books by Fogarty,Mumford [3], Kraft, Popov and Vinberg, Deudonne and Carrell, and most recently, Birkesand Richardson have been used. Example 1.1 was provided by Dave Anderson. Chapter2 was entirely taught by Dave Anderson as well. He mainly used [1] and [2]. The lattercontains more results with non-algebraically closed fields.

3

CHAPTER 1

Moduli spaces

To parametrize algebraic objects of certain type, need to set up a moduli problem; i.e. acontravariant functor

F ∶ Var Ð→ Setswhere F(pt) = the set of objects we want to parametrize and F(S) is the set of familiesof objects over S. Note that this F is not uniquely determined by the objects we want toparametrize.

Recall that M is a fine moduli space if F is represented by M , i.e. F(S) = Mor(S,M) forany variety S. Also recall that existence of a fine moduli space M is equivalent to existenceof a universal object above X ↦M where by a universal object we mean

F(S) = {F(η)(X) ∶ η ∶ S Ð→M}is satisfied for any S.

Remark. Note that existence of a nontrivial fibration Y over some S implies that the finemoduli space does not exist.

Definition 1. A variety M is called a coarse moduli space for F if

(1) F(pt) ≅M (as a set).

(2) F Ð→Mor(−,M) is a morphism of functors.

(3) Any other morphism of functors F Ð→Mor(−,N) factors (uniquely; already followssince it is already given on points) through some morphism M Ð→ N of varieties.

Remark. A fine moduli space is a coarse moduli space.

1. An example

Let V be an n-dimensional vector space, or simply kn. The objects we want to classify arelinear maps V TÐ→ V . We say two such objects T ∶ V Ð→ V and G′ ∶ V Ð→ V are equivalent ifthere is g ∈ GL(V ) such that

VT //

g

��

V

g

��

VT ′ // V

4

1. AN EXAMPLE 5

or that T and T ′ are conjugate. The moduli problem will be

F(S) = {(E,T ) ∶ E Ð→ S vector bundle of rank n , ET //

@@@@@@@ E

��

S

}/ ≅ .

Here isomorphism classes are given by commutative diagrams

VT1 //

g

��

V

g

��

VT2 // V

.

We will see that

All endomorphismsÐ→ No moduli spacesemisimple endomorphismsÐ→ coarse moduli space

cyclic endomorphismsÐ→ Fine moduli space

Proposition 1. There is no fine moduli space for this functor.

Proof. Consider E = A1 × V over S = A1 and let T ∶ E Ð→ E be given via

(t, v)↦ (t,

⎛⎜⎜⎜⎜⎜⎝

0 t 0 ⋯0 0 t ⋯

⋱0 ⋯ 0 t0 0 ⋯ 0

⎞⎟⎟⎟⎟⎟⎠

v)

For all t ≠ 0 these matrices are equivalent by conjugation by diag(1, t, t2,⋯, tn−1) and thematrix T ∣t=0 is not equivalent to them. �

1.1. Cyclic endomorphisms. T ∶ V Ð→ V is cyclic if v, Tv,⋯, Tn−1v form a basis of Vfor some v ∈ V . There would be two ways of defining the families here: (1) T ∶ E Ð→ E overS is locally cyclic if Ts is cyclic for any s ∈ S. (2) To require that T is globally cyclic.

1.2. Semisimple endormophisms.

F(pt) = { semisimple endomorphisms T ∶ V Ð→ V }where semisimple means diagonalizable, and let’s say char(k) /∣ n and

F(S) = {(E,T ) ∶ E Ð→ S vector bundle , ET //

@@@@@@@ E

��

S

semisimple endomorphisms }/ ≅

1. AN EXAMPLE 6

Theorem 1.1. (1) M = An is a coarse moduli space for F . here the isomorphismbetween F(pt) and An is given by

Tcharac.polyÐ→ (c1,⋯, cn)

where xn + c1xn−1 +⋯ + cn is the characteristic polynomial .

(2) There is not fine moduli space for F .

Proof. For (1) need to check conditions (2) and (3) of the definition. Given a familyof endomorphisms

E

��???????T // E

���������

S

over S, want to check that the induced map f ∶ S charÐ→ An is a morphism. It is enough todo this locally on S: Cover S by affine opens {Si} where E ≅ Si × Vj is the trivial bundleover Si. Over Si choose a global basis, b1,⋯, bn of global sections for E. In this basis T isgiven by an n×n matrix with entries in k[Si]. Taking the characteristic polynomial of thismatrix we see that f ∣Si is a morphism. To check condition (3) suppose F Ð→Mor(−,N) isa morphism of functors. Want to show this factors through a morphism M Ð→ N . Considerthe following family: S = An and E = An × V . Fix a basis (λ1,⋯, λn) for kn and assume

T = (λ1 00 λn

)

and apply our morphism to this family.

E

��

S = An

λ=(λ1,⋯,λn)↦(e(λ),⋯,e(λ))��

Snequiv// N

αi // k

M

where ei = ( the i-th characteristic polynomial )(−1)i.

Case 1: N is affine. Choose generators α1,⋯, αn ∈ k[N]. Pull them back to S. The pull-backs are symmetric functions on S; i.e. polynomials in e1(λ),⋯, en(λn). Therefore factorsthrough M .

1. AN EXAMPLE 7

Case 2: General N . We consider the special case of the natural action of G = Sn onX = An.1 In general cover N by affines and use the fact that e1(λ),⋯, en(λ) generate thefield of Sn-invariant rational functions on S = An. (Check!)

For second part assume the contrary that M = An is a fine moduli space (by uniquenessof coarse moduli space). Then in particular M carries a universal family; i.e. there is avector bundle E Ð→ An and a semisimple endomorphism T ∶ E Ð→ E over An such that thecharacteristic polynomial of T is xn + a1x

n−1 + ⋯ + an where a1,⋯, an are the coordinatefunctions on An. We claim that such a family cannot exist. Choose g ∈ k[t1,⋯, tn] suchthat E is trivial over the principal open subset U = {g ≠ 0} in An and 0 ∈ U i.e. g(0) ≠ 0.Over U choose a basis of global sections and write the matrix as (fij)n×n where fij ∈ k[U] =k[a1,⋯, an, g−1]. On the other hand T0 is nilpotent and semisimple, hence zero. Thusfij(0) = 0 for all i, j. Thus det(T ) vanishes to order n at zero. But detT = (−1)nan. Byour assumption on char(T ) if n ≥ 2, this is a contradiction. �

Question: What happens if n = 1? (Hint: consider the way we introduced families for cyclicendomorphisms. The answer depends on specific function F we consider, i.e. the type offamilies of semisimple endomorphisms we allow over a given bases.)

Example 1.1. Look at the moduli problemM1,1 of families of 1 pointed elliptic curvesmod isomorphisms.2 Claim:

(1) M1,1 ≅ A1 is the coarse moduli space (Use j-invariants machinery - See Hartshorne).

(2) There is no fine moduli space; there is a family X Ð→ S with all fibers isomorphicbut tnontrivial. Let E be an elliptic curve Γ ⊆ E ×E ×E. And Γ = {(x, y, x + y)}.And look at first projection.

▸ Exercise 1. Show this is NOT an example (that the fibration is nontrivial).

▸ Exercise 2. Fix an elliptic curve E, and look at the fibration over P1 ×0,∞ P1

given by E × P1 ∪E×{0},E×{∞} E × P1 an gluing on zero via identity and on ∞ viainverse. This has only 4 sections, and therefore is nontrivial.

⌟

1We have supplied a proof of this case in the course of proving that Spec k[X]G is a categorical quotientfor the G-action on X, where G is linearly reductive group, acting on an affine space X.

2This example was given by David Anderson, at the beginning of his class.

CHAPTER 2

Crash course on linear algebraic groups

Definition 2. An algebraic group is a group object in the category of varieties; i.e. G is avariety and morphisms are G ×G Ð→ G and G Ð→ G and has a point e ∈ G. In the schemeylanguage, the functor hG ∶ Sch/k Ð→ Set factors though the category of groups.

Definition 3. A linear algebraic group is an algebraic group G that is an affine variety.

Right away this tells you something about the coordinate ring of it! But let’s do an examplefirst:

Example 0.2.Gm = k∗ = Spec(k[x,x−1])

. ⌟

Note: The coordinate rings become k[G] have the structure of a commutative Hopf algebra:δ ∶ k[G]Ð→ k[G]⊗ k[G] the co-multiplication of the the algebra and c ∶ k[G]Ð→ k[G] is theantipade map and there is also a counit ε ∶ k[G]Ð→ k.

Example 0.3. For the multiplicative group:

δ ∶ k[x,x−1]Ð→ k[x,x−1]⊗ k[x,x−1]x↦ x⊗ x

and

c ∶ k[x,x−1]Ð→ k[x,x−1]x↦ x−1

and ε ∶ x↦ 1. ⌟▸ Exercise 3. Work out δ, c, ε for Ga (Ga = k with multiplication).

▸ Exercise 4. Gln = Spec k[xij]det.

▸ Exercise 5. Any Zariski closed subgroup of Gln; On and Sln and Bn ⊂ Gln the subgroupof upper-triangular matrices (called the Borel subgroups). Dn ⊆ Bn ⊂ Gln the subgroup ofdiagonal matrices (maximal torus) and Un ⊂ Bn ⊂ GLn strictly upper-triganluar matrices(1’s on the diagonal) (unipotent subgroup).

8

1. DIAGONALIZABLE GROUPS AND CHARACTERS 9

Proposition 2. Every linear algebraic group can be embedded in some Gln. The slogan isthat affine ⇔ linear.

Definition 4. A (rational) representation of G is a homomorphism G Ð→ Gl(V ) where Vis a vector space (homomorphism of algebraic groups is a map of the varieties that is alsoa map of groups).

A representation V is an irreducible one if there is no 0 ⊊W ⊊ V with G.W ⊆W .

Remark. You can also talk about representations where V is infinite. And there is animportant example of it. But always assume locally finite: i.e. for v ∈ V there is a fintiedimenional W where v ∈W ⊊ V which is G-stable (G.W ⊆W ).

Example 0.4. G acts on k[G] by g.f(x) = f(g−1x) for all x ∈ G.1 ⌟

Lemma 1. If W ⊊ k[G] is any finite dimensional subspace, then there is a subspace W ⊆V ⊆ k[G] such that V is finite-dimensional and G.V = V .

Proof. Without loss of generality assume W = span{f}. Write δ(f) = ∑imi ⊗ fi ∈k[G ×G]. δ corresponds to G ×GÐ→ G via (g1, g2)↦ (g−1

1 g2). Then for every g ∈ G

(g.f)(x) = f(g−1x) =∑i

mi(g)fi(x)

So g.f = ∑imi(g)fi. So g.f ∈ span{f1,⋯, fn}. Then span{g.f ∶ g ∈ G} does the trick. �

Proof of proposition. Take generators f1,⋯, fn of k[G]. By lemma assume theyare a basis for a G-stable subspace of k[G]. As in the lemma have mij ∈ k[G] such thatg.fi = ∑jmij(g)fj . and the map is k[xij]det Ð→ k[G] via xij ↦mij . And it remains to showthat this is surjective. Because

fi(g−1) = fi(g−1e) =∑j

mij(g)fj(e)

see fi = ∑j fj(e)mij and mij ’s generate k[G] so the map is surjective. �

1. Diagonalizable groups and characters

Dn ⊆ GLn the group of diagonal matrices. Note that k[Dn] ≅ k[x±1 ,⋯, x±n] which is isomor-phic to the group algebra k[Zn].

Definition 5. A character of an algebraic group G is a homomorphism χ ∶ GÐ→ Gm.

1The author has been struggling to figure this out in the language of groups schemes!

1. DIAGONALIZABLE GROUPS AND CHARACTERS 10

The set of charactersX(G) = Homalg.gp.(G,Gm)

is an abelian group under pointwise multiplication. (Warning: people intend to write theoperation of this group additively sometimes!)

χ1.χ2(g) = χ1(g)χ2(g).

Example 1.1. X(Gm) = Z via id ↦ 1 and a character of Gm is given via χ ∶ z ↦ zn ingeneral. ⌟

Note that Dn ≅ (Gm)n and hence χ(Dn) ≅ Zn (in of course a non-canonical way) andtherefore another way of writing the isomorphism in the beginning is k[Dn] = k[X(Dn)].Moral: Dn has lots of characters in contrast to Sln which has no nontrivial characters.

Definition 6. A linear algebraic group is diagonalizable if its isomorphic to a closedsubgroup of some Dn and if it also happens to be connected then it is called a torus.

Diagonalizable groups are very useful for the following structure theorem:

Proposition 3. For a linear algebraic group D the following are equivalent:

(1) D is diagonalizable.

(2) X(D) is finitely generated abelian group and k[D] ≅ k[X(D)].

(3) Every (finite dimensional, rational) representation of D is isomorphic to a directsum of 1-dimensional representations.

(4) D ≅ (k∗)r ×A for some finite abelian group A.

Corollary 1. T is a torus iff T ≅ (k∗)r (assuming algebraically closedness). (Spec(F3[x])(x)is not connected.)

Remark. In (3) if a given representation V you can write V = ⊕χ∈X(D)Vχ where

Vχ = {v ∈ V ∶ g.v = χ(g)v,∀g ∈D}and these latter are called weight spaces.

Remark. In (4) we certainly need the condition that A has no p-torsion when char(k) = p.

Remark. These tori are essentially the only groups with semi-simple representation theoryindependent of character of the field (linearly reductive condition).

Example 1.2 (of (3)). T = (k∗)2 ⊆ Gl2 and acting on M2×2 by conjugation. Then

T = {(z1 00 z2

)} is generated as an abelian group (has basis as Z-module) given by

χ1(g) = z1, χ2(g) = z2.

1. DIAGONALIZABLE GROUPS AND CHARACTERS 11

Note that g.(a bc d

) = ( a z1z−12 b

z−11 z2c d

) and

M2×2 = k.

χ1χ−12

³¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹µ

(0 10 0) ⊕k.

χ−11 χ2

³¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹µ

(0 01 0) ⊕k.

0

³¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹µ

(∗ 00 ∗) .

of weights χ1χ−12 and χ−1

1 χ2 and 0. ⌟

Example 1.3 (of (4)). X(D) = Z⊕ Z/2Z then if we work backwards to find the alge-braic group itself then

k[X(D)] ≅ k[x,x−1, y]/(y2 − 1)shows that D = Spec k[X(D)] = Gm × µ2. Note that µ2 is not reduced if char(k) = 2. ⌟

Proof. See Springer. We leave 2 ⇒4 ⇒1 ⇒2 and 3 ⇒1 as exercise. Remains to prove2 ⇒3. Given a representation D

ϕÐ→ GL(V ) choose a basis for V so

ϕ(g) = (aij(g))for some aij ∈ k[D] = k[X(D)] so

aij =∑χ

cχijχ.

Write ϕ(g) = ∑χ χ(g)Aχ where Aχ = (cχij). Will show Aχ ∈ End(V ) is projection onto Vχ.

Claim: For χ,ψ ∈X(D) have

AχAψ = δχ,ψAχ (orthogonal independence).

In fact since ϕ(gh) = ϕ(g)ϕ(h) get

∑η∈X(D)

η(gh)Aη =∑B

⎛⎝ ∑AχAψ=B

χ(g)ψ(h)⎞⎠B.

So entry-wise∑η

cηijη(g)η(h) = ∑χ,ψ

bχ,ψij χ(g)ψ(h).

We now use the theorem of linear independence of character for the group D ×D and get

bχ,ψij = cηijδχ,ψ,η.

Finally∑

χ∈X(D)Aχ = 1

and this is because the left hand side is ∑χ χ(1)Aχ = ϕ(1). So V = ⊕Vχ where Vχ =im(Aχ). �

Dual to characters: Called the co-characters also, are define by

2. REDUCTIVE GROUPS 12

Definition 7. For an algebriac group G a one-parameter subgroup is λ ∶ Gm Ð→ G. Y (G) =Homal.gp.(Gm,G) is also a group.

There is always a natural pairing

X(G) × Y (G)Ð→ Zvia (χ,λ)↦ χ ○ λ ∈ Hom(Gm,Gm).▸ Exercise 6. When G is a torus this is a perfect pairing (i.e. Y (G) = HomZ(X(G),Z)).

2. Reductive groups

Let k = k and G linear algebraic group. An element x ∈ G is semisimple if there is a faithful(injective) representation ϕ ∶ GÐ→ Gln such that

ρ(x) = (∗ 00 ∗)

n×n.

x ∈ G is unipotent if there is ρ ∶ GÐ→ Gln such that ρ(x) = (1 ∗0 1).

Theorem 2.1 (Jordan decomposition). For any x ∈ G there are unique elements xs, xu ∈ Gsuch that

x = xsxu = xuxsand xs is semisimple and xu is unipotent. Moreover, any homomorphism ϕ ∶ G Ð→ Hpreserves ss + unipotent parts, i.e. ϕ(x)s = ϕ(xs) and ϕ(x)u = ϕ(xu).

Proof. See Humphreys, section 15.3. �

Definition 8. G is unipotent if all x ∈ G are unipotent.

Remark. Any image or subgroup of a unipotent G is unipotent.

Definition 9. G is solvable if G = G0 ⊇ G1 ⊇ ⋯ ⊇ terminates in {e}, where Gi = [Gi−1,Gi−1]is the commutator group.

Note: For any algebraic group G, (G,G) is generated by all ghg−1h−1 and it is a closedsubgroup (not obvious).

Example 2.1. Un ⊆ GLn the subgroup of ( 1 ∗0 ∗ 1 ) elements is unipotent. It is also

solvable:⎛⎜⎝

1 ∗ ∗0 1 ∗0 0 1

⎞⎟⎠⊃⎛⎜⎝

1 0 ∗0 1 00 0 1

⎞⎟⎠⊃⎛⎜⎝

1 0 00 1 00 0 1

⎞⎟⎠

⌟

3. BOREL SUBGROUPS 13

Example 2.2. Bn{(∗ ∗0 ∗)} is solvable because (B,B) = U. ⌟

Remark. Any subgroup of a solvable group is solvable.

Generally unipotent group is solvable. The reason is

Proposition 4 (Lie - Kelchin). (char = 0?) If G is unipotent, then for any representationρ ∶ GÐ→ Gl(V ) there’s a basis of V such that ρ(G) ⊆ Un. Similarly if G is solvable, then forany representation ρ ∶ GÐ→ GL(V ) there is a basis of V such that ρ(G) ⊆ Bn.

3. Borel subgroups

Definition 10. A Borel subgroup of G is a maximal connected solvable (closed) subgroup.

Example 3.1. B = (∗ ∗0 ∗) ⊆ GLn. ⌟

Theorem 3.1 (Humphrey, 21.3). All Borel subgroups in G are conjugate.

Proof. Uses

(1)G/B ≅ { Borel subgroups }

is a projective variety.

(2) Borel’s fixed point theorem. (An action of a solvable group in a projective varietyhas a fixed point). This uses Lie-Kelchin.

�

Corollary 2. All maximal tori in G are conjugate.

The reason is that if T is solvable then T ⊆ B. Assume G is connected and nontrivial(≠ {e}. Then the radical of G is the maixmal connected solvable normal subgroup of G,R(G). Adding the word normal makes it unique! The unipotent radical is the maximalconnected unipotent normal subgroup of G, written as Ru(G).

Remark. Ru(G) ⊆ R(G).

Example 3.2. G = GLn then R(GLn) = k∗ = {scalara matrices }.1Ð→ k∗ Ð→ Gln Ð→ PGLn Ð→ 1

where the last term is a simple group. In particular from the above remark Ru(G) = 1here. ⌟

3. BOREL SUBGROUPS 14

Example 3.3. P = {(∗2×2 ∗2×2

0 ∗2×2)} ⊆ Gl4 has

Ru(P ) = {(I2×2 ∗0 I2×2

)}.

and

R(P ) = {⎛⎜⎜⎜⎝

∗ 0 ∗ ∗0 ∗ ∗ ∗0 0 ∗ 00 0 0 ∗

⎞⎟⎟⎟⎠}.

⌟Definition 11. G is semisimple if R(G) = {e}. G is reductive if Ru(G) = {e}.Remark. ss ⇒reductive.

Example 3.4. Sln, PGLn,Sln ×Slm are ss, and GLn, T are reductive. ⌟Remark. Z(Sln)µn is not connected.

Remark. If G is semisimple, then Z(G) is finite. Otherwise, Z(G)○ is a nontrivial solvablenormal subgroup.

Remark. If G is reductive then Z(G)○ = R(G) is a torus and (G,G) is semisimple.

Example 3.5. (GLn,GLn) = Sln. ⌟

Example 3.6. But R(B) = B) ≠ k∗ = Z(B). ⌟Remark. Any G, G/R(G) is semisimple and G/Ru(G) is reductive.

Example 3.7. Gln /k∗ = PGLn and B/U = D. ⌟

Up to finite quotient, semisimple groups are calassified to be

Sln,SO2n+1, Sp2n,SO2n

and the finite ones are G2, F4,E6,E7,E8.

CHAPTER 3

Action of algebraic groups

Let ϕ ∶ G ×X Ð→ X be the action (morphism) of an algebraic group G on variety X. Anorbit G.x of x is ϕ(G × {x}) and the stabilizer is StabG(x) = {g ∈ G ∶ g.x = x} which is aclosed subgroup of G.

Example 0.8. Any linear representation GÐ→ GL(V ) gives rise to a (linear) action ofG on V and an action of G on Gr(n, d) where n = dimV and 1 ≤ d ≤ n − 1. (Check thatthis is an algebraic action). ⌟

Definition 12. Given the action of an algebraic group G on an algebraic variety X, wesay that a morphism π ∶ X Ð→ Q is a categorical quotient for this action if it satisfies theuniversal property:

(1) π is constant on orbits, i.e. π(g.x) = π(x) for all x ∈X and g ∈ G.

(2) π′ ∶X Ð→ Q′ is another morphism constant on orbits then π′ factors through π

X

π��

π′

@@@@@@@

Q! // Q′

Example 0.9. The natural (permutation) action of G = Sn on An. Here π ∶ An Ð→ An

is the categorical quotient. ⌟

Remark. If the categorical quotient exists, it is unique.

1. Relation of coarse moduli spaces and categorical quotients

Let F be a moduli functor.

Definition 13. A family X over S (i.e. an element of F(S)) has a local universal propertyif for any family Y over T and any point t ∈ T there exists a Zariski open neighborhood Uof t in T and a morphism f ∶ U Ð→ S so that Y ∣U is the pullback of X via f .

Note: f may not be unique.

15

1. RELATION OF COARSE MODULI SPACES AND CATEGORICAL QUOTIENTS 16

Example 1.1.

F(S) = {(E,T ) ∶ E vector bundle over S ,T ∶ E Ð→ E endomorphism over S }

E =Mn × V

��

T // Mn × V

wwppppppppppp

S =Mn

with T (A,v) = (A,Av) for a generic matrix T = (xij) where xij are coordinate functionson Mn. Indeed if

E′ T ′ //

��

E′

~~||||||||

S′

is another family of endomorphism sand t ∈ S can trivialize E in a neighborhood of t overU , T ′ is given by a matrix (fij) where fij ∈ k[Mn]. Here we assume S′ affine. The desiremap f ∶ U Ð→ S =Mn is given by

u↦ (fij(u)) .⌟

Remark. Hilbert’s Theorem 90 shows that a vector bundle (trivial in etale topology) isZariski trivial.

Theorem 1.1. Suppose a family X over S has a local universal property for F . Moreover,assume that an algebraic group G is acting on S so that for s, t ∈ S, Xs ≅ Xt iff s, t are inthe same G-orbits. Then

(1) Morphisms of functors F Ð→ Mor(−,N) are in a natural bijective correspondencewith morphisms of varieties S Ð→ N , constant on orbits.

(2) If M is a coarse moduli space for F then M is a categorical quotient for theG-action on S with quotient map induced by F(S) Ð→ Mor(S,M) as the imageX ↦ π ∶ S Ð→M .

(3) Conversely suppose the G-action on S admits a categorical quotient π ∶ S Ð→ Qthen (Q,π) is a coarse moduli space for F if and only if Q is an “orbit space”, i.e.G-orbits in S are precisely the fibers of π.

Example 1.2. If S = Mn and G = Gln acting on S by conjugation as in previousexample, we know that

⎛⎜⎜⎜⎝

0 1 ⋯ 0⋮

0 0 ⋯ 10 0 ⋯ 0

⎞⎟⎟⎟⎠

1. RELATION OF COARSE MODULI SPACES AND CATEGORICAL QUOTIENTS 17

does not have a closed orbit, since the 0 matrix lies in the closure. Thus cannot have anorbit space for this action. ⌟

Proof. (1) Suppose we are given F Ð→ Mor(−.N). The image of X ∈ F(S) is amorphism S Ð→ N . This is constant on orbits since any two equivalence Xs,Xt

are mapped to the same point in N . Conversely, suppose a morphism f ∶ S Ð→ Nconstant on orbits is given. We want to use it to define a morphism α ∶ T Ð→ N forany family Y over T . Define α locally on T : Cover T by opens Ui such that overeach Ui, Y ∣Ui is obtained from X via a morphism βi ∶ Ui Ð→ S (this is our localuniversal property). Then define αi ∶ Ui Ð→ N as the composition of βi ∶ Ui Ð→ Sand f ∶ S Ð→ N . Now we check that αi and αj coincide on Ui ∩Uj . This will allowus to patch them together to a single map α ∶ T Ð→ N . In fact, αi(t) = αj(t) forany t ∈ Ui ∩ Uj because Yt ≅ Xβi(t) ≅ Xβj(t) so βj(t) and βi(t) are mapped via fto the same point in N . Check that the resulting maps α ∶ T Ð→ N give rise to amorphism of functors F Ð→Mor(−,N).

(2) SinceM is a coarse moduli space for F by the condition of local universal propertyand the correspondence of isomorphism classes of fibers of X and G-orbits it isclear the π sends each G-orbit in S to a point in M , i.e. is constant on G-orbits.If f ∶ S Ð→ N is constant on orbits, then we want to show that f factors throughπ. This follows form (1) and the definition of coarse moduli space.

(3) To be a coarse moduli space, Q has to be in one-to-one correspondence with F(pt)which is G/S, the set of G-orbits. Thus if Q is a coarse moduli space then it is anorbit space. The rest follows from (1).

�

Example 1.3. Let S = An and G = Sn, acting by permutations. This is related tothe functor of semisimple endomorphisms. The categorical quotient π ∶ An Ð→ An given by(x1,⋯, xn) Ð→ (t − x1)⋯(t − xn) is the orbit space (Generally the categorical quotient forany action of a finite group on an affine variety is always an orbit space. In this case everypoint is properly stable in the sense which will defined later. ⌟

CHAPTER 4

Categorical quotient of action of algebraic groupson affine spaces

Now we want to construct the categorical quotient for a given action of an algebraic groupG on an algebraic variety X. Assume for now that X is affine.

Observation 1: Suppose π ∶X Ð→ Y is a categorical quotient and Y is affine. The universalproperty says that the image of

π∗ ∶ k[Y ]Ð→ k[X]

is k[X]G = the ring of invariant functions on X. By universal property π is dominant(Check!). So π∗ is injective. We conclude that π is induced by the inclusion k[X]G ↪k[X].

Conclusion: If π exists and Y is affine then π∗ is the inclusion of k[Y ] = k[X]G in k[X]and this should in particular be a finitely generated k-algebra. (Hilbert’s 14-th problem -Nagata and Weyl: Not true in general but if G is reductive then this is the case.)

Here is an example when k[X]G is finitely generated but π ∶ X Ð→ Y is not the categoricalquotient.

Example 0.4. X = Gln and G is a subgroup of Gln consisting of matrices of the form

g = (Aa×a 0Bb×a Cb×b

)

where n = a + b is a fixed partition of n. Then G acts on X = Gln by multiplication on theleft g.M = gM . Let π ∶X Ð→ Gr(n, a) be the mapping

M ↦ span of first a rows of M

Observe that the fibers of π are precisely the G-orbits in X. Now π is the categoricalquotient for the G-action on X: say X Ð→ N is constant on G-orbits, then on the level ofpoints we get a map ψ ∶ Gr(n, a)Ð→ N . To show that ψ is a morphism consider the standardaffine patching of Gr(n, a). We construct a section si1,⋯,ia of π on each Ui1,⋯,ia . Say for

18

1. HILBERT’S 14-TH PROBLEM 19

simplicity (i1,⋯, ia) = (1,⋯, a). Then the section is

⎛⎜⎜⎜⎝

1 0 ⋯ 0 a1,a+1 ⋯ a1,n

0 1 ⋯ 0 a2,a+1 ⋯ a2,n

⋱0 0 ⋯ 1 aa,a+1 ⋯ aa,n

⎞⎟⎟⎟⎠↦

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 ⋯ 0 a1,a+1 ⋯ a1,n

0 1 ⋯ 0 a2,a+1 ⋯ a2,n

⋱0 0 ⋯ 1 aa,a+1 ⋯ aa,n0 0 ⋯ 0 1

⋱0 0 ⋯ 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

But k[X]G = k[Gr(n, a)] = k finitely generated but there is no affine quotient. ⌟

1. Hilbert’s 14-th problem

Let X be affine, and G act on X. Is k[X]G a finitely generated k-algebra?

We’ll show that the answer is affirmative if G is reductive. The special case of this isG = Sl2 acting on An+1 = the space of binary forms of degree n. SL2 acts by coordinatechanges:

f(x, y) = a0xn +⋯ + anyn ∼ (a0,⋯, an) ∈ An+1 g.↦ f((x, y)g−1).

The question is to find k[An]Sl2 (for instance, find the fundamental invariant, i.e. thegenerators of the algebraic of invariants). The question of computing the fundamentalinvariants is still open. Known cases are:

n ring of invariants1 k[a0, a1]Sl2 = k2 k[a0, a1, a2]Sl2 = k[D] ∶D = discriminant = a2

1 − 4a0a2

3 k[a0, a1, a2, a3]Sl2 = k[∆] ∶ ∆ = discriminant ,deg∆ = 4)4 k[a0, a1, a2, a3, a4]Sl2 = k[f2, f3] ∶ deg fi = i

Sylvester 5 4 fundamental invariants of degrees 4, 8, 12, 186 5 fundamental invariants of degrees 2, 4, 6, 10, 15

1986 7 29 fundamental invariants of degrees 4 (6), 8 (3),12 (6), 14 (4), 16 (2), 18 (9), 20, 22, 26, 30

1967 8 9 invariants of degrees 2, 3, 4, 5, 6, 7, 8, 9, 102010 9 92 fundamental invariants

10 106 fundamental invariants

What matters from the geometric point of view is that the embedding k[a0,⋯, an]Sl2 ↪k[a0,⋯, an] gives rise to a morphism

a ∈_

��

An+1

π��

(f1(a),⋯, fN(a)) ∈ Q ⊆ AN

2. REDUCTIVE AND LINEARLY REDUCTIVE GROUPS 20

where f0,⋯, fN are the generators of k[a0,⋯, an]Sl2 .

Lemma 2. N ≥ dimTπ(0)(Q).

Proof. Let M be the maximal ideal of π(0) in k[a0,⋯, an]Sl2 . Suppose f1,⋯, fNgenerate k[Q] as a k-algebra. After subtracting f(π(0)) from fi we may assume eachfi ∈M . Then f1,⋯, fN generate M/M2 as a k-vector space. �

Using this lemma, V. Kac showed that N ≥ p(n − 2) where p(x) = number of partition ofx. Hardy and Ramanujan showed

p(x) ∼ e√

2πx/3

4√

3x.

The general results are:

Theorem 1.1 (Gordan, 1868). k[a1,⋯, an]Sl2 is finitely generated.

Theorem 1.2 (Hilbert, 1890). Nullestellensatz and Hilbert’s basis theorem ⇒a new non-constructive proof.

The following theorem will motivate the next section.

Theorem 1.3. If G is a linearly reductive group acting on an affine variety X. Then k[X]Gis finitely generated k-algebra.

2. Reductive and linearly reductive groups

Definition 14. G is linearly reductive if every finite dimensional linear representation ofG decomposes as a direct sum of irreducibles.

Recall,

Theorem 2.1 (Maschke’s). Finite groups are linearly reductive.

Proof outline. Suppose GÐ→ Gl(V ) is a finite dimensional representation of a finitegroup G, and W ⊂ V is an invariant subspace. It is enough to show that W has a G-invariant complement. Choose some complement to W in V (not necessarily G-invariant)and let p ∶ V Ð→W be the projection along W ′′. Now average p over G to the G-equivariantmapping

pav(v) =1∣G∣ ∑g∈G

g−1pg(v).

Check that pav ∣W = id∣W . The conclusion is the W ′ ∶= kerpav is a G-invariant complementof W . �

2. REDUCTIVE AND LINEARLY REDUCTIVE GROUPS 21

Theorem 2.2. In characteristic zero, a linear algebraic group G is linearly reductive iff itis reductive.

Theorem 2.3 (Nagata). If G is a linearly reductive group action on an affine variety X.Then k[X]G is a finitely generated k-algebra.

Notions of reductivity:

(1) Reductive: the unipotent radical is trivial.

(2) Linear reductive: every finite dimensional subrepresentation is completely re-ducible.

(3) Geometrically reductive (due Nagata): For any finite dimensional representationG Ð→ Gl(V ) and 0 ≠ v ∈ V G there is a homogeneous polynomial F ∈ k[V ]G suchthat F (v) ≠ 0 and F (0) = 0.

(4) For any G-action on an affine variety X, k[X]G is a finitely generated k-algebra.

Theorem 2.4. In characteristic zero, all of the above four notions are equivalent. Incharacteristic p, (1), (3) and (4) are equivalent.

Remark. (2) ⇒(3) is easy: span(v) has a complement W ⊂ V . Let F ∶ V Ð→ spac(v) = A1

be the projection along W . Then F is a G-invariant linear form and F (v) = v ≠ 0.

Our goal is to show how (2) implies (4).

2.1. Reynolds operators. This is a way of averaging over G. Let G Ð→ Gl(V ) be afinite dimensional representation. Write V = V G ⊕W and PV ∶ V Ð→ V G be the projectionto V G along W . Note that here W and PV are unique. In fact W is the direct sum of thenon-trivial irreducible sub-representations of V .

Lemma 3. Suppose V is a finite dimensional G-module and V1 is a G-submodule. ThenPV (x) = PV1(x) for all x ∈ V .

Proof. Write V = V1 ⊕ V2 and V1 = V G1 ⊕W1 and V2 = V G

2 ⊕W2 as above. AndV G = V G

1 ⊕ V G2 . If x ∈ V then we can write is as x = x0 + x1 + x2 ∈ V G ⊕W1 ⊕W2. So

PV (x) = x0. If x ∈ V1 then x0 ∈ V G1 and x2 = 0 and we still have PV1(x) = x0. �

A G-moduleM is called rational or locally finite dimensional if a linear action of G on vectorspace M , where any finite collection of elements m1,⋯,mN ∈M lies in a finite-dimensionalG-invariant subspace V ofM . For instance, if G acts on an affine variety X thenM = k[X]is a rational G-module.

Corollary 3. Let M be a rational G-module. Then PM is well-defined.

2. REDUCTIVE AND LINEARLY REDUCTIVE GROUPS 22

Proof. To define PM(m) for m ∈ M , place m into a finite dimensional G-invariantV ⊆M and define PM(m) ∶= PV (m). If V1 and V2 are both finite dimensional G-invariantsubspaces of M containing m. Then there is a finite dimensional G-invariant subspaceV ⊆M containing both V1 and V2 and then apply the lemma. �

Proposition 5. Suppose h ∶ M1 Ð→ M2 is a (G-equivariant) homomorphism of rationalG-modules. Then h ○ PM1 = PM2 ○ h.

Proof. By rationality we may assume that m ∈M1 whereM1 is finite dimensional andM2 = h(M1). Write M1 = MG

1 ⊕W1 ⊕ ⋯ ⊕Wr where each Wi is a non-trivial irreduciblerepresentation of G. Now we apply h. By Schur’s lemma h(Wi) ≅Wi or is (0). Thus MG

2 =h(MG

1 ). Moreover, any m ∈M1 can be written as m =m0+w1+⋯+wr ∈MG1 ⊕W1⊕⋯⊕Wr.

Then h(m) = h(m0) + h(w1) + ⋯ + h(wr) where h(m0) ∈ MG2 and h(wi) ∈ h(Wi) for all i.

ThusPM2 ○ h(m) = h(m0) = h ○ PM1(m).

�

Corollary 4. If R is a rational G-algebra and r0 ∈ RG. Then PR(r0r) = r0PR(r) for anyr ∈ R.

Proof. Use the proposition with M1 = M2 = R and h being the multiplication byr0. �

Corollary 5. Let J ⊂ RG be an ideal. Then JR ∩RG = J .

Proof. ⊇ is obvious. For ⊆ suppose x = j1r2 + ⋯jnrn ∈ RG where j1,⋯, jn ∈ J andr1,⋯, rn ∈ R. Apply PR and get

x = PR(x) = PR(j1r1) +⋯ + PR(jnrn) = j1PR(r1) +⋯ + jnPR(rn)which is in J . �

Corollary 6. Assume R is a rational G-algebra. If R is noetherian then RG is noetherian.

Proof. Let J1 ⊂ J2 ⊂ ⋯ ⊂ Jn ⊂ ⋯ be an ascending chain of ideals in RG. ThenJ1R ⊂ J2R ⊂ ⋯ ⊂ JnR ⊂ ⋯ is an ascending chain of ideals in R. Since R is noetherianJnR = Jn+1R = ⋯ for some n ≥ 1. Now JnR ∩ RG = Jn+1R ∩ RG = ⋯ and apply corollary5. �

Lemma 4. Suppose S = ⊕∞i=0Si is a graded k-algebra (satisfying SiSj ⊆ Si+j). LetM = ⊕∞i≥1Siand x1,⋯, xn be homogeneous elements of S of degree ≥ 1 (i.e. each xi lies in some Sj , j ≥ 1).Then the following are equivalent:

a) S = S0[x1,⋯, xn] i.e. S is generated by x1,⋯, xn as an S0-algebra.

b) M = ∑ni=1 Sxi, i.e. M is generated by x1,⋯, xn as an ideal.

2. REDUCTIVE AND LINEARLY REDUCTIVE GROUPS 23

c) M/M2 = ∑ni=1 S0xi, where xi = xi mod M2, i.e. M/M2 is spanned by x1,⋯, xn asan S0-module.

Proof. (a) ⇒(b): Suppose x ∈ M , and by (a) x = ∑d1,⋯,dn≥0 sd1,⋯,dnxd11 ⋯xdnn with

sd1,⋯,dn ∈ S0 for all indices. Reducing both sides mod M we see that s0,⋯,0 = 0 and (b)follows.

(b)⇒(c) is obvious: Every x ∈M can be written as x = ∑ni=1 sixi for some si ∈ S. ModuloM2

we can replace each si on the right hand side by si[0] ∈ S0 where si = si[0]+si[1]+⋯, si[k] ∈Sk. Thus in M/M2, x ∈ ∑ni=1 S0xi.

(c)⇒(a): Since M = ∑ni=1 S0xi+M2 we have M2 = ∑ni,j=1 S0xixj +M3. Arguing inductively,we see that

Md = ∑d1+⋯+dn=d

S0xd11 ⋯xdnn +Md+1 ∀d ≥ 0

and thusS = ∑

d1,⋯,dn≤dS0x

d11 ⋯xdnn +Md+1 ∀d ≥ 0.

Now say S′ = S0[x1,⋯, xn] and observe that S′ = S: clearly S′ +Md+1 = S for all d ≥ 0.But every a ∈ Si lies in S′ for any i ≥ 0. Indeed take d = i then there is b ∈ S′ such thata = b + element of M i+1. Thus a = b[i] ∈ S′ and this completes the proof. �

Proof of (2)⇒ (4) in theorem 2.4. Case 1: X = V is a vector space with linearG-action. Then R = k[X] = ⊕Ri where Ri is the set of homogeneous polynomials of degreed on V . Then S = RG = ⊕i≥0RGi is also a graded k-algebra. Indeed, if f = f0 +⋯+ fd lies inS, where fi ∈ Ri then,

f = PR(f) = PR(f0) +⋯ + PR(fd)= PR0(f0)´¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶

∈R0

+⋯ + PRd(fd)´¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶

∈Rd

.

Since R = ⊕i≥0Ri we conclude that fi = PRi(fi) ∈ S as desired.

We know that R is noetherian by Hilbert’s basis theorem. Hence by corollary 6 S isnoetherian. Thus M = ⊕i≥0RGi is a finitely generated ideal. By (b) ⇒(a) of previous lemmaS is a finitely generated k-algebra.

Case 2: In the general affine case, let V be a finite dimensional G-invariant subspace ofk[X] containing a set of algebra generators of k[X]. Let f1,⋯, fn be a basis of V . Define

F ∶X Ð→ An

x↦ (f1(x),⋯, fn(x)).

2. REDUCTIVE AND LINEARLY REDUCTIVE GROUPS 24

Note that the G-action on V makes F into a G-equivariant map. Thus there is a surjectiveG-equivariant morphism

h ∶

R

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µk[x1,⋯, xn]Ð→ k[X]

xi ↦ fi

Now k[X]G = Pk[X](k[X]) = Pk[X] ○ h(R) = h ○ PR(R) = h(RG). By Case 1, RG is finitelygenerated, and therefore so is k[X]G. �

In the situation of the above theorem, we will refer to the morphism π ∶X Ð→ Q as a quotientor quotient map. Here π is induced by inclusion k[X]G ↪ k[X] as observed above. We willshow that π is the categorial quotient.

Proposition 6.

a) π is onto.

b) If x1,⋯, xn are closed and G-invariant subvarieties of X then π(X1 ∩ ⋯ ∩Xn) =π(X1) ∩⋯ ∩ π(Xn).

c) If Y ⊂ X is a closed and G-invariant then π∣Y ∶ Y Ð→ π(Y ) is a quotient forG-action on Y .

d) If U ⊂ Q is an open subset U ≠ ∅, then k[π−1(U)]G = k[U].

Proof. We translate each part into algebraic language:

a’) R = k[X] and S = k[X]G = k[Q], point in Q corresponding to maximal idealJ ⊂ S. Want to show JR ≠ R. But JR∩S = J by lemma 4 implying the assertion.

b’) If I1,⋯, In ⊂ R are G-invariant ideals then

(n

∑m=1

Im) ∩ S =n

∑m=1

(Im ∩ S).

⊇ is clear. For the converse start with x = x1 +⋯ + xn ∈ S, xm ∈ Im. But

x = PR(x) =∑m

PR(xm) =∑m

PIm(xm)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶∈IGm=Im∩S

∈n

∑m=1

(Im ∩ S).

c’) For I ⊂ R a G-invariant ideal, (R/I)G = S/I ∩ S. But consider the G-equivarianthomomorphism R Ð→ R/I. Then

(R/I)G = PR/I(R/I) = PR/I(h(R)) = hPR(R) = h(S) = S/I ∩ S.

d ’) For any f ∈ S, we have that R[1/f]G = S[1/f]. But this is obvious since af ∈ R[ 1

f ]is G-invariant if and only if a is G-invariant.

3. STABILITY 25

�

Corollary 7. Each fiber of π contains exactly one closed G-orbit.

Proof. Suppose q ∈ Q and π−1(q) contains closedG-orbits O1 andO2. Then O1∩O2 = ∅is impossible. Because π(O1) ∩ π(O2) contains q. It remains to show that π−1(q) containsa closed G-orbit. Let x ∈ π−1(q) by part (a) of proposition 6 above. If Gx is closed, we aredone. Otherwise take x1 ∈ Gx ∖Gx. Then Gx1 ⊊ Gx by Chevalley’s theorem: Gx containsan open subset of Gx. If Gx1 is closed we are done. Otherwise, take x2 ∈ Gx1 ∖Gx1 andcontinue. Get a desceding chain of inclusions .

⋯ ⊊ Gx1 ⊊ Gx1 ⊊ Gx.By noetherian property, get a contradiciton. �

Theorem 2.5. Let G,X,π ∶ X Ð→ Q be as in the proposition. Then π is the categoricalquotient for G-action on X.

Proof. We need to show that the dotted arrow ψ is a morphism in the diagram

X

π��

ϕ

��@@@@@@@@

Qψ//___ Y

Here ψ maps q ∈ Q to ϕ(x) for an arbitrary choice of x ∈ π−1(q) having closed orbit G.x ⊆X.ψ is well-defined on points. If Y is affine it is clear that ψ is a morphism: for any f ∈ k[Y ],ϕ∗(f) is a G-invariant function on X hence in k[X]G and ψ is just dual to the inclusionϕ∗ ∶ k[Y ]↪ k[X]G ↪ k[X].In general, we shall show that ψ is a morphism locally in a neighborhood of some q ∈ Q.Let Y0 be an affine open in Y containing ψ(q) = ϕ(x). set Z =X ∖ϕ−1(Y0). By proposition6 (b), the is f ∈ k[X]G which separates Z from Gx, i.e. f ∣Z ≅ 0 and f(x) ≠ 0. replace Xby Xf , Q by Qf and Y by Y0. By proposition 6 (d), k[Xf ]G = k[Qf ]. So we have reducedto the case where Y is affine. �

3. Stability

Let G be a linear reductive algebriac group and V a linear representation of G. Untilnow we know that k[V ]G is a finitely generated k-algebra. We know that π ∶ V Ð→ Q, thecategorical quotient map separates closed G-orbits in V . But in general, it is hard to tellwhich orbits are closed. We want to study conditions for this.

Definition 15. x ∈ V is called unstable if x ∈ π−1π(0). Equivalently, h(x) = 0 for anyhomogeneous invariant polynomial h ∈ k[V ]G of degree ≥ 1. All other points x ∈ V arecalled semistable. x is called properly stable if StabG(x) is finite and G.x is closed in V .The loci of these subsets of V , are denote respectively by V unst, V ss, V s.

3. STABILITY 26

Lemma 5.

a) V unst is closed,

b) V ss is open,

c) V s is open.

Proof. (a) and (b) are obvious. For (c) let

ϕ ∶ G × V Ð→ V × V(g, x)↦ (x, g.x)

and ∆ be the diagonal. Then the fiber of ϕ over (x,x) is isomorphic to the stabilizerStabG(x). By the fiber dimension theorem,

{x ∈ V ∶ dim StabG(x) ≥ 1}is closed in V . Now V s = π−1(Q∖π(Z)). We know that π(Z) is closed by proposition 6(c).Thus Q ∖ π(Z) and consequently V s are open. �

We know that the set of points with closed orbits is not always open. An important exampleis that of G = Sln, acting on Mn by conjugation. Closed orbits are the orbits of diagonalmatrices, and this set is not open for all n ≥ 2.

3.1. A case study. G = Gm so our G-action on V is diagonalizable, i.e. in some basisof V , T ∈ Gm acts via

⎛⎜⎝

tw1 0⋱

0 twn

⎞⎟⎠

where w1,⋯,wn ∈ Z. Suppose x =⎛⎜⎝

x1

⋮xn

⎞⎟⎠is basic. Let us say that wi is a weight associated

to x if xi ≠ 0. Here StabG(x) being finite for the properly stable points is a redundantcondition if the G-action is not trivial.

Proposition 7. Let Wx = { weights associated to x } ⊂ Z. Then

(1) x is properly stable if and only if Wx contains both positive and negative integers.

(2) x is unstable if and only if either every w ∈ Wx is positive or every w ∈ Wx isnegative.

Lemma 6. Let h ∶H Ð→ G be a homomorphism of algebraic groups. Then h(H) is closed inG.

Proof. HW problem. �

3. STABILITY 27

Fixing g ∈ Gm, let ϕ ∶ Gm Ð→ V via g ↦ g.v be given. Embed V into projective spaceP(V × k) via (a1,⋯, an) ↦ (a1 ∶ ⋯ ∶ an ∶ 1). Then ϕ extends uniquely to a morphismϕ ∶ P1 Ð→ P(V × k) by the valuative criterion.

P1ϕ// P(V × k)

Gm//

?�

OO

V?�

OO

Explicitly speaking: ϕ ∶ t ↦ (tw1x1 ∶ ⋯ ∶ twnxn ∶ 1) or t ↦ (tw1−mx1 ∶ ⋯ ∶ twn−mxn ∶ t−m)where m = min{0} ∪ {wi ∶ xi ≠ 0}. Then ϕ extends to A1 = Gm ∪ {0} and similarly to theother A1 = Gm ∪ {∞}. If ϕ(0) is in V we define limtÐ→0ϕ(t) ∶= ϕ(0) otherwise we say thatthis limit does not exist. Similarly for limtÐ→∞ϕ(t).

Example 3.1. Consider the action of Gm on k5 via

t↦

⎛⎜⎜⎜⎜⎜⎝

t−2 0 0 0 00 t−2 0 0 00 0 1 0 00 0 0 t 00 0 0 0 t2

⎞⎟⎟⎟⎟⎟⎠

.

Then when lifting the action to P5 we find that

limtÐ→0

⎛⎜⎜⎜⎜⎜⎝

11111

⎞⎟⎟⎟⎟⎟⎠

= (1 ∶ 1 ∶ 0 ∶ 0 ∶ 0 ∶ 0) ∈ P5 ∖ V

so the limit does not exist for our original action. However

limtÐ→0

⎛⎜⎜⎜⎜⎜⎝

00111

⎞⎟⎟⎟⎟⎟⎠

= (0 ∶ 0 ∶ 1 ∶ 0 ∶ 0 ∶ 1) ∈ P5

so the limit is (0,0,1,0,0) ∈ k5. ⌟

3. STABILITY 28

Example 3.2. Consider t ↦⎛⎜⎝

t−1 0 00 t−1 00 0 t−1

⎞⎟⎠. Then the unstable and not-properly sta-

ble points are the same: V (x1, x2) ∪ V (x3). For t↦⎛⎜⎝

t−1 0 00 1 00 0 t

⎞⎟⎠,

V s = V ∖ V (x1) ∪ V (x3)V uns = V (x1, x2) ∪ V (x2, x3)

⌟

Proof of proposition 7. We embed V ⊂ Pn and assume all xi ≠ 0 (otherwise restrictV to a subspace). Also assume m = w1 ≤ w2 ≤ ⋯ ≤ wn = M and w1 = ⋯ = wr < wr+1 andws < ws+1 = ws+2 = ⋯ = wn. Then in Pn,

p = limtÐ→0

t.x =⎧⎪⎪⎪⎨⎪⎪⎪⎩

(x1 ∶ ⋯ ∶ xr ∶ 0 ∶ ⋯ ∶ 0) m < 0(x1 ∶ ⋯ ∶ xr ∶ 0 ∶ ⋯ ∶ 0 ∶ 1) m = 0(0 ∶ ⋯ ∶ 0 ∶ 1) m > 0

,

and similarly

q = limtÐ→∞

t.x =⎧⎪⎪⎪⎨⎪⎪⎪⎩

(0 ∶ ⋯ ∶ 0 ∶ xs+1 ∶ ⋯ ∶ xn ∶ 0) M > 0(0 ∶ ⋯ ∶ 0 ∶ xs+1 ∶ ⋯ ∶ xn ∶ 1) M = 0(0 ∶ ⋯ ∶ 0 ∶ 1) M < 0

.

Let X be the closure of Gm.x in Pn. Then X = Gm.x∪{p, q}. If p, q /∈ V then Gm.x = V ∩Xis closed in V . This happens if m < 0 and M > 0 proving sufficiency in (1). Note also thatin this case StabGm x is finite. Conversely if x is properly stable then StabGm(x) is finite.Hence (m,M) ≠ (0,0). Moreover Gmx is closed in V then p, q /∈ V and thus m < 0 andM > 0.

For (2) is x in unstable iff 0 ∈ Gmx iff p or q are (0 ∶ ⋯ ∶ 0 ∶ 1) iff m > 0 or M < 0. �

Our main tool for determining stability of points on an affine space is the Hilbert-Mumfordcriterion. Since this is a crucial theorem in geometric invariant theory we will devote aseparate chapter to the statements and proof of this theorem.

CHAPTER 5

Hilbert-Mumford Criterion

Hilbert-Mumford Criterion 0.1. Let G be a linearly reductive group acting on a vectorspace V . Then

(1) x ∈ V is properly stable if and only if x is properly stable for every 1-parametersubgroup ρ ∶ Gm ↪ G. Equivalently limtÐ→0 ρ(t).x does not exist for any ρ.

(2) x ∈ V is semistable if and only if it is semistable for any 1-parameter subgroupρ ∶ Gm ↪ G. Equivalently limtÐ→0 ρ(t).x ≠ 0 for all ρ.

� Note 0.3. For λ ∶ Gm Ð→ G if it is not injective, but λ ≠ 1 then λ factors throughλ ∶ Gm Ð→ Gm = Gm/ker(λ) λ′Ð→ G. If ker(λ) = µd then the weights of λ = d (weights of

λ′). Thus can always replace λ by λ′. Hence we can always assume that λ is injective. ⌟

Example 0.4. Let V = kn+1 be the space of binary forms of degree n,

f(x, y) = a0xn + a1x

n−1y +⋯ + anyn

with the natural action of G = Sl2 by coordinate changes. Up to conjugation, every 1-

parameter subgroup has the form t↦ (t 00 t−1). So the weights of the action on kn+1 are

t↦⎛⎜⎜⎜⎝

tn 0 ⋯ 00 tn−2 ⋯ 0

⋱0 0 ⋯ t−n

⎞⎟⎟⎟⎠.

The unstables are those with coefficient aj of xiyj being zero where; f(x, y) is a multipleof x⌈n/2⌉. Not properly stable if aj = 0 whenever j > i; f(x, y) is a multiple of x⌊n/2⌋. Forn = 4 we get M1 the coarse moduli space of curves of genus one. For n = 6 we get M2 thecoarse moduli space of curves of genus two. ⌟

Example 0.5. Let V = Mn and Gln is acting by conjugation. We know that forA ∈Mn, G.A is closed in Mn if and only if A is diagonalizable. We also know that k[V ]Gln

is generated by e1(A),⋯, en(A), the coefficients of the characteristic polynomial of A. A isunstable iff ei(A) = 0 for all i iff A has all eigenvalues zero, iff A is nilpotent. If A is properlystable then A is diagonalizable. What do we know about StabGLn(A)? All polynomials inA are in it. In particular it will be infinite. (Check that the stabilizer is also infinite for

29

5. HILBERT-MUMFORD CRITERION 30

the action of Sln as well.) So V s = ∅. In any case we know the coarse moduli space for thisaction so we do not need the Hilbert-Mumford criteria for study of the orbit space here. ⌟

Example 0.6. Let G = Sln and V = Mn ×Mn. One can think of the elements of Vas representations of the free algebra k{x, y} Ð→ Mn. The action of any one-parametersubgroup diagonalizes as

λ ∶ t↦ diag(tα1 ,⋯, tαn),∑i

αi = 0, α1 ≥ ⋯ ≥ αn

α1 = ⋯ = αr1 > αr1+1 = ⋯ = αr1+r2 > ⋯So λ(t)(aij)n×n = (tαi−αjaij), λ(t)(bij)n×n = (tαi−αjbij). If all weights associated with(A,B) are ≥ 0 then both A and B are block-trinagular, i.e. of the form

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

r1 × r1 ∗

r2 × r2

⋱

0 rs × rs

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

So (A,B) is properly stable iff they have no common invariant subspace. In fact there is a

Theorem 0.1 (Burnside). A collection S ⊂Mn(k) for algebraically closed field k of n × nmatrices, has a common invariant subspace if and only if S does not generate Mn as ak-algebra.

Let ϕ ∶ k{x, y} Ð→ Mn correspond to (A,B) via x ↦ A,y ↦ B. Then what we showed isthat ϕ is properly stable iff ϕ is irreducible. By Burnside’s theorem this is equivalent to ϕbeing surjective.

λ(t) has strictly positive weights associated to (A,B) iff (A,B) are strictly block-uppertriangular with blocks of size ri:

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0r1×r1 ∗

0r2×r2

⋱

0 0rs×rs

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

5. HILBERT-MUMFORD CRITERION 31

So (A,B) is unstable iff A and B are stricly upper trinagular in same basis. Equivalentlyevery monomial in A,B of degree > 0 is nilpotent. ⌟

Example 0.7. Let V = k10 be the space of degree three forms in three variables x, yand z. Let G = Sl3.

Lemma 7. If f(x, y, z) cuts out a smooth cubic in P2 then f(x, y, z) is properly stable.

In fact if f is smooth after a change of basis we can rewrite it in

Weierstrass form xy2 = (x − α1z)(x − α2z)(x − α3z) αi’s are distinct

or Hasse form x3 + y3 + z3 + cxyz = 0 c ≠ −1(??)We need to check that if f is not properly stable then f is singular. By the Hilbert-Mumford criterion, there is a one parameter subgroup λ ∶ Gm Ð→ Sl3 such that every weightof λ associated to f in non-negative. After we replace λ by gλg−1 for a suitable g ∈ Sl3 andf by g.f we may assume

λ(t) =⎛⎜⎝

tα 0 00 tβ 00 0 tγ

⎞⎟⎠, α + β + γ = 0, α ≥ β ≥ γ.

In particular α > 0 and γ < 0.

Case β ≥ 0: xz2, yz2 and z3 have weights α + 2γ, β + 2γ,3γ < 0. Thus they enter intof(x, y, z) with zero coefficient. Now Df(0 ∶ 0 ∶ 1) = 0, i.e. the curve f = 0 is singular at(0 ∶ 0 ∶ 1) (and passes through it).

Case β < 0: Then every nontrivial f should be a multiple of x (otherwise it will havenegative weights). Thus f(x, y, z) = xq(x, y, z) where q is of degree 2. Hence this curvef = 0 is not smooth. ⌟

Theorem 0.2 (Hilbert-Mumford criterion for tori - Richardson). Let T = Grm be a torus

acting on V = kn. If w ∈ Tv for some v,w ∈ V then there exists a one parameter subgroupλ ∶ Gm Ð→ T such that limtÐ→0 λ(t)v ∈ Tw.

Remark. The theorem as stated becomes false if T is replaced by an arbitrary linearlyreductive group. It is true if we assume that Tw is closed in V (cf. an example in HW #4).

Proof. Diagonalize the T -action on V in basis e1,⋯, en:

t↦⎛⎜⎝

χ1(t)⋱

0 χn(t)

⎞⎟⎠

where χ1,⋯, χn ∶ T Ð→ Gm are characters. Say v = (x1,⋯, xn)t. We may then assume

x1,⋯, xn ≠ 0. Moreover after rescaling e1,⋯, en we may assume v = (1,⋯,1)t.

5. HILBERT-MUMFORD CRITERION 32

Lemma 8. There is h ∈ T such that hw =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

1⋮10⋮0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⎫⎪⎪⎪⎬⎪⎪⎪⎭# = s

for some 0 ≤ s ≤ n − 1.

Proof of lemma. After permuting coordinates we may assume w′ = (a1,⋯, as,0,⋯,0)twhere a1,⋯, as ≠ 0. Now it is enough to show that there is h ∈ T such that χ1(h) =

a1,⋯, χs(h) = as. We know that w = (a1,⋯, as,0,⋯,0)t lies in the closure of Tv =

⎛⎜⎝

χ1(t)⋮

χn(t)

⎞⎟⎠

in V . Thus (a1,⋯, as)t lies in closure of

⎛⎜⎝

χ1(t)⋮

χs(t)

⎞⎟⎠in ks as t ranges over T , and this is in

Gsm. In other words (a1,⋯, as)

t is in the closure of the image of the homomorphism

T Ð→ Gsm

t↦⎛⎜⎝

χ1(t)⋮

χs(t)

⎞⎟⎠.

We know that this image is closed (by a homework problem). Thus⎛⎜⎝

a1

⋮as

⎞⎟⎠=

⎛⎜⎝

χ1(h)⋮

χs(h)

⎞⎟⎠for

some h ∈ T and thus

h−1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

a1

⋮as0⋮0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎝

1⋮10⋮0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

.

�

For T = Grm a character χ ∈X(T ) has the form td11 ⋯tdrr for d1,⋯, dr ∈ Z. We identify χ with

(d1,⋯, dr) and X(T ) with Zr.

Lemma 9. Let X(T ) be the group of characters of T and C = SpanQ≥0(χ1,⋯, χn) in X(T )⊗Q = Qr. Then C ∩ (−C) ⊂ SpanQ(χ1,⋯, χs).

Proof. Otherwise there are d1,⋯, dn, e1,⋯, en ∈ Z≥0 such that

χd11 ⋯χdnn = χ−e11 ⋯χ−enn

5. HILBERT-MUMFORD CRITERION 33

and di ≠ 0 for some i > s. Now χd1+e11 ⋯χdn+enn ≅ 1. In particular f(x1,⋯, xn) = xd1+e11 ⋯xdn+enn

is a T -invariant monomial on V . So since w ∈ Tv, f(w) = f(v). But f(v) = 1 andf(w) = 0. �

Suppose χj(t1,⋯, tr) = taj11 ⋯tajrr ,1 ≥ j ≥ n. We are looking for λ(t) = (tb1 ,⋯, tbr) ∶ Gm Ð→ T ,so that the action of λ on V is given via

t↦⎛⎜⎝

tb1a11+⋯+bra1r

⋱tb1an1+⋯+branr

⎞⎟⎠.

So by the identification X(T ) ≅ Zr, say χi ↦ pi = (ai1,⋯, air). Then λ corresponds to linearform ` ∶ (a1,⋯, ar)↦ b1a1 +⋯brar. And what we need is to pick ` such that

`(p1) = ⋯ = `(ps) = 0, `(ps+1),⋯, `(pn) > 0.

Let C = SpanQ≥0(p1,⋯, pn). We mod out Qr by SpanQ(p1,⋯, ps). Let π be the projection

Qr Ð→W = Qr/SpanQ(p1,⋯, ps).For C = SpanQ≥0(ps+1,⋯, pn) lemma above implies that C∩−C = (0) inW . We are reducedto showing that a linear functional ` ∶W Ð→ Q exists such that `(pi) > 0 for all i = s+1,⋯, n.Separation Theorem. Let X and Y be convex subsets of Rm, X compact and Y closedand X ∩ Y = ∅. Then there is a linear functional ` ∶ Rm Ð→ R and α,β ∈ R such that

`(y) ≤ α < β ≤ `(x),∀x ∈X,y ∈ Y.

In out situation, let CR = SpanR≥0(p1,⋯, pn). Then CR ∩ (−CR) = (0) by a Homework 3,problem 5. Let X be the convex hull of p1,⋯, pm,

X = {c1p1 +⋯ + cmpm ∶ c1,⋯, cm ≥ 0, c1 +⋯ + cm = 0}.From the construction of pi’s we have that

α1p1 +⋯ + αmpm = −(β1p1 +⋯ + βmpm), α1,⋯, αm, β1,⋯, βm ≥ 0

is only possible if α = ⋯ = αm = β1 = ⋯ = βm = 0. This shows that the assumption of theSeparation Theorem

X ∩ (−CR) = (0)holds, so there exist α1,⋯, αn ∈ R and α < β such that

`a = a1x1 +⋯ + anxn, satisfies `a(y) ≤ α < β < `a(x),∀y ∈ −CR, x ∈X.Since 0 ∈ −CR, 0 = `a(y) ≥ α. Thus β > 0. Now, since p1,⋯, pm ∈X we have

`a(o1),⋯, `a(pm) ≥ β > 0.

By continuity, `b(p1),⋯, `b(pm) > 0 for any b sufficiently close to a. Now choose b =(b1,⋯, bn) ∈ Qn and multiply b by a common denominator of b1,⋯, bn to get a linear form`Nb = N`b(x1,⋯, xn) such that

`Nb(pi) > 0∀i, and Nb ∈ Za.

5. HILBERT-MUMFORD CRITERION 34

�

Now we prove the Hilbert-Mumford criteria in the general form we restate here. Unfortu-nately the proof only works over the complex numbers, k = C.

Theorem 0.3. Let G Ð→ Gl(V ) be a linearly reductive group with a linear representationG Ð→ Gl(V ). For x ∈ V let Y ⊂ V be G-invariant and closed subset such that Y ∩G.x ≠ ∅.Then there is a one-parameter subgroup λ ∶ Gm Ð→ G such that limtÐ→0 λ(t).x ∈ Y .

Proof due to Richardson. By Hilbert-Mumford criteria for tori we may assumeY ∩ Sx = ∅ for every maximal torus S of G. We will show that Gx ∩ Y = in that case.

What we need is a singular value decomposition, G =MTM where M is compact and T isa maximal torus. For example for G = Gln we can put M = Un and T the group of diagonalnon-singular matrices. For all m ∈ M , (m−1Tm)x ∩ Y = ∅. Thus m−1(Tmx) ∩ Y = ∅implying

T (mx) ∩ Y = ∅.T.mx and Y are both closed and T -invariant in V . By proposition 6 there is fm ∈ k[V ]Tsuch that

fm(mx) = 1 and fm ≅ 0 on Y .Now let Um = {v ∈ V ∶ ∣fm(v)∣ ≠ 0}. ThenM.x is covered by opens {Um ∶m ∈M}. SinceMxis compact there is a finite subcover Um1 ,⋯, UmN . Now consider the continuous function

f(v) = ∣fm1(v)∣ +⋯ + ∣fmN (v)∣This is a continuous T -invariant non-negative function V Ð→ R, f ≅ 0 on Y and ∣f ∣ > 1/2 onT.Mx. Thus on TMx, we have ∣f ∣ ≥ 1/2. This shows that TMx ∩ Y = ∅ where the closureis this time taken in the analytic topology.

We will be done if we show that M.TMx is closed in V ; then

MTMx ⊂MTMx =MTMx

implyingGx ∩ Y =MTMx ∩ Y =MMTx ∩ Y =M(TMx ∩ Y ) = ∅.

This follows from the

Lemma 10. Let M Ð→ Gln(R) be a representation of a compact Lie group M . If Z ⊂ Rn isclosed then MZ is closed.

Proof. Suppose y ∈MZ. Want to show y ∈MZ. We need to show y ∈MZ. Let B bethe closed unit ball centered at Y . Then

y ∈MZ ∩B ⊂M(Z ∩MB).Since Z ∩MB is compact, M.(Z ∩MB) is compact and therefore closed. Thus

y ∈M(Z ∩MB) =M.(Z ∩MB) ⊂M.Z.

5. HILBERT-MUMFORD CRITERION 35

�

Note finally that since G.x is open in the Zariski topology the analytic closure of it is thesame as the Zariski closure. This completes the proof. �

Remark. The nature of the proof of such decomposition is analytic. This is what re-stricts us to the complex numbers. Likewise, singular value decomposition works for Slnor other proper subgroups of Gln but for other groups we may not necessarily have such adecomposition.

Remark. We will call theorem 0.3, Richardson’s theorem for further reference. The fol-lowing shows how this theorem implies the Hilbert-Mumford criteria in the complex case.Before that we state a non-trivial result that we will be using in the proof.

Theorem 0.4 (Matsushima). If G is a linearly reductive group over complex numbers,StabG(x) is reductive if and only if Gx is affine.

Proof of Hilbert-Mumford criteria in the complex case. Suppose x ∈ V isunstable, i.e. Gx

∈0. Then the existence of λ follows from Richardson’s theorem withY = {0}. The converse is obvious.

For the second part assume x is not properly stable. Then either Gx is not closed, orStabG(x) is infinite. In the former case let Y be the unique closed G-orbit in Gx and applyRichardson’s theorem to this Y . Say Gx is closed but the stabilizer is infinite. If we knowthat StabG(x) is reductive then there is a one-dimensional torus in it so we can take λ suchthat λ(Gm) ⊂ StabG(x). This non-trivial fact follows from 0.4.

Conversely, assume x is properly stable. We want to show that limtÐ→0 λ(t)x does not existfor any λ ∶ Gm Ð→ G. Otherwise, say x0 = limtÐ→0 λ(t).x. x0 ∈ Gx = Gx. Since x is properlystable, x0 is properly stable. But StabG(x0) ⊃ λ(Gm) hence StabG(x0) is infinite.

�

Remark. Mumford’s proof of Hilbert-Mumford criteria goes as follows[4]: Hilbert finds acurve C ⊂ G such that C.x ∩ Y ≠ ∅. The question is how to get a one-parameter subgroupout of this curve. Say C ⊂ G = GLn ⊂ Mn×n ⊂ Pn2

. The resolve the singularities of C isPn2

. We get Csmooth which contains a point p ??. Op(Csmooth) is a local ring, and we passto completion Op(Csmooth) ≅ k [[t]] which is a DVR and we can write every element as utα.This gives rise to a k((t))-point of G = GLn, i.e. an invertible matrix A = (aij(t)) whereaij(t) ∈ k((t)). Write A = te(bij(t)) where bij(t) ∈ k [[t]]. The Smith normal form theorem,implies that there are h1, h2 ∈ GLn(k [[t]]) such that

h1(bij(t)h2 =⎛⎜⎝

td1

⋱tdr

⎞⎟⎠

;d1 ≥ d2 ≥ ⋯ ≥ dr.

5. HILBERT-MUMFORD CRITERION 36

Then h1Ah2 =⎛⎜⎝

te1

⋱ter

⎞⎟⎠, ei = di − e. Now we use this to make a desired one-parameter

subgroup. This is what Hilbert did. Mumford extends this by substituting the Smithnormal form with analogous tools in the case of other groups.

Lemma 11. Let G be a linearly reductive group and G Ð→ Gl(V ) a linear representation.Let X ⊂ V be a closed G-invariant irreducible, subvariety and finally π ∶ X Ð→ X//G thecategorical quotient. Assume Xps ≠ ∅. Then

(1) dimX//G = dim(X) − dimG,

(2) k(X)G = ffk[X]G ∶= k(X//G).

Proof. For x ∈ Xps ⊂open X, π−1π(x) = Gx. Every irreducible component of Gx hasdimension dimG, proving first claim by fiber dimension theorem. For the second assertionthe nontrivial case is the inclusion ⊂. Choose f1,⋯, fn ∈ k(X)G as generators of the fieldextension k ⊂ k(X)G. Consider the rational map ϕ ∶X −−� An. via x↦ (f1(x),⋯, fn(x)).Let Z be the indeterminacy locus of ϕ. Then Z is closed and G-invariant subset of X.

Claim: π(Z) ⊊ X//G. Indeed, both U = X ∖ Z and Xps are G-invariant dense open in X.If x ∈ U ∩Xps then Gx and Z are both closed G-invariant in V and Z ∩Gx = ∅. As weshowed earlier, π(Z) and π(Gx) are closed in X//G and π(Z) ∩ π(Gx) = ∅. Thus there is0 ≠ f ∈ k[X]G such that f ≅ 0 on Z. Now replace X by Xf = {x ∈ X ∶ f ≠ 0} and X//G by(X//G)f : ϕ ∶ Xf Ð→ An is regular. By the universal property of the categorical quotient, ϕfactors through π∣Xf ∶Xf Ð→ (X//G)f via ψ ∶ (X//G)f Ð→ An. In other words,

f1,⋯, fn ∈ k[(X//G)f ] = k[X//G][ 1f] = k[X]G[ 1

f].

Therefore f1,⋯, fn ∈ ffk[X]G. �

Remark. We did not need finite generation of k(X)G although it is the case here.

CHAPTER 6

Categorical quotients for projective varieties

Let X ⊂ P(V ) be a projective, G-invariant subvariety where G Ð→ Gl(V ) is a linear re-ductive group. We want to construct the categorical quotient in this situation. It will beProj k[Xaff]G.Recall that given a graded k-algebra R, with R0 = k an affine open (Proj R)F ≅ϕ Spec(Rf)0

via p↦ Rf ∩(Rf)0 for all homogeneous prime ideals p ⊂ R, such that ⊕i≥1Ri /⊂ . The inversemapping is just the contraction q↦ q ∩R for any prime q ∈ Spec(Rf)0,

q ⊂ k[ gfn

∶ f ∈ Rm,m = deg(f).n, n ∈ Z≥0].

Definition 16. Let G be a linear reductive group and G Ð→ Gl(V ) a given finite dimen-sional represenation. X ⊂ P(V ) be a G-invariant closed irreducible projective variety. Let

Xss//G ∶= Q = Proj k[Xaff]G,where Xaff ⊂ V is the affine cone over X.

The inclusion k[Xaff]G ↪ k[Xaff] induces a rational map

π ∶X −−� Q.

In concrete terms, if f0,⋯, fN are k-algebra generators of k[V ]G such that fi is homogeneousof degree di then

π ∶X −−� PNd0,⋯,dNx↦ [f0(x) ∶ ⋯ ∶ fN(x)].

The unstable, semistable and properly stable loci descend from Xaff to X. In fact thefollowing are equivalent

(1) π is not defined at x ∈X(2) f0(x) = ⋯ = fN(x) = 0

(3) f(x) = 0 for all homogeneous f ∈ k[Xaff]G of degree ≥ 1.

(4) f(x) = f(0) for any f ∈ k[Xaff]G,

(5) xaff ∈Xaff being unstable, and

37

6. CATEGORICAL QUOTIENTS FOR PROJECTIVE VARIETIES 38

(6) x ∈Xunst.

Lemma 12. π is regular on Xss.

Let R = k[V ]G = ⊕∞i=0Ri where Ri = k[V ]G is the space of degree i, G-invariant polynomialson V . For f ∈ Rd(d ≥ 1), let Uf = (Proj R)f as above. This is an affine subvariety ofQ = Proj R. Now let

Xf = {x ∈X ∶ f(x) ≠ 0}be a G-invariant open affine subvariety of X.

(Xaff)ss

��

Xf

��

⊂ Xss

��

// Y

Uf ⊂ Xss//G

;;xx

xx

x

Lemma 13. Let x ∈Xf ⊂X ⊂ P(V ) and xaff ∈ (Xaff)f ⊂Xaff ⊂ V . Then

(1) x is unstable (respectively properly stably), in Xf if and only if xaff is unstable(respectively properly stable) in Xaff.

(2) The categorical quotient for the G-action on Xf and π coincide on Xssf .

Proposition 8. π ∶Xss Ð→Xss//G is the categorical quotient ofr the G-action on Xss.

Proof. Clearly π is constant on orbits. Now suppose ϕ ∶ Xss Ð→ Y is constant onorbits. We need to show that ϕ factor through

Xss

��

ϕ

##FFFFFFFFF

Xss//G //___ Y

.

Cover Xss//G by affine opens Uf as above. Over each Uf we have

Xssf

π

��

ϕ

@@@@@@@@

Ufψ//___ Y

,

where π is the categorical quotient. Thus ψf exists and is unique. By uniqueness, ψf ∶Uf Ð→ Y patch together to give

ψ ∶Xss//GÐ→ Y.

6. CATEGORICAL QUOTIENTS FOR PROJECTIVE VARIETIES 39

�

Let G be a linearly reductive group, G Ð→ Gl(V ) a linear representation and dimV = n.Given a homogeneous invariant f ∈ k[V ]G of degree d ≥ 1, set

(k[V ]f)0 = k[M

f∶ M is a monomial in x1,⋯, xn of degree d ].

Here x1,⋯, xn is a basis of V ∗. Embed

P(V )f ∶= Spec(k[V ]f)0 = {x ∈ P(V ) ∶ f(x) ≠ 0}

into AN using above generators. Here N = (n+d−1n−1

) is the number of monomials of degree din x1,⋯, xn.

Lemma 14. Suppose x ∈ P(V ), f(x) ≠ 0 and xaff ∈ V is a representative of x in V .

(1) x is semi-stable in AN if and only if xaff is semistable in V .

(2) x is properly stable in AN if and only if xaff is properly stable in V .

Proof. By Hilbert-Mumford criterion, we may assume G = Gm. Diagonalize the actionof Gm on V in a basis e1,⋯, en with associated characters t ↦ tw1 ,⋯, t ↦ twn . Withoutloss of generality assume xaff = e1 + ⋯ + em. Denote the dual basis by x1,⋯, xn. Weightsassociated to x are therefore {w1,⋯,wm} =W .

In An, x has coordinates

(yd1,⋯,dn = {1

f(xaff) dm+1 = ⋯ = dn = 00 otherwise

∶ d1,⋯, dn ≥ 0, d1 +⋯ + dn = d) .

Weights associated to x in An are d1w1 + ⋯, emwm = wd1,⋯,dm where d1,⋯, dm ≥ 0 andd1 +⋯+ dm = n. Now observe that all w1,⋯,wm > 0 if and only if all wd!,⋯,dm > 0 Similarlyfor ≥ 0, ≤ 0 and < 0. �

Remark. If G = T is a torus, the Hilbert-Mumford criterion can be restated in terms ofC(x) the covex hull of the set of weights in X(T )⊗R associated to x, χ1,⋯, χm. In fact xis unstable if and only if 0 /∈ C(X) and is properly stable if and only if 0 is in the interiorof C(x).

In the above proof T = Gm,X(T ) = Z.

W ′ =W +⋯ +W =W +d

is the set of characters associated to x ∈ An and the cover hull of W ′ will be

d(convex hyll of W ).Thus 0 ∈ C(x) if and only if 0 ∈ C(xaff) and 0 is in the interior of C(x) if and only if 0 isin the interior of C(xaff).

6. CATEGORICAL QUOTIENTS FOR PROJECTIVE VARIETIES 40

Theorem 0.5. Let G Ð→ Gl(V ) be given as above. X ⊂ P(V ) be a closed G-invariantirreducible subset and set

Xss//G ∶= Proj k[Xaff]G.Then

(1) π ∶X −−� Xss//G is regular on Xss.

(2) Xss//G is the categorical quotient for the G-action on Xss.

(3) If x ∈Xps then π(x) = π(y) precisely when y ∈ Gx.(4) If Xps ≠ ∅ then

dimXss//G = dimX − dimG.

(5) Xss//G is a projective variety.

For the proof we need the following

Lemma 15. Suppose x, y ∈ V ss. Then there is a homogenous h ∈ k[V ]G of degree ≥ 1 suchthat h(x) ≠ 0 and h(y) ≠ 0.

Proof. Case 1. Gx ∩Gy ≠ ∅. Say z ∈ Gx ∩Gy. Thenf(x) = f(y) = f(z),∀f ∈ k[V ]G.

Since x ∈ V ss there is a homogeneous h ∈ k[V ]G of degree ≥ 1, such that h(x) ≠ 0 and weare done.

Case 2. Gx, Gy and {0} are disjoint closed G-invariant subsets of V . Hence

π(Gx) = π(x), π(Gy) = π(y), and π(0)are distinct. Here π ∶ V Ð→ V //G is the categorical quotient. In other words, there isf ∈ k[V ]G such that f(0) = 0 but f(x) ≠ 0 and f(y) ≠ 0. Write f = f1 +⋯ + fn, where fi’sare homogenous of degree i. Clearly each fi ∈ k[V ]G. Since f(x) ≠ 0, fi(x) ≠ 0 for some i.Similarly fj(y) ≠ 0 for some j. Now consider three possibilities: if fi(y) ≠ 0 take h = fi, iffj(x) ≠ 0 then take h = fj . If fi(y) = 0 and fj(x) = 0 take h = f ji + f ij . �

Proof. (0), (1) and (4) were proved last time. For (2) say x, y ∈Xf for some homoge-neous f ∈ k[V ]G. Indeed let

Z1 = Gxaff ∪Gyaff, Z2 = {0}.Since xaff, yaff are semistable, Z1 ∩Z2 = ∅. Thus there is f ∈ k[V ]G such that

f(0) = 0, f(x) ≠ 0, f(y) ≠ 0.

We can choose f homogeneous by lemma 15. In Xf we know that π(x) = π(t) impliesy ∈ Gx becasuse x ∈Xps

f . To compute dimXss//G in (3) we pass to open subsets

Xssf

πfÐ→Xssf //G

6. CATEGORICAL QUOTIENTS FOR PROJECTIVE VARIETIES 41

since Xpsf ≠ ∅ know that

dimXss//G = dimXf //G = dimXf − dimG = dimX − dimG.

�

Remark. Look at x ∈ P(V )ss mapping to π(x) ∈ P(V )ss//G. If x ∈ P(V )ps then Gx isclosed in P(V )ss but not in P(V ). Indeed choose a one-parameter subgroup λ ∶ Gm Ð→ G.Diagonalize λ so that for

x = (x1 ∶ ⋯ ∶ xn) ∈ P(V ), λ(t).x = (tw1x1 ∶ ⋯ ∶ twnxn).We may assume x1,⋯, xm ≠ 0 and xm+1 = ⋯ = xn = 0. Since x is properly stable, wi > 0,wm < 0. Assume

w1 ≥ ⋯ ≥ wm,w1 = ⋯ = ws > ws+1.

Then limtÐ→∞ λ(t).x = (x1 ∶ ⋯ ∶ xs ∶ 0 ∶ ⋯ ∶ 0) = y ∈ P(V ). Note that

⎛⎜⎝

tw1

⋱twn

⎞⎟⎠

and

⎛⎜⎜⎜⎜⎜⎜⎜⎝

1⋱

1tws+1−w1

⋱twn−w1

⎞⎟⎟⎟⎟⎟⎟⎟⎠

act in the same way on P(V ). We have y ∈ Gx and in fact y ∈ P(V )unst, the reason is thatweights of λ associated to y are w1,⋯,ws > 0. On the other hand every point in Gx isproperly stable. Thus y ∈ Gx ∖Gx and this Gx is not closed in P(V ).

CHAPTER 7

Application - Construction of the moduli of elliptic curves

1. Review on curves

We review some of the theory of curves in algebraic geometry and fix notations. So let Xbe a curve, i.e. a smooth projective irreducible variety of dimension one and let K be itscanonical line bundle. Recall that Op(X) is a discrete valuation ring

νp ∶ Op(X)∗ Ð→ N

which assigns to f its order of vanishing at p. One can extend νp to k(X)∗ Ð→ Z. We saythat f has a zero at p or order (or multiplicity) d if νp(f) = d ≥ 0, and a pole of order d ifνp(f) = −d < 0.

Recall that divisor on X are formal finite sums D = ∑p∈X np[p]. The divisor associated tof ∈ k(X)∗ is

div(f) = ∑p∈X

νp(f)[p].

We know that deg(div(f)) = 0 where deg(∑p∈E np[p]) = ∑p∈X np.

Associated to a divisor D there is a k-vector space

L(D) ∶= {f ∈ k(X)∗ ∶ div(f) +D ≥ 0} ∪ {0}

of dimension `(D), and Riemann-Roch theorem says

`(D) − `(K −D) = degD − g + 1.

Immediate consequences to recall are the cases

(1) D = 0; since the only regular functions on X are constants, this implies the `(K) =g, the dimension of the space of regular differential forms on X.

(2) D =K implies degK = 2g − 2.

Linearly equivalence: Recall that any two meromorphic differential forms differ by a rationalfunction coefficient. In other words K = div(w) for meromorphic function ω is independentof the choice of this function up to linear equivalence.

42

1. REVIEW ON CURVES 43

Case of elliptic curves: Here g = 1, `(K) = 1,deg(K) = 0. In fact K is linearly equivalenceto an effective divisor and we can let K = 0. Then Riemann-Roch can be restated as

`(D) − `(−D) = degD + 1 − g = degD.

So

`(D) = {degD degD > 00 degD < 0.

Also note that L(D) ≠ (0) if and only if D is linearly equivalent to an effective divisor. Soif degD = 0 then

`(D) = { 1 if D = div(f), for some f0 otherwise.

In particular D + div(1/f) ≥ 0 and D + div(g) ≥ 0 for any meromorphic f, g. So

div(g) − div(1/f) ≥ 0

so div(g.f) ≥ 0 and this means f.g is a constant.

Lemma 16. Choose a point A on X. Then every divisor D of degree 0 on X can be uniquelywritten as [p] − [A] for some (unique) point p ∈X.

Proof. Let D be a divisor of degree zero on X. Then deg(D + [A]) = 1 and

`(D + [A]) = 1.

So take 0 ≠ f ∈ L(D + [A]). Still deg(D + [A] + div(f)) = 1 and if effective, hence

D + [A] + div(f) = [p]

for some p ∈X. Thus D + div(f) = [p]− [A], i.e. D is linearly equivalenct to [p]− [A]. Foruniqueness if

[P ] − [A] ∼ [Q] − [A]then [P ] + div(f) = [Q] for some f ∈ k(X)∗. Thus f ∈∶ ([P ]). Also 1 ∈ L([P ]). Since`([P ]) = 1, f is a scalar multiple of 1, i.e a constant. We conclude that [Q] = [P ]+div(f) =[P ]. �

Let P,Q ∈X, thenD = [P ] + [Q] − 2[A] = [R] − [A]

by the above lemma. (P,Q)↦ R defines an operation X ×X ⊕Ð→X which is a commutativegroup on points of X.

Theorem 1.1. The group operations are given by morphisms, i.e. X is an algebraic group,with operations as above. A is the origin for this group structure.

2. ELLIPTIC CURVES AS RAMIFIED COVERINGS OF THE PROJECTIVE LINE 44

2. Elliptic curves as ramified coverings of the projective line

If f ∶ C1 Ð→ C2 is morphism of smooth curves of degree d = [k(C1) ∶ k(C2)] then recall theHurwitz formula

2 − 2g(C1) = d(2 − 2g(C2)) − ∑p∈C1

(ep − 1)

where ep is the ramification index of f at p

ep = ordp(f∗z)for a choice of uniformizing parameter z on C2 at f(p).

Now let E be an elliptic curve A ∈ E. L(A) is spanned by 1 as a k-vector space. Choose abasis {1, f} of L(2A) ⊃ L(A) with

f ∶ E −−� A1.

If we complete f to a regular map f ∶ E Ð→ P1 we get f−1(∞) = A. If λ ∈ A1 then the divisorof poles of f − λ is 2[A] so the divisor of zeroes of f − λ has degree 2. Thus

f ∶ E Ð→ P1

is a 2 ∶ 1 covering. By Hurwitz formula the number of ramification points of f is then

2(2 − 2g(P1)) − 2 − 2g(E) = 4.

Note that f can be replaced by f ′ = αf + β for any α,β ∈ k,α ≠ 0. This corresponds tomoving the ramification points by automorphism of P1 preserving ∞.

Conversely, any 2 ∶ 1 covering ϕ ∶ E Ð→ P1 can be obtained by composing f with someautomorphism P1 Ð→ P1. Indeed by Hurwitz formula ϕ has 4 ramification points. Aftercomposing with an automorphism of P1 we may assume one of them is ∞. The abusingthe same notation ϕ for ϕ ∶ E Ð→ A1 we have

div(ϕ) + 2[A] ≥ 0

so ϕ ∈ L(2[A]) and therefore ϕ = αf + β for some α,β ∈ k,α ≠ 0. This shows that ϕ and fare related by an automorphism of P1.

Theorem 2.1. (1) Every elliptic curves E is isomorphic to a smooth cubic curve inP2.

(2) If C is a smooth cubic curve in P2, it is a PGL3-translate of a Weierstrass cubic

y2z − f(x, z) = 0

for some binary cubic form f(x, z) with three distinct roots all different from (1 ∶ 0).

(3) A smooth cubic curve in P2 has genus 1.

(4) Two smooth cubics in P2 are isomorphism as abstract C1 and C2 are isomorphicas abstract curves if and only if they are PGL3 translates.

2. ELLIPTIC CURVES AS RAMIFIED COVERINGS OF THE PROJECTIVE LINE 45

Proof. (1) Choose bases 1 in L([A]), {1, f} in L(2[A]) and {1, f, g} in L(3[A]).Then the seven functions

1, f, g, f2, fg, f3, g2

in L(6[A]) are linearly dependent:

a1 + a2f +⋯ + a7g2 = 0.

In other words the map ψ ∶ E Ð→ P2 via p↦ [1 ∶ f(p) ∶ g(p)] has its image in cubiccurve C cut out by homogenization of the above equation.

Claim 1: deg(ψ) = 1.

k(E)

deg 2

��deg d

��deg 3

��5555555555555555

k(f, g)

{{vvvvvvvvv

##HHHHHHHHH

k(f) k(g)

Since d divides 2 and 3 we conclude that d = 1.

Claim 2: C is smooth. Assume p ∈ C is a singular point. Project from p to anyP1 ⊂ P2 which does not pass through p. This gives a degree one map C Ð→ P1.Composing with ψ ∶ E Ð→ C also of degree one we obtain a degree one map (i.e.a birational isomorphism E Ð→ P1). This is impossible because g(E) = 1 andg(P1) = 0.

We conclude from the above two claims that ψ is an isomorphism between E anda smooth cubic curve C.

(2) We may assume that C passes through A = (0 ∶ 1 ∶ 0) with is an inflection point ofC, i.e. C has a ‘double tangent’ at A. In other words A is the intersection of Cwith the Hessian curve given by

det( ∂2F

∂xi∂xj) .

We can make a change of coordinates such that the tangent line at A is z = 0.Then the equation F (x, y, z) of C has

(a) no y3 term (because A ∈ C),

(b) no y2x term (sing the tangent line at A is z = 0),

(c) no yx2 term (because of the double tangency),

(d) a non-zero y2z term (since z = 0 is the tangent line).

3. THE MODULI PROBLEM AND ITS SOLUTION 46

SoF = y2z + yzf1(x, z) + f3(x, z)

where fi is a binary form of degree i. After exchanging y by y + 12f1(x, z) we can

remove the middle term and get the equation

y2z = f(x, z)as desired. Since C is smooth, f(x, z) cannot be divisible by z and f(x, z) hasdistinct roots. If (a ∶ b) is a multiple root then (a ∶ 0 ∶ b) is a singular point of C.

(3) In view of part (2) project from A = (0 ∶ 1 ∶ 0) to the line y = 0. Tangent line to Cat p ∈ C is

∂F

∂x(p)x + ∂F

∂y(p)y + ∂F

∂z(p)z = 0

This passes through A if and only if ∂F∂y (p) = 0. Here F = y2z − f(x, y). Now

observe that π ∶ C Ð→ P1 is a 2:1 covering with precisely four ramification pointsand the claim follows from Hurwitz formula.

(4) By part (2) assume that C1 and C2 are both in Weierstrass form

C1 ∶ y2z = f1(x, z)C2 ∶ y2z = f2(x, z).

Suppose ϕ1 ∶ E Ð→ C1 and ϕ2 ∶ E Ð→ C2 are isomorphisms. Then, consideringprojections to the (x ∶ z) line we may assume ϕ1(A) = ϕ2(A) = (0 ∶ 1 ∶ 0). Thenthere is g ∈ PGL2 taking roots of f1 to roots of f2. This shows that C1,C2 arePGL3-translates. Note that

ϕ∗i (x/z) ∈ L(2[A]), ϕi(y/z) ∈ L(3[A]), i ∶ 1,2.Writing

ϕ∗2(x/z) = αϕ∗1(x/z) + β, and ϕ∗2(y/z) = c1ϕ∗1(y/z) + c2ϕ∗1(x/z) + c3.We obtain an element of PGL3 taking C1 to C2.

�

3. The moduli problem and its solution

Our original moduli problem is this:

F1,1(S) = {(p ∶X Ð→ S,A ∶ S Ð→X) ∶ satisfying (1), (2) and (3) below }

(1) p ∶ E Ð→ S is a proper smooth morphism,

(2) A ∶ S Ð→ E is a section,

(3) all geometric fibers of S are elliptic curves.

3. THE MODULI PROBLEM AND ITS SOLUTION 47

We introduce another functor

F1(S) = {p ∶X Ð→ S ∶ satisfying (1), (2) and (3) below }

(1) p ∶ E Ð→ S is a proper smooth morphism,

(2) There is an etale cover {Ui} of S and a section si ∶ Ui Ð→X ∣Ui over each Ui,

(3) all geometric fibers of S are elliptic curves.

The two functors coincide over a point but they are distinct in general. Now view P9 asthe space of plane cubic curves in P2 and let U ⊂ P9 be the open subset of smooth cubiccurves.

Theorem 3.1. The family of all smooth cubic curves over U ⊂ P9 has an etale local universalproperty for F1 (i.e. the local universal property of 13 with the opens taken to be etale opens).

Proof. The only thing that has remained to be proved for this assertion is to showthat there is an etale cover of U with local sections. But this is given by intersecting withthe Hessian curve. �

We conclude thatM1 = U//SL3

is a coarse moduli space for F1 if it is a categorical quotient and an orbit space.

Note that U = (P9)ps for aciton of SL3 (or equivalently for PGL3). Because of this U//SL3

is an orbit space. We know that

dim(U//SL3) = dimU − dim SL3 = 1

and in factM1 = A1 = U//SL3 ⊂ (P9)ss/SL3 = P1.

There is a moduli space for the moduli functor F1,1 as well. Indeed,

Theorem 3.2. THe following family has a local universal property ofr F1,1 in Zariskitopology.

S = {(a, b, c) ∶ (4a3 + 27b2)c = 1}.

Proof. There is a fibration

E = {(x ∶ y ∶ z) ∶ y2z = x3 + axz2 + bz3} ⊂ P2 × S pÐ→ S

via (x ∶ y ∶ z) ↦ (x, y, z) and the section is (a, b, c) ↦ ((0 ∶ 1 ∶ 0), (a, b, c)). Moreovers = (a1, b1, c1) and t = (a2, b2, c2) have isomorphic fibers

Es ∶ y2z = x3 + a1xz2 + b1z3

and Et ∶ y2z = x3 + a2xz2 + b2z3

3. THE MODULI PROBLEM AND ITS SOLUTION 48

if and only ifa2 = t2a1, b2 = t3b1

for some t ∈ Gm. Hint: in fact if α ∶ P2 Ð→ P 2 is the isomorphism of (C1, (0 ∶ 1 ∶ 0)) and(C2, (0 ∶ 1 ∶ 0)) given by

x↦ α11x + α12y + α13z

y ↦ α21x + α22y + α23z

z ↦ α31x + α32y + α33z

then αij = 0 if i ≠ j. So α has to take the tangent line z = 0 to C1 at (0 ∶ 1 ∶ 0) to thetangent line z = 0 to C2 at (0 ∶ 0 ∶ 1). Thus α31 = α32 = 0. Let A1,A2,A3 be the point oforder 2, the for y/z ∈ L(3[A]) we have

y/z(Ai) = 0, i = 1,2,3.

Thusy/z ∈ L(3[A] − [A1] − [A3])

which is one-dimensional. Thus α preserves y/z up to constant.

Note that each fibery2z − (x3 + axz2 + bz3) = 0

is a smooth cubic curve in P2. We can project this curve to P1 from (0 ∶ 1 ∶ 0). Theprojection is given by

(x ∶ y ∶ z)↦ (x ∶ z).This is generically 2:1 with four ramification points over (1 ∶ 0) and the 3 roots of x3 +axz2 + bz3. �

This time, our general machinery supplies us with a coarse moduli space

M1,1 = S//Gm.

In fact all points of S are property stable for our Gm-action in C3. This is because (a, b, c)is unstable if and only if either

a = b = 0 or c = 0and properly stable otherwise. Thus S//Gm will be an orbit space (each fiber of quotientmap S Ð→ S//Gm is exactly one orbit of Gm).

To realize S//Gm, we work out the ring of invariants

k[S]Gm = k[a, b, c]Gm/((4a3 + 27b2)c − 1).The ring of invariants is generated as a module by monomials

{aαbβcγ ∶ 2α + 3β − 6γ = 0}.It is easy to see then that this is the polynomial ring k[a3c, b2c]. So we have

k[S]Gm = k[a3c].

3. THE MODULI PROBLEM AND ITS SOLUTION 49

Remark. This invariant

a3c = a3

4a3 + 27b2is called the j-invariant of E ∶ y2z = x3 + axz2 + bz3. And we have a quotient map

SjÐ→ A1, via (a, b, c)↦ a3c.

CHAPTER 8

Construction of Mg

We give an outline of the construction of Mg when g ≥ 2 in this chapter.

1. Hilbert polynomial

For X ⊂ Pn, cut out by homogeneous ideal I ⊂ R = k[x0,⋯, xn], lethX(m) = dim(R/I)m.

Theorem 1.1 (Hilbert). There is a polynomial pX(m) such that

hX(m) = pX(m)for all large enough m >> 0.

Remark. pX(m) carries a lot of information about X ↪ Pn. In particular the leading termof PX(m) is d

s!ms where s is the dimension of X and d is the degree deg(X).

2. The Hilbert scheme

Consider the moduli problem that assigns to each scheme S, the set F(S) of flat properfamilies

X ⊂ Pn × Sover S such that π−1(s) has Hilbert polynomial p(m) as a subvariety of Pn.Theorem 2.1 (Grothendieck). There exists a fine moduli space, Hilbn,p for F .

3. Our moduli problem

Now fix g ≥ 2. Let C be a curve of genus g (smooth, projective). Choose a canonical divisorK on C. Let D = 3K be the divisor. Then by Riemann-Roch

deg(K) = 2g − 2 > 0,deg(3K) = 6g − 6⇒ `(3K) = 5g − 5.

Let f0,⋯, fn be a basis of L(3K), where n = 5g−6. Then there is a corresponding Veroneseembedding

ϕ ∶ C Ð→ Pn

p↦ (f0(p) ∶ ⋯ ∶ fn(p).

50

3. OUR MODULI PROBLEM 51

The Hilbert polynomial of ϕ(C) is

p(m) = (6m − 1)(g − 1).So we can think of ϕ(C) as a point of the Hilbert scheme Hilbn,p, denoted by [C, f0,⋯, fn].Note that PGLn+1 naturally acts on Pn and onHilbn,p, sending [C, f0,⋯, fn] to [C, f ′0,⋯, f ′n]for another basis of L(3K).Theorem 3.1 (Mumford). There existsa linear representation V of PGLn+1 and a lin-earization

Hilbn,p ↪ P(V )of the PGLn+1-action on Hilbn,p such that the image of [C, f0,⋯, fn] is properly stable inP(V ) for any smooth C and any basis f0,⋯, fn of L(3K) as above.

The GIT-quotientHg//PGLn+1 =Mg

is the coarse moduli space for the functor of curves of genus g. Here

Hg = {[C,f0,⋯, fn]} ⊂Hilbn,p.

APPENDIX A

Toric varieties as GIT quotients

1. Review of toric varieties

We consider the case of action of G = Gm on An (which can be generalized to action ofarbitrary tori on the affine space). Say the action is given via

t(x1,⋯, xn) = (ta1x1,⋯, tanxn).The induced action on coordinate ring is given on monomials xm = xm1

1 ⋯xmnn by

t.xm = t∑aimixm.So the ring of invariant regular functions is generated by

{xm ∶∑aimi = 0}.So if we correspond the point (m1,⋯,mn) ∈ Zn≥0 to xm, then k[An]G corresponds to thelattice points

a⊥ ⊂ Zn≥0.

Example 1.1. Let G act on A3 by the weight vector a = (1,1,−2). And the ring ofinvariants is generated by

m1 = (2,0,1),m2 = (0,2,1),m3 = (1,1,1).Say u = xm1 , v = xm2 ,w = xm3 . This way it is easy to see that

k[An]G = k[u, v,w]/(uv −w2).⌟

The invariant monomials of the coordinate ring form a subset of M = Zm, which give a realcone C ⊂ Rm that is a semigroup (actually a monoid). We let k[C] denote the semigroupalgebra

k[C] = { ∑finite

cixm ∶ ci ∈ k,mi ∈ C}

Definition 17. Spec k[C] is an affine toric variety.

If k[C] is graded by Z≥0, Proj k[C] is covered by affine toric varieties and is an example ofprojective toric varieties.

52

2. GIT QUOTIENTS 53

Example 1.2. Let C be the first quadrant cone on R2. Here M = Z2. On the integrallattice C ∩M , consider the grading

deg ∶ Z2 Ð→ Z; (i, j)↦ i + 2j.

This induces a grading on the polynomial ring by

degxm = deg(m)for any monomial xm. Here is how to get the projective variety

Proj k[C]

from this. It is obviously covered by U1 = Spec(k[C][x−1])0 and U2 = Spec(k[C][y−1])0. Butcombinatorially, let e1 = (1,0) vector correspond to x and e2 = (0,1) correspond to y. LetD = ⟨C ∪ {−e1}⟩ and let E be the line segment corresponding to the degree 0 piece in thisupper half plane. Then

(k[C][x−1])0 = k[E]and we can do the same thing with the half plane of first and fourth quadrants. The gluingis along

(k[C][x−1, y−1])0 = k[u±1].⌟

2. GIT quotients

Let G act on An and L be a line bundle. Then for any linearization of the action of G onL we have

An//G = Proj R, R ∶= ⊕∞d=0Γ(X,L⊗d)G.For instance let L be the trivial line bundle on An, then the induced action of a one-parameter subgroup

t(x1,⋯, xn) = (ta1x1,⋯, tanxn)is by a choice of character tb, and is explicitly given by

t.(x1,⋯, xn, xn+1) = (ta1x1,⋯, tanxn, tbxn+1)and the induced action on Γ(X,L) is

(t.f)(p) = tbf(t−1p).

Lemma 17. On monomial sections we have

t.xm = tb−∑aimi .xm for xm ∈ Γ(X,L)t.xm = tdb−∑aimi .xm for xm ∈ Γ(X,L⊗d).

Corollary 8. xm ∈ Γ(X,L⊗d)G if and only if db = ∑aimi.

2. GIT QUOTIENTS 54

Example 2.1. Say Gm is acting on A3 by a = (1,1,−2) and b = 1 from the abovenotation. Then the plane of

a.m = kgives the monomials in the k-grade piece Rk. The grading

deg ∶ Z3 Ð→ Z, m↦m.a

gives the GIT-quotient Proj k[C] = A3//G.

⌟

APPENDIX B

Moduli of representations of a quiver

To do.

55

APPENDIX C

Comments on homework problems

Problem 1 of set 1. Probably the easiest proof is to use the incidence X = {(Q,L) ∶L ⊂ Q} having projections to the space P

n(n+1)2

−1 of all quadrics and to Gr(n, d). �

Problem 1 of set 3. H acts on G. If g1, g2 ∈ G then Hg1 ≅ Hg2 iva h ↦ h(g−11 g2).

f(H) =H.e in G. So it is enough to show that there is a closedH-orbit in G. Use noetherianinduction for this: start with x! ∈ G. If Hx1 is not closed, then choose x2 ∈ Hx1 ∖Hx1. IfH2x2 is not closed, choose x3 ∈Hx2 ∖Hx2 and iterate. The sequence

⋯ ⊊Hx3 ⊊Hx2 ⊊Hx1

has to terminate. �

Problem 4 of set 3. For the second part we have

StabSL2(f0) = {(a bc d

) ∶ ad − bc = 1, (ax + by)n = xn}

= {(ζ 0c ζ−1) ∶ ζn = 1} = Ga ⋊ µn.

We conclude that StabSL2(f0) contains no Gm. �

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Bibliography

[1] Armand Borel. Linear algebraic groups, volume 126 of Graduate Texts in Mathematics. Springer-Verlag,New York, second edition, 1991.

[2] James E. Humphreys. Linear algebraic groups. Springer-Verlag, New York, 1975. Graduate Texts inMathematics, No. 21.

[3] D. Mumford, J. Fogarty, and F. Kirwan. Geometric invariant theory, volume 34 of Ergebnisse derMathematik und ihrer Grenzgebiete (2) [Results in Mathematics and Related Areas (2)]. Springer-Verlag,Berlin, third edition, 1994.

[4] P. E. Newstead. Introduction to moduli problems and orbit spaces, volume 51 of Tata Institute of Fun-damental Research Lectures on Mathematics and Physics. Tata Institute of Fundamental Research,Bombay, 1978.

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