Geometry of Families of CurvesFall 2012, taught by Joe Harris.

Contents

1 Parameter Spaces 2

2 Construction of the Hilbert Scheme 4

3 Geometry of Hilbert Schemes 5

4 Generalizations of the Hilbert Scheme Construction 5

5 Extraneous Components 6

6 Basic Properties of Reduced Hilbert Schemes 7

6.1 Digression on Picard Groups of Surfaces and Monodromy . . . . . . . . . . . . . . . 10

6.2 Linkage Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

7 General Theory of Hilbert Schemes 11

8 The Tangent Space of a Hilbert Scheme 14

9 Mumford’s Example 15

10 Generalizing These Techniques 17

10.1 Generalizing the Parametric Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 17

10.2 The Theory of Liaison (Linkage) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

11 Deformation Theory 19

12 Hurwitz Spaces 20

12.1 Aspects About the Geometry of H0d,g . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

13 Severi Varieties 23

14 Final Remarks Regarding Hilbert Schemes 24

15 Moduli Spaces 24

1

16 Examples of Stable Reduction 27

16.1 Variants of Stable Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

17 Geometry of Singular Curves 37

17.1 The δ-invariant of a Singularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

17.2 The Dual Graph of a Nodal Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

17.3 Planar Curve Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

18 Dualizing Sheaves of Curves 45

19 Kontsevich Spaces 47

20 Diaz’s Theorem 50

21 Unirationality of the Moduli Space in Small Genus 54

22 Divisors of the (Compactification of the) Moduli Space 56

1 Parameter Spaces

We’ll be interested in studying parameter spaces describing families of objects we’re interested in.Examples include:

• {subschemes, or subvarieties, of Pn with a given dimension, degree, and Hilbert polynomial}

• {sheaves or vector bundles on a given X}

• {schemes with given numerical invariants} modulo isomorphism

etc. Our goal is to identify these spaces with parts of a variety or scheme in a natural way. As asimple example, the set of hypersurfaces of degree d in Pn can be identified with PN = PH0(OPn(d)).

But in general, we need to specify what “natural” means. In the language of varieties, we meanthis: Given a family of objects

X B × Pn

B

(1.1)

with fibers schemes Xb ⊆ Pn of specified Hilbert polynomial p, we get a map

BφX−−→ Hp = {subschemes with Hilbert polynomial p}. (1.2)

2

We want to make Hp into a variety such that φX is regular.

In the world of schemes: A family means a flat family

X B × Pn

B

π (1.3)

We want that for all flat families X with Hilbert polynomial p, we have a map φπ : B → Hpcommuting with base change. Observe that if B = SpecC, then we get a bijection between pointsof Hp and subschemes of Pn with Hilbert polynomial p. If B is arbitrary and B′ is a point, thenφπ : b 7→ [Xb].

Given p, we have a functor {schemes} → {sets} given by

B 7→ {X ⊆ B × Pn flat over B with Hilbert polynomial p}. (1.4)

We say Hp is a fine parameter space if we have an isomorphism of functors

{families over B} ↔ Hom(B,Hp). (1.5)

Theorem 1.1. There exists a fine parameter space Hp,n for subschemes of Pn with Hilbert polyno-mial p.

As an example, the set of twisted cubics is given by maps P1 ↪→ P3 defined by [f0 : f1 : f2 : f3].But the corresponding morphism isn’t injective, for we could change the fi by an automorphism ofP1. And maps from a single curve isn’t useful in high genus.

Here is an alternate approach: C = V (Q1, Q2, Q3) and each quadric has specified coefficients.C is determined by span(Q1, Q2, Q3). We associate to C the vector space H0(IC,P3(2)), which wealso denote by I(C)2 ⊆ S2, where S is the graded ring K[W,X, Y, Z]. This vector space is a pointin G(3, 10). We obtain a map of sets {twisted cubics} → G(3, 10). The problem is that the imageis not closed: we can deform twisted cubics to get a nodal plane cubic with an embedded point atthe node.

Here’s another fact about fine parameter spaces: there exists a universal family, obtained from1 : H → H. This family is a

C H × Pn

H

(1.6)

such that for every family

X B × Pn

B

π (1.7)

3

we have X = C ×φπ B.

Goal: understand the geometry of parameter spaces, and use this to answer questions aboutthe schemes in question.

Examples of questions we could answer with the help of parameter spaces are:

• Given 12 lines `1, . . . , `12 ⊆ P3, how many twisted cubics meet all 12?

• Do there exist nonconstant families of twisted cubics over a complete base?

2 Construction of the Hilbert Scheme

Idea: associate to any X ⊆ Pn having Hilbert polynomial p its ideal I(X)m ⊆ Sm, a point inG(−, Sm). But can such an m always exist? It turns out it does. In fact it is required if we want Hto exist and be of finite type.

Theorem 2.1. Given p, there exists m0 such that for every m ≥ m0 and for every X ⊆ Pn withHilbert polynomial p, dim I(X)m =

(m+nn

)− p(m), and I(X)m determines I(X) up to saturation.

(Relevant background: flat limits, Geometry of Schemes II.3; cohomology and base change, 32646.7)

For Σ = {subschemes X of Pn with Hilbert polynomial p}, we have an embedding

Σ ↪→ G

((m+ n

n

)− p(m),

(m+ n

n

)), X 7→ I(X)m ⊆ Sm. (2.1)

We need to know that the image has an algebraic structure.

Theorem 2.2. The image is closed and has the structure of a scheme H satisfying the property ofbeing a fine parameter space.

To write down equations for the image, consider the map Λ⊗Skµk−→ Sm+k. We want rank(µk) ≤(

m+nn

)− p(m+ k). For U the universal bundle, this gives a determinental variety.

(Relevant text for next part: Chapter of of Moduli of Curves, Chapter 1 of Curves in ProjectiveSpace)

Recall H = Hp,n = {subsets X ⊆ Pn with Hilbert polynomial p}. We have to show that given

X B × Pn

B

(2.2)

flat with fibers of Hilbert polynomial p, we get a map B → H with b 7→ [I(Xb)m ⊆ Sm] ∈ G.The problem is with nonreduced points; this is where cohomology and base change is used.

4

3 Geometry of Hilbert Schemes

We will first consider the punctual Hilbert schemes

Hd,n = {subschemes of dimension 0 and degree d contained in Pn}. (3.1)

H contains the subset H0 of reduced schemes, which is equal to ((Pn)d\∆)/Sd, the set of distinctd-tuples of points.

Observation: if n ≥ 3 and d >> 0, then H has other, larger, components. As an example, forn = 3, look at subschemes Γ ⊆ Pn supported at a single point, and such that mk+1

p ⊆ Ip ⊆ mkp. mk

p

has codimension(k+2

3

), while mk+1

p has codimension(k+3

3

). The IΓ are of the form I = (mk+1, V )

for V ⊆ mk/mk+1, with d = deg V (I) =(k+2

3

)+ codimV . Choose codimV ∼ 1

2

(k+2

2

)∼ 1

4k2. The

dimension of the family of such schemes is ∼ 116k

4, but d ∼ 16k

3. So 116k

4 >> 3d for d large, implyingH has a component other than H0.

On the other hand, for n = 1, H = H0 = Pd, and for n = 2, H = H0 is smooth.

Now we’ll look at Hilbert schemes of curves. As an example,

H = H3m+1,3 ⊇ {twisted cubics} ⊇ Ht. (3.2)

We will see that Ht is irreducible of dimension 12; let H0 = Ht.

Proposition 3.1. H = H0 ∪ H1, where H1 = I(C : C = C0 ∪ {p}) for C0 a plane cubic andp ∈ P3 \ C0 a point.

H1 is irreducible of dimension 15, because the C0’s form a P9-bundle over (P3)∨, and then theC’s are fibered by the p’s.

Lemma 3.2. If C ⊆ P3 is of dimension 1 and degree 3, then pa(C) ≤ 1, with equality if and onlyif C is a plane cubic.

Proof. Look at a general plane section Γ = C ∩ H. Then letting h denote the Hilbert function,hC(m) ≥ hC(m− 1) + hΓ(m), because of the exact sequence

0→ OC(m− 1)→ OC(m)→ OΓ(m)→ 0. (3.3)

If Γ lies on a line, then hΓ(m) = 1, 2, 3, 3, . . ., otherwise hΓ(m) = 1, 3, 3, 3, . . .. Therefore hC(m) ≥1, 3, 6, 9, 12, . . .︸ ︷︷ ︸

3m

, implying pa(C) ≤ 1.

As a challenge problem, take C0 = V (Z, Y 4, Y 3X,Y 3W ), a triple line with a planar embeddedpoint. Is C0 ∈ H0?

4 Generalizations of the Hilbert Scheme Construction

• If Z ⊆ Pn is a closed subscheme, we can construct Hp,Z , the parameter space for subschemesof Z with Hilbert polynomial p. Consider those X with I(X) ⊇ I(Z), so we get a sub-Grassmannian.

5

• Given X and Y , we can parameterize {f : X → Y } as

{Γ : X × Y ↪→ Pn : p1 : X → X an isomorphism}, (4.1)

a quasi-projective scheme. (We might want to restrict the behavior of the embedded graph toget finite type.)

• We can consider the Hilbert scheme of nested pairs X ⊆ Y ⊆ Pn, as a subscheme of a flagmanifold.

• {X,Y, f : X → Y } can be thought of as a pair using the above graph construction.

• Relative Hilbert scheme: Given X π−→ Y , we can consider Hp,X/Y parameterizing subschemesof X contained in a fiber of Y .

As an example, this can be used as an aid to answer the question of how many nonconstantmaps f : C → D there can be, where C,D are smooth curves of genus g, h ≥ 2. Riemann-Hurwitzbounds the degree, so we can then fit the maps in a quasi-projective variety of finite type. Thisshows existence of a uniform bound in terms of g and h.

5 Extraneous Components

Recall the situation with H = H3m+1,3 = H0 ∪ H1, where H0 is the closure of the locus of twistedcubics, while H1 is the closure of the locus of C ∪ p for C a plane cubic and p ∈ P3 \ C.

Fact. H0 ∩ H1 is the closure of the locus of nodal plane cubics with a spatial embedded point atthe node.

In general, H = Hdm−g+1,n will have many “extraneous components” (ones which don’t containthe smooth irreducible nondegenerate curves). For example, there are components with generalmember C ∪ Γ where C is a plane curve of degree d and Γ is 0-dimensional of degree

(d−1

2

)− g.

Another example of extraneous components is given by ribbons. Take ` ⊆ P3 a line. X will bea subscheme of degree 2 supported on `. For example, consider a normal vector field to a line, andthen can pick a ribbon with corresponding planar Zariski tangent spaces.

(5.1)

Then I(X) = (X2, XY, Y 2, F (Z,W )X + G(Z,W )Y ) for F,G homogeneous of degree m. Wehave g(X) = −m, so in particular, for m ≥ 2, X cannot be a limit of reduced curves.

From now on, we will usually restrict to the non-extraneous components. The restricted Hilbertscheme H0 is a union of all of the irreducible components of H whose general point corresponds to areduced, irreducible, nondegenerate curve. However, extraneous components cannot necessarily beavoided.

6

As an example, we would like to know if H0 ⊆ H3m+1,3 is smooth. We can calculate the Zariskitangent space to H at a given point [C], so that H0 \ (H0 ∩H1) has tangent space of dimension 12,so that H is smooth at p. On H0 ∩H1, though, the tangent space to H has dimension 16; it is notclear whether this locus is smooth in H0. (It turns out to be.)

6 Basic Properties of Reduced Hilbert Schemes

Let H = Hdm−g+1,n (for now, n = 3), and let H0 be the restricted Hilbert scheme. We first look atthe example of d = 3, g = 0 (twisted cubics), and illustrate various methods of working with Hilbertschemes:

• The parametric approach: consider the scheme

Φ = {(F0, F1, F2, F3) : Fi a basis for H0(OP1(3))}/scalars o↪−→ P15. (6.1)

So Φ is irreducible of dimension 15. We have a map Φ→ H0 with fibers isomorphic to PGL2,which is 3-dimensional, so H0 is irreducible of dimension 12.

Examples of curves in H0: the curves

(6.2)

all lie in H: the first two lie on smooth quadrics of type (2, 1), and the third is a limit of curvesin the second family. What about I(X2, XY, Y 2)?

• Examination of varieties containing the curves: Consider

Σ = {(Q,C) : C ⊆ Q} ⊆ U × H, (6.3)

where H is the locus of twisted cubics, and U ⊆ P9 the locus of smooth quadrics. Σ projectsto U and H.To find the fibers of Σ→ U : the twisted cubics on a fixed quadric Q are curves of type (2, 1)and (1, 2), so the fibers are open in P5 q P5. Here P5 is the projectivization of (quadratics ⊗linears). So Σ has dimension 14.

The fibers of Σ→ H have dimension 2, since each C lies on a P2 of quadrics, and the smoothones are an open subset, so H0 has dimension 12. However, we haven’t proven irreducibility,since this method shows that Σ, and therefore H, has at most 2 components.

Actually Σ is irreducible, since (2, 1) and (1, 2) are interchangeable upon varying quadrics.

• Linkage: Say S, T are two surfaces of degrees s, t, and S ∩ T = C ∪D, curves of degrees c, dand genera g, h. We’ll assume S is smooth (this isn’t necessary). Then KS ∼ (s− 4)H for Hthe hyperplane section, and

7

2g − 2 = C.C +KS .C = C2 + (s− 4)c =⇒ C2 = 2g − 2− (s− 4)c. (6.4)

We also have

C.D = C.(tH − C) = tc− C2 = tc− (s− 4)c− (2g − 2). (6.5)

Finally, the last intersection is given by

D2 = D.(tH − C) = td− tc− (s− 4)c+ 2g − 2. (6.6)

By adjunction,

2h− 2 = D2 +K.D = D2 + (s− 4)d = (d− c)(s+ t− 4) + 2g − 2. (6.7)

In other words, we obtain

h− g = (d− c) · s+ t− 4

2. (6.8)

(Also d = st− c, of course.)Now consider

H

{(Q,Q′, C, `) : Q ∩Q′ = C ∪ `} Σ

G(1, 3) Hm+1,3

(6.9)

First, G(1, 3) is irreducible of dimension 4. The fibers of Σ → G(1, 3) are open subsets ofP6×P6, where P6 is the locus of quadrics containing a fixed `, so Σ is irreducible of dimension16. Finally, the fibers of Σ→ H are open in P2 × P2, so H0 is irreducible of dimension 12.

A harder example is d = 6, g = 3. For C ⊆ P3 of degree 6 and genus 3, we have

0→ H0(IC(2))→ H0(OP3(2)︸ ︷︷ ︸10

→ H0(OC(2))︸ ︷︷ ︸12−3+1=10

(6.10)

so H0(IC(2)) may be trivial. In degree 3, though, we have 20 and 16, so C lies on a P3 of cubicsurfaces. The residual curve is a twisted cubic.

We’ll now look at some general families of examples.

8

• g = 0 and d, n arbitrary: Look at H0 ⊆ Hmd+1,n, and consider

Ψ = {(F0, . . . , Fn) : Fi ∈ H0(OP1(d)), (6.11)Fi linearly independent, no common zeros, very ample}/scalars (6.12)

o↪−→ P(H0(OP1(d))⊕(n+1)) (6.13)

= P(d+1)(n+1)−1. (6.14)

Fix Me Formatting is awful right now. (1)

We have a natural map Ψ → H0 with fibers isomorphic to PGL2 of dimension 3, so H0 isirreducible of dimension (d+ 1)(n+ 1)− 4.

• Curves on (smooth) quadrics: (Curves on a quadric cone are limits of curves on smoothquadrics.) If Q ⊆ P3 is a smooth quadric, then Q ∼= P1 × P1, with two rulings by lines L,M .

M L

(6.15)

C ⊆ Q has type (a, b) if C ∼ aL + bM . That is, C is the zero locus of a bihomogeneouspolynomial of bidegree (a, b). If C has type (a, b), then degC = a + b. To get the genus, useadjunction: KQ = −2L− 2M , as a sum of pullbacks of the KP1 ’s; alternatively, it is −2H forH the hyperplane class in P3. This gives

g(C) =C.C +KQ.C

2+ 1 =

2ab− a− b2

+ 1 = (a− 1)(b− 1). (6.16)

Now use the incidence correspondence

U ⊆ P9 open

Φ =

{(Q,C) : C ⊆

(a,b)Q

}

H0

(6.17)

The fibers of Φ→ U have dimension (a+1)(b+1)−1, and there are two connected componentsif a 6= b (the fibers are irreducible if a = b). So Φ has dimension (a+ 1)(b+ 1) + 8. If a+ b ≥ 5,then Φ→ H0 has finite fibers, so dimH0 = (a+ 1)(b+ 1) + 8.

9

But does H0 has one or two components? We want to show that the monodromy on quadricsis transitive:

Lemma 6.1. The monodromy exchanges the two rulings, so Φ, and therefore H0, is irre-ducible.

Finally, if a, b ≥ 3, then this is an entire component of the Hilbert scheme, but this is not thecase if a = 1, 2 and b ≥ 4.

6.1 Digression on Picard Groups of Surfaces and Monodromy

If S is a smooth cubic surface, then S ∼= Bl{p1,...,p6}P2.

(6.18)

So we have

PicS = Z`⊕ Ze1 ⊕ · · · ⊕ Ze6, (6.19)

with `2 = 1, e2i = −1, and all other intersection pairings 0. Now Bl(P2) ↪→ S ⊆ P3 is given by

the linear system 3`− e1 − · · · − e6, and

KS = OS(−1) = −3`+ e1 + · · ·+ e6. (6.20)

The monodromy statement is that the monodromy acts on PicS as the full symmetry group ofthe lattice PicS preserving KS . This group has order 72 · 6! = 51840.

On the other hand, a very general surface S ⊆ P3 of degree m ≥ 4 has PicS = Z. The locuswhere PicS has rank at least 2 is a countable union of subvarieties.

6.2 Linkage Example

Recall the d = 6, g = 3 example. C ⊆ P3 may not lie on a quadric, but does lie on a cubic. If C lieson a quadric, then the type is (2, 4). Otherwise, C lies on a cubic. It turns out that a general suchcubic is smooth, and the residual curve of an intersection of two cubics is a twisted cubic.

In the first case, we get a 23-dimensional family of curves. In the second case, S ∩ S′ = C ∪ Tfor T a twisted cubic, T 2 = 1, T.KS = −3, T.C = 8. Now consider the space

10

{T} (dim 12)

Σ = {(S, S′, C, T )}

H0

(6.21)

For the fibers of Σ→ {T}, we have a map

H0(OP3(3))� H0(OT (3)) = H0(OP1(9))︸ ︷︷ ︸10

, (6.22)

so the fiber dimension is 9 + 9 = 18, implying Σ is irreducible of dimension 30. To get thefibers of Σ → H0, check that C lies on exactly four dimensions of cubics, so the fiber dimension is3 + 3 = 6, showing H0 for the second case is irreducible of dimension 24.

Finally, we claim the curves in the first case are limits of those in the second case, so H0 isirreducible.

7 General Theory of Hilbert Schemes

(A reference on the theorem on cohomology and base change is: Hartshorne, III 12.9 (?), or 32646.7.

For flat families, refer to Geometry of Schemes I 3, or (in connection with Hilbert schemes) 32648.)

Problem: Given a family of sheaves parameterized by B:

X

B(7.1)

that is, a sheaf F on X which is flat over B, let Xb = π−1(b) and Fb = F|Xb .Question: Do the spaces H0(Fb) fit together to form the fibers of a vector bundle or coherent

sheaf?

This turns out to not be the case, but there is a “best possible approximation” E = π∗F . Wehave maps φb : (π∗F)b → H0(Fb). Similarly for H i(Fb), we have Riπ∗F .

Examples: for E an elliptic curve and p ∈ E, let X = E × E → E, let ∆ be the diagonal andΓ = E × {p}.

11

Γ

∆

•

p (7.2)

Let F = OE×E(∆− Γ). This captures line bundles of degree 0 on E.

As another example, consider

p(−1,0)

q(1,0)

• •

• rt(0,t)

(7.3)

Here X = A1t × P2 and Γ = A1 × {p, q} ∪ {(t, rt)}.

(7.4)

Here F = IΓ(1).

12

Basic facts:

• The hi(Fb) are upper semicontinuous.

• In the open set U ⊆ B where h0(Fb) is constant, (π∗F)|U is locally free and φb is an isomor-phism for every b ∈ U .

• Where the hi(Fb) jump, they jump in adjacent pairs.

• When hi(Fb) and hi+1(Fb) both jump, the jump is reflected in Ri+1π∗F but not Riπ∗F .

(For the first example, π∗F = 0 but R1π∗F is a skyscraper sheaf supported at p.)

Idea/mnemonic: There is a complex of locally free sheaves

0→ K0 → K1 → K2 → · · · (7.5)

whose cohomology sheaves are the Riπ∗F . The kernel sheaves don’t recognize the drop in rank,but the cokernel sheaves do.

This relates to base change by

F F ′

X X×B B′

B B′

(7.6)

and identifying π∗F ′. φb occurs for B′ = {b}.This result is used to show that Hilbert schemes as constructed give a fine parameter space. A

map B → G(k, V ) is equivalent to inclusions

E V ⊗OB

B

φ

(7.7)

where E is locally free of rank dimV − k. So given

X B × Pn

B

π(7.8)

we get IX(m) ↪→ OB×Pn(m), then take direct images π∗IX(m) → Sm ⊗OB.

13

8 The Tangent Space of a Hilbert Scheme

The basic fact we’ll use is that H = Hp,n has the universal property of being a fine parameter space:for each scheme B, there is a natural bijection

{ X ⊆ B × Pn flat over B with Hilbert polynomial p} ↔ Mor(B,H).

Apply this to B = SpecC[ε]/(ε2), which we denote by I. For any scheme Z and point p ∈ Z, wehave

TpZ = {f : I→ Z : f(SpecC) = p}. (8.1)

So for every X ⊆ Pn, we have

T[X]H = {X ⊆ I× Pn flat over I such that X ∩ (SpecC× Pn) = X}. (8.2)

We call these first-order deformations of X.

To describe first-order deformations of X ⊆ Pn, do this locally (in An ⊆ Pn). Suppose X ⊆ I×Ansuch that X ∩ (SpecC× An) = X. Then X = V (I) for I ⊆ C[x1, . . . , xn, ε]/(ε

2).

Write I = {(fα + εgα)} for fα, gα ∈ C[x1, . . . , xn]. The condition that X ∩ (SpecC × An) = Xmeans that

{f ∈ C[x1, . . . , xn] : f + εg ∈ I for some g} = I(X). (8.3)

For flatness, if a module M over C[ε]/(ε2) is flat, then it preserves exactness of

0→ (ε) ↪→ C[ε]/(ε2)→ C→ 0. (8.4)

So C[x1, . . . , xn, ε]/I is flat if and only if εg ∈ I =⇒ g ∈ I(X). So for every f ∈ I(X), thereexists g ∈ C[x1, . . . , xn] such that f + εg ∈ I, and such a g is unique (mod I(X)). This means Xflat over I gives a homomorphism I(X)→ C[x1, . . . , xn]/I(X) by f 7→ g such that f + εg ∈ I.

We can patch these to get a global version: a first order deformation of X ⊆ Pn is a mapIX → OPn/IX = OX , so is a global section of Hom(IX ,OX) = NX/Pn . N is called the normalsheaf; when X is smooth, Np = TpPn/TpX.

We conclude that T[X]H = H0(NX/Pn).

For example, if X ⊆ Z ⊆ Pn are both smooth, we have an exact sequence

0→ NX/Z → NX/Pn → NZ/Pn |X → 0. (8.5)

As an application, if a, b ≥ 3, then the locus of smooth curves of type (a, b) on smooth quadricsurfaces is open in the appropriate H. Suppose C ⊆ Q ⊆ P3 is a smooth curve of type (a, b) on asmooth quadric. We haveQ ∼= P1×P1. WriteOQ(k, `) = π∗1OP1(k)⊗π∗2OP1(`). SoOQ(C) = OQ(a, b).Now use the exact sequence

14

0 NC/Q NC/P3 NQ/P3 |C 0

OC(a, b) OC(2) = OC(2, 2)

(8.6)

We need to show

dim{curves of type (a, b) on smooth quadrics}= dimT[C]H = h0(NC/P3).

The left hand side has been computed to be (a+ 1)(b+ 1)− 1 + 9. Subclaims:

1. h1(OC(a, b)) = 0.

2. h0(OC(a, b)) = (a+ 1)(b+ 1)− 1.

3. h0(OC(2, 2)) = 9.

For claim 1, we have the exact sequence

0→ OQ → OQ(a, b)res−−→ OC(a, b)→ 0. (8.7)

By Kunneth, h1(OQ(a, b)) = 0 if a, b ≥ −1. Alternatively, use Riemann-Roch. Also h2(OQ) = 0,so h1(OC(a, b)) = 0. And since h1(OQ) = 0, the above sequence is exact over global sections, so

h0(OC(a, b)) = h0(OQ(a, b))− h0(OQ) = (a+ 1)(b+ 1)− 1, (8.8)

showing claim 2.

For claim 3, use the exact sequence

0→ OQ(2− a, 2− b)→ OQ(2, 2)→ OC(2, 2)→ 0. (8.9)

Now h1(OQ(k, `)) = 0 for k, ` ≤ −1, so

h0(OC(2, 2)) = h0(OQ(2, 2)) = 9. (8.10)

9 Mumford’s Example

Here we encounter a situation where a component of the restricted Hilbert scheme H0 is everywherenonreduced. Consider d = 14, g = 24, and write H = H0

14m−23,3.

Suppose C ⊆ P3 is smooth of degree 14 and genus 24. Consider the restriction H0(OP3(m)) →H0(OC(m)). C can’t lie on a quadric (this is easy to check). Now check higher degrees.

We have degKC = 46, while for m ≥ 4, OC(m) has degree at least 56. So in this case, thedivisor is non-special, and Riemann-Roch implies h0(OC(m)) = h0(OP3(m) − 23. But for m = 3,h0(OC(3)) equals 19 or 20. So either C lies on a cubic or C doesn’t.

15

We check the second case first. Suppose C doesn’t lie on a cubic. Then C lies on two independentquartics S, S′. Write S ∩S′ = C ∪Q. By linkage, Q has degree 2 and genus 0, so Q is a plane conic.Consider Φ = {(C,Q)} mapping to {C} and {Q}. For the fibers of Φ → {Q}, the fibers are openin P25 × P25, since the quartics containing a given conic have codimension 9, or dimension 25. Thismeans the fiber dimension is 50. {Q} has dimension 8, so Φ is irreducible of dimension 58. Φ→ {C}has fiber dimension 2, so the locus of C not lying on a cubic is irreducible of dimension 56. (Weneed to know that H0(OP3(4))→ H0(OC(4)) is surjective; that is, h1(IC(4)) = 0.)

Now consider the case of C lying on a cubic S. We’ll assume S is smooth. C obviously can’t lieon a quartic, and we can check that C can’t lie on a quintic by linkage. But C must lie on a newsextic surface. Call such s sextic T . Now S ∩ T = C ∪ C ′, where C ′ has degree 4 and genus −1. C ′

could be either a union of two disjoint conics or a union of a line and a twisted cubic. (It is alsopossible for C ′ to be nonreduced, but this won’t occur generically.) We will conclude that H hasthree components:

• H0: C not lying on a cubic.

• H1: C lying on a cubic, and residual to two conics.

• H2: C lying on a cubic, and residual to a line and a twisted cubic.

The “pathological” component is H1, but we’ll also look at H2 to see how it differs from H1.

A conic Q on S has Q ∼ H − L for H the hyperplane class. Given two conics Q,Q′, they havethe same residual line, so Q,Q′ ∼ H − L. Now on S, C ∼ 6H −Q−Q′ ∼ 4H + 2L. We then have

h0(OC(3)) = 19 + h1(OC(3)) = 19 + h0(KC(−3)) = 19 + h0(OC(2L)) (9.1)

(since KC = (C +KS)|C = OC(3H + 2L)), which is at least 20.

Consider Σ = {(C,C ′)} mapping to {C} and {C ′}. Two conics impose independent conditionson cubics, so the space of cubics containing C ′ has dimension 19− 7− 7 = 5. Meanwhile, the spaceof sextics containing C ′ is 83−13−13 = 57. {C ′} has dimension 16, so Σ is irreducible of dimension78. Σ→ {C} has fiber dimension 23, so H1 is irreducible of dimension 56. In particular, H1 is nota limit of curves in H0.

Suppose C is smooth and C ∼ 4H + 2L ⊆ S ⊆ P3. Then we have

0 NC/S NC/P3 NS/P3 |C 0

OC(4H + 2L) OC(3)

(9.2)

NowKS = OS(−1),KC = (C+KS)|C , C2 = KC+OC(1), of degree 46+14 = 60, soOC(4H+2L)is nonspecial. The above exact sequence is exact on global sections, and h0(OC(4H+2L)) = 37 andh0(OC(3)) = 20, so dimTCH = 57.

In the H2 case, we can check that h0(OC(3)) = 19.

16

10 Generalizing These Techniques

A first approximation to the dimension of H at C is the dimension of TCH, which equals h0(NC/Pr).This can be further approximated by χ(NC/Pr). The Euler characteristic can actually be calculated:N = NC/Pr is a vector bundle of rank r − 1, and we have the exact sequence

0→ TC → TPr |C → NC/Pr → 0 (10.1)

with the tangent bundles having degrees 2−2g and (r+ 1)d, respectively, so c1(N ) = (r+ 1)d+2g − 2. Riemann-Roch implies

χ(N) = (r + 1)d+ 2g − 2− (r − 1)(g − 1) = (r + 1)d− (r − 3)(g − 1). (10.2)

This quantity will be called h(d, g, r), the expected dimension of H. In fact, a deformation theoryargument can be used to show that every component of H0 has dimension at least h(d, g, r).

Remark. Castelnuovo implies g ≤ π(d, r) ∼ d2

2(r−1) . In high degree and genus and r ≥ 4, this lowerbound is useless.

We say C ⊆ Pr is rigid if its deformations in Pr all arise from automorphisms of Pr; that is,PGLr+1 acting on the component of H containing C has a dense orbit. Question: Do there existrigid curves other than rational normal curves?

10.1 Generalizing the Parametric Approach

Here is an assertion for now: there exists a variety Mg parameterizing smooth abstract curves ofgenus g up to isomorphism, and Mg is irreducible of dimension 3g − 3 for g ≥ 2.

To parameterize curves in H, start with a given curve in Mg, then specify a map to Pr by a linebundle of degree d and an (r + 1)-tuple of sections. Let

Pd,g = {(C,L) : L ∈ PicdC} →Mg = {C}. (10.3)

The fiber dimension is g, so Pd,g is irreducible of dimension 4g − 3. Also let

grd = {(C,L, V ) : V r+1 ⊆ H0(L)} → Pd,g. (10.4)

Finally, H → grd with fibers isomorphic to PGLr+1, having dimension r2 + 2r.

The problem is to find the fibers of the map grd → Pd,g. Brill-Noether theory provides a answer:

dim grd ≥ dimPd,g − (r + 1)(g − d+ r). (10.5)

This implies dimH ≥ 4g − 3 − (r + 1)(d − g + r) + r2 + 2r, which exactly equals h(d, g, r).Furthermore, Brill-Noether implies any component ofH that dominatesMg has dimension h(d, g, r).

Conjecture 10.1. This is true for any component of H whose image in Mg has codimension lessthan g − 4.

17

10.2 The Theory of Liaison (Linkage)

We say that C,D are linked if C ∪ D is a complete intersection. If C is locally Cohen-Macaulay(CM) in P3, and S, T surfaces containing C with no common component, we define the residualcurve D in S∩T as D = V (Ann(IC/IS∩T )). Liaison is the equivalence relation generated by C ∼ Dwhen C,D are linked.

Theorem 10.2 (Gaeta). C is linked to a complete intersection if and only if C is arithmeticallyCM. That is, H0(OP3(m)) → H0(OC(m)) is surjective for every m. Equivalently, H1(IC(m)) = 0for every m.

Suppose S ∩ T = C ∪D, with C Cartier on S. Then use the sequence

0→ IS/P3(m)→ IC/P3(m)→ IC/S(m)→ 0. (10.6)

We have IS/P3(m) = OP3(m− s), and h1(OP3(m− s)) = h2(OP3(m− s)) = 0, so

H1(IC/P3(m)) = H1(IC/S(m)) (10.7)

= H1(OS(m)(−C)) (10.8)

= H1(KS(m− s+ 4)(−C)) (10.9)

= H1(OS(s+ t− 4−m)(−D))∗ (10.10)

= H1(ID/P3(s+ t− 4−m))∗. (10.11)

So the property of being arithmetically CM (or not) is preserved by liaison.

More generally, define

MC =⊕m

H1(IC/P3(m)) =⊕m

H0(OC(m))/(restrictions of polynomials of degree m). (10.12)

Then MC is a finite module over the ring S = C[X,Y, Z,W ]. Then MC is a liaison invariant upto possible twisting and dualizing. MC is called the Hartshorne-Rao module associated to C.

An example of non-arithmetically CM curves: If C = L ∪ L′ is a union of two skew lines, then

H1(IC(m)) =

{C m = 0

0 otherwise, (10.13)

so MC = C0. This is linked to a rational quartic curve by the intersection of a quadric and acubic. For such a curve, MC = C1.

Theorem 10.3 (Hartshorne-Rao). M defines a bijection between finite graded modules over S andliaison classes of C ⊆ P3.

As an example, take C to be the union of three pairwise skew lines. Then (MC)0 = C2, (MC)1 =C2, and all other graded pieces are zero. The module structure corresponds to a map H0(OP3(1))→

18

Hom(C2,C2). The locus where the determinant is zero is the dual of Q ⊇ L1 ∪ L2 ∪ L3. So twotriplets of skew lines are linked if and only if they lie on the same quadric.

Recall that sextic curves of genus 3 are linked to twisted cubics in a complete intersection oftwo cubics. So H6m−2,3 is unirational, and in particular, M3 unirational. Of course, unirationalityof M3 could have also been deduced from the plane quartic characterization.

11 Deformation Theory

References for this section are the book by Hartshorne and the paper by Vistoli.

Suppose X ⊆ Pn is any scheme. A deformation of X is an étale germ of a pointed scheme (B, b)and a subscheme X ⊂ Pn, flat over B ,with X ∩ ({b} × Pn) = X. so (X′, B′, b′) and (X, B, b) areequivalent if they agree on an étale neighborhood of the marked point.

Here are some other settings for deformations:

• f : X → Pn with X fixed: a deformation of f is a germ of f : B ×X → B with f |{b}×X = X.

• For E a vector bundle on X, a deformation of E is E on B ×X with E|{b}×X ∼= E.

• For X an abstract scheme, a deformation is X→ B along with φ : Xb∼−→ X.

Our goal is to describe a versal deformation space for X; that is, a deformation X → ∆ withX → 0 such that every deformation of X is a pullback of this one: given B → B with X → b, thereexists φ : B → ∆ with b 7→ 0. If we have uniqueness for first order deformations, meaning T0∆ isthe space of first order deformations, we say that X/∆ is miniversal.

To approach this ,we first describe the space of first order deformations. ForX ⊆ Pn a subscheme,this is H0(NX/Pn). For f : X → Pn, we get H0(f∗TPn). For E, it is H1(End(E)). In particular, if Eis a line bundle, this space is H1(OX). Finally, for abstract X, if X is smooth, this space is H1(TX).

The next issue is to see which first order deformations we can extend. Given a first orderdeformation X, we want to determine whether there exists

X X X′

SpecC SpecC[ε]/(ε2) SpecC[ε]/(ε3)

?∃

(11.1)

Let Def(X) denote the space of first order deformations of X. We have an addition in Def(X)because if A2

x,y, if X1 = V (x, y2) and X2 = V (x2, y), then X1 ∪X2 = V (x2, xy, y2), then restrict tothe sum of the two tangent vectors. This can be generalized since the union of two tangent vectorscontains all tangent vectors. So Def(X) is a vector space.

Now introduce a second vector space Obs(X), the “obstruction space”. We have a map φ2 :Obs → Sym2(Def∗), the space of homogeneous quadratic polynomials on Def, such that η ∈ Defextends to second order if and only if φ2(τ)(η) = 0 for every τ ∈ Obs.

For X ⊆ Pn, we have Obs = H1(NX/Pn). For f , it is given by H1(f∗TPn). For E, it isH2(End(E)). And for X abstract, it is H2(TX).

19

We get a sequence of maps φ3 : kerφ2 → Sym3(Def∗)/(imφ2), and more generally, φk :kerφk−1 → Symk(Def∗)/(imφ2, . . . , imφk−1). The images generate a homogeneous ideal I ⊆ Sym∗(Def∗),and η ∈ Def(X) is infinitely extendable if and only if η ∈ V (I).

Theorem 11.1 (Artin Approximation). Suppose we are given Xk → SpecC[ε]/(εk) with Xk ∩SpecC[ε]/(εk−1) = Xk−1. Then for each k, there exists an actual deformation over a smooth curvethat agrees with the Xk.

In particular, if X→ ∆ is miniversal, then the tangent cone to ∆ at 0 is V (I). This implies

dim ∆ = dimV (I) ≥ dim Def − dim Obs. (11.2)

So if X ⊆ Pn is a curve, we get

dimX H ≥ h0(N )− h1(N ) = χ(N ). (11.3)

As a special case, if Obs = 0, then ∆ is smooth and dim ∆ = dim Def. The problem is that theφk are too difficult to get a handle on.

As an example, recall H = H3m+1,3 the Hilbert scheme parameterizing twisted cubics. We haveH = H0 ∪ H1, where H0 is the closure of the locus of twisted cubics and H1 is the closure of thelocus of (plane cubic plus point)’s. Along the locus of twisted cubics (and in H0 \ H1), we haveh0(N ) = 12 and h1(N ) = 0, so for C ∈ H0 \ H1, C is a smooth point. But for C ∈ H0 ∩ H1, wehave h0(NC/P3) = 16 and h1(NC/P3) = 4. We conclude that dimTCH = 16.

We want to know whether H0 is smooth. Of course H isn’t, but only because the intersectionnecessitates it. To resolve this, Piene and Schlesinger calculated φ2 for C ∈ H0 ∩H1 and found (interms of x1, . . . , x16 for Def)

imφ2 = 〈x1x2, x1x3, x1x4, x1x5〉. (11.4)

So the tangent cone to H is V (x1)∩ V (x2, x3, x4, x5) This implies the two components are bothsmooth.

Remark. It is possible for Obs to be nonzero but all φk to be zero. For example, C = S ∩ T ⊆ P3.Then KC = OC(s+ t− 4) and NC/P3 = NC/S ⊕NC/T = O(t)⊕O(s). If either of s or t is at least4, then h1(N ) 6= 0. Meanwhile, if s = t, then H0 is open in G(2, H0(OP3(s))). The dimension of theGrassmannian is 2

(s+3

3

)− 4. But h0(NC/P3) = 2h0(OC(s)) = 2(

(s+3

3

)− 2). So no obstructions exist.

12 Hurwitz Spaces

Fix d and g, and let H0d,g be the space of simply branched covers f : C → P1 of degree d (simply

branched means that the branch divisor is reduced of degree b = 2d+2g−2, so C must be smooth).

We have a map H0d,g

π−→ Pb \∆, corresponding to the distinct set of b points in P1.

Claim. This map is finite, and can give H a structure of a variety such that the map is étale.

To find the preimage of a set of branch points, we find a cover of P1 with b points removed:

20

p0

×× ×

×γ1 γb

· · ·

(12.1)

Label the points over p0 in C by p1, . . . , pb. We then get σi ∈ Sd by the monodromy along γi.Simply branched forces the σi to be transpositions. Connectedness implys the monodromy groupgenerated by the σi is transitive. We also have

∏i σi = 1. So a fiber of π is given by

{(σ1, . . . , σb) : σi transpositions, 〈σ1, . . . , σb〉 = Sd,

∏σi = 1

}/(simultaneous conjugation).

(12.2)

We find that H0d,g is smooth of dimension 2d+ 2g − 2. Observe that this equals h(d, g, 1).

We can now understand abstract curves C using H0d,g:

Pb \∆

H0d,g

Mg

π

φ

(12.3)

If d ≥ 2g + 1 (and we can improve this), then φ is surjective.

Theorem 12.1 (Clebsch). H0d,g is connected, hence irreducible. Therefore Mg is irreducible.

To prove this, we want to determine the monodromy of π.

p0

× ×

γi γi+1

× ×

p0

γi+1γi

(12.4)

So (σ1, . . . , σb) (σ1, . . . , σi−1, σi+1, σ−1i+1σiσi+1, σi+2, . . . , σb).

(H0d,g is called the “small Hurwitz space”; it isn’t compact. Other conventions may be to mod

out by Aut(P1), or to mark the branch points.)

For π : H0d,g → P1 \∆, deg π is called the Hurwitz number.

21

Claim. Given any (σ1, . . . , σb) satisfying the appropriate conditions, then afer applying the abovetransformation, we may arrive at the normal form

(1 2), (1 2), . . . , (1 2)︸ ︷︷ ︸2g+2

, (2 3), (2 3), (3 4), (3 4), . . . (d− 1 d), (d− 1 d). (12.5)

(We can get rid of high numbers by (a b), (a b) 7→ −, (a b)(a c) 7→ (a c), (b c).)

12.1 Aspects About the Geometry of H0d,g

If you try to compactify by letting the branch points come together, the curve C can be very singular- we have little control over the result. Here is a better approach: as the branch points come together,

××

×

× ×

(12.6)

the family of curves and sections looks like

•

t = 0 ∆ (12.7)

We can then blow up at the intersection point to obtain

22

E (12.8)

Compactify H0d,g to the space Hd,g of admissible covers: covers of nodal curves of arithmetic

genus 0. The branch points still never come together.

We will also consider the Maroni stratification and divisor class theory.

13 Severi Varieties

Our goal here is to parameterize the set of maps f : C → P2 with C smooth of genus g and f ofdegree d. For now, we’ll focus on the image curves. Let PN denote the set of plane curves of degreed (specifically N =

(d+2

2

)− 1), and consider Vd,g parameterizing reduced, irreducible plane curves

of degree d and geometric genus g. We have Vd,g ⊆ V d,g ⊆ PN . Inside Vd,g, there is Ud,g consistingof nodal curves. Equivalently, write Ud,δ for curves having δ nodes, where δ =

(d−1

2

)− g.

Basic facts about Severi varieties:

• Ud,g is smooth of dimension N − δ = 3d+ g − 1 = h(d, g, 2).

• Ud,g is dense in Vd,g.

• Ud,g is irreducible.

We’ll be able to prove the first assertion. First consider the case δ = 1. Let

Φ = {(C, p) : C singular at p} ⊆ PN × P2. (13.1)

Then Φ has natural maps to V = Vd,1 and to P2. We can write

Φ =

(aij , x, y) :∑

aijxiyj︸ ︷︷ ︸

F

=∑

iaijxi−1j︸ ︷︷ ︸

G

=∑

jaijxiyj−1︸ ︷︷ ︸

H

= 0

. (13.2)

Our first claim is that if C has a node at p, then C is smooth at (C, p). To do this, we need tofind an invertible 3× 3 submatrix of the Jacobian. We’ll consider the submatrix

23

∂F∂x

∂G∂x

∂H∂x

∂F∂y

∂G∂y

∂H∂y

∂F∂a00

∂G∂a00

∂H∂a00

. (13.3)

Take p = (0, 0), so that a00 = a10 = a01 = 0. The above matrix becomes

0 a20 a11

0 a11 a02

1 0 0

. (13.4)

Because C has a node at p, det ( a20 a11a11 a02 ) 6= 0. We conclude that Φ is smooth of dimension N − 1at (C, p). In addition, this lets us conclude that π : Φ → PN is an immersion at (C, p), and theimage of the tangent space is a00 = 0. Hence if C is a plane curve with a node at p and no othersingularities, then V is smooth at C with tangent hyperplane {B : p ∈ B}.

Now consider the case of arbitrary δ. Now suppose C has nodes at p1, . . . , pδ and no othersingularities. There are then δ points of Φ lying over C. We have

(d−1

2

)− δ = g(C) = h0(K

C).

If C → C = V (f), then ω = g(x, y) dx∂f/∂y is a regular differential on C if and only if g has degree

at most d − 3 and g(pα) = 0 for α = 1, . . . , δ. Now H0(KC

) is the space of polynomials of degreeat most d − 3 vanishing at p1, . . . , pδ, having dimension at least

(d−1

2

)− δ. We force equality, so

p1, . . . , pδ impose independent conditions on polynomials of degree d−3, so a fortiori on polynomialsof degree d. This implies V d,1 has normal crossings at C. Now in a neighborhood of C, V d,δ is smoothof codimension δ in PN and has tangent space {B : p1, . . . , pδ} at C.

14 Final Remarks Regarding Hilbert Schemes

Let Hd,g,r denote the Hurwitz space of simply branched covers if r = 1, the Severi variety of nodalcurves if r = 2, and the Hilbert scheme of smooth curves otherwise. The expected dimension ish(d, g, r). For r = 1, 2, this scheme is always of the expected dimension, always smooth, and alwaysirreducible.

We would like to compactify these spaces. For r ≥ 2, there is an immediate compactification,but it is ugly. For r = 1, there is a nice compactification of admissible covers. One problem is tofind a better compactification if r ≥ 2. (Even Kontsevich spaces have issues.)

A conjecture is that Pic Hd,g,r ⊗ Q = 0 for r = 1, 2. This is possibly also true for r ≥ 3 if d islarge and for components of H dominating moduli.

15 Moduli Spaces

WE would hope to have a fine moduli space Mg for smooth projective curves of genus g. That is,we want a scheme Mg and, for every B, a natural bijection between families of curve of genus gover B and morphisms from B to Mg. Natural means commuting with base change. The propertyof being a fine moduli space would imply an isomorphism of functors Sch→ Set, F ∼= Mor(−,Mg)for F : B → {families}.

24

We’ll first look at the case g = 1. Given C of genus 1 and p ∈ C, |2p| give a map C → P1

which is a double cover branched at four points. After sending three of the points to 0, 1,∞ by anautomorphism of P1, we obtain the equation

Cλ : y2 = x(x− 1)(x− λ). (15.1)

A different choice of these points to send takes Cλ to Cλ′ for

λ′ ∈{λ,

1

λ, 1− λ, 1

1− λ,λ− 1

λ,

λ

λ− 1

}. (15.2)

S3 acts on U = P1 \ {0, 1,∞}, and set j(λ) = 256 (λ2−λ+1)3

(λ2(λ−1)2, a rational function invariant under

S3, so we get a map

U A1j

U/S3

j

∼ (15.3)

Take M1 = A1j , but M1 is not a fine moduli space! We do have that M1(C) is in biejction with

isomorphism classes of genus 1 curves, and also for a family C → B of curves of genus 1, we get amap B → M1 via y2 = (x − a)(x − b)(x − c)(x − d) for a, b, c, d ∈ OB(∆) (for ∆ some étale openset), or y2 = x(x− 1)(x− λ) for λ ∈ OB(∆), so we get a map ∆→ U

j−→ A1j .

We have obtained a natural transformation φ of functors F → Mor(−,M1), which is a bijectionon B = SpecC, but it is not an ismorphism of functors.

A problem is that j is triply ramified over j = 0. So if j is the j-function of an actual family,then all zeros of j must have order divisible by 4. Also all zeros of j − 1728 have order divisible by2. Finally, there’s another (global) obstruction to a map to M1 to come from a family.

The above shows the natural transformation isn’t surjective. It’s not injective, either: choose Eof genus 1 with a marked point. Let ι be multiplication by −1. Also let B′ → B be an unramifieddouble cover and consider (B′×E)/(τι)→ B′/(τ) = B. This is a nontrivial family of genus 1 curvesover B whose associated map B →M1 is constant.

Suppose F : Sch/C→ Set is a moduli functor. Then a scheme M is a Deligne-Mumford modulispace for F if:

• there exist natural transformations φ : F → Mor(−,M)

• φSpecC is a bijection

• φ is an isomorphism up to finite covers. This last point means:

– Given f : B →M, there is a finite cover i : B′ → B such that f ◦ i ∈ φB′ .– Given C → B and D → B such that φ(C) = φ(D), then there is a finite cover B′ such

that C ×B B′ is isomorphic to D ×B B′ over B.

25

Theorem 15.1 (Deligne-Mumford, ’69). There exists a Deligne-Mumford moduli space Mg forsmooth curves of genus g.

What we would like to know about Mg:

• Basic properties: Mg is irreducible of dimension 3g − 3 (if g ≥ 2).

• Can generalize to Mg,n, a Deligne-Mumford moduli space for n-pointed curves (curves withn distant ordered marked points)

• How do we handle the failure of Mg to be a fine moduli space?

• How is Mg constructed? Furthermore, how do we compactify Mg?

A family of n-pointed curves over B is C → B with sections σ1, . . . , σn having disjoint images.

To establish basic properties of Mg, we look at the map H0d,g

π−→ Mg. If d >> g (specificallyd ≥ 2g + 1), then π is surjective.

Given C of genus g, π−1(C) is an open subset of the set of rational functions f on C of degreed. To choose such a rational function, choose D = (f)∞ (there is a d-dimensional family of choices),and then choose f ∈ L(D) (Riemann-Roch implies the dimension is d− g + 1). So

dimπ−1(C) = 2d− g + 1 =⇒ dimMg = (2d+ 2g − 2)− (2d− g + 1) = 3g − 3. (15.4)

Remark. We also have dimMg,n = 3g− 3 +n, as long as Aut(C, p1, . . . , pn) is finite. That is, g ≥ 2,or g = 1 and n ≥ 1 or g = 0 and n ≥ 3,

To deal with the fact that Mg is not a fine moduli space (and there is not a universal family),there are three things that we can do:

• restrict to M0g, the subset of automorphism-free curves, an open subset whose complement

has codimension g − 2. M0g is a fine moduli space for automorphism-free curves.

• rigidify: for example consider curves with level structure, such as

{(C, σ1, . . . , σ2g)} where σi form a symplectic basis for H1(C,Z/m).

If m ≥ 3, then (C, σ1, . . . , σ2g) has no automorphisms. (We could also look at partial levelstructure.) There exists a fine moduli space M[m]

g for such objects, and this is a finite cover ofMg.

• Use stacks: enlarge the category of schemes to make Mg a fine moduli stack. (We won’t gointo detail here.)

26

Compactification of Mg

We want a projective Mg containing Mg as an open subset, sch that Mg is a moduli space for somelarger class of objects.

Recall the valuative criterion for properness: for ∆ a disc (or Spec of a DVR), ∆∗ the punctureddisc or generic fiber, then X is proper if and only if for all maps ∆∗ → X, there exists a uniqueextension ∆→ X.

Sow we want that for every family C∗ → ∆∗ of smooth curves, then after a finite base change∆∗ → ∆∗, there is a unique extension to a family C → ∆ of curves of this enlarged class.

A connected curve C is stable if C has only nodes as singularities and Aut(C) is finite.

Theorem 15.2. There exists a Deligne-Mumford moduli space Mg for the class of stable curves ofarithmetic genus g, and Mg is projective.

To do I think I missed this lecture. Try to fill it in. (2)

16 Examples of Stable Reduction

Suppose we have C → ∆ with C smooth and C0 having a cusp. After blowing up the cusp, we obtaina tacnode. Blowing up the tacnode gives E1 and Cν0 intersecting E2 at some point, then blow upthe triple point.

C0 Cν0E

2

E2

3

E1

2

Cν0 E3

6

3

2

E2

E1

Cν0

(16.1)

27

However, the special fiber is now highly nonreduced. The next step involves base change andnormalization.

E3

6

3

2

E2

E1

Cν0

∆t

•

•

•

(16.2)

If p ∈ D ⊆ C0 with D appearing in C0 with multiplicity 1, then (x, y, t) is given by y = t.Introduce s with s2 = t, then we get (x, y, s) : y = s2. This is a smooth branched cover of C whichis branched along D.

If D appears in C0 with multiplicity 2, we have t = y2. After a base change, we get s2 = y2, andthen have two sheets intersecting transversally along D. After normalizing, we get an unbrancheddouble cover.

If the multiplicity is 3, then we get t = y3, so s2 = y3 (a family of cusps), and then normalizinggives a smooth surface branched along D.

In summary, the effect of base change of order 2 and normalization is to replace C by a doublecover of C branched along the components of C with odd multiplicity. If D has odd multiplicity inC0, then the multiplicity is the same in the new family. But if D has even multiplicity in C0, thenthe preimage is a double cover of D, so the multiplicity is halved.

This statement generalizes to a base change of order p for p a prime (t = sp).

Applying base changes gives

E3

6

3

2

E2

E1

Cν0

2

E′3

3

3 E2

E′1

Cν0

E′′1

(16.3)

28

E′3 is a double cover of E3 branched at 2 points, so is rational. An unramified double cover ofP1 is two disjoint P1’s.

E′3

3

3 E2

E′1

Cν0

E′′1

3

E′′3

E′′2

E′1

Cν0

E′′′2

E′2

E′′1

(16.4)

E′′3 is a cyclic triple cover of E′3 branched at three points, so is isomorphic to y3 = x3 − 1, anelliptic curve with j = 0. This is true because C is smooth.

The final step is to blow down E′2, E′′2 , E′1, E′′1 . We arrive at the stable reduction

Cν0

E (16.5)

We say that E is an elliptic tail.

Next consider an example where C is smooth and C0 has a tacnode.

(16.6)

To resolve this, first blow up:

29

2 E1

Cν0

(16.7)

Now E1 intersects Cν0 at the node, so blow up again:

2 E1

E2 Cν0

4 (16.8)

Now perform two base changes of order 2 and normalization:

E′1

E′′1

E′2 Cν0

2

E′1

E′′1

E′′2 Cν0

(16.9)

Finally, blow down the rational tails:

E Cν0

(16.10)

We have E′2 ∼= P1, E′′2 an elliptic curve, and E = E′′2 , called an elliptic bridge. Look at the branchpoints of E → P1: we have j(E) = 1728 if C is smooth.

To do Figure this one out. (3)

We’ll also look at a triple point.

30

(16.11)

Blowing up gives

Cν0

P1

3

(16.12)

Finally, base change and normalization gives us the stable reduction

Cν0

E

(16.13)

with j(E) = 0.

Suppose we ended up with

31

C

P1

(16.14)

a semistable curve. We can still blow down because the P1 has self-intersection number −2. Weend up with

•

C

(16.15)

with local equation xy = t2. More generally, xy = tk arises from

C

P1’s

(16.16)

We also need to consider uniqueness of stable limits.

Finally, what if the curve is nonreduced? The last example we’ll look at is a family of plane quar-tics specializing to a double conic. That is, we have a quadric Q(X,Y, Z) and a quartic F (X,Y, Z),and we’ll consider the family V (Q2 + tF ) ⊆ ∆× P2.

32

××××××××

(16.17)

The total space is not smooth; it’s singular at t = 0 and Q = F = 0 (8 points). Blowing up thetotal space gives

2 (16.18)

with 8 P1’s as tails. Now we have a smooth total space and set-theoretic normal crossings. Wecan proceed as before:

C

(16.19)

C is a double cover of V (Q) ∼= P1 branched at 8 points. So it is a hyperelliptic curve of genus 3.Blowing down, we get the curve C as the special fiber of the stable reduction. Observe that C is ahyperelliptic limit of quartic curves.

If the general fiber is singular:

33

(16.20)

then normalize the general fiberFix Me Is this picture actually correct? (4)

(16.21)

Finally, in a situation such as

k

l (16.22)

if p - k`, then we have zp = xky` after base change. We get a singularity, which we would needto resolve.

34

If C → ∆ is a 1-parameter family of nodal curves and p ∈ C0 is a node of C0, then there existlocal coordinates on C near p given by C = V (xy− tk) for some k. For k ≥ 2, this is called an Ak−1

singularity.

For k = 2:

(16.23)

Fix Me Make this look better? (It will be tough!) (5)

Blowing up gives

C0E

(16.24)

For k = 3: we have xy = t3. The tangent cone is a union of two lines. Blowing up gives

C0

(16.25)

For k ≥ 4, blowing up gives

35

C0

•

(16.26)

This has local equation xy = tk−2, so we can repeat. We’ll end up with

C0

(16.27)

with k − 1 P1’s. This curve isn’t stable.

16.1 Variants of Stable Reduction

We say C is semistable if C is nodal and every smooth rational component of C meets the rest ofC in at least two points.

Theorem 16.1. Given any family C → ∆ with general fiber smooth, then after a finite base change,there exists C′ → ∆ with all fibers semistable and C′ smooth.

Given a family

C ∆× Pn

∆

(16.28)

with general fiber smooth:

Theorem 16.2. After a finite base change, there exists

36

C′ C

∆

(16.29)

such that all fibers of C′ → ∆ are nodal.

(Also observe that C′ → C, so this extends a family of curves in Pn.)

17 Geometry of Singular Curves

17.1 The δ-invariant of a Singularity

Suppose C is reduced with an isolated singularity at p. Let Cν be the normalization of C at p. Wehave a map Cν

π−→ C, and want to compare the arithmetic genera: that is, compare pa(Cν) withpa(C). To do this, we look at the sequence

0→ OC → π∗OCν → Fp → 0. (17.1)

Fp is a skyscraper sheaf supported at p, which depends on the singularity at p.

For p a node: Fp has rank 1.

Cν

q

r

C

p

(17.2)

In the above picture, f defined near q, r descends to C if and only if f(q) = f(r).

For p a cusp, the rank is 1. We need f ′(q) = 0.

37

q p

(17.3)

For p a tacnode, the rank is 2. We need f(q) = f(r) and f ′(q) = f ′(r).

q

r

C

p

(17.4)

Now suppose p is a planar triple point. Then we need f(q) = f(r) = f(s), but also a linear rela-tion between f ′(q), f ′(r), f ′(s) because the dimension of the tangent space to C at p has dimension2. So the rank of Fp is 3. (It would be 2 for a spatial triple point.)

q

r

s

p

(17.5)

Set δ(p) to be the rank of Fp (that is, h0(Fp)). Then χ(π∗OCν ) = χ(OC) + χ(Fp). Since π∗ isfinite, it preserves cohomology, so we obtain χ(π∗OCν ) = χ(OCν ) = 1 − pa(OCν ). Also χ(OC) =1− pa(OC). We conclude that pa(C) = pa(C

ν) + δ.

Now suppose C is nodal and connected, with δ nodes. Write C =⋃ki=1Ci with Ci nodal. Then

Cν =∐Cνi .

38

(17.6)

Then pa(C) = pa(Cν)+δ. Also, χ(OCqD) = χ(OC)+χ(OD), so pa(CqD) = pa(C)+pa(D)−1.

This implies

pa(C) = pa(Cν) + δ =

(∑pa(C

νi ))− k + 1 + δ =

(∑g(Ci)

)− k + 1 + δ. (17.7)

Therefore

∑g(Ci) = pa(C) + k − 1− δ (17.8)

which is less than or equal to pa(C).

17.2 The Dual Graph of a Nodal Curve

(17.9)

Given C, we construct a weighted graph ΓC with irreducible components of C corresponding tovertices of ΓC and nodes of C corresponding to edges of ΓC . For example, the dual graph of theabove curve is

39

• •

(17.10)

Make each vertex have weight equal to the genus of each component.

We can stratify Mg by the dual graph: set

MΓg = {C ∈Mg : ΓC = Γ} ⊂Mg, (17.11)

a locally closed subset.

As an example, take g = 2. The open stratum is

2 •2

(17.12)

The strata with 1 node are

•11

(17.13)

and

40

•

•

1

1

1 1

(17.14)

Those with 2 nodes are

•

(17.15)

and

1•

•

1

(17.16)

Those with 3 nodes are

•

•

(17.17)

41

and (the surface below consists of two spheres intersecting in three points)

• •

(17.18)

The specialization of strata is below. The marked points are disconnecting nodes; these allow usto constrain the possible limits of families of stable curves. The possibilities not already obstructeddo occur:

42

•

•

•

2

1

1

1

1

(17.19)

Now we determine the dimension of the stratum of type Γ. Say ni is the number of points onCνi lying over nodes. Then the number of parameters is

43

k∑i=1

(3gi − 3− ni) (17.20)

but we have

k∑i=1

ni = 2δ (17.21)

so the number of parameters equals

3

(k∑i=1

gi

)= 3k + 2δ = 3g + 3k − 3− 3δ − 3k + 2δ = 3g − 3δ. (17.22)

We conclude that the codimension of a stratum of curve with δ nodes is δ.

As a special case, the codimension 1 strata correspond to curves with a single node:

g − 1

α g − α (17.23)

The corresponding MΓg are called ∆0 and ∆α for α = 1, . . . , bg2c.

17.3 Planar Curve Singularities

(C, p) planar means dimTpC ≤ 2, so C is locally embeddable in a smooth surface. A deformationof C is

C C0 C

0 ∆

∼

(17.24)

modulo isomorphism over étale neighborhoods of 0 ∈ ∆; deformations of (C, p) are similar, butmodulo isomorphism of étale neighborhoods of p.

Theorem 17.1 (Reference: Hartshorne, Deformation Theory). Suppose C ⊆ A2 is reduced, p =(0, 0), and C = V (f) with f(x, y) ∈ C[x, y]. Introduce the Jacobian ideal

J =

(∂f

∂x,∂f

∂y

)⊆ C[x, y](x,y) = Op,A2 . (17.25)

(J is an ideal of finite index.) Then the miniversal deformation of (C, p) is simply

44

V (f + t1f1 + · · ·+ tkfk)� ∆t1,...,tk (17.26)

where fi is a basis for Op/J .

Here are some examples:

• For a node, f(x, y) = xy so J = (x, y), with versal deformation V (xy − t). So any familyC → B with C0 nodal at p is locally V (xy − α) for α ∈ O0,0. α is either 0 or locally tk. So inany family of nodal curves C → B, the locus of nodal fibers has codimension at most 1 in B.If we have equality and B is reduced, then we have a Cartier divisor.

• For a cusp, say V (y2 − x3), C = V (y2 − x3 − ax− b) is a versal deformation space C → ∆a,b.The fibers of C/∆ are: the fiber over (0, 0) is a cusp, the general fiber is smooth, and the fiberin the zero locus of 4a3 + 27b2 is nodal.

a

b

smooth

nodal

(17.27)

We get a map ∆a,b \ {(0, 0)} →Mg, which can be resolved by 3 blow-ups.

• For a tacnode, V (y2 − x4), we have C = V (y2 − x4 − ax2 − bx− c). For the fibers of C/∆, wecan have two nodes, a cusp, a single node, or a smooth curve. The cases of two nodes and acusp are eac curves in ∆a,b,c, while the nodal case is the discriminant surface.

• For a triple point, V (y3 − x3), C = V (y3 − x3 − axy − bx− cy − d) is a family over ∆a,b,c,d.

18 Dualizing Sheaves of Curves

We will extend the notion of the canonical bundle to possibly singular curves. We want Riemann-Roch, and we want families.

Suppose C is a reduced curve, and Cν π−→ C is its normalization. If p ∈ C is a singular point,and U is a neighborhood of p, then

ωC(U) =

meromorphic differentials ϕ on π−1(U) : ∀ f ∈ OC(U), p ∈ C,∑

q∈π−1(p)

Resq(fϕ) = 0

.

(18.1)

45

As an example, suppose p ∈ C is a node. ϕ can have at most simple poles at points lying overthe node, otherwise find a function vanishing to order 1 on one branch and arbitrarily high order onthe other branch; then the sum of the residues can’t be zero. Also, ϕ must have opposite residuesat these points.

If C = (y2 − x2), then dxy is such a differential. (We have t 7→ (t, t) and t 7→ (t,−t).)

Next, suppose p ∈ C is a cusp, such as y2 = x3. Then the map is t 7→ (t2, t3). Now a differentialcan have a double pole with residue zero, but ho higher order poles. We can take dt

t2, or dx

y again.

For p ∈ C a tacnode, y2 = x4: t 7→ (t, t2) and t 7→ (t,−t2). Again, we can have double poles butno triple poles. A valid differential is (dt

t2,−dt

t2), or dx

y . Another one is x dxy .

We observe that in these three cases:

• ω is locally free of rank 1.

• degω = degKCν + 2δ = 2g(C)− 2. Here g means arithmetic genus.

• h0(ω) = h0(Cν) + δ = g(C).

These hold whenever C has planar singularities, or more generally for local complete intersectionsingularities, or even Gorenstein singularities.

An example of a non-Gorenstein singularity is a spatial triple point. Then for U near the sin-gularity, ωC(U) consists of rational differentials on π−1(U) with simple poles at three points andresidues adding to 0. This fails the first two observations.

For C ⊆ S a smooth surface, if C is a smooth curve, then ωC = KS(C)|C . This formula applieseven if C is not smooth, as long as C is a Cartier divisor.

Suppose p ∈ C is a point of multiplicity m. Let S be the blow-up of S at p, and C be the propertransform of C in S.

E

C

S

π

S

(18.2)

C ∼ π∗(C)−mE, and KS

= π∗(KS)+E. We have E2 = −1 and E ·π∗(D) = 0 for every divisorD of S. So

46

2g(C)− 2 = C.C +KS.C = C.C −m2 + C.KS +m = 2g(C)− 2−m(m− 1). (18.3)

We conclude g(C) = g(C)−(m2

). For example, δ of a tacnode with a smooth component inter-

secting the node is(

32

)+(

22

)= 4.

To determine sections of ωC , relate to ωCby blowing up.

19 Kontsevich Spaces

Suppose X is a projective variety over C, and β ∈ H2(X;Z). Then

Mg(X,β) = {f : C → X : C nodal, Aut(f) finite, f∗([C]) = β}. (19.1)

Here Aut(f) finite means that for C0 ⊆ C any smooth rational component such that f |C0 isconstant, we must have that C0 ∩ C \ C0 consists of at least three points.

Remark. Taking X to be a point recovers Mg. More generally, for any X, β = 0 gives Mg ×X.

As a first example, consider M0(P2, 2), the Kontsevich space of “conic curves”. The Hilbertscheme of conics is P5. But the double line is not a Kontsevich stable map. So if P1 ft−→ P2 is givenby ft : [Ft, Gt, Ht] of smooth conics specializing to a double line, the specialization is not 2-1 ontoa line, We can associate a pair of branch points. So we get a map M0(P2, 2) → P5 which is anismorphism over the complement of the surface S ⊆ P5 of double lines, and blows up S: the fiberover a double line is ismorphic to P2.

Think of Mg(X,β) as an alternative compactification of the open U ⊆ H(X) parameterizingsmooth curves of genus g and class β, We want to find a relationship between Mg(X,β) and Hilbertschemes.

If X = Pn and β = d · `, write Mg(Pn, d) compactifying curves of degree d and genus g in Pn.As a next example, we’ll consider M0(P3, 2), If f : C → P3 is Kontsevich stable, then f must

be finite. Either the domain is P1 or a union of two P1’s. When we get a 2:1 map to a double linein the limit, we also get two branch points on the double line as a result.

But we don’t get a regular map to the Hilbert scheme! In H2m+1(P3), every point is a subschemeC ⊆ P3 lying on a unique plane. In the Kontsevich limit, we only see a 2:1 map onto a line, anddon’t see a particular plane.

We can introduce the Chow variety C2,1 of cycles of degree 2 and dimension 1 in P3. We geta map M0(P3, 2) → C2,1 taking a map to its image (with multiplicities indicated). We also have amap H2m+1(P3)→ C2,1 similarly. In fact, they are related birationally.

47

Γ

M0(P3, 2) H2m+1(P3)

C2,1

(19.2)

If we blow up the space of double lines, we get the space Γ.

As a third example, M1(P2, 3) includes the locus of smooth plane cubics, equal to U ⊆ P9 =H3,m(P2). We have a map M1(P2, 3)→ H3,m(P2) ∼= P9. In the cases

(19.3)

the resulting map is Kontsevich stable, so the map is locally an isomorphism. In the cuspidalcase, stable reduction gives

Cν0∼= P1

E (19.4)

(for E an elliptic tail). We obtain a map from this curve to the cuspidal curve taking E to apoint. This is a triple blow-up followed by a double blow-down. In the case of a triple line, we arriveat an elliptic curve mapping to a line with four branch points. This map is complicated!

Fact. M1(P2, 3) has three irreducible components!

Consider stable maps

48

P1

E

f

•

(19.5)

collapsing E to any point. Let’s count parameters: the image curve can be any singular cubic(8 parameters), and we specify where E maps to (1 parameter), and the j-invariant of E can beanything (1 parameter). We obtain a 10-dimensional family, larger than U .

Conversely, singular curves can’t specialize to smooth curves, so we get two different components(it’s an exercise to show that they’re not both specializations of a larger one).

The third component is given by

P1

•q

•p

P1E

•f(E)

(19.6)

The image curve has 5 + 2 = 7 parameters, and we need to specify (E, p, q) (2 parameters), sowe get a 9-dimensional family which is a separate component.

We’ve just seen that for plane cubics,H ∼= P9, but M1(P2, 3) has extraneous components. On theother hand,for twisted cubics, M0(P3, 3) is irreducible, but H3m+1(P3) has extraneous components.

Here is a problem: we can’t tell when a point in either H or Mg(X,β) lies in the closure of thecurves we’re interested in. As an example (we don’t know the answer): what stable maps f : C → P2

of degree d and genus g lie in the closure of the Severi variety?

A question from Bjorn Poonen: let C be a general hyperelliptic curve of genus g ≥ 2 with pa Weierstrass point. Does there exist q ∈ C and m > 2 with mq ∼ mp other than q anotherWeierstrass point?

Here are some ideas:

49

• If (C, p) has such a q, try to find a deformation of (C, p) losing q.

• Look at the monodromy on torsion points. If it’s transitive, it’s enough to show that therearen’t m2g − 1 such points.

20 Diaz’s Theorem

We would like to know what the largest dimensional complete subvariety of Mg is. We could alsoask a similar question, requiring that the variety pass through a general point on Mg.

Here is what we know:

• There exist complete curves in Mg through a general point (in fact, through any finite set ofpoints). This follows from the Satake compactification of Mg. Ms

g is a projective variety withMg as an open subset, but it is highly singular at the boundary and is not a moduli space. Avirtue is that for g ≥ 3, Ms

g \Mg has codimension 2.

• There exist complete families of large dimension when g >> 0. Here is a variant of Kodaira’sconstruction: Start with a curve C0 of genus h, and consider those C for which there existsf : C → C0 of degree 3, branched at only one point. This is contained in Mg for g = 3h− 1,and is a complete curve. (It’s a finite covering space of C0.)

Iterating this, given Σ ⊆ Mg complete of dimension m, we get Σ′ ⊆ M3g−1 complete ofdimension m + 1, where Σ′ is the set of C such that there exists f : C → C0 of degree 3branched at one point, and C0 ∈ Σ.

Theorem 20.1 (Diaz). If Σ ⊆Mg is complete of dimension m, then m ≤ g − 2.

Remark. The largest dimension m of complete Σ ⊆Mg has log3 g < m ≤ g−2. We also don’t knowif there exists a complete surface Σ ⊆Mg through a general point.

Here is Arbarello’s original proposed method of proof: we stratify Mg. Let

Γd ⊆Mg = {C : ∃ p ∈ C with r(dp) ≥ 1} (20.1)

and Γd = Γd \Γd−1. Alternatively, Γd is the set of C for which there exists f : C → P1 of degreed which is totally ramified at a point p ∈ C.

Arbarello proposed that Γd contains no complete curves. Now we have

Γ2 = Γ2 ⊆ Γ3 ⊆ Γ4 ⊆ · · · ⊆ Γg = Mg. (20.2)

(We have Γg = Mg since every curve has a Weierstrass point.)

If C → P1 is of degree d with a point of total ramification, then the number of other branchpoints is (2g − 2 + 2d)− (d− 1), so

dim Γd = # branch points− dimPGL2 = (2g + 2d− 2)− (d− 1) + 1− 3 = 2g + d− 3. (20.3)

So Γd is a hypersurface in Γd+1.

50

The proposed method of proving that Γd has no complete curves: given B ⊆ Γd complete, thenafter base change, we can assume:

• B is smooth

• There exists a universal family C → B

• There exists a section σ : B → C with r(dσ(b)) = 1 for every b ∈ B and r((d− 1)σ(b)) = 0 forevery b ∈ B.

σC

B

π

(20.4)

Look at E = π∗OC(dσ), a vector bundle of rank 2 on B. It is basepoint free, so we get

C PE∗

B

π

φ

P1-bundle(20.5)

Letting D be the rest of the branch locus of φ, we get

φ(σ)

D

(20.6)

What happens if D meets φ(σ)? We get a singular point of the cover. For example,

(1 2)(1 2 3 · · · d) = (2 · · · d), (20.7)

so we get a point with total ramification index d − 2. This requires a singularity (node). Thisforces a base point.

51

•

•(1 2)

(1 2 3 · · · d)

(20.8)

•

•

these cross

(20.9)

Here is the problem with this argument: D does not have to meet φ(σ). For there exist P1

bundles over B with a section σ and a disjoint curve D. Indeed, we don’t know whether Γd cancontain a complete curve.

But there do not exist P1 bundles over B with disjoint sections σ, τ and a curve D disjoint fromboth of them, except fo the trivial bundle and constant sections.

To see this, suppose we had

σ

τ

D

(20.10)

After a base change so that D splits into sections, we get

52

σ

τ

D

(20.11)

This forces D,σ, τ to be the union of at least three constant sections of trivial P1-bundles, sincea P1-bundle with at least 3 disjoint sections is trivial.

Diaz’s solution is to look at a different stratification

∆d = {C : ∃ f : C → P1 with deg f ≤ g and #f−1(0,∞) ≤ d}. (20.12)

Then we have

∆2 ⊆ ∆3 ⊆ · · · ⊆ ∆g = Mg. (20.13)

(∆g = Mg because, for example, we can take 0 to be a Weierstrass point and ∞ any otherbranch point of a function totally ramified of degree d at 0.)

Now ∆2 is the set of C for which there exists f : C → P1 totally ramified at two points. We havea component of ∆2 for each possible degree (up to g). Notice that ∆g has pure dimension 2g−3+d.

C

P1

degm

(20.14)

53

We have 2g+ 2m−2 branch points in total. There are 2m−d over 0,∞, so the number of otherbranch points is 2g + 2m− 2− (2m− d). We get 2g + d distinct branch points, so the dimension is2g − 3 + d.

Now suppose B ⊆ ∆d is complete. Then we have

∞

0

D

(20.15)

with D of degree 2g + d − 2 over B. By the above observation, D must meet either the 0 or∞ section. Say Ct

ft−→ P1 is branched at 0 and α(t). The monodromy over 0 is σ ∈ Sd, while themonodromy over α(t) is (1 2) after appropriate labelling. If 1 and 2 belong to different orbits of σ,then C0 is smooth, but #f−1

0 (0) < #f−1t (0). We end up in ∆d−1. If 1 and 2 belong to the same

orbit, then C0 is singular.

21 Unirationality of the Moduli Space in Small Genus

One important question is whether we can write down a general curve of genus g. In other words,does there exist a family C → B of curves of genus g, with B open in an affine (or projective) space,with the induced map φ : B →Mg dominant?

This is possible in low genus. We’ve already seen the genus 1 case.

• g = 2: All curves are hyperelliptic, so of the form

y2 =6∏i=1

(x− λi). (21.1)

All curves of genus 2 are expressible by choosing λi appropriately.

• g = 3: The non-hyperelliptic curves are plane quartics, so consider

∑i+j≤4

aijxiyj = 0 (21.2)

and let the aij vary, avoiding the hypersurface of singular quartics. The resulting image isdense in M3.

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• g = 4: f2(x, y, z) = g3(x, y, z) = 0 for general f2, g3 gives a complete intersection; the resultingimage is dominant in M4.

• g = 5: In general, the canonical curve is an intersection of three quadrics, so take three quadricsin P4.

• g = 6: CφK↪−−→ P5 is of degree 10, but not a complete intersection, so we need another method.

Brill-Noether theory says that a general C of genus 6 can be birationally embedded in P2 as asextic with four nodes. Furthermore, a general curve will have no three nodes collinear. Nowwe can take the four points to be

p1 = [1, 0, 0], p2 = [0, 1, 0], p3 = [0, 0, 1], p4 = [1, 1, 1]. (21.3)

So let V be the vector space of homogeneous F (X,Y, Z) of degree 6 vanishing to order atleast 2 at p1, . . . , p4. We have an open subset U ⊆ V consisting of those F which are nodal atp1, . . . , p4 and smooth otherwise. Then U →M6 is dominant.

• g = 7: Again, Brill-Noether theory implies that a general C can be embedded in P2 as a curveof degree 7 and 8 nodes. This means the Severi variety V7,7 = V 7,8 dominates M7.

Consider V 7,8 → (P2)8, the set of 8-tuples of points in P2 (specification of the nodes). Thegeneral fiber is P(92)−1−3·8 = P11. This means that V 7,8 is birationally a P11-bundle over(P2)8. Labelling the nodes would replace (P2)8 with (P2)8. This shows that V 7,8 is rational.Specifically, if

∆ = {(p,Γ) ∈ P2 × (P2)8 : p ∈ Γ}, (21.4)

then V 7,8 is birational to

P(

(π2)∗(π∗1OP2(7)⊗ I2

∆)). (21.5)

Therefore there exists U ⊆ V with U ↪→ A27 and a family C → U such that the map U →M7

is dominant.

• g = 8: Brill-Noether implies we can birationally embed a general curve in P2 as a degree 8 curvewith 13 nodes. Now we have V8,8 → (P2)13 with general fiber isomorphic to P(102 )−1−3·13 = P5,so the Severi variety is generically a P5-bundle over (P2)13. Again the Severi variety is rational.

• g = 9; embed as a octic with 12 nodes, so V8,9 → (P2)12 with general fiber P8.

• g = 10: take d = 9 and δ = 18. Then V9,10 → (P2)18, with general fiber P(112 )−1−3·18 = P0. SoV9,10 is actually birational to (P2)18.

• g = 11: if we try the same method, we will take d = 10, δ = 25, but then V10,11 does notdominate (P2)25! This means the 25 nodes must be in special position.

For the cases g = 11, 12, 13, 14, Sernesi, Ran, and Chang have used embeddings into P3 to showthat the Hilbert scheme was rational.

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φ : B →Mg dominant with B rational implies Mg is unirational. In particular, its plurigenerah0(Km

Mg) must be zero.

(Note: the genus 6-10 arguments actually had some holes. These were filled in by Arbarello andCornalba.)

One consequence of the above work for g ≤ 10 (14): the set of curves defined over Q is dense inMg. This is not known for g ≥ 15.

The existence of a family is equivalent to unirationality of Mg. (The other implication holdssince we can restrict to the automorphism-free locus, so that the map to Mg comes from a family.)

If X is smooth and projective, then these exists a dominant rational map Pn 99K X only ifh0(Km

X ) = 0 for every m > 0. (We can replace “smooth” by “having only canonical singularities”,which is satisfied by Mg.) So we want to know whether there exists an effective pluricanonicaldivisor on Mg. To do this, we study the divisor class theory of Mg, enough to identify some ofPic(Mg),KX , and classes of effective divisors.

22 Divisors of the (Compactification of the) Moduli Space

Divisors of Mg, or more generally subvarieties, come from geometric conditions. Here are someexamples:

• The locus ∆ of singular curves.

• Curves with special Weierstrass points: at least two give divisors.

• Curves with a semicanonical pencil (that is, there exists L with L2 = KC and h0(L) ≥ 2)gives a divisor.

• Brill-Noether loci: If ρ = ρ(r, d) = g − (r + 1)(g − d+ r) < 0, we can look at the set of C forwhich C has a grd. If ρ = −1, this gives a divisor.

• More generally, the set of C for which there exist p1, . . . , pk satisfying

h0

(∑i

mipi

)≥ r (22.1)

works.

Now suppose X is any variety, and L is a line bundle on X. For every map φ : B → C with Ba curve, we get a line bundle Lφ = φ∗L on B. If B′ α−→ B

φ−→ X, then Lφ◦α = α∗Lφ. Conversely, theassociation φ 7→ Lφ with the naturality condition determines L.

So a line bundle on Mg associates to every one-parameter family C π−→ B of stable curves adivisor class LC on B, such that for every B′ α−→ B, we have LC×BB′ = α∗LC . In this way, we canview a divisor class on Mg as a gadget that associates to a family of stable curves over B a divisorclass on B, which is compatible with base change.

Now we need to determine the variation of a family. Here are some methods:

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• Look at the Hodge bundle of C π−→ B. This bundle is given by E = π∗(ωC/B), a vector bundleof rank g on B.

•

•

C

B

π

(22.2)

We get a divisor class λ = c1(E) on B.

• Use the self-intersection number to determine κ = π∗(c1(ωC/B)2).

Given C → B, we’ll want to determine the intersection number of ∆ with the image in Mg

associated to this family. We’ll need a better understanding of ∆ to determine the scheme-theoreticintersection.

Recall that if p is a node of a curve C, then Def(C, p) = ∆t, with versal family xy− t. So around[C] ∈Mg, the divisor ∆ is locally (t). If C is stable with nodes p1, . . . , pδ, then we have

Def(C)→∏

Def(C, pi) =∏

∆ti =⇒ ∆ =(∏

ti

). (22.3)

We conclude that if C → B is a 1-parameter family of stable curves and φ : B → Mg is theassociated map, and b0 ∈ B is such that Cb0 has nodes p1, . . . , pδ with the local equation of C at pibeing xy − tmi ,

•

Cb0

(22.4)

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then we have

multb0(φ∗∆) =∑

mi. (22.5)

As an example, we’ll look at a general pencil of plane quartics. Take F,G to be two generalquartic polynomials in P2 and look at

C = V (t0F + t1G) ⊆ P1t × P2. (22.6)

Observe that all fibers Ct are stable, since the locus of quartics having a non-nodal singularityhas codimension 2. (We can further assume that all singular curves are irreducible with 1 node.)We would like to find the degrees of δ, λ, κ on P1.

Remark. C is a general divisor of type (1, 4), so is smooth by Bertini.

δ: Since C is smooth, each multiplicity is 1, implying that deg δ equals the number of nodes ofsingular fibers, which in turn equals the number of singular fibers.

Use Riemann-Hurwitz: given Xπ−→ B with B a smooth curve and X smooth of dimension n, let

F be the general fiber and let Γ = {b1, . . . , bδ} be the points at which the singular fibers lie over.

• (22.7)

Write X = π−1(B \ Γ) ∪ Xb1 ∪ · · · ∪ Xbδ . Each Xbi has a tubular neighborhood, which afterremoving Xbi , gives a S

1-bundle, therefore of Euler characteristic 0. We obtain

58

χ(X) = χ(π−1(B \ Γ)) +∑

χ(Xbi) (22.8)

= χ(B \ Γ)χ(F ) +∑

χ(Xbi) (22.9)

= χ(B)χ(F )− δχ(F ) +δ∑i=1

χ(Xbi) (22.10)

= χ(B)χ(F ) +∑b∈B

(χ(Xb)− χ(F )

). (22.11)

In our situation, we have δ singular fibers.

F : χ = −4 (22.12)

χ = −2− 1 = −3 (22.13)

59

We have χ(Xb) − χ(F ) = 1 for every nodal fiber, so the number of singular fibers equalsχ(C) − χ(B)χ(F ). Now χ(C) = χ(BlV (F,G)P2) = 3 + 16 = 19 since V (F,G) consists of 16 points.Also χ(B) = 2 and χ(F ) = −4, so there are 27 singular fibers.

Here is an alternative method that only works in P2. The locus of singular fibers is

V

(t0∂F

∂x+ t1

∂G

∂x, t0

∂F

∂y+ t1

∂G

∂y, t0

∂F

∂z+ t1

∂G

∂z

). (22.14)

κ: for Z ⊆ P1×P2, writeOZ(m,n) for (π∗1OP1(m)⊗π∗2OP2(n))|Z . We haveKP1×P2 = OP1×P2(−2,−3)so KC = KP1×P2(C)|C = OC(−1, 1). This means ωC/P1 = KC ⊗ π∗1K∗P1 = OC(1, 1).

So deg κ equals the triple intersection of divisors on P1×P2 of bidegrees (1, 4), (1, 1), (1, 1). Usingthe ring A(P1 × P2) = Z[α, β]/(α2, β3), we obtain

c21(ωC/P1) = (α+ 4β)(α+ β)(α+ β) = 9αβ2. (22.15)

Therefore deg κ = 9.

λ: We have

E = (π1)∗ωC/P1 = (π1)∗OC(1, 1) = OP1 ⊗ (π1)∗OC(0, 1) (22.16)

and (π1)∗OC(0, 1) is the trivial bundle of rank 3, so E = OP1(1)⊕3 implying deg c1(E) = 3.

Remark. δ can be broken up into δ = δ1 + · · · + δbg/2c, where δi is the contribution to the ithboundary component of Mg, using disconnecting nodes. We usually don’t do this, though.

For our next example, let S ⊆ P3 be a smooth quartic surface and C π−→ P1 be a general pencilof plane sections.

× × × × (22.17)

Let Γ be the intersection of the surface with the base line of the pencil. Then

60

C BlΓS P1 × S S

P1

α

π (22.18)

We can use genericity to show that every fiber is either smooth or irreducible with one node.(Or just look at general S.)

For the degree of δ, use Riemann-Hurwitz again. We have χ(B) = 2 and χ(F ) = −4. It turnsout that χ(S) = 24 (S is a K3 surface), so χ(C) = 28. Therefore deg δ = χ(C)− χ(B)χ(F ) = 36.

Alternatively, for S ⊆ P3, consider the Gauss map S[ ∂f∂x,...]

−−−−→ S∗ ⊆ (P3)∗. We want #(L∩S∗) forL a line. This equals degS∗.

Now introduce divisor classes on C ⊆ P1 × S: let η = π∗OP1(1) and ζ = α∗OS(1). Note thatC ∼ η + ζ in A(P1 × S). η is the class of a fiber, so η2 = 0. We have η · ζ = 4 since a plane sectionwith meet a fiber (a quartic curve) in four points. Finally, ζ2 = 4 as the intersection of S with ageneral line. Now

KP1×P3 = −2η − 4ζ =⇒ KP1×S = −2η =⇒ KC = (KP1×S + [C])|C = −η + ζ =⇒ ωC/P1 = η + ζ.(22.19)

We get κ = (η + ζ)2 = 12 as a result.

As for λ,

π∗ωC/P1 = π∗(π∗OP1(1)⊗ α∗OS(1)) = OP1(1)⊗ π∗αOS (1). (22.20)

π∗α∗OS(1) is a bundle of rank 1 with fiber over t ∈ P1 equal to H0(OCt(1)) = H0(OHt(1)). We

have an exact sequence

0→ OP1(−1)→ O⊕4P1 → π∗α

∗OS(1)→ 0 (22.21)

so c1 = 1, implying λ has degree 4.

Theorem 22.1. 1. PicMg ⊗Q is generated by λ, κ, δ0, . . . , δbg/2c.

2. If g ≥ 3, these satisfy the unique relation 12λ = κ+ δ.

3. αλ− βδ is ample if and only if α > 11β > 0.

4. KMg= 13λ− 2δ (at least for the moduli stack).

We will prove parts 2 and 4. 3 uses a technique not yet introduced, and 1 is very hard.

(Reference of how to get better understanding of stacks: Mumford, Picard groups of moduliproblems.)

Part 1 was proved by Harer, using topological methods. This uses the description of Mg viaTeichmüller space:

61

(22.22)

(22.23)

A Riemann surface with a pair-of-pants decomposition is specified by 2g− 2 pieces with variousgluings (6g − 6 real parameters in total). We can recover Mg = τg/Γg, where τg is the space above(which is contractible in C3g−3), and Γg is the mapping class group, which leaves the surface alone,but changes Teichmüller space. Harer tried to describe Γg.

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More recently, Arbarello and Cornalba gave a simpler proof, but one which was still heavilybased on topology and complex analysis.

Part 3 uses the notion of stability. This can be looked up in Moduli of Curves.

Parts 2 and 4 are applications of Grothendieck-Riemann-Roch.

After Serre duality, Riemann-Roch could be formulated as follows: for L a line bundle on C,χ(L) = c1(L)︸ ︷︷ ︸

d

−12 c1(TC)︸ ︷︷ ︸

2g−2

. More generally, if F is any coherent sheaf on C, then χ(F) = c1(F)− 12 ·

rank(F) · c1(TC).

Now if S is a surface and L is a line bundle, then

χ(L) =c2

1(L) + c1(TS)c1(L)

2+c2

1(TS) + c2(TS)

12. (22.24)

Hirzebruch generalized this further. Let X be a smooth variety of dimension n and F a sheaf onX having rank r. Then c(F) =

∏(1 +αi) where ck(F) is the kth elementary symmetric polynomial

of the αi. Now define ch =∑eαi . The homogeneous terms are symmetric, so expressible in terms

of the ci:

ch(F) = rank(F) + c1(F) +c2

1(F)− 2c2(F)

2+ · · · (22.25)

Also define the Todd class

Td(F) =∏ αi

1− e−αi= 1 +

c1(F)

1+c2

1(F) + c2(F)

12+ · · · (22.26)

Hirzebruch-Riemann-Roch states that if F is a coherent sheaf on X, then χ(F) = [ch(F) ·Td(Tx)]n.

Grothendieck put this in the relative setting. Suppose we have Xπ−→ B and F a coherent sheaf

on X.

B•(22.27)

Then we have

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∑i

(−1)ich(Riπ∗F) = π∗

(ch(F) · Td(TX)

Td(TB)

). (22.28)

Given a family C → B of stable curves (smooth for simplicity), write ω for c1(ωC/B). We wantλ = c1(π∗ωC/B). By GRR, R1π∗ωC/B is the structure sheaf, so we end up with

c1(π∗ωC/B) = π∗

((1 + ω +

1

2ω2 + · · ·

)(1− 1

2ω +

1

12ω2 + · · ·

))(22.29)

= π∗

(1 +

ω

2+ω2

12+ · · ·

)(22.30)

= g − 1 +κ

12. (22.31)

If there are singular fibers, we get a contribution of δ from c2 of the relative tangent bundle.This proves part 2.

To understand the canonical sheaf, if C ∈ Mg is automorphism-free, then TC(Mg) = H1(TC).Suppose C π−→ M0

g is the universal family of smooth automorphism-free curves. Then T (Mg) =R1π∗(TC/M0

g) so T ∗(Mg) = π∗(ω

2C/Mg

) by Kodaira-Serre duality. So KMg = π∗(ch(ω2) ·Td(ω∗)), andwe obtain KMg = 13λ. (We can make a correction of 13λ− 2δ for nodal curves.)

Recall the ample cone of aλ− bδ. We want to determine the effective cone:

a

b

effective

•KMg

ample slope Sg

(22.32)

Observe that if Sg < 132 , then Mg is of general type, so not unirational. If Sg > 13

2 , thenh0(Mk

g) = 0 for every k > 0, so Mg has negative Kodaira dimension.

We won’t take care of the issue that Mg is not smooth, but rather has canonical singularities.(To do this, analyze the local geometry of Mg.)

To show Sg has low slope, we need to exhibit effective divisors of Mg and calculate their class.

We might look at W , the set of C with a special Weierstrass point p ∈ C such that h0(OC((g−1)p)) ≥ 2. It turns out that the slope of [W ] is 9 +O(1

g ). This isn’t low enough.

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Another thing we might try is the set T of curves with a semicanonical pencil. This time theslope of [T ] equals 8 +O(1

g ), better but still not enough.

What does work is to look at the Brill-Noether divisor B ⊆Mg, equal to the set of C with a grdhaving ρ = −1.

As an example, consider C of genus g = 2k + 1 with a pencil of degree k + 1; that is, genus 3hyperelliptic curves. The result of calculation is that the slope of [B] is 6 + 12

g+1 . Hence Mg is ofgeneral type for g ≥ 24.

(Actually, this doesn’t always work. If g = p − 1, then such B doesn’t exist. Then the Petridivisor can be used. We won’t go over this.)

To do this, we first argue that [B] is a linear combination of the form aλ−b0δ0−· · ·−bbg/2cδbg/2c,and then calculate a and the bi by intersecting with test curves.

An example of a test curve is a pencil on a general K3 surface. A polarized K3 surface is Swith a (birational) embedding S ↪→ Pg of degree 2g − 2. For H ∼= Pg−1 a hyperplane section of Pg,S ∩H = C a canonical curve. Taking a pencil of hyperplane sections of S, we get a family of stablecurves of genus g

S BlΓS S

P1

π

α

(22.33)

Here Γ = S ∩ Pg−2, consisting of 2g − 2 points.

We have χ(S) = χ(S) + 2g − 2 = 2g + 22. But also δ = χ(S) − 2χ(C) = 6g + 18. NowωS/P1 = π∗OP1(1)⊗ α∗OPg(1), so E = π∗ωS/P1 = OP1(1)⊗ π∗(α∗OPg(1)), where π∗(α∗OPg(1)) is a

quotient of a trivial bundle of rank g + 1 by OP1(−1). We conclude that λ = degE = g + 1.

On the other hand, the degree of [B] on this family (that is, the number of fibers of the familythat have Brill-Noether number −1) is zero. By geometric Riemann-Roch, no hyperplane sectionhas a linear series with ρ = −1, because S does not contain many points with enough dependencerelations.

We conclude that the slope of [B] is 6g+18g+1 = 6 + 12

g+1 .

Remark. This is only a heuristic. We need to calculate coefficients of higher boundary components.

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To do. . .

2 1 (p. 9): Fix Me Formatting is awful right now.

2 2 (p. 27): I think I missed this lecture. Try to fill it in.

2 3 (p. 30): Figure this one out.

2 4 (p. 34): Fix Me Is this picture actually correct?

2 5 (p. 35): Fix Me Make this look better? (It will be tough!)

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