notes for math 282, geometry of algebraic curves · notes for math 282, geometry of algebraic...

NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES

AARON LANDESMAN

CONTENTS

1. Introduction 42. 9/2/15 52.1. Course Mechanics and Background 52.2. The Basics of curves, September 2 53. 9/4/15 63.1. Outline of Course 64. 9/9/15 84.1. Riemann Hurwitz 84.2. Equivalent characterizations of genus 84.3. Consequences of Riemann Roch 95. 9/11/15 106. 9/14/15 116.1. Curves of low genus 127. 9/16/15 137.1. Geometric Riemann-Roch 137.2. Applications of Geometric Riemann-Roch 148. 9/18/15 158.1. Introduction to parameter spaces 158.2. Moduli Spaces 169. 9/21/15 189.1. Hyperelliptic curves 1810. 9/23/15 2010.1. Hyperelliptic Curves 2010.2. Gonal Curves 2111. 9/25/15 2211.1. Curves of genus 5 2312. 9/28/15 2412.1. Canonical curves of genus 5 2412.2. Adjoint linear series 2513. 9/30/15 2713.1. Program for the remainder of the semester 2713.2. Adjoint linear series 2714. 10/2/15 2915. 10/5/15 3215.1. Castelnuovo’s Theorem 3316. 10/7/15 3417. 10/9/15 3618. 10/14/15 38

1

2 AARON LANDESMAN

19. 10/16/15 4119.1. Review 4119.2. Minimal Varieties 4220. 10/19/15 4421. 10/21/15 4722. 10/23/15 4922.1. Agenda and Review 4922.2. Resolutions of Projective Varieties and Green’s conjecture 5022.3. The Maximal Rank Conjecture 5223. 10/26/15 5324. 10/28/15 5524.1. Review 5525. 10/30/15 5725.1. Review 5725.2. Today: Brill Noether Theorem in dimension at least 3 5826. 11/2/15 6126.1. Review 6126.2. Hilbert Schemes 6127. 11/4/15 6427.1. Logistics 6427.2. Tangent Spaces in Brill Noether Theory 6427.3. Martens Theorem 6628. 11/6/15 6729. 10/9/15 7129.1. Agenda and Review 7130. 11/11/15 7530.1. Review 7530.2. The Genus 6 Canonical Model 7631. 11/13/15 7832. 11/16/15 8132.1. Logistics and Review 8132.2. A continuing study of genus 6 curves 8233. 11/18/15 8433.1. Logistics and Overview 8434. 11/20/15 8734.1. Plan, conventions, and review. 8734.2. An Upper bound on the dimension 8834.3. Proof of Existence for Brill Noether 8835. 11/23/15 9035.1. Review 9035.2. Inflectionary points of linear series 9236. 11/30/15 9336.1. Review and overview 9337. 12/2/15 9737.1. Overview 9737.2. Retraction 9737.3. Finishing the proof of Brill-Noether 97

NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 3

37.4. Further Questions 99

4 AARON LANDESMAN

1. INTRODUCTION

Joe Harris taught a course (Math 282) on algebraic curves at Harvard in Fall2015.

These are my “live-TEXed“ notes from the course. Conventions are as follows:Each lecture gets its own “chapter,” and appears in the table of contents with thedate.

Of course, these notes are not a faithful representation of the course, either in themathematics itself or in the quotes, jokes, and philosophical musings; in particular,the errors are my fault. By the same token, any virtues in the notes are to becredited to the lecturer and not the scribe. 1

Please email corrections to [email protected].

1This introduction has been adapted from Akhil Matthew’s introduction to his notes, with hispermission.

[email protected]


2. 9/2/15

2.1. Course Mechanics and Background.(1) Math 282, Algebraic Curves(2) CA Adrian(3) Text: ACGH, Volume 1(4) Four years ago, a similar course was taught, following ACGH. The idea

was: given a curve, what can we say about it. This is only half the story.Curves can appear in the abstract and in projective space. An importantpart of understanding curves is how they vary in flat families. The differ-ence between ACGH volumes 1 and 2, is that 1 deals with a fixed curveand 2 deals with families of curves. To learn more on families of curves,look at Moduli of Curves.

(5) Two major changes in the language since when the book was written: First,we will use cohomology, and second we will use schemes.

(6) Algebraic curves were first studied over the complex numbers. Some peo-ple studied complex analysis of Riemann Surfaces, and others studiedpolynomials in two variables.

Remark 2.1. We will use the language of smooth projective curves and compactRiemann surfaces interchangeably. We will assume all curves are over the complexnumbers.

The central problem of the course is

Question 2.2. What is a curve?

In the 19th century, a curve is a subset of Pn for some n.In the 20th century, a curve became an abstract curve, which exists indepen-

dently of any particular embedding in projective space.A similar perspective was adapted in group theory. Originally, people viewed

groups as subsets of GLn. Now, this is called representation theory.

Remark 2.3. In his textbook, Hartshorne says the goal of algebraic geometry is toclassify algebraic varieties. In the modern context, we can just specify the genus.However, in the 19th century, you would have to also specify the degree. We canthen ask, which pairs of d,g are realized as a curve. This is still not completelyknown. You can also specify a projective space, and then ask which curves can berealized in that projective space.

2.2. The Basics of curves, September 2.

Definition 2.4. Define the genus by g = 12 (1− χtopX).

Definition 2.5. An ordinary singular point of a curve of multiplicitym is a pointin which m branches of a curve meet transversely of a point. We can define thismore rigorously by saying that the completion of its local ring is isomorphic tok[x,y]/(f1 · · · fm) where fi are distinct linear functions in x,y.

Lemma 2.6. Let X be a curve. The following are equivalent:(1) The genus of X.(2) 1− χOX(3) 1

2 (degKX + 2)

6 AARON LANDESMAN

(4) 1− c, where c is the constant term of the Hilbert polynomial of C ⊂ Pr.(5) If C ∼= C0 ⊂ P2 of degree d with ordinary singular points of multiplicity mi,

then g(C) =(d−12

)−∑i

(mi2

)Definition 2.7. A divisor is an formal sum of the form D =

∑i nipi for ni ∈

Z,pi ∈ X. We sayD is effective ifD ≥ 0. We define the degree by degD =∑i ni.

For f a rational function on X, we define

div f = (f) =∑p

ordp(f) · p = (f)0 − (f)∞.

Remark 2.8. By the residue formula applied to the logarithmic derivative of arational function f, we have deg (f) = 0.

Definition 2.9. We say D ∼ E, or D is linearly equivalent to E if there exists arational function f with (f) = D− E. Effective divisors of degree d on X will benotated as Cd, which is by definition Cd/Sd.

Definition 2.10. Given D ∈ div X, we look at L(D) := {f ∈ K(X)× : (f) +D ≥ 0}.An alternative notation for L is H0. Since we can specify the polar part of sucha function, this is a finite dimensional vector space. This uses the fact that thereare no nonconstant meromorphic function. We define `(D) = dimL(D) and definer(D) = `(D) − 1.

Remark 2.11. If D ∼ E then L(D) ∼= L(E), as given by multiplication by f.

Definition 2.12. We define Picd(X) := Divisors of degree d/ ∼.

Remark 2.13. It turns out this set corresponds to the points of a variety.

Definition 2.14. Suppose ω is a rational 1-form, which looks locally like f(z) dz.We let (ω) =

∑ordp(ω)p. We define KX := (w) ∈ Pic2g−2(X).

3. 9/4/15

Note: there will be no class Monday or Friday.

3.1. Outline of Course.(1) This week: Basics of linear series(2) Starting 9/14, we’ll talk about curves of low genus and Castelnuovo’s the-

orem (gives an upper bound on the genus of a curve of degree d in projec-tive space)

(3) Brill-Noether Theory: This addresses the question “What can you say abouta general curve?” It makes sense to ask whether there is an open subset ofthe Hilbert scheme on which there is a uniformity of the correspondingcurves.

Remark 3.1. For the remainder of the course, we let X be a smooth projectivecurve. We let D =

∑i niPi and say D ∼ E if D is linearly equivalent to E. We let

KX be the divisor class of the canonical sheaf ω, of degree 2g− 2. Use L(D) forH0(C,D), let `(D) = h0(C,D) and r(D) = `(D) − 1.


The justification of looking at these linear systems is that we only allow polesat specified points. This gives us a finite dimensional vector space, which is some-thing quite manageable.

We next explain what this has to do with maps to projective space.

Remark 3.2. Given a map φ : X → Pr and let H∞ = V(x0) be a hyperplane cutout by the first coordinate of Pr. Then, the map can be given by [1, f1, . . . , fr], andD = φ−1H∞ and f1, . . . , fr ∈ L(D).

There are some issues with the above description. In particular, we will need toenforce that these maps are basepoint free. We will come back to this later.

Question 3.3. Given X of genus g and D a divisor of degree d on X, what can wesay about `(D).

Example 3.4. We know `(KX) = g.

Theorem 3.5. We have

`(D) = d− g+ 1+ `(KX −D)

Proof. Caution: there is a problem with the following proof. We have managed tosweet Serre duality under the rug. The problem is that we assumed D and K−Dwere effective.

Say D = p1 + · · ·+ pd, where pi are distinct points of X. Choose local coordi-nates zi around each pi. Any f ∈ L(D) can be written locally as aizi + h where his holomorphic. So, f is determined up to the addition of a constant function bya1, . . . ,ad. So, the question is: which d tuples arise as global principal parts offunctions. First, note that `(D) ≤ d+ 1.

Now, we ask, what is the obstruction to having a global rational function withthese polar parts. Here, we use that if we have a meromorphic 1-form on a curvethen the sum of the residues is 0 by stokes’ theorem. So, if ω ∈ L(KX), and f ∈L(D), then

∑i Respi(f ·ω) = 0.

Sayω(pi) = bidzi. So, plugging this in the above formula, we have∑i aibi =

0.So, we get g relations, but they might not be independent. The actual number

of relations we are getting is g− `(KX −D). So, `(D) ≤ d+ 1− g+ `(KX −D).The reverse inequality is not quite proved correctly, but to do it would be a lot

more work, so we give a heuristic argument. Now, we apply the same argumentto `(K−D). We have

`(K−D) ≤ 2g− 2− d+ 1− g+ `(D)

.Adding the two inequalities we have

`(D) + `(K−D) ≤ `(D) + `(K−D)

. So, we added two inequalities and got an equality, which means they were equal-ities to begin with. �

Suppose you are give a compact complex Riemann surface, how do you knowthere are any nonconstant meromorphic functions. This issue underlies the diffi-culty we are encountering in the proof above. In fact, in dimension at least 2, thereare compact complex manifolds with no meromorphic functions whatsoever.

8 AARON LANDESMAN

Remark 3.6. How should we define a canonical divisor on a singular curve? Wewill have to choose a canonical divisor on a singular curve so that it satisfies thecondition that sums of residues of f ·ω is 0.

Now, returning to the issue of obtaining a divisor D by a preimage of a hyper-plane, we look at locally.

Definition 3.7. Given a divisor D on Xwe define a sheaf OX(D) defined by

OX(D)(U)L := {f ∈ K(U)× : (f) +D ≥ 0 in U.}

When X is smooth, this corresponds to a line bundle.Observe that if D ∼ E,OX(D) ∼= OX(E), as given by multiplication by X. Addi-

tionally, define Picd to be line bundles of degree d, or equivalently, linear equiva-lence classes of degree d.

Finally, defineL(D) := H0(X,OX(D))

Remark 3.8. This construction is a special case of the fact that for locally factorialschemes which are regular in codimension 1, there is an isomorphism between thevector spaces of Weil and Cartier divisors.

Lemma 3.9. The collection of nondegenerate maps (maps whose image is not containedin a hyperplane are in bijection with pairs (V ,L) where L ∈ Picd(X),V ⊂ H0(L) withdimV = r+ 1, so that V is basepoint free.

Proof. See Vakil, chapter 16. �

4. 9/9/15

4.1. Riemann Hurwitz.

4.2. Equivalent characterizations of genus.(1) The dimension of the vector space of holomorphic 1 forms.(2) χ(OC) = 1− g(3) pC(m) = md+ 1− g(4) `(D) = d− g+ 1 for D >> 0.(5) Number of handles(6) b1(C) = 2g(7) degKC = 2g− 2(8) χ>(C) = 2− 2g.

Remark 4.1. It is easier to prove equivalence of these definitions in the algebraiccategory because in that setting we already have access to the algebraic functionscoming from projective space via an embedding.

Suppose π : C → B is a nonconstant map of compact Riemann surfaces ofdegree d. Then, for all but finitely many points in the target, the preimage consistsof d points. For all points p ∈ C, we can choose local coordinates so that π is of theform z 7→ zm, and whenm > 1 p is a ramified point.

Definition 4.2. Given a map π : C→ B as above, set vp(π) = m− 1 and define theramification divisor

R =∑p∈C

vp(π) · p(4.1)


and the branch divisor by

B =∑q∈B

∑p∈π−1(q)

· q(4.2)

We define the total ramification index as b := degR = degB

Theorem 4.3. (Riemann-Hurwitz) Suppose π : C→ B is a nonconstant map of compactRiemann surfaces of degree d. Then, 2g− 2 = d(2h− 2) + b.

Proof. We compare the degrees of ωC and π∗ωB. At a point of B, the pullback ofa from on Bwill pick up zeros equal to the ramification index. So,

KC = π∗KB + R(4.3)

Then, taking degrees, since degKC = 2g− 2, we obtain Riemann-Hurwitz. �

Remark 4.4. It’s also fun to prove this via topological Euler characteristic. You dothis by removing the ramification and branch locus. We then obtain an unramifiedcovering space, and use that on this locus, the Euler characteristic is multiplicative.Then, you add the points back in and deduce the formula.

4.3. Consequences of Riemann Roch.

Remark 4.5. Recall, nondegenerate maps C → Pr of degree d, modulo the actionof PGLr+1 are in bijection with pairs (L,V) so that L is a line bundle on C of degreed and V of dimension r+ 1 in H0(L) with no common zeros, i.e., no basepoints.

Given a line bundle and sections, we can choose φD(p) = [σ0(p), . . . ,σr(p)].We can also describe this intrinsically by Vp = {σ ∈ V : σ(p) = 0} . Then the map φsends φD : p 7→ Vp ∈ PV∨.

Definition 4.6. A grd is a line bundle of degree d and a vector space V of dimensionr+ 1 inside H0(L). Crucially, we do not assume that V is basepoint free.

Remark 4.7. If D = (L,V) is a basepoint free grd, we get a map C → Pr ∼= PV∨.The map being injective is equivalent to Vp+q = Vp ∩ Vq has codimension 2.Equivalently, Vp+q = {σ ∈ V : σ ≥ p+ q}. The map is an immersion if and only iffor all p ∈ C, V2p has codimension 2.

Lemma 4.8. A map φ is an embedding if and only if for all effective divisorsD ∈ C2 i.e.,of degree 2, VD has codimension 2.

Proof. This is Remark 4.7 �

Corollary 4.9. If degL ≥ 2g+1 thenφ|L| is an embedding, where |L| = (H0(L),H0(L)).

Proof. By Riemann Roch, degL ≥ 2g+ 1 =⇒ dimH0(L) = d− g+ 1. But, it’s alsotrue that degL(−D) ≥ 2g− 1 so H0(L) = d− g− 1. Therefore, any line bundle ofdegree 2g+ 1 or more gives rise to an embedding. �

Remark 4.10. We now want to find the best embedding, where it is easiest to dealwith the image. What is the lowest degree of an embedding we can find?

Lemma 4.11. Suppose D is a divisor of degree d ≥ g+ 3. Then, if D ∈ Cd general thenφ|O(D)| = φD is an embedding.

10 AARON LANDESMAN

Proof. Difficulty, we won’t do it now, but it will be helpful to know about theJacobian. �

Definition 4.12. A curve C is hyperelliptic if there exists a map π : C → P1 ofdegree 2. That is, C can be written as y2 =

∏2g+2i=1 (x− λi) on an open subset.

Exercise 4.13. Not every curve of genus at least 3 is hyperelliptic.

Example 4.14. In genus 2, the canonical sheaf makes C into a hyperelliptic curve.

Definition 4.15. For any curve C of genus g > 0, we have canonical map, definedto be the map associated to the dualizing sheaf.

Theorem 4.16. A curve C is hyperelliptic if and only if the canonical map is not anembedding.

Proof. Use the criterion for being a closed embedding, and Riemann Roch. Fur-thermore, use the characterization that being hyperelliptic is equivalent to havingan effective divisor of degree 2. �

5. 9/11/15

Today Joe is out, and Adrian is holding a review.

Remark 5.1. Deformations of abstract schemes are classified by H1(TC). Embed-ded deformations are classified by H0(NY/X).

Example 5.2. Let’s show there are non hyperelliptic curves. First, TCMg = H1(TC) =

H0(2KC). By Riemann-Roch, h0(2KC) = 3g− 3. This gives a proof that dimMg =3g− 3.

Exercise 5.3. Compute dimMg by first calculating the dimension of the space ofcovers of P1 of degree d.

Theorem 5.4. Let X be a smooth projective variety and Y ⊂ X be smooth of codimension1. Then, KY = (KX + Y)|Y .

Exercise 5.5. Let C be a smooth curve on P1 ×P1. Then, C is given by a bihomo-geneous polynomial of class (a,b).

Solution: We have KC = (KP1×P1 + C)|C. Then, Pic P1 × P1 ∼= ZH1 ⊕ZH2.So,

KC = KP1×P1 +C|C(5.1)

= (−2H1 − 2H2 + aH1 + bH2)|C(5.2)

= ((a− 2)H1 + (b− 2)H2)|C(5.3)

= (a− 2)b+ (b− 2)a(5.4)

= 2ab− 2a− 2b(5.5)

So, g = (a− 1)(b− 1).

Exercise 5.6. Compute the dimension of the space of twisted cubics in P3. (I.e., acomponent of the Hilbert scheme)

Solution: We have a map P1 → P3 of degree 3, which is nondegenerate. We canwrite any map sending [x,y] 7→ [fi(x,y)]. where fi is a homogeneous polynomial


of degree 3. This is 16 dimensional, we subtract 1 for scalars and subtract 3 for thePGL2 action.

Solution 2:

Lemma 5.7. If C is a twisted cubic, then C is contained in a quadric.

Proof. Note that OP3(2) has 10 dimensional cohomology, while OC(2) is 7 dimen-sional by Riemann-Roch. So, in fact, it lies on 3 quadrics. �

Now, consider (C,Q) so that C is contained in Q and this gives you what youwant.

¡++¿

6. 9/14/15

We will take the approach of interweaving new techniques with applications.Today’s new technique is the adjunction formula. We’ll give two proofs of theadjunction formula.

We’ll have the following setting: Let X be smooth varieties of dimension n.Every such scheme has a canonical bundle , which is an invertible sheaf of topdimensional forms KX = ∧nT∨X .

Suppose Y ⊂ X is a smooth divisor, i.e., codimension 1 subvariety. Recall wecan associated an invertible sheaf to the divisor Y, called L = OX(Y). There is alsoa section σ ∈ H0(L) corresponding to the regular function 1, with V(σ) = Y. Thekey fact is

Lemma 6.1. NY/X = L|Y = OX(Y)|Y = OY(Y).

Proof 1. Suppose we have a line bundle with total space L over X. Say p ∈ Y andq ∈ L where q is on the zero section over p. Then, TqL = TpX⊕ Lp. Then, thetangent space to σ, with V(σ) = Y, is the graph of a map TpX→ Lp. The kernel ofthis map is TpY, and by the conormal exact sequence, we have (NY/X)p = Lp. �

Here is an algebraic proof:

Proof 2. We have

N∨Y/X

∼= IY/I2Y = OY |Y = OX(−Y)|Y .(6.1)

And so, NY/X = OX(Y)|Y . �

We are now in a position to prove the adjunction formula. We have an exactsequence

0 TY TX|Y NY/X 0

Recall that in general if we have an exact sequence of vector spaces

0 An−1 Bn C1 0

then there is a natural isomorphism ∧nB ∼= ∧n−1A⊗C.Applying this to 6.1 we obtain

Theorem 6.2.

KY = KX|Y ⊗NY/X = KX(Y)|Y(6.2)

12 AARON LANDESMAN

Proof. This is immediate from applying 6.1 to the conormal exact sequence.�

Example 6.3. KPn = OPn(−n− 1). To see this, given Pn, we have homogeneouscoordinates z0, . . . , zn and on the affine open, we have inhomogeneous coordi-nates zi/z0. Look at dz1 ∧ · · ·dzn. This is holomorphic and nonzero in z0 6= 0.This has a pole of order n+ 1 along z0 = 0. Alternatively look at

dz1z1

∧ · · ·∧ dzn

zn(6.3)

is a meromorphic differential which has a pole on each of the coordinate hyper-planes, and is otherwise holomorphic.

Example 6.4. (1) If C ⊂ P2 is a smooth curve of degree d then by adjunctionOP2(C) = OP2(d) and so KC = OC(d − 3) which has degree (d − 3) · d.Therefore, the degree KX = 2g− 2 = (d− 3)d and so g =

(d−12

).

(2) If Q ⊂ P3 is a smooth quadric surface. Note Q ∼= P1 ×P1 → P3 via theSegre embedding. Then, the tangent plane to Q intersects Q at the twolines of the two rulings passing through the point. We say a curve C ⊂ Qhas bidegree (a,b) if C is the zero locus of a bihomogeneous polynomialof bidegree (a,b), i.e., if C meets lines of one ruling in a points and of theother ruling in b points.

(3) If C ⊂ Q ⊂ P3 is a smooth curve of bidegree (a,b). Then, deg(C) = a+ b,as can be seen by choosing the hyperplane to be a tangent hyperplane tothe quadric, whose intersection with the quadric surface is one line of eachruling. Now, we first compute KQ. We claim, KQ = OQ(−2,−2). First, Qis P1 ×P1, so to write a meromorphic 2 form on Q is the same as writingdown meromorphic 1 forms on both copies of P1 and the zeros are thepreimages of the zeros of the forms on P1. Then, we obtain a meromorphic2 form with poles on 2 lines in each ruling. So,

KQ = KP3 ⊗OP3(Q)|Q = OP3(−2)|Q(6.4)

Then, applying adjunction again,

KC = KQ ⊗OQ(C)|C(6.5)

= OQ(a− 2,b− 2)|C(6.6)

So, plugging in degrees, 2g− 2 = degKC = b(a− 2) + a(b− 2). Then,g = (a− 1)(b− 1). Now, if we take a curve with one of the bidegrees 1, wesee this is a rational curve, and so g = (a− 1)(b− 1).

6.1. Curves of low genus.

6.1.1. Genus 0. If C has genus 0, then C ∼= P1, as follows from Riemann-Roch (ora conic in P2 over arbitrary fields). Furthermore, there is only one line bundle onthe curve of any degree, because any two curves of the same degree are linearlyequivalent. Then, we can look at the complete linear system OP1(d) which embedsC as a rational normal curve of degree d ⊂ Pd.

This d is equal to the smallest possible degree of an irreducible nondegeneratecurve C ⊂ Pd, and any nondegenerate curve of that degree must be the rational


normal curve, as follows from Bezout’s theorem: If we had a curve of lower de-gree, d general points would lie in a hyperplane. Conversely, if we an irreducible,nondegenerate curve, d− 1 points of the curve span a codimension 2 linear space,and projection from that space will give an isomorphism of the curve with P1.

(1) How many surfaces or hypersurfaces does a curve lie on?(2) What is the normal bundle of a rational curve?(3) Many more open problems about rational normal curves.

6.1.2. Genus 1.(1) If we take d = 3, by Riemann Roch, this gives an embedding C → P2 as a

plane cubic. Conversely, a smooth plane cubic by adjunction is a genus 1curve.

(2) Now, take d = 4. This gives an embedding C → P3. If we have a cubicplane curve, it is the zero locus of a cubic polynomial. Does this curvelie on a quadric surface? There is a standard way of answering this usingthe following technique: We want to know whether H0(P3, IP3(2)) van-ish on C. Restricting this to the curve, and to see if the curve lies on aquadric surface, we want to check the kernel of the restriction map. Weknow H0(P3, IP3(2)) = 10 and H0(OC(2)) = 8. Therefore, the kernel is atleast two dimensional. In fact, this map has a two dimensional kernel. Sothis curve lies on a P1 of quadric surfaces. None of the quadrics containingthe curve are irreducible becauseC is irreducible, nondegenerate. Then, byBezout’s theorem, C is the intersection of two quadrics. We could also dothis by adjunction, and so Cmust be type (2, 2) on a quadric surface.

(3) For degree 5, we get an intersection of G(2, 5) with hypersurfaces in theplucker embedding. After this, the equations get even more complicated.

6.1.3. Genus 2. Let C be a curve of geometric genus 4. Take a divisor of degree 4,corresponding to a line bundle L. We get a map C → P2, where C has geometricgenus 2. By Riemann Roch, h0(L) = 3. Because and the image is of degree 4 in P2,but it cannot be smooth because all plane quartics have genus 3. In fact:

Exercise 6.5. For D of degree 4 on a curve C of genus 2, we can write p + q =

D−KC as an effective divisor for a unique divisor p+q onC. That isφD : C→ P2

is either a node if p 6= q or a cusp if p = q. The solution is just Riemann Roch.

7. 9/16/15

7.1. Geometric Riemann-Roch. New tool: Geometric Riemann-RochFor the moment being, we will assume

(1) Assume C is a non hyperelliptic curve of genus g(2) Assume D =

∑pi with pi distinct.

Let φ = φ|K| : C → Pg−1 be the canonical embedding. In coordinates: ifω1, . . . ,ωg are a basis for H0(K), then φ : p 7→ [ω1(p), . . . ,ωg(p)].

More intrinsically, Pg−1 = PH0(K)∨ and φ : p 7→ H0(K− p) ↪→ H0(K).

Theorem 7.1.

r(D) = d− 1− dimD

14 AARON LANDESMAN

Proof. Suppose p1, . . . ,pd ∈ C. Note that if n is the number of linear relations onthe points pi, so that n− d = dim spanp1, . . . ,pd − 1

`(K−D) = g− d+n = g− dim spanp1, . . . ,pd − 1

So, by Riemann-Roch, r(D) is the number of linear relations among the pointspi. So, if the pi are linearly independent then there does not exist a nonconstantmeromorphic function on Cwith simple poles at pi.

In general, φ : C → Pg−1 is some canonical map, which might not be an em-bedding if the curve is hyperelliptic. Let D be any effective divisor. In this case,define the span D = ∩φ−1(H)⊃DH ⊂ Pg−1.

If we take D = 2p we have to take the tangent line to the points, that is, theintersection of all planes tangent to the curve at that point. �

7.2. Applications of Geometric Riemann-Roch.

7.2.1. Genus 2, degree 5. SupposeD is a divisor class of degree 5 on a genus 2 curve.Then, φ : C → P3 embeds C as a quintic curve. Given an embedding, we can askabout the ideal of the embedding. Last time, we asked:

Question 7.2. What surfaces in P3 contain C?

To calculate the quadrics on which this surface lies, note that h0(OP3(2)) =

10,h0(OC(2)) = 10− 2+ 1.Then, the map

H0(OP3(2))→ H0(OC(2))

is in fact surjective, because if C were on two quadrics, being nondegenerate, itwould be degree at most 4. Therefore, the kernel is 1 dimensional, and the curvelies on a unique quadric. This quadric could be either a smooth quadric or a cone,and on the homework we’ll understand the distinction.

Now, let’s example the cubic surfaces containing C. We have

H0(OP3(3))→ H0(OC(3))

Now, C lies on 6 linearly independent cubics. However, we knew about 4 of thecubics C lies on from multiples of the quadratics. So, there are at least 2 linearlyindependent cubics, modulo the ideal generated by the quadratic polynomial Q.

Choose S a cubic containing C but not Q. What is S ∩Q. This is a curve ofdegree 6. Therefore, it must be the union C ∪ L, since L is degree 1. Here, weuse that complete intersections are unmixed (i.e., the resulting scheme structure isreduced). So, C is type (2, 3) or (3, 2) on the quadric.

Lemma 7.3. If M ⊂ Q is any line of type (1, 0) then M supC is S ∩Q. This gives abijective correspondence between cubics containing C but not Q and lines of the ruling ofS.

Proof. We have Q ∼= P1 × P1 → P3 by the Segre map. Homogeneous degree 1forms on P3 pull back to bidegree (1, 1) forms on Q. Therefore, there is a map

homogeneous polynomials of degree m on P3 → bihomogeneous polynomials of bidegree (m,m) on P3


Furthermore, this map is surjective, which can be proven by writing out the abovemap in coordinates. There’s another standard way of showing this, without re-sorting to coordinates. This map is the map on global sections associated to

0 OP3(m− 2) OP3(m) OQ(m) 0

is a surjection of sheaves. We’re asking whether it is surjective on global sections.Note that H1(L) = 0 for any line bundle on P3. Therefore, we have an associatedexact sequence on global sections. So, M ∪ C = S ∩Q. This shows that in factS1 ∩ S2 ∩Q = C. Alternatively, after we knew Cwas contained inQ, being quinticof genus 2, we would have known it was of type (2, 3), and we could have foundwhich cubics it lies on. �

For degree 6 line bundles of on genus 2 curves, the geometry gets much moresubtle, since in P2, when we found a surface containing the curve, the curve be-came a divisor on that surface.

7.2.2. Genus 4, non-hyperelliptic. Let’s focus on the canonical embedding. HereφK : C → P3 is a sextic curve, which is also the smallest degree embedding ofC in any projective space. Being non-hyperelliptic, it’s not expressible as a twosheeted cover of P1. So, there are no meromorphic functions of degree 2. How-ever, it is trigonal. We’ll see this shortly.

Looking at the usual exact sequences, we see

H0(OP3(2))→ H0(OC(2))

determines a surjective map, withC lying on a unique quadric surface, uniquenessfrom Bezout’s theorem.

Next, looking at cubics

H0(OP3(3))→ H0(OC(3))

The kernel is at least 5 dimensional, and so there is a cubic containing C but notQ.Then, S ∩Q is degree 6, hence equal to C. Conversely, by adjunction, any smoothcurve of the form S ∩Q = C is a canonical curve of genus 4. Now, our curve isrealized as a curve of type (3, 3) on Q ∼= P1 × P1. Now, by geometric RiemannRoch, we have three points on the curve which are linearly dependent. So, we’reasking whether the canonical curve contains three colinear points. That is, eachline in the quadric meets the line in a divisor of degree 3.

So, C is trigonal, and if Q is smooth, there exist 2 such maps, while if Q issingular, the curve is only trigonal in 1 way.

8. 9/18/15

8.1. Introduction to parameter spaces. Today: Jacobians. Application: everycurve of genus g admits a nondegenerate embedding C→ Pn of degree g+ 3.

One common question is when will there exist a function of degree d. In genus3 and 4, all curve have functions of degree 3, but this is not true in genus 5.

Jacobians are examples of parameter spaces. Fix C and let(1) Picard variety

Picd(C) = { Line bundles of degree d on C } .

16 AARON LANDESMAN

(2) Hilbert scheme

Hd,g,r = { curves C ⊂ Rr degree d, genus g }

(3) Moduli space of curves Mg.

Mg = { isomorphism classes of smooth projective curves of genus g }

This allows us to talk about the dimension of the family of line bundles on C.This is called the Picard variety or Jacobian.

8.2. Moduli Spaces. Here is a problem Joe got wrong on the quals, which led himto get a conditional in calculus, though it was actually intended by Barry Mazurto be a problem in algebraic geometry. The problem is to computer∫

dx√x2 + 1

The intention of Mazur was to look at y2 = x2 + 1 which is a hyperbola, andobserve this is a genus 0 curve, this curve is parameterized by t 7→ (

2t1−t2

, t2+11−t2

)which determines a map P1 → C yielding

∫dxy and when we pull it back to P1

it becomes∫R(t) dt and then we compute the integral of this rational function on

P1.The next integral that was considered was something like∫

dx√x3 + 1

which we can think of as∫dxy on the Riemann surface y2 = x3 + 1. This integral

isn’t quite well defined, since it depends on the path you take.Say we have a smooth projective curve C of genus g. Given p,q ∈ C, we want

to make sense of∫qp ∈ H0(K)∨/H1(C, Z) = J(C). which is called the Jacobian of

C. We can look at

H1(C, Z) H0(K)∨

Z2g Cg

Fix a base point p0 ∈ C. We get a map

C→ J

p 7→ ∫pp0

which is non canonical because it depends on a basepoint. More generally, for anydwe get a map µ.

µ : Cd → J

D =∑i

pi 7→∑∫pip0


Theorem 8.1. For D,D ′ ∈ Cd we have

D ∼ D ′ ⇐⇒ µ(D) = µ(D ′)

This implies that J ∼= Picd(C).

Proof. Idea: The forward direction is quite easy. If we start with D,D ′, the con-dition they be linearly equivalent means there is a family interpolating betweenthem. If they are sections of the same line bundle L, then we can take linear com-binations of D,D ′ parameterized by the two divisors. That is, there is a family{Dt}t∈P1 where D0 = D,D∞ = D ′ and

µ : P1 → J

t 7→ µ(Dt)

Now, since J is a complex torus, but the pullback of a form to P1 is 0, because P1

has no holomorphic 1 forms. Then, since 1 forms generate on J (in some sense, I’mnot sure exactly how) and so if they all pull back to 0 the map must be 0. �

Remark 8.2. The hard part of Theorem 8.1 was proven by Clebsch, even though itis known as Abel’s theorem

We have an isomorphism

Picd(C) ∼= J

noncanonically, which is why we will often want to write Picd(C) instead of J,though we won’t worry about that too much now.

Definition 8.3. Define

Wrd(C) ={L ∈ Picd(C) : h0(L) ≥ r+ 1

}Exercise 8.4. For L ∈ Picd(C) general, then h0(L) = max (0,d− g+ 1). Solu-tion: The idea is to equivalently formulate it as follows. If D ∈ Cd is a gen-eral effective divisor, we claim h0(L) = max (1,d− g+ 1). The proof of this,D =

∑di=1 pi. We start with h0(K) = g,h0(K − p1) = g − 1, and repeating we

see h0(K −D) = max(0,g − d), supposing we choose the pi so that they are asindependent as possible. Riemann-Roch then implies the statement. So, we onlyneed deal with the noneffective case. If d ≥ g the map µd is surjective. We onlyneed show if 0 ≤ d ≤ g then µd is generically 1-1. In particular,W0d ⊂ J is a closedsubvariety of dimension d. The proof of the statement for d ≤ g− 1 follows fromRiemann-Roch applied to the statement for d > g− 1.

Lemma 8.5. Say C is an arbitrary curve of genus g. If L ∈ Picg+3(C) is general, soh0(L) = 4, we get a map C→ P3. We claim this is an embedding.

Proof. If φL is not an embedding, then there exists an effective divisor of degree 2,this is equivalent to the existence of an effective divisor of degree 2 D = p+qwithh0(L(−D)) = h0(L) − 1 = 3. This would imply L(−D) ∈ W2g+1. Note that the setW2g+1 = K−W0g−3 by Riemann Roch. So, the existence of such a divisor D ∈ W02with h0(L(−D)) = 3 implying L(−D) ∈ K−W0g−3 which implies L ∈ W02 + (K−

W0g−3). So, the set of all such sums has dimension at most 2+ (g− 3) = g− 1. So,a general line bundle of degree g+ 3 will not lie on this g− 1 dimensional locus,and will therefore determine an embedding, i.e., be very ample. �

18 AARON LANDESMAN

Remark 8.6. We showed the above to show the existence of parameter spaces canallow us to prove theorems about the objects we’re concerned with. We’ve beenaccumulating many instances where we invoked these sorts of facts. On the home-work, we saw a line bundle of degree 5 determines something if and only if D isof the form D = 2K+ p, which is a one dimensional space of line bundles inside a2 dimensional space of line bundles.

On Monday we’ll talk about canonical embeddings of curves of genus 5 and 6.

9. 9/21/15

Today:(1) Hyperelliptic(2) Counting moduli(3) canonical curves of genus 5 and 6

9.1. Hyperelliptic curves. LetC be a hyperelliptic curve of genus g. Then, there isa map π : C→ P1 of degree 2. We have exactly 2g+2 branch pointsα1, . . . ,α2g+2.

Lemma 9.1. There exists a unique curve C with a map π : C → P1 branched atα1, . . . ,α2g+2.

Proof. Uniqueness is shown below via a topological argument. To show existence,we can write down C = V(y2 −

∏i(x− αi)), and specify a transition function to

the other chart, as in Vakil’s foundations of algebraic geometry. Alternatively, justcomplete this as a Riemann Surface. �

Given a Riemann sphere and 2g+ 2 points, draw arcs from a given branch pointto all other branch points. If we excise these arcs, we obtain two disjoint sheets.We now want to complete it. We can describe the surface complex analytically asfollows: Every time we go around a branch point the sheets are exchanged. Thisconstructs the complex analytic structure. The complex analytic structure comesfrom pulling back from P1.

Question 9.2. Suppose we have a three sheeted cover of the Riemann sphere sim-ply branched over 2g+ 4 points. How can we describe the structure of that Rie-mann surface?

The question is to describe the monodromy group at the branch points. Twoof them are exchanged, and there is one sheet containing the simple point. If weare given the branch points, to specify the cover, we only have to topologically de-scribe the cover, which is equivalent to describing the monodromy group aroundthe branch points. That is, the monodromy group is generated by transposition.We also know that loops around all branch points has a product which is trivial inthe fundamental group. So, we have transpositions τi ∈ S3, we know

∏i τi = id

and 〈τi〉 is transitive. Hence, 〈τi〉 = S3, or equivalently, there are at least twoτi. Furthermore, such τ1, . . . , τ2g+4 is determined up to simultaneous conjuga-tion, which corresponds to relabelling the sheets. We can now replace degree 3 bydegree d.

Now, if the map is of degree d, we would have α1, . . . ,α2g+2d−2 branch points,and we get a similar description. Here, we get similarly transpositions τi with 〈τi〉transitive and

∏i τi = id. This is the only relation because we know the funda-

mental group π1(S2 \{α1, . . . ,α2g+2d−2

}. Note that the order of this product


is important, and depends on the paths chosen between the points. The numberof solutions to this combinatorial problem is finite, and it has been worked out.This was calculated in degree 3 by Riemann and Hurwitz, and these numbers areRiemann-Hurwitz numbers.

Question 9.3. With the same setup as in Question 9.2 here is an open question:what is the number of covers with more complicated branching locus (i.e., non-simple branching locus)?

Remark 9.4. If d = 2 in the above question, Question 9.2, there is a unique suchRiemann surface, since there is a unique collection of such τi, namely τi = (12)for all i.

Question 9.5. What if we want to look at covers of curve other than P1. In fact,very few curves appear as branched curves of higher genus curves (as can be seenby counting moduli). We can ask a similar question for covers of elliptic curves:

What curves appear as branched covers of elliptic curves?However, going around a branched point is not the only way to get from one

sheet to another. One can also move around closed loops on the torus, not ho-mologous to the identity (i.e., around each of the two loops of the torus). So, alltogether, we have to specify two more generators, of the fundamental group ofthe torus, in addition to those of the branch points. We will get transpositions forthe branch points, and then two more arbitrary permutations corresponding tothe two loops. Call these τi,µ1,µ2. We get a relation

∏i τi = [µ1,µ2], since the

complement of the µi is a rectangle, and we want to go around the boundary ofthe rectangle.

Remark 9.6. The answer to this Question 9.5 is also open, and this question evencame up in Dawei Chen’s thesis.

Question 9.7. (Algebra Problem) Given α1, . . . ,αb, where b = 2g+ 4, we want acubic polynomial in ywith coefficients in C[x] of the form y3+α2(x)y

2+α1(x)y+α0(x) with discriminant

∏i(x−αi), and here it’s not clear that we can even make

such a curve, but we can do so via Riemann surfaces quite easily, and it’s equationwill necessarily have this form.

So, a hyperelliptic curve is either

V(y2 =

2g+2∏i=1

(x−αi)) or V(y2 =

2g+1∏i=1

(x−αi))

depending on whether the curve has a branch point at ∞ or not. So, in the firstcase, we add point q, r at ∞. To describe this map, we want to write down thecanonical map, and we can do this by writing down a differential.

We can write down the differential dx. We now ask, what is its divisor? In theplane, the divisor is the sum of the ramification points. Let pi be (αi, 0) ∈ C. Then,(dx) vanishes at the pi, so

(dx) =∑i

pi − 2q− 2r

where we computed the orders at infinity by symmetry, and degree considera-tions. To obtain a holomorphic function, we may note ( 1y ) =

∑pi+(g+ 1)(q+ r).

20 AARON LANDESMAN

We can now take products and note that(dx

y

)= (g− 1)(q+ r)

When g ≥ 2, this is holomorphic. We can generate all differentials by multiplyingby multiples of x. We will get g holomorphic differentials

dx

y,x dx

y, . . . ,

xg−1 dx

y

In this case, the canonical mapφK : C→ Pg−1 is the map given by[1, x, . . . , xg−1

].

In particular, the canonical map factors through the two sheeted cover as

(9.1)C P1

Pg−1

π

φK

φ|OP1

(g−1)

So, the canonical map is 2 : 1 onto the rational normal curve.Suppose we want to embed this curve in projective space. We’ll discuss this

next time.

Exercise 9.8. Every smooth curve admits an embedding of degree g+ 3. If C ishyperelliptic of genus g ≥ 2, then the smallest degree of an embedding C→ Pr isexactly g+ 3.

Remark 9.9. In some sense, hyperelliptic curves have the most linear series onthem, but they have the fewest embeddings.

10. 9/23/15

10.1. Hyperelliptic Curves. Recall from last time that ifC is hyperelliptic of genusg, then C is given by y2 = f(x) over A1 with deg(f) = 2g+ 2.

Note of course that such a curve is affine, but such an affine curve has a uniquecompletion to a smooth projective curve, by adding in two (or one) point at infin-ity, by taking its closure in P2 and resolving the singularities.

We can explicitly see that

H0(KC) = 〈dx

y, . . . ,

xg−1 dx

y〉

In other words, the canonical map factors through the hyperelliptic map to P1.

(10.1)C Pg−1

P1

φ

π νg−1

Definition 10.1. A line bundle L or divisor D or linear system |L| is special if

h0(K−D) 6= 0


Now, if we have a special divisor on a hyperelliptic curve, that means the di-visor is contained in the preimage of a hyperplane. So, a divisor D ∈ PicC, weknow that their image under the canonical map is linearly independent unless twopoints map to the same point.

So, by geometric Riemann Roch, the number of linear relations is exactly thenumber of pairs of points in D which map to the same point. That is, this is pre-cisely the multiple of the g12 that D contains.

In conclusion ifD is special, we can write |D| = r · g12 +D0, whereD0 is a fixedbase locus. That is, the map φD given by any special D factors through π, and, inparticular, there are no special, very ample linear series.

Lemma 10.2. The smallest degree of an embedding of a curve C→ Pr is g+ r. Note thathyperelliptic curves cannot be embedding into P2. But, if r ≥ 3, a general map will be anembedding.

Proof. Proved above. �

Exercise 10.3. Conversely, it is also true that every nonhyperelliptic curve has anembedding of degree less than g+ r.

10.2. Gonal Curves.

Definition 10.4. We say C is trigonal if there exists a degree 3 cover f : C→ P1.

Definition 10.5. We say C is tetragonal, pentagonal, hexagonal if there exists adegree 4,5,6 cover f : C→ P1.

Definition 10.6. Say C is k-gonal if there exists a g1k.

Warning 10.7. Be careful, it’s often unclear whether n-gonal curves are also n− 1gonal curves. That is, it is unclear whether we allow base points.

Remark 10.8. No one knows more about trigonal curves than Anand Patel.

Definition 10.9. The moduli space of genus g curves is

Mg = { isomorphism classes of smooth projective genus g curves}

Remark 10.10. The existence of Mg wasn’t proved until 1969, even though peoplehad been working with it a century earlier. For now, we’ll pretend we know whatwe mean by this.

Question 10.11. How do we calculate the dimension of Mg?

Question 10.12. Is Mg irreducible?

Understanding abstract curves is difficult, but we can understand the situationbetter by examining another moduli space:

Definition 10.13. Define the Hurwitz space

Hd,g ={(C, f) : C ∈Mg, f : C→ P1 of degree d with simple branching

}The one thing we can see for such a curve in Hd,g is the number of branch

points.

Lemma 10.14. dimMg = 3g− 3.

22 AARON LANDESMAN

Proof. Set b = 2d + 2g − 2 and the branch divisor will consist of an unorderedb tuple of distinct points. So, we obtain a map Hd,g → Pb \∆, where the imagecorresponds to a divisor of degree b on P1. We now endow Hd,g with the structureof an algebraic variety via the covering map to Pb. We next want to utilize theprojection map Hd,g →Mg by forgetting the map. Now, the dimension of Hd,g =b = 2d+ 2g− 2.

The question we now want to ask is, what is the fiber dimension of the fiber tothe map to Mg?

We can’t answer this in general, but when d > 2g− 2. We are now asking, howmany simply branched maps f : C → P1 of degree d are there? To specify such amap, we have to choose:

(1) A line bundle L ∈ Picd(C), which is g dimensional,(2) A pair of sections σ0,σ1 ∈ H0(L), up to multiplication (which is basepoint

free), which has dimension 2(d− g+ 1).

So, the dimension of the fiber is g+ 2(d− g+ 1) − 1.We should worry that Hd,g has the correct structure for both the projection

maps, but we won’t worry for now.So, the dimension of Mg is 2d+ 2g− 2− (2d− g+ 1) = 3g− 3. �

Remark 10.15. We can also think of Pb to be polynomials of degree b, and ∆ is thezero locus of the discriminant.

Exercise 10.16. Show that unordered b tuples of points without repetition corre-spond to Pb \∆.

Guess at a Solution: This is precisely symmetric functions in (P1)b, by the fun-damental theorem of symmetric functions.

Remark 10.17. The reason this method for computing dimMg worked so well wasbecause we introduced some auxiliary information that allows us to get a handleon the curve C.

Lemma 10.18. The dimension of the space of hyperelliptic curves is H2,g, which is 2g− 1dimensional

Proof. The map H2,g → { hyperelliptic curves } have three dimensional fibers.Therefore, the set of hyperelliptic curves is 2g− 1 = 2g+ 2− 3. �

Corollary 10.19. Not all genus g curves, for g ≥ 3, are hyperelliptic.

Proof. The dimension of the space of hyperelliptic curves is less than that of Mg.�

Next time: curves of genus 5 and 6.

11. 9/25/15

Today: Canonical curves of genus 5 and 6Monday: Adjoint seriesStarting Wednesday: Castelnuovo theory, chapter 3 in ACGHLet C be a smooth projective non-hyperelliptic genus 5 curve.


11.1. Curves of genus 5. We have a canonical map φ = φK : C → P4. We firstconsider which quadrics vanish on C.

H0(OP4(2))→ H0(OC(2))

which is a map from a 15 dimensional vector space to a 12 dimensional vectorspace, so the kernel is at least 3 dimensional. That, is C lies on at least threequadrics.

There are now two possibilities:

(1) So, it might be that curves are an intersection of three quadrics. In thissituation, C does not lie on any more quadrics, by the Noether af+bg+chtheorem.

(2) It can also be that C is a strict subset of ∩iQi.Let us now make four observations:

(1) Case 1 does occur because ifQi are three general quadrics, then by Bertini,C = ∩iQi is smooth, and by adjunction, we have KC = KP4(2+ 2+ 2) =OC(1).

(2) In case 1, C is not trigonal. If C were trigonal, then there exists a g13, so thecurve contains three collinear points. But, if we have three collinear points,every quadric would contain the line through those three points. So, if Cis trigonal, then it would not be an intersection of quadrics.

(3) If C has genus 0, C is a 1-sheeted cover of P1, but in genus 1 or 2, then Cis a two-sheeted cover of P1 (in genus 1, take any degree 2 series, in genus2, take the canonical linear series). General curves of genus 3 and 4 areexpressible as 3 sheeted covers of P1: In genus 3, we can project the planequartic away from a point, that is, the linear series K− P is a g13. In genus4, the rulings of a quadric cut out the g13 on C. It turns out general curvesof genus 5 are expressible as 4 sheeted covers of P1.

Note further that in genus 1 there is a 1 dimensional family of g12, ingenus 2, there is a finite number of g12. In general, in odd genus thereis a 1-dimensional family of such maps, and in even genus there is a 0-dimensional family of such maps.

(4) How many curves are complete intersections of quadrics in P4. Suchcurves arise as an open subset ofG(3,H0(OP4(2))) which is 3 · (15−3) = 36dimensional. The fibers of the map

U ={Λ ∈ G : C = ∩Q∈ΛQ is a smooth curve in P1

} →M5

Are PGL5. We can do a similar count to show the trigonal curves form a 10dimensional subvariety. Note that 3 · g− 3+ g2 − 1 = 36, so indeed, sinceMg is irreducible, a general curve will be expressible as such a completeintersection.

Quote from Joe: “We’ll answer the question of whether C is tetragonal, or I’mnot doing my job.“ We’ll see if he gets to it.

Question 11.1. Does the second case, in which C is not equal to an intersection ofquadrics, occur?

24 AARON LANDESMAN

Note that trigonal curves exist, and so they must be as in this second case. Startwith a trigonal curve of genus 5. Given |D| a g13, let’s look at |K−D|. By RiemannRoch, this is a g25. Note that if this curve has a base point, the curve has a g24, soit would either give a two sheeted cover of P1 (impossible since C is not hyper-elliptic) or a plane quartic (impossible since g = 5). Therefore, C → C0 ⊂ P2,and the image is not smooth, since a smooth plane quintic is genus 6. Let’s nowlook at |2D|, which is a g26. To see h0(2D) = 3, note that it is at most dimension3, by Riemann Roch, and at least 3 because the map Sym2H0(D) → H0(2D) isinjective, because the curve, when embedded in P1 is surjective, so does not lie onany quadrics.

Lemma 11.2. Now, if we take K− 2D = p+ q, we claim f(p) = f(q).

Proof. Note that 2(K −D) = K + p + q, so the canonical linear series KC is cutout on C by conics Q ⊂ P2 passing through R. To see this, a conic Q ⊂ P2, thepreimage f−1(Q) in C is a divisor of the form K+ p+ q. If r ∈ V(Q), we can writef−1(Q) = D+ p+ q where D ∼ K, and D is the residual 8 points of intersection ofQwith the curve. Furthermore, we get all canonical divisors in this way becauseChas genus 5 and we have a 5 dimensional space of conics passing through a pointin the plane. �

So, we now have a very interesting description of C: We can map

(11.1)

C C0 ⊂ P2

P4

φK

π

where the map π is given by quadratic polynomials vanishing at r. Now, we havethe C lies on a surface in P4. Let S = im P2 under the right map π. Then, wehave that the curve lies on a surface. This turns out to be the projection of theVeronese ν22 surface from a point. We can also say that we can resolve the mapP2 → S by blowing up at the point r. One way to describe this surface in P4 is totake a line and a plane conic living in complementary spaces spanning P4, take anisomorphism between them, and then take the union of the lines joining points onthe line and points on the conic.

Now, lines through r meets the curve at r and three other points. This deter-mines a trigonal map. So, the lines on the ruling of S meet the curve three times,and the intersection of the quadrics is precisely the surface S.

12. 9/28/15

Today: Adjoint series - appendix A Wednesday: Castelnuovo’s theorem - Chap-ter 3 But first canonical curves of genus 5

12.1. Canonical curves of genus 5. AssumeC has genus 5 and is not hyperelliptic.Let C → P4 be a canonical map. Then, C lies on three quadrics Qi. Last time, wesaw

(1) C = ∩iQi.(2) C ⊂ ∩iQi = S, where S is a cubic scroll.


As we saw last time, the second case corresponds to that were C is trigonal, athree sheeted cover of P1.

The first case corresponds to that where C is not trigonal, but C is expressibleas a 4 sheeted cover of P1.

Lemma 12.1. Every canonically embedded curve which is not trigonal admits a map ofdegree 4 to P1.

Proof. Let P2 ∼= { quadrics C ⊂ Q}. We can ask whether there are singular quadricsin this set. Inside P14 which is the space of all quadrics in P4, there is a quintichypersurface of singular quadrics. Singular quadrics in P4 are cones over quadricsurfaces, and so the cone over a quadric has two rulings by 2 planes. Two 2-planes from opposite rulings add up to a hyperplane section on the quadric cone.Altogether they intersect the curve in 8 points, so one of them must intersect in 4or fewer points. The intersection of the 2 plane and the curve is the intersectionwith the two other quadrics, which is 4 points. So, each 2 plane meets C in fourpoints. So, we get two g14’s on the curve and they are residual to each other (addto the canonical divisor).

In the rank 3 quadric case, it is a cone over a plane conic. In this case, we get ag14 which is semicanonical.

In the trigonal case, you can write down the quadrics containing the curve, andthey are all singular.

�

Question 12.2. Why do all g14’s come from quadrics in the above proof?Answer: If we have a canonical curve with 4 points D lying in a two plane Λ ,

look at the restriction map from{quadrics in P4 containing C

} → { quadrics in Λ ⊂ P2 ⊃ D}

There is a 1 dimensional family of conics containing D, and the above map issurjective. Therefore, the 2-planes on the quadric gives us all g14’s.

Remark 12.3. The converse, showing that all curves on the Segre cubic surface aretrigonal, we’ll see this from Castelnuovo theory.

12.2. Adjoint linear series. We’ll start by talking about plane curves, followingthree steps:

(1) Smooth curves C ⊂ P2. Even when the genus is a binomial coefficients,most curves are not plane curves. E.g., smooth plane quintics have dimen-sion 12 while all curves are genus 6 are 15 dimensional.

(2) Nodal curves C → C0 ⊂ P2. Given any smooth curve, there exists abirational embedding into the plane so that the image is a nodal planecurve.

(3) Arbitrary plane singularities.

12.2.1. Smooth plane curves. Choose homogeneous coordinates [X, Y,Z] on P2 andlet L∞ be the hyperplane Z = 0, and let x = X

Z ,y = YZ be affine coordinates on

P2 \ L∞ ∼= A2.Say C ⊂ P2 is a smooth curve of degree dwith C∩A2 the zero locus of f(x,y).Let’s further assume that [0, 1, 0] /∈ C, so the projection from [0, 1, 0], π : C →

P1, (x,y) 7→ x expresses C as a d sheeted cover of P1.

26 AARON LANDESMAN

Question 12.4. How do we write down all holomorphic differentials on C?

Let’s start with the differential dx, which is holomorphic in the affine plane.However, this will have poles on L∞. Now, by pulling back D∞ = L∞ ∩C. Then,(dx) = π−1(∞) = −2D∞. We have the poles of dx now, we must ask where (dx)is 0. We want to find a function with zeros at∞ and poles at the zeros of dx. Thesezeros are where ∂f(x,y)

∂x = 0.We know df|C = 0 since f is zero along the curve. But, we also know

∂f

∂xdx+

∂f

∂ydy = df|C = 0

Since C is smooth, we know ∂f∂x , ∂f∂y have no common zeros on C.

At a point pwhere dx = 0, we have ∂f∂y (p) = 0,∂f∂x (p) 6= 0, we have

ordp dx = ordp∂f

∂y

So, we can divide by ∂f∂y without introducing any new poles.

So, we consider the differential

ω0 =dx

fy, fy =

∂f

∂y

Now, ω0 is holomorphic and nonzero in the affine plane. Now, ∂f∂y is holomor-phic of degree d− 1, so the divisor must be (ω0) = (d− 3) ·D∞.

This is a very concrete form of the adjunction formula, since KC = OC(d− 3).We have a differential with zeros at infinity. If we multiply by a function, it

must be a polynomial in the affine plane, which will be holomorphic (even at∞)as long as n ≤ d− 3. So, consider{

g(x,y) dxfy

: degg ≤ d− 3}⊂ H0(KC)

Both sides have dimension(d−12

)and so the inclusion is an equality, and every

holomorphic differential is of this form.Here is a more modern proof of the above: Equivalently, by adjunction, KC ∼=

OC(d− 3). We have an exact sequence

(12.1) 0 OP2(−3) OP2(d− 3) KC 0

and since h1(OP2(−3)) = 0, we have that the map H0(OP2(d− 3)) → H0(KC) issurjective, and in fact defines an isomorphism.

12.2.2. Nodal Curves. Now, suppose we have C → C0 ⊂ P2 with C0 nodal. Forthe moment, suppose C0 is chosen so that at the nodes, the branches of C0 do nothave vertical tangent. The problem with the previous argument for smooth curvesis based on the assumption that 0 = df|C = ∂f

∂xdx+∂f∂ydy.

When C was smooth, these partials could not simultaneously vanish. How-ever, in the singular case, the partials vanish to order 1. We’re assuming that thenodal curve has no vertical tangents by our assumption, meaning that dx,dy arenonzero. (We assume this just for simplicity, we could do the same argumentwhere they do have zeros there.)


When we consider ω0 = dxfy

, we have ω0 will have simple poles at the pointsqi, ri, are the two preimages of each of the nodes pi so we must take g(x,y)ω0when g(x,y) had degree at most d− 3 and g vanishes at qi, ri. Now, if there areδ nodes, we impose δ conditions on g, so the dimension is at least

(d−12

)− δ, but

this is also the genus of the curve. Hence, we obtain all differentials in this way,and we also see that the nodes impose independent conditions on the vanishingof g.

13. 9/30/15

13.1. Program for the remainder of the semester. Through many examples, wehave some understanding that linear series are the key to understanding the ge-ometry of curves. We phrase this in terms of three questions, in increasing orderof depth and subtlety.

(1) What linear series grd may exist on a curve of genus g?(2) What very ample or birationally very ample linear series grd may exist on

a curve of genus g?(3) What linear series grd exist on a general curve?

Answers to above questions(1) Answered by Clifford’s theorem(2) Answered mostly (in the sense that it only answers the question of bira-

tionally very ample maps) by Castelnuovo’s theorem. For very ample grd’s,we know the answer for r = 3, which is relatively recent, r = 4 might beknown, but r ≥ 5 is still open.

(3) This is answered by Brill Noether theoryWe’re quite interested in relations between curves in the abstract and curves in

projective space. We’re really interested in embeddings of the curve in projectivespace, or birational embeddings, and much less so maps to projective space whicharen’t generically injective.

13.2. Adjoint linear series.

Remark 13.1. There’s a version of the following that applies to arbitrary reducedcurves (which is quite similar to the nodal case) and is written up in appendix Ain ACGH.

Recall the set up from last time: Suppose C ν−→ C0 ⊂ P2 is the normaliza-tion map of C, and C0 is a nodal plane curve of degree d. Say C0 has nodes atp1, . . . ,pδ, and let ν−1(pi) = qi ∪ ri. Set ∆ =

∑i(qi + ri) ∈ div C.

Choose coordinates [X, Y,Z] on P2 and let L∞ = V(Z), and x = X/Z,y = Y/Zbe coordinates on A2 ∼= P2 \L∞. LetD∞ = ν∗L∞. In A2, let us say thatC0 = V(f)where f(x,y) is a polynomial of degree d. Suppose further that at each node of thecurve, the tangent lines are not vertical, so that the divisors dx,dy have simplepoles at those points.

We now have two goals(1) Find an algorithmic way of writing down all differentials on C(2) If I have a divisor on C, how do we find the complete linear series associ-

ated to that divisor, algorithmically.

28 AARON LANDESMAN

Lemma 13.2. We can algorithmically write down all regular 1 forms by

W := ν∗g(x,y) dx

fy, degg ≤ d− 3,g(pi) = 0.

Proof 1. Start with dx. This has poles along L∞. To kill the poles at∞, we considerdxfy

where fy = ∂f∂y , because wherever dx has a zero fy does as well, at a point on

the curve. Therefore, polynomials of the form

W := ν∗g(x,y) dx

fy, degg ≤ d− 3

so long as we are away from the nodes. But at the nodes dx doesn’t vanish, whilefy does vanish. For this, we need g(pi) = 0. Then, by a dimension count, theinclusion

W → H0(KC)

is an isomorphism. �

We can also present a coordinate free proof.

Proof 2. Consider the fiber product

(13.1)

C S = Blp1,...,pδP2

C0 P2.

π

In S, we have exceptional divisors Ei and E =∑Ei. Let L be the pullback of the

class of a line in P2. Then, the Ei together with L generate the Picard group of S.Now,

C ∼ dL− 2E ∈ Pic(S)KS ∼ −3L+ E

By adjunction,

KC = (C+KS)|C = (d− 3)L− E|C = (d− 3)D∞ −∆

We can now show we get all differentials in the modern language of sheaves byusing the following exact sequence:(13.2)

0 IC((d− 3)L− E) ∼= O(−3L+ E) = KS OS((d− 3)L− E) OC((d− 3)L− E) ∼= KC 0

And we now use that h1(KS) = h1(OS) = 0 by Kodaira Serre duality, and so theinduced map on global sections is surjective. �

An additional fact is that

H0(OS(mL− E))→ H0(OC(mD∞ −∆))

is surjective for allm, and this is referred to as the completeness of adjoint series.Now, suppose we are given a divisor on C. Suppose we’re given a divisor B on

C of degree b. Assume that SuppB∩∆ = ∅, for notational convenience.

Question 13.3. Describe the complete linear series |B|.


Solution: Choose any plane curveG of degreem so that p1, . . . ,pδ ∈ G and B ⊂G. Now, consider ν∗G = B+∆+ R. Now, the divisor R has degree md− b− 2δ.Consider all plane curves of degree m = degG, with p1, . . . ,pδ ∈ H,R ⊂ H. Forany such H, we have ν∗H = ∆+ R+ S. Here R ∼ m ·D∞ −B−∆ and so

S ∼ mD∞ −∆− R ∼ B

So, this shows us how to find the complete linear series of B. If B = B+ − B−

a difference of effective divisors, we can choose G to contain B+ and H to containB−, and we still obtain the desired answer.

Remark 13.4. (History of genus) What was the genus to the ancients? Peoplelooked at lines and conics, and if you have any d points on a line, those points arethe intersections of a line and a curve of degree d. Any 2d points on a conic, the 2dpoints on a conic are in general the intersection of a plane curve of degree d and aconic. However, it is not true that 3d points on a cubic lie on the intersection witha curve of genus g. Rather, we will only find 3d− δ, where δ is some deficiency.Now, people defined genus as this deficiency. The idea that for each node δ, weget a 2 for 1 deal, in the sense that passing through a node drops the degree ofthe residual intersection by 2 but only imposes 1 linear condition, so the originaldefinition of genus was (

d− 1

2

)− δ

14. 10/2/15

Weierstrass pointsLet C be a compact Riemann surface of genus g ≥ 2.Application: Prove that automorphism group of such a Riemann surface is fi-

nite.


Sp = {− ordp(f) : f holomorphic on C \ p} .

This is a semigroup, and is called the Weierstrass semigroup of p ∈ C. The gapsequence Gp is N \ Sp. That is, the gap sequence is the order of poles that don’toccur.

Lemma 14.2. We have #Gp = g.

Proof. Observe

Sp ={m : h0(mp) > h0((m− 1)p)

}Gp =

{m : h0(mp) = h0((m− 1)p)

}

Now, let’s compare h0(mp) and h0(K−mp). By Riemann Roch, as we increasemby 1, precisely one of these two quantities changes. �

Lemma 14.3. For general p ∈ C, we have

Gp = {1, . . . ,g}Sp = {0,g+ 1,g+ 2, . . .}

Proof. Follows from Proposition 14.8. �

30 AARON LANDESMAN

TABLE 1. Chart ofm and gap sequence

m h0(mp) $hˆ0(K-mp)0 1 g1 1 g-12 1 or 2 g-2 or g-1...

......

m m-g+1 0

Proposition 14.4. There exists a finite set of points with a function with a pole of orderless than g+ 1 at p, and no other poles.

Proof. Follows from Proposition 14.8. �

Definition 14.5. A point p ∈ C is a Weierstrass point if the gap sequence is notGp = {1, . . . ,g}.

Definition 14.6. The weight of p ∈ C is the sum w(p) :=∑gi=1(ai − i).

Example 14.7. (1) For example, if the weight is 0, the cap sequence is 1, . . . ,g,and the point is not a Weierstrass point.

(2) If w = 1, then Gp = {1, . . . ,g− 1,g+ 1}(3) ifw = 2we haveGp = {1, . . . ,g− 1,g+ 2} or {1, . . . ,g− 2,g,g+ 1}, and the

former cannot happen in genus 3, because it is not a semigroup.

Proposition 14.8. For any C, we have∑p∈C

w(p) = g3 − g

Example 14.9. (1) If g = 2we haveφKC→ P1 of degree 2, and it has 6 branchpoints, which are the 6 Weierstrass points, each with weight 1.

(2) In g = 3, if C is hyperelliptic, we have 8 branched points of ΦK where thesemigroup is necessarily

Sp = {0, 2, 4, 6, 7, 8, 9, . . . , }Gp = {1, 3, 5}

and w = 3, which sums to 24 as claimed.(3) If g = 3 and C is not hyperelliptic. In this case, having a Weierstrass point,

we are precisely asking when h0(3p) > 1. Then, there is a line in planecontaining the divisor 3p, so we are asking for flex points of C. That is,by Geometric Riemann-Roch, the divisor 3p ⊂ C lies in a line, and p is aflex point of C ⊂ P2. The set of flex points are the points of intersectionwith the Hessian, and so the Hessian is a sextic plane curve, which meetsC of degree 4. Further, the intersection multiplicity of the point with theHessian is precisely the weight of the point. Overall, we get weight 4 · 6 =24, as desired.

Proposition 14.8. Let p ∈ C, and choose a local coordinate z around p. Choose abasis ω1, . . . ,ωg ⊂ H0(K). Write ωi = fi(z) dz. Being a Weierstrass point means


there is a differential vanishing to order g at p. We look at the Wronskian

∆ = det

f1 · · · fgf ′1 · · · f ′g...

. . ....

fg−11 · · · f

g−1g

This vanishes at p precisely when p is a Weierstrass point: To see this, note that

h0(gp) > 1 if and only if h0(K− gp) ≥ 1, which is the same as the existence of aholomorphic differential which vanishes to order g at p. So, Weierstrass points arezeros of ∆.

Next, we claim ∆ is a section of a power of the cotangent sheaf.Now, if we choose a different local coordinate, we want to see how the Wron-

skian changes over overlaps. Chose a different local coordinatew. The first row of

∆ is multiplied by ∂z∂w . The second row is multiplied by is multiplied by

(∂z∂w

)2,

plus some multiple of the first row. Hence, repeating this, the determinant is mul-

tiplied by(∂z∂w

)(g+12 ). Now, note that these are the transition functions for a power

of the cotangent bundle. Hence, ∆ ∈ Γ(K⊗(g+12 )). Hence, the number of zeros of ∆

is the degree of this line bundle. So, this is degree (2g− 2)(g+12

)= g3 − g.

It still remains to show the order of the Weierstrass function is equal to theweight of the Wronskian. This is left as an exercise, as is written in ACGH and3264, chapter 7. �

Remark 14.10. We don’t yet know that the Wronskian determinant is nonzero.However, in characteristic p, the Wronskian determinant actually can be 0. How-ever, this doesn’t happen in characteristic 0.

Remark 14.11. The maximum possible weight of a Weierstrass point correspondsto the semigroup which starts increasing as soon as possible. That is, Sp = {0, 2, 4, 6, . . . , 2g, 2g+ 1, 2g+ 2, . . .}which has gap sequenceGp = {1, 3, 5, 7, . . . , 2g− 1} which has weight

(g2

). This oc-

curs if and only if the curve is hyperelliptic. We further see there are 2g+ 2 suchpoints, and so these account for all Weierstrass points

Corollary 14.12. There are at least 2g+ 2 Weierstrass points, and exactly 2g+ 2 if andonly if C is hyperelliptic.

Proof. This is immediate from Remark 14.11. �

Question 14.13. Here are some open problems on Weierstrass points.(1) Which semigroups of Weierstrass points occur? For a long time it was

conjectured that every gap sequence occurs. However, Bookveits exhibiteda semigroup which cannot occur.

(2) Many of the deeper questions come from a variational point of view: Lookat Mg,1, the set of pointed curves, up to isomorphism. We have a stratifi-cation of this space, by associating to a point its Weierstrass gap sequence.We know that Weierstrass points occur in codimension 1. We can then ask,for a given semigroup, what is the codimension of the locus of points withthat gap sequence? We can ask if it is irreducible.

32 AARON LANDESMAN

Theorem 14.14. If C is a compact Riemann surface of genus g ≥ 2, then Aut(C) isfinite.

Proof. When C is hyperelliptic, we have an ad hoc argument: We can express C upto automorphism of P2, as a cover of P2 of order 2. Now, the automorphisms ofP1 fixing a set of 6 or more points is finite, so we’ll just cover the non-hyperellipticcase.

Now, we have strictly more than 2g+ 2Weierstrass points.

Lemma 14.15. If φ : C → C is any automorphism, the number of fixed points of φ is atmost 2g+ 2.

Proof. By Riemann Roch, we can find a meromorphic function f on C of degreeg+ 1 with g+ 1 poles and zeros. Consider f− φ∗f. This has poles where f haspoles or the pullback of f has poles. Hence, this difference has at most 2g + 2poles, hence at most 2g+ 2 zeros, which is precisely the number of fixed points.Therefore, the number of fixed points of φ is at most 2g+ 2. �

This suffices, because any automorphism fixing all the Weierstrass points mustbe the identity. �

15. 10/5/15

Here are some questions for the remainder of the course?(1) What linear series grd may exist on a curve of genus g?(2) What very ample or birationally very ample grd’s may exist? That is, for

which g, r,d does there exist(a) A smooth irreducible nondegenerate C ⊂ Pr of genus g and degree d(b) An irreducible nondegenerate C ⊂ Pr of degree d and geometric

genus g?(3) What grd’s exist on a general curve of genus g?

The problem of which very ample grd’s exist on smooth curves is still open, andthe most recent progress was made by a student of Harris in the last 10 years.

Riemann Roch says that grd is determined once 0 ≤ d ≤ 2g. So, the grd’s are onlyinteresting in the range 0 ≤ d ≤ 2g, at the parallelogram between d = 0,g = 0 andd = 2g− 1, r = g− 1. Then, Clifford’s theorem tells us all grd’s lie in the lower halfof this rectangle, that is, below th e line 2r = d.

Theorem 15.1. SayD has degree dwith 0 ≤ d ≤ 2g− 2. Then, r(D) ≤ d2 with equality

if and only if(1) d = r = 0,D = 0(2) d = 2g− 2, r = g− 1,D = K(3) C is hyperelliptic ad D = mg21.

Proof. Look at the multiplication map H0(D)⊗H0(K−D) → H0(K). Now, thinkof these in terms of projective spaces instead of vector spaces. Think of H0(D)

as a family of divisors parameterized by Pr, and H0(K−D) as parameterized byPg−d+r−1. Hence, we obtain

|D|× |K−D|→ |K|

Pr ×Pg−d+r−1 → Pg−1


This map is finite, because we’re asking how many ways we can express a canon-ical divisor as a sum of effective divisors in |D| and |K−D|. Hence, the dimensionof the source is at most the dimension of the target. Therefore, r+ g− d+ r− 1.So,

r+ g− d+ r− 1 ≤ g− 1implying r ≤ d/2. The equality will follow from our upcoming discussion ofCastelnuovo’s theorem. The strong form will be a way of saying when the multi-plication map is surjective (which will occur if and only if the curve is hyperellip-tic). �

Remark 15.2. Take C hyperelliptic and take D = r · g12 +D0 and D0 to be generalof degree d− 2r, so that these points are basepoints, and we can get all grd’s onhyperelliptic curves.

15.1. Castelnuovo’s Theorem. Assume C ⊂ Pr is a smooth irreducible, nonde-generate curve of degree d and genus g. First, observe that given r,d, we havethat g is bounded. Castelnuovo asks what the largest possible genus is. We won’tanswer which g occur. We’ll just find the largest.

Theorem 15.3. (This theorem will be stated next time, but it’s essentially a bound on thegenus as discussed above)

Proof idea. Here is the idea:The plan is to bound from below the dimensions of h0(OC(m)). We will do this

by bounding from below

h0(OC(m)) − h0(OC(m− 1)).

Whenm is large enough, we will have h0(OC(m)) = md−g+ 1, and by Riemann-Roch, we will know exactly what this dimension is. This will give us a bound fromabove on the genus.

We now carry out the proof.

Proof. Let Γ be a general hyperplane section of C ⊂ Pr. Let Γ be a general hyper-plane section of C ⊂ Pr. Consider

(15.1) 0 OC(m− 1) OC(m) OΓ (m) 0

Then,

(15.2) 0 H0(OC(m− 1)) H0(OC(m)) H0(OΓ (m))ρ

Therefore, H0(OC(m)) − H0(OC(m − 1)) = rk ρ, where rk ρ is colloquially thenumber of conditions imposed by Γ on H0(OC(m)). Hence, rk ρ is at least thenumber of conditions imposed by Γ imposed by H0(OPr(m)). as given by thecomposition of restrictions

(15.3) H0(OPr(m)) H0(OC(m)) H0(OΓ (m))

Say Γ imposes at least h conditions on H0(OC(m)). This is equivalent to havingp1, . . . ,ph ∈ Γ so that for all i, there is a σ ∈ H0(OC(m)) with σ(pj) = 0, j 6=i,σ(pi) 6= 0. Now, if we take a subspace im H0(OPr(m)) ⊂ H0(OC(m)).

34 AARON LANDESMAN

Now, we’ve moved away from the curve, and the question is:

Question 15.4. How many conditions do the d points p1, . . . ,pd ∈ Γ impose onhypersurfaces of degreem?

Now, we have Γ ⊂ H ∼= Pr−1 ∼= Pn, where n = r− 1. This consists of d distinctpoints.

So, the number of conditions imposed by Γ on H0(OPn(m)) is hΓ (m). wherehΓ (m) is the Hilbert function of Γ ⊂ Pn. We know naively that hΓ (m) ≥ min(d,m−1). Furthermore, this is the best we can do, because if all points of Γ are collinear,we have an equality.

We now want to say something about the geometry of a general hyperplanesection of a curve. We then obtain precisely the bound that the genus is at most(g−12

). This happens precisely when the curve we start with is a curve in P2 The

question that remains is what conditions we obtain on these lines when the curveis in Pt with t > 2. We then claim that for any curve in P2, a general hyperplanesection will not contain three collinear points.

The proof will be discussed next time. �

16. 10/7/15

Recall from last time that C → Pr be irreducible nondegenerate degree d andsmooth.

Remark 16.1. We assume C is smooth purely for notational convenience, but thefollowing will generalize to singular curves, and in general we can assume C ν−→C0 ⊂ Pr is a birational map and C0 may be singular.

Let Γ = C∩H be a general hyperplane section, intersecting C in d points.The key inequality

h0(OC(m)) − h0(OC(m− 1)) ≥The number of conditions imposed by Γ on hypersurfaces of degreem in H ∼= Pr−1

= hΓ (m)

where hΓ (m) is the Hilbert function, not the Hilbert polynomial, although for largem the two coincide and are equal to d.

So, we would like to bound hΓ (m) from below. From this, we deduce a lowerbound on h0(OC(m)), and so for m large, we will obtain from Riemann Roch,h0(OC(m)) = md− g+ 1, which yields an upper bound on the genus.

Question 16.2. In general, say Γ ⊂ Pn, with n = r− 1, is a collection of d points.What can we say about hΓ ?

Lemma 16.3. If the points are all in a line, the points impose min(m+ 1,d), which isachieved when Γ consists of d collinear points.

Proof. Let’s cover the case when d ≥ m+ 1. For any k ≤ m, we can find a hyper-surface of degree m containing p1, . . . ,pk ∈ Γ , but no other points of Γ . For eachpoint, we can choose a hyperplane containing 1 points and none of the others. So,intersecting these planes, we get a codimension k hyperplane. If the points arecolinear, we obtain equality. �


Remark 16.4. We have a converse to the above lemma, which says that if hΓ (m) =m+ 1 then the points must all be contained in a line. We can see this by choosingone of the hyperplanes to contain two points but no others.

Remark 16.5. If we plug in the minimum Hilbert function into our strategy, weobtain g =

(d−12

), which occurs precisely for plane curves.

So, if r ≥ 3 a general hyperplane section will not lie in a line, and so hΓ (m) willbe bigger than min(m+ 1,d). We now want to say something about the geometryof the general hyperplane section of an smooth nondegenerate curve.

Definition 16.6. A collection of d points in Pr are in linear general position if nosubset of k ≤ r points are linearly dependent.

Lemma 16.7. Under the above construction, Γ consists of d points in linear general posi-tion.

In the case of curves in P3, we want to show that there is a two parameter familyof hyperplanes containing three collinear points. Also, there’s only a two dimen-sional family of secant lines. So, we’re asking whether all secant lines to curvescan be three secant lines. This cannot happen because we know that curves canbe embedded into P2 with only nodes. The statement of the lemma is, however,much more general.

Remark 16.8. The proof we’ll see doesn’t work in characteristic p, and Joe isn’tsure whether it’s known whether this is true in characteristic p. In particular, theuniform position lemma is not true in characteristic p.

Proof. We have C ⊂ Pr. Let’s now introduce the family of hyperplane sections,and throw away all hyperplanes which are tangent to the curve, at least for now.Define

U = { hyperplanes H transverse to C } ⊂ (Pr)∨

Introduce the incidence correspondence

(16.1)

Φ Φ = {(H,p) ∈ U×C : p ∈ C∩H} C

(Pr)∨ U

α

β

Note that α is a d sheeted covering space, and is, in fact, a finite proper etale. It’sunramified because we removed the hyperplanes which are tangent. Above, Φ isthe closure ofΦ, containing hyperplanes with multiple points. It’s still flat, thoughnot necessarily unramified.

The next idea is to introduce the monodromy group. Fix a base hyperplane H0,as we vary the hyperplane, avoiding the tangent hyperplane, we can unambigu-ously follow the points. Hence, we obtain a map

π1(U,H0)→ Sd

Where Sd corresponds to permutations of H0 ∩C.

Question 16.9. What is the image G ⊂ Sd of π1(U)?

36 AARON LANDESMAN

Lemma 16.10. (Uniform Position Lemma) G = Sd.

Remark 16.11. We can now describe algebraically what this means for G = Sd.Look at the incidence correspondence

Φk = {(H,p1, . . . ,pk) : pi is distinct in H∩C}

We are claiming that Φk is irreducible. Strictly speaking, this only shows it isconnected, but since it’s also smooth, being a covering space of a smooth scheme,it is smooth, hence irreducible.

Proof. We’ll come back to this next time. �

Now, we can complete the proof of the main lemma of linear general position.We’ll just describe this in the case of curves in P3. The question is, can a generalhyperplane section of the curve contain 3 collinear points. Suppose every sectionhad three collinear points. We can vary the points so that they remain collinear.

More formally, let Φ ′k be the set

Φ ′k = {(H,p1, . . . ,pk) : p1, . . . ,pk are linearly dependent }

This is a closed subset. So, Φ ′k 6=⊂ Φk, implying dimΦ ′k < dimΦk = r, wherehere we are crucially using that the curve is nondegenerate. So the projection mapΦ ′k → U cannot dominate.

�

We can now deduce a better lower bound on hΓ .

17. 10/9/15

Let C ⊂ Pr be smooth nondegenerate curve of degree d and genus d.Let Γ = H∩C be a general hyperplane section. Today, we’ll

(1) prove the uniform position lemma(2) Use this to prove a lower bound on hΓ(3) Use this to derive an upper bound on g.

Lemma 17.1. (Uniform position lemma) Define

Φ ={(H,p) ∈ (Pr)∨ ×C : p ∈ H∩C

}Let Φ ⊂ Φ be the subset mapping to hyperplanes U ⊂ (Pr)∨ intersecting C in distinctpoints. Then, Φ → U is a d sheeted covering space. Fix a base point H0 ∈ U. We obtaina map

m : π1(U,H0)→ Aut(H0 ∩C)

Then,m is surjective, and so G = im (m) = Sd.

Proof. We will show(1) G is twice transitive(2) G contains a transposition

If G satisfies these two properties, it is the symmetric group, because it then con-tains all transpositions as given by conjugating the transposition by the twice tran-sitive action. We now show the above claims


(1) Define

Φ2 = {(H,p,q) ∈ U×C×C : p 6= q,p,q ∈ H∩C}

Consider the projection Φ2 → C×C \∆. We then see that all fibers of thismap are all irreducible open subsets of Pr−2 of the same dimension andhenceΦ2 is irreducible. This means G is twice transitive.

(2) Now, we want to show G contains a transposition. Choose a hyperplaneH0 which is simply tangent, so H0 ∈ (Pr)∨ \U, so

H0 ∩C = {p1, . . . ,pd−1}

and H0 is tangent to C at p1. Look at π : Φ → (Pr)∨ over an analytic(etale) neighborhood of H0 ∈ (Pr)∨, call it V . Note that at this point, themap is not a covering map at this point.

Remark 17.2. The existence of such a hyperplane depends on k havingcharacteristic 0. It is possible in characteristic p that every point is a flexpoint. (Or, every tangent line is a bitangent) This argument fails in char-acteristic p for this reason, and in this case the monodromy group will becontained in the alternating group, giving a counterexample to this lemmain characteristic p.

Now, when atH0, we have simple branching of π at the points p2, . . . ,pd−1and ramification of order 2 at p1. Then, p2, . . . ,pd−1 have unique analyticcontinuations q2, . . . ,qd−1, and there will be two points q,q ′ in a neigh-borhood of p1. Next, note that Φ is smooth. The fibers of the projectionmap Φ → C all have fibers isomorphic to Pr−1. Therefore, Φ is smooth.So, the preimage of U is still integral, because Φ → U is a covering map.Hence, we can draw an arc contained in Φ the analytic open set. Then, Φis smooth at (H0,p1), and soΦ∩ V1, where V1 corresponds to the compo-nent containing a neighborhood of (H0,p1), is still integral. In particular,we can join (H,q) and (H,q ′), which induces a transposition. The imageof this transposition γ in U is an element of π1(U,H0) which induces atransposition on H∩C.

�

Remark 17.3. The reference for this proof is Galois groups of enumerative prob-lems, duke math journal, around 1979.

Question 17.4. Clearly much of the above discussion was not special to smoothcurves. In particular, it generalizes to generically smooth curves. Does it general-ize even further?

So, we obtain from the uniform position lemma, proven last time, that for ageneric Γ hyperplane section of C ∈ Rr with r ≥ 3, the points of Γ will be ingeneral linear position.

Question 17.5. What sort of lower bound on hΓ can we deduce from the uniformposition lemma?

The idea will be to take a hyperplane containing 2 of the point, and if the pointsare in general position, it won’t contain a third point.

38 AARON LANDESMAN

Lemma 17.6. If Γ is a collection of d point in Pn in linear general position, then hΓ (m) ≥min(d,mn+ 1).

Proof. Take the case d ≥ mn+ 1. We claim, there exists a hypersurface X ⊂ Pn ofdegreem so that X contains p1, . . . , p̂i, . . . ,pmn+1 but not pi. This says the pointsimpose independent conditions, so the Hilbert function is at least mn+ 1. Now,group the points p1, . . . , p̂i, . . . ,pmn+1 intom sets Γα of n points. Take X = ∪αΓα.This does not contain pi because the points are in general linear position. �

Remark 17.7. Is the bound hΓ (m) ≥ min(d,mn+ 1) sharp? You might guess wecould do better if we use quadrics or cubics. However, this is in fact sharp.

Exercise 17.8. Find an example of a configuration of points Γ ⊂ Pn in linear gen-eral position for which hΓ (m) ≥ min(d,mn+ 1) is an equality.

Question 17.9. What are the curves which achieve the above upper bound onhyperplane sections.

18. 10/14/15

Recall that we have C → Pr a degree d genus g smooth nondegenerate curve.Let Γ = H∩C be a general hyperplane section. The basic inequality is

h0(OC(m)) − h0(OC(m− 1)) ≥ hΓ (m)

The uniform position lemma tells us that if Γ is in linear general position, then

hΓ (m) ≥ min(d,m(r− 1) + 1)

The basic lemma for showing this (letting r− 1 = n) then

Lemma 18.1. If Γ ⊂ Pn is a collection of d points in linear general position, then theinequality

hΓ (m) ≥ min(d,m(r− 1) + 1)

is sharp, and equality is achieved when Γ ⊂ C for C a rational normal curve.

Proof. So, once a hypersurface of degree m contains mn + 1 points of a rationalnormal curve, it will contain the whole curve by Bezout’s theorem. �

Let’s now compute the bound on the genus. Assuming d ≥ 2r− 1, we have

h0(OC) = 1

h0(OC(1)) ≥ r+ 1h0(OC(2)) ≥ 3rh0(OC(3)) ≥ 6r− 2

...

This continues until we obtain d < m(r− 1) + 1. Set

m0 = bd− 1r− 1

c

so writem0(r− 1) + 1+ εwith 0 ≤ ε ≤ r− 2.


Then,

h0(OC) = 1

h0(OC(1)) ≥ r+ 1h0(OC(2)) ≥ 3rh0(OC(3)) ≥ 6r− 2

...

h0(OC(m0)) ≥(m0 + 1

2

)(r− 1) +m0 + 1

...

h0(OC(m0 + k)) ≥(m0 + 1

2

)(r− 1) +m0 + 1+ kd

Applying Riemann Roch, for k large, we have

(m0 + k)d− g+ 1 = h0(OC(m0 + k) ≥

(m0 + 1

2

)(r− 1) +m0 + 1+ kd

Definition 18.2. A divisor D is special if H1(C,D) 6= 0, and is nonspecial other-wise.

Remark 18.3. We could have stopped at m0, since we can show h0(OC(m0)) isnonspecial. In general, this is the best bound we can give onm so that h0(OC(m))is nonspecial, although it’s an interesting question to ask what the least m is inparticular cases.

Cancelling kd, we have

m0d− g+ 1 = h0(OC(m0 + k) ≥

(m0 + 1

2

)(r− 1) +m0 + 1

So,

g ≤ m0(m0(r− 1) + 1+ ε) −(m0 + 1

2

)(r− 1) −m0

=

(m02

)(r− 1) + εm0


π(d, r) =(m02

)(r− 1) + εm0

where again, d = m0(r− 1) + ε, with 0 ≤ ε ≤ r.

Theorem 18.5. IfC is a curve of genus g and degree d in Pr, for r ≥ 3, then g ≤ π(r,d).

Proof. Completed above. �

The next step is to show there are curves of the maximal genus, called Casteln-uovo curves, and that they are particularly nice.

40 AARON LANDESMAN

TABLE 2. The function π(d, r)

d π m0r 0 1r+1 1 1r+2 2 1...

......

2r-2 r-2 12r-1 r-1 12r r+1 22r+1 r+3 2...

......

Example 18.6. Take r = 3.

(1) Take d = 2k+ 2,m0 = k, ε = 1. So, π(d, 3) =(k2

)+k = k2. This is precisely

the genus of a curve of type (k+ 1,k+ 1) on a quadric surface.Note, it makes sense that the curves achieving a bound can be found on

a quadric surface (the only way we can find the sharp equality on hΓ (m))is when the curve is a rational normal curve, i.e., is a conic in P2.

(2) Take d = 2d+ 1,m0 = k, ε = 0. Then, π(d, 3) = k(k− 1) which is the genusof a curve of type (k,k+ 1) on Q.

Let’s make a table for π(d, r).Asymptotically, we have

π(3, r) ∼d2

2r− 2

We see that there are curves of degree 2r in Pr+1 are canonical curves. We can alsosee that up to d = 2r, these bounds are a consequence of Clifford’s theorem.

We have g ≤ π(d, r). Equality implies two things. First, in order to have equal-ity, we must have equality of

hC(m) − hC(m− 1) ≤ h0(OC(m)) − h0(OC(m− 1)) ≥ hΓ (m)

throughout. That is, both inequalities must be equalities. So, we must have that Cis projectively normal In other words,

(18.1) H0(OPr(m)) H0(OC(m)) 0.

Remark 18.7. The word normal in projectively normal is intended. The surjectiv-ity condition is equivalent to saying that the cone over C in Pr+1 has a normalpoint at the vertex, assuming that C is smooth. If C is not smooth, the correspond-ing condition is called arithmetically Cohen Macaulay.

Corollary 18.8. (Noether’s theorem) If C ⊂ Pg−1 is canonical (meaning C is not hyper-elliptic), then the map

Sym•H0(K)→ ⊕H0(K⊗m)

is surjective.


Proof. This follows from C being extremal and hence the above maps

(18.2) H0(OPr(m)) H0(OC(m))

are surjective, which are the same as the maps

Sym•H0(K)→ ⊕H0(K⊗m).

�

A second consequence is that

hΓ (m) = min(d,m(r− 1) + 1)

Then, with one restriction, every linear combination of points with this minimalHilbert function hΓ = min(d,mn + 1) and d ≥ 2r − 1, then Γ lies on a rationalnormal curve B ⊂ Pr−1. In this case, B is the intersection of the quadrics Q ⊂Pr−1 containing Γ .

Now, by our homework, because the curve is projectively normal, (in partic-ular, linearly normal), then every quadric containing the hyperplane section isthe restriction of a quadric containing the curve. In this case, we can look at∩C⊂Q⊂PrQ. This must be a surface whose general hyperplane section is a rationalnormal curve.

Question 18.9. What are the surfaces in Pr whose hyperplane section is a rationalnormal curve?

19. 10/16/15

19.1. Review. Previously, we took C ⊂ Pr irreducible, nondegenerate of degreed and genus g, and found g ≤ π(d, r). We wondered whether this was sharp, andwhether we can characterize the curve achieving this maximum possible genus.

Lemma 19.1. (Castelnuovo’s Lemma) Let Γ ⊂ Pn be a configuration of d ≥ 2n + 3points in linear general position. Then, if hΓ (2) = 2n + 1, then Γ lies on a rationalnormal curve.

Proof. We may or may not prove this next week. �

Now, suppose Γ ⊂ Pn be of degree d in linear general position. Then, hΓ (m) ≥min(d,mn+ 1), as we saw before. So, the intersection of the quadrics containinga curve of maximum genus, so

g(C) = π(d, r)d ≥ 2r+ 1

Remark 19.2. Caution: The following does not hold when d = 2r.

then, ∩Q⊃CQ is a surface and S∩H is a rational normal curve.Today, we’ll talk about something which may seem to have little to do with the

above.

42 AARON LANDESMAN

19.2. Minimal Varieties. Say X ⊂ Pn is an irreducible, nondegenerate variety ofdimension k.

Question 19.3. What is the smallest possible degree of such a variety of such anX?

Answer: It’s not hard to obtain a lower bound. Suppose n ≥ 2. A generalhyperplane section is going to again be nondegenerate. So, if we continue cuttingdown the dimension, we will obtain an irreducible nondegenerate set of points:Say dimX = k. Observe X ∩H1 ∩ · · · ∩Hk. We then have a nondegenerate set ofpoints in Pn−k, and it takes n− k+ 1 points to span Pn−k Therefore, the degreeof Xmust be at least n− k+ 1.

Remark 19.4. We can also see the above by intersecting with only k − 1 hyper-planes and get a rational normal curve (also of degree n− k+ 1).

In some sense, minimal varieties are the simplest, smallest varieties.

Remark 19.5. There are a lot of interesting open problems about these minimalvarieties.

Say k = 2, so we’re looking for a surface of minimum possible degree. We willnow give a construction: In Pn, choose complimentary linear subspaces. Pk, Pl ⊂Pn, meaning that we we think of these as corresponding to V ,W ⊂ U, then V ⊕W = U. Caution, the k for Pk has nothing to do with k = dimX = 2.

Now, choose a rational normal curve in Pk and Pl, meaning a map

φ : P1 → Pk

φ ′ : P1 → Pl

where the images are C,C ′. We’re assuming k+ l = n− 1. Now, take X = Xk,l =

∪t∈P1φ(t),φ ′(t). This is an irreducible ruled surface. It’s nondegenerate becauseit contains the two rational normal curves.

Lemma 19.6. degX = n− 1.

Proof. To calculate degX, we can equivalently find degX∩H, where H is a hyper-plane containing Pk. By Bertini, for a general hyperplane section, X∩H is reduced.Set theoretically, the intersection has to be a curve. Now, the surface is a union oflines, all of which contain C. If we have any point not on C, then we contain thecomplete line. Therefore, the intersection consists of the union ofC and a collectionof lines of this ruling. The number of such lines is the number of points it meets C ′

in. It will meet C ′ in l points, because degC ′ = l. Therefore, X ∩H = C ∪li=1 Li,where Li are lines. So, degX = degX∩H = degC+ l = k+ l = n− 1. �

Remark 19.7. Here, we’re implicitly assuming k, l > 0, as if either were 0 wewould get a cone over a rational normal curve. When we do a calculation, we canthink of X as a curve in the grassmannian of lines. Hence, a map P1 → G(1,n).We can take X to be the preimage of a curve in the grassmannian. This will maponto a cone collapsing a curve to the vertex in the cone.

We can also show a general such surface, when k, l > 0 is the blow up of thiscone.

If k, l > 0, then this surface X is smooth, as can be seen in several ways. Forexample, we can calculate the tangent space at a given point.


These surfaces are called rational normal surface scrolls.

Example 19.8. (1) In P3, we can take X1,1 ⊂ P3. Choose two lines, and drawthe lines joining the points. This is a quadric surface.

(2) Consider X2,1 ⊂ P4. This is the union of lines joining a line in P4 and aconic in a plane.

(3) In P5, there are X3,1 and X2,2. It’s not obvious these are different, but, infact, they are, and they have different Hilbert polynomials.

Fact 19.9. (1) Abstractly, Xa,b ∼= Fa−b ∼= P(OP1(a)⊕OP1(b)).(2) We have a map P1 → G(1,n), so that the image is a rational normal curve

of degree n− 1 in G(1,n).(3) Take general linear forms Li,Mi to on Pn. If we look at the locus where

X =

{rk(L1 L2 · · · Ln−1M1 M2 · · · Mn−1

)< 1

}(4) We can now generalize this to any dimension k. Then, if we choose Pa1 , . . . , Pak ⊂

Pn to be complementary, so that∑k1 ai = n− k+ 1. Then, choose maps

parameterizing rational normal curves φi : P1 → Pai and set Xa1,...,ak =

∪t∈P1φ1(t), . . . ,φk(t). The first example is X1,1,1 ∈ P5. That is, we choosethree spanning lines in P5, and we draw the 2 plane spanned by triples ofpoints in P5. This three fold will have the minimum degree 3. Abstractly,this turns out to be the image of P1 × P2 → P5 under the Segre embed-ding, but we haven’t checked its a product.

Definition 19.10. A variety X ⊂ Pn is a minimal variety if X is irreducible, non-degenerate of degree n− dimX+ 1.

Theorem 19.11. If X in Pn is any minimal variety is either(1) X is a scroll(2) X is a quadric hypersurface (redundant in the case of surfaces).(3) X is a cone over the Veronese surface ν2(P2) of degree 4 in P5.

Proof. Not given. �

Another question we can ask about minimal surfaces is:

Question 19.12. What is the smallest possible Hilbert function? (In other words,how many hypersurfaces of a given degree can contain such a variety?)

Now, we can come back to the case of surfaces. For now, let’s omit the Veronesesurface. So, we’re just trying to find curves on a scroll. We would like to see ifwe can find a divisor class (i.e. curve) on a minimal surface whose genus achievesCastelnuovo’s bound.

Let X = Xk,l be a surface scroll. We will assume X is smooth. X is a P1 bundleover P1. So, the Picard group is generated by two elements, one which is a line ofthe ruling, and take the other to be a hyperplane class (which must be independentbecause it has nonzero intersection with the line class). So, let h be a hyperplanesection, f be a fiber. Then,

Pic(X) = Z〈h, f〉Writing out the intersection pairing, we haveThe next step is to find the canonical class:

44 AARON LANDESMAN

h fh n-1 1f 1 0

h fh n-1 1f 1 0

Lemma 19.13. KX ∼ −2h+ (n− 3)f.

Proof. Perhaps we’ll see next time. �

20. 10/19/15

Today and Wednesday, we’ll discuss outgrowths of Castelnuovo theory, and onFriday, we’ll start with Brill Noether Theory.

Recall the lemma from last time:

Lemma 20.1. Say Γ ⊂ Pn and d ≥ 2n + 3 points in linear general position. LethΓ (2) = 2n+ 1. Then, Γ is contained in a rational normal curve C in Pn.

Remark 20.2. The generalizations of Lemma 20.1 are what leads to further inves-tigation of the possible degree genus pairs of curves in Pr.

We know that ifC ⊂ Pr is irreducible smooth nondegenerate of degree d, genusg, we have g ≤ π(d, r).

Theorem 20.3. If C ⊂ Pr is irreducible, nondegenerate of degree d and genus g. Then,g ≤ π(d, r). If equality holds and d ≥ 2r+ 1 then C lies on a rational normal surfacescroll or a Veronese surface. Also, conversely, if S is a rational normal surface scroll and Cis a curve of class where d = m(r−1)+1+εwith 0 ≤ ε ≤ r−2 (m+1)h−(r−2−ε)for ε = 0 and C ∼ mh + f on such a surface, then equality is achieved (and a similarstatement holds for the Veronese surface).

Remark 20.4. The key is that when a curve is sitting in Pr, we know very littleabout its geometry. But, if we get the curve to sit on the surface, we know exactlyhow to translate the class of the curve into the genus. The key is to introduce

Proof. Say S is a rational normal surface scroll in Pr. Recall Pic(S) = Z〈h, f〉wheref is the class of a fiber and h is the hyperplane section.

Recall the intersection pairing: A priori, this generates the Picard group over Q.But, since this has determinant of absolute value 1, the Picard group can’t be anylarger than that generated by h, f. So, it is generated by h, f.

Now, write

KS = ah+ bf

and we apply adjunction to find a,b. If we apply adjunction to f, we have

−2 = 2g− 2 = f(KS + f) = a

so a = −2. Next, to find h, We know the hyperplane section is a rational normalcurve. We have

−2 = 2g− 2 = h(KS + h)


and by solving this, we conclude b = r− 3. So,

KS = −2h+ (r− 3)f

Remark 20.5. Let’s check this in the quadric surface in P3. By adjunction, thecanonical bundle is O(−2), which is precisely what we predicted.

So, if C ⊂ S has class ah+ bf (different a,b than those above) we see from theintersection table

degC = C · h = a(r− 1) + b

and, the genus of C is

g(C) =(ah+ bf) ((a− 2)h+ (b+ r− 3)f)

2+ 1

=

(a

2

)(r− 1) + (a− 1)(b− 1)

Remark 20.6. We can check the above when r = 3, and a curve in the basis h, fis a curve of type (a,a+ b), as we’re choosing a basis for the Picard group whosegenerators are a line of the ruling and a sum of the lines of the two rulings.

to find a curve of maximal genus, we write d = m(r− 1) + 1+ ε with 0 ≤ ε ≤r− 2. We take

C ∼ (m+ 1)h− (r− 2− ε)f

g =

(m

2

)(r− 1) +mε

The choice was unique when r = 3 except for possibly a confusing between tworulings of the quadric surface. The above is also unique except in one case ε = 0,in which case we have another solution given by C ∼ mh+ f.

Exercise 20.7. Check this on your own.

Finally, we should check when r = 5 and S is a Veronese surface. We can dothis because every curve on the Veronese surface has even degree, and it is theimage of some plane curve, which was can use to find the genus. This also yieldsequality g = π(d, r). �

Remark 20.8. If you had a surface and you wanted to bound the geometric genusof curves lying on the surfaceH0(K), you can use hyperplane sections to get lowerbounds on the hyperplane sections,

Remark 20.9. To exhibit curves which achieve the bound, we try to use Bertini’stheorem, to show that the linear system (m + 1)h − (r − 2 − ε)f, we take somevery singular curves which are unions of hyperplanes on the surface. We exhibitenough of the curves to show it’s basepoint free. Then, Bertini’s theorem tells usthere are smooth curves in these linear systems.

Remark 20.10. By Clifford’s theorem, you can’t have a curve of degree 2r withgenus bigger than r + 1. So, canonical curves are curves of maximum genus.However, canonical curves don’t satisfy d > 2r, so we can’t apply Castelnuovo’slemma. For instance, if we have 8 points in P3 which impose 7 conditions onquadrics, they could lie on the intersection of three quadrics. So, you really needthe 2n+ 3rd points for Castelnuovo’s lemma.

46 AARON LANDESMAN

But, now, Enriques made an interesting observation: Canonical curves lie onmany quadrics. In genus 5, a general canonical curve is the intersection of threequadrics.

Question 20.11. Is a canonical curve C ⊂ Pg−1 cut out by quadrics?

If not: that is, if there is p ∈ ∩Q⊃CQ but p /∈ C, we claim that C lies on aminimal degree surface. Then, we can go through the whole analysis of the boundon the genus. We can choose a general hyperplane H through p. Since p /∈ C, byBertini, such a hyperplane will intersect the curve transversely at {p1, . . . ,p2r}. Thesame analysis, by the uniform position lemma, implies Γ ∪ {p} is in linear generalposition. Essentially, this amounts to applying the uniform position lemma fromthe projection of the curve from the point C.

So, if we have g = π(d, r), which does hold for canonical curves, we obtain,hΓ∪{p} = 2r− 1, and so Γ ∪ {p} lies on a rational normal curve.

So, we obtain that ∩Q⊃CQ is a minimal rational surface. In particular, the in-tersection of the quadrics has to be a surface.

The above remark establishes the following, except for trigonality and planequintics.

Proposition 20.12. Let C be a canonical curve. If there is p ∈ ∩Q⊃CQ but p /∈ C, weclaim that C lies on a minimal degree surface. In particular the curve is trigonal or a planequintic.

Proof. To prove trigonality, if the curve is on a scroll, the fibers meet the curve inthree points. In the case that the curve lies on the Veronese surface, it is a planequintic. �

Theorem 20.13. If C ⊂ Pg−1 is a canonical curve, then either(1) C is cut out by quadrics(2) C is trigonal(3) g = 6 and C is a plane quintic.

Proof. This follows immediately from Proposition 20.12 �

Remark 20.14. We know trigonal curves will not be cut out by quadrics, becauseby geometric Riemann Roch, there will be three collinear points, and so it cannotbe cut out by quadrics, by Bezout.

Remark 20.15. Once you know that the ideal of the curve is generated by quadrics,you can ask what are the relations among the quadrics. That is, you can ask forthe next step of the resolution of the curve. Then, the relations are all linear, unlessC has a g14 or a g26. This was extended by Green and Lazarsfeld to a conjecture onthe resolution of the canonical curve, and Voisin proved this in the generic case.

Question 20.16. What if the curve does not lie on a rational normal surface scroll?

Then, Castelnuovo’s lemma characterizes curves with the smallest Hilbert func-tion. The next question is:

Question 20.17. What is the second smallest Hilbert function?

If we know this, we can start exhibiting degrees and genera below the boundwhich don’t occur.


21. 10/21/15

Recall, C ⊂ Pr is a smooth irreducible nondegenerate curve of degree d genusg, and Γ = H∩C. Then,

h0(OC(m)) − h0(OC(m− 1)) ≥ hΓ (m)

From these inequalities, we have g(C) ≤ π(d, r).

Remark 21.1. For today, the Veronese may be (incorrectly) referred to as a rationalnormal surface scroll.

We have

Lemma 21.2. Γ ⊂ Pn in linear general position of degree d. Then,(1) hΓ (m) ≥ min(d,mn+ 1) = h0(m)(2) If d ≥ 2n+ 3+ hΓ = h0 then Γ is contained in a rational normal curve.

Since we have determined the smallest Hilbert function corresponds to lyingon rational normal curve, this motivates the next question. In particular, we’d liketo know which curves have genus close to π(d, r) but not equal to π(d, r).

Question 21.3. What’s the second smallest Hilbert function?

We’ll see, in fact that there are gaps in the maximal genera below where themaximum is achieved, and we still don’t know exactly where those gaps are.

The argument for our bound π, we only looked at degree m curves which areunions of hyperplanes. However, in fact, we have a stronger condition. We’ll seewhat we can deduce for hyperplane sections with this stronger property.

Definition 21.4. A set of points Γ are in uniform position if for any two subsetsΓ ′, Γ ′′ ⊂ Γ so that |Γ ′| = |Γ ′′| then

hΓ ′ = hΓ ′′

Lemma 21.5. If Γ = C ∩H is a general hyperplane section of an irreducible C ⊂ Pr,then the points of Γ are in uniform position.

Example 21.6. Say r = 3 so we’re looking at curve in P3. Then, if we have acollection of point in a hyperplane H, then either no six points lie on a conic, orelse all points lie on a conic.

Proof. The idea is similar to that of linear general position: If three points were ona line, the all of Γ would lie on a line, contradicting nondegeneracy.

We have a similar situation here. If one set of 6 points lies on a conic, the we canswap out one point at a time, to see that every collection of 6 points lies on a conic.

Say U = (Pr)∗ \C∗, where C∗ means the subset of (Pr)∗ which are tangent ofC. The complement is the set of hyperplanes transverse to C. Define the incidencecorrespondence.

Φ = {(H,p) : p ∈ H∩C} ⊂ U×C

We now specialize to the case of curves lying on quadrics in P3. We can similarlyintroduce the set of sequences of points, which is

ψ = {(H,p1, . . . ,p6) : pi distinct in H∩C}

48 AARON LANDESMAN

and let

ψ ′ = {(H,p1, . . . ,p6) : pi lie on a conicH∩C}

If there exist 6 points p1, . . . ,p6 that don’t lie on a conic, we have ψ ′ ( ψ im-plies dimψ ′ < dimψ = r, because ψ is integral, as follows from the monodromystatement.

Therefore, general hyperplane sections can’t have some subset of points lyingon a conic and not all subsets not lying on a conic. �

Remark 21.7. So, the second smallest Hilbert function should correspond to thepoints lying on a cubic in P3. So, we currently have hΓ (m) ≥ min(d, 2m+ 1) inP3. If we replace the quadrics by a cubic, we would get a bound min(d, 3m+ 1),and there is a significant gap between these bounds.

Exercise 21.8. If we have some configuration of points in a plane and they don’t lieon a conic, then the next smallest Hilbert function is from those lying on a cubic.

First, we recall Castelnuovo’s theorem.

Theorem 21.9. (Castelnuovo) If Γ has minimal Hilbert function, then it has that becauseit lies on a rational normal curve B ⊂ Pn so that hΓ (m) = hB(m) = mn+ 1, withm ≤ d

n .

Recall, in general, if B ⊂ Pn is a curve of degree d and genus g, the Hilbertpolynomial of B is

pB(m) = d ·m− g+ 1

To minimize this, we first try to minimize d. The smallest possible degree is n,which is achieved by the rational normal curve. Now, we ask, which curve has thesecond smallest Hilbert function? The next smallest would have to have degreeat least n + 1. Then, the genus of the curve may be either 1, 0 as follows fromCastelnuovo’s theorem.

We conclude that the second smallest Hilbert function for hΓ of a collection Γ ofpoints in uniform position is achieved by points on an elliptic normal curve.

Remark 21.10. We’re guessing this on the basis of a suggestion: Configurations ofpoints with small Hilbert function have that small Hilbert function because theylie on a curve with small Hilbert function.

It takes some work to prove this is actually the case, since it might be that theydon’t lie on a curve, but it turns out that they do.

Proposition 21.11. If Γ is not contained in a rational normal curve, then

hΓ (m) =

m(n+ 1) if d > m(n+ 1)

m(n+ 1) − 1 if d = m(n+ 1)

d if d < m(n+ 1)

Proof. It follows from the comments preceding this, though those comments weren’treally proven. �

Corollary 21.12. If C 6⊂ a rational normal surface scroll, then we obtain that the genusis bounded by g ≤ π1(d, r) ∼ d2/2r


Proof. This follows from plugging in the inequalities from Proposition 21.11, weobtain g ≤ π1(d, r) ∼ d2/2r. We get this by computing the Hilbert function forlarge values ofm, and then using Riemann Roch to bound the genus. �

When we were analyzing curves on rational normal surface scrolls, we foundthe Picard group has rank 2, and so we used this to determine the possible generaof curves on these scrolls.

Let’s now describe what happens for curves in P3.

Proposition 21.13. In P3, for g > π1(d, 3) ∼ d2

6 , we have that g occurs if and only ifthere exist integers a,b so that a+ b = d and (a− 1)(b− 1) = g. If g ≤ π1(d, r), thenwe next obtain that the curve lies on a cubic or a higher degree curve, and every g occurs.

Proof. In the case that g ≤ π1(d, r) a cubic surface has Picard group of rank 7, andquartic has Picard groups of rank up to 19, and it turns out you can always solvethe system for each genus. �

Question 21.14. What about in higher dimensional space?

Proposition 21.15. In Pr, if g > π1(d, r) ∼ d2

2r , then the curve lies on S, and g occurs ifand only if there are a,b so that a(r− 1) + b = d, and

2g− 2 = (ah+ bf)((a− 2)h+ (b+ r− 3)f)

g =

(a

2

)(r− 1) + (a− 1)(b− 1)

in Pic(S).

Proof. The case g > π1(d, r) follows from prior analysis. �

Remark 21.16. In the case g ≤ π1(d, r). However, once r > 1 there are no delPezzo surfaces, and there are no surfaces of one more than the minimal degree,except for del Pezzo surfaces, but those die off after r = 9 or so, and the rank ofthe Picard group also decreases.

Look in the Montreal notes for more details.

Remark 21.17. If we drop the hypothesis of smoothness, and take the geometricgenus, we do get curve of every geometric genus up to π(d, r). The reason is thatin the linear system on a rational normal surface scroll S, we can find a curve classthat ache Ives the maximum, and there are smooth curves of this class. Further, inthis linear system, there exist irreducible nodal curves C ⊂ Swith δ nodes for any0 ≤ δ ≤ π(d, r).

Exercise 21.18. There exist irreducible plane curves with δ nodes for any δ between0 and

(d2

).

22. 10/23/15

22.1. Agenda and Review. Today(1) Green’s conjecture(2) Voisin’s theorem(3) The maximal rank conjecture (if time allows)

On Monday, we’ll move on to Brill Noether TheoryLet’s now recap what happened last time.

50 AARON LANDESMAN

Question 22.1. For which d,g, r does there exist a smooth nondegenerate C ⊂ Pr

of degree d and genus g?

This leads to the following question.

Question 22.2. What are the possible Hilbert functions of configurations of pointsin uniform position?

The question of all possible Hilbert functions has been answered by RichardStanley and others, which is solved by considering monomial ideals.

However, the question of uniform position points is more difficult. There areother interesting subschemes whose Hilbert functions we can ask about, such asHilbert functions of fat points (schemes supported at a point).

22.2. Resolutions of Projective Varieties and Green’s conjecture. Suppose X ⊂Pn is a variety or subscheme. We can look in I(X) ⊂ S = C[x0, . . . , xn]. SayI(X) = (f1, . . . , fk) and deg fi = ai.

Then, we have a resolution of the form

(22.1) · · · ⊕S(−bi) ⊕S(−ai) S S(X) 0

where S(−ai) correspond to the generators. Then, we can ask what relations existamong those generators. As a module the kernel of the map ⊕S(−ai) → S, de-termines the module of relations. This resolution terminates for a variety by theHilbert basis theorem. This resolution also tells you the Hilbert function.

Hence, understanding the Hilbert function is tantamount to describing the res-olution.

Example 22.3. Let X = C ⊂ P4 is a canonical curve of genus 5. We know C lies onthree quadrics. So, we have a resolution

(22.2) · · · ⊕S(−4)⊕2 S(−2)⊕3 S S(X) 0

in the case that the quadrics cut out the curve. In the case that the quadrics don’tcut out the curve, there’s an additional cubic in the ideal of the curve. There is alsoa linear relation among the quadrics in this case, but there’s only a quadratic rela-tion (that corresponding to QiQj −QjQi) in the case that the curve is a completeintersection of quadrics.

So, to understand a curve, we could try to understand the resolution of its ideal.The goal is to describe the resolution of canonical curves. SupposeC is a smooth

curve of genus g.

Say C is a smooth curve of genus g. Then, We have CφK−−→ Pg−1 ∼= PH0(K)∨.

and S = Sym•H0(K). Then, we introduce the canonical ring

R = ⊕∞m=0H

0(Km)

Now, by Noether’s theorem, the map S → R or equivalently, Sym•H0(K) →⊕mH0(Km) is surjective if and only if the curve is not hyperelliptic.

So, let’s assume C is not hyperelliptic. Then, we can ask about the resolution.By Enriques’ theorem, we can write

(22.3) · · · S(−2)⊕(g−22 ) S S(X) 0


in the case that the curve is further not trigonal, nor a plane quintic.This is equivalent to saying there does not exist a g13 or a g25.Now, here’s the next step, which will lead to Green’s conjecture. We can now

look at the linear relations among these quadrics.

(22.4)

· · · S(−3)⊕N S(−2)⊕(g−22 ) S S(X) 0

is exact if and only if C does not have a g14 or g26. Green’s conjecture naturallyextends this.

If L is a line bundle onC of degree d and dimension r = H0(L)− 1. If we assumeH0(L),H1(L) 6= 0 then d− 2r ≥ 0.Definition 22.4. The Clifford index of a linear series L which is a grd is d − 2r.Then, the Clifford index of a curve is the minimum over all line bundles L withh0(L),h1(L) 6= 0, other than O,K, of the Clifford index of L.

Remark 22.5. Observe, that

Cliff(C) = 0 ⇐⇒ C is hyperelliptic

Cliff(C) = 1 ⇐⇒ C is trigonal or a plane quintic

Cliff(C) = 2 ⇐⇒ C has a g13 or g25 or g37

Note for curves with g37, we get a map C→ P3, then the curve has a 4 secant line.Then, you can project from a point on the line, and we obtain that C is alreadytrigonal.

Here is a consequence of the Brill Noether Theorem.

Corollary 22.6. For a curve C general of genus g, then Cliff(C) = dg−22 e.Proof. Follows from Brill Noether, which we haven’t yet stated. �

Conjecture 22.7. (Green) This resolution of canonical curves, given by linear rela-tions at each step, is exact for Cliff(C) steps.

If you know the resolution remains linear half way through, you can get theremaining Koszul betti numbers from duality. So, you can get all betti numbers.

Then,

Theorem 22.8. (Voisin) Green’s conjecture is true for a general curve C.

Proof. Difficult �

Remark 22.9. Voisin does not prove that a curve with Clifford index dg−22 e has amaximally linear resolution. She just shows it for some open subset of those curve.

Remark 22.10. Here is the idea of proof of Voisin’s theorem. The set of curveswhich have a maximally linear resolution is an open condition: It’s saying that theranks of certain maps are as large as they can be. So, to prove Voisin’s theorem,you only have to exhibit a curve somewhere which have that sort of resolution.She does this for curves on a K3 surface. Her approach doesn’t seem like it willyield anything more in the direction of proving Green’s conjecture. So, she doesn’tprove which curves are excluded.

52 AARON LANDESMAN

22.3. The Maximal Rank Conjecture. Now, let’s look at the second graded pieceof an ideal of the curve I(C). We have a map

I(C)2 ⊗ S1 → I(C)3

We know the left hand side has dimension(g−22

)and S1 has dimension g. Then,

I(C)3 has dimension given by the kernel of

(22.5) 0 H0(IC(3)) H0(OPg−1(3)) H0(OC(3)) 0

The latter vector spaces have dimensions(g+23

), 5g−5, and so dim I(C)3 =

(g+33

)−

5g− 5. Then, if Green’s conjecture holds, the kernel of this map

I(C)2 ⊗ S1 → I(C)3

has maximal rank and generates the module of relations, and hence, Green’s con-jecture would imply S(−3)N1 generates the relations, and so N1 =

(g−22

)· g −(

g+23

)− (5g− 5) when C is not trigonal or a plane quintic or hyperelliptic.

So, now, we can ask the analogous question about curves in P3.

Question 22.11. Can we describe the resolution of an arbitrary embedding C →Pr?

The maximal rank conjecture is a solution to this for a general curve in Pr. Tospecify a map from a curve to projective space, you have to specify three pieces ofdata

(1) C general in Mg(2) L ∈Wrd general(3) Vr+1 ⊂ H0(L) general.

In this case

Fact 22.12. L2 is nonspecial.

This leads to the maximal rank conjecture.

Conjecture 22.13. The map

φm : H0(OPr(m))→ H0(OC(m))

which have dimensions(m+rr

)and md− g+ 1. Then, the conjecture is that this

map has maximal rank (the map is either injective or surjective).

If we believe this conjecture, we can say exactly what the Hilbert function is.Suppose we know φm has maximal rank. Then,

Conjecture 22.14. The second stage of the maximal rank conjecture says

I(C)m ⊗ S1 → I(C)m+1

has maximal rank, and we can continue following the resolution.

According to the conjecture, we can obtain the resolution of any general curve.Here is one little factoid.


23. 10/26/15

Question 23.1. What linear series may exist on a curve of genus g.

Answer: (Clifford) we must have d ≥ 2r in the range 0 ≤ d ≤ 2g− 2.

Question 23.2. What birationally very ample linear series may exist on a curve ofgenus g?

Answer: (Castelnuovo) we must have g ≤ π(d, r).If you fix d, and ask for the inequality on d, r, its traces out a quadratic curve in

the d, r plane.However, Castelnuovo curves are very nice, but also very special. This leads to

the main question of Brill-Noether theory.

Question 23.3. What linear series exist on a general curve of genus g?

Theorem 23.4. (Brill-Noether) Let C be a general curve of genus g. Then, C has a grd ifand only if

ρ := g− (r+ 1)(g− d+ r)

is ≥ 0.

Idea of Proof. Idea of proof: Look at the geometry of a curve, and then look at theJacobian of a curve, and introducing a bunch of subvarieties.

We’ll come back to the proof in a couple weeks, and start building up the theoryto give the proof now.

Fact 23.5. Recall,

Wrd :={L ∈ Picd(C) : h0(L) ≥ r+ 1

}Now, Wrd turns out to be a variety itself, and so, in fact, we can ask about generalpoints of it. We can also ask about its dimension and irreducibility, and so in.

It turns out

(1)

dimWrd = ρ

(2) If ρ > 0, thenWrd is irreducible.(3) If L ∈Wrd, if

(a) r ≥ 3, then L is very ample.(b) If r = 2, then φL is birational onto C0 ⊂ P2 is nodal(c) If r = 1, then φL → P1 is simply branched.

Remark 23.6. Our goal is to relate

Mg := { abstract smooth curves of genus g }

to

Hd,g,r := { curves C→ Pr irreducible nondegenerate smooth degree d and genus g curves}

When r ≥ 3, this Hd,g,r is an open subset of the Hilbert scheme. When, r = 2,we want to take the Severi variety.

54 AARON LANDESMAN

Definition 23.7. The Severi variety is

Vd,g ={C ⊂ P2 : irreducible degree d and geometric genus g.

}More precisely, it is a scheme whose k points are the above.

Then, when r = 1, we mean the Hurwitz space.

Definition 23.8. The Hurwitz space is

Hd,g ={f : C→ P1 : C is simply branched.

}More precisely, it is a scheme whose k points are the above.

Example 23.9. Let’s look at the case r = 1.

(23.1)

Hd,g

Mg (P1)b \∆ ∼= Pb \∆

where b = 2d + 2g − 2, and ∆ ∈ (P1)b is the diagonal while ∆ ⊂ Pb is thediscriminant locus. Again, the identification (P1)b ∼= Pb is by identifying divisorsof degree b with polynomials of degree b, modulo scalars. The right projectionsends a curve to its branch locus, and is a finite surjective map by a monodromyargument.

As a consequence,

dimHd,g = b

We can give Hd,g the structure of a complex manifold via the etale map to Pb \∆.When d is large compared to g, the map to Mg is dominant. Then,

Hd,g/PGL2 ={(C,D) : D is a g1d on C

}We have the map Hd/g/PGL2 → Mg is dominant with fibers of dimension

2d− g− 2. To see this, look at pairs (L,V) : V2 ⊂ H0(L). We have L ∈ Picd(C)which is g dimensional and V ∈ G(2,H0(L)) which is 2(d−g+ 1)− 4 dimensional.Putting these together, we see the fibers are of dimension 2d− g− 2. This tells usMg is 3g− 3 dimensional. So, we have that Hd,g/PGL2 has dimension 2d+ 2g− 5,and Mg has dimension 3g− 3. We guess that this is surjective when d+ 2g− 5 ≥3g− 3 implying

d ≥ g

2+ 1.

(1) If g = 0, we’re looking at rational curves, so d ≥ 1.(2) If g = 1, d ≥ 2. Here, there is a 1 dimensional family of g12’s.(3) If g = 2, then d ≥ 2, and C is a 2 sheeted cover of P1, in a unique way via

the canonical series. Here, there is a unique g12.(4) If g = 3, then C, if not hyperelliptic is a plane quartic, and K− p is a g13.

Here, there is a 1 dimensional family of g13’s.(5) If g = 4, d ≥ 3 as it is the intersection of a cubic and a quadric. Here there

are generals 2 g13’s.


(6) If g = 5, we have seen d ≥ 4, by the description of the complete intersec-tion of quadrics. (There is a 1 parameter family of g14’s.)

We see that in these low genus examples, the dimension of Wrd is indeed, ρ, asclaimed.

Example 23.10. Let’s look at the case when r = 2.(1) If g = 0, then d = 2.(2) If g = 1, then d = 3.(3) If g = 2, then d = 4. We see d ≥ 4 by Clifford’s theorem. Conversely,

if we take a general line bundle L ∈ Pic4(C) that L = KC + p + q hash0(L) = 3. Then, we deduce that φL : C→ P2 carries C to a nodal curve ofdegree 4, with 1 node. This maps p,q to the same point, but is otherwisean embedding.

(4) If g = 3, then d = 4, as general genus 3 curves are canonically embeddedby quadrics in P2.

(5) If g = 4, we can take a curve of type (3, 3) on a quadric. We can choosea point on the quadric from a point on the curve, you get an embeddinginto P2, except for the two lines on the quadric passing through the point,which are mapped to nodes in the plane.

Exercise 23.11. See what you can say when r = 2 and g = 5, 6. Genus 6 is a littletricky. See if you can compute cases when r = 3.

24. 10/28/15

24.1. Review. The general question is: what linear series exist on a general curveof genus g?

For which d,g, r does there exist a grd on C.Recall that in the case r = 1we introduced the Hurwitz space

(24.1)

Hd,g

Mg Pb \∆

as an incidence correspondence.Forα to be dominant, we obtain dimHd,g/PGL2 ≥dimMg, so we obtain d ≥ g

2 + 1.Here’s a somewhat difficult question:

Question 24.1. How do we give the Hurwitz space Hd,g the structure of a variety?

Next, recall we discussed the case r = 2. In this case we have handles on theplane curves. That is, we can access the coefficients of the polynomial cutting outthe curve in P2. Here, we look at the Severi variety

Vd,g = {C reduced and irreducible curve of geometric genus g and degree d} ⊂ P(d+22 )−1

Additionally, inside the Severi variety, we have

Vd,g ⊃ Ud,g :=

{C : C is nodal with δ nodes, whereδ =

(d− 2

2

)− g

}

56 AARON LANDESMAN

TABLE 3. My caption

F G H∂∂x 0 a20 a11∂∂y 0 a11 a02∂∂a00

1 0 0

The varietyUd,g is quite well behaved, but Vd,g is quite complicated. Ud,g will besmooth, while Vd,g is far from being smooth.

Remark 24.2. ACGH was divided into two parts: the geometry of fixed curvesand the varieties of families of curves. A good reference for the Hurwitz spaceand the Severi variety, see Moduli of Curves, by Harris and Morrison.

Fact 24.3. Here are two facts:Vd,g,Ud,g are locally closed subsets of PN where N =

(d+22

)− 1.

We have Ud,g ⊂ Vd,g is open and dense.

This is the analogous to the r = 1 case, when we say any branched cover canbe deformed to one with simple branching. In this case, it’s more of an exercisein deformation theory, but the fact is that any curve can be deformed to one withonly nodes, of the same geometric genus.

Theorem 24.4. Ud,g is smooth of dimension 3d+ g− 1.

Proof. Look at the incidence correspondence

(24.2)

Φ :={(C,p) ∈ PN ×P2 : p ∈ Csing

}

∆ ⊂ PN P2

where ∆ is the locus of singular curves. The fibers of P2 are linear space of dimen-sionN− 3. Hence, PN−3. So,Φ is irreducible and smooth of dimensionN− 1. So,∆ isN− 1 dimensional and irreducible. The fiber over ∆ is finite because a generalcurve has only finitely many singularities. Now, the three equations defining Φare very explicit.

F(a, x,y) =∑

aijxiyj

G(a, x,y) =∑

iaijxi−1yj

H(a, x,y) =∑

jaijxiyj−1

Say (C0,p0) ∈ Φ. Take p0 = (0, 0) in affine coordinates. Then, C0 = V(f) wheref =∑a0ijx

iyj = a20x2 + a11xy+ a02y

2 + · · · for a0ij ∈ k. Now, let’s consider thechart

Note,Φ is smooth being a PN−3 bundle over P2.Now, p0 is a node of C0 implies that Φ is smooth at (C0,p0) and the tangent

space to Φ at (C0,p0 is given by the vanishing of ∂∂x , ∂∂y , ∂

∂a00and

dπ : T(C0,p0)Φ→ TC0PN


is injective with image the hyperplane of curves containing p0.The converse is also true, but we’d have to consider the remaining rows. Now,

say C0 has δ nodes, called p1, . . . ,pδ, then neighborhoods of (C0,pi) ∈ Φ map toδ branches ∆1, . . . ,∆δ of ∆ at C0

Locally, Ud,g = ∩δi=1∆i. To conclude, we have to know whether the tangenthyperplanes intersect transversely or not. that is, we want to know that the tangentplanes to ∆ at a given singular point intersect transversely.

That is, we want to know if they are independent.But, the tangent hyperplanes to ∆i at C0,{

C ∈ PN : C 3 pi}

is transverse. That is, we claim p1, . . . ,pδ impose independent conditions on|OP2(d)|. But, we’ve seen that p1, . . . ,pδ impose independent conditions on |OP2(d−3)| by counting dimensions when we were describing nodal curves in the planeseveral weeks ago. Hence, they certainly impose independent conditions on |OP2(d)|.So, in a neighborhood of C0, we have that Ud,g is smooth of dimension

N− δ =

(d+ 2

2

)− 1− δ

=

((d− 1

2

)− δ

)− 1+ 3d

= 3d+ g− 1

�

Remark 24.5. ∆ has pretty large degree. It is degree something like 3d2, and notreally worth writing down.

Corollary 24.6. If a general curve of genus g has a g2d then d ≥ 23g+ 2.

Proof. We know dimUd,g/PGL3 ≥ dimMg. We see

dimUd,g/PGL3 ≥ dimMg

3d+ g− 9 ≥ 3g− 3

d ≥ 2

3g+ 2

�

Question 24.7. What happens when r ≥ 3?

In this case, we have no idea what happens. When r = 1, 2, we have very simpleequations for the curves or branch points. In P3, we don’t have a single definingequation, and we don’t even know how many equations define the curve. If wearbitrarily vary the equations, we won’t see a curve at all.

25. 10/30/15

25.1. Review. Brill Noether Theorem. Here’s a basic version. A general curve Cof genus g has a grd if and only if ρ = g− (r+ 1)(g− d+ r) ≥ 0, and in this casedimWrd(C) = ρ.

In the case r = 1we introduce the Hurwitz space of simply branched coveringsof P1, which is b := 2g+ 2d− 2 dimensional.

58 AARON LANDESMAN

We conclude that a general curve C has a g1d then d ≥ g2 + 1.

When r = 2, we can introduce the small Severi variety of curves in P2 with(d−12

)− g = δ nodes. We saw last time this has dimension 3d + g − 1. We can

conclude that a general curve has a g2d implies d ≥ 23g+ 2.

25.2. Today: Brill Noether Theorem in dimension at least 3. But, when r ≥ 3,there’s a natural parameter space for curves with a map to Pr called the Hilbertscheme.

H0d,g,r = {C ⊂ Pr : C is irreducible, smooth, nondegenerate degree d genus g }

Remark 25.1. In the r = 2 case, we were able to infinitesimally ask how many wayswe could deform the curve to first order and preserve the nodes. We don’t know ingeneral the dimension of H0d,g,r because it is intrinsically not well behaved. In ther = 1, 2 cases, the varieties are irreducible and an easy to write down dimension.For Hilbert schemes of curves in P3, the Hilbert scheme may have any number ofcomponents and be irreducible.

So, in this case, we’ll be able to use the Brill Noether theorem to describe, atleast in some cases, the Hilbert scheme.

Remark 25.2. For any r there is an expected dimension of the Hilbert scheme.When r = 3 the expected dimension of the Hilbert scheme is 4d. One of thethings Brill Noether theory says the dimension of the Hilbert scheme is 4d. Note,g doesn’t appear in P3, but in higher r, it does appear. Further, Brill Noethertheory implies this. However, we don’t know when this 4d occurs, or when thedimension is bigger.

Remark 25.3. Hopefully, today, we’ll see why the Brill Noether number is ρ =g− (r+ 1)(g− d+ r)

There are two reasons whyWrd is not a great object to look at.(1) It’s difficult to write down equations for Wrd because it’s a subvariety of

the Jacobian, which is complicated.(2) On the other hand, divisors on the curve of a given degree are quite ex-

plicit. If we want to relate moduli spaces, we’re looking not at invertiblesheaves, but rather actual linear series (and not complete linear series).

Recall:

Definition 25.4.

Wrd(C) ={L ∈ Picd(C) : h0(L) ≥ r+ 1

}⊂ Picd(C) ∼= J(C)

Crd ={D ∈ Cd : h0(O(D)) ≥ r+ 1

}Grd =

{(L,V) : L ∈ Picd(C),Vr+1 ⊂ H0(L)

}we have a map

u : Cd → Picd(C)

u−1(Wrd) = Drd

u(Crd) =Wrd


Remark 25.5.

Wrd ={L ∈ Picd : dimu−1(L) ≥ r

}and is different from Crd because there may be many divisors associated to a givenline bundle, in fact there is an r+ 1 dimensional vector space of them.

Remark 25.6. To describeWrd for the purposes of the homework, just describe thedimension and the irreducible components.

Lemma 25.7. Let r ≥ 0 and r ≥ d− g.

Wrd ⊂Wr−1d \Wrd

That is, no component of Wr−1d lies in Wrd. In other words, the Wrd give a stratificationof the Jacobian.

Question 25.8. The lemma says we can move an invertible sheaf slightly to reducethe number of sections by 1. How do we do this? This will be answered by thelemma.

Proof. Say L ∈Wrd \Wr+1d . That is, h0(L) = r+ 1. Consider the invertible sheavesof the form L(p − q), and we can specialize to L as p = q. The hypothesis thatr > d− g says

h0(K−D) > 0.

Choose p not in the base locus of |K−D|. So,

H0(K−D− p) = H0(K−D) − 1

H0(D+ p) = H0(D) = r+ 1

by Riemann Roch. Then, choose any point q not in he base locus of |D+ p|, we get

H0(D+ p− q) = H0(D+ p− 1) = r.

�

Let’s start by describing Crd.

Lemma 25.9. We can write down explicit defining equations for Crd in a neighborhood ofa divisor D = p1 + · · ·+ pd.

Remark 25.10. It may help to think of the proof for d ≤ g, and you can flip it tod ≥ g by applying Riemann Roch

Proof. Choose local coordinates zi for C in a neighborhood of each point pi. Then

h0(D) ≥ r+ 1 ⇐⇒ h0(K−D) ≥ g− d+ r

by geometric Riemann Roch. Letω1, . . . ,ωg be a basis for H0(K). Then, write

ωα = fα,i(zi)dzi

in a neighborhood of pi. Define the Brill-Noether matrix

M =

f1,1(z1) · · · fg,1(z1)...

. . ....

f1,d(zd) · · · fd,g(zd)

60 AARON LANDESMAN

Note, we may want to write the above matrix asω1(z1) · · · ωg(z1)...

. . ....

ω1(zd) · · · ωg(zd)

But the rows of the matrix are only defined up to scalars. We had to choose localcoordinates to settle this ambiguity, but this doesn’t change the rank.

The key observation is

H0(D) ≥ r+ 1 ⇐⇒ rk (M) ≤ d− r

That is, there are r linear relations among the rows of the matrix.Here,M(D) represents the evaluation map

H0(K)→ H0(K|D)

So, we’re looking for the locus where this matrix has at most rank d− r. That is,locally We have a map

(25.1)

Cd Md,g

Crd Md−rd,g

where Md,g is d× g matrices and Md−rd,g is d× g matrices of rank at most d− r.Then,Mka,b ⊂Ma,b is irreducible and has codimension (a− k)(b− k).

So, the dimension is (d− (d− r))(g− d+ r) = r(g− d+ r), so the codimensionMd−rd,g ⊂Md,g is codimension r(g− d+ r). The expected dimension is

dimCrd = d− r(g− d+ r)

This is also a lower bound on the dimension. �

Corollary 25.11. We have dimWrd ≥ d− r(g− d+ r) − r = g− (r+ 1)(g− d+ r),and d− r(g− d+ r) is the “expected dimension.”

Proof. This follows because the preimage under the map u is r dimensional. �

Remark 25.12. The key is saying that this is equal to the dimension, but we needto show that if this is a general curve, then this is the correct dimension.

Remark 25.13. The above estimate also gives us a way of giving a scheme structureto Crd, and then port that over to Wrd by taking the image. So, we can talk aboutWrd as a subscheme of the Jacobian variety. This is important because we’ll liketo describe the tangent spaces to Crd,Wrd scheme theoretically, defined by theseequations from the minors of local matrices.


26. 11/2/15

26.1. Review. Today, we’ll move forward on the discussion of Wrd,Crd. Take C tobe a smooth curve of genus g. Recall:

Wrd(C) :={L ∈ Picd(C) : h0(L) ≥ r+ 1

}Crd(C) :=

{D ∈ Cd : h0(L) ≥ r+ 1

}Grd(C) := {grd on C} =

{(L,V) : L ∈ Picd(C),Vr+1 ⊂ H0(L)

}Observe that if D =

∑di=1 pi is a reduced divisor, then in an analytic neighbor-

hood U of D ∈ Cd, we can write

Crd =

(q1, . . . ,qd) :

ω1(q1) · · · ωg(q1)...

. . ....

ω1(qd) · · · ωg(pd)

≤ d− r

={(q1, . . . ,qd) : rk

(H0(K)|q → H0(K|D)|q

)≤ d− r

}So, locally, we have a map

Uπ−→Md,g ⊃Md−rd,g

where Md,g is the set of d× g matrices, and Md−rd,g the subset of matrices of rankat most d− r.

Observe that Md−rd,g has codimension r(g − d + r) in Md,g. Therefore, U ∩π−1(Md−rd,g ) has “expected codimension” r(g−d+ r). So, the expected dimensionof Crd in Cd is

d− r(g− d+ r)

and every component of Crd has dimension at least d− r(g− d+ r) because thecodimension of the preimage is at most the codimension in the target.

26.2. Hilbert Schemes. Now, let’s discuss the relation between these Wrd and theHilbert scheme.

Remark 26.1. In the cases r = 1, 2, we were able to find the dimensions of theHurwitz space and the Severi variety, and deduce they had the correct dimension.But, things get a lot messier when you look at the Hilbert schemes in dimension atleast 3.

Now, we introduce global analogs ofWrd,Crd,Grd:


Wrd ={(C,L) : C smooth of genus g ,L ∈ Picd(C) : h0(L) ≥ r+ 1

} →Mg

Crd ={(C,D) : D ∈ Cd,C smooth of genus g ,L ∈ Picd(C) : h0(L) ≥ r+ 1

} →Mg

Grd ={(C,L,V) : C smooth of genus g ,L ∈ Picd(C),Vr+1 ⊂ H0(L)

} →Mg

62 AARON LANDESMAN

Further, we have maps

(26.1)

H0 Grd Mg

Wrd Mg,d = {(C,D)) : D ∈ Cd}

Crd

where the fibers of the first map are PGLr+1 and the fibers of the second map havedimension at least ρ.

Now, recall the Brill Noether matrixω1(q1) · · · ωg(q1)...

. . ....

ω1(qd) · · · ωg(pd)

Now, we have that the expected dimension of Grd is 3g− 3+ ρ. So, the expected

dimension of the Hilbert scheme

H0d,g,r = 3g− 3+ ρ+ r2 + 2r

4g− 3− (r+ 1)(g− d+ r) + r2 + 2r

Further, every component of H0d,g,r has at least this dimension.

Question 26.3. Do the the components of the Hilbert scheme other than thosecorresponding to smooth curves?

Harris doesn’t know of any specific components of lower dimension.Further, even Hilbert schemes of points are not well understood. There are

many components corresponding to nonreduced schemes. Iarabino found com-ponents entirely consisting of nonreduced schemes, which had dimension morethan 4d. More recently, Daniel Erman found components of the Hilbert scheme ofpoints with dimension less than 3d, which is the dimension of the smooth points.

Remark 26.4. How do we get an estimate on the dimension of a component of theHilbert scheme?

We can’t do it by writing down equations. For example, we can write downtwisted cubics as an element of G(3, 10), corresponding to three dimensional vec-tor spaces of quadrics whose zero locus is a twisted cubic. Unfortunately, this is21 dimensional, and the general set of three quadrics will intersect in 8 points.

The Hilbert scheme of twisted cubics is an interesting ground for exploration.

26.2.1. Estimating the dimension of H0d,g,r. Suppose we have a point C ⊂ Pr ofdegree d genus g. We want to know when we can deform a curve. We can’t reallysee the space of curves, but we can find the dimension of the tangent space tothe Hilbert scheme, which gives an upper bound on the dimension of the Hilbertscheme.

We want to estimate dimH0d,g,r in a neighborhood of C.So, our first approximation will be dim TCH = H0(C,NC/Pr).


Warning 26.5. Caution: There are components of the Hilbert scheme which areeverywhere nonreduced, such as smooth degree 24 genus 14 curves in P3. So, ourestimate won’t necessarily be correct, but it will be an upper bound.

Example 26.6. Let’s look at twisted cubics. We can look at three quadricsQ1,Q2,Q3,and translate that to

Q1 + εQ′1,Q2 + εQ ′2,Q3 + εQ ′3

This corresponds to a flat family.

C ⊂ Spec C[ε]/(ε2)×P3 → Spec C[ε]/(ε2)

and so this should be a module homomorphism. So, to conclude the cohomologyof the normal bundle is the same as a map

I(C)2 → (S(P3)/I(C)

)2

Qi 7→ Qi + εQ′i

Now, we can bound H0(C,NC/Pr) using Riemann Roch as follows:

Proposition 26.7. The dimension of the Hilbert scheme H0d,g,r is “expected” to be (r+1)d+ (r− 3)(g− 1).

Proof. First, we have an exact sequence

(26.2) 0 TC TPr |C NC/Pr 0

which implies NC/Pr = (r+ 1)d+ 2g− 2. So, we know the degree and the rank,and therefore we know the Euler characteristic. We’ll then make an a second ap-proximation that

h0(C,NC/Pr) = χ(C,NC/Pr) = (r+ 1)d+ 2g− 2− (r− 1)(g− 1)

= (r+ 1)d− (r− 3)(g− 1)

�

Remark 26.8. Note that to compute the expected dimension of the Hilbert schemeabove we made two estimations. These were both inequalities, and were in oppo-site dimensions, so we can’t get an actual bound form the above.

Example 26.9. Twisted cubics in P3 have dimension 12 which agree with theabove estimate. However, complete intersections, in general, will have higher di-mension. the first example of this being violated will be complete intersections ofquadrics and quartic. In fact, whenever the complete intersection has a quartic orhigher degree surface, we’ll be done.

Remark 26.10. We have two ways of estimating the dimension of the Hilbertscheme.

(1) The first way led us to the number

4g− 3− (r+ 1)(g− d+ r) + r2 + 2r

we got an expected number from the Brill-Noether type analysis.(2) The second way led us to

(r+ 1)d− (r− 3)(g− 1).

64 AARON LANDESMAN

Question 26.11. Which one is right?

In fact, the two estimates are the same.

27. 11/4/15

27.1. Logistics.(1) The homework will be up in a few days.(2) Read section IV.1 in ACGH. But, watch out for and ignore the argument

on page 161 to 162. For now, skip IV.2, IV.3, and read IV.4, IV.5.

27.2. Tangent Spaces in Brill Noether Theory. The framework is thatC is a smoothcurve of genus g, which are subvarieties

(27.1)

Crd Cd

Wrd Picd

u

where

Picd ∼= J =H0(K)∨

H1(C, Z)

where the Jacobian is defined by the map

u : D =∑i

pi 7→∑i

∫pip0

where this function above is dependent on choices of paths from p0 to pi and isreflected in the ambiguity of quotienting by H1(C, Z).

To understand these varieties, we will look at their tangent spaces.We have

TL Picd = H0(C,K)∨ ∼= H1(C,OC)

using Serre duality for the last step, and the description of the Jacobian for the firststep.

Also, for Cd, we have

TDCd = ⊕iTpiC = H0(C,K/K(−D))∨ ∼= H0(C,OC(D)/OC)

where the third term is the direct sum of the cotangent spaces at the points of D.

Remark 27.1. The middle description as ⊕iTpiC is only correct when the pointsare distinct. However, the third and fourth descriptions of TDCd is always correct,even when the points overlap.

Lemma 27.2. We have a natural isomorphism

TDCd = H0(C,OC(p)/OC)

and a pairing

H0(C,OC(p)/OC)⊗H0(K/K(−p))→ C

(f,ω) 7→ Resp(fw)


Proof. The first isomorphism was given above through

TDCd = ⊕iTpiC = H0(C,K/K(−D))∨ ∼= H0(C,OC(D)/OC)

The pairing is also an example of Serre duality. �

Now, recall,

u :∑i

pi 7→∑∫pip0

Therefore, by differentiating under the integral sign, we see

du =

ω1(p1) · · · ωg(p1)...

. . ....

ω1(pd) · · · ωg(pd)

Therefore,

D ∈ Crd ⇐⇒ rk (du) ≤ d− r

This implies that all fibers of u are reduced Pr’s. This is saying the corank of thedu is exactly the rank of the fiber, which is Pr, and so the tangent space must beexactly r dimensional.

Therefore,

Wrd = u(Crd)

Crd = u−1(Wrd)

By the above, this is also true at the scheme theoretic level. That is, we have

Tu(D)Wrd = du(TDC

rd)

Here, the key input is that the Brill Noether matrix, which defines Crd is also thedifferential of the map u.

Suppose L ∈ Picd and h0(L) = r+ 1, meaning

L ∈Wrd \Wr+1d

Question 27.3. If we slightly deform L in Picd, how does the space of global sec-tions vary?

Definition 27.4. A first order deformation of a line bundle L on a curve is a map

L̃→ Spec C[ε]/(ε2)×C

so that L|Spec C×C ∼= L.

Proposition 27.5. Vectors v ∈ TL Picd are in natural bijection with first order deforma-tions L̃ of L. Further, this correspondence can be made explicit.

Proof. See, for example, Hartshorne’s deformation theory textbook. �

We want to know,

Question 27.6. Is π∗L̃ locally free?

66 AARON LANDESMAN

We have a map

π∗L̃|SpecC → H0(L)

and we want to know when sections of H0(L) extend to an infinitesimal deforma-tion. Since we are working over the dual numbers, this is equivalent to π∗L̃ werelocally free.

Question 27.7. Given v ∈ TL Picd = H1(C,OC), which is equivalent to deforma-tions L̃ of L. Given σ ∈ H0(K), does σ extend to a section of L̃?

Answer: We have that σ extends if and only if

σ⊗ v ∈ H0(C,L)⊗H1(C,OC) 7→ 0 ∈ H1(C,L)

This implies that if L ∈Wrd \Wr+1d , and

v ∈ TL Picd = H1(C,OC)

then all sections

σ ∈ H0(C,L)

extend if and only if

H0(L)⊗ 〈v〉 7→ 0 ∈ H1(C,L)

Now, corresponding to the above, map, we have a map

H0(C,L)⊗H1(C,OC)→ H1(L)

or, equivalently, we can view this as a map

H0(C,L)⊗H1(C,L)∨ → H1(C,OC)∨

Now, using Serre duality, we can rewrite this as

H0(C,L)⊗H0(C,K⊗ L∨)µ0−−→ H0(C,K)

If we chase the diagrams, this map turns out to be multiplication.Next, from this, we obtain,

TLWrd = Ann(im µ0)

where the annihilator is a subset of H0(K)∨ annihilating the subset of H0(K).

27.3. Martens Theorem. This theorem will answer the question

Question 27.8. How large can dimWrd be?

Theorem 27.9. (Martens) Suppose 2 ≤ d ≤ g− 1. Then,

dimWrd ≤ d− 2r

and equality holds if and only if C is hyperelliptic.

Remark 27.10. This bound is really much larger than it should be, but it’s a goodexample of how we can use the first order analysis to understandWrd.


Proof. Consider the multiplication map

H0(C,L)⊗H0(C,K⊗ L∨)µ0−−→ H0(C,K)

Note, this is the same map we looked at when proving Clifford’s theorem. Weknow h0(D) = r+ 1,h0(K−D) = g−d+ r. Then, a given canonical divisor can bewritten in at most finitely many ways as a sum of divisors inH0(L) +H0(K⊗ L∨).Therefore,

dim im µ0 ≥ r+ (g− d+ r− 1) + 1 = g− d+ 2r

Therefore,

dimAnn (im µ0) ≤ d− 2rNow, we deal with the hyperelliptic case. If the curve is hyperelliptic, then to

describe a grd, we take r · g12 and add a fixed divisor, which has degree d− 2r.We’ll come back to the converse this Friday. �

Example 27.11. We want to look for a Wrd. To this, we can associate a residuallinear series. What is the first case where Wrd is not residual to a W0d. This is thecase ofW13 in genus 4. Suppose C is not hyperelliptic.

We have a canonical embedding C→ P3, which lies on a unique quadric, eithera smooth quadric or a quadric cone. We could describe the set of g13’s as lines onthe quadric surface. Therefore, we have two g13’s in the case that the curve lies ona smooth quadric surface or one if it lies on a quadric cone.

Question 27.12. IsW13 reduced in this case?

We’re looking at the following map:

H0(L)⊗H0(K⊗ L−1)→ H0(K)

We can see this map is an isomorphism. There are two cases(1) Case 1: The curve is on a smooth quadric. This corresponds to the case that

the product map is an isomorphism, and K⊗ L−1 6∼= L. Here, W13 consistsof two reduced points.

(2) Case 2: The curve is on a singular quadric. Then, K⊗ L−1 ∼= L, and so themultiplication map has a kernel because the line bundle is Serre dual toitself, and the kernel is given by the span of σ⊗ τ− τ⊗ σ. That is, if L ∼=K⊗ L−1, the multiplication map factors through Sym2H0(L). Hence, W13consists of 1 nonreduced point, because the tangent space is 1 dimensional.

Exercise 27.13. Show that this nonreduced scheme has degree 2. This would fol-low from the fact that the curves form a flat family, but can you prove it directly?

28. 11/6/15

Today we’ll discus deformations of line bundles with sections

Question 28.1. What is a parameter space?

Question 28.2. What does it mean to say that the variety Picd(C) “parameterizesline bundles of degree d on C”?

Remark 28.3. There are three levels on which we can answer this question.

68 AARON LANDESMAN

(1) We have a bijection{points of Picd

} → { line bundles of degree d on C }

where this bijection was defined by integration.(2) We have a correspondence respecting the way in which line bundles vary

in families. More specifically, let L be a family of line bundles, we have aline bundle L on B× C so that L|b×C is a line bundle of degree d on C,modulo line bundles on B. Then, the map

B→ Picd

b 7→ L|b×C

is a regular map.(3) “Picd represents the functor of families of line bundles of degree d.” More

precisely, we have an isomorphism of functors

{ schemes }→ { sets }

B 7→ { families of line bundles over B}

B 7→Mor(B, Picd)

In this case, family means a line bundle on the product whose restrictionsare line bundles of degree d on C.

Theorem 28.4. Picd represents the functor sending a scheme to families of line bundlesover B. In the same sense, Wrd \Wr+1d ⊂ Picd parameterizes families of line bundles onC with at least r+ 1 sections. More precisely, families of line bundles on C with at leastr+ 1 sections are line bundles L on B×C so that π∗L is locally free of rank r+ 1.

Proof. The proof is in ACGH, though, according to Joe, it’s fairly incomprehensiblethere. �

Definition 28.5. Define D := Spec C[ε]/(ε2).

Example 28.6. Take B = Spec C[ε]/(ε2). Then, Mor(B, Picd) is simply a tangentvector to Picd, and it can also be thought of as a “fuzzy line bundle” on C. Thissays,

TL Picd ={

line bundles L onD×C : L|Spec C×C ∼= L}

Let L ∈ Picd(C). Then,

TL Picd = { line bundles on D×C : L|C×C ∼= L}

To write down a line bundle, we can specify transition functions. So, choose anopen cover {Uα} of C, we trivialize the line bundle on each open set, and comparethe trivializations.

So, we can take gαβ ∈ O×C(Uαβ)where Uα ∩ Uβ := Uαβ. Then, a sectionσ ∈ H0(L) is given by a collection of functions{

sα ∈ O(Uα) : sα|Uαβ = gαβsβ

}We adapt the convention that gαβ = g−1βα.


Definition 28.7. A first order deformation of L is a line bundle L on D×C so thatL|Spec C×C ∼= L. We describe L via transition functions with respect to the opencover {D×Uα}.

Remark 28.8. We needn’t have hαβ ∈ OC(Uαβ) because gαβ are already invert-ible, so adding any multiple of εwill still be invertible.

Proposition 28.9. We have

TL Picd ∼= H1(C,OC)

Proof. Then, L is given by transition functions gαβ+εhαβ for some hαβ ∈ OC(Uαβ).Write hαβ = φαβ · gαβ so that

gαβ + εhαβ = gαβ(1+ εφαβ)

Note that gαβ satisfy the cocycle condition. Therefore,

1 = gαβ(1+ εφαβ)gβγ(1+ εφβγ)gγβ(1+ εφγα)

= 1+ ε(φαβ +φβγ +φγα)

which is then equivalent to {φαβ ∈ OC(Uαβ)

}is a cocycle in H1(C,OC). That is,

TL Picd ∼= H1(C,OC)

You have to check that if you choose an equivalent cocycle, you would get thesame line bundle.

�

Remark 28.10. We also have H0(K)∨ ∼= H1(O) ∼= TL Picd

Now, suppose we are at a point L ∈ Wrd \Wr+1d so that h0(C,L) = r+ 1. Ourgoal is to describe the tangent space toWrd at L.

Proposition 28.11. We have

TLWrd ={

first order deformations L ofL/D×C : every section σ ∈ H0(C,L) extends to a section of L.}

Further,

σ extends ⇐⇒ σ · v = 0

where σ · v ∈ H1(C,L) is multiplied via the cup product map

H1(C,OC)⊗H0(C,L)→ H1(C,L)

Proof. We want to describe, inside all tangent vectors to L, which is the space offirst order deformations, this is precisely the space of line bundles so that everysection extends.

We now want to describe when a particular section σ ∈ H0(C,L) extends to asection of L. Let v ∈ H1(C,OC). We want to answer

Question 28.12. Does σ extend to a section of the line bundle L corresponding tov.

70 AARON LANDESMAN

Let’s say L has transition functions gαβ and L has transition functions

gαβ(1+ εφαβ)

Say σ is given by a collection

{sα ∈ OC(Uα)}

where

sα = gαβsβ ∈ OC(Uαβ)

We want to know whether there exist a collection of sections

sα + εs ′α ∈ OD×C(D×Uα)

which satisfy the cocycle condition. By satisfying the cocycle condition, we meanthat we require

sα + εs ′α = gαβ(‘ + εφαβ)(sβ + εs ′β)

The key is to look at the term involving ε. That is, we obtain

s ′α = gαβ(φαβsβ + s ′β)

Multiplying this by gβα we get

gβαs′α = φαβsβs

′β

or, rearranging, we have

φαβsβ = −s ′β + gβαs′α

Now, we can realize this as a cup product of two cocycles. That is,

v ={φαβ

}∈ H1(C,OC)

Further,

σ = sβ ∈ H0(C,L)

and the product of these two cyclesφαβsβ is the cup product of these two cocyclesvσ ∈ H1(C,L). So, writing

φαβsβ = −s ′β + gβαs′α

is equivalent to the cup product vσ being a coboundary. In other words,

σ extends ⇐⇒ σ · v = 0

where σ · v ∈ H1(C,L) is multiplied via the cup product map

H1(C,OC)⊗H0(C,L)→ H1(C,L)

�

Finally, we wish to understand the tangent space toWrd.

Theorem 28.13. v ∈ TLWrd if and only if v ∈ Ann(im µ0) where

H0(C,L)⊗H0(C,K⊗ L∨)µ0−−→ H0(C,K)


Proof. By the preceding proposition, we have v ∈ TLWrd if and only if every σ ∈H0(L) extends. This is equivalent to

〈v〉 ⊗H0(C,L)→ 0 ∈ H1(C,L)

So, the relevant cup product map

H1(C,OC)⊗H0(C,K)→ H1(C,L)

Then, we can rewrite this as

H0(C,K)∨ ⊗H0(C,L)→ H0(C,K⊗ L∨)∨

and further by playing with duals, write this as the multiplication map

H0(C,L)⊗H0(C,K⊗ L∨)µ0−−→ H0(C,K)

Therefore, mapping to 0 is equivalent to the statement that v ∈ Ann(im µ0). �

On Monday, we’ll go over many examples.

Exercise 28.14. We’ve been describing explicit description of curves between genus1,2,3,4,5. For example, Brill Noether theory tells us a general curve is a threesheeted cover of P1. We knew that because the rulings of the quadric cut outpencils of degree 3.

The first case we don’t already know tells us that a general curve of genus 6 isexpressible as a 4 sheeted cover of P1.

29. 10/9/15

29.1. Agenda and Review.

Remark 29.1. There are only 9 classes left.

The agenda for the remainder of the semester is to:

(1) More about the schemes Crd,Wrd with examples.(2) Full statement of the Brill Noether Theorem and consequences. In particu-

lar, we’ll see what this tells us about the Hilbert scheme.(3) Proof of part of the Brill Noether theorem. This says that when ρ ≥ 0,

every general curve has a grd and when ρ < 0, the general curve does nothave a grd.

Recall from last time, let C be a smooth curve of genus g.Let L ∈Wrd \Wr+1d . We would like to describe

TLWrd = { first order deformations of L : all r+ 1 sections extend }

= Annim µ0

Here, we recall that first order deformations of L are in natural bijection with aline bundles L ∈ Pic(D× C) with L|Spec C×C ∼= L. Saying that sections extend isequivalent to π∗L is locally free, where π : D×C→ C is the projection map.

Also,

µ0 : H0(C,L)⊗H0(K⊗ L∨)→ H0(K)

72 AARON LANDESMAN

Proposition 29.2. Let D ∈ Crd \Cr+1d . The tangent space to Crd at a divisor D is

TDCrd = Ann im µ̃0

where

µ̃0 = ev ◦ µ0

H0(C,K) ev−→ H0(C,K/K(−D))

Proof. We have

(29.1)

Cd Crd

Picd Wrd

u

Then, taking tangent spaces of this diagram, we have

(29.2)

TDCrd∼= H0(K/K(−D))∨

TL Picd ∼= H0(K)∨

du

Here the map is simply the dual of the evaluation map

H0(C,K)→ H0(C,K/K(−D))

�

Corollary 29.3. We have dimWrd ≥ φ = g− (r+ 1)(g− d+ r) everywhere and µ0 isinjective if and only ifWrd is smooth of dimension ρ at L.

Proof. We knew the dimension bound from an incidence correspondence analysis.Further, if µ0 is injective then the dimension of the tangent space is equal to thisbound. Therefore, since the tangent space is an upper bound for the dimension,the inequality must be an equality. �

Remark 29.4. People often say you can prove that “using deformation theory.”What does this mean? One standard method is to use deformation theory to un-derstand the tangent space to some parameter space. Then, you can calculate thedimension of that tangent space by computing the cohomology group of somesheaf.

This is how 90% of deformation theory works.In 90% of the remaining applications of deformation theory, we write down a

condition, and analyze how deformations look over the dual numbers, and thencheck to see what holds on a given deformation. This is what we did last class.

The remaining 1% of deformation theory are trickier, and these occur when youget involved in higher order deformations.

We’ll see this later today, when we examine the multiplicity of points in Wrd.We’ll see this for genus 4 in class and in genus 6 on the homework.


Example 29.5. (1) Genus 4: We have Wrd = K −Wg−d+r−12g−2−d . The first inter-esting case this comes up in is W13 , not counting hyperelliptic curves. Wehave that C is not hyperelliptic, so

C→ P3

canonically embedded lies in a quadricQ ⊂ P3. Then, either the quadric issmooth, in which case there exist two g13’s on C. Call them L,M. These aredistinct. The sum of these g13’s is a hyperplane section of the curve. Thatis, L⊗M ∼= K and L 6∼= M. These are not isomorphic linear series becauseif a line of the opposite ruling cut a line of the same ruling we would haveto have a g23, violating Clifford’s theorem.

So, this is equivalent to saying L2 6∼= K.We will first need the following lemma.

Lemma 29.6. (Basepoint free pencil trick) Suppose we have σ, τ ∈ H0(C,L) andV(σ)∩V(τ) = ∅, that is, the subspace generated by σ, τ has no basepoints. Then,we have an exact sequence

(29.3) 0 ML−1 M⊕M M⊗ L 0

where the first map is given by multiplication by (−τ,σ) and the second map isgiven by multiplication by σ, τ.

Proof. Not too difficult, see, for instance, Saint-Donat’s article on Petri’stheorem. �

Lemma 29.7.

H0(C,L)⊗H0(C,M)µ−→0 H0(C,K)

is an isomorphism.

Proof. This is a map from the tensor product of two 2-dimensional spacesto a 4-dimensional space, so injectivity is equivalent to surjectivity is equiv-alent to being an isomorphism. The idea to prove injectivity is the base-point free pencil trick.

We have, from the basepoint free pencil trick,

(29.4) 0 KL−2 M⊕M K 0

Here, KL−2 6∼= O, so H0(C,KL−2) ∼= 0. This implies that the map

H0(C,M⊕M)→ H0(C,K)

is injective. �

Next, suppose we have that the quadric Q on which C lies is singular.Then, there is a unique g13 L so that L = K. Then, the image of the multipli-cation map

H0(C,M⊕M)→ H0(C,K)

74 AARON LANDESMAN

is a hyperplane inside H0(C,K), which is precisely H0(C,K − p), wherep is the singular cone point of the quadric. Then, W13 has 1 dimensionaltangent space at L. This tells usW13 consists of 1 nonreduced point.

In this situation, there does exist a first order deformation of the linebundle, at which both sections extend. The next question is:

Question 29.8. What is the multiplicity ofW13 at curves on singular quadrics?

We might guess that this is multiplicity 2, because for curve on a smoothquadric, it has degree 2. The challenge is to show thatW13 has multiplicityprecisely 2 for curves on singular quadrics.

Remark 29.9. This will follow from the proof of the Brill-Noether theorem,which will tell us that this scheme has multiplicity 2 using global aspectsof the geometry.

(2) Now, let’s consider genus 5. Suppose C is non hyperelliptic and non trig-onal. Then, C→ P4 where C = Q1 ∩Q2 ∩Q3 = ∩λ∈P2Qλ. So, we have ag14 on C. We obtain a map

W14(C)→ { singular quadrics containing C}L 7→ QL

and when these quadrics have rank 4, the quadric is a cone over a smoothquadric in P3, and so we get 2 g14’s, one for each of the rulings of thequadric in P3.

So, if the quadric Q has rank 4, then

L 6∼= KL−1

L2 6∼= K

Then, µ0 is injective andW14 is smooth of dimension 1 at L.But, if QL is rank 3, then W14 is singular at L with two dimensional

tangent space.We can now say the singular quadrics containing the curve forms a 1 di-

mensional family, because by Bertini the general member is smooth. Then,the general member accounts for a g14, that that is all g14’s.

We can also see this from a more general proposition (though it wasn’texplained much in class how to apply the following proposition).

Proposition 29.10. Say that C ⊂ Pn is any linearly normal curve. Say {Dλ} isa pencil of linearly equivalent divisors of degree d on C.

Then, dimDλ = n − h0(C,OC(1)(−D)) is the same for all λ. In otherwords, the hyperplanes containing the divisor is H0(C,OC(1)(−D)). Further-more, ∪λDλ is a rational normal scroll.

Proof. Not given. �

(3) Let’s now look at genus 6. TakeW14 ∼ K \W26 . From Brill Noether, ρ = 0 soa general curve C of genus 6will have only a finite number of g14s.

We’ll answer how many g14’s there are (5) in the next class.


Remark 29.11. The Hurwitz space H4,6 has dimension 18 and so H4,6modulo PGL2 has dimension 15 which has dimension equal to M6. So,we obtain a map

H4,6 →M6

which we expect to be dominant. We can prove this by exhibiting a pointin M6 which a finite nonempty fiber. So, if we have an isolated fiber ofthis map, it follows that the map is dominant. In other words, it suffices toexhibit a curve with a finite number of g14’s.

30. 11/11/15

30.1. Review. Today, we’ll start with Joe’s embarrassing confession regarding show-ing deformation theory for curves of genus 6.

Let’s recall what happened last time.Recall if C were a non hyperelliptic genus 4 curve then we looked at W13 ’s. We

had 2 cases.(1) If Q is smooth then there are 2 reduced g13’s L,M with L ⊗M ∼= K and

L 6=M. So,W13 consists of two reduced points.(2) If Q is singular, there exists a unique g13 corresponding to a semicanonical

L, i.e., L⊗2 ∼= K. In this case

µ0 : H0(C,L)⊗H0(C,K⊗ L−1)→ H0(C,K)

we obtain that this map factors through the 3 dimensional

(30.1)

H0(C,L)⊗H0(C,L) Sym2H0(C,L)

H0(C,K)

which is only 3 dimensional. Therefore, µ)9 has some kernel. So, W13 isa nonreduced point of multiplicity at least 2. We claimed, last time that ithas multiplicity exactly 2. The proof would be along the lines of the com-putation of the tangent space. We asked there when sections extended to adeformed vector bundle. We can go back to that here, and ask whether itextends to Spec C[ε]/ε3. If we show we can’t extend the first order defor-mation and sections to a second order deformations.

The embarrassing part is that Joe remembered having done this as agraduate student. Joe’s embarrassing confession is that he wasn’t able toreproduce that result after graduate school.

Exercise 30.1. Try seeing why this this W13 , using the method describedabove, multiplicity precisely 2.

After we see Brill Noether theory, it will follow that this has degree 2,but there should also be a deformation theoretic argument for this.

Remark 30.2. Since W13 is 0 dimensional but has 1 dimensional tangentspace, it must be Spec C[ε]/εk for some k. Therefore, if we could extend itto k but not k+ 1, it would have multiplicity precisely k.

76 AARON LANDESMAN

Remark 30.3. This is a general part of the philosophy of deformation theory: Ifyou only need second order deformation theory, you’re golden. If you have to gofurther, you’re screwed.

30.2. The Genus 6 Canonical Model. Now, let’s move on to studying curves ofgenus 6. So, let C be non hyperelliptic of genus 6. We have a canonical embedding

Cφ−→K P5. We obtain that C lies on 6 quadrics and has degree 10. Then, C =

∩Q⊃CQ, so long as C is not trigonal or a plane quintic.Then, ifC is general we have, by Brill Noether theoremW12 =W13 = ∅,W25 = ∅.

We can also show the above directly because these are only maps to P1, P2 via theSeveri variety and Hurwitz variety.

Example 30.4. If C ⊂ P2 is a plane quintic, we know that the Hilbert scheme ofplane quintics is P20 ∼= PV . Then, this space PV/PGL3 has dimension 20−8 = 12.But, dimM6 = 15, so the general curve will have emptyW25 .

Next, we note that W14 = K−W26 , which is 0 dimensional by the Brill Noethertheorem.

We will aim to prove this statement and also explain how to find all the g14’s ona general curve of genus 6.

Proposition 30.5. The space of g14’s on a genus 6 curve is 0 dimensional.

We’ll give two proofs.

Proof 1. Note the Hurwitz space H4,6 is 18 dimensional. So, H4,6/PGL2 is 15 di-mensional, and we have a map

H4,6/PGL2 →M6

is a map of two fifteen dimensional varieties. We want to show this map is domi-nant. To show this, it suffices to show it is generically finite. To show this, we onlyneed exhibit a single curve C ∈M6 so that the fiberW14(C) is finite.

So, let C0 ⊂ P2 be a plane sextic with nodes at p1, . . . ,p4 with no 3 collinear.Let C→ C0 be the normalization.

Question 30.6. what is the locus ofW14(C)?

By geometric Riemann Roch, takeD =∑i qi We haveD ∈ C14 if and only if the

qi fail to impose independent conditions on H0(C,K). We know

H0(C,K) ={g(x,y) dx

fy: degg = 3,g(pi) = 0

}That is, |K| is cut on C by cubics on P2 through p1 . . . ,p4 Now, equivalently, con-sider

S = Blp1,...,p4P2π−→ P2

Let

` = π∗ of the class of a line in P2

ei = class of the exceptional divisor over pi

e =

4∑i=1

ei


Then, let C be the proper transform of C0 in S. We obtain

C ∼ 6`− 2e

KS ∼ −3`+ 3e

KC ∼ (3`− e)|C

So, we have an exact sequence

(30.2) 0 OS(−3`+ e) OS(3`− e) KC 0

By Serre duality, H1(C,OS(−3`+ e)) = 0, so the second map is surjective. We canalso see this directly from our explicit description of H0(K) as cubics. Now, notethat the following are equivalent:

(1) D ∈ C14(2) qi fail to impose independent conditions on quadrics.(3) {p1, . . . ,p4,q1, . . . ,q4} fail to impose independent conditions on cubics.

Remark 30.7. Note that we’re going to assume these points are distinct.There are 2 ways to deal with this. Since C has no g13, this g14 has no base points.We can also do it by proving the following lemma scheme theoretically, mean-

ing that the 8 points need not be distinct. and it turns out that we will get the sameanswer.

Lemma 30.8. Let Γ = {p1, . . . ,p8} ⊂ P2. Then, Γ fails to impose independent conditionson cubics if and only if either

(1) 5 of the points are collinear or(2) all 8 lie on a conic.

Note that if 4 points are collinear, the cubic contains the line, so the 5th pointdoesn’t impose any additional conditions. Similarly, by Bezout, if the cubic con-tains 7 points on the conic, it contains the conic, so the 8th point doesn’t imposeany more conditions.

Proof. To prove this, we’ll need another lemma.

Lemma 30.9. 7 points fail to impose independent conditions on cubics if and only if 5 arecollinear.

Proof. If 7 points p1, . . . ,p7 fail to impose independent conditions on cubics, thenfor some point pi, any cubic containing pj for j 6= i must contain pi. Relabel thepoints so that i = 1. Let’s start by pairing up p2, . . . ,p7 into three pairs. So, wehave cubic which is a union of three lines, containing p1. So p1 is collinear with atleast two other points, say p2,p3. Let L be the line containing p1,p2,p3.

(1) Suppose L contains a 4th point, p4. Now pair up p2,p3,p4 on the line withp5,p6,p7 off the line. So, this is 3 lines, which is an reducible cubic, and soone of these three lines must equal L. So, L contains a 5th point and we’redone.

(2) To finish, suppose that L contains only 3 points. We’ll see this next time.�

�

�

78 AARON LANDESMAN

We can also see the above via the Severi variety. We’ll get a little extra informa-tion out of this proof than the one for the Hurwitz space.

Proof 2. Recall the Severi variety V6,6 of plane curves of degree 6 and geometricgenus 6, containing the subset U6,6 of sextics with four nodes. Inside U6,6 we havea subset W which is the set of 4 nodes in linear general position.(30.3)

V6,6 = { plane curves of degree 6 and geometric genus 6 }

U6,6 = { sextics with 4 nodes }

W = { 4 nodes in linear genera position, so that no three are collinear. }

Now, note that we have a map

U6,6 → {(p1,p2,p3,p4) ⊂ (P2)4 : not all 4 points are collinear}

Note not all 4 points can be collinear because then the curve would have degreeat least 2 · 4 = 8. Now, note that W is 23 dimensional. Then, W/PGL3 → M6are both 15 dimensional. Again, it suffices to exhibit a curve of genus 6 with afinite number of sextics with 4 nodes, in its preimage under the above map. Sincethose points not in general linear position form a closed subscheme, the map fromW → M6 cannot be dominant, so we only we will obtain that the general curvewill have non collinear nodes.

Remark 30.10. This will also imply that a general genus 6 curve can be realized asa plane sextic with 4 nodes.

We saw the existence of this curve in the above proof because having a W14 isthe same as having aW26 by Serre duality. �

31. 11/13/15

There are only 7 classes left. There is no homework this week. There will betwo more homework assignments, one due before thanksgiving and one due thelast week of classes. They’ll probably be due on Fridays.

Let’s come back to the Lemma from last time.

Lemma 31.1. If Γ is a collection of 8 points in P2, then Γ fails to impose independentconditions on cubics if and only if either

(1) Γ contains 5 collinear points or(2) Γ lies on a conic curve (i.e., all 8 points lie on a conic)

Proof. For this, we’ll need a sublemma, which we didn’t finish proving last time.

Lemma 31.2. If Γ consists of 7 points and it fails to impose independent conditions oncubics then Γ contains 5 collinear points.


Proof. The hypothesis says that for any one point, call it p1, we have that everycubic containing p2, . . . ,p7 also contains p1. Divide p2, . . . ,p7 into three pairs.Draw the lines through these pairs. We obtain that p1 is contained in this cubicand so p1 is collinear with two other points, say it is collinear with p2,p3.

Now, we have three collinear points p1,p2,p3. The remaining 4 points can beanywhere. We now have a bifurcation. Either there is a 4th points on L or not.We’ll use Lij for the line joining pi,pj.

(1) If there are 4 points on a line, say p1, . . . ,p4 lie on a line. Then, look at thethree lines respectively joining p2,p5 and p3,p6 and p4,p7. So, we musthave p1 on one of these three points, meaning one of the remaining threepoints lies on L. So, we have 5 points on L.

(2) Suppose there are precisely 3 points on L. Now, take the lines joiningp2,p4, call it L24 and L35 and L67. Therefore, p1 must lie on one of thesethree lines. We cannot have p1 on L24 or L35. Therefore, p1 lies on L67.Similarly, we can take L24 + L36 + L57. Therefore p1 also lies on L57. And,similarly, p1 lies on L47.

Therefore, L47,L57,L67 all contain p1, meaning that p1,p4,p5,p6,p7are 5 collinear points.

�

By the above sublemma, we may assume any 7 of the 8 points impose indepen-dent conditions. We define Lij := pipj.

So, any cubic containing 7 of the 8 points also contain the 8th point.Choose any three points p1,p2,p3 which are not collinear. Let Q be a conic

curve through the remaining 5 points. Now, look at the cubics L12 ∪ Q,L23 ∪Q,L13 ∪Q. We’re assuming p3 /∈ L12 does not lie on L12, so it lies onQ. Similarly,p1 ∈ Q and p2 ∈ Q. Hence, all 8 points lie in Q. �

Remark 31.3. This lemma is the beginning of an elementary branch of algebraicgeometry called interpolation.

In this context we’re asking about the map

(31.1) H0(P2, IΓ ) H0(P2,OP2(3)) H0(P2,OΓ (3))

and the above lemma is saying this map is surjective unless 5 points lie on a lineor 8 lie on a cubic.

That is, the hypothesis of the lemma says

h0(P2, IΓ (3)) = 3

The sublemma says for any Γ ′ ⊂ Γ do impose 7 independent conditions, thenh0(P2, IP2(3)) = 3.

Interpolation is a very broad, elementary subject. They can be fairly difficult,but can be very useful when you can prove them.

Now, let’s see what the lemma is good for. We’re trying to understand curvesof genus 6. Now, we’ll exhibit a curve of genus 6 and analyze its W14 and W26 .We’ll show the curve in question has only finitely many g14’s and finitely manyg26’s. Therefore, we’ll obtain that the curve we’ll see is “a general curve of genus6.”

80 AARON LANDESMAN

Proposition 31.4. There exists a curve C so that C has precisely 5 g14’s and these g14’s arereduced points ofW14(C).

Proof. Now, let’s return to genus 6. So, choose 4 points, p1, . . . ,p4 ∈ P2, so thatno three are collinear. In fact, any 4 points in linear general position can be carriedinto any other 4 points by PGL3. So, we may as well choose the points to be thecoordinate points and p4 = [1, 1, 1], if we like.

Now, let C0 ⊂ P2 be a general plane sextic which has some singularity at all pi.That is, C0 ∈ |I2Γ (6)|, where Γ is our set of 4 points. We claim

Lemma 31.5. We claim C has simple nodes at p1, . . . ,p4 and is otherwise smooth.

Proof. We can use Bertini’s theorem. A general member of a linear system issmooth outside the base locus of the linear series. We only need show the lin-ear series has no basepoints other than p1, . . . ,p4. That is, given 4 points and a5th point, we want to show we can find a sextic double at these 4 points and notpassing through a 5th point. We can just take some double lines and a conic to dothis. How do we see this curve is simply nodal, at the base points?

So, to see what happens, we can blow up the 4 points. Let

(31.2)

C S = Bl{p1,...,p4}P2

C0 P2

in PicS let ` be the pullback of a line, ei be the class of the exceptional divisor overpi and e =

∑i ei. Suppose we could apply Bertini’s theorem. Then, using Bernini

again, we see that the curve is smooth, which will meet the curve at two distinctpoints, hence it will be a node. So, it only remains to show there are no basepointson the exceptional divisors. To do this, we only need to exhibit some lines whicharen’t tangent to certain points. We can do this by just taking lines joining variouspoints. �

Now, we have

(31.3)

C S = Bl{p1,...,p4}P2

C0 P2

In this situation, we know |KC| is cut by cubic through the four points p1, . . . ,p4.We saw this explicitly, that

ω ∈ H0(KC)

ω =

{g(x,y) dx

fy: g(pi) = 0, degg = 3

}Say |D| is a g14 on C. Then, D =

∑i qi. A general divisor will have distinct points.

Question 31.6. What makes a divisor a g14?


This moves in a pencil if and only if it fails to impose independent conditions onKC. That is, we want to know whether q1, . . . ,q4 fail to impose independent con-ditions on |KC|. This is equivalent to whether p1, . . . ,p4,q1, . . . ,q4 fail to imposeindependent conditions on cubics.

We can now use the above Lemma 31.1, either all 8 of these points lie on a conicor 5 of them are collinear. In the latter case, We can’t have 3 points qi collinearwith 2 points pj because then the line would have intersection number 7 with thecurve, but it’s only a cubic. So, we would have q1, . . . ,q4 collinear with pi.

So, either(1) |D| is cut out on C by conics through p1, . . . ,p4 or(2) |D| is cut out on C by lines through pi for some i.

The Lemma tells us these are the only g14’s. Hence, the fiber dimension over theHurwitz moduli space M6 is finite at this curve, and so, there exist exactly 5 g14’son C.

Question 31.7. What about the tangent space Tu(D)W14?

Let’s say |D| is the linear series cut by conics. The other cases will be similar. Theseries residual |K−D| is cut by lines. That is, it is the g26 giving the map C → P2.In other words, D ∼ 2`− e in the Picard group of the blowup. Since

KS ∼ −3`+ e

C ∼ 6`− 2e

KC ∼ 3`− e

D ∼ 2`− e

K−D ∼ `

So, we have a map

H0(C,D)⊗H0(C,K−D)→ H0(C,K)

which is a map from a 2 dimensional vector space tensor a three dimensional vec-tor space to a 6 dimensional space. So, to check its an isomorphism, it suffices toshow its either injective or surjective. Now, this is precisely

H0(P2, Ip1,...,p4(2))⊗H0(P2,OC(1))→ H0(P2, Ip1,...,p4(3))

We can now forget the curve, and just look at things in the plane. The four pointsare a complete intersection of conics. So, the homogeneous ideal Ip1,...,p4 is gen-erated in degree 2 by conics, and so the map is surjective.

Therefore,W14(C) is reduced.

Exercise 31.8. Check that the other four points are reduced.

�

Next time, we’ll see special curves of degree 6 with nonreduced points at thesepoints.

32. 11/16/15

32.1. Logistics and Review. Six classes left!(1) Today, we’ll finish up with the discussion of curves of genus 6(2) We’ll see the full statement of Brill Noether theory for now.

82 AARON LANDESMAN

(3) We’ll discuss the plan for the rest of the semester.Recall that the idea from last time is that we wanted to write down a general

curve of genus 6. The curve we’re about to describe is a general curve of genus 6.We choose points p1, . . . ,p4 ∈ P2, so that no three are collinear. Then, we choosea general member of the linear series

C0 ∈ H0(P2, I2p1,...,p4(6))

which is a 16 = 28− 3 · 4 dimensional vector space. We will have C0 is nodal atp1, . . . ,p4 and otherwise smooth. Let

Cν−→ C0

be the normalization. We saw C has precisely 5 g14’s.(1) One is cut by conics containing p1, . . . ,p4.(2) Four are cut by lines containing a particular pi.

We checked last time that if D is a divisor cut by conics through the 4 points, then

H0(C,D)⊗H0(C,K−D)µ−→ H0(C,K)

is surjective. We saw this because, when we forgot the curve and looked in theplane, we saw that the ideal of the four points is generated in degree 2. Becausethe 4 points are a complete intersection, this follows from the Noether af + bgtheorem. Hence, the above map, corresponding to a point u(D) ∈W14 is a reducedpoint.

We can also say this in the following way: Introduce

(32.1)

C S := Blp1,...,p4P2

C0 P2.

Then,

KS ∼ −3`+ e

C ∼ 6`− 2e

KC ∼ 3`− e

So, here on S, the map

H0(C, 2`− e)⊗H0(C, `)→ H0(C, 3`− e)

This is an isomorphism on the curve because it is also an isomorphism on thesurface.

32.2. A continuing study of genus 6 curves. We can similarly compute that themultiplication map is an isomorphism for the other g14’s of lines through points.We can also do this in two ways:

(1) First, using the same trick, we want to check that the map

{ lines containing pi}⊗{

conics containing pj, j 6= i} → { cubics containing p1, . . . ,p4}

and check this is an isomorphism on P2.


(2) We can also see that there is no distinguished g14 in the following way.Recall

(32.2)

C S P5

C0 P2

φ|3`−e|

This realizes S as a del Pezzo surface in P5. The curve maps to its canon-ical embedding in P5. That is, the canonical model C ⊂ P5 lies on a delPezzo surface. Since the class of the curve is twice the class of the anti-canonical class, we see it is the complete intersection of a del Pezzo quinticsurface and a quadric hypersurface, once we check that the del Pezzo isprojectively normal.

Remark 32.1. We see genus 5 was the last time that a general curve isa complete intersection. The best description of genus 6 curves is as anintersection of a del Pezzo surface and a quadric. You can also describecanonical models of curves up through genus 10. Beyond that point, ourknowledge of canonical curves drops off. We don’t know much at all aboutthe geometry of a canonical curve in genus more than 10.

Lemma 32.2. Let S be a quintic del Pezzo surface in P5. Then, S contains 10lines.

Proof. The images of the 4 exceptional divisors determine 4 lines. We canalso consider a line in P2 passing through 2 of these points. This is alsomapped to a line in P5. The class of the line is 2` − ei − ej which hasintersection number 1with the class of the anti-canonical series 3`− e.

There is a simple way to draw these lines as well. with the four excep-tional divisors and Lij meeting Ei and Ej at a point.

�

Proposition 32.3. The curve C has 5 g26’s and 5 g14’s which are residual to eachother.

Proof. There exist 5 collections of 4 tuples of disjoint lines. For example,we can take L13,L23,L12,E4.

So, there exist five collections of 4 disjoint lines. We get five differentrepresentations of S as a 4 fold blow up of P2.

So, there exist five representations of C as a plane sextic with 4 nodes.�

Remark 32.4. There appeared to be four g14’s which were different in thatthey were cut by linear series, while one was cut by quadrics. However,that depended on choosing a realization of the surface as a blow up of 4points. However, we can blow down this surface at any of the five config-urations of 4 disjoint lines, and each of the 5 linear series becomes that cutby quadrics in one of these configurations. Because of this, we don’t havethe check the separate case in which the g14 is lines through a pi, to see it isa reduced point ofW14 .

84 AARON LANDESMAN

Remark 32.5. There are several different possible descriptions of the W14depending on the canonical model of a curve. Let’s assume the curve is nothyperelliptic or trigonal and is not bi-elliptic (mapping 2:1 onto an ellipticcurve or plane cubic). Generically, a curve will have 5 g14’s. However, thereexist curves with any number between 1 and 4 curves. We don’t want tolook for these curves on a del Pezzo surface, but rather, we’ll have to lookfor them on some surface in the same irreducible component of the Hilbertscheme of a del Pezzo surface.

Example 32.6. Here’s an example of a curve of genus 6. Let’s look at thecase when three points are collinear. We’ll see there are only 4 g14’s. Thishappens because, in effect, the g14’s cut by conics containing all 4 pointsalso contains the line L. This is precisely the same as the g14 cut by linescontaining the fourth point. We would expect in this case thatW14 is nonre-duced. We then get a map

H0(C,D)⊗H0(C,K−D)→ H0(C,K)

This is no longer an isomorphisms because all conics containing the firstthree points contain the line through the first three points. The image willthen be 5 dimensional. A conic through the 4 points will contain the line Lbut a cubic through them does not have to contain L. So, we see Tu(D)W

14

is 1 dimensional here. Then, we see the space consists of 4 points whichare reduced and 1 which has a 1 dimensional tangent space and this pointturns out to be a double point, although it requires some work to show ithas multiplicity exactly two.

When we take the blow up at these 4 points, we see the line through thethree points L has class `− e1 − e2 − e3 which has intersection number 0with the embedding class 3`− e, and so this line gets collapsed, and weobtain a singular del Pezzo quintic, with an ordinary double point.

The del Pezzo surface that contains the canonical curve is unique.

Question 32.7. As you vary the curve, how does the del Pezzo vary? Whenthe curve becomes bi-elliptic or trigonal, what happens to the curve?

For example, trigonal curves turn out to lie on the union of a quartic delPezzo in a hyperplane P4 ⊂ P5 union with a plane in P5.

Here is some food for thought, for the next class.

Question 32.8. Let’s observe(a) A curve of genus 0 has a unique g11.(b) A curve of genus 2 has a unique g12.(c) A general curve of genus 4 has two g13’s.(d) A general curve of genus 6 has five g14’s.

A general curve of genus 8 has a finite number of g15’s. How many?

Could it be that the answer is 14, and this traces out the Catalan num-bers?

33. 11/18/15

33.1. Logistics and Overview. Five classes remaining!


Today, Friday, Monday, and then Monday and Wednesday after thanksgiving.Today, we’ll see the full Brill-Noether Theorem

Definition 33.1. A scheme is connected in codimension 1 if between any twopoints, there exists a chain of components so that each pair of adjacent componentsmeet in codimension 1.

Theorem 33.2. (The full version of the Brill-Noether Theorem)Assume C is a general smooth curve of genus g over any algebraically closed field of

characteristic 0.(1)

dimWrd = dimGrd = ρ

where

ρ = g− (r+ 1)(g− d+ r)

and

dimCrd = ρ+ r.

In particular,

Wrd 6= ∅ ⇐⇒ ρ ≥ 0.(2) (Gieseker-Petri Theorem) For any invertible sheaf L on C,

µ0 : H0(C,L)⊗H0(C,K⊗ L∨)→ H0(C,K)

is injective.(3)

(Wrd)sing =Wr+1d

(Crd)sing = Cr+1d

(Grd)sing = ∅

(4) Set

Wg−1 ={L ∈ Picg−1 : h0(C,L) > 0

}= im u(Cg−1) ⊂ Picg−1

where u : C→ Pic1 is the Abel-Jacobi map. Define

θ :=[Wg−1

]∈ H2(J, Z)

the theta divisor in Picg−1. Then,

[Wrd] =

(r∏i=1

i!

(g− d+ r+ i)!

)θg−ρ.

Here are some special cases:(a) In particular, if ρ = 0, since θg = g! ∈ Z ∼= H2g(J, Z) we have

#Wrd = g!

r∏i=1

i!

(g− d+ r+ i)!.

(b) In particular, by the previous part,Wrd is smooth for a general curve, and aslong as there are only finitely many, this is the number, counting multiplicity.

86 AARON LANDESMAN

(c) In particular, if ρ = 0, r = 1, so g = 2k,d = k+ 1, we have

#W1d =(2k)!

k!(k+ 1)!

(5) If ρ > 0 then Wrd is connected in codimension 1. By the first and third parts,Wrd is irreducible. If ρ = 0, the monodromy of Wrd over Mg is transitive. It isthe symmetric group unless d = g− 1 in which case it is a wreath product. Inparticular, Wrd is irreducible.

(6) Let L ∈Wrd be general. Then,L is very ample if r ≥ 3φL : C→ P2 is birational and the image is nodal if r = 2φL : C→ P1 is simply branched if r = 1

(7) Let H be the open subset of the Hilbert scheme corresponding to smooth nonde-generate curves of degree d and genus g in Pr If ρ ≥ 0 there exists a uniquecomponent of H dominating Mg. When ρ < 0 there is no such component.

Proof. Here are some of the ideas of the proof: Parts 4, 5, and half of 1 follow froma Porteous’ formula calculation. There are in fact two ways to describe this as aPorteous’ formula calculation.

(1) We can say

D ∈ CrD ⇐⇒ rk(H0(C,K)→ H0(C,K|D)

)≤ d− r

Then, we can think of this map as given by a d×gmatrix, and those vectorspaces vary algebraically with the choice of the divisor. We introduce vec-tor bundles Eg (meaning E has rank g) and Fd on Cd where E = H0(C,K)is the trivial bundle of rank g and FD = H0(C,K|D). Then, defineφ : E→ Fto be the evaluation map. By Porteous’ formula, when the locus at whichthe matrix drops rank, the class of this locus is determined by the Chernclasses of the vector bundle. This cohomology is calculated in the coho-mology ring of Crd. You can also calculate the Chern classes of E becauseit is the trivial bundle, and you can calculate F by Grothendieck RiemannRoch. See Chapter 8 of ACGH for a proof. This allows us to determine theclass in part 4. Similarly, it lets us find the class as in part 1. This also tellsus the dimension of Crd is at least ρ+ r everywhere. That is, it determinesthe existence part of part 1. We can then set up an analogous calculationon Pic.

One can then do a parallel setup to find the class ofWrd.(2) Here is another way to carry out Porteous’ formula. On Picd, fix D =

p1 + · · ·+ pm for m � 0. We can form the vector bundle E on Picd. Wehave fibers

EL = H0(C,L(D))

and, we can “fit these together to form” a vector bundle E of rankm+ d−g+ 1. We can also form F by piecing together the fibers

FL = H0(C,L(D)|D).


Then, the locus

Wrd ={L ∈ Picd : rk φ|D ≤ m+ d− g− r

}⊂ Picd

Remark 33.3. However, it is slightly more tricky to construct these vectorbundles than those in the previous part. And, in this case, it’s not hard tosee

E∨ ⊗ F ∼= Hom(E, F)

is an ample vector bundle. This means that we can deduce the nonemp-tyness of the degeneracy locus: According to Fulton and Lazarsfeld, thedegeneracy locus is nonempty when E∨ ⊗ F is ample. This nonemptynesscan be found in chapter 7 of ACGH.

However, Fulton and Lazarsfeld showed that when E∨ ⊗ F is ample,the degeneracy locus are always connected in codimension 1 when theexpected dimension of the degeneracy locus is positive.

�

Exercise 33.4. In the case of genus 2, 4, 6we can not only count the number of g1r ’s,but in fact identify them. For example, in genus 6, we can identify the g14’s as the 4pencils through a point and the pencil of conics through 5 points. In genus 8, canyou find the 14 g15’s?

As my own guess, perhaps we can use that it has degree 14 and lies on a thegrassmannian G(2, 6).

On Friday, we’ll talk about the proofs of the remaining parts.

34. 11/20/15

34.1. Plan, conventions, and review.

34.1.1. Plan. There are only 4 classes left!Today, we’ll discuss the proof of Brill Noether theorem. Next class, we’ll learn

about plucker points and inflection points of linear series, and after thanksgiving,we’ll come back to the proof of Brill Noether.

34.1.2. Conventions. We’ll adapt the convention that if the dimension of a schemeis positive then the scheme is nonempty.

34.1.3. Review. We talked last time about the existence part of Brill Noether. Moregenerally, we discussed the case dimWrd ≥ ρ. This was proved by a Porteouscalculating in the late 1960s by Kleiman and Laksov, and, simultaneously, Kempf.

Mumford had the idea and told the two parties above, and these two partiesgot the result at the same time.

Remark 34.1. This lower bound dimWrd ≥ ρ holds for every single curve. How-ever, the reverse inequality only holds for a general curve.

88 AARON LANDESMAN

34.2. An Upper bound on the dimension. The other half of the proof involves avariational element, in the sense that we’ll have to choose general curves.

We want to show that dimWrd ≤ ρ and is empty if ρ < 0.To prove this upper bound, it is sufficient to exhibit a single smooth curve C of

genus gwithWrd(C) = ρ.

Remark 34.2. However, no one has ever written down a Brill-Noether generalcurve in arbitrary genus.

It’s not hard to write down special curves like plane curves of degree 9 with 5double points. However, a general curve of degree 23 shouldn’t have a g29, so thiswon’t be such a curve.

34.3. Proof of Existence for Brill Noether. Every proof of Brill Noether which hasever been proposed involves a specialization argument.

We’ll look at the proof originally suggested by Castelnuovo. Castelnuovo saidwe can’t write down a smooth curve, but we can look for singular curves, withonly relatively simple singularities. In this case, the geometric genus will be lower,hence simpler.

Castelnuovo suggested a g-nodal curve of arithmetic genus g. That is, we takeP1 with g pairs of points identified. So, a linear series on this curve can be pulledback to a linear series on P1, which are well understood.

So, here is the idea:Suppose we have a grd on C0. Then, we can pull it back to a grd on P1. On

P1, there’s a unique invertible sheaf of degree d, namely OP1(d) with an r + 1dimensional subspace of H0(OP1(d)).

To describe all grd’s on P1, we take

P1φ|O

P1(d)|

−−−−−−→ Pd

Then, a grd on C is simply a sub-linear series. That is, it is a linear space Λ ⊂ Pd ofdimension d− r− 1. So, the grd is{

H∩P1 : Λ ⊂ H}

So, the space of grd’s is simply the grassmannian G(d− r,d+ 1).Therefore the grd’s on C0 correspond to a grd, D on P1 with the property that, if

pi,qi are two preimages of a node under the normalization map

pi ∈ D ⇐⇒ qi ∈ D, for all D ∈ D

Translating this to the language of hyperplanes, we’re saying that for all hyper-planes H ⊃ Λ, we have

pi ∈ H ⇐⇒ qi ∈ H

Equivalently,

pi ∈ Λ,pi

or, equivalently,

pi,qi ∩Λ 6= ∅

Where a,bmeans the linear space spanned by a and b.


Therefore, grd’s on C0 are equivalent to

{Λ ∈ G(d− r,d+ 1) : Λ∩ Li 6= ∅, i = 1, . . . ,g}

So, we’re looking for planes meeting a collection of lines, which is really just aproblem in Schubert calculus.

Let’s first examine the dimension count.

Lemma 34.3. The “expected dimension of

{Λ ∈ G(d− r,d+ 1) : Λ∩ Li 6= ∅, i = 1, . . . ,g}

is ρ = g − (r + 1)(g − d + r), and this is an upper bound on the dimension of suchconfigurations of lines.

Proof. Let

Σ(L) = {Λ : Λ∩ L 6= ∅}

We will first count the dimension of this.We’ll call G := G(d− r,d+ 1). To find the dimension, we look at the incidence

correspondence

Φ := {(p,Λ) : p ∈ Λ} ⊂ L×G

(34.1)

Φ

L Σ(L) ⊂ G

Then, the left map has fibers isomorphic to G(d− r− 1,d) and so the dimension is

dimΦ = (r+ 1)(d− r− 1) + 1

= (r+ 1)(d− r) − r

Hence, since dimG(d− r+ 1,d+ 1) = (r+ 1)(d− r), we have that Σ(L) ⊂ G hascodimension r.

Now, if we have g lines, this has expected dimension

∩Σ(L) = (r+ 1)(d− r) − rg

= g− (r+ 1)(g− d+ r)

= ρ

�

Remark 34.4. There are two reasons why we are not finished with the proof:(1) However, the problem is that these lines are not general, and so we can’t

just say we’re done by Kleiman’s theorem. This is because the lines arechosen specially to intersect two points on the rational normal curve.

(2) The dimension ofWrd is upper semicontinuous in families. We’d like to saythat dimWrd, is upper semicontinuous. So, if the dimension on the specialfiber is equal to the maximum ρ, then it must be equal to ρ everywhere.

However, this is sitting inside Picd(C0), which is not a compact variety.The upper semicontinuity of dimension only applies to proper varieties.

We have Picd(C0) ∼= (C×)g. If we have a family Picd(C/∆) it might bethat the limit of grd’s on Ct may not be a grd on C0. A locally free sheaf of

90 AARON LANDESMAN

rank 1 on the general fiber could specialize to a torsion free sheaf of rank 1which may not be locally free.

Kleiman and Altman dealt with the second problem They showed the only way agrd can fail to be a grd on C0 is if it acquires basepoints at the nodes.

We can then handle this, because if a limit of grd has basepoints at δ of the nodesr1, . . . , rδ, We then have a linear series of degree δ lower, but of dimension r on thepartial normalization of C0 at the δ nodes.

Then, the dimension of the family of such limits is at most ρ(d− δ, r,g− δ) < ρ.So, the locus of grd’s whose limits are not grds has strictly smaller dimension.

So, it is sufficient to show dimWrd(C0) = ρ.That is, we only need to show the cycles

Σ(Li) ⊂ G(d− r,d+ 1)intersect properly.

Example 34.5. Here are some examples of the problem so far:(1) To show there does not exist a g12 on a general curve C of genus 3, we’re

saying that we can embed P1 by the complete linear series. Then, g12’s onthat curve correspond to points, and the condition is that the point lie onthree chords. But, three general chords to a conic are not concurrent. So,we’re done.

(2) To show there does not exist a g13 on a general curve C of genus 5, we canspecialize C to C0, a curve with 5 nodes. we embed the normalization ofC1 as a twisted cubic. Now, if we take 5 general chords to a twisted cubic,no line meets all 5. No line in P3 meets all 5.

35. 11/23/15

35.1. Review. There are only three classes left: Today and Monday and Wednes-day next week.

Recall the set up from last time.Consider a family

C→ ∆

with Ct smooth of genus g for t 6= 0 and C0 is g-nodal.That is, C0 ∼= P1/ {pi ∼ qi}

gi=1.

Question 35.1. How do we know we construct such a family?

We start with C and choose 2g general points and form C0. Then, we argue thatwe can find a family whose special fiber is C0 and whose general fiber is smooth.Then, general fiber of the family will satisfy the Brill-Noether theorem, by uppersemicontinuity of dimension ofWrd in families. That is,

dim (Wrd(C0)) ≤ ρ =⇒ dim (Wrd(Ct)) ≤ ρfor general t.

However, there is an issue that this is not proper over the disk. So, it’s possiblewe could have a component of Wrd(Ct) that appears over the special fiber thatdisappears on the general member. However, as we saw last time, Kleiman andAltman bounded the dimension of such components, and so we can still make theabove hold true for a general such family.


Now, given Wrd(C0), we can pull a linear series back to the normalization P1.We can then pull this back to a linear series D on P1 of degree d and dimension r,so that pi ∈ D ⇐⇒ qi ∈ D for all D ∈ D, 1 ≤ i ≤ g.

Then, the subspace of dimension r+ 1 corresponds to the projection away froma plane Λ ∼= Pd−r−1, so that there is a line Li joining pi and qi on the curve andmeeting Λ.

Recall the notation

Λ ∈ ∩iΣ(Li) ⊂ G(d− r,d+ 1).where Σ(Li) is a Schubert cycle. Recall the subspace of lines meeting Li is codi-mension r, so the subspace of g lines meeting Λ is codimension gr, which turnsout to be precisely ρ, as we saw last time.

Hence, the question is reduced to:

Proposition 35.2. For general chords Li to a rational normal curve, the cycles Σ(Li)intersect properly.

Remark 35.3. We’ll focus on the case when ρ is negative and show that Wrd of ageneral curve is empty.

Example 35.4. Recall the examples from last time:(1) “A general curveC of genus 3 is not hyperelliptic” amounts to the assertion

that when we embed C as a conic in P2, three general curve to a conic inP2 are no concurrent.

(2) “A general curve C of genus 5 is not trigonal“ amounts to the assertionthat if L1, . . . ,L5 ⊂ P3 are general chords to a twisted cubic, then no lineL ⊂ P3 meets all 5 such lines. Here, we see the problem arising for the firsttime. This time, not every chord is a chord to a twisted cubic.

Here is a sneaky way to do this, which was not the original proof, but itwas the second proof.

Remark 35.5. Note, g nodal curves only have dimension 2g − 3, insidethe moduli space of dimension 3g− 3. However, their behavior in generalmirrors that of a general curve with respect to the Brill Noether theorem.

The sneaky way is this:

Lemma 35.6. You can have three chords to a conic that are concurrent, but youcan’t have three tangent lines which are concurrent.

Similarly, we can’t five tangent lines to a twisted cubic which meets a line L.

Proof. If we had three concurrent tangent lines, projection from that com-mon point would give a 2 : 1 map from the conic to P1 with three branchpoints.

If there were a line meeting all five tangents to a twisted cubic wouldexpress the twisted cubic as a three sheeted cover of P1, and projectionfrom the line will have 5 ramification points, whereas by Riemann Hur-witz, such a map could only have 4 ramification points. �

So, following the idea of Lemma 35.6, the idea of proof of Proposition 35.2, isto specialize g general chords to a rational normal curve C ⊂ Pd to g arbitrarytangent lines.

92 AARON LANDESMAN

Proposition 35.7. For any g tangent lines Li to C ⊂ Pr, we have ∩iΣ(Li) is proper inG(d− r,d+ 1).

To prove this, we’ll need to generalize Riemann Hurwitz via the Plucker for-mula. But first, we’ll need to talk about inflectionary points of linear series.

35.2. Inflectionary points of linear series. Now, we’ll move to a more general setup, but keep in mind that we’ll only be applying this to P1.

Lemma 35.8. Let C be an arbitrary smooth curve of genus g. Let D = (L,V) be a grd onC. Then,

# {ordp σ : σ ∈ V \ {0}} = r+ 1.

Proof. There exists a basis for V consisting of sections with distinct orders of van-ishing at p. To construct this basis, replace a pair of sections with the same van-ishing order by two sections, one with the same order, and one with one higherorder. �

Definition 35.9. Given a linear series D = (L,V) on a smooth curve C of genus g.Define the vanishing sequence of D at p

0 ≤ a0 < a1 < · · ·aris the sequence of r+ 1 orders of vanishing from Lemma 35.8. We will alternativelynotate this as

ai(V ,p)

ai(D,p)

Set αi = ai − i, so that

0 ≤ α0 ≤ α1 ≤ · · · ≤ αr.This is then the ramification sequence of D at p. We also notate this sequence asα(D,p).

Proposition 35.10. For any D on C, we have

α(D,p) = 0

(meaning that αi(D,p) = 0 for all i) for p ∈ C general.

Remark 35.11. This is false in characteristic p, so we will have to work a bit toprove it.

Proof. Saying α0 > 0 is equivalent to p being a base point of D. Define φ := φD :C → Pr. Then, α1 > 0, given α0 = 0, this is equivalent to dφp = 0. Then, ifα1 = α0 = 0, to say α2 > 0 is equivalent to φ(p) being a flex point of φ(C). Inother words, the order of contact of Cwith its tangent line at p is at least 3. �

Definition 35.12. Let p ∈ C. Then, p is an inflectionary point or ramification pointof D if

α(D,p) 6= 0

Example 35.13. Take a smooth curve C ⊂ P3. Start with a line not meeting thecurve at p. Then, take a line meeting it but not tangent. Then, take a line tan-gent. This will yield a ramification sequence at such a point for the linear seriesembedding the curve in P3.


Theorem 35.14. Set

α(D,p) =∑i

αi(D,p)

Then, ∑i

α(D,p) <∞.

For D an arbitrary grd on an arbitrary smooth curve C, we have∑p∈C

α(D,p) = (r+ 1)d+ r(r+ 1)(g− 1)

Proof. We’ll possibly come back to this next time, but it’s also in 3264, and we didit in intersection theory last semester. �

Let’s see why this implies Brill Noether.

Proof of Proposition 35.7. We’ll look at the case that the expected dimension is neg-ative. That is, when (r + 1)(g − r) − rg = ρ < 0, we’ll show ∩Σ(Li) = ∅. Thegeneral statement is similar.

If Λ ∈ ∩Σ(Li), where Λ is a d − r − 1 plane in Pd which meets Li for all i,consider the linear series D, which is a grd on P1, cut by hyperplanes containingΛ. For each of the points pi, we have α1(D,pi) > 0 because we’re taking our linesto be the tangent lines at pi. In particular, α(D,pi) ≥ r. So, we must have that thetotal ramification index at these points, is

gr ≤ (r+ 1)d− r(r+ 1)

since g = 0. Hence, ρ ≥ 0. �

36. 11/30/15

36.1. Review and overview. Today is the penultimate class. There are 2 classesleft! Today, we’ll wrap up the proof of the Brill-Noether theorem. On Wednesday,we’ll talk about the consequences and open problems.

Here is a review of where we were in the proof of the Brill-Noether theorem.(1) Specialize to a g-nodal curve. More accurately, take C0 = P1/ ∼ where

pi ∼ qi for i = 1, . . . ,g, where the resulting curve has g nodes and normal-ization P1. This curve can be deformed to a smooth curve.

(2) We reduce the general statement to one on this g nodal curve. Note, thePicard variety on g-nodal curves of degree d is not proper. We can see thisbecause we can pull it back to P1, and then choose an isomorphism of thefibers over the two identified points, which is Gm. The gth power of thisis not proper.

(3) We want to describe the grd’s on C0. Embed P1 → Pd as a rational normalcurve of degree d, X ⊂ Pd. So, grd’s on C0 are in bijection with{

Λ ∼= Pd−r−1 ⊂ Pd : Λ∩ piqi 6= ∅}

For L ⊂ Pd a line, we set

Σ(L) := {Λ : Λ∩ L 6= ∅} ⊂ G(d− r,d+ 1).(4) By the above, reduce Brill-Noether to the following:

94 AARON LANDESMAN

Goal 36.1. If L1, . . . ,Lg are general chords toX ⊂ Pg, then ∩Σ(Li) is proper.That is,

dim∩Σ(Li) = (r+ 1)(d− r) − gr = ρ

This is our main goal for today.

Remark 36.2. Castelnuovo introduced the above proof technique, not in order toprove the Brill-Noether theorem, but instead to count the number of grd’s whenρ = 0. His aim was to use Schubert calculus to count the number of such grd’s.

It was later realized it is always true that g tangent lines will always have theexpected dimension. It is easier to deal with tangent lines because the statementis always true for them, because the other statement is only about a general secantline, where as this holds for all tangent lines. So, the idea is to show our goal forarbitrary tangent lines to X.

Recall the set up from last time. Given a grd with D = (L,V), on a smooth curveC of genus g and p ∈ C, let

{ai = ai(D,p)} = {ordp σ : σ ∈ V}where

0 ≤ a0 < a1 < · · · < ar ≤ dSet αi = ai − iwith

0 ≤ α0 ≤ α1 ≤ · · · ≤ αr ≤ d− rWe say a point p is an inflectionary point of D if α 6= 0.

Set

w = w(D,p) =∑i

αi(D,p)

to be the total weight of the inflectionary point p.The main theorem from last time we want to apply is the Plucker formula:

Proposition 36.3 (Plucker Formula). For any grd D on C we have∑p∈C

w(D,p) = (r+ 1)d+ r(r+ 1)(g− 1)

Proof. Let’s look locally on the curve. Take a point p ∈ C and an open set aroundit which both

(1) trivializes the line bundle L so that sections of L are regular functions(2) has a local coordinate z on C around the point p.

Take a basis σ0, . . . ,σr of H0(C,L) and write σi = fi(z). Consider the Wronskian

det

f0 · · · frf ′0 · · · f ′r...

. . ....

f(r)0 · · · f

(r)r

Write v = (f0, . . . , fr) be a vector valued function. Note that the Wronskian isequal to

v∧ v ′ ∧ · · ·∧ v(r)


Lemma 36.4. We claim

w(D,p) = ordpW

and in particular,W 6= 0.

Proof. First, let us show W 6= 0. Suppose it were. Then, the r derivatives of v andv are everywhere linearly dependent. Let’s start by identifying the last stage inthis wedge product which is nonzero. That is, say v∧ v ′ ∧ · · ·∧ v(k−1) 6= 0 butv∧ v ′ ∧ · · ·∧ v(k) = 0. Taking the derivative, the only nonzero term is

v∧ v ′ ∧ · · ·∧ v(k−1) ∧ v(k+1)

So, if v∧ v ′ ∧ · · ·∧ v(k) = 0. v∧ v ′ ∧ · · ·∧ v(k−1) ∧ v(k+1) = 0. Hence,

v(k+1) ∈ 〈v, v ′, . . . , v(k−1)〉.

Continuing this process, we see that

v(k+2) ∈ 〈v, v ′, . . . , v(k−1)〉.

and in general, by induction

v(k+t) ∈ 〈v, v ′, . . . , v(k−1)〉.

for all t. So, in the analytic world, integrating this, the whole curve must lie in thisk plane. I.e., if all the derivatives of a curve lie in a fixed subspace the image of amap must also lie in that subspace. This tells us the map is mapping us into a kplane. This contradicts nondegeneracy of the grd. �

Remark 36.5. Note, this is false in characteristic p. This fails in characteristic pbecause the statement that all derivative lie in the span of a k-plane does not meanthe curve itself lies in a k-plane.

At a point p ∈ Cwith ai = ai(D,p), up to a unit we can write our functions as

v = (za0 , za1 , . . . , zar)

Writing out the Wronskian, we have

det

za0 zar · · · zar

za0−1 za1−1 · · · zar−1

......

. . ....

za0−r za1−r · · · zar−r

The first nonzero derivative ofW is the

∑i(ai − i)th derivative.

We now come to proving the Plucker formula. Observe that W is a functiondefined locally. This W depends on the choices of our trivialization for L and thelocal coordinate we chose. We want to describe howW depends on these choices.

Question 36.6. How doesW depend on the choice of trivialization and local coor-dinate?

That is, if we chose a different trivialization differing by a nonzero functiong = gαβ, and a different coordinate ζwith ∂z

∂ζ = j = jαβ.

96 AARON LANDESMAN

The first row just changes by g. The second row changes by two terms. Thefirst g ′ times the first row, which is dependent on the first row. The second is gtimes the second row. Therefore, each row, module the span of the previous rowis multiplied by g. Therefore, in this caseW is multiplied by gr+1.

Now, how does W depend on the choice of local coordinate? In this case, the

total matrix is multiplied by j(r+12 ). Therefore, if we change by a function g and a

coordinate j, we obtain thatW is a well defined global section of

L⊗r+1 ⊗ K(r+12 )

C

which has degree

(r+ 1)d+

(r+ 1

2

)(2g− 2) = (r+ 1)d+ r(r+ 1)(g− 1).

Therefore, the number of zeros ofW is, therefore, equal to both∑p∈C

w(D,p) = (r+ 1)d+ r(r+ 1)(g− 1).

�

Remark 36.7. Think of Proposition 36.3 as an extension of Riemann-Hurwitz. Ob-serve, Riemann-Hurwitz is simply this expression in the case r = 1.

Next, we will see how to deduce the weakest form of Brill-Noether. (We actuallyare repeating this from last time.)

Proposition 36.8. Say L1, . . . ,Lg are any g tangent lines to X at pi. If ρ < 0, then∩Σ(Li) is empty.

Proof. This is quite immediate from the plucker formula. If Λ ∈ ∩Σ(Li) and D isthe corresponding grd on X ∼= P1, then the ramifications

α(D,pi) ≥ (0, 1, . . . , 1)

In particular, the total weight

w(D,pi) ≥ r

and so by the Plucker formula,

rg ≤ (r+ 1)d− r(r+ 1)

which precisely means ρ ≥ 0. Hence, contrapositively, when ρ < 0, there cannotexist any such Λ. �

Remark 36.9. To set up for Wednesday, we’d like to see how to go from this crudestatement: When ρ < 0 thenWrd is empty, to the stronger statement on the dimen-sion.

Next time, we’d also like to talk about the consequences of the Brill-Noethertheorem for studying Hilbert schemes.

There will be no homework after homework 9.


37. 12/2/15

37.1. Overview. This is the last class! Today, we’ll do three things(1) Make a retraction(2) Finish Brill-Noether(3) Discuss further questions

37.2. Retraction. Previously, Joe stated that no one had explicitly written down asmooth curve of genus g satisfying the Brill-Noether statement.

Recently, a paper has been published on the arxiv by Arbarello et al doing ex-actly this. Here is the construction:

Theorem 37.1. Take nine points p1, . . . ,p9 which are general. Let C ∈ P2 be a curvewith degree 3g that has a singularity of multiplicity (the order of vanishing of the definingequation, which generically looks like smooth branches crossing transversely at that point)g at p1, . . . ,p8 and a singularity of multiplicity g− 1 at p9.

Then, if E ⊂ P2 is the unique cubic containing p1, . . . ,p9, consider

OE(3)(−p1, . . . ,p9),

which is a line bundle of degree 0 on E. If this line bundle is not torsion of order ≤ g, thenthe normalization of C is Brill-Noether Petri general.

Proof. First, let us compute the geometric genus of g. By the formulas we’ve de-rived, we see the genus is(3g− 1

2

)− 8

(g

2

)−

(g− 1

2

)=1

2((3g− 1)(3g− 2) − 8g(g− 1) − (g− 1)(g− 2))

= g

We have to make an argument that a general such curve is smooth in the blowup. That is, in the linear system of the blow up divisor, a general curve is smoothof genus g. Further, the dimension of the linear system of such curves is

dim |C| = g.

The rest of the proof is in the paper. �

Corollary 37.2. Brill-Noether holds.

Proof. Using the above theorem, we have exhibited a given Brill Noether gen-eral curve, and hence, by upper semicontinuity, a general genus g curve is Brill-Noether general. �

37.3. Finishing the proof of Brill-Noether. We have shown in previous classeswe discussed the following. Let C ⊂ Pd be a rational normal curve, let L1, . . . ,Lgare arbitrary tangent lines to C, and let

Σ(L) = {Λ : Λ∩ L 6= ∅} ⊂ G(d− r,d+ 1) =: G.

We have showed by the Plucker formula that if ρ < 0, meaning

rg > (r+ 1)(d− r)

or equivalently ∑codim (Σ(Li ⊂ G) > dimG

98 AARON LANDESMAN

we have

∩iΣ(Li) = ∅

We then conclude that for ρ < 0 a general curve of genus g has no grd. We want toconclude that for a general curve C, we have dimWrd = ρ.

Remark 37.3. Here is how we strengthen the result we have so far. So far, wehave seen that Schubert cycle associated to tangent lines of a rational normal curvealways intersect dimensionally properly. We would like to extend this to otherSchubert cycles.

Definition 37.4. We define the osculating flag to a rational normal curve C ⊂ Pd

at p ∈ C to be a collection of subspaces

V0 ⊂ V1 ⊂ · · · ⊂ Vd−1 ⊂ Pd

where

V0 = p

V1 = 2p

V2 = 3p

...

Vd−1 = dp

Here, V2 is characterized by mp(H · C) ≥ 3 if and only if V2 ⊂ H, and V2 is theosculating 2 plane to C at p.

There are likely typos in the remainder of this subsection.

Theorem 37.5. Let C ⊂ Pd be a rational normal curve and p1, . . . ,pδ be any points.Let Vi be the osculating flag to C at pi. Then, the intersection of any Schubert cycles Σai

∩iΣai(Vi) ⊂ G(d− r,d+ 1)is proper.

Proof. This follows immediately from the Plucker formula. If Λ ∈ G(d− r,d+ 1)is a linear subspace Λ ∼= Pd−r−1 ⊂ Pd. Then, D = grd on C cut by hyperplanesH ⊃ Λ. Then, for any point p ∈ Cwith osculating flag V the total inflection weight

w(D,p) = maxΛ∈Σa(V)

(codimΣa ⊂ G(d− r,d+ 1)

That is, if Λ satisfies a certain osculating condition then the corresponding weighthas inflection weight at least equal to the sum of the indices in that Schubert cycle.Applying the plucker formula, we obtain that for any collection of Schubert cycles,Σai(Vi) defined relative to osculating flags to C, if∑

codim(Σai) > dimG(d− r,d+ 1)

then

∩iΣai(Vi) = ∅.

�

Corollary 37.6. The Brill Noether theorem holds (even for ρ > 0).


Proof. Then, the dimension of Wrd, we may note that the simple ramification lo-cus is just a hyperplane section. That is, to argue Wrd has dimension ρ and notlarger, then if Wrd had larger dimension, we could then impose the condition oframification at ρ+ 1 hyperplanes. If the dimension ofWrd were ρ+ 1 or more, thatintersection would be nonempty. So, these osculating flags allow us to add extracodimension 1 conditions. This lets us go from the inductive base case we saw lasttime to the general case about the dimension ofWrd.

�

37.4. Further Questions. First, as a basic consequence of the Brill-Noether theo-rem, including the Fulton-Lazarsfeld irreducibility statement we have the follow-ing.

Define

H := Hd,g,r

be the open subset of the Hilbert scheme parameterizing curves of degree d andgenus g in Pr. Officially, it is the open subset of

Hmd−g+1(Pr)

parameterizing smooth curves.

Lemma 37.7. Let r ≥ 3. Whenever ρ ≥ 0, there exists a unique irreducible componentof H dominating Mg.

Further, this component has the “expected dimension” equal to

3g+ 3+ ρ+ (r+ 1)2 − 1

where the 3g+ 3 = dimMg specified the abstract curve, ρ is the dimension of the choiceof line bundle, and (r+ 1)2 − 1 is the choice of sections for the embedding. Note,

3g+ 3+ ρ+ (r+ 1)2 − 1 = (r+ 1)d− (r− 3)g

= χ(NC/Pr)

Proof. This follows from Brill-Noether. First, suppose ρ > 0. On a general curve ofgenus g, there are very ample grd’s. Further, by Fulton and Lazarsfeld’s contribu-tion to Brill-Noether, we have a map

H→Mg

and the fibers are PGLr+1 bundles over Wr+1d . Hence, over a general point, thefibers of this map are irreducible. Since the map is also flat, the source will beirreducible.

When ρ = 0, we need a separate argument, here we’ll have to explain why themonodromy group is transitive. �

Definition 37.8. We call the unique irreducible component of H dominating Mgthe principle component, which we notate as H0 which implicitly depends ond,g, r.

There are now two directions of further study:(1) Understand the geometry of curves in the principle component of the Hilbert

scheme. We know quite a few things about this. For example,(a)

100 AARON LANDESMAN

Question 37.9. What equations define this curve?This is conjecturally answered by the maximal rank conjecture, whichbasically says that the trick we used to describe curves in low degreeand genus works in general. That is, the map

H0(OPr(m))→ H0(OC(m))

has maximal rank, which tells us the Hilbert function of such a curve,and hence is a first step to defining the ideal of such a curve.

(b)

Question 37.10. What is the inflectionary and secant plane behavior.That is, what sort of inflection points are we going to see? If you thinkabout the line of argument we made above to finish the Brill-Noethertheorem you will understand the following really good exercise:

Exercise 37.11. A general curve on the principle component has onlysimple ramification.

If you look at a curve in P3, we would expect that a curve has finitelymany 4 secant lines and no 5 secant lines.

Question 37.12. For a general curve in P3, are there always finitelymany 4-secant lines and no 5-secant lines.It’s 1 condition for a secant line to meet the curve. Hence, it is 4 condi-tions for a line to meet the curve 4 times and 5 conditions to meetit 5 times. We then get the above “expected dimensions” becausedimG(2, 4) = 4 of lines in P3.

(c) Flexibility (or interpolation):

Question 37.13. Given a general collection of points in Pr is there acurve containing the expected number of points.

(d) What is the geometry of the principle component itself? That is, whatis

Pic(H0)?

There is a conjecture that Pic(H0) is torsion.(2) What about the case that ρ < 0? If ρ < 0, Brill Noether tells us there is no

component of the Hilbert scheme that dominates moduli. However, we dostill have an expected dimension of the Hilbert scheme. We can ask

Question 37.14. When do components of the Hilbert scheme still have theexpected dimension when ρ < 0?

We know the “vast majority” of the components of the Hilbert schemedo have the expected dimension. However, there do seem to be a certainrange of g, r,dwhere the Hilbert scheme still have the expected dimension.We we formulate the following vague conjecture.

Conjecture 37.15. If H0 is any component of the Hilbert scheme H andthe image of H in Mg under the projection map has codimension less thang− 4, then H0 has the expected dimension.

notes for math 282, geometry of algebraic curves · notes for math 282, geometry of algebraic...

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