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Home , Adjunction formula, Canonical bundle, Cubic surface, Curve, Cusp (singularity), Dimension of a scheme

NOTES FOR MATH 282, OF ALGEBRAIC

AARON LANDESMAN

CONTENTS 1. Introduction 4 2. 9/2/15 5 2.1. Course Mechanics and Background 5 2.2. The Basics of curves, September 2 5 3. 9/4/15 6 3.1. Outline of Course 6 4. 9/9/15 8 4.1. Hurwitz 8 4.2. Equivalent characterizations of 8 4.3. Consequences of Riemann Roch 9 5. 9/11/15 10 6. 9/14/15 11 6.1. Curves of low genus 12 7. 9/16/15 13 7.1. Geometric Riemann-Roch 13 7.2. Applications of Geometric Riemann-Roch 14 8. 9/18/15 15 8.1. Introduction to spaces 15 8.2. Moduli Spaces 16 9. 9/21/15 18 9.1. Hyperelliptic curves 18 10. 9/23/15 20 10.1. Hyperelliptic Curves 20 10.2. Gonal Curves 21 11. 9/25/15 22 11.1. Curves of genus 5 23 12. 9/28/15 24 12.1. Canonical curves of genus 5 24 12.2. Adjoint linear series 25 13. 9/30/15 27 13.1. Program for the remainder of the semester 27 13.2. Adjoint linear series 27 14. 10/2/15 29 15. 10/5/15 32 15.1. Castelnuovo’s Theorem 33 16. 10/7/15 34 17. 10/9/15 36 18. 10/14/15 38 1 2 AARON LANDESMAN

19. 10/16/15 41 19.1. Review 41 19.2. Minimal Varieties 42 20. 10/19/15 44 21. 10/21/15 47 22. 10/23/15 49 22.1. Agenda and Review 49 22.2. Resolutions of Projective Varieties and Green’s conjecture 50 22.3. The Maximal Rank Conjecture 52 23. 10/26/15 53 24. 10/28/15 55 24.1. Review 55 25. 10/30/15 57 25.1. Review 57 25.2. Today: Brill Noether Theorem in at least 3 58 26. 11/2/15 61 26.1. Review 61 26.2. Hilbert Schemes 61 27. 11/4/15 64 27.1. Logistics 64 27.2. Spaces in Brill Noether Theory 64 27.3. Martens Theorem 66 28. 11/6/15 67 29. 10/9/15 71 29.1. Agenda and Review 71 30. 11/11/15 75 30.1. Review 75 30.2. The Genus 6 Canonical Model 76 31. 11/13/15 78 32. 11/16/15 81 32.1. Logistics and Review 81 32.2. A continuing study of genus 6 curves 82 33. 11/18/15 84 33.1. Logistics and Overview 84 34. 11/20/15 87 34.1. Plan, conventions, and review. 87 34.2. An Upper bound on the dimension 88 34.3. Proof of Existence for Brill Noether 88 35. 11/23/15 90 35.1. Review 90 35.2. Inflectionary points of linear series 92 36. 11/30/15 93 36.1. Review and overview 93 37. 12/2/15 97 37.1. Overview 97 37.2. Retraction 97 37.3. 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 Fall 2015. 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 the date. Of course, these notes are not a faithful representation of the course, either in the 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 be credited 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 his permission. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 5

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 , what can we say about it. This is only half the story. Curves can appear in the abstract and in . An important part 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 curve and 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 , and second we will use schemes. (6) Algebraic curves were first studied over the complex . Some peo- ple studied of Riemann , and others studied in two variables. Remark 2.1. We will use the language of smooth projective curves and compact Riemann surfaces interchangeably. We will assume all curves are over the complex numbers. 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 in projective space. A similar perspective was adapted in 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 is to classify 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 can then ask, which pairs of d, g are realized as a curve. This is still not completely known. You can also specify a projective space, and then ask which curves can be realized in that projective space. 2.2. The Basics of curves, September 2. 1 Definition 2.4. Define the genus by g = 2 (1 − χtopX). Definition 2.5. An ordinary singular of a curve of multiplicity m is a point in which m branches of a curve meet transversely of a point. We can define this more rigorously by saying that the completion of its local ring is isomorphic to k[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 1 (3) 2 (deg KX + 2) 6 AARON LANDESMAN

(4) 1 − c, where c is the constant term of the Hilbert of C ⊂ Pr. ∼ 2 (5) If C = C0 ⊂ P of degree d with ordinary singular points of multiplicity mi, d−1 mi then g(C) = 2 − i 2

Definition 2.7. A divisor isP an formal sum of the form D = i nipi for ni ∈ Z, ∈ X. We say D is effective if D ≥ 0. We define the degree by deg D = i ni. For f a rational on X, we define P P div f = (f) = ordp(f) · p = (f)0 − (f) p X . ∞ Remark 2.8. By the residue formula applied to the logarithmic of a f, we have deg (f) = 0. Definition 2.9. We say D ∼ E, or D is linearly equivalent to E if there exists a rational function f with (f) = D − E. Effective divisors of degree d on X will be d notated as Cd, which is by definition C /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 such a function, this is a finite dimensional . This uses the fact that there are no nonconstant . We define `(D) = dim L(D) and define r(D) = `(D) − 1. Remark 2.11. If D ∼ E then L(D) =∼ L(E), as given by by f.

Definition 2.12. We define Picd(X) := Divisors of degree d/ ∼. Remark 2.13. It turns out this corresponds to the points of a variety. Definition 2.14. Suppose ω is a rational 1-form, which looks locally like f(z) dz. 2g−2 We let (ω) = ordp(ω)p. We define KX := (w) ∈ Pic (X). P 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 about a general curve?” It makes sense to ask whether there is an open subset of the Hilbert on which there is a uniformity of the corresponding curves. Remark 3.1. For the remainder of the course, we let X be a smooth projective curve. 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 ω, of degree 2g − 2. Use L(D) for P H0(C, D), let `(D) = h0(C, D) and r(D) = `(D) − 1. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 7

The justification of looking at these linear systems is that we only allow poles at 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. r Remark 3.2. Given a map φ : X P and let H = V(x0) be a cut r out by the first coordinate of P . Then, the map can be given by [1, f1, ... , fr], and −1 ∞ D = φ H and f1, ... , fr ∈ L(D→).

There are∞ some issues with the above description. In particular, we will need to enforce 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 we say 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 to sweet Serre under the rug. The problem is that we assumed D and K − D were effective. Say D = p1 + ··· + pd, where pi are distinct points of X. Choose local coordi- nates z around each p . Any f ∈ L(D) can be written locally as ai + h where h i i zi is holomorphic. So, f is determined up to the of a constant function by a1, ... , ad. So, the question is: which d tuples arise as global principal parts of functions. First, note that `(D) ≤ d + 1. Now, we ask, what is the obstruction to having a global rational function with these polar parts. Here, we use that if we have a meromorphic 1-form on a curve then 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. Say ω(pi) = bidzi. So, plugging this in the above formula, we have i aibi = 0. P So, we get g relations, but they might not be independent. The actualP 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 argument to `(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 , how do you know there 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, there are compact complex with no meromorphic functions whatsoever. 8 AARON LANDESMAN

Remark 3.6. How should we define a canonical divisor on a singular curve? We will have to choose a canonical divisor on a singular curve so that it satisfies the condition 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- , we look at locally.

Definition 3.7. Given a divisor D on X we 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 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, define 0 L(D) := H (X, OX(D)) Remark 3.8. This construction is a special case of the fact that for locally factorial schemes which are regular in 1, there is an between the vector spaces of Weil and Cartier divisors. Lemma 3.9. The collection of nondegenerate maps (maps whose is not contained in a hyperplane are in with pairs (V, L) where L ∈ Picd(X), V ⊂ H0(L) with dim V = 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) deg KC = 2g − 2 (8) χ>(C) = 2 − 2g. Remark 4.1. It is easier to prove equivalence of these definitions in the algebraic category because in that setting we already have access to the algebraic functions coming from projective space via an embedding. Suppose π : C B is a nonconstant map of compact Riemann surfaces of degree d. Then, for all but finitely many points in the target, the preimage consists of d points. For all points→ p ∈ C, we can choose local coordinates so that π is of the form z 7 zm, and when m > 1 p is a ramified point.

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

(4.1) R =→ vp(π) · p p∈C X NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 9 and the branch divisor by   (4.2) B =   · q q∈B −1 X p∈πX(q) We define the total ramification index as b := deg R = deg B Theorem 4.3. (Riemann-Hurwitz) Suppose π : C B is a nonconstant map of compact Riemann 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 of a from on B will pick up zeros equal to the ramification index. So, ∗ (4.3) KC = π KB + R

Then, taking degrees, since deg KC = 2g − 2, we obtain Riemann-Hurwitz.  Remark 4.4. It’s also fun to prove this via topological Euler . You do this by removing the ramification and branch . We then obtain an unramified , 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 action of PGLr+1 are in bijection with pairs (L, V) so that L is a on C of degree 0 d and V of dimension r + 1 in H (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 φ ∨ sends φD : p 7 Vp ∈ PV . r Definition 4.6. A gd is a line bundle of degree d and a vector space V of dimension r + 1 inside H→0(L). Crucially, we do not assume that V is basepoint free. r r ∼ ∨ Remark 4.7. If D = (L, V) is a basepoint free gd, we get a map C P = 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 if for all p ∈ C, V2p has codimension 2.

Lemma 4.8. A map φ is an embedding if and only if for all effective divisors D ∈ C2 i.e., of degree 2, VD has codimension 2. Proof. This is Remark 4.7  0 0 Corollary 4.9. If deg L ≥ 2g + 1 then φ|L| is an embedding, where |L| = (H (L), H (L)). Proof. By Riemann Roch, deg L ≥ 2g + 1 = dim H0(L) = d − g + 1. But, it’s also true that deg L(−D) ≥ 2g − 1 so H0(L) = d − g − 1. Therefore, any line bundle of degree 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 deal with 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 then φ|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 the Jacobian.  Definition 4.12. A curve C is hyperelliptic if there exists a map π : C P1 of 2 2g+2 degree 2. That is, C can be written as y = i=1 (x − λi) on an open subset. → Exercise 4.13. Not every curve of genus atQ least 3 is hyperelliptic. Example 4.14. In genus 2, the canonical sheaf makes C into a . Definition 4.15. For any curve C of genus g > 0, we have canonical map, defined to 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 an embedding. Proof. Use the criterion for being a closed embedding, and Riemann Roch. Fur- thermore, use the characterization that being hyperelliptic is equivalent to having an effective divisor of degree 2. 

5. 9/11/15 Today Joe is out, and Adrian is holding a review. 1 Remark 5.1. Deformations of abstract schemes are classified by H (TC). Embed- 0 ded deformations are classified by H (NY/X). 1 Example 5.2. Let’s show there are non hyperelliptic curves. First, TCMg = H (TC) = 0 0 H (2KC). By Riemann-Roch, h (2KC) = 3g − 3. This gives a proof that dim Mg = 3g − 3.

Exercise 5.3. Compute dim Mg by first calculating the dimension of the space of covers of P1 of degree d. Theorem 5.4. Let X be a smooth and Y ⊂ X be smooth of codimension 1. 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). 1 1 ∼ Solution: We have KC = (KP1×P1 + C)|C. Then, Pic P × P = ZH1 ⊕ ZH2. So,

(5.1) KC = KP1×P1 + C|C (5.2) = (−2H1 − 2H2 + aH1 + bH2)|C

(5.3) = ((a − 2)H1 + (b − 2)H2)|C (5.4) = (a − 2)b + (b − 2)a (5.5) = 2ab − 2a − 2b So, g = (a − 1)(b − 1). Exercise 5.6. Compute the dimension of the space of twisted cubics in P3. (I.e., a component of the Hilbert scheme) Solution: We have a map P1 P3 of degree 3, which is nondegenerate. We can write any map sending [x, y] 7 [fi(x, y)]. where fi is a homogeneous polynomial → → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 11 of degree 3. This is 16 dimensional, we subtract 1 for scalars and subtract 3 for the PGL2 action. Solution 2: Lemma 5.7. If C is a , then C is contained in a .

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 .  Now, consider (C, Q) so that C is contained in Q and this gives you what you want. ¡++¿

6. 9/14/15 We will take the approach of interweaving new techniques with applications. Today’s new technique is the . We’ll give two proofs of the adjunction formula. We’ll have the following setting: Let X be smooth varieties of dimension n. Every such scheme has a , which is an invertible sheaf of top n ∨ dimensional forms KX = ∧ TX . Suppose Y ⊂ X is a smooth divisor, i.e., codimension 1 subvariety. Recall we can associated an invertible sheaf to the divisor Y, called L = OX(Y). There is also a section σ ∈ H0(L) corresponding to the regular function 1, with V(σ) = Y. The key 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 and q ∈ L where q is on the zero section over p. Then, TqL = TpX ⊕ Lp. Then, the to σ, with V(σ) = Y, is the graph of a map TpX Lp. The kernel of this map is TpY, and by the conormal exact sequence, we have (NY/X)p = Lp.  Here is an algebraic proof: → Proof 2. We have ∨ ∼ 2 NY/X = IY /IY = OY|Y = OX(−Y)|Y.(6.1)

And so, NY/X = OX(Y)|Y.  We are now in a to prove the adjunction formula. We have an exact sequence

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.

(6.2) KY = KX|Y ⊗ NY/X = KX(Y)|Y 12 AARON LANDESMAN

Proof. This is immediate from applying 6.1 to the conormal exact sequence.  n Example 6.3. KPn = OPn (−n − 1). To see this, given P , we have 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

dz1 dzn (6.3) ∧ ··· ∧ z1 zn 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 adjunction OP2 (C) = OP2 (d) and so KC = OC(d − 3) which has degree (d − 3) · d. d−1 Therefore, the degree KX = 2g − 2 = (d − 3)d and so g = 2 . (2) If Q ⊂ P3 is a smooth quadric surface. Note Q =∼ P1 × P1 P3 via the Segre embedding. Then, the tangent plane to Q intersects Q at the two lines of the two rulings passing through the point. We say a→ curve C ⊂ Q has bidegree (a, b) if C is the zero locus of a bihomogeneous polynomial of bidegree (a, b), i.e., if C meets lines of one ruling in a points and of the other 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 to the quadric, whose intersection with the quadric surface is one line of each ruling. Now, we first compute KQ. We claim, KQ = OQ(−2, −2). First, Q is P1 × P1, so to write a meromorphic 2 form on Q is the same as writing down meromorphic 1 forms on both copies of P1 and the zeros are the preimages of the zeros of the forms on P1. Then, we obtain a meromorphic 2 form with poles on 2 lines in each ruling. So,

(6.4) KQ = KP3 ⊗ OP3 (Q)|Q = OP3 (−2)|Q Then, applying adjunction again,

(6.5) KC = KQ ⊗ OQ(C)|C

(6.6) = OQ(a − 2, b − 2)|C

So, plugging in degrees, 2g − 2 = deg KC = b(a − 2) + a(b − 2). Then, g = (a − 1)(b − 1). Now, if we take a curve with one of the bidegrees 1, we see 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 (or a conic in P2 over arbitrary fields). Furthermore, there is only one line bundle on the curve of any degree, because any two curves of the same degree are linearly equivalent. Then, we can look at the complete linear system OP1 (d) which embeds C as a rational curve of degree d ⊂ Pd. This d is equal to the smallest possible degree of an irreducible nondegenerate curve C ⊂ Pd, and any nondegenerate curve of that degree must be the rational NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 13 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 from that space will give an isomorphism of the curve with P1. (1) How many surfaces or 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 1 curve. → (2) Now, take d = 4. This gives an embedding C P3. If we have a cubic , it is the zero locus of a cubic polynomial. Does this curve lie on a quadric surface? There is a standard way→ of answering this using 0 3 the following technique: We want to know whether H (P , IP3 (2)) van- ish on C. Restricting this to the curve, and to see if the curve lies on a quadric surface, we want to check the kernel of the restriction map. We 0 3 0 know H (P , IP3 (2)) = 10 and H (OC(2)) = 8. Therefore, the kernel is at least two dimensional. In fact, this map has a two dimensional kernel. So this curve lies on a P1 of quadric surfaces. None of the quadrics containing the curve are irreducible because C is irreducible, nondegenerate. Then, by Bezout’s theorem, C is the intersection of two quadrics. We could also do this by adjunction, and so C must be type (2, 2) on a quadric surface. (3) For degree 5, we get an intersection of G(2, 5) with hypersurfaces in the plucker embedding. After this, the get even more complicated.

6.1.3. Genus 2. Let C be a curve of 4. Take a divisor of degree 4, corresponding to a line bundle L. We get a map C P2, where C has geometric genus 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 = 2 D − KC as an effective divisor for a unique divisor p + q on C. That is φD : C P is either a node if p 6= q or a if p = q. The solution is just Riemann Roch. → 7. 9/16/15 7.1. Geometric Riemann-Roch. New tool: Geometric Riemann-Roch For the moment being, we will assume (1) Assume C is a non hyperelliptic curve of genus g (2) Assume D = pi with pi distinct. g−1 Let φ = φ|K| : CP P be the canonical embedding. In coordinates: if 0 ω1, ... , ωg are a for H (K), then φ : p 7 [ω1(p), ... , ωg(p)]. More intrinsically, P→g−1 = PH0(K)∨ and φ : p 7 H0(K − p) , H0(K). → Theorem 7.1. → → r(D) = d − 1 − dim D 14 AARON LANDESMAN

Proof. Suppose p1, ... , pd ∈ C. Note that if n is the number of linear relations on the 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 points pi. So, if the pi are linearly independent then there does not exist a nonconstant meromorphic function on C with 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, g−1 define the span D = →∩φ−1(H)⊃DH ⊂ P . If we take D = 2p we have to take the tangent line to the points, that is, the intersection of all planes tangent to the curve at that point. 

7.2. Applications of Geometric Riemann-Roch.

7.2.1. Genus 2, degree 5. Suppose D 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 ask about the ideal of the embedding. Last time, we asked: → Question 7.2. What surfaces in P3 contain C? 0 To calculate the quadrics on which this surface lies, note that h (OP3 (2)) = 0 10, h (OC(2)) = 10 − 2 + 1. Then, the map

0 0 H (OP3 (2)) H (OC(2)) is in fact surjective, because if C were on two quadrics, being nondegenerate, it → would be degree at most 4. Therefore, the kernel is 1 dimensional, and the curve lies 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

0 0 H (OP3 (3)) H (OC(3)) Now, C lies on 6 linearly independent cubics. However, we knew about 4 of the → cubics C lies on from multiples of the quadratics. So, there are at least 2 linearly independent 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 of degree 6. Therefore, it must be the union C ∪ L, since L is degree 1. Here, we use that complete intersections are unmixed (i.e., the resulting scheme structure is reduced). 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 sup C is S ∩ Q. This gives a bijective correspondence between cubics containing C but not Q and lines of the ruling of S.

Proof. We have Q =∼ P1 × P1 P3 by the Segre map. Homogeneous degree 1 forms 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

→ NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 15

Furthermore, this map is surjective, which can be proven by writing out the above map 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 associated exact sequence on global sections. So, M ∪ C = S ∩ Q. This shows that in fact S1 ∩ S2 ∩ Q = C. Alternatively, after we knew C was contained in Q, being quintic of genus 2, we would have known it was of type (2, 3), and we could have found which cubics it lies on.  For degree 6 line bundles of on genus 2 curves, the geometry gets much more subtle, 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. Here 3 φK : C P is a sextic curve, which is also the smallest degree embedding of C in any projective space. Being non-hyperelliptic, it’s not expressible as a two 1 sheeted→ cover of P . 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 0 0 H (OP3 (2)) H (OC(2)) determines a surjective map, with C lying on a unique quadric surface, uniqueness from Bezout’s theorem. → Next, looking at cubics 0 0 H (OP3 (3)) H (OC(3)) The kernel is at least 5 dimensional, and so there is a cubic containing C but not Q. Then, S ∩ Q is degree 6, hence equal to C→. Conversely, by adjunction, any smooth curve of the form S ∩ Q = C is a canonical curve of genus 4. Now, our curve is realized as a curve of type (3, 3) on Q =∼ P1 × P1. Now, by geometric Riemann Roch, we have three points on the curve which are linearly dependent. So, we’re asking whether the canonical curve contains three colinear points. That is, each line 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 is singular, the curve is only trigonal in 1 way.

8. 9/18/15 8.1. Introduction to parameter spaces. Today: Jacobians. Application: every curve 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 genus 3 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 r Hd,g,r = { curves C ⊂ R degree d, genus g }

(3) 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 him to get a conditional in calculus, though it was actually intended by Barry Mazur to be a problem in algebraic geometry. The problem is to computer dx √ x2 + 1 Z The intention of Mazur was to look at y2 = x2 + 1 which is a , and  2  t 7 2t t +1 observe this is a genus 0 curve, this curve is parameterized by 1−t2 , 1−t2 1 dx 1 which determines a map P C yielding y and when we pull it back to P R(t) dt → it becomes and then we compute theR of this rational function on P1. → R The next integral that was considered was something like dx √ x3 + 1 Z dx 2 3 which we can think of as y on the y = x + 1. This integral isn’t quite well defined, since it depends on the you take. R Say we have a smooth projective curve C of genus g. Given p, q ∈ C, we want q 0 ∨ to make sense of p ∈ H (K) /H1(C, Z) = J(C). which is called the Jacobian of C. We can look at R 0 ∨ H1(C, Z) H (K)

Z2g Cg

Fix a base point p0 ∈ C. We get a map C J p p 7 → 0 Zp which is non canonical because it depends→ on a basepoint. More generally, for any d we get a map µ.

µ : Cd J pi D = pi →7 0 i p X X Z 0 Theorem 8.1. For D, D ∈ Cd we have → D ∼ D0 µ(D) = µ(D0)

⇐⇒ NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 17

This implies that J =∼ Picd(C). Proof. Idea: The forward direction is quite easy. If we start with D, D0, the con- dition they be linearly equivalent means there is a family interpolating between them. If they are sections of the same line bundle L, then we can take linear com- binations of D, D0 parameterized by the two divisors. That is, there is a family 0 {Dt}t∈P1 where D0 = D, D = D and 1 ∞ µ: P J t 7 µ(Dt) → Now, since J is a complex , 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’m not 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 it is 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 r d 0 Wd(C) = L ∈ Pic (C): h (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 0 0 D = i=1 pi. We start with h (K) = g, h (K − p1) = g − 1, and repeating we see h0(K − D) = max(0, g − d), supposing we choose the p so that they are as P i independent as possible. Riemann-Roch then implies the statement. So, we only need deal with the noneffective case. If d ≥ g the map µd is surjective. We only 0 need show if 0 ≤ d ≤ g then µd is generically 1-1. In particular, Wd ⊂ J is a closed subvariety of dimension d. The proof of the statement for d ≤ g − 1 follows from Riemann-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, so h0(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 + q with 0 0 2 h (L(−D)) = h (L) − 1 = 3. This would imply L(−D) ∈ Wg+1. Note that the set 2 0 0 Wg+1 = K − Wg−3 by Riemann Roch. So, the existence of such a divisor D ∈ W2 0 0 0 with h (L(−D)) = 3 implying L(−D) ∈ K − Wg−3 which implies L ∈ W2 + (K − 0 Wg−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 can allow us to prove theorems about the objects we’re concerned with. We’ve been accumulating 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 is of the form D = 2K + p, which is a one dimensional space of line bundles inside a 2 dimensional space of line bundles. On Monday we’ll talk about canonical 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. Let C be a hyperelliptic curve of genus g. Then, there is 1 a map π : C P of degree 2. We have exactly 2g + 2 branch points α1, ... , α2g+2. Lemma 9.1. There exists a unique curve C with a map π : C P1 branched at → α1, ... , α2g+2. Proof. Uniqueness is shown below via a topological argument. To→ show existence, 2 we can write down C = V(y − i(x − αi)), and specify a transition function to the other chart, as in Vakil’s foundations of algebraic geometry. Alternatively, just Q complete this as a Riemann Surface.  Given a Riemann and 2g + 2 points, draw arcs from a given to 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 as follows: Every time we go around a branch point the sheets are exchanged. This constructs the complex analytic structure. The complex analytic structure comes from pulling back from P1. Question 9.2. Suppose we have a three sheeted cover of the sim- ply branched over 2g + 4 points. How can we describe the structure of that Rie- mann surface? The question is to describe the group at the branch points. Two of them are exchanged, and there is one sheet containing the simple point. If we are 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 around the 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 in the fundamental group. So, we have transpositions τi ∈ S3, we know i τi = id and hτii is transitive. Hence, hτii = S3, or equivalently, there are at least two Q τi. Furthermore, such τ1, ... , τ2g+4 is determined up to simultaneous conjuga- tion, which corresponds to relabelling the sheets. We can now replace degree 3 by degree 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 hτii transitive and i τi = id. This is the only relation because we know the funda- mental group π (S2 \ α , ... , α . Note that the order of this product Q1 1 2g+2d−2  NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 19 is important, and depends on the paths chosen between the points. The number of 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 are Riemann-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 such Riemann 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 seen by 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, all together, we have to specify two more generators, of the fundamental group of the torus, in addition to those of the branch points. We will get transpositions for the branch points, and then two more arbitrary corresponding to the 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 of the rectangle. Q Remark 9.6. The answer to this Question 9.5 is also open, and this question even came up in Dawei Chen’s thesis.

Question 9.7. (Algebra Problem) Given α1, ... , αb, where b = 2g + 4, we want a 3 2 cubic polynomial in y with coefficients in C[x] of the form y + α2(x)y + α1(x)y + α0(x) with 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 will necessarily have thisQ form. So, a hyperelliptic curve is either 2g+2 2g+1 2 2 V(y = (x − αi)) or V(y = (x − αi)) i=1 i=1 Y Y depending on whether the curve has a branch point at or not. So, in the first case, we add point q, r at . To describe this map, we want to write down the canonical 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 the plane, the divisor is the sum of the ramification points. Let pi be (αi, 0) ∈ C. Then, (dx) vanishes at the pi, so

(dx) = pi − 2q − 2r i X where we computed the orders at infinity by , and degree considera- 1 tions. To obtain a , we may note ( y ) = pi + (g + 1)(q + r). P 20 AARON LANDESMAN

We can now take products and note that  dx  = (g − 1)(q + r) y When g ≥ 2, this is holomorphic. We can generate all differentials by multiplying by multiples of x. We will get g holomorphic differentials dx x dx xg−1 dx , , ... , y y y g−1  g−1 In this case, the canonical map φK : C P is the map given by 1, x, ... , x . In particular, the canonical map factors through the two sheeted cover as → C π P1 (9.1) φK φ|O (g−1) P1 Pg−1

So, the canonical map is 2 : 1 onto the . 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 is hyperelliptic of genus g ≥ 2, then the smallest degree of an embedding C Pr is exactly g + 3. Remark 9.9. In some sense, hyperelliptic curves have the most linear series→ on them, but they have the fewest embeddings.

10. 9/23/15 10.1. Hyperelliptic Curves. Recall from last time that if C is hyperelliptic of genus g, 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 unique completion 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 dx xg−1 dx H0(K ) = h , ... , i C y y In other words, the canonical map factors through the hyperelliptic map to P1.

φ C Pg−1 (10.1) π νg−1

P1

Definition 10.1. A line bundle L or divisor D or linear system |L| is special if h0(K − D) 6= 0 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 21

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 ∈ Pic C, we know that their image under the canonical map is linearly independent unless two points map to the same point. So, by geometric Riemann Roch, the number of linear relations is exactly the number of pairs of points in D which map to the same point. That is, this is pre- 1 cisely the multiple of the g2 that D contains. 1 In conclusion if D is special, we can write |D| = r · g2 + D0, where D0 is a fixed base locus. That is, the map φD given by any special D factors through π, and, in particular, 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 that hyperelliptic curves cannot be embedding into P2. But, if r ≥ 3, a general map will be an embedding. → Proof. Proved above.  Exercise 10.3. Conversely, it is also true that every nonhyperelliptic curve has an embedding 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 a degree 4,5,6 cover f : C P1. → Definition 10.6. Say C is k-gonal if there exists a g1. → k Warning 10.7. Be careful, it’s often unclear whether n-gonal curves are also n − 1 gonal 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 people had been working with it a century earlier. For now, we’ll pretend we know what we 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 situation better by examining another moduli space: Definition 10.13. Define the Hurwitz space 1 Hd,g = (C, f): C ∈ Mg, f : C P of degree d with simple branching

The one thing we can see for such a curve in H is the number of branch → d,g points.

Lemma 10.14. dim Mg = 3g − 3. 22 AARON LANDESMAN

Proof. Set b = 2d + 2g − 2 and the branch divisor will consist of an unordered b b tuple of distinct points. So, we obtain a map Hd,g P \ ∆, where the image 1 corresponds to a divisor of degree b on P . We now endow Hd,g with the structure of an via the covering map to Pb.→ We next want to utilize the projection 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 to the map to Mg? We can’t answer this in general, but when d > 2g − 2. We are now asking, how many simply branched maps f : C P1 of degree d are there? To specify such a map, we have to choose: (1) A line bundle L ∈ Picd(C), which→ is g dimensional, 0 (2) A pair of sections σ0, σ1 ∈ H (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 the zero 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 dim Mg worked so well was because we introduced some auxiliary information that allows us to get a handle on the curve C.

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

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 6 Monday: Adjoint series Starting Wednesday: Castelnuovo theory, chapter 3 in ACGH Let C be a smooth projective non-hyperelliptic genus 5 curve. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 23

4 11.1. Curves of genus 5. We have a canonical map φ = φK : C P . We first consider which quadrics vanish on C. → 0 0 H (OP4 (2)) H (OC(2)) which is a map from a 15 dimensional vector space to a 12 dimensional vector → space, so the kernel is at least 3 dimensional. That, is C lies on at least three quadrics. There are now two possibilities: (1) So, it might be that curves are an intersection of three quadrics. In this situation, C does not lie on any more quadrics, by the Noether af+bg+ch theorem. (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 if Qi are three general quadrics, then by Bertini, C = ∩iQi is smooth, and by adjunction, we have KC = KP4 (2 + 2 + 2) = OC(1). 1 (2) In case 1, C is not trigonal. If C were trigonal, then there exists a g3, so the curve contains three collinear points. But, if we have three collinear points, every quadric would contain the line through those three points. So, if C is 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 C is a two-sheeted cover of P1 (in genus 1, take any degree 2 series, in genus 2, take the canonical linear series). General curves of genus 3 and 4 are expressible as 3 sheeted covers of P1: In genus 3, we can project the plane 1 quartic away from a point, that is, the linear series K − P is a g3. In genus 1 4, the rulings of a quadric cut out the g3 on C. It turns out general curves of genus 5 are expressible as 4 sheeted covers of P1. 1 Note further that in genus 1 there is a 1 dimensional family of g2, in 1 genus 2, there is a finite number of g2. In general, in odd genus there is 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. Such 0 curves arise as an open subset of G(3, H (OP4 (2))) which is 3 · (15 − 3) = 36 dimensional. The fibers of the map

1 U = Λ ∈ G : C = ∩Q∈ΛQ is a smooth curve in P M5

PGL Are 5. We can do a similar count to show the trigonal→ curves form a 10 dimensional subvariety. Note that 3 · g − 3 + g2 − 1 = 36, so indeed, since Mg is irreducible, a general curve will be expressible as such a . Quote from Joe: “We’ll answer the question of whether C is tetragonal, or I’m not 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 of quadrics, occur? 24 AARON LANDESMAN

Note that trigonal curves exist, and so they must be as in this second case. Start 1 with a trigonal curve of genus 5. Given |D| a g3, let’s look at |K − D|. By Riemann 2 2 Roch, this is a g5. Note that if this curve has a base point, the curve has a g4, so it would either give a two sheeted cover of P1 (impossible since C is not hyper- 2 elliptic) or a plane quartic (impossible since g = 5). Therefore, C C0 ⊂ P , and the image is not smooth, since a smooth plane quintic is genus 6. Let’s now 2 0 look at |2D|, which is a g6. To see h (2D) = 3, note that it is at most→ dimension 3, by Riemann Roch, and at least 3 because the map Sym2H0(D) H0(2D) is injective, because the curve, when embedded in P1 is surjective, so does not lie on any 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 cut out on C by conics Q ⊂ P2 passing through R. To see this, a conic Q ⊂ P2, the preimage f−1(Q) in C is a divisor of the form K + p + q. If r ∈ V(Q), we can write f−1(Q) = D + p + q where D ∼ K, and D is the residual 8 points of intersection of Q with the curve. Furthermore, we get all canonical divisors in this way because C has genus 5 and we have a 5 dimensional space of conics passing through a point in the plane.  So, we now have a very interesting description of C: We can map

2 C C0 ⊂ P (11.1) φK π P4 where the map π is given by quadratic polynomials vanishing at r. Now, we have the C lies on a surface in P4. Let S = im P2 under the right map π. Then, we have that the curve lies on a surface. This turns out to be the projection of the 2 Veronese ν2 surface from a point. We can also say that we can resolve the map P2 S by at the point r. One way to describe this surface in P4 is to take a line and a plane conic living in complementary spaces spanning P4, take an isomorphism→ between them, and then take the union of the lines joining points on the 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. Assume C has genus 5 and is not hyperelliptic. 4 Let C P be a canonical map. Then, C lies on three quadrics Qi. Last time, we saw

(1)→C = ∩iQi. (2) C ⊂ ∩iQi = S, where S is a cubic scroll. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 25

As we saw last time, the second case corresponds to that were C is trigonal, a three sheeted cover of P1. The first case corresponds to that where C is not trigonal, but C is expressible as a 4 sheeted cover of P1. Lemma 12.1. Every canonically embedded curve which is not trigonal admits a map of degree 4 to P1. Proof. Let P2 =∼ { quadrics C ⊂ Q}. We can ask whether there are singular quadrics in this set. Inside P14 which is the space of all quadrics in P4, there is a quintic of singular quadrics. Singular quadrics in P4 are cones over quadric surfaces, 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 4 or fewer points. The intersection of the 2 plane and the curve is the intersection with the two other quadrics, which is 4 points. So, each 2 plane meets C in four 1 points. So, we get two g4’s on the curve and they are residual to each other (add to the canonical divisor). In the rank 3 quadric case, it is a cone over a plane conic. In this case, we get a 1 g4 which is semicanonical. In the trigonal case, you can write down the quadrics containing the curve, and they are all singular.  1 Question 12.2. Why do all g4’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 is → 1 surjective. Therefore, the 2-planes on the quadric gives us all g4’s. Remark 12.3. The converse, showing that all curves on the Segre are trigonal, we’ll see this from Castelnuovo theory. 12.2. Adjoint linear series. We’ll start by talking about plane curves, following three 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 a birational embedding into the plane so that the image is a nodal plane curve. → (3) Arbitrary plane singularities. 12.2.1. Smooth plane curves. Choose homogeneous coordinates [X, Y, Z] on P2 and X Y let L be the hyperplane Z = 0, and let x = Z , y = Z be affine coordinates on P2 \ L =∼ A2. Say∞ C ⊂ P2 is a smooth curve of degree d with 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. These ∂f(x,y) ∞ zeros are where∞ ∂x = 0. We know df|C = 0 since f is zero along the∞ curve. But, we also know ∂f ∂f dx + dy = df| = 0 ∂x ∂y C ∂f ∂f Since C is smooth, we know ∂x , ∂y have no common zeros on C. ∂f ∂f At a point p where dx = 0, we have ∂y (p) = 0, ∂x (p) 6= 0, we have ∂f ord dx = ord p p ∂y ∂f So, we can divide by ∂y without introducing any new poles. So, we consider the differential dx ∂f ω0 = , fy = fy ∂y ∂f Now, ω0 is holomorphic and nonzero in the affine plane. Now, ∂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) dx ∞ : deg g ≤ d − 3 ⊂ H0(K ) f C y d−1 Both sides have dimension 2 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

1 0 0 and since h (OP2 (−3)) = 0, we have that the map H (OP2 (d − 3)) H (KC) is surjective, and in fact defines an isomorphism.

2 → 12.2.2. Nodal Curves. Now, suppose we have C C0 ⊂ P with C0 nodal. For the moment, suppose C0 is chosen so that at the nodes, the branches of C0 do not have vertical tangent. The problem with the previous→ argument for smooth curves ∂f ∂f is based on the assumption that 0 = df|C = ∂x dx + ∂y dy. 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 the nodal curve has no vertical by our assumption, meaning that dx, dy are nonzero. (We assume this just for simplicity, we could do the same argument where they do have zeros there.) NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 27

ω = dx ω When we consider 0 fy , we have 0 will have simple poles at the points qi, ri, are the two preimages of each of the nodes pi so we must take g(x, y)ω0 when g(x, y) had degree at most d − 3 and g vanishes at qi, ri. Now, if there are d−1 δ nodes, we impose δ conditions on g, so the dimension is at least 2 − 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 vanishing of g.

13. 9/30/15 13.1. Program for the remainder of the semester. Through many examples, we have 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 order of depth and subtlety. r (1) What linear series gd may exist on a curve of genus g? r (2) What very ample or birationally very ample linear series gd may exist on a curve of genus g? r (3) What linear series gd 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- r tionally very ample maps) by Castelnuovo’s theorem. For very ample gd’s, we know the answer for r = 3, which is relatively recent, r = 4 might be known, but r ≥ 5 is still open. (3) This is answered by Brill Noether theory We’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 projective space, or birational embeddings, and much less so maps to projective space which aren’t generically injective.

13.2. Adjoint linear series. Remark 13.1. There’s a version of the following that applies to arbitrary reduced curves (which is quite similar to the nodal case) and is written up in appendix A in ACGH. ν 2 Recall the set up from last time: Suppose C − C0 ⊂ P is the normaliza- tion map of C, and C0 is a nodal plane curve of degree d. Say C0 has nodes at −1 p1, ... , pδ, and let ν (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/Z 2 ∼ 2 ∗P 2 be coordinates on A = P \ L . Let D = ν L . In A , let us say that C0 = V(f) where f(x, y) is a polynomial of degree d. Suppose∞ further that at each node of the curve, the tangent lines are not∞ vertical,∞ so that∞ the divisors dx, dy have simple poles 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

∗ g(x, y) dx W := ν , deg g ≤ d − 3, g(pi) = 0. fy Proof 1. Start with dx. This has poles along L . To kill the poles at , we consider dx f = ∂f dx f fy where y ∂y , because wherever has a zero y does as well, at a point on the curve. Therefore, polynomials of the form∞ ∞ g(x, y) dx W := ν∗ , deg g ≤ d − 3 fy so long as we are away from the nodes. But at the nodes dx doesn’t vanish, while fy does vanish. For this, we need g(pi) = 0. Then, by a dimension count, the inclusion 0 W H (KC) is an isomorphism.  → We can also present a coordinate free proof. Proof 2. Consider the fiber product

2 C S = Blp1,...,pδ P (13.1) π

2 C0 P .

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 E together with L generate the of S. i P 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 by using 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

1 1 And we now use that h (KS) = h (OS) = 0 by Kodaira , and so the induced map on global sections is surjective.  An additional fact is that 0 0 H (OS(mL − E)) H (OC(mD − ∆)) m completeness of adjoint series is surjective for all , and this is referred to as the ∞ . Now, suppose we are given a divisor→ on C. Suppose we’re given a divisor B on C of degree b. Assume that Supp B ∩ ∆ = ∅, for notational convenience. Question 13.3. Describe the complete linear series |B|. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 29

Solution: Choose any plane curve G of degree m 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 = deg G, with p1, ... , pδ ∈ H, R ⊂ H. For any such H, we have ν∗H = ∆ + R + S. Here R ∼ m · D − B − ∆ and so S ∼ mD − ∆ − R ∼ B ∞ B B = B+ − B− So, this shows us how to find the complete∞ linear series of . If a difference of effective divisors, we can choose G to contain B+ and H to contain B−, and we still obtain the desired answer. Remark 13.4. (History of genus) What was the genus to the ancients? People looked at lines and conics, and if you have any d points on a line, those points are the intersections of a line and a curve of degree d. Any 2d points on a conic, the 2d points on a conic are in general the intersection of a plane curve of degree d and a conic. However, it is not true that 3d points on a cubic lie on the intersection with a 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, we get a 2 for 1 deal, in the sense that passing through a node drops the degree of the residual intersection by 2 but only imposes 1 linear condition, so the original definition of genus was d − 1 − δ 2

14. 10/2/15 Weierstrass points Let C be a compact Riemann surface of genus g ≥ 2. Application: Prove that group of such a Riemann surface is fi- nite. Definition 14.1. Define

Sp = {− ordp(f): f holomorphic on C \ p} . This is a semigroup, and is called the Weierstrass semigroup of p ∈ C. The gap sequence Gp is N \ Sp. That is, the gap sequence is the order of poles that don’t occur.

Lemma 14.2. We have #Gp = g. Proof. Observe 0 0 Sp = m : h (mp) > h ((m − 1)p)

0 0 Gp = m : h (mp) = h ((m − 1)p)

Now, let’s compare h0(mp) and h0(K − mp). By Riemann Roch, as we increase m by 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 of m and gap sequence

m h0(mp) $hˆ0(K-mp) 0 1 g 1 1 g-1 2 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 order less than g + 1 at p, and no other poles.

Proof. Follows from Proposition 14.8. 

Definition 14.5. A point p ∈ C is a if the gap sequence is not Gp = {1, ... , g}. g Definition 14.6. The weight of p ∈ C is the sum w(p) := i=1(ai − i). Example 14.7. (1) For example, if the weight is 0, theP 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) if w = 2 we have Gp = {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

w(p) = g3 − g p∈C X 1 Example 14.9. (1) If g = 2 we have φKC P of degree 2, and it has 6 branch points, 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 the semigroup 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 plane containing 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 a flex point of C ⊂ P2. The set of flex points are the points of intersection with the Hessian, and so the Hessian is a sextic plane curve, which meets C of degree 4. Further, the intersection multiplicity of the point with the Hessian 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 a 0 basis ω1, ... , ωg ⊂ H (K). Write ωi = fi(z) dz. Being a Weierstrass point means NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 31 there is a differential vanishing to order g at p. We look at the Wronskian   f1 ··· fg 0 0  f1 ··· fg  ∆ =   det  . .. .   . . .  g−1 g−1 f1 ··· fg 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 a holomorphic differential which vanishes to order g at p. So, Weierstrass points are zeros 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 coordinate w. The first row of 2 ∂z  ∂z  ∆ is multiplied by ∂w . The second row is multiplied by is multiplied by ∂w , plus some multiple of the first row. Hence, repeating this, the is mul- g+1  ∂z ( 2 ) tiplied by ∂w . Now, note that these are the transition functions for a power ⊗ g+1 of the . Hence, ∆ ∈ Γ(K ( 2 )). Hence, the number of zeros of ∆ g+1 3 is the degree of this line bundle. So, this is degree (2g − 2) 2 = g − g. It still remains to show the order of the Weierstrass function is equal to the weight of the Wronskian. This is left as an exercise, as is written in ACGH and 3264, 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 corresponds to the semigroup which starts increasing as soon as possible. That is, Sp = {0, 2, 4, 6, ... , 2g, 2g + 1, 2g + 2, ...} g which has gap sequence Gp = {1, 3, 5, 7, ... , 2g − 1} which has weight 2 . This oc- curs if and only if the curve is hyperelliptic. We further see there are 2g + 2 such points, and so these account for all Weierstrass points Corollary 14.12. There are at least 2g + 2 Weierstrass points, and exactly 2g + 2 if and only 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 exhibited a semigroup which cannot occur. (2) Many of the deeper questions come from a variational point of view: Look at 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 with that 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) is finite. Proof. When C is hyperelliptic, we have an ad hoc argument: We can express C up to automorphism of P2, as a cover of P2 of order 2. Now, the of P1 fixing a set of 6 or more points is finite, so we’ll just cover the non-hyperelliptic case. Now, we have strictly more than 2g + 2 Weierstrass points. Lemma 14.15. If φ : C C is any automorphism, the number of fixed points of φ is at most 2g + 2. Proof. By Riemann Roch,→ we can find a meromorphic function f on C of degree g + 1 with g + 1 poles and zeros. Consider f − φ∗f. This has poles where f has poles or the pullback of f has poles. Hence, this difference has at most 2g + 2 poles, 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 must be the identity. 

15. 10/5/15 Here are some questions for the remainder of the course? r (1) What linear series gd may exist on a curve of genus g? r (2) What very ample or birationally very ample gd’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? r (3) What gd’s exist on a general curve of genus g? r The problem of which very ample gd’s exist on smooth curves is still open, and the most recent progress was made by a student of Harris in the last 10 years. r r Riemann Roch says that gd is determined once 0 ≤ d ≤ 2g. So, the gd’s are only interesting in the range 0 ≤ d ≤ 2g, at the parallelogram between d = 0, g = 0 and r d = 2g − 1, r = g − 1. Then, Clifford’s theorem tells us all gd’s lie in the lower half of this rectangle, that is, below th e line 2r = d. d Theorem 15.1. Say D has degree d with 0 ≤ d ≤ 2g − 2. Then, r(D) ≤ 2 with equality if and only if (1) d = r = 0, D = 0 (2) d = 2g − 2, r = g − 1, D = K 2 (3) C is hyperelliptic ad D = mg1. Proof. Look at the multiplication map H0(D) ⊗ H0(K − D) H0(K). Now, think of these in terms of projective spaces instead of vector spaces. Think of H0(D) r 0 as a family of divisors parameterized by P , and H (K − D→) as parameterized by Pg−d+r−1. Hence, we obtain |D| × |K − D| |K| Pr × Pg−d+r−1 Pg−1 → → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 33

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 dimension of the source is at most the dimension of the target. Therefore, r + g − d + r − 1. So, r + g − d + r − 1 ≤ g − 1 implying r ≤ d/2. The equality will follow from our upcoming discussion of Castelnuovo’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).  1 Remark 15.2. Take C hyperelliptic and take D = r · g2 + D0 and D0 to be general r of degree d − 2r, so that these points are basepoints, and we can get all gd’s on hyperelliptic 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 have that g is bounded. Castelnuovo asks what the largest possible genus is. We won’t answer 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 the genus as discussed above) Proof idea. Here is the idea: 0 The plan is to bound from below the of h (OC(m)). We will do this by bounding from below 0 0 h (OC(m)) − h (OC(m − 1)). 0 When m is large enough, we will have h (OC(m)) = md − g + 1, and by Riemann- Roch, we will know exactly what this dimension is. This will give us a bound from above 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, 0 0 ρ 0 (15.2) 0 H (OC(m − 1)) H (OC(m)) H (OΓ (m))

0 0 Therefore, H (OC(m)) − H (OC(m − 1)) = rk ρ, where rk ρ is colloquially the 0 number of conditions imposed by Γ on H (OC(m)). Hence, rk ρ is at least the 0 number of conditions imposed by Γ imposed by H (OPr (m)). as given by the composition of restrictions

0 0 0 (15.3) H (OPr (m)) H (OC(m)) H (OΓ (m))

0 Say Γ imposes at least h conditions on H (OC(m)). This is equivalent to having 0 p1, ... , ph ∈ Γ so that for all i, there is a σ ∈ H (OC(m)) with σ(pj) = 0, j 6= 0 0 i, σ(pi) 6= 0. Now, if we take a subspace im H (OPr (m)) ⊂ H (OC(m)). Now, we’ve moved away from the curve, and the question is: 34 AARON LANDESMAN

Question 15.4. How many conditions do the d points p1, ... , pd ∈ Γ impose on hypersurfaces of degree m?

Now, we have Γ ⊂ H =∼ Pr−1 =∼ Pn, where n = r − 1. This consists of d distinct points. 0 So, the number of conditions imposed by Γ on H (OPn (m)) is hΓ (m). where n hΓ (m) is the Hilbert function of Γ ⊂ P . 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 hyperplane section of a curve. We then obtain precisely the bound that the genus is at most g−1 2 2 . This happens precisely when the curve we start with is a curve in P The question that remains is what conditions we obtain on these lines when the curve is in Pt with t > 2. We then claim that for any curve in P2, a general hyperplane section 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 and smooth. → Remark 16.1. We assume C is smooth purely for notational convenience, but the ν following will generalize to singular curves, and in general we can assume C − r C0 ⊂ P 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

0 0 h (OC(m)) − h (OC(m − 1)) ≥ The number of conditions imposed by Γ on hypersurfaces of degree m in H =∼ Pr−1

= hΓ (m) where hΓ (m) is the Hilbert function, not the Hilbert polynomial, although for large m the two coincide and are equal to d. So, we would like to bound hΓ (m) from below. From this, we deduce a lower 0 bound on h (OC(m)), and so for m large, we will obtain from Riemann Roch, 0 h (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 is achieved 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 each point, 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 are colinear, we obtain equality.  NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 35

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 choosing one of the to contain two points but no others. Remark 16.5. If we plug in the minimum Hilbert function into our strategy, we d−1 obtain g = 2 , which occurs precisely for plane curves.

So, if r ≥ 3 a general hyperplane section will not lie in a line, and so hΓ (m) will be bigger than min(m + 1, d). We now want to say something about the geometry of the general hyperplane section of an smooth nondegenerate curve. Definition 16.6. A collection of d points in Pr are in linear if no subset 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 family of 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 curves can be three secant lines. This cannot happen because we know that curves can be 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’t sure whether it’s known whether this is true in characteristic p. In particular, the uniform 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 α Φ Φ = {(H, p) ∈ U × C : p ∈ C ∩ H} C (16.1) β

(Pr)∨ U

Note that α is a d sheeted covering space, and is, in fact, a finite proper etale. It’s unramified because we removed the hyperplanes which are tangent. Above, Φ is the closure of Φ, containing hyperplanes with multiple points. It’s still flat, though not 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)?

Lemma 16.10. (Uniform Position Lemma) G = Sd. 36 AARON LANDESMAN

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 is connected, but since it’s also smooth, being a covering space of a , 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 general hyperplane section of the curve contain 3 collinear points. Suppose every section had three collinear points. We can vary the points so that they remain collinear. 0 More formally, let Φk be the set 0 Φk = {(H, p1, ... , pk) : p1, ... , pk are linearly dependent } 0 0 This is a closed subset. So, Φk 6=⊂ Φk, implying dim Φk < dim Φk = r, where here we are crucially using that the curve is nondegenerate. So the projection map 0 Φk 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 distinct points. Then, Φ U is a d sheeted covering space. Fix a base point H0 ∈ U. We obtain a 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 , 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 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 37

(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 this map are all irreducible open subsets of Pr−2 of the same dimension and hence Φ2 is irreducible. This→ means G is twice transitive. (2) Now, we want to show G contains a transposition. Choose a hyperplane r ∨ H0 which is simply tangent, so H0 ∈ (P ) \ U, so

H0 ∩ C = {p1, ... , pd−1} r ∨ and H0 is tangent to C at p1. Look at π : Φ (P ) over an analytic r ∨ (etale) neighborhood of H0 ∈ (P ) , call it V. Note that at this point, the map is not a covering map at this point. → Remark 17.2. The existence of such a hyperplane depends on k having characteristic 0. It is possible in characteristic p that every point is a flex point. (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 be contained in the alternating group, giving a counterexample to this lemma in characteristic p.

Now, when at H0, we have simple branching of π at the points p2, ... , pd−1 and ramification of order 2 at p1. Then, p2, ... , pd−1 have unique analytic 0 continuations 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 projection map 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, Φ is smooth at (H0, p1), and so Φ ∩ 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, q0), which induces a transposition. The image of this transposition γ in U is an element of π1(U, H0) which induces a transposition 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 smooth curves. 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 a generic Γ hyperplane section of C ∈ Rr with r ≥ 3, the points of Γ will be in general linear position.

Question 17.5. What sort of lower bound on hΓ can we deduce from the uniform position lemma? The idea will be to take a hyperplane containing 2 of the point, and if the points are in general position, it won’t contain a third point. 38 AARON LANDESMAN

n Lemma 17.6. If Γ is a collection of d point in P 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 of degree m so that X contains p1, ... , p^i, ... , pmn+1 but not pi. This says the points impose independent conditions, so the Hilbert function is at least mn + 1. Now, group the points p1, ... , p^i, ... , pmn+1 into m sets Γα of n points. Take X = ∪αΓ α. 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 we could 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 on hyperplane 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 0 → 0 h (OC(m)) − h (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 the inequality

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 rational normal 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 0 h (OC) = 1 0 h (OC(1)) ≥ r + 1 0 h (OC(2)) ≥ 3r 0 h (OC(3)) ≥ 6r − 2 . . This continues until we obtain d < m(r − 1) + 1. Set d − 1 m = b c 0 r − 1 so write m0(r − 1) + 1 + ε with 0 ≤ ε ≤ r − 2. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 39

Then, 0 h (OC) = 1 0 h (OC(1)) ≥ r + 1 0 h (OC(2)) ≥ 3r 0 h (OC(3)) ≥ 6r − 2 . . m + 1 h0(O (m )) ≥ 0 (r − 1) + m + 1 C 0 2 0 . . m + 1 h0(O (m + k)) ≥ 0 (r − 1) + m + 1 + kd C 0 2 0

Applying Riemann Roch, for k large, we have m + 1 (m + k)d − g + 1 = h0(O (m + k) ≥ 0 (r − 1) + m + 1 + kd 0 C 0 2 0 Definition 18.2. A divisor D is special if H1(C, D) 6= 0, and is nonspecial other- wise. 0 Remark 18.3. We could have stopped at m0, since we can show h (OC(m0)) is 0 nonspecial. In general, this is the best bound we can give on m so that h (OC(m)) is nonspecial, although it’s an interesting question to ask what the least m is in particular cases. Cancelling kd, we have

m + 1 m d − g + 1 = h0(O (m + k) ≥ 0 (r − 1) + m + 1 0 C 0 2 0 So, m + 1 g ≤ m (m (r − 1) + 1 + ε) − 0 (r − 1) − m 0 0 2 0 m  = 0 (r − 1) + εm 2 0 Definition 18.4. Define   m0 π(d, r) = (r − 1) + εm 2 0 where again, d = m0(r − 1) + ε, with 0 ≤ ε ≤ r. Theorem 18.5. If C 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 m0 r 0 1 r+1 1 1 r+2 2 1 ...... 2r-2 r-2 1 2r-1 r-1 1 2r r+1 2 2r+1 r+3 2 ......

Example 18.6. Take r = 3. k 2 (1) Take d = 2k + 2, m0 = k, ε = 1. So, π(d, 3) = 2 + k = k . 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 genus of a curve of type (k, k + 1) on Q. Let’s make a table for π(d, r). Asymptotically, we have d2 π(3, r) ∼ 2r − 2 We see that there are curves of degree 2r in Pr+1 are canonical curves. We can also see 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 0 0 hC(m) − hC(m − 1) ≤ h (OC(m)) − h (OC(m − 1)) ≥ hΓ (m) throughout. That is, both inequalities must be equalities. So, we must have that C is projectively normal In other words,

0 0 (18.1) H (OPr (m)) H (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 normal point at the , 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. → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 41

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

0 0 (18.2) H (OPr (m)) H (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 minimal Hilbert function hΓ = min(d, mn + 1) and d ≥ 2r − 1, then Γ lies on a rational normal 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 is the restriction of a quadric containing the curve. In this case, we can look at ∩C⊂Q⊂Pr Q. This must be a surface whose general hyperplane section is a rational normal curve.

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

19. 10/16/15 19.1. Review. Previously, we took C ⊂ Pr irreducible, nondegenerate of degree d and genus g, and found g ≤ π(d, r). We wondered whether this was sharp, and whether 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 + 3 points in linear general position. Then, if hΓ (2) = 2n + 1, then Γ lies on a rational normal curve.

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

n Now, suppose Γ ⊂ P 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 containing a 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 of dimension k. Question 19.3. What is the smallest possible degree of such a variety of such an X? Answer: It’s not hard to obtain a lower bound. Suppose n ≥ 2. A general hyperplane section is going to again be nondegenerate. So, if we continue cutting down the dimension, we will obtain an irreducible nondegenerate set of points: Say dim X = k. Observe X ∩ H1 ∩ · · · ∩ Hk. We then have a nondegenerate set of points in Pn−k, and it takes n − k + 1 points to span Pn−k Therefore, the degree of X must 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 minimal varieties. Say k = 2, so we’re looking for a surface of minimum possible degree. We will now 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 = dim X = 2. Now, choose a rational normal curve in Pk and Pl, meaning a map φ : P1 Pk φ0 : P1 Pl 0 → where the images are C, C . We’re assuming k + l = n − 1. Now, take X = Xk,l = 0 → ∪t∈P1 φ(t), φ (t). This is an irreducible . It’s nondegenerate because it contains the two rational normal curves. Lemma 19.6. deg X = n − 1. Proof. To calculate deg X, we can equivalently find deg X ∩ 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 of lines, all of which contain C. If we have any point not on C, then we contain the complete line. Therefore, the intersection consists of the union of C and a collection of lines of this ruling. The number of such lines is the number of points it meets C0 0 0 l in. It will meet C in l points, because deg C = l. Therefore, X ∩ H = C ∪i=1 Li, where Li are lines. So, deg X = deg X ∩ H = deg C + l = k + l = n − 1.  Remark 19.7. Here, we’re implicitly assuming k, l > 0, as if either were 0 we would get a cone over a rational normal curve. When we do a calculation, we can think of X as a curve in the 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 map onto 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 this cone. If k, l > 0, then this surface X is smooth, as can be seen in several ways. For example, we can calculate the tangent space at a given point. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 43

These surfaces are called rational normal surface scrolls. 3 3 Example 19.8. (1) In P , we can take X1,1 ⊂ P . Choose two lines, and draw the lines joining the points. This is a quadric surface. 4 4 (2) Consider X2,1 ⊂ P . This is the union of lines joining a line in P and a conic in a plane. 5 (3) In P , there are X3,1 and X2,2. It’s not obvious these are different, but, in fact, 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). n (3) Take general linear→ forms Li, Mi to on P . If we look at the locus where  L L ··· L  X = rk 1 2 n−1 < 1 M1 M2 ··· Mn−1

(4) We can now generalize this to any dimension k. Then, if we choose Pa1 , ... , Pak ⊂ n k P to be complementary, so that 1 ai = n − k + 1. Then, choose maps 1 ai parameterizing rational normal curves φi : P P and set Xa1,...,ak = P 5 ∪t∈P1 φ1(t), ... , φk(t). The first example is X1,1,1 ∈ P . That is, we choose three spanning lines in P5, and we draw the 2 plane→ spanned by triples of points 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 − dim X + 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). 2 5 (3) X is a cone over the ν2(P ) of degree 4 in P . 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 Veronese surface. So, we’re just trying to find curves on a scroll. We would like to see if we can find a divisor class (i.e. curve) on a minimal surface whose genus achieves Castelnuovo’s bound. 1 Let X = Xk,l be a surface scroll. We will assume X is smooth. X is a P bundle over P1. So, the Picard group is generated by two elements, one which is a line of the ruling, and take the other to be a hyperplane class (which must be independent because it has nonzero intersection with the line class). So, let h be a hyperplane section, f be a fiber. Then, Pic(X) = Zhh, fi Writing out the intersection pairing, we have The next step is to find the canonical class: 44 AARON LANDESMAN

h f h n-1 1 f 1 0

h f h n-1 1 f 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 on Friday, 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. Let n hΓ (2) = 2n + 1. Then, Γ is contained in a rational normal curve C in P . 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 if C ⊂ Pr is irreducible smooth nondegenerate of degree d, genus g, 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 surface scroll or a Veronese surface. Also, conversely, if S is a rational normal surface scroll and C is a curve of class where d = m(r − 1) + 1 + ε with 0 ≤ ε ≤ r − 2 (m + 1)h − (r − 2 − ε)f or ε = 0 and C ∼ mh + f on such a surface, then equality is achieved (and a similar statement holds for the Veronese surface). Remark 20.4. The key is that when a curve is sitting in Pr, we know very little about its geometry. But, if we get the curve to sit on the surface, we know exactly how 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) = Zhh, fi where f 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 any larger 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 normal curve. We have

−2 = 2g − 2 = h(KS + h) NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 45 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, the canonical 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 the intersection table deg C = C · h = a(r − 1) + b and, the genus of C is (ah + bf) ((a − 2)h + (b + r − 3)f) g(C) = + 1 2 a = (r − 1) + (a − 1)(b − 1) 2 Remark 20.6. We can check the above when r = 3, and a curve in the basis h, f is a curve of type (a, a + b), as we’re choosing a basis for the Picard group whose generators 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 m g = (r − 1) + mε 2 The choice was unique when r = 3 except for possibly a confusing between two rulings 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 do this because every curve on the Veronese surface has even degree, and it is the image of some plane curve, which was can use to find the genus. This also yields equality g = π(d, r).  Remark 20.8. If you had a surface and you wanted to bound the geometric genus of curves lying on the surface H0(K), you can use hyperplane sections to get lower bounds on the hyperplane sections, Remark 20.9. To exhibit curves which achieve the bound, we try to use Bertini’s theorem, to show that the linear system (m + 1)h − (r − 2 − ε)f, we take some very singular curves which are unions of hyperplanes on the surface. We exhibit enough of the curves to show it’s basepoint free. Then, Bertini’s theorem tells us there are smooth curves in these linear systems. Remark 20.10. By Clifford’s theorem, you can’t have a curve of degree 2r with genus 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’s lemma. For instance, if we have 8 points in P3 which impose 7 conditions on quadrics, they could lie on the intersection of three quadrics. So, you really need the 2n + 3rd points for Castelnuovo’s lemma. 46 AARON LANDESMAN

But, now, Enriques made an interesting observation: Canonical curves lie on many quadrics. In genus 5, a general canonical curve is the intersection of three quadrics. 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 a minimal degree surface. Then, we can go through the whole analysis of the bound on the genus. We can choose a general hyperplane H through p. Since p ∈/ C, by Bertini, such a hyperplane will intersect the curve transversely at {p1, ... , p2r}. The same analysis, by the uniform position lemma, implies Γ ∪ {p} is in linear general position. Essentially, this amounts to applying the uniform position lemma from the 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 . In particular, the in- tersection of the quadrics has to be a surface. The above remark establishes the following, except for trigonality and plane quintics.

Proposition 20.12. Let C be a canonical curve. If there is p ∈ ∩Q⊃CQ but p ∈/ C, we claim that C lies on a minimal degree surface. In particular the curve is trigonal or a plane quintic. Proof. To prove trigonality, if the curve is on a scroll, the fibers meet the curve in three points. In the case that the curve lies on the Veronese surface, it is a plane quintic.  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, because by geometric Riemann Roch, there will be three collinear points, and so it cannot be 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 for the next step of the resolution of the curve. Then, the relations are all linear, unless 1 2 C has a g4 or a g6. This was extended by Green and Lazarsfeld to a conjecture on the 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 bound which don’t occur. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 47

21. 10/21/15 Recall, C ⊂ Pr is a smooth irreducible nondegenerate curve of degree d genus g, and Γ = H ∩ C. Then, 0 0 h (OC(m)) − h (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 rational normal surface scroll. Recall Castelnuovo’s lemma: 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 and hΓ (2) = h0(2) then Γ is contained in a rational normal curve. Since we have determined the smallest Hilbert function corresponds to lying on rational normal curve, this motivates the next question. In particular, we’d like to 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 the maximum 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 are unions of hyperplanes. However, in fact, we have a stronger condition. We’ll see what 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 subsets Γ 0, Γ 00 ⊂ Γ so that |Γ 0| = |Γ 00| then

hΓ 0 = hΓ 00 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 a collection of point in a hyperplane H, then either no six points lie on a conic, or else all points lie on a conic. Proof. The idea is similar to that of linear general position: If three points were on a 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 can swap 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 of C. The complement is the set of hyperplanes transverse to C. Define the incidence correspondence. Φ = {(H, p): p ∈ H ∩ C} ⊂ U × C We now specialize to the case of curves lying on quadrics in P3. We can similarly introduce the set of sequences of points, which is

ψ = {(H, p1, ... , p6) : pi distinct in H ∩ C} 48 AARON LANDESMAN and let 0 ψ = {(H, p1, ... , p6) : pi lie on a conicH ∩ C} 0 If there exist 6 points p1, ... , p6 that don’t lie on a conic, we have ψ ( ψ im- plies dim ψ0 < dim ψ = r, because ψ is integral, as follows from the monodromy statement. Therefore, general hyperplane sections can’t have some subset of points lying on a conic and not all subsets not lying on a conic.  Remark 21.7. So, the second smallest Hilbert function should correspond to the 3 points lying on a cubic in P . So, we currently have hΓ (m) ≥ min(d, 2m + 1) in P3. 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 lie on 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 because n it lies on a rational normal curve B ⊂ P so that hΓ (m) = hB(m) = mn + 1, with d m ≤ n . Recall, in general, if B ⊂ Pn is a curve of degree d and genus g, the Hilbert polynomial 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 the second smallest Hilbert function? The next smallest would have to have degree at least n + 1. Then, the genus of the curve may be either 1, 0 as follows from Castelnuovo’s theorem. We conclude that the second smallest Hilbert function for hΓ of a collection Γ of points 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 of points with small Hilbert function have that small Hilbert function because they lie on a curve with small Hilbert function. It takes some work to prove this is actually the case, since it might be that they don’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 m(n + 1) if d > m(n + 1) hΓ (m) = 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’t really proven.  Corollary 21.12. If C 6⊂ a rational normal surface scroll, then we obtain that the genus 2 is bounded by g ≤ π1(d, r) ∼ d /2r NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 49

Proof. This follows from plugging in the inequalities from Proposition 21.11, we 2 obtain g ≤ π1(d, r) ∼ d /2r. We get this by computing the Hilbert function for large values of m, and then using Riemann Roch to bound the genus.  When we were analyzing curves on rational normal surface scrolls, we found the Picard group has rank 2, and so we used this to determine the possible genera of curves on these scrolls. Let’s now describe what happens for curves in P3.

3 d2 Proposition 21.13. In P , for g > π1(d, 3) ∼ 6 , we have that g occurs if and only if there exist a, b so that a + b = d and (a − 1)(b − 1) = g. If g ≤ π1(d, r), then we 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, and quartic has Picard groups of rank up to 19, and it turns out you can always solve the system for each genus.  Question 21.14. What about in higher dimensional space?

r d2 Proposition 21.15. In P , if g > π1(d, r) ∼ 2r , then the curve lies on S, and g occurs if and 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) a g = (r − 1) + (a − 1)(b − 1) 2 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 del Pezzo 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 of the Picard group also decreases. Look in the Montreal notes for more details. Remark 21.17. If we drop the hypothesis of , and take the geometric genus, we do get curve of every geometric genus up to π(d, r). The reason is that in the linear system on a rational normal surface scroll S, we can find a curve class that ache Ives the maximum, and there are smooth curves of this class. Further, in this linear system, there exist irreducible nodal curves C ⊂ S with δ nodes for any 0 ≤ δ ≤ π(d, r). Exercise 21.18. There exist irreducible plane curves with δ nodes for any δ between d 0 and 2 . 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 Theory Let’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 points in uniform position? The question of all possible Hilbert functions has been answered by Richard Stanley and others, which is solved by considering monomial ideals. However, the question of uniform position points is more difficult. There are other interesting subschemes whose Hilbert functions we can ask about, such as Hilbert functions of fat points (schemes supported at a point). 22.2. Resolutions of Projective Varieties and Green’s conjecture. Suppose X ⊂ n P is a variety or subscheme. We can look in I(X) ⊂ S = C[x0, ... , xn]. Say I(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 exist among 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 the Hilbert 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 on three 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’t cut out the curve, there’s an additional cubic in the ideal of the curve. There is also a 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 complete intersection 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. Suppose C is a smooth curve of genus g. φ Say C is a smooth curve of genus g. Then, We have C −−K Pg−1 =∼ PH0(K)∨. and S = Sym•H0(K). Then, we introduce the 0 m → R = ⊕m=0H (K ) Now, by Noether’s theorem, the map∞ S R or equivalently, Sym•H0(K) 0 m ⊕mH (K ) 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

⊕ g−2 (22.3) ··· S(−2) ( 2 ) S S(X) 0 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 51 in the case that the curve is further not trigonal, nor a plane quintic. 1 2 This is equivalent to saying there does not exist a g3 or a g5. Now, here’s the next step, which will lead to Green’s conjecture. We can now look at the linear relations among these quadrics.

⊕ g−2 (22.4) ··· S(−3)⊕N S(−2) ( 2 ) S S(X) 0

1 2 is exact if and only if C does not have a g4 or g6. Green’s conjecture naturally extends this. If L is a line bundle on C of degree d and dimension r = H0(L) − 1. If we assume H0(L), H1(L) 6= 0 then d − 2r ≥ 0. r Definition 22.4. The Clifford index of a linear series L which is a gd is d − 2r. Then, the Clifford index of a curve is the minimum over all line bundles L with h0(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 ⇐⇒ 1 2 3 Cliff(C) = 2 C has a g3 or g5 or g7 ⇐⇒ 3 3 Note for curves with g7, we get a map C P , then the curve has a 4 . Then, you can project from a⇐⇒ point on the line, and we obtain that C is already trigonal. → Here is a consequence of the Brill Noether Theorem. g−2 Corollary 22.6. For a curve C general of genus g, then Cliff(C) = d 2 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 the remaining 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  g−2 Remark 22.9. Voisin does not prove that a curve with Clifford index d 2 e has a maximally 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 curves which have a maximally linear resolution is an open condition: It’s saying that the ranks 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 . Her approach doesn’t seem like it will yield anything more in the direction of proving Green’s conjecture. So, she doesn’t prove which curves are excluded. 52 AARON LANDESMAN

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

I(C)2 ⊗ S1 I(C)3

g−2 S g We know the left hand side has dimension→ 2 and 1 has dimension . Then, I(C)3 has dimension given by the kernel of

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

g+2 g+3 The latter vector spaces have dimensions 3 , 5g − 5, and so dim I(C)3 = 3 − 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- N1 → g−2 jecture would imply S(−3) generates the relations, and so N1 = 2 · g − g+2 3 − (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? r → The maximal rank conjecture is a solution to this for a general curve in P . To specify a map from a curve to projective space, you have to specify three pieces of data

(1) C general in Mg r (2) L ∈ Wd 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 0 0 φm : H (OPr (m)) H (OC(m)) which have dimensions m+r and md − g + 1. Then, the conjecture is that this r → 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. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 53

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 of genus 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? r Theorem 23.4. (Brill-Noether) Let C be a general curve of genus g. Then, C has a gd if and 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 the Jacobian 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 theory to give the proof now. Fact 23.5. Recall,

r d 0 Wd := L ∈ Pic (C): h (L) ≥ r + 1 r Now, Wd turns out to be a variety itself, and so, in fact, we can ask about general points of it. We can also ask about its dimension and irreducibility, and so in. It turns out (1) r dim Wd = ρ r (2) If ρ > 0, then Wd is irreducible. r (3) If L ∈ Wd, if (a) r ≥ 3, then L is very ample. 2 (b) If r = 2, then φL is birational onto C0 ⊂ P is nodal 1 (c) If r = 1, then φL P is simply branched. Remark 23.6. Our goal is to relate → Mg := { abstract smooth curves of genus g } to r Hd,g,r := { curves C P 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 2 Vd,g = C ⊂ P : 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 1 Hd,g = f : C P : 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.

Hd,g (23.1) 1 ∼ b Mg (P )b \ ∆ = P \ ∆

1 b where b = 2d + 2g − 2, and ∆ ∈ (P )b is the diagonal while ∆ ⊂ P is the 1 ∼ b discriminant locus. Again, the identification (P )b = P is by identifying divisors of degree b with polynomials of degree b, modulo scalars. The right projection sends a curve to its branch locus, and is a finite surjective map by a monodromy argument. As a consequence,

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

1 Hd,g/PGL2 = (C, D): D is a gd 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 us Mg 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 g d ≥ + 1. 2 (1) If g = 0, we’re looking at rational curves, so d ≥ 1. 1 (2) If g = 1, d ≥ 2. Here, there is a 1 dimensional family of g2’s. (3) If g = 2, then d ≥ 2, and C is a 2 sheeted cover of P1, in a unique way via 1 the canonical series. Here, there is a unique g2. 1 (4) If g = 3, then C, if not hyperelliptic is a plane quartic, and K − p is a g3. 1 Here, there is a 1 dimensional family of g3’s. (5) If g = 4, d ≥ 3 as it is the intersection of a cubic and a quadric. Here there 1 are generals 2 g3’s. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 55

(6) If g = 5, we have seen d ≥ 4, by the description of the complete intersec- 1 tion of quadrics. (There is a 1 parameter family of g4’s.) r We see that in these low genus examples, the dimension of Wd is indeed, ρ, as claimed. 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, 4 if we take a general line bundle L ∈ Pic (C) that L = KC + p + q has 0 2 h (L) = 3. Then, we deduce that φL : C P carries C to a nodal curve of degree 4, with 1 node. This maps p, q to the same point, but is otherwise an embedding. → (4) If g = 3, then d = 4, as general genus 3 curves are canonically embedded by quadrics in P2. (5) If g = 4, we can take a curve of type (3, 3) on a quadric. We can choose a point on the quadric from a point on the curve, you get an embedding into 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 little tricky. 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 curve of genus g? r For which d, g, r does there exist a gd on C. Recall that in the case r = 1 we introduced the Hurwitz space

Hd,g (24.1)

b Mg P \ ∆ as an incidence correspondence.For α to be dominant, we obtain dim Hd,g/PGL2 ≥ g dim Mg, so we obtain d ≥ 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 the plane curves. That is, we can access the coefficients of the polynomial cutting out the curve in P2. Here, we look at the Severi variety (d+2)−1 Vd,g = {C reduced and irreducible curve of geometric genus g and degree d} ⊂ P 2 Additionally, inside the Severi variety, we have d − 2 V ⊃ U := C : C is nodal with δ nodes, whereδ = − g d,g d,g 2

The variety Ud,g is quite well behaved, but Vd,g is quite complicated. Ud,g will be smooth, while Vd,g is far from being smooth. 56 AARON LANDESMAN

TABLE 3. My caption

FGH ∂ ∂x 0 a20 a11 ∂ ∂y 0 a11 a02 ∂ 1 0 0 ∂a00

Remark 24.2. ACGH was divided into two parts: the geometry of fixed curves and the varieties of families of curves. A good reference for the Hurwitz space and the Severi variety, see Moduli of Curves, by Harris and Morrison. Fact 24.3. Here are two facts: N d+2 Vd,g, Ud,g are locally closed subsets of P where N = 2 − 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 can be deformed to one with simple branching. In this case, it’s more of an exercise in deformation theory, but the fact is that any curve can be deformed to one with only nodes, of the same geometric genus.

Theorem 24.4. Ud,g is smooth of dimension 3d + g − 1. Proof. Look at the incidence correspondence

N 2 Φ := (C, p) ∈ P × P : p ∈ Csing (24.2)  ∆ ⊂ PN P2 where ∆ is the locus of singular curves. The fibers of P2 are linear space of dimen- sion N − 3. Hence, PN−3. So, Φ is irreducible and smooth of dimension N − 1. So, ∆ is N − 1 dimensional and irreducible. The fiber over ∆ is finite because a general curve has only finitely many singularities. Now, the three equations defining Φ are very explicit.

i j F(a, x, y) = aijx y i−1 j G(a, x, y) = Xiaijx y i j−1 H(a, x, y) = X jaijx y

Say (C0, p0) ∈ Φ. Take p0 = (0, 0) in affineX coordinates. Then, C0 = V(f) where 0 i j 2 2 0 f = aijx y = a20x + a11xy + a02y + ··· for aij ∈ k. Now, let’s consider the chart P 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 (C , p is given by the vanishing of ∂ , ∂ , ∂ and 0 0 ∂x ∂y ∂a00 dπ : T Φ T N (C0,p0) C0 P is injective with image the hyperplane of curves containing p0. → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 57

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 to δ branches ∆1, ... , ∆δ of ∆ at C0 δ Locally, Ud,g = ∩i=1∆i. To conclude, we have to know whether the tangent hyperplanes intersect transversely or not. that is, we want to know that the tangent planes 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, N C ∈ P : 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 plane several 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 d + 2 N − δ = − 1 − δ 2 d − 1  = − δ − 1 + 3d 2 = 3d + g − 1  Remark 24.5. ∆ has pretty large degree. It is degree something like 3d2, and not really worth writing down. 2 2 Corollary 24.6. If a general curve of genus g has a gd then d ≥ 3 g + 2.

Proof. We know dim Ud,g/PGL3 ≥ dim Mg. We see

dim Ud,g/PGL3 ≥ dim Mg 3d + g − 9 ≥ 3g − 3 2 d ≥ g + 2 3  Question 24.7. What happens when r ≥ 3? In this case, we have no idea what happens. When r = 1, 2, we have very simple equations for the curves or branch points. In P3, we don’t have a single defining equation, and we don’t even know how many equations define the curve. If we arbitrarily 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 C r of genus g has a gd if and only if ρ = g − (r + 1)(g − d + r) ≥ 0, and in this case r dim Wd(C) = ρ. In the case r = 1 we introduce the Hurwitz space of simply branched coverings of P1, which is b := 2g + 2d − 2 dimensional. 1 g We conclude that a general curve C has a gd then d ≥ 2 + 1. 58 AARON LANDESMAN

When r = 2, we can introduce the small Severi variety of curves in P2 with d−1 2 − g = δ nodes. We saw last time this has dimension 3d + g − 1. We can 2 2 conclude that a general curve has a gd implies d ≥ 3 g + 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 Hilbert scheme.

0 r Hd,g,r = {C ⊂ P : 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 ways we could deform the curve to first order and preserve the nodes. We don’t know in 0 general the dimension of Hd,g,r because it is intrinsically not well behaved. In the r = 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 of components and be irreducible. So, in this case, we’ll be able to use the Brill Noether theorem to describe, at least 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 the things 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 Noether theory implies this. However, we don’t know when this 4d occurs, or when the dimension is bigger. Remark 25.3. Hopefully, today, we’ll see why the Brill Noether number is ρ = g − (r + 1)(g − d + r) r There are two reasons why Wd is not a great object to look at. r (1) It’s difficult to write down equations for Wd 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 invertible sheaves, but rather actual linear series (and not complete linear series). Recall: Definition 25.4. r d 0 d ∼ Wd(C) = L ∈ Pic (C): h (L) ≥ r + 1 ⊂ Pic (C) = J(C) r 0 Cd = D ∈ Cd : h (O(D)) ≥ r + 1 r d r+1 0 Gd = (L, V): L ∈ Pic (C), V ⊂ H (L) we have a map d u : Cd Pic (C) −1 r r u (Wd) = Dd →r r u(Cd) = Wd NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 59

Remark 25.5. r d −1 Wd = L ∈ Pic : dim u (L) ≥ r r and is different from Cd because there may be many divisors associated to a given line bundle, in fact there is an r + 1 dimensional vector space of them. r Remark 25.6. To describe Wd for the purposes of the homework, just describe the dimension and the irreducible components. Lemma 25.7. Let r ≥ 0 and r ≥ d − g. r r−1 r Wd ⊂ Wd \ Wd r−1 r r That is, no component of Wd lies in Wd. In other words, the Wd give a stratification of the Jacobian. Question 25.8. The lemma says we can move an invertible sheaf slightly to reduce the number of sections by 1. How do we do this? This will be answered by the lemma. r r+1 0 Proof. Say L ∈ Wd \ Wd . That is, h (L) = r + 1. Consider the invertible sheaves of the form L(p − q), and we can specialize to L as p = q. The hypothesis that r > 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.  r Let’s start by describing Cd. r Lemma 25.9. We can write down explicit defining equations for Cd in a neighborhood of a divisor D = p1 + ··· + pd. Remark 25.10. It may help to think of the proof for d ≤ g, and you can flip it to d ≥ 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 0 by geometric Riemann Roch. Let ω1, ... , ωg be a basis for H (K). Then, write ⇐⇒ ωα = fα,i(zi)dzi in a neighborhood of pi. Define the Brill-Noether matrix   f1,1(z1) ··· fg,1(z1)  . . .  M =  . .. .  f1,d(zd) ··· fd,g(zd) 60 AARON LANDESMAN

Note, we may want to write the above matrix as   ω1(z1) ··· ωg(z1)  . . .   . .. .  ω1(zd) ··· ωg(zd)

But the rows of the matrix are only defined up to scalars. We had to choose local coordinates 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

0 0 H (K) H (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

Cd Md,g (25.1)

r d−r Cd Md,g

d−r where Md,g is d × g matrices and Md,g is d × g matrices of rank at most d − r. k Then, Ma,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 codimension d−r Md,g ⊂ Md,g is codimension r(g − d + r). The expected dimension is

r dim Cd = d − r(g − d + r)

This is also a lower bound on the dimension. 

r Corollary 25.11. We have dim Wd ≥ 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 need to 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 structure r r to Cd, and then port that over to Wd by taking the image. So, we can talk about r Wd as a subscheme of the . This is important because we’ll like r r to describe the tangent spaces to Cd, Wd scheme theoretically, defined by these equations from the minors of local matrices. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 61

26. 11/2/15 r r 26.1. Review. Today, we’ll move forward on the discussion of Wd, Cd. Take C to be a smooth curve of genus g. Recall:

r d 0 Wd(C) := L ∈ Pic (C): h (L) ≥ r + 1 r 0 Cd(C) := D ∈ Cd : h (L) ≥ r + 1 r r d r+1 0 Gd(C) := {gd on C} = (L, V): L ∈ Pic (C), V ⊂ H (L)

d Observe that if D = i=1 pi is a reduced divisor, then in an analytic neighbor- hood U of D ∈ Cd, we can write P   ω1(q1) ··· ωg(q1) r  . . .  Cd = (q1, ... , qd):  . .. .  ≤ d − r  ω1(qd) ··· ωg(pd)   0 0  = (q1, ... , qd): rk H (K)|q H (K|D)|q ≤ d − r

So, locally, we have a map → π d−r U − Md,g ⊃ Md,g M d × g Md−r where d,g is the set of matrices,→ and d,g the subset of matrices of rank at most d − r. d−r Observe that Md,g has codimension r(g − d + r) in Md,g. Therefore, U ∩ −1 d−r π (Md,g ) has “expected codimension” r(g − d + r). So, the expected dimension r d of Cd in C is d − r(g − d + r)

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

r 26.2. Hilbert Schemes. Now, let’s discuss the relation between these Wd and the Hilbert scheme. Remark 26.1. In the cases r = 1, 2, we were able to find the dimensions of the Hurwitz 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 at least 3. r r r Now, we introduce global analogs of Wd, Cd, Gd: Definition 26.2. Define r d 0 Wd = (C, L): C smooth of genus g , L ∈ Pic (C): h (L) ≥ r + 1 Mg

Cr = (C D): D ∈ C C L ∈ d(C): h0(L) ≥ r + 1 M d , d, smooth of genus g , Pic → g

Gr = (C L V): C L ∈ d(C) Vr+1 ⊂ H0(L) M d , , smooth of genus g , Pic , g →

→ 62 AARON LANDESMAN

Further, we have maps

0 r H Gd Mg

(26.1) r Wd Mg,d = {(C, D)) : D ∈ Cd}

r Cd where the fibers of the first map are PGLr+1 and the fibers of the second map have dimension at least ρ. Now, recall the Brill Noether matrix   ω1(q1) ··· ωg(q1)  . . .   . .. .  ω1(qd) ··· ωg(pd) r Now, we have that the expected dimension of Gd is 3g − 3 + ρ. So, the expected dimension of the Hilbert scheme 0 2 Hd,g,r = 3g − 3 + ρ + r + 2r 4g − 3 − (r + 1)(g − d + r) + r2 + 2r

0 Further, every component of Hd,g,r has at least this dimension. Question 26.3. Do the the components of the Hilbert scheme other than those corresponding 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 more than 4d. More recently, Daniel Erman found components of the Hilbert scheme of points 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 the Hilbert scheme? We can’t do it by writing down equations. For example, we can write down twisted 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 is 21 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.

0 r 26.2.1. Estimating the dimension of Hd,g,r. Suppose we have a point C ⊂ P of degree d genus g. We want to know when we can deform a curve. We can’t really see the space of curves, but we can find the dimension of the tangent space to the Hilbert scheme, which gives an upper bound on the dimension of the Hilbert scheme. 0 We want to estimate dim Hd,g,r in a neighborhood of C. 0 So, our first approximation will be dim TCH = H (C, NC/Pr ). NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 63

Warning 26.5. Caution: There are components of the Hilbert scheme which are everywhere nonreduced, such as smooth degree 24 genus 14 curves in P3. So, our estimate 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 quadrics Q1, Q2, Q3, and translate that to 0 0 0 Q1 + εQ1, Q2 + εQ2, Q3 + εQ3 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 cohomology of the normal bundle is the same as a map →  3  I(C)2 S(P )/I(C) 2 0 Qi 7 Qi + εQ → i 0 Now, we can bound H (C, N r ) using Riemann Roch as follows: C/P→ 0 Proposition 26.7. The dimension of the Hilbert scheme Hd,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 0 h (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 scheme above 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 the above estimate. However, complete intersections, in general, will have higher di- mension. the first example of this being violated will be complete intersections of quadrics and quartic. In fact, whenever the complete intersection has a quartic or higher degree surface, we’ll be done. Remark 26.10. We have two ways of estimating the dimension of the Hilbert scheme. (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 that C is a smooth curve of genus g, which are subvarieties

r Cd Cd (27.1) u

r d Wd Pic where H0(K)∨ Picd =∼ J = H1(C, Z) where the Jacobian is defined by the map

pi u : D = pi 7 i i p0 X X Z where this function above is dependent on→ choices of paths from p0 to pi and is reflected in the ambiguity of quotienting by H1(C, Z). To understand these varieties, we will look at their tangent spaces. We have d 0 ∨ ∼ 1 TL Pic = H (C, K) = H (C, OC) using Serre duality for the last step, and the description of the Jacobian for the first step. Also, for Cd, we have 0 ∨ ∼ 0 TDCd = ⊕iTpi C = H (C, K/K(−D)) = H (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 ⊕iTpi C is only correct when the points are 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 0 TDCd = H (C, OC(p)/OC) and a pairing 0 0 H (C, OC(p)/OC) ⊗ H (K/K(−p)) C

(f, ω) 7 Resp(fw) → → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 65

Proof. The first isomorphism was given above through 0 ∨ ∼ 0 TDCd = ⊕iTpi C = H (C, K/K(−D)) = H (C, OC(D)/OC) The pairing is also an example of Serre duality.  Now, recall,

pi u : pi 7 i p0 X X Z Therefore, by differentiating under the integral→ sign, we see   ω1(p1) ··· ωg(p1)  . . .  du =  . .. .  ω1(pd) ··· ωg(pd) Therefore, r D ∈ Cd rk (du) ≤ d − r This implies that all fibers of u are reduced Pr’s. This is saying the corank of the du is exactly the rank of the fiber, which⇐⇒ is Pr, and so the tangent space must be exactly r dimensional. Therefore, r r Wd = u(Cd) r −1 r Cd = u (Wd) By the above, this is also true at the scheme theoretic level. That is, we have r r Tu(D)Wd = du(TDCd) r Here, the key input is that the Brill Noether matrix, which defines Cd is also the differential of the map u. Suppose L ∈ Picd and h0(L) = r + 1, meaning r r+1 L ∈ Wd \ Wd 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 L˜ Spec C[ε]/(ε2) × C ∼ so that L|Spec C×C = L. → d Proposition 27.5. Vectors v ∈ TL Pic are in natural bijection with first order deforma- tions L˜ 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∗L˜ locally free? 66 AARON LANDESMAN

We have a map 0 π∗L˜|Spec C H (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∗L˜ were locally free.

d 1 Question 27.7. Given v ∈ TL Pic = H (C, OC), which is equivalent to deforma- tions L˜ of L. Given σ ∈ H0(K), does σ extend to a section of L˜? Answer: We have that σ extends if and only if 0 1 1 σ ⊗ v ∈ H (C, L) ⊗ H (C, OC) 7 0 ∈ H (C, L)

This implies that if L ∈ Wr \ Wr+1, and d d → d 1 v ∈ TL Pic = H (C, OC) then all sections σ ∈ H0(C, L) extend if and only if H0(L) ⊗ hvi 7 0 ∈ H1(C, L) Now, corresponding to the above, map, we have a map → 0 1 1 H (C, L) ⊗ H (C, OC) H (L) or, equivalently, we can view this as a map → 0 1 ∨ 1 ∨ H (C, L) ⊗ H (C, L) H (C, OC) Now, using Serre duality, we can rewrite this as → µ H0(C, L) ⊗ H0(C, K ⊗ L∨) −−0 H0(C, K) If we chase the diagrams, this map turns out to be multiplication. Next, from this, we obtain, → r TLWd = 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 r Question 27.8. How large can dim Wd be? Theorem 27.9. (Martens) Suppose 2 ≤ d ≤ g − 1. Then, r dim Wd ≤ 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 good r example of how we can use the first order analysis to understand Wd. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 67

Proof. Consider the multiplication map µ H0(C, L) ⊗ H0(C, K ⊗ L∨) −−0 H0(C, K) Note, this is the same map we looked at when proving Clifford’s theorem. We know h0(D) = r + 1, h0(K − D) = g − d + r. Then,→ a given canonical divisor can be written in at most finitely many ways as a sum of divisors in H0(L) + H0(K ⊗ L∨). Therefore,

dim im µ0 ≥ r + (g − d + r − 1) + 1 = g − d + 2r Therefore,

dim Ann (im µ0) ≤ d − 2r Now, we deal with the hyperelliptic case. If the curve is hyperelliptic, then to r 1 describe a gd, we take r · g2 and add a fixed divisor, which has degree d − 2r. We’ll come back to the converse this Friday.  r Example 27.11. We want to look for a Wd. To this, we can associate a residual r 0 linear series. What is the first case where Wd is not residual to a Wd. This is the 1 case of W3 in genus 4. Suppose C is not hyperelliptic. We have a canonical embedding C P3, which lies on a unique quadric, either 1 a smooth quadric or a quadric cone. We could describe the set of g3’s as lines on 1 the quadric surface. Therefore, we have→ two g3’s in the case that the curve lies on a smooth quadric surface or one if it lies on a quadric cone. 1 Question 27.12. Is W3 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 −1 ∼ 1 the product map is an isomorphism, and K ⊗ L 6= L. Here, W3 consists of two reduced points. (2) Case 2: The curve is on a singular quadric. Then, K ⊗ L−1 =∼ L, and so the multiplication map has a kernel because the line bundle is Serre dual to itself, and the kernel is given by the span of σ ⊗ τ − τ ⊗ σ. That is, if L =∼ −1 2 0 1 K ⊗ L , the multiplication map factors through Sym H (L). Hence, W3 consists 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) “parameterizes line 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 a line 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 restrictions are line bundles of degree→ d on C.

Theorem 28.4. Picd represents the functor sending a scheme to families of line bundles r r+1 d over B. In the same sense, Wd \ Wd ⊂ Pic parameterizes families of line bundles on C with at least r + 1 sections. More precisely, families of line bundles on C with at least r + 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 incomprehensible there.  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 tangent vector to Picd, and it can also be thought of as a “fuzzy line bundle” on C. This says, d ∼ TL Pic = line bundles L onD × C : L|Spec C×C = L

Let L ∈ Picd(C). Then, d ∼ TL Pic = { line bundles on D × C : L|C×C = L} To write down a line bundle, we can specify transition functions. So, choose an open cover {Uα} of C, we trivialize the line bundle on each open set, and compare the trivializations. × So, we can take gαβ ∈ OC(Uαβ)where Uα ∩ Uβ := Uαβ. Then, a section σ ∈ H0(L) is given by a collection of functions

sα ∈ O(Uα): sα|Uαβ = gαβsβ

−1 We adapt the convention that gαβ = gβα. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 69

Definition 28.7. A first order deformation of L is a line bundle L on D × C so that ∼ L|Spec C×C = L. We describe L via transition functions with respect to the open cover {D × Uα}.

Remark 28.8. We needn’t have hαβ ∈ OC(Uαβ) because gαβ are already invert- ible, so adding any multiple of ε will still be invertible. Proposition 28.9. We have d ∼ 1 TL Pic = H (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) 1 is a cocycle in H (C, OC). That is, d ∼ 1 TL Pic = H (C, OC) You have to check that if you choose an equivalent cocycle, you would get the same line bundle.  0 ∨ ∼ 1 ∼ d Remark 28.10. We also have H (K) = H (O) = TL Pic r r+1 0 Now, suppose we are at a point L ∈ Wd \ Wd so that h (C, L) = r + 1. Our r goal is to describe the tangent space to Wd at L. Proposition 28.11. We have

r 0 TLWd = first order deformations L ofL/D × C : every section σ ∈ H (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 ⇐⇒ 1 0 1 H (C, OC) ⊗ H (C, L) H (C, L) Proof. We want to describe, inside all tangent vectors to L, which is the space of first order deformations, this is precisely the space→ of line bundles so that every section extends. We now want to describe when a particular section σ ∈ H0(C, L) extends to a 1 section of L. Let v ∈ H (C, OC). We want to answer Question 28.12. Does σ extend to a section of the line bundle L corresponding to v. 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 0 sα + εsα ∈ OD×C(D × Uα) which satisfy the cocycle condition. By satisfying the cocycle condition, we mean that we require 0 0 sα + εsα = gαβ(‘ + εφαβ)(sβ + εsβ) The key is to look at the term involving ε. That is, we obtain 0 0 sα = gαβ(φαβsβ + sβ)

Multiplying this by gβα we get 0 0 gβαsα = φαβsβsβ or, rearranging, we have 0 0 φαβsβ = −sβ + gβαsα Now, we can realize this as a cup product of two cocycles. That is, 1 v = φαβ ∈ H (C, OC) Further,  0 σ = sβ ∈ H (C, L) and the product of these two cycles φαβsβ is the cup product of these two cocycles vσ ∈ H1(C, L). So, writing 0 0 φαβsβ = −sβ + gβαsα 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 ⇐⇒ 1 0 1 H (C, OC) ⊗ H (C, L) H (C, L)  → r Finally, we wish to understand the tangent space to Wd. r Theorem 28.13. v ∈ TLWd if and only if v ∈ Ann(im µ0) where µ H0(C, L) ⊗ H0(C, K ⊗ L∨) −−0 H0(C, K)

→ NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 71

r Proof. By the preceding proposition, we have v ∈ TLWd if and only if every σ ∈ H0(L) extends. This is equivalent to

hvi ⊗ H0(C, L) 0 ∈ H1(C, L) So, the relevant cup product map → 1 0 1 H (C, OC) ⊗ H (C, K) H (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 → µ H0(C, L) ⊗ H0(C, K ⊗ L∨) −−0 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 genus 1,2,3,4,5. For example, Brill Noether theory tells us a general curve is a three sheeted cover of P1. We knew that because the rulings of the quadric cut out pencils of degree 3. The first case we don’t already know tells us that a general curve of genus 6 is expressible 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: r r (1) More about the schemes Cd, Wd 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, r every general curve has a gd and when ρ < 0, the general curve does not r have a gd. Recall from last time, let C be a smooth curve of genus g. r r+1 Let L ∈ Wd \ Wd . We would like to describe r TLWd = { 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 a ∼ line bundles L ∈ Pic(D × C) with L|Spec C×C = L. Saying that sections extend is equivalent to π∗L is locally free, where π : D × C C is the projection map. Also, 0 0 ∨ → 0 µ0 : H (C, L) ⊗ H (K ⊗ L ) H (K)

→ 72 AARON LANDESMAN

r r+1 r Proposition 29.2. Let D ∈ Cd \ Cd . The tangent space to Cd at a divisor D is r TDCd = Ann im µ˜ 0 where

µ˜ 0 = ev ◦ µ0

ev H0(C, K) − H0(C, K/K(−D))

Proof. We have → r Cd Cd

(29.1) u

d r Pic Wd

Then, taking tangent spaces of this diagram, we have

r ∼ 0 ∨ TDCd = H (K/K(−D))

(29.2) du d ∼ 0 ∨ TL Pic = H (K)

Here the map is simply the dual of the evaluation map

H0(C, K) H0(C, K/K(−D))

 → r Corollary 29.3. We have dim Wd ≥ φ = g − (r + 1)(g − d + r) everywhere and µ0 is r injective if and only if Wd 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 this bound. 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 the dimension of that tangent space by computing the cohomology group of some sheaf. 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 then check 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 you get involved in higher order deformations. r We’ll see this later today, when we examine the multiplicity of points in Wd. We’ll see this for genus 4 in class and in genus 6 on the homework. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 73

r g−d+r−1 Example 29.5. (1) Genus 4: We have Wd = K − W2g−2−d . The first inter- 1 esting case this comes up in is W3 , not counting hyperelliptic curves. We have that C is not hyperelliptic, so C P3 canonically embedded lies in a quadric Q ⊂ P3. Then, either the quadric is → 1 smooth, in which case there exist two g3’s on C. Call them L, M. These are 1 distinct. The sum of these g3’s is a hyperplane section of the curve. That is, L ⊗ M =∼ K and L 6=∼ M. These are not isomorphic linear series because if a line of the opposite ruling cut a line of the same ruling we would have 2 to have a g3, 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) and V(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 is given by multiplication by σ, τ. Proof. Not too difficult, see, for instance, Saint-Donat’s article on Petri’s theorem.  Lemma 29.7. 0 0 µ 0 H (C, L) ⊗ H (C, M) − 0 H (C, K) is an isomorphism. → Proof. This is a map from the tensor product of two 2-dimensional spaces to 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. 1 Then, there is a unique g3 L so that L = K. Then, the image of the multipli- cation map H0(C, M ⊕ M) H0(C, K) is a hyperplane inside H0(C, K), which is precisely H0(C, K − p), where → 1 p is the singular cone point of the quadric. Then, W3 has 1 dimensional 1 tangent space at L. This tells us W3 consists of 1 nonreduced point. 74 AARON LANDESMAN

In this situation, there does exist a first order deformation of the line bundle, at which both sections extend. The next question is: 1 Question 29.8. What is the multiplicity of W3 at curves on singular quadrics? We might guess that this is multiplicity 2, because for curve on a smooth 1 quadric, it has degree 2. The challenge is to show that W3 has multiplicity precisely 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 aspects of the geometry. (2) Now, let’s consider genus 5. Suppose C is non hyperelliptic and non trig- 4 onal. Then, C P where C = Q1 ∩ Q2 ∩ Q3 = ∩λ∈P2 Qλ. So, we have a 1 g4 on C. We obtain a map 1→ W4 (C) { singular quadrics containing C} L 7 QL and when these quadrics→ have rank 4, the quadric is a cone over a smooth 3 → 1 quadric in P , and so we get 2 g4’s, one for each of the rulings of the quadric in P3. So, if the quadric Q has rank 4, then L 6=∼ KL−1 L2 6=∼ K 1 Then, µ0 is injective and W4 is smooth of dimension 1 at L. 1 But, if QL is rank 3, then W4 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, 1 1 the general member accounts for a g4, that that is all g4’s. We can also see this from a more general proposition (though it wasn’t explained much in class how to apply the following proposition). n Proposition 29.10. Say that C ⊂ P is any linearly normal curve. Say {Dλ} is a pencil of linearly equivalent divisors of degree d on C. 0 Then, dim Dλ = n − h (C, OC(1)(−D)) is the same for all λ. In other 0 words, the hyperplanes containing the divisor is H (C, OC(1)(−D)). Further- more, ∪λDλ is a rational normal scroll. Proof. Not given.  1 2 (3) Let’s now look at genus 6. Take W4 ∼ K \ W6 . From Brill Noether, ρ = 0 so 1 a general curve C of genus 6 will have only a finite number of g4s. 1 We’ll answer how many g4’s there are (5) in the next class.

Remark 29.11. The Hurwitz space H4,6 has dimension 18 and so H4,6 modulo PGL2 has dimension 15 which has dimension equal to M6. So, we obtain a map

H4,6 M6

→ NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 75

which we expect to be dominant. We can prove this by exhibiting a point in M6 which a finite nonempty fiber. So, if we have an isolated fiber of this map, it follows that the map is dominant. In other words, it suffices to 1 exhibit a curve with a finite number of g4’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. 1 Recall if C were a non hyperelliptic genus 4 curve then we looked at W3 ’s. We had 2 cases. 1 ∼ (1) If Q is smooth then there are 2 reduced g3’s L, M with L ⊗ M = K and 1 L 6= M. So, W3 consists of two reduced points. 1 (2) If Q is singular, there exists a unique g3 corresponding to a semicanonical L, i.e., L⊗2 =∼ K. In this case 0 0 −1 0 µ0 : H (C, L) ⊗ H (C, K ⊗ L ) H (C, K) we obtain that this map factors through the 3 dimensional → H0(C, L) ⊗ H0(C, L) Sym2 H0(C, L) (30.1)

H0(C, K)

1 which is only 3 dimensional. Therefore, µ)9 has some kernel. So, W3 is a nonreduced point of multiplicity at least 2. We claimed, last time that it has 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 a deformed . We can go back to that here, and ask whether it extends 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 a graduate student. Joe’s embarrassing confession is that he wasn’t able to reproduce that result after graduate school.

1 Exercise 30.1. Try seeing why this this W3 , using the method described above, 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.

1 Remark 30.2. Since W3 is 0 dimensional but has 1 dimensional tangent space, it must be Spec C[ε]/εk for some k. Therefore, if we could extend it to k but not k + 1, it would have multiplicity precisely k. Remark 30.3. This is a general part of the philosophy of deformation theory: If you only need second order deformation theory, you’re golden. If you have to go further, you’re screwed. 76 AARON LANDESMAN

30.2. The Genus 6 Canonical Model. Now, let’s move on to studying curves of genus 6. So, let C be non hyperelliptic of genus 6. We have a canonical embedding φ 5 C − K P . 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. 1 1 2 Then,→ if C is general we have, by Brill Noether theorem W2 = W3 = ∅, W5 = ∅. We can also show the above directly because these are only maps to P1, P2 via the Severi variety and Hurwitz variety.

Example 30.4. If C ⊂ P2 is a plane quintic, we know that the Hilbert scheme of 20 ∼ plane quintics is P = PV. Then, this space PV/PGL3 has dimension 20 − 8 = 12. 2 But, dim M6 = 15, so the general curve will have empty W5 . 1 2 Next, we note that W4 = K − W6 , which is 0 dimensional by the Brill Noether theorem. 1 We will aim to prove this statement and also explain how to find all the g4’s on a general curve of genus 6. 1 Proposition 30.5. The space of g4’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 only 1 need exhibit a single curve C ∈ M6 so that the fiber W4 (C) is finite. 2 So, let C0 ⊂ P 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 of W1(C)? → 4 1 By geometric Riemann Roch, take D = i qi We have D ∈ C4 if and only if the q fail to impose independent conditions on H0(C, K). We know i P g(x, y) dx H0(C, K) = : deg g = 3, g(p ) = 0 f i y That is, |K| is cut on C by cubics on P2 through p1 ... , p4 Now, equivalently, con- sider

2 π 2 S = Blp1,...,p4 P − P Let → ` = π∗ of the class of a line in P2

ei = class of the exceptional divisor over pi 4 e = ei i=1 X NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 77

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

1 By Serre duality, H (C, OS(−3` + e)) = 0, so the second map is surjective. We can also see this directly from our explicit description of H0(K) as cubics. Now, note that the following are equivalent: 1 (1) D ∈ C4 (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. 1 1 There are 2 ways to deal with this. Since C has no g3, this g4 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 same answer. 2 Lemma 30.8. Let Γ = {p1, ... , p8} ⊂ P . Then, Γ fails to impose independent conditions on 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 point doesn’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 impose any 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 are collinear.

Proof. If 7 points p1, ... , p7 fail to impose independent conditions on cubics, then for some point pi, any cubic containing pj for j 6= i must contain pi. Relabel the points so that i = 1. Let’s start by pairing up p2, ... , p7 into three pairs. So, we have cubic which is a union of three lines, containing p1. So p1 is collinear with at least 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 with p5, p6, p7 off the line. So, this is 3 lines, which is an reducible cubic, and so one of these three lines must equal L. So, L contains a 5th point and we’re done. (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 geometric genus 6, containing the subset U6,6 of sextics with four nodes. Inside U6,6 we have a 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 2 4 U6,6 (p1, p2, p3, p4) ⊂ (P ) : not all 4 points are collinear Note not all 4 points can be collinear because then the curve would have degree → at least 2 · 4 = 8. Now, note that W is 23 dimensional. Then, W/PGL3 M6 are both 15 dimensional. Again, it suffices to exhibit a curve of genus 6 with a finite number of sextics with 4 nodes, in its preimage under the above map.→ Since those points not in general linear position form a closed subscheme, the map from W M6 cannot be dominant, so we only we will obtain that the general curve will have non collinear nodes. Remark→ 30.10. This will also imply that a general genus 6 curve can be realized as a plane sextic with 4 nodes. 1 We saw the existence of this curve in the above proof because having a W4 is 2 the same as having a W6 by Serre duality.  31. 11/13/15 There are only 7 classes left. There is no homework this week. There will be two more homework assignments, one due before thanksgiving and one due the last 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 independent conditions 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 on cubics then Γ contains 5 collinear points.

Proof. The hypothesis says that for any one point, call it p1, we have that every cubic 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 cubic and so p1 is collinear with two other points, say it is collinear with p2, p3. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 79

Now, we have three collinear points p1, p2, p3. The remaining 4 points can be anywhere. 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 the three lines respectively joining p2, p5 and p3, p6 and p4, p7. So, we must have p1 on one of these three points, meaning one of the remaining three points lies on L. So, we have 5 points on L. (2) Suppose there are precisely 3 points on L. Now, take the lines joining p2, p4, call it L24 and L35 and L67. Therefore, p1 must lie on one of these three 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, p7 are 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 on Q. 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 algebraic geometry called interpolation. In this context we’re asking about the map

0 2 0 0 2 (31.1) H (P , IΓ ) H (P2, OP2 (3)) H (P , OΓ (3)) and the above lemma is saying this map is surjective unless 5 points lie on a line or 8 lie on a cubic. That is, the hypothesis of the lemma says 0 2 h (P , IΓ (3)) = 3 The sublemma says for any Γ 0 ⊂ Γ do impose 7 independent conditions, then 0 2 h (P , 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 curves 1 2 of genus 6. Now, we’ll exhibit a curve of genus 6 and analyze its W4 and W6 . 1 We’ll show the curve in question has only finitely many g4’s and finitely many 2 g6’s. Therefore, we’ll obtain that the curve we’ll see is “a general curve of genus 6.” 1 1 Proposition 31.4. There exists a curve C so that C has precisely 5 g4’s and these g4’s are 1 reduced points of W4 (C). 2 Proof. Now, let’s return to genus 6. So, choose 4 points, p1, ... , p4 ∈ P , so that no three are collinear. In fact, any 4 points in linear general position can be carried 80 AARON LANDESMAN into any other 4 points by PGL3. So, we may as well choose the points to be the coordinate points and p4 = [1, 1, 1], if we like. 2 Now, let C0 ⊂ P be a general plane sextic which has some singularity at all pi. 2 That is, C0 ∈ |IΓ (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 is smooth 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 a 5th point, we want to show we can find a sextic double at these 4 points and not passing through a 5th point. We can just take some double lines and a conic to do this. 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

C S = Bl 2 {p1,...,p4}P (31.2)

2 C0 P in Pic S let ` be the pullback of a line, ei be the class of the exceptional divisor over pi 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 distinct points, henceP it will be a node. So, it only remains to show there are no basepoints on the exceptional divisors. To do this, we only need to exhibit some lines which aren’t tangent to certain points. We can do this by just taking lines joining various points.  Now, we have C S = Bl 2 {p1,...,p4}P (31.3)

2 C0 P

In this situation, we know |KC| is cut by cubic through the four points p1, ... , p4. We saw this explicitly, that 0 ω ∈ H (KC) g(x, y) dx ω = : g(p ) = 0, deg g = 3 f i y 1 Say |D| is a g4 on C. Then, D = i qi. A general divisor will have distinct points. 1 Question 31.6. What makes a divisorP a g4? This moves in a pencil if and only if it fails to impose independent conditions on KC. 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 impose independent conditions on cubics. We can now use the above Lemma 31.1, either all 8 of these points lie on a conic or 5 of them are collinear. In the latter case, We can’t have 3 points qi collinear NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 81 with 2 points pj because then the line would have intersection number 7 with the curve, 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. 1 The Lemma tells us these are the only g4’s. Hence, the fiber dimension over the 1 Hurwitz moduli space M6 is finite at this curve, and so, there exist exactly 5 g4’s on C. 1 Question 31.7. What about the tangent space Tu(D)W4 ? Let’s say |D| is the linear series cut by conics. The other cases will be similar. The 2 2 series residual |K − D| is cut by lines. That is, it is the g6 giving the map C P . 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 to show its either injective or surjective. Now, this is precisely 0 2 0 2 0 2 H (P , Ip1,...,p4 (2)) ⊗ H (P , OC(1)) H (P , Ip1,...,p4 (3)) We can now forget the curve, and just look at things in the plane. The four points → are 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. 1 Therefore, W4 (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 these points.

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. (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. 82 AARON LANDESMAN

2 We choose points p1, ... , p4 ∈ P , so that no three are collinear. Then, we choose a general member of the linear series

C ∈ H0(P2 I2 (6)) 0 , p1,...,p4 which is a 16 = 28 − 3 · 4 dimensional vector space. We will have C0 is nodal at p1, ... , p4 and otherwise smooth. Let ν C − C0

C 5 g1 be the normalization. We saw has precisely→ 4’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 µ H0(C, D) ⊗ H0(C, K − D) − H0(C, K) is surjective. We saw this because, when we forgot the curve and looked in the → plane, we saw that the ideal of the four points is generated in degree 2. Because the 4 points are a complete intersection, this follows from the Noether af + bg 1 theorem. Hence, the above map, corresponding to a point u(D) ∈ W4 is a reduced point. We can also say this in the following way: Introduce

2 C S := Blp1,...,p4 P (32.1)

2 C0 P .

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 the → surface.

32.2. A continuing study of genus 6 curves. We can similarly compute that the 1 multiplication map is an isomorphism for the other g4’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}

 P2 and check this is an isomorphism on . → NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 83

1 (2) We can also see that there is no distinguished g4 in the following way. Recall

φ|3`−e| C S P5 (32.2)

2 C0 P

This realizes S as a in P5. The curve maps to its canon- ical embedding in P5. That is, the canonical model C ⊂ P5 lies on a del Pezzo 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 quintic surface and a quadric hypersurface, once we check that the del Pezzo is projectively normal. Remark 32.1. We see genus 5 was the last time that a general curve is a complete intersection. The best description of genus 6 curves is as an intersection of a del Pezzo surface and a quadric. You can also describe canonical models of curves up through genus 10. Beyond that point, our knowledge of canonical curves drops off. We don’t know much at all about the 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 10 lines. Proof. The images of the 4 exceptional divisors determine 4 lines. We can also consider a line in P2 passing through 2 of these points. This is also 5 mapped to a line in P . The class of the line is 2` − ei − ej which has intersection number 1 with 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.  2 1 Proposition 32.3. The curve C has 5 g6’s and 5 g4’s which are residual to each other. 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 different representations of S as a 4 fold blow up of P2. So, there exist five representations of C as a plane sextic with 4 nodes.  1 Remark 32.4. There appeared to be four g4’s which were different in that they 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 4 points. 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 cut by quadrics in one of these configurations. Because of this, we don’t have 1 the check the separate case in which the g4 is lines through a pi, to see it is 1 a reduced point of W4 . 84 AARON LANDESMAN

1 Remark 32.5. There are several different possible descriptions of the W4 depending on the canonical model of a curve. Let’s assume the curve is not hyperelliptic or trigonal and is not bi-elliptic (mapping 2:1 onto an elliptic 1 curve or plane cubic). Generically, a curve will have 5 g4’s. However, there exist curves with any number between 1 and 4 curves. We don’t want to look for these curves on a del Pezzo surface, but rather, we’ll have to look for them on some surface in the same irreducible component of the Hilbert scheme of a del Pezzo surface. Example 32.6. Here’s an example of a curve of genus 6. Let’s look at the 1 case when three points are collinear. We’ll see there are only 4 g4’s. This 1 happens because, in effect, the g4’s cut by conics containing all 4 points 1 also contains the line L. This is precisely the same as the g4 cut by lines 1 containing the fourth point. We would expect in this case that W4 is nonre- duced. We then get a map H0(C, D) ⊗ H0(C, K − D) H0(C, K) This is no longer an because all conics containing the first three points contain the line through the→ first three points. The image will then be 5 dimensional. A conic through the 4 points will contain the line L 1 but a cubic through them does not have to contain L. So, we see Tu(D)W4 is 1 dimensional here. Then, we see the space consists of 4 points which are reduced and 1 which has a 1 dimensional tangent space and this point turns out to be a double point, although it requires some work to show it has multiplicity exactly two. When we take the blow up at these 4 points, we see the line through the three points L has class ` − e1 − e2 − e3 which has intersection number 0 with the embedding class 3` − e, and so this line gets collapsed, and we obtain 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? When the 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 del Pezzo 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 1 (a) A curve of genus 0 has a unique g1. 1 (b) A curve of genus 2 has a unique g2. 1 (c) A general curve of genus 4 has two g3’s. 1 (d) A general curve of genus 6 has five g4’s. 1 A general curve of genus 8 has a finite number of g5’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! NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 85

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 two points, there exists a chain of components so that each pair of adjacent components meet 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) r r dim Wd = dim Gd = ρ where ρ = g − (r + 1)(g − d + r) and r dim Cd = ρ + r. In particular, r Wd 6= ∅ ρ ≥ 0. (2) (Gieseker-Petri Theorem) For any invertible sheaf L on C, 0 0 ⇐⇒ ∨ 0 µ0 : H (C, L) ⊗ H (C, K ⊗ L ) H (C, K) is injective. (3) → r r+1 (Wd)sing = Wd r r+1 (Cd)sing = Cd r (Gd)sing = ∅ (4) Set g−1 0 Wg−1 = L ∈ Pic : h (C, L) > 0 g−1 = im u(Cg−1) ⊂ Pic where u : C Pic1 is the Abel-Jacobi map. Define   2 θ := Wg−1 ∈ H (J, Z) → the theta divisor in Picg−1. Then, ! r i! [Wr ] = θg−ρ. d (g − d + r + i)! i=1 Y Here are some special cases: (a) In particular, if ρ = 0, since θg = g! ∈ Z =∼ H2g(J, Z) we have r i! #Wr = g! . d (g − d + r + i)! i=1 Y r (b) In particular, by the previous part, Wd is smooth for a general curve, and as long 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 (2k)! #W1 = d k!(k + 1)! r (5) If ρ > 0 then Wd is connected in codimension 1. By the first and third parts, r r Wd is irreducible. If ρ = 0, the monodromy of Wd over Mg is transitive. It is the symmetric group unless d = g − 1 in which case it is a wreath product. In r particular, Wd is irreducible. r (6) Let L ∈ Wd be general. Then, L is very ample if r ≥ 3 2 φL : C P is birational and the image is nodal if r = 2 1 φL : C P 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 unique component 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 from a Porteous’ formula calculation. There are in fact two ways to describe this as a Porteous’ formula calculation. (1) We can say

r  0 0  D ∈ CD rk H (C, K) H (C, K|D) ≤ d − r

Then, we can think of this map as given by a d × g matrix, and those vector ⇐⇒ → spaces 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) 0 is the trivial bundle of rank g and FD = H (C, K|D). Then, define φ : E F to be the evaluation map. By Porteous’ formula, when the locus at which the matrix drops rank, the class of this locus is determined by the Chern→ classes of the vector bundle. This cohomology is calculated in the coho- r mology ring of Cd. You can also calculate the Chern classes of E because it is the trivial bundle, and you can calculate F by Grothendieck Riemann Roch. See Chapter 8 of ACGH for a proof. This allows us to determine the class in part 4. Similarly, it lets us find the class as in part 1. This also tells r us the dimension of Cd is at least ρ + r everywhere. That is, it determines the existence part of part 1. We can then set up an analogous calculation on Pic. r One can then do a setup to find the class of Wd. (2) Here is another way to carry out Porteous’ formula. On Picd, fix D = d p1 + ··· + pm for m  0. We can form the vector bundle E on Pic . We have fibers 0 EL = H (C, L(D)) and, we can “fit these together to form” a vector bundle E of rank m + d − g + 1. We can also form F by piecing together the fibers 0 FL = H (C, L(D)|D). NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 87

Then, the locus

r d d Wd = L ∈ Pic : rk φ|D ≤ m + d − g − r ⊂ Pic

Remark 33.3. However, it is slightly more tricky to construct these vector bundles than those in the previous part. And, in this case, it’s not hard to see

E∨ ⊗ F =∼ Hom(E, F)

is an ample vector bundle. This means that we can deduce the nonemp- tyness of the locus: According to Fulton and Lazarsfeld, the degeneracy locus is nonempty when E∨ ⊗ F is ample. This nonemptyness can 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 the expected dimension of the degeneracy locus is positive.



1 Exercise 33.4. In the case of genus 2, 4, 6 we can not only count the number of gr ’s, 1 but in fact identify them. For example, in genus 6, we can identify the g4’s as the 4 pencils through a point and the pencil of conics through 5 points. In genus 8, can 1 you find the 14 g5’s? As my own guess, perhaps we can use that it has degree 14 and lies on a the grassmannian 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 is positive then the scheme is nonempty.

34.1.3. Review. We talked last time about the existence part of Brill Noether. More r generally, we discussed the case dim Wd ≥ ρ. This was proved by a Porteous calculating in the late 1960s by Kleiman and Laksov, and, simultaneously, Kempf. Mumford had the idea and told the two parties above, and these two parties got the result at the same time.

r Remark 34.1. This lower bound dim Wd ≥ 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 a variational element, in the sense that we’ll have to choose general curves. r We want to show that dim Wd ≤ ρ and is empty if ρ < 0. To prove this upper bound, it is sufficient to exhibit a single smooth curve C of r genus g with Wd(C) = ρ. Remark 34.2. However, no one has ever written down a Brill-Noether general curve in arbitrary genus. It’s not hard to write down special curves like plane curves of degree 9 with 5 2 double points. However, a general curve of degree 23 shouldn’t have a g9, so this won’t be such a curve. 34.3. Proof of Existence for Brill Noether. Every proof of Brill Noether which has ever been proposed involves a specialization argument. We’ll look at the proof originally suggested by Castelnuovo. Castelnuovo said we can’t write down a smooth curve, but we can look for singular curves, with only relatively simple singularities. In this case, the geometric genus will be lower, hence simpler. Castelnuovo suggested a g-nodal curve of g. That is, we take P1 with g pairs of points identified. So, a linear series on this curve can be pulled back to a linear series on P1, which are well understood. So, here is the idea: r r 1 Suppose we have a gd on C0. Then, we can pull it back to a gd on P . On 1 P , there’s a unique invertible sheaf of degree d, namely OP1 (d) with an r + 1 0 dimensional subspace of H (OP1 (d)). r 1 To describe all gd’s on P , we take φ |O 1 (d)| P1 −−−−−−P Pd

r d Then, a gd on C is simply a sub-linear series. That is, it is a linear space Λ ⊂ P of r → dimension d − r − 1. So, the gd is H ∩ P1 : Λ ⊂ H

r So, the space of gd’s is simply the grassmannian G(d − r, d + 1). r r 1 Therefore the gd’s on C0 correspond to a gd, D on P 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, b means the linear space spanned by a and b. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 89

r Therefore, gd’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 a problem in . 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 such configurations 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

Φ (34.1) L Σ(L) ⊂ G

Then, the left map has fibers isomorphic to G(d − r − 1, d) and so the dimension is dim Φ = (r + 1)(d − r − 1) + 1 = (r + 1)(d − r) − r Hence, since dim G(d − r + 1, d + 1) = (r + 1)(d − r), we have that Σ(L) ⊂ G has codimension 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 are chosen specially to intersect two points on the rational normal curve. r (2) The dimension of Wd is upper semicontinuous in families. We’d like to say r that dim Wd, is upper semicontinuous. So, if the dimension on the special fiber is equal to the maximum ρ, then it must be equal to ρ everywhere. d However, this is sitting inside Pic (C0), which is not a compact variety. The upper semicontinuity of dimension only applies to proper varieties. d ∼ × g d We have Pic (C0) = (C ) . If we have a family Pic (C/∆) it might be r r that the limit of gd’s on Ct may not be a gd 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 1 which may not be locally free. Kleiman and Altman dealt with the second problem They showed the only way a r r gd can fail to be a gd on C0 is if it acquires basepoints at the nodes. r We can then handle this, because if a limit of gd has basepoints at δ of the nodes r1, ... , rδ, We then have a linear series of degree δ lower, but of dimension r on the partial 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. r r So, the locus of gd’s whose limits are not gds has strictly smaller dimension. r So, it is sufficient to show dim Wd(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 (1) To show there does not exist a g2 on a general curve C of genus 3, we’re 1 1 saying that we can embed P by the complete linear series. Then, g2’s on that curve correspond to points, and the condition is that the point lie on three chords. But, three general chords to a conic are not concurrent. So, we’re done. 1 (2) To show there does not exist a g3 on a general curve C of genus 5, we can specialize C to C0, a curve with 5 nodes. we embed the normalization of C1 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. ∼ 1 g → That is, C0 = P / {pi ∼ qi}i=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 that we 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 upper r semicontinuity of dimension of Wd in families. That is, r r dim (Wd(C0)) ≤ ρ = dim (Wd(Ct)) ≤ ρ for general t. However, there is an issue that this is⇒ not proper over the disk. So, it’s possible r we could have a component of Wd(Ct) that appears over the special fiber that disappears on the general member. However, as we saw last time, Kleiman and Altman bounded the dimension of such components, and so we can still make the above hold true for a general such family. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 91

r 1 Now, given Wd(C0), we can pull a linear series back to the normalization P . 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 from ∼ d−r−1 a plane Λ = P ⇐⇒ , so that there is a line Li joining pi and qi on the curve and meeting 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 turns out 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. r Remark 35.3. We’ll focus on the case when ρ is negative and show that Wd of a general curve is empty. Example 35.4. Recall the examples from last time: (1) “A general curve C 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 in P2 are no concurrent. (2) “A general curve C of genus 5 is not trigonal“ amounts to the assertion 3 that if L1, ... , L5 ⊂ P are general chords to a twisted cubic, then no line L ⊂ P3 meets all 5 such lines. Here, we see the problem arising for the first time. 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 it was the second proof. Remark 35.5. Note, g nodal curves only have dimension 2g − 3, inside the moduli space of dimension 3g − 3. However, their behavior in general mirrors 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 you can’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 branch points. If there were a line meeting all five tangents to a twisted cubic would express the twisted cubic as a three sheeted cover of P1, and projection from 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, is to specialize g general chords to a rational normal curve C ⊂ Pd to g arbitrary tangent lines. 92 AARON LANDESMAN

r Proposition 35.7. For any g tangent lines Li to C ⊂ P , we have ∩iΣ(Li) is proper in G(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 set up, but keep in mind that we’ll only be applying this to P1. r Lemma 35.8. Let C be an arbitrary smooth curve of genus g. Let D = (L, V) be a gd on C. 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 higher order.  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 < ··· ar is the sequence of r + 1 orders of vanishing from Lemma 35.8. We will alternatively notate 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 as α(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 to prove it.

Proof. Saying α0 > 0 is equivalent to p being a base point of D. Define φ := φD : r C P . Then, α1 > 0, given α0 = 0, this is equivalent to dφp = 0. Then, if α1 = α0 = 0, to say α2 > 0 is equivalent to φ(p) being a flex point of φ(C). In other→ words, the order of of C with its tangent line at p is at least 3.  Definition 35.12. Let p ∈ C. Then, p is an inflectionary point or ramification point of D if α(D, p) 6= 0 Example 35.13. Take a smooth curve C ⊂ P3. Start with a line not meeting the curve 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 series embedding the curve in P3. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 93

Theorem 35.14. Set

α(D, p) = αi(D, p) i X Then, α(D, p) < . i r X For D an arbitrary gd on an arbitrary smooth curve∞C, we have α(D, p) = (r + 1)d + r(r + 1)(g − 1) p∈C X Proof. We’ll possibly come back to this next time, but it’s also in 3264, and we did it in 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) = ∅. The general statement is similar. d If Λ ∈ ∩Σ(Li), where Λ is a d − r − 1 plane in P which meets Li for all i, r 1 consider the linear series D, which is a gd on P , cut by hyperplanes containing Λ. For each of the points pi, we have α1(D, pi) > 0 because we’re taking our lines to be the tangent lines at pi. In particular, α(D, pi) ≥ r. So, we must have that the total 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 classes left! 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 (1) Specialize to a g-nodal curve. More accurately, take C0 = P / ∼ 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, the Picard variety on g-nodal curves of degree d is not proper. We can see this because we can pull it back to P1, and then choose an isomorphism of the fibers over the two identified points, which is Gm. The gth power of this is not proper. r 1 d (3) We want to describe the gd’s on C0. Embed P P as a rational normal d r curve of degree d, X ⊂ P . So, gd’s on C0 are in bijection with ∼ d−r−1 d → Λ = P ⊂ P : 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

g Goal 36.1. If L1, ... , Lg are general chords to X ⊂ P , 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 to r prove the Brill-Noether theorem, but instead to count the number of gd’s when r ρ = 0. His aim was to use Schubert calculus to count the number of such gd’s. It was later realized it is always true that g tangent lines will always have the expected dimension. It is easier to deal with tangent lines because the statement is always true for them, because the other statement is only about a general secant line, where as this holds for all tangent lines. So, the idea is to show our goal for arbitrary tangent lines to X. r Recall the set up from last time. Given a gd with D = (L, V), on a smooth curve C of genus g and p ∈ C, let

{ai = ai(D, p)} = {ordp σ : σ ∈ V} where

0 ≤ a0 < a1 < ··· < ar ≤ d

Set αi = ai − i with

0 ≤ α0 ≤ α1 ≤ · · · ≤ αr ≤ d − r We say a point p is an inflectionary point of D if α 6= 0. Set

w = w(D, p) = αi(D, p) i X to be the total weight of the inflectionary point p. The main theorem from last time we want to apply is the Plucker formula: r Proposition 36.3 (Plucker Formula). For any gd D on C we have w(D, p) = (r + 1)d + r(r + 1)(g − 1) p∈C X Proof. Let’s look locally on the curve. Take a point p ∈ C and an open set around it 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. 0 Take a basis σ0, ... , σr of H (C, L) and write σi = fi(z). Consider the Wronskian

 f0 ··· fr  0 0  f0 ··· fr    det  . .. .   . . .  (r) (r) f0 ··· fr Write v = (f0, ... , fr) be a vector valued function. Note that the Wronskian is equal to v ∧ v0 ∧ ··· ∧ v(r) NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 95

Lemma 36.4. We claim

w(D, p) = ordp W and in particular, W 6= 0. Proof. First, let us show W 6= 0. Suppose it were. Then, the r of v and v are everywhere linearly dependent. Let’s start by identifying the last stage in this wedge product which is nonzero. That is, say v ∧ v0 ∧ ··· ∧ v(k−1) 6= 0 but v ∧ v0 ∧ ··· ∧ v(k) = 0. Taking the derivative, the only nonzero term is v ∧ v0 ∧ ··· ∧ v(k−1) ∧ v(k+1)

So, if v ∧ v0 ∧ ··· ∧ v(k) = 0. v ∧ v0 ∧ ··· ∧ v(k−1) ∧ v(k+1) = 0. Hence, v(k+1) ∈ hv, v0, ... , v(k−1)i. Continuing this process, we see that

v(k+2) ∈ hv, v0, ... , v(k−1)i. and in general, by induction

v(k+t) ∈ hv, v0, ... , v(k−1)i. for all t. So, in the analytic world, integrating this, the whole curve must lie in this k plane. I.e., if all the derivatives of a curve lie in a fixed subspace the image of a map must also lie in that subspace. This tells us the map is mapping us into a k r plane. This contradicts nondegeneracy of the gd. 

Remark 36.5. Note, this is false in characteristic p. This fails in characteristic p because the statement that all derivative lie in the span of a k-plane does not mean the curve itself lies in a k-plane.

At a point p ∈ C with ai = ai(D, p), up to a unit we can write our functions as v = (za0 , za1 , ... , zar ) Writing out the Wronskian, we have

 za0 zar ··· zar  za0−1 za1−1 ··· zar−1   det  . . . .   . . .. .  za0−r za1−r ··· zar−r

The first nonzero derivative of W is the i(ai − i)th derivative. We now come to proving the Plucker formula. Observe that W is a function defined locally. This W depends on theP choices of our trivialization for L and the local coordinate we chose. We want to describe how W depends on these choices. Question 36.6. How does W depend on the choice of trivialization and local coor- dinate? That is, if we chose a different trivialization differing by a nonzero function ∂z g = gαβ, and a different coordinate ζ with ∂ζ = j = jαβ. 96 AARON LANDESMAN

The first row just changes by g. The second row changes by two terms. The first g0 times the first row, which is dependent on the first row. The second is g times the second row. Therefore, each row, module the span of the previous row is multiplied by g. Therefore, in this case W is multiplied by gr+1. Now, how does W depend on the choice of local coordinate? In this case, the r+1 total matrix is multiplied by j( 2 ). Therefore, if we change by a function g and a coordinate j, we obtain that W is a well defined global section of

r+1 ⊗r+1 ( 2 ) L ⊗ KC which has degree r + 1 (r + 1)d + (2g − 2) = (r + 1)d + r(r + 1)(g − 1). 2 Therefore, the number of zeros of W is, therefore, equal to both

w(D, p) = (r + 1)d + r(r + 1)(g − 1). p∈C X 

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 actually are 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 is r ∼ 1 the corresponding gd on X = P , 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 cannot exist any such Λ. 

Remark 36.9. To set up for Wednesday, we’d like to see how to go from this crude r statement: When ρ < 0 then Wd is empty, to the stronger statement on the dimen- sion. Next time, we’d also like to talk about the consequences of the Brill-Noether theorem for studying Hilbert schemes. There will be no homework after homework 9. NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 97

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 a smooth 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: 2 Theorem 37.1. Take nine points p1, ... , p9 which are general. Let C ∈ P be a curve with degree 3g that has a singularity of multiplicity (the order of vanishing of the defining equation, which generically looks like smooth branches crossing transversely at that point) g at p1, ... , p8 and a singularity of multiplicity g − 1 at p9. 2 Then, if E ⊂ P 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, then the 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 g g − 1 1 − 8 − = ((3g − 1)(3g − 2) − 8g(g − 1) − (g − 1)(g − 2)) 2 2 2 2 = g We have to make an argument that a general such curve is smooth in the blow up. That is, in the linear system of the blow up divisor, a general curve is smooth of 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 classes d we discussed the following. Let C ⊂ P be a rational normal curve, let L1, ... , Lg are 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) > dim G X 98 AARON LANDESMAN we have

∩iΣ(Li) = ∅ r We then conclude that for ρ < 0 a general curve of genus g has no gd. We want to r conclude that for a general curve C, we have dim Wd = ρ. Remark 37.3. Here is how we strengthen the result we have so far. So far, we have seen that Schubert cycle associated to tangent lines of a rational normal curve always intersect dimensionally properly. We would like to extend this to other Schubert 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 d V0 ⊂ V1 ⊂ · · · ⊂ Vd−1 ⊂ P 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 the osculating 2 plane to C at p. There are likely typos in the remainder of this subsection. d Theorem 37.5. Let C ⊂ P 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) ∼ d−r−1 d r is a Λ = P ⊂ P . Then, D = gd on C cut by hyperplanes H ⊃ Λ. Then, for any point p ∈ C with osculating flag V the total inflection weight

w(D, p) = max (codimΣa ⊂ G(d − r, d + 1) Λ∈Σa(V) That is, if Λ satisfies a certain osculating condition then the corresponding weight has 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 ) > dim G(d − r, d + 1) then X

∩iΣai (Vi) = ∅.  Corollary 37.6. The Brill Noether theorem holds (even for ρ > 0). NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 99

r Proof. Then, the dimension of Wd, we may note that the simple ramification lo- r cus is just a hyperplane section. That is, to argue Wd has dimension ρ and not r larger, then if Wd had larger dimension, we could then impose the condition of r ramification at ρ + 1 hyperplanes. If the dimension of Wd were ρ + 1 or more, that intersection would be nonempty. So, these osculating flags allow us to add extra codimension 1 conditions. This lets us go from the inductive base case we saw last r time to the general case about the dimension of Wd.  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 and genus g in Pr. Officially, it is the open subset of r Hmd−g+1(P ) parameterizing smooth curves. Lemma 37.7. Let r ≥ 3. Whenever ρ ≥ 0, there exists a unique irreducible component of H dominating Mg. Further, this component has the “expected dimension” equal to 3g + 3 + ρ + (r + 1)2 − 1 where the 3g + 3 = dim Mg specified the abstract curve, ρ is the dimension of the choice of 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 of r genus g, there are very ample gd’s. Further, by Fulton and Lazarsfeld’s contribu- tion to Brill-Noether, we have a map

H Mg r+1 and the fibers are PGLr+1 bundles over Wd . Hence, over a general point, the fibers of this map are irreducible. Since→ the map is also flat, the source will be irreducible. When ρ = 0, we need a separate argument, here we’ll have to explain why the monodromy group is transitive. 

Definition 37.8. We call the unique irreducible component of H dominating Mg the principle component, which we notate as H0 which implicitly depends on d, 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, which basically says that the trick we used to describe curves in low degree and genus works in general. That is, the map 0 0 H (OPr (m)) H (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 think about the line of argument we made above to finish the Brill-Noether theorem you will understand the following really good exercise: Exercise 37.11. A general curve on the principle component has only simple ramification. If you look at a curve in P3, we would expect that a curve has finitely many 4 secant lines and no 5 secant lines. Question 37.12. For a general curve in P3, are there always finitely many 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 meet it 5 times. We then get the above “expected dimensions” because dim G(2, 4) = 4 of lines in P3. (c) Flexibility (or interpolation): Question 37.13. Given a general collection of points in Pr is there a curve containing the expected number of points. (d) What is the geometry of the principle component itself? That is, what is

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 do still have an expected dimension of the Hilbert scheme. We can ask Question 37.14. When do components of the Hilbert scheme still have the expected dimension when ρ < 0? We know the “vast majority” of the components of the Hilbert scheme do have the expected dimension. However, there do seem to be a certain range of g, r, d where 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 and the image of H in Mg under the projection map has codimension less than g − 4, then H0 has the expected dimension.