NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES

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1 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES AARON LANDESMAN CONTENTS 1. Introduction /2/ Course Mechanics and Background The Basics of curves, September /4/ Outline of Course /9/ Riemann Hurwitz Equivalent characterizations of genus Consequences of Riemann Roch /11/ /14/ Curves of low genus /16/ Geometric Riemann-Roch Applications of Geometric Riemann-Roch /18/ Introduction to parameter spaces Moduli Spaces /21/ Hyperelliptic curves /23/ Hyperelliptic Curves Gonal Curves /25/ Curves of genus /28/ Canonical curves of genus Adjoint linear series /30/ Program for the remainder of the semester Adjoint linear series /2/ /5/ Castelnuovo s Theorem /7/ /9/ /14/

2 2 AARON LANDESMAN /16/ Review Minimal Varieties /19/ /21/ /23/ Agenda and Review Resolutions of Projective Varieties and Green s conjecture The Maximal Rank Conjecture /26/ /28/ Review /30/ Review Today: Brill Noether Theorem in dimension at least /2/ Review Hilbert Schemes /4/ Logistics Tangent Spaces in Brill Noether Theory Martens Theorem /6/ /9/ Agenda and Review /11/ Review The Genus 6 Canonical Model /13/ /16/ Logistics and Review A continuing study of genus 6 curves /18/ Logistics and Overview /20/ Plan, conventions, and review An Upper bound on the dimension Proof of Existence for Brill Noether /23/ Review Inflectionary points of linear series /30/ Review and overview /2/ Overview Retraction Finishing the proof of Brill-Noether 97

3 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES Further Questions 99

4 4 AARON LANDESMAN 1. INTRODUCTION Joe Harris taught a course (Math 282) on algebraic curves at Harvard in Fall 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 mathematics 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 corrections to aaronlandesman@gmail.com. 1 This introduction has been adapted from Akhil Matthew s introduction to his notes, with his permission.

5 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES /2/ Course Mechanics and Background. (1) Math 282, Algebraic Curves (2) CA Adrian (3) Text: ACGH, Volume 1 (4) Four years ago, a similar course was taught, following ACGH. The idea was: given a curve, what can we say about it. This is only half the story. Curves can appear in the abstract and in projective space. An important part of understanding curves is how they vary in flat families. The difference 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 cohomology, and second we will use schemes. (6) Algebraic curves were first studied over the complex numbers. Some people studied complex analysis of Riemann Surfaces, and others studied polynomials 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 P n for some n. In the 20th century, a curve became an abstract curve, which exists independently of any particular embedding in projective space. A similar perspective was adapted in group theory. Originally, people viewed groups as subsets of GL n. Now, this is called representation theory. Remark 2.3. In his textbook, Hartshorne says the goal of algebraic geometry 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 The Basics of curves, September 2. Definition 2.4. Define the genus by g = 1 2 (1 χ topx). Definition 2.5. An ordinary singular point 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]/(f 1 f m ) where f i 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 χo X (3) 1 2 (deg K X + 2)

6 6 AARON LANDESMAN (4) 1 c, where c is the constant term of the Hilbert polynomial of C P r. (5) If C = C 0 P 2 of degree d with ordinary singular points of multiplicity m i, then g(c) = ( d 1) ) 2 i ( mi 2 Definition 2.7. A divisor is an formal sum of the form D = i n ip i for n i Z, p i X. We say D is effective if D 0. We define the degree by deg D = i n i. For f a rational function on X, we define div f = (f) = p ord p (f) p = (f) 0 (f). Remark 2.8. By the residue formula applied to the logarithmic derivative of a rational function 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 notated as C d, which is by definition C d /S d. Definition Given D div X, we look at L(D) := {f K(X) : (f) + D 0}. An alternative notation for L is H 0. Since we can specify the polar part of such a function, this is a finite dimensional vector space. This uses the fact that there are no nonconstant meromorphic function. We define l(d) = dim L(D) and define r(d) = l(d) 1. Remark If D E then L(D) = L(E), as given by multiplication by f. Definition We define Pic d (X) := Divisors of degree d/. Remark It turns out this set corresponds to the points of a variety. Definition Suppose ω is a rational 1-form, which looks locally like f(z) dz. We let (ω) = ord p (ω)p. We define K X := (w) Pic 2g 2 (X). 3. 9/4/15 Note: there will be no class Monday or Friday 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 theorem (gives an upper bound on the genus of a curve of degree d in projective 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 scheme 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 n ip i and say D E if D is linearly equivalent to E. We let K X be the divisor class of the canonical sheaf ω, of degree 2g 2. Use L(D) for H 0 (C, D), let l(d) = h 0 (C, D) and r(d) = l(d) 1.

7 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 something quite manageable. We next explain what this has to do with maps to projective space. Remark 3.2. Given a map φ : X P r and let H = V(x 0 ) be a hyperplane cut out by the first coordinate of P r. Then, the map can be given by [1, f 1,..., f r ], and D = φ 1 H and f 1,..., f r 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 l(d). Example 3.4. We know l(k X ) = g. Theorem 3.5. We have l(d) = d g l(k X D) Proof. Caution: there is a problem with the following proof. We have managed to sweet Serre duality under the rug. The problem is that we assumed D and K D were effective. Say D = p p d, where p i are distinct points of X. Choose local coordinates z i around each p i. Any f L(D) can be written locally as a i z i + h where h is holomorphic. So, f is determined up to the addition of a constant function by a 1,..., a d. So, the question is: which d tuples arise as global principal parts of functions. First, note that l(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(K X ), and f L(D), then i Res p i (f ω) = 0. Say ω(p i ) = b i dz i. So, plugging this in the above formula, we have i a ib i = 0. So, we get g relations, but they might not be independent. The actual number of relations we are getting is g l(k X D). So, l(d) d + 1 g + l(k X 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 l(k D). We have. Adding the two inequalities we have l(k D) 2g 2 d + 1 g + l(d) l(d) + l(k D) l(d) + l(k D). So, we added two inequalities and got an equality, which means they were equalities to begin with. Suppose you are give a compact complex Riemann surface, how do you know there are any nonconstant meromorphic functions. This issue underlies the difficulty we are encountering in the proof above. In fact, in dimension at least 2, there are compact complex manifolds with no meromorphic functions whatsoever.

8 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 hyperplane, we look at locally. Definition 3.7. Given a divisor D on X we define a sheaf O X (D) defined by O X (D)(U)L := {f K(U) : (f) + D 0 in U.} When X is smooth, this corresponds to a line bundle. Observe that if D E, O X (D) = O X (E), as given by multiplication by X. Additionally, define Pic d to be line bundles of degree d, or equivalently, linear equivalence classes of degree d. Finally, define L(D) := H 0 (X, O X (D)) Remark 3.8. This construction is a special case of the fact that for locally factorial schemes which are regular in codimension 1, there is an isomorphism between the vector spaces of Weil and Cartier divisors. Lemma 3.9. The collection of nondegenerate maps (maps whose image is not contained in a hyperplane are in bijection with pairs (V, L) where L Pic d (X), V H 0 (L) with dim V = r + 1, so that V is basepoint free. Proof. See Vakil, chapter Riemann Hurwitz. 4. 9/9/ Equivalent characterizations of genus. (1) The dimension of the vector space of holomorphic 1 forms. (2) χ(o C ) = 1 g (3) p C (m) = md + 1 g (4) l(d) = d g + 1 for D >> 0. (5) Number of handles (6) b 1 (C) = 2g (7) deg K C = 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 z m, and when m > 1 p is a ramified point. Definition 4.2. Given a map π : C B as above, set v p (π) = m 1 and define the ramification divisor (4.1) R = p C v p (π) p

9 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 9 and the branch divisor by (4.2) B = q B p π 1 (q) 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 ω C and π ω B. At a point of B, the pullback of a from on B will pick up zeros equal to the ramification index. So, (4.3) K C = π K B + R Then, taking degrees, since deg K C = 2g 2, we obtain Riemann-Hurwitz. Remark 4.4. It s also fun to prove this via topological Euler characteristic. You do this by removing the ramification and branch locus. We then obtain an unramified covering space, and use that on this locus, the Euler characteristic is multiplicative. Then, you add the points back in and deduce the formula Consequences of Riemann Roch. Remark 4.5. Recall, nondegenerate maps C P r of degree d, modulo the action of PGL r+1 are in bijection with pairs (L, V) so that L is a line bundle on C of degree d and V of dimension r + 1 in H 0 (L) with no common zeros, i.e., no basepoints. Given a line bundle and sections, we can choose φ D (p) = [σ 0 (p),..., σ r (p)]. We can also describe this intrinsically by V p = {σ V : σ(p) = 0}. Then the map φ sends φ D : p V p PV. Definition 4.6. A g r d 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. Remark 4.7. If D = (L, V) is a basepoint free g r d, we get a map C Pr = PV. The map being injective is equivalent to V p+q = V p V q has codimension 2. Equivalently, V p+q = {σ V : σ p + q}. The map is an immersion if and only if for all p C, V 2p has codimension 2. Lemma 4.8. A map φ is an embedding if and only if for all effective divisors D C 2 i.e., of degree 2, V D has codimension 2. Proof. This is Remark 4.7 Corollary 4.9. If deg L 2g + 1 then φ L is an embedding, where L = (H 0 (L), H 0 (L)). Proof. By Riemann Roch, deg L 2g + 1 = dim H 0 (L) = d g + 1. But, it s also true that deg L( D) 2g 1 so H 0 (L) = d g 1. Therefore, any line bundle of degree 2g + 1 or more gives rise to an embedding. Remark 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 Suppose D is a divisor of degree d g + 3. Then, if D C d general then φ O(D) = φ D is an embedding.

10 10 AARON LANDESMAN Proof. Difficulty, we won t do it now, but it will be helpful to know about the Jacobian. Definition A curve C is hyperelliptic if there exists a map π : C P 1 of degree 2. That is, C can be written as y 2 = 2g+2 i=1 (x λ i) on an open subset. Exercise Not every curve of genus at least 3 is hyperelliptic. Example In genus 2, the canonical sheaf makes C into a hyperelliptic curve. Definition For any curve C of genus g > 0, we have canonical map, defined to be the map associated to the dualizing sheaf. Theorem 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. Furthermore, use the characterization that being hyperelliptic is equivalent to having an effective divisor of degree /11/15 Today Joe is out, and Adrian is holding a review. Remark 5.1. Deformations of abstract schemes are classified by H 1 (T C ). Embedded deformations are classified by H 0 (N Y/X ). Example 5.2. Let s show there are non hyperelliptic curves. First, T C M g = H 1 (T C ) = H 0 (2K C ). By Riemann-Roch, h 0 (2K C ) = 3g 3. This gives a proof that dim M g = 3g 3. Exercise 5.3. Compute dim M g by first calculating the dimension of the space of covers of P 1 of degree d. Theorem 5.4. Let X be a smooth projective variety and Y X be smooth of codimension 1. Then, K Y = (K X + Y) Y. Exercise 5.5. Let C be a smooth curve on P 1 P 1. Then, C is given by a bihomogeneous polynomial of class (a, b). Solution: We have K C = (K P 1 P 1 + C) C. Then, Pic P 1 P 1 = ZH 1 ZH 2. So, (5.1) (5.2) (5.3) (5.4) (5.5) So, g = (a 1)(b 1). K C = K P 1 P 1 + C C = ( 2H 1 2H 2 + ah 1 + bh 2 ) C = ((a 2)H 1 + (b 2)H 2 ) C = (a 2)b + (b 2)a = 2ab 2a 2b Exercise 5.6. Compute the dimension of the space of twisted cubics in P 3. (I.e., a component of the Hilbert scheme) Solution: We have a map P 1 P 3 of degree 3, which is nondegenerate. We can write any map sending [x, y] [f i (x, y)]. where f i is a homogeneous polynomial

11 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 PGL 2 action. Solution 2: Lemma 5.7. If C is a twisted cubic, then C is contained in a quadric. Proof. Note that O P 3(2) has 10 dimensional cohomology, while O C (2) is 7 dimensional by Riemann-Roch. So, in fact, it lies on 3 quadrics. Now, consider (C, Q) so that C is contained in Q and this gives you what you want /14/15 We will take the approach of interweaving new techniques with applications. Today s new technique is the adjunction formula. We ll give two proofs of the adjunction formula. We ll have the following setting: Let X be smooth varieties of dimension n. Every such scheme has a canonical bundle, which is an invertible sheaf of top dimensional forms K X = n T X. 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 = O X (Y). There is also a section σ H 0 (L) corresponding to the regular function 1, with V(σ) = Y. The key fact is Lemma 6.1. N Y/X = L Y = O X (Y) Y = O Y (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, T q L = T p X L p. Then, the tangent space to σ, with V(σ) = Y, is the graph of a map T p X L p. The kernel of this map is T p Y, and by the conormal exact sequence, we have (N Y/X ) p = L p. Here is an algebraic proof: Proof 2. We have (6.1) And so, N Y/X = O X (Y) Y. N Y/X = I Y /I 2 Y = O Y Y = O X ( Y) Y. We are now in a position to prove the adjunction formula. We have an exact sequence 0 T Y T X Y N Y/X 0 Recall that in general if we have an exact sequence of vector spaces 0 A n 1 B n C 1 0 then there is a natural isomorphism n B = n 1 A C. Applying this to 6.1 we obtain Theorem 6.2. (6.2) K Y = K X Y N Y/X = K X (Y) Y

12 12 AARON LANDESMAN Proof. This is immediate from applying 6.1 to the conormal exact sequence. Example 6.3. K P n = O P n( n 1). To see this, given P n, we have homogeneous coordinates z 0,..., z n and on the affine open, we have inhomogeneous coordinates z i /z 0. Look at dz 1 dz n. This is holomorphic and nonzero in z 0 = 0. This has a pole of order n + 1 along z 0 = 0. Alternatively look at (6.3) dz 1 z 1 dz n z n is a meromorphic differential which has a pole on each of the coordinate hyperplanes, and is otherwise holomorphic. Example 6.4. (1) If C P 2 is a smooth curve of degree d then by adjunction O P 2(C) = O P 2(d) and so K C = O C (d 3) which has degree (d 3) d. Therefore, the degree K X = 2g 2 = (d 3)d and so g = ( d 1) 2. (2) If Q P 3 is a smooth quadric surface. Note Q = P 1 P 1 P 3 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 P 3 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 K Q. We claim, K Q = O Q ( 2, 2). First, Q is P 1 P 1, so to write a meromorphic 2 form on Q is the same as writing down meromorphic 1 forms on both copies of P 1 and the zeros are the preimages of the zeros of the forms on P 1. Then, we obtain a meromorphic 2 form with poles on 2 lines in each ruling. So, (6.4) (6.5) (6.6) K Q = K P 3 O P 3(Q) Q = O P 3( 2) Q Then, applying adjunction again, K C = K Q O Q (C) C = O Q (a 2, b 2) C So, plugging in degrees, 2g 2 = deg K C = 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) Curves of low genus Genus 0. If C has genus 0, then C = P 1, as follows from Riemann-Roch (or a conic in P 2 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 O P 1(d) which embeds C as a rational normal curve of degree d P d. This d is equal to the smallest possible degree of an irreducible nondegenerate curve C P d, and any nondegenerate curve of that degree must be the rational

13 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 13 normal curve, as follows from Bezout s theorem: If we had a curve of lower degree, d general points would lie in a hyperplane. Conversely, if we an irreducible, nondegenerate curve, d 1 points of the curve span a codimension 2 linear space, and projection from that space will give an isomorphism of the curve with P 1. (1) How many surfaces or hypersurfaces does a curve lie on? (2) What is the normal bundle of a rational curve? (3) Many more open problems about rational normal curves Genus 1. (1) If we take d = 3, by Riemann Roch, this gives an embedding C P 2 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 P 3. If we have a cubic plane curve, 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 the following technique: We want to know whether H 0 (P 3, I P 3(2)) vanish 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 know H 0 (P 3, I P 3(2)) = 10 and H 0 (O C (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 P 1 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 equations get even more complicated Genus 2. Let C be a curve of geometric genus 4. Take a divisor of degree 4, corresponding to a line bundle L. We get a map C P 2, where C has geometric genus 2. By Riemann Roch, h 0 (L) = 3. Because and the image is of degree 4 in P 2, but it cannot be smooth because all plane quartics have genus 3. In fact: Exercise 6.5. For D of degree 4 on a curve C of genus 2, we can write p + q = D K C as an effective divisor for a unique divisor p + q on C. That is φ D : C P 2 is either a node if p = q or a cusp if p = q. The solution is just Riemann Roch. 7. 9/16/ 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 = p i with p i distinct. Let φ = φ K : C P g 1 be the canonical embedding. In coordinates: if ω 1,..., ω g are a basis for H 0 (K), then φ : p [ω 1 (p),..., ω g (p)]. More intrinsically, P g 1 = PH 0 (K) and φ : p H 0 (K p) H 0 (K). Theorem 7.1. r(d) = d 1 dim D

14 14 AARON LANDESMAN Proof. Suppose p 1,..., p d C. Note that if n is the number of linear relations on the points p i, so that n d = dim spanp 1,..., p d 1 l(k D) = g d + n = g dim spanp 1,..., p d 1 So, by Riemann-Roch, r(d) is the number of linear relations among the points p i. So, if the p i are linearly independent then there does not exist a nonconstant meromorphic function on C with simple poles at p i. In general, φ : C P g 1 is some canonical map, which might not be an embedding if the curve is hyperelliptic. Let D be any effective divisor. In this case, define the span D = φ 1 (H) D H Pg 1. 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 Applications of Geometric Riemann-Roch Genus 2, degree 5. Suppose D is a divisor class of degree 5 on a genus 2 curve. Then, φ : C P 3 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 P 3 contain C? To calculate the quadrics on which this surface lies, note that h 0 (O P 3(2)) = 10, h 0 (O C (2)) = Then, the map H 0 (O P 3(2)) H 0 (O C (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 H 0 (O P 3(3)) H 0 (O C (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 = P 1 P 1 P 3 by the Segre map. Homogeneous degree 1 forms on P 3 pull back to bidegree (1, 1) forms on Q. Therefore, there is a map homogeneous polynomials of degree m on P 3 bihomogeneous polynomials of bidegree (m,m) on P 3

15 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 resorting to coordinates. This map is the map on global sections associated to 0 O P 3(m 2) O P 3(m) O Q (m) 0 is a surjection of sheaves. We re asking whether it is surjective on global sections. Note that H 1 (L) = 0 for any line bundle on P 3. Therefore, we have an associated exact sequence on global sections. So, M C = S Q. This shows that in fact S 1 S 2 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 P 2, when we found a surface containing the curve, the curve became a divisor on that surface Genus 4, non-hyperelliptic. Let s focus on the canonical embedding. Here φ K : C P 3 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 sheeted cover of P 1. So, there are no meromorphic functions of degree 2. However, it is trigonal. We ll see this shortly. Looking at the usual exact sequences, we see H 0 (O P 3(2)) H 0 (O C (2)) determines a surjective map, with C lying on a unique quadric surface, uniqueness from Bezout s theorem. Next, looking at cubics H 0 (O P 3(3)) H 0 (O C (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 = P 1 P 1. 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/ Introduction to parameter spaces. Today: Jacobians. Application: every curve of genus g admits a nondegenerate embedding C P n 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 Pic d (C) = { Line bundles of degree d on C }.

16 16 AARON LANDESMAN (2) Hilbert scheme H d,g,r = { curves C R r degree d, genus g } (3) Moduli space of curves M g. M g = { 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 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 x The intention of Mazur was to look at y 2 = x which is a hyperbola, ( and ) observe this is a genus 0 curve, this curve is parameterized by t 2t, t t 2 1 t 2 which determines a map P 1 C yielding dx y and when we pull it back to P 1 it becomes R(t) dt and then we compute the integral of this rational function on P 1. The next integral that was considered was something like dx x which we can think of as dx y on the Riemann surface y 2 = x This integral isn t quite well defined, since it depends on the path you take. Say we have a smooth projective curve C of genus g. Given p, q C, we want to make sense of q p H0 (K) /H 1 (C, Z) = J(C). which is called the Jacobian of C. We can look at H 1 (C, Z) H 0 (K) Z 2g C g Fix a base point p 0 C. We get a map C J p which is non canonical because it depends on a basepoint. More generally, for any d we get a map µ. D = i p p 0 µ : C d J p i p i p 0

17 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 17 Theorem 8.1. For D, D C d we have This implies that J = Pic d (C). D D µ(d) = µ(d ) Proof. Idea: The forward direction is quite easy. If we start with D, D, the condition 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 combinations of D, D parameterized by the two divisors. That is, there is a family {D t } t P 1 where D 0 = D, D = D and µ: P 1 J t µ(d t ) Now, since J is a complex torus, but the pullback of a form to P 1 is 0, because P 1 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 Pic d (C) = J noncanonically, which is why we will often want to write Pic d (C) instead of J, though we won t worry about that too much now. Definition 8.3. Define } Wd {L r (C) = Pic d (C) : h 0 (L) r + 1 Exercise 8.4. For L Pic d (C) general, then h 0 (L) = max (0, d g + 1). Solution: The idea is to equivalently formulate it as follows. If D C d is a general effective divisor, we claim h 0 (L) = max (1, d g + 1). The proof of this, D = d i=1 p i. We start with h 0 (K) = g, h 0 (K p 1 ) = g 1, and repeating we see h 0 (K D) = max(0, g d), supposing we choose the p i so that they are as independent as possible. Riemann-Roch then implies the statement. So, we only need deal with the noneffective case. If d g the map µ d is surjective. We only need show if 0 d g then µ d is generically 1-1. In particular, Wd 0 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 Pic g+3 (C) is general, so h 0 (L) = 4, we get a map C P 3. We claim this is an embedding. Proof. If φ L 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 h 0 (L( D)) = h 0 (L) 1 = 3. This would imply L( D) Wg+1 2. Note that the set W 2 g+1 = K W0 g 3 by Riemann Roch. So, the existence of such a divisor D W0 2 with h 0 (L( D)) = 3 implying L( D) K W 0 g 3 which implies L W0 2 + (K Wg 3 0 ). 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 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 homework, 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 embeddings of curves of genus 5 and /21/15 Today: (1) Hyperelliptic (2) Counting moduli (3) canonical curves of genus 5 and Hyperelliptic curves. Let C be a hyperelliptic curve of genus g. Then, there is a map π : C P 1 of degree 2. We have exactly 2g + 2 branch points α 1,..., α 2g+2. Lemma 9.1. There exists a unique curve C with a map π : C P 1 branched at α 1,..., α 2g+2. Proof. Uniqueness is shown below via a topological argument. To show existence, we can write down C = V(y 2 i (x α i)), and specify a transition function to the other chart, as in Vakil s foundations of algebraic geometry. Alternatively, just complete this as a Riemann Surface. Given a Riemann sphere and 2g + 2 points, draw arcs from a given branch point 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 P 1. Question 9.2. Suppose we have a three sheeted cover of the Riemann sphere simply branched over 2g + 4 points. How can we describe the structure of that Riemann surface? The question is to describe the monodromy 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 describe 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 τ i S 3, we know i τ i = id and τ i is transitive. Hence, τ i = S 3, or equivalently, there are at least two τ i. Furthermore, such τ 1,..., τ 2g+4 is determined up to simultaneous conjugation, 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 α 1,..., α 2g+2d 2 branch points, and we get a similar description. Here, we get similarly transpositions τ i with τ i transitive and i τ i = id. This is the only relation because we know the fundamental group π 1 (S 2 \ { α 1,..., α 2g+2d 2 }. Note that the order of this product

19 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., nonsimple 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 τ i, namely τ i = (12) for all i. Question 9.5. What if we want to look at covers of curve other than P 1. 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 homologous 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 permutations corresponding to the two loops. Call these τ i, µ 1, µ 2. We get a relation i τ i = [µ 1, µ 2 ], since the complement of the µ i is a rectangle, and we want to go around the boundary of the rectangle. 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 α 1,..., α b, where b = 2g + 4, we want a cubic polynomial in y with coefficients in C[x] of the form y 3 + α 2 (x)y 2 + α 1 (x)y + α 0 (x) with discriminant i (x α i), 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 equation will necessarily have this form. So, a hyperelliptic curve is either 2g+2 i=1 2g+1 V(y 2 = (x α i )) or V(y 2 = (x α i )) 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 p i be (α i, 0) C. Then, (dx) vanishes at the p i, so (dx) = i p i 2q 2r where we computed the orders at infinity by symmetry, and degree considerations. To obtain a holomorphic function, we may note ( 1 y ) = p i + (g + 1)(q + r). i=1

20 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 y, x dx y,..., xg 1 dx y In this case, the canonical map φ K : C P g 1 is the map given by [ 1, x,..., x g 1]. In particular, the canonical map factors through the two sheeted cover as (9.1) π C P 1 φ K φ O P g 1 P 1 (g 1) So, the canonical map is 2 : 1 onto the rational normal curve. Suppose we want to embed this curve in projective space. We ll discuss this next time. Exercise 9.8. Every smooth curve admits an embedding of degree g + 3. If C is hyperelliptic of genus g 2, then the smallest degree of an embedding C P r 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 /23/ Hyperelliptic Curves. Recall from last time that if C is hyperelliptic of genus g, then C is given by y 2 = f(x) over A 1 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 infinity, by taking its closure in P 2 and resolving the singularities. We can explicitly see that H 0 (K C ) = dx y,..., xg 1 dx y In other words, the canonical map factors through the hyperelliptic map to P 1. (10.1) C π φ ν g 1 P g 1 P 1 Definition A line bundle L or divisor D or linear system L is special if h 0 (K D) = 0

21 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES 21 Now, if we have a special divisor on a hyperelliptic curve, that means the divisor 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 precisely the multiple of the g 1 2 that D contains. In conclusion if D is special, we can write D = r g D 0, where D 0 is a fixed base locus. That is, the map φ D given by any special D factors through π, and, in particular, there are no special, very ample linear series. Lemma The smallest degree of an embedding of a curve C P r is g + r. Note that hyperelliptic curves cannot be embedding into P 2. But, if r 3, a general map will be an embedding. Proof. Proved above. Exercise Conversely, it is also true that every nonhyperelliptic curve has an embedding of degree less than g + r Gonal Curves. Definition We say C is trigonal if there exists a degree 3 cover f : C P 1. Definition We say C is tetragonal, pentagonal, hexagonal if there exists a degree 4,5,6 cover f : C P 1. Definition Say C is k-gonal if there exists a g 1 k. Warning 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 No one knows more about trigonal curves than Anand Patel. Definition The moduli space of genus g curves is M g = { isomorphism classes of smooth projective genus g curves} Remark The existence of M g 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 How do we calculate the dimension of M g? Question Is M g irreducible? Understanding abstract curves is difficult, but we can understand the situation better by examining another moduli space: Definition Define the Hurwitz space { } H d,g = (C, f) : C M g, f : C P 1 of degree d with simple branching The one thing we can see for such a curve in H d,g is the number of branch points. Lemma dim M g = 3g 3.

22 22 AARON LANDESMAN Proof. Set b = 2d + 2g 2 and the branch divisor will consist of an unordered b tuple of distinct points. So, we obtain a map H d,g P b \, where the image corresponds to a divisor of degree b on P 1. We now endow H d,g with the structure of an algebraic variety via the covering map to P b. We next want to utilize the projection map H d,g M g by forgetting the map. Now, the dimension of H d,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 M g? We can t answer this in general, but when d > 2g 2. We are now asking, how many simply branched maps f : C P 1 of degree d are there? To specify such a map, we have to choose: (1) A line bundle L Pic d (C), which is g dimensional, (2) A pair of sections σ 0, σ 1 H 0 (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 H d,g has the correct structure for both the projection maps, but we won t worry for now. So, the dimension of M g is 2d + 2g 2 (2d g + 1) = 3g 3. Remark We can also think of P b to be polynomials of degree b, and is the zero locus of the discriminant. Exercise Show that unordered b tuples of points without repetition correspond to P b \. Guess at a Solution: This is precisely symmetric functions in (P 1 ) b, by the fundamental theorem of symmetric functions. Remark The reason this method for computing dim M g worked so well was because we introduced some auxiliary information that allows us to get a handle on the curve C. Lemma The dimension of the space of hyperelliptic curves is H 2,g, which is 2g 1 dimensional Proof. The map H 2,g { hyperelliptic curves } have three dimensional fibers. Therefore, the set of hyperelliptic curves is 2g 1 = 2g Corollary Not all genus g curves, for g 3, are hyperelliptic. Proof. The dimension of the space of hyperelliptic curves is less than that of M g. Next time: curves of genus 5 and /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.

23 NOTES FOR MATH 282, GEOMETRY OF ALGEBRAIC CURVES Curves of genus 5. We have a canonical map φ = φ K : C P 4. We first consider which quadrics vanish on C. H 0 (O P 4(2)) H 0 (O C (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 i Q i. Let us now make four observations: (1) Case 1 does occur because if Q i are three general quadrics, then by Bertini, C = i Q i is smooth, and by adjunction, we have K C = K P 4( ) = O C (1). (2) In case 1, C is not trigonal. If C were trigonal, then there exists a g 1 3, 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 P 1, but in genus 1 or 2, then C is a two-sheeted cover of P 1 (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 P 1 : In genus 3, we can project the plane quartic away from a point, that is, the linear series K P is a g 1 3. In genus 4, the rulings of a quadric cut out the g 1 3 on C. It turns out general curves of genus 5 are expressible as 4 sheeted covers of P 1. Note further that in genus 1 there is a 1 dimensional family of g 1 2, in genus 2, there is a finite number of g 1 2. 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 P 4. Such curves arise as an open subset of G(3, H 0 (O P 4(2))) which is 3 (15 3) = 36 dimensional. The fibers of the map { } U = Λ G : C = Q Λ Q is a smooth curve in P 1 M 5 Are PGL 5. We can do a similar count to show the trigonal curves form a 10 dimensional subvariety. Note that 3 g 3 + g 2 1 = 36, so indeed, since M g is irreducible, a general curve will be expressible as such a complete intersection. 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 Does the second case, in which C is not equal to an intersection of quadrics, occur?

24 24 AARON LANDESMAN Note that trigonal curves exist, and so they must be as in this second case. Start with a trigonal curve of genus 5. Given D a g 1 3, let s look at K D. By Riemann Roch, this is a g 2 5. Note that if this curve has a base point, the curve has a g2 4, so it would either give a two sheeted cover of P 1 (impossible since C is not hyperelliptic) or a plane quartic (impossible since g = 5). Therefore, C C 0 P 2, and the image is not smooth, since a smooth plane quintic is genus 6. Let s now look at 2D, which is a g 2 6. To see h0 (2D) = 3, note that it is at most dimension 3, by Riemann Roch, and at least 3 because the map Sym 2 H 0 (D) H 0 (2D) is injective, because the curve, when embedded in P 1 is surjective, so does not lie on any quadrics. Lemma 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 K C is cut out on C by conics Q P 2 passing through R. To see this, a conic Q P 2, 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 (11.1) C C 0 P 2 φ K π P 4 where the map π is given by quadratic polynomials vanishing at r. Now, we have the C lies on a surface in P 4. Let S = im P 2 under the right map π. Then, we have that the curve lies on a surface. This turns out to be the projection of the Veronese ν 2 2 surface from a point. We can also say that we can resolve the map P 2 S by blowing up at the point r. One way to describe this surface in P 4 is to take a line and a plane conic living in complementary spaces spanning P 4, 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 determines 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 /28/15 Today: Adjoint series - appendix A Wednesday: Castelnuovo s theorem - Chapter 3 But first canonical curves of genus Canonical curves of genus 5. Assume C has genus 5 and is not hyperelliptic. Let C P 4 be a canonical map. Then, C lies on three quadrics Q i. Last time, we saw (1) C = i Q i. (2) C i Q i = S, where S is a cubic scroll.

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