CURVES NOTES Eric Larson Taught a Course On

CURVES NOTES

AARON LANDESMAN

1. INTRODUCTION Eric Larson taught a course on Curves to projective space at Stanford in Fall 2019. 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 email suggestions to [email protected]

2. CONVENTIONS Here are some conventions we will adapt throughout the notes. (1) I have a preference toward making any detail stated in class, which is not veriﬁed, into an exercise. This will mean there are many trivial exercises in the notes. When the exercise seems nontrivial, I will try to give a hint. Feel free to contact me if there are any exercises you do not know how to solve or other details which are unclear. (2) Throughout the notes, I will often include parenthetical remarks which describe things beyond the scope of the course. If some word or explanation is placed in quotation marks or in parentheses, and you don’t understand it, don’t worry about it! It’s more meant to give you a ﬂavor of how one might describe the 19th century ideas in this course in terms of 20th century algebraic geometry. newpage

1This introduction has been adapted from Akhil Matthew’s introduction to his notes, with his permission. 1 2 AARON LANDESMAN

3. 9/23/19 3.1. Goals of course. Goal 3.1. The goal throughout this term is to understand the geometry of curves in projective space. There are three overarching questions: (1) What are the degrees and genera of curves in projective space. I.e., what possible types of curves are there? (2) What is the geometry of the moduli space of such curves? (3) What is the geometry of an individual such curve? 3.2. Logistics. Ofﬁce hours will occur Monday and Tuesdays between 1:30 and 3. Grading for the class for undergrads will be based on solutions to homework problems which will be posed in lecture. Graduate students are encouraged to think about these homework problems, but do not need to turn anything in. To the extent that there is a textbook for this course (though we won’t be following it much) about half of what we’ll discuss can be found in Moduli of Curves by Harris and Morrison.

3.3. Maps from curves to projective space. Remark 3.2. For the entire duration of this class, we will always work over the complex numbers. Question 3.3. How does one describe a curve in projective space? Recall that for C an algebraic curve, a map C → Pr is the same as the 0 datum of a line bundle L on C together with x1, ... , xr+1 ∈ H (C, L ) which are basepoint free, i.e., there is no p ∈ C with xi(p) = 0 for all i. Question 3.4. How do we calculate H0(C, L )? This naturally connects to the topic of the dualizing sheaf.

3.4. The Dualizing sheaf. Let C be a smooth irreducible curve of genus g. Later, we will relax this assumption. Let p1, ... , pd ∈ C be distinct points. This is for ease of notation, and they can often coincide.

0 Question 3.5. What is dim H (C, OC (p1 + ··· + pd))?

A section of OC(p1 + ··· + pd) is determined by its principal parts at p1,..., pd. Up to the addition of a constant. Question 3.6. Which principal parts can be completed to a section? CURVES NOTES 3

0 0 If we have α ∈ H (KC) and σ ∈ H (OC(p1 + ··· + pd)) we must have ( · ) = ∑ Respi α σ 0. i

This condition is trivial if and only if α vanishes at p1, ... , pd. This yields the following lemma. Lemma 3.7. 0 0 0 dim H (OC (p1 + ··· + pd)) ≤ d + 1 − dim H (KC) − dim H (KC (−p1 − · · · − pd)) . Equivalently, 0 0 dim H (OC (p1 + ··· + pd)) − dim H (KC (−p1 − · · · − pd)) ≤ d + 1 − g. Theorem 3.8 (Riemann Roch and Serre Duality). The inequality in Lemma 3.7 is an equality. That is 0 0 dim H (OC (p1 + ··· + pd)) − dim H (KC (−p1 − · · · − pd)) = d + 1 − g. Remark 3.9. Riemann Roch is saying that the obstructions to constructing sections is precisely the condition that the residues when paired with any differential form sum to 0. Remark 3.10. Alternatively, we have a duality map 0 1 H (KC (−p1 − · · · − pd)) × H (OC (p1 + ··· + pd)) → C.(3.1)

We can then choose small disjoint neighborhoods ∆i of pi which pairwise do not intersect. Let U = C − {p1, ... , pd}. We can think of an element of 1 0 H (OC (p1 + ··· + pd)) by a 1-cocycle fi ∈ H O∆i−pi . Then, the pairing above from Serre duality (3.1) is given by ( { }) 7→ ( · ) g, fi ∑ Respi g fi . i Remark 3.11. We’ll primarily be interested in smooth curves, but they fit into a moduli space containing singular curves, so it will be useful to carry out many ideas for singular curves. Remark 3.12. The same logic as above applies when C is nodal, or at least has mild singularities if we replace the canonical bundle KC (which we were 1 ∈ H0( ) defining as ΩC) by ωC, the dualizing sheaf. So, the element α ωCe can be viewed as a meromorphic differential on the normalization Ce of C with the property that for any p ∈ C, Resp (α · f ) = 0 for any holomorphic f defined in a neighborhood of p. 4 AARON LANDESMAN

It’s important to note that the function f above is on the curve C, but the differential α is thought of as on the normalization. Or rather, if π : Ce → C is ∈ H0( ) the map from the normalization, we can view α π∗ωCe .

Question 3.13. What does ωC look like? 3.4.1. At smooth points. First, let’s answer this question at smooth points. i Start with a curve C and a local coordinate x. Then, α = ∑i aix dx. We have the condition that for any n ≥ 1, we must have

n−1 0 = Res αx = a−n.

Therefore, a−n = 0 for all n ≥ 1, we ﬁnd α must be regular. Remark 3.14. Hence, in the case C is smooth, we see that we exactly recover KC. 3.4.2. At nodes. Next, let’s answer this question Question 3.13 at nodal points. We start with local coordinates x and y at the two branches of the nodes, with xy = 0. We can write i i α = ∑(aix dx + biy dy). i Then, let’s investigate what the residue condition tells us. We certainly must have Res α = 0. So,

a−1 + b−1 = Res α = 0. Also, for any n ≥ 2

n−1 0 = Res α · x = a−n.

Similarly, b−n = 0. Therefore, the condition that Resp (α · f ) = 0 tells us α has at worst a simple pole along both branches with opposite residues.

Corollary 3.15. We have deg ωC = 2g − 2 for a nodal curve. Proof. This is true for a smooth curve. One can then use the explicit description of the dualizing sheaf above relating the curve to its normalization to see this statement also holds for nodal curves. Essentially, when we add a node, we increase the genus by 1, and we also increase the degree of ωC by 2.

Exercise 3.16. Fill in the details of this proof.

CURVES NOTES 5

3.5. Constructing sections of a line bundle. Suppose we have a curve C mapping to P2 by f so that f (C) has at worst nodal singularities. Let the image of C be F(x, y, z) = 0 with F homogeneous of degree d. We let Γ ⊂ f (C) denote the set of nodes of f (C) which are not nodes of C. Let’s now write down sections of ωC explicitly in terms of the equation F. On C, we have ∂F ∂F ∂F dx + dy + dz = dF = 0. ∂x ∂y ∂z Therefore, ∂F ∂F ∂F z dx + dy + dz = z(dF) = 0. ∂x ∂y ∂z We also have ∂F ∂F ∂F x + y + z = deg F · F. ∂x ∂y ∂z Therefore, we ﬁnd ∂F ∂F ∂F ∂F ∂F ∂F z dx + dy + dz = 0 = x + y + z dz. ∂x ∂y ∂z ∂x ∂y ∂z Rearranging, we ﬁnd y dz − z dy z dy − x dz = ∂F ∂F . ∂x ∂y By symmetry, we have y dz − z dy z dy − x dz x dy − y dx = = ∂F ∂F ∂F . ∂x ∂y ∂z or possibly a different one of the above three expressions, depending on ∂F whether ∂z is nonzero. This produces an expression which is a holomorphic differential. We’ll pick this up from here next class.

4. 9/25/19 We’ll start by recalling the basic setup from last time. We had a nodal curve C with a map f : C → P2. We called the image curve F(x, y, z) = 0. The nodes of C gave rise to nodes of F, but we may also have additional nodes. We call Γ the set of additional nodes of F. We showed y dz − z dy z dy − x dz x dy − y dx = = ∂F ∂F ∂F .(4.1) ∂x ∂y ∂z 6 AARON LANDESMAN

We note that this will be undeﬁned at the nodes. These expressions have degree −d − 3. We deﬁne 0 0 α : H (IΓ(d − 3)) → H (ωC).(4.2) G 7→ G · (4.1).(4.3) The upshot is the following. Lemma 4.1. The map α is an isomorphism and the nodes Γ of any connected plane curve (meaning that the source is connected; in other words, we should assume that normalizing that subset Γ of the nodes leaves the curve connected) of degree d impose independent conditions on polynomials of degree d − 3.

Proof. We know 2g − 2 = deg ωC = d(d − 3) − 2#Γ. Therefore, d − 1 #Γ = − g. 2

Question 4.2. How many sections are produced? That is, what is dim im α?

d−1 0 We have at least ( 2 ) − #Γ = g. Since dim H (ωC) ≤ g a priori, we have 0 dim H (ωC) = g and α is an isomorphism.

4.1. Sections of other line bundles. Start with an effective divisor D ⊂ Csm of degree m which is supported on the smooth locus of C. Pick a polynomial g of degree n for n ≥ d through D ∪ Γ. Let ∆ denote the residual, so ∆ ∪ D ∪ Γ is C ∩ V(g). Then, we obtain a map 0 0 H (IΓ∪∆(n)) → H (OC(D)) where the map is division by g, i.e., h 7→ h/g. An element lies in the kernel of the map if and only if C ⊂ V(h), i.e., F | h. In other words, we have an injection 0 0 0 H (IΓ∪∆(n))/F · H (O(n − d)) → H (OC(D)). Then, by Bezout’s theorem intersection V(h) ∩ C, and noting that the points of Γ occur with multiplicity 2, #∆ + #D + 2#Γ = nd, and so d − 1 #Γ ∪ ∆ = nd − m − − g 2 CURVES NOTES 7

for m = deg D. Then,

0 0 dim H (IΓ∪∆(n))/F · H (O(n − d)) n + 2 d − 1 n − d + 2 ≥ − nd − m − − g − 2 2 2 = m + 1 − g

Therefore, for D sufﬁciently large (i.e., sufﬁciently large degree on each component of C),

0 H (ωC(−D)) = 0. So, by Riemann Roch and Serre duality,

0 H (ωC(−D)) = 0. and so the inequality above is an equality and

0 0 0 dim H (OC(D)) = dim H (IΓ∪∆(n))/F · H (O(n − d)) = m + 1 − g. So, in this case β is an isomorphism. Since any line bundle can be written as D+ − D− where D+ is sufficiently large. Therefore, we have a procedure to 0 compute H (OC(D)) for any D. Exercise 4.3. Let C be the “smooth compactification” (meaning not that C is smooth, but rather that there are no new smooth singularities introduced by compactifying C) of the affine curve y2 = x3y + 30x3. Let D ⊂ C be the divisor (1, 6) + (2, 20) + (3, 45). (1) Give an explicit basis (in terms of rational functions of x and y) of 0 H (ωC). What are the local conditions at the singularities of C? Check your sections really do satisfy these conditions. 0 (2) Using Riemann Roch, calculate the dimension of H (OC(D)) 0 (3) Calculate H (OC(D)). Describe this in terms of rational functions in x and y. Warning 4.4. Be aware that C is not nodal. You will have to think about how what we did in class generalizes to curves with the type of singularities that C has. Also, you may want to use a computer algebra system to do some routine linear algebra.

Another application of the dualizing sheaf is the construction of M g. 8 AARON LANDESMAN

Deﬁnition 4.5. A map C → X for C a curve is stable if C is nodal and # AutX( f ) is ﬁnite and C C (4.4) X commutes. Lemma 4.6. A map f : C → X is stable if C is nodal and any rational contracted component has at least 3 nodes (special points on the normalization) and any elliptic contracted component has at least 1 node. Further, a map f : C → X from a nodal curve C is stable if and only if, for A an ∗ ∗ ⊗100 ample divisor on X, f A ⊗ ωC is ample (or ( f A ⊗ ωC) is very ample). Proof. Exercise 4.7. Fill this in.

Now, we can construct M g by taking the Hilbert scheme of curves of X in N ∗ ⊗100 P and consider the subvariety such that (OPN (1) A) |C = ( f A ⊗ ωC) . Then, M g is the quotient of this subscheme of the Hilbert scheme by PGLN+1.

Remark 4.8. We can similarly work with pointed curves (C, p1, ... , pn). Then, being stable amounts to the automorphisms fixing pi, or equivalently that rational contracted components have 3 special points, and in the Hilbert ∗ ⊗100 scheme condition we need to work with ( f A ⊗ (p1 + ··· + pn) ⊗ ωC) . 5. 9/27/19 We start by reviewing some things from the end of last class. Definition 5.1. A curve C is stable if C is nodal and Aut(C) is finite. We saw in the previous class: Lemma 5.2. An equivalent condition to Aut(C) being finite for a nodal curve is that both (1) every rational component as at least 3 nodes (on the normalization) and (2) any genus 1 component has at least 1 node. Proof. When a component of C has genus 0, that component has automorphism group PGL2 which is 3-dimensional. When g = 1, the automorphism group is a finite extension of a 1-dimensional group, hence the automorphism group is 1-dimensional. Marking either 3 or 1 nodes in either case CURVES NOTES 9

then reduces the dimension of the automorphism group by this amount, so it is then 0-dimensional. A further equivalent condition is: Lemma 5.3. An equivalent condition to Aut(C) being ﬁnite for a nodal curve is 4 that ωC is ample, or ωC is very ample.

Proof. The ﬁrst statement amounts to the fact that ωC is positive on each component, using the previous characterization Lemma 5.2. 1 3 To check the very ampleness condition, we need to check H (ωC(−p − q)) = 0. Equivalently, we want to show 0 −3 H (ωC (p + q)) = 0. −3 Then, ωC has degree at least −1 on each component, and then ωC (p + q) has degree at least −1 on each component, hence the above space of sections vanishes.

We can then deﬁne M g as the coarse moduli space of stable curves of genus g. We can construct this as those curves C in Hilb8g−8,g,7g−8, modulo ⊗4 the action of PGL7g−7, such that OC(1) = ωC . Then, Hilbd,g,n denoting the Hilbert scheme of degree d, genus g curves in Pn.

5.1. Construction of M g,n(X, β). We next construct a moduli space of stable maps to X of class β.

Deﬁnition 5.4. A map f : (C, p1,..., pn) → X is stable if C is nodal and Aut( f ) is ﬁnite. That is, we have

{p1,..., pn}

(5.1) C C

X and the automorphism group of f is the set of maps C → C in the above diagram making the diagram commute. Remark 5.5. Similarly to our characterization of stable curves above, we ﬁnd that having a ﬁnite automorphism group is equivalent to the statement that (1) any rational component contracted under f has at least 3 distinguished points and 10 AARON LANDESMAN

(2) any elliptic component contracted under f has at least 1 distinguished point. Further, for A an ample divisor on X, having ﬁnite automorphism group 4 ∗ 3 ∗ 3 is equivalent to ωC ⊗ ( f A) is ample and ωC ⊗ ( f A) is very ample. We’d like to deﬁne

M g,n(X, β)

as the moduli space of stable maps f : C → X such that f∗ [C] = β where β ∈ k−1 k−1 CH (X) for X of dimension k, or β ∈ H2(X), or β ∈ CH (X)/ { algebraic equivalence }. We’ll mostly take X to be projective space, and it won’t matter which sort of cohomology group we use. As before, one can construct this space as a subscheme of a Hilbert scheme, quotiented by a PGL action.

Example 5.6. We have M g,0(Spec C, 0) = M g. Question 5.7. Why is this a good deﬁnition?

We will focus on the case of M g. Remark 5.8. The key useful property justifying this definition of stable maps is that M g is proper. Equivalently, in terms of a valuative criterion, given a family of stable curves C → Spec K (for K the fraction field of a dvr) there exists a unique extension to a family of stable curves over Spec R, at least after base extension on the dvr. Topologically, one can think of this as a family of curves over a punctured disk, and extending this to the disk after taking a covering space of the disk. We won’t prove this, but we’ll give some examples and we’ll hopefully become friends with properness of M g. Example 5.9. Start with a smooth curve C and a pair of points p and q on C. We can glue p and q to obtain a nodal curve. Consider a family of curves as q approaches p. How do we fill this in to a family of stable curves? There is an obvious compactification where the central fiber is a cusp, but we want a stable central fiber. To see what happens, we start with our disk D and the trivial family C × D. We start with a horizontal section we are calling p, which is of the form D × p0, for p0 ∈ C. We also have a non-horizontal section q. Since the points we are identifying are coming together, we can blow up the point of intersection of p and q. This gives a new family of curves over D, where the central fiber is C ∪ P1 where these two components meet at one point. When we identify p and q, we get a family of nodal curves, where the central fiber CURVES NOTES 11

is C ∪ P1, where P1 meets C at one point and also has a node. Since P1 has three nodes, this curve is stable. Example 5.10. What if now p and q approach each other tangentially to second order, but the rest of the setup is the same as in our previous example. Then, we will blow up the intersection point of p and q in C × D. When we blow up, p and q meet transversely at the new component in the special fiber. We can then blow up again at the point of intersections. We now get C ∪ P1 ∪ P1 in the special fiber where C meets the first P1 at a point, and the first P1 meets the second P1 at a point. Then, the two sections pass through the second P1. We then identify the two sections p and q. After the identification, in the special fiber, we get C ∪ P1 ∪ P1 as above, but where the last P1 has a node. The middle P1 only has two marked points, so the family is not stable. The middle P1 has self intersection −2, so it can be contracted. To calculate the self intersections, we see C ∪ P1 ∪ P1 has self intersection 0, since it is linearly equivalent to the special fiber. Then, intersecting with the middle P1, we see its self intersection is −2 (since it intersects the other two components to order 1. We then contract this component, which is possible since its self intersection is −2. This contraction is a surface with an ordinary double point singularity. The special fiber then looks the same as in the previous example.

6. 9/30/19 Example 6.1. Start with a family of curves acquiring a cusp with smooth total space. Locally, in a neighborhood of the nodal point, the family looks like y2 = x3 + t. When we blow up at the singular point, the central ﬁber is the union of C and a copy of P1 with multiplicity 2. This multiplicity 2 is coming from the fact that y2 = x3 + t, and so t = y2 − x3. When we write x = αs, y = βs, then we see v · ∇F = α(2y) + β(3x2) = 2α2s + 3β3y2, which vanishes to order at least 2 in the direction v, and exactly 2 if α 6= 0. Now blow up this resulting surface at the point the double P1 meets C. Let Ce be the proper transform of C. This then has an exceptional divisor given by P1 with multiplicity 3. We can see this because the local equation looks like (y + x2)(y − x2)2 = 0, the latter being the equation for the double P1, and altogether this has multiplicity 3 since the lowest order term is y3. When we blow up this at the intersection of the double and triple P1, we get a new exceptional divisor isomorphic to P1 of degree 6 which meets the double P1 and triple P1 and Ce at distinct points. The local equation looks like (x − y)x2y3 which has order 6. We don’t quite have a nodal curve yet because the central ﬁber is not reduced. 12 AARON LANDESMAN

To make it reduced, we will need to make a base change. First, let’s try setting t = s2, and see what the resulting family looks like when we apply this and then further normalize. Say we start with x ∈ P1 where the exceptional P1 has order 2 and y ∈ P1where the exceptional P1 has order 3. Near x, since x lies on a component of multiplicity 2, the equation of the central fiber looks like t = z2 for z a local coordinate on the total space. We now set t = s2. We have s2 = z2 which can be rewritten as (s − z)(s + z). Hence, when we normalize x, has two preimages. On the other hand, near y, we obtain a local equation of the central fiber s2 = t = z3. This looks locally like a cuspidal curve times A1. When we normalize, the cusp is normalized, but have only 1 preimage. On the normalization, s vanishes to order 3. Therefore, the triple P1 remains a triple P1 after base change and normalization, while the double P1 has preimage given by two copies of P1, each of which occur with multiplicity 1. On the degree 6 P1, we now get a double cover of P1, which has multiplicity 3 which is branched over the intersection with Ce and the degree 3 P1. The next step is to make a base change by setting r3 = s and normalize again, and this will result in a nodal central fiber with a copy of P1 meeting Ce and 5 copies of P1 meeting it, each being reduced in the central fiber. When we apply the base change of order 3, the two P1’s of multiplicity 1 (call them red) and Ce persist, and we have a copy of P1 of order 3 (call it green) meeting Ce and the two P1’s, and then order 3 P1, and there is the single order 3 P1 (call it yellow) meeting that order 3 P1 just described. The yellow P1 become three copies of P1. The green P1, when normalized, is ramified over three points to order 3, and hence is a genus 1 curve with an automorphism of order 3. Therefore, it has j-invariant 0. The red P1 has self intersection −1, using that it is the central fiber minus the other things in the central fiber, which only meet with multiplicity 1. Therefore, we can contract the red P1’s. Similarly, the yellow P1s are −1 curves. Therefore, we can contract the yellow P1’s. So, we ultimately get a central fiber Ce meeting an elliptic curve E of j-invariant 0. This is our desired family of stable curves.

Remark 6.2. To spell out what was happening in the previous example when we were making base changes, we can say the following in general. When we make a base change of prime order p and normalize, we get a branched cover of the old fiber, branched over ∑i aiDi where ai ∈ {0, 1, ... , p − 1} with ai ≡ ai mod p and the old central fiber is ∑i aiDi. The monodromy around branch points are specified by the ai in our cyclic p-fold cover. CURVES NOTES 13

7. 10/2/19 It is a classical fact that every plane quartic curve has 28 bitangent lines. Suppose we have a pencil of plane quartic curves that degenerates to a double conic. Then, there are 8 nonreduced points in the base locus of this pencil.

Question 7.1. What is the limit of these 28 bitangents?

Example 7.2. Let’s compute the stable reduction of the above family. This problem is posed to show you the power of stable reduction. We have a family of plane quartics degenerating to a double conic. This is not stable because the conic is not reduced. Note that in this case, the total space is singular, because at a point in the base locus of the pencil, in a neighborhood of such a point, the surface looks like x2 − ty (for x corresponding to the equation of the double conic, and y the equation of a plane quartic in the pencil) which is an ordinary double point singularity when t = 0. Note that at non-basepoints on the conic, the local equation looks like x2 − t. So this surface has 8 singular points. When we blow up at these 8 singular points, we get a smooth surface, with a nonreduced (yellow) P1 of multiplicity 2 and 8 reduced (pink) P1’s meeting it. We then take an order 2 base change to make replace the yellow nonreduced P1 by a reduced curve. This is a double cover of P1 branched over 8 points, and so we obtain a hyperelliptic genus 3 curve. The 8 red P1 pink curves are −1 curves, so we can blow them down. Then, we obtain a family of smooth curves where the special ﬁber is a hyperelliptic genus 3 curve. The hyperelliptic curve can then be viewed as a double cover of the conic, branched over the 8 points of the base locus.

A line pq is bitangent on a smooth plane quartic if and only if 2(p + q) ∼ 0 KC. We note also that for smooth plane quartics, we have H (O(p + q)) = 1, since smooth plane quartics are not hyperelliptic (as the canonical map is not an embedding for hyperelliptic curves).

2 Proposition 7.3. Let C be a curve of genus g and L a line bundle with L = KC. Then, on a family of curves h0(L ) mod 2 is a deformation invariant. 14 AARON LANDESMAN

Proof. Pick a divisor D ⊂ C of large degree. We have a commutative diagram

0 H (L (D)|2D)

(7.1) 0 0 H (L (D)) H (L |D)

H0(L )

0 We have a quadratic form on H (L (D)|2D). We have a quadratic form (α, β) = ∑ Resp (α · β) . p∈D 0 0 Further, H (L |D) and H (L (D) are isotropic under this quadratic form. To 0 see this, if we multiply two elements of H (L |D), we get a holomorphic section, and so it has trivial residue. The claim for H0(L (D)) holds because if we take two meromorphic sections and multiply them, their product is again meromorphic, and so the sum of the residues of their poles vanishes. Then, H0(L ) is the intersection of these two isotropic subspaces. Then, the proposition follows from the fact that the orthogonal Grassmannian has two components, and which component a given isotropic lies in can be read off from the parity of its dimension of intersection with a ﬁxed isotropic subspace. In other words, as we deform the quadratic form, the parity of the dimension of their intersection is deformation invariant. Lemma 7.4. On a genus 3 hyperelliptic curve, the hyperelliptic involution is unique. Proof. Let C be a genus 3 curve. Suppose we had two hyperelliptic maps C → P1. This is the same as specifying a map C → P1 × P1. These hyperelliptic involutions are different, which implies that the map C → P1 × P1 is generically an embedding. The image of the curve is a (2, 2) curve on P1 × P1. So, the image is of genus 1. Since there are no maps from a genus 3 curve to a genus 1 curve, C cannot have two hyperelliptic maps. Letting p denote the image of p under the hyperelliptic involution. The limit of any bitangent is pq such that 2(p + q) ∼ K = p + p + q + q. So, we must have p + q ∼ p + q. Then, either p + q = p + q. Or, these are two different ﬁbers under the hyperelliptic map. I.e., p = q. If p = q¡ we would have h0(O(p + q)) = 2, which is not possible. Therefore, we must have 8 p = p, q = q. There are then 28 bitangents coming from the (2) pairs of points in the base locus. CURVES NOTES 15

This finishes the proof that there are 28 limits of bitangents in the special fiber. Note these can all be seen to be limits by using the fact shown above that the parity of h0(O(p + q)) is constant in families, and so the divisor of two points of intersection on the nonreduced conic deform to an effective degree 2 divisor, whose square is the canonical bundle. Remark 7.5. Given any family, here is the recipe for computing a stable reduction. (1) Blow up the family to make the total space smooth. (2) Blow up the family to make the central fiber (when reduced) nodal. (3) Base change to make the central fiber reduced. (4) blow down −1 and −2 P1’s in the central fiber. Exercise 7.6. What is the stable reduction of a family of curves with smooth total space acquiring a tacnode. Locally a tacnode looks like y2 = x4. Ge- ometrically, this has two branches intersecting at a point. These have the same tangent direction, but are otherwise as transverse as possible. In other words, there are two distinct branches meeting to exactly order 2 at a point.

8. 10/4/19 Today, we’ll discuss deformation theory of smooth maps. Let X → B be smooth. Question 8.1. How can we understand deformations of X over B? Let ∆ := Spec C [ε] / ε2. We’d like to understand the set of ﬂat extensions of X → B to X0 → B × ∆. X X0 (8.1)

B B × ∆. Everything we will say works equally well if we ﬁx a nontrivial deformation of B. The only part we are using is that the ideal of B in B × ∆ has square 0. 8.1. Constructing a bijection between deformations and extensions of differentials by the structure sheaf. The key observation is the following: Sup- pose we have an extension

X X0 (8.2)

B B × ∆. 16 AARON LANDESMAN

Then, associated to this extension, we have a short exact sequence

dε (8.3) 0 OX ΩX0/B|X ΩX/B 0

∨ where here OX can be viewed as the conormal bundle NX/X0 and the above sequence is the conormal exact sequence. Conversely, suppose we start with an extension

(8.4) 0 OX E ΩX/B 0

We have a map

d : OX → ΩX/B f 7→ d f . We construct the ﬁber product F as

F OX

(8.5) d p E ΩX/B.

Concretely,

F := {(e, x) ∈ E × OX : p(e) = dx} . This is a sheaf of rings on X with multiplication given by

(e1, x1)(e2, x2) = (x1e2 + x2e1, x1x2) . This is not just a sheaf of rings, but also a sheaf of C[ε]/ ε2 modules using ε (e, x) = (x · dε, 0) . From this, we can produce Spec F which is locally constructed as above and glued together locally on B. We obtain.

X Spec F (8.6)

B B × ∆.

Warning 8.2. F is not a sheaf of OX algebras, so there is no map Spec F → X, but we are still calling it “spec.” CURVES NOTES 17

There is a bijection between ﬁrst order deformations of X → B and extensions

(8.7) 0 OX • ΩX/B 0

Exercise 8.3. Show these maps are mutual inverses.

0 Lemma 8.4. There is a bijection between Ext (ΩX/B, OX) and automorphisms of any given deformation extending the trivial automorphism on X.

Proof. This follows from the definition of extensions. Remark 8.5. We have further equivalences between automorphisms of any given deformation extending the trivial automorphism on X, which are in 0 turn the same as infinitesimal automorphisms of X. Further, Ext (ΩX/B, OX) = 0 0 H (TX/B). So altogether we get a bijection between H (TX/B) and infinitesimal automorphisms of X.

1 Corollary 8.6. We have a bijection between Ext (ΩX/B, OX) and ﬁrst order deformations of X → B. 8.2. Lifting Deformations. n Deﬁnition 8.7. Let ∆n := Spec C[ε]/ (ε ).

Suppose we are given X and Xn ﬂat over B and B × ∆n and we consider deformations Xn+1

X Xn Xn+1 (8.8)

B B × ∆n B × ∆n+1.

Lemma 8.8. Associated to such a diagram, we have a “conormal-like” sequence

| (8.9) 0 Ω∆n+1/ Spec C ΩXn+1/B Xn ΩXn/B×∆n 0

Proof. To see where this is coming from, if X ⊂ Y, we have

∨ (8.10) 0 NX/Y ΩY|X ΩX 0 and for X → Y smooth, we have

(8.11) 0 ωY|X ΩX ΩX/Y 0. 18 AARON LANDESMAN

This sequence of differentials yields

(8.12) 0 Ω∆n ΩXn/B ΩXn/B×∆n 0 We get a diagram of sheaves | 0 Ω∆n+1 ΩXn+1/B Xn ΩXn/B×∆n 0 (8.13)

0 Ω∆n ΩXn/B ΩXn/B×∆n 0 | → The map ΩXn+1/B Xn ΩXn/B×∆n is coming as the composition of the latter maps in the conormal sequence and relative differential exact sequence. Conversely, suppose we have an extension p 0 Ω∆n+1 E ΩXn/B×∆n 0 (8.14)

0 Ω∆n ΩXn/B ΩXn/B×∆n 0 We then construct

F OXn

(8.15) d p E ΩXn/(B×∆n). This F can be expressed as

F := {(e, x) ∈ E × OXn : p(e) = dx} . This is a sheaf of rings by

(e1, x1) · (e2, x2) = (x1e2 + x2e1, x1x2) and is a sheaf of C [ε] / εn+1 modules by ε (e, x) = (ε · e + x · dε, εx) . We then have

X Xn Spec F (8.16)

B B × ∆n B × ∆n+1. arising from this data. Summarizing the above: CURVES NOTES 19

Proposition 8.9. There is a bijection between deformations

X Xn Xn+1 (8.17)

B B × ∆n B × ∆n+1. and extension diagrams p 0 Ω∆n+1 E ΩXn/B×∆n 0 (8.18)

0 Ω∆n ΩXn/B ΩXn/B×∆n 0 with inverse maps as constructed above. Proof. Exercise 8.10. Verify the maps constructed above are mutual inverses, completing the proof.

Corollary 8.11. There exists some Xn+1 ﬁtting into

X Xn Xn+1 (8.19)

B B × ∆n B × ∆n+1. if and only if the class of

(8.20) 0 Ω∆n ΩXn/B ΩXn/(B×∆n) 0 lies in the image of the natural map 1 → 1 Ext ΩXn/(B×∆n), Ω∆n+1 Ext ΩXn/(B×∆n), Ω∆n . This map sends E to the pushout P of

Ω∆n+1 E (8.21)

Ω∆n P.

Proof. Unwind the bijection shown in Proposition 8.9. 20 AARON LANDESMAN

9. 10/7/19 9.1. Interpretations of Ext groups. Recall that last class we had been examining diagrams of the form

0 X Xn Xn+1 0 (9.1)

0 B B × ∆n B × ∆n+1 0

and were asking when given Xn, a deformation of order n, can we extend it to a deformation of order n + 1 Xn+1 as above. We saw Xn lifts to Xn+1 if and only if (9.2) | ∈ 1( ) 0 Ω∆n Xn ΩXn/B ΩXn/(B×∆n) 0 Ext ΩXn/(B×∆n), Ω∆n

lies in the image of 1 → 1( ) Ext ΩXn/(B×∆n), Ω∆n+1 Ext ΩXn/(B×∆n), Ω∆n We have a short exact sequence

(9.3) 0 OX Ω∆n+1 Ω∆n 0.

This yields a short exact sequence (9.4) 1 1( ) 2( ) Ext ΩXn/(B×∆n), Ω∆n+1 Ext ΩXn/(B×∆n), Ω∆n Ext ΩXn/(B×∆n), OX

Then we can use 2( ) ' 2( ) Ext ΩXn/(B×∆n), OX Ext ΩX/B, OX .

This follows from the pushforward pullback adjunction from X → Xn, using that OX is supported on the reduced X, but the argument was omitted in class. We have proven the following:

0 Lemma 9.1. Ext (ΩX/B, OX) corresponds to infinitesimal automorphisms of X/B (automorphisms of a first order deformation). 1 Ext (ΩX/B, OX) corresponds to first order deformations of X/B. 2 Ext (ΩX/B, OX) corresponds to the obstructions to lifting a given nth order deformation to (n + 1)st order. CURVES NOTES 21

Example 9.2. Suppose X → B is a smooth curve of genus g with B = Spec C. 0 0 Then, Ext (ΩX, OX) = H (TX). So, infinitesimal automorphisms correspond to tangent vector fields. We have  3 if g = 0 0  h (TX) = 1 if g = 1 0 if g ≥ 2 1 1 Then, Ext (ΩX, OX) = H (TX). We have  0 if g = 0 1  h (TX) = 1 if g = 1 3g − 3 if g ≥ 2 2 2 Then, Ext (ΩX, OX) = H (TX) = 0 because X is a curve. 0 This implies Mg is smooth of dimension 3g − 3 for g ≥ 2. The H indicates 1 Mg is a Deligne Mumford stack, the H indicates the dimension, and the H2 indicates the smoothness (showing that first order deformations lift to arbitrarily higher order). 9.2. The Tangent Complex. We now want the above story to work when f : X → B is a family of curves with f not necessarily smooth. f Consider X −→ Y → Z. We get an exact sequence ∗ (9.5) 0 f ΩY/Z ΩX/Z ΩX/Y 0 To make this work for non-smooth f , we need an analog of Ω that induces an exact sequence in the above sense when f is not smooth. The usual procedure for doing this is the following. (1) Start with a functor F : A → B. (2) Embed in the derived category: Given an abelian category A such as modules or sheaves on a space X. We then map this into the derived category D(A ) by sending M to the complex (9.6) ··· 0 M 0 ··· (3) Find nice representatives of classes of objects in A where F is an exact functor. For example, we might take a complex of free objects. (4) Apply F term-wise. (5) From an exact sequence in A we obtain an exact triangle in D(A ). We now might want to take the sheaf of differentials and make it derived. Warning 9.3. Schemes or algebras are not an abelian category. 22 AARON LANDESMAN

Deﬁnition 9.4. Let ∆ be the category of nonempty ordered sets. The objects of ∆ are [n] := {0, ... , n} viewed as an ordered set, for n ≥ 0. The morphisms are order preserving maps. Exercise 9.5 (Easy exercise). Any order preserving map between ordered sets can be written as a composition of di’s and si’s with

di : [n] → [n + 1] (0, 1, 2, . . . , i − 1, i,..., n) 7→ (0, 1, . . . , i − 1, i + 1, . . . , n + 1) .

si : [n] → [n − 1] (0, 1, . . . , i, i + 1, . . . , n) 7→ (0, 1, . . . , i, i,..., n) .

Loosely speaking, di omits i and si repeats i.

The di and sj satisfy certain relations. Namely,

didj = dj−1di if i < j

disj = sj−1di if i < j (9.7) disj = id if i = j or i = j + 1

disj = sjdi−1 if i > j + 1

sisj = sj+1si if i ≤ j. Deﬁnition 9.6. A simplicial R-algebra is a functor ∆op → R − alg which includes the data of

(1) An R-algebra An for each n ≥ 0 (2) Maps di : An → An−1, si : An → Sn+1 satisfying the relations (10.1). (Note here the maps di and si go in “reverse” because we are working with ∆op instead of ∆. Remark 9.7. There is a forgetful functor from simplicial R-algebras to D(R) forgetting the R-algebra structure, by sending an object to A• where the n−1 i differential on degree n is given by ∑i=0 (−1) di, viewed as a map An → An−1. One much check this is indeed a differential. We now want to take our scheme X over Spec R and resolve it as a simplicial R-algebra. Deﬁnition 9.8. A free simplicial R-algebra A is a simplicial R-algebra A such that

(1) There is a set Xn such that An = R[Xn] is free (2) si(Xn) ⊂ Xn+1. CURVES NOTES 23

10. 10/9/19 Last time, we had deﬁned a notion of simplicial R-algebras, including all R-algebras. This is analogous to the way chain complexes contain all

R-modules. The inclusion is given by sending A 7→ (An)n∈Z≥0 with An = A and si = di = id. Lemma 10.1. Given an R-algebra A, we can construct a free simplicial resolution of A inductively (term by term).

sketch. The algorithmic procedure is that at each stage, we add new generators, at most one for each si of the previous generators (it will be one for each after accounting for the relations induced by the si, this is clearer in the examples below). We then compute the image in the chain complex and check if it is the desired kernel. If it is not the desired kernel, we add more generators yj so that d0yj maps to the kernel, and then dkyj = 0 for all k > 0.

To illuminate what is going on in the proof, let’s see some examples.

Remark 10.2. This is analogous to the construction of a projective resolution for modules over a ring.

Recall we had the following list of relations for the category ∆.

didj = dj−1di if i < j

disj = sj−1di if i < j (10.1) disj = id if i = j or i = j + 1

disj = sjdi−1 if i > j + 1

sisj = sj+1si if i ≤ j.

Example 10.3. Take R = C and A = C[x]. We start with C[x] in degree 0. We add C[x1] in degree 1 with x1 = s0x and d0x1 = d0s0x = x. Also, d1x1 = d1s0x = x, using the relations for simplicial R-algebras. Similarly, take C[x2] in degree 2 with x2 = x0s1 = s1x1. Then, we ﬁnd dix2 = x1 for all i. We can continue and take C[xi] in degree i with sjxi−1 = xi and djxi = xi−1. When we take the associated chain complex, we see that the maps are 0 from odd to even degrees and the identity from even to odd degrees.

Example 10.4. Take R = C[x] and A = C[x]/(x) = C. In degree 0, we can take A0 = R = C[x]. For the chain complex to be quasi-isomorphic to A in degree 0 in degree 1 we take R[y] with d0y = x and d1y = 0. In degree 2 we 24 AARON LANDESMAN

1 take R[y0, y1] with s0y = y0 and s1y = y1. We ﬁnd

d0y0 = d0s0y = id(y) = y

d1y0 = d1s0y = id(y) = y d2y0 = d2s0y = s0d1y = 0 and

d0y1 = x d1y1 = y d2y1 = y.

Remark 10.5. Observe that for a ≥ 1 a b a b a b a b a+b d(y0y1) = d0(y0y1) − d1(y0y1) = y x − y . These terms generate the ideal y(x − y) as an R[y]-module. We have a differential map d : R[y] → R sending 1 7→ 0 and ya 7→ xa for a ≥ 1. So, d is not a map of R-algebras. However, this is a map of R-modules. The cokernel of this map is C = R/(x). a b a b a+b Next, looking at d : R[y0, y1] → R[y]. This sends y0y1 7→ y x − y if b b a 6= 0. If a = 0, then y1 7→ x . In R[y], we see that y(x − y) ⊕ h1i lie in the kernel of d, and it is also the image of d. One can continue doing this in degree 3. In degree 3 we’ll need to add 6 generators for all the si’s of y0 and y1. We know 0 y0 := s1y0 = s1s0y = s0s0y = s0y0. 0 y1 := s2y0 = s2s0y = s0s1 = s0y1 0 y2 := s2y1 = s2s1y = s1s1y = s1y1. 0 0 0 It turns out we can take R[y0, y1, y2]. Altogether, we ﬁnd the di’s satisfy

0 0 0 y0 y1 y2 d0 y0 y1 x d1 y0 y1 y1 d2 y0 y0 y1 d3 0 y0 y1

0 Table 1: Table of di applied to the yj CURVES NOTES 25

y0 y1 d0 y x d1 y y d2 0 y

Table 2: differential from degree 2 to degree 1

So far we have started our resolution as 0 0 0 (10.2) R[y0, y1, y2] R[y0, y1] R[y] R 0 One can then keep going. . . Deﬁnition 10.6. Let S be an R-algebra. Take a free simplicial resolution of S which we call S ∈ ∆(R). Apply Ω•/R to obtain a simplicial S module {An}. These have di maps induced by the di on the resolution S of S. Then apply

• ⊗S S to obtain a simplicial S module {An ⊗Sn S}. These have the induced di. Given this corresponding S-module, we can apply the forgetful map i ∆(S) → C(S) sending {di} 7→ ∑i(−1) di. This produces a complex of S- • modules called the cotangent complex. We denote this ΩS/R ∈ D(S). Remark 10.7. This cotangent complex is independent of the choice of resolution. If we had two different resolutions, we would have a map between them inducing a quasi-isomorphism when we pass to chain complexes.

11. 10/11/19 Recall that we were discussing simplicial resolutions last time. We saw the following examples:

Example 11.1. Take R = C and A = C[x] and took a resolution with C[xi] in degree i so that xisjxi−1 and djxi = xi−1. Example 11.2. We took R = C[x], A = C[x]/(x). We had 0 0 0 (11.1) R[y0, y1, y2] R[y0, y1] R[y] R with yi = siy, d0y = x, d1y = 0, and we also had 0 y0 := s1y0 = s1s0y = s0s0y = s0y0 0 y1 := s2y0 = s2s0y = s0s1 = s0y1 0 y2 := s2y1 = s2s1y = s1s1y = s1y1. with 26 AARON LANDESMAN

0 0 0 y0 y1 y2 d0 y0 y1 x d1 y0 y1 y1 d2 y0 y0 y1 d3 0 y0 y1

0 Table 3: Table of di applied to the yj

We’d next like to compute the cotangent complex in both of these examples. Example 11.3. First, we compute the cotangent complex for Example 11.1. In degree i, the ring in degree i is C[xi] and the Kahler differentials are then C[xi] · dxi. We next need to turn this into a complex of C[x] modules. We have x act by xi. As a complex, this is given by

0 id 0 (11.2) C[xi] dxi C[xi] dxi C[x1] dxi C[x] dx with the differentials alternating between 0 and id. There is a quasi-isomorphism to

id 0 C[x2] dx2 C[x1] dx1 C[x] dx (11.3)

0 0 C[x] dx.

• We ﬁnd that ΩA/R = [ΩA/R]0. Remark 11.4. More generally, for A a smooth R-modules, the cotangent complex is concentrated in degree 0. Example 11.5. Let’s now compute the cotangent complex in Example 11.2. We have (11.4) 0 0 0 0 0 0 R[y0, y1, y2] · hdy0, dy1, dy2i R [y0, y1] · hdy0, dy1i R [y] · hdyi 0

When we tensor with A, we get the complex

0 0 0 0 (11.5) C · hdy0, dy1, dy2i Chdy0, dy1i Chdyi 0 CURVES NOTES 27

The second map is 0 because the differential before tensoring is x − 0, which becomes 0 after tensoring over R with A. The ﬁrst map is given by the matrix 1 0 0 0 0 −1 0 0 0 with respect to the bases dy0, dy1, dy2 and dy0, dy1. There is a quasi-isomorphism 0 0 0 Chdy0, dy1, dy2i Chdy0, dy1i C · hdyi 0 (11.6)

0 0 Chdyi 0

Exercise 11.6. Compute the cotangent complex for R = C[x] and A = C[x]/(x). Show that after taking Kahler differentials and tensoring with A, the resulting complex is exact in degree at least 2. Beginning of solution: Starting to compute the complex, we see that the degree 4 term has generators e, f , g, h with 0 0 0 e := s0y0 = s1y0 = s2y0 0 0 0 f := s3y0 = s0y1 = s1y1 0 0 0 g := s2y1 = s0y2 = s3y1 0 0 0 h := s1y2 = s2y2 = s3y2 Computing d, we ﬁnd

e f g h 0 0 0 d0 y0 y1 y2 x 0 0 0 0 d1 y0 y1 y2 y2 0 0 0 0 d2 y0 y1 y1 y2 0 0 0 0 d3 y0 y0 y1 y2 0 0 0 d4 0 y0 y1 y2

0 Table 4: Table of di applied to the yj

We should now compute that this complex is exact. It seems like a tedious calculation. When we compute the induced differential on relative sheaves of differentials over R and tensor with C[x]/(x) at this stage, we obtain the maps 0 0 0 (11.7) Chde, d f , dg, dhi Chdy0, dy1, dy2i Chdy0, dy1i 28 AARON LANDESMAN where the first map is given by 1 0 0 0 0 −1 and the second is given by   0 0 0 0 0 1 −1 0 0 0 0 0 0 so we see the image of the first map is dy1, and this is also the kernel of the second map. So the complex is indeed exact at this stage. Remark 11.7. We find

ΩA/R ' [Chdyi]1 . More generally when A = R/I is a complete intersection, we ﬁnd

h 2i ΩA/R ' I/I . 1 Once you check this for a point in A1, one can use a sequence of maps Spec A → Spec A1 → Spec A2 → · · · → R where each ring is deﬁned by a single equation. This equation gives a map to A1. That is we have

Spec A Spec A1 (11.8)

Spec C[x]/(x) A1.

And then thinking about how the cotangent complex ﬁts into an exact triangle tells us the above characterization holds for Spec A over Spec A1. One can then ﬁt these together using relevant exact triangles to obtain the cotangent complex for Spec A in Spec R. Remark 11.8. Here are the key properties of the cotangent complex which the examples illustrate: (1) Suppose we have

f X Y (11.9) Z. CURVES NOTES 29

We then obtain an exact triangle ∗ • • f ΩY/Z ΩX/Z (11.10) [1] • ΩX/Y. • (2) If f as in (11.9) is smooth, then Ω = [ΩX/Y] Xh/Y i 0 (3) If f is an lci embedding, Ω• = N∨ . X/Y X/Y 1 (4) The formation of the cotangent complex commutes with ﬂat base change. That is, given X → Y and Y0 → Y is a ﬂat map with

f X0 X (11.11)

Y0 Y a fiber square, then • ∗ • ΩX0/Y0 ' f ΩX/Y. (5) Given X → B and a deformation X X0 (11.12) B B × D where D = Spec C[ε]/ ε2, then 0 • Ext ΩX/B, OX corresponds to infinitesimal automorphisms of a given deformation 1 • Ext ΩX/B, OX parameterizes first order deformations of X/B 2 • Ext ΩX/B, OX parameterizes obstructions to lifting deformations from order i to order i + 1. h • i Example 11.9. Suppose X → B is smooth. Then ΩX/B = Ω . In this X/B 0 case i i Ext (ΩX/B, OX) = H (TX/B) . 30 AARON LANDESMAN

Suppose f : X → B is an lci embedding. Then, • = ∨ ΩX/B NX/B 1 . Then, i ∨ = i−1( ) Ext NX/B 1 , Ox H X, NX/B . Note here that it makes sense Ext0 = 0 because there are no inﬁnitesimal −1 automorphisms of an embedding. On the other side, H (NX/B) = 0, so this is consistent. Similarly, in degree 1, giving a deformations corresponds to a global section of the normal bundle, and geometrically, this corresponds to the deformation in the corresponding direction. 12. 10/14/19 Today, we’ll begin by discussing how the cotangent complex controls deformation theory. Lemma 12.1. Let C → B be a family of nodal curves. Then, the cotangent complex of C over B is • ΩC/B = [ΩC/B]0 . Proof. Away from the nodes of C, this holds because the family is smooth. In the neighborhood of a node, this family is, etale´ locally, C = Spec R[x, y]/(xy − t) → Spec R for t ∈ R. 2 Then, C factors through AR = Spec R[x, y]. So we have

f 2 C AR (12.1)

R. From this diagram, we obtain a factorization ∗ • ΩA2/R 0 f ΩC/R (12.2)

h ∨ i N A2 . C/ R 1 • Therefore, ΩC/R is the two term complex ∨ ∗ N 2 → f ΩA2/R. C/AR where the ﬁrst term lies in degree −1 and the second lies in degree 0. This map is injective, because it is so generically. CURVES NOTES 31

We therefore have • h ∨ ∗ i ΩC/R = coker N A2 → f ΩA2 /R = [ΩC/R]0 . C/ R R 0 Remark 12.2. This concentration of the cotangent complex in degree 0 holds more generally when the curve has planar singularities. The above lemma combined with the general package we have seen before shows the following: Corollary 12.3. In particular, for X a nodal curve, we have that 0 Ext (ΩX, OX) controls automorphisms of a ﬁrst order deformation. 1 Ext (ΩX, OX) controls ﬁrst order deformations. 2 Ext (ΩX, OX) 2 controls obstructions to deformations. By cohomological dimension reasons, Ext (ΩX, OX) = 0.

By applying the above corollary, we see M g is formally smooth, hence smooth. This yields the following corollary. 1 Corollary 12.4. M g is smooth with tangent space at X given by Ext (ΩX, OX).

12.1. Deformations of marked curves. Let’s next discuss M g,n. What if we mark points? Suppose we have a curve X with some collection of marked points Γ;. Suppose we want to deform the curve X as a marked curve. In other words, we deform the curve together with the sections. We can then glue C[x]/(x2) to each of the marked points, and then extend this to a deformation of the curve. In the case of a smooth curve, we obtain an 1 extension Ext (ΩX/B, OX). In the case of this extended curve, we want to pick a trivialization of the deformation at each of the points of Γ. Then, 1 a deformation corresponds to an element of Ext (ΩX/B, IΓ). There is a natural map 1 1 Ext (ΩX/B, IΓ) → Ext (ΩX/B, OX) sending a deformation of the curve with marked points to a deformation of the underlying curve. Here IΓ is the conormal bundle of Γ in X.

Corollary 12.5. The stack M g,n is smooth with tangent space at (X, Γ) given by 1 Ext (OX, IΓ). 32 AARON LANDESMAN

Remark 12.6. In the case the curve X is smooth, we have the following: Note 1 1 1 1 here Ext (ΩX/B, IΓ) = H (TX(−Γ)) and Ext (ΩX/B, OX) = H (TX). We have an exact sequence

(12.3) 0 TX(−Γ) TX TX|Γ 0. We get an exact sequence

0 1 1 (12.4) H (TX|Γ) H (TX (−Γ)) H (TX) 0. This makes sense because a deformation of the marked curve that doesn’t deform the curve is given by a deformation of the points, which is speciﬁed by a tangent vector. This is not true for nodal curves X because ΩX/B is not a vector bundle, so we’ll have another term in the spectral sequence for computing 1 Ext (ΩX/B, OX). Then, we have an exact sequence

(12.5) 0 OXsing ΩX ΩXe 0. In other words, dualizing the cotangent complex does not yield the “tangent complex” since we really have to take derived hom into OX. Example 12.7. Consider the map f : P1 → P2 t 7→ (s3, t2s, t3) These both live over Spec C. The cotangent complex of P2 over Spec C is the 2 1 • cotangent bundle of P , and similarly for P . Hence, we can compute Ω f as fitting in an exact triangle and is equivalent to the complex ∗ f ΩP2/ Spec C → ΩP1/ Spec C. This map is neither injective nor surjective. This complex therefore is not ∗ concentrated in a single degree. Note that ΩP1/ Spec C = O(−2). Next, f ΩP2 has first Chern class f ∗O(−3) = O(−9), because the map has degree 3. To compute the cohomology, we have a spectral sequence i j • =⇒ i−j • Ext H ΩP1/P2 , OP1 Ext ΩP1/P2 , OP1 . Using the spectral sequence, and letting p denote the point of P1 mapping to the singular point under f , we find 0( • ) = H ΩP1/P2 Op, CURVES NOTES 33

∗ Since the map f ΩP2 → ΩP1 has cokernel of degree 1, we have a surjection ∗ f ΩP2 → O(−3). (− ) −1( • ) = (− ) It therefore has kernel OP1 6 . It follows that H ΩP1/P2 OP1 6 . On the E2 page of the spectral sequence we have 0( 0( • ) ) 1( 0( • ) ) Ext H ΩP1/P2 , OP1 Ext H ΩP1/P2 , OP1 (12.6)

0( −1( • ) ) 1( −1( • ) ) Ext H ΩP1/P2 , OP1 Ext H ΩP1/P2 , OP1 These have dimensions 0 1 (12.7) 7 0 In particular, there are no differentials and the spectral sequence degenerates. Therefore, 2 M 0(P , 3) is smooth at [ f ] of dimension 8 = 7 + 1. Note that we are looking for Exti−j and so the anti-diagonals in the above chart converge to the Exti.

13. 10/16/19 Exercise 13.1. Start with V(xy) ⊂ P2 which is a disjoint union of two copies of P1 meeting at a point. Now, take V(uv) ⊂ P3 a disjoint union of two planes meeting at a point. We can map the each of the copies of P1 into each of the planes, so that the image is two conics meeting at a point, such that the conics are both tangent to the line of intersection of the planes. Show 3 M 0(P , 4) is smooth at f . 13.1. Maps to P1.

Deﬁnition 13.2. The Hurwitz space Hurd,g parameterizes C a smooth geometrically connected curve of genus g with a degree d (generically separable, 1 ◦ which is automatic in char 0) map C → P . We write Hurd,g for the open locus with simple branching. ◦ What is the tangent space to Hurd,g? We ﬁnd • ∗ Ω f = [ f ΩP1 → ΩC] ' ORam( f ) 34 AARON LANDESMAN

1 The tangent space is Ext ORam( f ), OC ' ⊕p∈Ram( f )C

Theorem 13.3. Mg is irreducible. Proof. The idea is to show ◦ (1) Hurd,g is surjective for d 0 and ◦ (2) Hurd,g is irreducible. Let’s address these in turn. (1) Let C be a smooth curve of genus g, L be a line bundle over C of large 4degree. Let V ⊂ H0(L ) be a general two dimensional space ◦ of sections. The following lemma implies the map Hurd,g → Mg is surjective. Lemma 13.4. For C → P1 via the linear system V has simple branching. Proof. We will compute dim V : C → P1 does not have simple branching . The branching can fail to be simple in three ways. Either: (a) there exists a point p ∈ C with V ∩ H0(L (−3p)) 6= 0 (b) there exists p ∈ C with V ⊂ H0(L (−p)) i.e., p is a basepoint (c) there exists p, q ∈ C with V ∩ H0(L (−2p − 2q)) 6= 0 In each of these possibilities, we will calculate the dimension of the possible two dimensional V. (a) The dimension is 1 + ((d − 3) + 1 − g − 1) + ((d + 1 − g) − 2) = 2d − 2g − 3 (given by the choice of point p and a choice of a line in H0(L (−3p)) and a choice of a line, and a choice of way to complete the line to a two dimensional space V). (b) 1 + 2((d − 1) + 1 − g − 2) = 2d − 2g − 3 (given by a choice of p and a 2-plane in H0(L (−p))) (c) 2 + ((d − 4) + 1 − g − 1) + ((d + 1 − g) − 2) = 2d − 2g − 3 (given by choice of p, q, a choice of line in H0(L (−2p − 2q)), and a choice of completing the line to a 2-dimensional space V). We now compare the above to dim Gr 2, H0(L ) = 2 (d + 1 − g − 2) = 2d − 2g − 2. ◦ ◦ (2) Now, we want to show Hurd,g is irreducible. We have a map Hurd,g → Symn P1 − ∆ where ∆ is the locus where two points come together and n = 2d + 2g − 2, the number of branch points. This target is irreducible. So we can show the source is irreducible by showing the map is ﬁnite etale´ and the monodromy group acts transitively on the ﬁber. Choose a basepoint p and branch points p1, ... , pn and let γi be simple loops around each of the pi. We have the relation γ1γ2 ··· γn = CURVES NOTES 35

1 since P1 is simply connected. The upshot is that the ﬁbers of π are indexed by sets n T := (γ1,..., γn) ∈ (transpositions in Sd) :

γ1γ2 ··· γn = 1, hγ1,..., γni acts transitively} /Sd,

where the quotient by Sd is via the conjugation action by relabelling the points pi. When we exchange pi and pi+1 the monodromy acts via Mi which sends −1 (γ1,..., γn) 7→ γ1,..., γi−1, γiγi+1γi , γi, γi+2,..., γn .

We obtain maps Mi : T → T.

Lemma 13.5. The span of the Mi act transitively on T Proof. We will do this by giving an algorithm taking some element of T and reducing it down to a nice form. Here is the algorithm: (a) Apply the Mi to move all transpositions not involving 1 to the end. (b) Suppose we have two adjacent transpositions (1a), (1b) for a 6= b. Then, we can send this to (ab)(1a). We then can push (ab) to the end using step 1 (c) We may assume that our sequence is (1a), (1a),..., (1a), other stuff not involving 1 . (d) There must be some element involving a in the stuff not involving −1 1 in order for the action to be transitive. We can then apply Mi to move the element involving a back. So our sequence will be of the form (1a), (1a),..., (1a), (ab) other stuff not involving 1 . with b 6= 1. Moving the 1a past ab, we get (1a), (1a),..., (1a), (1b), (1a), other stuff not involving 1 . and then (1a), (1a),..., (ab), (1a), (1a), other stuff not involving 1 . and moving it back we get (1a), (1a),..., (1a), (1b), (1b), (ab) other stuff not involving 1 . We then go back, and so we may assume we have only two (1a) and other things not involving 1. 36 AARON LANDESMAN

So, we can apply Mi to exchange two sheets in T, and conclude Mg is irreducible. 14. 10/18/19 Deﬁnition 14.1. The gonality of a curve C is the minimal degree of a map C → P1. The ﬁrst question we’ll address today is Question 14.2. What is the gonality of a general curve of genus g?

In other words, we have a map Hurd,g → Mg. The above question would be answered if we knew when this map is dominant. A first guess is that this is dominant when it could be dominant. We have dim Hurd,g = 2d + 2g − 2, since the cover is specified by the branch points, up to finite combinatorial data understood in terms of the monodromy of the branch points. Also, dim Mg = 3g − 3. There is a map

π : Hurd,g → Mg. So one might guess that π is dominant when 2d + 2g − 2 ≥ 3g − 3. This is not quite correct because if we compose with an automorphism of P1, we get the same curve in Mg. Conjecture 14.3. π is dominant if and only if 2d + 2g − 2 − 3 ≥ 3g − 3. Rearranging this, we want g + 2 d ≥ d e. 2 The map π can only be dominant if this inequality holds. Therefore, the g+2 gonality of a general curve of genus g is at least d 2 e. 1 g+2 Lemma 14.4. A general curve of genus g admits a map to P of degree d 2 e. Proof. We’ll start with a curve of genus g and a map of degree d to P1. We’ll then show that for a general deformation of the curve, we will also be able to deform the map along with the curve. We’ll do two cases depending on the parity of g. g+2 1 (1) Suppose g is even. Let d := d 2 e. Start with d copies of P , call them 1 1 X1, ... , Xd mapping to P as the identity on each P . Now, choose three points on X1 and X2 and glue them together. Then, choose three different points on X2 and three points on X3 and glue them. CURVES NOTES 37

In general, choose three points on Xi not already chosen and glue them to 3 points on Xi+1. We then obtain a connected nodal curve of genus 2(d − 1) = g. We just need to check that as we deform this curve, we can also deform the map to P1. The obstructions to lifting 1 ∗ deformations of C to deformations of f lie in Ext ( f ΩP1 → OC). In general f : X → Y obstructions to lifting a deformation of X 1 ∗ • to a deformation of f lie in Ext ( f ΩY, OX) coming from the exact triangle

• ∗ Ω f f ΩY (14.1)

ΩX

This yields an exact sequence (14.2) 0 ∗ 1 • 1 1 ∗ • E ( f ΩY, OX) Ext Ω f , OX Ext (ΩX, OX) Ext ( f ΩY, OX)

where the first term is deformations of the map fixing X and Y, the second is deformations of the map, the third is deformations of X, and the last is obstructions to deformations of the map fixing X and Y. 1 ∗ • If the obstruction Ext ( f ΩY, OX) vanishes, we see we can lift defor- 0 ∗ mations of X to deformations of the map. Note that Ext ( f ΩY, OX) corresponds to deformations (even though it is an Ext0 and not an 1 ∗ Ext ) because the f ΩY is really lying in degree −1 of the cotangent complex. However, 1 ∗ 1 1 ∗ Ext ( f ΩP1 , OC) = Ext (OC(−2), OC) = H ( f OP1 (2)). We’ll prove this by induction on d. When d = 1 it follows from 1 H (OP1 (2)) = 0. Now, let’s address the inductive step. We have an exact sequence

(14.3) 0 1 (2)(−Γ) 1 (2) (2) 0 OP OC∪ΓP OC

where Γ is the set of points where the last copy of P1 meets C, where C is the ﬁrst d − 1 copies of P1. Then, by induction, we see 1 1 H ( 1 (2)) = 0 since H ( (2)) = 0 and 1 (2)(−Γ) = 1 , OC∪ΓP OC OP OP (−1) which has vanishing H1. 38 AARON LANDESMAN

(2) Now we consider the case g is odd. Then we make d copies of P1 with g+2 1 d 2 e = d with g odd. We then glue the last copy of P at two points, and follow the same argument as above. Then, OP1 (2)(−Γ) = OP1 in the very last step, which still has vanishing H1. Question 14.5. Consider the curve C of genus 3 obtained by gluing two genus 1 curves E1 and E2 at two points. Is [C] ∈ M 3 in the closure of the hyperelliptic locus in M3?

To answer this question, we would like a compactification of Hurd,g. Recall 1 that Hurd,g = Mg(P , d). One possible compactification of this space is 1 M g(P , d). We can construct a stable map by contracting this curve to a point and attaching some P1’s mapping isomorphically. The problem is that this is a bad compactification. It doesn’t tell us anything about whether this curve is in the closure of the hyperelliptic locus. 1 The problem is that M g(P , d) → M g is dominant for every d, g. This is a compactification but it doesn’t tell you anything about the answer to this question. Another, quite naive compactification is to look at curves of genus g with a degree d map to P1 not contracting any components. The problem is that branch points approach each other, we obtain singularities of the source curve. The problem is that there is no map from this naive compactification to M g. Next time, we’ll discuss an alternate compactification that deals with this. Of course, if this were in the closure of hyperelliptics, it would have to contract a component. 15. 10/21/19 15.1. Admissible covers. At the end of last time, we were discussing the following question. Question 15.1. Let C be the union of two elliptic curves meeting at two points. Is C in the closure of the hyperelliptic locus in M 3? The key idea in order to construct a good compactification of hyperelliptic curves is similar to the idea for constructing the compactification of Mg,n. The idea is to never let the branch points collide. For example, if y2 = x(x − t)(x − 2t) · g(x). As t → 0, this acquires a cusp. Three of the branch points come together. But you can blow up that point in the total space of the target P1 × ∆, for ∆ the disk. But now, there is a natural question as to what the source curve is. It may be easiest to think about this in a complex analytic CURVES NOTES 39

fashion. We can think of the target P1 as a sphere, and we are pinching a loop γ on the sphere, turning it into a wedge of two spheres. Upstairs, in the cover, the number of loops over γ is the number of orbits of the monodromy element induced by γ. For example, if γ corresponds to (123)(45)(6) there are three loops over γ, one of which maps 3 to 1, one maps 2 to 1 and one maps 1 to 1. Example 15.2. Returning to the example y2 = x(x − t)(x − 2t) · g(x), the monodromy is a product of three transpositions (12)(12)(12) = (12). There- fore, in the resulting family γ has a single preimage which is ramified. So, the resulting cover over the blow up has four ramification points of order 2. The target of this limit is two rational curves meeting at a node. The source of the family limits to a curve of genus g − 1 and a curve of genus 1 mapping to the two rational curves. Both are separately ramified over the preimage of the node. The genus 1 curve is ramified over the three points corresponding to the terms x, x − t, and x − 2t and the node. In general, we have the following description of the limit. The ramification indices are equal along both branches at every node of the source. This compactification is called the space of admissible covers We have explained how the compactification looks set theoretically. It’s not obvious that the curve that fits in to the source is nodal. However, it is true that the source curve is always nodal. We’ll just verify this for a two-component curve, since it illustrates the key ideas.

Example 15.3. Suppose we have a curve with two components C1 and C2. Suppose we have k branch points limiting to a single point in the special ﬁber, corresponding to C1, and suppose there are b − k branch points on component C2 The genus of C1 is k + ∑ (a − 1) g = 1 − d + i i . 1 2

The ∑i(ai − 1) is the contribution from the ramification above the node. Similarly, b − k + ∑ (a − 1) g = 1 − d + i . 2 2 Suppose the fiber over the node has n (possibly non-reduced) points. Then, the genus of the union is at least g1 + g2 + n − 1 where n is the number of points where the two curves meet. There is equality if and only if the curve is nodal. We find b g ≥ g + g + n − 1 = g + g − 1 + d − (a − 1) = 1 − d + . 1 2 1 2 ∑ i 2 40 AARON LANDESMAN

b However, g = 1 − d + 2 using Riemann-Hurwitz, so the equality holds above and the cover must be nodal. Example 15.4. Suppose we take C → B where B is a rational curve with two components meeting at a point. Suppose over the node, C → B maps m : 1 with one preimage over the node. Locally, this looks like Spec C[u, v]/(uv) → Spec C[x, y]/(xy) with x 7→ um, y 7→ vm. We can include C[u, v]/(u, v) Spec C[u, v, t]/(uv − t) (15.1)

Spec C[x, y]/(xy) Spec C[x, y, t]/(xy − tm)

Remark 15.5. To answer our original question, if we have 8 branch points and we take 4 of them to degenerate to a single point, we see that two genus 1 curves meeting at two points does indeed lie in the closure of the hyperelliptic locus. In other words, if we have E1 and E2 meeting at p and q, we can map 1 1 E1 → P by O(p + q) and E2 → P by O(p + q). This yields a map E1 ∪p,q 1 1 E2 → P ∪ P , which is an admissible cover in the closure of the locus of hyperelliptic curves. Warning 15.6. Suppose we want to show some curve is not in the closure of the locus of hyperelliptic curves. We can’t just look to see if there is a map from our given curve to a tree of P1’s. The reason is that, given C, it might have been that you contracted some components of the admissible cover and then obtained C from that. So, one would need to show you cannot add 1-secant and 2-secant P1’s in such a way that there is an admissible covers map.

Example 15.7. Suppose we start with a genus 3 curve E1 ∪ E2 ∪ E3 with E1 meets E2 at a point and E2 meets E3 at a point. We know every genus 3 curve is trigonal. We can take the target B = B1 ∪ B2 ∪ B3 with Bi meeting Bi+1 at a point. We then can take E1 ∪ E2 mapping 2 : 1 to B1 ∪ B2 ramified over the node, and make E2 ∪ E3 mapping to B2 ∪ B3, which is unramified over the node. You can then attach a rational curve R3 meeting E2 at the other fiber over the node mapping to B3. You can also attach a rational curve R2 mapping to B2 meeting E3 at the point over the node not meeting E3. You then attach a rational curve R1 mapping to B1 meeting R2 over the nodal intersection B1 ∩ B2. Altogether E1 ∪ E2 ∪ E3 ∪ R1 ∪ R2 ∪ R3 is an admissible cover of B1 ∪ B2 ∪ B3. CURVES NOTES 41

Remark 15.8. In fact, the space of admissible covers is proper. The proof is essentially that you follow the points coming together, and you blow up just as you would in the process in stable reduction.

Exercise 15.9. Let C1 and C2 be two curves attached at a single point p. When is the union C1 ∪ C2 in the closure of the locus of trigonal curves? The condition should be in terms of Ci being trigonal or hyperelliptic, and whether p is a branch point of various types of the trigonal or hyperelliptic maps. Given necessary and sufficient conditions. 15.2. Existence of maps from curves to projective space. Here is a question that will occupy us for several of the future weeks. This is one of the main themes of the class. Question 15.10. When is there a curve of degree d and genus g in Pr? There are several ways to make this more precise. (1) When is there a smooth curve C of genus g with a nondegenerate map to Pr of degree d? (2) When is there a smooth curve C of genus g with a nondegenerate map to Pr of degree d that is an embedding? (3) When is there a smooth curve C of genus g with a nondegenerate map to Pr of degree d that is generically an embedding? (4) What if C is assumed to be a general curve? Remark 15.11. In this case, the first three questions above turn out to be quite similar. Hopefully, we’ll address the first three questions this week, and then we’ll spend multiple weeks on the last question.

16. 10/23/19 16.1. The uniform position principle. Let C ⊂ Pr be an integral nondegenerate (i.e., C does not lie in a hyperplane) curve. Let H ⊂ Pr be a hyperplane transverse to C. We obtain a map {(p, H) : p ∈ C ∩ H} (16.1)

(Pr)∨ C The left map is a degree d cover. Lemma 16.1 (Uniform position principle). The monodromy group of the left cover is Sd. 42 AARON LANDESMAN

Remark 16.2. This amounts to saying that there is no way to distinguish any subset of points of C ∩ H. For example, suppose r ≥ 3. Then, the general hyperplane section of C has no three collinear points. Proof. We can prove this claim in two steps. We will show the monodromy is doubly transitive and that it contains a transposition. To show the monodromy is doubly transitive, it is equivalent to show the incidence correspondence

{(p, q, H) : p, q ∈ C ∩ H, p 6= q} (16.2) ∨ (Pr) C × C − ∆.

is irreducible. We know C × C − ∆ is irreducible. The right projection is a Pr−2 bundle (this may use characteristic 0). To see the monodromy contains a simple transposition, we argue as follows. Let H be a hyperplane simply tangent to the curve at a single point. (This exists using that we are in characteristic 0, so not every tangent is a bitangent, and hence using a Bertini argument via projection from the tangent line, there is a hyperplane containing that tangent line which meets the curve elsewhere transversely.) Then, as we move in a small analytic disc around that point, the monodromy acts as a two cycle switching the two branches coming together at that point. 16.2. Clifford’s theorem. Question 16.3. When does there exist a map from a smooth genus g curve C → Pr of degree d? Let C be a curve of genus g and L be a line bundle on C. Question 16.4. What are the possible values of h0(L )? From Riemann roch, if d ≤ −1, then h0(L ) = 0. If d ≥ 2g − 1, we know what h0(L ) is by Riemann Roch. But there is some intermediate range with several possibilities. Namely, there is a parallelogram with vertices (d = −1, h0(L ) = 0), (g − 1, 0), (g − 1, g), (2g − 1, g) such that for d in the range −1 ≤ d ≤ 2g − 1, then h0(L ) lies in the above parallelogram. Clifford’s theorem says that in fact, h0(L ) actually lies in the lower half of the above parallelogram (lying on or above the line joining OC and KC. 0 d Theorem 16.5 (Clifford). For 0 ≤ d ≤ 2g − 2, then h (L ) ≤ 1 + 2 . Equality can hold for any d. Further, equality holds if and only if L = OC, KC, or C is CURVES NOTES 43

d hyperelliptic and L = 2 O(1). That is, in the last case, the resulting map is 2 : 1 onto a rational curve. Proof. Certainly, equality holds in the latter three cases. So, we just need to verify the inequality and that when equality holds, one of the conditions is satisﬁed. Lemma 16.6. For any effective line bundles L , M , then h0(L ⊗ M ) ≥ h0(L ) + h0(M ) − 1.

Proof. Let p1, ... , ph0(L )−1, q1, ... , qh0(M )−1 be general points on C. We will show there is a nonzero section of L ⊗ M vanishing at the above points. This will imply the lemma. But indeed, there is a section s of L vanishing at the pi and there is a section t of M vanishing at the qi. The s ⊗ t vanishes at the pi and qj. Note that if L or K ⊗ L ∨ are not effective, then Clifford’s theorem holds. Otherwise, we have h0(L ) + h0(K ⊗ L ∨) ≤ h0(K) + 1 = g + 1. Next, using Riemann Roch, h0(L ) − h0(K ⊗ L ∨) = d + 1 − g. Adding these two equations, 2 · h0(L ) ≤ d + 2. This is the statement of Clifford’s theorem. It only remains to analyze the equality case. We can only have equality hold in this when equality holds in Lemma 16.6. In other words, tracing through the argument of Lemma 16.6, we will have equality exactly when ∨ every section of KC is a product of a section of L and a section of K ⊗ L . Now, consider the map C → Pg−1. Choose a hyperplane which will meet the curve at 2g − 2 points. The above statement implies (using that every section is a pure tensor) implies that every hyperplane section can be parti- tioned so that one of the sets lie in the linear series for L and the complement lies in the linear series for K ⊗ L ∨. Therefore, the points in a divisor in L corresponds to a degree d divisor imposing h0(K) − h0(K ⊗ L ∨) conditions on hyperplanes (since h0(K ⊗ L ∨) is the dimension of the space of hyperplanes vanishing on our divisor). The above equality holding tells us this 0 0 ∨ 0 d dimension is h (K) − h (K ⊗ L ) = h (L ) − 1 = 2 . We therefore have d points imposing d/2 conditions on hyperplanes. This contradicts the uniform position principal if f were not an embedding, because any collection of d points will impose the same number of conditions on hyperplanes (as 44 AARON LANDESMAN

otherwise they would lie in a lower dimensional subspace). This generalizes Remark 16.2. This implies that f the canonical map is not an embedding, and hence the curve is hyperelliptic. Recall the reason for this: we are saying 0 that for all p there exists q with h (KC(−p − q)) = g − 1. By Riemann Roch, 0 h (OC(p + q)) = 2, so C is hyperelliptic. 17. 10/25/19 17.1. Castelnuovo’s bound: Existence of curves with birational maps to projective space. Question 17.1. When does there exist a curve of genus g with a degree d map to Pr that is a birational embedding? More speciﬁcally, for ﬁxed d and r, what is the maximum possible g? 0 The answer is known as Castelnuovo’s bound. The idea is that h (OC) = 1 0 and h (OC(k)) = dk + 1 − g if k 0, where OC(k) means the pullback of 0 0 OPr (k). We would like to bound the difference h (OC(k)) − h (OC(k − 1)). To bound the difference, we have the exact sequence of sheaves

(17.1) 0 OC(k − 1) OC(k) OC∩H(k) 0 for H ⊂ Pr a general hyperplane. We obtain a long exact sequence (17.2) 0 0 0 0 H (OC(k − 1)) H (OC(k)) H (OC∩H(k)) ···

0 0 From the map H (OPr (k)) → H (OC(k)) we obtain,

0 0 H (OPr (k)) H (OH(k)) (17.3)

0 0 0 H (OC(k − 1)) H (OC(k)) H (OC∩H(k)) and so 0 0 0 0 h (OC(k)) − h (OC(k − 1)) ≥ dim im H (OPr (k)) → H (OC∩H(k)) . Therefore, we similarly obtain 0 0 0 0 h (OC(k)) − h (OC(k − 1)) ≥ dim im H (OH(k)) → H (OC∩H(k)) .

r Lemma 17.2. Let p1, ... , pd ∈ P be in linear general position. Then, they impose at least min(d, kr + 1) conditions on polynomials of degree k. CURVES NOTES 45

Proof. It suffices to consider the case d = kr + 1. This is sufficient because if d < kr + 1, we can add in points until we obtain kr + 1 points, and then we obtain kr + 1 conditions, and so with only d points, we must have d conditions. Similarly, if d > kr + 1, then if the first d − 1 points impose kr + 1 conditions, d points must also impose that many conditions. So, we will now prove this lemma for d = kr + 1. It suffices to find a polynomial of degree k vanishing at p1,..., pd−1 and not at pd. We have points p1, ... , pr, pr+1, ... , p2r, ... , p(k−1)r+1, pkr = pd−1, pkr+1. So, p1, ... , pr span a hyperplane, and pr+1, ... , p2r span a hyperplane, and in general pjr+1, ... , pjr span a hyperplane Hj. Further, pd does not lie in H0 ··· Hk−1. So, H0 ··· Hk−1 vanishes at p1,..., pd−1 but not pd.

Remark 17.3. The above lemma is actually sharp. If we had d points p1, ... , pd 0 on a curve C, the number of conditions p1, ... , pd impose is at most h (OC(k)). 0 If C is a rational normal curve, then h (OC(k)) = kr + 1. So in fact it imposes exactly this many conditions by the lemma, and the lemma is sharp.

d−1 Proposition 17.4 (Castelnuovo’s bound). For m := b r−1 c, writing d = (r − 1)m + ε, for 1 ≤ ε ≤ r − 1, we have

m g ≤ (r − 1) + mε. 2

In particular the growth is on the order of d2/(2r − 2).

d−1 0 Proof. Write m := b r−1 c. Note that if k ≤ m, kr + 1 ≤ d. We know h (OC) = 1 and so by the lemma, Lemma 17.2 (which applies since the points are in linear general position by the uniform position principal)

0 h (OC(1)) ≥ 1 + r.

(using the lemma with r − 1 in place of r, as we are taking the hyperplane H which is a Pr−1)

0 h (OC(2)) ≥ 1 + r + (2r − 1) 0 h (OC(3)) ≥ 1 + r + (2r − 1) + (3r − 2) m 0 m + 1 h (OC(m)) ≥ ∑(ir − i + 1) = (r − 1) + m + 1. i=0 2 46 AARON LANDESMAN

Then, m + 1 h0(O (m + 1)) ≥ (r − 1) + m + 1 + d C 2 m + 1 h0(O (k)) ≥ (r − 1) + m + 1 + d(k − m). C 2 0 On the other hand, h (OC(k)) = kd + 1 − g. Putting these together, m + 1 kd + 1 − g ≥ (r − 1) + m + 1 + d(k − m). 2 for large k. Rearranging, we ﬁnd m + 1 g ≤ md − (r − 1) − m. 2 Writing d = (r − 1) + ε for 1 ≤ ε ≤ r − 1, we ﬁnd m + 1 g ≤ md − (r − 1) − m 2 m + 1 = m ((r − 1) m + ε) − (r − 1) − m 2 m = (r − 1) + mε. 2 Remark 17.5. In fact, Castelnuovo’s bound is sharp. For every d and r, there m r is a curve of this genus (r − 1)( 2 ) + mε in P . Example 17.6. Say r = 3. Then, this is saying ( m(m − 1) if d = 2m + 1 g ≤ m2 if 2m + 2 is achieved for C of type (m, m + 1) or (m + 1, m + 1) on a smooth quadric P1 × P1. Remark 17.7. More generally, equality is achieved for curves of type (m + 1) · H − (r − 2 − ε) · F on a rational normal surface scroll. Remark 17.8. In fact, these extremal curves on rational normal curves are in general smooth, so the maps C → Pr are even an embedding. Remark 17.9. The curves achieving Castelnuovo’s bound are very much not general in moduli because usually the gonality will be around half the genus, which is not true of the Castelnuovo curves, whose gonality is about m and genus is about m2. CURVES NOTES 47

17.2. Existence of maps from curves to projective space. Let C be a general curve of genus g. Question 17.10. Does there exist a nondegenerate map C → Pr of degree d? We’ll spend quite a bit of time working on this question. Given f : C → Pr, as we deform the curve C, we’d like to understand whether we can deform the map f with C. Obstructions to lifting deformations of C to deformations of f lie in 1 ∗ 1 ∗ Ext ( f ΩPr , OC) = H ( f TPr ). 1 ∗ So, we might expect the answer to be yes if and only if H ( f TPr ) = 0.

18. 10/28/19 Recall our goal last time was the following question: Question 18.1. Given a general curve C of genus g, does there exist a degree d nondegenerate map C → Pr? We saw last time that obstructions to lifting deformations of C to deforma- 1 ∗ tions of f lie in H ( f TPr ). Therefore, we would expect deformations to lift 1 ∗ when H ( f TPr ) = 0. We have an Euler exact sequence

⊕(r+1) (18.1) 0 OPr OPr (1) TPr 0

This implies c1(TPr ) = OPr (r + 1). We have ∗ χ( f TPr ) = (r + 1)d − r(g − 1). A naive guess might be that there exists a nondegenerate f : C → Pr with 1 ∗ H ( f TPr ) = 0 if and only if (r + 1)d − rg + r ≥ 0. 0 0 ∗ The map H (TPr ) ,→ H ( f TPr ) is an inclusion using that C is nondegenerate and there are no tangent vector fields fixing r + 2 points, using the Lefschetz trace formula which says any tangent field fixes at most r + 1 points, which is the first Chern class of TPr . The source has dimension 2 0 ∗ 2 (r + 1) − 1 using the Euler exact sequence Therefore, h ( f TPr ) ≥ r + 2r. A better guess is that there exists a map f : C → Pr which is nondegenerate 1 ∗ 2 of degree d with H ( f TPr ) = 0 if and only if (r + 1)d − rg + r ≥ r + 2r. This is saying (r + 1)d − rg − r(r + 1) ≥ 0. This prompts the following definition: Definition 18.2. Define ρ(d, g, r) := (r + 1)d − rg − r(r + 1). 48 AARON LANDESMAN

g lower bound on d for ρ(d, g, 2) ≥ 0 0 2 1 3 2 4 3 4 4 5 5 6 6 6

Table 5: When r = 2, what is the lower bound on g as a function of d

So, we guess there exists a degree d map when ρ(d, g, r) ≥ 0. Theorem 18.3 (Brill-Noether theorem). There exists a nondegenerate map f : C → Pr if and only if ρ(d, g, r) ≥ 0. To prove the if direction (if ρ ≥ 0 there is a nondegenerate map) it suffices to construct for every (d, g, r) with ρ(d, g, r) ≥ 0 a map f : C → Pr with 1 ∗ H ( f TPr ) = 0. First, let’s see what the bounds are when r = 2. If r = 2 and g = 0, then d ≥ 2. If g = 1, we find d ≥ 3. If g = 2, d ≥ 4. When g = 3, we find again that d ≥ 4. When g = 4, we find d ≥ 5. When g = 5, d ≥ 6 and when g = 6, d ≥ 6. To try and prove the Brill Noether theorem when r = 2, we need three sort of inductive moves. We would want to be able to go from (d, g, r) to (d + 1, g, r) to (d + 1, g + 1, r) and to (d + 1, g + 2, r) In other words, the lattice of allowable pairs (meaning ρ(d, g, r) ≥ 0) is generated by the rational normal curve and the three moves described above. So, our first goal is the following: ∗ r r Goal 18.4. Understand f TPr when f : P → P is a rational normal curve. We will do this by examining certain naturally defined subbundles of the tangent bundle. 18.1. The tangent pointing bundle. The setup is the following: Suppose we have a curve f : C ⊂ Pr and a linear space Λ with Λ ∩ r N f (Csing) = ∅. We have a projection map πΛ : P → P (for some N depending on the dimension of Λ) defined away from Λ. We’ll consider the vertical tangent space of the map πΛ. We can consider ∗ f TπΛ , the restriction of the vertical tangent space. You should think of the tangent pointing bundle as that on each point of the curve which points toward Λ. At every point of the curve, it is the tangent vectors such that if you were to follow that tangent vector in a line, you would hit Λ. CURVES NOTES 49

∗ Definition 18.5. The tangent pointing bundle TC→Λ ⊂ f TPr is the unique ∗ ∗ r subbundle whose restriction to C − Λ is f TπΛ ⊂ f TP . This existence of the extension follows from the valuative criterion for properness because given a vector bundle E → C we obtain a Grassmannian bundle G of rank r subbundles of E (meaning the quotient of the subbundle is a bundle) and the valuative criterion of properness says that a generic section of G → C extends to a section over all regular points of C. In particular, since ∗ ∗ r Λ did not intersect f (Csing) we can extend f TπΛ ⊂ f TP uniquely to C. Example 18.6. Take C ⊂ PV and Λ = p ∈/ C. Let H be a complementary hyperplane to p, meeting the curve in some points qi. Then, the underlying vector space V defining PV can be written as V = hHi ⊕ hpi. We can then construct the Euler field relative to p and H. That is, there is a Gm action on PV induced by

Gm × V → V λ(x, y) 7→ x + λy The differential at λ = 1 defines a vector field on PV called the Euler field associated to (p, H). The Euler field vanishes only on p ∪ H. This is because the fixed points of the action are precisely when x = 0 or y = 0. This yields a section of TC→Λ vanishing along C ∩ H. The upshot of this is r that TC→p ' OC(1), the pullback of O(1) from P .

19. 10/30/19 Last time, we were discussing pointing bundles, and we saw that if Λ = p, which is a point not on the curve, the pointing bundle to p is OC(1). Now, consider the case Λ = p ∈ f (C) and p is not a branch point of the map f and that p is the image of a unique point. Consider the Euler field for (p, H) which vanishes at f (C) ∩ H and at p. The Euler field also vanishes only to order 1 at p. Let us now calculate this order at p. To see this has a simple 0 at p, we may choose our coordinates so that H is the hyperplane at ∞ and p is the origin. The Euler field is then the r vector field on A that has value v at v. The upshot is that TC→p ' OC(1)(p) because we are looking for sections vanishing exactly along the intersection of f −1(H) and at p. Exercise 19.1. Let p ∈ f (C) the image of a unique point such that p is not a flex (i.e., the intersection of the tangent line to C at p with C has multiplicity exactly 2 at p) of f . Assume also that the map is unramified over p. Let L be the tangent line to f (C) at p, and assume also that L does not meet C at other points. Describe the bundle TC→L. 50 AARON LANDESMAN

∗ 1 Example 19.2. Now, we return to the goal of describing f TPr for f : P → Pr a rational normal curve. Let f : C = P1 → Pr be a rational normal curve. ⊕r Choose points p1, ... , pr. Consider i=1TC→pi . At a general point on the curve, the secant lines to each of the pi are independent, so at the generic point r ∗ ⊕ → r i=1TC→pi f TP . The first Chern classes of both of these bundles is (r + 1)r. The source bundle is a sum of r copies of O(1)(pi), so each has Chern class r + 1. The cokernel is therefore 0. This implies ∗ ⊕r f TPr ' O(r + 1) . 1 ∗ Corollary 19.3. In particular, H ( f TPr ) = 0. This is the base case of our Brill Noether theorem. The above description ∗ of f TPr will enable us to make an inductive argument. Let’s now give another example of using the tangent pointing bundle to describe the pullback of the tangent bundle. 1 r Example 19.4. Let f : L = P → P be a line. Choose r − 1 points p1, ... , pr−1 ⊕ Lr−1 ' general not on the curve. We have a map TL i=1 TL→pi Then, TL→pi OL(1). Also, TL ' OP1 (2). There is a map r−1 ∗ ⊕ ⊕ → r TL i=1 TL→pi f TP . Both have first Chern class r + 1, so this map is an isomorphism. ∗ r−1 ∗ In particular f TPr ' O(2) ⊕ O(1) , we have TL ⊂ f TPr . 19.1. Proving the existence part of the Brill Noether theorem. We want r 1 ∗ that for ρ ≥ 0, there exists f : C → P with H ( f TPr ) = 0. The base case is (r, 0, r) which we have shown in Example 19.2. We have three inductive steps (1) (d, g) 7→ (d + 1, g) (2) (d, g) 7→ (d + 1, g + 1) (3) (d, g) 7→ (d + r, g + r + 1). Let’s now tackle the first inductive step (d, g) 7→ (d + 1, g). Consider a curve C of degree d and genus g, and let D be the curve obtained by joining 1 ∗ C and a line L at a point p. By induction, H ( f |CTPr ) = 0. We have an exact sequence ∗ ∗ ∗ (19.1) 0 f |LTPr (−p) f TPr f |CTPr 0

1 ∗ 1 ∗ We know H ( f |CTPr ) = 0 by induction. Further, H ( f |LTPr (−p)) = 0 ∗ ⊕r−1 using that f |LTPr ' O(2) ⊕ O(1) . Here, the exact sequence is coming CURVES NOTES 51

from the fact that the kernel of the natural restriction map from tangent vectors on Pr restricted to D to tangent vectors on Pr restricted to C is tangent r 1 ∗ vectors on P restricted to L vanishing at p. This means H ( f TPr ) = 0. Next, let’s consider the second inductive step (d, g) 7→ (d + 1, g + 1). Now, we consider the curve D obtained by choosing two points on C and a 2-secant line joining p and q. We have an exact sequence

∗ ∗ ∗ (19.2) 0 f |LTPr (−p − q) f TPr f |CTPr 0

The H1 of the last term vanishes again by induction, and the H1 of the ﬁrst ∗ ⊕r−1 1 term vanishes again because f |LTPr ' O(2) ⊕ O(1) . Therefore, H of the middle term vanishes as well. We ﬁnally consider the third inductive step (d, g) 7→ (d + r, g + r + 1). Now, given a curve C, pick a collection of r + 2 general points Γ on C. Recall that the automorphism group of Pr acts transitively on sets of r + 2 points in general linear position. From this, we can construct a rational normal curve R passing through Γ. Now, we have a sequence

∗ ∗ ∗ (19.3) 0 f |RTPr (−Γ) f TPr f |CTPr 0

Exercise 19.5. Verify that this is indeed exact, where Γ is really viewed as a divisor on C.

1 ∗ We know H of the last term vanishes by induction. We know f |RTPr ' ⊕r ∗ ⊕r O(r + 1) . Therefore, f |RTPr (−Γ) ' O(r + 1) . This ﬁnishes the proof of the existence part of the Brill Noether theorem. Remark 19.6. The original proof was due to Kleiman and Laksov who do an intersection theory computation to compute nonemptiness. 19.2. A variant of the Brill Noether theorem involving marked points. Remark 19.7. One nice aspect about the above approach allows us to give a variant of the existence part of the Brill Noether theorem with marked points.

Let (C, p1,..., pn) be a general curve of genus g with n marked points. Let r q1,..., qn ∈ P be general points. Question 19.8. Does there exist a nondegenerate map of degree d f : C → Pr with f (pi) = qi for all i? 52 AARON LANDESMAN

r Given f : C → P with f (pi) = qi, we can ask what the obstructions are to lifting deformations of (C, p1,..., pn, q1,..., qn) to deformations of f ? These lie in 1 ∗ 1 ∗ r r Ext f ΩP , Ip1+···+pn ' H ( f TP (−p1 − · · · − pn) . So, just as we can study the Brill Noether theorem with marked points, by studying the cohomology of the pullback of the tangent bundle with a marked twist.

20. 11/1/19 Recall that last time we were trying to answer the question: r Question 20.1. Does there exist a map f : C → P with f (pi) = qi, for r (C, p1,..., pn) ∈ M g,n general and q1,..., qn ∈ P general? 1 ∗ Last time, we saw the answer is yes if H ( f TPr (−p1 − · · · − pn)) = 0. Let’s try and show this! Deﬁnition 20.2. Let E on C be a vector bundle. We say E satisﬁes interpolation if both (1) H1(C, E ) = 0 and (2) for any degree d > 0, there exists an effective Cartier divisor, D ⊂ C, of degree d such that H0(E (−D)) = 0 or H1(E (−D)) = 0.

Remark 20.3. The Euler characteristic tells us the difference of h0(E (−D)) − h1(E (−D)). Satisfying interpolation is essentially telling us that we can predict the cohomology of twists of E from the Euler characteristic. ∗ Lemma 20.4. Suppose f TPr satisfies interpolation. Then, for general points p1,..., pn on C, we have 1 ∗ H ( f TPr (−p1 − · · · − pn)) = 0 if and only if ∗ (r + 1)d − rg + r − rn = χ( f TPr (−p1 − · · · − pn)) ≥ 0. Proof. This follows from the definition of interpolation r Corollary 20.5. A map f : C → P with f (pi) = qi as in Question 20.1 exists if r ∗ there exists f : C → P for which f TPr satisfies interpolation and (r + 1)d − rg + r n ≤ b c. r CURVES NOTES 53

Remark 20.6. We will later see the “if” in the above statement can be up- graded to an “if and only if.” Proof. This follows from Lemma 20.4. Suppose ρ ≥ 0. We already know there exists a map f : C → Pr. We would ∗ like to say further that there exists an f such that f TPr satisfies interpolation. ∗ Corollary 20.7. If we knew f TPr this satisfies interpolation, the answer to Ques- tion 20.1 is yes if ρ ≥ 0 and (r + 1)d − rg + r n ≤ b c. r Proof. This is essentially a restatement of Corollary 20.5. Remark 20.8. Hopefully the above corollary justifies the name “interpolation.” The interpolation property controls the ability of maps of curves to interpolate through points. Here is now a generalization of the above definition of interpolation. Definition 20.9. Let E on C be a vector bundle of rank r. Let V ⊂ H0(E ) be a space of sections. We say (E , V) satisfies interpolation if (1) H1(C, E ) = 0 (2) For any non-negative degree d, there exists an effective Cartier divisor D of degree d with dim V ∩ h0 (E (−D)) = max {0, dim V − rd} .

Remark 20.10. Observe that by definition (and an easy computation of the Euler characteristic), E satisfies interpolation if and only if E , H0(C, E ) satisfies interpolation. ∗ We’d now like to show f TPr satisfies interpolation. We’ve already constructed f : C → Pr inductively last time. We’ll see inductively for the same ∗ curves we’ve constructed last time that f TPr already satisfies interpolation. We will first check the base case that f is a rational normal curve, and then move on to the three inductive steps. 20.1. Base case. Let f : C = P1 → Pr be a rational normal curve. We saw ∗ ⊕r last time that f TPr ' OP1 (r + 1) . This vector bundle satisfies interpolation, as we can see from the following lemma. In general, we have the following for when vector bundle on P1 satisfy interpolation: 1 Lemma 20.11. A vector bundle E ' ⊕iOP1 (ai) on P satisfies interpolation if and only if |ai − aj| ≤ 1 for all i, j and ai ≥ −1. 54 AARON LANDESMAN

Proof. To satisfy interpolation, we see that for any given integer n ≥ 0, the set of integers {ai − n} ≤ −1 or {ai − n} ≥ −1 and additionally ai ≥ −1 for all i (corresponding to the h1(E ) = 0 condition in the deﬁnition of interpolation. This implies the ai are within 1 of each other.

20.2. The setup for checking interpolation on reducible curves. Here is the general setup. Suppose we have E a vector bundle on a reducible curve X ∪Γ Y. We’ll think about this as being a reducible curve where X is a “simple” piece. This has no mathematical meaning, but there will be an asymmetry in X and Y, and so will be helpful conceptually. Typically X will be something like a line or a rational normal curve. We’ll assume we know what Cartier divisor to take on X to rig interpolation to hold. To this end, let ∆ ⊂ X − Γ 0 be an effective Cartier divisor such that H (E |X(−∆ − Γ)) = 0. Let

0 0 0 0 evX : H (E |X) → H (E |Γ) evY : H (E |Y) → H (E |Γ) denote the evaluation maps.

Lemma 20.12. In the above setup, E satisﬁes interpolation if

−1 0 0 V := evY evX H (E |X (−∆)) ⊂ H (E |Y)

satisﬁes interpolation and has dimension

χ (E (−∆)) = χ (E |Y) + χ (E |X (−∆ − Γ)) Proof. This follows from considering the exact sequence of sheaves

(20.1) 0 E |X(−Γ) E E |Y 0

and twisting appropriately by ∆ and other divisors on Y. The space V is isomorphic to

0 0 im H (E (−∆)) → H (E |Y .

We want to say this space of sections has the expected dimension and satisﬁes interpolation. So here we are taking the divisor we are considering and splitting it onto ∆ on X and the remainder of the divisor on Y. Exercise 20.13. Fill in the details of the above proof.

CURVES NOTES 55

20.3. First inductive step. In this step, we consider the inductive step (d, g) 7→ (d + 1, g). For this inductive step, we considered a nodal curve given by a smooth curve C meeting a line L at a point. We saw last time ∗ ⊕r−1 f |LTPr ' O(2) ⊕ O(1) .

Here, O(2) is canonically TL. We want to apply Lemma 20.12. We would 0 like to take ∆ in Lemma 20.12 to be two points so that H (E |X (−∆ − Γ) = 0. Here Γ is one point so we want ∆ to be (at least) two points. Observe that ∗ f |L(−∆) has a unique section whose value at p := Γ is TL|p. Let q be a point on L distinct from p. So, by Lemma 20.12, it suffices to show n 0 ∗ o σ ∈ H ( f |CTPr ) : σ|p ∈ TC→q|p satisfies interpolation and has dimension ∗ ∗ χ ( f |CTPr ) − χ (E |X (−∆ − Γ)) = χ ( f |CTPr ) − (r − 1). We’ll finish verifying this step next time.

21. 11/4/19 Recall that we’re in the middle of proving the Brill Noether existence theorem with marked points. Recall that a vector bundle E on a curve satisfies interpolation if (1) h1(E) = 0 (2) For all n ≥ 0, there exists an effective divisor D of degree n such that either h0(E(−D)) = 0 or h1(E(−D)) = 0. Note that in the above definition h0(E(−D)) = 0 when χ(E(−D)) ≤ 0 while h1(E(−D)) = 0 when χ(E(−D)) ≥ 0. Theorem 21.1. For ρ(d, g, r) = (r + 1)d − rg − r(r + 1) ≥ 0, there exists a stable r ∗ map f : C → P , and C a curve of genus g such that f TPr satisfies interpolation. We were proving this using three inductive steps. (1) (d, g) 7→ (d + 1, g) (2) (d, g) 7→ (d + 1, g + 1) (3) (d, g) 7→ (d + r, d + r + 1) The first step was obtained by joining C to a line at a point. The second was obtained by joining C to a line at two points. The third was obtained by joining a curve C to a rational normal curve meeting C in r + 2 points. Starting at rational normal curves, these three degenerations encompass the full Brill-Noether general range. 56 AARON LANDESMAN

Last time, we showed that in the first degeneration, in order to prove ∗ f TPr satisfies interpolation, it suffices to show

n 0 ∗ o σ ∈ H (C, f |CTPr ) : σ|p ∈ TC→q|p

satisfies interpolation. We’ll first make a new vector bundle on the curve (in terms of elementary modifications) that takes into account this data.

21.1. Modiﬁcations of vector bundles. Let X be a scheme, let E be a vector bundle on X and F ⊂ E a subbundle (so that the quotient is a bundle as well). Let D ⊂ X an effective Cartier divisor on X.

Deﬁnition 21.2. Deﬁne E[D → F] := ker (E → (E/F) |D) . In general this can be seen to be a vector bundle using that D has codimension 1. Remark 21.3. Observe

n 0 ∗ o 0 ∗ σ ∈ H (C, f |CTPr ) : σ|p ∈ TC→q|p = H (C, f |CTPr p → TC→q ).

21.2. Interpolation for the second degeneration. We’ll next try to understand what we’ll want for the second degeneration. We’ll try to understand ∗ interpolation of f TPr to C ∪ L where C meets L at the points p and q. We ∗ ⊕r−1 found in a previous class that f |LTPr = O(2) ⊕ O(1) . When we twist down by ∆, which is one point on the line L distinct from p and q, we get

∗ ⊕r−1 f |LTPr (−∆) = O(1) ⊕ O .

We can naturally identify the O(q) above with TL (−∆). We want to understand the restriction 0 ∗ r r H ( f |LTPr (−∆)) → TpP ⊕ TqP . We want to see whether this map is injective. Inside the source, we have TL(−∆) ' O(1) → TpL ⊕ TqL is surjective on global section. We now want to see what happens on the evaluations of elements of the sections of copies of O. We now can use this point ∆ to break this non-local behavior. We can limit ∆ to the point p. Then,

0 ∗ r r r im H ( f |LTPr (−∆)) → TqP ⊕ TpP = TqP ⊕ TpL.

∗ Then, for the degeneration we want to understand f |CTPr [p → q]. CURVES NOTES 57

21.3. Interpolation for the third degeneration. In the third case, we have a union C ∪ R for C a curve and R a rational normal curve meeting at r + 2 ∗ ⊕r points Γ. In this case, f |RTPr ' O (r + 1) . This yields an isomorphism 0 ∗ r H ( f |RTPr ) ' TΓP .

as can be seen using that on each factor the map OPr (r + 1) → OP1 (r + 1)|Γ induces an isomorphism on global sections. Note that here we did not need to twist down by any divisor on the rational normal curve. We now want to apply Lemma 20.12 to each of the three degenerations above. ∗ 21.4. Proof that f TPr satisﬁes interpolation. For every (d, g, r) with ρ(d, g, r) ≥ ∗ 0, we want to show f TPr satisﬁes interpolation. We proceed in the following steps. (1) First, apply the inverse of the third degeneration until we have d ≥ g + r. To see why this is possible, if d < g + r, then by Clifford’s theorem, r ≤ d/2. That is, d > 2r and g ≥ r + 1. So, when we apply this degeneration, we remain in the Brill Noether range ρ(d, g, r) ≥ 0.

Remark 21.4. One may think of this application of Clifford’s theorem as saying that the smallest special curve is a canonical curve. (2) We next induct on the statement that if f : C → Pr is embedded by a ∗ general line bundle of degree d ≥ g + r, then f TPr [p1 + ··· + pn → q] satisﬁes interpolation whenever d ≥ (r − 1) n + 1. Remark 21.5. We actually want the case of this when n = 0, but we’ll induct on this stronger statement. The inequality d ≥ (r − 1) n + 1 can be understood in terms of the exact sequence (21.1) ∗ ∗ 0 TC→q f TPr [p1 + ··· + pn → q] πq ◦ f TPr−1 (q − p1 − · · · − pn) 0

where πq denotes the projection from q. The inequality d ≥ (r − 1) n + 1 is equivalent to ∗ χ ( f TPr [p1 + ··· + pn → q]) χ T → ≤ b c. C q r

Exercise 21.6. Show the above inequality is necessary for satisfying interpolation. 58 AARON LANDESMAN

The above exact sequence gives us a way to do induction on r. (3) Prove the following lemma: Lemma 21.7. Given an exact sequence

(21.2) 0 L E Q 0

with (a) L nonspecial, i.e., H0(L) = 0, (b) Q satisfies interpolation (c) we have χ(Q) χ(L) − 1 ≤ ≤ χ(L) + 1 rk(Q) Then E satisfies interpolation. Sketch by picture. We need to show that if we twist down by effective divisors, we have at most one nonzero cohomology group. Look at the number line of degrees. There are three interesting values. χ(L) − 1, χ(L) and χ(L) + 1. The assumption is saying that χ(Q)/ rk Q falls somewhere in the range between χ(L) − 1 and χ(L) + 1. From the perspective of L, we’d like to understand where the cohomology of L goes from H0(L(−D)) = 0 to H1(L(−D)) = 0. We can deter- mine this transition by looking at whether χ(L) is positive or negative. So, whenever, χ(L) ≤ 0 we have h1(L(−D)) = 0 and when χ(L) ≥ 0, we have h0(L(−D)) = 0. Similarly by assumption for Q, we have h1(Q(−D)) = 0 whenever the degree of D is smaller than χ(Q)/ rk Q and h0(Q(−D)) = 0 whenever the degree of D is bigger than χ(Q)/ rk Q. Therefore, h1(E(−D)) = 0 whenever both h1(Q(−D)) = h1(L(−D)) = 0 and h0(E(−D)) = 0 whenever both h0(Q(−D)) = h0(L(−D)) = 0. By the assumptions we have made, we get that either h0(E(−D)) = 0 or h1(E(−D)) = 0. In our case, we find d ≤ (r − 1)n + 2r − 1 via elementary numerical ∗ calculations. If this is true, f TPr [p1 + ··· + pn → q] satisfies induction on r. So, we will be done when this inequality holds. (4) We may assume that d ≥ (r − 1) n + 2r. Now, we use our strategy of pulling off a 1-secant or 2-secant line. Let’s consider the 1-secant line ∗ 0 case. We are considering interpolation for f TPr [p1 + ··· + pn → q][p → q ] where p is the point of intersection and q0 is another point on the line. 0 Now degenerate by limiting q → q. Call p := pn+1. Then we are ∗ considering f TPr [p1 + ··· + pn+1 → q]. We replace d by d − 1 and n by n + 1 which still satisfy d ≥ (r − 1)n + 1. When we perform CURVES NOTES 59

this step enough, we will satisfy the inequality d ≤ (r − 1)n + 2r − 1. After reaching this bound, we will eventually reach r = 1.

22. 11/6/19 Today, and for the next several classes, we’ll discuss Brill-Noether nonexis- tence. If ρ(d, g, r) = (r + 1) d − rg − r (r + 1) < 0. Then, we’d like to show there does not exist a nondegenerate map f : C → Pr of degree d for C a general curve of genus g. The idea is to specialize to a singular curve and follow what the linear system V ⊂ H0(C, L) passes to in the specialization. This can be quite subtle, as we now see. Example 22.1. Consider a pencil of quadrics in P3 specializing to a quadric cone. We now intersect this family of quadrics with a smooth cubic surface through the cone point p in the special fiber. Our family of curves is Qt ∩ S, for Qt the pencil. A general curve is a smooth canonical curve of genus 4. However, the special fiber is a nodal genus 4 curve, with a node at p. This can be seen to be singular using the Jacobian criterion. Exercise 22.2. Show that the divisor associated to the ruling does not specialize to a Cartier divisor on the singular curve in the pencil. [This is the standard first example of a non-Cartier divisor, using that the quadric cone is not normal.] Show that if you blow up the point p in the total space of the family, so the reducible special fiber is the union of a hyperelliptic curve of genus 3 and a P1 meeting that curve at two points. The map of degree 3 on the general fiber then extends to the special fiber, and induces the hyperelliptic map on the genus 3 curve, and an isomorphism on the P1. 22.1. Compact type curves. The problem is that the line bundle in fact does 0 P1 not extend. This is related to the fact that PicQt∩S/P1 is not proper over . 0 Particularly, the fiber PicQ0∩S/ Spec C is not proper. There is a fix to the above example, which is to focus on compact type curves. Definition 22.3. A connected nodal curve C is compact type if the dual graph (of components of the curve C) is a tree. Recall that the dual graph of a curve is a graph whose vertices correspond to components of C and edges correspond to nodes. ' If C is compact type, then PicC/C ∏ components Ci PicCi/C For example, if C is a nodal curve with a single component, then PicC/C one can see the 60 AARON LANDESMAN

Jacobian is not proper, because there is an exact sequence

G (22.1) 0 m PicC/C PicCe/C 0 where Ce is the normalization of C.

22.2. Limit linear series. For the rest of this discussion we will assume our curves C have compact type. Even with this restriction to compact type curves, extending the linear system is still subtle.

Example 22.4. Consider a chain of elliptic curves E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5, where each Ei meets Ei+1 at a general point on Ei and Ei+1. This has genus g+3 5, and hence gonality at most b 2 c = 4. However, there are no finite maps from this curve to P1 of degree less than 10. The solution to this is to use admissible covers, which we saw before. 1 Suppose we started mapping E1 ∪ E2 to a broken chain of P ’s. They can meet at a ramification point. We cannot joint the third elliptic curve to the second at a ramification point because the nodal points were general, so they did not differ by a torsion point on the Jacobian of E2. We then continue by joining E4 to E3 at a ramification point and E5 to E4 at a non-ramification point. We then add in a bunch of copies of P1, and get a degree 4 map to a chain of 4 P1’s. Every time we add two elliptic curve components, the gonality will increase by 1. Example 22.5. Consider a family of curves over ∆ degenerating to a singular curve X where X = Y ∪ Z meeting at a single node p. Assume the total space of our family of curves is smooth. Suppose L 0 is a line bundle on the total space away from X. Because the relative Pic0 is proper, we can extend L to a line bundle on the total space. However, this is not unique, because the relative Picard scheme is not separated. In particular, O(Z) is Cartier on the total space. Therefore, L (Z) is another extension. Then, L (Z)|Y ' L |Y(p) and so the degree is one more than the degree of L |Y. On the other hand, L (Z)|Z ' L |Z(−p). Here the degree goes down by 1. You can see this because O(Z)|Z = O(−Y)|Z using that O(Z + Y)|Z is trivial, as Z + Y is the class of a fiber. So, for all degree distributions α = (y, z) such that y + z = d, there is an extension Lα such that deg Lα|Y = y, deg Lα|Z = z. For example,

L(y+1,z−1)(−Z) ' L(y,z)

L(y−1,z+1)(−Y) ' L(y,z) CURVES NOTES 61

The above isomorphisms can be rephrased as a map

L(y,z) → L(y±1,z∓1) given by multiplication by an equation of Z or Y respectively. Such a map is either 0 along Z or 0 along Y. We also have spaces of sections. Given V ⊂ H0(L ) on the general fiber of 0 dimension r + 1 on the general fiber, one obtains a limit Vα in H (Lα) for all degree distributions α. (This is, loosely speaking, constructed by pushing forward the line bundle, and considering a relative Grassmannian bundle in the pushforward. However, this is a problem because cohomology and base change might fail, so the pushforward may not be a vector bundle. But one can fix this by twisting up by a suitable relative effective Cartier divisor, and then twisting down.) Rephrasing the above on the central fiber, we have data

L(y+1,z−1)|Y ' L(y,z)|Y(p)

L(y+1,z−1)|Y ' L(y,z)|Y(−p) which mimics twisting by Z on the central ﬁber (but does not actually twist, because here we are only working on the singular ﬁber. We also have isomorphisms

L(y−1,z+1)|Y ' L(y,z)|Y(−p)

L(y−1,z+1)|Y ' L(y,z)|Y(p)

Deﬁnition 22.6. A prelimit linear series of degree d and dimension r + 1 on X = Y ∪p Z is for all degree distribution α = (y, z) with y + z = d the data of (1) Lα on X = Y ∪p Z such that deg Lα|Y = y, deg Lα|Z = z. 0 (2) Vα ⊂ H (Lα) of dimension r + 1 such that the following holds: we have isomorphisms

L(y+1,z−1)|Y ' L(y,z)|Y(p)

L(y+1,z−1)|Y ' L(y,z)|Y(−p) and

L(y−1,z+1)|Y ' L(y,z)|Y(−p)

L(y−1,z+1)|Y ' L(y,z)|Y(p) yielding a map

L(y,z) → L(y±1,z∓1) 62 AARON LANDESMAN

(which is either multiplying by a node or is the zero map) requiring that im V(y,z) ⊂ V(y±1,z∓1)

The problem with the above is that it is unwieldy and way too much information to keep track of. We will spend the next several classes pairing down this information to something more manageable. Next time, we’ll look at the map sending a prelimit linear series on C di sending (L, V) to ∏ Pic with (L | , V | ) ≤ ≤ Ci Ci/C (0,...,0,di,0,...,0) Ci (0,...,0,di,0,...,0) Ci 1 i n for a curve with n components, where the di appears in component i.

23. 11/8/19

Recall the setup from last time. We set X := Y ∪p Z be the union of two curves meeting at a point p.

Deﬁnition 23.1. A pre limit linear series (Lα, Vα) of degree d and dimension r + 1 on X is For all degree distributions α = (y, z) with y + z = d

(1) a line bundle Lα on X such that deg Lα|Y = y, deg Lα|Z = z and a 0 subspace Vα ⊂ H (Lα) of dimension r + 1 such that such that (1)

L(y+1,z−1)|Y ' L(y,z)|Y(p) and L(y+1,z−1)|Z ' L(y,z)|Z(−p)

iZ : L(y,z) → L(y+1,z−1)iY : L(y,z) → L(y−1,z+1) such that iZ V(y,z) ⊂ V(y+1,z−1)iY V(y,z) ⊂ V(y−1,z+1)

Remark 23.2. On Y, we obtain

L(y,z)|Y L(y,z)|Y(p) | (23.1) iZ Y

L(y+1,z−1)|Y CURVES NOTES 63 where the top map is an injection given by 1 ∈ H0(O(p)) and on Z we obtain

L(y,z)|Y L(y,z)|Y(−p) | (23.2) iZ Z

L(y+1,z−1)|Y where here the top map is 0. Goal 23.3. We want to understand the set theoretic image of the map {pre limit linear series} → {deg d, dim r + 1 linear series on Y } × {deg d, dim r + 1 linear series on Z} (Lα, Vα) 7→ L(d,0)|Y, V(d,0)|Y , L(0,d)|Z, V(0,d)|Z Deﬁnition 23.4. Given (Lα, Vα) we deﬁne (LY, VY) := L(d,0)|Y, V(d,0)|Y as the Y aspect and (LZ, VZ) := L(0,d)|Z, V(0,d)|Z as the Z aspect. We make the following observations.

(1) Note that V(d,0) ' V(d,0)|Y = VY and similarly for Z because the kernel is the sections supported away from Y or Z. (2) We have injections

V(y,z)|Y ,→ V(d,0)|Y = VY

V(y,z)|Z ,→ V(d,0)|Z = VZ so we get a map

V(y,z) ,→ VY ⊕ VZ.

Question 23.5. When can I choose V(y,z) for all y, z compatibly? First,

L(y,z)|Y ' LY(−zp)L(y,z)|Z ' LZ(−yp)

(3) We have V(y,z)|Y ⊂ VY(−zp) and V(y,z)|Z ⊂ VZ(−yp). So, after doing some linear algebra, in order to answer the question of when we can choose V(y,z) compatibly turns out to be the following necessary and sufﬁcient condition: (23.3) ( r + 1 if all sections of bothVY(−zp), VZ(−yp) vanish at p dim VY(−zp) + dim VZ(−yp) ≥ r otherwise 64 AARON LANDESMAN

Deﬁnition 23.6. For V ⊂ H0(L) of dimension r + 1 on an irreducible curve C. Let p ∈ C be a smooth point. Then, the vanishing sequence of V at p is

0 ≤ a0(V, p) < a1(V, p) < ··· < ar(V, p) is the set of possible orders of vanishing of nonzero sections in V at p. Remark 23.7. We have

dim V(−αp) = i ⇐⇒ ar−1(V, p) < α ≤ ar+1−i(V, p). Corollary 23.8. The upshot is that (23.3) is equivalent to ai Vy, p + ar−i (VZ, p) ≥ d for all i = 0, . . . , r. Proof. Let’s do one direction. If (23.3) holds for all y + z = d, it holds in particular for y = ar−i(VZ, p). Then dim VZ(−yp) = i + 1 and the evaluation map is nonzero. This implies dim VY(−(d − y)p) ≥ r + 1 − i. and so d − y ≤ ai(VY, p). Therefore,

d = d − y + y ≤ ai(VY, p) + ar−i(VZ, p). Deﬁnition 23.9. An (Eisenbud-Harris) limit linear series is the data of

((LY, VY) , (LZ, VZ)) on X of degree d and dimension r + 1 is d d (1) line bundles LY ∈ Pic (Y), LZ ∈ Pic (Z), and subspaces VY ⊂ 0 0 H (LY), VZ ⊂ H (LZ) both of dimension r + 1 such that for all 0 ≤ i ≤ r we have

ai(VY, p) + ar−i(VZ, p) ≤ d Remark 23.10. This is very helpful because it is so much less information than prelimit linear series. Having these vanishing conditions is also a very strong condition. Remark 23.11. In fact, we have only defined limit linear series set theoretically, but one can define a scheme structure on pre limit linear series and define a corresponding scheme structure on limit linear series as the image of the corresponding map (in certain Grassmannian bundles over Picard schemes) Remark 23.12. Both prelimit linear series and limit linear series fit into r proper families with Gd(C). in a degenerating family. CURVES NOTES 65

Exercise 23.13. Show when r = 1 that every limit linear series is the limit of linear series. In other words, when r = 1, show the above proper family has no components concentrated on the special fiber. Hint: Use admissible covers. The above exercise is an example of a “regeneration theorem.” Definition 23.14. A limit linear series is called refined if the inequality

ai(VY, p) + ar−i(VZ, p) ≤ d in the definition of limit linear series is an equality for all i. Remark 23.15. If a limit linear series is refined, then the limit linear series determines the prelimit linear series, at least set theoretically. We’ll now begin to answer the following question: Question 23.16. What does it mean geometrically for a limit linear series to be refined? As a slogan, the answer is that “no ramification points on the general fiber limit to p.” In other words, if we have a line bundle L on the generic fiber of a family and V ⊂ H0(L). There are certain ramification points, and these limit to ramification points on the special fiber. Being refined turns out to mean that none of the ramification points limit to the node p on the special fiber. So next we have to answer what ramification points of a linear series are. This should mean that sections vanish to higher order than one would naively expect. 0 Definition 23.17. A linear series V ⊂ H (L) is ramified at p if ar(V, p) > r. We can wrap the above into a count: # ramification points with multiplicity of V

r+1 r+1 is deg P L where P L|x = L|(r+1)x. On Monday, we’ll compute this + degree, using the Plucker formula. It turns out that deg Pr 1L is the locus where r + 1 sections fail to be linearly independent.

24. 11/11/19 We’ll continue answering the question we asked at the end of last time: Question 24.1. What does it mean geometrically for a limit linear series to be reﬁned? 66 AARON LANDESMAN

Consider a family of curves degenerating to a binary curve. On the general fiber we have some collection of ramification points (i.e., there is a nonzero section that vanishes to order r + 1, or equivalently the sequence of dimensions of sections of L vanishing at kp is not minimal.) Recall we defined a r+1 bundle of principal parts of L , P L whose restriction to x is L|(r+1)x.

r+1 r+1 Lemma 24.2 (Plucker formula). We have c1 P L = (r + 1) d + ( 2 )(2g − 2) . Proof. We have a sequence of bundles with maps Pr+1L → PrL → · · · → P2L → P1L .

The ﬁber of P2L → P1L over x is the quotient of sections of L vanishing at x to sections of L vanishing at x to order 2. This is the same as Ω|x. Overall, we ﬁnd ker P2L → P1L ' L ⊗ Ω. Similarly, ker PiL → Pi−1L ' L ⊗ Ω⊗i−1.

Putting this together, we ﬁnd

r+1 c1 P L = (r + 1) c1(L ) + (r + (r − 1) + ··· + 3 + 2 + 1) c1(Ω) r + 1 = (r + 1) d + (2g − 2) . 2 Next, we’d like to know how many ramiﬁcation points specialize into each component on the special ﬁber.

Remark 24.3. Recall that L(y,z) is the unique extension of L with degree y, z on the central fiber. For π the structure map, we take V(y,z) ⊂ π∗L(y,z) a locally free subsheaf. Ramification points of the general fiber yield ramification points of every V(y,z) (on the special fiber. Further, isolated ramification points of any V(y,z) generize to ramification points of the general fiber. Proposition 24.4. A limit linear series is refined if and only if no ramification points limit to nodes.

Proof. Say the two components of the special fiber have genera g1 and g2 with g1 + g2 = g, and the components meet at p. We can then find that the number of total ramification points is r + 1 (r + 1)d + (2g − 2) − (the multiplicity of the ramification at p) 2 1 CURVES NOTES 67

To compute this multiplicity of ramiﬁcation at p, look at a basis of sections ta0 + ··· , ta1 + ··· , ... , tar + ··· . We’ll also need to compute the derivatives of these to understand the multiplicity of the differentials. Looking at the leading terms, we have a matrix  a a −1 a −2  t 0 a0t 0 a0(a0 − 1)t 0 ··· ta1 a ta1−1 a (a − 1)ta1−2 ···  1 1 1   . . .. .   . . . .  a a −1 a −2 t r art r ar(ar − 1)t r ··· This determinant is   1 a0 a0(a0 − 1) ··· ( − ) ··· a +a +···+a −1−2···−r 1 a1 a1 a1 1  t 0 1 r det  . . . .   . . .. .  1 ar ar(ar − 1) ··· This determinant is the same as a Vandermonde determinant  2  1 a0 a0 ··· 2 1 a1 a ···  1  = a − a 6= det  . . .. .  ∏ i j 0.  . . . .  i

We can now blow up the surface at p. We’ll assume p is a smooth point of the total space. We can iterate this blowing up (similarly to the process of semistable reduction) to make all ramification points not pass through nodes in the central fiber. Hence, after blowing up, we can assume that any given family of linear series has a refined limit. However, we may not be able to make a modification so that all families of linear series have a refined limit. We might have to make different numbers of blow ups for different linear systems. So any single family of linear series can be assumed to have a limit, but we cannot assume all linear series have a refined limit.

25. 11/13/19 Today, we’ll address the following question: Question 25.1. When is a limit linear series the limit of linear series? In other words, if we have a family of curves degenerating to a curve of compact type Y ∪p Z, when is a linear series on the special fiber the limit of a linear series on the general fiber. The idea is to construct a scheme parameterizing limit linear series and show its relative dimension is at least ρ(g, d, r). Here relative dimension means the difference of dimensions is ρ(g, d, r). So, if we have a linear series at which the relative dimension is ρ, (so the total dimension of each component is ρ + 1 and the dimension of the fiber is also ρ, it must come as a limit from the general fiber, just by dimension considerations. We are therefore led to the question: Question 25.2. How do we construct a scheme parameterizing linear series? Example 25.3. Let C be a smooth curve of genus g. Let d and r be integers. r We want to construct a scheme Gd(C) whose points parameterize linear series. We can construct this by considering

d PicC/C × C

(25.1) π

d PicC/C The source has a universal line bundle L (the Poincare´ bundle). We’d like to take the pushforward, and try to make the cohomology of the fiber equal to the fiber of the cohomology. Of course, cohomology and base change may not apply, so we can’t quite do this. To fix this, take D ⊂ C a divisor of large CURVES NOTES 69

d degree N. Then, π∗L (D) is a vector bundle on PicC/C Now, consider the Grassmannian

Gr (r + 1, π∗L (D)) .

Inside of this Grassmannian, consider those r + 1 planes vanishing along d. d The Grassmannian has dimension dim PicC/C +(r + 1)(d + N + 1 − g − r − 1) = g + (r + 1)(d + N + 1 − g − r − 1) = ρ + (r + 1)N. Now, the subscheme of r + 1 planes vanishing along D is defined in the Grassmannian bundle by (r + 1)N equations. Therefore, every component of this subspace has dimension ≥ ρ(g, r, d) where ρ(g, r, d) = (r + 1)d − rg − r(r + 1). This construction works on an irreducible curve, but now we want to imitate this on a reducible curve. Say we have a relative curve C → B over a trait B whose generic fiber is smooth and special fiber is a reducible curve Y ∪p Z. Choose a collection of points ai on Y and bi on Z. Let D and E be (d,0) ' (0,d) divisors of large degree N and M. Observe PicC /B Pic since we can twist by the component Z d times. Consider

(d,0) × → (d,0) PicC /B BC PicC /B which is isomorphic to

(0,d) × → (0,d) PicC /B BC PicC /B .

We have LY a universal line bundle on the ﬁrst bidegree (d, 0) component and LZ a universal line bundle on the bidegree (0, d) component. We want to use both of these line bundles. Consider

Gr (r + 1, π∗LY(D + E)) × Gr (r + 1, π∗LZ(D + E)) . We’ll spell out the identification of these two Grassmannian bundles by giving bases of each which are identified. Let Fr(a + 1, V) denote the bundle of projective frames of dimension a + 1 in V, which means parameterizes collections of a + 1 points in PV fiber by fiber. The fibers look like GL(V) modulo the group of diagonal matrices. Let σ1, ... , σr+1 be the universal framing points in

Fr (r + 1, π∗LY(D + E))

and let τ1,..., τr+1 be the universal framing points in

Fr (r + 1, π∗LZ (D + E)) . 70 AARON LANDESMAN

We have a projection

Fr (r + 1, π∗LY(D + E)) × Fr (r + 1, π∗LZ (D + E))

(25.2) s

Gr (r + 1, π∗LY(D + E)) × Gr (r + 1, π∗LZ(D + E)) . The space of limit linear series we are looking for is a subscheme deﬁned by b − a (1) σiα r i = τr−iβ i . (2) σi|D = 0 (3) τi|E = 0. Remark 25.4. To understand why the ﬁrst equation makes sense α and br−i β are constant sections of OC (Y), OC (Z). Observe σiα is a section of a LY(D + E)(br−iY) which is isomorphic to LZ(d + E)(aiZ) in which τr−iβ i lives. This isomorphism comes from the isomorphisms LZ ' LY(dY) and O(Y + Z) ' O, which are isomorphisms we have chosen to begin with, using ai + br−i = d. Question 25.5. What does this image under s look like?

On the general fiber, this is just saying σi = τr−i and the second and third conditions are saying that σi = τr−i vanish along D and E. This is analogous to the construction for smooth curve from Example 25.3. On the special fiber, we can think of α as a function vanishing along Y and β as a function vanishing along Z. Recall, α and β give local coordinates at p for C . The first a equation implies that β i | σi. Also, σi|Y vanishes to order ai at p. Similarly, the restriction τr−i|Z vanishes to order br−i at p while Even if ai = 0, σi|Z and τr−i|Y are determined by τr−i|Z and σi|Y respectively, using the equations. In particular, σi|Z = 0 unless ai = 0. Therefore, these σi|Y vanishing to order at least ai at p and τr−i|Z vanish to order br−i at p. These are precisely limit linear series. These to not yet give us refined limit linear series. Inside of the scheme defined by the above equations, we have an open subscheme where σi|Y vanishes to order ai and τr−i|Z vanishes to order at least br−i at p. We now want to think about what the fiber dimension of s. On the open subscheme for refined linear series, the fiber dimension of s can be understood as follows. On the general fiber, the fibers are GLr+1 /diagonal matrices which has dimension r(r + 1). On the special fiber, the fibers of s look like (upper triangular matrices/diagonal matrices)2 . CURVES NOTES 71

r+1 This has dimension ( 2 ) · 2 = r(r + 1). This again has dimension r(r + 1). We have proven the following lemma: Lemma 25.6. Every component of the image of the map s has relative dimension at least g + 2(r + 1)(d + N + M + 1 − g − 1 − (r + 1)(d + N + M + 1 − g − 1) − (r + 1)N − (r + 1)M − r(r + 1) = ρ. Proof. The first g is coming form the Picard scheme. The second term is coming from the dimensions of spaces of sections, with the −1 corresponding to projectivizations. The (r + 1)(d + N + M + 1 − g − 1) term is coming from the first set of equations 1. The second sets of equations define (r + 1)N and (r + 1)M equations, and the r(r + 1) conditions are defining the dimensions of the fibers of the map s. 26. 11/15/19 Today, we’ll try to understand the following question: Question 26.1. What is the image of r θ d {gd (pre)limit linear series } −→ ∏ PicX/C ⊂ PicX/C . components X⊂C In other words, what collections of line bundles appear as limits of linear series? So far, everything we’ve done works for curves of compact type, even though we’ve only been stating it for binary curves. However, the answer to this question will not be uniform for all compact type curves. Definition 26.2. A chain curve is a curve of compact type on which every component has at most 2 nodes. In other words, a chain curve looks like a chain of curves. The image of the map θ from prelimit linear series to Pic. is contained in n 0 o (L ) : for any degree distribution α of total degree d, h (Lα) ≥ r + 1 . Here α is a collection of integers, one for each component, which sum to d. Proposition 26.3. On a chain curve the image of θ is precisely n 0 o (L ) : for any degree distribution α of total degree d, h (Lα) ≥ r + 1 . Proof. Given a system of line bundles satisfying the above condition, we need to cook up a limit linear series. Given components Ci with nodes pi joining Ci and Ci+1 define ai := maxh0(L | )≥r+1−i α where the notation (α,d−α) Ci (α, d − α) means that there is a total degree α on components Cj with j ≤ i 72 AARON LANDESMAN

and total degree d − α for curves Cj for j > i. This ai will encapsulate the degree to the left of the node pi. Let bi = d − ar−i. This will encapsulate the degree to the right of the node pi. We’ll now construct a space of sections of L on Ci with ramiﬁcation at least ai. We have nodes p := pk−1 and q := pk on component Ck. For any i, j we’ll put bj of our degrees to the right of q and ai of our degrees to the left of p. This leaves d − ai − bj degrees on Ck. The bj and ai refer to what we are hoping to make be the ramiﬁcation sequence at p and q. There exists a degree 0( )| = + − ( − ) = + distribution to the right of q with h Lβ C≥k+1 r 1 r j j 1 and evaluation at q is surjective. Here, β is a distribution with total degree bj Here, C≥k+1 means the union components Ct of C with t ≥ k + 1. Note that if this were not an equality or the evaluation map were not surjective, we could increase α and so bj would not be maximal. 0( | ) ≥ + Also, on C≥k, for any such degree distribution, h Ld−ai−bj,β C≥k r 1 − i. Then, L −ai p − bjq is the restriction of L to Ck. Therefore, 0 | − − + ( + ) − = ( ) ≥ + − h L Ck ai p bjq j 1 1 h L(d−ai−bj,β) r 1 i. where the −1 is coming from the surjectivity of the evaluation map at q (it is one condition for the sections to agree at q). Equivalently, we have 0 | − − ≥ + − − h L Ck ai p bjq r 1 i j. Hence, we have reduced the proposition to the following claim: Lemma 26.4. If 0 | − − ≥ + − − h L Ck ( ai p biq) r 1 i j, there exists Λ ⊂ H0 (L ) such that dim Λ = r + 1 with dim Λ (−ai p) ≥ r + 1 − i dim Λ −bjq ≥ r + 1 − j. Proof. We can argue by induction on r. Let’s start with r = −1. This holds tautologically by taking Λ = 0. We now must check an inductive step. We would like to apply our induc- | (− ) tive hypothesis to the line bundle L Ck a0 p . We will apply our inductive hypothesis. Suppose there exists Λ ⊂ 0 | (− ) Λ H L Ck a0 p of dimension r. CURVES NOTES 73 | − (− ) We will apply our inductive hypothesis to L Ck ( a0 p) with L a0 p and the sequences of ai given by {a1 − a0,..., ar − a0} and the bj given by {b0,..., br−1}. We conclude that dim Λ (−(ai − a0)p) ≥ r − (i − 1) dim Λ −bjq ≥ r − j. If we consider ⊂ 0( | ) Λ H L Ck such that

dim Λ(−ai p) ≥ r + 1 − i dim Λ(−a0 p − bjq) ≥ r − j. 0 = Let j : maxdim Λ(−a0 p−bjq)=r−j j Note that the inequality is an equality for j = 0, because Λ is r dimensional, so it holds for some j. 0 0 By hypothesis, dim H L −a0 p − bj0 q ≥ r + 1 − j . Therefore, pick σ ∈/ Λ. Then, Λ + hσi satisﬁes the conclusion, which completes the induction step. Deﬁnition 26.5. On a chain curve, we call such a collection of line bundles 0 in the image of the map θ (one where the h (Lα) ≥ r + 1 for all α of total degree d) a limit r-positive line bundle. Let’s review the maps we have

{ pre limit linear series }

(26.1) { limit linear series }

{ limit r-positive line bundles }

r The first two fit into families with a Gd this is the space of line bundles together with a subspace of sections of dimension at least r + 1. The last fits r r into a family with Wd. This Wd is the space of line bundles with at least r + 1 sections, but we do not keep track of what the space of sections is. When the dimension of the space of limit linear series is correct, we can lift that limit linear series to the general fiber. This is the basic package we have to limiting linear series on curves via degeneration. Next time, we’ll apply this package to prove some nice theorems. 74 AARON LANDESMAN

27. WEEKOF 11/18 I was out this week so wasn’t able to take notes. But here is a brief overview of what was covered. To start, we showed Brill Noether non-existence, that r when ρ(g, r, d) < 0, a general curve of genus g has no gd’s. We did this by degenerating to a chain of elliptic curves, and showing there were no limit linear series on the resulting chain of elliptic curves unless ρ(g, r, d) ≥ 0. The next topic was to describe limit linear series on chains of elliptic curves in more detail. For C a chain of elliptic curves, we described a bijection r between components of Wd(C) and Young Tableau fillings of an (r + 1) × (g + r − d) grid. I.e., we filled the numbers of the grid with numbers from 1 to g (not every number needed to be used, but every number can be used at most once) so that it is increasing along rows and columns. Specifically, suppose the node joining component Ci to component Ci+1 of C is labeled k k by pi. If we had ramification indices ai and bi on the left and right nodes appearing in component Ci of C. If pi denotes the node of C joining Ci and Ci+1, we place in the ith column the set of values k so that

k Lk ' O(ai pk−1 + br−i pk).(27.1)

1 g To ease notation we also set ai = i and bi = i. Some useful conditions in k k the proof of this equivalence were that ai + br−i ≤ d, with equality if and k k k only if Lk ' O(ai pk−1 + br−i pk) and otherwise ai + br−i = d − 1. We also k+1 k k have ai + br−i = d. Therefore, we inductively found that ai = i + k − 1 − k k+1 (# times (27.1) holds for i on the first k components ) and br−i = d − ai . We then did some explicit examples in terms of young tableau to see how k k r these ai and bi can be computed and correspond to components of gd’s. We saw that fillings of Young Tableau were in bijection with components of r Wd(C), for C the above chain of elliptic curves. Following this, we wanted r to investigate more subtle properties of Wd(C), such as connectedness so we needed to examine the scheme structure. We saw that the scheme structure was reduced (it essentially amounted to intersecting pullbacks of points from r various elliptic curves on a product of elliptic curves), implying that Wd(C) r is reduced. We constructed Wd(C) as an intersection of spaces of sections corresponding to different components of PicC/k, and we didn’t completely justify why each of these components were reduced. But granting this, we r justified why intersections of these were reduced. We then saw Wd(C) is also connected when ρ > 0, which was a game about moving numbers in Young Tableau around to show you could get from any young Tableau to any other. Recall Young Tableau correspond to components, and being able to move from one to another by replacing numbers corresponds to an intersection CURVES NOTES 75

of components. We were then able to deduce geometric connectedness r r and geometric reducedness for the general Wd by knowing it for Wd(C) because the joint condition of being geometrically reduced and geometrically connected generizes in flat families (even though geometric connectedness r on its own does not). We also said that Wd is reduced for a general curve r because it has connected fibers (which are Grassmannian) over Gd. This r r wasn’t spelled out in class, but the point is that Wd(C) → Gd(C) is its own Stein factorization since the fibers are geometrically reduced and connected. In particular, the map induced an isomorphism on global sections. Since the target is reduced and connected and projective, it has a 1-dimensional space of global sections. Therefore the source does as well and so must be connected. Here was an exercise: 1 Exercise 27.1. For C a general curve of genus 5, we have shown that G4(C) 1 is a curve, i.e., ρ(5, 1, 4) = 1. What is the genus of the curve G4(C)? [I believe the answer to this exercise is 11; in the degeneration to the elliptic curve in terms of Young Tableau, there are 10 genus 1 components of the resulting space of limit linear series, and they meet in a pattern whose dual graph has a single cycle.] Finally, we began discussing the Gieseker-Petri theorem: r Theorem 27.2. For C a general curve of genus g we have that Gd(C) is smooth. Because we can do a blow up to make the ramification sections of a family not pass through a node of the special fiber, we argued that it suffices to show the space of refined limit linear series on a chain of curves of genus 0 or 1 is smooth.

28. 12/2/19 Recall that last time we were trying to establish the Gieseker-Petri theorem.

r Theorem 28.1 (Gieseker-Petri). For C general of genus g, we have that Gd(C) is smooth. Proof. Last time we saw it suffices to show the space of refined limit linear series on a chain curve which has components of genus at most 1 is smooth. Now, we want to analyze the space of refined limit linear series on such chain curves. Let C be a curve with components Ci and nodes pi joining Ci and Ci+1. The space of refined limit linear series is k−1 k ä ∏ linear series with ramification ai at pk−1, ai at pk . k k ramification sequences {ai } 76 AARON LANDESMAN

Therefore, it suffices to show the space of k−1 k { linear series on Ck with ramification ai at pk−1 and ai at pk} k−1 is smooth. To simplify notation, we rename Ci by C, pk−1 by p, pk by q, ai k k by ai, and ai by bi . So we want to show the space of linear series on C with ramification sequence ai at p and bj at q is smooth. Here C has genus 0 or 1. We now split into two cases 28.1. The genus of C is 0. In the genus 0 case, we want

n 0 0 Z := Λ ⊂ H (OP1 (d)) : dim Λ ∩ H (O(d)(−ai p)) = r + 1 − i,(28.1)

0 o (28.2) dim Λ ∩ H (O(d)(−biq)) = r + 1 − i

0 Inside V = H (OP1 (d)) we have two flags V ⊃ V(−p) ⊃ · · · ⊃ V(−dp) ⊃ 0V ⊃ V(−q) ⊃ · · · ⊃ V(−dq) ⊃ 0. These flags are transverse using Riemann Roch and the fact that the curve is P1. That is, V(−ap) ∩ V(−bq) = H0(O(d)(−ap − bq)) and we know this has codimension equal to the sum of the codimensions of V(−ap) and V(−bq), by Riemann Roch. Further, any two transverse flags can be taken to any other pair of transverse flags of the same dimension by the action of GL(V). This follows from the fact that GL(V) acts transitively on the space of bases of V. We now use Kleiman’s Criterion (also sometimes called Kleiman-Bertini). Theorem 28.2 (Kleinman’s Criterion). Let V, W ⊂ X be smooth varieties (characteristic 0 may be important here) and let G be a smooth group acting transitively on X. Then, V ∩ gW is also smooth for g ∈ G general. Remark 28.3. One can make a version of this work in characteristic p as well, but you will need some additional assumption about how G acts on the total space of the tangent bundle of the relevant varieties. Proof. The idea of the proof is to consider the incidence correspondence

{(v, w, g) : gv = w} (28.3)

G V × W We know V × W is smooth and Stab(v) is smooth for every V. Therefore, the source is smooth, and by generic smoothness, the left projection map is smooth. CURVES NOTES 77

Now, note, that the space Z deﬁned above in (28.4) can be expressed as an intersection in X := Gr(r + 1, H0 (O(d)) of the space of

n 0 0 o A := Λ ⊂ H (O(d)) : dim Λ ∩ H (O(d)(−ai p)) = r + 1 − i and a general translate of

n 0 0 o B := Λ ⊂ H (O(d)) : dim Λ ∩ H (O(d)(−bi p)) = r + 1 − i . We want to show A ∩ B is smooth. But we know A ∩ B is isomorphic to A ∩ gB, since the ﬂags are transverse. By Kleiman’s criterion, this intersection is smooth. This is what we wanted to show. 28.2. The genus of C is 1. We now proceed to genus 1. If d = 0, this is an easy problem, (there are no sections when d = 0 and L 6= O, and when L = O, the resulting space is the one point space which is smooth. So we will assume d > 0. We want to understand the space

n 0 0 Z := Λ ⊂ H (L ) : dim Λ ∩ H (L (−ai p)) = r + 1 − i,(28.4)

0 o dim Λ ∩ H (L (−biq)) = r + 1 − i .(28.5)

d We have Luniv over C × PicC/C and there is a projection map π : C × d d d PicC/C → PicC/C. We then have a vector bundle π∗Luniv on PicC/C, which is a vector bundle by Cohomology and base change, using our assumption that d > 0 and Riemann Roch. As before, we also have flags V ⊃ V(−p) ⊃ · · · ⊃ V(−dp) ⊃ 0V ⊃ V(−q) ⊃ · · · ⊃ V(−dq) ⊃ 0. If L (−ap − bq) 6' O, then the flags above are transverse and we can proceed as in the genus 0 case. d It turns out that the map Z → PicC/C is not smooth, and failure of the map to be smooth occurs at the points O(dp), O((d − 1)p + q), ... , O(ap + bq), ... , O(dq). We want to show, however, that the total space of Z over these points is smooth. The flags fail to be transverse at these points. How- ever, the failure of transversality is the simplest possible failure. We have two linear subspaces which should meet at {0}, but instead meet at a line. In the projectivization, this means the two subspaces meet at a point. Said another way, H0(L (−ap − bq)) 6= 0. or in other words, L ' ap + bq. Then, the two members of the flag intersect precisely at a reduced point using that we have a scheme theoretic description of this as the locus where h0(L (−ap − bq)) ≥ 1, and this is reduced, since 78 AARON LANDESMAN

after translating, the locus of degree 0 line bundles with a section is a reduced point, corresponding to the trivial bundle. The key point is that we can trivialize most of the basis using a GL(V) action. This trivializes every subspace except a two dimensional such space. We can also trivialize the space so that one of the flags is stationary, but the other flag has one subspace which is allowed to move. So, in an analytic disk around the line bundle L , we have two flags which are almost all stationary, but the smallest space in one of the flags is moving, and the intersection point of the moving line with the 1-dimensional space in the other flag has nonzero derivative. Let’s see smoothness in an example. Example 28.4. Let’s start with a flag of a pink point in an orange line in a yellow plane and a purple point in a blue line in a green plane. We think of the blue line as moving. Let’s look at the set of lines meeting the blue and orange each in one point. The central fiber becomes reducible as a union of two planes (it is either contained in a plane or contains the intersection point). However, we have a map { lines meeting the blue line and orange line each in one point } (28.6)

{ lines meeting the orange line and the green plane } The target is smooth being an intersection of transverse flags, and the map is an isomorphism. In general, generalizing the above example, we can fiber over an appropri- ate target, and the map will either be an isomorphism or a P1 bundle, and the source will be smooth. This finishes the proof in the genus 1 case. 29. 12/4/19 Today, we’ll discuss irreducibility of the space of maps from curves to projective space. Theorem 29.1. Suppose ρ(g, r, d) ≥ 0. Then, there is a unique irreducible compo- r nent of M g(P , d) whose general member is a nondegenerate map from a smooth curve of genus g and which dominates M g. We’ll spend the rest of today’s class proving this. First, we observe that we have already done the case ρ > 0. r 29.1. The case ρ > 0. We showed two classes ago that Gd(C) is connected, r for C a general curve. We also showed last class that Gd(C) is smooth. r Then, connected and smooth together imply irreducible. Therefore, Gd(C) is CURVES NOTES 79

r irreducible, and so the relative Gd over Mg is irreducible. The space of maps to projective space is then a dense open in a PGL bundle over this universal r Gd. The PGL bundle corresponds to choosing a basis for the sections, and the open condition corresponds to throwing out the basepoint locus. The r r open is dense because if a Gd has a base point, it comes from some Gd−1, which has lower dimension by the Brill Noether theorem.

r 29.2. The case ρ = 0. We can now assume ρ = 0. Hence, Gd(C) is a finite r set of reduced points. We then have a universal Gd over Mg which is a generically finite map. We essentially want to understand the monodromy r of the map π : Gd → Mg. We want to know that the monodromy acts transitively on the geometric generic fiber.

Remark 29.2. You can characterize the locus where this map π is ramiﬁed as the locus of (C, L) where H0(C, L) ⊗ H0(C, K ⊗ L∨) → H0(C, K) is not surjective.

The plan of attack is via limit linear series. Start with a chain of elliptic curves C with components C1, ... , Cg. Then, limit linear series are indexed by Tableau of size (r + 1) × (g + r − d). When we move around in the locus of chains of elliptic curves, there is no monodromy in this family. We’re going to have to leave the space of chains of elliptic curves and loop around something so that the tableau gets permuted. To understand the monodromy, we’ll move to the larger locus where we smooth one of the nodes. So, we have a chain of elliptic curves and one genus 2 curve. Now suppose we smooth Ck ∪ Ck+1 to a single genus 2 curve. There is discrete ramiﬁcation data on the Cj for j 6= k, which can’t change as we apply such smoothenings. Therefore, the only way to exchange two tableau by this sort of monodromy operation if the two tableau are related by switching the positions of i and i + 1 in the tableau. What we need to show is the following:

Proposition 29.3. If we have two tableau which are related by swapping k and k + 1, then as we move in the space of chain curves with g − 2 genus 1 components and one component of genus 2, the monodromy action permutes those two tableau. Here the tableau correspond to further degenerations of the genus 2 curve to two genus 1 curves.

Proof. To set up notation, let p be the left point of the genus 2 curve and q be the right point with ramiﬁcation indices ai ant p and bj at q. Suppose k starts in column x and k + 1 starts in column y. We must have x 6= y if we can 80 AARON LANDESMAN

hope to swap the two tableau. By our conditions on the ai and bj, We have ( d − 2 if i ∈/ {x, y} ai + br−i = d − 1 if i ∈ {x, y} .

The d − 2 is coming from two numbers summing to d − 1 on component k, summing to d across the intersection point of Ck and Ck+1 and summing to d − 1 across Ck+1, and then d − 2 = (d − 1) − d + (d − 1). The d − 1 is one more than d − 2. We can swap k and k + 1 exactly when they are both not in the same row and not in the same column. How do we express the condition that the boxes are also not in the same row. This is expressed by ax − ay 6= ±1, since if they were in the same row, they would be directly k adjacent to each other. Then, ai = i + k − 1 − # elements ≤ k in column i. We can then unwind the deﬁnition to see ax − ay = ±1. We now want that for C a general curve of genus 2, as p, q ∈ C vary, monodromy exchanges (L , p, q) : L (−ax p − (d − a − ax)q) , L −ay p − d − 1 − ag q are effective .

Note that we really needed to show L (−ai p − br−iq) is effective, but this has degree 2 when i 6= x, y. since all degree 2 divisors on genus 2 curves are effective. Hence, we now want to see the two line bundles above are effective. That is, we have L (−ax p − (d − 1 − ax)q) ' O(z) and L (−ay p − (d − 1 − ay)q) ' O(w). Note that if we know one of z or w then we know the other. We ﬁnd that w − z needs to be linearly equivalent to ax p − ay p + (d − 1 − ax)q − (d − q − ay)p = (ax − ay)(p − q). We need to show that if we have one solution to this, then we have two solutions. Indeed, any genus 2 curve is hyperelliptic, so has a unique hyperelliptic involution ι, and we then have ι(z) − ι(w) = w − z. We then use ax − ay = n with n ∈/ {−1, 0, 1} with w − z = n(p − q). So, we need to say that as we vary p and q, we can exchange (w, z) and (ι(w), ι(z)). Let’s now rephrase this problem. We have a generic double cover C × C → 0 PicC/C sending (z, w) 7→ z − w. This double cover is irreducible. We want to pull this back to the space of pairs of points p and q, and ask if the pullback remains irreducible. This maps is given by

0 ×[n] 0 C × C → PicC/C −−→ PicC/C (p, q) 7→ p − q 7→ n(p − q). CURVES NOTES 81

So, we want to understand irreducibility of the fiber product X C × C (29.1) 0 ×[n] 0 C × C PicC/C PicC/C Note that when n = 1, this fiber product would be reducible, since there 0 is a diagonal component. We now have a copy C → PicC/C mapping by z 7→ z − z. This copy of C is the branch locus. We just need to check that 0 ×[n] 0 the map C × C → PicC/C −−→ PicC/C does not contain C. This is because if one takes a fiber product of two covers branched over loci not contained in each other, then the fiber product is irreducible. The only way this C can “swallow itself up” by multiplication by n for |n| > 1 is when im(C) is an abelian subvariety. However, this image is birational to C, and so is not an abelian subvariety because it has genus > 1. Hence, the fiber product is irreducible. So, we have now shown that if two tableau are related by exchanging k and k + 1 then they are exchanged by monodromy. So, it only remains to relate two tableau by exchanging k and k + 1 for various values of k. Question 29.4. What is the full monodromy group associated to this map r π : Gd → Mg. This is the full symmetric group when r = 1, but is an open problem when r > 1. It might also be known for some low g. riga 30. 12/6/19 r Recall we were trying to prove that M g(P , d) has a unique component whose general member is a nondegenerate map from a smooth curve. To prove this, we were examining a degeneration to a chain of elliptic curves, and then tried to examine the monodromy of the associated cover, to show the monodromy group acts transitively. We saw last time that given a tableau, with k and k + 1 in different rows and different columns, we were able to swap k and k + 1. It remains to prove the following combinatorial lemma: Lemma 30.1. Given any tableau, there exist a sequence of moves swapping k and k + 1 appearing in different rows and different columns after which the tableau is the standard tableau (with numbers appearing in order when reading left to right and then top to bottom). Proof. Here is the recipe. Keep swapping the number in the first box not containing the right number with the number one smaller. 82 AARON LANDESMAN

Example 30.2. 1 4 2 5 3 6 goes to 1 3 2 5 4 6 goes to 1 2 3 5 4 6 goes to 1 2 3 4 5 6 At every step, we either make a number in the next box correct, or we get it closer to being correct. Therefore, this process terminates. This lemma implies the monodromy is transitive, and hence we conclude Theorem 29.1. r In particular, by the Brill Noether theorem, a component of M g(P , d) has a unique component whose general member is a nondegenerate map from a smooth curve, and such a component exists if and only if ρ(g, r, d) ≥ 0. r Deﬁnition 30.3. We call such curves in the unique component of M g(R , d) whose general member is a nondegenerate map from a smooth curve Brill- Noether curves, or BN-curves. Example 30.4. If you look at genus 10 degree 6 curves in P3, there is a component where hyperelliptic curve mapping 2 : 1 onto a rational normal curve. These would not be Brill-Noether curves. r 1 ∗ Example 30.5. If we have a curve f : C → P such that H ( f TPr ) = 0 then [ f ] is a Brill-Noether curve. A natural question is: CURVES NOTES 83

Question 30.6. What is the geometry of Brill-Noether general curves? We’ll discuss the problem of interpolation. This is related to the following question. Question 30.7. Does there exist a Brill-Noether curve of degree d and genus r g through n general points p1,..., pn ∈ P ? As a simple example, there is a conic passing through 5 points in P2. From the perspective of this class, here is how we might go about answering this r question. Start with a map f : C → P . Suppose we have point pi ∈ C r mapping to points qi ∈ P for 1 ≤ i ≤ n. r As we deform the points qi ∈ P , we see that obstructions to lifting deformations of the qi to (C, pi, f ) lie in 1 • 1 Ext (Ω f , Ip1,...,pn ) ' H Nf (−p1 − · · · − pn) . at least if we assume f is an immersion. Conjecture 30.8. The answer to Question 30.7 is yes if and only if χ Nf (−p1 − · · · − pn) ≥ 0.

The conjecture is true if Nf satisﬁes interpolation. Recall this means that when you twist down by a collection of points, either h0 = 0 or h1 = 0. We have an exact sequence

∗ (30.1) 0 TC f TPr Nf 0.

We have ∗ χ(Nf ) = χ( f TPr ) − χ(TC) = [(r + 1) d − rg + r] − [3 − 3g] = (r + 1) d − (r − 3)(g − 1) . So we see χ Nf (−p1 − · · · − pn) = (r + 1) d − (r − 3)(g − 1) Hence, the conjecture is equivalent to saying something satisﬁes interpolation when this value is nonnegative. Remark 30.9. There are counterexamples to Conjecture 30.8, even in P3. For example, take (d, g) = (5, 2) or (d, g) = (6, 4). The conjecture would say we can pass curves of degree d and genus g through n points if n ≤ 10 in the ﬁrst case or n ≤ 12 in the second case. 84 AARON LANDESMAN

0 However, consider the space H (OP3 (2)). This is a 10-dimensional vector space. Further, there is a restriction map 0 0 H (OP3 (2)) → H (OC(2)) . This restriction map is a map from a 10 dimensional vector space to a 9 0 dimensional vector space by Riemann Roch. Therefore, H (IC(2)) 6= 0, and so C must lie on a quadric. There is a 10 dimensional vector space of quadrics in P3. Therefore, it is impossible to ﬁnd a quadric passing through 10 general points, by a dimension count. Therefore, the curves C as above cannot pass through 10 general points. This gives a counterexample to Conjecture 30.8. Remark 30.10. In P3, Remark 30.9 gives the only 2 counterexamples to the conjecture. Also, in any projective space, all counterexamples have difference between the expected answer and the actual answer which is at most 3. For example, in the degree 6 genus 4 case, we expected the answer was that it would pass through 12 points, but in fact in only passed through 9 points, and 12 − 9 = 3. Example 30.11. If we take (d, g) = (r + 2, 2) in Pr, we have a counterexample. Every genus 2 curve is hyperelliptic. There is a unique hyperelliptic involution. We now take every pair of points exchanged under the hyperelliptic involution, connect them with a line. These lines sweep out a surface S. In P3, that surface S is a quadric. In general, this is a certain rational normal scroll. The surface S can only pass through a certain number of points. However, S can only pass through some bounded number of points, and this constrains the number of points on which C passes. When r is odd, you can have balanced scrolls. Then, S ' P1 × P1, while in even dimensional projective spaces it is isomorphic to a Hirzebruch surface F1 = PP1 (OP1 ⊕ OP1 (1)). This gives a counterexample to Conjecture 30.8 in P3 and in P5. 4 This gives an example when Nf fails to satisfy interpolation in P , although it is not a counterexample to Conjecture 30.8. What is going on that if you are looking through such curves through 9 points, you would expect that there is a 2-dimensional family of such curves, and so you would expect the 2-dimensional family of curves to sweep out a 3-fold. However, it only sweeps out a surface, which gives failure of interpolation for the normal bundle. Said another way, you would expect a Brill-Noether curve would pass through 9 points and meet a general line, but in fact there is no such curve, even though there are curves passing through 9 general points and meeting a general plane. Next, let’s look at the (6, 4) example which also has a friend in P5. CURVES NOTES 85

Example 30.12. Let’s take r = 2k + 1 to be odd and (d, g) = (4k + 2, 2k + 2) in P2k+1. This is a canonically embedded curve. That is, a Brill-Noether curve of this type is a canonical curve of genus 2k + 2. The gonality of C is k + 2. Here, there was division by 2, which is the relevance of why we wanted r to be odd, and hence the genus to be even. Choose k + 2 points in the ﬁber of a map C → P1. We have a divisor 0 0 D ⊂ C of degree k + 2. Then, h (O(D)) = 2 so h (KC(−D)) = k + 1 by Riemann Roch. In other words, these k + 2 points are dependent, and only span a Pk instead of a Pk+1. So, these ﬁbers sweep out a variety X with X ' P1 × Pk when X is general. But X can only pass through a certain number of points, and this gives a counterexample to interpolation when r = 3 and r = 5. The above examples are the only counterexample in P3, and the only example where the normal bundle fails to satisfy interpolation in P4. If d ≥ g + r, then the above counterexamples with (d, g) = (r + 2, 2) are the only counterexamples with r arbitrary.