CURVES NOTES Eric Larson Taught a Course On
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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 2 H (C, L ) which are basepoint free, i.e., there is no p 2 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 2 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 a 2 H (KC) and s 2 H (OC(p1 + ··· + pd)) we must have ( · ) = ∑ Respi a s 0. i This condition is trivial if and only if a 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 Di of pi which pairwise do not intersect. Let U = C − fp1, ... , pdg. We can think of an element of 1 0 H (OC (p1 + ··· + pd)) by a 1-cocycle fi 2 H ODi−pi . Then, the pairing above from Serre duality (3.1) is given by ( f g) 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 2 H0( ) defining as WC) by wC, the dualizing sheaf. So, the element a wCe can be viewed as a meromorphic differential on the normalization Ce of C with the property that for any p 2 C, Resp (a · 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 a is thought of as on the normalization. Or rather, if p : Ce ! C is 2 H0( ) the map from the normalization, we can view a p∗wCe . Question 3.13. What does wC 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, a = ∑i aix dx. We have the condition that for any n ≥ 1, we must have n−1 0 = Res ax = a−n. Therefore, a−n = 0 for all n ≥ 1, we find a 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 a = ∑(aix dx + biy dy). i Then, let’s investigate what the residue condition tells us. We certainly must have Res a = 0. So, a−1 + b−1 = Res a = 0. Also, for any n ≥ 2 n−1 0 = Res a · x = a−n. Similarly, b−n = 0. Therefore, the condition that Resp (a · f ) = 0 tells us a has at worst a simple pole along both branches with opposite residues. Corollary 3.15. We have deg wC = 2g − 2 for a nodal curve. Proof. This is true for a smooth curve. One can then use the explicit descrip- tion 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 wC 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 G ⊂ f (C) denote the set of nodes of f (C) which are not nodes of C. Let’s now write down sections of wC 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 G the set of additional nodes of F.