# Tuesday, August 23 > See [Mukai's book](https://www.math.ens.psl.eu/~benoist/refs/Mukai.pdf) ## An aside on branched covers of curves :::{.remark} A question from me: is every curve a branched cover of $\PP^1$ over some number of points? Consider maps $f: X\to \PP^1$ where $f(X) \not\subset \PP^{n-1}$ is not contained in a hyperplane. This biject with basepoint free linear systems -- let $\mcf$ be an invertible sheaf, then a linear system is any linear subspace $V \subseteq H^0(X; \mcf)$. Writing $V = \gens{f_i}_{i=0}^n$, the bijection is sending $p\mapsto \tv{f_0(p): \cdots : f_n(p)} \in \PP^{n}$. Since $\mcf$ is invertible, locally $\ro{\mcf}{U} \cong \OO_U$ -- this map is well-defined precisely when not all the $f_i(p)$ are zero, which is precisely the basepoint-free condition. A map $f:X\to \PP^1$ thus corresponds to two sections which don't simultaneously vanish. If $X$ is projective, it admits a very ample line bundle $\mcl$ where the base locus of $H^0(\mcl)$ is empty. One can now project away from any point outside of the curve to get a regular map factoring the projective embedding: \begin{tikzcd} X && {\PP^n} \\ \\ && {\PP^{n-1}} \arrow[hook, from=1-1, to=1-3] \arrow[dotted, from=1-3, to=3-3] \arrow[from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiXFxQUF5uIl0sWzIsMiwiXFxQUF57bi0xfSJdLFswLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMiwiIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV0sWzAsMl1d) The projection: \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-08-23_12-54.pdf_tex} }; \end{tikzpicture} One can continue to project until reaching $\PP^1$. ::: :::{.definition title="Gonality"} The **gonality** of a curve $X$ is the minimal degree of a map $X\to \PP^1$, where degree is the size of a generic fiber. Here a cover may have a ramification locus upstairs and a branch locus downstairs, which are small in the sense that they are algebraic subsets. ::: :::{.remark} Recall $X$ is hyperelliptic if it admits a 2-to-1 map $X\to \PP^1$, so has gonality 2. Gonality 1 curves are isomorphic to $\PP^1$, and gonality 3 are *trigonal*. ::: ## Course plan :::{.remark} Plan for the course: 1. GIT ($1/2$ to $2/3$ of the course). Goals: - Construct moduli of vector bundles on a curve, surface, etc. - Construct $\Mg$ (Riemann's moduli space, not complete, projective, affine, but quasiprojective), $\bar{\Mg}$ (Deligne-Mumford's moduli of stable curves), $\bar{\mcm}_{0,n}$, and $\bar{\mgn}(V)$ (Kontsevich's moduli of stable maps, to rigorously define Gromov-Witten invariants).[^dimensions] - Sources: - Mukai's book, An Introduction to Invariants and Moduli (this will be our primary source). He covers stable vector bundles on curves. - Mumford's book on GIT and his paper about stability for algebraic curves, although this is perhaps unnecessarily difficult! We'll need this for $\bar{\Mg}$. 2. Hodge theoretic approaches. Goals: - Construct $\Ag$ the moduli of abelian varieties. See [Birkenhake and Lange](https://link.springer.com/book/10.1007/978-3-662-06307-1). - Construct $F_{2d}$ the moduli of K3 surfaces. See Huybrechts. [^dimensions]: $\dim \Mg = 3g-3$ for $g\geq 2$. ::: ## Intro :::{.remark} Let $X$ be a genus $g$ smooth projective curve. Over $\CC$, projective implies compact, and non-projective is a Riemann surface with finitely many punctures. More generally, over $k=\kbar$ smooth means that $\dim \T_{X, x} = \dim X$ at every point $x\in X$. Note that $X^{\sing} \subseteq X$ is an algebraic and thus closed subset, so curves have finitely many singularities (nodes, cusps, etc). There is only one topological type of curve, but there are distinct algebraic and conformal structures (which turn out to be equivalent notions for curves). One can show $\mcm_g(\CC)$ is an orbifold of dimension $3g-3$, i.e. locally a quotient $M/G$ of a manifold by a finite group. Similarly $\mcm_g$ is a quasiprojective algebraic variety of dimension $3g-3$ with only quotient singularities. Mumford was the first to ask questions about its geometry, e.g. is it rational? ::: :::{.definition title="Rational and unirational varieties"} A variety $X\in \Alg\Var\slice k$ is **rational** if $X\birational \PP^n$, so there is a common open subset $X\contains U \subset \PP^n$.[^zar_huge] Equivalently, there is an isomorphism of rational functions \[ k(X) \cong k(\PP^n)\cong k(\AA^n) ,\] where the latter is comprised of quotients of polynomials. One can take $n=\dim X$, since if $N > n$ one can factor a dominant map $\PP^N \to X$ through a hyperplane $\PP^{N-1}\to X$ which is still dominant. $X$ is **unirational** if there is a dominant morphism $f: \PP^n \birational X$, so a map defined on an open subset whose image is dense. Equivalently, $X$ admits a parameterization by coordinates $x_1, \cdots, x_n$, so there is a rational parameterization.[^rat_uniq] [^rat_uniq]: Note that if $X$ is rational, this parameterization is unique. In this case, there is a degree $d$ finite extension $k(x_1,\cdots, x_n)$ over the pullback of $k(X)$. [^zar_huge]: Note that open sets in the Zariski topology are large. ::: :::{.question title="A motivating question in birational geometry, the Lüroth problem"} Is the converse true? I.e. if there is a finite extension $k(x_1,\cdots, x_n)$ over $k(X)$, is it true that $k(X) = k(y_1,\cdots, y_n)$? So does unirational imply rational? ::: :::{.remark} Lüroth proved this in dimension 1, and as a consequence of the classification of surfaces, the Italian school showed this in dimension 2. See the **Castelnuovo criterion**, which shows $X$ is rational iff $X$ is regular, i.e. $q \da h^1(X; \OO_X) = 0$ and $p_2 \da h^0(X; 2K_X) = 0$. ::: :::{.warnings} This is **false** in dimension 3. 3-4 counterexamples were given in the 70s/80s, first due to Iskovskih-Manin, a second due to Clemens-Griffith, and later due to Mumford. ::: :::{.exercise title="?"} Show that if $k \subsetneq K \subset k(X)$, then $K$ is monogenic (generated by a single element). ::: :::{.proposition title="?"} $\Mg$ is rational for $g=2$. ::: :::{.proof title="Sketch"} Note that $3(g-1)=3$, and a genus 2 curve is a branched cover $X\to \PP^1$ ramified at 6 points $\ts{0,1,\infty, \lambda_1, \lambda_2, \lambda_3}$. This yields a dominant map $\AA^3\to \mcm_2$ which is finite-to-1 and defined up to the action of $S_6$. This is not defined if points collide, which corresponds to collapsing cycles in $X$, and is degree $6!$. Here we can write $X = V(y^2 = x(x-1)(x - \lambda_1) (x- \lambda_2)(x - \lambda_3)) \subseteq \AA^2\slice{\CC}$. If any \( \lambda_i = \lambda_j \) for $i\neq j$, one obtains a singularity locally modeled on the node $y^2=x^2$, which is the following over $\RR$: ![](figures/2022-08-23_13-49-10.png) Over $\CC$, this is two hyperplanes intersecting in a single point. We can thus write \[ \mcm_2 = \AA^3\smts{\lambda_i = \lambda_j \st i\neq j} / S^6 ,\] which is rational and unirational. ::: :::{.proposition title="?"} $\mcm_3$ is rational and unirational. ::: :::{.proof title="Sketch"} We need to show that a genus 3 curve can be parameterized by 6 parameters. Noting that a genus 3 curve is planar of degree 4, which suffices -- planar curves are given by polynomials $f_4(x_1, x_2, x_3) = \sum a_n x^n$, and these are the parameters. ::: :::{.remark} One of the classical Italian algebraic geometers (either Severi or Castelnuovo) "proved" the false statement that $\Mg$ is unirational for all $g$. In fact this is only true for $g\leq 9$. The idea is good though: any curve $X\embeds \PP^n$ can be projected to a curve in $X \to \PP^2$ with only finitely many nodes. The coordinates for the nodes can serve as parameters. Having a curve pass through given points is a linear condition, as is saying it is singular at a point (by computing partial derivatives). Being a *node* is not a linear condition -- instead, it is a quadratic algebraic condition coming from the vanishing of a $2\times 2$ determinant. It's also not clear that imposing singularity conditions locally are all independent, since singularities at some points can force singularities at others. Mumford proved that $\Mg$ is not unirational for $g\geq 24$, and is in fact *general type*, which is far from unirational. ::: :::{.remark} Next time: more general introduction, stable curves, a bit about Hodge theory, then starting Mukai's book. :::