Introduction

This year we will focus on the Abundance conjecture, which says, $$

Let $X/\mathbb{C}$ be a smooth projective manifold, and $K(X)$ is nef, then $K(X)$ is semi-ample.

and $c_1(X)$ lies in $H^{1,1}(X,\mathbb{C})\subset H^2(X,\mathbb{C})$.

Recall

• a line bundle is nef if for all proper curves $C\subset X$, the degree $\mathrm{deg}(L{|}_C)\ge 0$.

• a line bundle is semi-ample if there exists a morphism of $X\to {\mathbb{P}}^N$ where $L=f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^N}(1)$.

• semi-ample implies nef, but not the reverse.

• sloagan : cohomology determines geometry.

Let $L$ be a line bundle, when $(U_{ij},g_{ij})$ is the local data, we have

$$\left\{\frac{d{g}_{ij}}{g_{ij}}\right\}\in H^1(X,{\mathrm{\Omega }}_X^1)\cong H^{1,1}(X,\mathbb{C})$$

Our main theorem of the course is

Thm Aboundance conjecture is true for $\dim \le 3$.

Remark, in $\dim \le 2$ it is solved by Italian school. $\dim =3$ is by Kawamata, Miyaoka, Mori in the 80s.

Reference for the course: Flips and abundance for algebraic threefolds

Two Exercises

1. Semi-ample implies nef, find a counter example for the reverse, i.e. a nef line bundle which is not semi-ample.

2. Find $D,D^{\prime }$ two effective divisors such that their cohomology classes are the same $[D]=[D^{\prime }]$ where $D$ is semi-ample but $D^{\prime }$ is not.

Riemann-Roch Theorem

Today’s goal we will talke about Riemann-Roch theorem and Chern classes.

Let $X$ be a complex projective manifold and $E$ is a vector bundle on $X$. Now we want to assign certain cohomolgy classes to $E$, define the $k$-th Chern class $c_k(E)\in H^{2k}(X,\mathbb{Z})$ as

$$c_k(E)=(-1{)}^k\left[\frac{1}{k!}{\left(\frac{i}{2\pi }\right)}^k\mathrm{tr}\left(F^k\right)\right]$$

where $F$ is the curvature of the Chern connection on $E$. Don’t understand it? no worry we will define it in a Grothendieck way here for your convenience.

Grothendieck’s Axiomatic Definition Of Chern Classes

Define the total Chern class as

$$c(E)=\sum _{k=0}^{\dim X}{c}_k(E)\in H^{\ast }(X,\mathbb{Z}).$$

1. (Pullback Functoriality) For any morphism $f:X\to Y$, the $c_k(f^{\ast }E)=f^{\ast }{c}_k(E)$.

   This implies $c_k({\mathcal{O}}_X)=1_{k=0}$ by considering the diagram

   

2. For any short exact sequences

   $$0\to E^{\prime }\to E\to E^{″}\to 0$$

   we have

   $$c(E)=c(E^{\prime })\cdot c(E^{″})$$

   i.e. total Chern class is multiplicative.

3. If $E=\mathcal{O}(D)$ is a line bundle, we have

   $$c(\mathcal{O}(D))=1+[D]\in H^0+H^2$$

   where $c_1(D)=[D]$.

Now given any vector bundle $E$, you can always find ample $L$ that its certain negative power surjects onto $E$

$$\cdots \to L_2^{-n_2}\to L_1^{-n_1}\to E\to 0$$

then

$$c(E)=c(L_1{)}^{n_1}\cdot c(L_2{)}^{-n_2}\dots$$

Let $E$ be vector bundle of randk $r$, $Y=\mathbb{P}(E)={\mathrm{Proj}}_X{\mathrm{Sym}}^{\bullet }E$, then

Let $\zeta =c_1(\mathcal{O}(-1))$, then $H^{\ast }(Y,\mathbb{Z})$ is generated by $H^{\ast }(X,\mathbb{Z})$ and $1,\zeta ,{\zeta }^2,\dots ,{\zeta }^{r-1}$.

We have

$$-{\zeta }^r={\zeta }^{r-1}{c}_1(E)+\cdots +c_r(E)$$

Recommended Exercises

1. Verify the equality when $E$ is a line bundle. Note that $P(E)=X$ and what is $\mathcal{O}(1)$

2. Show that $c_k(E)=0$ if $k>min\{r,\dim X\}$.

3. Define the total Chern class of a manifold as $c(X):=c(T_X)$. Prove $c({\mathbb{P}}^n)=(1+\zeta {)}^{n+1}$, $\zeta$ is the hyperplane class.

4. Let $X\subset {\mathbb{P}}^n$ be a degree $d$ hypersurface, compute $c(X)$. Note the exact sequence

   $$0\to T_X\to T_{{\mathbb{P}}^n}{|}_X\to N_{X/{\mathbb{P}}^n}=\mathcal{O}(d){|}_X\to 0$$

Fact: $c_n(X)=\sum _i(-1{)}^i\dim H^i(X,\mathbb{Z})$, the top Chern class is the Euler characteristic.

Hirzebruch-Riemann-Roch Theorem

Let $X$ be a smooth projective variety, $L$ be a line bundle.

$$\chi (X,L)={\int }_X\mathrm{ch}(L)\cdot \mathrm{td}(X)$$

where

• $\mathrm{td}(X)=\prod _i\frac{c_i(T_X)}{1-\exp (-c_i(T_X))}$ is the Todd class,

  $$\mathrm{td}(T_X)=1+\frac{1}{2}{c}_1(T_X)+\frac{1}{12}(c_1^2(T_X)+c_2(T_X))+\frac{1}{24}{c}_1{c}_2+\dots$$

• $\mathrm{ch}(L)=e^{c_1(L)}=\sum _{m=0}^{\dim X}\frac{1}{m!}{c}_1(L{)}^m$ is the Chern character,

• $\chi (X,L)=\sum (-1{)}^i\dim H^i(X,L)$.

When $X$ is a curve, the todd class is $1+\frac{1}{2}(2-2g)=2-g$ and the Chern character is just the first Chern class, and the theorem reduces to

$$\chi (X,L)=\mathrm{deg}L+1-g.$$

Another corollary is for surfaces

$$\chi (X,L)=\chi (X,{\mathcal{O}}_X)+\frac{1}{2}{c}_1(L)\cdot (c_1(L)-c_1(K_X))$$

$$\chi (X,\mathcal{O}(D))=\chi ({\mathcal{O}}_X)+\frac{1}{2}D\cdot (D-K)$$

where $K_X$ is the canonical class.

Recommended Exercises

$$\chi (X,\mathcal{O}(m{K}_X))=\frac{2m^3-3m^2+m}{12}{c}_1(K_X{)}^3+\frac{m}{12}{c}_1(K_X)\cdot c_2(X)+\chi ({\mathcal{O}}_X)$$

Asymptotic Riemann-Roch

$X$ is a normal projective variety of dimension $n$, $D$ is a Cartier divisor and $E$ is a Weil divisor. Then $\chi (X,\mathcal{O}(D+E))=\frac{D^n}{n!}{m}^n+O(m^{n-1})$.

Proof

$X$ is smooth, it follows from the Hirzebruch-Riemann-Roch theorem. In general we can use a singular HRR or induction on dimension.

Let’s do a special case where $X$ has rational singularities.

$X$ has rational singularities if for some resolution of singularities $p:X^{\prime }\to X$ where $X^{\prime }$ smooth and $p$ birational,

$$R{p}_{\ast }{\mathcal{O}}_{X^{\prime }}={\mathcal{O}}_X.$$

Equivalently, $R^i{p}_{\ast }{\mathcal{O}}_{X^{\prime }}=0$ for $i>0$.

For example quotient, Du Val, terminal singularities are rational.

Claim (Exercise)

Let $X$ be rational, $L$ a line bundle, $p:X^{\prime }\to X$ resolution, then

$$H^i(X^{\prime },p^{\ast }L)\cong H^i(X,L)$$

(Hint: Leray Spectral Sequence)

$$\chi (X,L)=\chi (X^{\prime },p^{\ast }L)$$

Find an example where $H^1(X,L)\ne H^1(X^{\prime },f^{\ast }L)$.

Kodaira And Numerical Dimension

Let $X$ again be a normal projective variety whose dimension is $n$. $D$ a Cartier divisor. Define the Kodaira dimension $\kappa (D)$ as

$$\kappa (D)=\{\begin{array}{ll}-\mathrm{\infty }& \text{if }{h}^0(X,mD)=0\text{ for all }m>0\\ max\{\dim {\varphi }_{|kD|}(X)|k\in \mathbb{Z}\}& \text{otherwise}\end{array}$$

where ${\varphi }_{|kD|}:X\to \mathbb{P}{H}^0(X,\mathcal{O}(kD){)}^{\ast }$ defined by all section generators is a rational map.

D is nef (i.e. (D)|_C ), define

$$\mathcal{V}(X,D)=max\{k:D^k\ne 0\text{ in }\mathrm{Pic}(X)\otimes \mathbb{R}\}$$

Remark:

• $D^k:=c_1({\mathcal{O}}_X(D){)}^k\in H^{2k}(X,\mathbb{Z})$.

• $D^k\in A^k(X)$, asking that $D^k\cdot H^{n-k}\ne 0$ for all ample (equiv some) ample $H$.

Exercise:

• pick $0\le k\le n$, find $(X,L)$, where $\dim X=n$ and $K(L)=k,\mathcal{V}(L)=k$.

• $K(X,D),\mathcal{V}(X,D)\le \dim X$.

Reference

• Notions of Numerical Dimensions don’t coincide, Lesieutre.

Prop for a normal projective variety $X$, $\dim X=n$, $D$ Cartier

1. $\kappa (D)=\underset{m\to \mathrm{\infty }}{lim sup}\frac{\log h^0(X,\mathcal{O}(mD))}{\log m}$.

2. Suppose $\mathcal{O}(D)$ is semi-ample, then numerical dimension $\mathcal{V}(X,D)=\kappa (X,D)$.

3. Suppose $D$ is nef, then the numerical dimension $\mathcal{V}(X,D)\ge \kappa (X,D)$.

4. Suppose $D$ is nef and $\mathcal{V}=n$, then $\kappa =n$.

Question. For what pairs $j,k$ does there exists a divisor $D$ such that $\kappa (D)=j,\mathcal{V}(D)=k$?

Lecture 2

$X$ be a normal projective variety of dimension $n$, $D$ a Cartier divisor.

• $$\kappa (X,D)=\underset{n\to \mathrm{\infty }}{lim sup}\frac{\log h^0(X,\mathcal{O}(mD))}{\log m}$$

• $D$ semi-ample $\Rightarrow \mathcal{V}(X,D)=\kappa (X,D)$

• $D$ nef $\Rightarrow \mathcal{V}(X,D)\ge \kappa (X,D)$

• $D$ nef, $\mathcal{V}(X,D)=n\Rightarrow \kappa (X,D)=n$

If $f:X\to Y$ is a surjective morphism between projective varieties, then let’s say $G$ is a Cartier divisor on $Y$.

$$\kappa (G)=\kappa (f^{\ast }G)$$