Introduction
This year we will focus on the Abundance conjecture, which says, \(\newcommand{td}{\mathrm{td}} \newcommand{ch}{\mathrm{ch}}\)
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 \(\deg(L|_C)\ge 0\).
a line bundle is semi-ample if there exists a morphism of \(X\to \mathbb{P}^N\) where \(L=f^* \mathcal{O}_{\mathbb{P}^N}(1)\).
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
\[ \left\{ \frac{dg_{ij}}{g_{ij}} \right\} \in H^1(X,\Omega_X^1) \cong H^{1,1}(X,\mathbb{C}) \]
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
Semi-ample implies nef, find a counter example for the reverse, i.e. a nef line bundle which is not semi-ample.
Find \(D, D'\) two effective divisors such that their cohomology classes are the same \([D]=[D']\) where \(D\) is semi-ample but \(D'\) 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
\[ c(E) = \sum_{k=0}^{\dim X} c_k(E) \in H^*(X,\mathbb{Z}). \]
(Pullback Functoriality) For any morphism \(f: X\to Y\), the \(c_k(f^* E) = f^* c_k(E)\).
This implies \(c_k(\mathcal{O}_X) = 1_{k=0}\) by considering the diagram
For any short exact sequences
\[ 0\to E'\to E \to E'' \to 0\]
we have
\[ c(E) = c(E')\cdot c(E'') \]
i.e. total Chern class is multiplicative.
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 \(E\)
\[ \dots \to L_2^{-n_2} \to L_1^{-n_1} \to E \to 0 \]
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 \(H^*(Y,\mathbb{Z})\) is generated by \(H^*(X,\mathbb{Z})\) and \(1,\zeta,\zeta^2,\dots,\zeta^{r-1}\).
We have
\[ - \zeta^r = \zeta^{r-1} c_1(E) + \dots + c_r(E) \]
Recommended Exercises
Verify the equality when \(E\) is a line bundle. Note that \(P(E) = X\) and what is \(\mathcal{O}(1)\)
Show that \(c_k(E) = 0\) if \(k>\min\{ r, \dim X \}\).
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.
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.
\[ \chi(X,L) = \int_X \ch(L) \cdot \td(X) \]
where
\(\td(X) = \prod_i \frac{c_i(T_X)}{1-\exp(-c_i(T_X))}\) is the Todd class,
\[ \td(T_X) = 1 + \frac{1}{2} c_1(T_X) + \frac{1}{12} (c_1^2(T_X) + c_2(T_X)) + \frac{1}{24} c_1 c_2 + \dots \]
\(\ch(L) = e^{c_1(L)} = \sum_{m=0}^{\dim X} \frac{1}{m!}c_1(L)^m\) 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
\[ \chi(X,L) = \deg L + 1-g. \]
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
\[ \chi(X,\mathcal{O}(mK_X)) = \frac{2m^3-3m^2+m}{12} c_1(K_X)^3 + \frac{m}{12} c_1(K_X)\cdot c_2(X) + \chi(\mathcal{O}_X) \]
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 \(p: X'\to X\) where \(X'\) smooth and \(p\) birational,
\[ Rp_* \mathcal{O}_{X'} = \mathcal{O}_X. \]
Equivalently, \(R^ip_*\mathcal{O}_{X'}=0\) for \(i>0\).
For example quotient, Du Val, terminal singularities are rational.
Claim (Exercise)
Let \(X\) be rational, \(L\) a line bundle, \(p: X'\to X\) resolution, then
\[H^i (X', p^*L) \cong H^i(X,L) \]
(Hint: Leray Spectral Sequence)
\[ \chi(X,L) = \chi(X', p^*L) \]
Find an example where \(H^1(X,L)\neq H^1(X', f^*L)\).
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
\[ \kappa(D) = \begin{cases} -\infty & \text{if } h^0(X,mD) = 0 \text{ for all } m>0 \\ \max\{ \dim \phi_{|kD|}(X) | k\in\mathbb{Z}\} & \text{otherwise} \end{cases} \]
where \(\phi_{|kD|}: X\to \mathbb{P}H^0(X,\mathcal{O}(kD))^*\) defined by all section generators is a rational map.
D is nef (i.e. (D)|_C ), define
\[ \mathcal{V}(X,D) = \max \{ k : D^k \neq 0 \text{ in } \mathrm{Pic}(X) \otimes \mathbb{R}\} \]
Remark:
\(D^k := c_1(\mathcal{O}_X(D))^k \in H^{2k}(X,\mathbb{Z})\).
\(D^k \in A^k(X)\), asking that \(D^k\cdot H^{n-k}\neq 0\) 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
\(\kappa(D) = \limsup_{m\to \infty} \frac{\log h^0(X,\mathcal{O}(mD))}{\log m}\).
Suppose \(\mathcal{O}(D)\) is semi-ample, then numerical dimension \(\mathcal{V}(X,D) = \kappa(X,D)\).
Suppose \(D\) is nef, then the numerical dimension \(\mathcal{V}(X,D)\ge \kappa(X,D)\).
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.
\[\kappa(X,D) = \limsup_{n\to \infty} \frac{\log h^0(X,\mathcal{O}(mD))}{\log m}\]
\(D\) 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 \(Y\).
\[\kappa(G) = \kappa(f^* G)\]