# Tuesday, March 14

:::{.remark}
Today: Hirzebruch-Riemann-Roch and Chern classes.
Let $E\to X$ be a smooth $\CC\dash$vector bundle, then $\exists f: Y\to X$ such that $f^* E$ splits (as a smooth $\CC\dash$vector bundle) into a direct sum of line bundles, i.e. $f^* E\iso \bigoplus _{i\leq r} L_i$ where $r\da \rank E$.
One can ensure that $f^*: H^*(X; \ZZ)\injects H^*(Y; \ZZ)$ is injective.

This allows us to build $c_k(E)$ and thus $c_k(E)$ from the Chern roots $x_i \da c_1(L_i)$ of $E$.
Set $c_k(f^* E) \da \sigma_k(x_1,\cdots, x_r)$, the $k$th elementary symmetric polynomial in the $x_i$.
Note that $\prod_{i\leq r}(t + x_i) = \sum_{i\leq r} s_{r-i} x^i$ where e.g. 

- $s_1(x_1,\cdots, x_r) = \sum x_i$ 
- $s_2(x_1,\cdots, x_r) = \sum_{i < j} x_i x_j$
- $s_r(x_1,\cdots, x_r) = \prod x_i$

Define 
\[
c(E, t) \da \prod_{i\leq r} (t+x_i),\qquad c(E) \da c(E, 1) = \sum c_i
\]
where $c_i = H^{2i}(Y; \ZZ)$, and set $f^*(c_k E) \da c_k (f^* E)$; this uniquely defines $c_k(E)$.
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:::{.remark}
Note on proving the splitting principle: for $A\injects B\surjects C$ smooth vector bundles, putting a Hermitian metric on $B$ yields $A^\perp$ in $B$ and thus a (smooth) splitting.
Set $Y\da \ts{x\in X\st V_1 \subseteq V_2 \subseteq \cdots \subseteq V_r = E_x,\, \dim_\CC V_i = i}$, where $f: Y\to X$ by forgetting the flag.
Then $\dim Y = \dim X + \dim \Fl(\CC^r)$, then $f^* E$ admits a filtration $F^i$ where $F^1$ is a line bundle.
This yields SESs $F^{i-1} \injects F_i \surjects L_r$ which split.
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:::{.remark}
Define the total Chern character as $c(E) \da\sum_{i\leq r} e^{x_i}$.
Note that $\CC[t_1,\cdots, t_r]^{S_r} = \CC[s_1, \cdots, s_r]$, and 
\[
\sum_{i\leq r} \qty{\sum_{k\geq 0} { x_i^k \over k!} } 
&= r + (x_1+\cdots + x_r) + \qty{{x_1^2\over 2} + \cdots + {x_r^2\over 2}} + \cdots \\
&= r + c_1 + \qty{{c_1^2\over 2} - c_2} + \cdots
,\]
noting that e.g. $c_1^2 = \sum_i x_i^2 + 2 \sum_{i < j} x_i x_j$.
Define the total Todd class as 
\[
\Todd(E) \da \prod_i {1\over 1-e^{-x_i}} = 1 + {c_1\over 2} + {c_1^2 + c_2 \over 12} + {c_1c_2 \over 24}
.\]
Note this is holomorphic at each $x_i$ by L'Hopital, and moreover symmetric, and each term is a generating function for Bernoulli numbers.
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:::{.remark}
Recall that RR says that for a holomorphic line bundle of $L$, one can compute $\chi(L)$ in terms of $\deg L = \int c_1(L)$, a purely topological invariant.
The following theorem generalizes this:
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:::{.theorem title="Hirzebruch-Riemann-Roch"}
Let $E\to X$ be a holomorphic vector bundle over a compact complex manifold.
Defining $\chi(E) \da \sum (-1)^i h^0(X; E)$, 
\[
\chi(E) = \int_X \mathrm{Chern}(E) \Todd(\T_X )
\]
where the multiplication is in $H^*(X;\ZZ)$, noting that both classes are supported in $H^\even(X;\ZZ)$ and the integration means taking the top degree part in $H^{2\dim_\CC X}(X; \ZZ) \cong \ZZ$.
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:::{.remark}
Recovering RR: for $L\to X$ a line bundle on a curve, one has $\mathrm{Chern}(L) = e^{c_1(L)} = 1 + c_1(L)$ and $\Todd(T) = {c_1(T) \over 1 - e^{-c_1(T)}} = 1 + {c_1(T)\over 2}$.
Thus 
\[
\chi(L) = \int_X (1 + c_1(L) )\qty{1 + {c_1(T)\over 2}} = \int_X c_1(L) + {c_1(T)\over 2} = \deg L + {1\over 2}\int_X c_1(T)
.\]
Note that $c_1(T)$ is the fundamental class of the zeros of some section of $T$, i.e. the number of zeros of a vector field, which by Chern-Gauss-Bonnet yields $\int_X c_1(T) = \chi_\Top(X)$.
Thus 
\[
\chi(L) = \deg L + {1\over 2}\chi_\Top(X)
.\]
For a curve, $\chi_\Top(X) = 2-2g$, so $\chi(L) = \deg L + (1-g)$.
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:::{.remark}
Let $E\to S$ now be a line bundle over a surface, then $\chi(L) = h^0(L) - h^1(L) + h^2(L)$ since $L$ is a coherent sheaf and $\dim X = 2$.
HRR yields 
\[
\chi(L) = \int_S \qty{1 + c_1(L) + {c_2(L)\over 2!} }\qty{1 + {c_1(T)\over 2} + { c_1(T)^2 + c_2(T) \over 12}}
.\]
First consider the special case $L = \OO_S$ so $c_1(L) = 0$ and $\chi(\OO_S) = \int c_1(\T_S)^2 + c_2(\T_S)$.
By the splitting principle, if $\T_S = L_1 \oplus L_2$ then $\det \T_S = L_1 \tensor L_2$, so 
\[
c_1(\T_S) = c_1(\det \T_S) = c_1(L_1 \tensor L_2) = c_1(L_1) + c_1(L_2) = x_1 + x_2
.\]
Note also that $c_1(\det \T_S) = c_1(-K_S\dual) = -c_1(K_S)$ and so $c_1(\T_S) = K_S^2$.

For the second term, note that the top Chern class of $E$ is always the fundamental class of $V(s)$ for $s$ a generic smooth section of $E$.
In particular, $\int_S c_2(\T_S) = \chi_\Top(S)$ is the number of zeros of a smooth vector field.
This yields **Noether's formula**
\[
\chi(\OO_S) = {1\over 12}\qty{K_S^2 + \chi_\Top(S)}
.\]
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:::{.remark}
Let $X = \PP^2$, so $K = \OO(-3)$ and $\chi_\Top(X) = 3$, so $K^2 = (-3H)^2 = 9H^2 = 9$ so 
\[
\chi(\OO_{\PP^2}) = {1\over 12}\qty{ 9 + 3} = 1
.\]
Since $h^0(\OO) = 1$ and $h^1(\OO) = {1\over 2}\beta_1 = 0$, this yields $h^2(\OO) = 0$.
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:::{.remark}
RR for surfaces: 
\[
\chi(S, L) = \chi(\OO_S) + {1\over 2}(c_1(\T_S) c_1(L)) + {1\over 2}c_1(L))^2
.\]
Reworking this, note $c_1(\T_S) = -c_1(K_S)$ and thus 
\[
\chi(S, L) = \chi(\OO_S) + {1\over 2}\qty{L^2 - L\cdot K_S}
.\]
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:::{.remark}
Computing the Hodge diamond of a K3: recall $h^1(X) = 0$ and $K_X = \OO_X$.

- $H^0: 1$.
- $H^1: (0, 0)$. 
- $H^2: (1, N, 1)$ by Serre duality for some $N$.
- $H^3: (0, 0)$ by Poincare duality.
- $H^4: 1$ by Poincare duality.

Computing $N$: $\chi(\OO) = {1\over 12}(K^2 + \chi(X))$ where $\chi(\OO) = 1-0+1 = 2$ and $K^2 = 0$, this yields $\chi(X) = 24$ and $N = 22$.
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