# Friday, February 12

:::{.remark}
Last time: the splitting principle.
Suppose we have \( \bundle{E} = L_1 \oplus \cdots \oplus L_r \) and let $x_i \da c_i(L_i)$.
Then $c_k(\bundle{E})$ is the degree $2k$ part of \( \prod_{i=1}^r (1 + x_i ) \) where each $x_i$ is in degree $2$.
This is equal to $e_k(x_1, \cdots, x_r)$ where $e_k$ is the $k$th elementary symmetric polynomial.
:::

:::{.example title="?"}
For example,

- $e_1 = x_1 + \cdots x_r$.

- $e_2 = x_1 x_2 + x_1 x_3 + \cdots = \sum_{i < j} x_i x_j$

- $e_3 = \sum_{i<j<k} x_i x_j x_k$, etc.


:::

:::{.remark}
The theorem is that any symmetric polynomial is a polynomial in the $e_i$.
For example, $p_2 = \sum x_i^2$ can be written as $e_1^2 - 2e_2$.
Similarly, $p_3 = \sum x_i^3 = e_1^3 - 3e_1 e_2 -3e_3$
Note that the coefficients of these polynomials are important for representations of $S_n$, see *Schur polynomials*.
:::

:::{.remark}
Due to the splitting principle, we can pretend that $x_i = c_i(L_i)$ exists even when \( \bundle{E} \) doesn't split.
If \( \bundle{E} \to X \), the individual symbols $x_i$ don't exist, but we can write '
\[
x_1^3 + \cdots + x_r^3 = e_1^3 - 3e_1 e_2 - 3e_3 \da c_1(\bundle{E})^3 + 3c_1(\bundle{E}) c_2(\bundle{E}) + \cdots
,\] 
which is a well-defined element of $H^6(X; \ZZ)$.
So this polynomial defines a characteristic class of \( \bundle{E} \), and this can be done for any symmetric polynomial.
We can change basis in the space of symmetric polynomials to now define different characteristic classes.
:::

:::{.definition title="Chern Character"}
The **Chern character** is defined as 
\[
\ch(\bundle{E}) 
&\da \sum_{i=1}^r e^{x_i}\in H^*(X; \QQ) \\
&\da \sum_{i=1}^r \sum_{k=0}^{\infty } {x_i^k \over k!} \\
&= \sum_{k=0}^{\infty } {p_k(x_1, \cdots, x_r) \over k!} \\
&= \rank(\bundle{E}) + c_1(\bundle{E}) + { c_1(\bundle{E}) - c_2(\bundle{E}) \over 2!} + { c_1(\bundle{E})^3 - 3c_1(\bundle{E}) c_2(\bundle{E}) - 3 c_3(\bundle{E}) \over 3!} + \cdots \\
& \qquad \in H^0 + H^2 + H^4 + H^6 \\
&=\ch_0(\bundle{E}) + \ch_1(\bundle{E}) + \ch_2( \bundle{E} ) + \cdots, \\
&   \quad \ch_i(\bundle{E}) \in H^{2i}(X; \QQ) 
.\]
:::

:::{.definition title="Total Todd class"}
The **total Todd class** 
\[
\td(\bundle{E})
\da
\prod_{i=1}^r { x_i \over 1 - e^{-x_i} }
.\]

Note that
\[
{x_i \over 1 - e^{-x_i} } = 1 + {x_i \over 2} + {x_i^2 \over 12} + {x_i^4 \over 720} + \cdots = 1 + {x_i \over 2} + \sum_{i=1}^{\infty } { (-1)^{i-1} B_i \over (2i)! } x^{2i}
.\]
where L'Hopital shows that the derivative at $x_i = 0$ exists, so it's analytic at zero and the expansion makes sense, and the $B_i$ are Bernoulli numbers.
:::

:::{.remark title="Very important and useful!!"}
\( \ch(\bundle{E} \oplus \bundle{F}) = \ch(\bundle{E}) + \ch(\bundle{F}) \) and \( \ch( \bundle{E} \tensor \bundle{F} ) = \sum_{i,j} e^{x_i + y_j} = \ch( \bundle{E} ) \ch(\bundle{F} ) \)
using the fact that \( c_1(L_1 \tensor L_2) = c_1(L_1) c_1(L_2) \).
So $\ch$ is a "ring morphism" in the sense that it preserves multiplication $\tensor$ and addition $\oplus$, making the Chern character even better than the total Chern class.
:::

:::{.definition title="Todd Class"}
Let $X \in \Mfd_\CC$, then define the **Todd class** of $X$ as $\td_\CC(X) \da \td(TX)$ where $TX$ is viewed as a complex vector bundle.
If $X\in \Mfd_\RR$, define $\td_\RR = \td(TX \tensor_\RR \CC)$.
:::

## Section 5: Riemann-Roch and Generalizations

:::{.remark}
Let $X\in \Top$ and let \( \sheaf{F} \) be a sheaf of vector spaces.
Suppose $h^i(X; \sheaf{F}) \da \dim H^i(X; \sheaf{F}) < \infty$ for all $i$ and is equal to 0 for $i \gg 0$.
:::

:::{.definition title="Euler Characteristic of a Sheaf"}
The **Euler characteristic** of \( \sheaf{F} \) is defined as 
\[
\chi(X; \sheaf{F}) \da \chi(\sheaf{F}) \da \sum_{i=0}^{\infty } (-1)^i h_i(X; \sheaf{F} )
.\]
:::

:::{.warnings}
This is not always well-defined!
:::

:::{.example title="?"}
Let $X\in \Mfd_{\compact}$ and take \( \sheaf{F} \da \constantsheaf{\RR} \), we then have 
\[
\chi(X; \constantsheaf{\RR}) = h^0(X; \RR) - h^1(X; \RR) + \cdots = b_0 - b_1 + b_2 - \cdots \da \chi_{\Top}(X)
.\]

:::

:::{.example title="?"}
Let $X = \CC$ and take $\sheaf{F} \da \OO \da \OO^{\holomorphic}$ the sheaf of holomorphic functions.
We then have $h^{> 0}(X; \OO) = 0$, but $H^0(X; \OO)$ is the space of all holomorphic functions on $\CC$, making $\dim_\CC h^0(X; \OO)$ infinite.
:::

:::{.example title="?"}
Take $X = \PP^1$ with $\OO$ as above, $h^0(\PP^1; \OO) = 1$ since $\PP^1$ is compact and the maximum modulus principle applies, so the only global holomorphic functions are constant.
We can write $\PP^1 = \CC_1 \union \CC_2$ as a cover and $h^i(\CC, \OO) = 0$, so this is an acyclic cover and we can use it to compute $h^1(\PP^1; \OO)$ using Čech cohomology.
We have 

- $C^0(\PP^1; \OO) = \OO(\CC_1) \oplus \OO(\CC_2)$

- $C^1(\PP^1; \OO) = \OO(\CC_1 \intersect\CC_2) = \OO(\CC\units)$.

- The boundary map is given by
\[
\del_0: C^0 &\to C^1 \\
( f(z), g(z) ) &\mapsto g(1/z) - f(z)
\]
  and there are no triple intersections.

Is every holomorphic function on $\CC\units$ of the form $g(1/z) - f(z)$ with $f,g$ holomorphic on $\CC$.
The answer is yes, by Laurent expansion, and thus $h^1 = 0$.
We can thus compute $\chi(\PP^1; \OO) = 1-0 = 1$.

:::