# Monday, September 18

:::{.remark}
Today: computing $\Cl(X_\Sigma)$.
Write $T_N = (\cstar)^n \actson X_\Sigma$ for \( \Sigma \subseteq N_\RR \).
:::

:::{.theorem title="?"}
There is an exact sequence
\[
M &\to \Div_{T_N}( X_ \Sigma) = \bigoplus_{ \rho\in \Sigma(1)}\ZZ D_{ \rho} \to \Cl(X_ \Sigma) \to 0 \\
m &\mapsto \div(\chi^m),\qquad D \mapsto [D]
.\]
The first map is injective iff $\gens{u_ \rho\st \rho\in \Sigma(1)}_\RR = N_\RR$.
:::

:::{.remark}
This says the class group can be computed from only $T\dash$invariant divisors, and surjectivity here means any divisor is linearly equivalent to a toric divisor.
Moreover, $\Cl(X_ \Sigma)$ is a finitely-generated abelian group, which is not necessarily free.
:::

:::{.example title="?"}
What is $\Cl(\PP^2)$?
We have an exact sequence 
\[
0 \to \ZZ^2 &\to
\Div_T(\PP^2) = \gens{D_0, D_1, D_2}_\ZZ \to \Cl(\PP^2)\to 0 \\
(1, 0) &\mapsto \div(\chi^{1, 0}) = \sum v_{D_i}(\chi^{1, 0}) D_i = D_0 - D_2 \longrightarrow 0 \\
(0, 1) &\mapsto \div(\chi^{0, 1}) = \sum v_{D_i}(\chi^{0, 1}) D_i = D_1 - D_2 \longrightarrow 0
.\]
This yields $D_0 \sim D_2$ and $D_1\sim D_2$, so $\Cl(\PP^2) = \gens{D_1}_\ZZ = \gens{D_2}_\ZZ = \gens{D_3}_\ZZ$.
Writing $\Div_T(\PP^2) \cong \ZZ^3$ with $D_1 = (1, 0, 0), D_2 = (0, 1, 0), D_3 = (0, 0, 1)$ we similarly see 
\[
\Cl(\PP^2) = \ZZ^3 / \gens{(1,0,-1), (0, 1, -1)}_\ZZ\cong \ZZ^3/\ZZ^2 \cong \ZZ
\]
since this the denominator is a saturated lattice.
:::

:::{.example title="?"}
Compute $\Cl(\PP^n)$.
:::

## $\S$ 3.2: Cartier Divisors

:::{.remark}
Let \( \varphi: \abs{ \Sigma} \to \ZZ \) be a piecewise linear function on \( \Sigma \), i.e. \( \varphi \) is linear on each cone.
Recall Weil divisors are $\sum a_i D_i$ with $D_i$ prime divisors, and Cartier divisors are locally principal.
:::

:::{.theorem title="?"}
Assume $\gens{ \rho\st \rho\in \Sigma(1)}_\RR = N_\RR$.
There is an exact sequence 
\[
0 \to M \to \CDiv_{T_n}(X_ \Sigma) \to \Pic(X_ \Sigma)\to 0
.\]
:::

:::{.remark}
When is a Weil divisor Cartier?
:::

:::{.theorem title="?"}
A Weil divisor $D = \sum a_i D_{ \rho_i}$ is Cartier iff there exists an integral piecewise-linear function \( \varphi: \abs{ \Sigma} \to \RR \), so $\phi(N \intersect \abs{\Sigma}) \subseteq \ZZ$, such that \( \varphi(u_{ \rho_i}) = -a_i \) for each $i$.
:::

:::{.theorem title="?"}
$D = \sum a_i D_i$ is a principal Weil divisor iff $\phi$ is linear.
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:::{.remark}
Let $X_ \Sigma = \PP^2$, is $D_{ \rho_0}$ Cartier?
One has $D_{\rho_0} = 1 D_{\rho_0} + 0 D_{\rho_1} + 0 D_{\rho_2}$, so we need $\phi: \RR^2\to \RR$ where $\phi(1, 0) = -1$ and $\phi(0, 1) = \phi(-1, -1) = 0$.
One can take $\phi = \inp{\wait}{ (-1, 0)}$ and check $\phi(1, 0) = -1$ and $\phi(0, 1) = 0$.
So set $\phi(x, y) = -x$ on the first cone $\sigma_{01}$, and $\phi(x, y) = -x + y$ on the remaining cones by similarly checking $\inp{\wait}{(-1, 1)}$ on $\sigma_{20}$ and $\phi(x, y) = 0$ on $\sigma_{12}$ by checking $\inp{\wait}{(0, 0)}$ works.
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