# Wednesday, October 04


:::{.remark}
Recall that if $D\in \Div(X)$ with $D = \sum a_{\rho}D_{ \rho}$ is Cartier iff $\exists \phi: \abs{ \Sigma}\to \RR$ a PL function with $\phi(\rho) = -a_{ \rho}$ for every $\rho\in \Sigma(1)$.
Say $D$ is bpf if $\OO_X(D)$ has no base points, i.e. points $p$ with $s(p) = 0$ for all $s\in H^0(\OO_X(D))$.
If $D$ is bpf, there is a well-defined map $\psi: X\to \PP^n \da \PP H^0(\OO_X(D))\dual$ where $p\mapsto (f\mapsto f(p)$.
Note that sections $s: X\to \mcl$ correspond to rational maps $f:X\to \CC$, i.e. $f\in H^0(\OO_X(D))$ by, over $U$, writing $\pi\inv(U) = U\times \CC$ and setting $s = (x, f(x))$.
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:::{.theorem title="?"}
For $X$ proper and $\mcl\in \Pic(X)$, the global sections $H^0(\mcl)$ form a finite dimensional $\CC\dash$vector space.
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:::{.remark}
For $X = \PP^1$, we have $H^0(\OO_X) = \CC$ and $\PP(H^0(\OO_X)\dual) = \PP(\CC) = (\CC\smz)/\cstar = \pt$.
For toric varieties, $D$ is bpf if the corresponding PL function is convex.
So if $D = 1\ts{0} + 0\ts{\infty}$, then $D$ is Cartier since the corresponding graph is convex

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So $\OO_{\PP^1}$ is bpf.
One can check $\OO_X(\ts{0} + \ts{\infty}) = \OO_X(2)$.
Note that for $\PP^n$, there is a bijection $H^0(\OO_X(n)) \cong \kxn^{\homog}$.
So e.g. $\OO_X(-1)$ has no sections.
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:::{.definition title="?"}
$D$ is very ample if $D$ is bpf and $X\to \PP H^0(\OO_X(D))\dual$ is an embedding.
$D$ is ample if $kD$ is very ample for some $k$.
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:::{.remark}
$D\in \Div(X)$ is Cartier if $\phi$ is strictly convex.
E.g. $D = 0$ on $\PP^1$ is bpf and not ample since $\phi = 0$ is not strictly convex.
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