# Monday, September 11

## $\S 3.4$: Properness and Compactness.

:::{.remark}
Equip $X_\Sigma$ with the classical topology induced by $\CC^n$ -- if $X_\Sigma$ is smooth, it is naturally a complex manifold.
Question: when is $X_\Sigma$ compact in the classical topology?
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:::{.warnings}
In the Zariski topology, everything is compact!
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:::{.remark}
Zero-dimensional algebraic varieties of finite type can be covered by finitely many affine varieties.
More generally, any finite set of points is compact in the Zariski topology.
Why? Take any Zariski-open cover $\Union_\alpha U_\alpha$, and look at the complete of a single point $a_p$; this is an open set $U$.
Note that $U_\alpha^c$ is finitely many points $p_1, \cdots, p_n$. 
For each $p_i$, choose $U_{\alpha, p_i}$ containing $p_i$; then $\ts{U_{\alpha, p_i}}_{i\leq n}$ forms a finite subcover.
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:::{.exercise title="?"}
Show that $\AA^1$ is compact in the Zariski topology.
More generally, show this for $\AA^n$.
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:::{.remark}
Note that $\Sigma$ is compact in the analytic topology iff $\abs{\Sigma} = \RR^n$.
By considering fans, use this to check that $\AA^1$ is not compact but $\PP^1$ is compact.
Moreover, $\Sigma$ is a complete fan iff every 1-parameter subgroup $\cstar \actson (\cstar)^n \subseteq X_{ \Sigma}$ has a limit in $X_{ \Sigma}$ as $t\to 0$.
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:::{.remark}
An algebraic variety is proper if the morphism $X\to \pt$ is proper; this occurs iff $X$ is compact in the classical topology.
A toric variety $X_{ \Sigma}$ is proper iff $\Sigma$ is complete.
Note that projective implies proper, and although the converse holds for curves and surfaces, it does not hold generally.
A morphism $X\to Y$ is proper if it is universally closed -- for every $\psi: X\to Y$, take the fiber product: 

\[\begin{tikzcd}
	{X\fiberprod{Y} Z} && X \\
	\\
	Z && Y
	\arrow["\psi", from=3-1, to=3-3]
	\arrow["{\pi_Z}", from=1-1, to=3-1]
	\arrow["{\pi_X}", from=1-1, to=1-3]
	\arrow["f", from=1-3, to=3-3]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}\]
> [Link to Diagram](https://q.uiver.app/#q=WzAsNCxbMCwwLCJYXFxmaWJlcnByb2R7WX0gWiJdLFsyLDAsIlgiXSxbMiwyLCJZIl0sWzAsMiwiWiJdLFszLDIsIlxccHNpIl0sWzAsMywiXFxwaV9aIl0sWzAsMSwiXFxwaV9YIl0sWzEsMiwiZiJdLFswLDIsIiIsMCx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==)

Then $f$ is proper iff $\pi_X$ is a closed map for every such diagram, i.e. the images of closed sets are closed.
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:::{.theorem title="?"}
The following two definitions of properness are equivalent: 

1. $X$ is proper in the Zariski topology $\iff X$ is compact in the classical topology.
2. $X$ is proper $\iff X\to \pt$ is a proper morphism.

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:::{.exercise title="?"}
Show that $\AA^1$ is not proper using definition (2).

> Hint: take $Z = \PP^1$. Try taking a closed set which is not strictly contained in a fiber of $\AA^1\times \PP^1 \to \PP^1$, e.g. a diagonal $\ts{(x, x) \mid vx-u = 0}$ where $v, u$ are coordinates on $\AA^1\times \PP^1$. 

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