Compactness of the complex points of a $\mathbb{C}$-variety

Question

Let $X$ be a $\mathbb{C}$-variety and let $X_{cx}$ denote the topological space formed by its $\mathbb{C}$-points with the complex topology (i.e. the associated analytic space).

If $X$ is projective, then $X_{cx}$ is a compact topological space. I was thinking about the converse and found a GAGA-cite that $X_{cx}$ is compact iff $X$ is proper. This was not what I expected: If you draw for example the real points of the variety defined by $X^2+Y^2+Z^2=1$, you get a $2$-sphere which is nice and compact.

I can think about $\mathbb{C}$-points honestly only for curves since $\mathbb{R}^{\geq 4}$ is beyond of what I can imagine as a whole (and compactness is not a local property). Therefore, if the GAGA-cited theorem is true, there must be a big difference in relation to compactness between the real and the complex picture.

Why are there no non-trivial affine $X$ with $X_{cx}$ compact (a proper affine morphism is finite) and where is (morally) the difference to the real picture?

Answer 1

$\mathbb{C}$ is algebraically closed, which guarantees that affine varieties have points at infinity, and that there are ways to connect the points at infinity to the finite points. Seen from within affine space, these are ways to move a point off to infinity while staying on the variety. Hence the complex locus is unbounded and therefore noncompact. I'm assuming here the theorem, somewhat complicated to prove, that connectedness in Zariski and analytic topology is the same for complex varieties (and schemes).

In $\mathbb{R}^n$ the real locus can be bounded, because an affine variety may not have any real points at infinity, as in the example with the sphere $x^2+y^2+z^2=1$.

Answer 2

The complex points of a $\mathbb C$-scheme suffice to reconstruct the scheme.
More precisely there is an equivalence of categories between these schemes (more precisely integral schemes of finite type over $\mathbb C$) and classical varieties, obtained by sending the scheme $X$ to its set $X_{cx}$ of $\mathbb C$-points , endowed with the restriction of the structural sheaf: $\mathcal O_{X_{cx}}=O_X|X_{cx}$.

The root of this equivalence resides in two facts:
$\bullet$ The ring $\mathbb C[T_1,\ldots , T_n]$ is Jacobson, meaning that every prime ideal is the intersection of the maximal ideals containing it.
This is true for any polynomial ring but the following second fact necessitates algebraic closedness of $\mathbb C$:

$\bullet \bullet$Every maximal ideal $\mathfrak m \subset \mathbb C[T_1,\ldots , T_n]$ is given by a point $a=(a_1,\ldots, a_n)$: $\mathfrak m=(T_1-a_1,\ldots , T_n-a_n)$

This is completely false for $ \mathbb R[T_1,\ldots , T_n]$ and is the moral reason why the $\mathbb R$- points of a $\mathbb C$-scheme are vastly insufficient for determining the scheme: just think of the real points of $x_1^2+\ldots+x_n^2+x_{n+1}^2=0$ in $\mathbb P^n_\mathbb R$ [Actually there is literally nothing to think about :-)]

Answer 3

The following argument gives a rough idea for the non-compactness of affine $\mathbb{C}$-curves: let $X$ be such a curve and let $Y$ be its projective closure. Then $Y\setminus X$ is a finite set; take some point $x$ in this set. Chose a family $(A_i)_{i\in\mathbb{N}}$ of closed (in the fine topology) subsets $A_i\subseteq Y$ with the following properties: $A_i\supseteq A_{i+1}$, $A_i\neq A_{i+1}$, $\bigcap_i A_i =\{x\}$.

Then $(X\setminus A_i)_i$ is an open covering of $X$ that possesses no finite subcovering.