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It is possible to compute the Betti numbers of a smooth complex variety $X(\mathbb{C})$ by computing the cardinality of $X(\mathbb{F}_{p^n})$ for a prime $p$ with good reduction and a finite number of positive integers $n$; in other words, by brute force.

The above claim is wrong, so I'll phrase it the other way around. The Betti numbers of a smooth complex variety control the behavior of the number of points on $X(\mathbb{F}_{p^n})$; for example, for a smooth projective curve of genus $g$ we have $|\text{Card}(X(\mathbb{F}_q))| - q - 1| \le 2g \sqrt{q}$.

Generally I find the relationship between the arithmetic and topological properties of varieties surprising, although maybe this is a temporary kind of surprise that arithmetic geometers are used to. Another example: if $X$ is a curve, then whether the curvature of $X(\mathbb{C})$ is positive, zero, or negative determines whether $X(\mathbb{Q})$ is rationally parameterizable, a finitely generated group, or finite (unless it's empty).

2 added 449 characters in body

It is possible to compute the Betti numbers of a smooth complex variety $X(\mathbb{C})$ by computing the cardinality of $X(\mathbb{F}_{p^n})$ for a prime $p$ with good reduction and a finite number of positive integers $n$; in other words, by brute force.

Generally I find the relationship between the arithmetic and topological properties of varieties surprising, although maybe this is a temporary kind of surprise that arithmetic geometers are used to. Another example: if $X$ is a curve, then whether the curvature of $X(\mathbb{C})$ is positive, zero, or negative determines whether $X(\mathbb{Q})$ is rationally parameterizable, a finitely generated group, or finite (unless it's empty).

It is possible to compute the Betti numbers of a complex variety $X(\mathbb{C})$ by computing the cardinality of $X(\mathbb{F}_{p^n})$ for a prime $p$ with good reduction and a finite number of positive integers $n$; in other words, by brute force.