Let us set up the problem. Let $X/\mathbb{C}$ be a smooth projective variety. If $Y \subseteq X$ is a subvariety of codimension $k$, it gives rise to classes $[Y] \in H^{2k}(X, \mathbb{Z})$.
Question 1. How much cohomology do subvarieties see? In other words, what is the span of such these classes?
Theorem 2 (Hodge). For $X$ as above, there is a decomposition $$ H^k(X, \mathbb{C}) \cong \bigoplus_{p+q=k} H^{p,q}(X), \quad H^{p,q}(X) = H^q(X, \Omega^p). $$
This makes the restriction that $[Y]$ has to lie in $H^{k,k}(X)$.
Conjecture 3 (naive Hodge). These classes $[Y]$ integrally span $H^{2k}(X, \mathbb{Z}) \cap H^{k,k}(X)$.
This turns out to be false, but rationally this is still conjectured to be true. So let us define $$ \mathrm{Hdg}^k(X) = H^{2k}(X, \mathbb{Q}) \cap H^{k,k}(X). $$
Conjecture 4 (Hodge). These classes $[Y]$ rationally span $\mathrm{Hdg}^k(X)$.
We know very few cases of the Hodge conjecture. What we will focus on is a result that is a consequence of the Hodge conjecture.
Hodge loci#
Let us suppose that we have $f \colon X \to S$ a smooth projective morphism where $S/\mathbb{C}$ is connected, smooth and quasi-projective. We can now consider the Hodge classes $\mathrm{Hdg}^{2k}(X_s)$ in families, e.g., what its dimension is. We will see that there is a “generic dimension” and make the following definition.
Definition 5. We define the Hodge locus $\mathrm{Hdg}(f)$ of $f$ as the locus on $S$ where the dimension of $\mathrm{Hdg}^{2k}(X_s)$ is not minimal.
Proposition 6. The Hodge locus $\mathrm{Hdg}(f)$ is a countable union of analytic proper closed subsets.
Roughly speaking, on a small analytic neighborhood, we can look at the locus on which the flat extension of the section is Hodge. This can be checked to be an analytic condition.
Proposition 7. Assuming the Hodge conjecture, $\mathrm{Hdg}(f)$ is a countable union of closed subvarieties.
Here is a rough sketch of the proof. Using moduli theory, we can construct a scheme $C_M \to S$ parametrizing relative cycles on $\frac{1}{k}(Z_+ - Z_-)$ where $k, \deg(Z_+), \deg(Z_-) \le M$. Now there is a map from $C_M$ to the total space of the de Rham vector bundle, which is the de Rham realization. Taking the union over all $M$, we see that the Hodge locus is a countable union of varieties over $S$. Now the Hodge locus is union of the subvarieties that don’t dominate $S$.
The strategy of Bakker–Klingler–Tsimerman#
Theorem 8 (Cattani–Deligne–Kaplan). The Hodge locus is algebraic (unconditionally).
Later Bakker–Klingler–Tsimerman gave a simpler proof. The idea is that given $X \to S$, there is a space of Hodge structures $H(f)$ and a period map $$ \Phi_f \colon S \to H(f). $$ Here, note that $H(f)$ is generally analytic and not algebraic. Locally on $S$, if $\alpha$ is a cohomology class then there exists $$ H_\alpha \subseteq H(f) $$ which is the locus on which $H_\alpha$ is Hodge, and this is analytic. Then we have $$ \mathrm{Hdg}(f) = \bigcup \Phi_f^{-1}(H_\alpha) $$
Note that if $S$ was projective, we will be done by Chow’s theorem, which says that a closed analytic set of a projective variety is algebraic. This is what Cattani–Deligne–Kaplan in fact does. They compactify $S$ to a projective $\bar{S}$ and then “extend the period map” to conclude. The new approach is to use a variant of Chow’s theorem that does not assume compactness.
Theorem 9 (definable Chow). Let $X$ be a variety and let $Y \hookrightarrow X$ be a closed analytic subset that is definable. Then $Y$ is algebraic.
But what is definable? There is a large gap between algebraic geometry and analytic geometry. Definable geometry is somewhere in the middle, where we allow some analytic functions but not all.
Theorem 10 (Bakker–Klingler–Tsimerman). The moduli of Hodge structures $H(k)$ is definable, and the period map $\Phi_f$ is also definable.
Remark 11. It turns out that any analytic function is definable on small balls, and so Chow follows from definable Chow.