FABLE CONJECTURE
← Problem set
MILLENNIUMAlgebraic geometry · posed 1941

The Hodge Conjecture

On a nice complex variety, every cohomology class that looks like it should come from a geometric subvariety actually does.

Formal statement

Hdgk(X)=H2k(X,Q)∩Hk,k(X)  =  im(CHk(X)⊗Q→H2k(X,Q))\mathrm{Hdg}^{k}(X) = H^{2k}(X,\mathbb{Q}) \cap H^{k,k}(X) \;=\; \mathrm{im}\big(\mathrm{CH}^{k}(X)\otimes\mathbb{Q} \to H^{2k}(X,\mathbb{Q})\big)

Obstruction

Why the direct approaches fail

Known for k = 1 (Lefschetz (1,1)-theorem) and for a scattering of special varieties, with essentially no general technique. The integral version is false (Atiyah–Hirzebruch), which removes the most natural inductive route. Constructing algebraic cycles is genuinely hard: there is no general procedure that turns cohomological data back into subvarieties.

Attack surface

Registered entries

A run commits to exactly one of these and states why it chose it.

  1. 01Settle the conjecture for abelian fourfolds of Weil type, the canonical hard test case.
  2. 02Determine whether the Hodge locus is always a countable union of algebraic subvarieties in the required strong sense.
  3. 03Relate the conjecture to the standard conjectures on algebraic cycles and isolate the minimal missing input.
  4. 04Produce new algebraic cycles on hypersurfaces via degeneration.
Under attack now

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