Open Questions

Is every Hodge class on a smooth complex projective algebraic variety really a rational combination of the cohomology classes of actual algebraic subvarieties?

Citation Formats

General Reference

Mathematics Atlas. "Is every Hodge class on a smooth complex projective algebraic variety really a rational combination of the cohomology classes of actual algebraic subvarieties?." Accessed August 30, 2026. https://dev-mathematics.interactiveatlas.org/open-questions-index/hodge-conjecture-open-question.

APA Style

Mathematics Atlas. (n.d.). Is every Hodge class on a smooth complex projective algebraic variety really a rational combination of the cohomology classes of actual algebraic subvarieties?. Retrieved August 30, 2026, from https://dev-mathematics.interactiveatlas.org/open-questions-index/hodge-conjecture-open-question

BibTeX

@misc{mathematicsatlas-is-every-hodge-class-on-a-smooth-complex,
  author = {Mathematics Atlas},
  title = {Is every Hodge class on a smooth complex projective algebraic variety really a rational combination of the cohomology classes of actual algebraic subvarieties?},
  year = {2026},
  url = {https://dev-mathematics.interactiveatlas.org/open-questions-index/hodge-conjecture-open-question},
  note = {Accessed August 30, 2026}
}

Open Question

Proved only in codimension one, by the Lefschetz theorem on (1,1) classes. Higher codimension cases, including named examples such as the diagonal cycle's Kunneth components, remain open, and the natural integral version of the conjecture is known to be false.

What would resolve this A general proof for every codimension, or an explicit Hodge class shown not to be a combination of algebraic cycle classes.

OpenAlgebraic geometryClay Mathematics Institute

Cross-Tradition Connections

Question On

Hodge Conjecture

Source Clay Mathematics Institute

Sources

Clay Mathematics Institute

Clay Mathematics Institutehttps://www.claymath.org/wp-content/uploads/2022/06/hodge.pdfView the Source

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