Can MathGPT solve the Hodge conjecture?

No. No one has a proof, so no solver or AI can give one. MathGPT recognises the problem and shows the exact statement, its status and the established facts, labelled as reference information.

UnsolvedAlgebraic geometryMillennium Prize Problem

Hodge conjecture

Q: Is there a prize?

Prize of US \$1,000,000 offered by the Clay Mathematics Institute (Millennium Prize Problems, announced 24 May 2000).

On projective algebraic varieties, every Hodge class is a rational combination of algebraic cycles.

Exact statement

\operatorname{Hdg}^{k}(X)=H^{2k}(X,\mathbb Q)\cap H^{k,k}(X)\ \overset{?}{=}\ \operatorname{span}_{\mathbb Q}\{[Z]:Z\subset X\ \text{algebraic, codim }k\}

On a non-singular complex projective algebraic variety X, every rational cohomology class of type (k, k) is a rational linear combination of the classes of algebraic subvarieties of X.

Status

Open. Stated by W. V. D. Hodge (1950); known for divisors (codimension 1), open in general.

Posed: 1950, W. V. D. Hodge.

Prize of US $1,000,000 offered by the Clay Mathematics Institute (Millennium Prize Problems, announced 24 May 2000).

In plain words

A smooth projective variety is a shape cut out by polynomial equations. Topology attaches cohomology classes to it, and Hodge theory splits these classes into types $(p,q)$ . The classes of algebraic sub-shapes are always of type $(k,k)$ and rational.

The conjecture says the converse: every rational $(k,k)$ class comes from geometry built out of polynomial equations. It would mean the topology of these spaces is controlled by algebra in a precise way.

Key results so far

1924: Lefschetz (1,1) theorem: the conjecture holds for k = 1 (and, by duality, for k = dim X − 1).
1962: Atiyah and Hirzebruch show the version with integer coefficients is false.
2002: Voisin shows the natural extension to all compact Kähler manifolds is false.

What our solver can and cannot do

No computer program or AI can solve the Hodge conjecture by being asked. If you paste it into the MathGPT solver, you get this reference information, labelled as such, never an invented proof. The solver can still compute related special cases and examples. Open the solver.

Frequently asked questions

What does the Hodge conjecture say?: On a non-singular complex projective algebraic variety X, every rational cohomology class of type (k, k) is a rational linear combination of the classes of algebraic subvarieties of X.
Has the Hodge conjecture been solved?: Open. Stated by W. V. D. Hodge (1950); known for divisors (codimension 1), open in general.
Is there a prize?: Prize of US $1,000,000 offered by the Clay Mathematics Institute (Millennium Prize Problems, announced 24 May 2000).
Can MathGPT solve the Hodge conjecture?: No. No one has a proof, so no solver or AI can give one. MathGPT recognises the problem and shows the exact statement, its status and the established facts, labelled as reference information.

References

P. Deligne, 'The Hodge Conjecture', Clay Mathematics Institute (2000).
C. Voisin, Hodge Theory and Complex Algebraic Geometry I–II, Cambridge (2002–2003).

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