What is the answer to Hodge conjecture?

The current version of the Hodge Conjecture says: every Hodge class is algebraic. In other words, we can take any Hodge class, no matter how weird, and deform it into an honest subvariety.

Who proposed Hodge conjecture?

mathematician William Hodge
The conjecture was first formulated by British mathematician William Hodge in 1941, though it received little attention before he presented it in an address during the 1950 International Congress of Mathematicians, held in Cambridge, Mass., U.S. In 2000 it was designated one of the Millennium Problems, seven …

Does Tate conjecture imply Hodge conjecture?

Thus assuming Deligne’s conjecture that all Hodge classes are absolutely Hodge (which is known for abelian varieties and K3 surfaces), the Tate conjecture implies the Hodge conjecture.

Who proved the Poincaré conjecture?

Total citations10. To mathematicians, Grigori Perelman’s proof of the Poincaré conjecture qualifies at least as the Breakthrough of the Decade.

What is the Hodge conjecture in geometry?

In mathematics, the Hodge conjecture is a major unsolved problem in algebraic geometry that relates the algebraic topology of a non-singular complex algebraic variety to its subvarieties.

What is harmonic differential form according to Hodge conjecture?

I read from the Internet that according to the Hodge conjecture, a certain harmonic differential form in a projective, non-singular algebraic variety is a rational linear combination of the cohomology classes of algebraic cycles. Could anyone explain me what is that particular differential form?

Is the Hodge conjecture a Kähler manifold?

Statement of the Hodge conjecture. Because projective space carries a Kähler metric, the Fubini–Study metric, such a manifold is always a Kähler manifold. By Chow’s theorem, a projective complex manifold is also a smooth projective algebraic variety, that is, it is the zero set of a collection of homogeneous polynomials.

Is the cup product compatible with the Hodge decomposition?

(See Hodge theory for more details.) Taking wedge products of these harmonic representatives corresponds to the cup product in cohomology, so the cup product is compatible with the Hodge decomposition: