The Hodge conjecture

Q: On a shape cut out by polynomial equations, is every hole that passes Hodge's "looks algebraic" test really traced out by smaller shapes that are also cut out by polynomial equations?

Write down x^2 + y^2 = 1 and you have drawn a circle. Write down more equations in more letters and you draw surfaces, and then shapes with more dimensions than we can picture. Mathematicians have studied these shapes for centuries, and one of the first things they ask about any shape is: where are its holes? In 1950 the English mathematician W. V. D. Hodge noticed that holes in these equation-made shapes carry a hidden label, a bit like a colour. A few colours are special. Every hole that is traced out by a smaller equation-made shape inside has a special colour. Hodge asked whether the reverse is true: if a hole has the special colour, must it come from equations too? Seventy-five years later nobody knows. It is one of the seven Millennium Prize Problems, with one million dollars offered by the Clay Mathematics Institute for a proof or a disproof. It is also the most abstract of the seven, so this page leans on analogies, and it says plainly where each one breaks.

Q: Has the Hodge problem been solved?

Still open. The Clay Mathematics Institute lists the Hodge conjecture as unsolved, and no prize has been awarded. There is no accepted proof and no accepted counterexample for general smooth projective varieties. Accepted progress. Specialists have recently settled more special cases. For abelian fourfolds, Floccari and Fu's proof of the discriminant 1 case is accepted for publication in the Journal de Mathématiques Pures et Appliquées, while Markman's general Weil-class paper (arXiv 2502.03415) was still listed as a preprint when we checked. Combined with Moonen–Zarhin, these give all abelian fourfolds. Markman's ICM 2026 survey reports that abelian varieties of dimension up to 5 follow. A recent AI claim, not verified. On 6 October 2026 OpenAI released manuscripts in its public openai/math repository. One entry, "Family 032", claims a proof of the rational Hodge conjecture for every complex CM abelian variety, in every dimension and codimension, with corollaries for the Tate conjecture over finite fields and the Hodge standard conjecture for abelian varieties. The repository itself warns that results are at different stages of verification. It has not been through peer review, and no independent check of it has been published. Even if it is correct, it is a special case: CM abelian varieties are one family of shapes, and the Millennium Problem covers all smooth projective varieties. Self-published papers claiming the full conjecture also appear from time to time. None is peer reviewed or accepted.

On a shape cut out by polynomial equations, is every hole that passes Hodge's "looks algebraic" test really traced out by smaller shapes that are also cut out by polynomial equations?

$1,000,000 · still unsolved 20 min read, with things to play with

Write down $x^2 + y^2 = 1$ and you have drawn a circle. Write down more equations in more letters and you draw surfaces, and then shapes with more dimensions than we can picture. Mathematicians have studied these shapes for centuries, and one of the first things they ask about any shape is: where are its holes?

In 1950 the English mathematician W. V. D. Hodge noticed that holes in these equation-made shapes carry a hidden label, a bit like a colour. A few colours are special. Every hole that is traced out by a smaller equation-made shape inside has a special colour. Hodge asked whether the reverse is true: if a hole has the special colour, must it come from equations too?

Seventy-five years later nobody knows. It is one of the seven Millennium Prize Problems, with one million dollars offered by the Clay Mathematics Institute for a proof or a disproof. It is also the most abstract of the seven, so this page leans on analogies, and it says plainly where each one breaks.

1 · Play first

Trace the holes of a doughnut

Before the real thing, a warm-up that is fully solved. A doughnut surface (a torus) can be made from a square by gluing its left edge to its right edge and its top edge to its bottom edge. A straight line that leaves one side comes straight back in at the other.

Pick a class $(a, b)$ : a loop that goes $a$ times across and $b$ times up before it closes. The widget draws the straight loop of that class on the flat square and on the doughnut. Then add a second loop and count where they must cross.

Flat view: leave by one edge, come back in at the opposite edge.

Doughnut view: across = round the ring, up = round the tube. Dashed parts are behind.

Loop A: go across a times and up b times

Loop B: c across, d up

Loop A (2, 3): gcd(2, 3) = 1, so this is one simple closed loop: it never crosses itself.

Crossings of A and B: |a·d − b·c| = |2·0 − 3·1| = 3 Count the dots: 3.

What to notice

Try (2, 3), then (4, 6). Both lines have the same slope, but (4, 6) is not one loop: it falls apart into 2 copies of (2, 3). A class is one simple loop exactly when $\gcd(a,b) = 1$ .
Two straight loops of classes $(a,b)$ and $(c,d)$ cross exactly $|ad - bc|$ times. Check the dots against the formula.
Here every class is traced by honest loops. In the real problem the hard question is exactly whether that still works, and the answer is only hoped for with fractions.
Read the note under the widget: the torus is a warm-up, not a case of the open problem.

2 · The equation, piece by piece

Here is the conjecture in symbols. The first line is Hodge's discovery: the holes of a smooth equation-made shape split into labelled pieces. The second line picks out the special ones and asks whether they all come from smaller shapes. Tap any piece to see what it means.

\begin{aligned} \htmlClass{t0}{H^k(X,\mathbb{C})} &= \htmlClass{t1}{\bigoplus_{p+q=k}} \htmlClass{t2}{H^{p,q}(X)} \\[4pt] \htmlClass{t3}{\mathrm{Hdg}^p(X)} &= \htmlClass{t4}{H^{2p}(X,\mathbb{Q})} \htmlClass{t5}{\cap} \htmlClass{t6}{H^{p,p}(X)} \;\htmlClass{t7}{\overset{?}{=}}\; \htmlClass{t8}{\Big\{ \textstyle\sum_i q_i\,\mathrm{cl}(Z_i) \Big\}} \end{aligned}

H^k(X,\mathbb{C})

The k-th cohomology of X

$X$ is the shape: smooth, cut out by polynomial equations, and closed up so nothing escapes to infinity. $H^k$ is the bookkeeping of its $k$ -dimensional holes, counted with complex-number coefficients.

In one sentence: on a smooth projective complex variety, every rational class of type $(p,p)$ should be a rational combination of classes of subvarieties. Every word in that sentence matters, and the chapters below unpack them one at a time.

3 · What's going on

1Shapes made by equations

An algebraic variety is the set of points where some polynomial equations hold. One equation in two letters draws a curve, like the circle $x^2+y^2=1$ . One equation in three letters usually draws a surface, like the sphere $x^2+y^2+z^2=1$ . Add more letters and equations and you get shapes of any dimension.

The Hodge conjecture uses complex numbers, numbers of the form $a + bi$ with $i^2 = -1$ . Each complex number needs two real numbers to pin it down, so everything doubles. A complex "curve" is a surface to our eyes, two real dimensions. A complex surface has four real dimensions. The open cases start at complex dimension 4, which is 8 real dimensions.

Two more words from the official statement. Smooth (or non-singular) means the shape has no sharp points or self-crossings. Projective means it sits inside complex projective space, a way of adding "points at infinity" so the shape is closed up and nothing runs off the edge. Honest caveat: any picture you see of such a shape is a real slice or a shadow, never the object itself.

x^2 + y^2 = 1 \quad\longrightarrow\quad \text{a circle}

The simplest example of a shape made by an equation. The Hodge conjecture is about much bigger relatives.

2Holes, and the bookkeeping called cohomology

A doughnut has two essentially different loops: one around the "short way" through the hole, one around the "long way" round the ring. Any other loop on it can be slid into some combination of those two. That is what the widget shows: the class $(a,b)$ means " $a$ times one way, $b$ times the other".

Cohomology is the systematic bookkeeping for this, in every dimension. A shape has groups $H^0, H^1, H^2, \dots$ , one for each dimension, and the size of $H^k$ records how many independent $k$ -dimensional holes the shape has. You can count with whole numbers ( $\mathbb{Z}$ ), with fractions ( $\mathbb{Q}$ ) or with complex numbers ( $\mathbb{C}$ ), and those choices matter later.

Where the analogy breaks: "hole" is a stand-in. A cohomology class is better thought of as a way of measuring or detecting cycles, and most classes in high dimensions have no picture you could draw. Keep the doughnut in mind for intuition, but do not expect every class to look like a gap you could poke a finger through.

H^1(\text{torus},\mathbb{Z}) \cong \mathbb{Z}^2 \quad (\text{one number for each way round})

3Hodge's prism: splitting holes into types

Here is Hodge's discovery. On a smooth projective complex variety, each cohomology group with complex coefficients splits cleanly into pieces $H^{p,q}$ , with $p + q = k$ . It is a bit like a prism splitting white light into colours. The label $(p,q)$ records how a class behaves with respect to the complex directions: $p$ holomorphic ones and $q$ anti-holomorphic ones.

There is a mirror symmetry built in. Complex conjugation, which turns $a + bi$ into $a - bi$ , swaps the $(p,q)$ piece with the $(q,p)$ piece. So the pieces come in mirror pairs, except the balanced ones in the middle, $H^{p,p}$ , which are their own mirror image.

Where the analogy breaks: light is split by a physical prism, but here the splitting is a theorem that uses the complex structure of the shape. A shape that is only a real surface, like our doughnut drawn in the widget, has no such splitting at all. That is one reason the torus is only a warm-up.

H^k(X,\mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X), \qquad \overline{H^{p,q}} = H^{q,p}

4Hodge classes, and the bank-note test

Take a subvariety $Z$ inside $X$ : a smaller shape, also cut out by polynomial equations, with $p$ fewer complex dimensions. It traces out a hole, its class $\mathrm{cl}(Z)$ , in $H^{2p}$ . Two facts hold automatically. The class can be counted with whole numbers, so it is rational. And it lands in the balanced piece $H^{p,p}$ .

So every hole that comes from equations passes a test: rational, and of type $(p,p)$ . Classes that pass the test are called Hodge classes. The conjecture asks whether the test is perfect: is every Hodge class built from subvarieties?

Think of a bank note. Every genuine note passes the watermark check. The Hodge conjecture says that nothing which passes the check is forged. Where the analogy breaks: in the conjecture we are allowed to build a class from many subvarieties with fractional and negative weights, like paying with $\tfrac12$ of one note minus 3 of another. That freedom is essential, as the next chapter shows.

\mathrm{Hdg}^p(X) = H^{2p}(X,\mathbb{Q}) \cap H^{p,p}(X)

5Why fractions, and why "projective", really matter

Hodge first stated his question with whole-number coefficients. That version is false. In 1961 Michael Atiyah and Friedrich Hirzebruch used a tool called topological K-theory to find integral classes that pass the test but are not traced by subvarieties. Their examples were "torsion" classes, ones that vanish when multiplied by some whole number. In 1992 János Kollár found worse examples: a class that is not algebraic even though some whole-number multiple of it is.

So the modern statement uses fractions on purpose. Kollár's example is a perfect illustration: if $N$ copies of a class come from subvarieties, then with fractions the class itself is $\tfrac{1}{N}$ of that, and the rational conjecture is happy. Whole numbers can fail where fractions succeed.

The word "projective" also matters. In 1977 Steven Zucker showed there are complex tori (not projective) with Hodge classes that are not algebraic, and in 2002 Claire Voisin showed that natural extensions to compact Kähler manifolds, a wider family of smooth complex shapes, are false. Any proof has to use both the fractions and the equations in an essential way, which rules out many purely topological approaches.

N\cdot\alpha = \sum_i n_i\,\mathrm{cl}(Z_i) \;\Longrightarrow\; \alpha = \sum_i \tfrac{n_i}{N}\,\mathrm{cl}(Z_i)

With fractions you can always divide by N. With whole numbers you cannot, and that is where the integral version breaks.

6What is already known

The first case is older than the conjecture. In 1924 Solomon Lefschetz proved that every integral class of type $(1,1)$ in $H^2$ comes from divisors, which are subvarieties of codimension 1 (one complex dimension less than $X$ ). This is the Lefschetz $(1,1)$ theorem, and it settles the case $p = 1$ for every smooth projective variety. With a further result called hard Lefschetz, it settles every variety of dimension 3 or less, which is why Clay says the conjecture is known below dimension 4.

For abelian varieties, which are higher-dimensional cousins of the torus that are also cut out by equations, a lot is known. For general ones, and for products of elliptic curves, all Hodge classes come from divisors (Mattuck 1958, Tate 1965). In 1969 David Mumford found an abelian fourfold with Hodge classes that do not come from divisors, and in 1977 André Weil pointed to so-called Weil classes as the key test case.

Moonen and Zarhin showed in 1999 that for abelian varieties of dimension below 5, Weil classes were the only obstacle. In 2025 Eyal Markman posted a proof that Weil classes on abelian fourfolds of Weil type are algebraic (arXiv 2502.03415, still listed as a preprint when we checked), and Salvatore Floccari and Lie Fu gave a new proof of the discriminant 1 case, accepted at the Journal de Mathématiques Pures et Appliquées. Together with Moonen–Zarhin this gives the Hodge conjecture for all abelian fourfolds. In his ICM 2026 survey Markman states that it follows for abelian varieties of dimension up to 5.

p = 1:\quad \mathrm{Hdg}^1(X) = \{\textstyle\sum_i q_i\,\mathrm{cl}(D_i)\} \quad (\text{Lefschetz, 1924})

7What the doughnut can and cannot teach

The widget answers a baby version of Hodge's question: "which loop classes on a torus are traced by nice pieces?" The answer is complete. Every class $(a,b)$ is traced by straight loops. It is one simple loop exactly when $\gcd(a,b) = 1$ , and otherwise it is $\gcd(a,b)$ parallel copies. Two classes must cross $|ad - bc|$ times.

The real problem differs in three ways. First, the widget's loops are 1-dimensional holes on a real surface, with no $(p,q)$ splitting at all. Second, on the torus whole numbers always work, while in high dimensions they are known to fail. Third, the interesting Hodge classes on a complex torus of dimension 2 sit in $H^2$ , where Lefschetz already proved everything. The open cases begin in complex dimension 4.

What the warm-up does capture is the shape of the question: a list of all holes, a smaller list of holes that nice pieces can trace, and a test that might or might not tell the two lists apart. In the doughnut case the lists match and we can see why. The Hodge conjecture says they still match, with fractions, in every dimension.

(a,b)\cdot(c,d) = ad - bc

The intersection number on the torus, which the widget counts as crossings.

4 · Why it's so hard

If the conjecture is true, why has nobody proved it in 75 years? Pierre Deligne's official description for Clay lays out several walls. The common thread: knowing that a class passes the test tells you nothing about how to build the subvarieties that should trace it.

No machine for building subvarieties

In codimension 1, line bundles and their sections manufacture divisors on demand, which is how Lefschetz's case is proved. For codimension 2 and higher there is no known machine that turns a cohomology class into an actual subvariety.

Slicing does not work

A natural plan is induction: cut $X$ with hyperplanes and use the conjecture for the slices. Deligne explains that this needs control of objects called intermediate Jacobians, where the algebraic cycles can form a wildly large group, so the method is generally inapplicable.

You cannot even find the Hodge classes

In a family of varieties, no algorithm is known for deciding where a given class becomes of type $(p,p)$ . The candidates themselves are hard to locate.

The nearby versions are false

The integral version and the Kähler version both fail. A proof must use rational coefficients and projectivity in an essential way, which rules out many otherwise promising topological and analytic tools.

5 · What a solution would change

A foundation for motives

Alexander Grothendieck imagined a "universal cohomology" theory called motives. The Hodge conjecture is one of the pillars that would make that theory work, together with the standard conjectures.

Links to number theory

The conjecture sits beside the Tate conjecture, its counterpart for shapes over finite fields and number fields. Results for abelian varieties often pass between the two, through theorems of James Milne.

Understanding what equations can see

A proof would say that topology and algebra agree completely on the special holes: every one that looks algebraic is algebraic. There is no direct technology riding on it. Hodge numbers do appear in string theory, but we found no well-sourced claim that the conjecture itself has physical consequences.

6 · Where it stands

As of October 2026

Still open. The Clay Mathematics Institute lists the Hodge conjecture as unsolved, and no prize has been awarded. There is no accepted proof and no accepted counterexample for general smooth projective varieties.

Accepted progress. Specialists have recently settled more special cases. For abelian fourfolds, Floccari and Fu's proof of the discriminant 1 case is accepted for publication in the Journal de Mathématiques Pures et Appliquées, while Markman's general Weil-class paper (arXiv 2502.03415) was still listed as a preprint when we checked. Combined with Moonen–Zarhin, these give all abelian fourfolds. Markman's ICM 2026 survey reports that abelian varieties of dimension up to 5 follow.

A recent AI claim, not verified. On 6 October 2026 OpenAI released manuscripts in its public openai/math repository. One entry, "Family 032", claims a proof of the rational Hodge conjecture for every complex CM abelian variety, in every dimension and codimension, with corollaries for the Tate conjecture over finite fields and the Hodge standard conjecture for abelian varieties. The repository itself warns that results are at different stages of verification. It has not been through peer review, and no independent check of it has been published. Even if it is correct, it is a special case: CM abelian varieties are one family of shapes, and the Millennium Problem covers all smooth projective varieties.

Self-published papers claiming the full conjecture also appear from time to time. None is peer reviewed or accepted.

1924
Solomon Lefschetz proves the $(1,1)$ case. Every integral class of type $(1,1)$ in $H^2$ comes from divisors.
1930s–1940s
W. V. D. Hodge develops harmonic forms and the decomposition of cohomology into $(p,q)$ pieces.
1950
Hodge states the conjecture at the International Congress of Mathematicians, originally with whole-number coefficients.
1961
Atiyah and Hirzebruch show the integral version is false, using topological K-theory.
1969
Mumford finds an abelian fourfold with Hodge classes not generated by divisors. Grothendieck corrects Hodge's generalised conjecture.
1977
Weil singles out Weil classes as the key test case. Zucker shows projectivity cannot be dropped.
1992
Kollár finds a non-torsion counterexample to the integral version.
1995
Cattani, Deligne and Kaplan prove that Hodge loci in families are algebraic, as the conjecture predicts. Widely seen as the strongest evidence for it.
1999
Moonen and Zarhin show that for abelian varieties of dimension below 5, Weil classes are the only obstacle.
2002
Claire Voisin shows natural extensions to compact Kähler manifolds are false.
2025
Markman posts a proof that Weil classes on abelian fourfolds of Weil type are algebraic (preprint). Floccari and Fu give a new proof for discriminant 1 (accepted, JMPA). Together with Moonen–Zarhin, this covers all abelian fourfolds.
6 Oct 2026
OpenAI posts unverified manuscripts claiming the rational Hodge conjecture for CM abelian varieties, a special case.

Common mix-ups

MythIt is about the holes in a doughnut.

ActuallyLoops on a torus are a warm-up, and a fully solved one. The open cases are in complex dimension 4 and up, in degree $2p \ge 4$ .

MythEvery hole can be filled with whole pieces.

ActuallyThe whole-number version is false (Atiyah–Hirzebruch, Kollár, Totaro). Fractional combinations are an essential part of the conjecture, not a technicality.

MythIt holds for any smooth complex shape.

ActuallyIt fails for complex shapes that are not projective (Zucker, Voisin). Being cut out by polynomial equations really matters.

MythOpenAI solved the Hodge conjecture.

ActuallyIn October 2026 OpenAI claimed a special case, CM abelian varieties. The claim is unverified, and even if correct it would not settle the general problem.

MythNobody has made any progress.

ActuallyMany special cases are known: divisors, everything up to dimension 3, many abelian varieties, and since 2025 all abelian fourfolds.

7 · References

The Hodge Conjecture (problem page), Clay Mathematics Institute, 2026
The Hodge Conjecture (official problem description), Pierre Deligne, Clay Mathematics Institute, 2000
Millennium Prize Problems rules, Clay Mathematics Institute, 2026
Hodge conjecture, Wikipedia, 2026
Cycles on abelian 2n-folds of Weil type from secant sheaves on abelian n-folds, Eyal Markman, 2025
The Hodge conjecture for Weil fourfolds with discriminant 1 via singular OG6-varieties, Salvatore Floccari and Lie Fu, 2025
Secant sheaves and Weil classes on abelian varieties (ICM 2026 survey), Eyal Markman, 2025
openai/math repository, OpenAI, 2026
openai/math CONTENTS.md (Family 032), OpenAI, 2026
OpenAI Proves The Rational Hodge Conjecture For CM Abelian Varieties: Understand What This Means, OfficeChai, 2026

Next problemPoincaré: If every loop in a closed three-dimensional space can be shrunk to a point, must that space be the 3-sphere?