gpt 6 astra ultra attempt on Math-29

Deligne's Absolute Hodge Conjecture

Definitions and mathematical statement

A rational Hodge class $\alpha\in H^{2p}(X,{\mathbb{Q}}(p))$ becomes type $(0,0)$ in the twisted Hodge structure. Via algebraic de Rham comparison it has a conjugate on $X^\sigma$ for each $\sigma\in\operatorname{Aut}({\mathbb{C}})$. Absoluteness means \[ \forall\sigma\in\operatorname{Aut}({\mathbb{C}}),\qquad \alpha^\sigma\in H^{2p}(X^\sigma,{\mathbb{Q}}(p))\text{ is a Hodge class}. \] The conjecture requires this for every rational Hodge class on every smooth projective complex variety. Arbitrary field automorphisms of ${\mathbb{C}}$, not only continuous ones or complex conjugation, are quantified.

Formulation and concept references: [S1].

Short English statement

Does every rational Hodge class remain rational and of Hodge type after every automorphism of the complex numbers?

Research attempt

Scope, process, and checked context.

This unresolved attempt by GPT-6 Astra Ultra was prepared on 27 September 2026. The target is the assertion for every smooth projective complex variety, every rational Hodge class, and every field automorphism of $\mathbb C$. The attack is to build explicit certificates from already absolute classes and identify the information that such certificates cannot supply. The opposite track tests whether preservation of the algebraic Hodge filtration alone could force rationality.

Charles--Schnell [S1], Sections 2.2--2.4, supplies the comparison and functoriality conventions: algebraic cycle classes are absolute, and pullback, cup product and proper pushforward respect conjugation. Deligne's theorem in [E1] settles absoluteness of Hodge classes on abelian varieties. It does not assert the Hodge conjecture for those varieties or settle the present question for arbitrary $X$. These are external inputs, not results of the calculations below.

Separate the filtration from the rational lattice.

Temporarily use untwisted de Rham notation. Write $V_B=H^{2p}(X,\mathbb Q)$ and $V_{\rm dR}=H^{2p}_{\rm dR}(X/\mathbb C)$, and let $c_X:V_B\otimes\mathbb C\to V_{\rm dR}$ be comparison. For a field automorphism $\sigma$, algebraic base change gives a $\sigma$-semilinear map $s_\sigma$ between the de Rham spaces of $X$ and $X^\sigma$. It transports the Hodge filtration: \[ s_\sigma(F^pV_{\rm dR})=F^pH^{2p}_{\rm dR}(X^\sigma/\mathbb C). \] In the twisted notation of the statement, comparison must also include the Tate factor. It is the resulting conjugated class that must belong to the rational Betti space of $X^\sigma$. One cannot apply $\sigma$ entrywise to a rational Betti vector and declare that this is geometric conjugation: the two comparison maps lie between those operations.

Once a conjugated class is rational and belongs to $F^p$, its Hodge type is automatic. Indeed, after undoing the Tate twist, a rational class is real; if its components have first index at least $p$, complex conjugation forces second index at least $p$ too. In degree $2p$ only $(p,p)$ remains. Thus preservation of $F^p$ genuinely advances the argument, but leaves precisely the rationality of the conjugated comparison vector to prove.

Here is a linear-algebra stress test of that gap. Give the de Rham line $\mathbb Ce$ the full filtration and give its comparison rational structure the line $\mathbb Q\tau e$, with $\tau$ transcendental. The vector $\tau e$ is rational for this structure. An automorphism sending $\tau$ to $\tau+1$ exists: extend $\tau$ to a transcendence basis over $\overline{\mathbb Q}$, translate that variable, and extend the field isomorphism to an algebraic closure. Semilinear transport sends the vector to $(\tau+1)e$. Relative to the same comparison line its coefficient is $1+1/\tau$, which is not rational. The filtration was preserved throughout. This artificial line is not asserted to be the cohomology of a variety, and $\tau$ is not a substitute for a geometric Tate period. It isolates a missing compatibility condition, rather than giving a counterexample to Deligne's conjecture.

A complete elementary subcase.

Suppose the class is a rational linear combination of algebraic cycle classes. Write $\alpha=\sum_i a_i[Z_i]$, $a_i\in\mathbb Q$. Algebraic base change takes $Z_i$ to a codimension-$p$ cycle $Z_i^\sigma$ and commutes with its de Rham cycle class. The comparison convention therefore gives \[ \alpha^\sigma=\sum_i a_i[Z_i^\sigma]. \] Every term is rational and Hodge, for every $\sigma$, so $\alpha$ is absolute. In particular the Hodge conjecture would imply the target, but assuming that conjecture at this point would replace one open problem by a stronger input.

The codimension-one case can be completed using the classical Lefschetz $(1,1)$ theorem. Multiply a rational $(1,1)$ class by an integer to make it integral. The exponential sequence identifies such an integral class with the first Chern class of a holomorphic line bundle; on a projective variety GAGA makes the bundle algebraic. Conjugating that bundle proves the desired assertion, and division by the integer recovers the original class. At codimensions zero and $\dim X$, component and point classes give the same conclusion. No analogous general description of all intermediate Hodge classes is assumed.

Lemma 94.1: an explicit pairing certificate.

Let $d=\dim X$. Suppose $\beta_1,\ldots,\beta_r$ are absolute classes in $H^{2p}(X,\mathbb Q(p))$, and $\gamma_1,\ldots,\gamma_r$ are absolute classes in $H^{2d-2p}(X,\mathbb Q(d-p))$. Use the rationally normalized trace and set \[ G_{ji}=\operatorname{Tr}_X(\beta_i\smile\gamma_j),\qquad G\in\operatorname{GL}_r(\mathbb Q). \] For a rational Hodge class $\alpha$, put \[ b_j=\operatorname{Tr}_X(\alpha\smile\gamma_j),\quad c=G^{-1}b,\quad \alpha_0=\sum_i c_i\beta_i,\quad \delta=\alpha-\alpha_0. \] Then $\alpha_0$ is absolute, $\delta$ is rational Hodge and annihilates all the $\gamma_j$. If these pairings separate rational Hodge classes, then $\delta=0$ and $\alpha$ is absolute.

Proof. All entries of $G$ and $b$ are rational by the rational cup product and trace. Invertibility makes $c$ rational, so each $\alpha_0^\sigma=\sum_i c_i\beta_i^\sigma$ is rational Hodge. Subtracting gives a rational Hodge residual at the original embedding. The vector of its pairings is $b-Gc=0$. The additional separation hypothesis is exactly that this vanishing forces the residual to be zero. No information about the conjugate of the unknown residual was used.$\square$

For a checked instance take $X=\mathbb P^a\times\mathbb P^b$, with hyperplane classes $h,k$. The cohomology ring is $\mathbb Q[h,k]/(h^{a+1},k^{b+1})$. In codimension $p$ the nonzero monomials $h^ik^{p-i}$ form a basis. Their complementary classes are $h^{a-i}k^{b-p+i}$. The pairing matrix is the identity: an off-diagonal pair has an exponent exceeding $a$ or $b$, whereas a matching pair integrates to one. These algebraic classes separate the entire degree, so the certificate resolves all Hodge classes of this family. The check includes $p=0,a+b$ and the cases $a=0$ or $b=0$.

A route through abelian varieties and its precise limit.

Suppose there are algebraic correspondences inducing morphisms of rational Hodge structures \[ u:H^{2p}(X,\mathbb Q(p))\longrightarrow H^{2q}(A,\mathbb Q(q)), \qquad v:H^{2q}(A,\mathbb Q(q))\longrightarrow H^{2p}(X,\mathbb Q(p)), \] where $A$ is abelian, and suppose $v u(\alpha)=\alpha$. The class $u(\alpha)$ is Hodge and is absolute by Deligne's theorem. Functoriality of the conjugated correspondence then makes $v(u(\alpha))=\alpha$ absolute. This argument only needs a retraction on the class under consideration, not on the whole cohomology group. It supplies a concrete sufficient certificate which could be tested for particular geometric constructions.

The unsuccessful step is to assert that every Hodge class admits such a retraction. An arbitrary embedding of a rational vector space into another vector space is insufficient: the maps must be compatible with the Hodge structures and remain rational after geometric conjugation. Constructing $v$ merely by selecting a Betti basis assumes the compatibility that the attempt is trying to establish.

Likewise, spreading $X$ over a finitely generated field does not make a chosen Betti class a globally defined horizontal algebraic section. Monodromy can move it. Even when a section exists, an argument using transport from another fibre needs an absolute starting class and the hypotheses of a valid deformation theorem. The existence of a model over a small field alone supplies neither item.

Verification and next targets.

The certificate uses rational coefficients, so $\sigma$ fixes the coefficients but not the comparison matrices. Trace normalization includes the Tate twists in both complementary degrees. The toy example tests a logical implication only. The projective-product matrix can be verified symbolically from its ring relations; no numerical period approximation establishes rationality. All subcases above are classical consequences or explicit reorganizations of known facts; no novelty is claimed.

The first missing statement is that arbitrary rational Hodge classes are exhausted by certificates stable under every conjugation. Three specific next targets are to exhibit separating absolute complementary classes on a new family, construct an algebraic retraction to abelian cohomology for its primitive part, or control the residual $\delta$ under conjugated comparison maps. A putative counterexample must have nonzero residual beyond the available certificates and fail rationality for some field automorphism; the artificial line does not meet the geometric hypotheses.

Final status.

UNRESOLVED. The argument proves the pairing-certificate lemma, its projective-product application, and the conditional retraction criterion. It does not prove conjugate rationality for a general Hodge class and does not construct a geometric counterexample.

Sources

• S1 F. Charles and C. Schnell, Notes on absolute Hodge classes $2011$, Section 2.5, Conjecture 25. https://www.math.ens.psl.eu/~charles/AH.pdf. Formulation source.
• E1 P. Deligne (notes by J. S. Milne), Hodge cycles on abelian varieties, in Hodge Cycles, Motives, and Shimura Varieties, Lecture Notes in Mathematics 900 $1982$, pp. 9--100; revised electronic version hosted by J. S. Milne. https://www.jmilne.org/math/Documents/Deligne82.pdf. Primary source for the abelian-variety theorem; consulted 27 September 2026.