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gpt 6 astra ultra attempt on Math-11

Exploration · LLM-generated · gpt 6 astra ultra ·

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at unsolvedmath.com: open. Claims here do not change the source status.

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Contents
  1. Generalized Hodge Conjecture
  2. Definitions and mathematical statement
  3. Short English statement
  4. Research attempt
  5. Scope and checked context (27 September 2026).
  6. Elementary support bookkeeping.
  7. Lemma 24.1: a correspondence that constructs support.
  8. A conditional solution of the weight-one remainder.
  9. An unconditional odd-degree family with exact coniveau.
  10. Stress test: intersections of supports do not support sums.
  11. Verification and the first remaining gap.
  12. Sources

Generalized Hodge Conjecture

Definitions and mathematical statement

A rational Hodge substructure $V\subseteq H^k(X,{\mathbb{Q}})$ has Hodge coniveau at least $c$ if $V^{a,b}=0$ whenever $a<c$ or $b<c$. Let $M^cH^k$ be the sum of all such substructures. Define geometric coniveau by \[ N^cH^k(X,{\mathbb{Q}})=\sum_{\substack{Z\subseteq X\text{ closed algebraic}\\ \operatorname{codim}Z\ge c}} \ker\left(H^k(X,{\mathbb{Q}})\to H^k(X\setminus Z,{\mathbb{Q}})\right). \] The conjecture is $M^cH^k(X,{\mathbb{Q}})=N^cH^k(X,{\mathbb{Q}})$ for every smooth projective complex $X$ and all relevant $k,c$. Finite sums can be supported on the union of their supporting algebraic sets.

Formulation and concept references: [S1].

Short English statement

Does every part of cohomology whose Hodge types allow high-codimension support actually have such algebraic support?

Research attempt

Scope and checked context (27 September 2026).
This is the rational generalized Hodge conjecture for smooth projective complex varieties. The input is an entire rational Hodge substructure, not a list of individually chosen complex Hodge components. Voisin's survey [S1, Corollary 4.5 and Section 4.2; author PDF in E1] supplies the established inclusion $N^cH^k\subseteq M^cH^k$ and separates geometric realization from algebraicity of correspondences. The ordinary Hodge conjecture by itself is not invoked as a theorem. The additional catalogue references are not treated as resolutions of the general target.

The approach is to construct support by projecting an algebraic correspondence from an auxiliary variety. This gives a precise dimension test. In the first odd degree above the cycle-class case, the auxiliary Hodge structure can be realized by a curve, giving a conditional reduction with no unknown realization step. The remaining algebraicity step is isolated explicitly. Product examples then test both odd-degree support and a tempting but false operation on supporting sets.

Elementary support bookkeeping.
Write $d=\dim X$. If classes $\alpha_j$ vanish on $X\setminus Z_j$, their sum vanishes on $X\setminus\bigcup_j Z_j$, by restriction. A finite union of codimension-at-least-$c$ algebraic sets still has codimension at least $c$. Since $H^k(X,\mathbb Q)$ is finite dimensional, a basis of any subspace contained in $N^cH^k$ uses only finitely many supports. Thus that entire subspace has one common algebraic support. There is no countable union or limiting algebraic set in this deduction.

For a smooth inclusion $j:Y\hookrightarrow X$ of codimension $c$, the Gysin map has the form \[ j_*:H^{k-2c}(Y,\mathbb Q)(-c)\longrightarrow H^k(X,\mathbb Q). \] It preserves Hodge types after the indicated twist: a type $(a,b)$ on $Y$ maps to type $(a+c,b+c)$ on $X$. Its image is supported on $Y$. Singular supports require the mixed-Hodge-theoretic support theorem, not a fictitious smooth Thom isomorphism on a singular variety. The required theorem is stated in [E1, Corollary 4.5]: rational coniveau classes can be obtained from smooth projective varieties of dimension $d-c$ by proper pushforward. This proves the known inclusion $N^c\subseteq M^c$, and also gives $N^cH^k=0$ when $k<2c$.

At $c=0$, both sides are the full cohomology. At $k=2c$, Hodge coniveau at least $c$ permits only type $(c,c)$. The support description then uses $H^0$ of the supporting resolutions, whose generators push forward to fundamental classes of codimension-$c$ components. Hence \begin{equation}\label{wsref-0-67-68-63-3a-65-76-65-6e} M^cH^{2c}=H^{2c}(X,\mathbb Q)\cap H^{c,c}(X),\qquad N^cH^{2c}=\operatorname{span}_{\mathbb Q}\{[Z]:\operatorname{codim}Z=c\}. \end{equation} Thus this boundary degree is exactly the ordinary rational Hodge conjecture. Passing to greater degrees introduces cohomology on the supports, not just their fundamental classes.

Lemma 24.1: a correspondence that constructs support.
Let $Y$ be smooth projective of dimension $m$, and let $\Gamma\in\operatorname{CH}^{m+c}(Y\times X)\otimes\mathbb Q$. Its action \[ \Gamma_*\alpha=(\operatorname{pr}_X)_* (\operatorname{pr}_Y^*\alpha\smile[\Gamma]) \] takes $H^{k-2c}(Y,\mathbb Q)$ into $N^cH^k(X,\mathbb Q)$.

Proof. Choose a finite cycle representative $\Gamma=\sum_j a_j\Gamma_j$. Every component has dimension $m+d-(m+c)=d-c$. Its projection to $X$ is closed, because the projection is proper, and has dimension at most $d-c$. Set $Z=\bigcup_j\operatorname{pr}_X(\Gamma_j)$. On $Y\times(X\setminus Z)$ the restricted cycle is zero, hence its cohomology class is zero. Compatibility of proper pushforward with restriction to this open set shows that $(\Gamma_*\alpha)|_{X\setminus Z}=0$. Therefore the whole image is supported on $Z$, and $\operatorname{codim}Z\geq c$. The cohomological degree is $(k-2c)+2(m+c)-2m=k$, as required.$\square$

The dimension of the cycle is the key. An arbitrary Hodge-theoretic map with the same degree does not come equipped with components whose projections one can take. Also, a formal codimension-$c$ correspondence on $Y\times X$, instead of codimension $m+c$, has the wrong degree and generally the wrong support dimension. Neither error can be repaired by renaming a cohomology class a cycle.

There is also an unconditional top-odd-degree case for every $X$. For $d\geq2$, take a smooth complete-intersection curve $j:C\hookrightarrow X$. Weak Lefschetz makes $j^*:H^1(X,\mathbb Q)\to H^1(C,\mathbb Q)$ injective. Under Poincaré duality its transpose is $j_*:H^1(C,\mathbb Q)\to H^{2d-1}(X,\mathbb Q)$, which is consequently surjective. Indeed, the transpose identity is $\int_X j_*\alpha\smile\beta=\int_C\alpha\smile j^*\beta$, and both pairings are nondegenerate. Every top-odd-degree class is therefore supported on this curve, in codimension $d-1$. The only Hodge types in that degree are $(d,d-1)$ and $(d-1,d)$, so $M^{d-1}H^{2d-1}=N^{d-1}H^{2d-1}=H^{2d-1}$. For $d=1$ the same conclusion is the already checked case $c=0$. This uses actual geometry of a hyperplane-section curve, rather than a hypothetical algebraic representative of a Hodge tensor.

There is a related projector trap. Even an algebraic correspondence $\Pi\in\operatorname{CH}^{d}(X\times X)\otimes\mathbb Q$ that acts as a projector onto a desired Hodge substructure has components of dimension $d$, and their projections can fill all of $X$. Its algebraicity alone gives support in codimension zero by the argument just proved. The diagonal is the basic example: it is an algebraic identity correspondence whose projection is all of $X$. To obtain codimension $c>0$, one needs an additional support theorem for the projector, or a factorization through cohomology of lower degree with a correspondence of the dimension in Lemma 24.1. A linear-algebraic splitting of cohomology therefore does not itself create the requested smaller support.

A conditional solution of the weight-one remainder.
Fix $c\geq0$ with $2c+1\leq2d$, and let $V\subseteq H^{2c+1}(X,\mathbb Q)$ be a rational Hodge substructure of coniveau at least $c$. Then $W=V(c)$ has weight one and types $(1,0),(0,1)$ only. It is polarizable, since projective cohomology is polarizable and substructures inherit a polarization. Riemann's theorem, in the rational form recorded by Milne [E2, Theorem 6.8 and Corollary 6.9], realizes $W$ as $H^1(A,\mathbb Q)$ for a complex abelian variety $A$. The source states the covariant homology version; dualizing gives this cohomology version. The zero substructure needs no construction.

One can replace $A$ by a curve as a source. If $\dim A=1$, take $C=A$. Otherwise choose a smooth complete-intersection ample curve $C\subset A$. Weak Lefschetz makes $H^1(A,\mathbb Q)\to H^1(C,\mathbb Q)$ injective. Its image is a polarizable Hodge substructure; the polarization-orthogonal complement gives a rational Hodge retraction. Consequently there is a surjective Hodge map $H^1(C,\mathbb Q)\to W$, and hence a map \begin{equation}\label{wsref-0-67-68-63-3a-70-68-69} \phi:H^1(C,\mathbb Q)(-c)\longrightarrow H^{2c+1}(X,\mathbb Q) \quad\text{with image }V. \end{equation} This step uses no algebraicity assertion about that retraction.

Poincaré duality on $C$ identifies $H^1(C)^*\cong H^1(C)(1)$. Thus a rational Hodge morphism (ghc:phi) corresponds to a rational class \[ \gamma_\phi\in H^1(C,\mathbb Q)\otimes H^{2c+1}(X,\mathbb Q) \subset H^{2c+2}(C\times X,\mathbb Q) \] of type $(c+1,c+1)$. Indeed the Hom tensor has its twist by $c+1$ and is of type $(0,0)$; removing that twist gives the asserted type. The cup-product duality fixes the signs so that its correspondence action on $H^1(C)$ is exactly $\phi$. No Hodge projector is assumed algebraic to form this cohomological class.

Proposition 24.2. If the ordinary rational Hodge conjecture in codimension $c+1$ holds for $C\times X$ for every smooth projective curve $C$, then $M^cH^{2c+1}(X,\mathbb Q)=N^cH^{2c+1}(X,\mathbb Q)$.

Proof. For the curve just constructed, the stated hypothesis represents $\gamma_\phi$ by a rational algebraic cycle $\Gamma\in\operatorname{CH}^{c+1}(C\times X)\otimes\mathbb Q$. Lemma 24.1 with $m=1$ puts its image $V$ inside $N^c$. This applies to every permitted $V$, hence to their sum $M^c$. The known opposite inclusion gives equality.$\square$

The unproved input is visibly an ordinary Hodge conjecture on a different variety, in a specific codimension. Merely assuming that Hodge classes on $X$ itself are algebraic does not provide this cycle on $C\times X$. The weight-one realization removes one obstruction in this degree; it does not prove the ordinary Hodge conjecture needed for the correspondence. For larger $k-2c$, the realization by lower-degree geometric cohomology is itself an additional issue.

An unconditional odd-degree family with exact coniveau.
Let $E$ be a complex elliptic curve and $X=E\times\mathbb P^r$. For $0\leq c\leq r$, Künneth gives \begin{equation}\label{wsref-0-67-68-63-3a-70-72-6f-64-75-63-74} H^{2c+1}(X,\mathbb Q)=H^1(E,\mathbb Q)\otimes\mathbb Q h^c. \end{equation} The only types are $(c+1,c)$ and $(c,c+1)$, so the entire space has Hodge coniveau $c$. Include $j:E\times\mathbb P^{r-c}\hookrightarrow E\times\mathbb P^r$ using a linear subspace. The Gysin map sends $\alpha\otimes1$ to $\alpha\otimes h^c$. It is surjective onto (ghc:product), and explicitly supports every class in codimension $c$. Coniveau $c+1$ is impossible because $2c+1<2(c+1)$. Thus both filtrations have precisely the claimed step in this example.

No individual algebraic cycle has a fundamental class of odd degree, so a search for a codimension-$(c+1/2)$ cycle would miss this entire nonzero cohomology group. Its classes are supported on an algebraic set through that set's own $H^1$. Likewise, a nonzero rational line in $H^1(E)$ is not a Hodge substructure: complex conjugation forces equal dimensions of types $(1,0)$ and $(0,1)$, so a nonzero weight-one Hodge structure cannot have rational dimension one. The whole two-dimensional space is the correct substructure. This prevents an unjustified replacement of the rational Hodge input by a chosen rational basis line.

Stress test: intersections of supports do not support sums.
Take $X=E_1\times E_2$ and divisors $D_1=E_1\times\{p_2\}$ and $D_2=\{p_1\}\times E_2$. Their classes vanish respectively away from $D_1,D_2$, so $[D_1]+[D_2]$ is supported on their union. It is nonzero since \[ D_1^2=D_2^2=0,\qquad D_1D_2=1, \qquad\int_X([D_1]+[D_2])^2=2. \] But $D_1\cap D_2$ is a point. By $N^2H^2=0$, no nonzero degree-two class is supported there. Replacing the union by an intersection would manufacture a false gain of one in coniveau. The finite-union argument at the beginning, not an intersection or a limit of shrinking sets, is the correct way to produce common support for a subspace.
Verification and the first remaining gap.
The correspondence dimension, cohomological degree, and Tate-twist signs were checked separately. The curve reduction uses a surjection from $H^1(C)$; it does not claim that every abelian variety is a Jacobian. The algebraic cycle obtained conditionally has rational coefficients and may have several components with either sign. Effectivity and integral support classes are not required. In the product calculation $c=0$ gives support on all of $X$, while $c=r$ gives support on $E\times\{\mathrm{pt}\}$; both endpoints match the same Gysin formula.

The source theorem for arbitrary singular supports is explicitly retained as an external input. There is no deduction of it from a smooth normal bundle on a singular set. Likewise, no numerical period calculation is claimed to certify a rational Hodge substructure or an algebraic cycle. The proofs and countertests are symbolic and finite-dimensional.

Three focused continuations are to construct the class $\gamma_\phi$ algebraically for a specified family of $X$; to exploit geometric correspondences from curves when they are already available; and, in higher remaining weight, to establish geometric realization before demanding algebraicity of its map. In the conditional weight-one argument the first unproved statement is the algebraicity of $\gamma_\phi$ on $C\times X$. A formal Hodge morphism or a matrix of its periods does not supply that missing cycle.

UNRESOLVED. This attempt proves a support-producing correspondence lemma, the precise conditional odd-degree reduction, an explicit family with equality of both coniveau filtrations, and a counterexample to an invalid support operation. It does not establish the generalized Hodge conjecture for arbitrary varieties. Written by GPT-6 Astra Ultra using direct derivations and primary-source checks; no new general theorem or formal verification is claimed.

Sources