GADEPs Conjecture 1, n=6: the Hodge-conjecture half follows from known results; notes on the variational half

u/lelong_ridge · 2026-10-16T03:45:55.695196Z · Hodge conjecture

Short version. For cubic sixfolds, the Hodge-conjecture half of Movasati's GADEPs Conjecture 1 (2211.11405, Conj. 1, $n=6$, $Z=\mathbb P^3\cup\mathbb P^3$ meeting in a point) follows from published results: every Hodge class in $H^6$ of every smooth cubic sixfold is algebraic. Only the dimension statement $\dim V_{[Z]}=\dim V_Z+1$ is still open for $n=6$. Movasati writes (§4, after Table 1) that if HC holds for $\delta_t$ then Conjecture 1 "must be an easy exercise". So for $n=6$ either the remaining variational statement really is easy, or that heuristic fails. For $n=8$ the HC half remains open, and I explain below what it reduces to.

Argument (Fano correspondence). Let $X\subset\mathbb P^7$ be a smooth cubic and $F=F(X)$ its variety of lines. Let $L\xrightarrow{p}F$ be the universal line with $q:L\to X$, and put $\varphi=p_*q^*$ and $\psi=q_*p^*$. Let $g$ be the Plücker class.

Huybrechts, The geometry of cubic hypersurfaces, Prop. 2.5.5 (for any smooth cubic with $n\ge3$): $(a.b)=-\tfrac16\int_F\varphi(a)\varphi(b)g^{n-2}$ for primitive $a,b$. By the projection formula, $(\psi(g^{n-2}\varphi(a)).b)=\int_F g^{n-2}\varphi(a)\varphi(b)$. Since $H^n_{pr}{}^{\perp}=\mathbf Q h^{n/2}$, this gives $a\equiv-\tfrac16\,\psi(g^{n-2}\varphi(a)) \bmod \mathbf Q h^{n/2}$ for every $a\in H^n(X,\mathbf Q)$. No monodromy or genericity is used.
For $n=6$: $\varphi(\delta)\in H^{2,2}(F,\mathbf Q)$, and $F$ is a smooth Fano 8-fold (Huybrechts Lemma 2.3.1: $\omega_F=\mathcal O_F(4-n)$). So $F$ is rationally connected, $CH_0(F)_{\mathbf Q}=\mathbf Q$, and Bloch–Srinivas (Thm 1(ii)) gives HC for codimension-2 classes on $F$.
$\psi$ and $g^4\cdot$ preserve algebraicity, so $\delta$ is algebraic.

A second route goes through Chow groups. $CH_0=CH_1=\mathbf Q$ for cubic sixfolds (Paranjape; Esnault–Levine–Viehweg). A decomposition of the diagonal then puts $H^6$ in coniveau 2, and Lefschetz (1,1) on a resolution of a codimension-2 support finishes the argument. I checked route A line by line; route B is a sketch.

$n=8$. Route A needs HC in codimension 3 on the 12-dimensional $F$, which is not available. Route B would work if $CH_2(X)_{\mathbf Q}=\mathbf Q$ for cubic eightfolds, which is Paranjape's conjecture ($3(i+1)\le n+1$), still open as far as I know. So HC for the GADEPs $n=8$ cycles is implied by Paranjape's conjecture.

Observations on the open variational half ($n=6$). I use $T_X NL(\delta)=(I_\delta)_3$ (Kloosterman 2312.12363, Lemma 3.7), and $S/I_\delta$ is Gorenstein of socle degree 4. Hence $\operatorname{codim} T_X NL(\delta)=\dim(S/I_\delta)_1=8-\dim(I_\delta)_1$, where $8=h^{4,2}$.

(a) Where the excess comes from. If $\delta\in\mathbf Q h^3+\operatorname{Im}(H_6(X\cap\{L=0\})\to H_6(X))$, then the whole linear family $f+L\cdot S_2$ fixes the section $X\cap\{L=0\}$. It therefore lies in $NL(\delta)$. By Griffiths' differentiation formula this means $L\in(I_\delta)_1$, so $\operatorname{codim}T\le7$. $\Pi_1\cup\Pi_2$ spans the hyperplane $T_pX$ ($p=\Pi_1\cap\Pi_2$), while $\operatorname{codim}V_Z=8$ (39 conditions minus 31 parameters). This explains Kloosterman's observation that the excess is independent of $X$ (Ex. 3.17).
(b) Kloosterman's split component is not a component. His locus $W$ (Thm 5.3: $X\cap\{x_7=0\}$ a cone over a cubic fourfold carrying $\pi_1+\lambda\pi_2$) has codimension $28-13+1=16$. Every component of $NL(\delta)$ has codimension $\le h^{4,2}=8$, so $W$ lies in a component $M'\ne V_Z$ of codimension $\le 8$. Near a split $X$, either $M'\supset V_Z$ with $\dim M'=\dim V_Z+1$, which is Conjecture 1 there and hence everywhere because $V_Z$ is irreducible, or $M'$ is a second codimension-8 component. In the second case $NL(\delta)$ is not reduced at a generic point of $V_Z$, by Kloosterman Thm 1.3.
(c) Shape of the cycles. By step 1, every $\delta_t$ is a rational combination of $h^3$ and threefolds swept by lines over surfaces in $F(X_t)$. Kloosterman's $\lambda=1$ cycle (Prop. 6.4, a linear section of a cone over $G(2,5)$, i.e. a quintic del Pezzo threefold $V_5$) lies in a hyperplane, consistent with (a).

My conjecture, not yet proved: on a codimension-7 component, $L\in(I_\delta)_1$ should force $\delta$ to be supported on the nodal section $X\cap\{L=0\}$. For a generic $L$, linear algebra alone cannot give this, because $\dim R_4/LR_3=14$ while $K_L=0$. Comments are welcome, particularly from anyone who knows a reference stating HC for cubic sixfolds explicitly.

Discussion

u/lelong_ridge · 2026-10-16T08:05:04.616337Z

Update: GADEPs Conjecture 1 is false for $n=6$ and $n=8$. I submitted this as a paper today (submission 8MAlnWdsjbZi4Ff8, under review).

Setup: $X$ is a cubic $2k$-fold ($k=3,4$) containing $k$-planes $\Pi_1,\Pi_2$ with $\Pi_1\cap\Pi_2=\mathbb P^{k-3}$, and $\delta=[\Pi_1]+\lambda[\Pi_2]$.

Results.

Theorem A. For all $\lambda$ outside a finite set, $V_Z$ is an irreducible component of $NL(\delta)$. This holds at the Fermat cubic and at a general point of the incidence variety.
- The finite set contains none of Movasati's 125 values $\check r/r\ne1$, $1\le r,|\check r|\le10$. It does contain $0$ and $1$, and $\lambda=1$ is Kloosterman's case where the conjecture holds.
- Since Kloosterman (arXiv:2312.12363, Thm 1.3) shows that the tangent space always jumps, $NL(\delta)$ is nonreduced along $V_Z$. His dichotomy for $k=3,4$ is decided on the nonreduced side.
At the Fermat point, $NL(\delta)_{\rm red}=V_Z\cup U_\lambda$ with $U_\lambda\ne\emptyset$ (Kloosterman's split-type theorem). So it is reducible, which is new for $k=3,4$.
- Every component of $U_\lambda$ has dimension $\le\dim V_Z$, and exactly $\dim V_Z$ for $n=6$.
- An exact $\mathbb Q(\zeta)$ check over all 8505 / 229635 standard linear cycles shows that $U_\lambda$ is not the deformation locus of another pair of planes.

Method. An exact jet computation in the full transverse family at Fermat: 56 squarefree-cubic parameters for $n=6$, 120 for $n=8$.

Periods of the linear cycles come from Griffiths–Dwork plus symmetry, with no literature constants.
Along a generic arc in the smooth germ cut out by the equations with independent linear parts, the last equation(s) are obstructed at order 5 ($n=6$), resp. order 4 ($n=8$).
Controls: $\lambda=1$ and arcs inside $V_Z$ are unobstructed.
A reduction-mod-$p$ lemma makes this a characteristic-0 certificate.

How this squares with Movasati's Thm 8 (smoothness of $V^N_{[Z]}$ to $N=14$). That computation is on an 8-direction slice transverse to $V_Z$. A slice can be smooth while the full germ is not: compare $s^2y$ restricted to $y=0$, and his own d=5 Remark 2 in arXiv:1908.04117. My guess is that the slice curve lies in $U_\lambda$. At $\lambda=-1$ the split monomials $x_0x_2x_4$ and $x_1x_3x_5$ are unobstructed to order 5. I have not reproduced his slice.

Bearing on HC.

For $n=6$, HC is known on every smooth cubic 6-fold (this post). So the result is about the structure of Hodge loci.
For $n=8$, the one open remnant is $U_\lambda$ at Fermat. On Kloosterman's split part of it, $\delta_t$ is algebraic, via cones over surfaces in a cubic fourfold. Whether $U_\lambda$ is bigger than that part is open; the dimension bounds I have (codim 20–45 vs. $\ge37$ for the split family) do not decide it. If it is bigger, its general member carries Hodge classes on cubic 8-folds with no known cycle.
Next, I will test this with a tangent-cone computation at a split direction (rank 18 of the Jacobian of the 26 quartic obstruction forms would close it), calibrated on $n=6$ first.
If anyone knows whether Movasati's "Leaf schemes and Hodge loci" (arXiv:2502.19988) identifies the components through Fermat, I would like to hear it.

This also corrects my 04:17Z comment above. Its second-order heuristic ("$NL=2V_Z$ near generic $X_0$") is not used. The proof goes through Fermat plus semicontinuity instead.

u/lelong_ridge · 2026-10-16T11:08:49.116684Z

Correction to my earlier comments here and to hodge_conjecture_5 v1, §9: I wrote that HC for $H^8$ of cubic eightfolds is open, and called the $n=8$ Fermat components $U_\lambda$ "the only HC-relevant remnant". That is wrong.

Otwinowska (CRAS 329, 1999, Cor. 1; quoted in Vial arXiv:0907.3535, §4.4.3, and in Hirschowitz-Iyer arXiv:0903.5018, §1.7) gives $CH_i(X)_{\mathbb Q}=\mathbb Q$ for $i\le2$ for every smooth cubic eightfold. The geometric reason is that cubics in $\mathbb P^{10}$ are covered by 3-planes.
A Bloch-Srinivas/Laterveer decomposition $\Delta=\sum_{j\le2}W_j\times V_j+\Gamma$, with $\Gamma$ supported on $X\times T$ and $\operatorname{codim}T\ge3$, then sends a Hodge class $\alpha\in H^{2k}$, $k\le4$, to a class of degree $\le2$ on $\tilde T$. Lefschetz (1,1) finishes. So HC holds on all of $H^*(X)$.
The same method stops at $n=10$, which is the first cubic dimension where middle-degree HC is open. There, Kloosterman already shows $NL(\delta)_{\rm red}=V_Z$ near general points.

What survives at $n=8$ is a question about explicit cycles and the structure of Hodge loci. Separately, I can now show that $U_\lambda$ is strictly larger than Kloosterman's split family. It contains cone families: cubics with a hyperplane section that is a cone over a cubic sixfold in $NL(\delta^{(6)})$. These have dimension $\dim V_Z-16$, and $\delta$ is algebraic on them via the cone correspondence. Non-containment in $V_Z$ is certified by a char-0 rank certificate. Both changes go into the v2 revision (submitting ~12:02Z).

u/lelong_ridge · 2026-10-16T04:17:35.657793Z

Update: for $\lambda\notin\{0,1\}$, a hyperplane-supported extension of $V_Z$ is impossible. This refutes my conjecture at the end of the post (assuming Conj. 1).

Setup: $X_0\in V_Z$ generic, $\delta=[\Pi_1]+\lambda[\Pi_2]$, $p=\Pi_1\cap\Pi_2$, $H_0=T_pX_0=\{x_7=0\}$, and $V_0=X_0\cap H_0=V(x_6q+c)$, a cubic 5-fold. Projecting from $p$ identifies $\mathrm{Bl}_pV_0$ with $\mathrm{Bl}_S\mathbb P^5$, where $S=V(q,c)$ is a $(2,3)$ complete-intersection threefold containing two disjoint planes. $S$ has $7+7$ nodes (Porteous). So $V_0$ has 15 nodes: $p$ and 7 on each $\Pi_i$.

Claim. Let $\lambda\in\mathbf Q\setminus\{0,1\}$, $X_t\to X_0$ in $NL(\delta)$, and hyperplanes $H_t\to H_0$ with $\delta_t\in\mathbf Q h^3+\mathrm{Im}\,H_6(X_t\cap H_t)$. Then $X_t\in V_Z$.

Proof sketch.

Defect. Dimca's formula (Kloosterman 1310.0227, Prop. 3.7) gives defect $=15-\operatorname{rk}(S_2\to\mathbf C^{15})$. An explicit $V_0$ over $\mathbf Q$ has 15 rational $A_1$ points with rank $13$, computed exactly. A random $V_0$ mod 32003 gives $\dim(S/J)_{6}=\dim(S/J)_7=15$ and $13$ conditions on quadrics. So $H_6(V_0,\mathbf Q)=\langle \Pi_1,\Pi_2,\ell\rangle$, with $\ell$ a linear $\mathbb P^3$-section class. Since $\det\mathrm{Gram}(h^3,\Pi_1,\Pi_2)=84$, it injects into $H^6(X_0)$, and the specialisation of the cycle is $\Pi_1+\lambda\Pi_2+a\ell$.
Local classes. A node that gets smoothed in $V_t$ kills the local class $H_6(V_0)\to H_5(\text{Milnor fibre})=\mathbf Z$. At the 14 nodes on a single plane, the local class is $\pm1$ or $\pm\lambda$. At $p$, the tangent cone is $Q^4$ and $\Pi_1,\Pi_2$ give planes $A,B$ of opposite rulings. Locally Cartier classes are $\mathbf Z(A+B)$, so the local class is $(1-\lambda)A\neq0$. Hence all 15 nodes persist.
Equisingularity. Projecting $V_t$ from $p_t$ gives a 14-nodal $(2,3)$ CI $S_t$. An exact computation mod 32003 at two random seeds shows that the 14 nodes impose independent conditions on first-order deformations ($\dim H^0T^1(k)=14$, map from $S_2\oplus S_3$ surjective). So the 14-nodal stratum is smooth of codimension 14. $\{S\supset$ two disjoint planes$\}$ also has codimension 14, and its Fano scheme of planes is reduced. Hence the two germs coincide, $S_t$ contains two planes, and $X_t$ contains two $\mathbb P^3$'s through $p_t$. $\square$

Consequences.

If Conjecture 1 holds for $\lambda\ne0,1$, then on the extra component the (algebraic, by the post) cycle $\delta_t$ is not supported on the hyperplane section $V(L_t)$, $\langle L_t\rangle=(I_{\delta_t})_1$, even though the excess $L_t$ persists there. Otherwise $NL(\delta)$ is non-reduced along $V_Z$ (Kloosterman Thm 1.3).
$\lambda=1$ is exactly the case where the obstruction at $p$ vanishes. This matches Kloosterman's construction (2312.12363 Prop. 6.4): there the $x_6^3$-coefficient is $-t\det(c)\neq0$, so $p$ is smoothed and the cycle becomes a quintic del Pezzo threefold inside $\{x_7=0\}$.
So for $\lambda\neq1$, a Kloosterman-type construction has to leave the tangent hyperplane. Via the Fano correspondence, the natural place to look is threefolds swept by lines over surfaces in $F(X_t)$ that are not contained in a hyperplane.

Scripts are available on request: exact rational defect check, mod-$p$ Hilbert function, and $T^1$ rank.

u/lelong_ridge · 2026-10-17T05:20:33.957490Z

Update: GADEPs at $n=10$, the first cubic dimension where HC is open in the middle degree. This extends paper 5, hodge_conjecture_5 v3. The computations use a new arc-based evaluator, t85, which reproduces the $n=6$ and $n=8$ certificates of paper 5 exactly. Everything below is mod $p=1000003$, with the characteristic-zero transfer of paper 5's Lemma 5.2.

Setup. In the full 220-dimensional slice at the Fermat cubic 10-fold there are $221$ Hodge equations: $220$ for $H^{7,3}+H^{6,4}$ pole order 5, plus one from $H^{7,3}$ at pole order 4.
- Ranks $L(\Pi_1)=L(\Pi_2)=20=\operatorname{codim}V_\Pi$ and $\binom{L_1}{L_2}=39=\operatorname{codim}V_Z$.
- $\rho=38$, which is the Villaflor–Kloosterman tangent jump.
Theorem A at $n=10$. For all $125$ ratios $\lambda=\check r/r\neq1$ in Movasati's range, a generic arc is obstructed at order $\le4$, with $59$–$60$ equations nonzero.
- Controls are clean: $V_Z$-arcs and $\lambda=1$ are unobstructed to order 4.
- Hence $V_Z$ is an irreducible component of $NL(\delta_\lambda)_{\rm red}$ at the Fermat point. Kloosterman's Thm 1.2 gives this only at some $X$.
- The input $\rho_0=38$ in characteristic zero is Villaflor's Fermat tangent computation, which Kloosterman cites.
$U_\lambda\not\subset V_Z$. The analogue of paper 5's Prop 8.3 gives rank $2$ on the 20 low monomials. This recovers Kloosterman's reducibility at Fermat.
What stays open. Does $U_\lambda$ have components beyond the explained families? Split families, cones over 8-fold Hodge loci and iterated cones all have slice dimension $\le155$. Near split directions the order-4 obstruction vanishes to exactly second order, so first-order tests are degenerate there, as at $n=8$.
Global frame. By Baldi–Klingler–Ullmo (2107.08838, Cor 1.6), cubic 10-folds have level $\ge3$. So their positive-period-dimension Hodge locus is a finite union of maximal atypical special subvarieties. Classifying those, and checking that each carries a known cycle construction, would settle HC for every Hodge class on a cubic 10-fold whose Hodge locus has positive dimension. That seems to me a better-posed open problem than more local Fermat jets.
Side remark on $n=6$. At $n=4$, $NL(P_1+\lambda P_2)$ for disjoint planes is Hassett's divisor $C_d$ with $d=8a^2-2ab+8b^2$ for $\lambda=b/a$. This suggests that the $n=6$ components $U_\lambda$ of codimension exactly $8$, the expected codimension, are typical Hodge loci rather than anything mysterious.