Independent check of the character-lattice claims: verified, with one refinement (every triple, including $BCD$, has its own rank-2 relation lattice).
Setup. The eight weight forms as vectors in $\mathbb Z^4$ with coordinates $(u,v,w,t)$: $$a_1=(1,1,0,0),\ a_2=(0,0,1,1),\ b_1=(1,-1,0,0),\ b_2=(0,0,1,-1),\ c_1=(1,0,1,0),\ c_2=(1,0,-1,0),\ d_1=(0,1,0,1),\ d_2=(0,1,0,-1).$$ The $8\times 4$ weight matrix has rank $4$, and each of the six two-family submatrices also has rank $4$ - so "joint rank four although every pair has rank four" checks out at the level of the weight lattice.
The five rows. Each of $R_1,R_2,R_3,R_4,R_6$ is a genuine relation (each sums to the zero form), and:
- $\mathrm{rank}(R_1,R_2,R_3,R_4,R_6) = 4$, with Smith invariants $(1,1,1,1)$. Since the full relation lattice $\ker(\mathbb Z^8\to\mathbb Z^4)$ also has rank $4$, unit Smith invariants mean the five rows generate the entire integer relation lattice, not merely a full-rank sublattice. The strong form of the claim holds.
- $(R_1,R_2,R_3,R_4)$ alone: Smith invariants $(1,1,1,2)$, confirmed. The gap is completely explicit: the unique dependency among the five rows is $$R_1+R_2+R_3+R_4 = -2R_6,$$ so $R_6$ is a half-sum of the four-row system, and its class modulo $\langle R_1,\dots,R_4\rangle$ has order exactly $2$. This is the index-two gap in one line.
Refinement: all four triples are on equal footing. Your generating set has rows supported on $ABC$, $ABD$, $ACD$ only, but that is a property of the generating set, not of the lattice. Each of the four triples has a rank-2 integer relation lattice, with saturated bases:
- $ABC$: $R_1,\ R_2$;
- $ABD$: $R_3,\ R_4$;
- $ACD$: $R_6$ and $-a_1+a_2+c_2+d_2 = -(R_1+R_3+R_6)$;
- $BCD$: $b_1+b_2-c_1+d_1 = -(R_3+R_4+R_6)$ and $b_1-b_2-c_2+d_2 = -(R_2+R_3+R_6)$.
So the four exceptional triple-supported spaces in $H^4(ABCD)$ include a $BCD$ one supplied by genuine lattice elements (not just rational combinations) - consistent with your $52 = 28 + 4\cdot 4 + 8$ bookkeeping, which is at least arithmetically consistent.
Field arithmetic. All checks pass:
- $x=i\sqrt{3+\sqrt2}$ has minimal polynomial $t^4+6t^2+7$; $(x+y)^2=-6-2\sqrt7$, so $x+y$ has minimal polynomial $t^4+12t^2+8$;
- $t^4+6t^2+7=(t^2-x^2)(t^2-y^2)$, its splitting field is $N=\mathbb Q(x,y)$ of degree $8$, and its Galois group is generated by a $4$-cycle and a reflection - $D_4$ as claimed;
- the period matrices $\tau_\pm$ for both $(b,c)=(3,7)$ and $(6,8)$ are symmetric with positive-definite imaginary part (basic Riemann sanity only; I have not checked the polarization elements $1/(4X)$, $1/f'(X)$ or the CM action).
Not verified here. The dimension counts ($19=15+4$ for $A\times B\times C$; the $52$-dimensional decomposition) require the Hodge-theoretic dimension formulas, not just the lattice; I only confirm the arithmetic is consistent. The Moonen-Zarhin classification input [2, §4.1] I have not re-read for this check. And of course the lattice computation does not touch the actual problem - it controls the bookkeeping, but the exceptional $ABC$ correspondence still has to come from geometry.
Net: the character-lattice foundation of this reduction is exactly as stated, and slightly stronger (saturation is verified, and the $BCD$ lattice is explicit above).
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