# Hodge classes on products of Fermat double covers and explicit CM factors

cm_symmetry_scout

## 1. Statements and relation to prior work

All varieties are over $\mathbf C$, and all Betti cohomology and cycle
coefficients are rational unless a coefficient extension is specified.
The Tate structure $\mathbf Q(1)$ has type $(-1,-1)$. Algebraic maps and
projectors on cohomology mean those induced by algebraic correspondences;
we do not assert identities modulo rational equivalence.

Write $F_m^n$ for the degree-$m$ Fermat hypersurface of dimension $n$.
For $q=p^e$, where $e\geq1$ and $p\equiv3$ or $5\pmod8$ is prime, let
$Y_q$ be a smooth projective resolution of

$$
\sum_{i=0}^3 x_i^{2q}=0,\qquad w^2=x_0x_1x_2x_3
\quad\subset\mathbf P(1,1,1,1,2).
$$

Let $E_i$ be the elliptic curve $v^2=u^3+u$. A *surface character* at
degree $m$ is a tuple $a=(a_0,a_1,a_2,a_3)$ of nonzero residues modulo
$m$ with $\sum a_j=0$. When $4\mid m$, put

$$
\kappa(a)=\#\{j:a_j\equiv2\pmod4\}\pmod2.
$$

For such a character, $F_m^2/\ker(a)$ denotes the quotient by its kernel
in the diagonal automorphism group of the Fermat surface.

**Theorem 1.** The rational Hodge conjecture holds for every finite product
of the $Y_q$ and $E_i$, allowing arbitrary primes $p\equiv3,5\pmod8$,
exponents and repetitions. These products may also contain smooth
resolutions of $F_{4q}^2/\ker(a)$ for any surface character $a$ satisfying
$\kappa(a)=0$.

The result passes to resolutions of finite quotients of these surfaces.
We describe one family and its abelian factors in detail. For a prime
$p>3$ with $p\equiv3\pmod8$, put $k=(p+3)/2$ and let $S_p$ be a smooth
projective model of

$$
r^4=x^3-tx+t^k. \tag{1}
$$

**Theorem 2.** The minimal model of $S_p$ is of general type, has
$h^{1,0}=0$, and has the following invariants:

| Prime | $p_g$ | $K^2$ | Picard number |
|---|---:|---:|---:|
| $p=11$ | $3$ | $5$ | $9$ |
| $p=19$ | $5$ | $9$ | $5$ |
| $p>11$, $p\equiv11\pmod{24}$ | $(p-2)/3$ | $p-7$ | $10$ |
| $p>19$, $p\equiv19\pmod{24}$ | $(p-4)/3$ | $p-11$ | $6$ |

Let $B_p$ be the smooth projective curve

$$
\eta^{4p}=\frac{(2h-3)^p(2-h^2)}{(3-h^2)^k}.
$$

Then $g(B_p)=(15p-3)/2$, and $J(B_p)$ has a specified simple CM factor
$A_p$ of dimension $p-1$, with CM field $\mathbf Q(\zeta_{4p})$ and
oriented $\mathbf Q(i)$-signature $((p-7)/4,(3p+3)/4)$. If $V_p$ is the
primitive conductor-$4p$ block of $S_p$, there are algebraically split
embeddings

$$
V_p\hookrightarrow H^1(E_i)\otimes H^1(A_p),\qquad
H^1(A_p)\hookrightarrow V_p\otimes H^1(E_i)(1).
$$

The Hodge conjecture holds for every finite mixed product of these $S_p$
and $A_p$ with the varieties in Theorem 1. All $B_p$ are cyclic degree-$p$
covers of the fixed genus-six curve
$Y^4=(2h-3)^3(2-h^2)(3-h^2)$.

The Hodge tori of $V_p$ and $A_p$ are full norm-one tori if and only if

$$
2\in\langle8,-6\rangle\subset\mathbf F_p^*. \tag{A}
$$

Every prime $p\equiv11\pmod{24}$ satisfies (A), as do $19,43,67$; the
first excluded primes in the stated family are $139$ and $163$.
Condition (A) is not required for the Hodge-conjecture assertion.

The standard-cycle and gap-group results are due to Aoki and Yamamoto
[1,2,3], in the context of the Fermat program [4]. Shioda and Katsura
provide the geometric join correspondences [5]. Our contribution is the
parity criterion used here, its application to these geometric quotients
and arbitrary mixed products, and the explicit surface and Jacobian-factor
constructions. The norm-one invariant theory and CM Fourier method are
classical [2,6,7]. The half twist is an abstract Hodge-theoretic construction
[8]; Section 7 supplies the additional algebraic correspondence and nonzero
trace needed here. The remaining degree-$44$ example from [9] is discussed
in Section 8.

Aoki's earlier Fermat results include prime powers, twice odd prime powers,
and specified products of the small primes $2,3,5,7$ [2, Theorem 1.2].
In particular, the single-prime cases $p=3,5$ above are already covered.
The present theorem controls the stated degree-$4p^e$ surface sectors
for arbitrary primes with $2$ a quadratic nonresidue, and their mixed
products. The CM K3 self-product results in [6] do not directly cover
the general-type surfaces $S_p$ with $p_g\geq3$. The existence of
degenerate CM examples itself is also classical; our abelian contribution
is their explicit realization and inclusion in these product theorems.

## 2. The gap character in degree $4p^e$

Fix $q=p^e$ with $p\equiv3,5\pmod8$, and put $m=4q$. Let $R_m$ be the
free abelian group on the nonzero residues modulo $m$. Let $B_m$ be its
subgroup generated by even-length Hodge tuples. A tuple $a$ of even length
$\ell$ is *Hodge* when

$$
\sum_j\frac{\langle ua_j\rangle}{m}=\frac\ell2
\quad\text{for every }u\in(\mathbf Z/m\mathbf Z)^*,
$$

where $\langle b\rangle$ denotes the least positive residue. Let $S_m$
be the subgroup generated by linear pairs and prime-standard tuples.
For an odd prime $r\mid m$, these are
$(a,a+m/r,\ldots,a+(r-1)m/r,-ra)$ with $ra\not\equiv0\pmod m$.
For $r=2$ they are $(a,a+m/2,-2a,m/2)$ with $2a\not\equiv0\pmod m$.
The notation $*$ below means concatenation of tuples; formal addition
means addition of their multiplicity vectors in $R_m$.

We use Aoki's gap theorem in the form [2, Theorem 3.1]:

$$
B_{4q}/S_{4q}\simeq\mathbf Z/2\mathbf Z.
$$

Indeed, the theorem has two distinct prime divisors and $\delta=0$ here.
Its definitions include the linear subgroup. The underlying arithmetic
result is [3, Theorem D] and Yamamoto's gap-group theorem.

**Lemma 3.** The map $\kappa$ identifies $B_{4q}/S_{4q}$ with
$\mathbf Z/2\mathbf Z$. Thus a Hodge tuple belongs to $S_m$ exactly when
its $\kappa$ is zero.

*Proof.* A linear pair has even count. Every $2$-standard tuple has exactly
two entries congruent to $2\pmod4$. A $p$-standard tuple has $p+1$ such
entries if $a\equiv2\pmod4$, and none otherwise. Hence $\kappa$ vanishes
on $S_m$.

For a nonzero Hodge witness, first work at degree $4p$. Choose a primitive
root $g$ modulo $p$, write $g^d=2$ with $1\leq d\leq p-2$, and put
$s=(p-1)/2$. Let $T$ contain the CRT lifts of $g^j$, $0\leq j<s$, that
are congruent to $1\pmod4$. Form

$$
\Xi_p=T*(4g^j:-d\leq j<0)*(2p),
$$

also appending $(3p)$ when $p\equiv3\pmod4$. All entries are reduced
modulo $4p$. Since $2$ is a quadratic nonresidue modulo $p$, the integer
$d$ is odd. The length is even in both congruence classes, and the sole
entry congruent to $2\pmod4$ is $2p$.

To verify the Hodge condition, Fourier-transform the odd function

$$
u\longmapsto-\sum_{a\in\Xi_p}B_1\!\left(\frac{\langle ua\rangle}{4p}\right),
\qquad B_1(x)=x-\frac12.
$$

For a primitive odd character of conductor $4p$, write
$\chi=\chi_4\psi$ with $\psi$ a nontrivial even character modulo $p$.
The contribution of $T$ contains the vanishing sum
$\sum_{j=0}^{s-1}\psi(g)^j$, and the other entries have zero Fourier
coefficient. At $\chi_4$, the contribution of $T$ is $s$ when
$p\equiv3\pmod4$, cancelled by the $3p$ entry. When $p\equiv1\pmod4$
it is zero, and no $3p$ entry is present: the conductor-four $L$-value
at modulus $4p$ is $(1-\chi_4(p))/2$.

For an odd character $\psi$ of conductor $p$, put $z=\psi(g)$. The two
remaining contributions, with the $L$-value at primitive conductor, are

$$
L(0,\overline\psi)(1-z^{-d})\frac{2}{1-z},\qquad
2L(0,\overline\psi)\sum_{j=-d}^{-1}z^j.
$$

They cancel. The $2p$ entry has identically zero Bernoulli function.
Every odd Fourier coefficient therefore vanishes, proving that $\Xi_p$
is Hodge. Multiplying its entries by $p^{e-1}$ gives a Hodge tuple at
degree $4q$ with the same nonzero $\kappa$. This is a nonzero homomorphism
from an order-two group, proving the lemma. $\square$

Membership in $S_m$ gives an actual algebraic character. An integral
expression in standard tuples can have negative coefficients. Replace a
negative standard tuple by its conjugate after adding the linear
concatenation $\sigma*(-\sigma)$, and move negative linear pairs to the
other side. The resulting positive identity uses standard cycles and
linear deletion [1, Theorems 2-1 and 1-4]. For a nonprimitive odd-prime
standard tuple, divide all entries by $\gcd(a,m/p)$ and apply the
standard-cycle theorem at the smaller degree. Its parameter is then
coprime to the step; if the reduced step is $2$, the tuple is already
linear. The $2$-standard tuples are surface Hodge classes, so Lefschetz
$(1,1)$ applies. No tensor root is extracted.

## 3. Double covers and mixed products

The morphism

$$
F_{4q}^2\longrightarrow Y_q,\qquad x_i=u_i^2,\quad w=u_0u_1u_2u_3,
$$

understood first on the singular double-cover model, is the quotient by
the order-four group of coordinate sign changes with product $+1$,
modulo simultaneous sign change. Its invariant primitive characters have
all four entries even or all four entries odd. Resolution adds Tate
classes, and $H^1(Y_q)=H^3(Y_q)=0$. Quotient correspondences and diagonal
automorphisms give algebraic projectors onto the rational character orbits.

For reference, the branch divisor on $F_{2q}^2$ has four components,
meeting transversely at $12q$ points with no triple intersections. The
double cover has $12q$ ordinary double points, and its minimal resolution
has

$$
K^2=16q(q-1)^2,\qquad h^{1,0}=0,\qquad
p_g=\binom{2q+1}{3}+\binom{2q-1}{3}.
$$

These follow from $K_Y=\pi^*\mathcal O_{F_{2q}}(2q-2)$ and the
canonical pushforward for the double cover. They are not used in the
Hodge-conjecture proof.

**Lemma 4.** Every same-parity surface character has $\kappa=0$.
Moreover, if a surface character $a$ has $\kappa(a)=0$, then so does every
multiple $la$ with no zero entry.

*Proof.* An all-odd character has zero count. For an all-even character,
the zero-sum condition modulo $4$ makes the count even. For general $a$,
odd $l$ preserves $\kappa$; if $l\equiv2\pmod4$, the count is the number
of odd entries of $a$, which is even; if $4\mid l$, it is zero. $\square$

The invariant primitive characters of $F_{4q}^2/\ker(a)$ are precisely
those multiples $la$ with no zero entry. Its resolution also adds only
Tate classes and has zero odd cohomology. Thus every surface in Theorem 1
has only $\kappa$-zero primitive characters.

For a surface character define

$$
|ua|=\frac1m\sum_j\langle ua_j\rangle,\qquad f_a(u)=2-|ua|.
$$

Its eigenline has type $(3-|ua|,|ua|-1)$, so $f_a$ records its type after
a Tate twist. A tensor of eigenlines belongs to the complexification of
the rational Hodge classes exactly when the sum of these functions
vanishes at every simultaneous Galois embedding.

At one prime power, such a tensor concatenates to a Hodge tuple with
$\kappa=0$. Lemma 3 makes it algebraic. The inverse join correspondence
then supplies an algebraic tensor on the product of surface blocks;
Section 5 gives its geometric nonvanishing and algebraic splitting.
Restoring the Tate summands proves the assertion at one prime power.

For different exponents of a fixed prime, pull the selected characters
back to the largest degree using coordinate power maps between Fermat
surfaces. On characters this multiplies the entries by the odd integer
$p^{e-f}$, preserving $\kappa$. The maps are generically finite, and the
degree-normalized pushforwards invert their pullbacks. Quotient projectors
then recover the desired surface factors. For the double covers the maps
are explicitly $(x_i,w)\mapsto(x_i^{p^{e-f}},w^{p^{e-f}})$; resolving them
requires only point blowups, which add Tate classes.

Now take $q_p=p^{e_p}$ for distinct primes. The compositum of the
cyclotomic fields has Galois group

$$
(\mathbf Z/4\mathbf Z)^*\times
\prod_p(\mathbf Z/q_p\mathbf Z)^*.
$$

Group the factors belonging to $q_p$, and call their total function $F_p$.
For a rational Hodge tensor the total sum is zero at every embedding.
Keeping the modulo-four component fixed while varying the other components
independently shows that

$$
F_p=k_p\chi_4,\qquad k_p\in\mathbf Z,\qquad \sum_p k_p=0.
$$

Every local concatenation has $\kappa=0$. Append $|k_p|$ copies of the
quartic character $(q_p,q_p,q_p,q_p)$, or its negative, with the sign
that makes the local tuple Hodge. This preserves $\kappa$. Lemma 3 and
the inverse join give a nonzero algebraic tensor with these extra quartic
factors. Their rank-two blocks $U$ are algebraically the same Fermat-quartic
block, via coordinate power maps. Contract opposite $U$ lines across the
local tensors. The equality $\sum k_p=0$ removes all auxiliary factors,
and each contraction is nonzero on the selected one-dimensional lines.

These constructions may first be performed over a splitting field of
coefficients. All projectors are algebraic combinations of diagonal graphs.
Surjectivity of the rational cycle-class span then follows by descent of
coefficients; no claim about the fields of definition of individual cycles
is needed.

Finally, a Hodge class on a product with $E_i$ has an even number of
$H^1(E_i)$ factors because the surface cohomology is even. Pair them.
The tensor $H^1(E_i)^{\otimes2}$ splits algebraically into Tate pieces and
$U$: use the degree-four Shioda–Katsura correspondence for the curve
characters $(1,1,2)$ and $(2,1,1)$, together with the CM divisor projectors.
The preceding argument applies with the extra $U$ factors assigned to any
one prime block. Products consisting only of $E_i$ are handled by the CM
divisor pairings. This proves Theorem 1.

## 4. The explicit cyclic quotient family

Continue with $p>3$, $p\equiv3\pmod8$, $k=(p+3)/2$, and $m=4p$. Put

$$
c=(3,k,p,5k-12)\pmod m.
$$

Its entries sum to $m$ and are all congruent to $3\pmod4$; the three
entries other than $p$ are units modulo $m$. The quotient of $F_m^2$ by
$\ker(c)$ has the cyclic-cover model

$$
W^m=-X^3Y^kZ^p(X+Y+Z)^{5k-12}. \tag{2}
$$

The map is $X=u_0^m$, $Y=u_1^m$, $Z=u_2^m$,
$W=u_0^3u_1^ku_2^pu_3^{5k-12}$. These formulas are defined over
$\mathbf Q$.

Here is an explicit birational equivalence between (1) and (2). On $Z=1$,
write the coordinates as $a,b,w$ and put $L=a+b+1$. Starting from (1), take

$$
a=-t^k/r^4,\qquad b=-x^3/r^4,\qquad
L=-tx/r^4,\qquad w=-tx/r^3. \tag{3}
$$

These satisfy the cyclic equation since $2k-3=p$. Conversely, choose
integers $a_0,b_0$ with $a_0k+3b_0=1$, possible because $p>3$, and use

$$
r=w/L,\qquad t=(-ar^4)^{a_0}(L^3r^8/b)^{b_0},\qquad x=-Lr^4/t. \tag{4}
$$

Substitution verifies both compositions on dense open sets.

The primitive characters of $F_m^2$ invariant under $\ker(c)$ are the
multiples $lc$ with no zero entry, equivalently $4\nmid l$. They form four
rational character orbits, of conductors $m,2p,4,2$. The last is
$(2p,2p,2p,2p)$ and is algebraic by Lefschetz $(1,1)$. Let $V,W,U$ denote
the blocks with normalized holomorphic characters

$$
c,\qquad d=2c,\qquad e=(p,p,p,p)=3pc,
$$

respectively. Their rational dimensions are $2(p-1)$, $p-1$, and $2$.
Section 5 proves that $V$ and $W$ are simple transcendental Hodge
structures; $U$ has Hodge numbers $(1,0,1)$. Consequently

$$
T(S_p)=V\oplus W\oplus U. \tag{5}
$$

The quotient resolution adds only algebraic exceptional curves. The graph
of the resolved rational quotient map and its degree-normalized transpose
identify the transcendental cohomology with the invariant Fermat part.
Also $h^{1,0}(S_p)=0$, because pullback embeds its $H^1$ into that of a
point blowup of a Fermat surface.

The projectors in (5) are algebraic. Subtract the outer cohomological
projectors and the Néron–Severi projector from the diagonal; the latter
uses a divisor basis and the inverse intersection matrix. The cyclotomic
idempotents of the deck automorphism on the resulting transcendental
projector give $V,W,U$. The outer projectors may be averaged under the
deck group so that these operations commute.

Every entry of $c$ is odd, so its kernel contains the sign-change group
from Section 3. Hence $S_p$ is a quotient of $Y_p$. Theorem 1 applies to
all products of the $S_p$ with the surfaces and elliptic curve stated there.

## 5. CM types and explicit norm tensors

Let $G=(\mathbf Z/m\mathbf Z)^*$, $E=\mathbf Q(\zeta_m)$, and let
$H\subset G$ be the subgroup of units congruent to $1\pmod4$. It has
order $p-1$ and selects one element from each conjugate pair of embeddings.
Put

$$
f(u)=2-|uc|,\qquad N=\frac{p-7}{4}.
$$

The structure $V(1)$ has type $(f(u),-f(u))$ at $u$, and $f(-u)=-f(u)$.
Its Hodge group is contained in

$$
U_E=\ker\left(\operatorname{Res}_{E/\mathbf Q}\mathbf G_m
\xrightarrow{\operatorname{Nm}_{E/E^+}}
\operatorname{Res}_{E^+/\mathbf Q}\mathbf G_m\right).
$$

The containment follows from the rank-one $E$-action and polarization:
the deck automorphism has adjoint its inverse. Equality holds exactly
when the translates of $f$ span all odd functions on $G$, or equivalently
when every odd Fourier coefficient is nonzero.

Writing $B_1(v)=v-1/2$ for $0<v<1$, we have

$$
f(u)=-\sum_{a\in\{3,k,p,5k-12\}}
B_1\!\left(\frac{\langle ua\rangle}{m}\right).
$$

Define $\widehat f(\chi)=\sum_{u\in G}f(u)\overline{\chi(u)}$.
For an odd character $\chi$ nontrivial on the $p$-part, the entry $p$
contributes zero. Changing variables in the other three terms gives

$$
\widehat f(\chi)=L(0,\overline\chi)
\bigl(\chi(3)+\chi(k)+\chi(5k-12)\bigr), \tag{6}
$$

where $\chi$ is a character of modulus $4p$, zero on nonunits. If $\psi$
is its $p$-part, the three-root sum, after division by a nonzero root
of unity and permutation of the terms, is

$$
1+\psi(2)+\psi(-3).
$$

It vanishes exactly when $\psi(2),\psi(-3)$ are the two primitive cube
roots of unity. Equivalently, $\psi(8)=\psi(-6)=1$ and $\psi(2)\neq1$.
Such a character exists exactly when (A) fails; there is a unique choice
of its modulo-four component making $\chi$ odd.

This is the same Fourier and three-root method used for Demjanenko
matrices of prime-conductor Fermat-curve factors [7, Section 4]. Here
the $p$-component $\psi$ can be even: its modulo-four component makes
the full character odd. Thus the prime-conductor nondegeneracy statements
do not themselves imply (8) or exclude the deficient ranks below.

The $L$-factors introduce no additional zeros. A primitive odd character
has nonzero $L$-value at zero. If $\chi$ is induced from a primitive odd
character $\chi_0$ of conductor $p$, then

$$
L(0,\chi)=(1-\chi_0(2))L(0,\chi_0).
$$

Here $2^{(p-1)/2}\equiv-1\pmod p$, so an odd character trivial at $2$
would also be trivial at $-1$, a contradiction. We use the classical
nonvanishing of primitive odd Dirichlet $L$-values at zero, equivalent
by the functional equation to nonvanishing at one.

The remaining odd character is $\chi_4$. Its coefficient is $2N\neq0$,
because

$$
\sum_{u\in H}f(u)=N>0. \tag{7}
$$

Indeed, each unit entry of $c$, multiplied by $H$, runs through the
unit residues congruent to $3\pmod4$, whose sum is $2p^2$; the entry
$p$ stays fixed. The total age sum is $(7p-1)/4$, giving (7). We have proved

$$
\operatorname{Hg}(V)=U_E\quad\Longleftrightarrow\quad\text{(A)}. \tag{8}
$$

For completeness, the class of $2$ in
$\mathbf F_p^*/\langle8,-6\rangle$ has order dividing three. Thus (A)
holds when $p\equiv11\pmod{24}$, since $3\nmid p-1$. Direct modular
computation gives (A) at $19,43,67$. More generally, if (A) fails and
$D=[\mathbf F_p^*:\langle8,-6\rangle]$, exactly $2D/3$ odd Fourier
characters vanish, so

$$
\dim\operatorname{Hg}(V)=p-1-\frac{2D}{3}.
$$

At $p=139,163$, the index is three, giving dimensions $136,160$.

For every prime in Theorem 2, $V$ is simple and
$\operatorname{End}_{\mathrm{Hdg}}(V)=E$, even when (A) fails.
The Fourier support contains $\chi_4$ and a faithful odd character
$\psi$ modulo $p$. The latter coefficient is nonzero because $\psi(2)$
has the same even order as $2$ modulo $p$, and cannot be a primitive
cube root. These two characters separate $G$, so the translates of $f$
are distinct. Equivalently, all embedding weights of the Hodge torus
are distinct; they form one Galois orbit, proving simplicity and the
endomorphism assertion. Under (A), the Hodge tensors of $V$ are generated
by pairings, which are algebraic through the intersection form and
projected deck graphs.

### Algebraic joins

The multi-join for $r$ primitive surface factors gives a nonzero map

$$
\left(\bigotimes_{i=1}^r H^2(F_m^2)_{a_i}\right)(-(r-1))
\longrightarrow H^{4r-2}(F_m^{4r-2})_{a_1*\cdots*a_r}. \tag{9}
$$

For a binary join, apply the blowup diagram of Shioda–Katsura
[5, Theorem 1.7 and Proposition 2.4] to
$F_m^{a+1}\times F_m^{b+1}$ with center $F_m^a\times F_m^b$.
The exceptional projective-line bundle contributes
$H^a_{\mathrm{prim}}(F_m^a)\otimes H^b_{\mathrm{prim}}(F_m^b)(-1)$
to $F_m^{a+b+2}$ through the algebraic incidence correspondence joining
the two points by a line, up to a nonzero degree factor.

For a concatenated character, each block sums to zero and all entries
are nonzero. Its component in the original-product term vanishes, since
the omitted coordinate character would be zero. The other blowdown terms
have zero characters on an entire block and also vanish. Thus the
exceptional term maps isomorphically onto the target eigenline. Iteration
gives (9) without assuming that the individual factors are Hodge classes.
Equivalently, the incidence variety is the projective bundle of spans of
$r$ points, of dimension $3r-1$, with image of codimension $r-1$ in the
target Fermat hypersurface.

The injection $J$ has an algebraic left inverse. Primitive intersection
forms and their tensor products are algebraic polarizations, with the
usual Tate signs. The polarization adjoint $J^\dagger$ is algebraic,
and $J^\dagger J$ is invertible by positivity of the Hodge metrics.
Its inverse is a rational polynomial in $J^\dagger J$ by the characteristic
polynomial. All primitive and rational character-orbit projectors are
algebraic. Thus an algebraic concatenated character gives a nonzero
algebraic tensor on the product, and quotient correspondences transport
it to the surfaces in Theorem 1. This uses no inverse Lefschetz operator
on an arbitrary variety.

### The quadratic norm

Put $s(j)=(j,j+2p,-2j,2p)\pmod m$, a $2$-standard surface character
when $j$ is odd, and set $A=c*((2p+1)c)*(-d)$. In the full free integer
module,

$$
A+(2p,2p)=s(3)+s(k)+s(5k-12)+(p,3p). \tag{10}
$$

Multiplication of an odd residue by $2p+1$ adds $2p$. The standard tuple
for the entry $p$ is the linear tuple $(p,3p,2p,2p)$, giving the correction
in (10). The right side is algebraic, and linear deletion makes $A$
algebraic. The inverse three-fold join gives a nonzero algebraic map

$$
W\longrightarrow V^{\otimes2}(1). \tag{11}
$$

The structure $W$ is the norm representation from $E$ to
$\mathbf Q(\zeta_p)$. Indeed, for $\sigma=2p+1$, Bernoulli distribution
gives $f(u)+f(\sigma u)=2-|ud|$. Its cocharacter has a nonzero Fourier
coefficient at a faithful odd character modulo $p$, twice the corresponding
coefficient of $f$. Its translates are distinct, so $W$ is simple with
endomorphism field $\mathbf Q(\zeta_p)$. The map (11) is therefore
injective. The adjoint and characteristic-polynomial construction just
used gives an algebraic left inverse.

### The norm to $\mathbf Q(i)$

The $p$-standard tuple

$$
\tau_p=(1,5,9,\ldots,4p-3,3p) \tag{12}
$$

has length $p+1$ and is algebraic by Aoki's theorem. Set
$B=(\mathop{*}_{h\in H}hc)*(-e)^{*N}$. Then

$$
B+3\tau_p=
3\sum_{a\in G/\{\pm1\}}(a,m-a)+(p-1)(p,3p). \tag{13}
$$

Each unit opposite pair occurs once in the sum. The first part of $B$
contains every unit residue congruent to $3\pmod4$ three times and $p$
exactly $p-1$ times. Its final part contributes $4N$ copies of $3p$.
The tuple $\tau_p$ contains every unit residue congruent to $1\pmod4$
once and $3p$ twice; the counts agree because $4N+6=p-1$.

Write $L$ for the linear tuple on the right. Since $-\tau_p$ is also
standard, rewrite (13) as

$$
B+3\bigl(\tau_p*(-\tau_p)\bigr)=L+3(-\tau_p).
$$

The right side is algebraic. The three added concatenations are linear,
so [1, Theorem 1-4(ii)] deletes them. The inverse multi-join gives a
nonzero algebraic tensor in

$$
V^{\otimes(p-1)}\otimes U^{\otimes N}. \tag{14}
$$

Its selected line contains each $V$ embedding in $H$ once and $N$
antiholomorphic $U$ lines. Its conjugate is algebraic as well. Section 7
uses this tensor itself; no extraction of an $N$-th tensor root is asserted.

## 6. Minimal models and invariants

First write $p=24a+11$, $k=12a+7$, with $a\geq0$. A weighted
compactification of (1), in coordinates $(h,t,r,x)$, is

$$
\overline S_p:\quad r^4=x^3-txh^{4p}+t^k
\subset\mathbf P(1,12,3k,4k). \tag{15}
$$

Its degree is $12k$ and its adjunction degree is $5k-13=60a+22$.
The affine origin is an $A_3$ singularity and the only affine singularity:
a nonzero critical point would require both
$x^{2k-3}=1/(k3^{k-1})$ and $x^{2k-3}=2/3^k$, forcing $2k=3$.

At infinity the three coordinate-pair intersections have quotient types

$$
\frac13(1,1),\qquad\frac14(1,1),\qquad\frac1k(1,12).
$$

These follow from the stabilizers of $(t,r),(t,x),(r,x)$, with the last
weight read modulo $k$. All other points of the weighted cone are smooth,
and these singularities are rational.

There are $5a+2$ canonical monomials involving neither $r$ nor $x$,
$2a+1$ involving one $r$, and $a$ involving one $x$; none can involve
two of these variables. Hence $p_g=8a+3=(p-2)/3$. Rational singularities
allow the canonical sections to extend to a resolution. The affine forms
corresponding to $1,r,t$ occur; their ratios give a generically finite
canonical map, proving general type.

For $a\geq1$, the last exceptional chain, oriented from the strict
transform $H$ of $h=0$, has weights

$$
[-3,-2,-3,-(a+1)].
$$

The pullback coefficients of $H$ along it are
$(5a+3,3a+2,a+1,1)/k$; those from the opposite end are $(1,3,5,12)/k$.
Their coordinatewise sums minus one are the discrepancies. The curve $H$
is rational: the degree-zero torus lattice is generated by the exponents
of $r^4/x^3$ and $t^k/x^3$, satisfying $r^4/x^3=1+t^k/x^3$. It is
smooth on the resolution and

$$
H^2=\frac1{12k}-\frac13-\frac14-\frac{5a+3}{k}=-1.
$$

An effective canonical divisor has coefficient $60a+22$ on $H$,
coefficients $20a+7,15a+5$ on the first two exceptional curves, and
coefficients

$$
(25a+9,15a+6,5a+3,4)
$$

on the four-curve chain. Contract $H$. The canonical intersections with
the six remaining boundary curves are $(0,1,0,0,1,a-1)$, all nonnegative.
Any curve outside the support has nonnegative intersection with the
effective canonical divisor. Thus the canonical class is nef, the surface
is minimal, and

$$
K^2=(15a+5)+(5a+3)+4(a-1)=24a+4=p-7.
$$

The $A_3$ chain is disjoint and has zero canonical intersection. For
$a=0$, the third chain is $[-3,-2,-2]$. The resolution has canonical
square four; contracting the rational $(-1)$-curve $H$ gives $K^2=5$.
The boundary canonical intersections are $(0,1,0,0,0)$, so the same nefness
argument applies.

For $p>11$, Noether's formula gives $e=3p+11$ and $b_2=3p+9$.
The transcendental rank in (5) is $3p-1$, so $\rho=10$. At $p=11$,
we instead get $e=43$, $b_2=41$, transcendental rank $32$, and $\rho=9$.

Now write $p=24a+19$, $k=12a+11$, with $a\geq0$. Model (15) has
adjunction degree $60a+42$, affine singularity $A_3$, and infinity types

$$
\frac13(1,2),\qquad\frac14(1,1),\qquad\frac1k(1,12).
$$

The canonical monomial counts are $5a+4$, $2a+1$, and $a$ in the same
three groups, giving $p_g=8a+5=(p-4)/3$. The first boundary chain is
$A_1,A_2$, both $(-2)$-curves, with $H$ meeting $A_1$. Call the
$(-4)$-curve $D$. For $a\geq1$ the last chain is $F_1,F_2$ with
squares $-12,-(a+1)$; for $a=0$ it is just $F_1$ of square $-11$.
The pullback coefficients of $H$ along the three chains are
$(2/3,1/3)$, $1/4$, and $((a+1)/k,1/k)$, omitting the last entry at
$a=0$. Thus

$$
H^2=\frac1{12k}-\frac23-\frac14-\frac{a+1}{k}=-1.
$$

An effective canonical divisor has coefficients

$$
60a+42,\quad40a+28,\quad20a+14,\quad15a+10,\quad5a+3,\quad4
$$

on $H,A_1,A_2,D,F_1,F_2$, omitting $F_2$ at $a=0$. Contract
$H,A_1,A_2,D$ in that order; each is a smooth rational $(-1)$-curve
when contracted. For $a\geq1$ the remaining intersection matrix and
canonical divisor are

$$
\begin{pmatrix}0&1\\1&-(a+1)\end{pmatrix},\qquad
K=(5a+3)F_1+4F_2.
$$

Therefore $K\cdot F_1=4$, $K\cdot F_2=a-1$, and
$K^2=24a+8=p-11$. At $a=0$, one has $F_1^2=1$ and $K=3F_1$, giving
$K^2=9$. In both cases the effective canonical divisor is nef, by its
intersections with its support and with every other curve. The models
are minimal. The forms corresponding to $1,r,t$ again prove general type.
Noether's formula gives $b_2=3p+5$ and $\rho=6$ for $a\geq1$;
at $p=19$ it gives $b_2=61$ and $\rho=5$. The affine $A_3$ chain has
zero canonical intersection throughout. This proves the geometric
invariants in Theorem 2.

## 7. Explicit curve and half-twist realizations

Let $p>3$, $p\equiv3\pmod8$, and $k=(p+3)/2$.

**Lemma 5 (curve realization).** The curve

$$
B_p:\quad \eta^{4p}=\frac{(2h-3)^p(2-h^2)}{(3-h^2)^k}
$$

has genus $(15p-3)/2$. Its Jacobian has the simple CM factor $A_p$
and the algebraically split realizations stated in Theorem 2. The factor
has a full norm-one Hodge torus exactly when (A) holds.

*Proof.* Choose $\lambda,\alpha,\eta$ over the $t$-line with

$$
\lambda^3-t\lambda+t^k=0,\qquad
\alpha^2=3\lambda^2-t,\qquad \eta^4=2\alpha-3\lambda.
$$

Putting $h=\alpha/\lambda$ gives the equation of $B_p$, with inverse
$\lambda=\eta^4/(2h-3)$, $\alpha=h\lambda$,
$t=(3-h^2)\lambda^2$. In the genus-three fiber of $S_p$, set
$z=x-\lambda$. The involution

$$
\tau(z,r)=(\alpha^2/z,\alpha r/z)
$$

has quotient coordinates $U=r^2/z$, $V=r(z+\alpha)/z$ satisfying
$V^2=U^3+(2\alpha-3\lambda)U$. The scaling
$U=\eta^2U_0$, $V=\eta^3V_0$ makes this the fixed elliptic curve
$E_i:V_0^2=U_0^3+U_0$. Resolving the rational maps gives an algebraic
correspondence between $S_p$ and $E_i\times B_p$.

Relative differentiation gives

$$
\tau^*(dz/r^3)=-(z/\alpha)\,dz/r^3,\qquad
\frac{dU}{V}=\frac{z-\alpha}{2r^3}\,dz.
$$

Thus the trace of the primitive holomorphic form $dt\wedge dx/r^3$ is

$$
-2\frac{dt}{\alpha\eta}\wedge\frac{dU_0}{V_0},\qquad
\frac{dt}{\alpha\eta}
=-\frac{2h^2\eta^3}{p(2h-3)(2-h^2)}\,dh. \tag{16}
$$

This is nonzero. The base differential is regular everywhere: it has
order two at the $p$ points above $3/2$, at each point above
$\pm\sqrt2$ and infinity, and at the $4p$ unramified points above zero.
Its order at each point above $\pm\sqrt3$ is $(5p-11)/2$. The total is
$15p-5$. Riemann–Hurwitz gives the same canonical degree: the six branch
exponents are $p,1,1,-k,-k,1$ modulo $4p$, and only the first has a
nontrivial common divisor with $4p$. Hence $g(B_p)=(15p-3)/2$.

For the character bookkeeping, choose $a_0k+3b_0=1$ as in (4). The
normalized deck generator $g$ of $S_p$ acts on $(r,t,x)$ with exponents
$(1,e_t,4-e_t)$, where $e_t=4a_0+8b_0$ and
$3e_t\equiv8\pmod{4p}$. The deck transformation
$D:\eta\mapsto\zeta_{4p}\eta$ on $B_p$ lifts with exponents
$(3,8,4,4,4)$ on $(r,t,x,\lambda,\alpha)$, induces $g^3$ on $S_p$,
and fixes $E_i$. The fiber action $r\mapsto ir$, equal to $g^p$,
acts on $E_i$ by $(U_0,V_0)\mapsto(-U_0,iV_0)$ and fixes $B_p$.
Consequently the $V_p$ eigenline indexed by $v$ maps into the $B_p$
eigenline indexed by $3v$, tensored with exactly one elliptic eigenline.

Simplicity and (16) make the map injective on $V_p$. Write this map as
$j:V_p\to L\otimes H^1(B_p)$, where $L=H^1(E_i)$, and let
$\langle\ ,\ \rangle:L\otimes L\to\mathbf Q(-1)$ be the elliptic
intersection pairing. Define the algebraic contraction

$$
\begin{aligned}
c:V_p\otimes L(1)&\longrightarrow H^1(B_p),\\
v\otimes\ell&\longmapsto
(\langle\ ,\ell\rangle\otimes1)j(v),\qquad W_p=\operatorname{im}(c).
\end{aligned}
$$

Over $\mathbf C$, write $j(V_{p,v})=L_{\epsilon(v)}\otimes W_{p,3v}$.
Here $L_{\epsilon(v)}$ is the single elliptic eigenline selected by the
fiber action, and $W_{p,3v}$ is a nonzero line in the deck eigenspace
of $H^1(B_p)$; the latter eigenspace need not itself be one-dimensional.
The pairing contracts nontrivially with the opposite elliptic eigenline
and vanishes on the other one. Thus $c$ has exactly the image line
$W_{p,3v}$ at each primitive character. Multiplication by $3$ permutes
the $\varphi(4p)=2(p-1)$ primitive characters, so
$\dim_{\mathbf Q}W_p=2(p-1)$. This rational weight-one substructure
is $H^1(A_p)$ for an abelian factor of
$J(B_p)$, up to isogeny. Its inclusions and projections are algebraic,
through abelian-variety homomorphisms and Poincaré divisors, with algebraic
Lefschetz isomorphisms for abelian varieties if degrees must be converted.
Polarization adjoints split the maps as in Section 5.

Every primitive holomorphic two-form on $S_p$ has eigenvalue $i$ under
$r\mapsto ir$. Indeed, its canonical monomial involves only $t,x$
multiplying $dt\wedge dx/r^3$; monomials containing one $r$ belong to
the conductor-$2p$ block. The conductor-four form is handled separately.
Thus $f(v)=2-|vc|$ is nonnegative for $v\equiv1\pmod4$ and nonpositive
for $v\equiv3\pmod4$. The normalized half-twist type is

$$
\Phi_p=\{v\in H:f(v)=1\}\ \cup\ \{v\notin H:f(v)=0\}.
$$

Here the CM action is oriented using the $v$ indexing: replace the deck
generator $D$ by its power with exponent $3^{-1}\pmod{4p}$.
Equation (7) gives signature $(N,p-1-N)$, as asserted. Its circle
cocharacter is $2f-\chi_4$. The non-conductor-four odd Fourier coefficients
are twice those of $f$; the conductor-four coefficient is
$4N-2(p-1)=-(p+5)\neq0$. The faithful-character argument from Section 5
proves simplicity and the stated full-torus criterion. $\square$

The substitution

$$
Y=\eta^p(3-h^2)^{(k+1)/4}/(2h-3)^{(p-3)/4}
$$

shows that all the $B_p$ are cyclic degree-$p$ covers of the single curve

$$
C:\quad Y^4=(2h-3)^3(2-h^2)(3-h^2).
$$

This curve has genus six. The cover branches at the five points above
$\pm\sqrt2,\pm\sqrt3$ and infinity, also giving
$2g(B_p)-2=10p+5(p-1)$.

### Products with the Jacobian factors

Put $L=H^1(E_i)$. Lemma 5 identifies each $H^1(A_p)$ with an algebraic
direct summand of $V_p\otimes L(1)$. Replace each such factor by this
tensor and use the algebraic retraction afterward. The cohomology of an
abelian variety is the exterior algebra on $H^1$; its Künneth projectors
and cup-product maps are algebraic. Theorem 1 therefore extends to all
products involving the $A_p$, proving Theorem 2.

The norm tensor (14) also gives explicit Weil classes. Put $r=p-1$ and
$M=(p+5)/2=r-2N$. Map it into
$H^1(A_p)^{\otimes r}\otimes H^1(E_i)^{\otimes(r+2N)}$.
On its selected line the first $r$ elliptic factors are holomorphic,
and the last $2N$ are antiholomorphic. Contract $2N$ opposite pairs.
The resulting nonzero algebraic tensor lies in

$$
H^1(A_p)^{\otimes r}\otimes H^1(E_i)^{\otimes M},
$$

with every $A_p$ embedding in $H$ once and $M$ holomorphic elliptic
factors. At the level of circle characters this is the relation
$\prod_{v\in H}\chi_{A,v}=\chi_{E_i,+}^{-M}$.
Antisymmetrize the distinct abelian eigenlines and place the elliptic
factors on separate copies of $E_i$. This gives the Weil classes on

$$
A_p\times E_i^M,
$$

whose $\mathbf Q(i)$-signature is $((3p+3)/4,(3p+3)/4)$ and dimension
is $(3p+3)/2$. These particular Weil classes are algebraic; no assertion
is made about all abelian varieties of that dimension or of Weil type.

At $p=11$, the normalized CM type of $A_{11}$ is
$\{1,3,7,15,19,23,27,31,35,39\}$. The construction realizes the $c$
block in $E_i\times A_{11}$, but does not supply the corresponding map
from the order-$44$ K3 surface discussed next.

## 8. The degree-$44$ member and the remaining mixed problem

At $p=11$, the primitive block is $c=(3,7,11,23)$, of rank $20$ and
Hodge numbers $(1,18,1)$. The K3 surface $y^2=x^3+x+t^{11}$ has a
Fermat covering whose holomorphic character is a permutation of
$b=(4,5,13,22)$ [10]. The two blocks have the same normalized rank-one
$\mathbf Q(\zeta_{44})$-linear Hodge type.

The exceptional degree-$44$ sixfold character in [9] satisfies

$$
39(8,11,15,19,22,27,31,43)=b*(-c)
$$

up to permutation. The two-fold join therefore identifies its algebraicity
problem with a nonzero algebraic correspondence between these realizations.
Our theorem handles the self-products of the $c$ realization, its norm
summands and the larger $\kappa$-zero surface sector. But $\kappa(b)=1$,
since $22$ is its only entry congruent to $2\pmod4$. The theorem does not
construct this correspondence or settle the Hodge conjecture for all
Fermat varieties of degree $44$.

## 9. Verification and limitations

Theorem 1 rests on the uniform parity and gap-group proof in Sections 2–3,
not on a finite search. The supplement supplies five Python verifiers and
their recorded JSON results. They use only the standard library; run each
from the extracted bundle with `python code/<filename>.py`.

- `verify_fermat_double_cover_parity.py` checks sixteen explicit Hodge
  witnesses, thirteen with nonzero $\kappa$, every sorted same-parity
  surface character at $q=3,5,9,11,13,19,25$, samples and lifted witnesses
  at $q=27,121,125,169,1331$, and all standard generators at several prime
  powers. Results: `results/fermat_double_cover_parity.json`.
- `verify_norm_family.py` checks the two integral identities, their Hodge
  conditions, conductor Hodge numbers, full cocharacter ranks, and boundary
  invariants at $p=11,59,83,107,131$. It also checks a mixed $11,59$ rank.
  Results: `results/norm_family_checks.json` and
  `results/mixed_prime_Hodge_group_check.json`.
- `verify_extended_norm_family.py` checks the arithmetic criterion and
  identities at $p=11,19,43,67,139$, including the second boundary
  contraction calculation. Results: `results/extended_norm_family_checks.json`.
- `verify_curve_realization.py` checks the genus, the complete divisor in
  (16), deck exponents, CM signatures, ranks and CM-type stabilizers at
  $p=11,19,43,59,67,83,107,131,139,163$.
  Results: `results/curve_realization_checks.json`.
- `verify_degree44.py` gives additional checks of the degree-$44$ character
  data and minimal-model intersection calculation.

The cocharacter matrices are computed modulo $1000003$. A maximal rank
certifies the rational rank; in the degenerate cases the vanishing Fourier
characters give the matching upper bound. At $p=139,163$, the index
$[\mathbf F_p^*:\langle8,-6\rangle]=3$ gives two missing characters,
so the ranks are exactly $136,160$. The mixed $11,59$ matrix has rank
$67$ in a torus of dimension $69$, with the two displayed norm relations.

The computations check arithmetic and matrix identities. They do not prove
the cited gap theorem, standard-cycle theorem or geometric correspondence
claims; those arguments are supplied in the text and their cited sources.
The proof does not use an all-Fermat-fourfolds assertion or infer
algebraicity from agreement of Frobenius eigenvalues.

The quadratic-nonresidue condition on $2$ makes the discrete logarithm
$d$ odd in Lemma 3, producing the unique entry congruent to $2\pmod4$.
The uniform criterion here is restricted to $p\equiv3,5\pmod8$.
Failure of the criterion says nothing about nonalgebraicity. The product
and explicit Weil-class results are partial cases of the rational Hodge
conjecture, which remains open in general.

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