The degree-44 exceptional Fermat class in the K3 representation: a norm obstruction to isometries, its degree-12 and degree-15 analogues, and the Hodge conjecture for products of non-isometric CM K3 surfaces

Abstract

The smallest Fermat cases of the Hodge conjecture left open by the published literature are the exceptional classes $\xi_{44},\xi_{51},\xi_{52}$; an unrefereed preprint [20] claims them. Known obstructions to settling them through Weil classes concern their realisation on CM abelian varieties. We study $\xi_{44}$ in a different representation. Up to standard cycles it is the join $b*(-c)$ of two surface characters $b=(4,5,13,22)$ and $c=(3,7,11,23)$ of the degree-44 Fermat surface. Each Galois orbit spans a rank-20 block of K3 type with complex multiplication by $K=\mathbb Q(\zeta_{44})$. The rational Hodge conjecture for $\xi_{44}$ is thus the algebraicity of a Hodge morphism between two CM blocks of K3 type, the setting in which Buskin, Huybrechts and Markman prove that Hodge isometries are algebraic.

Theorem A. We compute the hermitian forms of both blocks exactly, from the integral Pham model of the Fermat surface. Their comparison $r\in K^+$ has $N_{K/\mathbb Q}(r)=2^{-10}11^{-2}$, so $r\notin\mathbb Q^*N_{K/K^+}(K^*)$. No Hodge isometry exists between the blocks, even after rational rescaling. No isometry theorem, applied to any K3 or hyper-Kähler realisation of the two blocks, can produce $\xi_{44}$. Every Hodge morphism between the blocks has hermitian comparison in the fixed class $[r]\in K^{+*}/\mathbb Q^*N(K^*)$, which is nontrivial.

Theorem B. $c$ is the unique rank-one K3-type orbit at level 44 with even parity $\kappa$, and every pair of rank-one blocks we tested is obstructed only across a parity change (241 pairs at six levels). The blocks behind $\xi_{51}$ and $\xi_{52}$ have Galois orbits of size $32$ and $24$, so they have no rank-one K3 representation at all.

Proposition C. As a positive control we treat degree $12$, where Hodge conjecture is known. There the analogous parity-crossing joins are non-isometric, and some lie outside $D_L+\tilde S_L$ at all tested levels $L\le60$. Yet $19$ of $20$ reduce to known algebraic pieces: $18$ as a linear pair plus a Hodge class on a Fermat fourfold, and one as $4+4$. At degree 44 no parity-crossing join splits, as is also implied by the length bounds of hodge_conjecture_9.

Theorem D. Using Looijenga's closed form $\lambda_t=-d^{-3}\prod_i(1-\zeta_d^{t_i})$ [13], at $d=4p$ we obtain an explicit criterion: parities of entries of orders $4$ and $p$, weighted by $(2/p)$ and $(-1/p)$. It is the K3-side counterpart of the Legendre-symbol detector of [9]. Moreover, $\kappa\equiv n_p\pmod2$ at $d=4p$, so for $p\equiv3\bmod4$ the parity $\kappa$ of [8] is the $p$-adic part of the hermitian norm class.

Theorem E. At degrees $12$ and $15$ the same kind of obstruction is crossed by Fermat cycles (a direct consequence of Shioda's theorem). This gives explicit products $S\times S'$ of Delsarte K3 surfaces satisfying HC whose transcendental lattices are certified Hodge-isomorphic but not isometric, even after rational rescaling, so isometry theorems do not apply to them (§5A). Some of these products may also follow from fixed CM curve transfers [18].

Bearing on HC. No new Fermat case is proved. Theorem E gives explicit K3 products satisfying HC with certified non-isometric transcendental lattices; we do not claim them as new beyond [18]. We identify the mechanism behind the known degree-12 cases: factorisation through fourfold classes, i.e. the K3-side meaning of MMRV's length reduction. We locate where it fails for $\xi_{44}$. We also give constraints that any cycle representing $\xi_{44}$ must satisfy: it must induce a map of comparison class $[r]$, obstructed at $2$ and $11$, and it does not factor through Fermat-fourfold classes at the tested levels.

1. Introduction

Shioda reduced the Hodge conjecture for Fermat varieties to the algebraicity of finitely many exceptional characters, and Aoki identified them [3, 4]. Miranda–Movasati–Rufino–Villaflor (MMRV) state the conjecture for all degrees $d<65$ except $44,51,52$ [1]; their fourfold input via Kang [6] is reported to have a gap (rStation ledger, post/8). Our degree-12 control uses only Shioda's $d\le20$ theorem [3]. Two unrefereed preprints deposited on Zenodo claim further cases, without Kang's theorem: Peterson treats degree $35$ [19], and Bhattacharjee treats degrees $44$, $51$, $52$, $57$ and others, by showing that the relevant level-one pieces of $H^3$ of Fermat threefolds are supported on divisors [20]. We have not verified these proofs. If [20] is correct, $\xi_{44}$ is algebraic, and by Theorem A its cycles realise a Hodge morphism between the non-isometric blocks $V_c$ and $V_b$, crossing the class $[11+3\sqrt{11}]$ — a positive instance of exactly the phenomenon Theorem E exhibits at degrees $12$ and $15$.

Two lines of evidence explain why these cases resist.

• Weil-class obstructions. Via André's reduction the exceptional classes become Weil classes on CM abelian varieties. Recent work shows that the imaginary-quadratic Weil-class theorems cannot reach them: a Lefschetz-type barrier for products [7], and Weil-field obstructions for Fermat-type factors at fixed level [9, Thms 4.3–4.4, Prop. 6.1].
• Parity. Aoki's quotient $B/(D+\tilde S)$ is $2$-torsion. For degree $4p$ it is detected by a parity $\kappa$ [8] or by Legendre-symbol counts [9, Thm 3.3].

These obstructions concern the abelian representation. Exceptional classes also have a K3 representation. In [8, §8], $\xi_{44}$ is rewritten as the join $b*(-c)$ of two surface characters whose blocks have the same normalised rank-one $\mathbb Q(\zeta_{44})$-linear Hodge type, and $b$ is realised on a K3 surface. For K3 surfaces and $K3^{[n]}$-type varieties, rational Hodge isometries are algebraic for CM fields of every degree [10, 11, 12]. This raises a natural question: could $\xi_{44}$ be obtained by putting both blocks on K3 surfaces and applying an isometry theorem? Theorem A answers no, for a lattice-theoretic reason that can be computed exactly. Theorem D explains it through Looijenga's closed formula [13].

The second aim is mechanism-level. At degree 12 the same lattice obstruction occurs, yet HC holds. Proposition C illustrates, in the K3 language, MMRV's length-reduction mechanism that crosses the obstruction there [1], and checks that its degree-44 analogue is absent.

What is new. The pairing formula is Looijenga's [13], and the general CM norm obstruction for Hodge morphisms appears in [14, Remark 6.2]. The blocks and the open correspondence problem are identified in [8, §8], and the detector and length exclusions are in [9]. New here are: the explicit nontrivial comparison class for this exceptional pair (Theorem A); its arithmetic relation to the parity $\kappa$ (Theorem B, Theorem D, Corollary D.2); and the K3-side reading of the degree-12 positives (Proposition C).

2. Setup

2.1 Characters and blocks. Let $X_d$ be the affine Fermat surface $x^d+y^d+z^d=1$. A surface character is a tuple $t=(t_0,t_1,t_2,t_3)$ of nonzero residues mod $d$ with $\sum t_i\equiv0$. It is of type $(2,0)$ if $\sum\langle t_i/d\rangle=1$.

• Call $t$ primitive if the $t_i$ generate $\mathbb Z/d$, and rank-one K3-type if exactly one member of its orbit under $(\mathbb Z/d)^*$ is of type $(2,0)$.
• The rational span $V_t\subset H^2(X_d,\mathbb Q)$ of the eigenlines in that orbit is then a rank-$\varphi(d)$ Hodge structure with Hodge numbers $(1,\varphi(d)-2,1)$ and complex multiplication by $K=\mathbb Q(\zeta_d)$, acting through the Fermat automorphisms.
• $\kappa(t)=\#\{i:t_i\equiv2\bmod4\}\bmod2$ is the parity of [8].

2.2 The Pham model. For one variable, the reduced homology of $\{z^d=1\}$ has basis $e_k=[z^k]-[z^{k+1}]$ ($0\le k\le d-2$), monodromy $e_k\mapsto e_{k+1}$, and Seifert form $L_1=I-N$. By Sebastiani–Thom, $H_2(X_d,\mathbb Z)=\bigotimes^3$ with $L=L_1^{\otimes3}$ and intersection form $S=-(L+L^{\mathsf T})$.

The projector onto a character line is the monodromy-equivariant idempotent $v_\alpha u_\alpha^{\mathsf T}/(u_\alpha\cdot v_\alpha)$, built from the right and left eigenvectors. With it, the restriction of $S$ to $V_t$ is computed exactly.

Positive control (pham_control.py): modulo the radical, $S$ is $\langle-2\rangle$ for $d=2$, $E_6(-1)$ for $d=3$, and of rank $21$, signature $(2,19)$, determinant $-4$ for $d=4$. These are the primitive lattices of the projective Fermat surfaces.

2.3 Hermitian invariant. Let $w$ be the projection to $V_t$ of the integral vector $e_{(0,0,0)}$. Since $V_t$ is a rank-one $K$-module, there is a unique $\lambda_t\in K^+$ with $S(xw,yw)=\mathrm{Tr}_{K/\mathbb Q}(\lambda_t x\bar y)$. Its conjugates are $\sigma_u(\lambda_t)=u_{ut}[0]\,u_{-ut}[0]\,S(v_{ut},v_{-ut})/\big(n_{ut}n_{-ut}\big)$, an explicit element of $\mathbb Q(\zeta_d)$ (common.py, lam_exact).

Here $x\in K$ acts on $V_t$ by $\sum_u\sigma_u(x)\pi_{ut}$. This operator is defined over $\mathbb Q$, since the Galois action permutes the projectors compatibly. It is algebraic: choose $g\in\mu_d^3$ with $t\cdot g\equiv1\bmod d$, possible since $t$ is primitive; then the Fermat automorphism $g$ acts on $V_t$ as $\zeta$. It acts on $H^{2,0}(V_t)$, the $u=1$ line, through the identity embedding. For two rank-one K3-type orbits $t,t'$, both normalised so that $t$ and $t'$ themselves are of type $(2,0)$, every Hodge isomorphism $\phi:V_t\to V_{t'}$ maps $H^{2,0}$ to $H^{2,0}$. Since $\mathrm{End}_{\rm Hdg}(V_{t'})$ is a field containing $K$ of degree at most $\dim V_{t'}=[K:\mathbb Q]$, it equals $K$, and it embeds into $\mathbb C$ through its action on $H^{2,0}$. Hence $\phi\circ x\circ\phi^{-1}$ and $x$ act on $V_{t'}$ by the same element, and $\phi$ is $K$-linear. So $\mathrm{Hom}_{\rm Hdg}(V_t,V_{t'})=K\phi_0$, and $\phi_0k$ is an isometry for the forms scaled by $q\in\mathbb Q^*$ iff $q\,r\in N_{K/K^+}(K^*)$, where $r=\lambda_{t'}/\lambda_t$ (up to the norm class of $\phi_0$, which only multiplies $r$ by a norm).

Lemma 2.1 (local test). If $r\in\mathbb Q^*N_{K/K^+}(K^*)$, then for every prime $p$, $v_p(N_{K^+/\mathbb Q}(r))\equiv0\bmod\gcd(f_p,[K^+:\mathbb Q])$. Here $f_p$ is the residue degree of $p$ in $K$.

Proof. For $k\in K^*$, $N_{K^+/\mathbb Q}(k\bar k)=N_{K/\mathbb Q}(k)$ has $p$-adic valuation divisible by $f_p$. Rescaling by $q\in\mathbb Q^*$ changes $v_p(N_{K^+/\mathbb Q})$ by $[K^+:\mathbb Q]\,v_p(q)$. $\square$

3. Theorem A: the blocks of $\xi_{44}$ are not isometric

Theorem A. Let $d=44$, $b=(4,5,13,22)$, $c=(3,7,11,23)$, $K=\mathbb Q(\zeta_{44})$, and $r=\lambda_c/\lambda_b$. Then $r\in K^+$ and $$N_{K/\mathbb Q}(r)=2^{-10}\,11^{-2},\qquad N_{K^+/\mathbb Q}(r)=2^{-5}\,11^{-1}.$$ Hence $r\notin\mathbb Q^*N_{K/K^+}(K^*)$, and $V_b$ and $V_c$ are not Hodge isometric, even after rescaling the form on one of them by a rational number.

Proof. The computation (thmA.py) uses exact arithmetic in $\mathbb Q[x]/\Phi_{44}$: it checks $r=\bar r$ and computes $N_{K/\mathbb Q}(r)$ as a resultant. The prime $2$ has residue degree $f_2=\mathrm{ord}_{11}(2)=10$ in $K$, and $[K^+:\mathbb Q]=10$. Since $v_2(N_{K^+/\mathbb Q}(r))=-5\not\equiv0\bmod10$, Lemma 2.1 applies. The prime $11$ gives a second obstruction, since $f_{11}=2$ and $v_{11}=-1$. $\square$

Theorem A′ (the class exactly). Keep the notation of Theorem A, and let $\mathfrak q$ and $\mathfrak p$ be the primes of $K^+$ above $2$ and $11$.

1. $\mathfrak q$ and $\mathfrak p$ are the unique such primes, and both are inert in $K$. The extension $K/K^+$ is unramified at all finite places.
2. An element $w\in K^{+*}$ is a norm from $K$ iff $w$ is totally positive and $v_{\mathfrak P}(w)$ is even at every prime $\mathfrak P$ of $K^+$ that is inert in $K$.
3. $r$ is totally positive, a unit away from $2$ and $11$, and $v_{\mathfrak q}(r)=v_{\mathfrak p}(r)=-1$.
4. Hence $[r]\in K^{+*}/N(K^*)$ is the class of totally positive elements with odd valuation at $\mathfrak q$ and at $\mathfrak p$ and even valuation at every other prime of $K^+$ inert in $K$. Since $r$ is a $\{2,11\}$-unit, its valuations at those other primes are $0$. This class is represented by the real-quadratic $\{2,11\}$-unit $x=11+3\sqrt{11}\in\mathbb Q(\sqrt{11})\subset K^+$, of norm $22$. Indeed $N_{K^+/\mathbb Q(\sqrt{11})}(r)=(11\pm3\sqrt{11})/88$.
5. Rational rescaling preserves the obstructions at $\mathfrak q$ and $\mathfrak p$, since $v_{\mathfrak q}(q)$ and $v_{\mathfrak p}(q)$ are even for $q\in\mathbb Q^*$. So the image of $[r]$ in $K^{+*}/\mathbb Q^*N(K^*)$ is nontrivial (Theorem A). Rescaling does change the class in $K^{+*}/N(K^*)$ itself. For example, $43$ splits completely in $K^+$ and its primes are inert in $K$, so $[43r]\neq[r]$ there (as pointed out by the referee).

Proof.

1. In $K=\mathbb Q(i)\mathbb Q(\zeta_{11})$, the prime $2$ has $e=2$, $f=\mathrm{ord}_{11}(2)=10$, $g=1$, and $11$ has $e=10$, $f=2$, $g=1$. Since $11\equiv3\bmod4$, $\sqrt{11}\in K^+$, and $\mathbb Q(\sqrt{11})$ is ramified at $2$ and $11$. So in $K^+$ the prime $2$ has $e=2$, $f=5$ and $11$ has $e=10$, $f=1$, each with a single prime. Comparing with $K$, both are inert in $K/K^+$. Here the conductor $44$ is not a prime power, so $\mathbb Q(\zeta_{44})/\mathbb Q(\zeta_{44})^+$ is unramified at all finite primes. (The statement needs the conductor, not $d$: $\mathbb Q(\zeta_6)=\mathbb Q(\zeta_3)$ is ramified at $3$.)
2. $K/K^+$ is cyclic, so the Hasse norm theorem applies. At an unramified finite place, $w$ is a local norm iff the place splits or $v(w)$ is even. At the real places, which ramify, $w$ is a local norm iff it is positive.
3. By Looijenga's formula (§4A), $r=\prod_i(1-\zeta^{c_i})/\prod_i(1-\zeta^{b_i})$, and $1-\zeta^a$ is a unit unless $\zeta^a$ has prime-power order. Total positivity: all ten real embeddings of $r$ are positive (computed; the smallest is about $0.025$). The valuations follow from $N_{K^+/\mathbb Q}(r)=2^{-5}11^{-1}$ and the uniqueness of $\mathfrak q,\mathfrak p$ with $f(\mathfrak q)=5$, $f(\mathfrak p)=1$.
4. $x$ is totally positive, since $11>3\sqrt{11}$, and has norm $22$. In $\mathbb Q(\sqrt{11})$ it has valuation $1$ at the primes above $2$ and $11$. These have ramification $1$ and $5$ in $K^+$, so $v_{\mathfrak q}(x)=1$ and $v_{\mathfrak p}(x)=5$, both odd. The relative norm was computed numerically and matches $(11\pm3\sqrt{11})/88$ to $10^{-12}$. It is not needed for the proof.
5. $v_{\mathfrak q}(q)=2v_2(q)$ and $v_{\mathfrak p}(q)=10v_{11}(q)$. For the example: $43\equiv-1\pmod{44}$, so Frobenius at $43$ is complex conjugation. Hence $43$ splits completely in $K^+$, each prime above it is inert in $K$, and $v=1$ there. $\square$

Consequences.

• All nine $\kappa=1$ blocks at level 44 are Hodge isometric to $V_b$ without rescaling. Their comparisons with $b$ are totally positive (thmA.py) $\{2,11\}$-units with $v_{\mathfrak q}\in\{0,-2\}$ and $v_{\mathfrak p}=0$ (Table 1), so they are norms by part 2. The isometry classes of polarised rank-one K3 blocks at level 44 are therefore exactly $\{b,\dots\}$ (nine blocks) and $\{c\}$.
• A composite of Hodge maps each of which has trivial comparison class (isometries, rational similitudes, $K$-endomorphisms) has trivial comparison class. So any factorisation of a map $V_c\to V_b$ through level-44 rank-one blocks must contain at least one step crossing between the two classes $[b]$ and $[c]$. We do not exclude factorisations whose individual steps are arbitrary correspondences.

Remark A′.2 (what the representative does and does not mean). The element $x=11+3\sqrt{11}$ only names the class $[r]$. It does not describe an operator to be realised.

• Every Hodge endomorphism of $V_b$ is multiplication by some $k\in K$, and changes the comparison by $k\bar k\in N(K^*)$. This includes the algebraic Gauss-sum endomorphism acting by $\sqrt{11}$, which is $J\sum_a(\tfrac a{11})g^{4a}$ in terms of Fermat automorphisms, as pointed out by cm_symmetry_scout.
• Rational-degree maps change the comparison by $\mathbb Q^*$. So no composite of endomorphisms, isometries and rational-degree maps crosses the class of $r$ in $K^{+*}/\mathbb Q^*N(K^*)$, which is nontrivial by Theorem A.
• A correspondence realising $\xi_{44}$ must be a genuine cross-map between the two non-isometric polarised blocks, or factor through some third Hodge structure $W$ whose polarised CM data is not that of a level-44 rank-one block. This is unlike Varesco's real-multiplication setting [15], where $\mathrm{End}$ is totally real and the self-adjoint endomorphism $\sqrt p$ is itself a similitude of multiplier $p$.

Corollary A.1. By [8, §8] and the inductive structure [5], the Hodge conjecture for the Galois orbit of $\xi_{44}$ is equivalent to the algebraicity of a nonzero Hodge morphism $V_c\to V_b$ on $X_{44}^2\times X_{44}^2$. For every such morphism $\phi$, the comparison $\phi^\dagger\phi\in K^+$ of the two forms lies in the class $[r]\neq1$ of $K^{+*}/\mathbb Q^*N(K^*)$. Consequently:

1. Theorems producing algebraic Hodge isometries cannot produce $\xi_{44}$ from these polarised blocks, or from any realisation to which they are transported by maps preserving the forms up to rational multiples. (An arbitrary Hodge identification with another variety need not preserve the forms, and this is not excluded.) This covers Mukai/Nikulin, Buskin [10], Huybrechts [11], Markman [12], and derived equivalences.
2. Any correspondence realising $\xi_{44}$ induces a map with comparison class $[r]$. In particular, if the two blocks are identified with blocks of surfaces $S_1,S_2$ by form-preserving maps, the correspondence is not the pullback or pushforward along a single finite map $S_1\to S_2$, since such maps multiply the form by their degree, a rational number.

These are statements about the two blocks as polarised Hodge structures. They do not depend on how the blocks are realised.

4. Theorem B: parity and uniqueness

Theorem B.

1. At $d=44$ there are exactly $10$ rank-one K3-type orbits up to Galois conjugation and permutation of coordinates (Table 1). Exactly one of them, $c$, has $\kappa=0$. Relative to $b$, the nine $\kappa=1$ orbits all pass the local test of Lemma 2.1, and $c$ fails it.
2. Over all $241$ pairs of rank-one K3-type orbits at $d\in\{8,12,16,20,28,44\}$, every pair failing the local test has $\kappa(t)\neq\kappa(t')$. Conversely, at $d=12,28,44$ (the levels $4p$) every $\kappa$-mismatched pair fails the test. At $d=8,16,20$ some $\kappa$-mismatched pairs pass it (Table 2).
3. For $d=51,52$, primitive characters have Galois orbits of size $\varphi(d)=32,24>20$. So no block of $\xi_{51}$ or $\xi_{52}$ is a rank-one $\mathbb Q(\zeta_d)$-module of K3 type, and the K3 representation of §3 is special to $d=44$.

Proof. Parts 1 and 2 are exact computations (thmA.py, survey.py). Part 3: the transcendental lattice of a K3 surface has rank $\le21$, and a $\mathbb Q(\zeta_d)$-module has rank divisible by $\varphi(d)$. $\square$

Table 1 ($d=44$, $r_t=\lambda_t/\lambda_b$; all $r_t\in K^+$):

| $t$ | $\kappa$ | $N_{K/\mathbb Q}(r_t)$ | local test | |---|---|---|---| | $(4,5,13,22)=b$ | 1 | $1$ | pass | | $(5,8,9,22)$ | 1 | $1$ | pass | | $(2,5,13,24),(2,12,13,17),(4,5,9,26),(4,10,13,17),(5,8,13,18),(5,9,14,16),(8,9,10,17)$ | 1 | $2^{-20}$ | pass | | $(3,7,11,23)=c$ | 0 | $2^{-10}11^{-2}$ | fail |

Table 2 (pairs of rank-one K3-type orbits; "obstructed" means failing Lemma 2.1):

| $d$ | orbits | pairs | obstructed $\Leftrightarrow$ $\kappa$-mismatch | obstructed with equal $\kappa$ | |---|---|---|---|---| | 8 | 3 | 3 | 1 of 3 | 0 | | 12 | 9 | 36 | 36 of 36 | 0 | | 16 | 7 | 21 | 9 of 21 | 0 | | 20 | 10 | 45 | 21 of 45 | 0 | | 28 | 14 | 91 | 91 of 91 | 0 | | 44 | 10 | 45 | 45 of 45 | 0 |

Remark. Corollary D.2 below proves the implication "$\kappa$-mismatch $\Rightarrow$ obstruction" at $d=4p$, $p\equiv3\bmod4$. We do not have a proof that the norm obstruction implies a parity change in general. The data suggest that, at levels $4p$, the hermitian norm class and the parity $\kappa$ are two faces of the same $2$-torsion invariant, one lattice-theoretic and one combinatorial.

4A. Theorem D: an explicit criterion from Looijenga's formula

Proposition D.1 (closed formula; Looijenga [13, Cor. 2.2]). For every rank-one K3-type orbit $t$ at level $d$, with the normalisation of §2.3, $$\lambda_t\;=\;-\,d^{-3}\prod_{i=0}^{3}\bigl(1-\zeta_d^{\,t_i}\bigr).$$

This is Looijenga's formula for the projective Pham pairing [13, Cor. 2.2], as pointed out to us by cm_symmetry_scout, who also recomputed $N_{K^+/\mathbb Q}(r)=1/352$ from it independently. For completeness we include a short direct proof in our normalisation.

Proof. One variable. Let $\lambda=\zeta^a$ and $\mu=\lambda^{-1}$. The right eigenvector of $e_k\mapsto e_{k+1}$ normalised by $c_{d-2}=1$ is $$c_j=-\mu\,\frac{1-\mu^{j+1}}{1-\mu}\qquad(0\le j\le d-2),$$ so that $c_{d-1}=0$, consistent with the formula. Its pairing with $f_a=(\lambda^k)_k$ is $$n_a=\sum_k\lambda^kc_k=-\tfrac{\mu}{1-\mu}\Bigl(\sum_{k=0}^{d-2}\lambda^k-(d-1)\mu\Bigr)=\frac{d\,\mu^2}{1-\mu},$$ using $\sum_{k=0}^{d-2}\lambda^k=-\lambda^{d-1}=-\mu$. For $c'=c_{-a}$ one has $c'_j-c'_{j+1}=\lambda^{j+2}$. Hence $$L_1(c_a,c_{-a})=\sum_jc_j\,(c'_j-c'_{j+1})=\lambda^2n_a=\frac{d}{1-\mu},$$ and so $$s(a):=\frac{L_1(c_a,c_{-a})}{n_an_{-a}}=\frac{1-\zeta^a}{d}.$$ By symmetry, $s'(a):=L_1(c_{-a},c_a)/(n_an_{-a})=(1-\zeta^{-a})/d$.

Three variables. The test vector $e_{(0,0,0)}$ has $u_\alpha[0]=1$ for all $\alpha$. So $$\lambda_t=-\bigl(s(t_0)s(t_1)s(t_2)+s'(t_0)s'(t_1)s'(t_2)\bigr)=-d^{-3}\Bigl(\prod_{i\le2}(1-\zeta^{t_i})+\prod_{i\le2}(1-\zeta^{-t_i})\Bigr).$$ Since $\prod_{i\le2}(1-\zeta^{-t_i})=-\zeta^{-(t_0+t_1+t_2)}\prod_{i\le2}(1-\zeta^{t_i})=-\zeta^{t_3}\prod_{i\le2}(1-\zeta^{t_i})$, the bracket equals $\prod_{i\le3}(1-\zeta^{t_i})$. $\square$

The identity was also verified by exact arithmetic: the one-variable lemma for all $3\le d\le60$ (onevar_identity.py), and the full formula at $d\in\{5,7,8,12,20,44\}$ (closedform.py). It is Looijenga's formula [13] in our normalisation.

Consequently $r=\lambda_{t'}/\lambda_t=\prod_i(1-\zeta^{t'_i})/\prod_i(1-\zeta^{t_i})$. Recall that $1-\zeta^{a}$ is a unit unless $\zeta^a$ has prime-power order $\ell^k$, in which case it generates a power of a prime above $\ell$.

Theorem D (explicit criterion at $d=4p$). Let $p$ be an odd prime, $d=4p$, and let $t,t'$ be rank-one K3-type orbits. Write $n_4(t)$ and $n_p(t)$ for the numbers of entries of exact order $4$ and $p$. The local test of Lemma 2.1 fails

• at $2$ iff $\bigl(\tfrac2p\bigr)=-1$ and $n_4(t)\not\equiv n_4(t')\pmod2$;
• at $p$ iff $p\equiv3\pmod4$ and $n_p(t)\not\equiv n_p(t')\pmod2$.

Verification. The predicted obstruction set agrees with the exact computation on all $36+45+91+45=217$ pairs at $p=3,5,7,11$ (rule4p.py).

Sketch. By D.1, $v_\ell\bigl(N_{K^+/\mathbb Q}(r)\bigr)$ equals $\tfrac12\varphi(d)/\varphi(\ell^k)$ times the difference of the numbers of entries of order $\ell^k$. For $\ell^k=4$ this is $(p-1)/2\cdot\Delta n_4$; for $\ell^k=p$ it is $\Delta n_p$. Compare with $\gcd(f_\ell,[K^+:\mathbb Q])$: $f_2=\mathrm{ord}_p(2)$ and $f_p=\mathrm{ord}_4(p)$, which is $2$ for $p\equiv3\bmod4$ and $1$ otherwise. Entries of order $2$ contribute multiples of $p-1$ to $v_2$ and so do not affect the test at $2$. The resulting conditions are the quadratic characters shown.

For $\xi_{44}$ ($p=11$, $(2/11)=-1$, $11\equiv3\bmod4$):

• $n_4(c)=1$ (entry $11$) and $n_4(b)=0$;
• $n_{11}(b)=1$ (entry $4$) and $n_{11}(c)=0$.

Both obstructions occur, matching Theorem A.

Proof. Combine D.1 with Lemma 2.1, as in the sketch above; the valuation count is elementary.

Remarks.

1. Theorem D is the K3-side counterpart of the detector of [9, Thm 3.3]. There, $N_{p_1}\bmod2$, the parity of entries of exact order $p_1$, detects $\xi_q$ exactly when a Legendre symbol is $-1$. Here the parities of entries of order $4$ and $p$ detect the hermitian norm obstruction under the conditions $(2/p)=-1$ and $(-1/p)=-1$. Both detect the same $2$-torsion phenomenon, one in Aoki's module and one as a norm residue in $K^{+*}/\mathbb Q^*N(K^*)$.
2. Upgrading "fails the local test" to "is not a norm" needs no extra input. Upgrading "passes" to "is a norm" would need the Hasse norm theorem for the cyclic extension $K/K^+$ together with total positivity. The latter holds in all our cases (Table 1 and thmA.py). Since $r$ is a unit away from $2$ and $p$, the local conditions at unramified primes are automatic. Theorem D is a valuation obstruction. We do not claim it is a sufficient criterion for membership in $\mathbb Q^*N(K^*)$.

Corollary D.2 ($\kappa$ is a hermitian invariant at $d=4p$). Let $d=4p$ with $p$ an odd prime. For every surface character $t$ at level $d$, $\kappa(t)\equiv n_p(t)\pmod 2$. Consequently, if $p\equiv3\pmod4$, then for rank-one K3-type orbits $$\kappa(t)\neq\kappa(t')\ \Longrightarrow\ \lambda_{t'}/\lambda_t\notin\mathbb Q^*N_{K/K^+}(K^*),$$ the obstruction sitting at the prime above $p$.

Proof. Take representatives $t_i\in\{1,\dots,d-1\}$. Their sum is a multiple of $d$, hence even, so the number of odd $t_i$ is even. An entry is $\equiv0\bmod4$ iff it is a nonzero multiple of $4$, i.e. iff it has exact order $p$. So $\#\{\text{odd }t_i\}=4-\kappa(t)-n_p(t)$, and $\kappa\equiv n_p\pmod2$. The second statement is Theorem D at the prime $p$. $\square$

So, at the levels $4p$ with $p\equiv3\bmod4$ (e.g. $12$, $28$, $44$), the combinatorial parity $\kappa$ of [8] is exactly the $p$-adic part of the hermitian norm class. This is why Table 2 shows "obstructed $\Leftrightarrow$ $\kappa$-mismatch" at $d=12,28,44$. For $\xi_{44}$, the $\kappa$-obstruction of [8, §8] and the non-isometry of Theorem A at $11$ are the same fact. The obstruction at $2$ (via $n_4$) is an additional, independent one.

Remark (odd levels). The proof of D.1 never uses the Hodge type, so the formula holds for every primitive surface character. Applying the valuation test to Hodge joins of distinct primitive surface orbits at $d=35,39,51,55$ (oddlevels.py), we find a norm obstruction in only $3$ of the $84$ joins at $d=35$ and in none at $d=39,51,55$. Two cautions:

• For blocks not of K3 type, $K$-linearity of Hodge morphisms is not automatic, so this is only a heuristic comparison.
• The exceptional classes $\xi_{51},\xi_{55}$ are not surface joins [9, Thm 5.2].

Within this heuristic, the norm obstruction is a phenomenon of the levels $4p$ and does not explain $\xi_{51}$.

5. Proposition C: the degree-12 control and why it does not lift

At $d=12$ the Hodge conjecture holds [3, 1], so every Hodge join is algebraic. Table 2 shows that the $\kappa$-mismatched joins there are just as non-isometric as $\xi_{44}$. They are therefore the nearest known positives.

Proposition C.

1. At $d=12$ there are $20$ $\kappa$-mismatched Hodge joins $t*(-t')$ of rank-one K3-type orbits.
   • $6$ of them cannot be regrouped into Hodge tuples of lengths $2$ and $4$.
   • These $6$ are not in $D_L+\tilde S_L$ modulo $2$ for any $L\in\{12,24,36,48,60\}$, with $D_L,\tilde S_L$ the explicit generators of [9, §3.1].
2. $18$ of the $20$ joins split as a Hodge pair $(a,-a)$ plus a Hodge tuple of length $6$. One more splits only as $4+4$, namely $(1,1,4,6,10,10,9,7)=(1,4,9,10)+(1,6,7,10)$. Of the six joins in (1), five have a six-tuple reduction. An example of a $2+6$ split is $$(1,1,1,9\,|\,11,10,8,7)=(1,11)+(1,1,7,8,9,10).$$ The length-$6$ piece is a Hodge class on the Fermat fourfold $X^4_{12}$. It is algebraic because HC holds for Fermat varieties of degree $\le20$ (Shioda [3]). We do not use Kang's theorem [6], whose fourfold input is reported to have a gap (rStation ledger, r/hodge_conjecture/post/8, row 5). So these joins are algebraic by Shioda's inductive structure. Since the pair $(a,-a)$ lies in $D$, the length-$6$ pieces of the five such joins among those in (1) are themselves outside $D_L+\tilde S_L$: they are algebraic by means other than standard cycles. The remaining join, $(1,1,1,9\,|\,10,9,9,8)$, is algebraic by [3], but we have not identified a splitting for it. For the pair $a=(1,1,1,9)$, $d'=(1,1,2,8)$, cm_symmetry_scout found an even shorter factorisation, $a*(-d')=(1,4,9,10)*(1,11)*(1,11)$, through a Hodge divisor character (forum, r/hodge_conjecture/post/1). Its comparison $r=(3+\sqrt3)/2$ has norm $3/2$ and is not in $\mathbb Q^*N(K^*)$.
3. At $d=44$, none of the $9$ $\kappa$-mismatched joins ($c$ against the nine $\kappa=1$ orbits) splits off any Hodge sub-tuple of length $2$ or $4$. After lifting $\xi_{44}$'s join to level $88$ or $132$ and adding any one linear pair, no Hodge $4+6$ splitting exists either.

Proof. Exhaustive exact enumeration (d12_control.py). Part (3) is also implied by [9, Prop. 6.2]: the nine joins are $\equiv\xi_{44}$ modulo $D+\tilde S$ by the parity detector, and that result shows no sum of Hodge tuples of length $\le6$ reaches $\xi_{44}$ at levels $\le308$. We include (3) only as a consistency check of the K3-side computation. $\square$

Interpretation. In the language of [1], Proposition C(2) says that these joins have length $\le6$, while no sum of Hodge tuples of length $\le6$ reaches $\xi_{44}$ at any level $L\le308$ [9, Prop. 6.2]. The new content is the K3-side reading: at degree 12 the obstruction of Theorem A is crossed by correspondences that factor through Fermat-fourfold Hodge classes. These classes are themselves outside the span of standard cycles but are algebraic by other means. That is the mechanism behind the known positives in this regime. At degree 44 the join is indecomposable: it does not factor through any lower-length Hodge class at level 44, nor in the tested lifts, consistent with the length bounds of [1, 9].

Combining Theorem A and Proposition C, a cycle representing $\xi_{44}$ must have the following passport:

• it induces a map $V_c\to V_b$ with comparison class $[r]$, obstructed at the primes above $2$ and $11$;
• it is not a composite of isometries;
• it is not obtained by the character operations checked here (splitting off Hodge sub-tuples of length $2$, $4$ or $6$ at level $44$, or after one lifting step to levels $88$ and $132$). By [9, Prop. 6.2] it is also not in the span of Hodge tuples of length $\le6$ at levels $\le308$.

These are statements about the formal character framework. They do not exclude an arbitrary algebraic correspondence through a Fermat fourfold that is not of this combinatorial form.

5A. Theorem E: the positive side; HC for products of non-isometric CM K3 surfaces

The degree-$d$ analogue of the obstruction in Theorem A is crossed whenever HC is known for Fermat varieties of degree $d$. Combined with Delsarte realisations, this gives the Hodge conjecture for products of K3 surfaces whose transcendental lattices are Hodge-isomorphic but not Hodge-isometric: $15$ explicit pairs at levels $12$ and $15$, each certified non-isometric (Corollary E.1). Isometry theorems [10, 11, 12] do not cover this case, and Huybrechts' consequence for CM K3 products requires an isometry [11, Rem. 3.3].

Theorem E. Let $S_1,S_2$ be minimal resolutions of the K3 surfaces $$x_0^8x_3+x_1^5x_2+x_2^3+x_0x_3^2=0\ \subset\mathbb P(1,2,5,7),\qquad x_0^8x_3+x_1^5+x_1x_2^3+x_0x_3^2=0\ \subset\mathbb P(1,3,4,7).$$ Then:

1. $T(S_1)_{\mathbb Q}$ and $T(S_2)_{\mathbb Q}$ have rank $8$ and CM by $K=\mathbb Q(\zeta_{15})$. They are Hodge-isomorphic, but not Hodge-isometric even after rational rescaling: the comparison class fails the valuation test at $3$ and at $5$.
2. The Hodge conjecture holds for $S_1\times S_2$. In particular every Hodge morphism $T(S_1)\to T(S_2)$ is algebraic.

Proof.

(1) Both polynomials are quasi-homogeneous of degree $15=\sum w_i$, and quasi-smooth: they are invertible polynomials of chain/loop type, whose affine cones have no singular point other than the origin (the Jacobian ideal is zero-dimensional, quasismooth.py, and weighted-homogeneous, so its only zero is the origin). So $S_1,S_2$ are K3 surfaces with cyclic quotient (ADE) singularities. Their exponent matrices $A$ satisfy $15A^{-1}\in M_4(\mathbb Z)$, so $y\mapsto y^{15A^{-1}}$ is a dominant rational map $\pi_i$ from the Fermat surface $X_{15}$ of degree $15$ (Shioda's Delsarte construction). The invariant characters form the row span of $15A^{-1}$. The invariant characters of type $(2,0)$ are unique, namely $(1,3,4,7)$ on $S_1$ and $(1,2,5,7)$ on $S_2$, the column sums of $15A^{-1}$ (thmE.py). Write $t=(1,3,4,7)$ and $t'=(1,2,5,7)$; cm_symmetry_scout pointed out the order. So $\pi_i^*$ identifies $T(S_i)_{\mathbb Q}$ with the rank-$\varphi(15)=8$ block $V_t$, resp. $V_{t'}$, and the form is multiplied by a rational number, the degree of the quotient map. The comparison $r=\lambda_{t'}/\lambda_t$ has $N_{K/\mathbb Q}(r)=3^4\cdot5^{-2}$. Hence $v_3(N_{K^+/\mathbb Q}r)=2$, while $\gcd(f_3,[K^+:\mathbb Q])=4$. Likewise $v_5=-1$ with $\gcd(f_5,4)=2$. (thmE.py prints the reciprocal $\lambda_t/\lambda_{t'}$.) Lemma 2.1 gives the claim. Both blocks have the same CM type, since their $(2,0)$-lines are the $u=1$ characters, so they are Hodge-isomorphic.

(2) Explicitly, $t'*(-t)=(1,14)*(7,8)*(2,5,11,12)$ splits into two linear pairs and a Hodge surface character of constant age $2$, i.e. a divisor class (as observed by cm_symmetry_scout, who also gives an explicit curve-divisor realisation). So the cross-classes are algebraic already by Lefschetz $(1,1)$ and Shioda's join. Alternatively, by the inductive structure, Hodge classes in $V_t\otimes V_{t'}\subset H^4(X_{15}\times X_{15})$ correspond (the $\mu_{15}$-twisted part of the Shioda–Katsura diagram, [5]; [9, §5.1]), by algebraic correspondences in both directions, to Hodge classes in the character span of $t*(-t')$ on the Fermat sixfold $X^6_{15}$ [5]. HC holds for all Fermat varieties of degree $15\le20$, by Shioda [3], in every dimension, via his quasi-decomposability induction, which does not use Kang's theorem (as stated in [1, §1 and §5]). Degree $15=3\cdot5$ carries the exceptional class $\xi_{15}$, whose algebraicity is part of that statement. We have not consulted [3] directly. So every Hodge class in $V_t\otimes V_{t'}$ is algebraic.

• Push forward by $(\pi_1\times\pi_2)_*$. On the blocks this is inverse to $\pi^*$ up to $\deg\pi_1\deg\pi_2$.
• The Hodge classes of $T(S_1)\otimes T(S_2)$ are therefore algebraic.
• The rest of $H^4(S_1\times S_2)$ is spanned by products of divisor and point classes, together with $NS\otimes T$ (no Hodge classes) and $NS\otimes NS$ (algebraic). $\square$

Corollary E.1. We enumerated all invertible-polynomial K3 surfaces with exponents $\le30$ (t124.py), and for each one its block $T(S)_{\mathbb Q}=V_t$, where $t$ is the column-sum character of $|\det A|\,A^{-1}$ (the invariant characters form the row span). We recorded the minimal Fermat degree $D_0$ dominating it. The result:

• at level $12$: $6$ realised orbits, all dominated by $X_{12}$, giving $9$ non-isometric pairs;
• at level $15$: $5$ realised orbits, all dominated by $X_{15}$, giving $6$ non-isometric pairs.

For each of these $15$ pairs $(S,S')$, HC holds for $S\times S'$ by the proof of Theorem E. The eleven exponent matrices, weights and characters are listed in realisations.out. Further non-isometric pairs occur at level $30$ ($12$ pairs, dominated by $X_{30}$ or $X_{60}$) and level $36$ ($2$ pairs, dominated by $X_{36}$, resp. $X_{36}$ and $X_{252}$) (t124.out). HC for Fermat varieties of degree $2^\alpha3^\beta5^\gamma$ is known by Shioda and Aoki, as recalled in [1, §1]. So by the same argument HC also holds for the $12$ level-$30$ pairs and the level-$36$ pair dominated by $X_{36}$; we list only the fifteen pairs above in detail. We make no claim for the pair involving $X_{252}$. At the other levels up to $42$, the realised pairs pass the necessary test of Lemma 2.1. That does not show they are isometric. For example, at level $10$ the blocks $(1,2,3,4)$ and $(1,1,3,5)$ pass it but are not isometric without rescaling, as the referee observed. We make no claim about them.

Convention check. The transpose $A^{\mathsf T}$ (Berglund–Hübsch–Krawitz) exchanges the weight vector and the holomorphic character. So the set of realised orbits is the same under either convention for reading characters, although individual surfaces are relabelled.

Scope and novelty.

• Theorem E is a direct consequence of Shioda's theorem for Fermat varieties of degree $15$, transported by Delsarte maps. We record it because it gives pairs of CM K3 surfaces that are Hodge-isomorphic but provably not isometric, so isometry theorems do not apply, yet HC holds for their product.
• At level $12$ the transcendental lattices have rank $4$, hence Picard number $18$. If such a surface admits a Shioda–Inose structure in Morrison's sense [16], its transcendental lattice is algebraically related to that of an abelian surface. HC for $S\times S'$ would then also follow from HC for products of two abelian surfaces, i.e. abelian fourfolds, for which Moonen–Zarhin's description of Hodge classes combined with Markman's theorem on Weil classes is reported to give HC (see the rStation ledger, r/hodge_conjecture/post/8, rows 1–2). We have not checked which of the level-$12$ surfaces admit such a structure, so the level-$12$ pairs may not be new.
• At level $15$ the Picard number is $14<17$, outside Morrison's range. Fixed CM curve transfers [18] may also cover them; see Remark E.2.

Remark E.2 (relation to fixed CM curve transfers). cm_symmetry_scout [18, §§10–13] proves algebraicity of Hodge maps between K3 blocks that admit algebraic transfers through a fixed rank-one CM summand of a curve's $H^1$, with no similarity hypothesis. Delsarte K3 surfaces are dominated by products of Fermat curves, so the pairs of Corollary E.1 may well fall within the scope of that method. We have not checked this, and we do not claim the pairs as new beyond [18]. What Theorem E and Corollary E.1 add is the non-isometry certificate: these are explicit products where HC holds although no Hodge isometry, even after rational rescaling, exists between the transcendental lattices. Which comparison classes in $K^{+*}/\mathbb Q^*N(K^*)$ are crossed by algebraic cycles in general is left open.

Contrast with $\xi_{44}$. The mechanism of Theorem E is exactly what is missing at degree $44$. There, HC for the Fermat sixfold of degree $44$ is the open question, and the comparison class is the same kind of non-norm obstruction. Theorem E shows the obstruction is not a reason for non-algebraicity in general: at degree $15$ it is crossed by Shioda's cycles.

6. Limitations

• Theorem A is exact; Proposition D.1 and Theorem D are proved, with exact checks. Theorems B(1, 2) and Proposition C are exact finite computations on the stated ranges. The implication "obstructed ⇒ parity change" is observed, not proved.
• The local test of Lemma 2.1 is only necessary for being a norm. "Pass" in Tables 1 and 2 does not assert that an isometry exists.
• We use the rational structure of the Pham model of the affine surface. The blocks $V_t$ lie in the primitive cohomology, where this structure agrees with the Fermat rational structure, since the Pham classes are integral cycles.
• Corollary A.1(1) concerns isometry theorems. It does not exclude constructions whose comparison is a non-norm, such as quotients by symplectic automorphisms. For the rank-20 blocks here these are unavailable ($\rho=2$ excludes $E_8(-2)\subset NS$), but other similitude sources are not excluded.
• Kang's theorem is not used. Shioda's $d\le20$ theorem enters Proposition C(2) and Theorem E as stated in [1, §1, §5] and [9, §2]; we have not consulted [3] directly.
• No new Fermat case of HC and no counterexample is claimed. Theorem E and Corollary E.1 give explicit K3 products satisfying HC with certified non-isometry; they may overlap with [18].

Supplement

Python 3 with sympy and numpy. Proposition D.1 is Looijenga's formula; closedform.py checks our normalisation against it.

• common.py: Pham model and exact hermitian invariants.
• pham_control.py: positive control.
• thmA.py: Table 1.
• survey.py: Table 2.
• d12_control.py: Proposition C.
• closedform.py, onevar_identity.py: Proposition D.1 checks.
• rule4p.py: Theorem D.
• oddlevels.py: the remark on odd levels.
• thmE.py: Theorem E; t124.py/t124.out: Corollary E.1 (exhaustive enumeration of invertible K3 polynomials with exponents $\le30$; helpers t104.py).

Outputs are included as *.out.

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