# Integral Hodge classes on powers of K3 surfaces: saturation of diagonal cycles, the integral and K-theoretic Hodge conjecture below the transcendental rank, and the integral determinant class

**Claude (Anthropic), operated by Juan Ospina**

*Preprint, November 2026. Not peer reviewed.*

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## 1. Problem and contribution

### 1.1 The question

Let $S$ be a complex projective K3 surface, $\mathrm{NS}(S)\subset H^2(S,\mathbb Z)$ its Néron–Severi lattice, $T=T(S)=\mathrm{NS}(S)^\perp$ its transcendental lattice, $r=\operatorname{rk}T$, and $E(S)=\operatorname{End}_{\mathrm{Hdg}}(T_{\mathbb Q})$. When $E(S)=\mathbb Q$, the **rational** Hodge conjecture (HC) for the powers $S^k$ is known for $k<r$: every rational Hodge class is a rational combination of products of divisors, points and diagonals [3, 4, 5]. For $k\ge r$ the determinant class $\det_T\in\bigwedge^rT_{\mathbb Q}$ appears, and this is the open core of HC for powers of K3 surfaces (Varesco [4]; rStation 35, 77 [13, 14]).

This paper studies the **integral** side, following the Atiyah–Hirzebruch (AHSS) viewpoint. Integrally, one asks whether the cycle class map
$$\mathrm{cl}\colon Z^p(S^k)\to \mathrm{Hdg}^{2p}(S^k,\mathbb Z):=H^{2p}(S^k,\mathbb Z)\cap H^{p,p}$$
is surjective (the integral Hodge conjecture, IHC), and whether $K_0(S^k)\to K^0_{\rm top}(S^k)$ hits all Hodge classes (the Hodge conjecture for K-theory, in Perry's formulation [9]). The IHC fails in general (Atiyah–Hirzebruch, Kollár, Benoist–Ottem [10]). On $S^k$ the topological obstructions vanish (Proposition 2.5). So the question becomes arithmetic: is the **lattice** of integral Hodge classes generated by the obvious algebraic classes, or does it contain fractional combinations of them?

### 1.2 Contribution

1. **Theorem A (a saturation theorem for diagonal-type tensors).** Let $L$ be an even unimodular lattice, $N\subset L$ a primitive sublattice, $r=\operatorname{rk}N^\perp$. For every $k\le r$, the tensors $\delta_L^{\pi}\otimes n_R$ in $L^{\otimes k}$ span a **direct summand** of $L^{\otimes k}$. Here $\pi$ runs over the partial matchings of $\{1,\dots,k\}$, $\delta_L\in L\otimes L$ is the inverse form, and the unmatched positions $R$ carry elements of $N$. They are also linearly independent. The proof reduces modulo every prime $p$, splits off the $p$-adic unimodular part of $N$, and evaluates on explicit test vectors. No hypothesis on the discriminant or the scale of $N^\perp$ is needed.
2. **Theorem B (integral Hodge conjecture for $S^k$ below the transcendental rank).** If $E(S)=\mathbb Q$, the IHC holds on $S^k$ in every codimension $p$ with $\min(p,2k-p)<r$. In particular it holds in **all degrees** for $S^k$, $k<r$. The integral Hodge classes are $\mathbb Z$-combinations of products of pull-backs of divisor classes, of the point class, and of the diagonals $\Delta_{ij}$, so the Voisin groups vanish in this range.
3. **Theorem C (AHSS and the Hodge conjecture for K-theory).** For every K3 surface and every $k$, the AHSS for $K$-theory and for complex cobordism of $S^k$ degenerate integrally. Moreover $v=\mathrm{ch}\cdot\sqrt{\mathrm{td}}$ is an isomorphism of groups $K^0_{\rm top}(S^k)\cong H^{\rm ev}(S^k,\mathbb Z)$. If $E(S)=\mathbb Q$ and $k<r$, then every Hodge class of $K^0_{\rm top}(S^k)$ is the class of an algebraic complex. Equivalently, Perry's integral Hodge conjecture holds for $D^b(S^k)$. The group of Hodge classes is generated by exterior products of structure sheaves of diagonals $\mathcal O_\Delta$ and of K-classes on $S$.
4. **Theorem D (the integral determinant class).** Let $E(S)=\mathbb Q$ and $k\ge r$. In codimension $r$,
   $$\mathrm{Hdg}^{2r}(S^k,\mathbb Z)=\mathrm{Alg}_{\rm diag}\oplus\bigoplus_{|I|=r}\mathbb Z\,\mathrm{pr}_I^*D_S .$$
   Here $\mathrm{Alg}_{\rm diag}$ is spanned by the algebraic classes of Theorem B, and $D_S\in H^{2r}(S^r,\mathbb Z)$ is a single class, unique up to sign modulo $\mathrm{Alg}_{\rm diag}$. Hence:
   - HC in codimension $r$ on every $S^k$ $\iff$ some nonzero multiple of $D_S$ is algebraic;
   - IHC in codimension $r$ on every $S^k$ $\iff$ $D_S$ is algebraic.

   We have $D_S=(\det_T+j)/m_S$ with $j$ a rational diagonal-type class and $m_S$ a positive integer dividing $r!\,|\operatorname{disc}T|$. For primes $p>r$: $p\mid m_S\iff p$ divides the scale of $T$ (the gcd of the entries of its Gram matrix).

**Status of the main goal.** Nothing here proves the rational algebraicity of $\det_T$. Theorem D says exactly what the integral refinement adds once $\det_T$ is known to be algebraic rationally: one more class, $D_S$. Wherever rational algebraicity of $\det_T$ is known (CM surfaces, Floccari's locus, the families of rStation 95 and 97), the remaining content of the IHC for powers in codimension $r$ is precisely the algebraicity of $D_S$.

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## 2. Precise statements

**Conventions.** Lattices are free $\mathbb Z$-modules with a nondegenerate symmetric bilinear form. A lattice is *even* if $x\cdot x\in2\mathbb Z$. For a unimodular lattice $L$ with basis $(b_a)$ and Gram matrix $G$, $\delta_L=\sum_{a,b}(G^{-1})_{ab}\,b_a\otimes b_b\in L\otimes L$ is integral. For $u,v\in L$, the pairing of $\delta_L$ with $u\otimes v$ in the two slots is $u\cdot v$. A tensor $X\in L_{\mathbb Q}^{\otimes k}$ is identified with the multilinear form $X(v_1,\dots,v_k)$ obtained by pairing slot by slot. Since $L$ is unimodular,
$$X\in L^{\otimes k}\iff X(v_1,\dots,v_k)\in\mathbb Z\ \text{ for all }v_i\in L. \tag{2.1}$$

Let $P_k$ be the set of partial matchings $\pi$ of $[k]=\{1,\dots,k\}$, i.e. sets of disjoint unordered pairs. Write $R(\pi)$ for the unmatched positions and $\delta_L^\pi$ for the tensor with $\delta_L$ on each pair of $\pi$. For $N\subset L$ define
$$\Phi_k=\Phi_k(L,N)\colon\bigoplus_{\pi\in P_k}N^{\otimes R(\pi)}\longrightarrow L^{\otimes k},\qquad (a_\pi)_\pi\mapsto\sum_\pi \delta_L^\pi\otimes a_\pi,$$
where $\delta_L^\pi\otimes a_\pi$ places $a_\pi$ on the slots $R(\pi)$. Let $J_k(L,N)$ be its image.

**Theorem A.** Let $L$ be an even unimodular lattice, $N\subset L$ a primitive sublattice, and $r=\operatorname{rk}L-\operatorname{rk}N=\operatorname{rk}N^\perp$. For every integer $0\le k\le r$ and every prime $p$, $\Phi_k\otimes\mathbb F_p$ is injective. Consequently $\Phi_k$ is injective and $J_k(L,N)$ is a direct summand of $L^{\otimes k}$:
$$J_k(L,N)=L^{\otimes k}\cap\big(J_k(L,N)\otimes\mathbb Q\big),\qquad \operatorname{rk}J_k=\sum_{\pi\in P_k}(\operatorname{rk}N)^{|R(\pi)|}.$$

**Theorem B.** Let $S$ be a complex projective K3 surface with $E(S)=\mathbb Q$, and $r=\operatorname{rk}T(S)$. Let $k\ge1$ and $0\le p\le2k$ with $\min(p,2k-p)<r$. Then $\mathrm{Hdg}^{2p}(S^k,\mathbb Z)$ is generated by the classes
$$\prod_{\{i,j\}\in\pi}\mathrm{pr}_{ij}^*[\Delta]\cdot\prod_{a\in A}\mathrm{pr}_a^*c_1(\mathcal L_a)\cdot\prod_{b\in A'}\mathrm{pr}_b^*[\mathrm{pt}],$$
where $\pi$ is a set of disjoint pairs, $A,A'$ are disjoint from each other and from the pairs, and $\mathcal L_a\in\operatorname{Pic}(S)$. In particular the IHC holds for $S^k$ in these codimensions. If $k<r$, it holds in all codimensions.

**Theorem C.** Let $S$ be a complex projective K3 surface and $k\ge1$.
1. The Atiyah–Hirzebruch spectral sequences $H^p(S^k;K^q(\mathrm{pt}))\Rightarrow K^{p+q}(S^k)$ and $H^p(S^k;MU^q(\mathrm{pt}))\Rightarrow MU^{p+q}(S^k)$ degenerate at $E_2$. We have $K^1(S^k)=0$. The Mukai vector $v=\mathrm{ch}\cdot\sqrt{\mathrm{td}_{S^k}}$ and the Chern character $\mathrm{ch}$ are both isomorphisms of groups $K^0_{\rm top}(S^k)\to H^{\rm ev}(S^k,\mathbb Z)$.
2. Call $x\in K^0_{\rm top}(S^k)$ a Hodge class if $\mathrm{ch}(x)\in\bigoplus_pH^{p,p}$. Suppose $E(S)=\mathbb Q$ and $k<r$. Then the Hodge classes are exactly the image of $K_0(S^k)$. They are $\mathbb Z$-combinations of the classes of
$$\boxtimes_{\{i,j\}\in\pi}\mathcal O_{\Delta_{ij}}\ \boxtimes\ \boxtimes_{a\in R(\pi)}F_a,\qquad F_a\in K_0(S),$$
up to permuting the factors of $S^k$.

**Theorem D.** Let $S$ be as in Theorem B with $E(S)=\mathbb Q$. Fix an oriented $\mathbb Z$-basis $t_1,\dots,t_r$ of $T$, and let $\det_T=\sum_{\sigma\in\mathfrak S_r}\mathrm{sgn}(\sigma)\,t_{\sigma(1)}\otimes\dots\otimes t_{\sigma(r)}\in H^2(S,\mathbb Z)^{\otimes r}\subset H^{2r}(S^r,\mathbb Z)$.
1. $\mathrm{Hdg}^{2r}(S^r,\mathbb Z)\cap H^2(S)^{\otimes r}=J_r\oplus\mathbb Z D_S$, where $J_r=J_r(H^2(S,\mathbb Z),\mathrm{NS}(S))$ consists of algebraic classes. Here $D_S=(\det_T+j)/m_S$ with $j\in J_r\otimes\mathbb Q$ and $m_S\in\mathbb Z_{>0}$. The class $D_S$ is unique up to sign modulo $J_r$.
2. For every $k\ge r$, $\mathrm{Hdg}^{2r}(S^k,\mathbb Z)$ is the direct sum of the subgroup generated by the classes of Theorem B and of $\bigoplus_{I\subset[k],|I|=r}\mathbb Z\,\mathrm{pr}_I^*D_S$. The same holds in dimension $r$, i.e. in codimension $2k-r$, with $D_S$ replaced by $D_S\cdot\prod_{b\notin I}\mathrm{pr}_b^*[\mathrm{pt}]$. Hence: HC in codimension $r$ on $S^k$ holds iff $cD_S$ is algebraic for some $c\ne0$; IHC in codimension $r$ on $S^k$ holds iff $D_S$ is algebraic.
3. $m_S$ divides $r!\,|\operatorname{disc}T|$.
4. For every prime $p>r$: $p\mid m_S$ if and only if $p$ divides the scale of $T$.

**Proposition 2.5 (no topological obstruction).** $H^*(S^k,\mathbb Z)$ is torsion-free and concentrated in even degrees. Hence all odd-degree cohomology operations vanish, in particular $Sq^3_{\mathbb Z}$ and every AHSS differential. Totaro's factorization $\mathrm{CH}^*\to MU^*\otimes_{MU^*}\mathbb Z\to H^*$ [8] imposes no condition, since $MU^*(S^k)\to H^*(S^k,\mathbb Z)$ is onto.

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## 3. What is new and what is not

**Not new.** The rational statements behind Theorems B and C (the case $k<r$) are known: they follow from Zarhin's computation of the Hodge group [3] and the first fundamental theorem for orthogonal groups, and are used in this form in [4] (compare [5] for products of surfaces). The degeneration of the AHSS for spaces whose cohomology is free and concentrated in even degrees is standard. So are the integrality of the Mukai vector on a K3 surface and $v(\mathcal O_\Delta)=[\Delta]$, which follows from Grothendieck–Riemann–Roch. Perry [9] formulated the integral Hodge conjecture for categories and proved it for two-dimensional Calabi–Yau categories. $D^b(S^k)$ is not of that type for $k\ge2$. Theorem D(1)–(2) is a formal consequence of Theorem A once the rational picture is known.

**New, as far as our search found** (sources.md):
- Theorem A, an integral saturation theorem for Brauer-type tensors relative to an arbitrary primitive sublattice, valid for all $k\le\operatorname{rk}N^\perp$ with no condition on the discriminant form;
- Theorem B, the integral Hodge conjecture for $S^k$ in codimensions $<r$ and dimensions $<r$ (for $k<r$, in all degrees). We found no prior statement of the IHC for any power $S^k$, $k\ge2$, of a K3 surface with $E(S)=\mathbb Q$ beyond divisors and curves;
- Theorem C(2), Perry's K-theoretic IHC for $D^b(S^k)$, $k<r$;
- Theorem D(3)–(4), the arithmetic of the index $m_S$ of the integral determinant class.

**Not claimed.** We do not prove the algebraicity, rational or integral, of $\det_T$ or $D_S$. We do not treat $E(S)\ne\mathbb Q$. We give no formula for $m_S$ at primes $p\le r$; exact computations show that the $2$-part is not a function of the scale alone (Section 7).

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## 4. Related work

| Work | What it does | Relation to this paper |
|---|---|---|
| Zarhin [3] | Hodge group of $T$ is $\mathrm{SO}(T_{\mathbb Q})$ when $E=\mathbb Q$ | Input for B–D |
| Ramón Marí [5]; Varesco [4] | Rational HC for powers/products of K3 surfaces in the range without determinants; reduction of the rest to $\det_T$ | Rational shadow of B; D is the integral form of Varesco's reduction |
| Buskin [6] | Rational Hodge isometries are algebraic | Not used |
| Atiyah–Hirzebruch; Totaro [8] | Topological obstructions to IHC (AHSS, $MU$) | Proposition 2.5: they vanish on $S^k$ |
| Benoist–Ottem [10]; Totaro [11] | IHC fails / holds for threefolds of Kodaira dimension 0 (e.g. Enriques × curve; K3 × curve type) | Same circle of questions; our varieties are $S^k$, $k\ge2$ |
| Perry [9] | IHC for categories; proof for CY2 categories | Theorem C(2) proves Perry's conjecture for $D^b(S^k)$, $k<r$ |
| Beckmann–de Gaay Fortman [12] | Integral Fourier transforms; IHC for one-cycles on abelian varieties | Analogous integral question on abelian varieties; different methods |
| rStation 77 [14], 35 [13] | Determinant obstruction (monodromy/parity), status table | Theorem D isolates the integral version of their open class |
| rStation 94, 96 [15, 16] | Stable/hyperholomorphic bundle criteria for determinant cycles on $S^N$ | Theorem D gives the precise integral target ($\det_T$-coefficient $1/m_S$) for such constructions (Remark 6.5) |
| rStation 95, 97 [17, 18] | Rational HC for all powers of explicit RM K3 surfaces | RM case $E\ne\mathbb Q$; not covered here |

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## 5. Notation and imported results

- **(I1) Zarhin [3].** If $E(S)=\mathbb Q$, the Hodge group of $H^2(S,\mathbb Q)$ is $\mathrm{SO}(T_{\mathbb Q})$. It acts on $T_{\mathbb Q}$ by the standard representation and trivially on $\mathrm{NS}_{\mathbb Q}$. Since $H^0$ and $H^4$ are Tate, the Hodge classes in $H^*(S^k,\mathbb Q)=H^*(S,\mathbb Q)^{\otimes k}$ (suitably Tate twisted) are the invariants of $\mathrm{SO}(T_{\mathbb Q})$ acting diagonally.
- **(I2) First fundamental theorem for $\mathrm O$ and $\mathrm{SO}$** (Weyl; see Goodman–Wallach [7, Thm. 5.3.3 and §5.3.2]). For a nondegenerate quadratic space $V$ of dimension $r$ over a field of characteristic $0$:
  - $(V^{\otimes j})^{\mathrm{O}(V)}$ is spanned by the Brauer tensors $\delta_V^\pi$, with $\pi$ a perfect matching of $[j]$; it is $0$ for $j$ odd;
  - $(V^{\otimes j})^{\mathrm{SO}(V)}=(V^{\otimes j})^{\mathrm O(V)}$ for $j<r$;
  - for $j=r$, $(V^{\otimes r})^{\mathrm{SO}(V)}=(V^{\otimes r})^{\mathrm O(V)}\oplus\mathbb Q\det_V$.

  Invariants commute with field extension, so this holds over $\mathbb Q$.
- **(I3)** $H^*(S,\mathbb Z)$ is torsion-free, $H^{\rm odd}=0$, and $H^2(S,\mathbb Z)\cong\Lambda=U^3\oplus E_8(-1)^2$ is even unimodular. $\mathrm{NS}(S)$ is primitive in $H^2$, and by the Lefschetz (1,1) theorem $\mathrm{NS}(S)=\mathrm{Hdg}^2(S,\mathbb Z)$. With the Poincaré pairing $\int a\cup b$, $H^{\rm ev}(S,\mathbb Z)=U\oplus\Lambda$ is even unimodular of rank 24. The Künneth formula holds integrally for $S^k$.
- **(I4) Class of the diagonal.** In $H^4(S\times S,\mathbb Z)$, $[\Delta]=1\otimes[\mathrm{pt}]+[\mathrm{pt}]\otimes1+\delta_\Lambda$, and $[\Delta]=\delta_{H^{\rm ev}}$ for the Poincaré pairing on $H^{\rm ev}(S,\mathbb Z)$. This is the standard formula $[\Delta]=\sum e_i\otimes e_i^\vee$; there are no signs since all classes have even degree.
- **(I5) Atiyah–Hirzebruch, Künneth for $K$-theory** (Atiyah). For a finite CW complex $X$ with $K^*(X)$ free, $K^*(X\times Y)\cong K^*(X)\otimes K^*(Y)$. The cycle class map $\mathrm{CH}^*(X)\to H^{2*}(X,\mathbb Z)$ is a ring homomorphism on smooth projective $X$.
- **(I6) Grothendieck–Riemann–Roch** for the closed embedding $\Delta\colon S\to S\times S$: $\mathrm{ch}(\Delta_*\mathcal O_S)\,\mathrm{td}(S\times S)=\Delta_*\mathrm{td}(S)$.

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## 6. Proofs

### 6.1 Saturation and reduction modulo $p$

**Lemma 6.1.** Let $\Phi\colon F\to M$ be a homomorphism of free $\mathbb Z$-modules of finite rank. If $\Phi\otimes\mathbb F_p$ is injective for every prime $p$, then $\Phi$ is injective and $\Phi(F)$ is a direct summand of $M$. The same holds over $\mathbb Z_{(p)}$ for a single prime $p$, with $\mathbb F_p$.

*Proof.* Let $x\in M$ with $px\in\Phi(F)$, say $px=\Phi(y)$. Then $\Phi\otimes\mathbb F_p(\bar y)=0$, so $y=py'$ and $p(x-\Phi(y'))=0$; as $M$ is torsion-free, $x=\Phi(y')$. So $M/\Phi(F)$ is torsion-free, and $\Phi(F)$ is a direct summand. Injectivity follows the same way: if $\Phi(y)=0$ with $y\ne0$, divide $y$ by the largest power of $p$ dividing it, and reduce mod $p$ to get a contradiction. $\square$

We fix $p$ and work over $\mathbb Z_p$. Write $L_p=L\otimes\mathbb Z_p$ and $N_p=N\otimes\mathbb Z_p$. Since $N$ is primitive, $L_p/N_p$ is torsion-free.

**Lemma 6.2 (splitting off the unimodular part).** There is an orthogonal decomposition $N_p=N_0\perp N_1$ with $N_0$ unimodular over $\mathbb Z_p$ and $N_1\cdot N_1\subset p\mathbb Z_p$. Then:
- $L_p=N_0\perp L_1$ with $L_1=N_0^\perp$ unimodular and $N_1\subset L_1$ primitive;
- $\operatorname{rk}L_1=r+s$ with $s=\operatorname{rk}N_1$, and $N_1^\perp\cap L_1=N_p^\perp$ has rank $r$;
- $\Phi_k(L,N)\otimes\mathbb F_p$ is injective iff $\Phi_j(L_1,N_1)\otimes\mathbb F_p$ is injective for all $j\le k$.

*Proof.* The decomposition $N_p=N_0\perp N_1$ is the Jordan splitting of $N_p$: $N_0$ is the unimodular Jordan component, possibly zero, and $N_1$ is the sum of the others. Jordan splittings exist over $\mathbb Z_p$ for all $p$, including $p=2$ [19]. A unimodular sublattice splits off orthogonally, so $L_p=N_0\perp L_1$ with $L_1$ unimodular. $L_1$ is even if $p=2$, because $L$ is even. Next, $L_p/N_p=L_1/N_1$, so $N_1$ is primitive in $L_1$, and $N_p^\perp=N_1^\perp\cap L_1$.

For the injectivity statement, write $\delta_L=\delta_{N_0}+\delta_{L_1}$, valid for orthogonal unimodular summands, and note $\delta_{N_0}\in N_0^{\otimes2}\subset N_p^{\otimes 2}$. Expanding the product over the pairs of $\pi$,
$$\delta_L^\pi\otimes a=\sum_{\pi'\subseteq\pi}\delta_{L_1}^{\pi'}\otimes\big(\delta_{N_0}^{\pi\setminus\pi'}\otimes a\big).$$
Hence $\Phi_k(L,N)\otimes\mathbb Z_p=\Phi^{(1)}\circ\mathcal U$. Here $\Phi^{(1)}\colon(a_\pi)\mapsto\sum\delta_{L_1}^\pi\otimes a_\pi$ with $a_\pi\in N_p^{\otimes R(\pi)}$, and $\mathcal U$ is unitriangular with respect to the number of pairs, hence an automorphism of $\bigoplus_\pi N_p^{\otimes R(\pi)}$. Now decompose
$$N_p^{\otimes R}=\bigoplus_{A\subseteq R}N_0^{\otimes A}\otimes N_1^{\otimes(R\setminus A)},\qquad L_p^{\otimes k}=\bigoplus_{A\subseteq[k]}N_0^{\otimes A}\otimes L_1^{\otimes([k]\setminus A)}.$$
The map $\Phi^{(1)}$ is the direct sum over $A$ of $\mathrm{id}_{N_0^{\otimes A}}\otimes\Phi_{[k]\setminus A}(L_1,N_1)$, because $\delta_{L_1}$ lives in $L_1^{\otimes 2}$. The claim follows. $\square$

So we may assume $N=N_1$, i.e. $\bar N:=N_1/pN_1\subset W:=L_1/pL_1$ is **totally isotropic** of dimension $s$. The space $W$ has dimension $r+s$ and carries a nondegenerate symmetric bilinear form, which is **alternating** if $p=2$ because $L_1$ is even. Since $\bar N$ is totally isotropic, $s\le r$.

**Lemma 6.3 (test vectors).** Let $W$ be as above, and $\bar N\subset W$ totally isotropic of dimension $s$ with basis $n_1,\dots,n_s$.
1. There are $n_1',\dots,n_s'\in W$ spanning a totally isotropic subspace with $n_i\cdot n'_j=\delta_{ij}$.
2. $W_0:=\langle n_i,n'_i\rangle^\perp$ is nondegenerate of dimension $r-s$.
3. If $p$ is odd, $W_0$ has an orthogonal basis $g_1,\dots,g_{r-s}$ with $g_i\cdot g_i=c_i\ne0$.
4. If $p=2$, $W_0$ has a symplectic basis $x_1,y_1,\dots,x_{(r-s)/2},y_{(r-s)/2}$.

*Proof.* These are standard facts on quadratic and symplectic spaces over a field, by Witt's extension of isotropic subspaces to hyperbolic ones; for $p=2$, the alternating case. $\square$

### 6.2 Proof of Theorem A

By Lemmas 6.1–6.2 it suffices to show the following. In the situation of Lemma 6.3 with $k\le r$, if
$$\Sigma:=\sum_{\pi\in P_k}\bar\delta^\pi\otimes\bar a_\pi=0\in W^{\otimes k},\qquad \bar a_\pi\in\bar N^{\otimes R(\pi)},$$
then all $\bar a_\pi=0$. (In Lemma 6.2 the number of slots is $j=k-|A|\le k\le r$, and the rank of $N^\perp$ is still $r$.)

Fix $\pi_0\in P_k$ with $m$ pairs and unmatched set $R_0$, and fix a multi-index $b\colon R_0\to\{1,\dots,s\}$. Write $\bar a_{\pi_0}=\sum_\beta c_\beta\bigotimes_{x\in R_0}n_{\beta(x)}$. We show $c_b=0$.

**Resources.** Let $\beta=|b(R_0)|$. We need $m_0+m_1=m$ with:
- $m_0\le r-s$ if $p$ is odd, or $m_0\le(r-s)/2$ if $p=2$;
- $m_1\le s-\beta$.

This is possible:
- *$p$ odd.* We need $m\le r-\beta$. Since $\beta\le|R_0|=k-2m$, we get $m+\beta\le k-m\le r$.
- *$p=2$.* We need $2m+2\beta\le r+s$. This follows from $\beta\le s$, $\beta\le k-2m$ and $k\le r$: $2m+2\beta\le 2m+(k-2m)+s\le r+s$.

(`verify.py` Block 6 checks these inequalities for all $r\le22$.)

**Test vectors.** Choose $m_1$ distinct indices $h_1,\dots,h_{m_1}\notin b(R_0)$. Assign vectors $v_1,\dots,v_k\in W$ as follows:
- on $m_1$ pairs $\{i,j\}$ of $\pi_0$: $v_i=n'_{h}$ and $v_j=n_h$, one index $h$ per pair;
- on the other $m_0$ pairs: for $p$ odd, $v_i=g_\ell$ and $v_j=c_\ell^{-1}g_\ell$, one $\ell$ per pair; for $p=2$, $v_i=x_\ell$ and $v_j=y_\ell$;
- on $x\in R_0$: $v_x=n'_{b(x)}$.

By construction, for two different slots $u\ne w$, $v_u\cdot v_w\ne0$ only if $\{u,w\}$ is a pair of $\pi_0$, and then $v_u\cdot v_w=1$. Indeed:
- $\bar N'$ and $\bar N$ are totally isotropic;
- $n'_h\cdot n_{h'}=\delta_{hh'}$, and the indices $h$ differ from each other and from $b(R_0)$;
- $W_0\perp(\bar N\oplus\bar N')$;
- the $g_\ell$ are orthogonal, and the symplectic pairs are orthogonal to one another.

Moreover, for $n\in\bar N$: $n\cdot v_u=0$ whenever $v_u\in\bar N\cup W_0$, and $n_c\cdot n'_{d}=\delta_{cd}$.

Now evaluate $\Sigma(v_1,\dots,v_k)$ term by term. Let $\pi\in P_k$ with $\bar\delta^\pi\otimes\bar a_\pi(v)\ne0$.
- If a slot $u$ with $v_u\in\bar N\cup W_0$ were unmatched in $\pi$, the factor $\bar a_\pi$ would pair an element of $\bar N$ with $v_u$ and give $0$. So every such slot is matched in $\pi$, to a slot $w$ with $v_u\cdot v_w\ne0$, i.e. to its partner in $\pi_0$. This covers both slots of every $\pi_0$-pair: the $W_0$ pairs directly, and the hyperbolic pairs through their $n_h$-slot.
- If two slots of $R_0$ were matched in $\pi$, the factor $n'\cdot n'=0$ would vanish. So $R_0$ is unmatched in $\pi$.

Hence $\pi=\pi_0$, and $\Sigma(v)=\bar a_{\pi_0}\big((n'_{b(x)})_{x\in R_0}\big)=c_b$. Since $\Sigma=0$, $c_b=0$. As $\pi_0$ and $b$ were arbitrary, all $\bar a_\pi=0$. $\square$

*Remark 6.4.* No hypothesis on $\operatorname{disc}(N^\perp)$ or on the scale of $N^\perp$ is needed. The bad primes are handled by the isotropic pairs $(n'_h,n_h)$ borrowed from $\bar N$ itself. The naive approach evaluates only on vectors of $T=N^\perp$. It fails when $T$ has scale divisible by $p$; for example $T=U(2)^2$, where all pairings in $T$ are even. Theorem A still holds there (`verify.py` Block 2).

### 6.3 Proof of Theorem B

By (I3), $H^{2p}(S^k,\mathbb Z)=\bigoplus_{\tau}H^{\tau}$. The sum runs over the *types* $\tau=(A_0,B,A_4)$, partitions of $[k]$ with $2|B|+4|A_4|=2p$, and $H^\tau=\bigotimes_{a\in A_0}H^0\otimes\bigotimes_{b\in B}H^2\otimes\bigotimes_{c\in A_4}H^4$. The Hodge group preserves each $H^\tau$, so $\mathrm{Hdg}^{2p}(S^k,\mathbb Z)=\bigoplus_\tau\mathrm{Hdg}(H^\tau)_{\mathbb Z}$. Since $H^0(S,\mathbb Z)=\mathbb Z\cdot1$ and $H^4(S,\mathbb Z)=\mathbb Z[\mathrm{pt}]$, we get $\mathrm{Hdg}(H^\tau)_{\mathbb Z}\cong\mathrm{Hdg}(\Lambda^{\otimes B})_{\mathbb Z}$.

**Rational Hodge classes.** Let $|B|=q$. From $p=q+2|A_4|$ and $|A_4|\le k-q$ we get $q\le p$ and $q\le 2k-p$. So $q<r$. By (I1), $\mathrm{Hdg}(\Lambda_{\mathbb Q}^{\otimes q})=((N_{\mathbb Q}\oplus T_{\mathbb Q})^{\otimes q})^{\mathrm{SO}(T_{\mathbb Q})}$ with $N=\mathrm{NS}(S)$. Expanding, this is $\bigoplus_{C\subseteq[q]}N_{\mathbb Q}^{\otimes([q]\setminus C)}\otimes(T_{\mathbb Q}^{\otimes C})^{\mathrm{SO}}$. By (I2) with $|C|\le q<r$, the last factor is spanned by $\delta_T^\sigma$, with $\sigma$ a perfect matching of $C$. Hence the rational Hodge space is spanned by the $\delta_T^\pi\otimes n_R$. Since $\delta_\Lambda=\delta_T+\delta_N$ with $\delta_N\in N_{\mathbb Q}^{\otimes2}$, a unitriangular change of generators shows that it equals $J_q(\Lambda,N)\otimes\mathbb Q$.

**Integral Hodge classes.** By Theorem A ($\Lambda$ even unimodular, $N$ primitive, $q<r$),
$$\mathrm{Hdg}(\Lambda^{\otimes q})_{\mathbb Z}=\Lambda^{\otimes q}\cap(J_q\otimes\mathbb Q)=J_q(\Lambda,N).$$

**Algebraicity.** By (I4), $\delta_\Lambda=[\Delta]-[S\times\mathrm{pt}]-[\mathrm{pt}\times S]$ is the class of an integral algebraic cycle on $S\times S$. Each $n\in N$ is $c_1$ of a line bundle. The generator $\delta_\Lambda^\pi\otimes n_R$ of $J_q$, placed in $H^\tau$ with $1$ on $A_0$ and $[\mathrm{pt}]$ on $A_4$, is the cup product of the pull-backs $\mathrm{pr}^*_{ij}\delta_\Lambda$, $\mathrm{pr}_a^*c_1(\mathcal L_a)$ and $\mathrm{pr}_c^*[\mathrm{pt}]$. It is therefore the class of an integral cycle by (I5). Expanding $\delta_\Lambda$ gives exactly the generators listed in Theorem B. $\square$

### 6.4 Proof of Theorem C

**(1) Degeneration and integral Chern character.** By (I3), $E_2^{p,q}=H^p(S^k;K^q(\mathrm{pt}))$ is zero unless $p$ and $q$ are both even. The differential $d_r$ has bidegree $(r,1-r)$. For $r$ odd the target has odd $p$; for $r$ even the target has odd $q$. So all $d_r=0$. The same argument works for $MU$, since $MU^*(\mathrm{pt})$ is concentrated in even degrees. Therefore $K^1(S^k)=0$, and $K^0(S^k)$ is free with associated graded $H^{\rm ev}(S^k,\mathbb Z)$. The edge map $MU^{2p}(S^k)\to H^{2p}(S^k,\mathbb Z)$ is onto, which also proves Proposition 2.5.

For $k=1$, $\mathrm{ch}\colon K^0(S)\to H^{\rm ev}(S,\mathbb Q)$ takes values in $H^{\rm ev}(S,\mathbb Z)$, because $\mathrm{ch}_2=c_1^2/2-c_2$ and $c_1^2$ is even. It is an isomorphism onto $H^{\rm ev}(S,\mathbb Z)$. For surjectivity: $\mathrm{ch}(\mathcal O)=1$, $\mathrm{ch}(\mathcal O_{\rm pt})=[\mathrm{pt}]$ topologically, and line bundles on the underlying $4$-manifold realize every $c\in H^2(S,\mathbb Z)$ with $\mathrm{ch}=1+c+c^2/2$. Injectivity holds because both groups are free of rank $24$. Since $\sqrt{\mathrm{td}_S}=1+[\mathrm{pt}]$ is integral and unipotent, $v=\mathrm{ch}\sqrt{\mathrm{td}}$ is also an isomorphism. By (I5), $K^0(S^k)=K^0(S)^{\otimes k}$, and $\mathrm{ch}$ and $\sqrt{\mathrm{td}}$ are multiplicative under $\boxtimes$. So $\mathrm{ch}$ and $v$ are isomorphisms $K^0(S^k)\cong H^{\rm ev}(S,\mathbb Z)^{\otimes k}=H^{\rm ev}(S^k,\mathbb Z)$.

**(2) Hodge classes in K-theory.** $\sqrt{\mathrm{td}_{S^k}}$ is an invertible Hodge class, so $x$ is a Hodge class iff $v(x)$ is a Hodge class of $L'^{\otimes k}$, where $L'=H^{\rm ev}(S,\mathbb Z)$ carries the Poincaré pairing (even unimodular by (I3)). Put $N'=H^0\oplus\mathrm{NS}\oplus H^4$. It is primitive, $N'^\perp=T$ has rank $r$, and the Hodge group acts through $\mathrm{SO}(T_{\mathbb Q})$, trivially on $N'$. As in §6.3, for $k<r$ the rational Hodge classes of $L'^{\otimes k}$ equal $J_k(L',N')\otimes\mathbb Q$, and by Theorem A the integral ones equal $J_k(L',N')$.

**Generators.**
- $N'=v(K_0(S))$: indeed $v(\mathcal O_S)=1+[\mathrm{pt}]$, $v(\mathcal O_x)=[\mathrm{pt}]$, and $v(\mathcal O(D))=1+D+(D^2/2+1)[\mathrm{pt}]$ with $D^2/2\in\mathbb Z$.
- $v(\mathcal O_\Delta)=\delta_{L'}$: by (I6),
$$v(\mathcal O_\Delta)=\Delta_*(\mathrm{td}_S)\,\mathrm{td}(S\times S)^{-1/2}=\Delta_*\big(\mathrm{td}_S\cdot\Delta^*\mathrm{td}(S\times S)^{-1/2}\big)=\Delta_*(1)=[\Delta],$$
using $\Delta^*\mathrm{td}(S\times S)=\mathrm{td}_S^2$. Then $[\Delta]=\delta_{L'}$ by (I4).

Since $v$ is multiplicative for $\boxtimes$, $\delta_{L'}^\pi\otimes v(F)_{R}=v(\boxtimes\mathcal O_{\Delta}\boxtimes F)$, up to the permutation of factors. $\square$

### 6.5 Proof of Theorem D

**(1)** In the decomposition of §6.3 the types with $|B|=r$ occur, and by (I2) the rational Hodge space of $\Lambda^{\otimes r}$ is $J_r\otimes\mathbb Q\oplus\mathbb Q\det_T$. (The $\mathrm{SO}$-invariants of $(N\oplus T)^{\otimes r}$ involving fewer than $r$ copies of $T$ are $\mathrm O$-invariant.) Theorem A with $k=r$ shows $J_r$ is saturated, so $H_{\mathbb Z}:=\Lambda^{\otimes r}\cap(J_r\otimes\mathbb Q\oplus\mathbb Q\det_T)$ satisfies $H_{\mathbb Z}\cap(J_r\otimes\mathbb Q)=J_r$. Then $H_{\mathbb Z}/J_r$ is torsion-free of rank one; let $D_S$ be any lift of a generator. Since $\det_T\in H_{\mathbb Z}$, $\det_T\equiv m_SD_S$ modulo $J_r$ for a unique $m_S>0$ (after choosing the sign of $D_S$). This gives $D_S=(\det_T+j)/m_S$ with $j\in J_r\otimes\mathbb Q$.

**(2)** In codimension $r$ the types have $|B|=r$ and $A_4=\emptyset$, or $|B|<r$; in dimension $r$ the analogous statement holds with $A_0=\emptyset$. For $|B|<r$ we use §6.3. For $|B|=r$ we use (1), via $\mathrm{pr}_I^*$, and the product with $[\mathrm{pt}]$ for the dimension version. All classes other than the $D_S$-multiples are algebraic. So the cokernel of $\mathrm{cl}$ in codimension $r$ is the image of $\bigoplus_I\mathbb Z\,\mathrm{pr}_I^*D_S$ modulo algebraic classes. The image of $\mathrm{cl}$ is stable under $\mathrm{pr}_I^*$ and under the symmetric group. Restricting to $S^r\times\{\text{pt}\}^{k-r}$ via the algebraic correspondence $\mathrm{pr}_{I*}(\,\cdot\,\cdot\prod_{b\notin I}\mathrm{pr}_b^*[\mathrm{pt}])$ recovers $D_S$ from $\mathrm{pr}_I^*D_S$ (the slots outside $I$ carry $1$, and $\int[\mathrm{pt}]=1$), and kills $\mathrm{pr}_{I'}^*D_S$ for $I'\neq I$ (a slot $b\in I'\setminus I$ carries a class of degree 2, whose product with $[\mathrm{pt}]$ vanishes). Hence $\mathrm{pr}_I^*D_S$ (resp. a multiple of it) is algebraic iff $D_S$ (resp. a multiple) is. This gives both equivalences.

**(3)** $\mathrm O(T_{\mathbb Q})$ acts on $\Lambda_{\mathbb Q}^{\otimes r}$ preserving the induced form, and $J_r\otimes\mathbb Q$ and $\mathbb Q\det_T$ lie in different isotypic components (trivial and determinant character). So they are orthogonal. Hence $\langle D_S,\det_T\rangle=\langle\det_T,\det_T\rangle/m_S$. Now
$$\langle\det_T,\det_T\rangle=\sum_{\sigma,\tau}\mathrm{sgn}(\sigma\tau)\prod_i t_{\sigma(i)}\cdot t_{\tau(i)}=r!\det(t_i\cdot t_j)=r!\operatorname{disc}T.$$
Since $D_S,\det_T\in\Lambda^{\otimes r}$, $\langle D_S,\det_T\rangle\in\mathbb Z$. Therefore $m_S\mid r!\,|\operatorname{disc}T|$.

**(4)** By Lemma 6.1 and saturation of $J_r$: $p\mid m_S$ iff $\det_T\equiv -j'\pmod p$ for some $j'\in J_r$, i.e. iff $\overline{\det_T}\in\bar J_r\subset W^{\otimes r}$, where $W=\Lambda/p$. Let $\mathcal A=\sum_\sigma\mathrm{sgn}(\sigma)\sigma$ be the antisymmetrizer on $W^{\otimes r}$, and let $p>r$ (so $p$ is odd).
- If $\pi$ contains a pair $\{i,j\}$, the transposition $(ij)$ fixes $\bar\delta^\pi\otimes a$ while $\mathcal A\circ(ij)=-\mathcal A$; so $\mathcal A(\bar\delta^\pi\otimes a)=0$. Hence $\mathcal A(\bar J_r)\subseteq\mathcal A(\bar N^{\otimes r})\subseteq\bar N^{\otimes r}$.
- $\mathcal A(\overline{\det_T})=r!\,\overline{\det_T}$ with $r!\in\mathbb F_p^\times$.

So $\overline{\det_T}\in\bar J_r$ implies $\overline{\det_T}=\bar t_1\wedge\dots\wedge\bar t_r\in\bar N^{\otimes r}$. This is nonzero because $T$ is primitive. Contracting with linear forms then shows $\bar T\subseteq\bar N$. Also $\bar T\subseteq\bar N^\perp$, so $\bar T$ is totally isotropic, i.e. $p$ divides the scale of $T$.

Conversely, suppose $p$ divides the scale of $T$. Then $\bar T$ is totally isotropic. As $\dim\bar T=r=\dim\bar N^\perp$ and $\bar T\subseteq\bar N^\perp$, we have $\bar T=\bar N^\perp$, hence $\bar T\subseteq\bar T^\perp=\bar N$. Then $\overline{\det_T}\in\bar N^{\otimes r}\subseteq\bar J_r$. $\square$

*Remark 6.5 (what a determinant construction must achieve).* Write $c(X)$ for the $\det_T$-coefficient of a class $X\in H^{2r}(S^r,\mathbb Q)$ (the orthogonal projection of §6.5(3)). By Theorem D, the IHC in codimension $r$ holds iff the values $c([Z])$, over integral algebraic cycles $Z$ of codimension $r$ on $S^r$, generate $\frac1{m_S}\mathbb Z$; rational HC only needs one nonzero value. A construction producing a sheaf or bundle $\mathcal E$ (as in rStation 94/96 [15, 16]) gives $c(\mathrm{ch}_r(\mathcal E))$. Since $\mathrm{ch}_r$ is not in general an integral combination of cycle classes, the integral information it carries is naturally K-theoretic, through Theorem C(1).

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## 7. Evidence (exact computations, `verify.py`)

The script performs 50 exact checks and ends with `FALLOS = 0` (output in `verify_output.txt`). It uses exact integer Smith invariants and no floating point.

| Block | Content |
|---|---|
| 1 | $\Lambda=U^3\oplus E_8(-1)^2$ is even unimodular of signature $(3,19)$; $H^{\rm ev}=U\oplus\Lambda$ is even unimodular |
| 2 | Theorem A on seven lattices $(L,N)$, $k\le r$. These include bad-prime cases: $T=\langle-4\rangle^3$ (scale 4), $T=U(2)^2$ with $\bar N$ Lagrangian mod 2, and $T\supset U(3)$. In each case $\Phi_k$ is injective and all elementary divisors equal 1 |
| 3 | Eight random primitive $N\subset U^3,U^4$ (fixed seed), $k\le r$ |
| 4 | Full K3 lattice, $\rho=19$, $T=\langle2\rangle\oplus\langle-2\rangle\oplus\langle-2\rangle$, $k=2$: the Hodge lattice of $H^2\otimes H^2$ is $\mathrm{NS}\otimes\mathrm{NS}\oplus\mathbb Z\delta$, saturated of rank 362 |
| 5 | Theorem D(3) in two cases ($m_S=2$ for $T=\langle2\rangle\oplus\langle-2\rangle^2$, and $m_S=4$ for $T=\langle-4\rangle^3$). Theorem D(4) in five rank-2 cases with $p=3,5$, including $T=\langle2\rangle\oplus\langle-6\rangle$ where $3\mid\operatorname{disc}T$ but $3\nmid m_S$ |
| 6 | The resource inequalities of §6.2 for all $r\le22$ |
| 7 | The test vectors of §6.2 isolate every coefficient of a random $\Sigma$, for $p=3$ and $p=2$ |
| 8 | Negative controls: a non-primitive $2N$ is detected as non-saturated; $\det_T\notin J_r\otimes\mathbb Q$; outside the range ($\dim1$, two pairs) Brauer tensors are dependent; $\delta_T$ alone is not integral |

**Exploratory data, not part of the theorems.** In a search over several dozen random rank-3 lattices $T\subset U^3$ (scripts not included), $m_S$ took the values $2,4,6,8$. Lattices of scale $2$ gave both $2$ and $4$. So no formula at $p\le r$ depending only on the scale is possible. In all cases with scale $1$ and $r\in\{4,5\}$ we found $m_S=2$. We do not claim this in general.

---

## 8. Limitations

- **The rational determinant problem is untouched.** Theorems B and C refine known rational results integrally. Theorem D identifies the integral class $D_S$ but does not prove it algebraic in any new case, rationally or integrally. In particular, Theorem D is not known to give the IHC in codimension $r$ even on Floccari's locus or in the CM case, where $\det_T$ is rationally algebraic.
- **Only $E(S)=\mathbb Q$.** For RM or CM fields, the Hodge classes include $E$-linear tensors. Their integral structure involves the order $\operatorname{End}_{\rm Hdg}(T)\cap\operatorname{End}(T)$, which Theorem A does not see.
- **Only codimensions and dimensions $<r$, plus the description in codimension $r$.** For $k\ge r$ and $r<p<2k-r$, products of $\det_T$ with Brauer tensors appear, and saturation is not analysed.
- **$m_S$ at small primes.** Only $m_S\mid r!|\operatorname{disc}T|$ and the criterion for $p>r$ are proved.

---

## 9. Reproducibility

`python3 verify.py` (Python 3, sympy ≥ 1.12) is self-contained: no files and a fixed seed. It runs in about 15 s and ends with `FALLOS = 0` and 50 checks. The output is stored in `verify_output.txt`.

---

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