Tate Conjecture for Algebraic Cycles

Tate Conjecture for Algebraic Cycles

OpenAI's claim

The manuscripts claim algebraicity of rational Hodge classes on CM abelian varieties, with consequences for generalized Hodge classes and finite-field Tate classes. The cited manuscript argues for that result; its relation to this problem is specified below.

Mathematical statement of the claimed result

Mathematical claim 1.

A complex abelian variety $A$ is CM if $\operatorname{End}(A)\otimes\mathbb Q$ contains a commutative semisimple algebra of dimension $2\dim A$. For every such $A$ and every $p\ge0$, OpenAI claims \[ \operatorname{CH}^p(A)_{\mathbb Q}\longrightarrow H^{2p}(A,\mathbb Q)\cap H^{p,p}(A) \quad\text{is surjective}. \] This also applies to arbitrary finite products of CM abelian varieties, including their powers. If $W\subset H^k(A,\mathbb Q)$ is a rational Hodge substructure and $W(r)$ is effective (all Hodge bidegrees are nonnegative), it claims a closed algebraic $Z\subset A$ of codimension at least $r$ with $W\subset\ker[H^k(A,\mathbb Q)\to H^k(A\setminus Z,\mathbb Q)]$. The manuscript additionally asserts that, for every abelian variety $A/\mathbb F_q$, every $r\ge0$ and prime $\ell\ne\operatorname{char}\mathbb F_q$, \[ \operatorname{CH}^r(A)\otimes\mathbb Q_\ell\longrightarrow H^{2r}_{\mathrm{\acute et}}(A_{\overline{\mathbb F}_q},\mathbb Q_\ell(r))^{\operatorname{Gal}(\overline{\mathbb F}_q/\mathbb F_q)} \quad\text{is surjective}. \] Numerical and $\ell$-adic homological equivalence are claimed to agree on rational cycles. Over any algebraically closed field, for an abelian $g$-fold and rational ample class $h$, the form $(-1)^r\deg(xyh^{g-2r})$ is claimed positive definite on the real extension of $\ker[h^{g-2r+1}:A^r_{\mathrm{num}}\to A^{g-r+1}_{\mathrm{num}}]$ for $0\le r\le g/2$.

Outline of the argument

Manuscript 1 (family 032, 2026-09-30): The argument realizes four-factor balanced tensors using unitary theta constructions. Elementary switches between equal-sum lists, followed by algebraic polarization contractions and descent of coefficients, build arbitrary balanced CM tensors from these elementary cycle classes.

Formal verification

Family 032 has no Lean scope page in this snapshot; no formalization of these manuscript results is listed.

This site has not executed the Lean files or checked their dependencies. These descriptions report the scope of the linked declarations, rather than a new verification of the proofs.

Status caveat

OpenAI attributes these results to an internal model. Its release notice cautions that results lacking full formalization can contain mistakes. This site has not independently assessed the proofs; a solved label records the scope of the reported claim.

Sources

• OpenAI, The rational Hodge conjecture for CM abelian varieties, 2026-09-30; citation key OAI:The-rational-Hodge-conjecture-for-CM-abelian-varieties-September-30-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:cm; Statement source: cc:ghc; Statement source: cc:tate; Statement source: cc:standard.
• OpenAI catalogue entry.

Tate Conjecture for Algebraic Cycles

Definitions and mathematical statement

Let $\bar X=X\times_k\bar k$, $G_k=\operatorname{Gal}(\bar k/k)$, and $\ell\ne\operatorname{char}k$. The Chow group $\operatorname{CH}^i(X)$ consists of codimension-$i$ cycles modulo rational equivalence. The Tate twist ${\mathbb{Q}}_\ell(i)$ is the $i$-fold tensor power of the cyclotomic representation. The assertion is surjectivity of \[ \operatorname{cl}_\ell:\operatorname{CH}^i(X)\otimes{\mathbb{Q}}_\ell \longrightarrow H^{2i}_{\mathrm{\acute et}}(\bar X,{\mathbb{Q}}_\ell(i))^{G_k} \] for every smooth geometrically irreducible projective $X/k$, every $i$, and every $k$ finitely generated over its prime field. The domain uses cycles over $k$, and the coefficients are ${\mathbb{Q}}_\ell$, not ${\mathbb{Z}}_\ell$.

Formulation and concept references: [S1].

Short English statement

Are all arithmetic-symmetry-invariant even cohomology classes generated by algebraic cycles defined over the base field?

Research attempt

Scope and source check (27 September 2026).

The target is surjectivity over the stated base field, for all codimensions and finitely generated fields. It is not the additional conjecture that homological and numerical equivalence agree. Nor may the rational coefficient field be replaced by an integral lattice. Li's formulation [S1] and Milne's account [E1] distinguish these assertions. Kahn [S2, S4] explicitly describes three incomplete approaches for surfaces over finite fields. Milne [S3] discusses failure of an integral Tate assertion for abelian varieties; that does not disprove the displayed rational map. None of these references is treated as a general solution.

The approach here separates three tasks often mixed together: finding geometric algebraic classes, descending classes to the base field, and certifying that enough independent classes have been found. The descent task admits an exact argument. The cycle-construction task remains open. For finite fields, a Frobenius calculation locates the relevant eigenspaces but also exposes a failure of counting eigenvalues alone. The elementary deductions below are not claimed as new results in the theory of cycles.

Boundary cases that fix the coefficient convention.

Write $d=\dim X$ and $V_i=H^{2i}_{\mathrm{\acute et}}(\bar X,\mathbb Q_\ell(i))$. For $i=0$, geometric irreducibility gives $V_0=\mathbb Q_\ell$, generated by $[X]$. For $i=d$, the trace identifies $V_d$ with $\mathbb Q_\ell$ with trivial Galois action. Any closed point $P$ has finite degree $n=[k(P):k]>0$, and its cycle class has trace $n$. Thus $[P]/n$ generates the target. No $k$-rational point is required. Codimensions outside $0\leq i\leq d$ contribute nothing.

This proves all codimensions for curves, but not divisors on arbitrary higher-dimensional varieties. More generally, for $X=\prod_{j=1}^s\mathbb P_k^{n_j}$ the projective-space cohomology and Künneth formulas give \[ H^{2i}(\bar X,\mathbb Q_\ell(i)) =\bigoplus_{\substack{a_1+\cdots+a_s=i\\0\leq a_j\leq n_j}} \mathbb Q_\ell h_1^{a_1}\cdots h_s^{a_s}. \] Each monomial is the class of a product of linear subspaces over $k$. Thus the selected map is surjective for this complete family. This calculation supplies a check on twists and degrees; a general projective variety does not have this cohomology presentation.

Lemma 15.1: finite-extension descent of an invariant class.

Let $L/k$ be finite Galois, with group $\Gamma$ and degree $n$. Suppose $v\in V_i^{G_k}$ and that $v$ is the class of $z\in\operatorname{CH}^i(X_L)\otimes\mathbb Q_\ell$ under the geometric identification of cohomology. Then \begin{equation}\label{wsref-0-74-63-3a-74-72-61-63-65} z_0=\frac1n f_*z\in\operatorname{CH}^i(X)\otimes\mathbb Q_\ell, \qquad\operatorname{cl}_\ell(z_0)=v, \end{equation} where $f:X_L\to X$ is the finite projection.

Proof. Finite flat pullback and proper pushforward satisfy $f^*f_*z=\sum_{\sigma\in\Gamma}\sigma z$. After applying the cycle map, its right side is $\sum_\sigma\sigma v=nv$, because the original class is fixed by all of $G_k$. The geometric pullback identifies the target cohomology spaces, so division by $n$ proves the claim. This argument is linear over $\mathbb Q_\ell$ and does not require the coefficients of $z$ to be rational numbers or its individual components to be invariant. $\square$

In particular, surjectivity over a finite Galois extension implies surjectivity over $k$: first restrict a $G_k$-invariant class, produce its cycle over $L$, and apply the lemma. A non-Galois separable extension is handled by pulling its cycle to a Galois closure. The denominators are legitimate even when $\ell\mid n$, since the coefficients are a field.

There is a qualification in imperfect characteristic. A cycle over $\bar k$ may require a purely inseparable extension in its field of definition, so it is not enough to say that every cycle descends to a finite separable extension. Finite purely inseparable base change is a universal homeomorphism and does not change geometric $\ell$-adic cohomology for $\ell\ne\operatorname{char}k$. For its finite flat projection of degree $p^e$, both push--pull composites on cycle groups multiply by $p^e$: there is a unique component above each component, and the residue degree times its flat multiplicity is $p^e$. Equivalently, the geometric fiber has that multiplicity. Dividing by $p^e$ gives the same class-descent statement. Combining this with separable descent handles a general finite algebraic field of definition. These are the usual cycle-map push--pull properties [E2], not an assumption that $k$ is perfect.

Every individual algebraic cycle has such a finite field of definition, and a finite linear combination has a common one. It follows that if a $G_k$-fixed class is known to lie in the geometric algebraic span, then it is represented by a cycle over $k$ with $\mathbb Q_\ell$ coefficients. This isolates the real gap: invariance has not been shown to imply membership in that geometric span.

The invariance assumption in Lemma 15.1 cannot be omitted. In the simple permutation representation with basis $e_1,e_2$ and an involution interchanging them, averaging $e_1$ gives $(e_1+e_2)/2$, not $e_1$. The anti-invariant vector $e_1-e_2$ is annihilated. Thus an orbit sum of a cycle does not preserve an arbitrary geometric class; it preserves the given class precisely when that class was already invariant. Also, averaging is performed on a known cycle representative. Writing the same averaging operator on cohomology only projects onto invariants and gives no procedure to lift an invariant vector through the cycle map. This distinction prevents the descent lemma from being read as an existence proof in disguise.

An actual integral countertest: the anisotropic conic.

Take $k=\mathbb Q$ and the smooth projective conic \[ X:\ x^2+y^2+z^2=0\subset\mathbb P^2_{\mathbb Q}. \] It has no real point. A closed point of odd degree would have residue field of odd degree over $\mathbb Q$, hence at least one real embedding; applying that embedding would give a real point of $X$. Therefore every closed point has even degree. The point $[1:i:0]$ over $\mathbb Q(i)$ yields a closed point of degree two, so the subgroup of degrees of zero-cycles is exactly $2\mathbb Z$.

Consequently, for the coefficient-replaced version of the displayed map, \[ \operatorname{CH}^1(X)\otimes\mathbb Z_2 \longrightarrow H^2(\bar X,\mathbb Z_2(1))^{G_{\mathbb Q}} \cong\mathbb Z_2, \] the image is $2\mathbb Z_2$, and the map is not surjective. With $\mathbb Q_2$ coefficients it is surjective, by dividing the degree-two class by two. This proves that the field and coefficient qualifications are doing real work. It is a counterexample only to that integral replacement over the fixed base field, not to the rational target or every convention called an integral Tate conjecture.

Finite-field attack: separate invariants from eigenvalue counts.

Now suppose $k=\mathbb F_q$. Let $F$ denote geometric Frobenius on untwisted $H^{2i}$ and put $T=q^{-i}F$ on $V_i$. This is the action after the Tate twist; geometric Frobenius acts on $\mathbb Q_\ell(1)$ by $q^{-1}$. Fixed classes over $\mathbb F_{q^m}$ are $\ker(T^m-I)$. The Weil eigenvalue bounds alone do not make all eigenvalues of $T$ roots of unity: algebraic numbers of complex absolute value one need not be algebraic integers after this normalization.

The following linear algebra requires no semisimplicity. Over a splitting field, on the generalized $\lambda$-eigenspace, $T^m-I$ is invertible unless $\lambda^m=1$. In the latter case \[ x^m-1=(x-\lambda)Q(x),\qquad Q(\lambda)=m\lambda^{m-1}\ne0. \] Thus $Q(T)$ is invertible on that generalized eigenspace and $\ker(T^m-I)=\ker(T-\lambda I)$ there. Generalized eigenvectors are not accidentally fixed by passing to a finite extension. Let $M$ be the least common multiple of the orders of the finitely many root-of-unity eigenvalues, taking $M=1$ if there are none. Then \begin{equation}\label{wsref-0-74-63-3a-73-74-61-62-69-6c-69-7a-65} \bigcup_{m\geq1}\ker(T^m-I)=\ker(T^M-I). \end{equation} The equality descends to $\mathbb Q_\ell$, since both sides are defined over it. The union is a vector space: a common multiple of two extension degrees fixes sums of the corresponding vectors.

This is a useful finite-dimensional reduction for a fixed variety, but it constructs no cycles spanning the stabilized space. It also does not identify its dimension with the total multiplicity of the relevant roots in the characteristic polynomial. For example, \[ T_1=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad T_2=\begin{pmatrix}1&1\\0&1\end{pmatrix} \] have the same characteristic polynomial and the same traces of every positive power. Yet their fixed spaces have dimensions two and one, respectively, and $T_2^m-I=m(T_2-I)$ has rank one for every $m>0$. These are test representations, not asserted examples of geometric Frobenius. They show that characteristic-polynomial data alone do not justify replacing a fixed-space calculation by a multiplicity count. Milne [E1, Section 1] keeps the relevant semisimplicity condition separate.

A finite intersection certificate.

Suppose an upper bound $r$ for $\dim V_i^{G_k}$ is established, and one finds cycles $z_1,\ldots,z_r$ of codimension $i$ and $w_1,\ldots,w_r$ of codimension $d-i$, all over $k$, such that \begin{equation}\label{wsref-0-74-63-3a-6d-61-74-72-69-78} B_{ab}=\deg(z_a\cdot w_b),\qquad \det B\ne0. \end{equation} Then the Tate map in codimension $i$ is surjective. Indeed, a relation $\sum_a c_a\operatorname{cl}_\ell(z_a)=0$ pairs with each $w_b$ to give $\sum_a c_aB_{ab}=0$. Invertibility forces every $c_a=0$. These $r$ invariant classes therefore fill a space whose dimension is at most $r$. The products are Chow intersection products on the smooth variety; no unproved transversality of chosen representatives is needed.

For finite fields, the multiplicity of $1$ in the characteristic polynomial of $T$ is always an upper bound, so it can be used in this sufficient certificate without assuming semisimplicity. Finding that many paired cycles would itself force equality of the dimensions. But a universal supply of complementary cycles with an invertible pairing is extra work. Even a nondegenerate full pairing need not restrict nondegenerately to invariant subspaces: for $T_2$ above and its contragredient, the fixed vectors are $e_1$ and $e_2^*$, whose pairing is zero. Therefore the certificate is sufficient, not an automatic consequence of duality.

Verification, first gap, and outcome.

The descent proof was checked at degrees divisible by $\ell$ and at purely inseparable degrees. The conic provides a concrete failure when the permitted denominators are removed. The Frobenius argument was checked on a nontrivial Jordan block rather than assuming diagonalizability. Invariant dimensions, intersection ranks, and eigenvalue multiplicities were kept distinct. No finite point count or cycle search is being reported as a proof for all varieties.

Three next targets are to construct cycles spanning the stabilized Frobenius space for a specified finite-field family; to obtain actual intersection matrices meeting (tc:matrix); and to identify the geometric algebraic span for Galois-invariant classes over other finitely generated fields. These cannot be replaced by averaging a class that has no known cycle representative. The first unresolved step is exactly that geometric algebraicity, before the proved descent lemma applies.

UNRESOLVED. The output proves descent and certification lemmas, elementary complete subcases, and explicit obstructions to integral and spectral shortcuts. It does not construct the missing cycles in general. Written by GPT-6 Astra Ultra with primary-source scope checks and direct algebraic verification; no formal proof certification is claimed.

Sources

• S1 Li, What is the Tate conjecture?, map $2$ and Conjecture 1 $2013$. https://www.math.columbia.edu/~chaoli/docs/TateConjecture.html. Formulation source. Consulted 24 September 2026: Cycle-class map and Conjecture 1.
• S2 An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn $2025$. https://arxiv.org/abs/2505.07337. Further reference; not a proof endorsement.
• S3 On the integral Tate conjecture for abelian varieties — J. S. Milne (September 8, 2025). https://arxiv.org/abs/2509.06707. Further reference; not a proof endorsement.
• S4 An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn, full text. https://arxiv.org/html/2505.07337v1. Further reference; not a proof endorsement.
• E1 J. S. Milne, The Tate conjecture over finite fields (AIM talk), revised 7 May 2008, especially Section 1. https://www.jmilne.org/math/articles/2007e.pdf. Primary author account separating cycle-surjectivity, numerical equivalence, and semisimplicity; consulted 27 September 2026.
• E2 The Stacks Project, Chow Homology, proper pushforward, flat pullback, and their compatibility. https://stacks.math.columbia.edu/tag/02P3. Also Topological invariance of the small étale site: https://stacks.math.columbia.edu/tag/04DY. References for finite-extension push--pull and purely inseparable invariance; consulted 27 September 2026.

Tate Conjecture for Algebraic Cycles

Definitions and mathematical statement

Let $\bar X=X\times_k\bar k$, $G_k=\operatorname{Gal}(\bar k/k)$, and $\ell\ne\operatorname{char}k$. The Chow group $\operatorname{CH}^i(X)$ consists of codimension-$i$ cycles modulo rational equivalence. The Tate twist ${\mathbb{Q}}_\ell(i)$ is the $i$-fold tensor power of the cyclotomic representation. The assertion is surjectivity of \[ \operatorname{cl}_\ell:\operatorname{CH}^i(X)\otimes{\mathbb{Q}}_\ell \longrightarrow H^{2i}_{\mathrm{\acute et}}(\bar X,{\mathbb{Q}}_\ell(i))^{G_k} \] for every smooth geometrically irreducible projective $X/k$, every $i$, and every $k$ finitely generated over its prime field. The domain uses cycles over $k$, and the coefficients are ${\mathbb{Q}}_\ell$, not ${\mathbb{Z}}_\ell$.

Formulation and concept references: [S1].

Short English statement

Are all arithmetic-symmetry-invariant even cohomology classes generated by algebraic cycles defined over the base field?

Research attempt

Current frontier.

Research check: 27 September 2026. The divisor theorem for abelian varieties is an established special case; it does not supply all higher-codimension classes on arbitrary varieties. The distinction between cycle-class surjectivity and the stronger package including semisimplicity is explicit in Milne, Section 1, especially Theorem 1.4. [S1], [E1]. Kahn's 2025 surface manuscript gives a reformulation and identifies a step that does not yet work in Section 3.2. Milne's September 2025 note concerns failure with integral coefficients, not a counterexample to the rational statement here. [S4], [S3]. We investigate a finite-dimensional certificate and the descent needed to make its cycles live over the stipulated base field.

Attack route.

One route is to lift invariant classes from special fibers; it requires algebraic cycles to lift, not merely their cohomology. Another is to build cycles until their intersection matrix exhausts the Frobenius eigenspace. A third is to find cycles after a field extension and descend them. The second and third routes can be combined without assuming the conjectural semisimplicity one is trying to exploit. The finite-field criterion below is stronger than the selected surjectivity statement; this strength is kept explicit.

Attempt: descent of an already algebraic invariant class.

Let $L/k$ be finite Galois, $p:X_L\to X$ the projection, and $\alpha\in H^{2i}(\bar X,{\mathbb{Q}}_\ell(i))^{G_k}$. Suppose that $\alpha=\operatorname{cl}_\ell(z)$ for $z\in\operatorname{CH}^i(X_L)\otimes{\mathbb{Q}}_\ell$. Then the cycle \[ z_0=\frac{1}{[L:k]}p_*z\in\operatorname{CH}^i(X)\otimes{\mathbb{Q}}_\ell \tag{15.1} \] has class $\alpha$. Indeed, on geometric cohomology, pullback of the pushforward is $\sum_{\sigma\in\operatorname{Gal}(L/k)}\sigma$; every summand fixes $\alpha$. Geometric base change identifies the target cohomology groups, so division by $[L:k]$ proves the claim. This uses rational $\ell$-adic coefficients even when $\ell$ divides the degree. It is not an integral argument. The push-pull and cycle-class compatibilities used here are those of the formulation and of the finite-field discussion in [E1].

Consequently, if a $G_k$-invariant class is represented by a finite combination of cycles defined over a finite separable extension, its field of definition is not an additional obstruction: take a Galois closure and apply $15.1$. This does not construct the cycles. In positive characteristic, this particular lemma is stated only for separable extensions; no unmentioned passage through a purely inseparable extension is needed for it.

Attempt: a certificate that also controls Jordan blocks.

Now specialize to $k={\mathbb{F}}_q$ and write $n=\dim X$, \[ V=H^{2i}(\bar X,{\mathbb{Q}}_\ell(i)),\qquad W=H^{2n-2i}(\bar X,{\mathbb{Q}}_\ell(n-i)). \] Let $F$ be geometric Frobenius acting on these twisted spaces. Poincare duality gives a nondegenerate $F$-invariant pairing $\langle\ ,\ \rangle:V\times W\to{\mathbb{Q}}_\ell$. Let $r$ be the algebraic multiplicity of $1$ in the characteristic polynomial of $F$ on $V$. Assume we have cycles over $k$, \[ z_1,\ldots,z_r\in\operatorname{CH}^i(X)_{{\mathbb{Q}}},\qquad w_1,\ldots,w_r\in\operatorname{CH}^{n-i}(X)_{{\mathbb{Q}}}, \] such that the rational matrix \[ D_{ba}=\deg(z_a\cdot w_b)\quad(1\leq a,b\leq r) \quad\hbox{satisfies}\quad\det D\ne0. \tag{15.2} \] Then the selected Tate assertion holds in codimension $i$ for $X$, and $F$ acts as the identity on its entire generalized $1$-eigenspace.

Here is a proof, including the dimension issue. The classes $v_a=\operatorname{cl}(z_a)$ lie in $V^F$. Pairing a relation $\sum_a t_av_a=0$ with all $\operatorname{cl}(w_b)$ yields $Dt=0$, so they are independent. On the other hand $\dim V^F\leq r$, because the fixed space is contained in the generalized $1$-eigenspace. Thus $\dim V^F=r$ and these classes span it. Equality with the generalized eigenspace excludes all nontrivial Jordan blocks at $1$. No semisimplicity was used as an input. If $r=0$, the fixed space is zero and the assertion is vacuous.

More explicitly, the algebraic correspondence \[ \Pi=\sum_{a,b}(D^{-1})_{ab}\,w_b\times z_a \in\operatorname{CH}^n(X\times X)_{{\mathbb{Q}}} \tag{15.3} \] acts on $V$ by $\Pi_*(v)=\sum_{a,b}v_a(D^{-1})_{ab}\langle v,\operatorname{cl}(w_b)\rangle$. It is the identity on the span of the $v_a$, has image that span, and therefore satisfies $\Pi_*^2=\Pi_*$ on $V$. In fact this particular product-cycle formula gives an idempotent even modulo rational equivalence. The composition of $w_d\times z_c$ followed by $w_b\times z_a$ is $D_{bc}w_d\times z_a$: in the middle factor the intersection is a zero-cycle whose pushforward is its degree. Summing with the coefficients in $15.3$ gives the matrix product $D^{-1}DD^{-1}=D^{-1}$, hence $\Pi\circ\Pi=\Pi$ in $\operatorname{CH}^n(X\times X)_{{\mathbb{Q}}}$. No assertion that arbitrary cohomological projectors lift to Chow projectors is used; this one is built from the assumed dual cycle lists. The connection with the finite-field Tate and semisimplicity conditions is discussed in [E1], Theorem 1.4; no historical novelty is asserted.

Attempt: what approximation would actually suffice.

Let $A\subseteq V$ be the ${\mathbb{Q}}_\ell$-span of all cycle classes over $k$. Because $V$ is finite dimensional, $A$ is a closed subspace. Thus an actual sequence $a_j\in A$ converging $\ell$-adically to $\alpha$ would establish $\alpha\in A$. To see closedness directly, choose a basis of $A$, extend it to one of $V$, and write $A$ as the common kernel of the remaining continuous coordinate functionals. The subtlety is not this limit; it is producing the approximants from actual cycles, with compatible accuracy and the correct base field.

Stress test.

Two elementary models obstruct tempting replacements for that missing construction. On the lattice ${\mathbb{Z}}_\ell$, multiplication by $1+\ell$ fixes every class modulo $\ell$, but its fixed space on ${\mathbb{Q}}_\ell$ is zero. Mod-$\ell$ invariance is therefore not a certificate of a rational Tate class. Independently, take \[ F=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad F^\vee=(F^{-1})^{\mathsf T}. \] The fixed line of $F$ is spanned by $e_1$; the fixed line of $F^\vee$ is spanned by the second coordinate functional. Their pairing is zero. All eigenvalues equal $1$, yet the invariant pairing is degenerate and $15.2$ with $r=2$ is impossible. This refutes the purely linear-algebraic inference from eigenvalues of the correct weight to semisimplicity. It is not asserted to be the cohomology of a counterexample variety.

If cycles are first constructed over an extension, their individual classes need not be fixed by $G_k$; averaging can kill noninvariant components and lower rank. Formula $15.1$ avoids that loss only for the specified invariant class already represented there. Finally, neither point counts determining $r$ nor a positive numerical intersection form produces $r$ independent algebraic classes. For finitely generated fields other than finite fields, the one-operator certificate does not replace the full $G_k$-invariance condition. These are genuine limits of the route.

Outcome.

Outcome: Conditional reduction. A nondegenerate cycle-intersection matrix of the full Frobenius multiplicity would prove the finite-field case under consideration and supply an explicit algebraic idempotent correspondence. An invariant class already algebraic over a finite separable extension descends with rational coefficients. Neither step establishes the required abundance of cycles, and the matrix criterion deliberately includes an additional semisimplicity conclusion. It is not claimed equivalent to, or weaker than, the selected conjecture. Historical novelty of these auxiliary observations is not asserted.

Sources

• S1 Li, What is the Tate conjecture?, map $2$ and Conjecture 1 $2013$. https://www.math.columbia.edu/~chaoli/docs/TateConjecture.html. Catalogue formulation source. Consulted 24 September 2026: Cycle-class map and Conjecture 1.
• S2 An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn $2025$. https://arxiv.org/abs/2505.07337. Catalogue research/status reference; not a proof endorsement.
• S3 On the integral Tate conjecture for abelian varieties — J. S. Milne (September 8, 2025). https://arxiv.org/abs/2509.06707. Catalogue research/status reference; not a proof endorsement.
• S4 An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn, full text. https://arxiv.org/html/2505.07337v1. Catalogue research/status reference; not a proof endorsement.
• E1 James S. Milne, The Tate conjecture over finite fields, version 2.1, 7 November 2007, Section 1 and Theorem 1.4. https://www.jmilne.org/math/articles/2007e.pdf. Added research source. Cycle surjectivity, intersection pairing and the stronger semisimplicity package. Consulted 27 September 2026.