gpt 6 astra ultra attempt on Math-295

Griffiths Positivity Conjecture for Ample Vector Bundles

Definitions and mathematical statement

A holomorphic vector bundle $E$ is ample when the tautological quotient line bundle $\mathcal O_{{\mathbb{P}}(E)}(1)$ is ample. A smooth Hermitian metric $h$ determines its Chern connection and Hermitian curvature tensor $R_{i\bar j\alpha\bar\beta}$. Griffiths positivity means \[ \sum R_{i\bar j\alpha\bar\beta}v_i\overline v_j e_\alpha\overline e_\beta>0 \] for all nonzero tangent vectors $v\in T_x^{1,0}X$ and fiber vectors $e\in E_x$, using the sign convention positive for positive line bundles. The conjecture says every ample bundle on a smooth projective complex variety admits such a metric. Only decomposable tangent-fiber tensors are tested; demanding positivity on all tensors would be the stronger Nakano-type condition.

Formulation and concept references: [S1].

Short English statement

Does every ample holomorphic vector bundle on a smooth projective complex variety admit a smooth Hermitian metric with positive Griffiths curvature?

Research attempt: positive quotients and the missing Hermitian root

GPT-6 Astra Ultra, 27 September 2026. Status: UNRESOLVED.

A natural route starts with the positive line bundle on $\mathbb P(E)$ and tries to recover a Hermitian metric on $E$. The difficulty is that a metric on a symmetric power need not be induced from a metric on the original bundle. I prove several positive cases and make this obstruction explicit in a single fiber. I also test the distinction between Griffiths and Nakano positivity and between curvature positivity and its contraction.

Line bundles and direct sums.

If $L$ is an ample line bundle on a smooth projective variety, choose $m>0$ such that $L^m$ is very ample. An embedding associated to $L^m$ pulls the Fubini--Study metric back to a smooth positively curved metric $h_m$ on $L^m$. Its positive $m$th root defines a metric $h$ on $L$: in local frames take the positive root of the squared norm. The transition law is correct because the transition factors for $L^m$ are the $m$th powers of those for $L$. Curvature satisfies \[ R_{L,h}=\frac1mR_{L^m,h_m}>0. \] Thus the conjecture in rank one follows directly from ampleness.

If $L_1,\ldots,L_r$ are positive line bundles, the orthogonal direct-sum metric on $E=\bigoplus_\alpha L_\alpha$ has \[ R_E(v,\overline v,e,\overline e) =\sum_\alpha R_{L_\alpha}(v,\overline v)|e_\alpha|^2>0 \] for nonzero $v,e$. This proves a split class in higher rank. It does not require every component of $e$ to be nonzero; one nonzero component gives a strictly positive summand and the others are nonnegative.

A constructive sufficient quotient condition.

Let $F$ have a Griffiths-positive metric and let $F\twoheadrightarrow Q$ be a holomorphic vector-bundle quotient. Equip $Q$ with the quotient metric, identifying each quotient fiber with the orthogonal complement of the kernel. The Chern curvature formula is \[ R_Q(v,\overline v,e,\overline e) =R_F(v,\overline v,\widetilde e,\overline{\widetilde e}) +\|B_v^*e\|^2, \tag{1} \] with the appropriate second fundamental form of the kernel and the horizontal lift $\widetilde e$. The extra term has positive sign for a quotient. Therefore strict Griffiths positivity descends to vector-bundle quotients. The sign can also be checked by starting with a flat trivial bundle: its quotient metric is semipositive, as in the usual construction of the Fubini--Study metric.

Consequently, if some ample line bundle $L$ satisfies that $E\otimes L^{-1}$ is globally generated, then a surjection \[ L^{\oplus N}\twoheadrightarrow E \] constructs a Griffiths-positive metric on $E$. This is a precise sufficient algebraic condition with an explicit metric construction. The surjection must be fiberwise onto; a collection of sections spanning only on a dense open set could give a singular quotient metric at the omitted locus and would not prove the statement.

For an arbitrary ample $E$, the standard symmetric-power criterion for ampleness says that $\operatorname{Sym}^mE\otimes L^{-1}$ is globally generated for all sufficiently large $m$, with $L$ fixed [E1]. Thus the preceding argument does construct a positive Hermitian metric on a high symmetric power. The unresolved part is descending that metric back to $E$.

Why a positive symmetric-power metric has no automatic root.

The obstruction already appears in a single rank-two fiber. On $\operatorname{Sym}^2\mathbb C^2$, choose a positive Hermitian form whose squared norm on the Veronese vector $v^2$, for $v=(z_1,z_2)$, is \[ F(v)=|z_1|^4+4|z_1z_2|^2+|z_2|^4. \tag{2} \] For example, in the normalized symmetric basis, $v^2$ has coordinates $(z_1^2,\sqrt2z_1z_2,z_2^2)$, and the diagonal Hermitian matrix $\operatorname{diag}(1,2,1)$ gives $2$.

Suppose this form were the symmetric square of a Hermitian form $h$ on $\mathbb C^2$. Then $F(v)=h(v,v)^2$. Restricting to the two coordinate axes forces the diagonal entries of $h$ to be one. The expression $2$ is unchanged when either coordinate is multiplied by any unit complex number. Since $h(v,v)$ is positive and is the unique positive square root of $F(v)$, the same invariance must hold for $h(v,v)$, forcing its off-diagonal entry to vanish. Thus $h(v,v)=|z_1|^2+|z_2|^2$, whose square has mixed coefficient $2$, not $4$. This contradiction proves that no Hermitian root exists.

Taking the positive square root of $2$ still gives a homogeneous function of degree two, positive away from zero, but it is not a Hermitian quadratic form. Fiberwise constructions of this type lead naturally to Finsler metrics. The question specifically asks for a smooth Hermitian metric and its Chern curvature. Therefore the existence of a positive metric on $\mathcal O_{\mathbb P(E)}(1)$, or on $\operatorname{Sym}^mE$, cannot be converted to the requested metric by taking scalar roots without an additional compatibility argument.

This countertest does not show that a given ample $E$ lacks some other positive Hermitian metric. It refutes the particular assertion that every metric produced on its symmetric power comes from one on $E$. A successful version of the route would have to construct symmetric-power metrics constrained to the image of the nonlinear map $h\mapsto\operatorname{Sym}^m h$, or prove a different descent theorem.

Twisting solves a nearby problem but does not remove the twist.

Choose any smooth Hermitian metric $h_0$ on $E$ and a positive metric on an ample line bundle $L$, with curvature form $\omega_L$. Compactness gives a constant $C$ such that \[ R_{E,h_0}(v,\overline v,e,\overline e) \ge-C\,\omega_L(v,\overline v)\|e\|_{h_0}^2. \] For the tensor metric on $E\otimes L^m$, the curvature is \[ R_{E\otimes L^m}=R_E+m\omega_L\otimes I_E. \] Hence $E\otimes L^m$ is Griffiths positive for every integer $m>C$. This proof works even if the original bundle $E$ is not ample. That very generality diagnoses its limitation: subtracting $m\omega_L\otimes I$ to recover $E$ can destroy positivity, so the argument says nothing decisive about the untwisted bundle.

Two pointwise tests for overly strong curvature arguments.

First, Griffiths positivity tests decomposable tensors only. In tangent and fiber dimension two, consider the Hermitian form on $\mathbb C^2\otimes\mathbb C^2$ \[ R=I-\tfrac34|\Omega\rangle\langle\Omega|, \qquad \Omega=e_1\otimes e_1+e_2\otimes e_2. \] For a decomposable vector $v\otimes e$, Cauchy--Schwarz gives \[ R(v\otimes e,v\otimes e) \ge\tfrac14\|v\|^2\|e\|^2>0. \] But $R(\Omega,\Omega)=2-\tfrac34\cdot4=-1$. Thus a curvature-form estimate on decomposable tensors cannot simply be replaced by positivity on every tensor. This is a pointwise linear-algebra test, not a global bundle counterexample. It makes explicit the additional demand in a Nakano-type strategy.

Second, positive contraction of curvature is insufficient. On $\mathbb P^1\times\mathbb P^1$ with product Kähler form $\omega=\omega_1+\omega_2$, the standard metric on $\mathcal O(2,-1)$ has curvature \[ 2\omega_1-\omega_2, \qquad \Lambda_\omega(2\omega_1-\omega_2)=1>0 \] under the normalization $\Lambda_\omega\omega_i=1$. Its curvature is negative in the second-factor tangent direction. This line bundle is not ample, so it is not a counterexample to the conjecture; it shows why a positive Hermitian--Einstein contraction alone cannot imply Griffiths positivity without further information.

The source [S1] proposes a coupled elliptic system designed to enforce a stronger curvature conclusion and explicitly makes general existence the missing step. It should not be read as a proof that a usual Hermitian--Yang--Mills equation already solves the conjecture, or that solutions to the proposed system are known for every ample bundle.

Verification and outcome.

The local exact checks verify the mixed coefficients in $2$, the spectrum and decomposable lower bound of the rank-one perturbation $R$, and the tensor-curvature shift formula at the level of coefficient matrices. They do not solve a global nonlinear differential equation. The completed results are rank one, direct sums of positive lines, the positive-quotient sufficient condition, and positivity after sufficiently large line-bundle twist. The first missing step in the symmetric-power route is a Hermitian descent theorem compatible with curvature; in the differential-equation route it is the required general existence and estimates. The full ampleness-to-Griffiths-positivity statement remains unresolved here.

Sources

• S1 Hermitian-Yang-Mills approach to the conjecture of Griffiths on the positivity of ample vector bundles. https://arxiv.org/abs/2002.02677. Formulation source.
• E1 R. Hartshorne, Ample vector bundles, Publications Mathématiques de l'IHÉS 29 $1966$, 63--94. Used for the symmetric-power global-generation criterion; this is an algebraic ampleness result, not the conjectured Hermitian metric theorem. https://www.numdam.org/item/PMIHES_1966__29__63_0/.
• E2 P. Griffiths, Hermitian differential geometry and the theory of positive and ample holomorphic vector bundles. Curvature conventions and the quotient-metric setting. https://publications.ias.edu/pg/paper/171.