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The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and S.T. Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Yau-Donaldson flow

$$\frac{\partial h_t}{\partial t}=-2h_t(\Lambda F_{h_t}-\lambda Id)$$ to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow

The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and Shing Tung Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Yau-Donaldson flow

$$\frac{\partial h_t}{\partial t}=-2h_t(\Lambda F_{h_t}-\lambda Id)$$ to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow

The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and S.T. Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Donaldson flow to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow

The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and S.T. Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Yau-Donaldson flow

$$\frac{\partial h_t}{\partial t}=-2h_t(\Lambda F_{h_t}-\lambda Id)$$ to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow

Some additional comments to @diverietti answer,

The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and S.T. Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Donaldson flow to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow

The positivity of holomorphic bisectional curvature is as same as Griffiths positivity of the holomorphic tangent bundle

Y.T. Siu and S.T. Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture saying that every compact K"ahler manifold with positive holomorphic bisectional curvature is biholomorphic to the projective space

Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see

F. Campana and H. Flenner, A characterization of ample vector bundles on a curve. Math. Ann. 287 (1990), no. 4, 571–575.

Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric

A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that

$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$

in the sense of currents

But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$.

The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known )

Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric

Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 (1970) 125–180

Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths

singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases

A singular Hermitian vector bundle $(E, h)$ which is positively curved in the sense of Griffiths is weakly positive

We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity?

There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive )

See Paun survey paper

The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Donaldson flow to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow