Griffiths-positive metric

Question

How to find a Griffiths positive metric on an ample vector bundle?

Answer 1

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

Answer 2

As soon as the base manifold has dimension greater than one, the existence of Griffiths positive metrics on an ample vector bundle is not known (and even in the one dimensional case, it is not obvious: it's a theorem!). Anyway this existence is conjectural (Griffiths' conjecture).

On the other hand, if a hermitian vector bundle $E\to X$ has positive Griffiths curvature, then it is ample. This is quite straightforward, since any hermitian metric on $E$ induces in a natural way a (quotient) hermitian metric on the tautological line bundle $\mathcal O_{E}(1)\to\mathbb P(E)$ over the projectivized manifold of hyperplanes of $E$. With such a metric, the curvature of $\mathcal O_{E}(1)$ can be computed in terms of the curvature of $E$ and one can see that if $E$ is Griffiths positive then $\mathcal O_{E}(1)$ has strictly positive curvature. Thus, by the Kodaira projectivity criterion, $\mathcal O_{E}(1)$ is ample and therefore, by definition, $E$ is ample.

The point is that not every metric on $\mathcal O_{E}(1)$ comes from a metric from $E$. So, if $E$ is ample, that is if $\mathcal O_{E}(1)$ is ample, then $\mathcal O_{E}(1)$ admits a positively curved metric but then one doesn't know how to produce a metric on $E$ with the desired positivity properties.

If you want to see it from another point of view, the difficulty is that if $E$ is ample one is able to construct Griffiths positive metrics on some high symmetric power $S^mE$ of $E$, but, except when the rank of $E$ is one, we don't know how to extract $m$-th roots of such metrics in order to get one on $E$.