7
$\begingroup$

Let $\mathcal E\to X$ be a stable vector bundle over a polarized projective manifold $(X,\omega)$. It is well-known, that in this case $\mathcal E$ admits Hermitian-Einstein metric, i.e., a metric $h$, such that (may be up to some constants)

$$ {\rm tr}_\omega F(\mathcal E, h)=\lambda \rm Id, $$ where

  • $F(\mathcal E, h)\in \Lambda^{1,1}(T^*M)\otimes \rm End(\mathcal E)$ is the curvature of Chern connection on $(\mathcal E, h)$
  • ${\rm tr}_\omega\colon \Lambda^{1,1}(T^*M)\otimes \rm End(\mathcal E)\to \rm End(\mathcal E)$ is a contraction with $\omega$
  • $\lambda=\int_X c_1(\mathcal E)\wedge\omega^{n-1}/\int_X \omega^n$.

Assume additionally that $\mathcal E$ admits a metric $h_0$ such that the corresponding curvature $F(\mathcal E, h_0)$ is Griffiths non-negative, i.e., for any $v\in T_x^{1,0}X$, $e\in \mathcal E_x$, $h_0\bigl(F(\mathcal E, h_0)(v,\bar v)e,\bar e\bigr)\ge 0$.

Question: Is it true that under this additional non-negativity assumption the Hermitian-Einstein metric $h$ is also Griffiths non-negative?

Upd Oct 5. In the present form the question might seem not well-motivated, so I would like to make two points.

1) Why one should expect this to be true?
Hermitian-Einstein metric can be constructed as limits of a heat-type flow (see, e.g., Yum-Tong Siu, Lectures on Hermitian-Einstein metrics for stable bundles and Kahler-Einstein metrics, 1986). On the other hand, various geometric flows are known to preserve certain positivity conditions. For example, Ricci flow preserves positivity of curvature operator. So the question can be resolved by studying which positivity conditions are preserved by the relevant heat flow.

2) Why one might care?
Existence of Hermitian-Einstein metric implies Kobayashi-Lubke inequality relating $c_2(\mathcal E)$, $c_1(\mathcal E)$ and $\omega$. At the same time, for a Griffiths non-negative vector bundle there is a whole family of Fulton-Lazarsfeld inequalities, involving higher Chern classes. It is reasonable to expect, that once $\mathcal E$ admits a metric that is both Hermitian-Einstein and positive, then stronger inequalities hold. This in turn can help in various computations involving Hirzebruch-Riemann-Roch formula.

$\endgroup$
  • 1
    $\begingroup$ For the first part of your question:We want to run a heat flow to get Hermitian Einstein. Finding initial metric is important to make reference metric to run the flow, so for instance for twisted Kahler Einstein along some fibration like Iitaka, existence of such initial metric correspond to some positivity of metric on moduli part of Fujino-Mori canonial bunle formula and also existence of Zariski decomposition . read Hajime Tsuji's paper..... $\endgroup$ – user21574 Jan 24 '17 at 22:29
  • 1
    $\begingroup$ ...... In this case we can take such initial metric to be a canonical metric on moduli space of Hermitian Einstein metric, like Weil-Petersson metric which is semi-positive. But we can not take any inital metric , we must choose those metrics with vanishing Lelong number, due to Demailly. Read some papers related to Kahler Ricci flow with initial metric start with a metric with vanishing Lelong number. $\endgroup$ – user21574 Jan 24 '17 at 22:30
  • 1
    $\begingroup$ As an additional related comment if the metric $h$ admit hermitian-Einstein metric $\Lambda_\omega F_h=\lambda I$ , then constant $\lambda$ can not be negative or zero(life here is different with Kahler-Einsein metric ). The reason is due to following equality $\Delta |s|^2=|\nabla s|^2-<(\Lambda_\omega F_h)s,s>\geq 0$ where $s$ is a section and hence $|s|$ is bounded. But maximum principle for subharmonic functions, so $|\nabla s|^2=\lambda |s|^2$, so if the coefficient $\lambda$ in $\Lambda_\omega F_h=\lambda I$ be negative or zero, all sections are parallel or zero $\endgroup$ – user21574 May 16 '17 at 0:37
  • 1
    $\begingroup$ Of course Hermitian-Einstein metrics could have negative constant $\lambda$ (take line bundle $\mathcal O(-1)$ over the projective space). You have just proved that in the case it is negative, there are no holomorphic sections. $\endgroup$ – Yury Ustinovskiy May 16 '17 at 22:44
  • 1
    $\begingroup$ In fact the corollary of previous comment is that in the case of singular hermitian-Einstein metric (well defined!) then we can get uniqueness of solution when $\lambda>0$ . But this is very special case for uniqueness $\endgroup$ – user21574 May 17 '17 at 0:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.