5
$\begingroup$

Let $X$ be a compact, connected, Kähler manifold, of dimension $d$, with Hermitian metric $\omega$; let $E$ be a vector bundle on $X$ of rank $r\geq2$.

By [1] definition 3.1.2:

A line bundle $L$ over $X$ is said numerically effective (nef, for short) if for any $\epsilon>0$ there exists a smooth Hermitian metric $h_{\epsilon}$ on $L$ such that \begin{equation*} \Omega_{h_{\epsilon}}(L)\geq-\epsilon\omega; \end{equation*} that is the curvature form $\Omega_{h_{\epsilon}}(L)$ of the Chern connection on $L$ (with respect to $h_{\epsilon}$) can have an arbitrary negative part. $E$ is nef if the tautological bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$ is nef.

By [1] definition 3.1.3:

$E$ is $1$-nef if for any $\epsilon>0$ there exists a Hermitian metric $h_{\epsilon}$ on $E$ such that $\Omega_{h_{\epsilon}}(E)\geq-\epsilon\omega$.

By [1] proposition 3.2.4, the $1$-nef bundles $E$ over $X$ are nef; but the inverse is unknown in general; excepted for:

  1. $d=1$ (i.e. algebraic curves), see [1] theorem 3.3.1;
  2. on toric and Abelian varieties $E\otimes\det E$ is $1$-nef;
  3. tangent bundle $TX$ of $X$, where it is nef and $d\in\{2,3\}$;

(2) is justified in [1] at page 113, (3) follows by [2] theorems 6.1, 7.1 and [1] proposition 3.2.4.

Question: Are there other examples of manifolds over which the nef bundles are $1$-nef? Or is there an example of nef not $1$-nef bundle over some manifold $X$?


[1] M. A. A. De Cataldo - Singular Hermitian metrics on vector bundles, J. reine ang. Math. 502 (1998) 93-122

[2] J.-P. Demailly, T. Peternell, M. Schneider - Compact complex manifolds with numerically effective tangent bundles, J. Algebraic Geom. 3 (1994) 295-345

$\endgroup$
  • $\begingroup$ We have 2-nef bundle, 3-nef bundle, ... $n$-nef bundle for any positive integer $n$ see Ugo Bruzzo, Beatriz Graña Otero, Metrics on semistable and numerically effective Higgs bundles, J. reine ang. Math. 612 (2007) 59-79 . I think when a VB is stable in the sense of Mumford, 1-nef bundle=nef bundle $\endgroup$ – user21574 Dec 8 '17 at 18:53
  • $\begingroup$ Yes, in general I can consider $t$-nef bundles for $t\in\{1,...,\min\{d,r\}\}$ on $X$; $t$-nef is $s$-nef for $s\in\{1,...,t\}$; nef bundles are semistables in Mumford sense (it's a corollary of [2] theorem 2.18), and it turns out that $1$-nef bundles are semistables. But all this does not answer to my question! $\endgroup$ – Armando j18eos Dec 9 '17 at 8:18
  • $\begingroup$ Interesting comment (with correct citation, I was careless to give the correct citation, sorry), First of all, that singular hermitian metric is not well defined for Vector bundles in general. You may assume that $h$ is negatively curved in the sense of Griffiths then you can define such singular hermitian metric on a VB .but I will think about it more. But I am not an expert, you may as Jean Pierre Demailly directly by email see also mathoverflow.net/questions/238825/… $\endgroup$ – user21574 Dec 9 '17 at 8:23
  • $\begingroup$ Bruzzo with Biswas, Graña Otero, Gurjar, Hernández Ruipérez and others, extended the notions of nefness and $t$-nefness to Higgs bundles, respectively, on smooth projective varieties over algebrically closed fields of characteristic $0$ (H-nef) and connected, compact Kähler manifolds ($t$-H-nef). They proved that ($t$-)nef implies ($t$-)H-nef but the vice versa does not hold. $\endgroup$ – Armando j18eos Dec 9 '17 at 8:27
  • $\begingroup$ I was wrong: the nef bundles with dual nef (numerically flat bundles, for short nflat) are semistables!, of course, it turns out that the $1$-nflat bundles ($1$-nef bundles with dual $1$-nef) are semistables. $\endgroup$ – Armando j18eos Dec 10 '17 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.