0
$\begingroup$

Let $X$ be a compact connected Kähler manifold, of dimension $d\geq3$, with Hermitian metric $\omega$; let $E$ be a vector bundle on $X$ of rank $r\geq2$.

By [1] definition 1.2:

A line bundle $L$ on $X$ 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.

By [1] definition 1.9:

$E$ is nef if the tautological bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$ is nef. $E$ is numerically flat (nflat, for short) if $E$ and $E^{\vee}$ are both nef.

In the proof of theorem 1.18, one assumes $E$ nflat; let $F$ be a reflexive subsheaf of $E$ of minimal rank $p$ of degree $0$, one proves that $\det F$ is a line bundle and by [1] lemma 1.20 $F$ is a subbundle of $E$. By assumption, $F$ is a ($\omega$-)stable subbundle of $E$.

The authors state that $F^{\vee}$ is a locally free quotient sheaf of $E^{\vee}$!

Question: Why does this hold?

I understand that $\left(E_{\displaystyle/F}\right)^{\vee}\equiv\mathcal{Q}^{\vee}$ is a reflexive subsheaf of $E^{\vee}$ (cfr. [2] proposition V.5.18 or this answer), on an open subset $U$ of $X$ it is locally free, so $\mathcal{E}xt^1_{\mathcal{O}_X}\left(\mathcal{Q},\mathcal{O}_X\right)_{|U}=0$ and $F^{\vee}_{|U}$ is a locally free quotient sheaf of $E^{\vee}_{|U}$.

Remark: For the cases of complex curves and Kähler surfaces, the previous statement holds by [2] corollary V.5.20; because, without change the names, $\textrm{codim}(X\setminus U)\geq3$.


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

[2] S. Kobayashi (1987) Differential Geometry of Complex Vector Bundles, Iwanami Shoten Publishers and Princeton University Press

$\endgroup$
1
$\begingroup$

If $F$ is a subbundle of $E$, then one has a natural surjection of vector bundles $E^*\to F^*\to 0$ induced by the restriction map. This is equivalent to saying the $F^*$ is a quotient bundle of $E^*$ (by the kernel of the map above, which is a vector subbundle of $E^*$), or if you prefer, $\mathcal O_X(F^*)$ is a locally free quotient of $\mathcal O_X(E^*)$.

$\endgroup$
  • $\begingroup$ Thank you: your answer suggested me another solution, which I printed as another answer. $\endgroup$ – Armando j18eos Sep 14 '18 at 15:49

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.