2
$\begingroup$

Let $X$ be a projective variety over an algebraically closed field $k$ and $D^b(X)$ be the derived category of bounded complexes of coherent sheaves on $X$. Let $S$ be the Serre functor on $D^b(X)$. An object $P$ of $D^b(X)$ is called a point-like object if it satisfies the following three conditions.

  1. $S(P)\cong P[d]$ for some integer $d$,
  2. Hom$(P,P[i])=0$ for $i<0$,
  3. Hom$(P,P)\cong k$.

One of the results of Bondal and Orlov claims that if $X$ is a smooth projective variety over an algebraically closed field $k$ and suppose that $\omega_X$ or $\omega_X^*$ is ample where $\omega_X$ is the canonical sheaf. Then the point-like objects in $D^b(X)$ are the objects which are isomorphic to $\mathcal{O}_x[m]$, where $x$ is a closed point of $X$ and $m$ is an integer.

Of course the above result no longer holds if neither $\omega_X$ nor $\omega_X^*$ is ample. For example if $X$ is an abelian variety, then $L[m]$ is also a point-like object where $L$ is an invertible sheaf (or any simple sheaf) on $X$ and $m$ is an integer.

My question is: for an abelian variety $X$ over an algebraically closed field $k$, does it exist point-like objects other than $L[m]$ for a simple sheaf $L$?

Improve this question
$\endgroup$
5
  • $\begingroup$ do you know what happens for X an elliptic curve? I guess P has to be a sheaf, by homological dimension of D(X), but it's not clear to me why it must be a line bundle. $\endgroup$
    – Yosemite Sam
    Commented Nov 2, 2016 at 2:34
  • $\begingroup$ @YosemiteSam I'm still not sure why $P$ must be a sheaf up to shift also it seems so. Could you give more reason? $\endgroup$
    – Zhaoting Wei
    Commented Nov 2, 2016 at 3:34
  • 1
    $\begingroup$ Any autoequivalence of $D(X)$ takes a point-like object to a point-like object. And in case of abelian variety there are many autoequivalences, some of them take line bundles to complexes with several cohomology sheaves. $\endgroup$
    – Sasha
    Commented Nov 2, 2016 at 6:57
  • $\begingroup$ @YosemiteSam By the way, even when X is an elliptic curve, P does not need to be a line bundle --- any indecomposable vector bundle would do. $\endgroup$
    – t3suji
    Commented Nov 2, 2016 at 14:16
  • 1
    $\begingroup$ @ZhaotingWei Cor 3.15 of Huybrechts tells you that any E in D(C) splits as a direct sum of Ei[i] with Ei a sheaf. If E is pointlike and has two summands Ei[i], Ej[j] then the identity of both would contribute to Hom(P,P) which would have to have dimension at least two. i.e. P must be a shifted sheaf. And now t3suji's and Sasha's comments make it clear it's a nightmare in general. Maybe an alternative question is what are the pointlike objects when the (anti)canonical is almost ample, like big&nef. $\endgroup$
    – Yosemite Sam
    Commented Nov 2, 2016 at 14:55

0

Reset to default

You must log in to answer this question.