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Let $X$ be a smooth projective variety of dimension $n$. Let $D$ be a smooth divisor of $X$. Let $i:D\hookrightarrow X$ be the inclusion. Let $H$ be an ample line bundle on $X$.

Let $E$ be a vector bundle of rank $r$ on $X$. Consider $E|_D$. Suppose I know that $E|_D$ is $\mu_{i^*H}$ semistable, does it imply that $E$ is $\mu_H$ semistable?

Thanks in advance!

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  • $\begingroup$ @gsvr, can you give me a reference? $\endgroup$
    – gradstudent
    Dec 30, 2015 at 13:46
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    $\begingroup$ What's happening here if $X$ is a high genus curve? $\endgroup$
    – Allen Knutson
    Dec 30, 2015 at 14:21

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First of all, as Allen Knutson remarks, $X$ should have dimension at least two.

Now the Mehta-Ramanathan theorem tells you that if $\mathcal E$ is $H$-semistable and $D \in |mH|$ is general, with $m$ sufficiently large, then $\mathcal E \big|_D$ will likewise be $H\big|_D$-semistable. But you are asking for the reverse direction. This is actually much easier. Namely, assume that $\mathcal E$ is not $H$-semistable, and let $\mathcal F \subset \mathcal E$ be a (saturated) destabilizing subsheaf. Then $\mathcal F\big|_D \subset \mathcal E\big|_D$ will still be destabilizing. Here you do not even need $m$ to be large (but $|mH|$ should be basepoint-free and $D$ should be general, to avoid problems when $\mathcal F$ is not locally free). On the other hand, if $D$ is not numerically equivalent to a multiple of $H$, the statement you are asking for definitely fails.

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  • $\begingroup$ I don't understand the statement that $|mH|$ should be basepoint free and $D$ should be general. For any $D\in|mH|$, if $E$ is locally free or even torsion free on $X$, slope of $E|_D$ is $m$ times the slope of $E$ right. How is basepoint free necessary? $\endgroup$
    – gradstudent
    Feb 17, 2016 at 17:42
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    $\begingroup$ The restriction of a torsion-free sheaf to a smooth Cartier divisor need not be torsion-free. Then you need to divide out the torsion and this might mess up your slope. If $D$ is in general position, this does not happen. In any case, I was not making a precise statement, I was just pointing out the fact that here one has to be careful. $\endgroup$
    – pgraf
    Feb 19, 2016 at 17:43

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