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!



If the restriction of a vector bundle to a divisor is semi stable, then is the vector bundle itself semistable?
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.


