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Let $X$ be a projective variety over the filed of complex number and $X^s$ be its smooth locus with $X \setminus X^s$ has codimension at least $2$. Let $E$ be a reflexive sheaf on $X^s$. Assume that $i_*E$ is semistable with respect to an ample divisor $A$ on $X$, where $i: X^s \to X$ be the inclusion. Let $\pi: Y \to X$ be a resolution of singularities of $X$.

Question: Is $\pi^*(i_*E)$ semistable with respect to some ample divisor in $Y$ ?

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  • $\begingroup$ In many cases the sheaf $\pi^*(i_*E)$ is even not torsion free. $\endgroup$
    – Sasha
    Commented Dec 5, 2022 at 10:03
  • $\begingroup$ What about if $E$ is itself locally free on $X^s$ and $X$ is normal ? $\endgroup$
    – LAPRAS
    Commented Dec 5, 2022 at 10:22
  • $\begingroup$ Even in this case. $\endgroup$
    – Sasha
    Commented Dec 5, 2022 at 10:30
  • $\begingroup$ If we consider the sheaf $(\pi^*(i_*E))/Tor$, then can we say about its semistability ? $\endgroup$
    – LAPRAS
    Commented Dec 5, 2022 at 13:15
  • $\begingroup$ Which semistability you are interested in? $\endgroup$
    – Sasha
    Commented Dec 5, 2022 at 13:26

1 Answer 1

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I assume that the resolution $\pi \colon Y \to X$ is the identity over $X^s$, so that $X^s \subset Y$. Then the sheaf $$ E_Y := \pi^*(i_\ast E) / \mathrm{torsion} $$ is always slope semistable with respect to $\pi^*A$.

Indeed, let $F \subset E_Y$ be a reflexive subsheaf. Then $F\vert_{X^s}$ is a reflexive subsheaf in $E$ and $$ F_X := i_\ast(F\vert_{X_s}) \subset i_\ast E $$ is a reflexive subsheaf in $i_\ast E$. Then we have $$ \mu(F) = \frac{1}{r(F)} c_1(F) \cdot (\pi^*A)^{n-1} = \frac{1}{r(F_X)} c_1(F_X) \cdot A^{n-1} = \mu(F_X) $$ and similarly $$ \mu(E_Y) = \frac{1}{r(E_Y)} c_1(E_Y) \cdot (\pi^*A)^{n-1} = \frac{1}{r(i_\ast E)} c_1(i_\ast E) \cdot A^{n-1} = \mu(i_\ast E). $$ Since $i_\ast E$ is slope semistable, we have $\mu(F_X) \le \mu(i_\ast E)$, hence $\mu(F) \le \mu(E_Y)$, and hence $E_Y$ is semistable.

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$
    – LAPRAS
    Commented Dec 5, 2022 at 15:57
  • $\begingroup$ Is $\pi^*A$ ample ? $\endgroup$
    – LAPRAS
    Commented Dec 5, 2022 at 17:11
  • $\begingroup$ Of course it is not. $\endgroup$
    – Sasha
    Commented Dec 5, 2022 at 19:19
  • $\begingroup$ But I want semistability with respect to an ample divisor. I think $\pi^*A - \sum a_iE_i$ is ample for some $a_i$. Here $E_i's$ are components of the exceptional divisor. Can we say something with respect to this ample divisor ? $\endgroup$
    – LAPRAS
    Commented Dec 5, 2022 at 19:53
  • $\begingroup$ You can try to look at all semisimple subscheaves $F \subset E_Y$ of the same slope $\pi^*A$ as $E_Y$ and try to find $a_i$ such that their $(\pi^*A - \sum a_iE_i)$-slope is less than that. $\endgroup$
    – Sasha
    Commented Dec 6, 2022 at 5:11

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