# open embedding of moduli of stable pair

Let $S$ smooth projective surfac and $X=K_S$ the total space of canonical bundle and $\bar{X}=Proj(Sym(K_S^{\vee}\oplus \mathcal{O}_S))$ be the projective completion of $X$. I have seen in the literature that there is an open immersion of the corresponding moduli of stable pairs: $$P_n(X,i_*\beta)\subseteq P_n(\bar{X},i_*\beta)$$ where $i$ is the inclusion of $S$ as the zero section and $\beta$ the homology class of a curve in $S$. I want to see the prove of this. Can any body point out how to prove this?

thank yoy very much

Let $f:X\to \overline X$ be the open immersion, $i:S\to X$ the zero section, $\beta\in H_2(S,\mathbb Z)$ a curve class. Then to a stable pair $$[s:\mathcal O_X\to F]\in P_n(X,i_\ast \beta)$$ you can attach $$\overline s:\,\mathcal O_{\overline X} \longrightarrow f_\ast \mathcal O_X\overset{f_\ast s}{\longrightarrow}f_\ast F,$$ which is a stable pair in $P_n(\overline X,f_\ast i_\ast\beta)$ since $f_\ast F$ is pure (it has the same support as $F$, which is pure) and the cokernel of $\overline s$ has finite support (the same as the support of the cokernel of $s$). I leave it to you to write such association in families. This is an open immersion, since for a point $$[\mathcal O_{\overline X}\to E]\in P_n(\overline X,f_\ast i_\ast\beta),$$ the condition of belonging to $P_n(X, i_\ast\beta)$ simply means Supp $E\subset X$, which is an open condition.
Added. Let $s:\mathcal O_{X\times P}\to \mathscr F$ be the universal stable pair over $X\times P$, with $P=P_n(X, i_\ast\beta)$. Then the closed subscheme $V:=\textrm{Supp }\mathscr F\hookrightarrow X\times P$ is proper over $P$. On the other hand, the projection $\overline X\times P\to P$ is separated. This implies the map $$u:V\hookrightarrow X\times P\xrightarrow{f\,\times\, \textrm{id}_P}\overline X\times P$$ is proper. On the other hand, since $\mathscr F$ is coherent, it is a pushforward from a coherent sheaf on $V$ [see Lemma $29.9.7$]. Then $(f\times\textrm{id}_P)_\ast\mathscr F$ is also a pushforward from $V$, and since $u$ is proper it is coherent.
• Thank you very much @Andrea Ricolfi. I try to work it for families. The problem is in family $f_*F$ does not necessarily coherent because the support does not need to be proper relative to base scheme parametrizing the family. – Rizal Afgani Sep 9 '17 at 14:29
• In fact $f_\ast F$ is coherent even though $f$ is not proper, see my expanded answer. – Andrea Ricolfi Sep 9 '17 at 16:45
• How can you conclude that the support $V$ of $\mathcal{F}$ is proper? – Rizal Afgani Sep 9 '17 at 19:34