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Let X be a symplectic complex manifold of dimension $2n $, i.e. there exists a non degenerate holomorphic 2-form $\sigma $ such that $ H^0 (X,\Omega^2)=\mathbb {C}\cdot\sigma $. Suppose that there exists a submanifold $ P\subset X $ such that $P\cong\mathbb {P}^n $. Let $\tilde {X} $ be the blow up of X along P. Using the Euler sequence and the symplectic structure $\sigma $, we can show that the projection of the exceptional divisor $ D=\mathbb {P}(\mathcal {N}_{P/X}) $ on $\mathbb {P}^n $ is isomorphic to the projective bundle $\mathbb {P}(\Omega_P) $. Hence we can identify it with the incidence variety $$ \{(x, H)\,|\, x\in H\}\subset\mathbb {P}^n\times(\mathbb {P}^n)^*.$$ So we can project on the dual projective space $(\mathbb {P}^n)^*$ and define a blow down $\tilde {X}\to X'$. The Mukai flop is then the birational map $ X---> X'$ obtained by composing the blow up and the blow down.

My question is about the existence of the blow down: why does it exist? Which are the hypothesis we must check to say that a contraction is a blow down? Reference are very welcome!

Thank you very much!

User3773
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    The used contraction criterion here is usually referred to as (Moishezon-)Nakano-Fujiki-criterion. The only reference I know by heart is Fujiki, On the blowing down of analytic spaces. If no one is quicker than me, I will provide more details, but I can't promise that this will happen very soon. – Ben Dec 25 '14 at 19:07
  • Dear Ben thank you very much! I'll try to get and use that paper..but anyway a solution will be very welcome also everytime (at least it will be a test for my argument). – User3773 Dec 25 '14 at 19:32

1 Answers1

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Thanks to Ben's comment, I have found the following solution, please to leave a comment if there are some imprecisions. The Corollary of the Theorem 2' (at the bottom of page 502) in the paper suggested by Ben can be rephrased as follows:

let $X$ be a complex manifold, $A$ a complex submanifold of codimension $1$ in $X$ and $f:A\to A'$ a fiber bundle over a complex manifold $A'$ with every fiber $F$ connected. If both $\;\mathcal{N}^*_{A/X}|_F\;$ and $\;\left(\mathcal{N}_{A/X}|_F\otimes K_F\right)^*\;$ are ample (here $K_F$ if the canonical of $F$), then there exists a unique (up to isomorphism) complex manifold $X'$ and a proper surjective map $f':X\to X'$ such that the pair $(X',f')$ is the blow down of $X$ along $f$. In particular, if $F\cong\mathbb{P}^n$, then only the ampleness of $\;\mathcal{N}^*_{A/X}|_F\;$ can be assumed.

Using the notation of the question, we must apply this result with $X=\tilde{X}$, $A=D$ and $A'=(\mathbb{P}^n)^*$. Since $D$ is a divisor, it is of codimension $1$ and we can use the adjunction formula: $$ \mathcal{N}^*_{A/X}=\mathcal{O}_A(-A). $$ When we restrict to $F=\mathbb{P}^n$, we get $\:\mathcal{N}^*_{A/X}|_F=\mathcal{O}(-1)$ and so we have done.

User3773
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    My apologies. I provided the 'wrong' reference. What you're doing is fine, except that from the cited corollary you don't get smoothness of $X'$. Fortunately, I just found the appropriate article(s). In On the inverse of monodical transformations by Nakano, there is a theorem proven which I will state below. However, there is a gap an an unnecessary condition. This was recognised and fixed by Nakano and Fujiki in Supplement to "On the inverse of monodical transformations". From both articles we get...(continued below) – Ben Dec 30 '14 at 15:58
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    For $X$ a complex manifold of dimension $\geq 3$ and a codimension one submanifold $A$, which admits a locally trivial fibration $f\colon A\to A'$ onto a smooth manifold $A'$, the fibres being isomorphic to $\mathbb{P}^{r}$ with $r\geq 1$, for a contraction $X\to X'$ of $A$ to $A'$ along $f$ onto a smooth(!) manifold $X'$ to exist it is necessary and sufficient that the conormal bundle of $A$ in $X$ is, restricted to a fibre, isomorphic to $\mathcal{O}{\mathbb{P}^r}(1)$. In particular, your argumentation remains valid since you noticed $\mathcal{N}{A|X}^* = \mathcal{O}_{A/A'}(-1)$ globally. – Ben Dec 30 '14 at 16:13
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    Edit to the above: there is a dual too much... It's $\mathcal{N}^*{A|X} = \mathcal{O}{A/A'}(1)$ of course. This means that there has to be something wrong. I'll have a look at it. – Ben Dec 30 '14 at 17:32
  • I've read the right criterion. I still don't know why the quantity he calls $[S]$ is our conormal bundle, anyway he states that this bundle must be equal to $\mathcal{O}(1)^{-1}$ when restricted to the fibre. Since the fibre is a projective space the last line bundle is just the tautological one, i.e. $\mathcal{O}(-1)$, and so there is no problem. What do you think about? – User3773 Dec 31 '14 at 12:46
  • Well, the ideal sheaf of the Cartier divisor $A$ is $\mathcal{J} = \mathcal{O}X(-A)$, hence the conormal bundle of $A$ in $X$ is $\mathcal{J}/\mathcal{J}^2 = \mathcal{O}_X(-A)|_A$ and this is the inverse of $\mathcal{O}_X(A)|_A =: [A]$. Put together, the assumption of the theorem is $\mathcal{N}{A|X}^* (=-[A]) = \mathcal{O}(1)$ after restriction to fibres. This I don't see right now. – Ben Dec 31 '14 at 15:54
  • I was recalling my argument. When I calculated the restriction to the fibre, I didn't consider the structure of $D$. In the same way as I claimed that $D={(x,H),|,x\in H}\subset\mathbb{P}^n\times(\mathbb{P}^n)^*$, it is natural to see $D$ like the fibre bundle over $\mathbb{P}^n$ with typical fibre equal to the dual projective space. So the restriction $\mathcal{O}_D(-D)|_F$ is like the tautological bundle over the dual projective space, i.e. the hyperplane bundle over the projective space:this is what we want. It is not rigorous, but intuitively it seems to work. What do you think? – User3773 Jan 02 '15 at 16:14