Question about the construction of Mukai flop 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!
 A: 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.
