I'm trying to understand some aspects of Oguiso's example of a Calabi-Yau threefold of Picard number 2, described in the paper ["Automorphism groups of Calabi-Yau manifolds of Picard number two"](http://arxiv.org/abs/1206.1649).


Take $X \subset \mathbb P^3 \times \mathbb P^3$ to be the intersection of general hypersurfaces of bidegree $(1,1)$, $(1,1)$, and $(2,2)$. The claim is that the projections $p\_i : X \to \mathbb P^3$ have degree $2$ (clear enough), and contract "$(2(2+1)+2)^3 = 8^3$" rational curves. I'm not sure how to arrive at this count. Where does this number come from, and what can be said about the points of $\mathbb P^3$ over which there are positive-dimensional fibers?



Counting contracted curves
I set up the enumerative computation in a manner similar to Serge Lvovski, but I also get a different answer than the one in Oguiso's paper. Serge Lvovski's answer gives $4\times 4\times 6=96$ rather than $8^3$. My answer gives $120$, rather than $8^3$.
The discrepancy may be due to the same issue as in my previous comment. 


Let $A$ and $B$ be the two global sections of $\mathcal{O}(1,1)$ and let $C$ be the global section of $\mathcal{O}(2,2)$. Push forward by projection to the second $\mathbb{P}^3$ factor, i.e., consider $A$ and $B$ as global sections of the rank $4$ vector bundle $V:=H^0(\mathbb{P}^3,\mathcal{O}(1))\otimes\_{\mathbb{C}} \mathcal{O}(1)$. 
 These two sections give a coherent subsheaf $S \cong \mathcal{O}\oplus \mathcal{O}$ of $V$. The issue in my comment above is that the cokernel $V/S$ fails to be locally free at $4$ points of $\mathbb{P}^3$.


What is the image $I$ of the map $S\otimes\_{\mathcal{O}} V \to \text{Sym}^2 V$, or equivalently, what is the kernel $K$? Of course the image of $\bigwedge^2 S$ inside $S\otimes\_{\mathcal{O}} S \subset S\otimes\_{\mathcal{O}}V$ is in the kernel $K$. I claim this is the entire kernel. Since the kernel is a torsion-free subsheaf of a locally free sheaf, it can be computed at the generic point of $\mathbb{P}^3$, where the claim is basic linear algebra (in fact the same linear algebra works on the open subset that is the complement of the $4$ points where $V/S$ is not locally free). So the image $I$ is a locally free sheaf of rank $7$, and it is a coherent subsheaf of $\text{Sym}^2V$. As above, the cokernel $\text{Sym}^2 V/I$ fails to be locally free at the $4$ bad points.


Now form the direct sum of $I$ with the rank $1$ subbundle $T\cong \mathcal{O}$ of $\text{Sym}^2V$ 
generated by the global section $C$. 
Inclusion of each summand into $\text{Sym}^2 V$ gives a map of locally free sheaves, $$I\oplus T \to \text{Sym}^2 V$$ The rank of the domain is $7+1=8$ and the rank of the target is $10$. By Thom-Porteous, the expected codimension of the degeneracy locus is $(10-7)(8-7) = 3$, i.e., we expect finitely many points of $\mathbb{P}^3$ where the linear subsystem of $H^0(\mathbb{P}^3,O(2))$ generated by $A$, $B$ and $C$ is not codimension $2$. Moreover, Thom-Porteous predicts the total length of this degeneracy locus is the degree three coefficient of the power series $c\_t(\text{Sym}^2 S)/c\_t(I\oplus T)$. 


Of course $\text{Sym}^2 V$ is just $\mathcal{O}(2)^{\oplus 10}$, which has total Chern class $$ c\_t(\mathcal{O}(2)^{\oplus 10}) = c\_t(\mathcal{O}(2))^{10} = (1+2H)^{10} = 1 + 20H + 180H^2 + 960 H^3,$$ where $H$ is $c\_1(\mathcal{O}(1))$ on $\mathbb{P}^3$. Similarly, $c\_t(I\oplus T) = c\_t(I)c\_t(T) = c\_t(I)\cdot 1$, since $T\cong \mathcal{O}$. Since also $K \cong \mathcal{O}$, $c\_t(I)$ equals $c\_t(S\otimes V)$, which is just $$c\_t(I\oplus T) = (1+H)^8 = 1 + 8H + 28H^2 + 56 H^3.$$ Thus the power series we need is $$ c\_t(\text{Sym}^2 V)/c\_t(I\oplus T) = 1 + 12H + 56H^2 + 120H^3.$$ This is how I derive the number 120.


Of course we already know our $4$ bad points will be in the degeneracy locus, and they should probably have multiplicity $>1$ since the linear subsystem has codimension $5$ at each of these $4$ points. So there should be at most 116 points (and almost certainly fewer) where the fiber is a line. It is difficult for me to reconcile this with $8^3$.


$\textbf{Edit}$. I just computed the length at each of the $4$ bad points is $10$: the scheme structure of the degeneracy locus at each bad point is cut out by precisely the third power of the maximal ideal of the point. So the total contribution of the $4$ bad points to the total length is $4\times 10 = 40$. Thus I am getting that there are $4$ fibers that are smooth plane conics, and also $80$ positive-dimensional fibers that are lines. I do not understand how Oguiso computes $8^3$ positive-dimensional fibers. 


$\textbf{Second Edit}$. I think I've reconstructed Oguiso's way of getting $8^3$. The computation does have a mistake, but the method is a reasonable one. Choose homogeneous coordinates $X\_0$, $X\_1$, $X\_2$, $X\_3$ on the first factor $\mathbb{P}^3$. Then we can write $A$, $B$ and $C$ as $$ A = A\_0X\_0 + A\_1X\_1 + A\_2X\_2 + A\_3X\_3,$$ 
$$ B = B\_0X\_0 + B\_1X\_1 + B\_2X\_2 + B\_3X\_3,$$ 
$$ C = \sum\_{0\leq i \leq j \leq 3} C\_{i,j}X\_iX\_j, $$ where the $A\_i$ and $B\_j$ are global sections of $\mathcal{O}(1)$, and the $C\_{i,j}$ are global sections of $\mathcal{O}(2)$. Now we can use $A$ to "eliminate" $A\_3X\_3$. Similarly, using $A\_3B-B\_3A$, we can use this to "eliminate" $(A\_3B\_2-A\_2B\_3)X\_2$. Thus, if we multiply $C$ by $A\_3^2(A\_3B\_2-A\_2B\_3)^2$, then every instance of $X\_3$ occurs as $A\_3X\_3$, and every instance of $X\_2$ occurs as $(A\_3B\_2-A\_2B\_3)X\_2$. Thus, modulo the homogeneous ideal generated by $A$ and $A\_3B-B\_3A$, we have
$$
A\_3^2(A\_3B\_2-A\_2B\_3)^2C \equiv F\_{0,0}X\_0^2 + F\_{0,1}X\_0X\_1 + F\_{1,1}X\_1^2,
$$
where now $F\_{i,j}$ is a linear combination of the coefficients $A\_3^2(A\_3B\_2-A\_1B\_3)^2C\_{i,j}$. Of course each of these coefficients is a global section of $\mathcal{O}(8)$. Wherever $F\_{0,0}$, $F\_{0,1}$ and $F\_{1,1}$ are all zero, the expression above is in the homogeneous ideal generated by $A$ and $B$, thus the common zero locus of $A$, $B$ and this expression has dimension $\geq 1$ in the fiber $\mathbb{P}^3$. If the common zero locus of $F\_{0,0}$, $F\_{0,1}$ and $F\_{1,1}$ were finite, it would have length $8^3$ by Bezout. Of course this common zero locus is not finite; the expression is zero on the hyperplane where $A\_3$ equals $0$. 


