For two intersecting $D6$ branes $a$ and $b$ wrapped around a $6$ dimensional torus $T^6 = T^2 \times T^2 \times T^2$ specified by

$$ \textrm{D6-brane a:}\, (l_1^a,l_2^a,l_3^a) $$

$$ \textrm{D6-brane b:}\, (l_1^b,l_2^b,l_3^b) $$

where the $l_i$ are th directions of the D-brane on the $i$th $T^2$ it can be shown that the total intersection number is given by the product of the $3$ intersection numbers $\sharp(l_i^a,l_i^b)$ on the $i$th $T^2$

$$ I_{ab} = \prod\limits_{i=1}^3 = \sharp(l_i^a,l_i^b) $$

with

$$ \sharp(l_i^a,l_i^b) = det\left( \begin{array}{c} l_i^a \\ l_i^a \\ \end{array} \right) $$

The magnitude of the three factors can geometrically be visualized as the area of the parallelogram defined by $l_i^a$ and $l_i^b$ whereas the sign of the angle needed to align the fist direction with the second.

What is the approach to calculate the intersection number, if the six dimensional torus is replaced by a manifold that can not be factored that easily, such as for example a CY?