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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?

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    $\begingroup$ Your question seems to be how one calculates the intersection number of two 3-dimensional submanifolds on an oriented 6-manifold, in particular, on a CY 3-fold. That's a somewhat vague question: how are the manifold and its submanifolds specified? But you might consult the famous mirror symmetry paper of Candelas-de la Ossa-Greene-Parkes; they need four cycles that form a symplectic basis for the 3rd homology of the mirror quintic. $\endgroup$
    – Tim Perutz
    Commented Mar 2, 2014 at 15:27
  • $\begingroup$ @TimPerutz thanks, do you have a link to that paper? $\endgroup$
    – Dilaton
    Commented Mar 3, 2014 at 10:18

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