= Solution
Put $L=L_1\cap L_2$, and let $j:L\hookrightarrow M$ be inclusion. The <transverse intersection theorem> makes $L$ a smooth <submanifold> without boundary, with
$$
T_xL=T_xL_1\cap T_xL_2,\qquad
\operatorname{codim}_{\mathbb R}L
=\operatorname{codim}_{\mathbb R}L_1+\operatorname{codim}_{\mathbb R}L_2.
$$
It is compact because it is a closed subset of either compact input <submanifold>.
Write $\nu_r=TM|_{L_r}/TL_r$. On $L$, consider the bundle map
$$
TM|_L\longrightarrow\nu_1|_L\oplus\nu_2|_L,\qquad
v\longmapsto(v\bmod TL_1,\ v\bmod TL_2).
$$
Its kernel is $TL$, and it is onto because $TL_1+TL_2=TM$ at every intersection point. Thus the <normal bundle of a transverse intersection> has the actual bundle isomorphism
$$
\nu(j)\cong\nu_1|_L\oplus\nu_2|_L.
$$
Transfer the direct-sum <complex structure> $J_1\oplus J_2$ through this isomorphism. This supplies a <complex structure> on the <normal bundle> itself, not merely after stabilization.
Geometrically, multiply the two classes by forming their external product in $M\times M$ and pulling it back along the diagonal $\Delta:M\to M\times M$. The transversality hypothesis is precisely what makes the inverse image of $L_1\times L_2$ under $\Delta$ the <submanifold> $L$. The <normal bundle> of this inverse image is the direct sum just calculated. Multiplication of the corresponding <Thom classes> agrees with the <Thom class> of this direct sum, so the <complex cobordism product represented by transverse intersections> is
$$
\boxed{[L_1,i_1,\nu_1]*[L_2,i_2,\nu_2]
=[L_1\cap L_2,\ j,\ \nu_1|_L\oplus\nu_2|_L].}
$$
If the inputs have complex normal ranks $a,b$, the product has degree $2(a+b)$. If they are disjoint, the representative is empty and the product is zero.
Back to article page
