self-intersection of lagrangian submanifold

Question

Let's consider lagrangian submanifold $X$ in symplectic manifold $M$. Is it true that self-intersection index of $X$ is equal to the Euler characteristic $\chi(X)$? Can we construct (not canonical) isomorphism between tangent bundle $TX$ and normal bundle $NX$?

Answer 1

First let us prove $TX \cong NX$. Let $J$ be an $\omega$-compatible almost complex structure on $(M, \omega)$, so that $$g(v, w) = \omega(v, Jw)$$ defines a metric on $M$ and $\omega(Jv, Jw) = \omega(v, w)$. Now if $v \in TX$, we have that $$g(v, Jv) = \omega(v, J^2 v) = - \omega(v, v) = 0,$$ so that $Jv \in NX$. Hence we have an isomorphism $$J|_{TX}: TX \xrightarrow{~\cong~} NX.$$

Now from differential topology we know that $$e(NX)[X] = X \cdot X.$$ Since $NX \cong TX$, it follows that $e(NX)[X] = e(TX)[X] = \chi(X)$ and hence $$\chi(X) = X \cdot X.$$