Why the Euler characteristics of a compact connected lagrangian submanifold of $\mathbb{R}^4$ is zero? Let's consider space $\mathbb{R}^4$ with the standard symplectic structure and let $L\subseteq \mathbb{R}^4$ be a compact connected embedded submanifold. There is a fact that if $L$ is lagrangian submanifold, then it is a torus, or, equivalently in this case, $\chi(L)=0$. I saw this fact in many books and articles, but all proofs I could find are either not very detailed or too difficult for me. More precisely, I am interested in these facts: why, if $L$ is otientable, then $\chi(L)=0$; why, if $L$ is not orientable, then $\chi(L)\equiv 0 (\text{mod 2})$. Both facts as I understood are easy, but in all proofs I saw there are some moments I do not understand.
I will give some information about my knowledge and problems that I have reading proofs of this fact.
I know about Weinstein's neighborhood theorem. I understood that I have to use that some neighborhood of $L$ is symplectomorphic to some neighborhood of zero-section in cotangent bundle $T^* L$. Also, many proofs use isomorphism $TL≅T^* L≅NL$. I understand that these isomorphisms exist but do not know why it is useful here. Then, I know about Euler characteristics as a number of zeros (with signes or modulo two) of a generic vector field on a manifold. Also, I am a bit familiar with the intersection index of two submanifolds, so as I understand Euler characteristics of submanifold is the intersection index of the zero-section in its tangent bundle with itself.
It seems to me that I have all instruments to proof these facts but concrete steps I do not see. Could you help me please?
 A: Calculate the homological intersection number of L with a pushoff. If the pushoff is using a section of the normal bundle (which is isomorphic to the tangent bundle) then you get minus the Euler characteristic because that's the homological intersection of a section of the tangent bundle with the zero section. However, you can also pushoff using a translation in $\mathbb{C}^n$, which displaces L from itself if you translate far enough because L is compact and hence bounded. This gives intersection number zero. Now notice that if L is orientable then homological intersection number can be computed using any pushoff by standard results in differential topology (it's just $[L]^2$). If it's not orientable then this still works mod 2. In fact, as observed by Audin, using the Pontryagin square instead of $[L]^2$, you get the result modulo 4 in the non-orientable setting. In fact, you don't need the Weinstein neighbourhood theorem, only the fact that normal and tangent bundles are isomorphic (which is strictly easier: it just uses linear algebra rather than Moser's argument).
The same strategy tells you the Euler characteristic equals $[L]^2$ in general (not just in $\mathbb{C}^n$).
