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A statement: The self-intersection index of lagrangian submanifold $M \subset X$ is equal to Euler characteristic $\chi(M)$. How I should oriented $X$?

Let's consider some example. The null-section $M^2$ of cotangent bundle $T^{*}\!M^2$ is lagrangian submanifold of $T^{*}\!M$. Let $(q_1,q_2)$ be local coordinate on $M$ and $(p_1,p_2)$ is natural coordnate on $T^{*}\!M$. The local forms $\text{d}\,q_1 \wedge \text{d}\,q_2 \wedge \text{d}\,p_1 \wedge \text{d}\,p_2$ define the natural orientation of
$T^{*}\!M$. As I think, for this orientation the self-intersection index $M$ in $T^{*}\!M$ is equal to $\chi(M)$.

The natural symplectic structure on $T^{*}\!M$ is $\omega^2 = \text{d}\,p_1 \wedge \text{d}\,q_1 + \text{d}\,p_2 \wedge \text{d}\,q_2$.

The symplectic form also define orientation $$ \omega^2 \wedge \omega^2 = 2\,\text{d}\,p_1 \wedge \text{d}\,q_1 \wedge \text{d}\,p_2 \wedge \text{d}\,q_2 $$

So, the orientation induced by symplectic strucuture is not natural. And I should oriented $X$ by $(-1)\omega^2 \wedge \omega^2$.

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If you are dealing with lagrangians inside $X$, then it is clear that you are assuming the existence of a symplectic form, let us say $\omega\in\Omega^2(X)$.

Assume that the dimension of $X$ is the even number $2n$. If $f$ is a smooth function on $X$ such that $f(p)>0$ for every $p\in X$, and $\mathrm{Vol}(X)\in\Omega^{2n}(X)$ is any volume form, then $f\cdot \mathrm{Vol}(X)$ is also a volume form: there is no uniqueness or "natural" choice.

The most famous choices for volume forms for $X$ are $\omega\wedge\cdots\wedge\omega$ and $\frac{1}{n!}\omega\wedge\cdots\wedge\omega$, where the latter is more commonly called the Liouville volume (not always, though).

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