Let $M$ be closed orientable $2n$-manifold, where $n$ is odd. It is well known that the $\mathbb Z$-module $H^\bullet(M;\mathbb Z)$ has graded-commutative multiplication and $H^{2n}(M;\mathbb Z)\simeq\mathbb Z$. So (by Poincaré duality) there is a skew-symmetric quadratic form $[-]\smile[-]$ on $H_n(M;\mathbb Z)$.

Clearly, for any closed orientable $n$-submanifold $S\subset M$ we have $[S]\smile[S]=0$. On the over hand, this can be computed as the Euler class of the normal bundle $NS$. Indeed, this Euler class is equal to the sum of intersection points $S\cap S'$, where $S'$ is “another copy of $S$ in general position with the first one”.

**Question:** *how can we see the fact $[S]\smile[S]=0$ geometrically in terms of the normal bundle?*

It seems that, vice versa, any orientable rank $n$ vector bundle $E$ over a closed orientable $n$-manifold $S$ can be used to make closed orientable $2n$-manifold $M$, so that $E\subset M$ becomes a tubular neighborhood of its zero section $S\subset E$.
*Is this correct?*

*Do we always have the property $e(E)=0$ for $n$, $E$, $S$ as above?*

I think I can prove just that $w_n(E)=0$ using splitting principle (recall that $e\underset2\equiv w_n$). Also the fact $e(TS)=0$ is well-known, but how can we handle the case of arbitrary $E$?