# Intersection form of $2n$-manifold for odd $n$

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 $$E$$?

The following argument can be phrased in terms of cohomology (it amounts to a proof that the Euler class of an odd-rank bundle is 2-torsion) but here is a purely intersection-theoretic phrasing.

It suffices to show that if $$E \to S$$ is any vector bundle with $$\text{rank}(E) = \dim S = 2k+1$$, then a generic section has no zeroes when counted with sign. As you know (and by a standard argument), the signed count of such zeroes is independent of the choice of generic section. If $$\phi$$ is such a section, write the oriented zero set as $$Z(\phi)$$.

Pick a generic section $$\phi$$. Then $$Z(-\phi)$$ can be identified with $$Z(\phi)$$ with the opposite orientation, because the negation map $$\Bbb R^{2k+1} \to \Bbb R^{2k+1}$$ is orientation-reversing. Thus $$\# Z(\phi) = \# Z(-\phi) = -\# Z(\phi),$$ the first equality because the count is independent of the choice of generic section, the second by the observation about orientations.

Thus $$\# [S \cap S] = \# Z(\phi)$$ is an integer equal to its negative, thus zero.

As for your follow-up questions: if $$E \to S$$ is any vector bundle over closed base, it may be realized as the normal bundle of an embedding of $$S$$ into a closed manifold by taking the fiberwise one-point compactification of $$E$$ into a sphere bundle over $$S$$.

In general, if $$E$$ has rank $$2k+1$$, the class $$e(E) \in H^{2k+1}(S;\Bbb Z)$$ is 2-torsion. If $$E$$ has rank equal to the dimension of $$S$$ and $$S$$ is oriented, this implies $$e(E) = 0$$. If $$E$$ has smaller rank than the dimension of $$S$$, or if $$S$$ is non-orientable, it is possible for this class to be non-zero.

• okay, the argument is purely trivial! thank you! Sep 12, 2022 at 21:42
• @AndreyRyabichev No problem. Now that I think of it, by thinking of $BGL_{2k+1}$ as an infinite-dimensional manifold, applying the geometric argument in the universal case proves that a rank 2k+1 bundle over any reasonable space has $e(E)$ 2-torsion. (Of course, there are also purely algebraic arguments.)
– mme
Sep 12, 2022 at 21:48