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 $E$?


1 Answer 1


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.

  • $\begingroup$ okay, the argument is purely trivial! thank you! $\endgroup$ Sep 12, 2022 at 21:42
  • $\begingroup$ @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.) $\endgroup$
    – mme
    Sep 12, 2022 at 21:48

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