# Under what condition is a fiber bundle cobordant to the trivial bundle?

Let $E$ be the total space of a fiber bundle with base $B$ and fiber $F$, where $B$ and $F$ are smooth manifolds.
Under what condition is $E$ unoriented cobordant to $B\times F$?

And what happens if we take $B$, $F$, and $E$ oriented. Is $E$ oriented cobordant to $B\times F$?

Edit : As mentioned in the comment, there are classical counterexamples, due to Atiyah, in the oriented case when the base is not simply connected.

Furthermore Dold showed that the unoriented cobrodism ring is a polynomial ring over $\mathbb{Z}/2$ with generator the $\mathbb{R}\mathbb{P}^{i}$ for $i$ even, and some fiber bundle with base $\mathbb{R}\mathbb{P}^{2^r-1}$ and fiber $\mathbb{C}\mathbb{P}^{s2^r}$ with $s,r\geq 1$. see http://www.map.mpim-bonn.mpg.de/Unoriented_bordism or http://link.springer.com/article/10.1007%2FBF01473868. Those bundles have $\mathbb{R}\mathbb{P}^{2^r-1}$ as base spaces which are null-bordant, so they provide counterexample.

However, in both those cases, counterexamples (seem to) come from the fact that the base space is not simply connected. In fact, signature is known to be multiplicative in the simply-connected case, see http://www.maths.ed.ac.uk/~aar/papers/chs.pdf.

So I will try to reformulate my question with those facts in mind :

Given that $B$ is simply connected, is there some reasonable hypothesis on $F$ assuring that $E$ is cobordant to $B\times F$?

• Does anyone know if the question "cobordant over $B$" is different? – Jesse C. McKeown Jun 20 '16 at 16:59
• I think the answer to this question might provide an example where "cobordant" and "cobordant over B" are different. – S. Douteau Jun 20 '16 at 17:05
• For oriented cobordisms there are classical couterexamples due to Atiyah: the signature need not be multiplicative for fiber bundles, see maths.ed.ac.uk/~aar/papers/atiyahsgn.pdf even though it is multiplicative for products. – Igor Belegradek Jun 21 '16 at 1:16
• Yes there are classical counterexamples for oriented cobordisms. However, the signature is multiplicative when the action of the fundamental group of the base is trivial on the cohomology of the fiber. – David C Jun 21 '16 at 12:16

1. Sphere bundles of vector bundles. If $$E \to B$$ is a vector bundle and $$S(E) \to B$$ is the associated sphere bundle, then $$S(E)$$ bounds the disc bundle $$D(E)$$, so it is nullcobordant. On the other hand, $$S^{n-1}\times B$$ is nullcobordant as $$S^{n-1}$$ is, so $$S(E)$$ and $$S^{n-1}\times B$$ are cobordant. Provided $$E$$ is an orientable bundle, the same is true in oriented cobordism.
2. Finite covering spaces. If $$F \to E \xrightarrow{\pi} B$$ is a finite covering space (i.e. $$F$$ is finite), then $$TE \cong \pi^*TB$$. It follows that the Stiefel-Whitney numbers of $$E$$ are just the Stiefel-Whitney numbers of $$B$$ multiplied by $$\deg \pi = |F|$$. So $$E$$ has the same Stiefel-Whitney numbers as the disjoint union of $$|F|$$ copies of $$B$$ (Stiefel-Whitney numbers add under disjoint sum), which is nothing but $$F\times B$$.
A similar consideration with Pontryagin numbers shows that if $$E$$ and $$B$$ are oriented, then the same conclusion holds in oriented cobordism.
3. Principal bundles. If $$G \to E \xrightarrow{\pi} B$$ is a principal $$G$$-bundle, and $$\dim G \geq 1$$, then $$E$$ is nullcobordant. To see this, note that $$TE \cong V\oplus H$$ where $$V$$ and $$H$$ are the vertical and horizontal spaces respectively. Now $$V\cong E\times\mathfrak{g}$$ while $$H \cong \pi^*TB$$, so $$TE \cong (E\times\mathfrak{g})\oplus\pi^*TB$$. In particular, $$w(TE) = w((E\times\mathfrak{g})\oplus\pi^*TB) = w(\pi^*TB) = \pi^*w(TB).$$ As $$\dim E = \dim B + \dim G > \dim B$$, all the Stiefel-Whitney numbers of $$E$$ vanish, so it is nullcobordant. On the other hand, $$G\times B$$ is nullcobordant because $$G$$ is (Lie groups are parallelisable, so all their Stiefel-Whitney numbers are zero). Again, provided that $$E$$ and $$B$$ are orientable, the same is true in oriented cobordism (the Pontryagin numbers of $$E$$ are also zero).