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I want to understand whether orientable surface bundles over the circle, i.e. with orientable total space, are always trivial, so I though I would revive an old post and ask for a few clarifications, but don't have enough credits, so I'll post them here instead.

  1. What restrictions are placed on the structure group by orientability of the total sapce? (Also for more general fiber bundles.)

  2. Is the argument (and affirmative answer) given to the old post the same for general fiber bundles over $\mathbb{S}^1$, as opposed to vector bundles?

I'll outline Ma Ming's answer (which I didn't fully understand) here for self-containedness:

Let $E \rightarrow \mathbb{S}^1$ be an $SL(n)$ (vector) bundle. Its classification depends on the homotopy class of $\mathbb{S}^0 \rightarrow SL(n)$ which is trivial, so $E$ is trivial.

Thanks!

Community
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student
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  • To explain Ma Ming's answer in more detail: Consider $S^1$ to be two closed intervals glued together at their endpoints. In order to construct a bundle $E\to S^1$, we just need information about how the fibers at those endpoints $S^0$ are glued together (relative to each other), which is the same as giving a map in $[S^0; SL(n)] = \pi_0 SL(n)$. The general, rigorous setup is called the clutching map; see wikipedia or Allen Hatcher's book on vector bundles, for example. – anomaly Sep 19 '14 at 21:15
  • Ma Ming's answer is about bundles of oriented vector spaces. It doesn't tell you anything about surface bundles. – Qiaochu Yuan Sep 20 '14 at 01:01
  • However, I just found a Theorem (Steenrod, Topology of fibre bundles 18.5) for classification of bundles over $\mathbb{S}^n$, which should allow for a similar approach to that of Ma Ming: The equivalence classes of bundles over $\mathbb{S}^n$ with group $G$ is in 1-1 correspondence with equivalence classes of elements of $\pi_{n-1} (G)$ under operations of $\pi_0 (G)$. – student Sep 20 '14 at 11:16
  • @student: yes, the much more general fact is that bundles over $X$ with structure group $G$ are in one-to-one correspondence with homotopy classes of maps from $X$ to the classifying space $BG$. When $X$ is the suspension $SY$ of some other space $Y$ then we have $[X, BG] \cong [Y, \Omega BG] \cong [Y, G]$ and that reproduces the case of spheres with $Y = S^{n-1}$. (I am being a bit sloppy about basepoints.) – Qiaochu Yuan Sep 22 '14 at 05:56

1 Answers1

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No; in fact there are many interesting nontrivial such bundles. Here are some details.

Surface bundles over a circle $S^1$ with fiber an orientable surface $\Sigma_g$ can be constructed as follows. Let $f : \Sigma_g \to \Sigma_g$ be a diffeomorphism. The quotient of the product $\Sigma_g \times [0, 1]$ by the equivalence relation

$$(x, 0) \sim (f(x), 1)$$

defines a $3$-manifold called the mapping torus $M_f$ of $f$, with a map to $S^1$ coming from projecting to the second coordinate. All fiber bundles over $S^1$ arise in this way. I believe it is moreover the case that the diffeomorphism class of the total space depends only on the class of $f$ in the mapping class group $\text{MCG}(\Sigma_g) \cong \pi_0 \text{Diff}(\Sigma_g)$ and that $M_f$ is orientable iff $f$ is orientation-preserving, hence iff its image in the mapping class group lies in the orientation-preserving subgroup $\text{MCG}^{+}(\Sigma_g)$ of the mapping class group.

The easiest nontrivial case to understand here is $g = 1$, where we get torus bundles. Here the orientation-preserving mapping class group is

$$\text{MCG}^{+}(\Sigma_1) \cong \text{SL}_2(\mathbb{Z})$$

so any non-identity element of $\text{SL}_2(\mathbb{Z})$ gives rise to a nontrivial orientable $\Sigma_1$-bundle over $S^1$. You get $3$-manifolds exhibiting three of the eight Thurston geometries this way. For $g \ge 2$ see Nielsen-Thurston classification.

Qiaochu Yuan
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  • Dear Qiaochu Yuan, could you please provide a reference for the fact that every fibre bundle over $S^1$ arises this way? – Eduardo Longa Mar 02 '23 at 21:36