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Which manifolds are parallelizable?

This question is all about vector bundles isomorphic to trivial ones (for the sake of completeness I've collected definitions I refer to below). Let us agree to call trivial those bundles too.

Question What does characterize manifolds with a trivial tangent bundle?

At first I would have thought that such a manifold must be somewhat "dull", like a Euclidean space. But let us consider the unit circle $\mathbb{S}^1$ and the unit sphere $\mathbb{S}^2$. The first is not a Euclidean space and still has a trivial tangent bundle. The second has not a trivial tangent bundle, for the existance of a vector bundle isomorphism $F \colon \rm{T}\mathbb{S}^2 \to \mathbb{S}^2\times \mathbb{R}^2$ would contradict the Hairy ball theorem. (cfr. Spivak A comprehensive introduction to differential geometry, chapter 3).

This little example induces me to think that there must be more behind this. Can someone add something?

A quick review of definitions:

A vector bundle is a five-tuple $(E, B, V, \pi)$ where $E, B$ are topological spaces, $V$ is a vector space, $\pi\colon E \to B$ is surjective and continuous, every fibre of $\pi$ has a vector space structure and a local triviality condition is satisfied (for every $p\in B$ there exists a neighborhood $U$ and a homeomorphism $t \colon \pi^{-1}(U) \to U \times V$ whose restriction to each fiber is a linear isomorphism). The vector bundle $(B \times V, B, V, \pi_1)$ is called trivial.

If $(E, B, V, \pi)$ and $(E', B', V', \pi')$ are vector bundles, a pair of continuous functions $(\tilde{f}, f)$ such that the following diagram commutes

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} E & \ra{\tilde{f}} & E' \\ \da{\pi} & & \da{\pi'} \\ B & \ra{f} & B' \\ \end{array} $$ and the restriction of $\tilde{f}$ to every fiber is linear is called a vector bundle map. A bijective vector bundle map whose inverse is still a vector bundle map is called a vector bundle isomorphism.

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Giuseppe Negro
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    I think the word for this is parallelizable. – Dylan Moreland Jul 07 '11 at 16:24
  • @Dylan @Theo: Ok! Now I know which word to look for. Thank you! – Giuseppe Negro Jul 07 '11 at 16:30
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    @Theo: that's not entirely true. H spaces are known to have trivial tangent sphere bundle, but apparently they can have non-trivial tangent bundles. – Willie Wong Jul 07 '11 at 16:34
  • In any case, let's wait and see if Ryan Budney will pop over and comment on this question. – Willie Wong Jul 07 '11 at 16:36
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    @Willie: Thanks! I knew there was something not quite right here... So let me restate the correct part again to avoid spreading misinformation: Lie groups are parallelizable as can be seen by identifying the Lie algebra with the left-invariant vector fields. – t.b. Jul 07 '11 at 16:37
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    This looks awfully similar to a question I asked a while ago. So in case you haven't seen it yet: http://math.stackexchange.com/questions/46297/which-manifolds-are-parallelizable – Marek Jul 07 '11 at 16:37
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    @Marek: In fact I had not seen your question. Mine is a complete duplicate of yours so I'm going to delete it in the next two hours. Thank you for pointing this out! – Giuseppe Negro Jul 07 '11 at 16:58
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    @dissonance: Why delete it? It would be sufficient if the question is closed as a duplicate (I just cast my vote). – t.b. Jul 07 '11 at 17:09
  • @Theo: I voted to close too. I didn't know of this option in case of duplicate questions! – Giuseppe Negro Jul 07 '11 at 17:21
  • I've closed the question. – Zev Chonoles Jul 07 '11 at 17:48
  • One advantage about duplicates is that, for example, if another user comes along and knows about the notion of a trivial tangent bundle but does not know the terminology "parallelizable", having your question here as a duplicate (rather than deleted) will allow him/her to more easily find the answer. – Willie Wong Jul 12 '11 at 15:03

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