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Let $M$ be a connected smooth manifold, and assume that it is parallelisable; that is, its tangent bundle is trivial. Does $M$ admit an H space structure? That is, does there exist a smooth map $\mu:M\times M\to M$ and an identity element $e$ satisfying $\mu(e,x)=\mu(x,e)=x$?

The motivation for asking is the following: given any Lie group $G$, its tangent bundle is trivial. What about the converse? It's hard enough coming up with a parallelisable manifold that is not a Lie group ($\mathbb{S}^7$ is such an example). The best idea I've heard is to think about quotients of Lie groups by discrete subgroups, but the few examples I've tried weren't parallelisable in the end.

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The answer is no. It's easy to come up with non-Lie Group manifold with trivial tangent bundles: any 3-manifold that's not itself a Lie group, like the connect sum of two 3-dimensional lens spaces. There are many obstructions to being an h-space. H-spaces which are manifolds do not need to have trivial tangent bundles. This was answered in an earlier thread. mathoverflow.net/questions/51593/… –  Ryan Budney Jan 22 '11 at 21:21
@Ryan: why not put that as an answer. –  Willie Wong Jan 23 '11 at 0:17
Thanks, Ryan. I'll take a look. –  Jordan Watts Jan 23 '11 at 2:40

1 Answer 1

up vote 8 down vote accepted

Ryan Budney's comment pretty much killed the question, but anyway...

Let $X_{m,n} = S^{2m}\times S^{2n+1}$, with $m\le n$ (strictly) positive integers.

Lemma: $X=X_{m,n}$ is parallelisable.

Proof: this follows from playing around with vector bundles, the key facts being that $TX = \pi^*(TS^{2m}) \oplus \pi^*(TS^{2n+1})$ and that trivial bundles are natural. More precisely, the second factor of $X$ has Euler number 0, so one can split off a trivial 1-bundle from its tangent bundle, pull it back through the projection and see it as the pull-back of a trivial bundle over the first factor. So $TX = \pi^*(TS^{2m}\oplus \varepsilon) \oplus V$, and the first summand is trivial. Now one can split off a trivial bundle of rank 2 from the first factor and use it to trivialise the second factor.

Lemma: $X$ is not an $H$-space.

Proof: the cohomology ring (say with $\mathbb{F}_2$ coefficients) of an $H$-space is a finite-dimensional commutative Hopf algebra, therefore it's generated in odd dimensions. But, in our case, $H^*(X)$ has $H^{2m}(X)$ as the first nontrivial group.

This gives a nice family of simply connected counterexamples to your question, the smallest of which is $S^2\times S^3$. Notice how, actually, there is no simply connected counterexample in dimension one, two, three (thanks to Perelman) and four (e.g. because the Euler number of a simply connected 4-manifold is strictly positive).

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very nice, I'll be sure to remember that argument! –  Olivier Bégassat Mar 28 '11 at 15:53
I'm less familiar with H-spaces, but ignoring that, this is beautiful. Thanks! –  Jordan Watts Apr 4 '11 at 23:28

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