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.



Does a Trivial Tangent Bundle Induce a Multiplication?
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).


