Conditions for a smooth manifold to admit the structure of a Lie group

Question

As we know, Lie group is a special smooth manifold. I want to find some geometric property, which is only satisfied by the Lie group. I only found one property: parallelizability. Can you show me something more?

Answer 1

If $G$ is any topological group, the connected component $G_0$ containing the identity is a closed normal subgroup and the other connected components are cosets. So a necessary condition for a space to admit the structure of a topological group is that all of its connected components be homeomorphic. In the case of a smooth manifold $G_0$ is also open, so the space of components $G/G_0$ is discrete. It follows that a smooth manifold $X$ admits the structure of a Lie group iff all of its components are homeomorphic and any one of them admits the structure of a Lie group. In other words, it is no real loss of generality to assume that $X$ is connected.

There are lots of deep necessary conditions on the topology of $X$ for $X$ to admit the structure of a Lie group. See for instance Theorem 2.7 of these notes which asserts the following:

a) $\pi_1(X)$ is a finitely generated abelian group.
b) $\pi_2(X)$ is the trivial group.
c) $\pi_3(X)$ is a finitely generated free abelian group.

There is much more to say here, although others are more qualified than I to say it.

Answer 2

To add some conditions on the topology, if a compact manifold $M$ admits the structure of an abelian Lie group, then $M$ must be isomorphic to $(S^1)^n$.

More generally, if $M$ admits the structure of a Lie group, then $M$ must have the same cohomology ring with $\mathbb{Q}$ coefficients as a product of odd spheres. If $M$ admits the structure of a nonabelian Lie group, then $S^3$ must appear in the product of odd spheres. Further, such an $M$ must be rationally elliptic, which puts VERY strong constraints on the rational homotopy groups.

The universal cover of $M$ can only have $2,3,$ or $5$ torsion in its cohomology.

A noncompact Lie group must be diffeomorphic to $\mathbb{R}^k\times G$ where $G$ is a compact Lie group. (I'm not positive about this line. Certainly a noncompact simply connected group must be diffeomorphic to a product like this, with $G$ simply connected).

On the other hand, I think it could be VERY difficult to come up with sufficient topological conditions to be a Lie group. For example, the space $SO(7)\times S^7$ is diffeomorphic to $SO(8)$, and hence carries a Lie group structure. However, as mentioned by others, $S^7$ itself does not carry a Lie group structure.

Answer 3

The cohomology of the manifold has to be a Hopf Algebra. More generally, the manifold has to be an H-space. This implies the unit sphere bundle is fibrewise homotopy-trivial (see comments below). Earlier I thought manifold + h-space implied triviality of the tangent bundle but that's not clear.

A manifold that's an H-space is called an "H-manifold" in the literature. Sometimes this is ambiguous since if H is a group, "H-manifold" could mean "manifold with an action of the group H" so beware when searching the literature.

I believe there is a well-known obstruction theory for H-manifolds to be Lie groups and many algebraic topologists know a pile of examples off the top of their heads. Unfortunately, if I ever did know such a pile of examples I've forgotten!