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?
Asked
Active
Viewed 2,255 times
22
-
1The fundamental group of each connected component is abelian. – t.b. Jan 08 '11 at 11:23
-
4Each connected component of a Lie group admits the structure of a Lie group :) – Mariano Suárez-Álvarez Jan 08 '11 at 14:19
-
8parallelizability is not only satisfied by Lie groups. For example, all H-spaces are parallelizable but not every H-space is a Lie group (e.g. $S^7$). – Eric O. Korman Jan 08 '11 at 18:33
-
Ya,I only want some conditions to drop some mainfolds from the Lie group,the iff condition is too difficult. – Strongart Jan 09 '11 at 11:04
3 Answers
21
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.
Pete L. Clark
- 97,892
-
2The problem of which spheres are Lie groups isn't too hard, actually. It's easy to show that an abelian compact Lie group is a Torus and a nonabelian compact lie group has nontrivial $H^3$. The problem of which spheres are H-spaces, however, turned out to be a very difficult and rich question. – Jason DeVito - on hiatus Jan 08 '11 at 19:04
-
@Jason: thanks. I think this is the second time you've told me this. – Pete L. Clark Jan 08 '11 at 19:32
-
Yes,the structure theorem is very powerful,and some group in the topology also can do this job. – Strongart Jan 11 '11 at 10:56
14
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.
Jason DeVito - on hiatus
- 50,257
-
The line you're uncertain about is almost true, barring the trivial fact that you need to multiply by some discrete set if $\pi_0$ of your Lie group is infinite. There's a proof in Hochschild's "Structure of Lie groups" that doesn't use any assumptions on the Lie group. – Oct 04 '15 at 01:42
4
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!
Ryan Budney
- 22,958
-
How does one show that a manifold which is an $H$-space is parallelizable? – Mariano Suárez-Álvarez Jan 09 '11 at 02:07
-
-
Thanks. $TM$ being fiber homotpy equivalent to a trivial bundle is the same thing as being parallelizable? – Mariano Suárez-Álvarez Jan 09 '11 at 04:36
-
@Mariano: sorry, I don't know the answer to that. I earlier thought the answer was known but the more I look into it the less solid information I've been able to find. I'll revise my answer. – Ryan Budney Jan 09 '11 at 22:17
-
@Mariano: Igor answers your question here: http://mathoverflow.net/questions/51593/fibrewise-homotopy-equivalence-of-unit-sphere-bundles-vs-isomorphism-of-tangent-b – Ryan Budney Jan 10 '11 at 02:35