Lie Groups and Manifolds - MathOverflow

Lie Groups and Manifolds

2009-11-12T21:45:53Z

I'm trying to get a better handle on the relation between Lie groups and the Manifolds they correspond to. Firstly, is the relationship injective? that is, does each Lie group correspond to a unique manifold? Or are all the manifolds corresponding to a particular group homeomorphic?

Also, what formal form does the relationship take? I can intuitively understand the relationship between, say, $SO(3)$ and $S^2$ by thinking about rotating the sphere into itself, but what how does this generalize to a more general group or manifold.

Answer by Mariano Suárez-Alvarez for Lie Groups and Manifolds

2009-11-12T21:52:57Z

There are manifolds which are groups in many ways. A very simple example is $\mathbb{R}^3$, which is an abelian Lie group in the obvious way, and a nilpotent group when seen as the set of upper triangular unipotent $3\times 3$ matrices, that is, the set of $3\times 3$ matrices which are upper triangular and have ones along the diagonal.

When the dimension is larger, things get `worse' (or better, depending on your persective) There are uncountably many Lie group structures on $\mathbb R^n$ for large $n$ (at least $8$, if I recall correctly)

Answer by Danny Calegari for Lie Groups and Manifolds

2009-11-12T22:45:12Z

$SO(3)$ is homeomorphic to $RP^3$, not to $S^2$. The relationship between $SO(3)$ and $S^2$ is that $SO(3)$ is the group of (orientation-preserving) isometries of $S^2$ in its round metric. If $M$ is any Riemannian manifold, the group of isometries of $M$ is a Lie group (this is an old theorem of Kobayashi (edit: I mean Myers-Steenrod; see comments).

Answer by fuzzytron for Lie Groups and Manifolds

2009-11-12T23:48:52Z

Also, what formal form does the relationship take? I can intuitively understand the relationship between, say, SO(3) and S2 by thinking about rotating the sphere into itself, but what how does this generalize to a more general group or manifold.

The relationship you're describing here is called group action - you have a homomorphism $g$ from $SO(3)$ to a subgroup of automorphisms on (the standard embedding of) $S^2$. In other words, for every rotation in $SO(3)$ you have a mapping of $S^2$ to itself; this correspondence commutes with composition. However, the existence of a homomorphism does not mean that $S^2$ and $SO(3)$ are the same. In particular, $g$ is not an isomorphism: there are "more" rotations than there are mappings of the sphere to itself. In fact, there is no Lie group isomorphic to $S^2$, i.e., there is no group operation that makes $S^2$ a Lie group (this fact follows from the "hairy ball theorem").

Answer by Jason DeVito for Lie Groups and Manifolds

2009-11-13T02:25:41Z

To add a bit,

There are also many examples of compact manifolds with multiple group structures.

As a quick example, first recall that $SU(2)$ is the collection of all $A \in M_2(\mathbb{C})$ with $A\overline{A}^t = Id$ and $det(A) = 1$. It is a Lie group (which is actually diffeomorphic to $S^3$.)

The manifold $S^1\times SU(2)$ has (at least) 2 group structures. The first is simply the product. The second is isomorphic to the Lie group $U(2)$, those matrices $A\in M_2(\mathbb{C})$ such that $ A\overline{A}^t = Id$ (no extra condition no the determinant).

For another example, recall that $SO(n)$ is the Lie group consisting of all $A\in M_n(\mathbb{R})$ such that $AA^t = Id $. Then $SO(3)\times SU(2)$ is diffeomorphic to $SO(4)$ but the group structures are different.

Answer by Gian Maria Dall'Ara for Lie Groups and Manifolds

2009-11-14T10:24:45Z

I add that there are a lot of manifolds which does not admit a Lie group structure. A nice obstruction is that the fundamental group should be abelian. This is true even for topological groups. So there's no way to put a topological group structure on surfaces of genus higher than 1. This can be easily understood by inspection of the map $\gamma(t)\sigma(r)$ for $\gamma$ and $\sigma$ two loops based at the identity. An obstruction in the smooth category (if I remember correctly) is the fact that if $H^1(G)$ is trivial than $H^3(G)$ must be non trivial (maybe $\dim G>0$), showing that $S^0$, $S^1$ and $S^3\cong SU^2$ are the only spheres which can be lie groups. They are the units of the only associative division algebras over the reals.

Answer by Gian Maria Dall'Ara for Lie Groups and Manifolds

2009-11-19T16:39:54Z

I realized that one very fundamental geometric constraint on the underlying manifold of a Lie group which wasn't mentioned is that every such manifold is parallelizable, i.e. the tangent bundle is globally trivial. This is very easily seen by choosing a basis for the tangent space at the identity and moving it around with group translations. This, together with the "hairy ball theorem" gives you the non-existence of lie group structures on even dimensional spheres ($\dim>0$).

Answer by sinbad for Lie Groups and Manifolds

2009-11-23T10:55:41Z

I like to add one pt. every lie group,as manifold, is orientable and has Euler charactersitic 0.

Answer by Amin for Lie Groups and Manifolds

2011-11-01T10:24:10Z

Just to comment on the relation between S^2 and SO(3) : there is indeed, for symmetric spaces, a natural correspondence between them and Lie groups, which in particular gives the S^2-SO(3) pair. A symmetric space is roughly a connected manifold M with global symmetries s_x at each point (satisfying certain properties). Now define G(M) to be the group generated by even products of symmetries. Then one can show (using Palais's theorem) that this is a Lie group, which is connected and acting transitively on the symmetric space. Obviously, if you are in the Riemannian context, this will be a Lie group of isometries. Further, it's the 'smallest' subgroup of the isometry group transitive and stable under the involution given by conjugation by symmetry at any base point, which is thus a sort of uniqueness.