# Lie Groups and Manifolds

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.

• What do you mean by "does each Lie group correspond to a unique manifold"? Every Lie group is per definition a manifold. Unique in what sense? Commented Nov 12, 2009 at 22:01
• I'm reminded of an old joke. Q: What's the physicist's definition of a group? A: A Lie group without the manifold structure.
– HJRW
Commented Nov 12, 2009 at 22:51
• @lwassink: SO(3) is definitely not homeomorphic to S^2. I know because I made the same mistake a few months ago! The intuitive reason that SO(3) and S^2 can't be homeomorphic is that SO(3) is a 3-dimensional manifold, while S^2 is 2-dimensional. To see why SO(3) is 3-dimensional, imagine a sphere with a dot on it; hold it so the dot is on top. There are three independent ways to rotate the sphere. First, you can move the dot towards or away from your body. Second, you can move the dot to the left or the right. Third, you can spin the sphere around the dot, so the dot doesn't move. Commented Nov 13, 2009 at 1:16
• I think the question Iwassink was asking is the following. Let G be a group which admits a manifold structure making it a Lie group. Is it possible that it admits another non-homeomorphic manifold structure with respect to which it is also a Lie group? This seems like a completely reasonable question to me and, though I would guess the answer's no, I wouldn't know how to prove it. Commented Nov 13, 2009 at 2:20
• If you don't make any extra assumptions on your group Hugh, a change of topology can give you two different manifold structures. Take a 1 and a 2-dimensional vector space over the reals. As groups, they're isomorphic. So if you allow a change-of-topology you can think of R as a lie group in two different ways. It's kind of silly. Commented Nov 13, 2009 at 3:12

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.

• I wanted to add something slightly more exotic. $SO(8)$ is diffeomorphic to $SO(7)\times S^7$, but $S^7$ doesn't have ANY group structure (though it ALMOST does: there is a multiplication which satisfies all of the usual group axioms EXCEPT associativity. Google search Cayley numbers for more information). Commented Nov 16, 2009 at 1:46
• This is very very pretty. I'll have to think about these. Are there any lower dimensional examples? Commented Nov 23, 2009 at 17:36
• As mentioned below, R^2 has at least 2 different Lie group structures. The smallest (in terms of dimension) compact example is the one above with SU(2)x S^1 and U(2) (which is dimension 4). This is because for dim = 1 or 2, all compact Lie groups are isomorphic to products of circles. In dimension 3, The only compact Lie groups are T^3, SU(2), and SO(3). (This can be proved by classifying all semisimple Lie algebras of dim <=3). All 3 are pairwise nondiffeomorphic (nonhomotopy equivalent in fact) as pi_1(T^3) = Z^3, pi_1(SU(2)) = 0, pi_1(SO(3)) = Z/2Z (in fact, SU(2) double covers SO(3) ) Commented Nov 23, 2009 at 20:46

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)

• Even R^2 is also homeomorphic to the 2-dimensional solvable Lie group Aff(R) of affine transformations of the line [which in fact is isometric to the hyperbolic plane]. Another Lie group structure on R^3 is given by the solvable group Sol = R^2 semidirect R, where the action of R on R^2 is by the diagonal matrix with diagonal entries (e^t, e^-t). Commented Nov 12, 2009 at 22:32
• Yet another natural Lie group structure on $\mathbb{R}^3$ is as the universal cover of $SL_2(\mathbb{R})$.
– Ben
Commented May 5, 2015 at 4:07
• I realize I'm nitpicking a century late to the party :) but shouldn't the group in the example of Tom Church be $\mathrm{Aff}^{+}(\mathbb{R})$ (which is connected) instead of $\mathrm{Aff}(\mathbb{R})$? Commented Sep 16, 2021 at 22:23

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.

• The fact that other spheres do not have group structures can actually be seen in the topological category, using the fact that any topological group has a classifying space (en.wikipedia.org/wiki/Classifying_space), whose loopspace is homotopy-equivalent to the group. Cohomology operations can then be used to show that no other spheres can be loopspaces. Commented Nov 14, 2009 at 19:51
• Very interesting! What about the relation with division algebras? Is it incidental or there is something deeper? Commented Nov 15, 2009 at 8:45

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").

$SO(3)$ 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).

• The isometry group is a Lie group by the Myers-Steenrod theorem (which in fact is non-trivial). Was this assumed first by Kobayashi, or what do you mean by "an old theorem of Kobayashi"? Commented Nov 12, 2009 at 22:54
• You're right! I was confusing this with a theorem of Kobayashi that if $P$ is a $G$-structure on a compact manifold, where the Lie algebra of $G$ is elliptic, then the group of automorphisms of $P$ is a Lie group. Commented Nov 12, 2009 at 23:16
• So if I understand correctly, the group of isometries of a smooth manifold is a Lie group, but it need not be diffeomorphic to the manifold whose isometry group it is. Is there ever a case where a Lie group is both the isometry group of a manifold and diffeomorphic to that same manifold? Commented Nov 13, 2009 at 7:07
• The real line acting on itself by translations should do the trick Commented Nov 13, 2009 at 7:31
• Well, the real line acting on itself by translation makes up the identity componenet of the isometry group in this case - but the full isometry group is disconnected. For more examples like this, if you pick any connected centerless Lie group G at all and pick a "generic" left-invariant metric, I'd be willing to bet that the identity componenet of the isometry group is isomorphic to G. (But I don't know this for a fact) Commented Nov 15, 2009 at 22:20

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.

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 ($\dim>0$).

• You don't need the "hairy ball theorem". In fact, from what you said it follows that the Euler characteristic $\chi$ is zero for any Lie group. This number is 2 for even spheres and hence no group structure on them. Commented Nov 21, 2009 at 20:51

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

• Doesn't the real line have Euler characteristic equal to 1? Maybe you want to assume compact, so that there is a maximal torus. I think that would do it. Commented Nov 27, 2009 at 4:36
• Well, the better statement is that every (connected) Lie group is either contractible or has Euler characteristic 0. Every Lie group deformation retracts onto its maximal compact subgroup, and hence, the homotopy type of a Lie group is that of a compact Lie group. Any compact Lie group of dimension > 0 has euler characteristic 0, but a point, of course, has euler characteristic 1. Commented Dec 1, 2009 at 19:28