Introduction to Lie Groups and Lie Algebras September 21, 2004

Gabriel C. Drummond-Cole

November 30, 2004

From now on all of my Lie groups are connected. If G is a Lie group then 𝔤 is the Lie algebra corresponding to G, also written T_{1}G and Lie(G).

There is a bracket [,], and I won’t remind you of how it’s defined, but it’s bilinear, skewsymmetric, and satisfies the Jacobi identity. There are relations to the group multiplication. There is the Ad action Ad e^{tx} = 1+tad x+… for x ∈ 𝔤, which defines the ad x action 𝔤 → 𝔤 by ad x(y) = [x, y].

You have the formula e^{x}e^{y} = e^{f}(x, y). Here f = x + y + \frac{1}{2}[x, y] + …, where the other terms have x, y, and the commutator. The coefficients are universal. This is called the Campbell Hausdorff formula and we haven’t actually proved it.

Now, what next? I want to continue the story about the relation between Lie groups and Lie algebras. Say we have ϕ : G_{1} → G_{2}; then there is a corresponding map ϕ_{∗} : 𝔤_{1} → 𝔤_{2} which preserves [,]. That is, Hom(G_{1}, G_{2}) ↪ Hom(𝔤_{1}, 𝔤_{2}). Is this equality? It’s pretty clear that a lift would be unique.

If you know the map at the level of Lie algebras, it uniquely defines the map in a neighborhood of the identity in the Lie group, which then generates the group, giving uniqueness.

In general, though, not everything can be lifted. Let G_{1} = S^{1}, G_{2} = ℝ. Then the Lie algebras are ℝ with zero bracket. There is a map (the identity) on the level of Lie algebras, but a Lie group morphism (which must be induced by t → αt : ℝ → ℝ) should satisfy 2πα = 0. Then it’s the zero morphism.

Is there any way to fix this?

Theorem 1 If G_{1} is simply connected then Hom(G_{1}, G_{2}) = Hom(𝔤_{1}, 𝔤_{2}).

The reason we’re trying to do this is because Lie algebras are easier to work with. The other type of question is the question of subgroups. Namely, suppose H ⊂ G (a Lie subgroup, I don’t consider any others). Then of course you can immediately form the corresponding Lie algebras, and you have an inclusion 𝔥 ⊂ 𝔤. This is not only an embedding of vector spaces, but also preserves the commutator.

Think of inclusion as a group morphism; then this induces an algebra morphism. Can we do it the other way around? Does a Lie subalgebra correspond to a Lie subgroup?

The answer again is unfortunately no. Not every Lie subalgebra can be related to a Lie subgroup. Say your G is ℝ^{2}, and 𝔥 is a subspace spanned by (1, α). Then H = ℝ(1, a).

Look instead at T^{2}, where 𝔤 = ℝ^{2} with the trivial bracket. Then let 𝔥 = ℝ(1, a) with a irrational. At the level of Lie groups, this covers an everywhere dense subset of the torus. This is then not a closed one-dimensional submanifold.

To accomodate for this kind of problem we have the definition

Definition 1 An immersed (analytic, etc.) subgroup in G is a subset H ⊂ G such that

1. H is a subgroup of G

2. H is almost a submanifold; it is the image of an immersion: H = Im(H̃ ↪ G) for an immersion i of a Lie group H̃. That is, it is a map of manifolds.

So that’s almost a Lie subgroup, but not quite. So you should be talking about these more general subgroups. Does this solve the problem? The answer is yes. Here is a theorem:

Theorem 2 There is a bijection between connected immersed subgroups H ⊂ G and subalgebras 𝔥 ⊂ 𝔤. The correspondence comes from H → 𝔥 = Lie(H) = T_{1}H.

There is also a third fundamental theorem which we’ll discuss next time. Let me note that theorem one easily follows from theorem two, and the other way around as well.

The third theorem will tell us that an abstract Lie algebra is the Lie algebra of some Lie group.

Here’s the proof of the first theorem from the second. Suppose f : 𝔤_{1} → 𝔤_{2}. We want to show it lifts to ϕ : G_{1} → G_{2} with ϕ_{∗} = f.

Define G = G_{1} × G_{2}. Then the Lie algebra of G is 𝔤_{1} × 𝔤_{2}. Then this contains 𝔥 = {(x, f(x))}. I claim that this is a subalgebra. It’s a subspace, and [(x, f(x)), (y, f(y))] = ([x, y], [f(x), f(y)]) = ([x, y], f([x, y])). So that’s a very simple but quite useful trick. Then there is a corresponding subgroup H ⊂ G_{1} × G_{2}. It’s the image of some embedding Im(H̃ ↪ G_{1}×G_{2}). Look at the projection p : G_{1}×G_{2} → G_{1}. Apply this projection to H or if you prefer to H̃. Then I claim that H̃ ↪ G_{1} ×G_{2} → G_{1} is a covering map, that is, at the level of tangent spaces is an isomorphism. This follows from the fact that projection 𝔤_{1} → 𝔤_{1} × 𝔤_{2} → 𝔤_{1} is an isomorphism.

You also need to check that it’s surjective. This is because it’s a group morphism covering a neighborhood of the identity. So it’s a covering map. So leaving apart some stupid topological problem, a surjective map which is an isomorphism at the level of covering spaces is a covering map. Since G_{1} is simply connected this is just a diffeomorphism. Then we can follow the inverse map and then the other projection.

So define ϕ as composition of G_{1} → H ⊂ G_{1} × G_{2} →^{p_{2}} G_{2}. You have to check that it is a group morphism, but that basicly follows because H is a subgroup.

I still haven’t proved theorem 2. To do this I need to ask how many of you are familiar with distribution. Not all, so I’ll give you the method but I won’t go into detail.

Let me start by the following remark. Assume you have a Lie group which corresponds to the subalgebra 𝔥. Then at every point p ∈ H, T_{p}H = (T_{1}H)p = 𝔥 · p = (R_{p^{−1}})_{∗}𝔥.

Now at every point in G, define 𝒟_{p} = 𝔥·p ⊂ T_{p}G. This is called a distribution, and generalizes the picture of a differential equation as a direction field. This simple argument shows that the Lie group should have the property that the tangent space is (T_{1}H)p. If H exists, then T_{p}H = 𝒟_{p}. So H is called an integral manifold of the distribution 𝒟_{p}.

Lemma 1 For every point g ∈ G, there is loally an integral manifold containing g, namely H^{0} · g, where H^{0} = exp in some neighborhood of 0 in 𝔥.

After we prove this we’re basically done. There are at least two or three proofs. I know two.

1. You can use the Frobenius thoerem, that tells you that 𝒟_{p} is integrable. It’s generated by right-invariant vector fields corresponding to elements of 𝔥, and 𝔥 is closed under [,] If I take ge^{tx} for x ∈ 𝔥, then this curve is the integral curve for a right invariant vector field corresponding to g. In particular, its tangent vector at every point will lie in this distribution. Thus, e^{tx} · g lies in an integral manifold containing g. Now it shows that elements of form H^{0} · g are in the integral manifold containing g. Comparing dimensions you find out they are the same.

2. There is an alternative proof which doesn’t use Frobenius at all, but instead uses the Campbell Hausdorff formula. They are actually closely related but I’d rather not spend time on this.

Now that I’ve proved the lemma, you have G, 𝔤, and 𝔥. You create out of it a distribution. Locally there is always a manifold. In particular near the identity this is exp(𝔥). So just take an integral manifold and continue it as far as you can. Now take H to be the maximal connected integral manifold containing 1. This doesn’t have to be a submanifold in G. But by arguments having nothing to do with Lie theory you can show that this is the image of an injective map between manifolds. I’m not going to show this because it is so believable.

The only thing which remains to be shown is why this thing is a subgroup. The points on this integral manifold will be exactly of for exp(t_{1}x_{1}). This is how you extend. Then an element will be in H only if it is the product of finitely many things of form exp(t_{i}x_{i}) for x_{i} ∈ 𝔥. So H is a subgroup, and that is basically the end.

So that essentially proves these two theorems. In the last two minutes let me give you some obvious corrolaries. There are obvious ways to get morphisms from a Lie group to another, if the first is simply connected. For example, where you’re supposed to show the map SU(2) → SO(3) is a covering map, instead of using a messy fromula it suffices to define ϕ_{∗} : 𝔰𝔲(2) → 𝔰𝔬(3) and check that this is a morphism of Lie algebras.

For example, if the basis is

(σ_{x} = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , σ_{y} = ( \begin{matrix}0 & i \\ i & 0 \end{matrix} ) , σ_{z} = ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} )

Then you take σ_{i} → 2T_{i}, where these latter are the infinitessimal rotations.

There is a similar statement for normal groups.

Theorem 3 Say H is a connected Lie subgroup of G. Then H is normal if and only if ad xh ∈ 𝔥 for all x ∈ 𝔤. That is, [x, h] ∈ 𝔥. Such an 𝔥 is called an ideal in 𝔤.

If H is normal then Lie(G/H) = 𝔤/𝔥.

This theorem is relatively easy and the time is up so let’s stop here.