Lie Subalgebras and Lie Subgroups

This post will be shorter than usual. I thought it might be fun to write up some neat small results that have come up in my classes. I'll start today by talking about Lie subalgebras and Lie subgroups.

Recall that we can use the group structure of a Lie group GG to define a product on TeGT_eG (the tangent space to the identity). We call TeGT_eG with this product structure the Lie algebra of GG, and denote it g\g. The Lie algebra encodes a lot of significant information about the group - the Baker-Campbell-Haussdorf formula lets us relate the group product to the Lie bracket (at least in the image of the exponential map).

A Lie subgroup HGH \subseteq G is a Lie group HH along with an injective Lie group homomorphism ι:HG\iota:H \inj G. The differential of this homomorphism gives us a map between their Lie algebras dι:hgd\iota: \h \to \g. The image of dιd\iota is a Lie subalgebra of g\g (i.e. a linear subspace which is closed under the Lie bracket). This gives us a nice way of associating Lie subalgebras of g\g to Lie subgroups of GG.

A natural follow-up question to ask is whether this correspondence works the other way as well: given a Lie subalgebra hg\h \subseteq \g, does it necessarily come from a Lie subgroup i:HGi:H \inj G? It turns out that the answer is yes! The proof is pretty neat, and not too long, although that's largely because I'll use a powerful theorem without proof.

The general idea of the proof is fairly intuitive. We can view the subalgebra hg\h \subseteq \g as a linear subspace of TeGT_eG. Using left-multiplication, we can translate this subspace to get a subspace of TxGT_xG for all xGx \in G. Then, we can essentially "integrate up" these planes to get a submanifold which is tangent to these planes. To make this argument more formal, we will look at distributions, which are just assignments of planes to each point in a manifold.

A kk-plane distribution on a manifold MnM^n is a rank-kk subbundle of the tangent bundle TMTM. Explicitly, this means that for each point xMx \in M, we assign a kk-dimensional subspace ΔxTxM\Delta_x \subseteq T_xM, and we make these choices in a smooth way. We denote the distribution by Δ\Delta. Now, suppose we have a submanifold NMN \subseteq M such that for every xNx \in N, Δx\Delta_x is the tangent space to NN at xx. In this case, we call NN an integral manifold of Δ\Delta.

We call a distribution Δ\Delta involutive if for any vector fields X,YX,Y whose vectors all lie in Δ\Delta, then the Lie bracket [X,Y][X,Y] also lies in Δ\Delta. Frobenius' Theorem tells us that if a distribution is involutive, then we can find a unique maximal integral manifold passing through any point xMx \in M (Frobenius' theorem is actually stronger than this, but this is enough for us). This is great for us!

We can use Frobenius' theorem to find our subgroup HH. Suppose we have a Lie subalgebra h\h. We can construct a distribution Δ\Delta be defining Δx:=dLxh\Delta_x := dL_x\h. Since h\h is a Lie subalgebra, it is closed under the Lie bracket. So Δ\Delta is involutive. Thus, we can find a maximal integral manifold of Δ\Delta passing through the identity ee. Suggestively, we'll call this submanifold HH.

Now, we just need to show that HH is a subgroup. This sounds like it might be difficult, but there's actually a clever trick that makes it really easy!. Let hHh \in H. Consider the translated submanifold h1Hh^{-1}H. h1Hh^{-1}H is an integral manifold of h1Δh^{-1}\Delta. Since we constructed Δ\Delta by left-translating a subspace of TeGT_eG, Δ\Delta must be left-invariant. So h1Δh^{-1}\Delta is just Δ\Delta. Thus, h1Hh^{-1}H is a maximal integral submanifold of Δ\Delta. And since hHh \in H, h1h=eh1Hh^{-1}h = e \in h^{-1}H. By uniqueness of maximal integral submanifolds, we conclude that h1H=Hh^{-1}H = H. Thus, HH is a subgroup.

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