## 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 $G$ to define a product on $T_eG$ (the tangent space to the identity). We call $T_eG$ with this product structure the *Lie algebra* of $G$, and denote it $\g$. 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 $H \subseteq G$ is a Lie group $H$ along with an injective Lie group homomorphism $\iota:H \inj G$. The differential of this homomorphism gives us a map between their Lie algebras $d\iota: \h \to \g$. The image of $d\iota$ is a Lie subalgebra of $\g$ (i.e. a linear subspace which is closed under the Lie bracket). This gives us a nice way of associating Lie subalgebras of $\g$ to Lie subgroups of $G$.

A natural follow-up question to ask is whether this correspondence works the other way as well: given a Lie subalgebra $\h \subseteq \g$, does it necessarily come from a Lie subgroup $i:H \inj G$? 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 $\h \subseteq \g$ as a linear subspace of $T_eG$. Using left-multiplication, we can translate this subspace to get a subspace of $T_xG$ for all $x \in G$. 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 $k$-plane distribution on a manifold $M^n$ is a rank-$k$ subbundle of the tangent bundle $TM$. Explicitly, this means that for each point $x \in M$, we assign a $k$-dimensional subspace $\Delta_x \subseteq T_xM$, and we make these choices in a smooth way. We denote the distribution by $\Delta$. Now, suppose we have a submanifold $N \subseteq M$ such that for every $x \in N$, $\Delta_x$ is the tangent space to $N$ at $x$. In this case, we call $N$ an *integral manifold* of $\Delta$.

We call a distribution $\Delta$ *involutive* if for any vector fields $X,Y$ whose vectors all lie in $\Delta$, then the Lie bracket $[X,Y]$ also lies in $\Delta$. Frobenius' Theorem tells us that if a distribution is involutive, then we can find a unique maximal integral manifold passing through any point $x \in M$ (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 $H$. Suppose we have a Lie subalgebra $\h$. We can construct a distribution $\Delta$ be defining $\Delta_x := dL_x\h$. Since $\h$ is a Lie subalgebra, it is closed under the Lie bracket. So $\Delta$ is involutive. Thus, we can find a maximal integral manifold of $\Delta$ passing through the identity $e$. Suggestively, we'll call this submanifold $H$.

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