Lie group topology

I'll assume you have a basic understanding of general topology -- if not, consult the topology articles here. Most of the abstract stuff and "weird" cases are not really important, because it is easy to see that Lie groups are manifolds.

We need to be careful while studying the topology of Lie groups, because we already have an intuitive picture of a Lie group, and we need to be careful to prove all the things we just "believe" to be true.

The main point of the topology of a Lie group is that the group elements define the "flows" on the manifold. What this means is that left-multiplication is a homeomorphism, and it's not absurd to say that inversion is a homeomorphism, because it represents a "reflection" of the manifold. That these conditions make sense is confirmed by looking at the proofs of the following "obvious" facts.

(1) In a connected group, a neighbourhood of the identity generates the entire group, i.e. $H\le G\land H\in N(1)\implies H=G$ for connected $G$.

Let's think about why this is true. Why does $H$ need to be a neighbourhood -- why must it contain an open set containing the identity? Suppose instead we just knew it contained a set $Q$ that looked like this:

Well, $H$ still contains the orange point, but we cannot say it contains the purple point, because it's perfectly happy not containing it -- it's not like we have some vertical element in the Lie group that if you multiplied to some point in $Q$, you'd get the purple point. But instead if $Q$ was an open neighbourhood of the identity:

Then the purple point has to be in $H$, because $Q$ contains flows in "all directions" on the group. To actually prove that every point will be contained in $H$ -- well, we know that the point is (will eventually be) that $H$ is the connected component of $G$ (and since $G$ is connected, $H=G$) -- let's just show that $H$ is both open and closed, i.e. nothing in $H$ touches its exterior, and nothing in its exterior touches $H$. Here's the proof:

• Nothing in $H$ touches anything -- Suppose $\exists x\in H, x\in\mathrm{cl}(H')$. Then $xQ$ contains a point in $H'$.
• Nothing outside $H$ touches it -- Suppose $\exists x\in H', x\in\mathrm{cl}(H)$. Then $xQ$ contains a point in $H$, so $x$ must be in $H$.

We're really just formalising the notion of "translating $Q$ to its edges to extend $H$ further and further". The key fact we've used here is, of course, that left-multiplication is a homeomorphism, so $xQ$ is still an open set.

(2) The connected component of the identity is a subgroup.

The idea is that taking two elements $g,h$ of the connected component, their product should remain in the connected component. Once again, this follows from the continuity of left-multiplication -- considering the action of left-multiplication by $g$ on the connected component, its continuity implies that the image must remain connected.

(3) If a subgroup contains a neighbourhood of the identity, it contains the connected component of the identity.

Corollary to (1) and (2).

(4) The connected component of the identity is a normal subgroup.

Conjugation is a continuous map.

(5) Open subgroups are closed.

Corollary to (3). Alternate proof: the complement is the union of some cosets, which are open sets too. A weaker theorem can be made of closed sets -- closed subgroups with finite index are open.

What this means: any open subgroup is a union of connected components.

(6) Intuition for compact subgroups

How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $W$ of the identity by observing that $xW$ must be in the subgroup? The idea is that these translations of $W$ form an open cover of the group, if it has a finite subcover, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $W$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact.

(7) A compact, connected Abelian Lie group is a torus.

This is a generalisation of "a finite Abelian group is the direct product of cyclic groups".

The idea behind the proof is that in the Abelian case, the exponential map is a homomorphism from the Lie algebra to the Lie group, but the Lie algebra cannot detect compactness in the Lie group -- the kernel of the exponential map can. We know from our study of the exponential map that it has a discrete kernel, and in the Abelian case is surjective -- thus the Lie group is homeomorphic to $\mathbb{R}^n/\mathbb{Z}^n$, which is an $n$-torus.

(8) A connected Abelian Lie group is a cylinder (direct product of a torus and an affine space)

Analogous to above, except $\mathbb{R}^m/\mathbb{Z}^n$ where $m\ge n$.