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 $H$.

**(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 $m\ge n$.**

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