Connected Lie group is second countable?

Question

I know this is true from various sources, unfortunately none of them give the full proof. I already have a start:

Let $G$ be connected Lie Group. Choose $K$ to be any compact neighbourhood of the identity $e$. Then the interior $\text{int}(K)$ is an open neighbourhood of $e$. Now take $V = \text{int}(K) \cap \text{int}(K)^{-1}$, then $V$ is still an open neighbourhood of $e$ and $V \subset K$.

Consider $H = \bigcup_{n \in \mathbb{N}}V^n$. This is an open subgroup of $G$. But its complement $H^C = \bigcup_{g \neq e \in G} gH$ is also open. Because $G$ is connected and $H$ is nonempty, $H = G$. In particular,

$$ G = \bigcup_{n \in \mathbb{N}}V^n = \bigcup_{n \in \mathbb{N}}K^n$$

I'm not sure how to proceed from here, though. I imagine one could take $K$ such that it is homeomorphic to a closed disk in $\mathbb{R}^n$, then this disk is second countable and thus $K$ is second countable. But even then I don't know how this translates to the entire space being second countable?

Intuitively I would assume that taking one may construct a countable dense subset of $G$ and translate the countable around the identity to each element of this subset... but I don't think that's enough?

Answer 1

The proof follows from the following lemma.

Lemma: The identity component $G_e$ of a Lie group is second countable.

Proof: Since the Lie algebra $\mathfrak{g}$ of $G$ is second countable, then so is its $k$-fold image $\exp(\mathfrak{g})\cdots \exp(\mathfrak{g})$ in $G$ for each $k ∈ \mathbb{N}$. Now $G_e$ is the countable union over $k ∈ \mathbb{N}$ of these sets.

It follows from the Lemma that a Lie group is second countable if and only if it has at most countably many components.

Answer 2

I don't think you need to go to the Lie algebra in order to show this.

You already showed $G=\bigcup_{n\in\mathbb{N}}K^n$ where each $K^n$ is compact (since $K$ is compact and $K^n$ is the image of $K\times\cdots\times K$ under the continuous map $G^n\to G:\left(g_1,\dots,g_n\right)\mapsto g_1\cdots g_n$). Now, take some open identity neighbourhood $U$ in $G$ which is second countable (which exists since $G$ is locally diffeomorphic to $\mathbb{R}^n$). We show there are $\left\{g_n\right\}_{n\in\mathbb{N}}$ such that $G=\bigcup_{n\in\mathbb{N}}g_n U$. This is enough since each $g_nU$ is second countable (since it is homeomorphic to $U$, since $x\mapsto g_nx$ is a continuous map with a continuous inverse $x\mapsto g_n^{-1}x$) and the union is countable (and therefore the union of the bases of $g_nU$ is a countable basis of $G$).

For every $n\in\mathbb{N}$, the compact set $K^n$ is covered by $\left\{kU\right\}_{k\in K^n}$, so we have a finite subcover $$K^n=\bigcup_{i=1}^{m_n}k_iU.$$ But $G=\bigcup_{n\in\mathbb{N}}K^n$, so $$G=\bigcup_{n\in\mathbb{N}}\bigcup_{i=1}^{m_n}k_iU,$$ which is a countable union.

This proof is essentially from Hilgert and Neeb's "Structure and Geometry of Lie Groups".