Group Theory/Topological groups

Definition (topological group):

A topological group $G$ is a group whose underlying set is endowed with a topology such that

1. the group law is a continuous function $G\times G\to G$ and
2. inversion is a continuous function $G\to G$.

Thus, a topological group is a group with structure in the category of topological spaces.

Proposition (every topological group is a uniform space):

Let $G$ be a topological group, and let $(U_{\alpha })_{}$ be a neighbourhood system of its identity. Then the sets

${\displaystyle }$

form an entourage system whose induced topology is identical to the topology of $G$.

Proof: $\Box$

Theorem (Birkhoff‒Kakutani theorem):

Proposition (the connected component of the identity of a topological group is one of its normal subgroups):

Let $G$ be a topological group, and let $K\subseteq G$ be the connected component of its identity. Then $K\trianglelefteq G$.

Proof: $\Box$

Proposition (each locally compact topological group is the disjoint union of translates of one of its σ-compact open subgroups):

Let $G$ be a locally compact topological group. Then there exists a σ-compact open subgroup $H\leq G$, from which we may of course deduce that

$G=\bigsqcup _{g\in S}gH$,

where $S$ is a set that contains one element of each left coset of $H$ (the squared union symbol indicating that the union is disjoint). Moreover, each left coset $gH$ of $H$ is σ-compact and open.

Proof: We shall denote the identity of $G$ by $e$. Let $K$ be a compact neighbourhood of $e$. We set $L:=K\cup K^{-1}$. Since the image of a compact set via a continuous map is compact and the union of two compact sets is compact, $L$ is a compact neighbourhood of $e$. Moreover, induction, the fact that the product of two compact sets is compact and the fact that the image of a compact set via a continuous map is compact (applied to the continuous group law map) yield that all the sets $L^{2},L^{3},\ldots$ are compact. Yet, the group

$H:=\langle L\rangle \leq G$,

ie. the group generated by the elements of $L$, is the union of these sets, hence σ-compact.

It remains to show that $H$ is open. To this end, we may use that since $L$ is a neighbourhood of $e$, there exists an open set $V\subseteq L$ such that $e\in L$. Since multiplication by a group element is an isomorphism, the set $hV$ are open in $H$ whenever $h\in H$. Hence,

$H=\bigcup _{h\in H}hV$

is open.

Finally, $gH$ is open and σ-compact because multiplication by $g$ is an automorphism of $G$ in the category of topological spaces, whence it preserves openness and compactness. $\Box$