# Group Theory/Topological groups

Definition (topological group):

A topological group ${\displaystyle G}$ is a group whose underlying set is endowed with a topology such that

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

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

Proposition (every topological group is a uniform space):

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

${\displaystyle }$

form an entourage system whose induced topology is identical to the topology of ${\displaystyle G}$.

Proof: ${\displaystyle \Box }$

Theorem (Birkhoff‒Kakutani theorem):

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

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

Proof: ${\displaystyle \Box }$

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

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

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

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

Proof: We shall denote the identity of ${\displaystyle G}$ by ${\displaystyle e}$. Let ${\displaystyle K}$ be a compact neighbourhood of ${\displaystyle e}$. We set ${\displaystyle L:=K\cup K^{-1}}$. Since the image of a compact set via a continuous map is compact and the union of two compact sets is compact, ${\displaystyle L}$ is a compact neighbourhood of ${\displaystyle e}$. 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 ${\displaystyle L^{2},L^{3},\ldots }$ are compact. Yet, the group

${\displaystyle H:=\langle L\rangle \leq G}$,

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

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

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

is open.

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