## An Introduction to Topological Groups

We’re going to talk about an interesting merger of abstract algebra and topology, namely topological groups.

Now you might be wondering, why study topological algebraic structures? Topology and abstract algebra are hard enough! Well as a matter of fact they pop up all the time. Most notably, in calculus, you were actually working with the topological field $\mathbb{R}$. The topology of $\mathbb{R}$ as well as its algebraic properties were used to prove statements about differentiation and integration. There’s no better place to start than with definitions, so let’s define a topological group and introduce some of its basic properties.

#### The Definition of A Topological Group and Some Consequences

Definition: A topological group $G$ is a group $G$ endowed with a topology such that the group operation and inverse map

\begin{aligned} G \times G \to G &\qquad (g,h) \mapsto gh \\ G \to G &\qquad g \mapsto g^{-1} \end{aligned}

are continuous maps. Here we give $G \times G$ the product topology. By convention, if $G$ is finite it’s given the discrete topology.

Now a nice property of groups is that we can move between any two elements in the group by translation (i.e. multiplication). We would hope that if $U \subseteq G$ is open, then $gU$ and $Ug$ are open for any $g \in G$. This is indeed true, and to see it start by fixing $g \in G$. Then the map $x \mapsto (g,x)$ is obviously continuous. Composing with the group operation, the map $x \mapsto (g,x) \mapsto gx$ is continuous. This composition is said to be a left-translation or more precisely left-translation by $g$. Since $x \mapsto (g^{-1},x) \mapsto g^{-1}x$ is a continuous inverse, left-translation is a homeomorphism. There is a symmetric argument for right-translation. So, the above tells us in particular that if $U$ is open, then so is $gU$ and $Ug$. We’ve neglected the inverse map $g \to g^{-1}$ up until now, but it has an analogous property as the above translation maps do. It’s a continuous involution, so is a homeomorphism. Therefore if $U$ is open, then the image of $U$ under the inverse map is open, and we call it the inverse of $U$ and denote it by $U^{-1} = \{g^{-1}|g \in U\}$.

Another topological property which is important for topological groups is homogeneity. If $X$ is a topological space, then $X$ is homogeneous if for any $x,y \in X$ there is a homeomorphism $f:X \to X$ such that $f(x) = y$. Intuitively, this says that the topology of $X$ looks the same locally any any point $x \in X$. If we look at a topological group $G$, then it’s homogeneous because if $g,h \in G$, then the left-translation map $x \mapsto hg^{-1}x$ sends $g$ to $h$. A consequence of this is that if we have a local basis at the identity $e$, then it determines a local basis everywhere and therefore a basis for the entire topology. Moreover, homogeneity says that it suffices to verify local topological properties of $G$ only at the identity $e$. Indeed, if we have such a property at $e$ we can then translate this property everywhere in $G$. We will see an example of this below.

#### Symmetric Neighborhoods and The Equivalence of $T_{1}$ and $T_{2}$

An interesting property of topological groups is that the $T_{1}$ and $T_{2}$ (Hausdorff) separation axioms are equivalent! In the following we will prove this statement, but we will need a small technical proposition first which is incredibly useful beyond its introduction here. This proposition is about symmetric neighborhoods. Firstly, a set $S \subseteq G$ is called symmetric if $S = S^{-1}$. In other words, the set is closed under inverses. Now for the proposition:

Proposition: Let $G$ be a topological group.

i. Every neighborhood $U$ of the identity contains a neighborhood $V$ of the identity such that $VV \subseteq U$.

ii. Every neighborhood $U$ of the identity contains a symmetric neighborhood $V$ of the identity.

In particular, every neighborhood $U$ of the identity contains a symmetric neighborhood $V$ of the identity such that $VV \subseteq U$.

proof: Let us first prove i. We may assume $U$ is open because if not, we take its interior. The restriction of the group operation $\varphi: U \times U \to G$ is continuous, so $\varphi^{-1}(U)$ is an open set containing $(e,e)$ of the form $\varphi^{-1}(U) = V_{1} \times V_{2}$ for some open neighborhoods $V_{1}$ and $V_{2}$ of $e$. If we set $V = V_{1} \cap V_{2}$, then $V$ is a neighborhood of $e$ such that $\varphi(V \times V) = VV \subseteq U$. To prove ii note that $U$ and $U^{-1}$ are both neighborhoods of the identity. Then $V = U \cap U^{-1}$ is as well and by definition is symmetric. This $V$ is the desired open set. To prove the last statement, apply i and then ii.

Note that by our homogeneity discussion above, the proposition holds for any $g \in G$, not just the identity. Now that we have the technical proposition, let’s state and prove the equivalence of $T_{1}$ and $T_{2}$.

Theorem: Let $G$ be a topological group. Then $G$ is $T_{2}$ if and only if it’s $T_{1}$.

proof: The forward implication is clear since every $T_{2}$ space is $T_{1}$. To prove the inverse implication, let $g,h \in G$. Since $G$ is $T_{1}$ we can find a neighborhood $U$ about the identity disjoint from $gh^{-1}$, and we may assume $U$ is open by replacing it with its interior. By the proposition, there is a symmetric open neighborhood $V$, about the identity, contained in $U$, and such that $VV \subseteq U$. Again, we may assume $V$ is open by replacing it with its interior. We claim the open sets $gV$ and $hV$ are disjoint. If they are not, we have $gv = hv'$ for some $v,v' \in G$ which implies $gh^{-1} = v'v^{-1} \in VV \subset U$, a contradiction. Hence $G$ is $T_{2}$.

It’s a very easy exercise, that I leave to you, to show these equivalent statements are further equivalent to the conditions of the identity being closed and every singleton of $G$ being closed. It’s also a fun exercise to show that the general linear group over the reals $GL_{n}(\mathbb{R})$ is a topological group with respect to matrix multiplication and the Euclidean topology (here you view matrices as $n^{2}$ dimensional vectors in $\mathbb{R}^{n^{2}}$), and that its special linear group $SL_{n}(\mathbb{R})$ is a closed subgroup. I leave this to you as well if you want a slightly more difficult exercise. That’s all, thank you for reading!

#### References

Fourier Analysis on Number Fields – Dinakar Ramakrishnan, Robert J. Valenza