An Introduction to Topological Groups

Hey there, and welcome back! Today we’re going to talk about an interesting merger of abstract algebra and topology, namely topological groups. Last week a colleague asked me if there were any mathematical topics which were used in class but that I was never previously taught. While I do self study a lot, I had never come across algebraic structures that were also topologized, and I now use these frequently in class. So, I thought it would be nice to give a little introduction about 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, when you were taking calculus, you were actually working with the topological field $\mathbb{R}$ . You were using the topology of $\mathbb{R}$ as well as its algebraic properties as a field to prove statements about differentiation and integration as make some computations. It is sometimes beneficial to simplify the algebraic structure which a topology is endowed on, and this is one of the many reasons why we study topological groups/rings/modules/vector spaces rather than just topological fields. However, other deep fields of mathematics deal with other topologized algebraic structures for a variety of reasons.

Now 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!