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!


 

 

 

References

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

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