# The Unapologetic Mathematician

## Topological Groups

Now we’ve said a lot about the category $\mathbf{Top}$ of topological spaces and continuous maps between them. In particular we’ve seen that it’s complete and cocomplete — it has all limits and colimits. But we’ve still yet to see any good examples of topological spaces. That will change soon.

First, though, I want to point out something we can do with these limits: we can define topological groups. Specifically, a topological group is a group object in the category of topological spaces. That is, it’s a topological space $G$ along with continuous functions $m:G\times G\rightarrow G$, $e:\{*\}\rightarrow G$, and $i:G\rightarrow G$ that satisfy the usual commutative diagrams. A morphism of topological groups is then just like a homomorphism of groups, but by a continuous function between the underlying topological spaces.

Alternately we can think of it as a group to which we’ve added a topology so that the group operations are continuous. But as we’ve seen, a topological structure feels a bit floppier than a group structure, so it’s not really as easy to think of a “topology object” in a category. So we’ll start with $\mathbf{Top}$ and take group objects in there.

Now it turns out that every topological group is a uniform space in at least two ways. We can declare the set $E_U=\{(x,y)|xy^{-1}\in U\}$ to be an entourage for any neighborhood $U$ of the identity, along with any subset of $G\times G$ containing such an $E_U$. Since any neighborhood of $e$ contains $e$ itself, each $E_U$ must contain the diagonal $\{(x,x)\}$. The intersection $E_U\cap E_V$ is the entourage $E_{U\cap V}$, and so this collection is closed under intersections.

To see that $\bar{E}_U$ is an entourage, we must consider the inversion map. Any neighborhood $N$ of the identity contains an open set $U$. Then the preimage $i^{-1}(U)$ is just the “reflection” that sends each element of $U$ to its inverse, which must thus be open. The reflection of $N$ contains the reflection of $U$, and is thus a neighborhood of the identity. Then $\bar{E}_U=\{(x,y)|yx^{-1}=(xy^{-1})^{-1}\in U\}$ is the same as $E_{i^{-1}(U)}$.

Now, why must there be a “half-size” entourage? We’ll need to construct a half-size neighborhood of the identity. That is, a neighborhood $V$ so that the product of any two elements of $V$ lands in the neighborhood $U$. Then $(x,y)$ and $(y,z)$ in $E_V$ means that $xy^{-1}$ and $yz^{-1}$ are in $V$, and thus their product $xz^{-1}$ is in $U$, so $(x,z)\in E_U$.

To construct this neighborhood $V$ let’s start by assuming $U$ is an open neighborhood by passing to an open subset of our neighborhood if necessary. Then its preimage $m^{-1}(U)$ is open in $G\times G$ by the continuity of $m$, and $U\times G$ and $G\times U$ will be open by the way we built the product topology. The intersection of these will be the collection of pairs $(x,y)\in G\times G$ with both $x$ and $y$ in $U$, and whose product also lands in $U$, and will be open as a finite intersection of open sets. We can project this set of pairs onto its first or second factor, and take the intersection of these two projections to get the open set $V$ which is our half-size neighborhood.

The uniform structure we have constructed is called the right uniformity on $G$ because if we take any element $a\in G$ the function from $G$ to itself define by right multiplication by $a$ $x\mapsto m(x,a)$ — is uniformly continuous. Indeed, right multiplication sends an entourage $E_U=\{(x,y)|xy^{-1}\in U\}$ to itself, since the pair $(xa,ya)$ satisfies $xa(ya)^{-1}=xaa^{-1}y^{-1}=xy^{-1}\in U$. Left multiplication, on the other hand, sends a pair $(x,y)$ in $E_U$ to $(ax,ay)$, for which we have $ax(ay)^{-1}=axy^{-1}a^{-1}\in aUa^{-1}$. Thus to an entourage $E_U$ we can pick the entourage $E_{a^{-1}Ua}$. So left multiplication is also uniformly continuous, but not quite as easily. We could go through the same procedure to define the left uniformity which again swaps the roles of left and right multiplication. Note that the left and right uniformities need not be the same collection of entourages, but they define the same topology.

Still, this doesn’t tell us how to get our hands on any topological groups to begin with, so here’s a way to do just that: start with an ordered group. That is, a set with the structures of both a group and a partial order so that if $a\leq b$ then $ga\leq gb$ and $ag\leq bg$. Using this translation invariance we can determine the order just by knowing which elements lie above the identity, for then $a\leq b$ if and only if $e\leq a^{-1}b$. The elements $x$ with $1\leq x$ form what we call the positive cone $G^+$.

We can now use this to define a topology by declaring the positive cone to be closed. Then we’d like our translations to be homeomorphisms, so for each $a$ the set of $x$ with $a\leq x$ must also be closed. Similarly we want inversion to be a homeomorphism, and since it reverses the order we find that for each $a$ the set of $x$ with $x\leq a$ is closed. And then we can use the complements of all these as a subbase to generate a topology. This topology will in fact be uniform by everything we’ve done above.

And, finally, one specific example. The field $\mathbb{Q}$ of rational numbers is an ordered group if we forget the multiplication. And thus we get a uniform topology on it, generated by the subbase of half-infinite sets. Specifically, for each rational number $a$ the set $(a,\infty)$ of all $x\in\mathbb{Q}$ with $a and the set $(-\infty,a)$ of all $x\in\mathbb{Q}$ with $x are declared open, and they generate the topology. A neighborhood of $0\in\mathbb{Q}$ will be any subset which contains one of the form $(-a,a)$. Since the group is abelian, both the left and the right uniformities coincide. For each rational number $a$ we have an entourage $E_a=\{(x,y)|-a. That is, a pair of rational numbers are in $E_a$ if they differ by less than $a$.

November 27, 2007 Posted by | Group theory, Topology | 2 Comments