Topological Group Reflections: turning a group with topology into a topological group, Part I

The main complicating factor with the quotient topology on the fundamental group $\pi_{1}^{qtop}(X,x_0)$ is that it often fails to be a topological group. More generally, if $M$ is a topological monoid and the quotient space $G=M/\mathord{\sim}$ is group by a relation that preserves the operation, then $G$ need not be a topological group. From one viewpoint, this could be considered a disappointing fact. So what causes this failure to happen? It turns out that this is a fairly deep topological question that I don’t have a complete answer too. It is partially due to some issues with the general topological category.

At this point, some folks may be inclined to point out that this failure doesn’t occur if one works internal to some “nicer” category like the category of k-spaces. This is a fair point. I agree that there are many situations in which the k-space category is more suitable and that one can obtain a group-object for $\pi_1$ in k-spaces. However, most folks who are adamant about forcing things into a “convenient” category (typically in AT) are uncomfortable outside of such categories and are not used to dealing with objects where the actual topology matters. To insist that a convenient coreflection is the only reasonable “fix” is short-sighted and closes the door to good mathematics. It turns out that forcing things into the k-space category loses too much information that is actually important to topologized fundamental groups and what they are good at detecting. If I had only stuck with k-spaces and been closed off to considering more challenging topological options, I wouldn’t have come up with the mathematics for what has (weirdly) become my most popular paper and led to surprising applications in topological group theory.

In this post, I’m going to talk about the actual “topological fix” to the quotient topology of $\pi_{1}^{qtop}(X,x_0)$ . We’re going to decide if there exists a functorial topology on $\pi_1(X,x_0)$ that gives a topological group while retaining as much of the original information of the quotient topology as possible. Along the way, we’ll explore some interesting mathematical structures and categorical topology.

Let’s approach this from a very general viewpoint. Say I have a group $G$ with identity element $e\in G$ and some topology $T$ on $G$ . The pair $(G,T)$ might already be a topological group. If so, great! Leave it alone. But maybe it’s not. Maybe the group operation doesn’t interact nicely with the topology at all. Specifically, the group operation $\mu:G\times G\to G$ , $\mu(a,b)= ab$ and group inversion $i:G\to G$ , $i(a)= a^{-1}$ are NOT assumed to be continuous. But let’s see if we can modify the topology $T$ , with minimal information loss, so that we end up with a topological group. The underlying group will be identical to the original group. It’s only the topology that’s going to change.

Let $\sigma:G\times G\to G$ be the function $\sigma(a,b)=ab^{-1}$ . Notice that if $\sigma$ is continuous, then inversion $i(b)=\sigma(e,b)=b^{-1}$ is continuous and group multiplication $\mu(a,b)=\sigma(a,i(b))=ab$ is continuous. Thus $G$ is a topological group. Conversely, if $G$ is a topological group, then $\sigma(a,b)=\mu(a,i(b))$ is continuous.

Proposition 1. Let $G$ be a group with topology. Then the following are equivalent:

$G$ is a topological group ,
$\sigma:G\times G\to G$ is continuous,
$\sigma:G\times G\to G$ is a quotient map,
$\sigma:G\times G\to G$ is a continuous open surjection.

Proof. We’ve already noted the equivalence of 1. and 2. Certainly 4 $\Rightarrow$ 3 $\Rightarrow$ 2 from basic topology. That 2 implies 4 is an introductory excercise in topological group theory. $\square$

Since the function $\sigma$ tells us, in a sense, how close $T$ is to being a topological group, we can “force” it to be continuous by changing the topology of the codomain.

Given a group $G$ equipped with topology $T_0$ , let $c(G)$ be the same group as $G$ but equipped with the quotient topology inherited from $\sigma$ , i.e. so that $\sigma:G\times G\to c(G)$ is a quotient map. Let $T_1$ denote this quotient topology. Thus $U\in T_1$ if and only if $\sigma^{-1}(U)$ is open in $G\times G$ . The identity function $G\to c(G)$ can be defined by $\sigma(g,e)=g$ and is is continuous. Hence, the topology $T_1$ of $c(G)$ is coarser than that of $G$ . Also, according to Proposition 1, we have $c(G)=G$ if and only if $G$ is a topological group.

So we fixed it right? Isn’t $c(G)$ a topological group? Well…that’s not immediately clear. We’d have to check that $\sigma:c(G)\times c(G)\to c(G)$ is continuous and there does not seem to be an easy way to prove this. We only know that $\sigma:G\times G\to c(G)$ is continuous. Even so, it seems that we have at least moved in the right direction since we removed some open sets from $T_0$ to get the topology $T_1$ but not too many open sets.

So let’s do it again. Let $c^2(G)$ be the same group as $G$ but equipped with the quotient topology inherited from $\sigma$ with domain $c(G)\times c(G)$ , i.e. so that $\sigma:c(G)\times c(G)\to c^2(G)$ is a quotient map. Let $T_2$ denote this quotient topology. Now, $c(G)$ is a topological group if and only if $c^2(G)=c(G)$ . If this is the case, we may stop. If not, we try it again.

Since $G\mapsto c(G)$ is a general operation we can perform for any group with any topology, we could define $c^{n+1}(G)=c(c^{n}(G))$ recursively for all $n\geq 1$ where $T_n$ denotes the topology of $c^{n}(G)$ . We are left with proper inclusions of topologies $T_0\supseteq T_1\supseteq T_2\supseteq T_3\supseteq \cdots$ . This sequence will stabilize if and only if $c^{n}(G)$ is a topological group for some $n$ . However, there is nothing promising that this will happen. So regular induction may fail us… We cry a few tears and then remember that there is, in fact, a way to continue this process to persist past infinity. This is where we are ever so grateful for

transfinite induction.

We have so far defined the induction for finite ordinals. The first infinite ordinal is $\omega=\{0,1,2,3,\dots\}$ . Let $c^{\omega}(G)$ be the same group as $G$ but with topology $T_{\omega}=\bigcap_{n=0}^{\infty}T_n$ . Notice that $T_n\supseteq T_{\omega}$ for all $n\in\omega$ . The reason that this intersection makes sense (besides being a natural choice) is because the descending sequence $T_0\supseteq T_1\supseteq T_2\supseteq T_3\supseteq \cdots$ of topologies is telling us that to end up with a topological group we must remove the open sets from $T_0\backslash T_1$ and then from $T_1\backslash T_2$ and so on. If we remove all of these open sets, we’re only left with those in $\bigcap_{n\geq 0}T_n$ .

Most transfinite induction arguments break into two cases – the successor ordinal case (a single step) and limit ordinal case (moving past infinitely many steps). The construction above provides a template for how to do the general transfinite induction: assume $c^{0}(G)=G$ . Suppose $\lambda$ is an ordinal and $c^{\kappa}(G)$ has been defined as the group $G$ with topology $T_{\kappa}$ for every ordinal $\kappa<\lambda$ .

Successor ordinal case: Suppose $\lambda=\kappa+1$ for ordinal $\kappa$ . Since $c^{\kappa}(G)$ is assumed to be defined, we may let $c^{\lambda+1}(G)=c(c^{\kappa}(G))$ , that is, the same underlying group as $G$ but with the quotient topology inherited from $\sigma$ with domain $c^{\kappa}(G)\times c^{\kappa}(G)$ , i.e. so that $\sigma: c^{\kappa}(G)\times c^{\kappa}(G)\to c^{\lambda}(G)$ is a quotient map. Let $T_{\lambda}$ denote the topology of $c^{\lambda}(G)$ .

Limit ordinal case: Suppose $\lambda$ is a limit ordinal. Our induction hypothesis is that $c^{\kappa}(C)$ is defined with topology $T_{\kappa}$ for all $\kappa<\lambda$ . Thus we let $c^{\lambda}(G)$ be the same group as $G$ but with the topology $T_{\lambda}=\bigcap_{\kappa<\lambda}T_{\kappa}$ .

This transfinite induction results in a nested sequence of topologies all on the same group $G$ :

$T_0\supseteq T_1\supseteq T_2 \supseteq \cdots T_{\omega}\supseteq T_{\omega+1}\supseteq T_{\omega+2}\supseteq\cdots\supseteq T_{\omega+\omega}\supseteq\cdots\supseteq T_{\omega^{\omega}}\supseteq\cdots T_{\omega_1}\supseteq \cdots$

where I interjected some other infinite ordinals for good measure. But this sequences goes on for all ordinals.

Proposition 2: $c^{\lambda}(G)$ is a topological group if and only if $T_{\kappa}=T_{\kappa+1}$ .

Proof. If $c^{\kappa}(G)$ is a topological group, then $\sigma: c^{\kappa}(G)\times c^{\kappa}(G)\to c^{\kappa}(G)$ is continuous and is therefore a quotient map. But quotient topologies are unique and the above construction gives that $\sigma: c^{\kappa}(G)\times c^{\kappa}(G)\to c^{\kappa+1}(G)$ is a quotient map. Hence, $T_{\kappa}=T_{\kappa+1}$ . Conversely, suppse $T_{\kappa}=T_{\kappa+1}$ . But $\sigma: c^{\kappa}(G)\times c^{\kappa}(G)\to c^{\kappa+1}(G)$ is continuous by construction and since we are assuming $c^{\kappa}(G)=c^{\kappa+1}(G)$ , we make this replacement in the codomain to see that $\sigma: c^{\kappa}(G)\times c^{\kappa}(G)\to c^{\kappa}(G)$ is continuous. Hence, $c^{\kappa}(G)$ is a topological group. $\square$

This proposition tells us that the transfinite sequence $\{c^{\kappa}(G)\}_{\kappa}$ of groups with topology can only stabilize if it stabilizes at a topological group. So must it stabilize?

Proposition 3: The transfinite sequence $\{c^{\kappa}(G)\}_{\kappa}$ of groups with topology stabilizes at some ordinal $\kappa$ .

Proof. Here, we have to appeal to cardinality. Suppose that the transfinite sequence of topologies $\{T_{\kappa}\}_{\kappa}$ never stabilizes. This means for each ordinal $\kappa$ , $T_{\kappa+1}$ is a proper subset of $T_{\kappa}$ and all of the sets $T_{\kappa}\backslash T_{\kappa+1}$ are pairwise-disjoint in the original topology $T_0$ . In other words, for each ordinal $\kappa$ , we removed some at least one set $U_{\kappa}\in T_{\kappa}\backslash T_{\kappa+1}$ as we proceeded through the induction.

But the ordinal numbers go really far…too far. If you give me any set like $T_0$ , there exists an ordinal number $\lambda$ with a cardinality greater than that of $T_0$ . But the function $f:\lambda\to T_0$ , $f(\kappa)=U_{\kappa}$ is an injection and this is a contradiction because we can’t inject $\lambda$ into $T_0$ since we found $\lambda$ so that $|\lambda|>|T_0|$ . $\square$

Putting everything together, we get the following.

Theorem 4: The transfinite sequence of topologies $\{c^{\kappa}(G)\}_{\kappa}$ stabilizes at some ordinal. In particular, the transfinite sequence $\{c^{\kappa}(G)\}_{\kappa}$ of groups with topology stabilizes at a topological group, denoted $\tau(G)$ .

The construction $G\mapsto \tau(G)$ takes in a group with topology and outputs a topological group in “the most efficient way possible.” Formally, the topology of $\tau(G)$ is the finest topology on the group that is (1) coarser than that of $G$ and (2) which makes the group a topological group.

Did we really need to go through all this trouble to build $\tau(G)$ ? Well…there are other ways we could choose to show that topology of $\tau(G)$ exists but none of them are going to have simple descriptions. Because, remember, a union of topologies need not be a topology. So we can’t build the topology from below. We really do have to carefully wittle down the topology of $G$ without removing too much information.

My bolded claim is not exactly proved yet, but this and many other things we can say about the construction become easier to prove once we show that $\tau$ is a functor. We’ll talk about maps in Part II.

4 Responses to Topological Group Reflections: turning a group with topology into a topological group, Part I

David Roberts says:

September 20, 2024 at 1:21 am

In your proof of Proposition 3, you can take lambda to be the Hartogs’ number of the set T_0, which is the smallest ordinal (or even just well-ordering, as in the original treatment) with no injection to T_0. And this can, by the way, be constructed without AC 🙂

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