# The Birkhoff–Kakatuni theorem

November 09, 2023

The purpose of this post is to explore a little a very cool metrization theorem for topological groups. (By the way, for some reason I felt like aiming this post a bit lower than some of my others.) It’s a fun bit of general topology, which I’m partial to: topology was the first subject where I really felt like math was something I could love doing. For a while some of my internet handles were references to the Urysohn lemma, a beautiful result that will come up here for us.

Anyway, here’s the statement of the theorem:

Theorem (Birkhoff, Kakatuni). Suppose $$G$$ is a topological group. Then $$G$$ is metrizable—and in fact admits a left-invariant metric—if and only if $$G$$ is Hausdorff and first countable.

## 1. Background

Let’s unpack that a lot. Groups and topological spaces I assume you know. (By the way, if you don’t and you’d like to be invited a little further into these blog posts, feel free to shoot me an email!) A topological group is simply a group $$G$$ which is at the same time a topological space and for which the operations of multiplication (i.e. the function $$G \times G \to G$$ sending $$(g,h)$$ to $$gh$$) and inversion (the map $$G \to G$$ sending $$g$$ to $$g^{-1}$$) are continuous.

A metric on a set $$X$$ is a function $$d\colon X \times X \to \mathbb{R}$$ which satisfies the following rules:

• (Positive Definite) We have that $$d(x,y) \ge 0$$ for all points $$x$$ and $$y$$ in $$X$$, and if $$d(x,y) = 0$$, then we conclude that $$x = y$$.
• (Symmetric) $$d(x,y) = d(y,x)$$.
• (Triangle Inequality) For all $$x$$, $$y$$ and $$z$$ in $$X$$, the “length of the third side”, that is, $$d(x, z)$$ is not more than the sum of the lengths of the first two. In symbols, $$d(x, z) \le d(x, y) + d(y, z)$$.

If $$d$$ is a metric on $$X$$, we can define the metric “ball” of radius $$r > 0$$ about a point $$x$$. This is the set $B_r(x) = \{ y \in X : d(x,y) < r \}$ of points in $$X$$ at distance strictly smaller than $$r$$ from $$x$$. A metric generates a topology on $$X$$ by the rule that a set $$U \subset X$$ is open when it has the property that for each point $$x \in U$$, there exists $$\epsilon > 0$$ such that $$U$$ contains $$B_\epsilon(x)$$. In other words, if $$x$$ is a point of $$U$$, then $$U$$ actually contains all the points sufficiently close to $$x$$ as well. Put more succinctly, we say that the metric balls form a basis for the topology on $$X$$.

A topological space $$X$$ is metrizable if there exists a metric (which is not uniquely determined) on the set $$X$$ for which metric balls form a basis for the given topology on $$X$$. In other words, the metric topology and the original topology are the same. So while a metric space comes equipped with a metric—it’s part of the data—a metrizable space does not. So while the classes of metric and metrizable spaces are the same, somehow the latter are more mysterious.

For groups $$G$$, a metric $$d\colon G \times G \to \mathbb{R}$$ is left invariant if it is, uhh, left invariant under the (left) action of $$G$$ on itself. That is, for all $$g$$, $$x$$ and $$y \in G$$, we have $$d(gx, gy) = d(x, y)$$. Geometrically, one can phrase this as saying that the left action of $$G$$ on itself is by isometries, metrically rigid motions.

As a geometric group theorist, I love thinking about groups as symmetries of geometric objects: concretely what this means is that if I have a group $$G$$, I’m going to want a metric space $$X$$ and an action of $$G$$ on the space $$X$$ by isometries to go with it. A left-invariant metric on $$G$$ itself provides me with just such an $$X$$, namely $$G$$ itself.

We’re already at the point where we can make a cute observation about the theorem: a priori one might think that there exist groups which are metrizable as topological spaces, but for which it’s not possible to find such a metric which actually agrees with the group structure—that is, is left invariant. Not so, says the Birkhoff–Kakatuni theorem!

## 2. The conditions of the theorem are necessary

Here’s another cute observation: if $$X$$ is a metric space, then it is Hausdorff, which let me remind you, means that if $$x$$ and $$y$$ are distinct points of $$X$$, then there exist open sets $$U$$ and $$V$$ containing $$x$$ and $$y$$ respectively, which are still disjoint. It requires you to mispronounce the name, but you might think that the points $$x$$ and $$y$$ are housed within $$U$$ and $$V$$, and disjointness says that these houses are off of each other—“housed off”.

A topological space $$X$$ is first countable if every point $$x$$ has a countable neighborhood basis. That is, there is a collection of open sets $$U_n$$ which we can index by the natural numbers such that if $$V$$ is an open set containing $$x$$, then $$V$$ contains $$U_n$$ for some $$n$$. This condition is also necessary: Notice that if an open set $$V$$ contains $$x$$, then by the definition of the metric topology it contains $$B_\epsilon(x)$$ for some sufficiently small $$\epsilon > 0$$. But then there exists $$N \in \mathbb{N}$$ so large that $$\frac{1}{N} < \epsilon$$. Therefore $$V$$ also contains $$B_{\frac{1}{N}}(x)$$, and the collection of balls of radius $$\frac{1}{n}$$ is already indexed by the natural numbers (sorry, for me zero is not a natural number).

## 3. The Urysohn lemma

I want to take a small diversion to state and prove the Urysohn Lemma, because I think it is so cute and it turns out to help us prove the theorem.

Lemma (Urysohn). A topological space $$X$$ is “normal” if and only if each pair of closed subsets of $$X$$ is separated by a continuous function.

Let’s make sense of the statement. A space $$X$$ is normal (terrible name, but we’re stuck with it) if whenever $$A$$ and $$B$$ are disjoint closed subsets of $$X$$, there exist open sets $$U \supset A$$ and $$V \supset B$$ which are still disjoint. In Hausdorff spaces, points are closed sets, so this is like amping up the definition of a Hausdorff space.

Let’s make the quick observation that if $$X$$ is normal, $$A \subset X$$ is closed, and $$U \supset A$$ is open, then we can squish in an open set $$V$$ and a closed set $$B$$ in there so that $$A \subset V \subset B \subset U$$. Indeed, to see this, observe that $$X - U$$ is closed, since $$U$$ is open, and it is clearly disjoint from $$A$$. By normality, there exist disjoint open neighborhoods $$V \supset A$$ and $$W \supset X - U$$, and the set $$B = X - W$$ is a closed set fitting into the chain $$A \subset V \subset B \subset U$$.

Anyway, $$A$$ and $$B$$ are separated by a continuous function if there exists $$f\colon X \to [0,1]$$ such that for all $$a \in A$$, we have $$f(a) = 0$$ and for all $$b \in B$$, we have $$f(b) = 1$$. Now a priori the condition of being separated by a continuous function is stronger than normality applied to $$A$$ and $$B$$, since it implies it (consider the preimages of the open intervals $$(\frac{-1}{4},\frac{1}{4})$$ and $$(\frac{3}{4}, 1\frac{1}{4})$$ as the required disjoint open sets) but is not implied by the normality hypothesis.

That’s what’s so cool about Urysohn’s lemma! As if by magic, it produces a continuous real-valued function from thin air. Here’s how.

### 3.1. Proof of the Urysohn lemma

We proceed inductively on the “dyadic rationals” between zero and one: i.e. each fraction between 0 and 1 which can be put in lowest terms as $$\frac{k}{2^n}$$. So $$0$$ is $$\frac{0}{2^0}$$, $$1$$ is $$\frac{1}{2^0}$$, and then there’s $$\frac{1}{2}$$, then $$\frac{1}{2^2} = \frac{1}{4}$$ and $$\frac{3}{2^2} = \frac{3}{4}$$, and so on. The induction is on $$n$$. The point is that at each step after the base case of $$n = 0$$, between every two existing dyadic rationals, there is exactly one new one.

Anyway, we inductively start with $$C(0) = A$$ and $$U(1) = X - B$$. Using the “quick observation” above, we fit in a pair of sets $$C(0) \subset U(\frac{1}{2}) \subset C(\frac{1}{2}) \subset U(1)$$, where $$C(\frac{1}{2})$$ is closed and $$U(\frac{1}{2})$$ is open. Then we fit in $$U(\frac{1}{4})$$ and $$C(\frac{1}{4})$$ in between $$C(0)$$ and $$U(\frac{1}{2})$$, also $$U(\frac{3}{4})$$ and $$C(\frac{3}{4})$$ between $$C(\frac{1}{2})$$ and $$U(1)$$, and then so on and so forth.

That’s a lot! What does it buy us? Well the idea is that we think of these nested sets topographically like level sets of a function $$f$$, and then we click our heels together three times or something, and the wished-for function appears.

In actual math, define a function $$f\colon X \to [0, 1]$$ by the rule that $$f(x) = \inf \{ r : x \in U(r) \}$$, and $$f(x) = 1$$ if there is no $$r$$ for which $$x \in U(r)$$. Here of course $$\inf$$ means infimum or greatest lower bound—think of it as being “like” the minimum, but allowing us to recognize that by using binary decimal expansion I can, for example, get closer and closer and closer to $$\frac{1}{3}$$, even though $$\frac{1}{3}$$ is not a dyadic rational number, and there will in general be some actual point $$x$$ satisfying $$f(x) = \frac{1}{3}$$.

It’s pretty easy to show that $$f$$ is continuous by playing the sets $$U(r)$$ and $$C(r)$$ off of each other cleverly, and it’s obvious that $$f(a) = 0$$ for all $$a \in A$$ and $$f(b) = 1$$ for all $$b \in B$$.

### 3.2. What if we had a continuous function?

Let’s start returning back to the setting of topological groups, and suppose that we had a left-invariant metric on $$G$$. Then consider the function $$\|\cdot\|\colon G \to \mathbb{R}$$ defined as $$\|g\| = d(1, g)$$, where $$1$$ denotes the identity of $$G$$.

This function has the properties that it is continuous, takes the value $$0$$ only on $$1$$, is symmetric in the sense that $$\|g^{-1}\| = \|g\|$$ and is “submultiplicative” in the sense that $\|gh\| = d(1, gh) \le d(1, g) + d(g, gh) = d(1, g) + d(1, h) = \|g\| + \|h\|.$

Such a norm on $$G$$ would immediately yield a left-invariant metric on $$G$$ by the rule that $$d(g, h) = \|g^{-1}h\|$$. This metric would be continuous since $$\|\cdot\|$$ was continuous. It will generate the topology, one can show, as soon as the norm balls form a neighborhood basis for $$1$$: that is, a set $$U \subset G$$ containing $$1$$ is open if and only if it contains all $$g$$ with $$\|g\| < \epsilon$$ for $$\epsilon > 0$$ sufficiently small.

The idea for the rest of the proof will be to try and construct such a function.

## 4. The necessary conditions are sufficient

Let’s start to prove the harder direction of the Birkhoff–Kakatuni theorem. Notice that because the inversion map $$g \mapsto g^{-1}$$ is continuous, it is a homeomorphism. If $$U$$ is a neighborhood of $$1$$, then so is $$U^{-1}$$ and $$U \cap U^{-1}$$. This latter set is additionally symmetric in the sense that if $$g \in U \cap U^{-1}$$, so is $$g^{-1}$$.

So let $$G = U_1 \supset U_2 \supset \cdots$$ be a symmetric, countable neighborhood basis of $$1$$. One exists after a little finagling by the previous observation and the assumption that $$G$$ is first countable. Since $$G$$ is Hausdorff, one can show that the intersection $$\bigcap_{n = 1}^\infty U_n$$ is the single point $$\{1\}$$.

Notice that the left action of $$G$$ on itself is by homeomorphisms, so if $$U$$ is open in $$G$$, so is $$gU$$. Since the union of open sets is open, if $$U$$ is open and $$A$$ is any set in $$G$$, the set $$AU = \{ ag : a \in A, g \in U \}$$ is open, as is, for similar reasons, $$UA$$.

Since multiplication $$G \times G \to G$$ is continuous, the preimage of the open neighborhood $$U_n$$ of $$1$$ under multiplication is open and contains $$(1,1)$$, and therefore contains $$U_m \times U_m$$ for some $$m$$. In other words, for each $$n$$ there exists $$m$$ such that $$U_m U_m \subset U_n$$. Thus we can and will assume that given our neighborhood basis of the identity, we have that $$U_{n+1} U_{n+1} \subset U_n$$.

This lets us play the Urysohn game: let’s reindex so that what was $$U_n$$ is now $$U_{\frac{1}{2^n}}$$, and define $$U_{\frac{k}{2^n}}$$ according to the binary expansion of $$\frac{k}{2^n}$$. For example, $$\frac{5}{8} = \frac{1}{2} + \frac{1}{8}$$, and we let $$U_{\frac{5}{8}} = U_{\frac{1}{8}} U_{\frac{1}{2}}$$. Since groups are not always commutative (abelian), it’s important that we choose a convention for left versus right here, so we’ll say that we put the biggest denominators on the left. It’s pretty easy to show that as the dyadic rationals increase, the corresponding subsets get larger, using our hypothesis that $$U_{\frac{1}{2^{n+1}}} U_{\frac{1}{2^{n+1}}} \subset U_{\frac{1}{2^n}}$$.

This lets us define a function $$f\colon G \to [0,1]$$. I’d like to try $$f(g) = \inf\{ \frac{k}{2^n} : g \in U_{\frac{k}{2^n}}\}$$ if this set is nonempty, and $$f(g) = 1$$ if not. Interestingly, this function, while likely continuous (we’ll show that in a second), is not quite good enough to be a norm. For one thing it’s no longer obviously symmetric, and submultiplicativity is also not obvious to me.

Instead, apparently the standard trick is to consider the function $$F = 1 - f$$. It has a unique maximum, namely, $$1$$ at $$1$$, the sets $$F^{-1}(1 - \frac{1}{2^n})$$ form a neighborhood basis of $$1$$, and the following mysterious property:

• For every $$\epsilon > 0$$, there exists $$U \ni 1$$ open such that for all $$x \in G$$, we have $$|F(gx) - F(x)| \le \epsilon$$ for all $$g \in U$$.

Despite me calling it mysterious, this property is not hard to prove: just let $$\frac{1}{2^n}$$ be smaller than $$\epsilon$$ and take $$U = U_{\frac{1}{2^n}}$$. Then $$f(gx)$$, one can show, is at most equal to the binary expansion of $$f(x)$$ to the $n$th decimal place, plus $$\frac{1}{2^n}$$. In other words, the error between $$F(gx)$$ and $$F(x)$$ is at most $$\frac{1}{2^n}$$, because their first $$n-1$$ binary digits agree.

Anyway, just (“just”) define $$d\colon G \times G \to \mathbb{R}$$ by the rule that $$d(g, h) = \sup_{x \in G} | F(g^{-1}x) - F(h^{-1}x) |$$. (As usual $$\sup$$ means supremum, or least upper bound.) It’s getting close to my bedtime, so I’ll let you verify that this gives a metric on $$G$$ (not too hard), which is left invariant and generates the topology on $$G$$. Pretty neat!