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!

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).

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.

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!