A sad fact that every topologist should be surprised about at some point in their career is that many apparently “nice” CW complexes are not actually metrizable. The purpose of this post is to introduce CW complexes and talk about metrizing them.
The $n$-sphere and $(n+1)$-disc
Before we get to the definition of a CW complex, which I think is one of the neatest bits of “synthetic” topology out there, we need to meet the building blocks of CW complexes: the $n$-sphere and the $(n+1)$-disc.
In calculus and linear algebra, we meet the spaces $\mathbb{R}^n$: starting with $\mathbb{R}$, the real line, which we just talked about, and then Cartesian products thereof.
It’s always interesting to me to remember that $\mathbb{R}^n$ with the Euclidean metric is just a model of Euclidean geometry, a map and not the territory. Nothing in classical geometry required that every point in space be specified by a tuple of coordinates.
Anyway, $\mathbb{R}^n$ comes equipped with the Euclidean norm $\|\cdot\|$, defined as \[ \|\vec x\| = \sqrt{x_1^2 + \cdots + x_n^2}, \] where $\vec x$ denotes the coordinate tuple $(x_1,\ldots,x_n)$.
The set of points $\vec x$ in $\mathbb{R}^{n+1}$ satisfying $\|\vec x\| = 1$ is the unit $n$-sphere, $S^n$. There is a unit $0$-sphere in $\mathbb{R}$ comprising the points $1$ and $-1$, then a unit circle in $\mathbb{R}^2$, a unit $2$-sphere in $\mathbb{R}^3$, the brain-breaking three-dimensional sphere in $\mathbb{R}^4$, and so on. If instead we ask that $\|\vec x \| \le 1$, we get the unit $(n+1)$-disc $D^{n+1}$ in $\mathbb{R}^{n+1}$. Thus for example the unit circle in $\mathbb{R}^2$ does not contain the points it surrounds, while the unit $2$-disc does.
Every point of $D^{n+1}$ may be written as $t \vec x = (tx_0, \ldots, tx_n)$ with $t$ in $[0,1]$ and $\vec x$ a point of the unit $n$-sphere. This representation is unique except for at the origin where any choice of $\vec x$ paired with $t = 0$ will do.
Metric spaces
The spaces $\mathbb{R}^{n+1}$, $D^{n+1}$ and $S^n$ are examples of metric spaces. A metric on a set $X$ is a function $d\colon X \times X \to \mathbb{R}$ satisfying the following properties
- $d(x,y) \ge 0$ and $d(x,y) = 0$ implies that $x = y$. Intuitively this makes sense: the “distance” between two points should be nonzero when the points are different, and a negative distance doesn’t make much sense.
- $d(x,y) = d(y,x)$. This also makes sense: turning the ruler around shouldn’t change its measurement. (There are interesting generalizations of the concept of a metric space where this doesn’t hold! And intuitively there are arguments to be made against this axiom too: for example, maybe as the crow flies it doesn’t matter whether you’re going from point A to point B, but things like hills or a strong wind could make it experientially easier to go one way than the other.)
- $d(x, z) \le d(x, y) + d(y, z)$. This is known as the “triangle inequality” and is the most important of the axioms. Maybe you remember from a geometry class that even the long side of a triangle with straight line sides is always shorter than the sum of the other two sides. Intuitively you might think it’s faster to go straight to your destination $z$ from home at $x$ than to detour through $y$—at worst $y$ could be “on the way” and then it wouldn’t take you any longer.
The spaces above are metric spaces: the Euclidean norm on $\mathbb{R}^{n+1}$ yields a metric $d \colon \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}$ defined as $d(\vec x,\vec y) = \| \vec x - \vec y\|$, where the subtraction is done “pointwise”. Since both $D^{n+1}$ and $S^{n}$ are subsets of $\mathbb{R}^{n+1}$, restricting the domain of definition of the metric yields a metric. This metric is slightly better for the disc than for the sphere, for the reason that the disc is convex: it contains the straight lines between any two points. Points on this straight line make the triangle inequality above an equality, for which mathematicians say that straight lines are geodesic in $\mathbb{R}^{n+1}$.
Anyway, because the sphere $S^{n}$ never contains a straight line, the Euclidean distance on $\mathbb{R}^{n+1}$ restricted to $S^{n}$ is not geodesic. The round metric, which I believe I mentioned a few posts ago, corrects this problem.
A metric space deserves the name “space” because it comes equipped with a topology, where the sets $B_{\epsilon}(x) = \{ y : d(x,y) < \epsilon \}$ of points $y$ at distance less than $\epsilon$ from $x$ are declared “open”.
Metrizability
Topological spaces in general can sometimes feel a little arbitrary: the rules of what defines a “topology”, while at an appropriate level of generality, allow all kinds of strange behavior. Metric spaces are a bit better behaved. They are, for instance, Hausdorff and first countable, and they come equipped with many many maps to the real numbers $\mathbb{R}$, one of the fundamental mathematical objects. So, given a space, one often wants to know if it is metrizable. That is, does there exist a metric $d \colon X \times X \to \mathbb{R}$ such that the metric topology on $X$ and whatever topology it walked in with have the same open sets?
Being metrizable already implies the existance of the nice properties, but it is qualitatively weaker than being metric. That is, given a metrizable space, it doesn't follow that one can easily put to hand the “right” metric to look at it with.
CW complexes
One way to build new topological spaces from old ones is to take some shapes and glue. For us, the shapes will be discs, and we’ll glue pretty arbitrarily, but always along the boundary spheres in such a way that we can build up our space “one dimension at a time”.
Formally, imagine beginning with a bunch of discs, sorted by dimension but otherwise fairly arbitrary. Let’s let $e^{n}_{i}$ denote an \(n\)-dimensional disc, where $i$ is drawn from some indexing set $I$. We'll say that our CW complex is countable if $I$ can be taken to be the counting numbers $\mathbb{N}$. If we have zero \(n\)-dimensional discs for $n$ above some threshold $N$, then we'll say our CW complex, whatever it is, is \(N\)-dimensional.
Here's the guts of the definition: imagine laying out your \(0\)-dimensional discs (or \(0\)-cells for short), then connect them up with \(1\)-cells in some way, then layer on the \(2\)-cells so that the circle boundary of each \(2\)-dimensional disc drizzles over what you have already in any way you'd like, and then continue with the higher-dimensional cells.
Explicitly, define $X^{0}$ to be a discrete set of points, one for each \(0\)-cell. Then supposing $X^{n}$ has been defined, define $X^{n+1}$ by taking a map $f^{n+1}_{i}\colon S^{n} \to X^{n}$, one for each \((n + 1)\)-cell, forming the disjoint union \[ \coprod_{i} e_i^{n+1} \sqcup X^n \] and taking the quotient by identifying each point on the boundary circle of $e^{n+1}_{i}$ with its image under $f^{n+1}_{i}$.
The name CW
As far as I understand it, the name CW is an abbreviation for two topological facts about the complex. One of them, “closure-finiteness” requires that the closure of each open \(n\)-cell contains only finitely many cells, all of dimension less than \(n\). This follows by construction from the compactness of $S^{n-1}$.
The other one, “weak topology” is confusing, since “weak” with respect to topologies now typically means the exact opposite of what it means in this context. What’s relevant is that a set is closed in $X$ if and only if it meets each closed cell in a closed set.
Not all CW complexes are metrizable
Maybe the simplest example is the following: begin with $e_{0}^{0}, e_{1}^{0}, \ldots$ a countable collection of \(0\)-cells and a countable collection $e_{1}^{1}, e_{2}^{1}, \ldots$ a countable collection of \(1\)-cells. The boundary “\(0\)-sphere” of each \(1\)-cell is just a two-point set; so attach the \(1\)-cell by sending one end of $e_{i}^{1}$ to $e_{0}^{0}$ and the other to $e_{i}^{0}$. The picture is a little spiky star-shaped graph, with countably infinitely many edges coming out of a single vertex. Call this $X$.
Although I haven’t talked extensively—and I don’t want to—about the topology on $X$, it should be intuitively clear that any open neighborhood of the center vertex contains points from each edge. So, supposing $X$ were a metric space, we could choose a sequence of points in $X$, one from each $e_{i}^{1}$, such that the \(n\)th point is at distance $\frac{1}{n}$ from $e_{0}^{0}$. These points converge in the sense of the previous post to $e_{0}^{0}$ in the metric. But! notice that the collection of these points actually meets each closed cell in a closed set, namely a single point. Since the a sequence converges—if at all—to a limit point of the sequence thought of as a set, this is a contradiction. Therefore $X$ is not metrizable.
Continuous injections into metric spaces
I want to prove the following:
Theorem. Every countable CW complex of dimension $N$ admits a continuous injection into $\mathbb{R}^{k(N)}$ for some increasing function $k(N)$ of $N$.
It turns out that if $X = X^{N}$ is additionally locally finite, the map we construct is an embedding, but I won't prove it. When the map is an embedding (i.e. a homeomorphism onto its image), our previous observation allows us to give $X$ the structure of a metric space. When we have merely a continuous injection, we still have a metric on the set $X$, but the metric topology and the CW topology may differ, as the previous example shows. What is true from the construction is that the metric topology is “weaker” or “coarser”, in the sense that it has fewer open sets.
The function $k(n)$ we will use may be defined equivalently as $k(n) = \frac{1}{2}(n+1)(n+2)$ or as the function satisfying $k(0) = 1$ and $k(n) = k(n-1) + n + 1$. The optimal function turns out to be $k(n) = 2n + 1$ for Big Theorem reasons whose proof I haven’t attempted to understand.
The construction is inductive and explicit: points of $e^{n+1}_{i}$ may be written as $t\vec v$ for $t \in [0,1]$ and $\vec v \in S^{n}$. If $f^{n+1}_{i} \colon S^{n} \to X^{n}$ is the attaching map and $\iota_{n} \colon X^{n} \to \mathbb{R}^{k(n)}$ is the function inductively supposed to exist, let $\vec v_{k(n) + 1}$ denote the point of $\mathbb{R}^{k(n+1)}$ whose \((k(n) + 1)\)st coordinate is $1$ and all other coordinates are $0$.
The function on $e^{n+1}_{i}$ defined as $t\vec v \mapsto (t\iota_{n}f^{n+1}_{i}(\vec v), (1-t)t\vec v, (1-t))$ has the property that it agrees with $\iota_{n}$ for points in the boundary \(n\)-sphere of $e^{n+1}_{i}$ and is injective. To make distinct \(n+1\)-cells have distinct image, we will add to this function the function $t\vec v \mapsto 2 i (1 - t) \vec v_{k(n) + 1}$. (The $2$ is there because the diameter of the unit disc is $2$.)
Therefore we may inductively define $\iota_{n+1}$ as agreeing with $\iota_{n}$ on $X^{n}$ and otherwise being the sum of the functions above. Checking continuity is straightforward.