# How can uncountably many closed smooth 4-manifolds be presented by an essentially countable alphabet (Kirby diagrams)?

A smooth, closed 4-manifold admits a handle decomposition which is specified completely by its Kirby diagram. A Kirby diagram, up to isotopy, can be seen as a labelled morphism in the tangle category. Now the tangle category is finitely presented (it is the free ribbon category on one generator), or in other words, one could devise an alphabet and a language in which to write down Kirby diagrams as words.

It seems that it's not known whether there are closed 4-manifolds with uncountably many smooth structures. I don't understand that. It seems like there can only be countably many closed smooth 4-manifolds in total since they can be represented by words in a finite alphabet!

Now I'm aware that there are uncountably many exotic $\mathbb{R}^4$s and these usually require an infinite number of 3-handles or so. I can see that a language containing infinitely large words can be uncountably big. But my question is about closed manifolds, and they should have finite Kirby diagrams.

## 1 Answer

There are not uncountably many closed smooth 4-manifolds. The fact that they can all be represented by a finite handlebody diagram is essentially a proof (plus the fact that there are only countably many isotopy classes of knots on a 3-manifold). Alternatively, in 4 dimensions, DIFF = PL, and so every closed smooth 4-manifold is given by a finite combinatorial triangulation, which are countable.

You can only get uncountably many smooth structures when you work with noncompact 4-manifolds.