# Horrible sets and blowups in Hubbard's Teichmuller theory

Edit: I can rephrase this question this way: When blowing up every point in the $x$-axis in $\mathbb{C}^2$ by means of an inverse limit of finite blowups, how can anything be 'left over'? The horrible set described by Hubbard must consist of points on the axis that are not blown up, but every point on the axis is blown up on the way to the inverse limit, since the index set is a directed set. It's not like the inverse limit only considers finitely many blowups at a time; every point in the inverse limit lives in every blowup. Why is there a horrible set?

## Original Question:

I've had this question for some time. In Hubbard's Teichmuller theory book, on page 9, he describes an ugly complex 2-manifold that is not second countable. He constructs it by taking $\mathbb{C}^2$ and blowing up every point along the axis $\mathbb{C} \times 0$. More specifically, he considers all blowups of finitely many points along this axis, and takes their inverse limit under the natural projection maps from one blowup to another which has a strictly smaller subset of blowup points. He notes that there is a natural map $p$ from the blown-up space to $\mathbb{C}^2$.

He denotes $p^{-1}(\mathbb{C} \times 0)$ by $Y$ and claims that $Y$ consists of one copy of projective space for each point of the complex line plus a 'horrible set'. Here's my question:

Where does the horrible set come from?

Because every point in the inverse limit is given by $a=\mathop{\Pi} \limits_{\alpha \in J} a_\alpha \in \mathop{\Pi} \limits_{\alpha \in J} X_\alpha$ with $p_{\alpha\beta}(a_\alpha)=a_\beta$ whenever $\alpha<\beta$ (here the $X_\alpha$ are all the blowups with partial order given by inclusion of the blow up points and the p maps are the natural projections). So, any point in the inverse limit has a coordinate $a_0$ in the non-blown up space $X_0=\mathbb{C}^2$ (the minimal element in the ordering). There is an $\alpha$ corresponding to the blowup of the point $a_0$, and in $X_\alpha$, we must have that $a_\alpha$ is some element of the copy of projective space created in the blowup. This means that our point is in the 'nice' set, and not in the horrible set. So where does the horrible set come from?

## Example as written in Hubbard

A 2-dimensional complex manifold that is not second countable

We will describe a connected complex manifold of dimension 2 that is not second countable. This manifold is a close analog of Example 1.3.1 [which was the classic non-second countable surface], but the elementary "cut and past" approach used there doesn't work so well in higher dimensions, so we will instead use a description in terms of blow-ups. (For blow-ups, see Hatshorne, Shafarevich; see page 30 of Thurston's Three-Dimensional Geometry and Topology for an informal introduction. But readers who don't know about blow-ups can skip this example; it has no further applications in this book).

In $\mathbb{C}^2$, we will blow up every point of $\mathbb{C}\times${0}. More specifically, for any finite subset $Z\subset \mathbb{C}$, we denote by $$\widetilde{\mathbb{C}_Z^2}$$

the blow-up of $\mathbb{C}^2$ at all the points of $Z$, and set $X:=\mathop{\lim}\limits_{\mathop{\leftarrow}\limits_Z} \widetilde{\mathbb{C}_Z^2}$, where the finite subsets are partially ordered by inclusion. (For inverse limits, see Hatcher). There is a natural map $p:\widetilde{\mathbb{C}_Z^2} \rightarrow \mathbb{C}^2$.

This space $X$ is not a manifold, the inverse image $Y:=p^{-1}(\mathbb{C}\times${0}) consists of the disjoint union $Y_1$ of uncountable many copies $\mathbb{P}^1_z$ of $\mathbb{P}^1$, one for every $z\in \mathbb{C}$, and some horrible set $Y_2=Y-Y_1$. The set $Y_2$ is not closed; it accumulates on exactly one point of each $\mathbb{P}_z^1$, namely, the point corresponding to the horizontal direction through $z$.

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This is not quite the same as your question, but may be Riemann-Zariski space is helpful. By definition, the latter is the "cohomology classes" you obtain when taking limit of the cohomology groups of finite blowups of a given compact Kahler surface. –  nono Nov 13 '12 at 15:13
@nono I think the key idea he uses is that we only blow up in one direction, so that most elements of the blowups (i,e. all the ones not pointing in that direction) are not close to each other. Is there a version of the Riemann Zariski space that only looks at blowups on a lower-dimensional subset? –  Brian Rushton Nov 13 '12 at 23:19
It does seem as if there is no "horrible set". However, if one removes from $X$ the "strict transform" of $\mathbb{C} \times 0$, i.e., the set of horizontal directions on each $\mathbb{P}^1_z$, then one does get a complex manifold which is not second countable so the example still works. –  ulrich Nov 18 '12 at 6:57