Morgan Shalen compactification of $\mathbb C^2$

Question

I'm reading the Otal's survey on the compactification of Morgan Shalen. (available here)

He claims in an example (page 8) that the compactification of $\mathbb C^2$ is $S^4$, which sounds completely natural, but I'm not able to understand why, and if I do the calculations I obtain a different object.

Here the definition:

Let $X$ be an affine algebraic set and take a finite (or countable) generator set $F$ of the ring of the regular functions on $X$. In the example I'm interested in $X=\mathbb C^2$ and the coordinates $z,w$ are my generating set.

Let $[0,\infty)^F$ and denote by $\mathbb P^F$ its projectivized, with $\pi$ the natural projection. In my example, $F$ has two elements so $\mathbb P^F$ is a closed segment.

Define $\theta_0:X\to [0,\infty)^F$ by $\theta_0(x)=(\log(|f(x)|+2))_{f\in F}$ and

$\theta=\pi\circ\theta_0$.

Let $\hat X$ be the one-point compactification of $X$.

The MS compactification of $X$, w.r.t. $F$ is the closure of the graphic of $\theta(X)$ in $\hat X\times\mathbb P^F$

In the example the function $\theta$ is $$\theta(z,w)=\dfrac{\log(|z|+2)}{\log(|w|+2)}$$

By studying the level sets of such function it seems to me that the compactification of $\mathbb C^2$ is a singular object, while in the survey is claimed that it is $S^4$.

In particular the closure of a level set in $\hat X\times \mathbb P^F$ is the one-point compactification of the level set itself. Level sets are $|z|+2=(|w|+2)^c$ so at infinity they are of the form $T^2\times[a,\infty)$ whose one-point compactification is the cone over $T^2$, which is singular. As level sets are disjoint this would show that the compactification of $\mathbb C^2$ is not a manifold.

My question is: Is there any place where i can find the proof that the MS compactification of $\mathbb C^2$ is $S^4$? Or, can anyone give some hint?

Answer 1

Not true. Let $S^1$ be the unit circle in $\mathbb C$. Let $\Delta$ be the closure, in the projective plane, of the quadrant $Q=[log\ 2,\infty)\times [log\ 2,\infty)\subset\mathbb R^2$. It is topologically a $2$-simplex.

Map the product $S^1\times S^1\times\Delta$ to the one-point compactification of $\mathbb C\times \mathbb C$ by giving a proper map from the dense open subset $S^1\times S^1\times Q$ to $\mathbb C\times \mathbb C$, namely $$ (a,b,(s,t))\mapsto ((e^s-2)a,(e^t-2)b). $$ Map $S^1\times S^1\times\Delta$ also to the projective segment $\mathbb P^F$ by $$ (a,b,(s,t))\mapsto t/s. $$ The combined map $S^1\times S^1\times\Delta\to \mathbb C\times \mathbb C\times \mathbb P^F$ displays the space you are asking about as a quotient of $S^1\times S^1\times\Delta$. Now look at which points have been identified. For a point $p$ in the interior of the ``infinity'' side of $\Delta$, $S^1\times S^1\times p$ goes to one point, and a neighborhood of that point in the quotient space looks like the product of an open interval and a cone on $S^1\times S^1$. Thus a neighborhood looks like a cone on the suspension of $S^1\times S^1$, and is not a manifold.

Answer 2

I will not answer probably to the question but i want to report my computation here, since maybe you can help me.

I perfectly know that the compactification is relative to a complex variety, but for a moment let's forget that and we try to do the same construction for $\mathbb{R}^2$. We consider the generating set $F=\{x,y\}$ given by coordinates and we consider the closure of the image of the function $$\theta:\mathbb{R}^2 \rightarrow S^2 \times \mathbb{P}^F, \hspace{10pt} \theta(x,y)=((x,y),[\log(|x|+2):\log(|y|+2)]).$$

If we want to close this image, i think we need to add only point to each level set of the function $$\varphi(x,y)=\frac{\log(|y|+2)}{\log(|x|+2)}$$ and two more points $\{\infty\} \times [0:1],\{\infty\} \times [1:0]$. This is equivalent to determine a suitable identification of the boundary of the disk $D^2$, as reported in this picture https://drive.google.com/file/d/0B16wiL4cSf4zbVpRcVNzaGJTYmc/view?usp=sharing.

More precisely, the points with the same color are identified on the boundary. In this way you are adding the whole space $\mathbb{P}^F \cong [0,\infty]$ and identifying the upper-right quarter of the boundary with the other ones following the identification drawn in picture. So, the space $\overline{X}^F$ should be a singular quotient of the 2-sphere $S^2$.

If we define the projection on the second factor $\pi_2:X^F \rightarrow \mathbb{P}^F$ we can consider the preimage $\pi_2^{-1}(a)$ of a point $a \in [0,\infty] \cong \mathbb{P}^F$. The picture reported below describes the topological behaviour of the preimage https://drive.google.com/file/d/0B16wiL4cSf4zRHhLTjE0Q2NnVjQ/view?usp=sharing.

Is this computation correct? Are there any mistakes in my reasoning?I met the same problems studying the MS compactification. In particular i do the same computation as above and i get something singular.

If there were no mistake in my computation, we could be able to describe at least the preimage $\pi_2^{-1}(a)$ also for the complex case. I think the result could be the following https://drive.google.com/file/d/0B16wiL4cSf4zUlJSVEZQSGYxcTg/view?usp=sharing. From this considerations, i can't see how to prove that $\overline{C^2}^F$ is homeomorphic to $S^4$.

Thanks in advance for your help.