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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?

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    $\begingroup$ Please do not "bump" the question by altering the subject line this way. Subject lines should only be used to indicate mathematical content. I'm rolling back. $\endgroup$ – Todd Trimble Apr 3 '15 at 12:57
  • $\begingroup$ I don't believe this statement. The space in question is not a manifold. It seems to be homeomorphic to the pushout $D^4\leftarrow S^3\to D^1$ where the first arrow is the inclusion of the boundary and the second is the map from the join of two circles to the join of two points (induced by mapping each circle to one of the points). Over the interior of the $1$-disk the map from the $3$-sphere is a fiber bundle with fiber the torus $S^1\times S^1$, so that a neighborhood (in the pushout space) of an interior point on the $1$-disk is the cone on the suspension of a torus. $\endgroup$ – Tom Goodwillie Apr 5 '15 at 17:24
  • $\begingroup$ In other words, it looks like you are right. I will elaborate this as an answer when I have time. $\endgroup$ – Tom Goodwillie Apr 5 '15 at 19:24
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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.

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  • $\begingroup$ Since the statement in the notes is not true, do you think there is an "easy" generalization of this construction valid for $\mathbb{C}^2$ to the more general case of $\mathbb{C}^n$? $\endgroup$ – ness1 Apr 6 '15 at 19:05
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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.

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