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Reading Perelman's preprint(1991) Alexandrov space II now. Got confused about the last section 6.4, which contains an example which indicate that the statement ".... manifold is diffeomorphic to the normal bundle over soul" in Cheeger-Gromoll's Soul Theory won't hold for Alexandrov spaces. I also read BBI's book(A course in Metric Geometry) (page 400-401) which also contains the description of this example. I am using the notation in BBI's book, the example goes as follows:

Let $\pi: K_0(\mathbb{CP}^2)\to K_0(\mathbb{CP}^1)$ be the projection. $\bar{B}_0(1)$ be the unit ball in $\mathbb{CP}^1$ (Note: here should be $K_0(\mathbb{CP}^1$), right?). Let $X^5=\pi^{-1}(\bar{B}_0(1))$. Take double of $X^5$ and it will be the example

The picture in my mind is $K_0(\mathbb{CP}^1)$ is sub-cone of $K_0(\mathbb{CP}^2)$, so the projection is the projection on the second factor if we write the coordinate in cone as $(t, x)$ for $t\in \mathbb R$ and $x\in \mathbb{CP}^2$.

My question is what is the topology of $X^5$?

1) Is $\bar{B}_0(1)$ a close ball? If so then $\bar{B}_0(1)$ will have boundary $\mathbb{CP}^1$, right? and $X^5$ will be a closed cone over $\mathbb{CP}^2$, right?

2) Is $X^5$ compact?

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    $\begingroup$ If you are interested in errata to BBI, it is now on my home page (see my profile). I included misprints that I see in this example. $\endgroup$
    – Sergei Ivanov
    Commented Apr 27, 2010 at 19:59

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No $X^5$ is not a cone over $CP^2$ and is not compact. The projection has nothing to do with the cone structure. In fact, it's better to forget about the cone structure altogether (until you ask what is the topology of the thing).

The spaces are just quotients of $\mathbb C^3$ and $\mathbb C^2$ by the standard circle action, and the projection is induced by the coordinate projection $\mathbb C^3\to \mathbb C^2$. For example, there is a whole half-line in the pre-image of the origin.

I suggest you visualize a similar construction with $\mathbb R$ in place of $\mathbb C$ or lower the dimensions by 1 (or both) to see what is going on.

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  • $\begingroup$ I see, so $X^5= \{(z_1, z_2, z_3| |z_1|^2+|z_2|^2\le 1, z_1, z_2, z_3\in \mathbb C\}/ \{S^1 action\}$, right? $\endgroup$
    – J. GE
    Commented Apr 21, 2010 at 21:57
  • $\begingroup$ @MG: yes. And indeed, too many misprints there. I'll try to find the latest errata (it used to be on YuB's site but somehow disappeared). $\endgroup$
    – Sergei Ivanov
    Commented Apr 22, 2010 at 6:50

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