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Many manifolds can be obtained from gluing the boundary of a ball. For example, $\mathbb{RP}(2)$ is obtained from gluing the two edges of a bi-gon (2-ball). Or, lens spaces are obtained from a 3-cell whose boundary consists of two $p$-gons, which are then glued under a rotation by $q$ edges.

Question: Is there a similar construction for the complex projective plane $\mathbb{CP}(2)$? Ideally, is there a way to divide the boundary of the 4-ball into two identical 3-cells, which are then glued under some symmetry? Or is there at least such a gluing construction with $4$ or more 3-cells?

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    $\begingroup$ One very weak way to interpret the general question: "Is there some surjection $B^4 \to \Bbb{CP}^2$ which is a homeomorphism on the interior and finite-to-one on the boundary?" $\endgroup$
    – mme
    Commented Aug 22 at 17:31
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    $\begingroup$ All closed connected compact n-manifolds can be glued from a ball. If the manifold is PL, take a triangulation and knock out codimension-1 simplices to get a cell complex with a single n-cell , then its boundary consists of pairs of simplices glued together. In the topological case there are results about the existence of handle structures for which one may do a similar thing. $\endgroup$
    – Ian Agol
    Commented Sep 5 at 15:29
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    $\begingroup$ @IanAgol Yes, that makes sense. I wanted to go the opposite direction, hoping to get a simple "cellulation" for $\mathbb{CP}(2)$ from the answer to this question. In the mean time I think I found one, a 4-cell whose boundary consists of 6 3-cells that are glued pairwise. $\endgroup$
    – Andi Bauer
    Commented Sep 8 at 10:12

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There was a gap in my first answer to the first question, discussed in the comments. I hope it's clear that this doesn't constitute a full answer to OP's question, which remains interesting.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way.

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

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  • $\begingroup$ Thanks a lot for your answer! I can't reproduce your cell counting: For what I have in mind, you'd have one 4-cell, one 3-cell, and the number of 0-cells, 1-cells, and 2-cells depends on the cellulation of the 2-sphere boundary of the 3-sphere, and also on the symmetry we use for identification, doesn't it? $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 14:10
  • $\begingroup$ For example, in the lens space example in one dimension lower, the number of 0-cells and 1-cells depends on the identification/symmetry of the two boundary $p$-gons. For $L(4;1)$ you get a single 1-cell and a single 0-cell, and for $L(4;2)$ (by which I mean an overly complicated cell-gluing construction of $L(2;1)$ with twice as many cells) you get two 1-cells and two 0-cells. Of course there's no Euler characteristic in 3D, I'm just trying to argue that the number of 0-cells, 1-cells, and 2-cells will depend on the identification/symmetry. $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 14:32
  • $\begingroup$ Is there a way to see that the Euler characteristic is still independent on the 2-sphere cellulation and its symmetry? (And am I assuming correctly that you might have just mistyped "two 3-cells" instead of "one 3-cell"?) $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 14:34
  • $\begingroup$ And a last clarification question: The non-cellular construction you're describing is the one where we attach a 2-sphere to the boundary of a 4-ball, corresponding to a CW decomposition with one 0-cell, one 2-cell, and one 4-cell, correct? That's the same as in @MikhailKatz answer, right? $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 14:40
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    $\begingroup$ Cool thanks! I didn't mean to restrict to involutive symmetries, but it's still good to know that it can't work with an involution. $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 17:05
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There is a famous Hopf fibration of an odd-dimensional sphere $S^{2n-1}$. If you quotient out the orbits of the boundary of the $2n$-ball, you get the complex projective $n$-space.

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  • $\begingroup$ Thanks! Can you elaborate on how this answers the question? What you're describing sounds like the CW decomposition of $\mathbb{CP}(2)$ with a 4-ball attached to a 2-sphere. I was looking for a way to glue the boundary points of the 4-ball in a pairwise manner. $\endgroup$
    – Andi Bauer
    Commented Aug 22 at 14:19
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    $\begingroup$ You can't glue them in a pairwise manner, but if you identify each fiber (i.e., a circle) of the Hopf fibration to a point, you get the complex projective space. $\endgroup$
    – Mikhail Katz
    Commented Aug 22 at 14:21

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