9

I was asked to prove that $$\mathbb{CP}^{2}\#\overline{\mathbb{CP}^{2}}\not\cong \mathbb{S}^{2}\times \mathbb{S}^{2}$$ as fibre bundles over $\mathbb{S}^{2}$ with fibre $\mathbb{S}^{2}$. Since the above connected sum is as manifolds instead of as bundles, it is not clear to me how to prove it via elementary methods.

Here are some thoughts:

1) Can I prove them non-homotopic as manifolds? If they are homotopic as fibre bundles, then they should be homotopic as manifolds as well. Therefore, since all the algebraic invariants on the right hand side is relatively easy to compute, I should be able to solve the problem. However, it is not clear to me how the underlying holomorphic structure of $\overline{\mathbb{CP}^{2}}$ influence the structure. For example, since $\mathbb{CP}^{2}, \mathbb{S}^{2},\overline{\mathbb{CP}^{2}}$ are all simply connected, we cannot get any information from $\pi_{1}$. For $\pi_{2}$ we know $\pi_{2}(\mathbb{S}^{2}\times \mathbb{S}^{2})=\pi_{2}(\mathbb{S}^{2})\times \pi_{2}(\mathbb{S}^{2})=\mathbb{Z}\times \mathbb{Z}$ via Hurewitz's theorem. But computing $$\pi_{2}(\mathbb{CP}^{2}\#\overline{\mathbb{CP}^{2}})$$ seem to be nontrivial even if one use Ryan Budney's method (https://mathoverflow.net/questions/93282/homotopy-groups-of-connected-sums) because they are both simply connected. We can also compute the homology, but it is not clear to me what kind of space is $\mathbb{CP}^{2}-\mathbb{D}^{4}$. If we view $\mathbb{CP}^{2}\cong \mathbb{CP}^{1}\cup \mathbb{C}^{2}$ with the attaching map given by $\mathbb{S}^{3}\rightarrow \mathbb{S}^{2}$, then remove $\mathbb{D}^{4}$ seems to be giving us $\mathbb{CP}^{1}$ back. But this is not rigorous.

Using the hints at here (Computing the homology and cohomology of connected sum) it seems at $H_{2}$ level they again coincide: the result is $\mathbb{Z}\oplus \mathbb{Z}$.

The most important thing is I noticed I did not use the conjugate relationship in the above arguments at all. So there must be something missing here. I guess I can try to compute $\pi_{3}$ and $\pi_{4}$ as well. Since they are CW Complexes, they must differ at some point otherwise it would violate Whitehead's thoerem.

2) Can I prove this via cohomological methods like Chern class? Again, I do not know how to compute it...

Community
  • 1
Bombyx mori
  • 19,638
  • 6
  • 52
  • 112
  • Could you describe how this connected sum of projective spaces forms a $S^{2}$-bundle over $S^{2}$? As you remarked, the manifolds in question are simply connected, so in fact - since I assume you computed homology - you already know the $\pi_{2}$ group by Hurewicz theorem. It's clear it's enough to prove that these manifolds are non-homotopic, but as you observed the usual invariants seem to fail. I tried to compute the intersection form in cohomology but in seems that even those agree. I'll be happy to see some solution. – Piotr Pstrągowski Apr 20 '13 at 11:54
  • @PiotrPstragowski: $\mathbb{CP}^{2}$ itself is already an projective line bundle over $\mathbb{CP}^{1}$. So I think maybe the connected sum can "fuse" the fibre on two bundles and give a new bundle with a twisted fibre. But this is only a rough guess, I do not know. This is why I can only do the usual computation in above. – Bombyx mori Apr 20 '13 at 18:58
  • The complex projective plane is not a projective line bundle over $\mathbb{CP}^{1} \simeq S^{2}$. The long exact sequence of homotopy tells you this is impossible. – Piotr Pstrągowski Apr 21 '13 at 00:19
  • @PiotrPstragowski: I have not think of this over, I thought by the inclusion map $\mathbb{CP}^{1}\rightarrow \mathbb{CP}^{2}$ I can find the fibre over $\mathbb{CP}^{1}$ by $(\lambda u,\lambda v,\lambda w)\rightarrow (\lambda u,\lambda v)$. But this map cannot map $(0,0,1)$ to anywhere and I feel something is wrong. – Bombyx mori Apr 21 '13 at 00:30
  • @PiotrPstragowski: May I ask which fibration over $\mathbb{CP}^{1}$ you are talking about? – Bombyx mori Apr 21 '13 at 00:32
  • I'm just saying there are no fibrations $\mathbb{CP}^{1} \rightarrow \mathbb{CP}^{2} \rightarrow \mathbb{CP}^{1}$. – Piotr Pstrągowski Apr 21 '13 at 11:21
  • @PiotrPstragowski: I see. I need to think about it. – Bombyx mori Apr 21 '13 at 16:18

1 Answers1

5

It seems that the intersection form on second cohomology is indeed enough to distinguish these 4-manifolds, as long as you use integer coefficients. (I'm not sure why I did not think of that earlier, since by Freedman's results this is the first thing one should check.)

All cohomology is with $\mathbb{Z}$ coefficients. Your manifolds are orientable, so a choice of orientation on each of them induces an intersection form

$H^{2}(M) \times H^{2}(M) \rightarrow H^{4}(M) \rightarrow \mathbb{Z}$

given by cup product and then evaluation on the fundamental class. (See: Wikipedia - Intersection form).

The intersection form of $\mathbb{CP}^{2}$#$(-\mathbb{CP}^{2})$ is, up to a sign, given by

$\left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)$,

intuitively since the square of the generator of $H^{2}(\mathbb{CP}^{2})$ is equal to $1$ and square of the generator of $H^{2}(-\mathbb{CP}^{2})$ is -1 because of the reversed orientation.

On the other hand, the intersection form of $S^{2} \times S^{2}$ is, up to a sign, given by

$\left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)$,

since if $h_{1}, h_{2}$ are the generators of $H^{2}(S^{2})$ of the first, resp. second copy of $S^{2}$, then $h_{1} \cup h_{2}$ is the generating $4$-cell of $H^{4}(S^{2} \times S^{2})$, but $h_{i}^{2} = 0$ for dimensional reasons. Since a general element of $H^{2}(S^{2} \times S^{2})$ can be written as $ah_{1} + bh_{2}$ with $a, b \in \mathbb{Z}$ and

$(ah_{1} + bh_{2})^{2} = a^{2}h_{1} ^{2} + b^{2} h_{2}^{2} + 2ab h_{1} h_{2} = 2ab$

one sees that there is no element of $H^{2}(S^{2} \times S^{2})$ that squares to an odd number, in stark contrast with the intersection form on $\mathbb{CP}^{2}$#$(-\mathbb{CP}^{2})$. This proves that these two manifolds are not even homotopy equivalent, so in particular not diffeomorphic.

Piotr Pstrągowski
  • 5,283
  • Let me think about this for a while. I am not that familiar with intersection forms, so need to review.... – Bombyx mori Apr 20 '13 at 19:00
  • I think you have already hinted at a (slightly) easier way to tell these manifolds apart: the cup product structure (indeed, a complete invariant of (the homotopy clas of) simply connected 4-manifolds). –  Apr 20 '13 at 19:00
  • @SteveD: Thanks! I looked it up and this seems to be right. I still do not really understand why $H^{2}(\overline{\mathbb{CP}^{2}})$ has cup product negative -1, though. – Bombyx mori Apr 20 '13 at 23:36
  • The intersection form is really just a cup product in disguise. The point is that multiplying elements from $H^{2}$ gives you actual cohomology classes in $H^{4}$, not numbers. It is the choice of a generator of $H^{4}$ (ie. choice of an orientation) that tells you how to convert these classes into numbers. This is why the intersection form on $H^{2}(\mathbb{CP}^{2})$ is positive and negative in $H^{2}(-\mathbb{CP}^{2})$. – Piotr Pstrągowski Apr 21 '13 at 00:16
  • 3
    Right, but the intersection forms you gave are in fact conjugate in $GL(2,\mathbb{Q})$. It just seems easier - at least to me - to remark that for $S^2\times S^2$, no element of $H^2$ squares to a generator of $H^4$, while this does hold true in the other manifold. –  Apr 21 '13 at 00:23
  • @SteveD: May I ask why $GL(2,\mathbb{Q})$ is involved? I did not thought about that... – Bombyx mori Apr 21 '13 at 00:31