Let $M$ be a closed, smooth $4$-manifold with integral cohomology ring isomorphic to that of $\mathbb{CP}^2$, is it diffeomorphic to it?
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$\begingroup$ Do you have any reason to believe that this would be true? $\endgroup$– user2520938Commented Sep 21, 2020 at 11:27
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1$\begingroup$ If the manifold has a Morse function with three critical points, then by Eells-Kuiper it is homotopy equivalent to $CP^2$. In particular, it is simply connected and then by Freedman homeomorphic to $CP^2$. The question of exotic smooth structures on $CP^2$ is open, albeit a conjecture of Fintushel and Stern says there are infinitely many. $\endgroup$– ThiKuCommented Sep 21, 2020 at 11:33
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2 Answers
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No. If $\Sigma$ is any homology 4-sphere with non-trivial fundamental group, $\mathbb{CP}^2 \# \Sigma$ is a homology $\mathbb{CP}^2$ with non-trivial fundamental group. (Here $\#$ denotes connected sum.) There are many examples of homology 4-spheres: for instance, Kervaire produced many examples in his paper Smooth homology spheres and their fundamental groups.**
Whether there is any simply connected 4-manifold with the same homology as $\mathbb{CP}^2$ but that is not diffeomorphic to it, is an important open question, which is somewhat related to the 4-dimensional smooth Poincaré conjecture.
** The main statement is about homology $n$-spheres for $n$ strictly larger than 4, but the remark at the end of the introduction mentions that the constructive part of the proof works for $n = 4$ as well.
answered Sep 21, 2020 at 11:29
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$\begingroup$ Sorry, i do not understand your example since I required the space to be a 4-dimensional manifold. $\endgroup$– Nick LCommented Sep 21, 2020 at 11:32
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2$\begingroup$ @MarcoGolla Do you know if in addition we say X is irreducible and co-homology ring isomorphc to CP^2 then is it homeomorphic to CP^2? $\endgroup$ Commented Sep 21, 2020 at 12:34
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1$\begingroup$ I don't know off hand. I think it's a good question, but I also think it's hard to check irreducibility. (Note that irreducibility must be defined with more care than in the 3-d case, since there might be non-trivial homotopy 4-spheres that are invertible.) $\endgroup$ Commented Sep 21, 2020 at 13:30
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2$\begingroup$ @KevinCasto: almost. Freedman's theorem implies that if X is a simply-connected smooth 4-manifold with the same cohomology as $\mathbb{CP}^2$, then it is homeomorphic to it. (However, there is a simply-connected, non-smoothable topological 4-manifold with the same cohomology as $\mathbb{CP}^2$.) $\endgroup$ Commented Sep 21, 2020 at 17:47
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This was originally going to be a comment to Marco Gallo's answer, but got too long. I figured since this provides an alternative solution to the problem, I'll post it as an answer.
An alternative answer to this question are the fake projective planes, these are algebraic surfaces which have the same cohomology as $\mathbb{CP}^2$, but nontrivial fundamental groups. They have universal cover the complex unit ball which topologically is just $\mathbb{R}^4$.
I offer this additional answer since they provide examples very different to Marco Galla's answer and answer a question of Anubhav Mukherjee's in the comment. Either they are irreducible or have an irreducible summand that is a cohomology $\mathbb{CP}^2$.
Suppose $X$ is one of the fake projective planes. If $X$ is irreducible, we're done. If $X$ is reducible, then it admits a connect sum decomposition $X = Y \# \Sigma$ where $Y$ is a cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ that is not a homotopy $S^4$. Again if $Y$ is irreducible and not homeomorphic to $\mathbb{CP}^2$, we're done. If $Y$ is reducible, repeat this process again until you get either an irreducible cohomology $\mathbb{CP}^2$. This process must terminate by compactness.
This gives a connect sum decomposition which we'll write as $X = Y \# \Sigma$ (slightly abusing notation here) where $Y$ is an irreducible cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ which is the connect sum of all of the homology $S^4$'s we found. If $Y$ is not $\mathbb{CP}^2$, we're done.
Now we see that $Y$ can not be $\mathbb{CP}^2$. Suppose $Y = \mathbb{CP}^2$ and consider the universal cover $f: \tilde{\Sigma} \rightarrow \Sigma$. Take a point $p \in \Sigma$ and it's inverse image $f^{-1}(p)$. We can now construct a cover $f^*: Y \# |f^{-1}(p)|\mathbb{CP}^2 \rightarrow \Sigma \# \mathbb{CP}^2 = X$ by connect summing a $\mathbb{CP}^2$ at each point in $f^{-1}(p)$. This covering space is simply connected since it's summands are simply connected and so it is the universal cover. This universal covering space has 2nd homology coming from the $\mathbb{CP}^2$ summands. This contradicts $X$ having universal cover $\mathbb{R}^4$. Thus $Y$ is not $\mathbb{CP}^2$ and is an irreducible cohomology $\mathbb{CP}^2$.
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$\begingroup$ Can you please tell me the description of such algebraic surfaces which are fake projective spaces? Or any references? $\endgroup$ Commented Sep 23, 2020 at 2:02
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6$\begingroup$ Kai, the fake projective planes only have the rational cohomology of $CP^2$, they have torsion in $H_1$. See Theorem 9.1 arxiv.org/abs/math/0512115v2 $\endgroup$– Ian AgolCommented Sep 23, 2020 at 3:25
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$\begingroup$ Fake projective planes are irreducible. More generally, aspherical manifolds of dimension at least three are always irreducible. $\endgroup$ Commented Sep 24, 2020 at 23:21