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Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Can the universal cover of $X$ be contractible?

It semms to me that the answer is negative when $X$ is simply connected using results of Freedman and Donaldson. Is anything known when $X$ is not simply connected? Donaldson proved that also in this case the intersection form is diagonalizable over $\mathbb Z$.

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I don't understand your second paragraph: if $X$ is simply connected, and has non-trivial intersection form, then $X$ is its own universal cover, so it is not contractible. There is no need for appealing to Freedman or Donaldson. –  Ian Agol Dec 27 '11 at 3:38
    
You're right: this second paragraph makes no sense, as it stands. In fact I was thinking of the following question: Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Then the intersection form is diagonalizable over $\mathbb Z$. Does it follow that $X$ is homeomorphic to a connected sum of a manifold $Y$ with trivial intersection form and a finite number of $\mathbb C \mathbb P^2$-s (with direct or reversed orientation)? If this was the case, one could relate the universal cover of $X$ to that of $Y$ and try to show that the first is not contractible. –  Matei Jan 5 '12 at 13:28
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1 Answer

In algebraic geometry, there are examples of "fake projective planes", which in this context means smooth complex surfaces of general type with the same cohomology ring as the complex projective plane. It is known that the universal cover of such spaces is the complex hyperbolic ball. So the answer to your question is yes. (The first such fake projetive plane was shown to exist by Mumford.)

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I don't know of examples where the cohomology ring is that of the connected sum of n projetive planes for n bigger than 1. For small n I think one could look for hyperbolic 4-manifolds, but hyperbolic examples cannot exist for large n. –  Peter Kronheimer Dec 15 '11 at 14:20
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The previous comment is off-base, because hyperbolic 4-manifolds have signature zero and cannot be definite if their intersection form is non-trivial. –  Peter Kronheimer Dec 15 '11 at 14:45
    
Thank you very much for your answer! The situation is more complex than I was imagining. I would be interested indeed to know whether also examples with $n$ bigger than $1$ exist. –  Matei Dec 15 '11 at 21:02
    
I don't think I know any examples with n bigger than 1. I agree that this is an interesting question. –  Peter Kronheimer Dec 15 '11 at 22:11
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