This might be well known for algebraic topologist. So I am looking for an explicit example of a 4 dimensional manifold with fundamental group isomorphic to the rationals $\mathbb Q$.
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8$\begingroup$ What do you mean by "explicit"? You can take presentation complex for $Q$, immerse it in $R^4$ and then pull-back regular neighborhood. The result is your manifold. $\endgroup$– MishaCommented Jul 1, 2013 at 13:37
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$\begingroup$ See [this stackexchange question and answers.][1] [1]: math.stackexchange.com/questions/36775/… $\endgroup$– Igor RivinCommented Jul 1, 2013 at 17:12
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1 Answer
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Since any compact manifold has the homotopy type of a finite CW-complex (see this MathOverflow question: Are non-PL manifolds CW-complexes?) and $\mathbb{Q}$ is not finitely presented, the manifold $X$ you are looking for is necessarily non-compact.
An explicit construction of a non-compact three-manifold $M$ with $\pi_1(M)=\mathbb{Q}$ can be found in the paper
B. Evans and L. Moser: Solvable Fundamental Groups of Compact 3-Manifolds, Transactions of the American Mathematical Society 168 (1972), see in particular page 209.
Now it suffices to take $X=M \times \mathbb{R}$.
answered Jul 1, 2013 at 14:09