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Does there exist a closed 4-manifold $X$ such that $\pi_2(X)$ contains torsion elements?

And, if so, does there exist a closed 4-manifold $X$ such that $\pi_2(X)\neq 0$ but $\pi_2(X)\otimes \mathbb{Q}=0$? What about compact 4-manifolds with boundary?

This seems to be a basic question about the topology of 4-manifolds, but I can't seem to find a quick answer to any part of this question or find any existing work on this in the literature. May I ask if there is any previous work on this question, or if there is a short answer? Any hint either in the smooth category or the topological category would be greatly appreciated. Thanks a lot!

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    $\begingroup$ If $X$ is a closed $4$-manifold such that $\pi_2(X)$ has torsion, then $\pi_1(X)$ is infinite. Indeed, if $\pi_1(X)$ is finite, its universal cover $Y$ is compact, and $\pi_2(X)\cong\pi_2(Y)\cong H_2(Y)\cong H^2(Y)\cong Hom(H_2(Y),\mathbb Z)$ using Hurewicz, Poincare duality and universal coefficients. Note that $Hom$ is finitelty generated free abelian. $\endgroup$
    – Igor Belegradek
    Commented Mar 15 at 2:25
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    $\begingroup$ Remark 6.3 of this paper seems to indicate the question was open for closed 4-manifolds as of 2008: arxiv.org/pdf/0802.0995.pdf $\endgroup$
    – Anthony Conway
    Commented Mar 15 at 2:57
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    $\begingroup$ Remark 6.3 is worded as if the authors are unsure about the exact relationship to $H^2(\pi,\mathbb Z\pi)$ being free abelian. What is the precise relationship between the two questions? As in my previous comment $\pi_2(X)$ is isomorphic to $H^2_c(Y)$, the compact support cohomology of the universal cover. If $Y$ is contractible, then $H^2_c(Y)$ is isomorphic to $H^2(\pi,\mathbb Z\pi)$ where $\pi=\pi_1(X)$, but I don't know if this holds in general. Maybe the classifying map $X\to B\pi$ can be used somehow. $\endgroup$
    – Igor Belegradek
    Commented Mar 15 at 15:31
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    $\begingroup$ Essentially $H^2(\pi;\mathbb{Z}\pi)$ injects into $H^2_c(\widetilde{X})$. Indeed one has $\pi_2(X)=\pi_2(\widetilde{X})=H_2(\widetilde{X})=H_2(X;\mathbb{Z}\pi)=H^2(X;\mathbb{Z}\pi)=H^2_c(\widetilde{X})$ and as mentioned in their sequence (3.1), the UCSS shows that $H^2(\pi;\mathbb{Z}\pi)$ injects into $H^2(X;\mathbb{Z}\pi)=H^2_c(\widetilde{X})$. $\endgroup$
    – Anthony Conway
    Commented Mar 15 at 18:58

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