Timeline for Does there exist a closed 4-manifold whose $\pi_2$ contains torsion elements
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| when toggle format | what | by | license | comment | |
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| Mar 15 at 18:58 | comment | added | Anthony Conway | 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 $H^2(X;\mathbb{Z}\pi)=H^2_c(\widetilde{X})$. | |
| Mar 15 at 15:31 | comment | added | Igor Belegradek | 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. | |
| Mar 15 at 2:57 | comment | added | Anthony Conway | 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 | |
| Mar 15 at 2:25 | comment | added | Igor Belegradek | 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. | |
| S Mar 15 at 1:06 | review | First questions | |||
| Mar 15 at 1:51 | |||||
| S Mar 15 at 1:06 | history | asked | Boyu Zhang | CC BY-SA 4.0 |