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Sep 27, 2023 at 5:42 vote accept Jordi Daura
Sep 26, 2023 at 1:20 answer added Ian Agol timeline score: 3
Sep 23, 2023 at 7:54 comment added Jordi Daura @IanAgol The motivation of this question comes precisely from the paper of Cappell, Weinberger and Yan you linked. I was trying to prove that if the center of the fundamental group is large (for example, $\mathcal{Z}(\pi_1(M))\cong \mathbb{Z}^{n-1}$ where $n$ is the dimension of $M$) then I have an action of the torus $T^{rank\mathcal{Z}(\pi_1(M))}$ on $M$.
Sep 22, 2023 at 18:16 comment added Ian Agol Not sure if relevant, but such an example has universal cover Euclidean space. doi.org/10.1016/0040-9383(75)90034-8 Also, there exists aspherical manifolds with non-trivial center and no circle action, disproving a conjecture of Conner-Raymond. doi.org/10.1112/jtopol/jtt023 One could ask a weaker question though: If $Z(\pi_1(M)) \neq 0$, then does $\pi_1(M)/\mathbb{Z}$ look like an $n-1$-dimensional aspherical orbifold group? If so, one might be able to rule out $\mathbb{Z}$ center.
Sep 21, 2023 at 15:19 comment added Moishe Kohan I am sure there is no trivial reason and a proof would be hard and require new ideas.
Sep 21, 2023 at 8:47 comment added Jordi Daura Thanks for the suggestion. I edited the title accordingly so the positive answer is what I would like to be true, that there always exists an element of infinite order on the inner automorphism group (unless $\pi_1$ is abelian).
Sep 21, 2023 at 8:44 history edited Jordi Daura CC BY-SA 4.0
edited title
Sep 20, 2023 at 9:17 comment added YCor It might be a bit confusing that you ask the question in one direction in the title ("can it be periodic") and the opposite direction in the body (is there an element of infinite order"), so it's unclear what "a positive answer" should be. Anyway I think it's an open question as well in this context.
S Sep 20, 2023 at 9:03 review First questions
Sep 20, 2023 at 10:32
S Sep 20, 2023 at 9:03 history asked Jordi Daura CC BY-SA 4.0