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Mar 13, 2021 at 0:48 comment added Tyrone The result I quoted predates Cohen-Moore-Neisendorfer and is due to Toda. I didn't think Neisendorff could prove the CMN stuff in his text (I see now he did but left out the fun parts). Amusingly you also have an easy non-H-space counterexample a dimension lower in $S^2$.
Mar 12, 2021 at 2:41 comment added Nicholas Kuhn @Tyrone You have the Cohen-Neisendorfer-Moore odd prime exponent theorem stated incorrectly and it is stronger than you say: $S^{2n+1}$ has exponent $p^n$. (The $n=1$ theorem is due to Selick, and grounds an inductive proof.) You are right that I forgot that $S^3$ is an H-space, so my conjecture was a bit too quick.
Mar 11, 2021 at 16:16 comment added Tyrone I should add that $S^3$ also has $p$-primary exponent $p^2$ for any odd prime $p$ (this is due to Toda). In general, for $p$ an odd prime, the $p$-primary exponent of $S^{2n+1}$ is $p^{2n}$. Since $S^{2n+1}$ is an H-space when localised at an odd prime, its degree $p^{2n}$ self-map annihilates all $p$-torsion in $\pi_*S^{2n+1}$.
Mar 11, 2021 at 15:58 comment added Tyrone I think the conjecture already fails for $S^3$ which has 2-primary exponent $4$ (this is a result of James).
Mar 11, 2021 at 15:50 comment added Nicholas Kuhn I would even conjecture that, if a simply connected finite complex $X$ has nontrival mod $p$ homology, then any self map of $X$ that is not null after localization at $p$ will induce a nontrivial homomorphism infinitely often on the p primary part of $\pi_*(X)$.
Mar 11, 2021 at 15:42 history answered Nicholas Kuhn CC BY-SA 4.0