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Feb 22, 2022 at 22:33 answer added Fernando Muro timeline score: 5
Feb 22, 2022 at 22:25 answer added Tim Campion timeline score: 6
Feb 22, 2022 at 21:03 comment added Tim Campion Also, Prasit's Main Theorem 1.3 in the paper you linked seems to do the sort of thing you want, doesn't it? It gives a lower bound for the associativity of $M(p^r)$ when $r$ is large. If I'm reading it right, it says that as $r \to \infty$, the associativity level $n$ of $M(p^r)$ (with $p$ fixed) increases to $\infty$ as well. So a fortiori, the same is true after truncation. Are you really hoping that $n$ actually reaches $\infty$ at some finite $r$ or something?
Feb 22, 2022 at 20:56 comment added Tim Campion Would you also be interested in higher associativity / commutativity of truncated Moore spaces? Not that I have any insight either way, but if you're ultimately interested in something about $\pi$-finite spaces, then maybe truncated Moore spaces are relevant too?
Feb 22, 2022 at 20:12 comment added KotelKanim @FernandoMuro. This much is clear to me, but thanks for making it explicit. I am mainly interested in results that are uniform in $p$ and $d$ and allow arbitrarily large $r$. So for $d=1$, the question is what can be said about $M(2^r,1)$.
Feb 22, 2022 at 15:17 comment added Fernando Muro The reason why you don't get anything for $p=2$, $r=1$, and any $d\geq 1$ is the same as in mathoverflow.net/questions/87919/…
Feb 22, 2022 at 14:25 comment added Fernando Muro For $p$ odd, $r\geq 1$, and $d=1$ the truncation is the same as for $d=0$ hence you get an $E_\infty$-structure. The only non-trivial case would then be $p=2$.
Feb 22, 2022 at 14:16 history edited KotelKanim CC BY-SA 4.0
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Feb 21, 2022 at 15:20 history asked KotelKanim CC BY-SA 4.0