I only know through stories that mod 3 moore spectrum is not associative. I do not know of any proof. I have been informed that Toda had proved it in the paper "Extended $p^{th}$ power". I was not able to follow it. Can anybody give me a proof that mod 3 Moore spectrum is not associative?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ This is also Lemma 6.2 in Toda's On spectra realizing exterior parts of the Steenrod algebra (sciencedirect.com/science/article/pii/0040938371900176), which you might find helpful. (I don't know if this proof is any different from the one in the Toda reference you mention, which I don't have available.) $\endgroup$– Eric PetersonCommented Jan 15, 2013 at 20:47
-
2$\begingroup$ That paper of Toda's is a little concentrated. The nature of the answer to this question will probably depends pretty heavily on whether you know what a Massey product is. You can show that there must be a map from $M$ to $H\mathbb{Z}/3$ which preserves the unit and multiplication. The resulting map $H_* M \to H_* H\mathbb{Z}/3$ has as image a square-zero class in the dual Steenrod algebra ($\tau_0$) whose triple Massey product $\langle \tau_0, \tau_0, \tau_0 \rangle$ is not in the image. $\endgroup$– Tyler LawsonCommented Jan 16, 2013 at 4:42
-
$\begingroup$ Hi @TylerLawson. Could you say a bit more about the shape of the argument? You are saying that if $M$ had an $A_{\infty}$-structure, and the above map is induced from an $A_{\infty}$-map then for all elements in the image, their triple Massey products (a representative of which is given by the $A_{\infty}$-structure) must always be in the image because - is this correct? $\endgroup$– Elden ElmantoCommented Jul 11, 2014 at 20:39
Add a comment
|