Is the mod-2 Moore spectrum a retract of a shift of its tensor square?

Question

Let $M_p(i)$ be the mod $p^i$ Moore spectrum, i.e. the cofiber of $p^i: \mathbb S \to \mathbb S$. Upper and lower bounds on the $n$ for which $M_p(i)$ admits an $A_n$ structure are known, cf. Bhattacharya. I gather from this that $M_p(i)$ admits at least an $A_2$ structure for all primes $p$ and $i \in \mathbb N$, except for the mod-2 Moore spectrum $M_2(1)$, which does not admit an $A_2$ structure.

One consequence of a spectrum $X$ having an $A_2$ structure is that $X$ is a retract of $X\wedge X$. If $M_2(1)$ were a retract of $M_2(1) \wedge M_2(1)$, then the retract map would be an $A_2$ structure, so that can't happen.

But the Spanier-Whitehead dual of $M_p(i)$ is $\Sigma^{-1} M_p(i)$, so by a triangle equation we have that $M_p(i)$ is always a retract of $\Sigma^{-1} M_p(i)^{\wedge 3}$.

So it seems like there is conflicting evidence for the resolution of the following

Question: Is the mod-2 Moore spectrum $M_2(1)$ a retract of $\Sigma^n M_2(1) \wedge M_2(1)$ for some $n \in \mathbb Z$?

Answer 1

The mod $2$ cohomology of $S^0/2 \wedge S^0/2$ is a $\mathbf{F}_2$-vector space on generators in degrees 0, 1, 1, and 2. The classes in degrees 0 and 2 are connected by a nontrivial $\mathrm{Sq}^2$, so you cannot split $S^0/2$ off (any shift of) $S^0/2 \wedge S^0/2$. The topological version of this statement is the fact that there is a cofiber sequence $$S^1 \xrightarrow{2 \vee \eta} S^1 \vee S^0/2 \to S^0/2 \wedge S^0/2,$$ where the map $S^1\to S^0/2$ is given by composing $\eta:S^1\to S^0$ with the inclusion of the bottom cell of $S^0/2$. However, $S^0/p$ does split off $S^0/p \wedge S^0/p$ for $p$ odd, because the top cell (in dimension 2) is not connected to the bottom cell.