What is the homotopy type of the smash power of Moore spectra $(S/2)^{\otimes n}$?

Question

Let $S/2$ be the mod $2$ Moore spectrum, and let $n \in \mathbb N$.

Question: What is the homotopy type of the $n$th smash power $(S/2)^{\otimes n}$?

Notes:

• When $p$ is odd, we have $S/p \otimes S/p = S/p \oplus \Sigma S/p$, and so when $p$ is odd, we have $(S/p)^{\otimes n} = (S/p) \otimes \otimes^{n-1} (S\oplus \Sigma S) = \oplus_{i = 0}^{n-1}\binom{n-1}{i} \Sigma^i S/p$.

• But for $p = 2$, the above formula fails for $n = 2$. In other words, $2 \neq 0$ as an endomorphism of $S/2$ (though $4 = 0$). We can see that $S/2 \neq (S/2) \oplus (\Sigma S/2)$ because $H^\ast(S/2;\mathbb F_2)$ has a $Sq^1$, so by the Cartan formula $H^\ast(S/2 \otimes S/2; \mathbb F_2)$ has a $Sq^2$, but $(S/2) \oplus (\Sigma S/2)$ doesn't have a $Sq^2$.

• I seem to remember that there is a formula for $n = 3$ involving $S/\eta$. If this is the case, then just by looking at mod $2$ homology it would have to be something like $(S/2)^{\otimes 3} = (S/2) \otimes ((S/\eta) \oplus \Sigma S \oplus \Sigma S)$, but I'm not sure how to convince myself that this is actually the case.

This question is related to another old question of mine. And a couple of similar questions has been asked before, but the answers only go so far as showing that $S/2 \otimes S/2$ does not split as $S/2 \oplus \Sigma S/2$ -- no positive results about understanding $(S/2)^{\otimes n}$ are mentioned.

Answer 1

Firstly, as shown in my answer to your `old question', arbitrary smash products of copies of $S/2$ will decompose into wedges of indecomposable spectra of an infinite number of homotopy types (even after suspensions).

At the prime 2, one can use idempotents in $\mathbb Z_2[\Sigma_3]$ to show any spectrum of the form $X \wedge X \wedge X$ decomposes as a wedge of the form $Y \vee Y \vee Z$.

When $X=S/2$, $Y = \Sigma S/2$ (which is likely what you meant) and $Z$ will have the same mod 2 cohomology (as an $A$-module) as $S/\eta \wedge S/2$. (Likely this cohomology module pins down the homotopy type.)

Added later ...

More generally, one can use known information about the modular representation theory of the symmetric groups to show the following: Suppose the mod 2 cohomology of $X$ is two dimensional. The idempotents in the group ring $\mathbb Z_2[\Sigma_n]$ can be used to (naturally) decompose $X^{\wedge n}$ into a wedge decomposition of the form $$ X^{\wedge n} \simeq \bigvee_{i=0}^{[(n-1)/2]} a_n(i) X_n(i)$$ where $a_n(0)=1$ but the other $a_n(i)$'s might be hard to compute. [That indexing set corresponds to 2-regular Young diagrams with two columns: pairs $(i,j)$ with $0 \leq i < j$ and $i+j=n$.]

If $X$ was $\mathbb RP^2$, $\mathbb CP^2$, $\mathbb HP^2$, or the Octonian projective plane, then the spectrum $X_{2^k}(0)$ will have $Sq^{2^{k+c}}$ acting nontrivially, with $c=0,1,2,3$ corresponding to the four examples. I don't know if these pieces further decompose or are atomic. Maybe $H^*(X_n(i); \mathbb Z/2)$ doesn't decompose as a direct sum as an $A$-module in our four examples.