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 similar question 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.

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.

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

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 similar question 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.