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Consider an element $e \in \pi_n^s(S^0)$, in the stable homotopy groups of sphere. Let $C$ denote the spectrum which is cone of $e$, i.e. $C$ fits in the cofiber sequence $$ S^n \to S^0 \to C.$$
$\textbf{Question:}$ I want to know under what condition $e$ viewed as a self map of $C^{\wedge k}$ (the $k$-fold smash product of C) is null-homotopic, in other words how do we detect if $e \wedge 1_{C^{\wedge k}}$ is homotopically trivial.

Let me expand on the case when $k=1$.

We have a cofiber sequence $$S^n \wedge C \to S^0 \wedge C \to C \wedge C $$ where the left most map is $e \wedge 1_C$. If $e \wedge 1_C$ is trivial then the $C$ splits of $C \wedge C$.

I believe that if $e \cup_1 e \neq 0$ then $e \wedge 1_C$ is non-trivial. The reason being (I think) $e \cup_1 e$ is the non-trivial attaching map for the top-cell of $C \wedge C/\Sigma_2$ to the $0$-cell. Let me know if I am stating something incorrect. I think the converse should be true as well.

$\textbf{Question:}$ If $e \cup_1 e \neq 0$ then can we say that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k>1$ as well?

Further let me give an example, let $e = 2: S^0 \to S^0$. We know that $2$ is a non-trivial as a self-map of $M_2$ (the cone of $2$), the reason being $2 \cup_1 2 = \eta$.

$\textbf{Question:}$ Is $ 2 \wedge 1_{M_2^{\wedge k}}$ non-trivial for all $k$?

I believe the answer should be yes. Does anyone know of a proof?

$\textbf{Question:}$ Is there an example of an $e$ such that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k <n$ but homotopically trivial for $k = n$, where n > 1?

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    $\begingroup$ Just starting to realize that considering $C \wedge DC$ as a ring might just resolve the issue for $k>1$. $\endgroup$
    – Prasit
    Commented Feb 26, 2015 at 2:48
  • $\begingroup$ I always though 2 was trivial on $M_2$... Very interesting. $\endgroup$
    – user43326
    Commented Feb 26, 2015 at 21:02
  • $\begingroup$ So, if I understood correctly, we have $[MZ/2,MZ/2]\cong Z/4$ and not $Z/2\oplus Z/2$? $\endgroup$
    – user43326
    Commented Feb 27, 2015 at 9:54
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    $\begingroup$ Yes that is correct. $\endgroup$
    – Prasit
    Commented Feb 28, 2015 at 11:49

1 Answer 1

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The answer to your third question is "yes".

A little more generally, if $Sq^{n+1}$ acts nontrivially in the mod $2$ cohomology of $C$, then $Sq^{(k+1)(n+1)}$ acts nontrivially in the mod $2$ cohomology of $C^{\wedge k+1}$, by the Cartan formula. If $e$ smashed with the identity on $C^{\wedge k}$ were null-homotopic, then its mapping cone, which is $C^{\wedge k+1}$, would split in a way that is not compatible with this nontrivial cohomology operation. A similar argument works for $e = \alpha_1$, using the Steenrod operations in mod $p$ cohomology.

I think this kind of argument goes back to Michael Barratt. There is a Cartan formula for secondary cohomology operations, due to Leif Kristensen, which might give similar results for some classes $e$ of Adams filtration $2$, but I have not looked at the details.

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  • $\begingroup$ Thats a very concrete answer. BTW I had some alternate proof in mind. Let me know if I am making a mistake here. If we view $C \wedge DC$ as a ring spectra, say $R$, then the question boils down to whether $e \neq 0 \in \pi_*R$ implies $e_n = e \wedge 1 \wedge \ldots \wedge 1$ is nonzero. And the answer should be yes as $e$ which is nonzero factors through $e_n$. In fact $e = m \circ e_n$ where m is $n$-fold multiplication. $\endgroup$
    – Prasit
    Commented May 17, 2015 at 2:50
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    $\begingroup$ @Prasit: OK, now I understand how you use that $C \wedge DC \simeq F(C, C)$ is a ring spectrum, with multiplication corresponding to composition. So you have proved that the answer to your fourth question is "no". More generally, for any essential (= not null-homotopic) map $e : X \to Y$ the smash product of $e$ with $a$ copies of the identity on $X$ and $b$ copies of the identity on $Y$ will be essential, where we assume that $X$ is finite (= dualizable) if $a>0$, and that $Y$ is finite if $b>0$. $\endgroup$ Commented May 17, 2015 at 11:08
  • $\begingroup$ I am now curious if the proof that you mentioned for the case when $e = 2$, can be massaged to answer the fourth question of mine. $\endgroup$
    – Prasit
    Commented May 17, 2015 at 18:42

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