Timeline for When does a map in the stable homotopy group gets killed when smashed with cone of itself?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2015 at 11:21 | vote | accept | Prasit | ||
| May 17, 2015 at 18:42 | comment | added | Prasit | 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. | |
| May 17, 2015 at 11:08 | comment | added | John Rognes | @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 $b>0$. | |
| May 17, 2015 at 2:50 | comment | added | Prasit | 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. | |
| May 16, 2015 at 7:39 | history | answered | John Rognes | CC BY-SA 3.0 |