Timeline for The second stable homotopy group
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2023 at 16:49 | history | edited | Leo | CC BY-SA 4.0 |
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| Mar 21, 2023 at 16:33 | vote | accept | Leo | ||
| Mar 21, 2023 at 10:27 | comment | added | Tom Goodwillie | I do not think that your proposed splitting map can be a homomorphism, because there cannot be a natural splitting. If there were, then the map $A\hat{\otimes }A\to A\wedge A$ in Fernando Muro's answer would have a natural right inverse $s$ that is a homomorphism. But by universal example one can work out what the natural maps $t:A\hat A\to A\hat{\otimes }A$ are, and none of them does the job. | |
| Mar 20, 2023 at 17:43 | answer | added | Fernando Muro | timeline score: 11 | |
| Mar 19, 2023 at 20:18 | comment | added | Achim Krause | @ChrisSchommer-Pries, I presume your $\wedge$ is a $\vee$. Unfortunately, the suspension of $h$ splits, using the James splitting and the observation that the Hopf invariant of $\eta$ is $1$. So the suspension spectrum of the cofiber of $S^2\to \Omega S^2\vee S^2$ just splits into a bunch of spheres. | |
| Mar 19, 2023 at 18:18 | comment | added | Chris Schommer-Pries | @AchimKrause what happens if you attach a 3-cell to $\Omega S^2 \wedge S^2$ by $(h, 2)$ where "h" is the adjoint of the hopf map. This is an unstable analog of what you suggest. Does it split in this case? | |
| Mar 19, 2023 at 14:50 | history | edited | Leo | CC BY-SA 4.0 |
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| Mar 19, 2023 at 14:37 | comment | added | Leo | @AchimKrause Thanks! Indeed, if my argument in the updated version of my question is correct, it doesn't work for spectrum $X$. | |
| Mar 19, 2023 at 14:33 | history | edited | Leo | CC BY-SA 4.0 |
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| Mar 18, 2023 at 19:00 | comment | added | Achim Krause | Here's a small observation: For a $1$-connective spectrum, the corresponding sequence does not necessarily split, as can be seen by attaching a $3$-cell to $\mathbb{S}^1\oplus \mathbb{S}^2$ along $(\eta,2)$. So if this holds for spaces, it is an unstable phenomenon, so the Adams spectral sequence is probably not helpful (on its own). | |
| Mar 18, 2023 at 18:15 | comment | added | Leo | @მამუკა ჯიბლაძე Hello! I decided to remove the same question on math.stackexchange, so let me address your question here. Differentials of the type $H_2(X,\pi^s_i)\to H_0(X,\pi^s_{i+1})$ are trivial due to the canonical splitting $\mathbb{S}_{\bullet}(X)=\pi^s_{\bullet}\oplus\pi^s_{\bullet}(X)$. The other differential $H_3(X,\pi^s_0)\to H_1(X,\pi^s_1)$ is the composition of the mod-2 reduction and the dual of $\mathrm{Sq}^2$, which is trivial due to the dimensional reason. | |
| Mar 18, 2023 at 18:00 | comment | added | Leo | @JohnPalmieri Thank you for the suggestion. Let me stick to mathoverflow and remove that question. | |
| Mar 18, 2023 at 16:57 | comment | added | John Palmieri | I think it's better to post a question to just one of mathoverflow and math.stackexchange, not to both at the same time: math.stackexchange.com/questions/4661599/…. | |
| Mar 18, 2023 at 16:22 | history | edited | Leo | CC BY-SA 4.0 |
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| Mar 18, 2023 at 15:53 | history | asked | Leo | CC BY-SA 4.0 |