Timeline for The homotopy of universal Thom spectrum
Current License: CC BY-SA 3.0
14 events
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| Jun 9, 2014 at 9:48 | comment | added | Neil Strickland | @Justin: thanks, I was just confused; it is the $p$'th space of the $E_2$ operad that is relevant, not the $E_p$ operad. | |
| Jun 9, 2014 at 8:22 | history | edited | Justin Noel | CC BY-SA 3.0 |
There was a mistake: Neil Strickland's observation about the contractibility of holds for odd primes
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| Jun 8, 2014 at 15:13 | comment | added | Justin Noel | @Prasit: I did not fully understand your comment, but I will reiterate how my answer relates to your question. The case in your question is $MSL_1(S_p)$ whose $\pi_0$ is torsion-free. Since I suspected you meant to ask about $MGL_1(S_p)$ I included that case. That spectrum is universal in the sense that it is contractible and hence terminal. I am claiming that the homotopy groups of $MG$ are $\bZ/p$-modules and calculating them is equivalent to calculating the homology as as a comodule over the dual Steenrod algebra. I did not claim that the latter problem was easy; it is just algebraic. | |
| Jun 8, 2014 at 13:46 | comment | added | Prasit | In the particular case that I asked in the question, there is a map from $K\mathbb{Z}/p \to MG$ which is a $A_2$ which also implies $MG$ is wedge sum of $K\mathbb{Z}/p$. @Justin: From what you said it is cear that $MG$ or $\mathcal{M}$ is wedge summand of $K\mathbb{Z}/p$, one for each of $\mathbb{Z}/p$ basis of the homotopy. However, however it is not clear what the homotopy of $MG$ or $\mathcal{M}$ is. I would like to conclude that the homotopy of $\mathcal{M}$ is the same $S^0_p$ as $\mathbb{Z}/p$. But I do not know if it is true. | |
| Jun 8, 2014 at 11:08 | comment | added | Justin Noel | @NeilStrickland: This is an unpublished result of Hopkins and Mahowald. It appears as Thm 4.12 here: nullplug.org/publications/p-torsion.pdf . | |
| Jun 8, 2014 at 11:07 | comment | added | Justin Noel | @Prasit: $R$ is a wedge of suspensions $H\mathbb{Z}/p$'s so you get one $\mathbb{Z}/p$ in the homotopy groups for each summand and for each such summand you get a copy of the dual Steenrod algebra in the homology. | |
| Jun 8, 2014 at 10:25 | comment | added | Neil Strickland | @Justin: you say that $E_2$ is enough instead of $E_\infty$. Do you have a reference for that? I think I know an argument for $E_p$, but not for $E_2$. | |
| Jun 8, 2014 at 3:27 | comment | added | Prasit | How do one show that "The generators of these copies of the dual Steenrod algebra are in 1-1 correspondence with generators of $\pi_*(R)$ as a graded $\mathbb{Z}/p$-vector space." | |
| Jun 7, 2014 at 20:41 | vote | accept | Prasit | ||
| Jun 7, 2014 at 11:36 | history | edited | Justin Noel | CC BY-SA 3.0 |
Further fill out the answer following helpful comments from Neil Strickland.
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| Jun 7, 2014 at 10:35 | comment | added | Justin Noel | Thanks Neil! I was in a bit of a rush. I will complete my response and add your comments. | |
| Jun 7, 2014 at 7:27 | comment | added | Neil Strickland | We could instead use an intermediate space $G$ with $\pi_0(G)=1+p\mathbb{Z}_p$ and then we would get an $E_\infty$ Thom spectrum $\mathcal{M}$ with $\pi_0(\mathcal{M})=\mathbb{Z}/p$. Any such $E_\infty$ ring spectrum is known to be a wedge of suspensions of $H\mathbb{Z}/p$, so if you could calculate the homology, you would have a collapsing Adams spectral sequence for the homotopy. | |
| Jun 7, 2014 at 7:22 | comment | added | Neil Strickland | In the original question $GL_1(S^0_p)$ was defined to be a connected component, so the tensor product should be over $\mathbb{Z}$ and the answer is $\mathbb{Z}_p$. If we instead take $GL_1(S^0_p)$ to be the union of invertible components, then we get what you wrote, but $\mathbb{Z}_p^\times$ is acting trivially on $\mathbb{Z}$ and by multiplication on $\mathbb{Z}_p$, so the tensor product is zero. | |
| Jun 7, 2014 at 5:38 | history | answered | Justin Noel | CC BY-SA 3.0 |