Timeline for Endomorphism ring spectrum of the Eilenberg-MacLane spectrum
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 8, 2017 at 18:56 | vote | accept | Martin Frankland | ||
| May 8, 2017 at 18:44 | answer | added | Tyler Lawson | timeline score: 32 | |
| May 8, 2017 at 18:20 | comment | added | Martin Frankland | For $p=2$, one version of the statement can be found in [H.J. Baues, The algebra of secondary cohomology operations; Theorem 4.6.5], though phrased in a different language. | |
| May 8, 2017 at 18:03 | comment | added | Martin Frankland | @Denis Nardin: No, thank you for your comments! They're helping me clear up the situation. Let me look for a reference to a more precise statement of "known not to be..." | |
| May 8, 2017 at 18:01 | comment | added | Denis Nardin | @Martin Oh now I understand the problem. Of course the map $R\to R$ is not $R$-linear in the example at hand. I still believe that it factors through that map when $R$ is $E_\infty$, unless you have a reference that it doesn't, although I don't see an immediate proof... Sorry for being so dense :) | |
| May 8, 2017 at 17:44 | comment | added | Martin Frankland | Yes, I think that's what I had in mind. For $R$ an $E_{\infty}$ ring spectrum, I don't see that $S$-linear endomorphisms $R \to R$ would be $R$-linear, so I don't see why the composition map $\mathrm{End}_S(R) \wedge_S \mathrm{End}_S(R) \to \mathrm{End}_S(R)$ would factor through the natural map $\mathrm{End}_S(R) \wedge_S \mathrm{End}_S(R) \to \mathrm{End}_S(R) \wedge_R \mathrm{End}_S(R)$. Are you claiming that it does? | |
| May 8, 2017 at 17:36 | comment | added | Tom Goodwillie | When you have a ring map $k\to R$, do you say that $R$ is an $k$-algebra? I think that most of us do not, unless $k$ is commutative and its image is in the center of $R$. In other words, by a $k$-algebra (for a commutative ring $k$) we mean a $k$-module equipped with a $k$-bilinear multiplication. At least I am guessing that Martin would say this, and that that's what Denis needs to know to clear up the confusion. | |
| May 8, 2017 at 17:28 | comment | added | Denis Nardin | I am afraid I do not understand. I am claiming $\mathrm{End}(H\mathbb{F}_p)$ is, in a canonical way, an $H\mathbb{F}_p$-algebra (and more generally that $\mathrm{End}_S(R)$ has a canonical $R$-algebra structure), so I don't understand how can $\mathrm{End}(H\mathbb{F}_p)$ be "known" not to be an algebra over $H\mathbb{F}_p$. What am I missing? | |
| May 8, 2017 at 17:26 | comment | added | Martin Frankland | Well, $\mathrm{End}_S(H\mathbb{F}_p)$ is certainly an $H\mathbb{F}_p$-module. The issue is about composition of endomorphisms being "$H\mathbb{F}_p$-linear in both variables". Does that clarify the ambiguity? | |
| May 8, 2017 at 17:20 | comment | added | Denis Nardin | Precisely. In fact for any $E_1$-ring spectrum $R$, an $R$-module structure on $M$ is the same thing as an $E_1$-ring map $R\to \mathrm{End}(M)$ (if I recall correctly this is how Lurie defines the $E_1$-structure on $\mathrm{End}(M)$). | |
| May 8, 2017 at 17:20 | history | edited | Martin Frankland | CC BY-SA 3.0 |
added 4 characters in body
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| May 8, 2017 at 17:18 | comment | added | Martin Frankland | @Denis Nardin: You mean the analogue of the map in plain algebra that would send a scalar $s$ to the module endomorphism "multiply by $s$", right? | |
| May 8, 2017 at 17:07 | comment | added | Denis Nardin | I am a bit confused: since $H\mathbb{F}_p$ is a $H\mathbb{F}_p$-module spectrum, then there is a map of $E_1$-algebras $H\mathbb{F}_p\to \mathrm{End}(H\mathbb{F}_p)$. | |
| May 8, 2017 at 17:01 | history | asked | Martin Frankland | CC BY-SA 3.0 |