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Aug 25 at 4:40 comment added მამუკა ჯიბლაძე Unfortunately several links have died, I wonder if it is possible to revive them...
Sep 12, 2018 at 14:50 history edited skd CC BY-SA 4.0
updated broken link
Mar 1, 2018 at 23:23 comment added Dylan Wilson @skd but the fact that you have an E_2-map before taking Thom spectra means you're at least E_2 but you could be more (on accident), and then it takes some argument to show that you're at most E_2... but ok, I'll stop quibbling
Mar 1, 2018 at 22:47 comment added skd @DylanWilson (i): I think I mean at best, since it's not E_3 (unless my understanding of English is subpar); (ii): it's not relevant, just thought I'd mention that fun fact :P ; (iii): I do mean h(n-1). You're right, of course: it isn't much of a reduction; reducing the reduction is the most interesting part of the whole story.
Mar 1, 2018 at 22:45 comment added Dylan Wilson also (for the OP), aren't there lots of silly things "above" the sphere? like if $X$ is an $E_{\infty}$-space then $\Sigma^{\infty}_+X \to S^0$ by killing $X$ seems to work.
Mar 1, 2018 at 22:40 comment added Dylan Wilson @skd penultimate paragraph: (i) maybe you mean "at least" not "at best" E_2, (ii) I don't see how the E_2-structure etc. is relevant?, and (iii) at the end you mean "h(n-1)" not "h(n+1)", and "reduces to" is technically correct but not much of a reduction... :)
Mar 1, 2018 at 14:09 history edited skd CC BY-SA 3.0
added some references
Oct 23, 2017 at 8:03 vote accept მამუკა ჯიბლაძე
Jul 12, 2017 at 20:50 comment added მამუკა ჯიბლაძე @skd I don't even know what integral K-theory is, I meant just complex K-theory. Its Chern character goes to HQ, and HZ is somewhere nearby, so I just lack vision of how are they situated wrt each other...
Jul 12, 2017 at 20:36 comment added skd @მამუკაჯიბლაძე I don't know, but that's an interesting question. By "global" K, do you mean integral K-theory?
Jul 12, 2017 at 20:35 comment added skd @NicholasKuhn Thanks for pointing that out; I should've said that the restriction of the source to $\Sigma^n_+ \mathbf{RP}^\infty$ is nullhomotopic for every $n$.
Jul 12, 2017 at 19:58 comment added Nicholas Kuhn The comment about the Kahn-Priddy map (aka the transfer associated to the inclusion of the trivial group into the group of order 2) says something that is very very wrong: each of those maps is definitely not nullhomotopic. Also, restricted to the unit, it is multiplication by 2, so it isn't a map of ring spectra, as it isn't unital.
Jul 12, 2017 at 9:13 comment added მამუკა ჯიბლაძე @skd OK here is the first question that was born in me during reading your exciting answer. There are few "global" ring spectra that I see on your map - S, MSpin, MString, MU, KO, TMF, HQ, and those mysterious X(n). Can you tell how MSpin, MString, etc. and those X(n) relate to each other? Could you also indicate where do things like global K, HZ reside? Are there any others? There must also be a global such guy at each height.
Jul 12, 2017 at 9:03 comment added მამუკა ჯიბლაძე @RobertBruner Exciting! Would not it be shocking if every ring spectrum has an algebraic version (something lake AEll(S), AMU(S), ..., AS(S)) and they all sit above S repeating the whole pattern! (Sorry for shameless wishful thinking)
Jul 12, 2017 at 5:43 comment added Robert Bruner There is a very interesting ring above S: the Waldhausen K-theory, A(*) = K(S) augments onto S.
Jul 11, 2017 at 22:45 comment added მამუკა ჯიბლაძე Great! Is it also a map of ring spectra?
Jul 11, 2017 at 22:36 comment added skd @მამუკაჯიბლაძე Sorry; I'd understood the question differently. There are nontrivial maps $X\to S$, of course. For instance, there's a map $\Sigma^\infty_+\mathbf{RP}^\infty\to S$, called the Kan-Priddy map. (This is defined via the composition of the maps $\mathbf{RP}^{n-1}_+ \to O(n)$ sending a line to the reflection it defines and $O(n)\to \Omega^n S^n$, which adjuncts to $\Sigma^n_+ \mathbf{RP}^{n-1}\to S^n$. Each of these maps is nullhomotopic --- but when we send $n\to \infty$, we get a nontrivial map $\Sigma^\infty_+ \mathbf{RP}^\infty\to S$.)
Jul 11, 2017 at 22:35 comment added skd @RobertBruner Thanks! I'll edit that into my notes.
Jul 11, 2017 at 22:00 comment added მამუკა ჯიბლაძე And yes, most probably after having digested your information I will come up with some further questions, thanks.
Jul 11, 2017 at 21:59 comment added მამუკა ჯიბლაძე Concerning "above $S$" - $\mathbb Z$ is initial in rings but some other rings still may have homomorphisms to it, right?
Jul 11, 2017 at 21:56 comment added მამუკა ჯიბლაძე Many thanks for your brilliantly illuminating answer (for me, so it is hardly flattering, as it is not so difficult to illuminate somebody as submerged into darkness as me). And even more thanks for many links to extremely informative sources. I will try to study them.
Jul 11, 2017 at 18:34 history edited skd CC BY-SA 3.0
fixed typos
Jul 11, 2017 at 16:27 comment added Robert Bruner In your notes you say it is unfortunate that a lot of people use ASS to refer to the Adams spectral sequence. Actually, a lot of us scrupulously say Adams ss to avoid this ugly abbreviation.
Jul 11, 2017 at 16:22 history edited skd CC BY-SA 3.0
added references
Jul 11, 2017 at 16:13 history answered skd CC BY-SA 3.0