Timeline for Is there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?
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17 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2011 at 14:00 | comment | added | Akhil Mathew | Interesting. Thanks for letting me know. | |
| Dec 11, 2011 at 1:23 | comment | added | Dylan Wilson | @Akhil: According to him this representability question will be the subject of DAG XIV. Though I'm sure that his elliptic cohomology paper/book will be equally as enlightening. | |
| Dec 11, 2011 at 0:51 | comment | added | Akhil Mathew | @Dylan: This sounds very interesting, and I'm looking forward to hearing more when Lurie posts this stuff. Perhaps you're referring to the forthcoming exposition on elliptic cohomology referred to in his survey? (Thanks for the kind words about my web-page, though I'm afraid I understand almost nothing about DAG at this point!) | |
| Dec 10, 2011 at 23:58 | comment | added | Dylan Wilson | (I apologize for the my vagueness and unclear exposition- I'm working under the assumption that Akhil has pretty much seen or at least glanced at many of Lurie's papers, mainly due to his wonderful webpage: people.fas.harvard.edu/~amathew/dag.html ) :) | |
| Dec 10, 2011 at 23:55 | comment | added | Dylan Wilson | (contd) the thing doing the representing would have to be like an infinite-dimensional artin stack... and I actually haven't seen the exact hypotheses on Lurie's representability theorem, so I don't know if you could apply it in this case. He says he's trying to post this before the new year, but until then the only reference I know is his original thesis, but this treats the case where E_infty rings are replaced by SCRs, so I'm not sure if the same proof applies. There also finite-dimensionality hypotheses... | |
| Dec 10, 2011 at 23:54 | comment | added | Dylan Wilson | @Akhil: Yes that's the one. The one with the more obvious title seems to just be an announcement of results. On another note, it sounds like you're expressing a purely formal groupy way of constructing MU. I don't think such a thing is written down but I'm tempted to think one could attempt the same idea as in Lurie's work on elliptic cohomology: define what should be a "formal group" or "derived formal group" over an E_infty ring, then define the obvious functor, show that it's represented by something reasonable, and take global sections. The only part that I think would be awkward is that | |
| Dec 10, 2011 at 23:44 | comment | added | Akhil Mathew | (contd.) Yet $MU$ is universal among complex-oriented cohomology theories, and the fact that its formal group law is the universal one seems more than mere coincidence. But of course I'm a beginner to this field and my opinions should be taken very seriously... | |
| Dec 10, 2011 at 23:42 | comment | added | Akhil Mathew | @David: Thanks for doing that! I've been trying to get somewhere in Lurie's notes, but I think his proof is essentially a slightly fancier restatement of the usual proof (i.e. the Adams spectral sequence computation, except that he only carries it out at $p = 2$). In particular, one of the things that feels somehow "morally wrong" about it (to me, at least) is that the result seems to fall out by coincidence: that is, $\pi_* MU$ is computed to be a polynomial ring that happens to match the Lazard ring, and the map $L \to \pi_* MU$ is checked on indecomposables to match. | |
| Dec 10, 2011 at 23:39 | comment | added | Akhil Mathew | @Eric: Thanks for your additional comments, which I enjoyed reading. | |
| Dec 10, 2011 at 23:38 | comment | added | Akhil Mathew | @Dylan: Do you mean "Elementary proofs of some results in cobordism theory"? I haven't read that, but I thought the original one didn't have any proofs. | |
| Dec 10, 2011 at 20:32 | comment | added | Eric Peterson | ... Really, for all its successes and results, stable homotopy theory is still in these senses a young field. @David: Those are exactly the notes that spurred Akhil's question, referenced in the comment thread above. High level for sure, but not high enough! :) | |
| Dec 10, 2011 at 20:31 | comment | added | Eric Peterson | I don't think there is something that primal yet available; instead, we have this deep sea of computations that permit many guesses, some of which turn out to be true. For instance, MU is designed to be universal among complex-oriented spectra with a choice of coordinate on the formal group. You could hope that the constraint "among complex-oriented spectra" is empty, and instead MU carries a coord. universal among all formal groups. That this is true is another shocking freeness result, but we don't yet have a construction of a spectrum that starts with that data from scratch, so to speak... | |
| Dec 10, 2011 at 19:41 | comment | added | David White | @Akhil: My academic brother thinks about things like this a lot, and I've sent him the link to your question. Perhaps this will finally convince him to join MO! Anyway, I think he once told me something about Lurie having a nice clever proof of this fact about $\pi_*MU$ and if it's Lurie I'm sure some good high level explanation is given. Here is the best link I can find, to some lecture notes my brother has worked through last year. You can just change the "Lecture8" to "Lecture7" or another number to see the setup: math.harvard.edu/~lurie/252xnotes/Lecture8.pdf | |
| Dec 10, 2011 at 17:55 | comment | added | Dylan Wilson | It might not add any insight, but quillens original proof of this fact used no homotopy theory, except to know that the coefficient ring was of finite type... You should read his paper in any case | |
| Dec 10, 2011 at 16:26 | comment | added | Akhil Mathew | Thanks. I guess I'm looking for something really primal here. For instance, I don't know a high-concept explanation that $\pi_* MU$ should be the Lazard ring! As far as I know, that goes through the Steenrod algebra and the Adams spectral sequence. | |
| Dec 10, 2011 at 5:02 | history | edited | Eric Peterson | CC BY-SA 3.0 |
added 416 characters in body
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| Dec 10, 2011 at 4:55 | history | answered | Eric Peterson | CC BY-SA 3.0 |