Timeline for Model structure of commutative dg-algebras inside all dg-algebras
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2010 at 12:29 | comment | added | Jacob Lurie | The homotopy category definitely doesn't embed fully faithfully. Maybe the easiest example (working over a field C) is to take the usual polynomial ring R = C[x,y]. In commutative dgas this is already cofibrant, so [R,A] = H_0(A) x H_0(A) for any cdga A. In associative dgas a cofibrant replacement for R is the free associative algebra on x,y and z in hom. degree 1 with dz=xy-yx. So if A happens to be a cdga you get [R,A] = H_0(A) x H_0(A) x H_1(A). | |
| Nov 12, 2009 at 14:29 | comment | added | Mark Hovey | I would definitely mean A/A[A,A]A, yes, since we have to get a CDGA out of our DGA A, so we need to mod out by the differential ideal generated by the commutators. The reference you gave refers to the derived functors of this as "higher abelianization", but as you say the discussion is very brief, and there are no references to anywhere it is discussed in depth. Tantalizing! Thanks for the reference. | |
| Nov 11, 2009 at 23:32 | comment | added | Vladimir Dotsenko | You mean A/A[A,A]A, rather than A/[A,A], right? (If we mod out the linear span of commutators, this does not really make thinks commutative [and derived functors compute cyclic homology, I believe].) The derived functors of this are briefly discussed (for some examples) in arxiv.org/abs/math/0610410. | |
| Nov 11, 2009 at 20:44 | comment | added | Tyler Lawson | Secondly, I would be skeptical about CDGAs being a full subcategory of the homotopy category of DGAs, as (from characteristic p considerations) I would have expected there to be multiple inequivalent commutative DGAs whose cofibrant replacements (as associative DGAs) became equivalent. But in characteristic zero there are no Steenrod operations and so I'm not sure how one might find some invariant not detected on the associative level. | |
| Nov 11, 2009 at 15:28 | history | answered | Mark Hovey | CC BY-SA 2.5 |