Timeline for A canonical way to kill a subset of cohomology in a dg-algebra: via $A_\infty$-algebras? References?
Current License: CC BY-SA 3.0
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| Jan 28, 2013 at 6:57 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Jan 26, 2013 at 11:51 | comment | added | Akhil Mathew | This paper of Tate gives examples of killing cohomology classes in cdgas: projecteuclid.org/…. | |
| Jan 26, 2013 at 10:25 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Jan 26, 2013 at 7:25 | comment | added | Mikhail Bondarko | I have no particular set of generators in $S$; $S$ is a 'large' subgroup in $H^\ast(A)$. Also see the update of the question. Thank you very much! | |
| Jan 26, 2013 at 7:23 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
I updated the question keeping in mind my discussion with Tyler Lawson.
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| Jan 26, 2013 at 5:28 | comment | added | Tyler Lawson | If you are in a good setting (maybe you have something commutative enough, with an ideal given by a regular sequence) then you can often construct a quotient that looks nice on the cohomology level, but there may multiple inequivalent ways to do it. | |
| Jan 26, 2013 at 5:22 | comment | added | Tyler Lawson | Mikhail: The shortest quotient would be to take some kind of homotopy pushout; the category of modules would then be some kind of category of dg $A$-modules $M$ equipped with, for each $s \in S$, a chain homotopy from multiplication-by-$s$ to zero. But this only has a "versal" property, and it depends on the actual set of generators (e.g. if you use different generators or include redundant generators in $S$) you get a different and likely inequivalent algebra. This has kind of lousy properties. My own feeling is that quotients are just hard in the derived setting. | |
| Jan 26, 2013 at 5:06 | comment | added | Mikhail Bondarko | Also, I am rather interested in the category of modules over the algebra obtained than in the algebra itself. | |
| Jan 26, 2013 at 5:02 | comment | added | Mikhail Bondarko | I would expect the quotient to have some kind of universal property; yet I am not sure that this is possible. I don't know much about these things; any hints or references would be very welcome! What is 'the best' quotient in your opinion? | |
| Jan 26, 2013 at 4:56 | comment | added | Tyler Lawson | Are you looking for the resulting map on cohomology to be some kind of quotient map? Are you looking for the quotient to have some kind of universal property? Are you looking for something else? For many definitions of "quotient" there is almost always a nontrivial moduli space of them. | |
| Jan 26, 2013 at 4:51 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Jan 26, 2013 at 4:41 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |