Timeline for Unifying "cohomology groups classify extensions" theorems
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22 events
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| Nov 7, 2022 at 7:40 | comment | added | David Corwin | This much older question is related: mathoverflow.net/questions/118044/… As mentioned in the answer there, (1)/(2) and (3)/(4) are related by a universal coefficient exact sequence. | |
| Nov 9, 2019 at 16:54 | vote | accept | CommunityBot | moved from User.Id=30211 by developer User.Id=481663 | |
| Nov 9, 2019 at 15:40 | answer | added | skd | timeline score: 17 | |
| Nov 9, 2019 at 14:53 | answer | added | Tim Porter | timeline score: 3 | |
| Nov 9, 2019 at 4:48 | comment | added | Phil Tosteson | I don't know the details, but the old-school approach is to define cohomology for algebras over a monad (or triple) using canonical resolutions. I think Barr and Beck gave an interpretation of $H^1$ or $H^2$ in this context. According to its introduction, the book "Simplicial methods and the interpretation of “triple” cohomology" gives a Yoneda style description of all of the $H^i$. Probably from a modern perspective these results can be reinterpreted in terms of the cotangent complex as Denis suggests. | |
| Nov 9, 2019 at 2:10 | answer | added | Student | timeline score: 1 | |
| Nov 9, 2019 at 1:26 | comment | added | user30211 | @JulianRosen thanks! fixed. | |
| Nov 9, 2019 at 1:25 | history | edited | user30211 | CC BY-SA 4.0 |
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| Nov 9, 2019 at 1:25 | comment | added | Julian Rosen | In (1), I think you want $0\to N\to L\to M\to 0$. | |
| Nov 9, 2019 at 0:52 | history | edited | user30211 | CC BY-SA 4.0 |
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| Nov 8, 2019 at 23:36 | comment | added | user30211 | @tj_ thanks- I will fix this tonight | |
| Nov 8, 2019 at 21:53 | comment | added | skd | @Qfwfq and OP: OK, I'll write an answer shortly. Sorry for the brevity. | |
| Nov 8, 2019 at 20:01 | history | edited | user30211 | CC BY-SA 4.0 |
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| Nov 8, 2019 at 18:58 | comment | added | Paul Siegel | In analysis: K-homology classifies extensions of C* algebras. | |
| Nov 8, 2019 at 17:52 | comment | added | Denis Nardin | Yeah, I think the answer is basically the general theory of the cotangent complex (and the observation that cohomology = certain Ext groups), but someone should write down a proper explanation of it | |
| Nov 8, 2019 at 16:57 | comment | added | user30211 | @skd. Can you explain a bit more? What about the quillen cohomology example, which is from the derived category of rings (one's choice of simplicial rings / DGAs / $E-\infty$ ring spectra)? Maybe I see... we are considering the composition of the derived functor of $\text{Hom}_R(-, N)$ and the derived abelianization functor. And I still need to work out why we get extensions of $G$ in group cohomology and not extensions of $R[G]$. | |
| Nov 8, 2019 at 16:56 | comment | added | Qfwfq | @skd: what's the explanation for the cohomology degree being 1 for 1) and 2 for 3) while both classify "two step" extensions? Is the H^2 figuring in 3) actually an Ext^1 in the appropriate abelian category? (You may have already answered this in your comment but I'm not knowledgeable in homotopy theory..) | |
| Nov 8, 2019 at 16:42 | comment | added | Qfwfq | Also, 2) and 3) are related by being Ext^k in categories of sheaves on a space: Spec(R) for 2), and BG for 3). | |
| Nov 8, 2019 at 16:32 | comment | added | skd | These are just Ext's in some abelian category, i.e., cohomology of derived Hom in some derived category. For instance, group cohomology H^i(G, M) is Ext^i(Z, M) in Z[G]-modules. Cohomology is just extensions: for example, if HZ denotes the Eilenberg-Maclane spectrum, then H^n(X; Z) = [X, K(Z,n)] = [X_+, S^n HZ], where X_+ is the suspension spectrum of X. This can be thought of as "Ext^n(X_+, HZ)" in spectra, but it's really just pi_{-n} of the mapping spectrum Map(X_+, HZ). Not sure if there's anything more general that can be said. | |
| Nov 8, 2019 at 16:32 | history | edited | user30211 |
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| Nov 8, 2019 at 16:26 | history | edited | user30211 | CC BY-SA 4.0 |
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| Nov 8, 2019 at 16:13 | history | asked | user30211 | CC BY-SA 4.0 |