Timeline for Graded commutativity of cup in Hochschild cohomology
Current License: CC BY-SA 2.5
13 events
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| Dec 12, 2009 at 20:47 | comment | added | Mariano Suárez-Álvarez | You should check, for example, Hilton-Stammbach's book on homological algebra, where everything is explained in detail. Sadly, the comments in MO are forcefully too short and, after a little while, do not get jsMath-love, so they are not a good medium for explaining this. :/ | |
| Dec 12, 2009 at 20:39 | comment | added | darij grinberg | Yes, you did, by your definition of the action on $M^{\mathbb{ad}}$. Anyway, my knowledge of group cohomology is rather... trivial (done a few exercises in Lang), and I'm more interested in the general case of Hochschild cohomology (and Harrison as well). | |
| Dec 12, 2009 at 19:49 | comment | added | Mariano Suárez-Álvarez | No, that is not the differential in the usual bar complex for group cohomology (did I give you that impression?) | |
| Dec 12, 2009 at 19:45 | comment | added | darij grinberg | This is puzzling me. Is the $\delta$ in group cohomology really defined by $\left(\delta f\right) \left(a_1 , ... , a_{n+1}\right) = a_1 f\left(a_2 , ... , a_{n+1}\right)$ $ + \sum\limits_{i=1}^{n} f\left(a_1 , ... , a_{i-1} , a_{i}a_{i+1} , a_{i+2} , ... , a_{n+1}\right) + \left(-1\right)^{n+1} a_{n+1} f\left(a_1 , ... , a_n\right) a_{n+1}^{-1}$ ? This was new to me. Anyway, I don't care too much about the group case; thanks a lot for the help in the case of general algebras. | |
| Dec 12, 2009 at 16:00 | comment | added | Mariano Suárez-Álvarez | If $M$ is a $\mathbb ZG$-bimodule, you can turn it into a left $G$-module $M^{\mathrm{ad}}$ with action $g\cdot m = gmg^{-1}$. Then it is true that $H^\bullet(\mathbb ZG,M)\cong H^\bullet(G, M^{\mathbb{ad}})$. This is explained, for example, in Cartan-Eilenberg, Chapter X, section 4. | |
| Dec 12, 2009 at 15:46 | comment | added | darij grinberg | Sorry, the hom doesn't go to $\mathbb{Z}$, but to the group module $M$ (which is turned into a $\mathbb{Z}G$-bimodule as follows: The left action of $\mathbb{Z}G$ on $M$ is the linearization of the group action that we started with; the right action of $\mathbb{Z}G$ on $M$ is the linearization of $mg=m$ for every $m\in M$ and $g\in G$). | |
| Dec 12, 2009 at 15:38 | comment | added | darij grinberg | And, if I am not mistaken, these isomorphisms commute with $\delta$. | |
| Dec 12, 2009 at 15:37 | comment | added | darij grinberg | Okay, I was imprecise as always. What I meant is: In group cohomology (written the inhomogenous way), a n-cochain is an element of $\mathrm{Hom}_{\mathrm{Set}}\left(G^n,\mathbb{Z}\right)$. (If it's already here that I am wrong, sorry.) By the universal property of the free $\mathbb{Z}$-module, $\mathrm{Hom}_{\mathrm{Set}}\left(G^n,\mathbb{Z}\right)\cong\mathrm{Hom}_{\mathbb{Z}}\left(\mathbb{Z}G^n,\mathbb{Z}\right)$ canonically. Finally, $\mathbb{Z}G^n\cong\left(\mathbb{Z}G\right)^{\otimes n}$. | |
| Dec 12, 2009 at 15:34 | comment | added | Mariano Suárez-Álvarez | What exactly do you mean by their being the "same"? If you make a concrete statement, it would be easier to know hwo to say no :P | |
| Dec 12, 2009 at 15:31 | vote | accept | darij grinberg | ||
| Dec 12, 2009 at 15:09 | comment | added | darij grinberg | Sorry, I should have been more precise when talking about group cohomology. But if we allow $k$ to be an arbitrary commutative ring (rather than a field), then the group cohomology is the Hochschild cohomology over $kG$ for $k=\mathbb Z$, right? | |
| Dec 12, 2009 at 14:50 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
edited body; added 331 characters in body
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| Dec 12, 2009 at 14:33 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |