Timeline for Describe the second cohomology group $H^2(Z_n \times Z_n. k^*)$.
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2022 at 14:16 | comment | added | Sasha | Yes, I think you are correct about the sign. | |
| Sep 13, 2022 at 12:06 | comment | added | Hetong Xu | Thank you! (Since when I'm checking $d_1 \circ d_2=0$ in your notation, there is a quite subtle sign issue. Maybe the second column in $d_2$ should be $(1-t_1, -(1-t_2))^{t}$ or conversely?) | |
| Sep 13, 2022 at 11:49 | comment | added | Sasha | Yes, precisely. | |
| Sep 13, 2022 at 10:20 | comment | added | Hetong Xu | Thank you! So we can also regard it as the total complex of a double complex $M_1^{\bullet} \otimes_{\mathbb{Z}} M_2^{\bullet}$? (So we can get the description of the differential map in the new complex.) | |
| Sep 13, 2022 at 9:17 | comment | added | Sasha | If you have two rings $R_1$ and $R_2$ and two modules $M_1$ over $R_1$ and $M_2$ over $R_2$, then $M_1 \otimes M_2$ is naturally a module over $R_1 \otimes R_2$ (tensor products are over the integers). Similarly, if you have two complexes $M_1^\bullet$ and $M_2^\bullet$ their tensor product is the complex with $n$-th term $\oplus_{i+j = n} M_1^i \otimes M_2^j$. This is what I mean by tensor square of resolutions. | |
| Sep 12, 2022 at 8:51 | comment | added | Hetong Xu | Sorry for such a late comment. Yet I'm a little bit confused at the term "tensor square" of two resolutions. Is there an operation called "tensor square of resolutions" and here is a particular example? Or this is not actually a mathematical notion? Thank you! | |
| Oct 8, 2010 at 9:29 | vote | accept | unknown | ||
| Sep 15, 2010 at 7:17 | vote | accept | unknown | ||
| Sep 15, 2010 at 7:17 | |||||
| Aug 26, 2010 at 11:53 | history | answered | Sasha | CC BY-SA 2.5 |