Timeline for How to show twisted complexes over a DG category is again a DG category?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2014 at 15:15 | answer | added | Quel | timeline score: 5 | |
| Mar 3, 2014 at 23:18 | comment | added | Fernando Muro | I don't see any problem with degrees here, $d_Af$ will have degree $l-1$ in $A$ and $k-1=l-1+i-j$ in the twisted DG-category. | |
| Mar 3, 2014 at 16:21 | comment | added | Xingting | Take the simplest case that all $q_{ij}=0$, then $df=d_Af$. For the graded Leibniz rule, you need to check that $d(fg)=d(f)g+(-1)^{|f|}fd(g)$. Note that the degree of f now is $k=l+j-i$. So if you use the differential in the original $A$, the degree of $f$ is $l$. So there is a a difference. I do not know how to fix it. | |
| Mar 3, 2014 at 11:23 | comment | added | Fernando Muro | Can you maybe explain why you think there are problems in the definition of the differential? | |
| Mar 3, 2014 at 5:44 | review | First posts | |||
| Mar 3, 2014 at 6:29 | |||||
| Mar 3, 2014 at 5:28 | history | asked | Xingting | CC BY-SA 3.0 |