Timeline for Filtered ring giving rise to a graded-commutative ring
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2011 at 17:48 | comment | added | Will Sawin | If you know $F^0$ to $F^{i-1}$, you can use this to find out the minimum necessary for $F^i$. The minimum is most likely to give you the most interesting ring, so go with that. Then calculate the minimum for $F^{i+1}$.... If $F^0$ is well-behaved enough, you get a graded ring with the appropriate commutator property. | |
| Sep 20, 2011 at 19:09 | comment | added | euklid345 | I'm sorry, you are right, in the graded commutative case this question is more subtle than I thought. My dumb answer only works if you already have a $\mathbb{Z}/2\mathbb{Z}$ grading. | |
| Sep 20, 2011 at 18:40 | comment | added | Pierre | ... I mean when one of $i$ or $j$ is even, sorry! but you get the idea. | |
| Sep 20, 2011 at 18:40 | comment | added | Pierre | I agree, there's no graded-commutator available. Of course you could state $F^i F^j - F^j F^i \subset F^{i+j+1}$ when $i$ and $j$ are both even, and $F^i F^j + F^j F^j \subset F^{i+j+1}$ in the other cases. But that's paraphrasing. | |
| Sep 20, 2011 at 18:06 | comment | added | Mariano Suárez-Álvarez | euklid345: if the algebra is not graded to begin with, there is no graded-commutator to compute! :) | |
| Sep 20, 2011 at 17:59 | comment | added | euklid345 | Of course, $[,]$ is the graded commutator. You were asking for well-kown and simple... | |
| Sep 20, 2011 at 17:24 | comment | added | Pierre | That would make it commutative, rather than graded-commutative. Unless the meaning you give to the square brackets depends on i and j -- that would be an option, but it's hardly better than saying "$gr R$ is graded-commutative". I was hoping for something natural, somehow. | |
| Sep 20, 2011 at 16:14 | history | answered | euklid345 | CC BY-SA 3.0 |