Timeline for Compact expression for triples of subsets with total sum zero
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2021 at 14:51 | comment | added | Fedor Petrov | Yes, but it seems to be unavoidable, because the answer really depends on arithmetic properties of $|G|$ | |
| Nov 21, 2021 at 14:19 | comment | added | Vlad Matei | But now it still seems delicate since we sum over partitions and we have to be careful which parts in our partition of $k$ share a common factor with $|G|$. | |
| Nov 21, 2021 at 14:16 | comment | added | Vlad Matei | Hi Fedor! Great idea. To formalize what you wrote in general you need to write the elementary symmetric sum in terms of the power sum. We have the the green boxed formula ( for f_X(m)) math.stackexchange.com/questions/3983165/… | |
| Nov 21, 2021 at 11:11 | comment | added | Fedor Petrov | (continuation) Usually everything cancels, you need only the unit character and the characters which are unit on the subgroup of cubes. | |
| Nov 21, 2021 at 11:11 | comment | added | Fedor Petrov | Let $G$ have multiplicative notation, let $\chi$ run over all characters on $G$, then the number of your triples is $n^{-1}(\sum_{|A|=k}\chi(\prod(A))^3$. For computing the sum $\sum_{|A|=k} \chi(\prod(A))$, you may use inclusion-exclusion: say, for $k=3$ you write in the group algebra: $\sum_{|A|=3} \prod(A)=\frac16 (\sum_{g\in G} g)^3-\frac12\sum_{g\in G} g^2\cdot \sum_{h\in G} h+\frac13\sum_{g\in G} g^3$, and apply to this equation the character $\chi$ (extended to the group algebra). | |
| Nov 21, 2021 at 8:59 | history | asked | Vlad Matei | CC BY-SA 4.0 |