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S Apr 28, 2023 at 23:42 history bounty ended Marcos
S Apr 28, 2023 at 23:42 history notice removed Marcos
Apr 28, 2023 at 23:42 vote accept Marcos
Apr 27, 2023 at 21:34 answer added Hung-Hsun Yu timeline score: 2
Apr 23, 2023 at 21:56 history edited Marcos CC BY-SA 4.0
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S Apr 21, 2023 at 22:14 history bounty started Marcos
S Apr 21, 2023 at 22:14 history notice added Marcos Authoritative reference needed
Apr 20, 2023 at 15:02 history edited Marcos CC BY-SA 4.0
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Apr 20, 2023 at 13:22 comment added მამუკა ჯიბლაძე @Marcos Sure. It is in Mathematica. ker[n_]:=If[n==1,{s[1]},Sort@Map[#.Table[s[i],{i,n}]&,NullSpace[Mod[Table[Which[Last[p]==i,Binomial[i,First[p]],First[p]==i,Binomial[n-i+1,Last[p]-i],True,0],{p,Subsets[Range[n],{2}]},{i,n}],2],Modulus->2]]]
Apr 20, 2023 at 11:41 comment added Marcos @მამუკაჯიბლაძე I know its been a while, but can you give me the code you used for making those computations? It would help me a lot!
Mar 25, 2023 at 16:23 comment added Marcos @მამუკაჯიბლაძე I'll try to prove some of this facts and see what I came with. Thanks!!!
Mar 25, 2023 at 16:21 comment added მამუკა ჯიბლაძე No idea. It might be tricky to actually prove this. One more empirical observation: number of summands in the basis vector of the kernel of $f_n$ seems to be $2^j-2$ where $j$ is the number of 1s in the binary expansion of $n+1$
Mar 25, 2023 at 16:19 comment added Marcos @მამუკაჯიბლაძე However, proving those claims seems dificultad, do you have any idea?
Mar 25, 2023 at 16:17 comment added Marcos @მამუკაჯიბლაძე Thanks, that is interesting, since I want to compute the kernel to find a homology group. And It turns out that the dimension of the image is $1$ iff $n=2^{m+1}-1$ and zero otherwise. Hence, It seems that most homology groups have rank $1$.
Mar 25, 2023 at 15:42 comment added მამუკა ჯიბლაძე Additionally, for $k=m$, i. e. for $n=2^{m+1}-1$ the kernel is spanned by $\sigma_{2^m}$ and it seems that in no other cases does it contain any single $\sigma$, only nontrivial sums.
Mar 25, 2023 at 15:36 comment added მამუკა ჯიბლაძე Empirically, the kernel is spanned by $\sigma_{2^k}$ and $\sigma_{2^m}$ for $n=2^m+2^k-1$, $k<m$, and is 1-dimensional for all other $n$ (checked for $n\leqslant256$)
Mar 25, 2023 at 15:01 comment added მამუკა ჯიბლაძე Here are the spanning vectors of kernels for some small $n$:$$\begin{array}{rl}n&\ker\\\hline6&\sigma_1+\sigma_2+\sigma_3+\sigma_4+\sigma_5+\sigma_6\\7&\sigma_4\\8&\sigma_1,\sigma_8\\9&\sigma_2,\sigma_8\\10&\sigma_1+\sigma_2+\sigma_3+\sigma_8+\sigma_9+\sigma_{10}\\11&\sigma_4,\sigma_8\\12&\sigma_1+\sigma_4+\sigma_5+\sigma_8+\sigma_9+\sigma_{12}\\13&\sigma_2+\sigma_4+\sigma_6+\sigma_8+\sigma_{10}+\sigma_{12}\\14&\sum_{i=1}^{14}\sigma_i\\15&\sigma_8\\16&\sigma_1,\sigma_{16}\\17&\sigma_2,\sigma_{16}\\18&\sigma_1+\sigma_2+\sigma_3+\sigma_{16}+\sigma_{17}+\sigma_{18}\end{array}$$
Mar 25, 2023 at 14:44 comment added მამუკა ჯიბლაძე For $n=6$, the kernel is 1-dimensional, spanned by $\sum_{i=1}^6\sigma_i$, since\begin{align*}f(\sigma_1)&=\sigma_{13}+\sigma_{15}\\f(\sigma_2)&=\sigma_{23}+\sigma_{26}\\f(\sigma_3)&=\sigma_{13}+\sigma_{23}\\f(\sigma_4)&=\sigma_{45}+\sigma_{46}\\f(\sigma_5)&=\sigma_{15}+\sigma_{45}\\f(\sigma_6)&=\sigma_{26}+\sigma_{46}\end{align*}In general, kernels seem to be either 1- or 2-dimensional: they are 2-dimensional for $n=2,4,5,8,9,11,16,17,19,23,32,33,35,39,47$ and 1-dimensional for all other $n<50$
Mar 25, 2023 at 13:48 history edited Marcos CC BY-SA 4.0
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Mar 25, 2023 at 9:04 comment added Marcos @FriedrichKnop I've edited the question, hope it helps!
Mar 25, 2023 at 9:03 history edited Marcos CC BY-SA 4.0
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Mar 25, 2023 at 7:42 comment added Friedrich Knop What happens for small $n$?
Mar 24, 2023 at 19:08 history edited Marcos CC BY-SA 4.0
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Mar 23, 2023 at 17:18 history asked Marcos CC BY-SA 4.0