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Sep 24 at 2:47 history edited Hung-Hsun Yu CC BY-SA 4.0
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May 2, 2023 at 8:34 comment added Marcos @მამუკაჯიბლაძე I've tried to generalize this question in a more general setting using your idea of $2$-cocycle in mathoverflow.net/questions/445969/….
Apr 29, 2023 at 7:16 comment added მამუკა ჯიბლაძე Nice! An (obvious) observation: your $F$ is a 2-cocycle. I suspect this is not a coincidence - @Marcos mentions in a comment that all this has to do with computing second (co?)homology of something...
Apr 29, 2023 at 5:49 comment added Hung-Hsun Yu Thanks for catching the typo! The notation $[x^ay^bz^c]f(x,y,z)$ is the abbreviation for the coefficients of $x^ay^bz^c$ in $f(x,y,z)$.
Apr 29, 2023 at 5:48 history edited Hung-Hsun Yu CC BY-SA 4.0
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Apr 28, 2023 at 23:42 history bounty ended Marcos
Apr 28, 2023 at 23:42 vote accept Marcos
Apr 28, 2023 at 9:45 comment added Marcos Moreover, I don't folow what $[x^ay^bz^c]F(x+z,y)+F(z,y)=1$ means. The $[x^ay^bz^c]$ is multiplication by $x^ay^bz^c$? Because in such case the equation feels weird.
Apr 28, 2023 at 9:28 comment added Marcos Also, the second sumand should be $\sum_{i<j<n+1}\binom{n+1-i}{j-i}x^iy^{n+1-j}z^{j-i}$
Apr 28, 2023 at 8:59 comment added Marcos Thanks!! Just a couple of things. If I understood your proof correctly the set you defined as $S$ should be $I$. Also, what does the notation $a\oplus b$ mean? The reason why I care about this map is because it appears as the differential map used to compute the homolgy of a certain group, so I wasn't aware of the polynomial interpretation of the formula you gave.
Apr 27, 2023 at 21:57 review Late answers
Apr 27, 2023 at 22:59
Apr 27, 2023 at 21:34 history answered Hung-Hsun Yu CC BY-SA 4.0