@ Michael Albanese. Journal of Pure and Applied Algebra 221 (2017) 1134-1145

For $n=3$ it seems to be true as it says something about the areas of the sides of a parallelepiped.

For the interpretation with central extensions the action on the modules needs to be trivial.

@YCor: Of the two, Borel was less inclined to make jokes. And I like your joke about humor.

@Martin Rubey: Recursion works better: rel ={{1, 1}, {1, 2}, {1, 10}, {2, 2}, {3, 2}, {3, 3}, {3, 4}, {4, 4}, {5, 4}, {5, 5}, {5, 6}, {6, 6}, {7, 6}, {7, 7}, {7, 8}, {8, 8}, {9, 8}, {9, 9}, {9, 10}, {10, 10}}; max = Max[rel]; Do[down[i] = Select[rel, Max[#] == i &], {i, max}]; checkdown[f_List] := AllTrue[down[Length[f]], MemberQ[rel, f[[#1]]] &]; num[{}] := Sum[num[{i}], {i, max}]; num[f_List] := If[checkdown[f],If[Length[f] == max,1,Sum[num[Append[f, i]], {i, max}]],0]; num[{}] gives 10030.

@Martin Rubey: In Mathematica I put endoQ[f_List]:=AllTrue[rel,MemberQ[rel,f[[#1]]]&] then rel = {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}, {1, 2}, {3, 2}, {3, 4}, {5, 4}, {5, 6}, {1, 6}} then Length[Select[Tuples[Range[6], 6], endoQ[#] &]]. One must not start with 0.

But I would describe it as {1, 2}, {3, 2}, {3, 4}, {5, 4}, {5, 6}, {1, 6}, a cycle with directions alternating.

The 234 is confirmed by my program.

@Pierre MATSUMI.This is not the way it is done. If you agree, you should accept the answer.

No, that is false. In my answer $g$ is irreducible, but $(g)$ is not prime. Could you rephrase your question about `the relation' ?

If $g$ is prime, then the answer is yes. One shows that multiplication by $g$ on $S[X]/(f)$ is injective and thus has a nontrivial determinant. And a matrix times its adjoint is the determinant times the identity matrix.

I know that Elsevier mathematical journals convert to XML.

The fiber is $P/B$.

Notice the homomorphism $S^{r+1}(E)\to E\otimes S^r(E)$ that sends $f$ to $x\otimes \frac{\partial f}{\partial x}+y\otimes \frac{\partial f}{\partial y}$.