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Brundan and Kleshchev showed that if $k$ is a field of characteristic $p$, then the group ring $kS_n$ of the symmetric group $S_n$ admits an integer grading which is nontrivial in the sense that it is not concentrated in degree 0. This grading is explicit, in the sense that Brundan-Kleshchev give an explicit isomorphism of $kS_n$ with a specific graded ring. However, on the occasions that I've tried to work out examples of this grading by tracing through the isomorphism, I have found it extremely hard going. (Caveat: I am definitely an amateur at this!)

Question 1: Are examples of the Brundan-Kleshchev grading worked out anywhere?

In the above question I think I am looking for particular choices of $n$ or $p$ or both, where the gradings of specific elements of $kS_n$ are written out.

Question 2: Are there any alternative ways to describe or compute the Brundan-Kleshchev grading on $kS_n$?

(n.b. Brundan and Kleshchev give gradings on a more general family of algebras than just $kS_n$ for $\mathrm{char}(k)=p$, but I am just asking about this special case.)

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  • $\begingroup$ "non-trivial grading" in what? Usually "non-trivial" means the grading is not concentrated in degree zero. An obvious non-grading of $kG$ is precisely graded by $G$, and hence also by the abelian quotient of $G$ $\mathbf{Z}/2\mathbf{Z}$— you would certainly not call this "hidden". Do you mean in $\mathbf{Z}$ (or a torsion-free abelian group)? $\endgroup$
    – YCor
    Apr 22, 2021 at 16:01
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    $\begingroup$ @YCor The hidden grading is an integer grading, nontrivial meaning that it’s not concentrated in degree 0. $\endgroup$
    – Richard Hepworth
    Apr 22, 2021 at 17:27
  • $\begingroup$ Is $p$ assumed to be $\le n$? if it's true for arbitrary large $p$ and fixed $n$, I'd guess it's also true in characteristic zero. $\endgroup$
    – YCor
    Apr 22, 2021 at 17:33
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    $\begingroup$ @YCor: There is no assumption relating $n$ and $p$, and the result also holds in characteristic 0. $\endgroup$
    – Richard Hepworth
    Apr 22, 2021 at 18:30
  • $\begingroup$ For anyone confused by this, it seems that semisimple algebras like matrix algebras can have nontrivial gradings - we can grade the algebra of $n \times n$ matrices so that matrices with unique nonvanishing entry in row $i$ and column $j$ have grade $i-j$. But probably these are more interesting once we have $p\leq n$. $\endgroup$
    – Will Sawin
    Apr 23, 2021 at 2:20

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