Consider the group algebra of the symmetric group $\mathbb{C}S_n$. Then there is a corresponding Lie algebra $\mathfrak{L}(S_n)$ defined by $$[\sigma, \tau] = \sigma\circ\tau - \tau\circ\sigma,$$ where $\sigma, \tau \in S_n$. The structure of $\mathfrak{L}(S_n)$ in terms of simple factors has been considered in this post. One can also ask the same question for the Lie subalgebra of $\mathfrak{L}(S_n)$ generated by transpositions, which was considered in this post.

Now, since there is a $\mathbb{Z}_2$ grading of $\mathbb{C}S_n$, one can also define a Lie superalgebra $s\mathfrak{L}(S_n)$ on it by replacing the commutators with anti-commutators $$\{\sigma, \tau\} = \sigma\circ\tau + \tau\circ\sigma,$$ for all $\sigma, \tau \in S_n^{(1)}$, where $S_n^{(1)}$ is the odd part of the symmetric group, and all other commutators remain unchanged. Now we have similar questions: what is the structure of $s\mathfrak{L}(S_n)$ in terms of simple Lie superalgebras? What is the subalgebra of $s\mathfrak{L}(S_n)$ generated by transpositions?

My attempt is for $n=3$, $s\mathfrak{L}(S_n) \cong \mathfrak{gl}(1|1) \oplus \mathfrak{gl}(1|0) \oplus \mathfrak{gl}(0|1)$, while the subalgebra generated by transpositions is $\mathfrak{sl}(1|1) \oplus \mathfrak{gl}(1|0) \oplus \mathfrak{gl}(0|1)$. I think in general $s\mathfrak{L}(S_n)$ should be very similar to $\mathfrak{L}(S_n)$, but it might be much harder to determine the subalgebra generated by transpositions.

  • $\begingroup$ It sounds like what you need is the structure of the Schur superalgebra: arxiv.org/abs/1209.6327 $\endgroup$
    – Buzz
    Aug 11, 2022 at 2:18

1 Answer 1


Inspired by this post, Jonathan Kujawa and I wrote a paper answering this question. The paper is now posted on the arXiv (https://arxiv.org/abs/2310.01555).

First, as an associative superalgebra, $\mathbb{C}S_n$ is a direct sum of simple matrix superalgebras of types M and Q, i.e., of the forms $\mathfrak{gl}(m|m)$ and $\mathfrak{q}(n)$. The number of summands of type M (resp. of type Q) is equal to the number of equivalence classes of partitions $\lambda \vdash n$ under the relation generated by $\lambda \sim \lambda'$ (where $\lambda'$ is the transpose or conjugate partition) such that $\lambda \neq \lambda'$ (resp. $\lambda = \lambda'$). In other words, the number of summands of type M is one half the number of non-symmetric partitions of $n$, and the number of summands of type Q is equal to the number of symmetric partitions of $n$.

The structure of $\mathbb{C}S_n$ as an associative superalgebra immediately yields the structure as a Lie superalgebra under the (super)commutator.

The Lie subsubperalgebra generated by the transpositions ends up being equal to the (direct) sum of $\mathfrak{D}(\mathbb{C}S_n)$ (the derived subsuperalgebra of the Lie superalgebra $\mathbb{C}S_n$) and the one-dimensional odd subspace spanned by $T_n = \sum_{1 \leq i < j \leq n} (i,j)$, the sum in $\mathbb{C}S_n$ of all transpositions.

  • $\begingroup$ This is very awesome! It's quite surprising to me that the Lie subsuperalgebra generated by transpositions contains almost everything except for the centers. $\endgroup$ Oct 7, 2023 at 5:40

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