# What is the Lie superalgebra generated by permutations?

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.

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

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.