Define a Lie bracket on the group algebra of the permutation group $S_n$ in the following way: $$[\sigma, \tau] = \sigma\circ\tau - \tau\circ\sigma,$$ where $\sigma, \tau \in S_n$, and the multiplication on permutations is defined as composition. My question is, what is the dimension of the Lie subalgebra generated by transpositions, i.e. $(ij)$? My conjecture is that the dimension is given by $C_n - \lfloor \frac{n}{2} \rfloor$, where $C_n$ is the Catalan number. Is this correct and what is the proof?

For example, when $n=3$, using the cycle notation, we have $$ [(12),(23)] = (132) - (123) \\ [(23),(31)] = (132) - (123) \\ [(31),(12)] = (132) - (123) \\ $$ and $$ [(12), (132) - (123)] = 2((23) - (13)), \text{etc.} $$ Therefore this algebra is $4 = C_3 - 1$ dimensional.

If this conjecture is correct, the result should not be hard to generalize to the Lie subalgebras generated by $S_k$ for $k<n$, which should be related to A214015 and A026820.

Update: As pointed in the comment, this conjecture is wrong. It fails from $n=6$, where there is an unexpected $\mathfrak{so}(16)$ piece in the Lie algebra.

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    $\begingroup$ Dimension is what you say for $n=4$ and $n=5$, however for $n=6$ it is $249$. $\endgroup$ Sep 26, 2020 at 19:20
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    $\begingroup$ I used GAP: first tr:=List(Combinations([1..6],2),p->(p[1],p[2])); then G:=Group(tr); then A:=GroupRing(Rationals,G); then emb:=Embedding(G,A); then L:=LieAlgebra(A); then S:=Subalgebra(L,List(tr,x->LieObject(emb(x)))); and finally Dimension(S); $\endgroup$ Sep 26, 2020 at 20:55
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    $\begingroup$ I just ask about LieCenter(S); and SemiSimpleType(LieDerivedSubalgebra(S)); assuming that it is reductive (which it turns out to be: if it would not be the case, GAP would fail on determining the semisimple type). Mine is finally done with the structure, it is $\mathfrak{sl}(5)\oplus\mathfrak{sl}(9)\oplus\mathfrak{sl}(5)\oplus\mathfrak{so}(16)$ (plus 1-dimensional center). $\endgroup$ Sep 27, 2020 at 15:58
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    $\begingroup$ The following paper (in french) gives an explicit description of this Lie algebra (see Theorem A) : I. Marin, L'algèbre de Lie des transpositions ( arXiv:math/0502119). $\endgroup$
    – Adrien
    Oct 13, 2020 at 8:30
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    $\begingroup$ @Adrien you should post this as an answer. Now it is hidden behind the 'show 11 more comments' button $\endgroup$
    – Vincent
    Oct 13, 2020 at 15:08

1 Answer 1


In "L'algèbre de Lie des transpositions" (arXiv:math/0502119 ), Ivan Marin shows the Lie algebra generated by transpositions is the product of a 1 dimensional Lie algebra, and of a semi-simple Lie algebra, and provides an explicit decomposition of the latter as a direct sum of special linear, symplectic and orthogonal Lie algebras. See theorem A and section 5.

  • $\begingroup$ Thanks a lot! This is exactly the result I need. $\endgroup$ Oct 14, 2020 at 3:54

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