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.

`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);`

$\endgroup$`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$7more comments