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](https://oeis.org/A214015) and [A026820](https://oeis.org/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.



What's the dimension of the Lie algebra generated by transpositions on $n$ objects?
In "L'algèbre de Lie des transpositions" ([arXiv:math/0502119](https://arxiv.org/abs/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.


