# What's the dimension of the Lie algebra generated by transpositions on $n$ objects?

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, 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.

• Dimension is what you say for $n=4$ and $n=5$, however for $n=6$ it is $249$. – მამუკა ჯიბლაძე Sep 26 '20 at 19:20
• I used GAP: first tr:=List(Combinations([1..6],2),p->(p,p)); 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); – მამუკა ჯიბლაძე Sep 26 '20 at 20:55
• 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). – მამუკა ჯიბლაძე Sep 27 '20 at 15:58
• 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). – Adrien Oct 13 '20 at 8:30
• @Adrien you should post this as an answer. Now it is hidden behind the 'show 11 more comments' button – Vincent Oct 13 '20 at 15:08

## 1 Answer

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.

• Thanks a lot! This is exactly the result I need. – WunderNatur Oct 14 '20 at 3:54