12
$\begingroup$

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.

$\endgroup$
12
  • 4
    $\begingroup$ Dimension is what you say for $n=4$ and $n=5$, however for $n=6$ it is $249$. $\endgroup$ – მამუკა ჯიბლაძე Sep 26 '20 at 19:20
  • 4
    $\begingroup$ 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); $\endgroup$ – მამუკა ჯიბლაძე Sep 26 '20 at 20:55
  • 3
    $\begingroup$ 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). $\endgroup$ – მამუკა ჯიბლაძე Sep 27 '20 at 15:58
  • 3
    $\begingroup$ 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 '20 at 8:30
  • 2
    $\begingroup$ @Adrien you should post this as an answer. Now it is hidden behind the 'show 11 more comments' button $\endgroup$ – Vincent Oct 13 '20 at 15:08
9
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! This is exactly the result I need. $\endgroup$ – WunderNatur Oct 14 '20 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.