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Consider the Lie algebra $\mathfrak{g}$ span by $x_1,\ldots,x_n,y_1,\ldots,y_n$ with the commutator

$\left[y_i,x_i\right]=y_i, \left[x_i,x_{i+1}\right]=y_i, \left[y_i,x_{i+1}\right]=-y_i,\quad 1\leq i\leq n$

and the other brackets not mentioned are zero and the indices are mod $n$ (for instance when $n=3$ we have $\left[x_3,x_1\right]=y_3=\left[x_1,y_3\right]$).

We have that $g$ is solvable, $Z(\mathfrak{g})=\langle\sum_{i=1}^n(x_i+y_i)\rangle$ and $\mathfrak{h}=\langle y_i,i=1,\ldots,n\rangle$ is abelian ideal of $g$.

This Lie algebra appears in completely integrable systems, it gives a Hamiltonian formulation of the relativistic Toda lattice. I want to know if this is a very known solvable Lie algebra and what can other properties can say about it?

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  • $\begingroup$ If $n=2$ it's not well-defined (because $[x_1,x_2]=y_1$ and $[x_2,x_1]=y_2$ is absurd), so assume $n\ge 3$. Then set $X_i=x_i+y_{i-1}$. Then in the new basis $X_1,\dots,X_n,y_1,\dots,y_n$, the nonzero brackets are $[X_i,y_i]=-y_i$ and $[X_i,y_{i-1}]=y_{i-1}$. This is a semidirect product of two $n$-dimensional abelian Lie algebras, the first acting on the second with $n$-weights $w_1\dots,w_n$, which are dependent with only one relation ($\sum w_i=0$). Hence unlike what you claim, its center is 1-dimensional, generated by $\sum_i X_i=(\sum_i x_i+\sum_i y_i)$. $\endgroup$
    – YCor
    Apr 2, 2015 at 5:27
  • $\begingroup$ I know this Lie algebra as the direct product of a 1-dimensional abelian Lie algebra and a $2n-1$-dimensional Lie algebra called $\mathfrak{sol}_{2n-1}$, usually defined as the semidirect product of an $n$-dimensional ideal by the $(n-1)$-dimensional of diagonal matrices of trace 0. (PS: in my previous comment I mean "$n$ weights", not "$n$-weights".) $\endgroup$
    – YCor
    Apr 2, 2015 at 10:05
  • $\begingroup$ The Lie group $Sol_{2n-1}$ and its Lie algebra are "very known", also in geometry. For example, $Sol_{2n-1}$ has a quadratic Dehn function for $n\ge 3$ (C. Pittet). $\endgroup$ Apr 2, 2015 at 12:00

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