In my work, I came across the following Lie algebra of vector fields: $$ \mathfrak{L}_q = \textrm{vect}_{\mathbb{R}}\left\{\partial_{x_1},\ldots,\partial_{x_m},\varepsilon_ix_j\partial_{x_i} - \varepsilon_jx_i\partial_{x_j},\quad\textrm{for}\quad 1\leq i<j\leq m\right\}, \;\textrm{where}\;\varepsilon_i=\left\{\begin{array}{rl}+1&1\leq i\leq p\\ -1 &p+1\leq i\leq m \end{array}\right.. $$ Obviously $\mathfrak{L}_q$ is $\frac{m}{2}(m+1)$-dimensional, is composed of an abelian ideal $\mathfrak{I}=\textrm{vect}_{\mathbb{R}}\left\{\partial_{x_1},\ldots,\partial_{x_m}\right\}$, and a subalgebra $\mathfrak{l}=\textrm{vect}_{\mathbb{R}}\left\{\varepsilon_ix_j\partial_{x_i} - \varepsilon_jx_i\partial_{x_j}\right\}$ which isomorphic to $\mathfrak{so}(p,q)$ (with $p+q=m$).
Question: I am trying to characterize that Lie algebra among all $\frac{m}{2}(m+1)$-dimensional Lie algebras.
Partial answer: So far the conjecture I have is the following:
$\mathfrak{L}$ a $\frac{m}{2}(m+1)$-dimensional Lie algebra is isomorphic to $\mathfrak{L}_q$ if and only if :
- $\mathfrak{L}=\mathfrak{I}\oplus\mathfrak{l}$, where $\mathfrak{I}$ is a $m$-dimensional abelian ideal, and $\mathfrak{l}$ is a subalgebra;
- $\mathfrak{l}$ is isomorphic to $\mathfrak{so}(p,q)$;
- The action of $\mathfrak{l}$ on $\mathfrak{I}$ is a faithful representation of $\mathfrak{so}(p,q)$.
My proof strategy is to show that $\mathfrak{L}$ admits a basis which has the same commutativity relations than $\mathfrak{L}_q$. The point I did not manage to solve is to prove the existence of a basis of $\mathfrak{I}$ that satisfies the commutativity relations with $\mathfrak{l}$.
Additional question: Does there exist an algebraic criterion to check if $\mathfrak{l}$ is isomorphic to $\mathfrak{so}(p,q)$?