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Consider the real Lie algebra $\mathfrak{so}_{p,q}(\mathbb{R})$ with $p+q=m$, we label the elements of the algebra $X_{i,j}$ for $1\leq i<j\leq m$. Let $Y$ be a set of $m$ vectors.

Can I always find a basis $(Y_1,\cdots,Y_m)$ of $Y$ such that

$$ [X_{i,j},Y_i]= -Y_j\quad \text{and}\quad [X_{i,j},Y_j] =\pm Y_i $$

and of course $[X_{i,j},Y_k]=0$ when $k\neq i,j$ ? The $\pm$ sign depend on the ratio of the eigenvalues of the underlaying quadratic form.

Observe that the above relations mean that $Y_i$ and $Y_j$ are eigenvectors of $[X_{i,j},[X_{i,j},\cdot]]$.

Thank you!

Btw do you have some references about real Lie algebras without going throught the complex field ?

Athena
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  • I do not understand what exactly you label as $X_{i,j}$. Are these supposed to be certain generators (some of which would be redundant, because the dimension of the Lie algebra is certainly smaller than $m^2$)? Do you maybe have a specific matrix representation of that Lie algebra in mind -- then you should write it down, please. – Torsten Schoeneberg Jan 28 '20 at 17:24
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    Thks for yr reply, there is actually $\frac{m(m-1)}{2}$ basis elements as I took $i<j$. Element of the algebra can be expressed as $X=\begin{pmatrix}A&B\-B^T&D\end{pmatrix}$ where $A+A^T=0$ and $D+D^T=0$ ($A\in \mathfrak{so}p(\mathbb{R})$, $B\in M{p,q}(\mathbb{R})$ and $D\in \mathfrak{so}q(\mathbb{R})$). Let $E{i,j}$ be the matrix matrix whose entry are zeros except for the $(i,j)$th which is $+1$. The notation $X_{i,j}$ stands for one of the element basis : $$ E_{i,j}-E_{j,i},\ 1\leq i<j\leq p \ E_{i,j}, \ 1\leq i\leq p,\ p+1\leq j\leq m \ E_{i,j}-E_{j,i},\ p+1\leq i<j\leq m $$ – Athena Jan 28 '20 at 18:39

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