2
$\begingroup$

I'm trying to show that the Lie algebra $\cal{L}$ generated by the matrices \begin{equation} t_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & +1 & 0 \end{pmatrix} \quad t_2 = \begin{pmatrix} 0 & 0 & x \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \quad t_3 = \begin{pmatrix} 0 & -x & 0 \\ +1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ \end{equation} with $x \in \mathbb{R}$ and $x \ne 1$ is isomorphic to $\mathfrak{so}(3)$. The corresponding commutation relations are \begin{equation} [t_1, t_2] = t_3 \quad [t_2, t_3] = x \, t_1 \quad [t_3, t_1] = t_2 \end{equation} Specifically, I'm looking for a matrix $A$ such that \begin{equation} \epsilon_{ijk}=\sum_{n,m,s} A_{in}\,A_{jm}\,(A^{-1})_{sk}\,f_{nms} \end{equation} with $f_{nms}$ denoting ${\cal L}$'s structure constants \begin{equation} f_{2,3,1} = x = -f_{3,2,1} \qquad f_{3,1,2} = f_{1,2,3} = 1 = -f_{1,3,2} = -f_{2,1,3} \end{equation} and the Levi-Civita symbol $\epsilon_{ijk}$, the structure constants of $\mathfrak{so}(3)$. How do I construct $A$?

$\endgroup$
1
$\begingroup$

Hint: try correcting the relations rather than looking for an explicit isomorphism. More specifically, the first and last relation (which are the same as for $\mathfrak{so}(3)$) remain unchanged if you multiply $t_2$ and $t_3$ by the same factor. Now choose that factor in such a way that you "correct" the middle relation.

$\endgroup$
  • $\begingroup$ Thanks, your hint helped me to clear up my confusion. It turns out, that $A$ is the diagonal matrix $diag(1,1/\sqrt{x},1/\sqrt{x})$. $\endgroup$ – user71769 Mar 19 '18 at 20:14

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.