Let $sl_2$ be the 3-dimensional simple Lie algebra with a basis $\{x,y,h\}$ such that $[h,x]=2x$, $[h,y]=-2y$, $[x,y]=h$. Consider another Lie algebra $L$ with a basis $\{e_1,e_2,e_3\}$ such that $[e_1,e_2]=e_3$, $[e_2,e_3]=e_1$, $[e_3,e_1]=e_2$.

If we let the base field be the complex field $C$, then $sl_2$ and $L$ are isomorphic as Lie algebras. And I tried to construct the specific isomorphism but failed. So I want to know the form of this isomorphism ? Thanks in advance.

  • 1
    $\begingroup$ Just compute the isomorphism $\phi$ directly. It is a non-singular matrix of size $3$. Write down the equations in the coefficients of $\phi$ and solve them. You will need that some coefficient is non-real, because it is false over the real numbers - see here. $\endgroup$ – Dietrich Burde Apr 9 '18 at 10:34

Let $\phi\colon \mathfrak{sl}_2(\mathbb{C})\rightarrow \mathfrak{so}_3(\mathbb{C})$ be a bijective linear map satisfying $\phi([a,b])=\{\phi(a),\phi(b)\}$ for all $a,b\in \mathfrak{sl}_2(\mathbb{C})$. Here the curly Lie brackets are the ones of the second Lie algebra. Then $\phi\circ {\rm ad}(a)={\rm Ad}(\phi(a))\circ \phi$ on the level of matrices, where ${\rm Ad}$ denotes the adjoint representation of the second Lie algebra. Solving the equations of this matrix equation in the coefficients of $\phi$ gives, among other possibilities, $$ \phi=\begin{pmatrix} i & i & 0 \cr 0 & 0 & 2i \cr 1 & -1 & 0\end{pmatrix}. $$ Since $\det(\phi)=-4$, this linear map is bijective.

  • $\begingroup$ I really appreciate this. Thank you. $\endgroup$ – XSR Apr 10 '18 at 0:58
  • $\begingroup$ You are welcome. If you want you can accept this answer. $\endgroup$ – Dietrich Burde Apr 13 '18 at 18:06

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