Lie algebra isomorphism between ${\rm sl}(2,{\bf C})$ and ${\bf so}(3,\Bbb C)$

Solution 1:

Some hints: you have introduced a basis $\{e,f,h\}$ for the Lie algebra $sl(2,\mathbb C)$; all you need is a suitable basis $\{e_i\}$ in $\mathbb C^{3}$ and an isomorphism $\phi:\mathbb C^{3}\rightarrow sl(2,\mathbb C)$ of vector spaces s.t. $\phi(e_i\wedge e_j)=[\phi(e_i),\phi(e_j)]$. Can you find such basis? Try with the simplest one... Then you should define the isomorphism $\phi$ simply as $\phi(e_i)=...$ (choose the right element of the basis for $sl(2,\mathbb C)$: you need to preserve compatibility with brackets) and extend it $\mathbb C$-linearly.

Solution 2:

To Learner,

I am trying to find a isomorphism from ${\bf R}^3$ to $sl(2,{\bf R})$. But from the comments of Avitus, I realized that I should find a suitable basis in ${\bf C}^3$.

Note that $\{ e,f,h\} $ of $sl(2,{\bf R})$ in original post is a basis in $sl(2,{\bf C})$ : And these elements satisfy the relation $$\ast\ [e,f]=h,\ [e,h]=-2e,\ [f,h]=2f $$

So remaining thing is to make a basis in ${\bf C}^3$ : I do not know who will find through $reasonable\ way$. My way is try and error, i.e., just trying.

(1) If $v_k=\sum_{i=1}^3 a^k_i e_i,\ a_i^k\in {\bf C}$ where $e_i$ is a canonical basis in ${\bf R}^3$, then let $\phi (v_1)=e,\ \phi (v_2)=f,\ \phi (v_3)=h$ so that we have an equation for $a^k_i$. Then we solve it (I believe that this way would work).

(2) Another way : Note that from $\ast$ we have $$ [e+f,h]=-2(e-f) $$ And $$ [e+f,e-f]=-2h $$

That is we have $v=e+f,\ w=h,\ x=e-f $ s.t. they are basis for $sl(2, {\bf R})$ and $$ [v,w]=-2x,\ [w,x]=2v,\ [x,v]=2w $$

This relation is similar to $(\{ e_i\},\wedge )$ so that we let $$ \phi(v)=ae_1,\ \phi (w)=be_2,\ \phi (x)=ce_3 $$ So from calculation we have $$\frac{1}{2} (a,b,c)=( \pm i,\pm i,1)$$