Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$ I think that this is an exercise. I can not find a solution.
We can define Lie bracket multiplication on $\mathbb{C}^3$ : $$ x\wedge y $$ where $x=(x_1,x_2, x_3)$, $y=
(y_1,y_2,y_3)$, and $\wedge $ is the wedge product we know.
Consider the Lie algebra $\mathfrak {sl}(2,\mathbb{C})= \{ X\in M_2(\mathbb{C}) \mid\ {\rm Trace} (X) =0\}$
and $$
e= \left(
  \begin{array}{cc}
    0 & 1 \\
    0 & 0 \\
  \end{array}
\right),\ f= \left(
  \begin{array}{cc}
    0 & 0 \\
    1 & 0 \\
  \end{array}
\right),\ h= \left(
  \begin{array}{cc}
    1 & 0 \\
    0 & -1 \\
  \end{array}
\right). $$ Note that $$ [e,f]=h,\ [e,h]=-2e,\
[f,h]=2f.$$
Here, the problem is to find an explicit isomorphism between $\mathfrak {sl}(2,\mathbb{ C})$ and $\mathbb{C}^3$.
Thank you in advance.

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.
