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.
 A: 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.
A: 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
    $$\frac{1}{2} (a,b,c)=( \pm i,\pm i,1)$$
A: Here is a description of the isomorphism $\mathfrak{sl}_2(\mathbb C)\cong\mathfrak{so}_3(\mathbb C)$ which is perhaps more enlightening:
In general given a semisimple Lie algebra $\mathfrak g$ we have the adjoint representation $\mathrm{Ad}\colon\mathfrak g\to\mathfrak{gl}(\mathfrak g)$, sending $X$ to $\mathrm{ad}\ X\colon Y\mapsto [XY]$. In fact, since $[\mathfrak g\mathfrak g]=\mathfrak g$, the homomorphism $\mathrm{Ad}$ lands in $\mathfrak{sl}(\mathfrak g)$. Moreover, $\mathrm{Ad}$ is injective since its kernel is $\mathfrak z(\mathfrak g)=0$.
Note that $\mathfrak{g}$ comes with a non-degenerate pairing (the Killing form) $K\colon\mathfrak g\times\mathfrak g\to\mathbb C$, given by $K(X,Y):=\mathrm{tr}(\mathrm{ad}\ X\circ\mathrm{ad}\ Y)$. Then we see that $K([XY],Z)=K(X,[YZ])$, so $\mathrm{ad}\ X$ is orthogonal with respect to the pairing $K(-,-)$. Thus, the adjoint representation is in fact an injective homomorphism $\mathfrak g\hookrightarrow\mathfrak{so}(\mathfrak g)$.

When $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$, this gives an isomorphism, since $\dim\mathfrak g=\dim\mathfrak{so}(\mathfrak g)=3$.


Note: This argument does not work over $\mathbb R$, since not all non-degenerate bilinear pairings $V\times V\to\mathbb R$ on a $\mathbb R$-vector space $V$ are equivalent to each other. For instance, $(x,y)*(z,w):=xz+yw$ and $(x,y)*'(z,w):=xz-yw$ are not equivalent. Bilinear forms are classified by its signature $(p,q)$, corresponding to the bilinear form on $\mathbb R^{p+q}$ given by the symmetric matrix $\begin{pmatrix}1_{p\times p}\\&-1_{q\times q}\end{pmatrix}$. Correspondingly, we have Lie algebras $\mathfrak{so}_{p,q}$, which only become $\mathfrak{so}_{p+q}$ over $\mathbb C$.
One can check that the Killing form $K(-,-)$ on $\mathfrak{sl}_2(\mathbb R)$ has signature $(2,1)$, so $\mathfrak{sl}_2(\mathbb R)\cong\mathfrak{so}_{2,1}(\mathbb R)$, not $\mathfrak{so}_3(\mathbb R)$.
