I'm still working on a Lie algebra isomorphism from the Lie algebra of $SL_2(\mathbb C)$ into the Lie algebra of $O(3, \mathbb C)$.
It has been suggested to me to use linear combinations of coefficients but I'm not sure what I'm doing is right since I got that all coefficients are equal to zero. Of course they are not all zero since the resulting map wouldn't be an isomorphism.
Please can someone tell me what I'm doing wrong?
Let $b_1,b_2,b_3$ be the basis of $\mathfrak{sl}_2$ and $B_1,B_2,B_3$ the basis of $\mathfrak o$:
$$ b_1 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} \text{ and } b_2 = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, b_3 = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix} $$
and a basis for $\mathfrak o$ is given by
$$ B_1 = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}, B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{pmatrix}, B_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{pmatrix}$$
so that
$(\ast)$ $[b_1,b_2]=b_3, [b_1,b_3]=-2b_1$, $[b_2,b_3]=2b_2$ and
$(\ast \ast)$ $[B_1,B_2]=B_3, [B_1,B_3]=-B_2, [B_2,B_3]=B_1$.
If $\varphi: \mathfrak{sl}_2\to \mathfrak o$ denotes the isomorphism $(\ast)$ translates into the following equations we want to satisfy:
$$ [\varphi(b_1),\varphi(b_2)]=\varphi(b_3)$$
$$ [\varphi(b_1),\varphi(b_3)]=-2\varphi(b_1) $$
$$[\varphi(b_2),\varphi(b_3)]=2\varphi(b_2)$$
Representing the image using the following coefficients
$$\varphi(b_1) = a_1 B_1 + a_2 B_2 + a_3 B_3$$ $$\varphi(b_2) = c_1 B_1 + c_2 B_2 + c_3 B_3$$ $$\varphi(b_3) = d_1 B_1 + d_2 B_2 + d_3 B_3$$
and using that the bracket is bilinear and that $[x,x]=0$ and that $[x,y]=-[y,x]$ and the equation $(\ast \ast)$ I got the following system of equations:
$$ [\varphi(b_1),\varphi(b_2)] = (a_1 c_2 - a_2 c_1)B_3 + (a_1 c_3 - a_3c_1)(-B_2) + (a_2 c_3 - a_3 c_2)B_1 = varphi(b_3) = d_1 B_1 + d_2 B_2 + d_3 B_3$$
$$ (a_1d_2 - a_2d_1)B_3 + (a_1d_3 - a_3 d_1)(-B_2) + (a_2 d_3 - a_3 d_2)B_1 = -2(a_1B_1+a_2B_2+a_3B_3) $$
$$(c_1d_2 - c_2d_1)B_3 + (c_1d_3 - c_3d_1)(-B_2) + (c_2d_3 - c_3d_2)B_1 = 2(c_1B_1 + c_2B_2 + c_3B_3)$$
Then I sorted the equations into three systems $"B_1", "B_2" , "B_2"$ like e.g. "B_1":
$$\begin{array}{ccccc} a_2c_3 &- a_3 c_2 &- d_1 & = &0\\ a_2 d_3 &-a_3d_2 &+ 2a_1 &=&0\\ c_2d_3 &-c_3d_2 &-2c_1 &= &0 \end{array}$$
(I'm omitting the other two to avoid making the question too long)
Then I collected the second equation in each of the three systems to get
$$ \begin{pmatrix}2&d_3&-d_2\\ -d_3&2 &d_1\\ d_2 & -d_1 & 2 \end{pmatrix}\begin{pmatrix}a_1 \\ a_2 \\ a_3\end{pmatrix}=0$$
and using the third rows
$$ \begin{pmatrix}-2&d_3&-d_2\\ -d_3&-2 &d_1\\ d_2 & -d_1 & -2 \end{pmatrix}\begin{pmatrix}c_1 \\ c_2 \\ c_3\end{pmatrix}=0$$
From which I deduced $a_1 = - c_1, a_2 = -c_2, a_3 = -c_3$. I substituted this into the first equation of each system which resulted in $d_1=d_2=d_3 =0$ from which it follows that all the coefficients are zero.
I put a lot of work into finding this isomorphism already and I'd appreciate any help towards finding it.
I think that your method looks sound, but the conclusion $a_1=-c_1$, $a_2=-c_2$, $a_3=-c_3$ is wrong. The matrices being multiplied by the $a$'s and $c$'s are not opposites: they are opposite along the main diagonal but the same off it. So the $a$'s and $c$'s won't be opposite.
Addendum: by your method, you should arrive at equations that any correct isomorphism (or actually any lie algebra homomorphism) would satisfy, but you should not expect your equations to have unique solutions. $\mathfrak{o}$ has many isomorphisms to itself (for example, for any fixed $P\in O(3,\mathbb{C})$, the map $X\mapsto PXP^{-1}$ is an isomorphism of $\mathfrak{o}$ to itself), so there will be many isomorphisms from $\mathfrak{sl}_2$ to $\mathfrak{o}$, and thus your equations will have many solutions.