Prove that $\mathfrak{su}(2)$ is not isomorphic to $\mathfrak{sl}(2,\mathbb R)$ I would like to prove that the two algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb R)$ are not isomorphic as Lie algebras.
I started considering the following basis for $\mathfrak{su}(2)$:
\begin{equation}
\sigma_1=\frac{1}{2}
\begin{pmatrix} 
i &0\\
0 &-i\\
\end{pmatrix}
\quad
\sigma_2=\frac{1}{2}
\begin{pmatrix} 
0 &i\\
i &0\\
\end{pmatrix}
\quad
\sigma_3=\frac{1}{2}
\begin{pmatrix} 
0 &-1\\
1 &0\\
\end{pmatrix}
\end{equation}
and for  $\mathfrak{sl}(2,\mathbb R)$:
\begin{equation}
A_1=\frac{1}{2}
\begin{pmatrix} 
1 &0\\
0 &-1\\
\end{pmatrix}
\quad
A_2=\frac{1}{2}
\begin{pmatrix} 
0 &1\\
0 &0\\
\end{pmatrix}
\quad
A_3=\frac{1}{2}
\begin{pmatrix} 
0 &0\\
1 &0\\
\end{pmatrix}
\end{equation}
It is possible to check that for  the $\sigma$s the following commutation relation holds:
\begin{equation}
[\sigma_i,\sigma_j]=\epsilon_{ijk}\sigma_k.
\end{equation}
While if we consider:
\begin{equation}
[A_1,A_2]=A_2
\end{equation}
Then it seems that I can't find any isomorphism between these 2 basis.
Now I would like to generalize the argument to every basis, does anyone have any idea on how to proceed?
 A: There are several ways to see that the two Lie algebras are not isomorphic. First, $\mathfrak{sl}(2)$ has a $2$-dimensional subalgebra $\langle A_1,A_2\rangle$, but $\mathfrak{su}(2)$ has not - see above. Secondly, the Killing forms are not equivalent, hence the Lie algebras cannot be isomorphic, see this duplicate:
The Killing forms of $\mathfrak{su}(n)$ and $\mathfrak{sl}(n,\Bbb R)$ are not isomorphic (as real Lie algebras)
Furthermore, this question has appeared already here:
Showing the Lie Algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are not isomorphic.
A: Looking at a general basis will not help much, I guess, but what you have done can be completed: Note that the subspace $U = \def\<#1>{\left<#1\right>}\<A_1, A_2>$ of $\def\sl{\mathfrak{sl}(2, \mathbf R)}\sl$ generated by $A_1$ and $A_2$ is a subalgebra, as $[A_1, A_2] = A_2 \in U$. So $\sl$ has a two-dimensional subalgebra. We will prove that $\def\su{\mathfrak{su}(2,\mathbf C)}\su$ does not have any. Define $\phi \colon \su \to \mathbf R^3$ by $\phi(\sigma_i) = e_i$, where $e_i$ denotes the $i$-th standard unit vector. Then we have 
$$ \phi([\sigma_i, \sigma_j]) = \phi(\epsilon_{ijk}\sigma_k)=\epsilon_{ijk}e_k = e_i \times e_j = \phi(\sigma_i) \times \phi(\sigma_j) $$
where $\epsilon_{ijk}$ is the Levi-Civita symbol and $\times$ denotes the usual cross product. Hence 
$$ \phi([A, B]) = \phi(A) \times \phi(B),\qquad A,B \in \su $$
by linearity which implies that $\phi$ is an isomorphism of algebras of $\su$ with $(\mathbf R^3, \times)$. Now let $A, B \in \su$ be linearly independent, then $\phi(A)$ and $\phi(B)$ span a two-dimensional linear subspace $V$ of $\mathbf R^3$, hence $\phi(A) \times \phi(B)$ is a non-zero vector orthogonal to $V$; that is, $\phi([A,B]) \not\in V$, giving - as $\phi$ is invertible - $[A,B]\not\in \phi^{-1}[V] = \<A,B>$. So $A, B$ do not generate a two-dimensional subalgebra. As $A$ and $B$ were arbitrary, there isn't any. 
Therefore $\su$ and $\sl$ aren't isomorphic.
