TITLE: Prove that $\mathfrak{su}(2)$ is not isomorphic to $\mathfrak{sl}(2,\mathbb R)$
QUESTION [10 upvotes]: 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?

REPLY [4 votes]: 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.