Dynkin diagrams classify real semisimple Lie algebras

Both algebras are three-dimensional over $\mathbb{R}$ and are semisimple. The split algebra has the basis $e=E_{12},f=E_{21},h=\mathrm{diag}(1,-1)$, and its form in [L2] is nondegenerate because for nonzero traceless $X$ one has $K(X,X^T)=4\mathrm{tr}(X{X}^T)>0$. For $\mathfrak{s}\mathfrak{u}(2)$, let $M_a$ be the matrix of ${\mathrm{ad}}_{A_a}$ in the basis of [L3]. The displayed brackets give $K(A_a,A_b)=\mathrm{tr}(M_a{M}_b)=-8{\delta }_{ab}$, so its Killing form is negative definite and nondegenerate. Cartan's criterion [L1] gives semisimplicity in both cases.

The element $e=E_{12}\in {\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is nonzero and ${\mathrm{ad}}_e$ is nilpotent: ${\mathrm{ad}}_e(e)=0$, ${\mathrm{ad}}_e(h)=-2e$, ${\mathrm{ad}}_e(f)=h$, ${\mathrm{ad}}_e^2(f)=-2e$, ${\mathrm{ad}}_e^3(f)=0$. In $\mathfrak{s}\mathfrak{u}(2)$, if $X={\sum }_a{x}_a{A}_a\mathrm{\ne }0$ then ${\mathrm{ad}}_X={\sum }_a{x}_a{M}_a$ is a nonzero real skew-symmetric operator in the basis $(A_1,A_2,A_3)$; a nonzero skew-symmetric operator is not nilpotent, because a nilpotent operator $S$ satisfies $\mathrm{tr}(S^2)=0$ while $\mathrm{tr}(S^2)=-{\sum }_{i,j}{s}_{ij}^2<0$ for real skew $S\mathrm{\ne }0$. Hence $\mathfrak{s}\mathfrak{u}(2)$ has no nonzero element with nilpotent adjoint.

An isomorphism of Lie algebras carries elements with nilpotent adjoint to elements with nilpotent adjoint, since ${\mathrm{ad}}_{\phi (X)}=\phi {\mathrm{ad}}_X{\phi }^{-1}$; by step 1.2 the algebras ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ and $\mathfrak{s}\mathfrak{u}(2)$ are therefore not isomorphic.

The real basis $e,f,h$ of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is a complex basis of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. For the compact algebra, $A_3=ih$, $A_2=e-f$, and $A_1=i(e+f)$, so conversely $h=-i{A}_3$, $e=(A_2-i{A}_1)\mathrm{/}2$, and $f=(-A_2-i{A}_1)\mathrm{/}2$; hence $A_1,A_2,A_3$ are also a complex basis of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. Complexifying either inclusion therefore gives ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. By [L4] both complexifications have diagram $A_1$, while step 2.1 shows the real algebras are not isomorphic. This is the required counterexample.