The root-space decomposition classifies real semisimple Lie algebras with no extra data

Define the two real Lie algebras ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})=\{A\in M_2(\mathbb{R}):\mathrm{tr}A=0\}$ and ${\mathfrak{s}\mathfrak{u}}_2=\{A\in M_2(\mathbb{C}):A^{\ast }=-A,\mathrm{tr}A=0\}$, each under the commutator bracket. Both are closed under the bracket and are three-dimensional over $\mathbb{R}$: ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ has basis $e,f,h$ as in the special linear case, while the general element of ${\mathfrak{s}\mathfrak{u}}_2$ is $\left(\begin{array}{cc}i\alpha & \beta \\ -\bar{\beta }& -i\alpha \end{array}\right)$ with $\alpha \in \mathbb{R}$, $\beta \in \mathbb{C}$.

Both real algebras are semisimple. For ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$, the adjoint matrices in the real basis $(h,e,f)$ give the Killing matrix $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ from [L1], whose determinant is nonzero. For ${\mathfrak{s}\mathfrak{u}}_2$, use the real basis $(ih,e-f,i(e+f))$. The same bracket computation gives the diagonal Killing matrix $\mathrm{diag}(-8,-8,-8)$ in this basis. Thus both Killing forms are nondegenerate, and [L1] makes both algebras semisimple.

Both complexify to ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. For ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ this is clear from the real basis $e,f,h$. For ${\mathfrak{s}\mathfrak{u}}_2$, the three real matrices $\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$ belong to ${\mathfrak{s}\mathfrak{u}}_2$ and are linearly independent over $\mathbb{C}$, so the complex span of ${\mathfrak{s}\mathfrak{u}}_2$ is the three-dimensional space of traceless complex matrices. Hence both real algebras have the same complexification and therefore the same root-space decomposition over $\mathbb{C}h$.

They are not isomorphic: an isomorphism of real Lie algebras preserves nilpotent elements, since it conjugates adjoint operators. The element $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ is a nonzero nilpotent element of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$, because ${\mathrm{ad}}_e$ is nilpotent and nonzero. On the other hand ${\mathfrak{s}\mathfrak{u}}_2$ has no nonzero nilpotent element: every $A\in {\mathfrak{s}\mathfrak{u}}_2$ is normal, hence diagonalisable over $\mathbb{C}$ with purely imaginary eigenvalues $i{\theta }_1,i{\theta }_2$, and the eigenvalues of ${\mathrm{ad}}_A$ on the complexification are the differences $i({\theta }_j-{\theta }_k)$; if all of them vanished then ${\theta }_1={\theta }_2$, so $A$ would be a scalar multiple of the identity and then $\mathrm{tr}A=0$ forces $A=0$.

Consequently the common complexification and its root-space decomposition do not determine the real semisimple Lie algebra: ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ and ${\mathfrak{s}\mathfrak{u}}_2$ are non-isomorphic real semisimple Lie algebras with the same complexification ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$, whose root decomposition is that of the previous facts. Real forms therefore require extra data, and the statement is false.