= Solution
Write $\mathfrak g_{\mathbb C}=L(G_{\mathbb C})$ and $\mathfrak h=L(H)$. A root is a nonzero weight $\alpha\in\mathfrak h^*$ of the adjoint action. Its <root space> is
$$
\mathfrak g_\alpha=\{Y\in\mathfrak g_{\mathbb C}:[Z,Y]=\alpha(Z)Y\text{ for all }Z\in\mathfrak h\}.
$$
At the <group> level these weights come from nontrivial characters of the Cartan <subgroup> acting by conjugation. The Cartan <root-space decomposition> is
$$
\mathfrak g_{\mathbb C}=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha.
$$
The zero-<weight space> is $\mathfrak h$; for the complexification of a <compact> <group> the algebra is reductive, with any centre contained in this zero-<weight space>.
For the product in question,
$$
G_{\mathbb C}=SL_2(\mathbb C)\times SL_2(\mathbb C),\qquad
\mathfrak g_{\mathbb C}=\mathfrak{sl}_2(\mathbb C)\oplus\mathfrak{sl}_2(\mathbb C).
$$
Take the diagonal <Cartan subalgebra> generated by $H_1=(\operatorname{diag}(1,-1),0)$ and $H_2=(0,\operatorname{diag}(1,-1))$. On $Z=xH_1+yH_2$, define $\alpha_1(Z)=2x$ and $\alpha_2(Z)=2y$. The roots are exactly
$$
\boxed{\Phi=\{\alpha_1,-\alpha_1,\alpha_2,-\alpha_2\}}.
$$
If $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb C(E,0)$, $\mathbb C(F,0)$, $\mathbb C(0,E)$, and $\mathbb C(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the <root system> $A_1\times A_1$.
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