Assume this is all over $\mathbb{C}$. Given (semisimple, finite-dimensional) Lie algebras $\mathfrak{g}_1$ and $\mathfrak{g}_2$ with Cartan subalgebras $\mathfrak{h}_1$ and $\mathfrak{h}_2$, why is the root system $\Phi$ of $\mathfrak{g}:=\mathfrak{g}_1\oplus \mathfrak{g}_2$ with respect to $\mathfrak{h}:=\mathfrak{h}_1\oplus \mathfrak{h}_2$ equal to the union of the respective root systems, $\Phi=\Phi_1\sqcup\Phi_2$?
My main concern is the following: Define
$\Phi(\mathfrak{h}\subset \mathfrak{g}) :=\lbrace \alpha\in \mathfrak{h}^*-\lbrace 0\rbrace\mid \mathfrak{g}_\alpha \neq 0\rbrace$, where $\mathfrak{g}_\alpha = \lbrace X\in \mathfrak{g}\mid \forall H\in \mathfrak{h},\:\text{ad}_H(X)=\alpha(H)X\rbrace$. It is clear to me that $\Phi_1\sqcup\Phi_2\subseteq \Phi$, where $\Phi_1\hookrightarrow\Phi$ by $\alpha\mapsto \alpha\oplus 0$, i.e. the map $\mathfrak{h}_1\oplus \mathfrak{h}_2\to \mathbb{C}$ given by $(H_1,H_2)\mapsto \alpha(H_1)$. Similar for $\Phi_2$. However, I do not understand why we should not be able to have roots of the form $\alpha_1\oplus \alpha_2$, $\alpha_i\in \Phi_i$, where none of the $\alpha_i$ are $0$. They should be interpreted as the maps $(H_1,H_2)\mapsto \alpha_1(H_1)+\alpha_2(H_2)$. They are clearly linear, meaning $\alpha_1\oplus\alpha_2\in \mathfrak{h}^*=(\mathfrak{h}_1\oplus \mathfrak{h}_2)^*=\mathfrak{h}_1^*\oplus \mathfrak{h}_2^*$, and since the Lie bracket of the direct sum is just the coordinatewise Lie brackets, I should think we get $\mathfrak{g}_{\alpha_1\oplus \alpha_2}=\lbrace (X,Y)\in \mathfrak{g}_1\oplus \mathfrak{g}_2 \mid X\in \mathfrak{g}_{1,\alpha_1},\:Y\in \mathfrak{g}_{2,\alpha_2}\rbrace$, which is non-zero by our choice of $\alpha_1,\alpha_2$.
But if this was true, the roots wouldn't decompose into two orthogonal subsets, and the whole idea of classifying Dynkin diagrams falls apart, and so on and so forth. What am I missing?
Best Answer
Seems like this was answered in comments:
On an element of the proposed space $\lbrace (X,Y)\in \mathfrak{g}_1\oplus \mathfrak{g}_2 \mid X\in \mathfrak{g}_{1,\alpha_1},\:Y\in \mathfrak{g}_{2,\alpha_2}\rbrace$, a general $H = (H_1,H_2) \in \mathfrak h$ operates via
$$(X,Y) \mapsto (\alpha_1(H_1)X, \alpha_2(H_2)Y).$$
That is not the same as
$$(X,Y) \mapsto ((\alpha_1(H_1)+\alpha_2(H_2))X, (\alpha_1(H_1)+\alpha_2(H_2))Y)$$
but only this would be an operation which would make that space deserve the notation "$\mathfrak{g}_{\alpha_1 \oplus \alpha_2}$".