Irreducibility and connected Dynkin diagrams

Suppose first that $\mathrm{\Phi }\mathrm{\ne }\mathrm{\varnothing }$. If $\mathrm{\Gamma }$ is disconnected, partition its vertices as $\mathrm{\Delta }=S\bigsqcup T$ into two nonempty unions of connected components. Then $(S,T)=0$ by [L3], and $U=\mathrm{span}S$, $W=\mathrm{span}T$ are nonzero orthogonal subspaces with $E=U\oplus W$ by [L2]. Every root has support in just one side. Otherwise, after replacing a root by its negative if necessary, choose a positive root $\beta$ of least height whose support meets both $S$ and $T$. It is not simple, so [L4] writes $\beta =\gamma +\delta$ for positive roots $\gamma ,\delta$. Minimality makes each summand supported on one side, and because $\beta$ is mixed they lie on opposite sides; hence $(\gamma ,\delta )=0$. Reflection in $\gamma$ then gives $s_{\gamma }(\beta )=\beta -\frac{2(\beta ,\gamma )}{(\gamma ,\gamma )}\gamma =\delta -\gamma \in \mathrm{\Phi },$ but $\delta -\gamma$ has nonzero simple-root coefficients of both signs, contradicting [L2]. Thus $\mathrm{\Phi }=(\mathrm{\Phi }\cap U)\bigsqcup (\mathrm{\Phi }\cap W)$ is an orthogonal splitting with both parts nonempty, and $\mathrm{\Phi }$ is reducible.

If $\mathrm{\Phi }$ is reducible, write $\mathrm{\Phi }={\mathrm{\Phi }}_1\bigsqcup {\mathrm{\Phi }}_2$ as an orthogonal union of nonempty subsystems spanning orthogonal nonzero subspaces by [L1]. Put ${\mathrm{\Delta }}_i=\mathrm{\Delta }\cap {\mathrm{\Phi }}_i$. Every simple root belongs to exactly one ${\mathrm{\Phi }}_i$, so $\mathrm{\Delta }={\mathrm{\Delta }}_1\bigsqcup {\mathrm{\Delta }}_2$. Each ${\mathrm{\Delta }}_i$ is nonempty: choose a positive root $\beta \in {\mathrm{\Phi }}_i$ and expand it in the basis $\mathrm{\Delta }$ using [L2]; orthogonal projection to the other component, together with linear independence of the simple roots there, forces all coefficients from ${\mathrm{\Delta }}_{3-i}$ to vanish, while $\beta \mathrm{\ne }0$ leaves a coefficient from ${\mathrm{\Delta }}_i$. Since $({\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2)=0$, no edge of $\mathrm{\Gamma }$ joins ${\mathrm{\Delta }}_1$ to ${\mathrm{\Delta }}_2$, and $\mathrm{\Gamma }$ is disconnected.

For $\mathrm{\Phi }\mathrm{\ne }\mathrm{\varnothing }$, steps 1.1 and 1.2 prove the two implications by contraposition, so irreducibility is equivalent to connectedness of $\mathrm{\Gamma }$. For $\mathrm{\Phi }=\mathrm{\varnothing }$, the spanning axiom gives $E=0$ and the base and diagram are empty; the zero space has no splitting into two nonzero subspaces, so this root system is irreducible by [L1]. Conversely an empty diagram gives $E=0$ by [L2], hence $\mathrm{\Phi }=\mathrm{\varnothing }$. Thus in all ranks irreducibility is equivalent to the diagram being empty or connected.