Unique irreducible decomposition

Define a graph $G$ with vertex set $\mathrm{\Phi }$, two distinct vertices $\alpha ,\beta$ being joined by an edge exactly when $(\alpha ,\beta )\mathrm{\ne }0$. Let $C_1,\dots ,C_m$ be the connected components of $G$, with $m=0$ if $\mathrm{\Phi }$ is empty, so that $\mathrm{\Phi }=C_1\bigsqcup \cdots \bigsqcup C_m$ and every $C_i$ is nonempty.

If $\alpha \in C_i$ and $\beta \in C_j$ with $i\mathrm{\ne }j$, then $(\alpha ,\beta )=0$, since otherwise an edge would join the two vertices and they would lie in one component. Consequently $\mathrm{span}{C}_i\perp \mathrm{span}{C}_j$ for $i\mathrm{\ne }j$, and $E=\mathrm{span}\mathrm{\Phi }=\mathrm{span}{C}_1\oplus \cdots \oplus \mathrm{span}{C}_m$ is an orthogonal direct sum.

For uniqueness, let $\mathrm{\Phi }={\mathrm{\Psi }}_1\bigsqcup \cdots \bigsqcup {\mathrm{\Psi }}_k$ with each ${\mathrm{\Psi }}_j$ a reduced crystallographic root system in $F_j=\mathrm{span}{\mathrm{\Psi }}_j$, the $F_j$ pairwise orthogonal, and $E={⨁}_j{F}_j$. If $\alpha \in {\mathrm{\Psi }}_j$ and $\beta \in {\mathrm{\Psi }}_{j^{\mathrm{\prime }}}$ with $j\mathrm{\ne }{j}^{\mathrm{\prime }}$ then $(\alpha ,\beta )=0$ because $F_j\perp F_{j^{\mathrm{\prime }}}$; hence no edge of $G$ joins distinct parts, and each connected component $C_i$ of $G$ is contained in a single ${\mathrm{\Psi }}_j$.

For each $i$ one has $C_i=\mathrm{\Phi }\cap \mathrm{span}{C}_i$. Indeed, if $\gamma \in \mathrm{\Phi }\cap \mathrm{span}{C}_i$ then $\gamma ={\sum }_{\alpha \in C_i}{c}_{\alpha }\alpha$; if $\gamma \mathrm{\notin }{C}_i$ then $(\gamma ,\alpha )=0$ for every $\alpha \in C_i$ by step 2.1 applied to the components, whence $(\gamma ,\gamma )={\sum }_{\alpha }{c}_{\alpha }(\gamma ,\alpha )=0$ and $\gamma =0$, contradicting $0\mathrm{\notin }\mathrm{\Phi }$.

Each $C_i$ is a reduced crystallographic root system in $E_i=\mathrm{span}{C}_i$: this is [L3] applied to $V=E_i$, whose intersection with $\mathrm{\Phi }$ is $C_i$ by step 3.1, and $C_i$ spans $E_i$ by definition. Moreover $C_i$ is irreducible: if $C_i=(C_i\cap U)\bigsqcup (C_i\cap V)$ came from an orthogonal decomposition $E_i=U\oplus V$ with both summands nonzero, then no edge of $G$ would join a vertex in $C_i\cap U$ to a vertex in $C_i\cap V$, so the graph $G$ restricted to $C_i$ would be disconnected, contradicting that $C_i$ is a component of $G$.

Conversely each ${\mathrm{\Psi }}_j$ is a union of components: if $\gamma \in {\mathrm{\Psi }}_j$ then by the argument of step 3.1 applied inside the subsystem ${\mathrm{\Psi }}_j$, the whole component $C_i\ni \gamma$ of the graph $G$ lies in ${\mathrm{\Psi }}_j$, since a root of $\mathrm{\Phi }$ nonorthogonal to $\gamma$ must lie in $F_j$ (it is orthogonal to every other $F_{j^{\mathrm{\prime }}}$). Hence ${\mathrm{\Psi }}_j=\bigcup \{C_i:C_i\subseteq {\mathrm{\Psi }}_j\}$.

Steps 3.1 and 4.1 exhibit $\mathrm{\Phi }$ as the disjoint union of the irreducible root systems $C_1,\dots ,C_m$, whose spans are pairwise orthogonal and span $E$; this is the asserted decomposition.

If each ${\mathrm{\Psi }}_j$ is irreducible, discard all empty ${\mathrm{\Psi }}_j$ (whose spans are zero). Each remaining ${\mathrm{\Psi }}_j$ is by step 4.2 a nonempty union of components, and by step 4.1 each component is irreducible; an irreducible root system cannot be the orthogonal disjoint union of two nonempty root subsystems, so ${\mathrm{\Psi }}_j$ contains exactly one component. Therefore the components $C_1,\dots ,C_m$ are a permutation of the nonempty parts ${\mathrm{\Psi }}_j$, and the decomposition into nonempty components is unique up to order. If $\mathrm{\Phi }=\mathrm{\varnothing }$, every ${\mathrm{\Psi }}_j$ is empty and deleting them leaves exactly the empty decomposition with $E=0$.