Existence and uniqueness of the highest root

The set ${\mathrm{\Phi }}^{+}$ is finite and nonempty: since $\mathrm{\Phi }$ is nonempty, choose a root and its negative, exactly one of which is positive. Hence the root order is a partial order, and a comparison chain of positive roots is finite because heights strictly increase along it; hence ${\mathrm{\Phi }}^{+}$ has a maximal element $\theta$, i.e. a positive root such that $\theta \le \gamma$ for no positive root $\gamma \mathrm{\ne }\theta$.

Let $W_0$ be the subgroup of the orthogonal group generated by the simple reflections; every root is in its orbit of a simple root. The generators are $s_{\alpha }$ for $\alpha \in \mathrm{\Delta }$. Let $\gamma ={\sum }_i{n}_i{\alpha }_i$, $n_i\ge 0$, be a positive root that is not simple. Then $0<(\gamma ,\gamma )={\sum }_i{n}_i(\gamma ,{\alpha }_i)$ produces an index $i$ with $(\gamma ,{\alpha }_i)>0$; put $k=\frac{2(\gamma ,{\alpha }_i)}{({\alpha }_i,{\alpha }_i)}>0$, so that the reflected root $s_{{\alpha }_i}(\gamma )=\gamma -k{\alpha }_i$ lies in $\mathrm{\Phi }$ by [L1]. Its simple-root coordinates are those of $\gamma$ except the $i$-th, which is $n_i-k$. If $n_i-k<0$, the one-sign assertion of [L2] forces every other coordinate, which is unchanged and nonnegative, to vanish. Thus $s_{{\alpha }_i}(\gamma )$ is a negative multiple of ${\alpha }_i$, hence equal to $-{\alpha }_i$ by the reducedness clause of [L1]; applying $s_{{\alpha }_i}$ again would then give $s_{{\alpha }_i}(-{\alpha }_i)={\alpha }_i=\gamma$, contrary to the choice of $\gamma$. Therefore $s_{{\alpha }_i}(\gamma )$ is a positive root, of height $\mathrm{ht}(\gamma )-k<\mathrm{ht}(\gamma )$. Iterating this strict descent, which stays at height $\ge 1$ while the root is positive, must end at a simple root, since the argument would strictly lower the height of any nonsimple positive root; a simple root has height $1$. Hence $\gamma =w(\delta )$ for a simple root $\delta$ and a product $w$ of simple reflections. Negative roots are negatives of positive ones, and $-\delta =s_{\delta }(\delta )$.

The maximal root $\theta$ is dominant: $(\theta ,{\alpha }_i)\ge 0$ for every simple root ${\alpha }_i$. Indeed, if $(\theta ,{\alpha }_i)<0$, then the positive roots $\theta ,{\alpha }_i$ cannot be proportional: reducedness would force equality, giving a positive pairing. Thus $\theta +{\alpha }_i\in \mathrm{\Phi }$ by [L3] applied to the nonproportional pair $({\alpha }_i,\theta )$; this root is positive (a sum of positive roots) and $\theta +{\alpha }_i>\theta$ because the difference is the simple root ${\alpha }_i$, contradicting maximality of $\theta$.

The Dynkin diagram of $\mathrm{\Phi }$, with vertices $\mathrm{\Delta }$ and an edge between ${\alpha }_i,{\alpha }_j$ when $({\alpha }_i,{\alpha }_j)\mathrm{\ne }0$, is connected. Otherwise $\mathrm{\Delta }=S\bigsqcup T$ with $({\alpha }_i,{\alpha }_j)=0$ for all $i\in S$, $j\in T$; then each $s_{{\alpha }_i}$, $i\in S$, fixes every ${\alpha }_j$, $j\in T$, and vice versa, so the subgroup generated by the two families is their commuting product $W_S{W}_T$ and $W_0\mathrm{\Delta }=(W_{S}S)\bigsqcup (W_{T}T)$ by step 1.2; the spans of $W_{S}S\subseteq \mathrm{span}S$ and $W_{T}T\subseteq \mathrm{span}T$ are nonzero, orthogonal, and span $E$, so $\mathrm{\Phi }$ would be reducible, contradicting [L5].

The support of $\theta$ is all of $\mathrm{\Delta }$: write $\theta ={\sum }_i{n}_i{\alpha }_i$ with $n_i\ge 0$. If $n_i=0$ for some $i$, then by step 2.1 and [L4] $0\le (\theta ,{\alpha }_i)={\sum }_{j:n_j>0}{n}_j({\alpha }_j,{\alpha }_i)\le 0$, so $({\alpha }_j,{\alpha }_i)=0$ for every $j$ with $n_j>0$, that is, no vertex outside the nonempty support of $\theta$ is adjacent in the diagram to any vertex of the support; this contradicts the connectedness of step 2.2.

At least one simple root pairs strictly positively with $\theta$. Indeed step 3.1 writes $\theta ={\sum }_i{n}_i{\alpha }_i$ with every $n_i>0$, while step 2.1 gives $(\theta ,{\alpha }_i)\ge 0$ for every $i$. Since $0<(\theta ,\theta )={\sum }_i{n}_i(\theta ,{\alpha }_i),$ not all these nonnegative pairings can vanish.

Uniqueness: let ${\theta }^{\mathrm{\prime }}$ be a second highest root. Steps 2.1 through 3.1 apply equally to ${\theta }^{\mathrm{\prime }}$, so ${\theta }^{\mathrm{\prime }}={\sum }_i{c}_i{\alpha }_i$ with every $c_i>0$. Together with steps 2.1 and 4.1 this gives $(\theta ,{\theta }^{\mathrm{\prime }})={\sum }_i{c}_i(\theta ,{\alpha }_i)>0.$ If $\theta$ and ${\theta }^{\mathrm{\prime }}$ were proportional, reducedness and positivity would already force $\theta ={\theta }^{\mathrm{\prime }}$. Otherwise [L3] gives $\theta -{\theta }^{\mathrm{\prime }}\in \mathrm{\Phi }$; this root is positive, in which case ${\theta }^{\mathrm{\prime }}<\theta$ and ${\theta }^{\mathrm{\prime }}$ is not maximal, or negative, in which case $\theta <{\theta }^{\mathrm{\prime }}$ and $\theta$ is not maximal. Both alternatives are impossible, so ${\theta }^{\mathrm{\prime }}=\theta$.

The dominance statement for all positive roots follows because $(\theta ,\gamma )={\sum }_i{c}_i(\theta ,{\alpha }_i)\ge 0$ whenever $\gamma ={\sum }_i{c}_i{\alpha }_i$ is positive with $c_i\ge 0$, using step 2.1. Every positive root lies below a maximal root by finiteness; uniqueness makes that maximal root $\theta$, so $\theta$ is also the greatest positive root. In rank one $\mathrm{\Phi }=\{\alpha ,-\alpha \}$, and $\theta =\alpha$. The empty system in $E=0$ remains irreducible under the library definition but is expressly excluded here; it has no highest root. Dominance need not be strict on every simple root: in $A_3$, $\theta ={\epsilon }_1-{\epsilon }_4$ pairs to zero with ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. This completes the proof.