Highest-root dominance and the fundamental alcove

Let $v$ be the regular vector defining ${\mathrm{\Phi }}^{+}$. Its orthogonal projection to $E_i$ has the same nonzero pairings with roots in ${\mathrm{\Phi }}_i$, so ${\mathrm{\Phi }}_i^{+}={\mathrm{\Phi }}_i\cap {\mathrm{\Phi }}^{+}$ is a positive system on ${\mathrm{\Phi }}_i$. Its simple roots are ${\mathrm{\Delta }}_i=\mathrm{\Delta }\cap {\mathrm{\Phi }}_i$: if $\gamma \in {\mathrm{\Phi }}_i^{+}$ is a sum $\gamma =\beta +\delta$ of positive roots of $\mathrm{\Phi }$, project the equality to each $E_j$ with $j\mathrm{\ne }i$. Each positive root lies in one component; if exactly one summand lies in $E_j$, its projection is nonzero, and if both lie in $E_j$, their sum is nonzero because $B(v,\beta +\delta )>0$. Thus neither summand lies outside $E_i$. Decomposability in ${\mathrm{\Phi }}_i$ is therefore equivalent to decomposability in $\mathrm{\Phi }$. Finally, each root in ${\mathrm{\Phi }}_i$ has only ${\mathrm{\Delta }}_i$ coordinates in the basis $\mathrm{\Delta }$, so ${\mathrm{\Delta }}_i$ spans $E_i$; it is linearly independent as a subset of $\mathrm{\Delta }$. Hence the simple-root basis splits into bases ${\mathrm{\Delta }}_i$ of the $E_i$.

Fix $i$. Write ${\theta }_i={\sum }_{s\in S_i}{n}_{i,s}{\alpha }_s$ with $n_{i,s}\in {\mathbb{Z}}_{\ge 0}$, not all zero, by [F2]. If its support $T=\{s:n_{i,s}>0\}$ were proper, then for every $s\in S_i\setminus T$, dominance from [F3] and nonpositivity [F7] would give $0\le B({\theta }_i,{\alpha }_s)={\sum }_{t\in T}{n}_{i,t}B({\alpha }_t,{\alpha }_s)\le 0.$ Thus every simple root in $T$ is orthogonal to every simple root in $S_i\setminus T$, so by [F8] the connected Dynkin graph would be disconnected. Hence $T=S_i$, and every $n_{i,s}>0$. For $\gamma \in {\mathrm{\Phi }}_i^{+}$, the finite set $P_{\gamma }:=\{\eta \in {\mathrm{\Phi }}_i^{+}:\gamma \le \eta \}$ is nonempty. It has a maximal element $\eta$ because it is finite. If $\eta \le {\eta }^{\mathrm{\prime }}$ for a positive root ${\eta }^{\mathrm{\prime }}$, then transitivity gives $\gamma \le {\eta }^{\mathrm{\prime }}$, so ${\eta }^{\mathrm{\prime }}\in P_{\gamma }$ and maximality forces ${\eta }^{\mathrm{\prime }}=\eta$. Thus $\eta$ is maximal among all positive roots, and uniqueness of the highest root in [F3] gives $\eta ={\theta }_i$. By the root-order definition in [F3], ${\theta }_i-\gamma ={\sum }_{s\in S_i}{m}_s{\alpha }_s$ with $m_s\in {\mathbb{Z}}_{\ge 0}$. For $x\in C$, each $B(x,{\alpha }_s)>0$, so $B(x,\gamma )>0$ by [F2] and $B(x,{\theta }_i)-B(x,\gamma )={\sum }_{s\in S_i}{m}_{s}B(x,{\alpha }_s)\ge 0.$ The dominance inequality $B({\theta }_i,\gamma )\ge 0$ is [F3]. If $B(x,{\theta }_i)<1$, the displayed bound gives $B(x,\gamma )<1$ as well.

Fix $i$, write $n=\mathrm{\mid }{S}_i\mathrm{\mid }\ge 1$, and define the linear map $T_i:E_i\to {\mathbb{R}}^{S_i}$ by $T_i(x{)}_s=n_{i,s}B(x,{\alpha }_s)$. Its kernel is zero: if every coordinate vanishes, then $x$ is orthogonal to the basis ${\mathrm{\Delta }}_i$, hence to all of $E_i$, and positive definiteness gives $x=0$. The domain and codomain both have dimension $n$, so $T_i$ is bijective. Its coordinate functionals are continuous by Cauchy–Schwarz, so $A_i=T_i^{-1}(\{u:u_s>0,{\sum }_s{u}_s<1\})$ is open. It is nonempty, since $u_s=1\mathrm{/}(2n)$ gives a point of it, and it is convex; it is exactly the interior of the closed coordinate simplex below. Its closure is $T_i^{-1}(\{u:u_s\ge 0,{\sum }_s{u}_s\le 1\})$: continuity gives one inclusion; for any $u$ in that closed set, put $x=T_i^{-1}(u)$ and $x^{\ast }=T_i^{-1}(u^{\ast })$, where $u_s^{\ast }=1\mathrm{/}(2n)$. Then $x_t=(1-t)x+t{x}^{\ast }$ lies in $A_i$ for $0<t\le 1$, and $\mathrm{\parallel }{x}_t-x{\mathrm{\parallel }}_B=t\mathrm{\parallel }{x}^{\ast }-x{\mathrm{\parallel }}_B\to 0$, proving the other inclusion. The closed coordinate simplex is the convex hull of $0$ and the standard coordinate vectors $e_s$, which are affinely independent; its inverse image is therefore a geometric simplex. Every point in the closure is ${\sum }_s{u}_s{T}_i^{-1}(e_s)$ with $0\le u_s\le 1$, so by [F9] its norm is at most ${\sum }_s\mathrm{\parallel }{T}_i^{-1}(e_s){\mathrm{\parallel }}_B$. Its barycentric coordinates are $u_s$ and $u_0=1-{\sum }_s{u}_s$. Each of the $n+1$ equalities $u_s=0$ or $u_0=0$ cuts out an $(n-1)$-dimensional face; every boundary point satisfies one of these equalities, so these are exactly the facets. These coordinate facets correspond to $H_{{\alpha }_s,0}$ and $H_{{\theta }_i,1}$.

On $A_i$, step 1.2 gives $0<B(x,\gamma )<1$ for every positive root $\gamma \in {\mathrm{\Phi }}_i^{+}$; for a negative root $-\gamma$ it gives $-1<B(x,-\gamma )<0$. Hence no integer level $H_{\beta ,k}$ for $\beta \in {\mathrm{\Phi }}_i$ meets $A_i$. For nonempty $J$ equal to a singleton $\{i\}$ or to all components $I$, the set $A_J:={\prod }_{j\in J}{A}_j$ is nonempty, convex, and avoids every wall of the arrangement on $E_J:={⨁}_{j\in J}{E}_j$. Let $D_J$ be the connected component of that wall complement containing a point $x\in A_J$. For each simple root ${\alpha }_s$ in these components, the continuous function $y\mapsto B(y,{\alpha }_s)$ has no zero on $D_J$ and is positive at $x$, so its negative and positive preimages cannot both be nonempty, since they would separate $D_J$; it therefore stays positive throughout $D_J$. For each $j\in J$, the function $y\mapsto B(y,{\theta }_j)-1$ likewise has no zero on $D_J$ and is negative at $x$, so it stays negative. Thus $D_J\subseteq A_J$. Since $A_J$ is connected and meets $D_J$, maximality of the connected component gives $A_J\subseteq D_J$, hence $A_J=D_J$. Taking singleton $J$ proves that each $A_i$ is an alcove; taking $J=I$ proves that $A$ is an alcove for the full arrangement when $I$ is nonempty. By [F4] these components are open, as claimed.

If $\mathrm{\Phi }=\mathrm{\varnothing }$, then $E=0$ by [F1], the wall arrangement is empty, and its complement has the single component $\{0\}$. If ${\mathrm{\Phi }}_i=\{\pm {\alpha }_s\}$ is of type $A_1$, its positive roots contain only its simple root ${\alpha }_s$, so ${\theta }_i={\alpha }_s$ and the two inequalities defining $A_i$ are exactly $0<B(x,{\alpha }_s)<1$. The product statement already established applies independently to every component, including any collection of $A_1$ factors. The proof constructs only finite component decompositions, finite coordinate vectors, and finite sums; no axiom of choice is used.