Intersection of root subcomplexes and purity under convexity

(Cone and realization intersections.) If $x\in c[Y]\cap c[Z]$, then $x\in c[F]$ for some face $F\in Y$ and $x\in c[F^{\mathrm{\prime }}]$ for some face $F^{\mathrm{\prime }}\in Z$. By [F1], $x\in c[F\cap F^{\mathrm{\prime }}]$, and $F\cap F^{\mathrm{\prime }}$ is a common face, so $x\in c[Y\cap Z]$. The reverse inclusion follows from $Y\cap Z\subseteq Y,Z$. Intersecting this cone equality with $S^{n-1}$ gives the realization equality. If there is no common vertex, every common face is empty, so the cone intersection is $c[\mathrm{\varnothing }]=\{0\}$ and its sphere section is empty.

(Closed face cones.) The empty-face cone is $\{0\}$ and is closed. Let $F=\{v_1,\dots ,v_m\}$ be a nonempty face, $U=\mathrm{span}(F)$, and let $G=(⟨v_i,v_j⟩{)}_{i,j}$ be its Gram matrix. For any nonzero coefficient vector $a$, $a^{\mathsf{T}}Ga=\mathrm{\parallel }{\sum }_i{a}_i{v}_i{\mathrm{\parallel }}^2>0$ by independence, so $G$ is invertible. For $x\in U$, its unique coordinate vector in the basis $F$ is $G^{-1}(⟨x,v_j⟩{)}_j$, whose coordinates are continuous by [F3]. Hence $c[F]$ is the intersection of the closed subspace $U$ with the inverse images of the closed half-line $[0,\mathrm{\infty })$ under these coordinate maps; it is closed in $V$. Since $X(c)$ has finitely many faces, $c[Y]$, $c[Z]$, and $C$ are finite unions or intersections of closed face cones and are closed.

(Cone and spherical convexity.) Every nonzero vector of $C$ is a nonnegative combination of positive-root vertices, so the common open-half-space functional in [F1] is strictly positive on it; in particular $K$ contains no antipodal pair. If $C$ is convex, the segment between any $u,v\in K$ lies in $C$ and avoids $0$; normalizing that segment gives the shorter great-circle arc, so $K$ is geodesically convex. Conversely, suppose $K$ is geodesically convex. For nonzero $x,y\in C$, write $x=ru$, $y=sv$ with $r,s>0$ and $u,v\in K$. If $u=v$, then $x+y\in C$. Otherwise the normalized positive combination $(ru+sv)\mathrm{/}\mathrm{\parallel }ru+sv\mathrm{\parallel }$ lies on the shorter arc from $u$ to $v$, hence in $K$, so again $x+y\in C$. Thus $C$ is closed under addition and nonnegative scaling, and is convex.

(Full span of each maximal simplex.) Let $L=\mathrm{span}(C)$. Since $Y\cap Z$ has a vertex, $L\mathrm{\ne }\{0\}$. Choose a maximal simplex $F$ of $Y\cap Z$; it is nonempty. Suppose $U:=\mathrm{span}(F)$ is a proper subspace of $L$. The point $x:={\sum }_{v\in F}v$ has strictly positive coordinates in the independent list $F$. The common vertices span $L$ by step 1.1, so some common vertex $q$ lies outside $U$. For $0<t\le 1$, convexity gives $x_t=(1-t)x+tq\in C$, and $x_t\mathrm{\notin }U$. Take $t_j=1\mathrm{/}j$ for $j\ge 2$. There are finitely many faces of $Y\cap Z$, so one face $F^{\mathrm{\prime }}$ has $x_{t_j}\in c[F^{\mathrm{\prime }}]$ for infinitely many $j$. Along that subsequence $x_{t_j}\to x$, and step 1.2 makes $c[F^{\mathrm{\prime }}]$ closed; hence $x\in c[F^{\mathrm{\prime }}]$. Since $x\in c[F]$, [F1] gives $x\in c[F\cap F^{\mathrm{\prime }}]$. The coordinates of $x$ in the independent family $F$ are all strictly positive, so uniqueness of those coordinates forces $F\subseteq F^{\mathrm{\prime }}$. Maximality gives $F=F^{\mathrm{\prime }}$, contradicting $x_{t_j}\mathrm{\notin }U$. Therefore $\mathrm{span}(F)=L$.

(Empty and one-vertex cases; dimensions.) If $Y\cap Z$ has no vertex, its sole face is $\mathrm{\varnothing }$, step 1.1 gives the empty realization, and the nonempty hypothesis of (2) fails. If it has exactly one vertex $v$, its only nonempty face is $\{v\}$; then $C=c[\{v\}]$ is a ray, its span is $\mathrm{span}(v)$, and the unique maximal simplex spans it. In the general nonempty case, step 2.1 gives $\mathrm{span}(F)=L$ for every maximal simplex. By [F4], $\mathrm{\mid }F\mathrm{\mid }=\dim L$ and $\dim F=\dim L-1$; hence all maximal simplices have the same dimension, as claimed.

The span of $K$ equals $L$: every nonzero point of $C$ is a positive scalar multiple of its normalization in $K$, and $K\subseteq C$. This also verifies the span formulation for the single-vertex case and completes (2)–(3).