The factorization criterion, linear independence of the faces, and the geometric simplicial structure of X(sigma)

By [F1], $M(c)=V$ and ${\mathrm{\ell }}_T(c)=n$. For every $a\in {\mathrm{\Phi }}_{+}$, [F2] gives $t_a$ with $\rho (t_a)=R(a)$, and [F8] gives its moved line $\mathbb{R}a$. Apply the subspace-restriction theorem [F5] to this line inside $M(c)$; its orthogonal restriction is the unique reflection with normal $a$, so Carter's formula gives $t_a{\le }_{T}c$. Thus every positive-root reflection lies below $c$.

Let $q=t_1\cdots t_m$ be a reduced reflection factorization. Each factor $t_r$ lies below $q$: if $q=x{t}_{r}y$, then $t_{r}q=(t_{r}x{t}_r)y$ is a product of $m-1$ reflections, so the triangle inequality forces ${\mathrm{\ell }}_T(t_{r}q)=m-1$. If $r<s$, the product $t_r{t}_s$ has reflection length $2$ when $t_r\mathrm{\ne }{t}_s$: their canonical images are distinct involutions with distinct moved lines, so $\rho (t_r)\rho (t_s)\mathrm{\ne }I$; its determinant is $+1$, whereas every reflection has determinant $-1$. Thus $t_r{t}_s$ is neither the identity nor a reflection, and its reflection length is $2$. Write $q=x{t}_{r}y{t}_{s}z$. Then $(t_r{t}_s{)}^{-1}q=t_s{t}_{r}x{t}_{r}y{t}_{s}z=(t_s{t}_{r}x{t}_r{t}_s)(t_{s}y{t}_s)z,$ a product of $m-2$ reflections. The triangle inequality gives the reverse lower bound $m-2$, so $t_r{t}_s{\le }_{T}q$.

Let $q:=R(a_k)\cdots R(a_1)$. If ${\mathrm{\ell }}_T(R(a_1)\cdots R(a_k)c)=n-k$, then ${\mathrm{\ell }}_T(q)\le k$ and [F1], [F3] give $n={\mathrm{\ell }}_T(c)\le {\mathrm{\ell }}_T(q)+{\mathrm{\ell }}_T(q^{-1}c)\le k+(n-k)=n\mathrm{.}$ Thus ${\mathrm{\ell }}_T(q)=k$ and $q{\le }_{T}c$, so the factorization of $q$ is reduced. For $i>j$, step 1.2 gives $R(a_i)R(a_j){\le }_{T}q{\le }_{T}c$. Since $R(a_i){\le }_{T}c$ by step 1.1, ${\mathrm{\ell }}_T(R(a_i)c)=n-1$; also ${\mathrm{\ell }}_T(R(a_j)R(a_i)c)=n-2$. Hence $R(a_j){\le }_{T}R(a_i)c$. By [F2] and [F8] the moved line of $R(a_j)$ is $\mathbb{R}{a}_j$; by [F3] and [F4] it lies in $M(R(a_i)c)=F(R(a_i)c{)}^{\perp }$. The vector $\mu (a_i)$ spans that fixed line by [F6], so $\mu (a_i)\cdot a_j=0$.

Conversely, suppose $\mu (a_i)\cdot a_j=0$ for all $i>j$. The matrix $D=(\mu (a_i)\cdot a_j{)}_{i,j=1}^k$ is upper triangular with diagonal $1$ by [F6]; therefore both the roots $a_i$ and the vectors $\mu (a_i)$ are linearly independent, so $k\le n$. For $k=0$ the conclusion is ${\mathrm{\ell }}_T(c)=n$ by [F3], so assume $k\ge 1$. Define $q_{k+1}=1$. Descending on $i$, assume $q_{i+1}:=R(a_k)\cdots R(a_{i+1}){\le }_{T}c$ with length $k-i$. Each factor $R(a_j)$, $j>i$, is below $q_{i+1}$ by the first claim of step 1.2. Complements reverse order: if $u{\le }_{T}v{\le }_{T}w$, transitivity gives $u{\le }_{T}w$; writing $v=ux$ and $w=vy$ with additive reflection lengths then gives ${\mathrm{\ell }}_T(xy)={\mathrm{\ell }}_T(x)+{\mathrm{\ell }}_T(y)$, and conjugation invariance gives ${\mathrm{\ell }}_T(y^{-1}xy)={\mathrm{\ell }}_T(x)$, hence $y=v^{-1}w{\le }_T{u}^{-1}w=xy$. Put $v:=q_{i+1}^{-1}c$. For every $j>i$, complement reversal gives $v{\le }_{T}R(a_j)c$, and [F6] puts $\mu (a_j)$ in $F(R(a_j)c)\subseteq F(v)$. These $k-i$ independent vectors form a basis of $F(v)$: Carter's formula gives $\dim F(v)=n-{\mathrm{\ell }}_T(v)=n-(n-k+i)=k-i\mathrm{.}$ The assumed zero pairings put $a_i$ in $F(v{)}^{\perp }=M(v)$ by [F4]. Restricting $\rho (v)$ to the line $\mathbb{R}{a}_i$ gives the orthogonal reflection $R(a_i)=\rho (t_{a_i})$ by [F2]; the restriction theorem [F5] and Carter's formula therefore give $R(a_i){\le }_{T}v$. Write $v=R(a_i)z$ with ${\mathrm{\ell }}_T(v)=1+{\mathrm{\ell }}_T(z)$. Then $c=q_{i+1}R(a_i)z$, whose displayed $k-i+1+{\mathrm{\ell }}_T(z)=n$ reflection factors force the factorization to be reduced; consequently $q_i:=q_{i+1}R(a_i){\le }_{T}c$ and ${\mathrm{\ell }}_T(q_i)=k-i+1$. At $i=1$ this yields ${\mathrm{\ell }}_T(q)=k$ and ${\mathrm{\ell }}_T(q^{-1}c)=n-k$. This proves (1).

For $i<j$, the two-root instance of (1) says $R(a_j)R(a_i){\le }_{T}c⟺\mu (a_j)\cdot a_i=0,$ since the product of two distinct positive-root reflections has length $2$ by step 1.2. If $F=\{a_1<\cdots <a_k\}$ is a simplex, every pair is an edge by [F10], so these pairwise equivalences make $D=(\mu (a_i)\cdot a_j)$ upper triangular with diagonal $1$. Pairing a vanishing combination ${\sum }_j{\lambda }_j{a}_j=0$ with each $\mu (a_i)$ gives $D\lambda =0$; hence every ${\lambda }_j=0$. Thus the vertices of every nonempty face are linearly independent. Conversely, if the reverse product of an increasing tuple lies below $c$ and has length $k$, step 1.2 applied to its reduced factorization shows each pair product is below $c$, so the tuple is a simplex. The empty tuple is the empty simplex by [F10].

Let $f_0:={\sum }_{s\in S}{\beta }_s$, with the dual vectors from [F13]. Then $B(f_0,{\alpha }_s)=1$ for every simple root; since each positive root is a nonzero nonnegative combination of simple roots by [F7], every positive root has positive pairing with $f_0$. Also [F7] gives positive pairing with every $f\in {\mathcal{C}}^{\circ }$ by the chamber transfer [F17]. For a nonempty face $F$, its independent unit roots have a positive-definite Gram matrix with diagonal $1$. By [F12], $c[F]\cap S^{n-1}$ is the associated spherical simplex of dimension $\mathrm{\mid }F\mathrm{\mid }-1$. The independence of the cone generators makes $c[F]$ pointed and simplicial. For the empty face, [F10] gives $c[\mathrm{\varnothing }]=\{0\}$ and its sphere section is empty.

The roots of $P_{\sigma }$ lie in $M(\sigma )$: if $a\in P_{\sigma }$, then $t_a{\le }_T\sigma$ by definition; [F2] identifies its linear reflection, [F8] gives $M(t_a)=\mathbb{R}a$, and [F3] gives $M(t_a)\subseteq M(\sigma )$. Thus a face of $X(\sigma )$ has at most $\dim M(\sigma )={\mathrm{\ell }}_T(\sigma )$ vertices by step 3.1 and Carter's formula. If $\sigma =1$, then $P_{\sigma }=\mathrm{\varnothing }$ and $X(1)=\{\mathrm{\varnothing }\}$ has dimension $-1$. If $n=1$ and $\sigma \mathrm{\ne }1$, the rank-one data give $W=\{1,c\}$, $\sigma =c$, and $P_{\sigma }=\{{\alpha }_1\}$; hence $X(\sigma )$ has one vertex and dimension $0={\mathrm{\ell }}_T(\sigma )-1$. If $n\ge 2$ and $\sigma \mathrm{\ne }1$, [F9] gives $\mathrm{\Delta }=\{{\delta }_1,\dots ,{\delta }_k\}\subseteq P_{\sigma }$ and ${\epsilon }_i=R({\delta }_1)\cdots R({\delta }_{i-1}){\delta }_i$. Each $R({\delta }_j)$ lies in the reflection subgroup $W_{\sigma }$ of [F9], so every ${\epsilon }_i$ is a root of that subgroup; its positivity from [F9] puts it in $P_{\sigma }$. Hence the increasing reordering ${\theta }_1<\cdots <{\theta }_k$ lies in $P_{\sigma }$. Its reverse product is below $c$ with length $k={\mathrm{\ell }}_T(\sigma )$ by [F9], so step 3.1 makes it a $k$-vertex simplex. Therefore $\dim X(\sigma )=k-1$. Finiteness follows from finiteness of ${\mathrm{\Phi }}_{+}$ in [F1], and the full-subcomplex and $X(\sigma ,\rho )$ claims follow from [F10].

Let $K_i:=X(\sigma ,{\tau }_i)$. By [F10], $K_i$ has vertices ${\tau }_1,\dots ,{\tau }_i$. Any new simplex of $K_{i+1}$ must contain the new vertex ${\tau }_{i+1}$. Its other vertices form a face $B$ of $K_i$, and the two-root criterion of step 3.1 says each such vertex $b$ is joined to ${\tau }_{i+1}$ exactly when $\mu ({\tau }_{i+1})\cdot b=0$. Conversely, any face $B$ of $K_i$ with all vertices in this hyperplane gives a simplex $B\cup \{{\tau }_{i+1}\}$. This includes $B=\mathrm{\varnothing }$, proving the cone-over-link description and the nested inclusions.

Put $K_0:=\{\mathrm{\varnothing }\}$ and induct on $i$ to prove the cone intersection formula for all faces of $K_i$. At $i=0$ both cones equal $\{0\}$. Suppose the formula holds at $i$ and consider faces of $K_{i+1}$. If both are in $K_i$, use induction. Otherwise each new face has the form $B\cup \{{\tau }_{i+1}\}$ with $B\in K_i$ and every vertex of $B$ orthogonal to $\mu ({\tau }_{i+1})$, by step 4.3. For a cone point in such a face, its coefficient on ${\tau }_{i+1}$ is its pairing with $\mu ({\tau }_{i+1})$, because $\mu ({\tau }_{i+1})\cdot {\tau }_{i+1}=1$ by [F6] and its pairings with the base vertices are zero. Every old vertex ${\tau }_j$, $j\le i$, has $\mu ({\tau }_{i+1})\cdot {\tau }_j\le 0$ by [F6]. Therefore a cone on an old face intersects a cone with apex ${\tau }_{i+1}$ only where the apex coefficient is zero; there the induction hypothesis identifies the intersection with the cone on the common base face. For two faces both containing the apex, equality of a common cone point gives equality of its apex coefficients after pairing with $\mu ({\tau }_{i+1})$, and then equality of the base cone points; induction identifies their base intersection. In each case the intersection is exactly the cone on the common face. Since every face of $X(\sigma )$ belongs to $K_t$, this proves the formula in (4).

Define the normalized cone map on a barycentric point of the geometric realization by $\iota (({\lambda }_v{)}_{v\in V(F)}):=\frac{{\sum }_v{\lambda }_{v}v}{{\parallel {\sum }_v{\lambda }_{v}v\parallel }_B},{\lambda }_v\ge 0,{\sum }_v{\lambda }_v=1.$ The denominator is nonzero because $B(f_0,v)>0$ for every positive root by step 4.1. Its restriction to each simplex is a homeomorphism onto the corresponding spherical simplex by [F12], and it is continuous globally by the weak-topology definition [F11]. If two such images agree, the two positive combinations lie on the same ray; step 5.1 puts that ray in the cone on the common face, and linear independence of that face plus the barycentric sum-one condition makes the original points equal. Thus $\iota$ is a continuous bijection onto $\mathrm{\mid }X(\sigma )\mathrm{\mid }$. By [F1] and step 4.2, $X(\sigma )$ is a finite abstract simplicial complex, so its ordinary realization is compact by [F15]. If $A$ is closed in that realization, [F15] makes $A$ compact; the open-cover argument in [F15] makes $\iota [A]$ compact, and it is closed in the Hausdorff sphere by [F15] and [F16]. Thus $\iota$ is a closed continuous bijection onto its image and therefore a topological embedding. No Choice is used: the only compactness input is [F15], whose finite-complex proof reduces to finite-dimensional Heine-Borel.