Finite noncrossing intervals are lattices, independently of the Coxeter element

(Rank one.) Suppose $S=\{s\}$. By [F8], $W$ has at most two elements; the map $s\mapsto -1$ to the group $\{1,-1\}$ satisfies the presentation, so $W=\{1,s\}$. The group is abelian, so $T=\{s\}$, and the bipartite element is $\gamma =s$. By [F7], $\mathrm{\Phi }=\{e_s,-e_s\}$, ${\mathrm{\Phi }}_{+}=\{e_s\}$, the reflection with normal $e_s$ is $s$, and $M(s)=\mathbb{R}{e}_s$. Thus $P_1=\mathrm{\varnothing }$, $P_s=\{e_s\}$, and $X(s)$ has one vertex $e_s$, while $X(1)$ has empty realization. If either $a=1$ or $b=1$, then $a\wedge b=1$ and both identities hold. Otherwise $a=b=s$, whose only common lower bounds are $1,s$, so $a\wedge b=s$. Its unique maximal simplex is $F=\{e_s\}$ and its reverse reflection product is $R(e_s)=s$. Thus $M(s)=\mathrm{span}(F)=\mathrm{span}(\mathrm{\mid }X(s)\mathrm{\mid })$ and $P_s=P_s\cap P_s$. This proves every clause of (1) in rank one.

(Identity and empty intersections in rank at least two.) Assume $\mathrm{\mid }S\mathrm{\mid }\ge 2$. If $a=1$ or $b=1$, the only element below $1$ is $1$, so $a\wedge b=1$; by [F2], $M(1)=\{0\}$, $P_1=\mathrm{\varnothing }$, and $\mathrm{\mid }X(1)\mathrm{\mid }=\mathrm{\varnothing }$, so both displayed identities hold. Now suppose $a,b\mathrm{\ne }1$ and $P_a\cap P_b=\mathrm{\varnothing }$. Any common lower bound $\tau$ has $P_{\tau }\subseteq P_a\cap P_b=\mathrm{\varnothing }$ by transitivity. By [F2], $M(\tau )=\mathrm{span}(P_{\tau })=\{0\}$, so ${\mathrm{\ell }}_T(\tau )=0$ by [F1] and $\tau =1$. Thus $1$ is the greatest common lower bound and both identities again hold. To see that this case occurs, take $m(s,t)=3$, $\gamma =st$, $a=s$, $b=t$. By [F9], $st$ has order $3$, so it is neither the identity nor a reflection, since every reflection is conjugate to a simple involution. As it is a product of two reflections, ${\mathrm{\ell }}_T(\gamma )=2$. Also $s^{-1}\gamma =t$ and $t^{-1}\gamma =tst$ are reflections, so $s,t{\le }_T\gamma$. By [F10], $P_s=\{e_s\}$ and $P_t=\{e_t\}$, which are distinct because the simple roots are linearly independent. Thus $P_s\cap P_t=\mathrm{\varnothing }$.

(The nonempty common face in rank at least two.) Assume $\mathrm{\mid }S\mathrm{\mid }\ge 2$ and $P_a\cap P_b\mathrm{\ne }\mathrm{\varnothing }$, set $Y=X(a)$, $Z=X(b)$, and let $C=c[Y]\cap c[Z]$. By [F4], $c[X(a)]=c[P_a]$ and $c[X(b)]=c[P_b]$; each is a positive cone, so $C$ is convex. The common-root set gives a common vertex, and [F4] identifies $\mathrm{\mid }Y\cap Z\mathrm{\mid }=\mathrm{\mid }Y\mathrm{\mid }\cap \mathrm{\mid }Z\mathrm{\mid }$. Choose a maximal simplex $F=\{v_1<\cdots <v_r\}$ of $Y\cap Z$. It is a simplex of $X(\gamma )$, so [F3] gives $\sigma =R(v_r)\cdots R(v_1){\le }_T\gamma$ and ${\mathrm{\ell }}_T(\sigma )=r$. By [F5] and the purity conclusion of the preceding item, $M(\sigma )=\mathrm{span}(F)=L:=\mathrm{span}(C)$. Since $C\subseteq c[X(a)]=c[P_a]$ and $\mathrm{span}(P_a)=M(a)$ by [F2], one has $M(\sigma )\subseteq M(a)$; similarly $M(\sigma )\subseteq M(b)$. With $\sigma ,a,b{\le }_T\gamma$, rigidity [F1] gives $\sigma {\le }_{T}a,b$. Now $P_{\sigma }\subseteq P_a\cap P_b$ by transitivity. Conversely, each $\alpha \in P_a\cap P_b$ is a common vertex of $Y$ and $Z$, hence belongs to $\mathrm{\mid }Y\cap Z\mathrm{\mid }\subseteq C$ and to $L=M(\sigma )$. Thus $M(t_{\alpha })=\mathrm{span}(\alpha )\subseteq M(\sigma )$; since $t_{\alpha },\sigma {\le }_T\gamma$, rigidity gives $t_{\alpha }{\le }_T\sigma$, so $\alpha \in P_{\sigma }$. Hence $P_{\sigma }=P_a\cap P_b$. If $\tau {\le }_{T}a,b$, then $P_{\tau }\subseteq P_{\sigma }$, so $M(\tau )=\mathrm{span}(P_{\tau })\subseteq M(\sigma )$ by [F2]; rigidity gives $\tau {\le }_T\sigma$. Therefore $\sigma =a\wedge b$. Finally, $\mathrm{span}(\mathrm{\mid }Y\mathrm{\mid }\cap \mathrm{\mid }Z\mathrm{\mid })=\mathrm{span}(C)$ because every nonzero point of the cone normalizes into its sphere section. This proves all nonempty-case identities.

(Joins.) Let $U=\{x\in [1,\gamma ]:a{\le }_{T}x,b{\le }_{T}x\}$. It is nonempty because $\gamma \in U$, and finite because $W$ is finite. Iterating the binary meet established in steps 1.1–1.3 gives the greatest lower bound $m$ of $U$. Since $a$ and $b$ are lower bounds of every member of $U$, they satisfy $a,b{\le }_{T}m$; and $m{\le }_{T}x$ for every common upper bound $x$. Thus $m$ is the least common upper bound, $a\vee b$. By induction on cardinality, the iterated binary meet of any nonempty finite subset is its greatest lower bound: this is immediate for a singleton, and adjoining one element replaces the existing meet $m$ by $m\wedge x$. This proves (2).

(Conjugacy and independence of $c$.) For any finite-type $W$ and Coxeter elements $c,c^{\mathrm{\prime }}$, [F11] gives $c^{\mathrm{\prime }}=wc{w}^{-1}$ for some $w\in W$. Conjugation maps $T$ bijectively to itself, so it preserves ${\mathrm{\ell }}_T$ and ${\le }_T$; its inverse is conjugation by $w^{-1}$. Hence it is an order isomorphism of the two intervals. Also $\rho (wx{w}^{-1})-\mathrm{i}\mathrm{d}=\rho (w)(\rho (x)-\mathrm{i}\mathrm{d})\rho (w{)}^{-1}$, so $M(wx{w}^{-1})=\rho (w)M(x)$. An order isomorphism preserves greatest lower bounds and least upper bounds by their defining universal properties, and therefore is a lattice isomorphism once the bipartite interval is known to be a lattice. This proves (4) and transfers (1)–(2) to every Coxeter element in the connected case.

(Reducible systems.) If $S=\mathrm{\varnothing }$, then $W=\{1\}$, the interval and the empty product are both one-element lattices. Otherwise use the component decomposition and interval identity [F6]. Each $W_{S_i}$ is a connected finite-type Coxeter group, so its noncrossing interval is a finite lattice by steps 1.1–1.3, 2.1, and 3.1. Componentwise meets and joins make the finite product a lattice. This proves (3) and completes the theorem.