Sortable Projections and Finite Cambrian Lattices

Let $A\mathrm{\ne }\mathrm{\varnothing }$ be a set of c-sortable elements and put $I:={\bigcap }_{a\in A}N(a^{-1})$. Since $W$ is finite, $I$ is finite. In each noncommutative generalized rank-two subsystem, [F6]-[F7] put all the traces $N(a^{-1})$ at the same c-oriented end of its angular order; their intersection is therefore empty, the allowed terminal singleton, or an initial segment in that fixed order. In a zero-orientation subsystem each trace is empty or a singleton, so their intersection is again empty or a singleton. Thus $I$ satisfies [F8], so [F8] gives a unique $u\in W$ with $N(u)=I$; put $m:=u^{-1}$, so $N(m^{-1})=I$. For every $a\in A$, $N(m^{-1})\subseteq N(a^{-1})$, so $m{\le }_{R}a$ by [F10]. If $v$ is any lower bound of $A$, then $N(v^{-1})\subseteq I=N(m^{-1})$, so $v{\le }_{R}m$; therefore $m=\bigwedge A$ and the displayed inversion-set identity holds. The same rank-two traces show $m$ is c-aligned, hence c-sortable by [F6].

Fix $s\in S$ and put $D_s:=\{u:s\mathrm{̸}{\le }_{R}u\}$. By [F14], $\mathrm{\ell }(su)=\mathrm{\ell }(u)+1$ for $u\in D_s$ and $\mathrm{\ell }(su)=\mathrm{\ell }(u)-1$ otherwise. Thus $u\mapsto su$ sends $D_s$ into $\{w:s{\le }_{R}w\}$. Conversely, if $w{\ge }_{R}s$, write $w=su$ with $\mathrm{\ell }(w)=1+\mathrm{\ell }(u)$. If $u{\ge }_{R}s$, then [F14] gives $\mathrm{\ell }(su)=\mathrm{\ell }(u)-1$, contradicting this equality, so $u\in D_s$; hence the map is onto. If $u,v\in D_s$ and $u{\le }_{R}v$, write $v=ux$ with additive length. Then $sv=(su)x$ and $\mathrm{\ell }(sv)=\mathrm{\ell }(su)+\mathrm{\ell }(x)$, so $su{\le }_{R}sv$. Conversely, if $su{\le }_{R}sv$, write $sv=(su)x$ with additive length; cancellation gives $v=ux$, and the ascent identities give $\mathrm{\ell }(v)=\mathrm{\ell }(u)+\mathrm{\ell }(x)$, so $u{\le }_{R}v$. Therefore left multiplication by $s$ is an order isomorphism from $D_s$ onto $\{w:s{\le }_{R}w\}$.

Suppose $u{⋖}_{R}v$, $u\mathrm{̸}{\ge }_{R}s$ and $v{\ge }_{R}s$. Then $e_s\in N(v^{-1})\setminus N(u^{-1})$ by [F10]. The inclusion in [F10] and the one-length rise across a cover imply that this difference has one root, so it equals $\{e_s\}$. Write $v=ur$ with $r\in S$ by [F10] and [F14]. By [F11], the unique new prefix root in $N(v^{-1})$ is $\rho (u)e_r$, hence $\rho (u)e_r=e_s$ and $ur{u}^{-1}=s$. Using $r^2=1$ from [F20], $sv=s(ur)=(ur{u}^{-1})(ur)=u=vr$. Thus $s$ is a cover reflection of $v$.

We prove join preservation by lexicographic induction on $(\mathrm{\mid }S\mathrm{\mid },\mathrm{\ell }(x\vee y))$. If $\mathrm{\mid }S\mathrm{\mid }=0$ or $\mathrm{\ell }(x\vee y)=0$, then $W=\{1\}$ or $x=y=1$, respectively, and the identity holds. For every other pair, assume it holds for all pairs of smaller lexicographic measure. This is the induction hypothesis.

Let $A\mathrm{\ne }\mathrm{\varnothing }$ be c-sortable and set $w:=\bigvee A$, which exists by [F12]. For every $a\in A$, $a={\pi }_c(a){\le }_R{\pi }_c(w)$ by [F4]-[F5], so ${\pi }_c(w)$ is an upper bound of $A$ and $w{\le }_R{\pi }_c(w)$. Since ${\pi }_c(w){\le }_{R}w$ by [F4], we get $w={\pi }_c(w)$; hence $w$ is c-sortable. Together with step 1.1 this proves (2).

If $X,Y\mathrm{̸}{\ge }_{R}s$, their rank-one parabolic prefixes are both $1$. By join preservation of the parabolic prefix map in [F13], $Z:=X\vee Y$ also has rank-one prefix $1$, so $Z\mathrm{̸}{\ge }_{R}s$. Step 1.2 shows $sZ$ is an upper bound of $sX,sY$. If $w$ is any common upper bound of $sX,sY$, then $w{\ge }_{R}s$ and we may write $w=s{w}^{\mathrm{\prime }}$ with $w^{\mathrm{\prime }}:=sw$ and $\mathrm{\ell }(w)=1+\mathrm{\ell }(w^{\mathrm{\prime }})$. If $w^{\mathrm{\prime }}{\ge }_{R}s$, [F14] would instead give $\mathrm{\ell }(s{w}^{\mathrm{\prime }})=\mathrm{\ell }(w^{\mathrm{\prime }})-1$, contradicting $s{w}^{\mathrm{\prime }}=w$ and that length equality; hence $w^{\mathrm{\prime }}\mathrm{̸}{\ge }_{R}s$. Step 1.2 now gives $X,Y{\le }_R{w}^{\mathrm{\prime }}$, hence $Z{\le }_R{w}^{\mathrm{\prime }}$, and the same order isomorphism gives $sZ{\le }_{R}w$. Therefore $sX\vee sY=s(X\vee Y)$.

By monotonicity, ${\pi }_c(x\wedge y){\le }_R{\pi }_c(x)\wedge {\pi }_c(y)$. The right side is c-sortable by step 1.1 and is below $x\wedge y$ because ${\pi }_c(x){\le }_{R}x$ and ${\pi }_c(y){\le }_{R}y$. Since ${\pi }_c(x\wedge y)$ is the greatest c-sortable element below $x\wedge y$ by [F4], the reverse inequality holds. Thus ${\pi }_c(x\wedge y)={\pi }_c(x)\wedge {\pi }_c(y)$.

Suppose $x,y\mathrm{̸}{\ge }_{R}s$. The rank-one case of [F13] shows $x\vee y\mathrm{̸}{\ge }_{R}s$, so [F2] computes all projections in $W_J$, where $J=S\setminus \{s\}$. The prefix join identity $(x\vee y{)}_J=x_J{\vee }_{W_J}{y}_J$ from [F13] and the induction hypothesis from step 1.4 applied in the lower-rank parabolic give ${\pi }_c(x\vee y)={\pi }_{c^{\mathrm{\prime }}}(x_J{\vee }_{W_J}{y}_J)={\pi }_{c^{\mathrm{\prime }}}(x_J){\vee }_{W_J}{\pi }_{c^{\mathrm{\prime }}}(y_J)={\pi }_c(x){\vee }_{W_J}{\pi }_c(y)$. These last two outputs lie in $W_J$ by [F2], and their join in $W$ equals their join in $W_J$: if $a,b\in W_J$, then F13,(3) gives $a,b{\le }_R(a{\vee }_{W}b{)}_J$ and $(a{\vee }_{W}b{)}_J=a{\vee }_{W_J}b$, while $(a{\vee }_{W}b{)}_J{\le }_{R}a{\vee }_{W}b$; hence $a{\vee }_{W}b=(a{\vee }_{W}b{)}_J$. Thus the displayed value is ${\pi }_c(x){\vee }_W{\pi }_c(y)$. Here $x_J,y_J$ are their actual $W_J$-prefixes; no identity claim about them is needed.

Let $y\mathrm{̸}{\ge }_{R}s$ and put $z:=s\vee y$. In a saturated chain from $y$ to $z$, take the first cover $u{⋖}_{R}v$ whose upper element satisfies $v{\ge }_{R}s$. Its lower element is not above $s$, so step 1.3 gives $v=su$ and $s$ is a cover reflection of $v$. Since $v$ is a common upper bound of both $s$ and $y$, leastness gives $z{\le }_{R}v$; the chain gives $v{\le }_{R}z$, so $v=z$ and $s$ is a cover reflection of $z$. Let $q:={\pi }_c(z)$. The one-letter element $s$ is c-sortable, so ${\pi }_c(s)=s$ by [F5]; monotonicity gives $s{\le }_{R}q$. Since $sz=u\mathrm{̸}{\ge }_{R}s$ and ${\pi }_c(sz){\le }_{R}sz$ by [F5], ${\pi }_c(sz)\mathrm{̸}{\ge }_{R}s$ and therefore ${\pi }_c(sz)\mathrm{\ne }q$. By [F18], $zC\subseteq {\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(q)$ but $(sz)C\subseteq ̸{\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(q)$. Step 1.3 gives $sz=zr$ for some $r\in S$, so $s=zr{z}^{-1}$; by [F11], $\rho (z)e_r=\pm e_s$, and [F21] shows that $rC$ is the adjacent chamber to $C$ across $H_{e_r}$: $C\cap H_{e_r}$ is a facet, the reflection fixes it pointwise and exchanges its sides, and the arrangement is W-invariant. Applying $z$ gives that $zC$ and $zrC$ are adjacent across $z{H}_{e_r}=H_{\rho (z)e_r}=H_s$ by [F11]. The cone is the intersection of the skip-root halfspaces [F3] and a union of closed chambers [F4], so their common facet lies in its boundary. The skip-root set is finite, and each defining hyperplane distinct from $H_s$ intersects $H_s$ in a proper subspace. Start at a relative-interior point of the facet. For each such hyperplane still containing the point, perturb within $H_s$ in a direction outside that hyperplane; a sufficiently small perturbation stays in the relative interior and preserves the nonzero evaluations for hyperplanes already avoided. Finite iteration yields a point outside all those intersections. At this point a defining skip-root inequality is an equality, and its hyperplane must be $H_s$. By [F17], the skip root normal to this wall is either $e_s$ or $-e_s$. Since $s{\le }_{R}z$, [F10] gives $\rho (z^{-1})e_s\in {\mathrm{\Phi }}_{-}$. For $p\in C^{\circ }$, invariance of $B$ gives $B(\rho (z)p,e_s)=B(p,\rho (z^{-1})e_s)<0$ by [F16]-[F17]. Thus the included chamber is on the negative side of $H_s$, so the inward skip-root normal is $-e_s$. By [F19], $-e_s=-{\beta }_s$ in the negative skip basis means $s\in \mathrm{C}\mathrm{o}\mathrm{v}(q)$, so $s$ is a cover reflection of $q$.

Suppose $x,y{\ge }_{R}s$ for an initial letter $s$ of $c$, and put $X:=sx$, $Y:=sy$. Then $X,Y\mathrm{̸}{\ge }_{R}s$ by [F14]. Applying step 2.2 to $X,Y$ gives $x\vee y=s(X\vee Y)$; since $s^2=1$ by [F20], this is equivalent to $X\vee Y=s(x\vee y)$, whose length is $\mathrm{\ell }(x\vee y)-1$. By the recursion [F2], ${\pi }_c(x)=s{\pi }_{scs}(X)$, ${\pi }_c(y)=s{\pi }_{scs}(Y)$ and ${\pi }_c(x\vee y)=s{\pi }_{scs}(s(x\vee y))$. The induction hypothesis from step 1.4 for $(X,Y)$ in the rotated system $scs$ gives ${\pi }_{scs}(X\vee Y)={\pi }_{scs}(X)\vee {\pi }_{scs}(Y)$; these two projection values are not above $s$ because they lie below $X,Y$ by [F5]. Applying step 2.2 again yields ${\pi }_c(x)\vee {\pi }_c(y)=s({\pi }_{scs}(X)\vee {\pi }_{scs}(Y))=s{\pi }_{scs}(X\vee Y)={\pi }_c(x\vee y)$.

By the initial cover decomposition [F19], $q=s\vee q_J$ for $J=S\setminus \{s\}$. Parabolic compatibility [F18] gives $q_J={\pi }_{c^{\mathrm{\prime }}}(z_J)$, and join preservation of prefixes [F13] gives $z_J=(s\vee y{)}_J=s_J\vee y_J=y_J$, since $s_J=1$. The rank-drop branch of [F2] gives ${\pi }_c(y)={\pi }_{c^{\mathrm{\prime }}}(y_J)$; hence $q_J={\pi }_c(y)$ and ${\pi }_c(s\vee y)=s\vee {\pi }_c(y)={\pi }_c(s)\vee {\pi }_c(y)$. This proves (3).

In the mixed case, assume $x{\ge }_{R}s$ and $y\mathrm{̸}{\ge }_{R}s$, and put $z:=s\vee y$. Then $x,z{\ge }_{R}s$. Since $x\vee y$ is an upper bound of $s$ and $y$, $z{\le }_{R}x\vee y$; since $y{\le }_{R}z$, also $x\vee y{\le }_{R}x\vee z$, so $x\vee z=x\vee y$. The both-above case 3.1, whose induction step uses the shorter join $s(x\vee z)$, now gives ${\pi }_c(x\vee y)={\pi }_c(x)\vee {\pi }_c(z)$. By step 3.2, ${\pi }_c(z)={\pi }_c(s)\vee {\pi }_c(y)$; monotonicity and $s{\le }_{R}x$ give ${\pi }_c(s){\le }_R{\pi }_c(x)$, hence ${\pi }_c(x\vee y)={\pi }_c(x)\vee {\pi }_c(y)$. This proves the join identity in every case.

Define $\phi :W\mathrm{/}{\sim }_c\to {\pi }_c(W)$ by $\phi ([x{]}_c)={\pi }_c(x)$. It is well defined and injective by the definition of ${\sim }_c$, and surjective by the definition of ${\pi }_c(W)$. By [F5], every image is c-sortable and every c-sortable element is fixed, so ${\pi }_c(W)$ is exactly the c-sortable sublattice. The quotient order is defined by $[x{]}_c{\le }_c[y{]}_c$ exactly when ${\pi }_c(x){\le }_R{\pi }_c(y)$, so $\phi$ is an order isomorphism; the identities proved in steps 2.3, 2.4, 3.1 and 4.1 make it a lattice isomorphism. Equality of ${\pi }_c$-images is preserved by both meet and join, so by [F15] and the definition [F1], ${\sim }_c$ is a lattice congruence, the proposed class operations are representative-independent, and $W\mathrm{/}{\sim }_c$ is a lattice. The quotient map $p_c$ is surjective and preserves meet and join by those operations. All sets and inductions used here are finite, and no Choice is used.

Fix $J=S\setminus \{s\}$ and let $w=w_J\cdot {}^{J}w$ be the length-additive parabolic factorization. The prefix inversion formula in [F4] and opposition in [F5] give $N(((w{w}_0{)}_J{)}^{-1})=N((w{w}_0{)}^{-1})\cap {\mathrm{\Phi }}_{J,+}=N(w_0{w}^{-1})\cap {\mathrm{\Phi }}_{J,+}={\mathrm{\Phi }}_{J,+}\setminus (N(w^{-1})\cap {\mathrm{\Phi }}_{J,+})={\mathrm{\Phi }}_{J,+}\setminus N(w_J^{-1})$. Applying opposition inside $W_J$ gives $N(w_0(J)w_J^{-1})={\mathrm{\Phi }}_{J,+}\setminus N(w_J^{-1})=N((w_J{w}_0(J){)}^{-1})$. Both $(w{w}_0{)}_J$ and $w_J{w}_0(J)$ lie in $W_J$, so equality of their inverse inversion sets gives equality of the elements by the order criterion and antisymmetry in [F6]. Hence $(w{w}_0{)}_J=w_J{w}_0(J)$.

Suppose $s$ is final in $c$ and $\mathrm{\ell }(sw)<\mathrm{\ell }(w)$; set $q:={\pi }_c(w)$ and $v:={\pi }_{cs}(w_J)$. By [F3] and [F10], both $s$ and $v$ are $c$-sortable and lie below $w$, since $s{\le }_{R}w$ and $v{\le }_R{w}_J{\le }_{R}w$. Their join $x:=s\vee v$ is $c$-sortable by [F14] and below $w$, so $x{\le }_{R}q$ by [F11]. Thus $q{\ge }_{R}s$; the terminal cover decomposition [F12] gives $q=s\vee q_J$. The prefix $q_J$ is $cs$-sortable by [F10] and $q_J{\le }_R{w}_J$ by [F4], hence $q_J{\le }_{R}v$ by [F11] applied inside $W_J$. Therefore $q=s\vee q_J{\le }_{R}s\vee v=x$, and with $x{\le }_{R}q$ this proves ${\pi }_c(w)=s\vee {\pi }_{cs}(w_J)$.

We prove by lexicographic induction on $(\mathrm{\mid }S\mathrm{\mid },\mathrm{\ell }(x))$ that whenever $x{\le }_{R}y$ and ${\pi }_c(x)={\pi }_c(y)$, one has $u_c(x)=u_c(y)$. If $S=\mathrm{\varnothing }$, then $W=\{1\}$ and this holds; at every positive-rank pair assume it holds for all smaller measures, and fix an initial letter $s$ of $c$.

Define $\tau (w):=w{w}_0$, so $u_c=\tau \circ {\pi }_{c^{-1}}\circ \tau$. The map $\tau$ reverses right weak order: if $y=xv$ with additive length, then $x{w}_0=y{w}_0(w_0{v}^{-1}{w}_0)$ and $\mathrm{\ell }(w_0{v}^{-1}{w}_0)=\mathrm{\ell }(v)$ by [F5], so $y{w}_0{\le }_{R}x{w}_0$ by [F6]; since ${\tau }^2$ is the identity, this is an order anti-isomorphism. Therefore $u_c$ is order-preserving by [F13]. Since ${\pi }_{c^{-1}}(w{w}_0){\le }_{R}w{w}_0$ by [F9], applying $\tau$ gives $w{\le }_R{u}_c(w)$. Finally, $u_c(u_c(w))=\tau ({\pi }_{c^{-1}}({\pi }_{c^{-1}}(w{w}_0)))=\tau ({\pi }_{c^{-1}}(w{w}_0))=u_c(w)$ by idempotence in [F9].