The Kreweras complement of [1,c], and the type-A model by noncrossing set partitions

(Rank of a Coxeter element.) Write $c=s_{i_1}\cdots s_{i_n}$, where each simple reflection occurs once. Its simple-root normals, in the reverse list, are independent unit roots. Clause (2) of [F1] applied to this reversed list gives ${\mathrm{\ell }}_T(c)=n$. If $n=0$, $c=1$ and the same equality is immediate.

(The group-theoretic complement maps the interval to itself.) The set $T$ is invariant under conjugation: conjugation permutes its defining conjugates of simple reflections. Conjugating a shortest reflection factorization and then conjugating back shows ${\mathrm{\ell }}_T(gw{g}^{-1})={\mathrm{\ell }}_T(w)$ for all $g,w\in W$. If $w{\le }_{T}c$, then ${\mathrm{\ell }}_T(c)={\mathrm{\ell }}_T(w)+{\mathrm{\ell }}_T(w^{-1}c)$, so ${\mathrm{\ell }}_T(K(w))=n-{\mathrm{\ell }}_T(w)$. Also $K(w{)}^{-1}c=c^{-1}wc$, whose reflection length is ${\mathrm{\ell }}_T(w)$. Therefore ${\mathrm{\ell }}_T(K(w))+{\mathrm{\ell }}_T(K(w{)}^{-1}c)=n={\mathrm{\ell }}_T(c)$ and $K(w){\le }_{T}c$. Thus $K(w)$ is in the interval. The map is injective, and the interval is finite because $W$ is finite, so it is onto. Direct multiplication gives $K(K(w))=c^{-1}wc$ and $wK(w)=c$.

(The reflection set in type A.) For $N=1$ there are no simple reflections and no transpositions. For $N\ge 2$, conjugating any simple reflection $s_i=(ii+1)$ by $g\in S_N$ gives $(g(i)g(i+1))$ (Conjugating a cycle relabels each entry: $g(a_1\dots a_k)g^{-1}=(g(a_1)\dots g(a_k))$), so every reflection is a transposition. Conversely, for any transposition $(ab)$ choose a permutation $g$ with $g(1)=a$ and $g(2)=b$; then $g{s}_1{g}^{-1}=(ab)$, so every transposition is a reflection. A right multiplication by a transposition changes the number of cycles by exactly one: if its two labels lie in one cycle, it cuts that cycle at those labels into two; if they lie in different cycles, it joins the cycles. Thus any expression of $w$ as $r$ transpositions must have $r\ge N-\mathrm{\#}\{\text{cycles of }w\}$, since reaching the identity requires increasing the cycle count to $N$ one step at a time. Conversely each cycle $(a_1{a}_2\cdots a_m)$ is $(a_1{a}_m)(a_1{a}_{m-1})\cdots (a_1{a}_2)$, a product of $m-1$ transpositions. Multiplying these expressions over the disjoint cycles gives the matching upper bound and proves the formula. It also covers $N=1$, where the identity is the empty product.

(An interval block exists.) Every noncrossing partition with at least two blocks has a block consisting of consecutive vertices in the original cyclic order. A singleton block suffices. Otherwise choose a block $B$ minimizing $\max B-\min B$ in the linear order $1<\cdots <N$. If $B$ is not a linear interval, there are successive elements $b<b^{\mathrm{\prime }}$ of $B$ and a label $x$ with $b<x<b^{\mathrm{\prime }}$. Let $D$ be the block containing $x$. Any $y\in D$ outside $(b,b^{\mathrm{\prime }})$ would make the chords $b{b}^{\mathrm{\prime }}$ and $xy$ have alternating endpoints and therefore cross, contradicting disjointness of the block hulls. Hence $D\subseteq (b,b^{\mathrm{\prime }})$, so $\max D-\min D<b^{\mathrm{\prime }}-b\le \max B-\min B$, contrary to minimality. Thus $B$ is a linear interval, and hence consecutive in the original cyclic order.

(Order reversal and lattice duality.) If $u{\le }_{T}v$, then ${\mathrm{\ell }}_T(v)={\mathrm{\ell }}_T(u)+{\mathrm{\ell }}_T(u^{-1}v)$. The conjugation invariance just proved and $v(u^{-1}v)v^{-1}=v{u}^{-1}$ give ${\mathrm{\ell }}_T(K(v{)}^{-1}K(u))={\mathrm{\ell }}_T(c^{-1}v{u}^{-1}c)={\mathrm{\ell }}_T(u^{-1}v)={\mathrm{\ell }}_T(v)-{\mathrm{\ell }}_T(u)={\mathrm{\ell }}_T(K(u))-{\mathrm{\ell }}_T(K(v))\mathrm{.}$ Hence $K(v){\le }_{T}K(u)$. An order-reversing bijection of a lattice carries every least upper bound to a greatest lower bound and vice versa, by the defining universal properties. The lattice hypothesis is supplied by [F5].

(Noncrossing increasing cycles lie below $c$: singleton removal.) Define $w_{\pi }$ to be the product of the cyclically increasing cycles on the blocks of $\pi$. We prove $w_{\pi }{\le }_{T}c$ by induction on $N$. The one-block partition gives $w_{\pi }=c$. If $\pi$ has a singleton block $\{a\}$, remove it to obtain a noncrossing partition ${\pi }^{\mathrm{\prime }}$ on the remaining cyclically ordered set $Y$ of size $N-1$, with long cycle $c_Y$ and permutation $w^{\mathrm{\prime }}$. By induction, ${\mathrm{\ell }}_Y(w^{\mathrm{\prime }})+{\mathrm{\ell }}_Y(w^{\mathrm{\prime }-1}{c}_Y)=N-2$. Regard these permutations as fixing $a$ in $S_N$, and let $p$ be the predecessor of $a$ in the cyclic order. For $\tau =(pa)$, direct evaluation on the labels gives $c=c_Y\tau$. Since $w^{\mathrm{\prime }}$ fixes $a$, $q=w^{\mathrm{\prime }-1}{c}_Y$ also fixes $a$; multiplying $q$ on the right by $\tau$ joins the singleton cycle $\{a\}$ to the cycle containing $p$. Hence ${\mathrm{\ell }}_N(w_{\pi })={\mathrm{\ell }}_Y(w^{\mathrm{\prime }})$ and ${\mathrm{\ell }}_N(w_{\pi }^{-1}c)={\mathrm{\ell }}_Y(w^{\mathrm{\prime }-1}{c}_Y)+1$. Their sum is $N-1={\mathrm{\ell }}_N(c)$, proving $w_{\pi }{\le }_{T}c$. This includes the discrete partition and the cases $N\le 2$.

(Elements below $c$ have noncrossing increasing cycles.) Induct on $d=N-1-{\mathrm{\ell }}_T(w)$ for $w{\le }_{T}c$. If $d=0$, then $w=c$. If $d>0$, take a shortest transposition factorization $w^{-1}c=t_1\cdots t_d$. A shortest factorization of $w$ followed by this one is a shortest factorization of $c$, so its prefix $x=w{t}_1$ satisfies $w{\le }_{T}x{\le }_{T}c$ and ${\mathrm{\ell }}_T(x)={\mathrm{\ell }}_T(w)+1$. By induction, $\pi (x)$ is noncrossing and its cycles are cyclically increasing. Since right multiplication by $t_1$ lowers reflection length by one, step 1.3 shows that $t_1$ splits one cycle of $x$. If that cycle is $(b_1{b}_2\cdots b_r)$ in cyclic order and $t_1=(b_i{b}_j)$ with $i<j$, the two resulting cycles have supports and cyclic orders $(b_i,b_{j+1},\dots ,b_r,b_1,\dots ,b_{i-1})\text{and}(b_j,b_{i+1},\dots ,b_{j-1}),$ with singleton cycles interpreted as fixed points. Both are cyclically increasing. Their convex hulls lie on opposite sides of the chord $b_i{b}_j$ and meet its line only at distinct endpoints, so are disjoint. Every other block hull was disjoint from the old block hull and remains disjoint from its two sub-hulls. Thus $\pi (w)$ is noncrossing and every cycle is cyclically increasing.

(Noncrossing increasing cycles lie below $c$: interval-block contraction.) Now suppose $\pi$ has no singleton blocks and at least two blocks. By step 1.4 it has a consecutive block $B=\{a,a+1,\dots ,a+m-1\}$ in cyclic order, with $m\ge 2$. Contract $B$ to a single label $a$ to obtain a cyclically ordered set $Y$ of size $N-m+1$ and a noncrossing partition ${\pi }^{\mathrm{\prime }\mathrm{\prime }}$ whose block at $a$ is the singleton $\{a\}$. Let $c^{\mathrm{\prime }\mathrm{\prime }}$ be the long cycle on $Y$ and $w^{\mathrm{\prime }\mathrm{\prime }}=w_{{\pi }^{\mathrm{\prime }\mathrm{\prime }}}$, so $w^{\mathrm{\prime }\mathrm{\prime }}$ fixes $a$. Write ${\gamma }_B=(aa+1\cdots a+m-1)$. Extending permutations of $Y$ to fix the deleted labels gives $w_{\pi }={\gamma }_B{w}^{\mathrm{\prime }\mathrm{\prime }}$ and $c=c^{\mathrm{\prime }\mathrm{\prime }}{\gamma }_B$, with $w^{\mathrm{\prime }\mathrm{\prime }}$ commuting with ${\gamma }_B$. By induction, ${\mathrm{\ell }}_Y(w^{\mathrm{\prime }\mathrm{\prime }})+{\mathrm{\ell }}_Y(w^{\mathrm{\prime }\mathrm{\prime }-1}{c}^{\mathrm{\prime }\mathrm{\prime }})=\mathrm{\mid }Y\mathrm{\mid }-1=N-m$. The disjoint-cycle formula gives ${\mathrm{\ell }}_N(w_{\pi })=(m-1)+{\mathrm{\ell }}_Y(w^{\mathrm{\prime }\mathrm{\prime }})$, while $w_{\pi }^{-1}c=w^{\mathrm{\prime }\mathrm{\prime }-1}{\gamma }_B^{-1}{c}^{\mathrm{\prime }\mathrm{\prime }}{\gamma }_B={\gamma }_B^{-1}(w^{\mathrm{\prime }\mathrm{\prime }-1}{c}^{\mathrm{\prime }\mathrm{\prime }}){\gamma }_B,$ so conjugation invariance and the cycle formula give ${\mathrm{\ell }}_N(w_{\pi }^{-1}c)={\mathrm{\ell }}_Y(w^{\mathrm{\prime }\mathrm{\prime }-1}{c}^{\mathrm{\prime }\mathrm{\prime }})$. The two lengths sum to $(m-1)+(N-m)=N-1={\mathrm{\ell }}_N(c)$. Therefore $w_{\pi }{\le }_{T}c$. The one-block case was handled in step 2.2; these cases exhaust all partitions.

(The partition map is a bijection and preserves order.) Steps 2.2, 2.3 and 3.1 show that each noncrossing partition has an interval element $w_{\pi }$ and every interval element arises this way. Its cycle supports determine each of its cyclically increasing cycles, so this correspondence is bijective. If $u{\le }_{T}v$, choose a shortest transposition factorization of $u^{-1}v$ and append it to a shortest factorization of $u$. Every prefix is shortest, giving a chain in $[1,c]$ from $u$ to $v$ whose steps multiply on the right by a transposition and raise length by one. By step 1.3 each step joins two cycles, so $\pi (u)$ refines $\pi (v)$. Conversely suppose $\pi (u)$ refines $\pi (v)$. For each block $B$ of $\pi (v)$, restrict $v$ to its cyclically increasing cycle $v_B$ and let $u_B$ be the product of the cycles of $u$ supported in $B$. The induced partition $\pi (u){\mathrm{\mid }}_B$ is noncrossing, and its cycles remain cyclically increasing in the induced cyclic order on $B$. By steps 2.2, 2.3 and 3.1 applied to the labels in $B$, $u_B{\le }_T{v}_B$. The blocks $B$ are disjoint, and the cycle formula in step 1.3 gives additivity of reflection length across these supports for $u$, $v$, and $u^{-1}v$. Summing ${\mathrm{\ell }}_B(v_B)={\mathrm{\ell }}_B(u_B)+{\mathrm{\ell }}_B(u_B^{-1}{v}_B)$ over all $B$ gives ${\mathrm{\ell }}_T(v)={\mathrm{\ell }}_T(u)+{\mathrm{\ell }}_T(u^{-1}v)$, hence $u{\le }_{T}v$. Thus the bijection is an order isomorphism.

(The region partition is the classical complement.) For $N=1$, the sole black and white blocks are singletons, $c=K(1)=1$, and all assertions in (4) hold, with $\widehat{0}=\widehat{1}$. Assume $N\ge 2$. Draw the convex hull edges of each black block of $\pi$ in the alternating $2N$-gon. These noncrossing chords cut the disk into polygonal regions. Group white vertices lying in the same region. Each region is a convex polygonal cell of the dissection by noncrossing chords, so grouping its white vertices gives a noncrossing partition. Any compatible white block must lie in one region, since a segment joining vertices in different regions crosses a black block edge. Hence this region partition is the coarsest interleaving partner, namely $K_{\mathrm{c}\mathrm{l}}(\pi )$. Let $\alpha =w_{\pi }$, and label $d_i$ as the white vertex in the gap after $b_i$. Tracing the boundary of the region at $d_i$ to the next white vertex passes black vertex $b_{i+1}$ and then follows the boundary edge of its black block back to its predecessor $b_j$, where $j={\alpha }^{-1}(i+1)$. Thus the successor permutation of white vertices within their regions is $q(i)={\alpha }^{-1}(i+1)={\alpha }^{-1}c(i)$. Its cycles are exactly the white blocks of $K_{\mathrm{c}\mathrm{l}}(\pi )$, so $q=w_{K_{\mathrm{c}\mathrm{l}}(\pi )}={\alpha }^{-1}c$ and $\alpha q=c$. Since $K(w)=w^{-1}c$, this proves $\pi (K(w))=K_{\mathrm{c}\mathrm{l}}(\pi (w))$. By step 2.1 and the order isomorphism, $K_{\mathrm{c}\mathrm{l}}$ is an order-reversing bijection, and its square is relabeling by $c^{-1}$, namely $i\mapsto i-1$. From steps 1.2 and 1.3, ${\mathrm{\ell }}_T(K(w))=(N-1)-(N-\mathrm{\mid }\pi (w)\mathrm{\mid })=\mathrm{\mid }\pi (w)\mathrm{\mid }-1$. Applying the cycle formula to $K(w)$ gives $\mathrm{\mid }{K}_{\mathrm{c}\mathrm{l}}(\pi )\mathrm{\mid }=N+1-\mathrm{\mid }\pi \mathrm{\mid }$. If two labels belonged to a common block of both $\pi$ and $K_{\mathrm{c}\mathrm{l}}(\pi )$, the corresponding black and white chords would have alternating endpoints and cross; hence their only common lower bound in refinement order is $\widehat{0}$, giving $\pi \wedge K_{\mathrm{c}\mathrm{l}}(\pi )=\widehat{0}$. Each cycle of $w_{\pi }$ and $w_{K_{\mathrm{c}\mathrm{l}}(\pi )}$ stays inside a class of the equivalence relation generated by their block memberships. Thus each permutation preserves every equivalence class setwise, so their product $c$ also preserves each class setwise. Since $c$ is transitive, the only such class is the whole label set. Every common upper bound is consequently $\widehat{1}$, so $\pi \vee K_{\mathrm{c}\mathrm{l}}(\pi )=\widehat{1}$.

The general complement proof uses only finite reflection length, conjugation invariance of its defining set, and the lattice property of the interval. The type-A model uses the Coxeter presentation and permutations only; it invokes no Lie-theoretic root system, crystallographic hypothesis, finite classification, or Choice. No enumeration of the general finite-type interval is asserted.