Noncrossing Partition Lattices and Kreweras Complements — Examples

(The fourteen interval elements.) By [F3], ${\mathrm{\ell }}_T(w)=4-\mathrm{\#}\{\text{cycles of }w\}$ and $w{\le }_{T}c$ exactly when its cycles are cyclically increasing and its support partition is noncrossing. Length $0$ gives only $1$. Length $1$ gives all six transpositions; each has one pair block and two singleton blocks, so is noncrossing, and its 2-cycle is cyclically increasing. Length $2$ means two cycles, hence either a 3-cycle and a fixed point or two transpositions. For each of the four 3-element supports, exactly one orientation is cyclically increasing, giving $(123),(124),(134),(234)$; each support partition is noncrossing. Of the three double transpositions, $(12)(34)$ and $(14)(23)$ have noncrossing pair blocks, while $(13)(24)$ has crossing pair blocks. Length $3$ means a single 4-cycle; only $(1234)$ is cyclically increasing in the stated order. These cases exhaust the possible cycle counts and give precisely the list in (1).

(Direct complement products.) For an involution $w$, $K(w)=wc$. Applying $c=(1234)$ on the right first gives $K(12)=(234)$, $K(23)=(134)$, $K(34)=(124)$, $K(13)=(12)(34)$, $K(24)=(14)(23)$, and $K(14)=(123)$. The same multiplication gives $K((12)(34))=(24)$ and $K((14)(23))=(13)$. For the 3-cycles, multiplying their inverses by $c$ gives $K(123)=(34)$, $K(124)=(23)$, $K(134)=(12)$, and $K(234)=(14)$; also $K(1)=c$ and $K(c)=1$. This is the full list in Statement (3).

(The fifteen partitions.) By block sizes, the set partitions of four labels consist of one partition with one block, six with three blocks, seven with two blocks, and one with four blocks, for a total of fifteen. A partition with one or four blocks is noncrossing. The six three-block partitions have one pair and two singletons, so are noncrossing. Among the seven two-block partitions, the four triple-plus-singleton partitions are noncrossing; the three pairings are $\{1,2\}\mid \{3,4\}$, $\{1,4\}\mid \{2,3\}$, and $\{1,3\}\mid \{2,4\}$, of which only the last has alternating endpoints. This proves the unique crossing claim. The first-step list has fourteen elements, all with noncrossing cyclically increasing cycles; [F3] says each noncrossing partition has a unique such interval permutation. Thus the supports in (1) give exactly the fourteen noncrossing partitions.

(Order, square, and block counts.) [F4] gives that $K$ is an order-reversing bijection of the interval, $K^2(w)=c^{-1}wc$, and ${\mathrm{\ell }}_T(K(w))=3-{\mathrm{\ell }}_T(w)$. Conjugating a cycle by $c^{-1}$ relabels each entry by $1\mapsto 4\mapsto 3\mapsto 2\mapsto 1$, so the support partition rotates as stated. Since ${\mathrm{\ell }}_T(w)=4-\mathrm{\mid }\pi (w)\mathrm{\mid }$ and ${\mathrm{\ell }}_T(K(w))=4-\mathrm{\mid }\pi (K(w))\mathrm{\mid }$ by [F3], the length complement gives $\mathrm{\mid }\pi (K(w))\mathrm{\mid }=5-\mathrm{\mid }\pi (w)\mathrm{\mid }$.

(The dihedral group and its reflections.) The defining relations give $c^m=1$, $t=sc$, and $scs=c^{-1}$. The order of $c$ is exactly $m$: if $m\ge 3$, let $R(i)=i+1$ and $J(i)=-i$ on $\mathbb{Z}\mathrm{/}m\mathbb{Z}$. Since $J^2=1$ and $JRJ=R^{-1}$, the assignment $s\mapsto J$, $t\mapsto JR$ satisfies $s^2=t^2=1$ and $st\mapsto R$, so [F1] gives a homomorphism with the image of $c$ of order $m$. If $m=2$, map $s,t$ to the independent coordinate flips of $(\mathbb{Z}\mathrm{/}2\mathbb{Z}{)}^2$; these are commuting involutions, satisfy the defining relations, and their product has order $2$. Since $c^m=1$ in $W$, in both cases $c$ has exact order $m$. Now $W=⟨s,c⟩$ and $s{c}^k=c^{-k}s$, so every word reduces to $c^k$ or $c^{k}s$, with $k$ taken modulo $m$. These at most $2m$ forms are distinct: the $c^k$ are distinct by the exact order just proved, the $c^{k}s$ are distinct by cancellation, and the two families are separated by the homomorphism $\epsilon :W\to \{1,-1\}$ with $\epsilon (s)=\epsilon (t)=-1$, which exists by [F1] because both simple generators map to $-1$ and $st$ maps to $1$. Thus $\mathrm{\mid }W\mathrm{\mid }=2m$. The reflection set is exactly $T=\{c^{k}s:0\le k<m\}$. Every conjugate of $s$ or $t$ has $\epsilon =-1$ and is therefore in this list. Conversely, for every integer $j$, $c^{j}s{c}^{-j}=c^{2j}s$ and, since $t=sc$, $c^{j}t{c}^{-j}=c^{2j-1}s$. The exponents $2j$ and $2j-1$ cover all residues modulo $m$, so every $c^{k}s$ is a conjugate of a simple reflection. In particular $\mathrm{\mid }T\mathrm{\mid }=m$.

(The Coxeter form and its plane action.) Put $\theta =\pi \mathrm{/}m$ and $q=\cos \theta$. For $x=a{e}_s+b{e}_t$, [F3] and [F5] give $B(x,x)=a^2-2qab+b^2=(a-qb{)}^2+{\sin }^2\theta b^2$. Since $0<\theta \le \pi \mathrm{/}2<\pi$, [F5] gives $\sin \theta >0$, so this is positive for every nonzero $(a,b)$. In orthonormal coordinates take $e_s=(1,0)$ and $e_t=(-\cos \theta ,\sin \theta )$. The reflection formula $R(v)=I-2v{v}^{\mathsf{T}}$ and [F5] give $\rho (c)=\rho (s)\rho (t)=\left(\begin{array}{cc}\cos (2\theta )& -\sin (2\theta )\\ \sin (2\theta )& \cos (2\theta )\end{array}\right)\mathrm{.}$ Thus $c$ rotates this plane through $2\pi \mathrm{/}m$. The order calculation in step 1.1 proves finite type, including $m=2$, where $s,t$ commute and the diagram consists of two isolated vertices.

(Reflection lengths.) Every $r\in T$ is nonidentity and is itself a reflection, so ${\mathrm{\ell }}_T(r)=1$. Each nonidentity rotation $c^k$ is not in $T$ by the sign $\epsilon$, and $c^k=(c^{k}s)s$ is a product of two reflections; hence ${\mathrm{\ell }}_T(c^k)=2$. In particular $c$ has length $2$.

(The interval below $c$.) The identity and $c$ lie below $c$. For $r_k=c^{k}s\in T$, $r_k$ is a conjugate of an involutory simple reflection, so $r_k^{-1}=r_k$, and $r_k^{-1}c=r_{k}c=c^{k}sc=c^{k-1}s=r_{k-1}\in T,$ so ${\mathrm{\ell }}_T(r_k)+{\mathrm{\ell }}_T(r_k^{-1}c)=2={\mathrm{\ell }}_T(c)$ and every reflection lies below $c$. If $c^k$ is a rotation other than $1$ or $c$, then $c^{1-k}$ is also a nonidentity rotation, so ${\mathrm{\ell }}_T(c^k)+{\mathrm{\ell }}_T((c^k{)}^{-1}c)=2+2=4\mathrm{\ne }2$. These are all group elements by step 1.1, proving the interval formula. When $m=2$ there are no rotations other than $1,c$, so the same argument covers that case.

(Lattice operations.) The interval in step 3.1 has bottom $1$, top $c$, and $m$ distinct reflections of equal length $1$ between them. Distinct reflections are incomparable by [F2], since each has the same length and a strict absolute-order comparison would require positive length increase. Thus two distinct reflections have only $1$ as common lower bound and only $c$ as common upper bound; operations with $1$ and $c$ are forced by their bottom/top roles. This proves the displayed lattice operations directly and verifies the finite-type lattice conclusion in this example.

(Kreweras action on reflections.) By [F4], $K$ is an order-reversing bijection; its explicit action is $K(1)=c$, $K(c)=1$, and $K(r_k)=r_{k-1}$ by step 3.1. It therefore cycles through all $m$ reflections. Direct multiplication gives $K^2(w)=c^{-1}wc$ and $c^{-1}{r}_{k}c=c^{k-2}s=r_{k-2}$. By [F6] and step 1.2, this conjugation rotates each reflection axis through $-2\pi \mathrm{/}m$ in the displayed orientation (an angle of magnitude $2\pi \mathrm{/}m$); for $m=2$, a rotation through $\pi$ fixes every unoriented axis. Iterating $k\mapsto k-2$ returns to $k$ exactly when $m$ divides $2j$. The least positive such $j$ is $m$ for odd $m$ and $m\mathrm{/}2$ for even $m$. Hence $K^2$ has one cycle for odd $m$, two cycles for even $m$, and is the identity on $T$ when $m=2$.

(The interval below the crossing element.) By [F1], ${\mathrm{\ell }}_T(x)=4-2=2$. If $t$ is a transposition, then $t^{-1}x=tx$. For $t=(13)$ or $(24)$, $tx$ is the other transposition, so ${\mathrm{\ell }}_T(t^{-1}x)=1$ and $t{\le }_{T}x$. The remaining four transpositions join the two cycles of $x$; explicitly, $(12)x=(1324)$, $(14)x=(1342)$, $(23)x=(1243)$, and $(34)x=(1423)$, each a 4-cycle of reflection length $3$. None is below $x$. If $u{\le }_{T}x$, [F2] gives ${\mathrm{\ell }}_T(u)\le 2$; length $0$ forces $u=1$, and length $2$ forces ${\mathrm{\ell }}_T(u^{-1}x)=0$, hence $u=x$. Therefore the displayed four elements are the entire interval. Its two distinct atoms have meet $1$ and join $x$, so it is the Boolean lattice on two generators.

(The two maximal elements.) By [F1], the identity, the three transpositions, and the two 3-cycles are exactly the elements of $S_3$, with reflection lengths $0,1,2$, respectively. The products are $s_1{s}_2=(123)$ and $s_2{s}_1=(132)$. They are distinct and have equal length, so [F2] makes them incomparable. No element has length greater than $2$, so each is maximal. A common upper bound would have to be strictly above one of these distinct maximal elements; hence none exists and $\mathrm{Abs}(S_3)$ has no join for this pair.

(The crossing obstruction.) The diagonals joining $1$ to $3$ and $2$ to $4$ cross in the square, so the support partition of $x$ is crossing. Also direct right-to-left multiplication gives $x^{-1}c=(1432)$, which has length $3$ by [F1], while ${\mathrm{\ell }}_T(c)=3$ and ${\mathrm{\ell }}_T(x)=2$. Thus ${\mathrm{\ell }}_T(x)+{\mathrm{\ell }}_T(x^{-1}c)=5\mathrm{\ne }3={\mathrm{\ell }}_T(c)$, so $x\mathrm{̸}{\le }_{T}c$ by [F2]. This verifies directly the exclusion predicted by the type-A criterion.

(The Coxeter intervals are five-element lattices.) Fix either 3-cycle $d$. For $d=(123)$, the products $td$ for $t=(12),(23),(13)$ are $(23),(13),(12)$, respectively; for $d=(132)$ they are $(13),(12),(23)$. Hence every $t\in T$ satisfies ${\mathrm{\ell }}_T(d)=1+{\mathrm{\ell }}_T(t^{-1}d)=2$ and lies below $d$. If $w{\le }_{T}d$ has length $2$, [F2] forces ${\mathrm{\ell }}_T(w^{-1}d)=0$, so $w=d$. Therefore $[1,d{]}_{{\le }_T}=\{1\}\cup T\cup \{d\}$. Its three transpositions are incomparable atoms, any two have meet $1$ and join $d$, so this interval is a five-element lattice. Finally, $s_1(s_1{s}_2)s_1=s_2{s}_1$, so the two Coxeter elements are conjugate.