Weak Order, Inversions, and Lattice Operations — Examples

The six elements: the type-$A$ isomorphism and inversion-length formula [F3,F4] give $\mathrm{\mid }W\mathrm{\mid }=6$, $\mathrm{\ell }(s)=\mathrm{\ell }(t)=1$, $\mathrm{\ell }(st)=\mathrm{\ell }(ts)=2$ and $\mathrm{\ell }(sts)=3$. The values $1,s,t,st,ts,sts$ are distinct: $1$ has length $0$ by [F2], the isomorphism separates $s$ and $t$, elements of different lengths are distinct, and $st=ts$ would imply $sts=t$ after right multiplication by $s$, contradicting their lengths. From the presentation [F1], $(st{)}^3=1$ and $s^2=t^2=1$; hence $(sts)(tst)=1$, while $tst$ is an involution. Therefore $sts=tst=:w_0$, and the six displayed elements exhaust $W$.

The $A_2$ root set and reflection actions: put $c:=\cos (\pi \mathrm{/}3)$. By [F18,F19], $2c^2-1=\cos (2\pi \mathrm{/}3)=-c$, so $(2c-1)(c+1)=0$. Since $\pi \mathrm{/}3<\pi \mathrm{/}2$, [F20] and [F21] give $c=\cos (\pi \mathrm{/}3)>\cos (\pi \mathrm{/}2)=0$, hence $c=1\mathrm{/}2$. From the Coxeter form and reflection formula [F11,F12], $\rho (s)e_s=-e_s$, $\rho (s)e_t=e_t+e_s$, $\rho (t)e_t=-e_t$, and $\rho (t)e_s=e_s+e_t$; thus $\rho (s)(e_s+e_t)=e_t$ and $\rho (t)(e_s+e_t)=e_s$. These actions and linearity show that $\mathcal{R}:=\{\pm e_s,\pm e_t,\pm (e_s+e_t)\}$ is invariant under both generators. Since $W$ is generated by $S$ [F2] and $\rho$ is a homomorphism [F13], every $\rho (w)$ preserves $\mathcal{R}$. The root-orbit definition [F14] gives $\mathrm{\Phi }\subseteq \mathcal{R}$; conversely $e_s,e_t$ are roots, $-e_s=\rho (s)e_s$, $-e_t=\rho (t)e_t$, $e_s+e_t=\rho (s)e_t$, and $-(e_s+e_t)=\rho (s)\rho (t)e_t$, so $\mathrm{\Phi }=\mathcal{R}$. The positive cone and sign theorem [F15] then give ${\mathrm{\Phi }}_{+}=\{{\alpha }_s,{\alpha }_t,{\alpha }_s+{\alpha }_t\}$, where ${\alpha }_s=e_s$ and ${\alpha }_t=e_t$.

Covers: the cover characterization [F7] says a right cover is right multiplication by a simple generator with length rise one. Multiplying the six values of step 1.1 by $s,t$ gives the rises $1s=s$, $1t=t$, $st=s\cdot t$, $ts=t\cdot s$, $w_0=st\cdot s$, and $w_0=ts\cdot t$; each raises length by one by [F4]. The other six products are $ss=1$, $tt=1$, $(st)t=s$, $(ts)s=t$, $w_{0}s=st$, and $w_{0}t=ts$, none of which raises length. Hence the right covers are exactly the six displayed in the statement.

The asymmetry: $st=s\cdot t$ and $\mathrm{\ell }(st)=\mathrm{\ell }(s)+\mathrm{\ell }(t)$, so $s{\le }_{R}st$ by [F5]. If $s{\le }_{L}st$, [F6] gives $st=xs$ and $\mathrm{\ell }(st)=\mathrm{\ell }(x)+\mathrm{\ell }(s)$; right multiplication by $s$ gives $x=sts=w_0$, contradicting $2=\mathrm{\ell }(st)=\mathrm{\ell }(x)+1=4$. Likewise $st=s\cdot t$ with $\mathrm{\ell }(st)=\mathrm{\ell }(s)+\mathrm{\ell }(t)$ gives $t{\le }_{L}st$, while $t{\le }_{R}st$ would force $x=t^{-1}st=tst=w_0$ and $2=\mathrm{\ell }(st)=1+3$, impossible. Inversion is an order isomorphism by [F9], consistent with these comparisons.

The six inversion sets: by the definition [F16] and prefix-root formula [F17], the reduced expressions from step 1.1 and the roots from step 1.2 give $N(1)=\mathrm{\varnothing }$, $N(s^{-1})=N(s)=\{{\alpha }_s\}$, $N(t^{-1})=N(t)=\{{\alpha }_t\}$, $N((st{)}^{-1})=N(ts)=\{{\alpha }_s,\rho (s){\alpha }_t\}=\{{\alpha }_s,{\alpha }_s+{\alpha }_t\}$, and $N((ts{)}^{-1})=N(st)=\{{\alpha }_t,\rho (t){\alpha }_s\}=\{{\alpha }_t,{\alpha }_s+{\alpha }_t\}$. For $w_0=sts=tst$, the same formula gives $N(w_0^{-1})=\{{\alpha }_s,\rho (s){\alpha }_t,\rho (st){\alpha }_s\}=\{{\alpha }_s,{\alpha }_s+{\alpha }_t,{\alpha }_t\}={\mathrm{\Phi }}_{+}$, since $\rho (st){\alpha }_s=\rho (s)({\alpha }_s+{\alpha }_t)={\alpha }_t$.

The right order: the cover chains of step 2.1 and the chain characterization in [F7] give exactly the $17$ relations: $1{\le }_{R}v$ for all six $v$; $s{\le }_{R}s,st,w_0$; $t{\le }_{R}t,ts,w_0$; $st{\le }_{R}st,w_0$; $ts{\le }_{R}ts,w_0$; and $w_0{\le }_R{w}_0$. The remaining $19$ of the $36$ ordered pairs do not satisfy $u{\le }_{R}v$; eight have incomparable entries, and eleven are reversals of strict comparisons. The down-sets are $\downarrow 1=\{1\}$, $\downarrow s=\{1,s\}$, $\downarrow t=\{1,t\}$, $\downarrow st=\{1,s,st\}$, $\downarrow ts=\{1,t,ts\}$ and $\downarrow w_0=W$; the up-sets are $\uparrow 1=W$, $\uparrow s=\{s,st,w_0\}$, $\uparrow t=\{t,ts,w_0\}$, $\uparrow st=\{st,w_0\}$, $\uparrow ts=\{ts,w_0\}$ and $\uparrow w_0=\{w_0\}$. The minimum $1$ and poset properties used here are in [F9].

The order isomorphism: inversion is an order isomorphism by [F9], so left weak order is the image of the right order under $w\mapsto w^{-1}$. In particular, $s{\le }_{R}st$ but $s\mathrm{̸}{\le }_{L}st$, while $t{\le }_{L}st$ but $t\mathrm{̸}{\le }_{R}st$, as shown in step 2.2.

Meets: intersections of down-sets from step 3.1 are $\{1\}$ for the four cross pairs $(s,t),(s,ts),(st,t),(st,ts)$; they are the down-set of the smaller element for pairs in one chain, and the down-set of the other element when one is $w_0$. By [A1] their greatest elements are the greatest common lower bounds, so the cross-pair meets are $1$, the meet along a chain is its smaller element, and $w_0\wedge v=v$ for $v\in \{s,t,st,ts\}$. Each listed value lies below both elements and dominates every common lower bound, the universal property in [F22]. For the empty meet every element is a lower bound, and the maximum $w_0$ gives $\bigwedge \mathrm{\varnothing }=w_0$, in agreement with [F10].

Joins: intersections of up-sets from step 3.1 are $\{w_0\}$ for the four cross pairs $(s,t),(s,ts),(st,t),(st,ts)$ and for every pair containing $w_0$; on a chain, the intersection is the up-set of the larger element. By [A1] their least elements are the least common upper bounds, so every cross-pair join and every join involving $w_0$ is $w_0$, and joins along a chain are its larger element. Each listed value is an upper bound and lies below every common upper bound, the universal property in [F22]. For the empty join every element is an upper bound, and the minimum $1$ gives $\bigvee \mathrm{\varnothing }=1$, in agreement with [F10].

The criterion on all $36$ pairs: write $\gamma ={\alpha }_s+{\alpha }_t$. The six sets of step 2.3 are $\mathrm{\varnothing }$, $\{{\alpha }_s\}$, $\{{\alpha }_t\}$, $\{{\alpha }_s,\gamma \}$, $\{{\alpha }_t,\gamma \}$, and $\{{\alpha }_s,{\alpha }_t,\gamma \}$. Their inclusion relations consist of the six reflexive pairs; the five strict inclusions from $\mathrm{\varnothing }$ to every nonempty set; the four inclusions from each singleton to its containing intermediate set and to the full set; and the two inclusions from the intermediate sets to the full set, for $17$ relations total. The remaining $19$ ordered pairs do not satisfy the directed inclusion $N(u^{-1})\subseteq N(v^{-1})$. Comparing these $17$ inclusions and $19$ failures with the corresponding right-order relations and failures from step 3.1 proves, in both directions, $u{\le }_{R}v⟺N(u^{-1})\subseteq N(v^{-1})$ on all $36$ pairs by [F8]. No Choice is used.

Every reduced expression is alternating and every element of $W$ has exactly one reduced expression. If a reduced expression had equal adjacent letters, [F2] would delete them and shorten a word for the same element, a contradiction; thus every reduced expression is alternating by [A1]. Conversely every alternating word is reduced by [F8]. If two reduced expressions have the same value, their lengths both equal the length of that element by [F1], so they have the same length $q$. For $q=0$ both are the empty word; for $q\ge 1$, [A1] says each is $a_q$ or $b_q$. If their first letters agree then the words are identical. If $q=2k$ is even and the first letters differ, equality would give $(st{)}^k=(ts{)}^k=(st{)}^{-k}$, using $s^2=t^2=1$ from [F12], and hence $(st{)}^q=1$, contrary to [F3]. If $q=2k+1$ is odd, equality $(st{)}^{k}s=(ts{)}^{k}t$ gives $(st{)}^k=(st{)}^{-k}ts=(st{)}^{-k-1}$ after right multiplication by $s$, again forcing $(st{)}^q=1$, contrary to [F3]. Thus reduced expressions are unique; every element has one by the definition of $\mathrm{\ell }$ in [F1].

The generators are incomparable in both orders. If $s{\le }_{R}t$, then [F4] gives $t=sx$ and $\mathrm{\ell }(t)=\mathrm{\ell }(s)+\mathrm{\ell }(x)=1+\mathrm{\ell }(x)$ by [F7]. Since $\mathrm{\ell }(t)=1$, $\mathrm{\ell }(x)=0$, so $x=1$ by [F1] and [F13], contradicting $s\mathrm{\ne }t$ in [F3]. Interchanging $s,t$ excludes $t{\le }_{R}s$. If $s{\le }_{L}t$, then [F11] gives $t=xs$ and $\mathrm{\ell }(t)=\mathrm{\ell }(x)+\mathrm{\ell }(s)=\mathrm{\ell }(x)+1$; the same length-zero argument gives $x=1$ and $t=s$, a contradiction. Interchanging $s,t$ excludes $t{\le }_{L}s$.

The powers $(st{)}^k$, $k\in \mathbb{Z}$, are pairwise distinct, and $W$ is infinite. For $k\ge 1$, $(st{)}^k$ is the value $a_{2k}$; for $k\le -1$, $(st{)}^k=(ts{)}^{-k}$ is the value $b_{-2k}$ by [F12]; and $(st{)}^0=1$. If $(st{)}^i=(st{)}^j$ for $i\mathrm{\ne }j$, cancellation gives $(st{)}^{i-j}=1$, contrary to the infinite order in [F3]. Thus the powers are pairwise distinct and form an infinite subset of $W$.

For every $v\in W$, $[1,v{]}_R$ is the finite chain of values of the prefixes of its unique reduced expression; in particular $[1,sts{]}_R=\{1,s,st,sts\}$ and $[1,ts{]}_R=\{1,t,ts\}$. Let $p_j$ be the prefix of length $j$ of the unique reduced expression of $v$. Each prefix is alternating and hence reduced by [F8], so prefixes of different lengths have different values by [F1]. By [F5], $u{\le }_{R}v$ holds exactly when the reduced expression of $u$ is a prefix of this unique expression of $v$; thus the elements of $[1,v{]}_R$ are exactly the prefix values, ordered by prefix inclusion. They form a finite chain with $\mathrm{\ell }(v)+1$ elements. The displayed intervals follow because $sts$ and $ts$ are alternating reduced words.

The set $\{s,t\}$ has no upper bound in either weak order, so its right join $s\vee t$ does not exist and $W$ is not a lattice. If $z$ were a right upper bound, $s{\le }_{R}z$ and $t{\le }_{R}z$ would give, by [F5], reduced expressions of $z$ beginning with $s$ and with $t$. This contradicts the uniqueness in step 1.1. If $z$ were a left upper bound, [F16] would give reduced expressions of $z$ ending in $s$ and in $t$, the same contradiction. Therefore there is no upper bound in either order; by [F10] a right join must be an upper bound, so $s\vee t$ does not exist. Since a lattice has a join for every pair by [F15], $W$ is not a lattice.

If $\mathrm{\varnothing }\mathrm{\ne }X\subseteq [1,v{]}_R$, then $X$ has a meet and a join: they are its least and greatest elements in the finite chain of step 2.2. The least element is a lower bound of $X$, and every lower bound is below it because it belongs to $X$; hence it is $\bigwedge X$ by [F10]. Dually, the greatest element is an upper bound of $X$ and is below every upper bound, so it is $\bigvee X$. In particular, $s\wedge st=s$ and $s\vee st=st$ in $[1,sts{]}_R$.

The interval translation isomorphism: $s{\le }_{R}sts$ by step 2.2, and $s^{-1}(sts)=s(sts)=ts$ by [F12]. Hence [F9] gives the order isomorphism $x\mapsto sx$ from $[1,ts{]}_R$ to $[s,sts{]}_R$. Using the displayed interval of step 2.2, its values are $s,st,sts$, so $[s,sts{]}_R=\{s,st,sts\}$.

Every nonempty subset of $[1,sts{]}_R$ has a join, while the empty subset also has join $1$. Every nonempty such subset is bounded above by $sts$, so [F6] supplies its join. For the empty subset, every element is an upper bound by vacuity; [F13] gives $\mathrm{\ell }(1)=0$, and [F4] gives $1{\le }_{R}w$ for every $w\in W$ by writing $w=1\cdot w$ with $\mathrm{\ell }(w)=\mathrm{\ell }(1)+\mathrm{\ell }(w)$. Thus $1$ is the least upper bound of the empty set by [F10]. The operations on nonempty subsets were explicitly computed from finite chains in step 3.1, and the empty join was proved directly, so no Axiom of Choice is used. In contrast, step 2.3 shows that $\{s,t\}$ has no join; hence the boundedness hypothesis for nonempty subsets in Weak order is a meet-semilattice, finite Coxeter groups are lattices, and joins of simple reflections exist exactly for finite parabolics (1) cannot be dropped.

The $A_2$ data: $W=\{1,s,t,st,ts,w_0\}$ with the lengths of Fact [F2], and the six inversion sets $N(w^{-1})$ of Fact [F3]; in particular, no other subset of ${\mathrm{\Phi }}_{+}$ displayed below occurs as an inversion set $N(w^{-1})$.

Failure of the join equality: $s\vee t=w_0$ by [F4] and $N(w_0^{-1})={\mathrm{\Phi }}_{+}=\{{\alpha }_s,{\alpha }_t,{\alpha }_s+{\alpha }_t\}$ by step 1.1, while $N(s^{-1})\cup N(t^{-1})=\{{\alpha }_s\}\cup \{{\alpha }_t\}=\{{\alpha }_s,{\alpha }_t\}$. The union $\{{\alpha }_s,{\alpha }_t\}$ is not among the six sets of step 1.1, so it is not the inversion set $N(w^{-1})$ of any element $w\in W$; in particular $N((s\vee t{)}^{-1})={\mathrm{\Phi }}_{+}\mathrm{\ne }\{{\alpha }_s,{\alpha }_t\}=N(s^{-1})\cup N(t^{-1})$, refuting the join half of the displayed statement.