Coxeter, Artin, and Hecke Interfaces — Examples

The braid pairs of the type-$A_{n-1}$ matrix are the pairs of alternating words $\{s_i{s}_{i+1}{s}_i,s_{i+1}{s}_i{s}_{i+1}\}$ for $\mathrm{\mid }i-j\mathrm{\mid }=1$ and the commuting pairs $\{s_i{s}_j,s_j{s}_i\}$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$; their images under $f(s_i):=(ii+1)$ are equal in $S_n$ by the presentation of [F6]. Hence [F2] gives a homomorphism ${\pi }_n:G_n\to S_n$ with ${\pi }_n({\sigma }_i)=(ii+1)$. Its image is a subgroup of $S_n$ containing the adjacent transpositions, which generate $S_n$ by [F7], so the image is all of $S_n$ by [F10]: ${\pi }_n$ is surjective.

The monoid isomorphism $G_n^{+}\cong B_n^{+}$: define $\phi :G_n^{+}\to B_n^{+}$ on generators by $\phi ({\sigma }_i):={\bar{\sigma }}_i$. The braid pairs of the type-$A$ matrix map to the two defining word pairs of $B_n^{+}$ of [F9], whose two members are ${\equiv }^{+}$-equivalent in $B_n^{+}$, so [F1] gives a monoid homomorphism $\phi$. Conversely the elements $a_i:={\sigma }_i\in G_n^{+}$ satisfy $a_i{a}_{i+1}{a}_i=a_{i+1}{a}_i{a}_{i+1}$ and $a_i{a}_j=a_j{a}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$, because the corresponding alternating words are ${\equiv }^{+}$-equivalent in $A^{+}(S,m)$ and hence equal in $G_n^{+}$; so [F9] gives a monoid homomorphism $\psi :B_n^{+}\to G_n^{+}$ with $\psi ({\bar{\sigma }}_i)={\sigma }_i$. The composites $\phi \circ \psi$ and $\psi \circ \phi$ fix every generator, hence are the respective identities by the uniqueness clauses of [F1] and [F9]; thus $\phi$ and $\psi$ are mutually inverse isomorphisms.

The same assignment $f(s_i):=(ii+1)$ has equal values on the two words of every braid pair by [F6], so [F1] with $M=S_n$ gives a monoid homomorphism ${\pi }_n^{+}:G_n^{+}\to S_n$ with ${\pi }_n^{+}({\sigma }_i)=(ii+1)$. The composites ${\pi }_n\circ \gamma$ and ${\pi }_n^{+}$ are monoid homomorphisms $G_n^{+}\to S_n$ agreeing on every generator ${\sigma }_i$, so they are equal by the uniqueness clause of [F1]: ${\pi }_n\circ \gamma ={\pi }_n^{+}$.

The map $b:S_n\to G_n^{+}$ is well defined by the type-$A$ isomorphism of the given data: its input $w\in S_n$ corresponds to the unique element of $W$ with the same name, and [F3] applies. It satisfies ${\pi }_n^{+}(b_w)=w$ by [F3], and ${\pi }_n\circ \gamma \circ b={\mathrm{i}\mathrm{d}}_{S_n}$ because ${\pi }_n\circ \gamma ={\pi }_n^{+}$ by step 2.1 and ${\pi }_n^{+}\circ b=\mathrm{i}\mathrm{d}$ by [F4]; in particular $b$ is injective by [F4] and is a set-theoretic section of both ${\pi }_n^{+}$ and ${\pi }_n\circ \gamma$.

For $n\ge 3$, the element $s_1{s}_2$ permutes the first three symbols as in $S_3$ and fixes the others; it corresponds to $(12)(23)=(123)$, whose one-line form is $[2,3,1,4,\dots ,n]$ (with no tail when $n=3$); the pairs $(1,3)$ and $(2,3)$ are its only inversions: $(1,2)$ is not an inversion and all pairs involving the increasing tail contribute none, so $\mathrm{inv}(s_1{s}_2)=2$ by [F8]. By the given type-$A$ clause $\mathrm{\ell }(s_1{s}_2)=2$ while $\mathrm{\ell }(s_1)=\mathrm{\ell }(s_2)=1$, so $\mathrm{\ell }(s_1{s}_2)=\mathrm{\ell }(s_1)+\mathrm{\ell }(s_2)$ and [F5] gives $b_{s_1}{b}_{s_2}=b_{s_1{s}_2}$; explicitly both sides equal ${\sigma }_1{\sigma }_2$, and no braid move is needed for this pair.

For $n=3$, the two words $s_1{s}_2{s}_1$ and $s_2{s}_1{s}_2$ both act as the transposition $(13)$, whose one-line form $[3,2,1]$ has all three pairs as inversions, so $\mathrm{inv}(w_0)=3$ by [F8]; with $\mathrm{\ell }(w_0)=3$ by the given type-$A$ clause, both words have length $\mathrm{\ell }(w_0)$ and hence are reduced expressions of $w_0$. Step 3.1 therefore gives $b_{w_0}$ as the product along either word, and the two products are equal in $G_3$ because ${\sigma }_1{\sigma }_2{\sigma }_1$ and ${\sigma }_2{\sigma }_1{\sigma }_2$ are the two words of a braid pair of $A(S,m)$, hence equal in $G_3^{+}$ and in $G_3$.

Scope and choice: only presentation-level maps, the positive monoid isomorphism and the displayed finite computations are proved; no topological model, no Garside or lattice property, and no embedding of $G_n^{+}$ into $G_n$ or $B_n^{+}$ into $B_n$ is asserted. All constructions are given on generators of explicitly presented monoids and groups, and no choice is used.

The rank-one data: since a braid pair requires two distinct letters, the braid-pair set of $(\{s\},m)$ is empty by [F1], so ${\equiv }^{+}$ is the diagonal and $A^{+}=S^{\ast }$ is the free monoid on the single generator $s$, with elements $[s^k]$ for $k\ge 0$ and $[s^i]=[s^j]$ exactly when $i=j$, because $L([s^k])=k$ by [F3]. The relator set of $W$ is $\{s^2\}$, so $W=⟨s\mid s^2=1⟩=\{1,s\}$ with $\mathrm{\ell }(1)=0$ and $\mathrm{\ell }(s)=1$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action).

The positive lifts are $b_1=[\epsilon ]$ (the empty word is a reduced expression of $1$ because $\mathrm{\ell }(1)=0$) and $b_s=[s]={\sigma }_s$ (the one-letter word is a reduced expression of $s$ because $\mathrm{\ell }(s)=1$), by the definition of $b$ and step 1.1.

But $b_s{b}_s=[s][s]=[ss]$ by the product rule of [F1], while $b_{s\cdot s}=b_{s^2}=b_1=[\epsilon ]$ since $s^2=1$ in $W$; the two are different elements of $A^{+}$, because $L([ss])=2\mathrm{\ne }0=L([\epsilon ])$ by [F3]. Hence $b_s{b}_s\mathrm{\ne }{b}_{s\cdot s}$, so the assignment $b$ is not a monoid homomorphism: it fails to preserve the product $s\cdot s=1$.

The same defect appears at the group level: $(\gamma \circ b)(s{)}^2=\gamma ([ss])={\sigma }_s^2$ while $(\gamma \circ b)(s^2)=\gamma ([\epsilon ])=1_A$, and ${\sigma }_s^2\mathrm{\ne }{1}_A$ because $\mathrm{deg}({\sigma }_s^2)=2\mathrm{\ne }0=\mathrm{deg}(1_A)$ by [F3] and [F4], a group homomorphism preserving the identity. Thus $\gamma \circ b$ is not a group homomorphism $W\to A$.

The failure is not an artefact of the rank-one computation: for every finite Coxeter matrix with $S\mathrm{\ne }\mathrm{\varnothing }$ and every $s\in S$, $[ss]$ is the class of the word $ss$ in $A^{+}(S,m)$, and its length satisfies $L([ss])=2\mathrm{\ne }0=L([\epsilon ])$ by [F3]; combined with $s^2=1$ in $W$ and $\mathrm{\ell }(s)=1$ (Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action) this gives $b_s{b}_s\mathrm{\ne }{b}_{s^2}$ for the ambient monoid, and no parabolic reduction is needed. The general theorem records that $b_u{b}_v=b_{uv}$ holds exactly on the pairs with $\mathrm{\ell }(uv)=\mathrm{\ell }(u)+\mathrm{\ell }(v)$ by [F5], so the refuted statement is obtained by dropping that length hypothesis.

Scope and choice: this counterexample is a finite computation in the rank-one system plus the stated supplier clauses; it constructs no topological model, claims nothing about embeddings of $A^{+}$ into $A$, and uses no choice.

The four relations of part 1 are the relations [F1]–[F4] specialized to $v_s=v$: the normalized relation is the defining relation $T^2=(v-v^{-1})T+1$, and the other three read $S^2=(Q-1)S+Q$, $(H^{-}{)}^2=(v^{-1}-v)H^{-}+1$ and $(T^{\mathrm{E}\mathrm{W}}+1)(T^{\mathrm{E}\mathrm{W}}-q^{-2})=0$ with $Q=q^2$. The interconversions are [F5] at $v_s=v$. In rank one the standard basis of The standard basis of the generic Hecke algebra and base change is $\{1,T\}$, and $\{1,S\}$, $\{1,H^{-}\}$, $\{1,T^{\mathrm{E}\mathrm{W}}\}$ are obtained by fixing $1$ and scaling the other basis vector $T$ by the units $q$, $-1$, $-q^{-1}$, respectively. For each such unit $a$, the $R$-linear map $1\mapsto 1$, $T\mapsto aT$ has inverse $1\mapsto 1$, $T\mapsto a^{-1}T$, hence carries a basis to a basis by [F7].

For the rank-two check, each of the three substitutions multiplies every generator by one constant: $S_s\mapsto v$, $H_s\mapsto -1$, $T_s^{\mathrm{E}\mathrm{W}}\mapsto -v^{-1}$. Hence an alternating product of $m$ factors acquires the same constant factor, $v^m$, $(-1{)}^m$ or $(-v^{-1}{)}^m$, on both sides of the braid relation, so the braid relation is preserved by each substitution; this is exactly the content of the corresponding change of generators in [F2]–[F5]. For the diagonal formulas, the substituted standard-basis element is the product of the images of the generators along a reduced expression, $S_w=S_{s_1}\cdots S_{s_k}=v^k{T}_{s_1}\cdots T_{s_k}=v^{\mathrm{\ell }(w)}{T}_w$, and similarly $H_w^{-}=(-1{)}^{\mathrm{\ell }(w)}{T}_w$ and $T_w^{\mathrm{E}\mathrm{W}}=(-v^{-1}{)}^{\mathrm{\ell }(w)}{T}_w$, using the product rule and independence of the standard basis in Reduced-word independence of T_w and the length-multiplication rules (1).

For the rank-one Kazhdan–Lusztig conversion, set $T^{\mathrm{K}\mathrm{L}}:=-q^{-1}T$, which is part 1's $T^{\mathrm{E}\mathrm{W}}$; since $T=-q{T}^{\mathrm{K}\mathrm{L}}$ and the relation for $T$ holds, the substitution is consistent, and $T^{\mathrm{K}\mathrm{L}}$ satisfies $(T^{\mathrm{K}\mathrm{L}}{)}^2=q^{-2}{T}^2=q^{-2}((v-v^{-1})T+1)=(q^{-2}-1)T^{\mathrm{K}\mathrm{L}}+q^{-2}$ by the computation of step 1.1. With $H^{\mathrm{K}\mathrm{L}}:=q{T}^{\mathrm{K}\mathrm{L}}=-T=H^{-}$ one gets $(H^{\mathrm{K}\mathrm{L}}{)}^2=(H^{-}{)}^2=(v^{-1}-v)H^{-}+1=(q^{-1}-q)H^{\mathrm{K}\mathrm{L}}+1$, the rank-one case of the recorded conversion [F6] with $\mathrm{\ell }(s)=1$.

Scope and choice: this example compares four quadratic normalizations and verifies the exponent conversions of the standard bases in one rank-two family; it constructs no canonical basis, no Kazhdan–Lusztig polynomial and no positivity or bar-invariance statement, all of which remain in their designated proof homes, and the general coefficient conversion [F6] is supplied by the seam lemma. Every computation is a substitution in $R=\mathbb{Z}[v^{\pm 1}]$, and no choice is used.

The representation is faithful: $\rho$ is injective by clause (3) of The root-length criterion and faithfulness of the canonical reflection representation, applied to the system $S=\{s,t\}$, $m(s,t)=\mathrm{\infty }$. Hence $\rho$ is a faithful canonical realization of the system, and by clause (4)(a) of Hecke and Lie seam contract compatibility: the Artin-to-Hecke map, normalization conversions, root-length matching, and the reflection-faithfulness boundary the canonical construction is a realization in the sense of Elias–Williamson, with $\mathfrak{h}=V$, ${\alpha }_s^{\vee }=e_s$ and ${\alpha }_s$ the functional $2B(-,e_s)$.

The radical: since $m(s,t)=\mathrm{\infty }$ one has $B(e_s,e_s)=B(e_t,e_t)=1$ and $B(e_s,e_t)=-1$ (The real Coxeter form, its radical, reflections, and form-preserving maps). Hence $B(e_s+e_t,e_s)=1-1=0$ and $B(e_s+e_t,e_t)=-1+1=0$, so $\mathbb{R}(e_s+e_t)\subseteq \mathrm{rad}(B)$; for $v=a{e}_s+b{e}_t$, one has $B(v,e_s)=a-b$ and $B(v,e_t)=b-a$. Thus $v$ lies in the radical exactly when $a=b$. The radical is therefore the one-dimensional line $\mathrm{rad}(B)=\mathbb{R}(e_s+e_t)$, and is a hyperplane.

The radical is fixed pointwise by $W$: for $a=e_s$ or $e_t$ and $v\in \mathrm{rad}(B)$ one has $B(v,a)=0$, so the reflection formula gives $r_a(v)=v$ (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (2)); a composite of finitely many maps fixing a vector fixes it, and every element of $W$ is a finite product of the generators $s,t$ and their inverses by [F3]. Hence every element of $W$, in particular $\rho ((st{)}^k)$ for $k\mathrm{\ne }0$, fixes $\mathrm{rad}(B)$ pointwise.

There are infinitely many reflections: by part (4) of The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness, $st$ has infinite order in $W$. Put $r:=st$; since $s{r}^{-1}=sts=rs$, induction gives $s{r}^{-m}=r^{m}s$ and hence $r^{m}s{r}^{-m}=r^{2m}s$ for every $m\in \mathbb{Z}$. Each $r^{2m}s$ lies in the reflection set $T=\{ws{w}^{-1}:w\in W,s\in S\}$ (The canonical reflection homomorphism, roots, reflections, and the positive cone (2)), and the elements $r^{2m}s$, $m\in \mathbb{Z}$, are pairwise distinct: if $r^{2m}s=r^{2m^{\mathrm{\prime }}}s$ then right multiplication by $s$ gives $r^{2m}=r^{2m^{\mathrm{\prime }}}$, hence $r^{2(m-m^{\mathrm{\prime }})}=1$ and $m=m^{\mathrm{\prime }}$ because $r$ has infinite order. So $W$ has infinitely many reflections. Moreover $(st{)}^k$ for $k\mathrm{\ne }0$ has infinite order, whereas every conjugate of $s$ or $t$ is an involution; hence these powers are not reflections.

The matrices of the two generators in the basis $(e_s,e_t)$ follow from the reflection formula with $2c(s,t)=2$ for $m=\mathrm{\infty }$: $r_s(e_s)=-e_s$ and $r_s(e_t)=e_t+2e_s$, and $r_t(e_t)=-e_t$, $r_t(e_s)=e_s+2e_t$ (The real Coxeter form, its radical, reflections, and form-preserving maps, Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (3)). By step 1.2 the fixed space of $r_s$ is $\ker B(-,e_s)=\{a{e}_s+b{e}_t:a-b=0\}=\mathbb{R}(e_s+e_t)=\mathrm{rad}(B)$, and similarly $\mathrm{Fix}(r_t)=\ker B(-,e_t)=\mathrm{rad}(B)$: the two distinct reflections $r_s\mathrm{\ne }{r}_t$ have the same fixed hyperplane. In coordinates [F2] $[r_s]=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right),[r_t]=\left(\begin{array}{cc}1& 0\\ 2& -1\end{array}\right),$ differ in their $(1,1)$-entries, so indeed $r_s\mathrm{\ne }{r}_t$ while $\mathrm{Fix}(r_s)=\mathrm{Fix}(r_t)$.

Every full fixed space of codimension one of an element of $W$ equals $\mathrm{rad}(B)$. Every element of $W$ is $r^k$ or $r^{k}s$ for some $k\in \mathbb{Z}$ with $r:=st$: repeatedly deleting adjacent equal letters from a word in $s,t$ (using $s^2=t^2=1$) leaves an alternating word, and the alternating words are $(st{)}^k$, $(st{)}^{k}s$, $(ts{)}^k=(st{)}^{-k}$ and $(ts{)}^{k}t=(st{)}^{-k-1}s$. For every reflection $wa{w}^{-1}$ of $W$, with $a\in \{s,t\}$, its image is $\rho (wa{w}^{-1})=r_{\rho (w)e_a}$ with $\rho (w)e_a\in \mathrm{\Phi }$ (Descent of the reflection representation, unit root norms, and conjugation of reflections (4)); its fixed space is $\ker B(-,\alpha )$, a hyperplane (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (2)) containing $\mathrm{rad}(B)$ by step 1.3, hence equal to $\mathrm{rad}(B)$. For the elements $r^k$ write $A:=\rho (r)$, whose matrix is the product of the two matrices of step 2.1, $A=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)$; then $N:=A-I=\left(\begin{array}{cc}2& -2\\ 2& -2\end{array}\right)$ satisfies $N^2=0$, so $A^k=(I+N{)}^k=I+kN$ for every $k\in \mathbb{Z}$ (the case $k=-1$ being $(I+N{)}^{-1}=I-N$ when $N^2=0$), and $(I+kN)v=v$ with $k\mathrm{\ne }0$ exactly when $Nv=0$, i.e. exactly when $v\in \ker (A-I)=\{a{e}_s+b{e}_t:a=b\}=\mathrm{rad}(B)$; the identity $A^0=I$ fixes all of $V$, of codimension zero. For the elements $r^{k}s$ one has $(r^{k}s{)}^2=r^k(s{r}^{k}s)=r^k{r}^{-k}=1$ because $srs=r^{-1}$, so $\rho (r^{k}s)$ is an involution; it fixes $\mathrm{rad}(B)$ pointwise by step 1.3, while $\rho (r^{k}s)=A^k{r}_s=(I+kN)r_s=\left(\begin{array}{cc}-1-2k& 2+2k\\ -2k& 1+2k\end{array}\right)\mathrm{\ne }I$, so its fixed space is a proper subspace containing the hyperplane $\mathrm{rad}(B)$, hence equals $\mathrm{rad}(B)$. Therefore the only full fixed space of codimension one of an element of $W$ is $\mathrm{rad}(B)$.