Positive elementary ribbon elements

PDF p. 22: (5.2)–(5.3)

Let $\mathrm{\Gamma }$ be a spherical Coxeter diagram, meaning that its Coxeter group $W$ is finite, with vertex set ${\mathrm{\Gamma }}_0$ and associated Artin–Tits group $A$. Let $A^{+}$ be the submonoid of $A$ generated by the standard Artin generators. For $J\subseteq {\mathrm{\Gamma }}_0$, let $w_J$ be the longest element of the standard parabolic subgroup $W_J$, and let ${\mathrm{\Delta }}_J$ be its positive reduced-word lift in $A$.

Let $I\subseteq {\mathrm{\Gamma }}_0$ and $a\in I^c={\mathrm{\Gamma }}_0\setminus I$. Write $I+a=I\cup \{a\}$, and let $[I,a]$ be the vertex set of the connected component containing $a$ in the diagram induced on $I+a$. Write $[I,a]-a=[I,a]\setminus \{a\}$. The positive elementary ribbon element associated to $(a,I)$ is

${\mu }_{a,I}={\mathrm{\Delta }}_{[I,a]-a}^{-1}{\mathrm{\Delta }}_{[I,a]}\in A^{+}\mathrm{.}$

Equivalently, ${\mu }_{a,I}={\mathrm{\Delta }}_I^{-1}{\mathrm{\Delta }}_{I+a}$. Inverses and products in these expressions are taken in $A$.

Coxeter-system, roots, length and weak-order conventions

Subset and vertex-complement notation

Positive Artin and parabolic monoids

Connected component notation [I,a]

Longest parabolic elements and opposition permutation

The spherical Coxeter system has finite rank, since its distinct simple generators belong to the finite group $W$. Every standard parabolic subgroup is finite. Hence Longest parabolic elements and opposition permutation supplies each displayed longest element and its well-defined positive lift. The component and set operations are supplied by Connected component notation [I,a] and Subset and vertex-complement notation, and the quotient is defined in the group $A$.

Put $K=[I,a]$ and $J=K-a$. In the spherical Artin system on $K$, the positive lift of $w_J\in W_K$ is a simple element, hence a left divisor of ${\mathrm{\Delta }}_K$, by Independent spherical Artin Garside input. Thus ${\mathrm{\Delta }}_K={\mathrm{\Delta }}_{J}b$ for a positive element $b$ of that system. Mapping its standard generators to the corresponding generators of $A$ preserves the braid relations, so the same equality holds in $A$ and $b\in A^{+}$. Therefore ${\mathrm{\Delta }}_J^{-1}{\mathrm{\Delta }}_K=b\in A^{+}$.

Let $H=(I+a)\setminus K$. There are no diagram edges between $H$ and $K$, and $I=H\bigsqcup J$. The disconnected-factor result Finite parabolic longest elements and disconnected factors gives $w_I=w_H{w}_J$ and $w_{I+a}=w_H{w}_K$, with lengths additive in both products. By Reduced-word lifts and the prefix-order convention, ${\mathrm{\Delta }}_I={\mathrm{\Delta }}_H{\mathrm{\Delta }}_J$ and ${\mathrm{\Delta }}_{I+a}={\mathrm{\Delta }}_H{\mathrm{\Delta }}_K$. Group cancellation now gives ${\mathrm{\Delta }}_I^{-1}{\mathrm{\Delta }}_{I+a}={\mathrm{\Delta }}_J^{-1}{\mathrm{\Delta }}_K={\mu }_{a,I}$. The argument also covers empty $H$ or $J$, using the identity as the empty lift.