3. Roots, words and inversion sets

In this post we gather some technical results relating roots, words and inversion sets. This is all standard Coxeter group theory.

Definition 1: Let $(R,B)$ be a root system and let $\mathbf{w}=(s_{x_1},\ldots,s_{x_n})$ be a word. For each index $i\leq n$ we assign a root $$r_i(\mathbf{w})=s_{x_n}s_{x_{n-1}}\cdots s_{x_{i+1}}(x_i)$$
The proof of Lemma 3.1 below is actually quite easy and we leave it to the reader.

Lemma 3.1: Let $(R,B)$ be a root system and let $\mathbf{u}=(s_{x_1},\ldots,s_{x_n})$ be a word. Let $x\in B$ and let $$\mathbf{v}=(s_{x_1},\ldots,s_{x_n},s_x)$$ Then for each $1\leq i\leq n$ we have $$r_i(\mathbf{v})=x\ast r_i(\mathbf{u})$$ $\square$

The next lemma allows us to build inversion sets root by root.

Lemma 3.2: Let $(R,B)$ be a root system, $w\in W(R)$, and $x\in B$. If $x\notin I(w)$, then $$I(ws_x)=\{x\}\cup (x\ast I(w))$$ If $x\in I(w)$, then $$I(ws_x)=x\ast (I(w)-\{x\})$$
Proof: Suppose $x\notin I(w)$. Then since $ws_x(x)=w(-x)=-w(x)\in R^-(B)$ we have that $x\in I(ws_x)$. If $y\in I(ws_x)$, then $$ws_x(y)=w(s_x\ast y)\notin R^+(B)$$ Since $y\neq x$, $x\ast y\in R^+(B)$. Thus $x\ast y\in I(w)$. Conversely, if $y\in I(w)$, then $w(y)=ws_x(x\ast y)\notin R^+(B)$, hence $x\ast y\in I(ws_x)$. Thus $y\mapsto x\ast y$ is a bijection of $I(w)$ with $I(ws_x)-\{x\}$, and we are done. The second part is left to the reader. $\square$

Theorem 3.3 tells us why the roots we assign to these indices are important.

Theorem 3.3: Let $(R,B)$ be a root system and let $\mathbf{w}=(s_{x_1},\ldots,s_{x_n})$ be a reduced word for the element $$w=s_{x_1}\cdots s_{x_n}$$ Then $$I(w)=\{r_i(\mathbf{w})|1\leq i\leq n\}$$
Proof: We prove this by induction on $\ell(w)$. The base case is $\ell(w)=0$, in which case a reduced word for $w$ is empty and hence the result is true. For the induction step, we assume the result is true for $w$ with an arbitrary word $\mathbf{w}=(s_{x_1},\ldots,s_{x_n})$, let $x\in B-I(w)$, and prove the result for the element $ws_x$ and the word $$\mathbf{w}'=(s_{x_1},\ldots,s_{x_n},s_x)$$ We define unambiguously $$I_{ws_x}=\{r_i(\mathbf{w}')|1\leq i\leq n+1\}$$ By Lemma 3.1 above and the induction hypothesis, we have that $$I_{ws_x}=\{x\}\cup (x\ast I(w))$$ Hence by Lemma 3.2, $$I_{ws_x}=I(ws_x)$$ and the result follows. $\square$

Definition 2: Given an element $w\in W(R)$, denote by $R(w)$ the set of reduced words for $w$. Given a reduced word $\mathbf{w}=(s_{x_1},\ldots,s_{x_n})$ for the element $w$, we define a bijection $b_{\mathbf{w}}:I(w)\to [\ell(w)]$ by declaring that $b_{\mathbf{w}}(r_j(\mathbf{w}))=j$. Then define $$B(w)=\{b_{\mathbf{w}}|\mathbf{w}\in R(w)\}$$ These can be seen as linear orderings on the inversion set. These correspond bijectively to reduced words as we prove now.

Theorem 3.4: Let $(R,B)$ be a root system and let $w\in W(R)$. Then the map $R(w)\to B(w)$ given by $\mathbf{w}\mapsto b_{\mathbf{w}}$ is a bijection.

Proof: Clearly the map is surjective. We prove it is injective by induction on $\ell(w)$. This is clear if $\ell(w)=0$. Otherwise, suppose $\mathbf{u}\neq \mathbf{v}$ are reduced words for $w$. If $\mathbf{u}_{\ell(w)}\neq \mathbf{v}_{\ell(w)}$, then there exist distinct $x,x'\in B\cap I(w)$ such that $b_{\mathbf{u}}(x)=b_{\mathbf{v}}(x')=\ell(w)$, hence $b_{\mathbf{u}}\neq b_{\mathbf{v}}$. If instead $\mathbf{u}_{\ell(w)}=\mathbf{v}_{\ell(w)}$, then there exists $x\in B\cap I(w)$ such that $b_{\mathbf{u}}(x)=b_{\mathbf{v}}(x)=\ell(w)$. Define $$\mathbf{u}'=(\mathbf{u}_i)_{i < \ell(w)}$$ and $$\mathbf{v}'=(\mathbf{v}_i)_{i < \ell(w)}$$ Then $\mathbf{u}'\neq \mathbf{v'}$ and if $y\neq x$ then $$b_{\mathbf{u}}(y)=b_{\mathbf{u}'}(x\ast y)$$ and $$b_{\mathbf{v}}(y)=b_{\mathbf{v}'}(x\ast y)$$ By the induction hypothesis, $b_{\mathbf{u}'}(x\ast y)\neq b_{\mathbf{v}'}(x\ast y)$ for some $y\in I(w)$, hence $b_{\mathbf{u}}\neq b_{\mathbf{v}}$ and the result follows by induction. $\square$

Uniqueness of inversion sets is a very useful fact that we now prove.

Proposition 3.5: Let $(R,B)$ be a root system and let $u,v\in W(R)$. If $I(u)=I(v)$, then $u=v$.

Proof: We prove this by induction on $\ell(u)=\ell(v)$. If $\ell(u)=\ell(v)=0$ this is clear. Otherwise let $x\in I(u)\cap B$. Then $$I(us_x)=I(vs_x)=x\ast (I(u)-\{x\})$$ By the induction hypothesis, $us_x=vs_x$, hence $u=(us_x)s_x=(vs_x)s_x=v$ by induction. $\square$