ScalingStacks

Proof. Multiplying if necessary $\sigma^{\prime}$ and $\sigma^{\prime\prime}$ by a power of $c$, we can assume $\sigma$, $\sigma^{\prime}$ and $\sigma^{\prime\prime}$ are in $W_{n}$.

Let $\sigma^{\prime}=s_{a_{1}}\cdots s_{a_{m}}$ and $\sigma^{\prime\prime}=s_{a_{m+1}}\cdots s_{a_{d}}$ be two reduced decompositions. The Exchange Lemma [Hu, Theorem 5.8] shows that there is $i$ such that $\sigma s_{a}=s_{a_{1}}\cdots s_{a_{i-1}}s_{a_{i+1}}\cdots s_{a_{d}}$.

If $i>m$, then $\sigma^{\prime\prime}s_{a}=s_{a_{m+1}}\cdots s_{a_{i-1}}s_{a_{i+1}}\cdots s_{a_{d}}$ and this contradicts $\ell(\sigma^{\prime\prime}s_{a})>\ell(\sigma^{\prime\prime})$. So, $i\leq m$. We have $\sigma s_{a}=s_{a_{1}}\cdots s_{a_{i-1}}s_{a_{i+1}}\cdots s_{a_{m}}\sigma^{\prime\prime}$. We deduce that $\alpha^{\prime}=s_{a_{1}}\cdots s_{a_{i-1}}s_{a_{i+1}}\cdots s_{a_{m}}$ has length $m-1$ and the lemma follows. ∎

Proof. Consider a pair $(i,j)\in L(\sigma)$ with $1\leq i\leq n$ and such that $(i,j^{\prime}){\not\in}L(\sigma)$ and $(j^{\prime},j){\not\in}L(\sigma)$ for $i<j^{\prime}<j$. Given $j^{\prime}$ with $i<j^{\prime}<j$, we have $\sigma(i)<\sigma(j^{\prime})<\sigma(j)$, a contradiction. It follows that $j=i+1$. We have

We deduce by induction on $|\tilde{L}(\sigma)|$ that $\ell(\sigma)\leq|\tilde{L}(\sigma)|$.

We prove the statements on $\{\bigl(s_{a_{l}}\cdots s_{a_{r+1}}(i_{r}),s_{a_{l}}\cdots s_{a_{r+1}}(i_{r}+1)\bigr)\}_{1\leq r\leq l}$ by induction on $\ell(\sigma)$. By induction, the statements hold for $\sigma s_{a_{l},a_{l}+1}$. In particular, $\ell(\sigma s_{a_{l},a_{l}+1})=|\tilde{L}(\sigma s_{a_{l},a_{l}+1})|$. It follows that $\ell(\sigma)=\ell(\sigma s_{a_{l},a_{l}+1})+1>|\tilde{L}(\sigma s_{a_{l},a_{l}+1})|$. Assume $(i_{l},i_{l}+1){\not\in}L(\sigma)$. It follows that $L(\sigma s_{a_{l},a_{l}+1})=s_{a_{l},a_{l}+1}(L(\sigma))\coprod\bigl(\{(i_{l},i_{l}+1)\}+n{\mathbf{Z}}\bigr)$, hence $|\tilde{L}(\sigma)|<|\tilde{L}(\sigma s_{a_{l},a_{l}+1})|=\ell(\sigma s_{a_{l},a_{l}+1})=\ell(\sigma)-1$, a contradiction. It follows that $(i_{l},i_{l}+1)\in L(\sigma)$, hence

The last statement of the lemma follows now by induction.

Consider now $(i,j)\in L(\sigma)$. Up to translating $(i,j)$ diagonally by $n{\mathbf{Z}}$, we can assume there is $r$ such that $i=s_{a_{l}}\cdots s_{a_{r+1}}(i_{r})$ and $j=s_{a_{l}}\cdots s_{a_{r+1}}(i_{r}+1)$. So $\sigma s_{i,j}=c^{d}s_{a_{1}}\cdots s_{a_{r-1}}s_{a_{r+1}}\cdots s_{a_{l}}$, hence $\sigma s_{i,j}<\sigma$. The lemma follows. ∎

Proof. Consider $(j_{1},j_{2})\in L(\sigma)$ and let $s=s_{j_{1},j_{2}}$. Consider integers $i<j$ with $i-j{\not\in}n{\mathbf{Z}}$.

If $s(i)<s(j)$, then $(i,j)\in L(\sigma)$ if and only if $s(i,j)=(s(i),s(j))\in L(\sigma s)$.

Assume now $s(i)>s(j)$. We have three possibilities:

$\bullet\$ $i-j_{1}\in n{\mathbf{Z}}$, $j-j_{2}{\not\in}n{\mathbf{Z}}$: we have $(i,j)\in L(\sigma)$ if and only if $(i,j)\in L(\sigma s)$ or $\sigma(i)>\sigma(j)>\sigma s(i)$ (and then $(i,j){\not\in}L(\sigma s)$)

$\bullet\$ $i-j_{1}{\not\in}n{\mathbf{Z}}$, $j-j_{2}\in n{\mathbf{Z}}$: we have $(i,j)\in L(\sigma)$ if and only if $(i,j)\in L(\sigma s)$ or $\sigma s(j)>\sigma(i)>\sigma(j)$ (and then $(i,j){\not\in}L(\sigma s)$).

$\bullet\$ $i=j_{1}+nr$, $j=j_{2}+nr^{\prime}$ with $r,r^{\prime}\in{\mathbf{Z}}$: we have $(i,j)\in L(\sigma)$ if and only if $(i,j)\in L(\sigma s)$ or $\sigma(j_{1})-\sigma(j_{2})>n(r^{\prime}-r)>\sigma(j_{2})-\sigma(j_{1})$ (and then $(i,j){\not\in}L(\sigma s)$).

We deduce there is an injective map $a:L(\sigma s)\to L(\sigma)$ given by

and

Note that $a(L(\sigma s))\sqcup((j_{1},j_{2})+n{\mathbf{Z}})\subset L(\sigma)$.

Let us now prove the lemma. We have $\sigma=c^{d}w$ and $\sigma^{\prime}=c^{d^{\prime}}w^{\prime}\in\hat{{\mathfrak{S}}}_{n}$ for some $w,w^{\prime}\in W_{n}$. Assume $\sigma^{\prime}<\sigma$ and $\ell(\sigma^{\prime})=\ell(\sigma)-1$. We have $d=d^{\prime}$, $w^{\prime}<w$ and $\ell(w^{\prime})=\ell(w)-1$. It follows that there is a reduced decomposition $w=s_{a_{1}}\cdots s_{a_{l}}$ and $r\in\{1,\ldots,l\}$ such that $w^{\prime}=s_{a_{1}}\cdots s_{a_{r-1}}s_{a_{r+1}}\cdots s_{a_{l}}$. Let $j_{1}=s_{a_{l}}\cdots s_{a_{r+1}}(i_{r})$ and $j_{2}=s_{a_{l}}\cdots s_{a_{r+1}}(i_{r}+1)$. We have $(j_{1},j_{2})\in L(\sigma)$ and $\sigma^{\prime}=\sigma s_{j_{1},j_{2}}$ (Lemma 3.2.3).

The discussion above shows that $\{i\in{\mathbf{Z}}\ |\ j_{1}<i<j_{2},\ \sigma(j_{1})>\sigma(i)>\sigma(j_{2})\}=\emptyset$ and $\min\bigl(\frac{j_{2}-j_{1}}{n},\frac{\sigma(j_{1})-\sigma(j_{2})}{n}\bigr)<1$. The lemma follows. ∎