ScalingStacks

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

$$L(\sigma)=\bigl(\{(i,i+1)\}+n{\mathbf{Z}}\bigr)\coprod(s_{i,i+1},s_{i,i+1})(L(\sigma s_{i,i+1})).$$

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

$$L(\sigma)=s_{a_{l},a_{l}+1}(L(\sigma s_{i_{l},i_{l}+1}))\coprod\bigl(\{(i_{l},i_{l}+1)\}+n{\mathbf{Z}}\bigr).$$

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. ∎