ScalingStacks

Proof. Let $w\in W$ and $s\in S$ with $ws>w$. We have $d(T_{w}T_{s})=d(T_{ws})=\sum_{w^{\prime}<ws,\ \ell(w^{\prime})=\ell(w)}T_{w^{\prime}}$. We have [Hu, Theorem 5.10]

It follows that $d(T_{w}T_{s})=d(T_{w})T_{s}+T_{w}=d(T_{w})T_{s}+T_{w}d(T_{s})$.

Consider now $v\in W$ and $s\in S$ with $vs<v$. We have $d(T_{v})=d(T_{vs}T_{s})=d(T_{vs})T_{s}+T_{vs}$ by the result above. It follows that $d(T_{v})T_{s}+T_{v}d(T_{s})=T_{vs}T_{s}+T_{v}=0=d(T_{v}T_{s})$.

We deduce that $d(T_{w}T_{w^{\prime}})=d(T_{w})T_{w^{\prime}}+T_{w}d(T_{w^{\prime}})$ for all $w,w^{\prime}\in W$.

Since $d^{2}(T_{s})=0$ for $s\in S$, it follows that by induction that $d^{2}=0$. ∎

Proof. The first statement follows from Proposition 3.1.3. The second statement is a property of the multiplication of $T_{w}$’s.

For the third statement, let us assume $r<r^{\prime}$. We have $s_{i_{r+1}}\cdots s_{i_{r^{\prime}}}=s_{i_{r}}\cdots s_{i_{r^{\prime}-1}}$ reduced, hence $s_{i_{r}}s_{i_{r+1}}\cdots s_{i_{r^{\prime}}}$ is not reduced, a contradiction. ∎

Proof. Let $v\in W$. There is a unique decomposition $v=v^{\prime}v^{\prime\prime}$ where $\ell(v)=\ell(v^{\prime})+\ell(v^{\prime\prime})$, $v^{\prime\prime}\in W_{I}$ and $v^{\prime}\in W^{I}$.

We have $d(T_{v})=d(T_{v^{\prime}})T_{v^{\prime\prime}}+T_{v^{\prime}}d(T_{v^{\prime\prime}})$. If $u\in W$ and $u<v^{\prime}$, then $u{\not\in}w_{S}W_{I}$. It follows that

∎