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]

$$\{w^{\prime}\in W\ |w^{\prime}<ws,\ \ell(w^{\prime})=\ell(w)\}=\{w^{\prime\prime}s\ |\ w^{\prime\prime}<w,\ w^{\prime\prime}<w^{\prime\prime}s,\ \ell(w^{\prime\prime})=\ell(w)-1\}\sqcup\{w\}.$$

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 $d^{2}=0$. ∎