ScalingStacks

Proof. Define $w^{I}=w_{S}w_{I}$ and ${{}^{I}w}=w_{I}w_{S}$, so that ${{}^{I}w}\cdot w^{I}=1$. We have $w^{I}\in W^{I}$.

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 $v^{\prime}\in W^{I}$ [Hu, Proposition 1.10]. Furthermore, $\ell(v^{\prime})<\ell(w^{I})$ unless $v^{\prime}=w^{I}$. We have $t_{S,I}(T_{v})=\delta_{v^{\prime},w^{I}}T_{v^{\prime\prime}}$.

There is a unique decomposition $v^{\prime\prime}=v_{1}v_{2}$ with $\ell(v^{\prime\prime})=\ell(v_{1})+\ell(v_{2})$, $v_{2}\in W_{J}$ and $v_{1}$ has minimal length in $v^{\prime\prime}W_{J}$. We have $v=(v^{\prime}v_{1})v_{2}$ where $\ell(v)=\ell(v^{\prime}v_{1})+\ell(v_{2})$ and $v^{\prime}v_{1}$ has minimal length in $vW_{J}$. Furthermore, $v^{\prime}v_{1}=w^{J}$ if and only if $v^{\prime}=w^{I}$ and $v_{1}=w_{I}w_{J}$. It follows that

$$t_{I,J}\circ t_{S,I}(T_{v})=\delta_{v^{\prime},w^{I}}t_{I,J}(T_{v^{\prime\prime}})=\delta_{v^{\prime},w^{I}}\delta_{v_{1},w_{I}w_{J}}T_{v_{2}}=t_{S,J}(T_{v}).$$

This shows the first statement of the lemma.

We have $T_{v^{\prime\prime}}T_{x}\in H_{I}$, hence

$$t_{S,I}(T_{v}T_{x})=t_{S,I}(T_{v^{\prime}}(T_{v^{\prime\prime}}T_{x}))=\delta_{v^{\prime},w^{I}}T_{v^{\prime\prime}}T_{x}=t_{S,I}(T_{v})T_{x}.$$

This shows the second statement of the lemma.

Let $x^{\prime}=w^{I}\cdot x\cdot{{}^{I}w}$. We have $\ell(w^{I}\cdot x\cdot{{}^{I}w})=\ell(x)$. Since $T_{x^{\prime}}T_{v}$ is a linear combination of elements $T_{yz}$ with $y\leq x^{\prime}$ and $z\leq v$, it follows that if $v^{\prime}\neq w^{I}$, then $T_{x^{\prime}}T_{v^{\prime}}$ is a linear combination of elements $T_{w^{I}\cdot y\cdot{{}^{I}w}z}$ with $y\in W_{I}$ and $z{\not\in}w^{I}W_{I}$, hence of elements $T_{u}$ with $u{\not\in}w^{I}W_{I}$. So, if $v^{\prime}\neq w^{I}$, then $t_{S,I}(T_{x^{\prime}}T_{v})=0$.

Assume now $v^{\prime}=w^{I}$. We have $T_{x^{\prime}}T_{v}=T_{w^{I}\cdot x\cdot{{}^{I}w}}T_{w^{I}}T_{v^{\prime\prime}}=T_{w^{I}\cdot x}T_{v^{\prime\prime}}=T_{w^{I}}T_{x}T_{v^{\prime\prime}}$ because $\ell(x^{\prime}\cdot w^{I})=\ell(w^{I}\cdot x)=\ell(w^{I})+\ell(x)=\ell(x^{\prime})+\ell(w^{I})$. We deduce that $t_{S,I}(T_{x^{\prime}}T_{v})=T_{x}T_{v^{\prime\prime}}=T_{x}t_{S,I}(T_{v})$. This shows the third statement of the lemma.

Let $v_{0}\in W^{I}$. We have $\ell(w^{I})=\ell(w^{I}v_{0}^{-1})+\ell(v_{0})$. Let $v\in W^{I}$. Note that $T_{w^{I}v_{0}^{-1}}T_{v}=T_{w^{I}v_{0}^{-1}v}$ or $T_{w^{I}v_{0}^{-1}}T_{v}$ is a linear combination of $T_{w}$’s with $\ell(w)<\ell(w^{I}v_{0}^{-1})+\ell(v)$. It follows that if $t_{S,I}(T_{w^{I}v_{0}^{-1}}T_{v})=0$ if $\ell(v)<\ell(v_{0})$ or $\ell(v)=\ell(v_{0})$ and $v\neq v_{0}$. We have also $t_{S,I}(T_{w^{I}v_{0}^{-1}}T_{v_{0}})=1$.

Since $H$ is a free right $H_{I}$-module with basis $\{T_{v}\}_{v\in W^{I}}$, we deduce that $\hat{t}_{S,I}$ is surjective. Since $\hat{t}_{S,I}$ is an $R$-module morphism between free $R$-modules of the same finite rank, it follows that it is an isomorphism. This shows the fifth statement of the lemma.

Let $s\in S$ and $v\in W$. Let $s^{\prime}=w_{S}\cdot s\cdot w_{S}\in S$. If $v{\not\in}\{w_{S},w_{S}\cdot s\}$, then $t_{S,\emptyset}(T_{v}T_{s})=0$ and $t_{S,\emptyset}(T_{s^{\prime}}T_{v})=0$. If $v=w_{S}\cdot s$, then $T_{v}T_{s}=T_{w_{S}}=T_{s^{\prime}}T_{v}$. If $v=w_{S}$, then $t_{S,\emptyset}(T_{v}T_{s})=a_{s}=t_{S,\emptyset}(T_{s^{\prime}}T_{v})$. So, we have shown that $t_{S,\emptyset}(T_{v}T_{s})=t_{S,\emptyset}(T_{s^{\prime}}T_{v})$. It follows by induction on $\ell(w)$ that $t_{S,\emptyset}(T_{v}T_{w})=t_{S,\emptyset}(T_{w_{S}\cdot w\cdot w_{S}}T_{v})$ for all $w\in W$.

Consider now $h^{\prime}\in H$ commuting with $H_{I}$. Let $h^{\prime\prime}\in H_{I}$. We have

$$t_{I,\emptyset}(t_{S,I}(hh^{\prime})h^{\prime\prime})=t_{I,\emptyset}(t_{S,I}(hh^{\prime}h^{\prime\prime}))=t_{S,\emptyset}(hh^{\prime\prime}h^{\prime})=t_{S,\emptyset}(\iota_{S}(h^{\prime})hh^{\prime\prime})=\\ =t_{I,\emptyset}(t_{S,I}(\iota_{S}(h^{\prime})hh^{\prime\prime}))=t_{I,\emptyset}(t_{S,I}(\iota_{S}(h^{\prime})h)h^{\prime\prime}).$$

It follows that $\hat{t}_{I,\emptyset}(t_{S,I}(hh^{\prime}))=\hat{t}_{I,\emptyset}(t_{S,I}(\iota_{S}(h^{\prime})h))$, hence $t_{S,I}(hh^{\prime})=t_{S,I}(\iota_{S}(h^{\prime})h)$. This completes the proof of the lemma. ∎