ScalingStacks

Proof. Given the defining relations for ${\mathcal{U}}^{\bullet}$, the construction of the lemma does define (uniquely) a monoidal functor $H$.

Fix $l_{1},\ldots,l_{n}\in\{1,\ldots,3\}$. Given $i\in\{1,\ldots,n-1\}$ such that $l_{i}\leq l_{i+1}$, we put $\tilde{T}_{i}=a_{l_{1}}\cdots a_{l_{i-1}}\lambda_{l_{i},l_{i+1}}a_{l_{i+2}}\cdots a_{l_{n}}$. Note that $\tilde{T}_{i}\tilde{T}_{i+1}\tilde{T}_{i}$ is well-defined if and only if $l_{i}\leq l_{i+1}\leq l_{i+2}$, hence if and only if $\tilde{T}_{i+1}\tilde{T}_{i}\tilde{T}_{i+1}$ is well-defined. As a consequence, given $i_{1},\ldots,i_{r},j_{1},\ldots,j_{s}\in\{1,\ldots,n-1\}$ such that $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r}}$ and $\tilde{T}_{j_{1}}\cdots\tilde{T}_{j_{s}}$ are well-defined and $T_{i_{1}}\cdots T_{i_{r}}=T_{j_{1}}\cdots T_{j_{s}}$, then we have $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r}}=\tilde{T}_{j_{1}}\cdots\tilde{T}_{j_{s}}$. This shows the faithfulness of $H$.

Consider $i_{1},\ldots,i_{r}$ such that $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r}}$ is well-defined and non-zero. Let $w=s_{i_{1}}\cdots s_{i_{r}}\in{\mathfrak{S}}_{n}$. We show by induction on $r$ that given $(i,j)\in\tilde{L}(w)$, we have $l_{i}\leq l_{j}$.

Let $w^{\prime}=s_{i_{1}}\cdots s_{i_{r-1}}$. Put $d=i_{r}$ and $w^{\prime}=ws_{d}$. Since $T_{i_{1}}\cdots T_{i_{r}}\neq 0$, we have $r=\ell(w)$. We have $\tilde{L}(w)=\{(d,d+1)\}\coprod s_{d}(\tilde{L}(w^{\prime}))$ by Lemma 3.2.3. We have a well-defined map $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r-1}}$ from $a_{l_{1}}\cdots a_{l_{d-1}}a_{l_{d+1}}a_{l_{d}}a_{l_{d+2}}\cdots a_{l_{n}}$. It follows by induction that given $(i,j)\in\tilde{L}(w^{\prime})$, we have $l_{s_{d}(i)}\leq l_{s_{d}(j)}$. Since $\tilde{L}(w)=\{(d,d+1)\}\coprod s_{d}(\tilde{L}(w^{\prime}))$ (Lemma 3.2.3), we deduce that $l_{i}\leq l_{j}$ for all $(i,j)\in\tilde{L}(w)$.

Consider now $w\in{\mathfrak{S}}_{n}$ such that given $(i,j)\in\tilde{L}(w)$, we have $l_{i}\leq l_{j}$. Let $w=s_{i_{1}}\cdots s_{i_{r}}$ be a reduced decomposition of $w$. We show by induction on $r$ that $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r}}$ is well-defined. As before, we define $d$ and $w^{\prime}$. By induction on $r$, the element $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r-1}}$ gives a well-defined map from $a_{l_{1}}\cdots a_{l_{d-1}}a_{l_{d+1}}a_{l_{d}}a_{l_{d+2}}\cdots a_{l_{n}}$. Since $(d,d+1)\in\tilde{L}(w)$, it follows that $l_{d}\leq l_{d+1}$, hence $\tilde{T}_{d}$ is a well-defined map from $a_{l_{1}}\cdots a_{l_{n}}$. We deduce that $\tilde{T}_{i_{1}}\cdots\tilde{T}_{i_{r}}$. This shows that $T_{w}$ is in the image of $H$. ∎