By the double centralizer theory we know that as ${\text{End}}_{𝔘}\left(W\right)\otimes 𝔘$ modules, $W\cong \underset{\lambda }{⨁}{𝒵}^{\lambda }\otimes {\Lambda }_{\lambda },$ where ${𝒵}^{\lambda }$ are irreducible modules for ${\text{End}}_{𝔘}\left(W\right)$ and ${\Lambda }_{\lambda }$ are irreducible modules for $𝔘\text{.}$ By taking traces on both sides of this isomorphism we have $${\text{tr}}_q\left(b\right)=\text{Tr}\left(v^{-1}ub\right)=\sum _{\lambda \in \hat{W}}{\eta }_{\lambda }\left(b\right){\chi }^{\lambda }\left(v^{-1}u\right)=\sum _{\lambda \in \hat{W}}{\eta }_{\lambda }\left(b\right){\text{dim}}_q\left({\Lambda }_{\lambda }\right),$$ where ${\eta }_{\lambda }$ is the irreducible character of ${\text{End}}_{𝔘}\left(W\right)$ on the module ${𝒵}^{\lambda }$ and ${\chi }^{\lambda }$ is the irreducible character of the irreducible $𝔘\text{-module}$ ${\Lambda }_{\lambda }\text{.}$ Thus the trace of a minimal idempotent $p_{\mu }$ in the minimal ideal corresponding to $\mu$ is $$\text{wt}\left(\mu \right)={\text{tr}}_q\left(p_{\mu }\right)=\sum _{\lambda \in \hat{W}}{\eta }_{\lambda }\left(p_{\mu }\right){\text{dim}}_q\left({\Lambda }_{\lambda }\right)=\sum _{\lambda \in \hat{W}}{\delta }_{\lambda \mu }{\text{dim}}_q\left({\Lambda }_{\lambda }\right)={\text{dim}}_q\left({\Lambda }_{\mu }\right)\text{.}$$

$\square$

(a) Let $g\in 𝔘,$ $x\in V,$ $\varphi \in V^{*}\text{.}$ Then, by direct computation, $$\begin{array}{ccc}gě(x\otimes \varphi )& =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩g(\sum _i{e}_i\otimes e^i)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\Delta \left(g\right)(\sum _i{e}_i\otimes e^i)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\sum _{g,i}{g}_{\left(1\right)}{e}_i\otimes g_{\left(2\right)}{e}^i\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\sum _{i,j,k}\sum _g⟨g_{\left(1\right)}{e}_i,e^j⟩\\ & & \times ⟨g_{\left(2\right)}{e}^i,e_k⟩(e_j\otimes e^k)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\sum _{i,j,k}\sum _g⟨g_{\left(1\right)}{e}_i,e^j⟩\\ & & \times ⟨e^i,S\left(g_{\left(2\right)}\right)e_k⟩(e_j\otimes e^k)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\sum _{j,k}\sum _g⟨g_{\left(1\right)}S\left(g_{\left(2\right)}\right)e_k,e^j⟩(e_j\otimes e^k)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}ux⟩\sum _{j,k}⟨\epsilon \left(g\right)e_k,e^j⟩e_j\otimes e^k\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\epsilon \left(g\right)⟨\varphi ,v^{-1}ux⟩\sum _j{e}_j\otimes e^j\\ & =& \epsilon \left(g\right)ě(x\otimes \varphi ),\end{array}$$ where we are using the identity $\sum _g{g}_{\left(1\right)}S\left(g_{\left(2\right)}\right)=\epsilon \left(g\right)$ which follows from the definition of the antipode in a Hopf algebra. On the other hand, since $v$ is in the center of $𝔘$ and $u^{-1}xu=S^{-2}\left(x\right)$ for all $x\in 𝔘,$ $$\begin{array}{ccc}ěg(x\otimes \varphi )& =& ě(\sum _g{g}_{\left(1\right)}x\otimes g_{\left(2\right)}\varphi )\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\sum _{g,i}⟨g_{\left(2\right)}\varphi ,v^{-1}u{g}_{\left(1\right)}x⟩e_i\otimes e^i\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\sum _{g,i}⟨\varphi ,u{u}^{-1}S\left(g_{\left(2\right)}\right)v^{-1}u{g}_{\left(1\right)}x⟩e_i\otimes e^i\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\sum _{g,i}⟨\varphi ,v^{-1}u{S}^{-1}\left(g_{\left(2\right)}\right)g_{\left(1\right)}x⟩e_i\otimes e^i\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\sum _{g,i}⟨\varphi ,v^{-1}u\epsilon \left(g\right)x⟩e_i\otimes e^i\\ & =& \epsilon \left(g\right)ě(x\otimes \varphi ),\end{array}$$ where we are using the identity $\sum _g{S}^{-1}\left(g_{\left(2\right)}\right)g_{\left(1\right)}=\epsilon \left(g\right)$ which follows from the definition of the skew antipode in a Hopf algebra.

(b) This follows from the following easy computation. $$\begin{array}{ccc}{ě}^2(x\otimes \varphi )& =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}ě(⟨\varphi ,v^{-1}ux⟩\sum _i{e}_i\otimes e^i)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}ux⟩\sum _i⟨e^i,v^{-1}u{e}_i⟩\sum _j{e}_j\otimes e^j\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}ux⟩{\text{dim}}_q\left(V\right)\sum _j{e}_j\otimes e^j\\ & =& ě(x\otimes \varphi )\text{.}\end{array}$$

$\square$

Let $W=V^{\otimes (m-1)}\text{.}$ Let $\left\{w_t\right\}$ and $\left\{e_i\right\}$ be bases of $W$ and $V$ respectively and let $\left\{w^s\right\}$ and $\left\{e^i\right\}$ be dual bases in $W^{*}$ and $V^{*}$ respectively.

(a) Then $$\begin{array}{ccc}{ě}_{m}b{ě}_m(w_s\otimes e_i\otimes e^j)& =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨e^j,v^{-1}u{e}_i⟩\sum _k{ě}_{m}b(w_s\otimes e_k\otimes e^k)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨e^j,v^{-1}u{e}_i⟩\sum _{k,t,l}⟨b(w_s\otimes e_k),w^t\otimes e^l⟩{ě}_m(w_t\otimes e_l\otimes e^k)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨e^j,v^{-1}u{e}_i⟩\sum _{k,t,l,p}⟨b(w_s\otimes e_k),w^t\otimes e^l⟩⟨e^k,v^{-1}u{e}_l⟩\\ & & \times (w_t\otimes e_p\otimes e^p)\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨e^j,v^{-1}u{e}_i⟩\sum _{k,t,p}⟨(\text{id}\otimes v^{-1}u)b(w_s\otimes e_k),w^t\otimes e^k⟩\\ & & \times (w_t\otimes e_p\otimes e^p)\\ & =& {\epsilon }_{m-1}\left(b\right){ě}_m(w_s\otimes e_i\otimes e^j)\text{.}\end{array}$$ The remaining assertion follows since ${ě}_m$ commutes with elements of $\text{End}\left(W\right)\subseteq \text{End}(W\otimes V\otimes V^{*})\text{.}$

(b) The action of ${\epsilon }_{m-1}\left(a_{1}b{a}_2\right)$ on a basis element $w_j$ of $W$ satisfies $$\begin{array}{ccc}{\epsilon }_{m-1}\left(a_{1}b{a}_2\right)w_j& =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}\sum _{k,p}⟨(\text{id}\otimes v^{-1}u)(a_1\otimes \text{id})b(a_2\otimes \text{id})(w_j\otimes e_k),w^p\otimes e^k⟩w_p\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}{a}_1\sum _{k,p}⟨(\text{id}\otimes v^{-1}u)b(a_2{w}_j\otimes e_k),w^p\otimes e^k⟩w_p\\ & =& a_1{\epsilon }_{m-1}\left(b\right)a_2{w}_j\text{.}\end{array}$$

(c) Let $x\in 𝔘$ and let $\bar{x}\in \text{End}\left(W\right)$ be the endomorphism of $W$ determined by the action of $x$ on $W\text{.}$ Then, since $b\in {\text{End}}_{𝔘}(W\otimes V),$ $\bar{x}{ě}_{m}b{ě}_m={ě}_m\bar{x}b{ě}_m={ě}_{m}b\bar{x}{ě}_m={ě}_{m}b{ě}_m\bar{x}\text{.}$ This implies that $\bar{x}{\epsilon }_{m-1}\left(b\right){ě}_m={\epsilon }_{m-1}\left(b\right)\bar{x}{ě}_m\text{.}$ Since the map $\text{End}\left(W\right)\to \text{End}(W\otimes V\otimes V^{*})$ given by $a\mapsto a{ě}_m$ is injective, it follows that $\bar{x}{\epsilon }_{m-1}\left(b\right)={\epsilon }_{m-1}\left(b\right)\bar{x}\text{.}$

$\square$

(a) By the definition of the Markov trace and the fact that $v^{-1}u$ is a grouplike element of $𝔘,$ $${\text{mt}}_m\left(a\right)=\frac{{\text{Tr}}_q\left((v^{-1}u\otimes v^{-1}u)(a\otimes \text{id})\right)}{{\text{dim}}_q{\left(V\right)}^m},$$ by the definition of the quantum trace on $V^{\otimes m}\text{.}$ Since traces on tensor products of modules are the products of the individual traces we may write $${\text{mt}}_m\left(a\right)=\frac{\text{Tr}\left(v^{-1}ua\right)\text{Tr}\left(v^{-1}u\text{id}\right)}{{\text{dim}}_q{\left(V\right)}^m},$$ where the first Tr in the numerator is on $V^{\otimes (m-1)}$ and the second is on $V\text{.}$ Then, by the definition of quantum dimension, we get $${\text{mt}}_m\left(a\right)=\frac{\text{Tr}\left(v^{-1}ua\right){\text{dim}}_q\left(V\right)}{{\text{dim}}_q{\left(V\right)}^m}={\text{mt}}_{m-1}\left(a\right)\text{.}$$ In particular, ${\text{mt}}_m\left(1\right)={\text{tr}}_q\left({\text{id}}^{\otimes m}\right)/{\text{dim}}_q{\left(V\right)}^m=1\text{.}$

(b) Let $W=V^{\otimes (m-1)}\text{.}$ Let $\left\{w_s\right\}$ be a basis of $W$ and let $\left\{w^s\right\}$ be a dual basis in $W^{*}\text{.}$ Let $\left\{e_i\right\}$ be a basis of $V$ and let $\left\{e^i\right\}$ be a dual basis of $V^{*}\text{.}$ Let $b\in {𝒵}_m\text{.}$ Since the element $v^{-1}u$ is a grouplike element of $𝔘$ we have $$\begin{array}{ccc}{\text{dim}}_q\left(V\right){\text{tr}}_q\left({\epsilon }_{m-1}\left(b\right)\right)& =& {\text{dim}}_q\left(V\right)\sum _s⟨v^{-1}u{\epsilon }_{m-1}\left(b\right)w_s,w^s⟩\\ & =& \sum _{s,k}⟨(v^{-1}u\otimes \text{id})(\text{id}\otimes v^{-1}u)b(w_s\otimes e_k),w^s\otimes e^k⟩\\ & =& \sum _{s,k}⟨(v^{-1}u\otimes v^{-1}u)b(w_s\otimes e_k),w^s\otimes e^k⟩\\ & =& \sum _{s,k}⟨v^{-1}ub(w_s\otimes e_k),w^s\otimes e^k⟩\\ & =& {\text{tr}}_q\left(b\right),\end{array}$$ where the quantum trace on the left hand side of equation is the quantum trace on $V^{\otimes (m-1)}$ and the quantum trace on the right side of the equation is the quantum trace on $V^{\otimes m}\text{.}$ The statement follows by converting to Markov traces.

(c) Let ${ě}_m$ be the element of ${\text{End}}_{𝔘}(V^{\otimes m}\otimes V^{*})$ given by ${ě}_m={\text{id}}^{\otimes (m-1)}\otimes ě,$ where $ě$ is as in (3.2). Then, since $a\in {\text{End}}_{𝔘}\left(V^{\otimes (m-1)}\right),$ $a$ commutes with ${ě}_m$ and $$\begin{array}{ccc}{ě}_{m}a{Ř}_{m-1}{ě}_m& =& a{ě}_m{Ř}_{m-1}{ě}_m\\ & =& a({\text{id}}^{\otimes (m-1)}\otimes \left(ěŘě\right))\\ & =& a({\text{id}}^{\otimes (m-1)}\otimes {\epsilon }_1\left(Ř\right)ě)\\ & =& {\text{dim}}_q{\left(V\right)}^{-1}\alpha a{ě}_m\text{.}\end{array}$$ It follows that ${\epsilon }_{m-1}\left(a{Ř}_{m-1}\right)=\alpha a$ and thus that $${\text{mt}}_m\left(a{Ř}_{m-1}\right)={\text{mt}}_{m-1}\left({\epsilon }_{m-1}\left(a{Ř}_{m-1}\right)\right)={\text{dim}}_q{\left(V\right)}^{-1}\alpha {\text{mt}}_{m-1}\left(a\right)\text{.}$$

(d) This follows immediately from Lemma (3.1) and the definition of the Markov traces.

$\square$

(1) By Proposition (3.7)(a) it is enough to show that ${ě}_2Ř{ě}_2={\left({\text{dim}}_q\left(V\right)\right)}^{-1}v{\left(\lambda \right)}^{-1}{ě}_2$ as elements of ${\text{End}}_{𝔘}(V\otimes V\otimes V^{*})\text{.}$ Let $\left\{e_i\right\}$ be a basis of $V$ and let $\left\{e^i\right\}$ be a dual basis in $V^{*}\text{.}$ It follows from the identities (2.5), (2.6) and (2.7) that if $ℛ=\sum _i{a}_i\otimes b_i$ and $(S\otimes \text{id})\left(ℛ\right)={ℛ}^{-1}=\sum _j{c}_j\otimes d_j,$ then $$\sum _i{b}_i{S}^2\left(a_i\right)=\sum _j{d}_{j}S\left(c_j\right)=\sum _j{S}^{-1}\left(d_j\right)c_j=u^{-1}\text{.}$$ Let $x,y\in V$ and let $\varphi \in V^{*}\text{.}$ Then, $$\begin{array}{ccc}{ě}_2Ř{ě}_2(x\otimes y\otimes \varphi )& =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}uy⟩{ě}_2Ř\sum _{k}x\otimes e_k\otimes e^k\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}⟨\varphi ,v^{-1}uy⟩{ě}_2\sum _{k,i}{b}_i{e}_k\otimes a_{i}x\otimes e^k\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}uy⟩\sum _{k,i,l}⟨e^k,v^{-1}u{a}_{i}x⟩b_i{e}_k\otimes e_l\otimes e^l\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}uy⟩\sum _{i,l}\left(b_i{v}^{-1}u{a}_{i}x\right)\otimes e_l\otimes e^l\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}uy⟩\sum _{i,l}{b}_i{S}^2\left(a_i\right)v^{-1}ux\otimes e_l\otimes e^l\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-2}⟨\varphi ,v^{-1}uy⟩\sum _l{u}^{-1}{v}^{-1}ux\otimes e_l\otimes e^l\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}{ě}_2(v^{-1}x\otimes y\otimes \varphi )\\ & =& {\left({\text{dim}}_q\left(V\right)\right)}^{-1}v{\left(\lambda \right)}^{-1}{ě}_2(x\otimes y\otimes \varphi )\text{.}\end{array}$$

(2) follows immediately, since, by Proposition (2.14), the element $v=e^{-h\rho }u$ acts on an irreducible module ${\Lambda }_{\lambda }$ of highest weight $\lambda$ by the constant $q^{-⟨\lambda ,\lambda +2\rho ⟩}\text{.}$

$\square$

Step 1. Let $M\prime =({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)})\in {𝒯}^m$ and let $M″=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-1)})\text{.}$ Then $${\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)=\frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\mu }^{(m-1)}\right)}{E}_{M″M″}\ne 0\text{.}$$

		Proof.
	Suppose that $${\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)=\sum _{(U″,R″)\in {\Omega }^{m-1}}{a}_{U″R″}{E}_{U″R″}\in {𝒵}_{m-1},$$ for some constants $a_{U″R″}\in k\text{.}$ Suppose that $(S″,T″)\in {\Omega }^{m-1}$ and that $S″=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{(m-1)})\text{.}$ Then $$\begin{array}{ccc}{\text{mt}}_m\left(E_{S″T″}{E}_{M\prime M\prime }\right)& =& {\text{mt}}_m\left(E_{S″S″}{E}_{S″T″}{E}_{M\prime M\prime }\right)\\ & =& {\text{mt}}_m\left(E_{S″T″}{E}_{M\prime M\prime }{E}_{S″S″}\right)\\ & =& {\delta }_{S″M″}{\delta }_{T″M″}{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)\text{.}\end{array}$$ On the other hand, by Proposition (3.7b) $$\begin{array}{ccc}{\text{mt}}_{m-1}\left({\epsilon }_{m-1}\left(E_{S″T″}{E}_{M\prime M\prime }\right)\right)& =& {\text{mt}}_{m-1}\left(E_{S″T″}{\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)\right)\\ & =& {\text{mt}}_{m-1}\left(E_{S″T″}{\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)E_{S″S″}\right)\\ & =& {\text{mt}}_{m-1}\left(E_{S″T″}{a}_{S″T″}{E}_{T″S″}{E}_{S″S″}\right)\\ & =& a_{S″T″}{\text{wt}}_{m-1}\left({\sigma }^{(m-1)}\right)\text{.}\end{array}$$ By Proposition (3.7a), these two expressions are equal. Since the weights of the Markov trace are nonzero, it follows that $$a_{S″T″}=\{\begin{array}{cc}\frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\mu }^{(m-1)}\right)},& \text{if}\hspace{0.17em}S″=T″=M″,\\ 0,& \text{otherwise,}\end{array}$$ and, if $S″=T″=M″$ then $a_{S″T″}\ne 0\text{.}$ The formula for ${\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)$ follows. $\square$

Step 2. It follows from Proposition (1.4) that ${ě}_m$ has the form $$\begin{array}{cc}{ě}_m=\sum _{(S,T)\in {\Omega }_{m-1}^{m+1}}{\left({ě}_m\right)}_{ST}{E}_{ST},& \text{(3.13)}\end{array}$$ since, by its definition, ${ě}_m$ commutes with all elements of ${𝒵}_{m-1}\text{.}$

Step 3. Let $(S,T)\in {\Omega }^{m+1}\text{.}$ Suppose $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{(m+1)})$ and define $S\prime =({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ and $S″=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{(m-1)})\text{.}$ Define $T\prime$ and $T″$ analogously. Let $M\prime =({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)})\in {𝒯}^m$ and let $M″=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-1)})\text{.}$ Then, if ${\left({ě}_m\right)}_{SS}\ne 0,$ then $$\begin{array}{cc}{\left({ě}_m\right)}_{SS}=\frac{{\text{wt}}_m\left({\sigma }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\sigma }^{(m-1)}\right)},& \text{(3.14)}\end{array}$$ and $$\begin{array}{cc}{\left({ě}_m\right)}_{SM}{\left({ě}_m\right)}_{MS}=\frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right){\text{wt}}_m\left({\sigma }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}{\left({\sigma }^{(m-1)}\right)}^2}\text{.}& \text{(3.15)}\end{array}$$

		Proof.
	It follows from the path algebra definitions and (3.13) that $$E_{SS}{ě}_m{E}_{M\prime M\prime }{ě}_m{E}_{TT}={\delta }_{S″M″}{\delta }_{M″T″}\sum _M{\left({ě}_m\right)}_{SM}{\left({ě}_m\right)}_{MT}{E}_{ST},$$ where the sum is over all tableaux $M$ such that $M=({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)},{\sigma }^{(m+1)})\text{.}$ Since the Bratteli diagram is multiplicity free there is at most one such $M\text{.}$ Thus $$\begin{array}{cc}{E}_{SS}{ě}_m{E}_{M\prime M\prime }{ě}_m{E}_{TT}={\delta }_{S″M″}{\delta }_{M″T″}{\left({ě}_m\right)}_{SM}{\left({ě}_m\right)}_{MT}{E}_{ST}\text{.}& \text{(3.16)}\end{array}$$ Let $S$ and $M″$ be as above. Then $$\begin{array}{ccc}{E}_{SS}{\epsilon }_{m-1}\left(E_{M\prime M\prime }\right){ě}_m{E}_{TT}& =& \frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\mu }^{(m-1)}\right)}{E}_{SS}{E}_{M″M″}{ě}_m{E}_{TT}\\ & =& \frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\mu }^{(m-1)}\right)}{\delta }_{S″M″}{\delta }_{M″T″}{\left({ě}_m\right)}_{ST}{E}_{ST}\text{.}& \text{(3.17)}\end{array}$$ Since ${ě}_m{E}_{M\prime M\prime }{ě}_m={\epsilon }_{m-1}\left(E_{M\prime M\prime }\right){ě}_m,$ it follows that (3.16) and (3.17) are equal. Assuming that $S=T$ and that $S″=M″,$ i.e. ${\sigma }^{\left(i\right)}={\mu }^{\left(i\right)},$ for all $i\le m-1,$ this gives the following equation. $$\begin{array}{cc}{\left({ě}_m\right)}_{SM}{\left({ě}_m\right)}_{MS}=\frac{{\text{wt}}_m\left({\mu }^{\left(m\right)}\right)}{{\text{wt}}_{m-1}\left({\mu }^{(m-1)}\right)}{\left({ě}_m\right)}_{SS}\text{.}& \text{(3.18)}\end{array}$$ The formula in (3.14) follows by setting $M=$ S\text{.}$$ The formula in (3.15) now follows from (3.14) and (3.18) (recall that ${\mu }^{(m-1)}={\sigma }^{(m-1)}\text{).}$ $\square$

Step 4. For each $\lambda \in {\widehat{𝒵}}_{m-1}$ there exist $S$ such that $(S,S)\in {\Omega }_{\lambda }^{m+1}$ and ${\left({ě}_m\right)}_{SS}\ne 0\text{.}$

		Proof.
	Fix $\lambda \in {\widehat{𝒵}}_{m-1}$ and let $M$ be such that ${\mu }^{(m-1)}=\lambda \text{.}$ Assume that ${\left({ě}_m\right)}_{SS}=0$ for all $S$ such that $(S,S)\in {\Omega }_{\lambda }^{m+1}\text{.}$ Then ${\left({ě}_m\right)}_{SM}{\left({ě}_m\right)}_{MS}=0$ for all $S\text{.}$ So, by (3.16), $E_{SS}{ě}_m{E}_{M\prime M\prime }{ě}_m{E}_{SS}=0$ for all $S\text{.}$ This implies that ${\epsilon }_{m-1}\left(E_{M\prime M\prime }\right)=0$ which is a contradiction to Step 1. $\square$

Step 5. If $S$ is such that ${\left({ě}_m\right)}_{SS}\ne 0$ then ${\sigma }^{(m-1)}={\sigma }^{(m+1)}\text{.}$

		Proof.
	Let $S\in {𝒯}^{m+1}$ be a tableau such that ${\left({ě}_m\right)}_{SS}\ne 0\text{.}$ Then, as $𝔘\text{-modules,}$ ${\Lambda }_{{\sigma }^{(m+1)}}\cong E_{SS}b(V^{\otimes m}\otimes V^{*})$ for all $b\in {𝒞}_{m+1}$ such that $E_{SS}b\ne 0\text{.}$ In particular, since ${\left({ě}_m\right)}_{SS}\ne 0,$ $$E_{SS}{ě}_m{E}_{S\prime S\prime }{ě}_m=c_{S\prime }{E}_{SS}{E}_{S″S″}{ě}_m\ne 0$$ and we have that $$\begin{array}{ccc}{\Lambda }_{{\sigma }^{(m+1)}}& \cong & E_{SS}ěm{E}_{S\prime S\prime }{ě}_m(V^{\otimes m}\otimes V^{*})\\ & =& c_{S\prime }{E}_{SS}{E}_{S″S″}{ě}_m(V^{\otimes m}\otimes V^{*})\\ & \cong & E_{SS}(E_{S″S″}{V}^{\otimes (m-1)}\otimes {ě}_m(V\otimes V^{*}))\text{.}\end{array}$$ Since $E_{S″S″}{V}^{\otimes (m-1)}\cong {\Lambda }_{{\sigma }^{(m-1)}}$ and $e_m(V\otimes V^{*})\cong {\Lambda }_{\varnothing }$ it follows that ${\Lambda }_{{\sigma }^{(m+1)}}$ is isomorphic to an irreducible component in the tensor product ${\Lambda }_{{\sigma }^{(m-1)}}\otimes {\Lambda }_{\varnothing }\text{.}$ Thus ${\Lambda }_{{\sigma }^{(m+1)}}\cong {\Lambda }_{{\sigma }^{(m-1)}},$ and so ${\sigma }^{(m+1)}={\sigma }^{(m-1)}$ as elements of $\widehat{𝔘}\text{.}$ $\square$