Computation of Central Elements

## Computation of ${Z}_{V}^{\left(l\right)}$ when $V=L\left({\epsilon }_{1}\right)$

Let $𝔤={𝔰𝔬}_{2r+1}$ or $𝔤={𝔰𝔭}_{2r}$ or $𝔤={𝔰𝔬}_{2r}$, and let $V=L\left({\epsilon }_{1}\right)$, Let ${z}_{V}^{\left(l\right)}$ be the central elements in ${U}_{q}𝔤$ defined by $zV(l) = ε id⊗trV t+ 12 y l and let zV u = ∑ i∈ℤ≥0 zV(l) u-l.$ Then $π0 zV u + ε u - 12 = εu + 12 u+ 12 y u- 12 y -1 u- 12 δ u- 12 y u+ 12 y +1 u+ 12 δ σρ ∏ i=1 r u + hi + 12 u- hi+ 12 u+ hi - 12 u- hi - 12 .$

Proof.
By (5.20), as operators on $L\left(0\right)\otimes V$, $t = 12 ⟨ ε1 , ε1 + 2ρ ⟨ - ⟩ ε1 , ε1 + 2ρ + ⟨ 0, 0+ 2ρ =0 and 12 y +t l = 12 y l .$ So, as operators on $L\left(0\right)\otimes V$, $zV(l) = ε id⊗trV 12 y + t l = ε id⊗trV 12 y l = ε dim V 12 y l$ and, since $dim\left(V\right)=\epsilon +y$, $zV u = ∑ l∈ℤ≥0 zV(l) u-l = ∑ l∈ℤ≥0 ε dim V 12 y l u-l = ε dimV 1 1- 12 y u-1 .$ Therefore
 $zV u + εu - 12 = 1+εy 1- 12 y u-1 +εu - 12 = u+εyu+ εu - 12 u- 12 y u- 12 y = ε u 2 + 12 u + 12 ε y u + 14 y u- 12 y = εu + 12 u+ 12 y u- 12 y .$ 6.13
By the second identity in (2.6) and the definition of $\Phi$ in Theorem 5.2(a), $yk∈𝒲k-1 acts on L0 ⊗ V⊗k = L0 ⊗ V⊗k-1 ⊗ V as 12 y + t.$ If $L\left(\mu \right)$ is and irreducible ${U}_{q}𝔤$-module in $L\left(0\right)\otimes {V}^{\otimes k-1}$, then (5.20) gives that ${y}_{k}$ acts on the $L\left(\lambda \right)$ component of $L\left(\mu \right)\otimes V$ by the constant
 6.14
As in [
Naz, Theorem 2.6], the irreducible ${𝒲}_{k}$-module ${𝒲}_{k}^{\mu /0}={𝒲}_{k}^{\mu }$ has a basis $\left\{{v}_{T}\right\}$ indexed by up-down tableaux $T=\left({T}^{\left(0\right)},{T}^{\left(1\right)},\dots ,{T}^{\left(k\right)}\right)$, where ${T}^{\left(0\right)}=\varnothing$, ${T}^{\left(k\right)}=\mu$, and ${T}^{\left(i\right)}$ is a partition obtained from ${T}^{\left(i-1\right)}$ by adding or removing a box, and Thus $∏ i=1 k-1 u-yi-1 u+yi+1 u-yi 2 u+yi 2 u-yi+1 u-yi-1$ acts on the $L\left(\mu \right)\otimes {𝒲}_{k-1}^{\mu }$ isotypic component in the ${U}_{q}𝔤\otimes {𝒲}_{k-1}$-module decomposition
 $L0 ⊗ V⊗k-1 ≅ ⊕ μ Lμ ⊗ 𝒲 k-1 μ ,$ 6.15
by
 $∏ i=1 k-1 u+ c T(i) , T(i-1) -1 u+ c T(i), T(i-1) +1 u-c T(i), T(i-1) 2 u+c T(i), T(i-1) 2 u-c T(i), T(i-1) +1 u-c T(i), T(i-1) -1$ 6.16
for any up-down tableaux $T$ of length $k$ and shape $\mu$. If a box is added (or removed) at step $i$ and then removed (or added) at step $j$, then the $i$ and $j$ factora of this product cancel. Therefore, (
6.16) is equal to
 $∏ b∈μ u+ 12 y +cb -1 u+ 12 y +c b +1 u- 12 y -c b 2 u+ 12 y +c b 2 u- 12 y +c b +1 u- 12 y -c b -1$ 6.17
(see [
Naz, Lemma 3.8]). Simplifying one row at a time,
 $∏ b∈μ u+ 12 y +c b -1 u+ 12 y +c b +1 u+ 12 y +c b 2 = ∏ i=1 r u+ 12 y -i u+ 12 y+ μi -i+1 u+ 12 y +1 -i u+ 12 y + μi -i = u+ 12 y -r u+ 12 y +1 ∏ i=1 r u+ 12 y + μi -i +1 u+ 12 y + μi -i ,$ 6.18
if $\mu =\left({\mu }_{1},\dots ,{\mu }_{r}\right)$. It follows that (
6.17) is equal to
 $u+ 12 y -r u+ 12 y +1 u- 12 y -1 u- 12 y +r ∏ i=1 r u+ 12 y + μi -i +1 u+ 12 y + μi -i u- 12 y - μi-i u+ 12 y - μi -i +1 = u- 12 δ u+ 12 y +1 u- 12 y -1 u+ 12 δ evμ+ρ ∏ i=1 r u+ hi+ 12 u+ hi - 12 u- hi + 12 u- hi - 12$ 6.19
where since $evμ σρ h ε i ∨ = evμ+ρ h ε i ∨ = ⟨ μ+ρ, ε i ∨ ⟩ = μi + 12 y-2i+1 = 12 y + 12 -i$ and $12 y -r = 12 y-2r = 12 y-2r+δ - 12 δ = - 12 δ$ Combining (
6.13) and (6.19), the identity (4.14) gives that $zV u + εu - 12 = zV 1 + εu - 12 ∏ i=1 k-1 u+ yi -1 u+ yi +1 u- yi 2 u+ yi 2 u- yi +1 u- yi -1$ acts on the $L\left(\mu \right)\otimes {𝒲}_{k-1}^{\mu }$-component in the ${U}_{q}𝔤\otimes {𝒲}_{k-1}$-module decomposition in (6.15) by $εu + 12 u+ 12 y u- 12 y u- 12 δ u+ 12 y +1 u- 12 y -1 u+ 12 δ evμ+ρ ∏ i=1 r u+ hi+ 12 u+ hi - 12 u- hi + 12 u- hi - 12 .$ $\square$

To help illuminate the cancellation done in (6.18), we walk through the example where $\mu =\left(5,5,3,3,1,1\right)$, and so the contents of the boxes are $TABLEAU GOES HERE.$ In this example, the product over the boxes in the first row of the diagram is Thus, simplifying the product one row at a time, $∏ b∈μ x+cb -1 x+cb +1 x+cb x+cb = x-1 x+0 x+5 x+4 ⋅ x-2 x-1 x+4 x+3 ⋅ x-3 x-2 x+1 x+0 ⋅ x-4 x-3 x+0 x-1 ⋅ x-5 x-4 x-3 x-4 ⋅ x-6 x-5 x-4 x-5 = x-6 x+0 x+5 x+4 ⋅ x+4 x+3 ⋅ x+1 x+0 ⋅ x+0 x-1 ⋅ x-3 x-4 ⋅ x-4 x-5$ leading us to the identity

Let $𝔤={𝔰𝔬}_{2r+1}$ or $𝔤={𝔰𝔭}_{2r}$ or $𝔤={𝔰𝔬}_{2r}$, and let $V=L\left({\epsilon }_{1}\right)$ and $z=\epsilon {q}^{y}$, where Let ${Z}_{V}^{\left(l\right)}$ be the central elements in ${U}_{q}𝔤$ defined by $ZV(l) = ε id⊗qtrV z ℛ12 ℛ l$ and let $ZV(l) = ∑ l∈ℤ≥0 ZV(l) u-l and ZV- = ∑ l∈ℤ≥0 ZV(-l) u-l.$ Then $π0 ZV+ u - z q-q-1 - u 2 u 2 -1 = z q-q-1 u+q u-q-1 u-εqδ u+1 u-1 u-εq-δ σρ ∏ i=1 r u-ε L i -2 q-1 u-ε L i 2 q-1 u-ε L i -2 q u-ε L i 2 q .$

Proof.
As operators on $L\left(0\right)\otimes V$, $z ℛ12 ℛ = z q ⟨ ε1 , ε1 + 2ρ ⟨ - ⟩ ε1 , ε1 + 2ρ + ⟨ 0, 0+ 2ρ ⟩ =z and z ℛ12 ℛ l = zl.$ So, as operators on $L\left(0\right)$, $ZV(l) =ε id⊗ qtrV z ℛ12 ℛ l = zl ε dimq V$ and, since $\epsilon {dim}_{q}\left(V\right)=\frac{z-{z}^{-1}}{q-{q}^{-1}}+1$, $ZV+ = ∑ l∈ℤ≥0 ε dimq V zl u-l = ε dimq V 1 1-z u-1 = z-z-1+ q-q-1 q-q-1 1-z u-1 .$ A similar calculation yields $ZV+ u = z-z-1+ q-q-1 q-q-1 1-z u-1 and z-z-1 +q+q-1 q-q-1 1- z-1 u-1$ (see [BB], the last displayed equation in the proof of Lemma 7.4). Amazingly, as operators on $L\left(0\right)$,
 $ZV+ - ZV(0) + z q-q-1 - 1 u 2 -1 = z q-q-1 1- z-1 u-1 1-z u-1 u+q u-q-1 u+1 u-1$ 6.21
and
 $ZV- - ZV(0) + -z-1 q-q-1 - 1 u 2 -1 = -z-1 q-q-1 1-z u-1 1- z-1 u-1 u-q u+q-1 u+1 u-1 .$ 6.22

By (5.39), $Yk acts on L0 ⊗ V⊗k as z ℛ12ℛ.$ If $L\left(\mu \right)$ is an irreducible ${U}_{q}𝔤$-module in $L\left(0\right)\otimes {V}^{\otimes k-1}$, then (5.23) and (5.26) give that ${Y}_{k}$ acts on the $L\left(\lambda \right)$ component of $L\left(\mu \right)\otimes V$ by the constant where $c\left(\lambda /\mu \right)$ is computed in (6.14) (see [OR]). By induction, as in [OR, Theorem 6.3(b)], the irreducible ${W}_{k}$-module ${W}_{k}^{\mu /0}={W}_{k}^{\mu }$ has a basis $\left\{{v}_{T}\right\}$ indexed by up-down tableaux $T=\left({T}^{\left(0\right)},{T}^{\left(1\right)},\dots ,{T}^{\left(k\right)}\right)$, where ${T}^{\left(0\right)}=\varnothing$, ${T}^{\left(k\right)}=\mu$, and ${T}^{\left(i\right)}$ is a partition obtained from ${T}^{\left(i-1\right)}$ by adding or removing a box, and Thus $∏ i=1 k-1 u-Yi 2 u- q-2 Yi-1 u- q 2 Yi-1 u - Yi-1 2 u- q 2 Yi u- q-2 Yi$ acts on the $L\left(\mu \right)\otimes {W}_{k-1}^{\mu }$ isotypic component in the ${U}_{q}𝔤$-module decomposition

 $L0 ⊗ V⊗k-1 ≅ ⊕ μ Lμ ⊗ W k-1 μ ,$ 6.23
by
 $∏ i=1 k-1 u-ε q 2 c T(i) , T(i-1) 2 u- q-2 q -2 c T(i), T(i-1) u-ε q2 q -2c T(i), T(i-1) u-ε q -2c T(i), T(i-1) 2 u-ε q2 q -2c T(i), T(i-1) u-ε q-2 q 2c T(i), T(i-1)$ 6.24
for any up-down tableaux $T$ of length $k$ and shape $\mu$. If a box is added (or removed) at step $i$ and then removed (or added) at step $j$, then the $i$ and $j$ factora of this product cancel. Therefore, (
6.24) is equal to
 $∏ b∈μ u-z q 2cb 2 u-z-1 q -2cb+1 u-z-1 q -2cb-1 u-z-1 q -2cb 2 u-z q 2cb+1 u-z q 2cb-1$ 6.25
(see [
BB], the next to last displayed equation in the proof of Lemma 7.4). Simplifying one row at a time, $∏ b∈μ u-z-1 q -2cb-1 u-z-1 q -2cb+1 u-z-1 q -2cb u-z-1 q -2cb = ∏ i=1 r u-z-1 q-2-i u-z-1 q -2μi-i+1 u-z-1 q -2-i-1 u-z-1 q-2μi-i = u-z-1 q2r u-z-1 q2⋅0 ∏ i=1 r u-z-1 q -2μi-i+1 u-z-1 q -2μi-i$ if $\mu =\left({\mu }_{1},\dots ,{\mu }_{r}\right)$. It follows that (6.25) is equal to
 $u-z-1 q2r u-z-1 u-z u-z q-2r ∏ i=1 r u-z-1 q -2μi-i+1 u-z-1 q -2μi-i ⋅ u-z q2μi-i u-z q2μi-i+1 = u-ε qδ u-z-1 u-z u-ε qδ evμ+ρ ∏ i=1 r u-ε L i -2 q-1 u-ε L i 2 q-1 u-ε L i -2 q u-ε L i 2 q$ 6.26
where since $evμ+ρ L i 2 = evμ+ρ K 2 ε i ∨ = q ⟨ μ+ρ , 2 ε i ∨ ⟩ = q 2 μi + y-2i+1 = q y+1+2 μi-i = εz q 2 μi -i +1$ and $z-1 q2r = ε q-2r qδ q2r = ε qδ.$ Combining (
6.21) and (6.26), the identity (4.15) gives that $Zk+ - Zk(0) + z q-q-1 - 1 u2-1 = Z1+ - Z1(0) + z q-q-1 - 1 u2-1 ∏ i=1 k-1 u-Yi 2 u-q-2 Yi-1 u-q2 Yi-1 u- Yi-1 2 u-q2 Yi u-q-2 Yi ,$ acts on the $L\left(\mu \right)\otimes {W}_{k-1}^{\mu }$ isotypic component in the ${U}_{q}𝔤\otimes {W}_{k-1}$-module decomposition in (6.23) by $z q-q-1 1-z-1u-1 1-zu-1 u+q u-q-1 u+1 u-1 u-ε qδ u-z-1 u-z u-ε qδ evμ+ρ ∏ i=1 r u-ε L i -2 q-1 u-ε L i 2 q-1 u-ε L i -2 q u-ε L i 2 q = z q-q-1 u+q u-q-1 u+1 u-1 u-εqδ u-εq-δ evμ+ρ ∏ i=1 r u-ε L i -2 q-1 u-ε L i 2 q-1 u-ε L i -2 q u-ε L i 2 q .$

$\square$

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