Braid Group Action on Affine Yangian

Braid Group Action on Affine Yangian

Ryosuke KODERA

R. Kodera

Department of Mathematics, Graduate School of Science, Kobe University,
Kobe 657-8501, Japan<EMAIL_ADDRESS>http://www2.kobe-u.ac.jp/~kryosuke/

Received August 09, 2018, in final form February 27, 2019; Published online
March 16, 2019

We study braid group actions on Yangians associated with symmetrizable
Kac–Moody Lie algebras. As an application, we focus on the affine Yangian of
type A and use the action to prove that the image of the evaluation map
contains the diagonal Heisenberg algebra inside $\hat{\mathfrak{gl}}_{N}$.

affine Yangian; braid group action; evaluation map

17B10; 17B37; 17B67

##  1 Introduction

The purpose of this paper is to introduce braid group actions on Yangians and
study their properties. Moreover we give an application for the affine Yangian
of type A.

Braid group actions on Kac–Moody Lie algebras and their integrable
representations are classical. Lusztig initiated the study of braid group
actions on quantized enveloping algebras associated with Kac–Moody Lie
algebras. See [15, notes on Part VI] for historical remarks. They are
fundamental tools for the construction of the PBW bases of the quantized
enveloping algebras of finite type in [13, 14] (see also [8, 11, 16]). Then
Lusztig [12] used the PBW basis for the construction of the canonical basis in
ADE type. The braid group action is important also in affine type. Beck [1]
used it to construct the Drinfeld generators and proved the equivalence of two
presentations of the quantized affine algebra.

Another important family of quantum groups are Yangians which were introduced
by Drinfeld. The Yangian $Y(\mathfrak{g})$ associated with a simple Lie
algebra $\mathfrak{g}$ contains the universal enveloping algebra
$U(\mathfrak{g})$ as a subalgebra and it is easy to see that the braid group
action on $U(\mathfrak{g})$ extends to $Y(\mathfrak{g})$. Although this is a
trivial observation, it seems that properties of the braid group action on the
Yangian have not been seriously studied. (A different kind of braid group
actions on quantized affine algebras and Yangians are studied by
Ding–Khoroshkin [3]. It differs from the action introduced in this paper and
our results are independent of their work.)

Let us give a remark on a difference between the situations for the quantized
enveloping algebra and the Yangian. The Yangian $Y(\mathfrak{g})$ is known to
be an additive degeneration of the quantized enveloping algebra
$U_{q}(\hat{\mathfrak{g}})$ of affine type. The braid group acting on
$U_{q}(\hat{\mathfrak{g}})$ is of affine type, while one acting on
$Y(\mathfrak{g})$ is of finite type. Thus the symmetry of Yangian coming from
this consideration is smaller than that of quantized enveloping algebra given
by Lusztig.

We can define the Yangian associated with a symmetrizable Kac–Moody Lie
algebra $\mathfrak{g}$. Then we see that the braid group of $\mathfrak{g}$
acts on it. In this paper we mainly focus on the affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ of type A and the general case is
studied in Appendix A. The affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ is a two-parameter deformation of the
universal enveloping algebra of the universal central extension of
$\mathfrak{sl}_{N}\big{[}s,t^{\pm 1}\big{]}$, and is related to many
interesting objects; symmetry of the spin Calogero–Sutherland model [9, 17];
Schur–Weyl type duality for degenerate double affine Hecke algebra [4]; quiver
variety associated with the cyclic quiver [9, 18].

The affine Yangian $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ contains
$U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ as a subalgebra and hence it admits
an action of the braid group associated with $\hat{\mathfrak{sl}}_{N}$. We
give a formula for the action on generators of degree one. We use it to show
the compatibility of the braid group action and the coproduct on
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ introduced by Guay [5],
Guay–Nakajima–Wendlandt [6]. This compatibility result holds for the Yangian
of arbitrary finite or affine type except for $A_{1}^{(1)}$ and $A_{2}^{(2)}$
as discussed in Appendix A. See Propositions 3.12 and A.4 for precise
statements.

We hope to apply the braid group action to the study of structure theory and
representation theory of the affine Yangian. As a first step, we consider the
following in the second half of the paper.

Guay [5] introduced an evaluation map for the affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ whose target space is a certain
completion of $U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$. By its definition, the
image contains $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$, and we have expected
that it contains $U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$. We give an
affirmative answer to this question, under a certain assumption on the
parameters, by constructing elements of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ whose images by the evaluation map
coincide with generators of the diagonal Heisenberg algebra inside
$\hat{\mathfrak{gl}}_{N}$ (Theorem 4.18). To construct such elements, we use
the braid group action. Certainly it is desirable to lift the Heisenberg
subalgebra inside the affine Yangian and we will come back to this problem in
a future.

Our main result Theorem 4.18 implies that the pull-back of an irreducible
$\hat{\mathfrak{gl}}_{N}$-module by the evaluation map is irreducible as a
module of $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$. We determine the highest
weights of the evaluation modules in [10].

The plan of this paper is as follows. In Section 2, we define the affine
Yangian $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ and recall some automorphisms
and the coproduct. In Section 3, we introduce the braid group action and study
its properties. In particular, the compatibility with the coproduct is proved
in Section 3.4. We construct elements of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ whose images by the evaluation map
coincide with Heisenberg generators in Section 4. In Appendix A, we consider
the braid group action on the Yangian associated with a symmetrizable
Kac–Moody Lie algebra. Then we give a proof of the compatibility with the
coproduct when it is known to be well-defined.

##  2 Affine Yangian

###  2.1 Affine Yangian

Fix an integer $N\geq 3$ throughout the paper. We use the notation
$\\{x,y\\}=xy+yx$.

######  Definition 2.1.

The affine Yangian $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ is the algebra
over $\mathbb{C}$ generated by $x_{i,r}^{+}$, $x_{i,r}^{-}$, $h_{i,r}$
($i\in\mathbb{Z}/N\mathbb{Z}$, $r\in\mathbb{Z}_{\geq 0}$) with parameters
$\varepsilon_{1},\varepsilon_{2}\in\mathbb{C}$ subject to the relations:

$\displaystyle[h_{i,r},h_{j,s}]=0,\qquad\big{[}x_{i,r}^{+},x_{j,s}^{-}\big{]}=\delta_{ij}h_{i,r+s},\qquad\big{[}h_{i,0},x_{j,r}^{\pm}\big{]}=\pm
a_{ij}x_{j,r}^{\pm},$
$\displaystyle\big{[}h_{i,r+1},x_{j,s}^{\pm}\big{]}-\big{[}h_{i,r},x_{j,s+1}^{\pm}\big{]}=\pm
a_{ij}\dfrac{\varepsilon_{1}+\varepsilon_{2}}{2}\big{\\{}h_{i,r},x_{j,s}^{\pm}\big{\\}}-m_{ij}\dfrac{\varepsilon_{1}-\varepsilon_{2}}{2}\big{[}h_{i,r},x_{j,s}^{\pm}\big{]},$
$\displaystyle\big{[}x_{i,r+1}^{\pm},x_{j,s}^{\pm}\big{]}-\big{[}x_{i,r}^{\pm},x_{j,s+1}^{\pm}\big{]}=\pm
a_{ij}\dfrac{\varepsilon_{1}+\varepsilon_{2}}{2}\big{\\{}x_{i,r}^{\pm},x_{j,s}^{\pm}\big{\\}}-m_{ij}\dfrac{\varepsilon_{1}-\varepsilon_{2}}{2}\big{[}x_{i,r}^{\pm},x_{j,s}^{\pm}\big{]},$
$\displaystyle\sum_{w\in\mathfrak{S}_{1-a_{ij}}}\big{[}x_{i,r_{w(1)}}^{\pm},\big{[}x_{i,r_{w(2)}}^{\pm},\dots,\big{[}x_{i,r_{w(1-a_{ij})}}^{\pm},x_{j,s}^{\pm}\big{]}\dots\big{]}\big{]}=0,\qquad
i\neq j,$

where

$\displaystyle a_{ij}=\begin{cases}\hphantom{-}2&\text{if }i=j,\\\ -1&\text{if
}i=j\pm 1,\\\ \hphantom{-}0&\text{otherwise,}\end{cases}\qquad
m_{ij}=\begin{cases}\hphantom{-}1&\text{if }j=i-1,\\\ -1&\text{if }j=i+1,\\\
\hphantom{-}0&\text{otherwise.}\end{cases}$

######  Remark 2.2.

In this paper, we define
$\hat{\mathfrak{sl}}_{N}=\mathfrak{sl}_{N}\otimes\mathbb{C}\big{[}t,t^{-1}\big{]}\oplus\mathbb{C}c$
without the degree operator. In [6], the Yangian $Y_{\hbar}(\mathfrak{g})$ of
affine type is defined to be an algebra containing the degree operator $d$ and
one without $d$ is denoted by $Y_{\hbar}(\mathfrak{g}^{\prime})$. In
particular, the algebra defined in Definition 2.1 coincides with
$Y_{\hbar,\varepsilon}(\mathfrak{g}^{\prime})$ in the notation of [6,
Definition 7.1].

The subalgebra generated by $x_{i,0}^{+}$, $x_{i,0}^{-}$, $h_{i,0}$
($i\in\mathbb{Z}/N\mathbb{Z}$) is isomorphic to
$U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ (See [5, Theorem 6.1] for $N\geq 4$
and [7, Theorem 6.9] in general). We write $x_{i}^{\pm}=x_{i,0}^{\pm}$,
$h_{i}=h_{i,0}$ and identify them with the standard Chevalley generators of
$\hat{\mathfrak{sl}}_{N}$. Let $\\{\alpha_{i}\\}_{i\in\mathbb{Z}/N\mathbb{Z}}$
be the simple roots of $\hat{\mathfrak{sl}}_{N}$. The null root $\delta$ is
given by $\delta=\sum\limits_{i=0}^{N-1}\alpha_{i}$. Let
$\theta=\sum\limits_{i=1}^{N-1}\alpha_{i}$ be the highest root of
$\mathfrak{sl}_{N}$ and $h_{-\theta}$ the coroot corresponding to $-\theta$.
We denote by $\hat{\Delta}$, $\hat{\Delta}_{+}$, and
$\hat{\Delta}_{+}^{\mathrm{re}}$ the set of roots, positive roots, and
positive real roots for $\hat{\mathfrak{sl}}_{N}$, respectively. We need to
consider $\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d$ with the degree operator
to deal with the coproduct on $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$. Fix a
nondegenerate invariant symmetric bilinear form $(\,,\,)$ on
$\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d$ such that $(x_{i}^{+},x_{i}^{-})=1$
and denote the induced bilinear form on the dual of the Cartan subalgebra by
the same letter.

Let $\hat{W}$ be the Weyl group of $\hat{\mathfrak{sl}}_{N}$ generated by the
simple reflections $s_{i}$ ($i\in\mathbb{Z}/N\mathbb{Z}$). We denote by
$s_{\theta}$ the reflection corresponding to the highest root $\theta$. For
each $\alpha\in\hat{\Delta}$, we assign the translation element $t_{\alpha}$
in $\hat{W}$. For example, we have $s_{0}s_{\theta}=t_{\theta}$. The action of
$t_{\alpha}$ on the root lattice
$Q=\sum\limits_{i=1}^{N-1}\mathbb{Z}\alpha_{i}$ of $\mathfrak{sl}_{N}$ is
given by $t_{\alpha}(\lambda)=\lambda-(\alpha,\lambda)\delta$ for $\lambda\in
Q$.

Set $\hbar=\varepsilon_{1}+\varepsilon_{2}$ and
$\tilde{h}_{i,1}=h_{i,1}-\frac{\hbar}{2}h_{i}^{2}$. We can deduce the
following identities directly from the defining relations of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$.

######  Lemma 2.3.

We have

$\displaystyle\big{[}h_{i,1},x_{i}^{\pm}\big{]}=\pm
2x_{i,1}^{\pm}\pm\hbar\big{\\{}h_{i},x_{i}^{\pm}\big{\\}},\qquad\big{[}\tilde{h}_{i,1},x_{i}^{\pm}\big{]}=\pm
2x_{i,1}^{\pm},\qquad\big{[}x_{i,1}^{\pm},x_{i}^{\pm}\big{]}=\pm\hbar\big{(}x_{i}^{\pm}\big{)}^{2},$
$\displaystyle\big{[}h_{i,1},x_{j}^{\pm}\big{]}=\mp\left(x_{j,1}^{\pm}+\dfrac{\hbar}{2}\big{\\{}h_{i},x_{j}^{\pm}\big{\\}}-m_{ij}\dfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{j}^{\pm}\right)\qquad\text{if}\quad
i=j\pm 1.$

###  2.2 Automorphisms

We introduce algebra (anti-)automorphisms of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ which will be used later.

Let $\omega$ be the algebra anti-automorphism of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ defined by $x_{i,r}^{\pm}\mapsto
x_{i,r}^{\mp}$, $h_{i,r}\mapsto h_{i,r}$. Here anti-automorphism means that
the invertible operator $\omega$ satisfies $\omega(XY)=\omega(Y)\omega(X)$. It
is easy to check that it is well-defined. The following algebra automorphism
corresponds to a rotation of the Dynkin diagram.

######  Proposition 2.4 ([4, Lemma 3.5]).

The assignment

$\displaystyle
x_{i,r}^{\pm}\mapsto\sum_{s=0}^{r}\dbinom{r}{s}\varepsilon_{2}^{r-s}x_{i-1,s}^{\pm},\qquad
h_{i,r}\mapsto\sum_{s=0}^{r}\dbinom{r}{s}\varepsilon_{2}^{r-s}h_{i-1,s}$

gives an algebra automorphism $\rho$ of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$.

In particular we have
$\rho(x_{i,1}^{\pm})=x_{i-1,1}^{\pm}+\varepsilon_{2}x_{i-1}^{\pm}$.

######  Remark 2.5.

The relation between generators $X_{i,r}^{\pm}$, $H_{i,r}$ with parameters
$\lambda$, $\beta$ used in [4, 5] and ours are as follows:

$\displaystyle
X_{0,r}^{\pm}=\sum_{s=0}^{r}\dbinom{r}{s}\left(\dfrac{N}{4}(\varepsilon_{1}-\varepsilon_{2})\right)^{r-s}x_{0,s}^{\pm},\qquad
H_{0,r}=\sum_{s=0}^{r}\dbinom{r}{s}\left(\dfrac{N}{4}(\varepsilon_{1}-\varepsilon_{2})\right)^{r-s}h_{0,s},$
$\displaystyle
X_{i,r}^{\pm}=\sum_{s=0}^{r}\dbinom{r}{s}\left(\dfrac{i}{2}(\varepsilon_{1}-\varepsilon_{2})\right)^{r-s}x_{i,s}^{\pm},\qquad
H_{i,r}=\sum_{s=0}^{r}\dbinom{r}{s}\left(\dfrac{i}{2}(\varepsilon_{1}-\varepsilon_{2})\right)^{r-s}h_{i,s},\qquad
i\neq 0,$
$\displaystyle\lambda=\hbar,\qquad\beta=\frac{1}{2}\hbar-\dfrac{N}{4}(\varepsilon_{1}-\varepsilon_{2}).$

###  2.3 Coproduct

A formula for the coproduct on the affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ was stated in [5].
Guay–Nakajima–Wendlandt [6] gave a detailed proof of the well-definedness. To
recall it, we consider a bigger algebra
$Y\big{(}\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d\big{)}$, which is generated
by $x_{i,r}^{\pm}$, $h_{i,r}$, $d$ with defining relations given in [6,
equation (2.8)]. Moreover we need certain completions
$Y\big{(}\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d\big{)}\widehat{\otimes}Y\big{(}\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d\big{)}$
and
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}\widehat{\otimes}Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$
of the tensor products since the coproduct involves infinite sums. See [6,
Section 5] for the precise definition of the completion.

We define the half Casimir operator $\Omega_{+}$ as follows. Let $\\{u_{k}\\}$
be a $\mathbb{C}$-basis of the Cartan subalgebra of
$\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d$ and $\big{\\{}u^{k}\big{\\}}$ its
dual basis with respect to the nondegenerate bilinear form $(\,,\,)$. Let
$\big{\\{}x_{\alpha}^{(k)}\big{\\}}_{\alpha\in\hat{\Delta},1\leq
k\leq\operatorname{mult}\alpha}$ be a root vector basis satisfying
$\big{(}x_{\alpha}^{(k)},x_{-\alpha}^{(l)}\big{)}=\delta_{k,l}$. Here
$\operatorname{mult}\alpha$ denotes the dimension of the root space
corresponding to $\alpha$. We take simple root vectors as
$x_{\pm\alpha_{i}}^{(1)}=x_{i}^{\pm}$. Put

$\displaystyle\Omega_{+}=\sum_{k}u^{k}\otimes
u_{k}+\sum_{\begin{subarray}{c}\alpha\in\hat{\Delta}_{+}\\\ 1\leq
k\leq\operatorname{mult}\alpha\end{subarray}}x_{-\alpha}^{(k)}\otimes
x_{\alpha}^{(k)}$

as an element of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d\big{)}\widehat{\otimes}Y\big{(}\hat{\mathfrak{sl}}_{N}\oplus\mathbb{C}d\big{)}$.

Define a $\mathbb{C}$-linear operator $\square$ on
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ by $\square(X)=X\otimes 1+1\otimes
X$. Note that $\square$ is not an algebra homomorphism, but satisfies
$\square([X,Y])=[\square(X),\square(Y)]$.

######  Theorem 2.6 ([6, Definition 4.6, Theorem 4.9, Proposition 5.18,
Section 7]).

There exists an algebra homomorphism $\Delta\colon
Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}\to
Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}\widehat{\otimes}Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$
uniquely determined by

$\displaystyle\Delta(X)=\square(X)\qquad\text{for}\quad X=x_{i}^{\pm},h_{i},$
$\displaystyle\Delta\big{(}x_{i,1}^{+}\big{)}=\square(x_{i,1}^{+})-\hbar\big{[}1\otimes
x_{i}^{+},\Omega_{+}\big{]},$
$\displaystyle\Delta\big{(}x_{i,1}^{-}\big{)}=\square(x_{i,1}^{-})+\hbar\big{[}x_{i}^{-}\otimes
1,\Omega_{+}\big{]},$
$\displaystyle\Delta\big{(}\tilde{h}_{i,1}\big{)}=\square(\tilde{h}_{i,1})+\hbar[h_{i}\otimes
1,\Omega_{+}].$

######  Remark 2.7.

Explicitly we have

$\displaystyle[1\otimes x_{i}^{+},\Omega_{+}]=-h_{i}\otimes
x_{i}^{+}+\sum_{\begin{subarray}{c}\alpha\in\hat{\Delta}_{+}\\\ 1\leq
k\leq\operatorname{mult}\alpha\end{subarray}}x_{-\alpha}^{(k)}\otimes\big{[}x_{i}^{+},x_{\alpha}^{(k)}\big{]},$
$\displaystyle[x_{i}^{-}\otimes 1,\Omega_{+}]=x_{i}^{-}\otimes
h_{i}+\sum_{\begin{subarray}{c}\alpha\in\hat{\Delta}_{+}\\\ 1\leq
k\leq\operatorname{mult}\alpha\end{subarray}}\big{[}x_{i}^{-},x_{-\alpha}^{(k)}\big{]}\otimes
x_{\alpha}^{(k)},$ $\displaystyle[h_{i}\otimes
1,\Omega_{+}]=-\sum_{\alpha\in\hat{\Delta}_{+}^{\mathrm{re}}}(\alpha_{i},\alpha)\,x_{-\alpha}^{(1)}\otimes
x_{\alpha}^{(1)}.$

Hence the target of $\Delta$ is
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}\widehat{\otimes}Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$
without $d$.

##  3 Braid group action

We define automorphisms $T_{i}$ $(i\in\mathbb{Z}/N\mathbb{Z})$ of the affine
Yangian and study their properties.

###  3.1 Definition

Since the adjoint actions of $x_{i}^{\pm}$ on
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ are locally nilpotent, the operators
$\exp\operatorname{ad}x_{i}^{\pm}$ are well defined by

$\displaystyle\exp\operatorname{ad}x_{i}^{\pm}=\sum_{n=0}^{\infty}\dfrac{1}{n!}\big{(}\operatorname{ad}x_{i}^{\pm}\big{)}^{n}.$

These are algebra automorphisms of $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ as
$\operatorname{ad}x_{i}^{\pm}$ are derivations. We define an algebra
automorphism $T_{i}$ of $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ for each
$i\in\mathbb{Z}/N\mathbb{Z}$ by

$\displaystyle
T_{i}=\exp\operatorname{ad}x_{i}^{+}\exp\operatorname{ad}(-x_{i}^{-})\exp\operatorname{ad}x_{i}^{+}.$

This operator appears in [6] and is used to construct real root vectors of
Yangians.

###  3.2 Braid relations

A proof of the following proposition is exactly the same as one for the fact
that $\\{T_{i}\\}$ satisfy the braid relations as automorphisms of
$U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$. We give a proof for the sake of
completeness.

######  Proposition 3.1.

The operators $\\{T_{i}\\}$ satisfy the braid relations. That is, we have

$\displaystyle T_{i}T_{j}=T_{j}T_{i}\quad\text{if $a_{ij}=0$},\qquad
T_{i}T_{j}T_{i}=T_{j}T_{i}T_{j}\quad\text{if $a_{ij}=-1$}.$

For each $w\in\hat{W}$ with a reduced expression $w=s_{i_{1}}\cdots
s_{i_{l}}$, we can define an algebra automorphism $T_{w}$ of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ by $T_{w}=T_{i_{1}}\cdots T_{i_{l}}$
thanks to the braid relations.

Let us start the proof with some preparations. A proof of the following
formulas is straightforward.

######  Lemma 3.2.

Assume $a_{ij}=-1$. Then

  1. (i)

$\exp\operatorname{ad}x_{i}^{+}$ sends:

$\displaystyle x_{i}^{+}\mapsto x_{i}^{+},\qquad x_{j}^{+}\mapsto
x_{j}^{+}+\big{[}x_{i}^{+},x_{j}^{+}\big{]},\qquad x_{i}^{-}\mapsto
x_{i}^{-}+h_{i}-x_{i}^{+},\qquad x_{j}^{-}\mapsto x_{j}^{-},$ $\displaystyle
h_{i}\mapsto h_{i}-2x_{i}^{+},\qquad h_{j}\mapsto
h_{j}+x_{i}^{+},\qquad\big{[}x_{i}^{+},x_{j}^{+}\big{]}\mapsto\big{[}x_{i}^{+},x_{j}^{+}\big{]},$
$\displaystyle\big{[}x_{i}^{-},x_{j}^{-}\big{]}\mapsto\big{[}x_{i}^{-},x_{j}^{-}\big{]}+x_{j}^{-};$

  2. (ii)

$\exp\operatorname{ad}(-x_{i}^{-})$ sends:

$\displaystyle x_{i}^{+}\mapsto x_{i}^{+}+h_{i}-x_{i}^{-},\qquad
x_{j}^{+}\mapsto x_{j}^{+},\qquad x_{i}^{-}\mapsto x_{i}^{-},\qquad
x_{j}^{-}\mapsto x_{j}^{-}-[x_{i}^{-},x_{j}^{-}],$ $\displaystyle h_{i}\mapsto
h_{i}-2x_{i}^{-},\qquad h_{j}\mapsto
h_{j}+x_{i}^{-},\qquad\big{[}x_{i}^{+},x_{j}^{+}\big{]}\mapsto\big{[}x_{i}^{+},x_{j}^{+}\big{]}-x_{j}^{+},$
$\displaystyle\big{[}x_{i}^{-},x_{j}^{-}\big{]}\mapsto\big{[}x_{i}^{-},x_{j}^{-}\big{]}.$

The following two propositions are well known. Proposition 3.3 follows from
Lemma 3.2. Then Proposition 3.4 follows from Proposition 3.3.

######  Proposition 3.3.

We have

$\displaystyle T_{i}(x_{j}^{+})=\begin{cases}-x_{i}^{-}&\text{if $i=j$},\\\
\big{[}x_{i}^{+},x_{j}^{+}\big{]}&\text{if $a_{ij}=-1$},\\\ x_{j}^{+}&\text{if
$a_{ij}=0$},\end{cases}\qquad
T_{i}\big{(}x_{j}^{-}\big{)}=\begin{cases}-x_{i}^{+}&\text{if $i=j$},\\\
-\big{[}x_{i}^{-},x_{j}^{-}\big{]}&\text{if $a_{ij}=-1$},\\\
x_{j}^{-}&\text{if $a_{ij}=0$},\end{cases}$ $\displaystyle
T_{i}(h_{j})=\begin{cases}-h_{i}&\text{if $i=j$},\\\ h_{j}+h_{i}&\text{if
$a_{ij}=-1$},\\\ h_{j}&\text{if $a_{ij}=0$}.\end{cases}$

######  Proposition 3.4.

Assume $a_{ij}=-1$. Then we have

$\displaystyle T_{i}T_{j}\big{(}x_{i}^{+}\big{)}=x_{j}^{+},\qquad
T_{i}T_{j}\big{(}x_{i}^{-}\big{)}=x_{j}^{-},\qquad T_{i}T_{j}(h_{i})=h_{j}.$

###### Proof of Proposition 3.1.

We use that
$\varphi\circ\exp\operatorname{ad}X\circ\varphi^{-1}=\exp\operatorname{ad}\varphi(X)$
holds for any algebra automorphism $\varphi$. If $a_{ij}=0$, then we have

$\displaystyle
T_{i}T_{j}T_{i}^{-1}=\exp\operatorname{ad}T_{i}\big{(}x_{j}^{+}\big{)}\exp\operatorname{ad}T_{i}\big{(}{-}x_{j}^{-}\big{)}\exp\operatorname{ad}T_{i}\big{(}x_{j}^{+}\big{)}=T_{j}$

by the formulas in Proposition 3.3. If $a_{ij}=-1$, then we have

$\displaystyle
T_{i}T_{j}T_{i}T_{j}^{-1}T_{i}^{-1}=\exp\operatorname{ad}T_{i}T_{j}\big{(}x_{i}^{+}\big{)}\exp\operatorname{ad}T_{i}T_{j}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}T_{i}T_{j}\big{(}x_{i}^{+}\big{)}=T_{j}$

by the formulas in Proposition 3.4. The proof is complete. ∎

We give an alternative definition of $T_{i}$.

######  Proposition 3.5.

The operator $T_{i}$ coincides with
$\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}x_{i}^{+}\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}$.

###### Proof.

Put
$T_{i}^{\prime}=\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}x_{i}^{+}\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}$
for a while. Then we have

$\displaystyle
T_{i}T_{i}^{\prime}T_{i}^{-1}=\exp\operatorname{ad}T_{i}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}T_{i}\big{(}x_{i}^{+}\big{)}\exp\operatorname{ad}T_{i}\big{(}{-}x_{i}^{-}\big{)}=T_{i}$

by the formulas in Proposition 3.3. Hence the assertion is proved. ∎

Note that $T_{i}^{-1}$ is given by

$\displaystyle\exp\operatorname{ad}x_{i}^{-}\exp\operatorname{ad}\big{(}{-}x_{i}^{+}\big{)}\exp\operatorname{ad}x_{i}^{-}=\exp\operatorname{ad}\big{(}{-}x_{i}^{+}\big{)}\exp\operatorname{ad}x_{i}^{-}\exp\operatorname{ad}\big{(}{-}x_{i}^{+}\big{)}.$

######  Lemma 3.6.

We have $(i)$ $\omega\circ T_{i}=T_{i}\circ\omega$ and $(ii)$ $\rho\circ
T_{i}=T_{i-1}\circ\rho$.

###### Proof.

For the assertion (i), we use
$\omega(\exp\operatorname{ad}X(Y))=\exp\operatorname{ad}(-\omega(X))(\omega(Y))$.
Then we have $\omega(T_{i}(X))=T_{i}^{\prime}(\omega(X))$. Since we have
proved $T_{i}^{\prime}=T_{i}$, the assertion holds.

The assertion (ii) is obvious. ∎

Let $M$ be a $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$-module and assume that
$x_{i}^{\pm}$ acts on $M$ locally nilpotently. Then an automorphism
$T_{i}^{M}$ of $M$ is defined similarly. The following property is immediate.

######  Proposition 3.7.

Let $M$ be a $Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$-module and assume that
$x_{i}^{\pm}$ acts on $M$ locally nilpotently. Then for any $X\in
Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ and $m\in M$,

$\displaystyle T_{i}^{M}(Xm)=T_{i}(X)T_{i}^{M}(m)$

holds.

In fact, this is a general property for any associative algebra such that it
contains $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ as a subalgebra and the
operators $\exp\operatorname{ad}x_{i}^{\pm}$ are well-defined.

###  3.3 Action on generators

We compute the action of $T_{i}$ on $x_{j,1}^{\pm}$, $\tilde{h}_{j,1}$. We
will use the following formulas.

######  Lemma 3.8.

Assume $a_{ij}=-1$. Then

  1. (i)

$\exp\operatorname{ad}x_{i}^{+}$ sends:

$\displaystyle x_{i,1}^{+}\mapsto
x_{i,1}^{+}-\hbar\big{(}x_{i}^{+}\big{)}^{2},$ $\displaystyle
x_{i,1}^{-}\mapsto
x_{i,1}^{-}+h_{i,1}-x_{i,1}^{+}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+\hbar\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle x_{j,1}^{-}\mapsto x_{j,1}^{-},$
$\displaystyle\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}\mapsto\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}+x_{j,1}^{-},$
$\displaystyle\big{\\{}h_{i},x_{i}^{+}\big{\\}}\mapsto\big{\\{}h_{i},x_{i}^{+}\big{\\}}-4\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\big{(}x_{i}^{+}\big{)}^{2}\mapsto(x_{i}^{+})^{2};$

  2. (ii)

$\exp\operatorname{ad}(-x_{i}^{-})$ sends:

$\displaystyle x_{i,1}^{+}\mapsto
x_{i,1}^{+}+h_{i,1}-x_{i,1}^{-}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}+\hbar\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle x_{i,1}^{-}\mapsto
x_{i,1}^{-}-\hbar\big{(}x_{i}^{-}\big{)}^{2},$ $\displaystyle
x_{j,1}^{-}\mapsto x_{j,1}^{-}-\big{[}x_{i}^{-},x_{j,1}^{-}\big{]},$
$\displaystyle h_{i,1}\mapsto
h_{i,1}-2x_{i,1}^{-}-\hbar\big{\\{}h_{i},x_{i}^{-}\big{\\}}+3\hbar\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\big{\\{}h_{i},x_{i}^{+}\big{\\}}\mapsto\big{\\{}h_{i},x_{i}^{+}\big{\\}}-3\big{\\{}h_{i},x_{i}^{-}\big{\\}}-2\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}+2h_{i}^{2}+4\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\big{(}x_{i}^{+}\big{)}^{2}\mapsto\big{(}x_{i}^{+}\big{)}^{2}+h_{i}^{2}+\big{(}x_{i}^{-}\big{)}^{2}+\big{\\{}h_{i},x_{i}^{+}\big{\\}}-\big{\\{}h_{i},x_{i}^{-}\big{\\}}-\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}.$

###### Proof.

By a direct computation using the defining relations of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ with Lemma 2.3. For example, by

$\displaystyle\operatorname{ad}x_{i}^{+}\big{(}x_{i,1}^{-}\big{)}=h_{i,1},$
$\displaystyle\big{(}\operatorname{ad}x_{i}^{+}\big{)}^{2}\big{(}x_{i,1}^{-}\big{)}=\big{[}x_{i}^{+},h_{i,1}\big{]}=-2x_{i,1}^{+}-\hbar\big{\\{}h_{i},x_{i}^{+}\big{\\}},$
$\displaystyle\big{(}\operatorname{ad}x_{i}^{+}\big{)}^{3}\big{(}x_{i,1}^{-}\big{)}=-2\big{[}x_{i}^{+},x_{i,1}^{+}\big{]}-\hbar\big{\\{}\big{[}x_{i}^{+},h_{i}\big{]},x_{i}^{+}\big{\\}}=2\hbar\big{(}x_{i}^{+}\big{)}^{2}+4\hbar\big{(}x_{i}^{+}\big{)}^{2}=6\hbar\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\big{(}\operatorname{ad}x_{i}^{+}\big{)}^{4}\big{(}x_{i,1}^{-}\big{)}=0,$

we have

$\displaystyle\exp\operatorname{ad}x_{i}^{+}\big{(}x_{i,1}^{-}\big{)}=x_{i,1}^{-}+h_{i,1}-x_{i,1}^{+}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+\hbar\big{(}x_{i}^{+}\big{)}^{2}.$
∎

######  Proposition 3.9.

We have

$\displaystyle
T_{i}\big{(}x_{j,1}^{+}\big{)}=\begin{cases}-x_{i,1}^{-}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}&\text{if
$i=j$},\\\ \big{[}x_{i}^{+},x_{j,1}^{+}\big{]}&\text{if $a_{ij}=-1$},\\\
x_{j,1}^{+}&\text{if $a_{ij}=0$},\end{cases}$ $\displaystyle
T_{i}\big{(}x_{j,1}^{-}\big{)}=\begin{cases}-x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}&\text{if
$i=j$},\\\ -\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}&\text{if $a_{ij}=-1$},\\\
x_{j,1}^{-}&\text{if $a_{ij}=0$},\end{cases}$ $\displaystyle
T_{i}\big{(}\tilde{h}_{j,1}\big{)}=\begin{cases}-\tilde{h}_{i,1}-\hbar\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}&\text{if
$i=j$},\\\
\tilde{h}_{j,1}+\tilde{h}_{i,1}+\tfrac{\hbar}{2}\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}+m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}h_{i}&\text{if
$a_{ij}=-1$},\\\ \tilde{h}_{j,1}&\text{if $a_{ij}=0$}.\end{cases}$

###### Proof.

The formulas for $a_{ij}=0$ trivially hold. Let us consider the other cases.

We compute $T_{i}\big{(}x_{j,1}^{-}\big{)}$. First assume $i=j$. We have

$\displaystyle\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}x_{i}^{+}\big{(}x_{i,1}^{-}\big{)}=\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\big{(}x_{i,1}^{-}+h_{i,1}-x_{i,1}^{+}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+\hbar\big{(}x_{i}^{+}\big{)}^{2}\big{)}$
$\displaystyle\qquad{}=\big{(}x_{i,1}^{-}-\hbar(x_{i}^{-})^{2}\big{)}+\big{(}h_{i,1}-2x_{i,1}^{-}-\hbar\big{\\{}h_{i},x_{i}^{-}\big{\\}}+3\hbar\big{(}x_{i}^{-}\big{)}^{2}\big{)}$
$\displaystyle\qquad\quad{}-\big{(}x_{i,1}^{+}+h_{i,1}-x_{i,1}^{-}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}+\hbar\big{(}x_{i}^{-}\big{)}^{2}\big{)}$
$\displaystyle\qquad\quad{}-\tfrac{\hbar}{2}\big{(}\big{\\{}h_{i},x_{i}^{+}\big{\\}}-3\big{\\{}h_{i},x_{i}^{-}\big{\\}}-2\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}+2h_{i}^{2}+4\big{(}x_{i}^{-}\big{)}^{2}\big{)}$
$\displaystyle\qquad\quad{}+\hbar\big{(}\big{(}x_{i}^{+}\big{)}^{2}+h_{i}^{2}+\big{(}x_{i}^{-}\big{)}^{2}+\big{\\{}h_{i},x_{i}^{+}\big{\\}}-\big{\\{}h_{i},x_{i}^{-}\big{\\}}-\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}\big{)}$
$\displaystyle\qquad{}=-x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+\hbar\big{(}x_{i}^{+}\big{)}^{2}.$

Then we have

$\displaystyle\exp\operatorname{ad}x_{i}^{+}\big{(}{-}x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+\hbar\big{(}x_{i}^{+}\big{)}^{2}\big{)}$
$\displaystyle\qquad{}=-\big{(}x_{i,1}^{+}-\hbar\big{(}x_{i}^{+}\big{)}^{2}\big{)}+\tfrac{\hbar}{2}\big{(}\big{\\{}h_{i},x_{i}^{+}\big{\\}}-4\big{(}x_{i}^{+}\big{)}^{2}\big{)}+\hbar\big{(}x_{i}^{+}\big{)}^{2}=-x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}.$

Next assume $a_{ij}=-1$. Then we have

$\displaystyle
T_{i}\big{(}x_{j,1}^{-}\big{)}=\exp\operatorname{ad}x_{i}^{+}\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\big{(}x_{j,1}^{-}\big{)}=\exp\operatorname{ad}x_{i}^{+}\big{(}x_{j,1}^{-}-\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}\big{)}$
$\displaystyle\hphantom{T_{i}\big{(}x_{j,1}^{-}\big{)}}{}=x_{j,1}^{-}-\big{(}\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}+x_{j,1}^{-})=-\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}.$

The formulas for $T_{i}\big{(}x_{j,1}^{+}\big{)}$ are obtained by applying
$\omega$ to these results.

We compute $T_{i}(h_{j,1})$. First assume $i=j$. Then we have

$\displaystyle
T_{i}(h_{i,1})=T_{i}\big{(}\big{[}x_{i}^{+},x_{i,1}^{-}\big{]}\big{)}=\big{[}{-}x_{i}^{-},-x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}\big{]}=-h_{i,1}-\hbar\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}+\hbar
h_{i}^{2}.$

Next assume $a_{ij}=-1$. Then we have

$\displaystyle
T_{i}(h_{j,1})=T_{i}\big{(}\big{[}x_{j}^{+},x_{j,1}^{-}\big{]}\big{)}=\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},-\big{[}x_{i}^{-},x_{j,1}^{-}\big{]}\big{]}$
$\displaystyle\hphantom{T_{i}(h_{j,1})}{}=-\big{[}\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{i}^{-}\big{]},x_{j,1}^{-}\big{]}-\big{[}x_{i}^{-},\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{j,1}^{-}\big{]}\big{]}.$

Since we have

$\displaystyle\big{[}\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{i}^{-}\big{]},x_{j,1}^{-}\big{]}=\big{[}\big{[}h_{i},x_{j}^{+}\big{]},x_{j,1}^{-}\big{]}=\big{[}{-}x_{j}^{+},x_{j,1}^{-}\big{]}=-h_{j,1}$

and

$\displaystyle\big{[}x_{i}^{-},\big{[}[x_{i}^{+},x_{j}^{+}\big{]},x_{j,1}^{-}\big{]}\big{]}=\big{[}x_{i}^{-},\big{[}x_{i}^{+},h_{j,1}\big{]}\big{]}=\big{[}x_{i}^{-},x_{i,1}^{+}+\tfrac{\hbar}{2}\big{\\{}h_{j},x_{i}^{+}\big{\\}}-m_{ji}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{i}^{+}\big{]}$
$\displaystyle\hphantom{\big{[}x_{i}^{-},\big{[}[x_{i}^{+},x_{j}^{+}\big{]},x_{j,1}^{-}\big{]}\big{]}}{}=-h_{i,1}+\tfrac{\hbar}{2}\big{(}\big{\\{}\big{[}x_{i}^{-},h_{j}\big{]},x_{i}^{+}\big{\\}}+\big{\\{}h_{j},\big{[}x_{i}^{-},x_{i}^{+}\big{]}\big{\\}}\big{)}+m_{ji}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}h_{i}$
$\displaystyle\hphantom{\big{[}x_{i}^{-},\big{[}[x_{i}^{+},x_{j}^{+}\big{]},x_{j,1}^{-}\big{]}\big{]}}{}=-h_{i,1}-\tfrac{\hbar}{2}\big{\\{}x_{i}^{-},x_{i}^{+}\big{\\}}-\hbar
h_{i}h_{j}+\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}m_{ji}h_{i},$

we conclude

$\displaystyle
T_{i}(h_{j,1})=h_{j,1}+h_{i,1}+\tfrac{\hbar}{2}\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}+\hbar
h_{i}h_{j}+\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}m_{ij}h_{i}.$

We can easily obtain the formulas for $T_{i}(\tilde{h}_{j,1})$ from these
identities. ∎

######  Proposition 3.10.

Assume $a_{ij}=-1$. Then we have

$\displaystyle
T_{i}T_{j}\big{(}x_{i,1}^{+}\big{)}=x_{j,1}^{+}-m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{j}^{+}-\tfrac{\hbar}{2}\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},x_{i}^{-}\big{\\}},$
$\displaystyle
T_{i}T_{j}\big{(}x_{i,1}^{-}\big{)}=x_{j,1}^{-}-m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{j}^{-}-\tfrac{\hbar}{2}\big{\\{}x_{i}^{+},T_{i}\big{(}x_{j}^{-}\big{)}\big{\\}},$
$\displaystyle
T_{i}T_{j}(h_{i,1})=h_{j,1}-m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}h_{j}-\tfrac{\hbar}{2}\\{x_{i}^{+},x_{i}^{-}\\}+\tfrac{\hbar}{2}\big{\\{}T_{i}(x_{j}^{+}),T_{i}\big{(}x_{j}^{-}\big{)}\big{\\}}.$

###### Proof.

We show the first identity. Since we have

$\displaystyle
T_{i}T_{j}\big{(}x_{i,1}^{+}\big{)}=T_{i}\big{(}\big{[}x_{j}^{+},x_{i,1}^{+}\big{]}\big{)}=\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},-x_{i,1}^{-}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}\big{]},$

it follows from

$\displaystyle\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{i,1}^{-}\big{]}=\big{[}h_{i,1},x_{j}^{+}\big{]}=-\big{(}x_{j,1}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{j}^{+}\big{\\}}-m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{j}^{+}\big{)}$

and

$\displaystyle\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},\big{\\{}h_{i},x_{i}^{-}\big{\\}}\big{]}=\big{\\{}\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},h_{i}\big{]},x_{i}^{-}\big{\\}}+\big{\\{}h_{i},\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{i}^{-}\big{]}\big{\\}}$
$\displaystyle\hphantom{\big{[}\big{[}x_{i}^{+},x_{j}^{+}\big{]},\big{\\{}h_{i},x_{i}^{-}\big{\\}}\big{]}}{}=-\big{\\{}\big{[}x_{i}^{+},x_{j}^{+}\big{]},x_{i}^{-}\big{\\}}-\big{\\{}h_{i},x_{j}^{+}\big{\\}}=-\big{\\{}T_{i}(x_{j}^{+}),x_{i}^{-}\big{\\}}-\big{\\{}h_{i},x_{j}^{+}\big{\\}}.$

The second identity is obtained by applying $\omega$ to the first one. The
third identity follows from

$\displaystyle
T_{i}T_{j}(h_{i,1})=T_{i}T_{j}\big{(}\big{[}x_{i,1}^{+},x_{i}^{-}\big{]}\big{)}=\big{[}x_{j,1}^{+}-m_{ij}\tfrac{\varepsilon_{1}-\varepsilon_{2}}{2}x_{j}^{+}-\tfrac{\hbar}{2}\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},x_{i}^{-}\big{\\}},x_{j}^{-}\big{]}$

and

$\displaystyle\big{[}\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},x_{i}^{-}\big{\\}},x_{j}^{-}\big{]}=\big{\\{}\big{[}T_{i}(x_{j}^{+}),x_{j}^{-}\big{]},x_{i}^{-}\big{\\}}+\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},\big{[}x_{i}^{-},x_{j}^{-}\big{]}\big{\\}}$
$\displaystyle\hphantom{\big{[}\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},x_{i}^{-}\big{\\}},x_{j}^{-}\big{]}}{}=\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}-\big{\\{}T_{i}\big{(}x_{j}^{+}\big{)},T_{i}\big{(}x_{j}^{-}\big{)}\big{\\}}.$
∎

We give formulas for $T_{i}^{2}$. Proofs are straightforward.

######  Proposition 3.11.

We have $T_{i}^{2}(X)=X$ for $X=x_{j}^{\pm},x_{j,1}^{\pm}$ $(a_{ij}\neq-1)$,
$h_{k},h_{k,1}$ $(k\in\mathbb{Z}/N\mathbb{Z})$. We have $T_{i}^{2}(X)=-X$ for
$X=x_{j}^{\pm},x_{j,1}^{\pm}$ $(a_{ij}=-1)$. In particular, $T_{i}^{4}=1$.

###  3.4 Compatibility with the coproduct

The goal of this subsection is to prove the following proposition.

######  Proposition 3.12.

We have $\Delta\circ T_{i}=(T_{i}\otimes T_{i})\circ\Delta$.

We use the following lemma to prove Proposition 3.12.

######  Lemma 3.13.

We have $(T_{i}\otimes T_{i})\Omega_{+}=\Omega_{+}+x_{i}^{+}\otimes
x_{i}^{-}-x_{i}^{-}\otimes x_{i}^{+}$.

###### Proof.

Note that the bilinear form $(\,,\,)$ is $T_{i}$-invariant. Therefore, if
$\\{u_{k}\\}$ and $\big{\\{}u^{k}\big{\\}}$ are dual bases, then
$\\{T_{i}(u_{k})\\}$ and $\big{\\{}T_{i}\big{(}u^{k}\big{)}\big{\\}}$ are also
dual bases. Moreover, if we put

$\displaystyle
y_{\alpha}^{(k)}=\begin{cases}T_{i}\big{(}x_{s_{i}(\alpha)}^{(k)}\big{)}&\text{if
$\alpha\in\hat{\Delta}\setminus\\{\alpha_{i},-\alpha_{i}\\}$},\\\
x_{i}^{+}&\text{if $\alpha=\alpha_{i}$},\\\ x_{i}^{-}&\text{if
$\alpha=-\alpha_{i}$},\end{cases}$

then $\big{\\{}y_{\alpha}^{(k)}\big{\\}}$ also satisfy
$\big{(}y_{\alpha}^{(k)},y_{-\alpha}^{(l)}\big{)}=\delta_{k,l}$ and we have

$\displaystyle\big{\\{}T_{i}\big{(}x_{\alpha}^{(k)}\big{)}\big{\\}}_{\alpha\in\hat{\Delta}_{+},1\leq
k\leq\operatorname{mult}\alpha}=\big{\\{}y_{\alpha}^{(k)}\big{\\}}_{\alpha\in\hat{\Delta}_{+},1\leq
k\leq\operatorname{mult}\alpha}\cup\big{\\{}{-}x_{i}^{-}\big{\\}}\setminus\big{\\{}x_{i}^{+}\big{\\}},$
$\displaystyle\big{\\{}T_{i}\big{(}x_{-\alpha}^{(k)}\big{)}\big{\\}}_{\alpha\in\hat{\Delta}_{+},1\leq
k\leq\operatorname{mult}\alpha}=\big{\\{}y_{-\alpha}^{(k)}\big{\\}}_{\alpha\in\hat{\Delta}_{+},1\leq
k\leq\operatorname{mult}\alpha}\cup\big{\\{}{-}x_{i}^{+}\big{\\}}\setminus\big{\\{}x_{i}^{-}\big{\\}}.$

Hence we obtain

$\displaystyle(T_{i}\otimes
T_{i})\Omega_{+}-\Omega_{+}=\big{(}{-}x_{i}^{+}\big{)}\otimes\big{(}{-}x_{i}^{-}\big{)}-x_{i}^{-}\otimes
x_{i}^{+}.$ ∎

###### Proof of Proposition 3.12.

We use

$\displaystyle\big{[}\square\big{(}x_{i}^{+}\big{)},\Omega_{+}\big{]}=-x_{i}^{+}\otimes
h_{i},\qquad\big{[}\square\big{(}x_{i}^{-}\big{)},\Omega_{+}\big{]}=h_{i}\otimes
x_{i}^{-}$

(see [6, Lemma 4.2]).

For $X=x_{j}^{\pm},h_{j}$, both $\Delta T_{i}(X)$ and $(T_{i}\otimes
T_{i})\Delta(X)$ are equal to $\square\,T_{i}(X)$. Hence it is enough to show
$\Delta T_{i}\big{(}x_{j,1}^{+}\big{)}=(T_{i}\otimes
T_{i})\Delta\big{(}x_{j,1}^{+}\big{)}$ since
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ is generated by $x_{j}^{\pm}$,
$h_{j}$, $x_{j,1}^{+}$ ($j\in\mathbb{Z}/N\mathbb{Z}$).

First assume $i=j$. We claim that the both sides coincide with

$\displaystyle\square\,T_{i}\big{(}x_{i,1}^{+}\big{)}+\hbar\big{(}\big{[}1\otimes
x_{i}^{-},\Omega_{+}\big{]}+x_{i}^{-}\otimes h_{i}\big{)}.$

The left-hand side is

$\displaystyle\Delta\big{(}{-}x_{i,1}^{-}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}\big{)}$
$\displaystyle\qquad{}=-\big{(}\square\big{(}x_{i,1}^{-}\big{)}+\hbar\big{[}x_{i}^{-}\otimes
1,\Omega_{+}\big{]}\big{)}+\tfrac{\hbar}{2}\square\big{(}\big{\\{}h_{i},x_{i}^{-}\big{\\}}\big{)}+\hbar\big{(}h_{i}\otimes
x_{i}^{-}+x_{i}^{-}\otimes h_{i}\big{)}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{i,1}^{+}\big{)}+\hbar\big{(}{-}\big{[}x_{i}^{-}\otimes
1,\Omega_{+}\big{]}+h_{i}\otimes x_{i}^{-}+x_{i}^{-}\otimes h_{i}\big{)},$

hence we see the claim by $-\big{[}x_{i}^{-}\otimes
1,\Omega_{+}\big{]}+h_{i}\otimes x_{i}^{-}=\big{[}1\otimes
x_{i}^{-},\Omega_{+}\big{]}$. The right-hand side is

$\displaystyle(T_{i}\otimes
T_{i})\big{(}\square\big{(}x_{i,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{i}^{+},\Omega_{+}\big{]}\big{)}=\square\,T_{i}\big{(}x_{i,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{i}^{+}\big{)},(T_{i}\otimes T_{i})\Omega_{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{i,1}^{+}\big{)}+\hbar\big{[}1\otimes
x_{i}^{-},\Omega_{+}+x_{i}^{+}\otimes x_{i}^{-}-x_{i}^{-}\otimes
x_{i}^{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{i,1}^{+}\big{)}+\hbar\big{(}\big{[}1\otimes
x_{i}^{-},\Omega_{+}\big{]}+x_{i}^{-}\otimes h_{i}\big{)},$

hence the claim holds.

Next assume $a_{ij}=0$. Then the left-hand side is $\Delta(x_{j,1}^{+})$. The
right-hand side is

$\displaystyle(T_{i}\otimes
T_{i})\big{(}\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{)}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},(T_{i}\otimes T_{i})\Omega_{+}\big{]}$
$\displaystyle\qquad{}=\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}+x_{i}^{+}\otimes x_{i}^{-}-x_{i}^{-}\otimes
x_{i}^{+}\big{]}$
$\displaystyle\qquad{}=\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}=\Delta\big{(}x_{j,1}^{+}\big{)},$

hence the claim holds.

Finally assume $a_{ij}=-1$. We claim that the both sides coincide with

$\displaystyle\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{(}\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}-x_{i}^{+}\otimes
x_{j}^{+}\big{)}.$

The left-hand side is

$\displaystyle\Delta\big{(}\big{[}x_{i}^{+},x_{j,1}^{+}\big{]}\big{)}=\big{[}\square\big{(}x_{i}^{+}\big{)},\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{]}$
$\displaystyle\hphantom{\Delta\big{(}\big{[}x_{i}^{+},x_{j,1}^{+}\big{]}\big{)}}{}=\square\,T_{i}\big{(}x_{j,1}^{-}\big{)}-\hbar\big{[}\square\big{(}x_{i}^{+}\big{)},\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{]}.$

Then we see the claim by

$\displaystyle\big{[}\square\big{(}x_{i}^{+}\big{)},\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{]}=\big{[}\big{[}\square\big{(}x_{i}^{+}\big{)},1\otimes
x_{j}^{+}\big{]},\Omega_{+}\big{]}+\big{[}1\otimes
x_{j}^{+},\big{[}\square\big{(}x_{i}^{+}\big{)},\Omega_{+}\big{]}\big{]}$
$\displaystyle\qquad{}=\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}+\big{[}1\otimes
x_{j}^{+},-x_{i}^{+}\otimes h_{i}\big{]}=\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}-x_{i}^{+}\otimes x_{j}^{+}.$

The right-hand side is

$\displaystyle(T_{i}\otimes
T_{i})\big{(}\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{)}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},(T_{i}\otimes T_{i})\Omega_{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}+x_{i}^{+}\otimes
x_{i}^{-}-x_{i}^{-}\otimes x_{i}^{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{(}\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}-x_{i}^{+}\otimes
x_{j}^{+}\big{)},$

hence the claim holds. ∎

##  4 Heisenberg generators

###  4.1 Affine Lie algebra $\boldsymbol{\hat{\mathfrak{gl}}_{N}}$

Let $\mathfrak{gl}_{N}$ be the complex general linear Lie algebra consisting
of $N\times N$ matrices. We denote by $E_{i,j}$ the matrix unit with
$(i,j)$-th entry $1$, and set $\mathbf{1}=\sum\limits_{i=1}^{N}E_{i,i}$. The
indices $i$, $j$ of $E_{i,j}$ are regarded as elements of
$\mathbb{Z}/N\mathbb{Z}$. The transpose of an element $X$ of
$\mathfrak{gl}_{N}$ is denoted by ${}^{t}X$.

Let
$\hat{\mathfrak{gl}}_{N}=\mathfrak{gl}_{N}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}c$
be the affine Lie algebra whose Lie bracket is given by

$\displaystyle\big{[}X\otimes t^{r},Y\otimes t^{s}\big{]}=[X,Y]\otimes
t^{r+s}+r\delta_{r+s,0}\operatorname{tr}(XY)c,\qquad\text{$c$ is central}.$

We denote the element $X\otimes t^{s}$ by $X(s)$. We identify the generators
as usual:

$\displaystyle x_{0}^{+}=E_{N,1}(1),\qquad$ $\displaystyle
x_{0}^{-}=E_{1,N}(-1),\qquad$ $\displaystyle h_{0}=E_{N,N}-E_{1,1}+c,$
$\displaystyle x_{i}^{+}=E_{i,i+1},\qquad$ $\displaystyle
x_{i}^{-}=E_{i+1,i},\qquad$ $\displaystyle h_{i}=E_{i,i}-E_{i+1,i+1},\qquad
i\neq 0.$

We define automorphisms analogous to $\omega$ and $\rho$ for
$\hat{\mathfrak{gl}}_{N}$. Let $\omega$ be the anti-automorphism of
${U}\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$ defined by
$\omega(X(s))={}^{t}X(-s)$ and $\omega(c)=c$. The assignment

$\displaystyle x_{i}^{\pm}\mapsto x_{i-1}^{\pm},\qquad h_{i}\mapsto
h_{i-1},\qquad c\mapsto
c,\qquad\mathbf{1}(s)\mapsto\mathbf{1}(s)+\delta_{s,0}c$

gives an algebra automorphism $\rho$ of
$U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$.

######  Lemma 4.1.

We have
$\rho(E_{ij}(s))=E_{i-1,j-1}(s+\delta_{i,1}-\delta_{j,1})+\delta_{s,0}\delta_{i,1}\delta_{j,1}c$.

###### Proof.

We can show the identities for $i\neq j$ inductively from those for the
Chevalley generators. Then we can show

$\displaystyle\rho(h_{i}(s))=\begin{cases}h_{-\theta}(s)&\text{if $i=1$},\\\
h_{i-1}(s)&\text{otherwise}\end{cases}$ (4.1)

for $s\neq 0$. Indeed we have
$\rho(h_{1}(s))=\rho([E_{1,2}(s),E_{2,1}])=[E_{N,1}(s+1),E_{1,N}(-1)]=h_{-\theta}(s)$
for $i=1$. The other cases are similarly proved. The identity (4.1) will be
used later.

Let us consider the case $i=j$. The case $s=0$ is proved as follows. Note that
the identity $\mathbf{1}=\sum\limits_{i=1}^{N-1}ih_{i}+NE_{N,N}$ holds.
Applying $\rho$ to the both sides, we obtain

$\displaystyle\mathbf{1}+c=\sum_{i=1}^{N-1}ih_{i-1}+N\rho(E_{N,N}).$

The right-hand side is equal to

$\displaystyle\sum_{i=1}^{N-1}ih_{i}-Nh_{N-1}+c+N\rho(E_{N,N}),$

hence we have $\rho(E_{N,N})=E_{N,N}+h_{N-1}=E_{N-1,N-1}$. Then we can
inductively show

$\displaystyle\rho(E_{i,i})=\rho(h_{i}+E_{i+1,i+1})=h_{i-1}+E_{i,i}=E_{i-1,i-1}$

for $i=N-1,N-2,\ldots,2$. For $i=1$, we have
$\rho(E_{1,1})=\rho(h_{1}+E_{2,2})=h_{0}+E_{1,1}=E_{N,N}+c$. The case $s\neq
0$ is similarly proved by considering

$\displaystyle\mathbf{1}(s)=\sum_{i=1}^{N-1}ih_{i}(s)+NE_{N,N}(s)$

and (4.1). ∎

We similarly define an algebra automorphism

$\displaystyle
T_{i}=\exp\operatorname{ad}x_{i}^{+}\exp\operatorname{ad}\big{(}{-}x_{i}^{-}\big{)}\exp\operatorname{ad}x_{i}^{+}$

of $U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$.

######  Lemma 4.2.

We have $T_{i}(X(s))=T_{i}(X)(s)$ for $X\in\mathfrak{gl}_{N}$ if $i\neq 0$.

###### Proof.

Obvious from the definition of $T_{i}$ and the Lie bracket of
$\hat{\mathfrak{gl}}_{N}$. ∎

######  Lemma 4.3.

We have

$\displaystyle
T_{0}(E_{j,j}(s))=\begin{cases}E_{1,1}(s)-\delta_{s,0}c&\text{if }j=N,\\\
E_{N,N}(s)+\delta_{s,0}c&\text{if }j=1,\\\
E_{j,j}(s)&\text{otherwise}\end{cases}$

and

$\displaystyle T_{i}(E_{j,j}(s))=\begin{cases}E_{i+1,i+1}(s)&\text{if }j=i,\\\
E_{i,i}(s)&\text{if }j=i+1,\\\ E_{j,j}(s)\ &\text{otherwise},\end{cases}\qquad
i\neq 0.$

###### Proof.

We show the case $i\neq 0$. By Lemma 4.2, it is enough to consider the case
$s=0$. Apply $T_{i}$ to the identity
$\mathbf{1}=\sum\limits_{j=1}^{N-1}jh_{j}+NE_{N,N}$. Then the left-hand side
is $T_{i}(\mathbf{1})=\mathbf{1}$ and the right-hand side is

$\displaystyle
T_{i}\left(\sum_{j=1}^{N-1}jh_{j}+NE_{N,N}\right)=\begin{cases}\displaystyle\sum_{j=1}^{N-1}jh_{j}-Nh_{N-1}+NT_{N-1}(E_{N,N})&\text{if
$i=N-1$},\vspace{1mm}\\\
\displaystyle\sum_{j=1}^{N-1}jh_{j}+NT_{i}(E_{N,N})&\text{otherwise}.\end{cases}$

This shows

$\displaystyle T_{i}(E_{N,N})=\begin{cases}E_{N-1,N-1}&\text{if $i=N-1$},\\\
E_{N,N}&\text{otherwise}.\end{cases}$

Now let $i=1$. We can inductively show that
$E_{N-1,N-1},E_{N-2,N-2},\ldots,E_{3,3}$ are invariant under $T_{1}$ and

$\displaystyle
T_{1}(E_{2,2})=T_{1}(h_{2}+E_{3,3})=h_{2}+h_{1}+E_{3,3}=E_{1,1},$
$\displaystyle T_{1}(E_{1,1})=T_{1}(h_{1}+E_{2,2})=-h_{1}+E_{1,1}=E_{2,2}.$

Thus we have shown the assertion for $i=1$. Similarly we can show the other
cases.

The case $i=0$ is obtained from the case $i=1$ by applying $\rho$ and using
Lemma 4.1. ∎

######  Lemma 4.4.

We have $T_{0}T_{1}\cdots
T_{k-1}(x_{k}^{+})=\begin{cases}E_{N,k+1}(1)&\text{if $0\leq k\leq N-2$},\\\
-E_{N,1}(2)&\text{if $k=N-1$}.\end{cases}$

###### Proof.

First we prove the assertion for $0\leq k\leq N-2$ by induction on $k$. The
case $k=0$ is trivial. Assume that it holds for $k$, then

$\displaystyle T_{0}T_{1}\cdots
T_{k-1}T_{k}\big{(}x_{k+1}^{+}\big{)}=T_{0}T_{1}\cdots
T_{k-1}\big{(}\big{[}x_{k}^{+},x_{k+1}^{+}\big{]}\big{)}=\big{[}T_{0}T_{1}\cdots
T_{k-1}\big{(}x_{k}^{+}\big{)},x_{k+1}^{+}\big{]}$
$\displaystyle\hphantom{T_{0}T_{1}\cdots
T_{k-1}T_{k}\big{(}x_{k+1}^{+}\big{)}}{}=[E_{N,k+1}(1),E_{k+1,k+2}]=E_{N,k+2}(1).$

Next we prove the assertion for $k=N-1$. We have

$\displaystyle T_{0}T_{1}\cdots
T_{N-2}\big{(}x_{N-1}^{+}\big{)}=T_{0}T_{1}\cdots
T_{N-3}\big{(}\big{[}x_{N-2}^{+},x_{N-1}^{+}\big{]}\big{)}=\big{[}T_{0}T_{1}\cdots
T_{N-3}\big{(}x_{N-2}^{+}\big{)},T_{0}\big{(}x_{N-1}^{+}\big{)}\big{]}$
$\displaystyle\hphantom{T_{0}T_{1}\cdots
T_{N-2}\big{(}x_{N-1}^{+}\big{)}}{}=[E_{N,N-1}(1),-E_{N-1,1}(1)]=-E_{N,1}(2).$

The proof is complete. ∎

######  Lemma 4.5.

Let $i\leq j$. We have

$\displaystyle T_{i}T_{i+1}\cdots
T_{j-1}\big{(}x_{j}^{+}\big{)}=E_{i,j+1},\qquad T_{i}T_{i+1}\cdots
T_{j-1}\big{(}x_{j}^{-}\big{)}=E_{j+1,i},$ (4.2) $\displaystyle
T_{j}T_{j-1}\cdots T_{i+1}\big{(}x_{i}^{+}\big{)}=(-1)^{j-i}E_{i,j+1},\qquad
T_{j}T_{j-1}\cdots T_{i+1}\big{(}x_{i}^{-}\big{)}=(-1)^{j-i}E_{j+1,i}.$ (4.3)

###### Proof.

The assertion is easily proved by induction. ∎

######  Lemma 4.6.

We have

$\displaystyle T_{i}(E_{i+1,j})=E_{i,j}\qquad$ $\displaystyle\text{if $j+1\leq
i\leq N-1$ \ or \ $i+2\leq j$},$ (4.4) $\displaystyle
T_{i}(E_{j,i+1})=E_{j,i}\qquad$ $\displaystyle\text{if $1\leq i\leq j-2$ \ or
\ $j\leq i-1$},$ (4.5) $\displaystyle T_{i}(E_{i,1})=-E_{i+1,1}\qquad$
$\displaystyle\text{if $2\leq i\leq N-1$},$ (4.6) $\displaystyle
T_{i}(E_{2,i})=-E_{2,i+1}\qquad$ $\displaystyle\text{if $3\leq i\leq N-1$}.$
(4.7)

###### Proof.

The identity (4.5) is deduced from (4.4) by applying $\omega$. We use (4.2) to
show the other identities as

$\displaystyle
T_{i}(E_{i+1,j})=\begin{cases}T_{i}T_{j}T_{j+1}{\cdots}T_{i-1}\big{(}x_{i}^{-}\big{)}=T_{j}T_{j+1}{\cdots}T_{i-2}\big{(}x_{i-1}^{-}\big{)}=E_{i,j}&\text{if
$j+1\leq i\leq N-1$},\\\ T_{i}T_{i+1}T_{i+2}\cdots
T_{j-1}\big{(}x_{j}^{+}\big{)}=E_{i,j}&\text{if $i+2\leq j$},\end{cases}$
$\displaystyle T_{i}(E_{i,1})=T_{i}T_{1}T_{2}\cdots
T_{i-2}\big{(}x_{i-1}^{-}\big{)}=T_{1}T_{2}\cdots
T_{i-2}T_{i}\big{(}x_{i-1}^{-}\big{)}=T_{1}T_{2}\cdots
T_{i-2}\big{(}\big{[}x_{i-1}^{-},x_{i}^{-}\big{]}\big{)}$
$\displaystyle\hphantom{T_{i}(E_{i,1})}{}=\big{[}T_{1}T_{2}\cdots
T_{i-2}\big{(}x_{i-1}^{-}\big{)},x_{i}^{-}\big{]}=[E_{i,1},E_{i+1,i}]=-E_{i+1,1},$
$\displaystyle T_{i}(E_{2,i})=T_{i}T_{2}T_{3}\cdots
T_{i-2}\big{(}x_{i-1}^{+}\big{)}=T_{2}T_{3}\cdots
T_{i-2}T_{i}\big{(}x_{i-1}^{+}\big{)}=T_{2}T_{3}\cdots
T_{i-2}\big{(}\big{[}x_{i}^{+},x_{i-1}^{+}\big{]}\big{)}$
$\displaystyle\hphantom{T_{i}(E_{2,i})}{}=\big{[}x_{i}^{+},T_{2}T_{3}\cdots
T_{i-2}\big{(}x_{i-1}^{+}\big{)}\big{]}=[E_{i,i+1},E_{2,i}]=-E_{2,i+1}.$ ∎

###  4.2 Evaluation map

The evaluation map for the affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ was introduced by Guay [5]. It is an
affine analog of the well-known evaluation map from $Y(\mathfrak{sl}_{N})$ to
$U(\mathfrak{gl}_{N})$. Let $U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}_{{\rm
comp},-}$ be the completion of ${U}\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$
defined in [10, Definition 2.5]. From now on, we regard the central element
$c$ of $\hat{\mathfrak{gl}}_{N}$ as a complex number.

The following result can be deduced from a formula for
$\operatorname{ev}(H_{i,1})$ $(i\neq 0)$ where
$H_{i,1}=h_{i,1}+(i/2)(\varepsilon_{1}-\varepsilon_{2})h_{i}$, given in [5,
Section 6, pp. 462–463]. See [10] for details.

######  Theorem 4.7 ([5, 10]).

Assume $\hbar c=N\varepsilon_{2}$. Then there exists an algebra homomorphism
$\operatorname{ev}\colon Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}\to
U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}_{{\rm comp},-}$ uniquely determined by

$\displaystyle\operatorname{ev}(x_{i,0}^{+})=x_{i}^{+},\qquad\operatorname{ev}(x_{i,0}^{-})=x_{i}^{-},\qquad\operatorname{ev}(h_{i,0})=h_{i},$
$\displaystyle\operatorname{ev}(x_{i,1}^{+})=\begin{cases}(1+N\varepsilon_{2})x_{0}^{+}+\hbar\displaystyle\sum_{s\geq
0}\sum_{k=1}^{N}E_{k,1}(s+1)E_{N,k}(-s)\quad\text{if $i=0$},\\\
(1+i\varepsilon_{2})x_{i}^{+}+\hbar\displaystyle\sum_{s\geq
0}\bigg{(}\sum_{k=1}^{i}E_{k,i+1}(s)E_{i,k}(-s)\\\
\qquad{}+\displaystyle\sum_{k=i+1}^{N}E_{k,i+1}(s+1)E_{i,k}(-s-1)\bigg{)}\quad\text{if
$i\neq 0$},\end{cases}$
$\displaystyle\operatorname{ev}(x_{i,1}^{-})=\begin{cases}(1+N\varepsilon_{2})x_{0}^{-}+\hbar\displaystyle\sum_{s\geq
0}\sum_{k=1}^{N}E_{k,N}(s)E_{1,k}(-s-1)\quad\text{if $i=0$},\\\
(1+i\varepsilon_{2})x_{i}^{-}+\hbar\displaystyle\sum_{s\geq
0}\bigg{(}\sum_{k=1}^{i}E_{k,i}(s)E_{i+1,k}(-s)\\\
\qquad{}+\displaystyle\sum_{k=i+1}^{N}E_{k,i}(s+1)E_{i+1,k}(-s-1)\bigg{)}\quad\text{if
$i\neq 0$},\end{cases}$
$\displaystyle\operatorname{ev}(h_{i,1})=\begin{cases}(1+N\varepsilon_{2})h_{0}-\hbar
E_{N,N}(E_{1,1}-c)\\\ \qquad{}+\hbar\displaystyle\sum_{s\geq
0}\sum_{k=1}^{N}(E_{k,N}(s)E_{N,k}(-s)-E_{k,1}(s+1)E_{1,k}(-s-1))\quad\text{if
$i=0$},\\\ (1+i\varepsilon_{2})h_{i}-\hbar
E_{i,i}E_{i+1,i+1}+\hbar\displaystyle\sum_{s\geq
0}\bigg{(}\sum_{k=1}^{i}E_{k,i}(s)E_{i,k}(-s)\\\
\qquad{}+\displaystyle\sum_{k=i+1}^{N}E_{k,i}(s+1)E_{i,k}(-s-1)-\displaystyle\sum_{k=1}^{i}E_{k,i+1}(s)E_{i+1,k}(-s)\\\
\qquad{}-\displaystyle\sum_{k=i+1}^{N}E_{k,i+1}(s+1)E_{i+1,k}(-s-1)\bigg{)}\quad\text{if
$i\neq 0$}.\end{cases}$

We will use the following property.

######  Proposition 4.8.

We have

$\displaystyle\mathrm{(i)}\
\omega\circ\operatorname{ev}=\operatorname{ev}\circ\,\omega,\qquad\mathrm{(ii)}\
\rho\circ\operatorname{ev}=\operatorname{ev}\circ\,\rho,\qquad\mathrm{(iii)}\
T_{i}\circ\operatorname{ev}=\operatorname{ev}\circ\,T_{i}.$

###### Proof.

The assertion (i) is immediate from the definition of $\operatorname{ev}$. The
assertion (ii) is stated in [5, Section 6, p. 463] and a proof is given in
[10, Proposition 3.6]. Since $\operatorname{ev}$ is the identity on the
subalgebra $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ and $T_{i}$ is defined via
the generators of $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$, the assertion
(iii) holds. ∎

###  4.3 Construction of Heisenberg generators

We construct elements $a_{m}$ ($m\in\mathbb{Z}$) of the affine Yangian
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ such that
$\operatorname{ev}(a_{m})=\mathbf{1}(m)$ under the assumption
$\varepsilon_{2}\neq 0$.

First consider the case $m=0$.

######  Proposition 4.9.

We have

$\displaystyle\operatorname{ev}\left(\sum_{i=0}^{N-1}\tilde{h}_{i,1}\right)=\varepsilon_{2}\mathbf{1}+c+\tfrac{\hbar}{2}c^{2}.$

###### Proof.

Put

$\displaystyle A_{0}=\sum_{s\geq 0}\sum_{k=1}^{N}E_{k,N}(s)E_{N,k}(-s),\qquad
B_{0}=\sum_{s\geq 0}\sum_{k=1}^{N}E_{k,1}(s+1)E_{1,k}(-s-1),$ $\displaystyle
A_{i}=\sum_{s\geq
0}\left(\sum_{k=1}^{i}E_{k,i}(s)E_{i,k}(-s)+\sum_{k=i+1}^{N}E_{k,i}(s+1)E_{i,k}(-s-1)\right),\qquad
i\neq 0,$ $\displaystyle B_{i}=\sum_{s\geq
0}\left(\sum_{k=1}^{i}E_{k,i+1}(s)E_{i+1,k}(-s)+\sum_{k=i+1}^{N}E_{k,i+1}(s+1)E_{i+1,k}(-s-1)\right),\qquad
i\neq 0,$

so that

$\displaystyle\operatorname{ev}\big{(}\tilde{h}_{i,1}\big{)}=\begin{cases}(1+N\varepsilon_{2})h_{0}-\tfrac{\hbar}{2}\big{(}E_{N,N}^{2}+(E_{1,1}-c)^{2}\big{)}+\hbar(A_{0}-B_{0})&\text{if
$i=0$},\\\
(1+i\varepsilon_{2})h_{i}-\tfrac{\hbar}{2}\big{(}E_{i,i}^{2}+E_{i+1,i+1}^{2}\big{)}+\hbar(A_{i}-B_{i})&\text{otherwise}.\end{cases}$

Here we use

$\displaystyle-\hbar
E_{N,N}(E_{1,1}-c)-\tfrac{\hbar}{2}h_{0}^{2}=-\tfrac{\hbar}{2}\big{(}E_{N,N}^{2}+(E_{1,1}-c)^{2}\big{)},$
$\displaystyle-\hbar
E_{i,i}E_{i+1,i+1}-\tfrac{\hbar}{2}h_{i}^{2}=-\tfrac{\hbar}{2}\big{(}E_{i,i}^{2}+E_{i+1,i+1}^{2}\big{)},\qquad
i\neq 0.$

Then we have $A_{i+1}-B_{i}=E_{i+1,i+1}^{2}$ for $i=0,\ldots,N-1$, where we
regard $A_{N}=A_{0}$. Therefore

$\displaystyle\operatorname{ev}\left(\sum_{i=0}^{N-1}\tilde{h}_{i,1}\right)=c+\varepsilon_{2}\left(Nh_{0}+\sum_{i=1}^{N-1}ih_{i}\right)-\tfrac{\hbar}{2}\left(2\sum_{i=1}^{N}E_{i,i}^{2}-2cE_{1,1}+c^{2}\right)+\hbar\sum_{i=1}^{N}E_{i,i}^{2}$
$\displaystyle\hphantom{\operatorname{ev}\left(\sum_{i=0}^{N-1}\tilde{h}_{i,1}\right)}{}=c+\varepsilon_{2}\sum_{i=1}^{N-1}ih_{i}+N\varepsilon_{2}h_{0}+\hbar
cE_{1,1}-\tfrac{\hbar}{2}c^{2}.$

By the assumption $\hbar c=N\varepsilon_{2}$, we have

$\displaystyle N\varepsilon_{2}h_{0}+\hbar
cE_{1,1}-\tfrac{\hbar}{2}c^{2}=N\varepsilon_{2}(E_{N,N}+c)-\tfrac{\hbar}{2}c^{2}=N\varepsilon_{2}E_{N,N}+\tfrac{\hbar}{2}c^{2}.$

The assertion holds since we have
$\mathbf{1}=\sum\limits_{i=1}^{N-1}ih_{i}+NE_{N,N}$. ∎

Assume $\hbar c=N\varepsilon_{2}$ and $\varepsilon_{2}\neq 0$. Put

$\displaystyle
a_{0}=\dfrac{1}{\varepsilon_{2}}\left(\sum_{i=0}^{N-1}\tilde{h}_{i,1}-c-\tfrac{\hbar}{2}c^{2}\right).$

Then we have $\operatorname{ev}(a_{0})=\mathbf{1}$ by Proposition 4.9.

Next consider the case $m\geq 1$. For each $i\in\mathbb{Z}/N\mathbb{Z}$ and a
fixed $m$, define an element $w(i,m)$ of the affine Weyl group $\hat{W}$ by

$\displaystyle w(i,m)=t_{-\alpha_{i+1}}^{m-1}s_{i+1}s_{i+2}\cdots
s_{i-3}s_{i-2}.$

This element has the property $w(i,m)(\alpha_{i-1})=-\alpha_{i}+m\delta$.
Hence the elements

$\displaystyle\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1}^{+}\big{)}\big{]},\qquad\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1,1}^{+}\big{)}\big{]}$

have weight $m\delta$. We will see

$\displaystyle\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1}^{+}\big{)}\big{]}=(-1)^{m-1}\times\begin{cases}h_{-\theta}(m)&\text{if
$i=0$},\\\ h_{i}(m)&\text{otherwise}\end{cases}$

in Lemma 4.13, and will compute the value of

$\displaystyle\operatorname{ev}\big{(}\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1,1}^{+}\big{)}\big{]}\big{)}=\big{[}x_{i}^{+},T_{w(i,m)}\operatorname{ev}\big{(}x_{i-1,1}^{+}\big{)}\big{]}$

in Proposition 4.15. Then we will take the summation over $i$ in Proposition
4.16. The result will yield a construction of the elements $a_{m}$ in Theorem
4.18.

By Lemma 3.6(ii), $\rho\circ T_{w(i,m)}=T_{w(i-1,m)}\circ\rho$ holds. The case
$i=1$ will be important. Note that $w(1,1)=s_{2}s_{3}\cdots s_{N-1}$ and
$t_{-\alpha_{2}}=s_{2}s_{3}\cdots s_{0}s_{1}s_{0}\cdots s_{3}$.

######  Lemma 4.10.

We have

$\displaystyle
T_{t_{-\alpha_{2}}}(E_{i,i}(s))=\begin{cases}E_{1,1}(s)&\text{if $i=1$},\\\
E_{2,2}(s)+\delta_{s,0}c&\text{if $i=2$},\\\ E_{3,3}(s)-\delta_{s,0}c&\text{if
$i=3$},\\\ E_{i,i}(s)&\text{if $4\leq i\leq N$}\end{cases}$

for any $s\in\mathbb{Z}$.

###### Proof.

The assertion is easily proved by Lemma 4.3. ∎

######  Lemma 4.11.

We have

$\displaystyle T_{w(1,m)}(E_{1,1}(s))=E_{1,1}(s),\qquad
T_{w(1,m)}(E_{N,N}(-s))=E_{2,2}(-s)+\delta_{s,0}(m-1)c$

for any $s\in\mathbb{Z}$.

###### Proof.

The assertion follows from

$\displaystyle T_{2}T_{3}\cdots T_{N-1}(E_{1,1}(s))=E_{1,1}(s),\qquad
T_{2}T_{3}\cdots T_{N-1}(E_{N,N}(-s))=E_{2,2}(-s)$

and Lemma 4.10. ∎

######  Lemma 4.12.

We have

$\displaystyle
T_{w(i,m)}(x_{i-1}^{+})=(-1)^{m-1}\times\begin{cases}E_{1,N}(m-1)&\text{if
$i=0$},\\\ E_{i+1,i}(m)&\text{otherwise}.\end{cases}$

###### Proof.

We prove the assertion for $i=1$. The other cases are deduced from this case
by applying $\rho$ and Lemma 4.1. First assume $m=1$. Recall
$T_{w(1,1)}=T_{2}T_{3}\cdots T_{N-1}$. We have

$\displaystyle T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{0}^{+}\big{)}=T_{2}T_{3}\cdots
T_{N-2}\big{(}\big{[}x_{N-1}^{+},x_{0}^{+}\big{]}\big{)}=\big{[}T_{2}T_{3}\cdots
T_{N-2}\big{(}x_{N-1}^{+}\big{)},x_{0}^{+}\big{]}$
$\displaystyle\hphantom{T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{0}^{+}\big{)}}{}\overset{\text{by
(\ref{eq:increase})}}{=}[E_{2,N},E_{N,1}(1)]=E_{2,1}(1),$

hence the case $m=1$ is proved. Next consider the case $m\geq 2$. Since the
case $m=2$ yields the equality for general $m\geq 2$ inductively, it is enough
to prove

$\displaystyle T_{t_{-\alpha_{2}}}(E_{2,1}(1))=-E_{2,1}(2).$

Since $E_{2,1}(1)$ is invariant under $T_{i}$ for $3\leq i\leq N-1$, we have

$\displaystyle T_{t_{-\alpha_{2}}}(E_{2,1}(1))=T_{2}T_{3}\cdots
T_{0}T_{1}T_{0}(E_{2,1}(1)).$ (4.8)

We have

$\displaystyle
T_{0}T_{1}T_{0}(E_{2,1}(1))=T_{0}T_{1}T_{0}\big{(}T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{0}^{+}\big{)}\big{)}=T_{0}T_{1}T_{2}T_{3}\cdots
T_{N-2}\big{(}x_{N-1}^{+}\big{)}=-E_{N,1}(2).$

Here the second equality follows from the braid relations and Proposition 3.4,
and the last from Lemma 4.4. Then the right-hand side of (4.8) is

$\displaystyle T_{2}T_{3}\cdots T_{N-1}(-E_{N,1}(2))\overset{\text{by
Lemma~{}\ref{lem:higher}}}{=}-(T_{2}T_{3}\cdots
T_{N-1}(E_{N,1}))(2)\overset{\text{by (\ref{eq:TE-1})}}{=}-E_{2,1}(2),$

hence the assertion is proved. ∎

######  Lemma 4.13.

We have

$\displaystyle\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1}^{+}\big{)}\big{]}=(-1)^{m-1}\times\begin{cases}h_{-\theta}(m)&\text{if
$i=0$},\\\ h_{i}(m)&\text{otherwise}.\end{cases}$

###### Proof.

Immediate from Lemma 4.12. ∎

######  Lemma 4.14.

We have

$\displaystyle T_{w(1,m)}(E_{k,1}(s+1)E_{N,k}(-s))$
$\displaystyle\qquad{}=(-1)^{m-1}\times\begin{cases}E_{1,1}(s+1)E_{2,1}(-s+m-1)&\text{if
$k=1$},\\\ E_{3,1}(s-m+2)E_{2,3}(-s+2m-2)&\text{if $k=2$},\\\
E_{k+1,1}(s+1)E_{2,k+1}(-s+m-1)&\text{if $3\leq k\leq N-1$},\\\
E_{2,1}(s+m)\Big{(}E_{2,2}(-s)+\delta_{s,0}(m-1)c\Big{)}&\text{if
$k=N$}.\end{cases}$

###### Proof.

We prove

$\displaystyle T_{w(1,m)}(E_{k,1}(s+1))=\begin{cases}E_{1,1}(s+1)&\text{if
$k=1$},\\\ (-1)^{(m-1)(N+1)}(-E_{3,1}(s-m+2))&\text{if $k=2$},\\\
-E_{k+1,1}(s+1)&\text{if $3\leq k\leq N-1$},\\\
(-1)^{m-1}E_{2,1}(s+m)&\text{if $k=N$}\end{cases}$ (4.9)

and

$\displaystyle
T_{w(1,m)}(E_{N,k}(-s))=\begin{cases}(-1)^{m-1}E_{2,1}(-s+m-1)&\text{if
$k=1$},\\\ (-1)^{(m-1)N}(-E_{2,3}(-s+2m-2))&\text{if $k=2$},\\\
(-1)^{m-1}(-E_{2,k+1}(-s+m-1))&\text{if $3\leq k\leq N-1$},\\\
E_{2,2}(-s)+\delta_{s,0}(m-1)c&\text{if $k=N$}.\end{cases}$ (4.10)

The equalities (4.9) for $k=1$ and (4.10) for $k=N$ follow from Lemma 4.11.

Consider (4.9) for $k=N$ and (4.10) for $k=1$. Note that Lemma 4.12 for $i=1$
is nothing but (4.9) for $k=N$ and $s=0$. We can prove the other cases by
applying $[-,E_{1,1}(\pm s)]$ to this case as

$\displaystyle\big{[}T_{w(1,m)}(E_{N,1}(1)),E_{1,1}(\pm
s)\big{]}=T_{w(1,m)}\big{(}\big{[}E_{N,1}(1),E_{1,1}(\pm
s)\big{]}\big{)}=T_{w(1,m)}\big{(}E_{N,1}(1\pm s)\big{)}.$

Here we use the fact that $E_{1,1}(\pm s)$ is invariant under $T_{w(1,m)}$
proved in Lemma 4.11. In the sequel, we prove (4.9) and (4.10) for $2\leq
k\leq N-1$.

We prove (4.9). First we consider the case $m=1$. We prove

$\displaystyle T_{w(1,1)}\big{(}E_{k,1}(s+1)\big{)}=-E_{k+1,1}(s+1)$

for $2\leq k\leq N-1$ by induction on $k$. Since $T_{w(1,1)}=T_{2}T_{3}\cdots
T_{N-1}$ does not involve $T_{0}$, it is enough to prove
$T_{w(1,1)}(E_{k,1})=-E_{k+1,1}$ by Lemma 4.2. When $k=2$, we have

$\displaystyle T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{1}^{-}\big{)}=T_{2}\big{(}x_{1}^{-}\big{)}=-E_{3,1}.$

Assume that the assertion holds for $k$. Since we have
$E_{k+1,1}=-T_{k}(E_{k,1})$ by (4.6),

$\displaystyle T_{2}T_{3}\cdots T_{N-1}(E_{k+1,1})=-T_{2}T_{3}\cdots
T_{N-1}T_{k}(E_{k,1})=-T_{k+1}T_{2}T_{3}\cdots T_{N-1}(E_{k,1})$
$\displaystyle\hphantom{T_{2}T_{3}\cdots
T_{N-1}(E_{k+1,1})}{}=T_{k+1}(E_{k+1,1})\overset{\text{by
(\ref{eq:TE-3})}}{=}-E_{k+2,1}.$

Thus we have proved the case $m=1$. Next we consider the case $m\geq 2$. Since
the case $m=2$ yields the equality for general $m\geq 2$ inductively, it is
enough to prove:

$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{3,1}(s+1)\big{)}=(-1)^{N+1}E_{3,1}(s),$ (4.11)
$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{k+1,1}(s+1)\big{)}=E_{k+1,1}(s+1)\qquad\text{if
$3\leq k\leq N-1$}.$ (4.12)

If we prove the assertion for $s=0$, we can prove the other cases by applying
$[-,E_{1,1}(s)]$ to this case by using the fact that $E_{1,1}(s)$ is invariant
under $T_{t_{-\alpha_{2}}}$ proved in Lemma 4.10. We prove (4.11) for $s=0$.
Since we have $E_{3,1}=-T_{2}\big{(}x_{1}^{-}\big{)}$, the left-hand side of
(4.11) for $s=0$ is

$\displaystyle-T_{2}T_{3}\cdots T_{0}T_{1}T_{0}\cdots
T_{3}\big{(}T_{2}\big{(}x_{1}^{-}\big{)}(1)\big{)}\overset{\text{by
(\ref{eq:decrease})}}{=}-T_{2}T_{3}\cdots
T_{0}T_{1}T_{0}\big{(}(-1)^{N-2}E_{N,1}(1)\big{)}$
$\displaystyle\qquad{}=(-1)^{N-1}T_{2}T_{3}\cdots
T_{0}T_{1}T_{0}\big{(}x_{0}^{+}\big{)}=(-1)^{N}T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{1}^{-}\big{)}$
$\displaystyle\qquad{}=(-1)^{N}T_{2}\big{(}x_{1}^{-}\big{)}=(-1)^{N+1}E_{3,1}.$

We prove (4.12) for $s=0$ by backward induction on $k$. The case $k=N-1$ is
proved as

$\displaystyle T_{2}T_{3}\cdots T_{0}T_{1}T_{0}\cdots
T_{3}\big{(}x_{0}^{+}\big{)}=T_{2}T_{3}\cdots
T_{0}T_{1}\big{(}x_{N-1}^{+}\big{)}=T_{2}T_{3}\cdots
T_{N-2}\big{(}x_{0}^{+}\big{)}=x_{0}^{+}.$

Assume that the assertion holds for $k$. Since we have
$E_{k,1}=T_{k}(E_{k+1,1})$ by (4.4),

$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{k,1}(1)\big{)}=T_{t_{-\alpha_{2}}}T_{k}\big{(}E_{k+1,1}(1)\big{)}=T_{k}T_{t_{-\alpha_{2}}}\big{(}E_{k+1,1}(1)\big{)}=T_{k}\big{(}E_{k+1,1}(1)\big{)}=E_{k,1}(1).$

Here the second equality holds by $k\geq 3$. Thus we have proved (4.9).

We prove (4.10). First we consider the case $m=1$. We prove

$\displaystyle T_{w(1,1)}\big{(}E_{N,k}(-s)\big{)}=-E_{N,k+1}(-s)$

for $2\leq k\leq N-1$ by backward induction on $k$. By Lemma 4.2, it is enough
to prove $T_{w(1,1)}(E_{N,k})=-E_{2,k+1}$. When $k=N-1$, we have

$\displaystyle T_{2}T_{3}\cdots
T_{N-1}\big{(}x_{N-1}^{-}\big{)}=-T_{2}T_{3}\cdots
T_{N-2}\big{(}x_{N-1}^{+}\big{)}=-E_{2,N}$

by (4.2). Assume that the assertion holds for $k$. Since we have
$E_{N,k-1}=T_{k-1}(E_{N,k})$ by (4.5),

$\displaystyle T_{2}T_{3}\cdots
T_{N-1}\big{(}E_{N,k-1}\big{)}=T_{2}T_{3}\cdots
T_{N-1}T_{k-1}\big{(}E_{N,k}\big{)}=T_{k}T_{2}T_{3}\cdots
T_{N-1}\big{(}E_{N,k}\big{)}$ $\displaystyle\hphantom{T_{2}T_{3}\cdots
T_{N-1}\big{(}E_{N,k-1}\big{)}}{}=-T_{k}(E_{2,k+1})\overset{\text{by
(\ref{eq:TE-2})}}{=}-E_{2,k}.$

Thus we have proved the case $m=1$. Next we consider the case $m\geq 2$. Since
the case $m=2$ yields the equality for general $m\geq 2$ inductively, it is
enough to prove:

$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{2,3}(-s)\big{)}=(-1)^{N}E_{2,3}(-s+2),$ (4.13)
$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{2,k+1}(-s)\big{)}=-E_{2,k+1}(-s+1)\qquad\text{if
$3\leq k\leq N-1$}.$ (4.14)

If we prove the assertion for $s=0$, we can prove the other cases by applying
$[E_{2,2}(-s),-]$ to this case by using the fact that $E_{2,2}(-s)$ is
invariant under $T_{t_{-\alpha_{2}}}$ for $s\geq 1$ proved in Lemma 4.10. We
prove (4.13) for $s=0$. Since we have

$\displaystyle T_{N-1}T_{N-2}\cdots
T_{3}\big{(}x_{2}^{+}\big{)}=(-1)^{N-3}E_{2,N}$

by (4.3), and

$\displaystyle
T_{0}T_{1}T_{0}\big{(}E_{2,N}\big{)}=T_{1}T_{0}T_{1}\big{(}E_{2,N}\big{)}\overset{\text{by
(\ref{eq:TE-1})}}{=}T_{1}T_{0}\big{(}E_{1,N}\big{)}\overset{\text{by
Lemma~{}\ref{lem:0_to_k}}}{=}T_{1}\big{(}{-}E_{N,1}(2)\big{)},$

the left-hand side of (4.13) for $s=0$ is

$\displaystyle(-1)^{N-2}T_{2}T_{3}\cdots
T_{N-1}T_{1}\big{(}E_{N,1}(2)\big{)}=(-1)^{N-2}T_{2}T_{1}T_{3}\cdots
T_{N-1}\big{(}E_{N,1}(2)\big{)}$ $\displaystyle\qquad{}\overset{\text{by
(\ref{eq:TE-1})}}{=}(-1)^{N-2}T_{2}T_{1}\big{(}E_{3,1}\big{)}(2)=(-1)^{N-2}T_{2}T_{1}\big{(}{-}T_{2}\big{(}x_{1}^{-}\big{)}\big{)}(2)$
$\displaystyle\qquad{}=(-1)^{N-1}T_{2}\big{(}x_{2}^{-}\big{)}(2)=(-1)^{N}E_{2,3}(2).$

We prove (4.14) for $s=0$ by induction on $k$. Assume $k=3$. Since we have
$E_{2,4}=T_{2}\big{(}x_{3}^{+}\big{)}$, the left-hand side of (4.14) for $s=0$
is

$\displaystyle T_{2}T_{3}\cdots T_{0}T_{1}T_{0}\cdots
T_{3}\big{(}T_{2}\big{(}x_{3}^{+}\big{)}\big{)}=T_{2}T_{3}\cdots
T_{0}T_{1}T_{0}\cdots T_{4}\big{(}x_{2}^{+}\big{)}=T_{2}T_{3}\cdots
T_{0}T_{1}\big{(}x_{2}^{+}\big{)}$ $\displaystyle\qquad\overset{\text{by
Lemma~{}\ref{lem:0_to_k}}}{=}T_{2}T_{3}\cdots
T_{N-1}\big{(}E_{N,3}(1)\big{)}\overset{\text{by
(\ref{eq:TE-1})}}{=}T_{2}T_{3}(E_{4,3})(1)=-T_{2}\big{(}x_{3}^{+}\big{)}(1)=-E_{2,4}(1).$

Assume that the assertion holds for $k$. Since we have
$E_{2,k+2}=-T_{k+1}(E_{2,k+1})$ by (4.7), the left-hand side is

$\displaystyle
T_{t_{-\alpha_{2}}}\big{(}E_{2,k+2}\big{)}=-T_{t_{-\alpha_{2}}}T_{k+1}\big{(}E_{2,k+1}\big{)}=-T_{k+1}T_{t_{-\alpha_{2}}}\big{(}E_{2,k+1}\big{)}$
$\displaystyle\hphantom{T_{t_{-\alpha_{2}}}\big{(}E_{2,k+2}\big{)}}{}=T_{k+1}\big{(}E_{2,k+1}\big{)}(1)=-E_{2,k+2}(1).$

Here the second equality holds by $k\geq 3$. ∎

For a fixed $m\geq 1$, put

$\displaystyle S_{i,j}(p)=E_{i,j}(p)E_{j,i}(m-p).$

######  Proposition 4.15.

We have

$\displaystyle\big{[}x_{i}^{+},T_{w(i,m)}\operatorname{ev}\big{(}x_{i-1,1}^{+}\big{)}\big{]}$
$\displaystyle=(-1)^{m-1}\left(A_{i}+\hbar\sum_{k=1}^{N}(P_{i,k}-Q_{i,k})-\hbar\sum_{s=0}^{m-2}E_{i,i}(s+1)E_{i+1,i+1}(-s+m-1)\right),$
(4.15)

where

$\displaystyle
A_{i}=\begin{cases}(1+(N-1)\varepsilon_{2})h_{-\theta}(m)+(m-1)\hbar
ch_{-\theta}(m)+\hbar cE_{N,N}(m)&\text{if $i=0$},\\\
(1+N\varepsilon_{2})h_{1}(m)+(m-1)\hbar ch_{1}(m)&\text{if $i=1$},\\\
(1+(i-1)\varepsilon_{2})h_{i}(m)+(m-1)\hbar ch_{i}(m)+\hbar ch_{i}(m)&\text{if
$2\leq i\leq N-1$},\end{cases}$ $\displaystyle P_{i,k}=\sum_{s\geq
0}S_{k,i}(p(i,k)),\qquad Q_{i,k}=\sum_{s\geq 0}S_{k,i+1}(q(i,k)).$

The integers $p(i,k)$ and $q(i,k)$ are given by

$\displaystyle p(0,k)=\begin{cases}s+m-1&\text{if $k=1$},\\\ s-m+1&\text{if
$k=2$},\\\ s&\text{if $3\leq k\leq N-1$},\\\ s+1&\text{if $k=N$},\end{cases}$
$\displaystyle p(i,k)=\begin{cases}s&\text{if $1\leq k\leq i-1$},\\\
s+1&\text{if $k=i$},\\\ s+m&\text{if $k=i+1$},\\\ s-m+2&\text{if $k=i+2$},\\\
s+1&\text{if $i+3\leq k\leq N$},\end{cases}\qquad 1\leq i\leq N-2,$
$\displaystyle p(N-1,k)=\begin{cases}s-m+1&\text{if $k=1$},\\\ s&\text{if
$2\leq k\leq N-2$},\\\ s+1&\text{if $k=N-1$},\\\ s+m&\text{if
$k=N$},\end{cases}\qquad q(i,k)=\begin{cases}p(0,k)+1&\text{if $i=0$},\\\
p(i,k)&\text{otherwise}.\end{cases}$

###### Proof.

We prove the assertion for $i=1$. By Lemma 4.14, we have

$\displaystyle(-1)^{m-1}T_{w(1,m)}\operatorname{ev}\big{(}x_{0,1}^{+}\big{)}$
$\displaystyle\qquad{}=(-1)^{m-1}(1+N\varepsilon_{2})T_{w(1,m)}\big{(}x_{0}^{+}\big{)}+(-1)^{m-1}\hbar\sum_{s\geq
0}\sum_{k=1}^{N}T_{w(1,m)}\big{(}E_{k,1}(s+1)E_{N,k}(-s)\big{)}$
$\displaystyle\qquad{}=(-1)^{m-1}(1+N\varepsilon_{2})T_{w(1,m)}\big{(}x_{0}^{+}\big{)}+(m-1)\hbar
cE_{2,1}(m)$ $\displaystyle\qquad\quad{}+\hbar\sum_{s\geq
0}\bigg{(}E_{1,1}(s+1)E_{2,1}(-s+m-1)+E_{2,1}(s+m)E_{2,2}(-s)$
$\displaystyle\qquad\quad{}+E_{3,1}(s-m+2)E_{2,3}(-s+2m-2)+\sum_{k=4}^{N}E_{k,1}(s+1)E_{2,k}(-s+m-1)\bigg{)}.$

Therefore

$\displaystyle(-1)^{m-1}[x_{1}^{+},T_{w(1,m)}\operatorname{ev}(x_{0,1}^{+})]=(1+N\varepsilon_{2})h_{1}(m)+(m-1)\hbar
ch_{1}(m)$ $\displaystyle{}+\hbar\sum_{s\geq
0}\bigg{(}{-}S_{1,2}(s+1)+E_{1,1}(s+1)h_{1}(-s+m-1)+h_{1}(s+m)E_{2,2}(-s)+S_{2,1}(s+m)$
$\displaystyle{}-S_{3,2}(s-m+2)+S_{3,1}(s-m+2)+\sum_{k=4}^{N}\big{(}{-}S_{k,2}(s+1)+S_{k,1}(s+1)\big{)}\bigg{)}.$
(4.16)

Here we use Lemma 4.13. Since we have

$\displaystyle\sum_{s\geq
0}\big{(}E_{1,1}(s+1)h_{1}(-s+m-1)+h_{1}(s+m)E_{2,2}(-s)\big{)}$
$\displaystyle\qquad{}=\sum_{s\geq
0}\big{(}S_{1,1}(s+1)-S_{2,2}(s+m)\big{)}-\sum_{s=0}^{m-2}E_{1,1}(s+1)E_{2,2}(-s+m-1),$

the right-hand side of (4.16) is equal to

$\displaystyle(1+N\varepsilon_{2})h_{1}(m)+(m-1)\hbar ch_{1}(m)$
$\displaystyle\qquad{}+\hbar\sum_{s\geq
0}\bigg{(}\bigg{(}S_{1,1}(s+1)+S_{2,1}(s+m)+S_{3,1}(s-m+2)+\sum_{k=4}^{N}S_{k,1}(s+1)\bigg{)}$
$\displaystyle\qquad{}-\bigg{(}S_{1,2}(s+1)+S_{2,2}(s+m)+S_{3,2}(s-m+2)+\sum_{k=4}^{N}S_{k,2}(s+1)\bigg{)}\bigg{)}$
$\displaystyle\qquad{}-\hbar\sum_{s=0}^{m-2}E_{1,1}(s+1)E_{2,2}(-s+m-1).$

Hence the assertion holds for $i=1$. Then apply $\rho$ to (4.15) for $i=1$.
The left-hand side is

$\displaystyle\begin{split}&\rho\big{(}\big{[}x_{1}^{+},T_{w(1,m)}\operatorname{ev}\big{(}x_{0,1}^{+}\big{)}\big{]}\big{)}=\big{[}x_{0}^{+},T_{w(0,m)}\operatorname{ev}\big{(}x_{0,1}^{+}+\varepsilon_{2}x_{0}^{+}\big{)}\big{]}\\\
&\hphantom{\rho\big{(}\big{[}x_{1}^{+},T_{w(1,m)}\operatorname{ev}\big{(}x_{0,1}^{+}\big{)}\big{]}\big{)}}{}=\big{[}x_{0}^{+},T_{w(0,m)}\operatorname{ev}\big{(}x_{0,1}^{+}\big{)}\big{]}+\varepsilon_{2}h_{-\theta}(m).\end{split}$

Here we use Proposition 4.8(ii) and Lemma 4.13. Consider the right-hand side.
We can see

$\displaystyle\rho(A_{1})=A_{0}+\varepsilon_{2}h_{-\theta}(m)-\hbar
cE_{N,N}(m),$
$\displaystyle\rho(P_{1,k})=\begin{cases}P_{0,N}+E_{N,N}(m)c&\text{if
$k=1$},\\\ P_{0,k-1}&\text{if $2\leq k\leq
N$},\end{cases}\qquad\rho(Q_{1,k})=Q_{0,k-1},$
$\displaystyle\rho\big{(}E_{1,1}(s+1)E_{2,2}(-s+m-1)\big{)}=E_{N,N}(s+1)E_{1,1}(-s+m-1).$

Hence the assertion holds for $i=0$. Then apply $\rho$ to (4.15) for $i=0$.
Similarly the left-hand side is

$\displaystyle\rho\big{(}\big{[}x_{0}^{+},T_{w(0,m)}\operatorname{ev}\big{(}x_{N-1,1}^{+}\big{)}\big{]}\big{)}=\big{[}x_{N-1}^{+},T_{w(N-1,m)}\operatorname{ev}\big{(}x_{N-2,1}^{+}\big{)}\big{]}+\varepsilon_{2}h_{N-1}(m),$

and for the right-hand side, we can see

$\displaystyle\rho(A_{0})=A_{N-1}+\varepsilon_{2}h_{N-1}(m)+\hbar
cE_{N,N}(m),$
$\displaystyle\rho(P_{0,k})=P_{N-1,k-1},\qquad\rho(Q_{0,k})=\begin{cases}Q_{N-1,N}-E_{N,N}(m)c&\text{if
$k=1$},\\\ Q_{N-1,k-1}&\text{if $2\leq k\leq N$},\end{cases}$
$\displaystyle\rho\big{(}E_{N,N}(s+1)E_{1,1}(-s+m-1)\big{)}=E_{N-1,N-1}(s+1)E_{N,N}(-s+m-1).$

Note that $c$ never appears in the last equality as $-s+m-1$ cannot be $0$ for
$0\leq s\leq m-2$. Hence the assertion holds for $i=N-1$. Continuing this
process we prove the assertions for $i=N-2,N-3,\ldots,2$ since we have

$\displaystyle\rho(A_{i+1})=A_{i}+\varepsilon_{2}h_{i}(m),\qquad\rho(P_{i+1,k})=P_{i,k-1},\qquad\rho(Q_{i+1,k})=Q_{i,k-1},$
$\displaystyle\rho\big{(}E_{i+1,i+1}(s+1)E_{i+2,i+2}(-s+m-1)\big{)}=E_{i,i}(s+1)E_{i+1,i+1}(-s+m-1).$
∎

######  Proposition 4.16.

We have

$\displaystyle\operatorname{ev}\left((-1)^{m-1}\sum_{i=0}^{N-1}\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1,1}^{+}\big{)}\big{]}\right)=\varepsilon_{2}\mathbf{1}(m)+\hbar
R_{m},$ (4.17)

where

$\displaystyle
R_{m}=(-1)^{m-1}\sum_{i=0}^{N-1}x_{i}^{+}T_{w(i,m)}\big{(}x_{i-1}^{+}\big{)}+\sum_{p=1}^{m-1}\sum_{1\leq
i\leq j\leq N-1}h_{i}(p)h_{j}(m-p)$
$\displaystyle\hphantom{R_{m}=}{}-\sum_{p=-m+2}^{m-1}\bigg{(}E_{1,N}(p-1)E_{N,1}(m-p+1)+\sum_{i=0}^{N-2}E_{i+2,i+1}(p)E_{i+1,i+2}(m-p)\bigg{)}$
$\displaystyle\hphantom{R_{m}=}{}+\sum_{p=-m+2}^{0}\bigg{(}E_{1,N-1}(p-1)E_{N-1,1}(m-p+1)+E_{2,N}(p-1)E_{N,2}(m-p+1)$
$\displaystyle\hphantom{R_{m}=}{}+\sum_{i=0}^{N-3}E_{i+3,i+1}(p)E_{i+1,i+3}(m-p)\bigg{)}.$

######  Remark 4.17.

The point of the statement of Proposition 4.16 is as follows: although each
term
$\operatorname{ev}\big{(}\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1,1}^{+}\big{)}\big{]}\big{)}$
lies in the completion of $U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$, we obtain
$R_{m}$ an element of $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ as a remainder
term after cancellation.

###### Proof.

We use the notation in Proposition 4.15. We have

$\displaystyle\sum_{i=0}^{N-1}A_{i}=\varepsilon_{2}\left((N-1)h_{-\theta}+Nh_{1}+\sum_{i=2}^{N-1}(i-1)h_{i}\right)(m)+\hbar
c\left(E_{N,N}+\sum_{i=2}^{N-1}h_{i}\right)(m)$
$\displaystyle\hphantom{\sum_{i=0}^{N-1}A_{i}}{}=\varepsilon_{2}\left((N-1)h_{-\theta}+Nh_{1}+\sum_{i=2}^{N-1}(N+i-1)h_{i}+NE_{N,N}\right)(m)$
$\displaystyle\hphantom{\sum_{i=0}^{N-1}A_{i}}{}=\varepsilon_{2}\left(\sum_{i=1}^{N-1}ih_{i}+NE_{N,N}\right)(m)=\varepsilon_{2}\mathbf{1}(m).$

In the second equality we use the condition $\hbar c=N\varepsilon_{2}$. We
compute $\sum\limits_{k=1}^{N}(P_{i+1,k}-Q_{i,k})$ as follows:

$\displaystyle\sum_{k=1}^{N}(P_{0,k}-Q_{N-1,k})=-\sum_{s=0}^{2m-3}S_{1,N}(s-m+1)+\sum_{s=0}^{m-2}S_{2,N}(s-m+1)$
$\displaystyle\hphantom{\sum_{k=1}^{N}(P_{0,k}-Q_{N-1,k})=}{}+S_{N-1,N}(0)+\sum_{s=0}^{m-2}S_{N,N}(s+1),$
$\displaystyle\sum_{k=1}^{N}(P_{1,k}-Q_{0,k})=\sum_{s=0}^{m-2}S_{1,1}(s+1)-\sum_{s=0}^{2m-3}S_{2,1}(s-m+2)$
$\displaystyle\hphantom{\sum_{k=1}^{N}(P_{1,k}-Q_{0,k})=}{}+\sum_{s=0}^{m-2}S_{3,1}(s-m+2)+S_{N,1}(1),$
$\displaystyle\sum_{k=1}^{N}(P_{i+1,k}-Q_{i,k})=S_{i,i+1}(0)+\sum_{s=0}^{m-2}S_{i+1,i+1}(s+1)-\sum_{s=0}^{2m-3}S_{i+2,i+1}(s-m+2)$
$\displaystyle\hphantom{\sum_{k=1}^{N}(P_{i+1,k}-Q_{i,k})=}{}+\sum_{s=0}^{m-2}S_{i+3,i+1}(s-m+2)\qquad\text{for}\
1\leq i\leq N-3,$
$\displaystyle\sum_{k=1}^{N}(P_{N-1,k}-Q_{N-2,k})=\sum_{s=0}^{m-2}S_{1,N-1}(s-m+1)+S_{N-2,N-1}(0)$
$\displaystyle\hphantom{\sum_{k=1}^{N}(P_{N-1,k}-Q_{N-2,k})=}{}+\sum_{s=0}^{m-2}S_{N-1,N-1}(s+1)-\sum_{s=0}^{2m-3}S_{N,N-1}(s-m+2).$

Hence the assertion holds by

$\displaystyle(-1)^{m-1}x_{i}^{+}T_{w(i,m)}\big{(}x_{i-1}^{+}\big{)}=\begin{cases}S_{N,1}(1)&\text{if
$i=0$,}\\\ S_{i,i+1}(0)&\text{otherwise}\end{cases}$

and

$\displaystyle\sum_{i=0}^{N-1}\big{(}S_{i,i}(s+1)-E_{i,i}(s+1)E_{i+1,i+1}(-s+m-1)\big{)}$
$\displaystyle\qquad{}=\left(\sum_{i=1}^{N-1}E_{i,i}(s+1)h_{i}(-s+m-1)\right)+E_{N,N}(s+1)h_{-\theta}(-s+m-1)$
$\displaystyle\qquad{}=\sum_{i=1}^{N-1}(E_{i,i}-E_{N,N})(s+1)h_{i}(-s+m-1)=\sum_{i=1}^{N-1}\sum_{j\geq
i}h_{j}(s+1)h_{i}(-s+m-1).$ ∎

Applying $\omega$ to (4.17), we obtain

$\displaystyle\operatorname{ev}\left((-1)^{m-1}\sum_{i=0}^{N-1}\big{[}T_{w(i,m)}\big{(}x_{i-1,1}^{-}\big{)},x_{i}^{-}\big{]}\right)=\varepsilon_{2}\mathbf{1}(-m)+\hbar\omega(R_{m}).$

Now the following theorem has been proved.

######  Theorem 4.18.

Assume $\hbar c=N\varepsilon_{2}$ and $\varepsilon_{2}\neq 0$. Let $R_{m}$
$(m\geq 1)$ be the element of $U(\hat{\mathfrak{sl}}_{N})\subset
Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$ as in Proposition 4.16, and define for
$m\in\mathbb{Z}$,

$\displaystyle
a_{m}=\dfrac{1}{\varepsilon_{2}}\times\begin{cases}\displaystyle\left((-1)^{m-1}\sum_{i=0}^{N-1}\big{[}x_{i}^{+},T_{w(i,m)}\big{(}x_{i-1,1}^{+}\big{)}\big{]}-\hbar
R_{m}\right)&\text{if $m>0$},\\\
\displaystyle\left(\sum_{i=0}^{N-1}\tilde{h}_{i,1}-c-\tfrac{\hbar}{2}c^{2}\right)&\text{if
$m=0$},\\\
\displaystyle\left((-1)^{m-1}\sum_{i=0}^{N-1}\big{[}T_{w(i,-m)}\big{(}x_{i-1,1}^{-}\big{)},x_{i}^{-}\big{]}-\hbar\omega(R_{-m})\right)&\text{if
$m<0$}.\end{cases}$

Then we have $\operatorname{ev}(a_{m})=\mathbf{1}(m)$. In particular, the
image of the evaluation map $\operatorname{ev}$ contains
$U\big{(}\hat{\mathfrak{gl}}_{N}\big{)}$.

######  Corollary 4.19.

Assume $\hbar c=N\varepsilon_{2}$ and $\varepsilon_{2}\neq 0$. Then the pull-
back of an irreducible $\hat{\mathfrak{gl}}_{N}$-module by the evaluation map
$\operatorname{ev}$ is irreducible as a module of
$Y\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$.

######  Remark 4.20.

If $\varepsilon_{2}=0$, we cannot apply our construction. In fact, if we
assume $c\neq 0$, the condition $\hbar c=N\varepsilon_{2}$ and
$\varepsilon_{2}=0$ implies $\varepsilon_{1}=0$. The affine Yangian at
$\varepsilon_{1}=\varepsilon_{2}=0$ is isomorphic to the universal enveloping
algebra of the universal central extension of the Lie algebra
$\mathfrak{sl}_{N}\big{[}s,t^{\pm 1}\big{]}$. Moreover the evaluation map
becomes the genuine evaluation at $s=0$. In this situation, the image of the
evaluation map is $U\big{(}\hat{\mathfrak{sl}}_{N}\big{)}$, and hence it does
not contain the Heisenberg algebra generated by $\mathbf{1}(m)$
($m\in\mathbb{Z}$).

######  Remark 4.21.

We do not know whether the elements $a_{m}$ ($m\in\mathbb{Z}$) satisfy the
Heisenberg relations. The construction of the Heisenberg subalgebra inside the
affine Yangian will be left as a future work.

##  Appendix A General case

We use the notation
$(\operatorname{ad}X)^{(n)}(Y)=(\operatorname{ad}X)^{n}(Y)/n!$ for divided
power operators.

###  A.1 Yangian and braid group action

Let $(a_{ij})_{i,j\in I}$ be a symmetrizable generalized Cartan matrix and fix
integers $(d_{i})_{i\in I}$ such that $(d_{i}a_{ij})_{i,j\in I}$ is symmetric.
We denote by $\mathfrak{g}$ the corresponding Kac–Moody Lie algebra. Then the
Yangian $Y(\mathfrak{g})$ is defined to be generated by $x_{i,r}^{+}$,
$x_{i,r}^{-}$, $h_{i,r}$ ($i\in I$, $r\in\mathbb{Z}_{\geq 0}$) with a
parameter $\hbar\in\mathbb{C}$ subject to the relations:

$\displaystyle[h_{i,r},h_{j,s}]=0,\qquad\big{[}x_{i,r}^{+},x_{j,s}^{-}\big{]}=\delta_{ij}h_{i,r+s},\qquad\big{[}h_{i,0},x_{j,r}^{\pm}\big{]}=\pm
d_{i}a_{ij}x_{j,r}^{\pm},$
$\displaystyle\big{[}h_{i,r+1},x_{j,s}^{\pm}\big{]}-\big{[}h_{i,r},x_{j,s+1}^{\pm}\big{]}=\pm
d_{i}a_{ij}\tfrac{\hbar}{2}\big{\\{}h_{i,r},x_{j,s}^{\pm}\big{\\}},$
$\displaystyle\big{[}x_{i,r+1}^{\pm},x_{j,s}^{\pm}\big{]}-\big{[}x_{i,r}^{\pm},x_{j,s+1}^{\pm}\big{]}=\pm
d_{i}a_{ij}\tfrac{\hbar}{2}\big{\\{}x_{i,r}^{\pm},x_{j,s}^{\pm}\big{\\}},$
$\displaystyle\sum_{w\in\mathfrak{S}_{1-a_{ij}}}\big{[}x_{i,r_{w(1)}}^{\pm},\big{[}x_{i,r_{w(2)}}^{\pm},\dots,\big{[}x_{i,r_{w(1-a_{ij})}}^{\pm},x_{j,s}^{\pm}\big{]}\dots\big{]}\big{]}=0,\qquad
i\neq j.$

Set $x_{i}^{\pm}=x_{i,0}^{\pm}$, $h_{i}=h_{i,0}$. Then the standard Chevalley
generators of $\mathfrak{g}$ are identified with $d_{i}^{-1/2}x_{i}^{+}$,
$d_{i}^{-1/2}x_{i}^{-}$, $d_{i}^{-1}h_{i}$.

Following [6], we define

$\displaystyle
T_{i}=\exp\operatorname{ad}d_{i}^{-1/2}x_{i}^{+}\exp\operatorname{ad}\big{(}{-}d_{i}^{-1/2}x_{i}^{-}\big{)}\exp\operatorname{ad}d_{i}^{-1/2}x_{i}^{+}.$

In the sequel, we put $e_{i}=d_{i}^{-1/2}x_{i}^{+}$,
$f_{i}=d_{i}^{-1/2}x_{i}^{-}$.

######  Proposition A.1.

The operators $\\{T_{i}\\}$ satisfy the braid relations. That is, we have

$\displaystyle T_{i}T_{j}=T_{j}T_{i}\quad\text{if $a_{ij}=0$},\qquad
T_{i}T_{j}T_{i}=T_{j}T_{i}T_{j}\quad\text{if $a_{ij}=-1$},$ $\displaystyle
T_{i}T_{j}T_{i}T_{j}=T_{j}T_{i}T_{j}T_{i}\quad\text{if $a_{ij}=-2$},\qquad
T_{i}T_{j}T_{i}T_{j}T_{i}T_{j}=T_{j}T_{i}T_{j}T_{i}T_{j}T_{i}\quad\text{if
$a_{ij}=-3$}.$

###### Proof.

The identities follow from

$\displaystyle T_{i}(e_{j})=e_{j},\qquad$ $\displaystyle
T_{i}(f_{j})=f_{j},\qquad$ $\displaystyle a_{ij}=0,$ $\displaystyle
T_{i}T_{j}(e_{i})=e_{j},\qquad$ $\displaystyle T_{i}T_{j}(f_{i})=f_{j},\qquad$
$\displaystyle a_{ij}=-1,$ $\displaystyle T_{i}T_{j}T_{i}(e_{j})=e_{j},\qquad$
$\displaystyle T_{i}T_{j}T_{i}(f_{j})=f_{j},\qquad$ $\displaystyle a_{ij}=-2,$
$\displaystyle T_{i}T_{j}T_{i}T_{j}T_{i}(e_{j})=e_{j},\qquad$ $\displaystyle
T_{i}T_{j}T_{i}T_{j}T_{i}(f_{j})=f_{j},\qquad$ $\displaystyle a_{ij}=-3$

as in the proof of Proposition 3.1. ∎

######  Proposition A.2.

We have

$\displaystyle
T_{i}\big{(}x_{j}^{\pm}\big{)}=\begin{cases}-x_{i}^{\mp}&\text{if $i=j$},\\\
\big{(}{\pm}\operatorname{ad}d_{i}^{-1/2}x_{i}^{\pm}\big{)}^{(-a_{ij})}\big{(}x_{j}^{\pm}\big{)}&\text{if
$a_{ij}<0$},\\\ x_{j}^{\pm}&\text{if $a_{ij}=0$},\end{cases}\qquad
T_{i}(h_{j})=\begin{cases}-h_{i}&\text{if $i=j$},\\\
h_{j}-a_{ij}h_{i}&\text{if $a_{ij}<0$},\\\ h_{j}&\text{if
$a_{ij}=0$},\end{cases}$ $\displaystyle
T_{i}\big{(}x_{j,1}^{\pm}\big{)}=\begin{cases}-x_{i,1}^{\mp}+\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{\mp}\big{\\}}&\text{if
$i=j$},\\\
\big{(}{\pm}\operatorname{ad}d_{i}^{-1/2}x_{i}^{\pm}\big{)}^{(-a_{ij})}\big{(}x_{j,1}^{\pm}\big{)}&\text{if
$a_{ij}<0$},\\\ x_{j,1}^{\pm}&\text{if $a_{ij}=0$}.\end{cases}$

###### Proof.

The formulas for $T_{i}\big{(}x_{j}^{\pm}\big{)}$, $T_{i}(h_{j})$ are deduced
from well-known formulas for the Chevalley generators. We produce a
computation of $T_{i}\big{(}x_{j}^{-}\big{)}$ for $a_{ij}<0$ since the case
$T_{i}\big{(}x_{j,1}^{-}\big{)}$ for $a_{ij}<0$ is verified in a very similar
way. Put $m=-a_{ij}$. We have

$\displaystyle
T_{i}(x_{j}^{-})=\exp\operatorname{ad}e_{i}\exp\operatorname{ad}(-f_{i})\big{(}x_{j}^{-}\big{)}=\sum_{n=0}^{m}\dfrac{(-1)^{n}}{n!}\exp\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{n}\big{(}x_{j}^{-}\big{)}.$
(A.1)

We can prove

$\displaystyle\exp\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{n}\big{(}x_{j}^{-}\big{)}=\sum_{k=0}^{n}\binom{n}{k}\left(\prod_{l=k}^{n-1}(m-l)\right)(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}$
(A.2)

by induction on $n$. Indeed, if we assume the assertion for $n$, we have

$\displaystyle\exp\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{n+1}\big{(}x_{j}^{-}\big{)}=\operatorname{ad}\left(\exp\operatorname{ad}e_{i}(f_{i})\right)\big{(}\exp\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{n}\big{(}x_{j}^{-}\big{)}\big{)}$
(A.3)
$\displaystyle\hphantom{\exp\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{n+1}\big{(}x_{j}^{-}\big{)}}{}=\operatorname{ad}\big{(}f_{i}+d_{i}^{-1}h_{i}-e_{i}\big{)}\left(\sum_{k=0}^{n}\binom{n}{k}\left(\prod_{l=k}^{n-1}(m-l)\right)(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}\right).$

Since we have

$\displaystyle\operatorname{ad}h_{i}(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}=d_{i}(-2k+m)(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}$

and

$\displaystyle\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}=\sum_{s=0}^{k-1}(\operatorname{ad}f_{i})^{k-s-1}\big{(}\operatorname{ad}d_{i}^{-1}h_{i}\big{)}(\operatorname{ad}f_{i})^{s}\big{(}x_{j}^{-}\big{)}$
$\displaystyle\hphantom{\operatorname{ad}e_{i}(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}}{}=\sum_{s=0}^{k-1}(-2s+m)(\operatorname{ad}f_{i})^{k-1}\big{(}x_{j}^{-}\big{)}=k(m-k+1)(\operatorname{ad}f_{i})^{k-1}\big{(}x_{j}^{-}\big{)},$

the coefficient of $(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}$ in
(A.3) is

$\displaystyle\binom{n}{k-1}\left(\prod_{l=k-1}^{n-1}(m-l)\right)+\binom{n}{k}\left(\prod_{l=k}^{n-1}(m-l)\right)(m-2k)$
$\displaystyle\qquad{}-\binom{n}{k+1}\left(\prod_{l=k+1}^{n-1}(m-l)\right)(k+1)(m-k).$

It is easy to see that this is equal to

$\displaystyle\binom{n+1}{k}\left(\prod_{l=k}^{n}(m-l)\right).$

Thus (A.2) is proved. Then (A.1) is equal to

$\displaystyle\sum_{n=0}^{m}\dfrac{(-1)^{n}}{n!}\sum_{k=0}^{n}\binom{n}{k}\left(\prod_{l=k}^{n-1}(m-l)\right)(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}$
$\displaystyle\qquad{}=\sum_{k=0}^{m}\dfrac{(-1)^{k}}{k!}(\operatorname{ad}f_{i})^{k}\big{(}x_{j}^{-}\big{)}\times\left(\sum_{n=k}^{m}(-1)^{n-k}\binom{m-k}{n-k}\right)=\dfrac{(-1)^{m}}{m!}(\operatorname{ad}f_{i})^{m}\big{(}x_{j}^{-}\big{)}.$

The formula for $T_{i}\big{(}x_{i,1}^{-}\big{)}$ is proved in a way similar to
Proposition 3.9. We use

$\displaystyle\exp\operatorname{ad}e_{i}\big{(}x_{i,1}^{+}\big{)}=x_{i,1}^{+}-d_{i}^{1/2}\hbar\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}e_{i}\big{(}x_{i,1}^{-}\big{)}=x_{i,1}^{-}+d_{i}^{-1/2}h_{i,1}-x_{i,1}^{+}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}+d_{i}^{1/2}\hbar\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}e_{i}\big{(}\big{\\{}h_{i},x_{i}^{+}\big{\\}}\big{)}=\big{\\{}h_{i},x_{i}^{+}\big{\\}}-4d_{i}^{1/2}\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}e_{i}\big{(}\big{(}x_{i}^{+}\big{)}^{2}\big{)}=\big{(}x_{i}^{+}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}(-f_{i})\big{(}x_{i,1}^{+}\big{)}=x_{i,1}^{+}+d_{i}^{-1/2}h_{i,1}-x_{i,1}^{-}-\tfrac{\hbar}{2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}+d_{i}^{1/2}\hbar\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}(-f_{i})\big{(}x_{i,1}^{-}\big{)}=x_{i,1}^{-}-d_{i}^{1/2}\hbar\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}(-f_{i})(h_{i,1})=h_{i,1}-2d_{i}^{1/2}x_{i,1}^{-}-d_{i}^{1/2}\hbar\big{\\{}h_{i},x_{i}^{-}\big{\\}}+3d_{i}\hbar\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}(-f_{i})\big{(}\big{\\{}h_{i},x_{i}^{+}\big{\\}}\big{)}=\big{\\{}h_{i},x_{i}^{+}\big{\\}}-3\big{\\{}h_{i},x_{i}^{-}\big{\\}}-2d_{i}^{1/2}\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}$
$\displaystyle\hphantom{\exp\operatorname{ad}(-f_{i})\big{(}\big{\\{}h_{i},x_{i}^{+}\big{\\}}\big{)}=}{}+2d_{i}^{-1/2}h_{i}^{2}+4d_{i}^{1/2}\big{(}x_{i}^{-}\big{)}^{2},$
$\displaystyle\exp\operatorname{ad}(-f_{i})\big{(}\big{(}x_{i}^{+}\big{)}^{2}\big{)}=\big{(}x_{i}^{+}\big{)}^{2}+d_{i}^{-1}h_{i}^{2}+\big{(}x_{i}^{-}\big{)}^{2}+d_{i}^{-1/2}\big{\\{}h_{i},x_{i}^{+}\big{\\}}$
$\displaystyle\hphantom{\exp\operatorname{ad}(-f_{i})\big{(}\big{(}x_{i}^{+}\big{)}^{2}\big{)}=}{}-d_{i}^{-1/2}\big{\\{}h_{i},x_{i}^{-}\big{\\}}-\big{\\{}x_{i}^{+},x_{i}^{-}\big{\\}}.$

As we mentioned, $T_{i}\big{(}x_{j,1}^{-}\big{)}$ for $a_{ij}<0$ is computed
by replacing $x_{j}^{-}$ with $x_{j,1}^{-}$ in the argument for
$T_{i}\big{(}x_{j}^{-}\big{)}$. Then the formulas for
$T_{i}\big{(}x_{j,1}^{+}\big{)}$ are obtained from those for
$T_{i}\big{(}x_{j,1}^{-}\big{)}$ by applying $\omega$. ∎

######  Remark A.3.

In this appendix we impose no further condition on $(a_{ij})_{i,j\in I}$, and
hence on $\mathfrak{g}$, to study the braid group action. However the defining
relations of $Y(\mathfrak{g})$ given here may not produce a correct definition
of Yangian for some generalized Cartan matrix. For example, it is known that
for $Y(\hat{\mathfrak{sl}}_{2})$ the defining relations should be modified as
in [9, Definition 5.1] or [2, Section 1.2] (In [9], some relations are
missing). In [7, equation (2.1)], the authors suggest the following condition:

$\displaystyle\text{for all $i$ and $j$, if $a_{ij}\leq-2$ then $a_{ji}=-1$
holds}.$

See [7, Section 2, Lemma 4.2, Remark 5.15] for an explanation on some
evidences.

###  A.2 Compatibility with the coproduct

Assume that $\mathfrak{g}$ is a Kac–Moody Lie algebra of finite or affine type
except for $A_{1}^{(1)}$ and $A_{2}^{(2)}$. Then the coproduct $\Delta$ on
$Y(\mathfrak{g})$ is well-defined by the same formula as in Theorem 2.6 [6,
Definition 4.6, Theorem 4.9, Proposition 5.18]. Let us prove the main result
of this appendix.

######  Proposition A.4.

We have $\Delta\circ T_{i}=(T_{i}\otimes T_{i})\circ\Delta$.

###### Proof.

Lemma 3.13 holds in a general situation and we use it. It is enough to prove
$\Delta T_{i}\big{(}x_{j,1}^{+}\big{)}=(T_{i}\otimes
T_{i})\Delta\big{(}x_{j,1}^{+}\big{)}$ for $a_{ij}<0$ since the proofs
concerning the other generators are the same as in the proof of Proposition
3.12. We claim that the both sides coincide with

$\displaystyle\square\,T_{i}(x_{j,1}^{+})-\hbar\big{(}\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}-d_{i}^{1/2}x_{i}^{+}\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}\big{)}.$

The left-hand side is

$\displaystyle\Delta\big{(}(\operatorname{ad}e_{i})^{(-a_{ij})}\big{(}x_{j,1}^{+}\big{)}\big{)}=(\operatorname{ad}\square(e_{i}))^{(-a_{ij})}\big{(}\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{)}$
$\displaystyle=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\bigg{(}\big{[}(\operatorname{ad}\square(e_{i}))^{(-a_{ij})}\big{(}1\otimes
x_{j}^{+}\big{)},\Omega_{+}\big{]}$
$\displaystyle\quad{}+\sum_{n=1}^{-a_{ij}}\big{[}(\operatorname{ad}\square(e_{i}))^{(-a_{ij}-n)}\big{(}1\otimes
x_{j}^{+}\big{)},(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})\big{]}\bigg{)}$
$\displaystyle=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\bigg{(}\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}+\sum_{n=1}^{-a_{ij}}\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-n)}\big{(}x_{j}^{+}\big{)},(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})\big{]}\bigg{)}.$

We show

$\displaystyle\sum_{n=1}^{-a_{ij}}\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-n)}\big{(}x_{j}^{+}\big{)},(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})\big{]}=-d_{i}^{1/2}x_{i}^{+}\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}.$

Since we have

$\displaystyle(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})=\begin{cases}-d_{i}^{-1/2}x_{i}^{+}\otimes
h_{i}&\text{if $n=1$,}\\\ x_{i}^{+}\otimes x_{i}^{+}&\text{if $n=2$,}\\\
0&\text{if $n\geq 3$,}\end{cases}$

we see

$\displaystyle\begin{split}&\sum_{n=1}^{-a_{ij}}\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-n)}\big{(}x_{j}^{+}\big{)},(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})\big{]}\\\
&\qquad{}=\big{[}1\otimes x_{j}^{+},-d_{i}^{-1/2}x_{i}^{+}\otimes
h_{i}\big{]}=-d_{i}^{1/2}x_{i}^{+}\otimes x_{j}^{+}\end{split}$

if $a_{ij}=-1$, and

$\displaystyle\sum_{n=1}^{-a_{ij}}\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-n)}\big{(}x_{j}^{+}\big{)},(\operatorname{ad}\square(e_{i}))^{(n)}(\Omega_{+})\big{]}$
$\displaystyle\qquad{}=\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)},-d_{i}^{-1/2}x_{i}^{+}\otimes
h_{i}\big{]}+\big{[}1\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-2)}\big{(}x_{j}^{+}\big{)},x_{i}^{+}\otimes
x_{i}^{+}\big{]}$
$\displaystyle\qquad{}=d_{i}^{-1/2}\big{(}2d_{i}(-a_{ij}-1)+d_{i}a_{ij}\big{)}x_{i}^{+}\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}$
$\displaystyle\qquad\quad{}-d_{i}^{1/2}(-a_{ij}-1)x_{i}^{+}\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}$
$\displaystyle\qquad{}=-d_{i}^{1/2}x_{i}^{+}\otimes(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}$

if $a_{ij}\leq-2$, hence the claim holds. The right-hand side is

$\displaystyle(T_{i}\otimes
T_{i})\big{(}\square\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
x_{j}^{+},\Omega_{+}\big{]}\big{)}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\big{(}T_{i}\otimes
T_{i}\big{)}\Omega_{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}+x_{i}^{+}\otimes
x_{i}^{-}-x_{i}^{-}\otimes x_{i}^{+}\big{]}$
$\displaystyle\qquad{}=\square\,T_{i}\big{(}x_{j,1}^{+}\big{)}-\hbar\big{(}\big{[}1\otimes
T_{i}\big{(}x_{j}^{+}\big{)},\Omega_{+}\big{]}+x_{i}^{+}\otimes\big{[}T_{i}\big{(}x_{j}^{+}\big{)},x_{i}^{-}\big{]}-x_{i}^{-}\otimes\big{[}T_{i}(x_{j}^{+}),x_{i}^{+}\big{]}\big{)}.$

Then the desired identity follows from

$\displaystyle\big{[}T_{i}(x_{j}^{+}),x_{i}^{-}\big{]}=\left[\dfrac{1}{(-a_{ij})!}(\operatorname{ad}e_{i})^{-a_{ij}}\big{(}x_{j}^{+}\big{)},d_{i}^{1/2}f_{i}\right]$
$\displaystyle\hphantom{\big{[}T_{i}(x_{j}^{+}),x_{i}^{-}\big{]}}{}=\dfrac{1}{(-a_{ij})!}d_{i}^{1/2}\sum_{n=0}^{-a_{ij}-1}(\operatorname{ad}e_{i})^{-a_{ij}-n-1}\big{(}\operatorname{ad}d_{i}^{-1}h_{i}\big{)}(\operatorname{ad}e_{i})^{n}\big{(}x_{j}^{+}\big{)}$
$\displaystyle\hphantom{\big{[}T_{i}(x_{j}^{+}),x_{i}^{-}\big{]}}{}=\dfrac{1}{(-a_{ij})!}d_{i}^{1/2}\sum_{n=0}^{-a_{ij}-1}(2n+a_{ij})(\operatorname{ad}e_{i})^{-a_{ij}-1}\big{(}x_{j}^{+}\big{)}$
$\displaystyle\hphantom{\big{[}T_{i}(x_{j}^{+}),x_{i}^{-}\big{]}}{}=\dfrac{1}{(-a_{ij})!}d_{i}^{1/2}a_{ij}(\operatorname{ad}e_{i})^{-a_{ij}-1}\big{(}x_{j}^{+}\big{)}=-d_{i}^{1/2}(\operatorname{ad}e_{i})^{(-a_{ij}-1)}\big{(}x_{j}^{+}\big{)}$

and

$\displaystyle\big{[}T_{i}(x_{j}^{+}),x_{i}^{+}\big{]}=\left[\dfrac{1}{(-a_{ij})!}(\operatorname{ad}e_{i})^{-a_{ij}}\big{(}x_{j}^{+}\big{)},d_{i}^{1/2}e_{i}\right]=-\dfrac{1}{(-a_{ij})!}d_{i}^{1/2}(\operatorname{ad}e_{i})^{-a_{ij}+1}\big{(}x_{j}^{+}\big{)}=0.\\!\\!\\!\\!$
∎

### Acknowledgments

The author would like to thank Yoshihisa Saito for suggesting him to study
braid group action on affine Yangian. Discussions with him improved contents
of this paper. He also thanks Nicolas Guay for telling him the reference [3]
and the referees for helpful comments. This work was supported by JSPS KAKENHI
Grant Number 26287004, 17H06127, 18K13390, and The Kyoto University
Foundation.

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