11institutetext: Department of Mathematics, University of York, York YO10 5DD,
United Kingdom

# Yangian of the periplectic Lie superalgebra

Maxim Nazarov

###### Abstract

We study in detail the Yangian of the periplectic Lie superalgebra. For this
Yangian we verify an analogue of the Poincaré -Birkhoff -Witt Theorem.
Moreover we introduce a family of free generators of the centre of this
Yangian.

###### Keywords:

Hopf algebra, Lie superalgebra, Yangian

Subject Classifications 16T20, 17A70, 17B37

### 1

Let $\mathfrak{gl}_{{\hskip 0.5pt}M|N}$ be the general linear Lie superalgebra
over the complex field ${\mathbb{C}}\,$. We will assume that both $M$ and $N$
are positive integers. The Yangian of $\mathfrak{gl}_{{\hskip 0.5pt}M|N}$ was
introduced in N1 (18) by extending the definition of the Yangian
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip
0.5pt}N}{\hskip-0.5pt})$ of the Lie algebra $\mathfrak{gl}_{{\hskip
0.5pt}N}\,$, see for instance N4 (21). The Yangian of $\mathfrak{gl}_{{\hskip
0.5pt}M|N}$ is a deformation of the universal enveloping algebra of the
polynomial current Lie superalgebra $\mathfrak{gl}_{{\hskip 0.5pt}M|N}[u]$ in
the class of Hopf algebras, see N5 (22) for further details.

Let $\mathfrak{p}_{N}$ be the periplectic Lie superalgebra over
${\mathbb{C}}\,$. This is a subalgebra of $\mathfrak{gl}_{{\hskip 0.5pt}N|N}$
preserving a non-degenerate odd symmetric bilinear form on the
$\mathbb{Z}_{2}$-graded vector space ${\mathbb{C}}^{{\hskip 0.5pt}N|N}$. The
Lie superalgebra $\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ acts on
${\mathbb{C}}^{{\hskip 0.5pt}N|N}$ in the natural way. Equivalently, the Lie
superalgebra $\mathfrak{p}_{N}$ can be defined as a fixed point subalgebra of
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ with respect to a certain involutive
automorphism. This automorphism is denoted by $\omega\,$, see Section 3 for an
explicit definition of $\mathfrak{p}_{N}$ in these terms.

The corresponding Yangian $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ was briefly introduced in N2 (19). Note
that the construction of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ therein involved the periplectic Brauer
algebra which was later studied in by other authors, see for instance CP (6,
7, 12, 17).

According to a general scheme of D (9), the Yangian of the Lie superalgebra
$\mathfrak{p}_{N}$ cannot be introduced as a deformation of the universal
enveloping algebra of the polynomial current Lie superalgebra
$\mathfrak{p}_{N}[u]$ in the class of Hopf algebras. This is because by S (23)
the only supersymmetric $\mathfrak{p}_{N}$-invariant element of
$\mathfrak{p}_{N}^{{\hskip 0.5pt}\otimes 2}$ is zero. Consequently, there is
no natural Lie co-superalgebra structure on $\mathfrak{p}_{N}[u]\,$.

Remarkably, there is a natural Lie co-superalgebra structure on the twisted
polynomial current Lie superalgebra $\mathfrak{g}=\\{\,g{\hskip
0.5pt}(u)\in\mathfrak{gl}_{{\hskip 0.5pt}N|N}[u]:\omega{\hskip 0.5pt}({\hskip
0.5pt}g{\hskip 0.5pt}(u))=g{\hskip 0.5pt}(-{\hskip 0.5pt}u){\hskip
0.5pt}\\}\,.$ The Yangian $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is a deformation of the universal
enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})$ in the class of Hopf algebras. See N2 (19)
for further discussion of this fact.

The definition of the Yangian $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is given in our Section 10. It is based
on a new solution of the quantum Yang-Baxter equation constructed in N2 (19).
This solution is a rational function of two variables $u\,,{\hskip-0.5pt}v$
with values in the periplectic Brauer algebra. Unlike other rational solutions
discovered before N2 (19) this is not a function of only the difference $u-v$
of the variables, see our Section 5.

Our Theorem 1 is stated in N2 (19) without proof. We prove it here. We also
prove an analogue for $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of the Poincaré -Birkhoff -Witt Theorem
for $\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})\,$, see MM
(15). This analogue is our Theorem 2 which was also stated in N2 (19) without
proof. We sketch its proof here. Its Corollary 1 describes
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ as a vector
space explicitly.

Unlike its analogues AMR (2) for the Yangians of the symplectic and orthogonal
Lie algebras, Theorem 2 cannot be proved by using the method of N3 (20). The
reason for it is explained in our Section 11. We use the Diamond Lemma of B
(4) instead.

The Lie superalgebra $\mathfrak{p}_{N}$ is a subalgebra of $\mathfrak{g}\,$.
It consists of all those elements of $\mathfrak{g}$ which do not depend on the
variable $u\,$. So the universal enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is a Hopf subalgebra of
$\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})\,$. We also have
an embedding of $\operatorname{U}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip
0.5pt})$ to the Yangian $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ as a Hopf subalgebra, see Corollary 2
to our Theorem 2.

There also is a subalgebra of $\mathfrak{g}$ isomorphic to
$\mathfrak{gl}_{{\hskip 0.5pt}N}[u]\,$. It consists of all
$g(u)\in\mathfrak{g}$ taking values only in the even part of
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}\,$. Hence there is a Hopf subalgebra of
$\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip 0.5pt})$ isomorphic to
$\operatorname{U}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip 0.5pt}N}[u]{\hskip
0.5pt})\,$. We have an embedding of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{gl}_{{\hskip 0.5pt}N}{\hskip-0.5pt})$ to
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ as an
associative subalgebra, see Corollary 3. But it is not a Hopf algebra
embedding.

Our Theorem 3 yields an explicit description of the centre of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. We use an
analogue for $\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
of the quantum Berezinian N1 (18) for the Yangian of $\mathfrak{gl}_{{\hskip
0.5pt}M|N}\,$.

Irreducible finite-dimensional representations of the Lie superalgebra
$\mathfrak{p}_{N}$ were recently studied by several authors, see for instance
AGG (1, 3, 5, 8, 10, 11). It would be very interesting to study irreducible
finite-dimensional representations of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip 0.5pt}$. For the Yangian of
$\mathfrak{gl}_{{\hskip 0.5pt}M|N}$ this was done in Z (25) and more recently
in L (13, 14, 24).

### 2

We shall use the following general conventions. Let $\mathrm{A}$ and
$\mathrm{B}$ be associative $\mathbb{Z}_{{\hskip 0.5pt}2}$-graded algebras.
Their tensor product $\mathrm{A}\otimes\mathrm{B}$ is also an associative
$\mathbb{Z}_{2}$-graded algebra such that for any homogeneous elements
$X,X^{\prime}\in\mathrm{A}$ and $Y,Y^{\prime}\in\mathrm{B}$

$\displaystyle(X\otimes Y){\hskip 0.5pt}(X^{\prime}\otimes Y^{\prime})$
$\displaystyle=X{\hskip 0.5pt}X^{\prime}\otimes Y{\hskip
0.5pt}Y^{\prime}\,(-1)^{{\hskip 0.5pt}\deg X^{\prime}\deg Y},$
$\displaystyle\deg{\hskip 0.5pt}(X\otimes Y)$ $\displaystyle=\deg X+\deg
Y{\hskip 0.5pt}.$

Furthermore, for any two $\mathbb{Z}_{2}$-graded modules $U$ and $V$ over
$\mathrm{A}$ and $\mathrm{B}$ respectively, the vector space $U\otimes V$ is a
$\mathbb{Z}_{2}$-graded module over $\mathrm{A}\otimes\mathrm{B}$ such that
for any homogeneous elements $x\in U$ and $y\in V$

$\displaystyle(X\otimes Y){\hskip 0.5pt}({\hskip 0.5pt}x\otimes y{\hskip
0.5pt})$ $\displaystyle=X{\hskip 0.5pt}x\otimes Y{\hskip
0.5pt}y\,(-1)^{{\hskip 0.5pt}\deg x\,\deg Y},$ (1) $\displaystyle\deg\hskip
0.5pt({\hskip 0.5pt}x\otimes y{\hskip 0.5pt})$ $\displaystyle=\deg x+\deg
y\,.$ (2)

A homomorphism $\alpha:\mathrm{A}\to\mathrm{B}$ is a linear map such that
$\alpha{\hskip 0.5pt}(X\,X^{\prime})=\alpha{\hskip 0.5pt}(X)\,\alpha{\hskip
0.5pt}(X^{\prime})$ for all $X,X^{\prime}\in\mathrm{A}\,$. But an
antihomomorphism $\beta:\mathrm{A}\to\mathrm{B}$ is a linear map such that for
all homogeneous $X,X^{\prime}\in\mathrm{A}$

$\beta{\hskip 0.5pt}(X{\hskip 0.5pt}X^{\prime}{\hskip 0.5pt})=\beta{\hskip
0.5pt}(X^{\prime})\,\beta{\hskip 0.5pt}(X)\,(-1)^{{\hskip 0.5pt}\deg X\deg
X^{\prime}}.$ (3)

Let $n$ be any positive integer. If the algebra $\mathrm{A}$ is unital, let
$\iota_{{\hskip 0.5pt}p}$ be its embedding into the tensor product
$\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip 0.5pt}n}$ as the $p{\hskip
0.5pt}$-th tensor factor :

$\iota_{{\hskip 0.5pt}p}{\hskip 0.5pt}(X)=1^{\otimes{\hskip
0.5pt}(p-1)}\otimes X\otimes 1^{\otimes{\hskip
0.5pt}(n-p)}\quad\text{for}\quad p=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}n\,.$

We will also use various embeddings of
$\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip 0.5pt}m}$ into
$\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip 0.5pt}n}$ for $m=1{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}n$. For any choice of
$m$ pairwise distinct indices $p_{1}{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}p_{m}\in\\{{\hskip 0.5pt}1{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}n{\hskip 0.5pt}\\}$ and of an element
$X\in\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip 0.5pt}m}$ of the form
$X=X^{(1)}\otimes\ldots\otimes X^{(m)}$ we will denote

$X_{{\hskip 0.5pt}p_{1}\ldots{\hskip 0.5pt}p_{m}}={\hskip 0.5pt}\iota_{{\hskip
0.5pt}p_{1}}(X^{(1)}){\hskip 0.5pt}\ldots\,{\hskip 0.5pt}\iota_{{\hskip
0.5pt}p_{m}}(X^{(m)})\in\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip 0.5pt}n}.$

We will then extend the notation $X_{{\hskip 0.5pt}p_{1}\ldots{\hskip
0.5pt}p_{m}}$ to all elements $X\in\mathrm{A}^{{\hskip-0.5pt}\otimes{\hskip
0.5pt}m}$ by linearity.

### 3

Let the indices $i\,,j\,,k\,,l$ run through ${\hskip 0.5pt}\pm 1{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}\pm N\,$. Put
${\bar{\imath}\,}=0$ if $i>0$ and ${\bar{\imath}\,}=1$ if $i<0\,$. Now
consider the $\mathbb{Z}_{{\hskip 0.5pt}2}$-graded vector space
${\mathbb{C}}^{{\hskip 0.5pt}N|N}{\hskip 0.5pt}$. Let
$e_{i}\in{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ be an element of the standard
basis. The $\mathbb{Z}_{{\hskip 0.5pt}2}$-grading on ${\mathbb{C}}^{{\hskip
0.5pt}N|N}$ is defined by $\deg e_{i}={\bar{\imath}\,}\,$. We will be using
the bilinear form $\langle\ ,\,\rangle$ on ${\mathbb{C}}^{{\hskip 0.5pt}N|N}$
defined by setting $\langle\,e_{i},e_{j}\,\rangle=\delta_{\,i{\hskip
0.5pt},{\hskip 0.5pt}-j}$ for any indices $i$ and $j\,$. This form is clearly
symmetric.

Let $E_{{\hskip 0.5pt}ij}\in\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ be the standard matrix unit, defined by
$E_{{\hskip 0.5pt}ij}\,e_{{\hskip 0.5pt}k}=\delta_{{\hskip
0.5pt}jk}\,e_{{\hskip 0.5pt}i}\,$. The associative algebra
$\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ is
$\mathbb{Z}_{{\hskip 0.5pt}2}$-graded so that $\deg E_{{\hskip
0.5pt}ij}={\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}\,$. Hence
${\mathbb{C}}^{{\hskip 0.5pt}N|N}$ is a $\mathbb{Z}_{2}$-graded module over
$\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$. For any
$n$ we can also identify the tensor product $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}n}$ with the
algebra $\operatorname{End}\hskip 1.0pt(({\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}n})$ acting on the vector space
$({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}n}$ by repeatedly
using the conventions (1) and (2).

We can define an antiautomorphism $\tau$ of $\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ by mapping

$\tau:\,E_{{\hskip 0.5pt}ij}\mapsto E_{{\hskip
0.5pt}ji}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}}\,.$

Note that $\tau$ is not involutive, while $\tau^{{\hskip 0.5pt}2}$ is the
parity automorphism of $\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N}$

$E_{{\hskip 0.5pt}ij}\mapsto E_{{\hskip
0.5pt}ij}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}+{\hskip
0.5pt}{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (4)

We can also define an involutive automorphism $\pi$ of
$\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ by mapping

$\pi:\,E_{{\hskip 0.5pt}ij}\mapsto E_{\,-i{\hskip 0.5pt},{\hskip 0.5pt}-j}\,.$

The compositions $\tau\,\pi$ and $\pi\,\tau$ again differ by the automorphism
(4). Hence both $\tau\,\pi$ and $\pi\,\tau$ are involutive antiautomorphisms
of $\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$.
Explicitly,

$\tau\,\pi:\,E_{{\hskip 0.5pt}ij}\mapsto E_{\,-j{\hskip 0.5pt},{\hskip
0.5pt}-i}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$

Consider the Lie superalgebra $\mathfrak{gl}_{{\hskip 0.5pt}N|N}\,$. To avoid
confusion, denote by $e_{{\hskip 0.5pt}ij}$ the element of
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ corresponding to $E_{{\hskip
0.5pt}ij}\in\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$.
Then $\deg e_{{\hskip 0.5pt}ij}={\bar{\imath}\,}+{\hskip
0.5pt}{\bar{\jmath}\,}$ and

$[\,e_{{\hskip 0.5pt}ij}\,,e_{{\hskip 0.5pt}kl}\,]=\delta_{jk}\,e_{{\hskip
0.5pt}il}-\delta_{{\hskip 0.5pt}li}\,e_{{\hskip 0.5pt}kj}\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}\,.$ (5)

It follows that both $\pi$ and $-\,\tau$ are automorphisms of the Lie
superalgebra $\mathfrak{gl}_{{\hskip 0.5pt}N|N}\,$. Let $\omega=-\,\tau\,\pi$
be their composition. The automorphism $\omega$ is involutive. Further,

$\langle\,Z\,x{\hskip 0.5pt},y\,\rangle+\langle\,x{\hskip 0.5pt},\omega{\hskip
0.5pt}(Z)\,y\,\rangle(-1)^{{\hskip 0.5pt}\deg x\deg Z}=0$

for any homogeneous $x{\hskip 0.5pt},y\in{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ and
$Z\in\mathfrak{gl}_{{\hskip 0.5pt}N|N}\,$. The Lie superalgebra
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ acts on the vector space
${\mathbb{C}}^{{\hskip 0.5pt}N|N}$ via the above identification with
$\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$.

Now the periplectic Lie superalgebra $\mathfrak{p}_{N}$ is the fixed point
subalgebra of $\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ with respect to the
automorphism $\omega\,$. It can also be defined as the subalgebra of
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ preserving the form $\langle\ ,\,\rangle$
on ${\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$. This subalgebra is spanned by the
elements

$f_{{\hskip 0.5pt}ij}=e_{{\hskip 0.5pt}ij}+\omega{\hskip 0.5pt}(e_{{\hskip
0.5pt}ij})=e_{{\hskip 0.5pt}ij}-e_{{\hskip 0.5pt}-j{\hskip 0.5pt},{\hskip
0.5pt}-i}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (6)

### 4

Take the element of the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}$

$P=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes E_{{\hskip
0.5pt}ji}\,(-1)^{\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (7)

It acts on the vector space $({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}2}$ so that $e_{i}\otimes e_{j}\mapsto e_{j}\otimes
e_{i}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}}{\hskip
0.5pt}.$ Here we identify the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}$ with the
algebra $\operatorname{End}\hskip 1.0pt(({\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}2})$ using (1). Note that
$P^{{\hskip 0.5pt}2}=1\,$. Also note that $({\hskip
0.5pt}\tau\,\pi\otimes\tau\,\pi{\hskip 0.5pt})(P)=-\,P\,.$ Now let

$Q=(-\,\tau\,\pi\otimes{\mathrm{id}}\,)(P)=(\,{\mathrm{id}}\otimes\tau\,\pi{\hskip
0.5pt})(P)\,.$ (8)

Explicitly,

$Q=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes E_{\,-i{\hskip 0.5pt},{\hskip
0.5pt}-j}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+{\hskip 0.5pt}{\bar{\imath}\,}{\hskip 0.5pt}+{\hskip
0.5pt}{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (9)

The image of the action of $Q$ on $({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}2}$ is one-dimensional and is spanned by

$\sum_{i}\,e_{{\hskip 0.5pt}i}\otimes e_{\,-{\hskip
0.5pt}i}\,{(-1)}^{\,{\bar{\imath}\,}}\,.$ (10)

Here we regard $Q$ as an element of $\operatorname{End}\hskip
1.0pt(({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}2})$ by once again identifying the latter algebra with
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}2}$ using (1). Also note that

$P\,Q=-\,Q\,,\quad Q\,P=Q\quad\text{and}\quad Q^{{\hskip 0.5pt}2}=0\,.$ (11)

By (M, 17, Theorem 4.5) the supercommutant of the image of $\mathfrak{p}_{N}$
in $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}n}$ for any $n$ is generated by all elements
$P_{\,pq}$ and $Q_{\,pq}$ where $1\leqslant p<q\leqslant n\,$. Here we use the
standard comultplication on the universal enveloping algebra of
$\mathfrak{p}_{N}\,$.

### 5

Consider a function of complex variables $u\,,{\hskip-0.5pt}v$ with values in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}$

$R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)=1-\frac{P}{u-v}+\frac{Q}{u+v}\,.$

By (8) we have

$\displaystyle(\tau\,\pi\otimes 1)(R(u\,,{\hskip-0.5pt}v))$
$\displaystyle=R(u\,,-{\hskip 0.5pt}v)\,,$ (12)
$\displaystyle(1\otimes\tau\,\pi)(R(u\,,{\hskip-0.5pt}v))$
$\displaystyle=R(-{\hskip 0.5pt}u\,,{\hskip-0.5pt}v)\,.$ (13)

We also have

$R{\hskip 0.5pt}(u{\hskip 0.5pt},{\hskip-0.5pt}v)\,R{\hskip 0.5pt}(-{\hskip
0.5pt}u\,,-{\hskip 0.5pt}v)=1-\frac{1}{(u-v)^{{\hskip 0.5pt}2}}\,.$ (14)

Indeed, due to the relation $P^{{\hskip 0.5pt}2}=1$ and to (11)

$\displaystyle\left(1-\frac{P}{u-v}+\frac{Q}{u+v}\right)\left(1+\frac{P}{u-v}-\frac{Q}{u+v}\right)=$
$\displaystyle 1-\frac{P^{{\hskip 0.5pt}2}}{(u-v)^{{\hskip
0.5pt}2}}+\frac{P\,Q+Q\,P}{(u-v)(u+v)}-\frac{Q^{{\hskip
0.5pt}2}}{(u+v)^{{\hskip 0.5pt}2}}=1-\frac{1}{(u-v)^{{\hskip 0.5pt}2}}\,.$

###### Theorem 1

The rational function $R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ satisfies the
Yang-Baxter equation in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}({\hskip
0.5pt}u,v,w{\hskip 0.5pt})$

$R_{{\hskip 0.5pt}12}(u\,,{\hskip-0.5pt}v){\hskip 0.5pt}{\hskip
0.5pt}R_{{\hskip 0.5pt}13}(u\,,{\hskip-0.5pt}w){\hskip 0.5pt}{\hskip
0.5pt}R_{{\hskip 0.5pt}23}(v\,,{\hskip-0.5pt}w)=R_{{\hskip
0.5pt}23}(v\,,{\hskip-0.5pt}w){\hskip 0.5pt}{\hskip 0.5pt}R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}w){\hskip 0.5pt}{\hskip 0.5pt}R_{{\hskip
0.5pt}12}(u\,,{\hskip-0.5pt}v)\,.$ (15)

###### Proof

Using the definition of $R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ the equality in
(15) will follow from the relations in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}$ displayed
below :

$\displaystyle P_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}13}=P_{{\hskip
0.5pt}23}\,P_{{\hskip 0.5pt}12}=P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}23}\,,$
(16) $\displaystyle P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}12}=P_{{\hskip
0.5pt}12}\,P_{{\hskip 0.5pt}23}=P_{{\hskip 0.5pt}23}\,P_{{\hskip 0.5pt}13}\,,$
(17) $\displaystyle Q_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}=-\,P_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}12}=-\,Q_{{\hskip 0.5pt}13}\,P_{{\hskip
0.5pt}23}\,,$ (18) $\displaystyle Q_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}13}=-\,Q_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}23}=-\,P_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}13}\,,$ (19) $\displaystyle Q_{{\hskip
0.5pt}12}\,P_{{\hskip 0.5pt}13}=-\,Q_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}23}=-\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}\,,$ (20)
$\displaystyle P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}=-\,Q_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}12}=-\,Q_{{\hskip 0.5pt}23}\,P_{{\hskip
0.5pt}13}\,,$ (21) $\displaystyle P_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}13}=Q_{{\hskip 0.5pt}23}\,P_{{\hskip 0.5pt}12}=Q_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}13}\,,$ (22) $\displaystyle Q_{{\hskip
0.5pt}13}\,P_{{\hskip 0.5pt}12}=P_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}23}=Q_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}\,,$ (23) $\displaystyle
P_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}23}=P_{{\hskip
0.5pt}23}\,P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}12}\,,$ (24) $\displaystyle
P_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}=Q_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}12}\,,$ (25) $\displaystyle
Q_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}23}=P_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}\,,$ (26) $\displaystyle
Q_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}=0\,,$ (27)
$\displaystyle Q_{{\hskip 0.5pt}23}\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip
0.5pt}12}=0\,,$ (28) $\displaystyle P_{{\hskip 0.5pt}12}\,P_{{\hskip
0.5pt}13}\,Q_{{\hskip 0.5pt}23}=P_{{\hskip 0.5pt}23}\,Q_{{\hskip
0.5pt}13}\,P_{{\hskip 0.5pt}12}=Q_{{\hskip 0.5pt}12}\,P_{{\hskip
0.5pt}13}\,P_{{\hskip 0.5pt}23}=-\,Q_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}13}\,Q_{{\hskip 0.5pt}23}\,,$ (29) $\displaystyle Q_{{\hskip
0.5pt}23}\,P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}12}=P_{{\hskip
0.5pt}12}\,Q_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}23}=P_{{\hskip
0.5pt}23}\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}=-\,Q_{{\hskip
0.5pt}23}\,Q_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}\,.$ (30)

To prove (15), the relations (16) and (17) are used along with the identity

$\frac{1}{u-v}\,\frac{1}{u-w}-\frac{1}{u-v}\,\frac{1}{v-w}+\frac{1}{u-w}\,\frac{1}{v-w}=0$
(31)

which is easy to verify. The relations (18) and (19) are used with the
identity

$\frac{1}{u+v}\,\frac{1}{u+w}+\frac{1}{u+v}\,\frac{1}{v-w}-\frac{1}{u+w}\,\frac{1}{v-w}=0$

which is obtained from (31) by changing the sign of $u\,$. The relations (20)
and (21) are used with the identity

$\frac{1}{u+v}\,\frac{1}{u-w}+\frac{1}{u+v}\,\frac{1}{v+w}-\frac{1}{u-w}\,\frac{1}{v+w}=0$

which is obtained from (31) by changing the sign of $v\,$. The relations (22)
and (23) are used with the identity

$\frac{1}{u-v}\,\frac{1}{u+w}-\frac{1}{u-v}\,\frac{1}{v+w}+\frac{1}{u+w}\,\frac{1}{v+w}=0$

which is obtained from (31) by changing the sign of $w\,$. Finally, the
relations (29) and (30) are used along with another identity which is easy to
verify :

$\displaystyle\frac{1}{u-v}\,\frac{1}{u-w}\,\frac{1}{v+w}$
$\displaystyle-\frac{1}{u-v}\,\frac{1}{u+w}\,\frac{1}{v-w}\,{\hskip 0.5pt}+$
$\displaystyle\frac{1}{u+v}\,\frac{1}{u-w}\,\frac{1}{v-w}$
$\displaystyle-\frac{1}{u+v}\,\frac{1}{u+w}\,\frac{1}{v+w}=0\,.$

Let us verify the relations (16) to (23). The relations (16) and (17) follow
from the description of the action of $P$ on the vector space
$({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}2}$ as
provided in the beginning of Section 4. Due to (8), the relations (18) and
(19) can be obtained by applying the antiautomorphism $\tau\,\pi$ of
$\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$ to the
relations (16) and (17) relative to the first tensor factor of
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}$. Similarly, the relations (20) and (21)
can be obtained by applying $\tau\,\pi$ to the relations (16) and (17)
relative to the second tensor factor of $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}$. The
relations (22) and (23) can be obtained by applying $\tau\,\pi$ to the
relations (16) and (17) relative to the third tensor factor of
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}$.

Next let us verify the relations (24) to (30). The above mentioned description
of the action of $P$ on $({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}2}$ shows that either side of (24) equals
$P_{{\hskip 0.5pt}13}\,$. This description also implies (25) and (26). It also
implies that

$Q_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}=Q_{{\hskip
0.5pt}12}\,Q_{{\hskip 0.5pt}21}\,P_{{\hskip 0.5pt}13}=Q_{{\hskip
0.5pt}12}\,P_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}21}\,P_{{\hskip
0.5pt}13}=Q_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}12}\,P_{{\hskip
0.5pt}12}\,P_{{\hskip 0.5pt}13}=0$

where we also used (11). So we get the relation (27). Similarly, we get (28) :

$Q_{{\hskip 0.5pt}23}\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}12}=P_{{\hskip
0.5pt}13}\,Q_{{\hskip 0.5pt}21}\,Q_{{\hskip 0.5pt}12}=-\,P_{{\hskip
0.5pt}13}\,Q_{{\hskip 0.5pt}21}\,P_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}12}=-\,P_{{\hskip 0.5pt}13}\,P_{{\hskip 0.5pt}12}\,Q_{{\hskip
0.5pt}12}\,Q_{{\hskip 0.5pt}12}=0\,.$

Also similarly, each of the first three products in (29) is equal to
$-\,P_{{\hskip 0.5pt}13}\,Q_{{\hskip 0.5pt}23}\,$. By using the first equality
in (19) and then the second equality in (20),

$-\,Q_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}13}\,Q_{{\hskip
0.5pt}23}=Q_{{\hskip 0.5pt}12}\,P_{{\hskip 0.5pt}23}\,Q_{{\hskip
0.5pt}23}=-\,Q_{{\hskip 0.5pt}12}\,Q_{{\hskip 0.5pt}23}=-\,P_{{\hskip
0.5pt}13}\,Q_{{\hskip 0.5pt}23}\,.$

Hence we get the last equality in (29). The proofs of all three equalities in
(30) are very similar to those in (29) and are omitted here. ∎

### 6

We will first define the extended Yangian of the Lie superalgebra
$\mathfrak{p}_{N}\,$. This is a complex associative unital algebra
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ with a
countable set of generators

$T^{{\hskip 0.5pt}(r)}_{ij}\quad\text{where}\quad
r=1,2,\,\ldots\quad\text{and}\quad i\,,j=\pm\,1{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}\\!\pm N\,.$ (32)

The algebra $\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
is $\mathbb{Z}_{{\hskip 0.5pt}2}$-graded so that $\deg T_{ij}^{{\hskip
0.5pt}(r)}={\bar{\imath}\,}+{\bar{\jmath}\,}$ for all $r\,$. To write down
defining relations for the generators (32) of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ we will use the formal power series in
$u^{-1}$ with coefficients from $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$T_{ij}(u)=\delta_{{\hskip 0.5pt}ij}\cdot 1+T_{ij}^{{\hskip 0.5pt}(1)}{\hskip
0.5pt}u^{-1}+T_{ij}^{{\hskip 0.5pt}(2)}{\hskip 0.5pt}u^{-2}+\ldots\,.$ (33)

Now combine all the series (33) into a single element

$T(u)=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes
T_{ij}(u)\in(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip 0.5pt}[[u^{-1}]]\,.$ (34)

For any $n$ and any $p=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}n$ we will denote

$T_{p}(u)=(\iota_{p}\otimes{\mathrm{id}}){\hskip 0.5pt}({\hskip
0.5pt}T(u))\in(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}n}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,[[u^{-1}]]\,.$ (35)

By using this notation for $n=2$ the defining relations of
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ are

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,{\hskip
0.5pt}T_{1}(u)\,T_{2}(v)=T_{2}(v)\,T_{1}(u)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,.$ (36)

Namely, after multiplying each side of (36) by $u^{{\hskip 0.5pt}2}-v^{{\hskip
0.5pt}2}$ it becomes an equality of formal Laurent series in $u^{-1},v^{-1}$
with coefficients in $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. By expanding the relation (36) in
the basis of $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}$ consisting of

$E_{{\hskip 0.5pt}ij}\otimes E_{{\hskip 0.5pt}kl}\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}\quad\text{where}\quad
i\,,j\,,k\,,l=\pm\,1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}\\!\pm N$

while using the definitions (7) and (9), we get a collection of relations

$\displaystyle[\,T_{ij}(u){\hskip
0.5pt},T_{kl}(v)\,]=\frac{\,T_{kj}(u)\,T_{il}(v)-T_{kj}(v)\,T_{il}(u)}{u-v}\,(-1)^{{\hskip
0.5pt}\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{k}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{l}\,}\,+\,{\bar{k}\,}{\hskip
0.5pt}{\bar{l}\,}}$ $\displaystyle-\,\,\delta_{{\hskip 0.5pt}-i,{\hskip
0.5pt}k}\,\sum_{h}\,\frac{\,T_{{\hskip 0.5pt}hj}(u)\,T_{{\hskip
0.5pt}-h,{\hskip 0.5pt}l}(v)}{u+v}\,(-1)^{{\hskip
0.5pt}\,{\bar{h}\,}{\bar{l}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{l}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}}$
$\displaystyle+\,\,\delta_{{\hskip
0.5pt}j,-l}\,\,\sum_{h}\,\frac{\,T_{k,-h}(v)\,T_{ih}(u)}{u+v}\,(-1)^{{\hskip
0.5pt}\,{\bar{h}\,}{\bar{k}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{l}\,}+\,{\bar{h}\,}}$ (37)

where $h=\pm 1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}\pm N$ and the square brackets stand for the supercommutator.

By comparing (15) with (36), we obtain that for any $t\in{\mathbb{C}}$ the
assignment

$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip
0.5pt}[[u^{-1}]]\to(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}2}{\hskip 0.5pt}[[u^{-1}]]:\,T(u)\mapsto
R(u\,,t{\hskip 0.5pt})$

defines a representation $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$. We will denote this representation by
$\rho_{\,t}\,$. By the definitions (7) and (9), for any $r\geqslant 0$

$\rho_{\,t}:T_{ij}^{{\hskip 0.5pt}(r+1)}\mapsto(-{\hskip 0.5pt}E_{{\hskip
0.5pt}ji}+(-1)^{{\hskip 0.5pt}r}{\hskip 0.5pt}\tau\,\pi{\hskip
0.5pt}(E_{{\hskip 0.5pt}ji}))\,t^{\,r}{\hskip
0.5pt}{(-1)}^{\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$

### 7

Due to (36) for any formal power series $c{\hskip 0.5pt}(u)$ in $u^{-1}$ with
coefficients from ${\mathbb{C}}$ and leading term $1{\hskip 0.5pt}$, an
automorphism of the algebra $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ can be defined by

$T_{ij}(u)\mapsto c{\hskip 0.5pt}(u)\,T_{ij}(u)\,.$ (38)

Now for all $i$ and $j$ define a formal power series $T^{\,\prime}_{ij}(u)$ in
$u^{-1}$ with coefficients in the algebra $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by using the element inverse to (34) :

$T(u)^{-1}=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes T^{\,\prime}_{ij}(u)\,.$

###### Proposition 1

These mappings define two commuting antiautomorphisms of the algebra
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\\!:$

$\displaystyle T_{ij}(u)$ $\displaystyle\mapsto T_{ij}(-{\hskip 0.5pt}u)\,,$
(39) $\displaystyle T_{ij}(u)$ $\displaystyle\mapsto T^{\,\prime}_{ij}(u)\,.$
(40)

###### Proof

By using the notation (35) for $n=2$ we get the equalities

$\displaystyle T_{1}(u)\,T_{2}(v)$ $\displaystyle=\sum_{i,j,k,l}\,E_{{\hskip
0.5pt}ij}\otimes E_{{\hskip 0.5pt}kl}\otimes
T_{ij}(u)\,T_{kl}(v)\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}\,,$ (41) $\displaystyle
T_{2}(-v)\,T_{1}(-u)$ $\displaystyle=\sum_{i,j,k,l}\,E_{{\hskip
0.5pt}ij}\otimes E_{{\hskip 0.5pt}kl}\otimes T_{kl}(-v)\,T_{ij}(-u)\,.$ (42)

Due to (3) the antihomomorphism property of (39) follows from the relation

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes
1)\,T_{2}(-v)\,T_{1}(-u)=T_{1}(-u)\,T_{2}(-v)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)$

which is obtained from (36) by using (14). The antihomomorphism (39) is
clearly involutive and therefore bijective.

Similarly to (42), by again using the notation (35) for $n=2$ we get

$T_{2}(v)^{-1}\,T_{1}(u)^{-1}=\sum_{i,j,k,l}\,E_{{\hskip 0.5pt}ij}\otimes
E_{{\hskip 0.5pt}kl}\otimes T^{\,\prime}_{kl}(v)\,T^{\,\prime}_{ij}(u)\,.$

By comparing this with (41), the antihomomorphism property of (40) follows
from the relation

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1){\hskip
0.5pt}\,T_{2}(v)^{-1}\,T_{1}(u)^{-1}=T_{1}(u)^{-1}\,T_{2}(v)^{-1}{\hskip
0.5pt}(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)$

which is obtained by multiplying both sides of the defining relation (36) on
the left and right by $T_{2}(v)^{-1}$ and then by $T_{1}(u)^{-1}$.

The antihomomorphism (40) clearly commutes with (39). The bijectivity of the
antihomomorphism (40) will follow from Proposition 6 below. ∎

Furthermore, let us introduce the element of $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip 0.5pt}[[u^{-1}]]$

$\displaystyle T^{\,\ast{\hskip-0.5pt}}(u)=({\hskip
0.5pt}\tau\,\pi\otimes{\mathrm{id}}\,)(T(u))=$
$\displaystyle\sum_{i,j}\,E_{\,-j{\hskip 0.5pt},{\hskip 0.5pt}-i}\otimes
T_{ij}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes
T_{\,-j{\hskip 0.5pt},{\hskip 0.5pt}-i}{\hskip
0.5pt}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$

###### Proposition 2

An automorphism of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ commuting with (39) is defined by

$T_{ij}(u)\mapsto T_{\,-j{\hskip 0.5pt},{\hskip 0.5pt}-i}{\hskip
0.5pt}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (43)

###### Proof

The composition $\tau\,\pi$ is an antiautomorphism of the
$\mathbb{Z}_{2}$-graded associative algebra $\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}\,$. Therefore by applying the map
$\tau\,\pi\otimes\tau\,\pi\otimes{\mathrm{id}}$ to both sides of (36) and then
using (12),(13) we get the relation

$T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,T^{\,\ast{\hskip-0.5pt}}_{2}(v)\,(R{\hskip
0.5pt}(-{\hskip 0.5pt}u\,,-{\hskip 0.5pt}v)\otimes 1)=(R{\hskip
0.5pt}(-{\hskip 0.5pt}u\,,-{\hskip 0.5pt}v)\otimes
1)\,T^{\,\ast{\hskip-0.5pt}}_{2}(v)\,T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,.$ (44)

Here for $p=1,2$ we denote

$T^{\,\ast{\hskip-0.5pt}}_{p}(u)=\iota_{p}\otimes{\mathrm{id}}\,({\hskip
0.5pt}T^{\,\ast{\hskip-0.5pt}}(u))\in(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip
0.5pt}2}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,[[u^{-1}]]\,.$

Multiplying (44) on the left and right by $R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1$ and using (14) yields

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes
1)\,T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,T^{\,\ast{\hskip-0.5pt}}_{2}(v)=T^{\,\ast{\hskip-0.5pt}}_{2}(v)\,T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)$

which yields the homomorphism property of (43). This homomorphism is clearly
involutive, hence bijective. It clearly commutes with (39) as well. ∎

### 8

Let us multiply the relation (36) by $u+v$ and then set $v=-{\hskip
0.5pt}u\,$. We get

$(Q\otimes 1)\,{\hskip 0.5pt}T_{1}(u)\,T_{2}(-{\hskip 0.5pt}u)=T_{2}(-{\hskip
0.5pt}u)\,T_{1}(u)\,(Q\otimes 1)\,.$ (45)

Since the image of the action of $Q$ on $({\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}2}$ is one-dimensional, either
side of the last displayed relation must be equal to $Q\otimes Z(u)$ where
$Z(u)$ is a power series in $u^{-1}$ with coefficients from
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. It is
immediate that every coefficient of the series $Z(u)$ has $\mathbb{Z}_{{\hskip
0.5pt}2}$-degree $0$ and that the leading term of $Z(u)$ is $1\,$. Hence

$Z(u)=1+Z^{{\hskip 0.5pt}(1)}{\hskip 0.5pt}u^{-1}+Z^{{\hskip 0.5pt}(2)}{\hskip
0.5pt}u^{-2}+\ldots\,.$

###### Proposition 3

The elements $Z^{{\hskip 0.5pt}(1)},Z^{{\hskip 0.5pt}(2)}\ldots$ of the
algebra $\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ are
central.

###### Proof

We will work with the elements (35) where $n=3\,$. Using (36) we get

$\displaystyle(R_{{\hskip 0.5pt}13}(u\,,{\hskip-0.5pt}v)\,R_{{\hskip
0.5pt}23}(-{\hskip 0.5pt}u\,,{\hskip-0.5pt}v)\otimes
1)\,T_{1}(u)\,T_{2}(-{\hskip 0.5pt}u)\,T_{3}(v)=$ $\displaystyle
T_{3}(v)\,T_{1}(u)\,T_{2}(-{\hskip 0.5pt}u)\,(R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}v)\,R_{{\hskip 0.5pt}23}(-{\hskip
0.5pt}u\,,{\hskip-0.5pt}v)\otimes 1)\,.$ (46)

Due to (14) we have the identity in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}({\hskip
0.5pt}u\,,{\hskip-0.5pt}v{\hskip 0.5pt})$

$-{\hskip 0.5pt}R_{{\hskip 0.5pt}13}(u\,,-{\hskip 0.5pt}v)\,P_{{\hskip
0.5pt}12}\,R_{{\hskip 0.5pt}23}(-{\hskip 0.5pt}u\,,{\hskip-0.5pt}v)=-{\hskip
0.5pt}P_{{\hskip 0.5pt}12}{\hskip-0.5pt}\left(1-\frac{1}{(u+v)^{{\hskip
0.5pt}2}}\right).$

By applying to it the antiautomorphism $\tau\,\pi$ relative to the first
tensor factor of $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{\otimes{\hskip 0.5pt}3}$ and then using (8),(12) we get

$Q_{{\hskip 0.5pt}12}\,R_{{\hskip 0.5pt}13}(u\,,{\hskip-0.5pt}v)\,R_{{\hskip
0.5pt}23}(-{\hskip 0.5pt}u\,,{\hskip-0.5pt}v)=Q_{{\hskip
0.5pt}12}{\hskip-0.5pt}\left(1-\frac{1}{(u+v)^{{\hskip 0.5pt}2}}\right).$

Therefore by multiplying the relation (46) by $Q_{{\hskip 0.5pt}12}\otimes 1$
on the left and using the above definition of the series $Z(u)$ we obtain

$(Q_{12}\otimes Z(u)){\hskip 0.5pt}\,T_{3}(v)\left(1-\frac{1}{(u+v)^{{\hskip
0.5pt}2}}\right)=T_{3}(v)\,{\hskip 0.5pt}(Q_{12}\otimes
Z(u))\left(1-\frac{1}{(u+v)^{{\hskip 0.5pt}2}}\right).$

The last relation implies that any generator $T_{ij}^{{\hskip 0.5pt}(r)}$ of
the algebra $\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
commutes with all coefficients of the series $Z(u)\,$. ∎

By the definition of $Z(u)$ we have relations in $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{\otimes{\hskip
0.5pt}2}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,[[u^{-1}]]$

$\displaystyle(Q\otimes 1)\,{\hskip 0.5pt}T_{1}(u)\,T_{2}(-{\hskip 0.5pt}u)$
$\displaystyle=Q\otimes Z(u)\,,$ (47) $\displaystyle T_{2}(-{\hskip
0.5pt}u)\,T_{1}(u)\,(Q\otimes 1)$ $\displaystyle=Q\otimes Z(u)\,.$

Applying $\tau\,\pi\otimes{\mathrm{id}}\otimes{\mathrm{id}}$ to both sides of
these relations and then using (12) gives

$\displaystyle T^{\,\ast{\hskip-0.5pt}}_{{\hskip 0.5pt}1}(u)\,(-{\hskip
0.5pt}P\otimes 1)\,T_{2}(-{\hskip 0.5pt}u)$ $\displaystyle=-{\hskip
0.5pt}P\otimes Z(u)\,,$ $\displaystyle T_{2}(-{\hskip 0.5pt}u)\,(-{\hskip
0.5pt}P\otimes 1)\,T^{\,\ast{\hskip-0.5pt}}_{{\hskip 0.5pt}1}(u)$
$\displaystyle=-{\hskip 0.5pt}P\otimes Z(u)\,.$

Equivalently,

$\displaystyle T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,T_{1}(-{\hskip
0.5pt}u)\,(-{\hskip 0.5pt}P\otimes 1)$ $\displaystyle=-{\hskip 0.5pt}P\otimes
Z(u)\,,$ $\displaystyle(-{\hskip 0.5pt}P\otimes 1)\,T_{1}(-{\hskip
0.5pt}u)\,T^{\,\ast{\hskip-0.5pt}}_{1}(u)$ $\displaystyle=-{\hskip
0.5pt}P\otimes Z(u)\,.$

The latter two can be simpler written as relations in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,[[u^{-1}]]$

$\displaystyle T^{\,\ast{\hskip-0.5pt}}(u)\,T(-{\hskip 0.5pt}u)$
$\displaystyle=1\otimes Z(u)\,,$ (48) $\displaystyle T(-{\hskip
0.5pt}u)\,T^{\,\ast{\hskip-0.5pt}}(u)$ $\displaystyle=1\otimes Z(u)\,.$

Observe that the last two are also equivalent to each other due to Proposition
3. Explicitly, these last two relations can be rewritten respectively as

$\displaystyle\sum_{k}\,T_{{\hskip 0.5pt}-k{\hskip 0.5pt},{\hskip
0.5pt}-i}{\hskip 0.5pt}(u)\,T_{{\hskip 0.5pt}kj}(-{\hskip
0.5pt}u)\,{(-1)}^{\,{\bar{\jmath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{\jmath}\,}}$ $\displaystyle=\delta_{{\hskip
0.5pt}ij}\,Z(u)\,,$ (49) $\displaystyle\sum_{k}\,T_{ik}(-u)\,T_{{\hskip
0.5pt}-j{\hskip 0.5pt},{\hskip 0.5pt}-k}{\hskip
0.5pt}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{k}\,}+\,{\bar{k}\,}}$
$\displaystyle=\delta_{{\hskip 0.5pt}ij}\,Z(u)\,.$ (50)

###### Proposition 4

The antiautomorphism (40) maps $Z(u)\mapsto Z(u)^{-1}\,$.

###### Proof

Due to Proposition 3 the relation (47) can be rewritten as

$Q\otimes Z(u)^{-1}=(Q\otimes 1)\,{\hskip 0.5pt}T_{2}(-{\hskip
0.5pt}u)^{-1}\,T_{1}(u)^{-1}\,.$

The right hand side of the last displayed relation can also be obtained by
applying to the left hand side of (47) the antiautomorphism (40) relative to
the tensor factor $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Hence Proposition 4 follows from
(47). ∎

###### Proposition 5

The square of antiautomorphism (40) of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is given by

$T_{ij}(u)\mapsto Z(u)^{-1}{\hskip 0.5pt}Z(-{\hskip 0.5pt}u){\hskip
0.5pt}\,T_{ij}(u)\,.$

###### Proof

By (48)

$T(u)^{-1}=Z(-{\hskip 0.5pt}u)^{-1}\,T^{\,\ast{\hskip-0.5pt}}(-{\hskip
0.5pt}u)\,.$

Thus for any indices $i$ and $j$ we have the relation

$T^{\,\prime}_{ij}(u)=Z(-{\hskip 0.5pt}u)^{-1}\,T_{\,-j{\hskip 0.5pt},{\hskip
0.5pt}-i}{\hskip 0.5pt}(-{\hskip 0.5pt}u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}{\hskip 0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt}.$ (51)

By applying the antiautomorphism (40) to the right hand side of this relation,
and then using the same relation for $-{\hskip 0.5pt}j{\hskip 0.5pt},-{\hskip
0.5pt}i$ and $-{\hskip 0.5pt}u$ instead of $i\,,j$ and $u$ we get the series

$Z(u)^{-1}\,T_{ij}(u)\,Z(-{\hskip 0.5pt}u)=Z(u)^{-1}{\hskip 0.5pt}Z(-{\hskip
0.5pt}u){\hskip 0.5pt}\,T_{ij}(u)\,.$

Here we also used Propositions 3 and 4. ∎

###### Proposition 6

The automorphism (43) of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ maps $Z(u)\mapsto Z(-{\hskip
0.5pt}u)\,$.

###### Proof

By setting $j=i$ in (49) we get an explicit formula

$Z(u)=\sum_{k}\,T_{{\hskip 0.5pt}-k{\hskip 0.5pt},{\hskip 0.5pt}-i}{\hskip
0.5pt}(u)\,T_{{\hskip 0.5pt}ki}(-{\hskip
0.5pt}u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{\imath}\,}}\,.$ (52)

It follows from (52) that the automorphism $\eqref{T}$ maps $Z(u)$ to the sum

$\sum_{k}\,T_{{\hskip 0.5pt}ik}{\hskip 0.5pt}(u)\,T_{{\hskip 0.5pt}-i,{\hskip
0.5pt}-k}(-{\hskip 0.5pt}u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{k}\,}}\,.$

On the other hand, by setting $j=i$ in (50) we get another explicit formula

$Z(u)=\sum_{k}\,T_{{\hskip 0.5pt}ik}(-{\hskip 0.5pt}u)\,T_{{\hskip
0.5pt}-i{\hskip 0.5pt},{\hskip 0.5pt}-k}{\hskip
0.5pt}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{k}\,}}\,.$ (53)

Comparing the sum in (53) with the previous display completes our proof. ∎

### 9

There is a natural Hopf algebra structure on the extended Yangian
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. A
coassociative comultiplication homomorphism $\Delta:\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is defined by

$\Delta:\,T_{ij}(u)\,\mapsto\,\sum_{k}\,T_{ik}(u)\otimes
T_{kj}(u)\,{(-1)}^{{\hskip 0.5pt}({\hskip 0.5pt}{\bar{\imath}\,}{\hskip
0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})({\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})}$ (54)

where the tensor product is taken over the subalgebra ${\mathbb{C}}[[u^{-1}]]$
of $\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip
0.5pt}[[u^{-1}]]\,$. The counit homomorphism
$\varepsilon:\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to{\mathbb{C}}$ is defined by
$T_{ij}(u)\mapsto\delta_{ij}\,$. Further, the antipodal mapping
$\operatorname{S}:\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is just the antiautomorphism (40).
Justification of these definitions is very similar to that in the case of the
Yangian of the general linear Lie algebra $\mathfrak{gl}_{{\hskip
0.5pt}N}{\hskip 0.5pt}$, see (N4, 21, Section 4). Here we omit the details.

###### Proposition 7

For the formal power series $Z(u)$ in $u^{-1}$ we have

$\Delta:\,Z(u)\mapsto Z(u)\otimes
Z(u)\quad\text{and}\quad\varepsilon:\,Z(u)\mapsto 1\,.$ (55)

###### Proof

By using the above formula (52) with the index $k$ replaced by $-{\hskip
0.5pt}h\,$, and then using the definition (54) twice with $k$ replaced
respectively by $-{\hskip 0.5pt}k$ and $l\,$, the comultiplication
homomorphism $\Delta$ maps $Z(u)$ to the sum

$\displaystyle\sum_{h,k,l}\,(\,T_{{\hskip 0.5pt}h,{\hskip
0.5pt}-k}(u)\,T_{{\hskip 0.5pt}-h,{\hskip 0.5pt}l}(-{\hskip 0.5pt}u){\hskip
0.5pt})\otimes(\,T_{{\hskip 0.5pt}-k,-i}(u)\,T_{{\hskip 0.5pt}li}(-{\hskip
0.5pt}u){\hskip 0.5pt})\,{(-1)}^{\,{\bar{h}\,}{\hskip 0.5pt}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{k}\,}+\,{\bar{k}\,}{\hskip
0.5pt}{\bar{l}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}{\hskip
0.5pt}+\,{\bar{k}\,}}$ $\displaystyle=\ \sum_{k,l}\,\,(\,\delta_{{\hskip
0.5pt}kl}\,Z(u){\hskip 0.5pt})\otimes(\,T_{{\hskip 0.5pt}-k,-i}(u)\,T_{{\hskip
0.5pt}li}(-{\hskip 0.5pt}u){\hskip 0.5pt})\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{\imath}\,}{\hskip 0.5pt}+\,{\bar{k}\,}{\hskip
0.5pt}+\,{\bar{l}\,}}$ $\displaystyle=\ \sum_{k}\,\,Z(u)\otimes(\,T_{{\hskip
0.5pt}-k,-i}(u)\,T_{{\hskip 0.5pt}ki}(-{\hskip 0.5pt}u){\hskip
0.5pt})\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{\imath}\,}}=\,Z(u)\otimes Z(u)$

as needed. Here we also used the relation (49) with the indices $k{\hskip
0.5pt},l{\hskip 0.5pt},-{\hskip 0.5pt}h$ instead of $i{\hskip 0.5pt},j{\hskip
0.5pt},k$ respectively. Then we used the above formula (52) in its original
form.

Due to the definition of the counit homomorphism $\varepsilon\,$, the second
statement in (55) immediately follows from (52). ∎

Note that Proposition 4 also follows from (55), because the multiplication
$\mu:\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ and the unit mapping
$\delta:{\mathbb{C}}\to\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}):1\mapsto 1$ obey

$\mu\,({\hskip 0.5pt}\operatorname{S}\otimes\,{\mathrm{id}}{\hskip
0.5pt})\,\Delta=\delta\,{\hskip 0.5pt}\varepsilon\,.$

Indeed, by applying to the coefficients of the series $Z(u)$ the homomorphisms
at the two sides of the last displayed identity, we get the equality
$\operatorname{S}(Z(u))\,Z(u)=1\,$.

### 10

The Yangian $\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
of the Lie superalgebra $\mathfrak{p}_{N}$ is defined as the quotient of the
$\mathbb{Z}_{{\hskip 0.5pt}2}$-graded algebra $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by the relations

$Z^{{\hskip 0.5pt}(1)}=Z^{{\hskip 0.5pt}(2)}=\ldots=0\,.$

Hence $\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is a
complex associative unital algebra with the generators (32), subject to the
relations (37) and $Z(u)=1\,$. Here we still use the series (33). Explicit
formulas for $Z(u)$ in terms of these series are given by (52) and (53). The
$\mathbb{Z}_{{\hskip 0.5pt}2}$-grading on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is still defined by setting $\deg
T_{ij}^{{\hskip 0.5pt}(r)}={\bar{\imath}\,}+{\bar{\jmath}\,}$ for every $r\,$.

Due Propositions 4 and 7, the Hopf algebra structure descends from
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ to the
quotient $\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.
Due to Proposition 5, the antipodal map on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is involutive. In terms of the series
(32), this map is still defined by (40).

By $\eqref{TPij}$ the antipodal map on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ coincides with the composition of the
two commuting maps on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ defined by (39) and (43). These two
maps on $\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
commute by Proposition 2. The map (39) on $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ clearly descends to the quotient
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$, whereas
(43) descends to $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ due to Proposition 7.

Consider the universal enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of the Lie superalgebra
$\mathfrak{p}_{N}\,$. This is an associative algebra generated by the elements
$f_{{\hskip 0.5pt}ij}\in\mathfrak{p}_{N}\,$, see (6). The algebra
$\operatorname{U}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is
$\mathbb{Z}_{{\hskip 0.5pt}2}$-graded so that $\deg
f_{ij}={\bar{\imath}\,}+{\bar{\jmath}\,}\,$. The defining relations are

$\displaystyle f_{\,-j{\hskip 0.5pt},{\hskip 0.5pt}-i}=-\,f_{{\hskip
0.5pt}ij}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}{\hskip 0.5pt},$ $\displaystyle[\,f_{{\hskip
0.5pt}ij}\,,f_{{\hskip 0.5pt}kl}\,]=\delta_{{\hskip 0.5pt}jk}\,f_{{\hskip
0.5pt}il}-\delta_{{\hskip 0.5pt}li}\,f_{{\hskip 0.5pt}kj}\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}$ $\displaystyle-\,\delta_{-{\hskip
0.5pt}i,{\hskip 0.5pt}k}\,f_{{\hskip 0.5pt}-j,{\hskip
0.5pt}l}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}\\!-\delta_{{\hskip 0.5pt}j,-l}\,f_{{\hskip
0.5pt}i,{\hskip 0.5pt}-k}\,{(-1)}^{\,{\bar{k}\,}{\hskip
0.5pt}{\bar{l}\,}{\hskip 0.5pt}+\,{\bar{l}\,}}{\hskip 0.5pt},$

see (5). But now the square brackets stand for the supercommutator in
$\operatorname{U}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

On the other hand, by setting $Z(u)=1$ in (49) and taking the coefficients at
$u^{-1}$ we obtain the relations in the algebra $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$T_{\,-j{\hskip 0.5pt},{\hskip 0.5pt}-i}^{{\hskip 0.5pt}(1)}=T_{{\hskip
0.5pt}ij}^{{\hskip 0.5pt}(1)}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}{\hskip 0.5pt}+\,{\bar{\jmath}\,}}\,.$

Multiplying (37) by $u^{{\hskip 0.5pt}2}-v^{{\hskip 0.5pt}2}$ and then taking
the coefficients at $u\,v^{-1}$ yields

$\displaystyle[\,T_{{\hskip 0.5pt}ij}^{{\hskip 0.5pt}(1)}\,,T_{{\hskip
0.5pt}kl}^{{\hskip 0.5pt}(1)}\,]=(\,\delta_{{\hskip 0.5pt}kj}\,T_{{\hskip
0.5pt}il}^{{\hskip 0.5pt}(1)}-\delta_{{\hskip 0.5pt}il}\,T_{{\hskip
0.5pt}kj}^{{\hskip 0.5pt}(1)}\,)\,(-1)^{{\hskip
0.5pt}\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{k}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{l}\,}\,+\,{\bar{k}\,}{\hskip
0.5pt}{\bar{l}\,}}$ $\displaystyle-\,\delta_{-{\hskip 0.5pt}i,{\hskip
0.5pt}k}\,T_{{\hskip 0.5pt}-j,{\hskip 0.5pt}l}^{{\hskip
0.5pt}(1)}\,(-1)^{{\hskip 0.5pt}\,{\bar{\imath}\,}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}{\hskip 0.5pt}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}}\\!+\delta_{{\hskip 0.5pt}j,-l}\,T_{{\hskip
0.5pt}k,{\hskip 0.5pt}-i}^{{\hskip 0.5pt}(1)}\,(-1)^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{l}\,}{\hskip 0.5pt}+\,{\bar{\imath}\,}}{\hskip 0.5pt}.$

By comparing the last two displays with the above relations in
$\operatorname{U}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$, we deduce
that a homomorphism $\operatorname{U}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ can be defined by mapping

$-{\hskip 0.5pt}f_{{\hskip 0.5pt}ji}\,{(-1)}^{\,{\bar{\jmath}\,}}\mapsto
T_{ij}^{{\hskip 0.5pt}(1)}\,.$ (56)

Our Theorem 2 will imply that this homomorphism is an embedding. Moreover it
is an embedding of Hopf algebras, because for $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by our definitions

$\Delta:\,T_{ij}^{{\hskip 0.5pt}(1)}\mapsto T_{ij}^{{\hskip 0.5pt}(1)}\otimes
1+1\otimes T_{ij}^{{\hskip 0.5pt}(1)}{\hskip 0.5pt},\
\quad\varepsilon:\,T_{ij}^{{\hskip 0.5pt}(1)}\mapsto 0\,,\
\quad\operatorname{S}:\,T_{ij}^{{\hskip 0.5pt}(1)}\mapsto-\,T_{ij}^{{\hskip
0.5pt}(1)}{\hskip 0.5pt}.$

### 11

For any formal power series $c{\hskip 0.5pt}(u)$ in $u^{-1}$ with coefficients
from ${\mathbb{C}}$ and leading term $1{\hskip 0.5pt}$, the automorphism (38)
of the associative algebra $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ maps $Z(u)$ to

$c{\hskip 0.5pt}(u)\,c{\hskip 0.5pt}(-u)\,Z(u)\,.$

So (38) descends to $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ if and only if $c{\hskip
0.5pt}(u)\,c{\hskip 0.5pt}(-u)=1\,$. The special Yangian of ${\hskip
0.5pt}\mathfrak{p}_{N}$ is the fixed point subalgebra of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ relative to
all these automorphisms. We will denote this subalgebra of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by
$\operatorname{Z}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. It will
not be confused with the subalgebra of $\operatorname{U}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ consisting of all elements invariant
relative to the adoint action of $\mathfrak{p}_{N}\,$, since the latter
subalgebra is just ${\mathbb{C}}$ by (S, 23, Proposition 3).

The special Yangian $\operatorname{Z}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is moreover a Hopf subalgebra of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Indeed,
the anipodal map on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ preserves the subalgebra
$\operatorname{Z}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by the
definition of this subalgebra. Now let $c{\hskip 0.5pt}(u)$ by any formal
power series in $u^{-1}$ with coefficients from ${\mathbb{C}}$ and leading
term $1{\hskip 0.5pt}$, satisfying the condition $c{\hskip 0.5pt}(u)\,c{\hskip
0.5pt}(-u)=1\,$. By (54) the comultiplication on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ maps

$\displaystyle c{\hskip 0.5pt}(u)\,T_{ij}(u)\,\mapsto\,$
$\displaystyle\sum_{k}\,({\hskip 0.5pt}c{\hskip 0.5pt}(u)\,T_{ik}(u))\otimes
T_{kj}(u)\,{(-1)}^{{\hskip 0.5pt}({\hskip 0.5pt}{\bar{\imath}\,}{\hskip
0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})({\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})}=$
$\displaystyle\sum_{k}\,T_{ik}(u)\otimes({\hskip 0.5pt}c{\hskip
0.5pt}(u)\,T_{kj}(u))\,{(-1)}^{{\hskip 0.5pt}({\hskip
0.5pt}{\bar{\imath}\,}{\hskip 0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})({\hskip
0.5pt}{\bar{\jmath}\,}{\hskip 0.5pt}+\,{\bar{k}\,}{\hskip 0.5pt})}\,.$

Hence the comultiplication intertwines the action of the automorphism (38) on
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ with its
action on any one of the two tensor factors of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. It follows that the image of the
subalgebra $\operatorname{Z}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ under
comultiplication is contained in both $\operatorname{Z}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ and $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{Z}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Therefore this image is contained in
$\operatorname{Z}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\otimes\operatorname{Z}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ as needed.

We will conclude this section with an important observation. For any
$t\in{\mathbb{C}}$ consider the representation $\rho_{\,t}$ of
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ defined at
the end of Section 6. By using the formula (48) and the relations (12),(14)
this representation maps $Z(u)$ to

$({\hskip 0.5pt}\tau\,\pi\otimes{\mathrm{id}}\,)(R{\hskip 0.5pt}(u\,,t{\hskip
0.5pt}))\,R{\hskip 0.5pt}(-{\hskip 0.5pt}u\,,t{\hskip 0.5pt})=R{\hskip
0.5pt}(u\,,-\,t{\hskip 0.5pt})\,R{\hskip 0.5pt}(-{\hskip 0.5pt}u\,,t{\hskip
0.5pt})=1-\frac{1}{(u+t{\hskip 0.5pt})^{{\hskip 0.5pt}2}}\,.$

So the composition of $\rho_{\,t}$ with the automorphism (38) of
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ maps $Z(u)$
to

$c{\hskip 0.5pt}(u)\,c{\hskip 0.5pt}(-u)\left(1-\frac{1}{(u+t{\hskip
0.5pt})^{{\hskip 0.5pt}2}}\right)$

which may be equal to $1$ for some $c{\hskip 0.5pt}(u)$ if and only if
$t=0\,$. So the composition of $\rho_{\,t}$ with (38) may factor to a
representation of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ if and only if $t=0\,$. The choice of
$c{\hskip 0.5pt}(u)$ for that purpose is not unique then. For example, we can
choose

$c{\hskip 0.5pt}(u)=\frac{u}{u+1}\quad\text{or}\quad c{\hskip
0.5pt}(u)=\frac{u}{u-1}\,.$

However, the composition of the tensor product $\rho_{\,t}\otimes\rho_{{\hskip
0.5pt}-{\hskip 0.5pt}t}$ of representations of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ with an automorphism (38) may factor to
a representation of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ for any $t\in{\mathbb{C}}\,$. This is
because due to (55), the latter composition maps $Z(u)$ to

$c{\hskip 0.5pt}(u)\,c{\hskip 0.5pt}(-u)\left(1-\frac{1}{(u+t{\hskip
0.5pt})^{{\hskip 0.5pt}2}}\right)\left(1-\frac{1}{(u-t{\hskip 0.5pt})^{{\hskip
0.5pt}2}}\right).$

The choice of $c{\hskip 0.5pt}(u)$ for $\rho_{\,t}\otimes\rho_{{\hskip
0.5pt}-{\hskip 0.5pt}t}$ is still not unique. For example, we can choose

$c{\hskip 0.5pt}(u)=\frac{(u+t{\hskip 0.5pt})^{{\hskip 0.5pt}2}}{(u+t{\hskip
0.5pt})^{{\hskip 0.5pt}2}-1}\quad\text{or}\quad c{\hskip
0.5pt}(u)=\frac{(u-t{\hskip 0.5pt})^{{\hskip 0.5pt}2}}{(u-t{\hskip
0.5pt})^{{\hskip 0.5pt}2}-1}\,.$

### 12

There is a natural ascending $\mathbb{Z}{\hskip 0.5pt}$-filtration on the
associative algebra $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. It is defined by setting the degree
of $T^{{\hskip 0.5pt}(r)}_{ij}$ to $r$ for each $r\geqslant 1$ and all indices
$i\,,j\,$. Let $\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ be the $\mathbb{Z}{\hskip
0.5pt}$-graded algebra corresponding to this filtration. Denote by $X^{{\hskip
0.5pt}(r)}_{ij}$ the image of the generator $T^{{\hskip 0.5pt}(r)}_{ij}$ in
the degree $r$ component of $\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

The $\mathbb{Z}_{2}$-grading on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ descends to
$\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ so that $\deg X^{{\hskip
0.5pt}(r)}_{ij}={\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}\,$. For any
$r,s\geqslant 1$ by taking the coefficients at $u^{{\hskip 0.5pt}-r}{\hskip
0.5pt}v^{{\hskip 0.5pt}-s}$ in (37) we immediately obtain the supercommutation
relation in the $\mathbb{Z}_{2}$-graded algebra
$\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$[{\hskip 0.5pt}\,X^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}ij},{\hskip
0.5pt}X^{{\hskip 0.5pt}(s)}_{{\hskip 0.5pt}kl}\,]=0\,.$

For any indices $i\,,j$ introduce the series with coefficients in the algebra
$\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$X_{{\hskip 0.5pt}ij}(u)=\delta_{ij}\cdot 1+X_{{\hskip 0.5pt}ij}^{{\hskip
0.5pt}(1)}{\hskip 0.5pt}u^{-1}+X_{{\hskip 0.5pt}ij}^{{\hskip 0.5pt}(2)}{\hskip
0.5pt}u^{-2}+\ldots\,.$

Then the relations (49) with $Z(u)=1$ imply that for all $i\,,j$ we have

$\sum_{k}\,X_{{\hskip 0.5pt}-k{\hskip 0.5pt},{\hskip 0.5pt}-i{\hskip
0.5pt}}{\hskip 0.5pt}(u)\,X_{{\hskip 0.5pt}kj{\hskip 0.5pt}}(-{\hskip
0.5pt}u)\,{(-1)}^{\,{\bar{\jmath}\,}{\hskip
0.5pt}{\bar{k}\,}+\,{\bar{\jmath}\,}}=\delta_{{\hskip 0.5pt}ij}\,.$ (57)

The numerator in the first line of the defining relation (37) of
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ vanishes at
$v=u\,$. Further, if we multiply by $u+v$ the expressions in the second and in
the third line of (37) and then set $v=-\,u\,$, we will get

$-\,\delta_{{\hskip 0.5pt}-i,k}\,\delta_{{\hskip
0.5pt}j,-l}\,Z(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}}\quad\text{and}\quad\delta_{{\hskip
0.5pt}j,-l}\,\delta_{{\hskip
0.5pt}-i,k}\,Z(u)\,\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{l}\,}+\,{\bar{\imath}\,}}$

by using (49) and (50) respectively. The latter two expressions cancel each
other. So by using (49) and (50) the defining relation (37) can be rewritten
is an equality of formal power series in $u^{-1},v^{-1}$ with coefficients in
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

### 13

We will also employ another ascending $\mathbb{Z}{\hskip 0.5pt}$-filtration on
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. It is
defined by setting the degree of $T^{{\hskip 0.5pt}(r)}_{ij}$ to $r-1$ for any
$r\geqslant 1\,$. The corresponding $\mathbb{Z}{\hskip 0.5pt}$-graded algebra
will be denoted by $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Let $Y^{{\hskip 0.5pt}(r)}_{ij}$ be
the image of $T^{{\hskip 0.5pt}(r)}_{ij}$ in the degree $r-1$ component of
$\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

The $\mathbb{Z}_{2}$-grading on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ descends to
$\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ so that $\deg Y^{{\hskip
0.5pt}(r)}_{ij}={\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}\,$. For any
$r\geqslant 1$ by equating to $\delta_{{\hskip 0.5pt}ij}$ the left hand side
of (49) and then taking the coefficients at $u^{{\hskip 0.5pt}-r}$ we obtain
the relation in the algebra
$\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$Y_{{\hskip 0.5pt}-j{\hskip 0.5pt},{\hskip 0.5pt}-i}^{{\hskip
0.5pt}(r)}=-\,Y_{{\hskip 0.5pt}ij}^{{\hskip
0.5pt}(r)}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}+\,{\bar{\jmath}\,}{\hskip 0.5pt}+\,r}\,.$ (58)

For $r{\hskip 0.5pt},{\hskip-0.5pt}s\geqslant 1$ by taking coefficients at
$u^{{\hskip 0.5pt}-r}{\hskip 0.5pt}v^{{\hskip 0.5pt}-s}$ in (37) we obtain the
relation

$\displaystyle[\,\,Y^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}ij},{\hskip
0.5pt}Y^{{\hskip 0.5pt}(s)}_{{\hskip 0.5pt}kl}\,]\,=(\,\delta_{{\hskip
0.5pt}kj}\,Y^{{\hskip 0.5pt}(r+s-1)}_{{\hskip 0.5pt}il}-\delta_{{\hskip
0.5pt}il}\,Y^{{\hskip 0.5pt}(r+s-1)}_{{\hskip 0.5pt}kj}\,)\,(-1)^{{\hskip
0.5pt}\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{k}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{l}\,}\,+\,{\bar{k}\,}{\hskip
0.5pt}{\bar{l}\,}}\,+$ $\displaystyle\delta_{-{\hskip 0.5pt}i,{\hskip
0.5pt}k}\,Y_{{\hskip 0.5pt}-j,{\hskip 0.5pt}l}^{{\hskip
0.5pt}(r+s-1)}\,(-1)^{{\hskip 0.5pt}\,{\bar{\imath}\,}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}{\hskip 0.5pt}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}+\,r}-\,\delta_{{\hskip
0.5pt}j,-l}\,Y_{{\hskip 0.5pt}k,{\hskip 0.5pt}-i}^{{\hskip
0.5pt}(r+s-1)}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{\imath}\,}{\hskip 0.5pt}+\,r}\,.$ (59)

Here we used the equality to $1$ of the coefficient at $u^{{\hskip
0.5pt}-r}{\hskip 0.5pt}v^{{\hskip 0.5pt}-s}$ in the expansion of

$\frac{v^{\,1-r-s}-u^{{\hskip 0.5pt}1-r-s}}{u-v}$

as a polynomial in $u^{-1},v^{-1}\,$. We also used the relation (58) obtained
just above.

The structure of $\mathbb{Z}_{2}$-graded Hopf algebra on
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ descends to
$\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Due to the definition (54) the
comultiplication, the counit homomorphism and the antipodal antiautomorphism
for $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ are defined respectively by

$Y_{ij}^{{\hskip 0.5pt}(r)}\mapsto Y_{ij}^{{\hskip 0.5pt}(r)}\otimes
1+1\otimes Y_{ij}^{{\hskip 0.5pt}(r)}{\hskip 0.5pt},\ \quad Y_{ij}^{{\hskip
0.5pt}(r)}\mapsto 0\,,\ \quad Y_{ij}^{{\hskip
0.5pt}(r)}\mapsto-\,Y_{ij}^{{\hskip 0.5pt}(r)}{\hskip 0.5pt}.$ (60)

### 14

Now consider the twisted polynomial current Lie superalgebra $\mathfrak{g}$ as
defined in Section 1. This subalgebra of $\mathfrak{gl}_{{\hskip
0.5pt}N|N}[u]$ is spanned by the elements

$g^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}ij}=e_{{\hskip
0.5pt}ij}\,u^{r}{\hskip-0.5pt}+{\hskip 0.5pt}\omega{\hskip 0.5pt}(e_{{\hskip
0.5pt}ij}){\hskip 0.5pt}(-{\hskip 0.5pt}u)^{r}=e_{{\hskip
0.5pt}ij}\,u^{r}{\hskip-0.5pt}-e_{{\hskip 0.5pt}-j{\hskip 0.5pt},{\hskip
0.5pt}-i}\,u^{r}{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}{\hskip 0.5pt}+\,{\bar{\jmath}\,}{\hskip 0.5pt}+\,r}$

where $r=0,1,2,\,\ldots$ and $i\,,j=\pm\,1{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}\\!\pm N\,$. Here $g^{{\hskip
0.5pt}(0)}_{{\hskip 0.5pt}ij}=f_{{\hskip 0.5pt}ij}$ by (6). For $r,s\geqslant
0$

$\displaystyle g^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}-j{\hskip 0.5pt},{\hskip
0.5pt}-i{\hskip 0.5pt}}=-{\hskip 0.5pt}\,g^{{\hskip 0.5pt}(r)}_{{\hskip
0.5pt}ij}{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}{\hskip 0.5pt}+\,r}\,,$ (61)
$\displaystyle[\,g_{{\hskip 0.5pt}ij}^{{\hskip 0.5pt}(r)}\,,g_{{\hskip
0.5pt}kl}^{{\hskip 0.5pt}(s)}\,]=\delta_{{\hskip 0.5pt}jk}\,g_{{\hskip
0.5pt}il}^{{\hskip 0.5pt}(r+s)}-\delta_{{\hskip 0.5pt}li}\,g_{{\hskip
0.5pt}kj}^{{\hskip 0.5pt}(r+s)}\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}$ $\displaystyle-\,\delta_{-{\hskip
0.5pt}i,{\hskip 0.5pt}k}\,g_{{\hskip 0.5pt}-j,{\hskip 0.5pt}l}^{{\hskip
0.5pt}(r+s)}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}{\hskip 0.5pt}+\,r}\\!-\delta_{{\hskip
0.5pt}j,-l}\,g_{{\hskip 0.5pt}i,{\hskip 0.5pt}-k}^{{\hskip
0.5pt}(r+s)}\,{(-1)}^{\,{\bar{k}\,}{\hskip 0.5pt}{\bar{l}\,}{\hskip
0.5pt}+\,{\bar{l}\,}{\hskip 0.5pt}+\,s}\,.$ (62)

The universal enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})$ is generated by the elements $g^{{\hskip
0.5pt}(r)}_{{\hskip 0.5pt}ij}$ as an associative algebra. It is
$\mathbb{Z}_{2}$-graded so that the degrees of these generators are equal to
${\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}$ respectively. The defining
relations for these generators are (61) and (62) where the square brackets now
denote the supercommutator.

By comparing (61),(62) with (58),(59) above we deduce that the mapping

$-\,g^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}ji}\,(-1)^{\,{\bar{\jmath}\,}}\mapsto
Y_{ij}^{{\hskip 0.5pt}(r+1)}$ (63)

for $r\geqslant 0$ defines a homomorphism $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})\to\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of $\mathbb{Z}_{2}$-graded associative
algebras. Moreover, this is a Hopf algebra homomorphism by the definitions
(60).

Note that for any $t\in{\mathbb{C}}$ the image of $T_{ij}^{{\hskip
0.5pt}(r+1)}\in\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
under the representation $\rho_{\,t}$ coincides with the image of the left
hand side of (63) under the evaluation representation
$\operatorname{U}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip 0.5pt}N|N}[u]{\hskip
0.5pt})\mapsto\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N}$
defined by mapping $e_{{\hskip 0.5pt}ij}\,u^{r}\mapsto E_{{\hskip
0.5pt}ij}\,t^{\,r}\,$.

###### Theorem 2

The mapping (63) defines an isomorphism $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})\to\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of $\mathbb{Z}_{2}$-graded Hopf
algebras.

###### Proof

The homomorphism $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})\to\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ defined by (63) above is clearly
surjective. We have to prove that this homomorphism is injective as well.

Let $\mathcal{I}$ be the set of all triples $(i,j,r)$ with $r\geqslant 0\,$.
Choose any total ordering $<$ on this set. Also choose any subset
$\mathcal{J}\subset\mathcal{I}$ such that for $i\neq-{\hskip 0.5pt}j$ it
contains only one of the triples $(i,j,r)$ and $(-{\hskip 0.5pt}j{\hskip
0.5pt},-{\hskip 0.5pt}i{\hskip 0.5pt},r)\,$, while for $i=-{\hskip 0.5pt}j$ it
contains $(i,j,r)$ if and only if $\,{\bar{\imath}\,}{\hskip-0.5pt}+{\hskip
0.5pt}r$ is odd. Due to (61) the left hand sides of (63) corresponding to the
triples $(i,j,r)\in\mathcal{J}$ form a basis of the vector space
$\mathfrak{g}\,$.

Let $\mathcal{T}$ be the set of all products of $T_{ij}^{{\hskip 0.5pt}(r+1)}$
corresponding to $(i,j,r)\in\mathcal{J}$ taken in non-decreasing orders, such
that two adjacent triples can be the same $(i,j,r)$ only if
$\,{\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}=0$ in $\mathbb{Z}_{2}\,$.
We will prove that $\mathcal{T}$ is linearly independent in
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. This is
equivalent to the linear independence in
$\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of the set of products obtained from
$\mathcal{T}$ by replacing every factor $T_{ij}^{{\hskip 0.5pt}(r+1)}$ by the
corresponding $Y_{ij}^{{\hskip 0.5pt}(r+1)}\,$. Hence we will prove the
injectivity of the homomorphism defined by (63).

Let $\mathcal{S}$ be the set of all products of the generators
$T_{ij}^{{\hskip 0.5pt}(r+1)}$ of the algebra $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. Let us call a pair of triples
$(i,j,r)$ and $(k,l,s)$ _misordered_ if $(i,j,r)>(k,l,s)\,$, or
$(i,j,r)=(k,l,s)$ and $\,{\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}=1$ in
$\mathbb{Z}_{2}\,$. For any product $A\in\mathcal{S}$ let $\mu(A)$ be the
number of pairs of factors of $A$ corresponding to misordered pairs of
triples. The pairs of factors of $A$ counted by $\mu{\hskip 0.5pt}(A)$ need
not to be adjacent. Let $\nu{\hskip 0.5pt}(A)$ be the number of factors of $A$
corresponding to the triples which are not in $\mathcal{J}\,$.

Let $A$ and $B$ be any two products from the set $\mathcal{S}\,$. Let $a$ and
$b$ be the numbers of their factors respectively. Let $c$ and $d$ be
respectively the sums of the degrees of their factors relative to the
$\mathbb{Z}{\hskip 0.5pt}$-filtration on $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ from Section 13. We will write $A\prec
B$ if $c<d\,$, or $c=d$ but $a<b\,$, or $c=d$ and $a=b$ but $\nu{\hskip
0.5pt}(A)<\nu{\hskip 0.5pt}(B)\,$, or $A$ is a permutation of factors of $B$
and $\mu{\hskip 0.5pt}(A)<\mu{\hskip 0.5pt}(B)\,$. The relation denoted by
$\prec$ is a partial ordering on the set $\mathcal{S}{\hskip 0.5pt}$, with
every descending chain terminating. Notice that the condition $A\prec B$
implies that $XA\,Y\prec XB\,Y$ for any $X,Y\in\mathcal{S}\,$.

Let $S(u)$ be the element of the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})\otimes\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,[[u^{-1}]]$ inverse to
$T^{\,\ast{\hskip-0.5pt}}(-{\hskip 0.5pt}u)\,$. When determining the inverse
element here we do not use any relations in $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. The relation (48) with $Z(u)=1$ can
be now written as $S(u)=T(u)\,$. For any $i,j$ define a formal power series
$S_{ij}(u)$ in $u^{-1}$ with coefficients in $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$,

$S(u)=\sum_{i,j}\,E_{{\hskip 0.5pt}ij}\otimes S_{ij}(u)\,.\vspace{-4pt}$

Define the elements $S_{ij}^{{\hskip 0.5pt}(1)},S_{ij}^{{\hskip
0.5pt}(2)},\,\ldots\in\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ as the coefficients of the series
$S_{ij}(u)$ at $u^{-1},u^{-2},\,\ldots$ respectively. For any $r\geqslant 0$

$S^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}=T_{{\hskip 0.5pt}-j{\hskip
0.5pt},{\hskip 0.5pt}-i}^{{\hskip
0.5pt}(r+1)}\,{(-1)}^{\,{\bar{\imath}\,}{\hskip
0.5pt}{\bar{\jmath}\,}+\,{\bar{\jmath}\,}{\hskip 0.5pt}+\,r}+C$ (64)

where $C$ is a linear combination of products of two or more generators of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ with the sums
of the degrees of factors by the $\mathbb{Z}{\hskip 0.5pt}$-filtration from
Section 13 being less than $r\,$. These products occur in the linear
combination $C$ only if $r>0\,$.

Take any $(i,j,r)\notin\mathcal{J}\,$. If $i\neq-{\hskip 0.5pt}j\,$, then
$(-{\hskip 0.5pt}j{\hskip 0.5pt},-{\hskip 0.5pt}i{\hskip
0.5pt},r)\in\mathcal{J}\,$. In this case we will replace any factor
$T^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}$ of $B$ by the right hand side of
(64). But if $i=-{\hskip 0.5pt}j$ and
$\,{\bar{\imath}\,}{\hskip-0.5pt}+{\hskip 0.5pt}r$ is even, then we will
replace $T^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}$ by $C/2\,$. We will call
either replacement a _reduction_ of $B\,$. It does not change the value of
$B\in\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ and is
compatible with $\prec$ on $\mathcal{S}\,$, so that its result is a linear
combination of products $A\prec B\,$.

For any triples $(i,j,r)$ and $(k,l,s)$ the relation (37) gives an equality in
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$

$T^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}\,T^{{\hskip 0.5pt}(s+1)}_{{\hskip
0.5pt}kl}=T^{{\hskip 0.5pt}(s+1)}_{{\hskip 0.5pt}kl}\,T^{{\hskip
0.5pt}(r+1)}_{{\hskip 0.5pt}ij}\,{(-1)}^{{\hskip
0.5pt}(\,{\bar{\imath}\,}+\,{\bar{\jmath}\,}\,)({\hskip
0.5pt}{\bar{k}\,}+\,{\bar{l}\,}\,)}+D$ (65)

where $D$ is a linear combination of single generators of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of degree
$r+s$ by the $\mathbb{Z}{\hskip 0.5pt}$-filtration from Section 13, and of
products of two generators with the sums of the degrees of factors being equal
to $r+s-1\,$. Here we use the expansions

$\frac{1}{u-v}\,=u^{-{\hskip 0.5pt}1}+u^{-{\hskip
0.5pt}2}\,v\,+\,\ldots\quad\text{and}\quad\frac{1}{u+v}\,=u^{-{\hskip
0.5pt}1}-u^{-{\hskip 0.5pt}2}\,v\,+\,\ldots$

then equate the coefficients at $u^{-{\hskip 0.5pt}r-1}{\hskip 0.5pt}v^{-s-1}$
of the series at the two sides of (37).

If $(i,j,r)>(k,l,s)$ then we will replace any product of the corresponding two
adjacent factors of $B$ by the right hand side of (65). But if
$(i,j,r)=(k,l,s)$ and $\,{\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}=1$ in
$\mathbb{Z}_{2}$ then we will replace the product of the two factors of $B$ by
$D/2\,$. The value of $B\in\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ will not change then. Either
replacement will be called a _reduction_ of $B$ too. It is compatible with the
ordering $\prec$ on $\mathcal{S}$ as well.

When proving linear the independence of the set $\mathcal{T}$ in
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$, we may
assume that $(1,1,r)\notin\mathcal{J}$ for every $r\geqslant 0\,$. The
collection of relations $S^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}11}=T^{{\hskip
0.5pt}(r+1)}_{{\hskip 0.5pt}11}$ can be written as the equality
$S_{11}(u)=T_{11}(u)\,$. Using only the relation (48) in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})\otimes\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip 0.5pt}[[u^{-1}]]\,$, this
equality gives $Z(u)=1\,$. But the relation (48) follows from (36), see
Section 8. So the relation $Z(u)=1$ follows from (36) and from the relations
$S^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}11}=T^{{\hskip 0.5pt}(r+1)}_{{\hskip
0.5pt}11}$ for every $r\geqslant 0\,$. Under our assumption on the choice of
$\mathcal{J}{\hskip 0.5pt}$, the latter relations are used to define
reductions via (64).

Let now us multiply each side of the relation (36) by $P\otimes 1$ on the left
and on the right. Also exchange the variables $u$ and $v\,$. The result is the
relation

$(R{\hskip 0.5pt}(-{\hskip 0.5pt}u\,,-{\hskip 0.5pt}v)\otimes 1)\,{\hskip
0.5pt}T_{2}(v)\,T_{1}(u)=T_{1}(u)\,T_{2}(v)\,(R{\hskip 0.5pt}(-{\hskip
0.5pt}u\,,-{\hskip 0.5pt}v)\otimes 1)$ (66)

since by (11)

$P\,R{\hskip 0.5pt}(v\,,{\hskip-0.5pt}u)\otimes 1)\,P=R{\hskip 0.5pt}(-{\hskip
0.5pt}u\,,-{\hskip 0.5pt}v)\,.$ (67)

But (66) can also be obtained by multiplying each side of (36) on the left and
on the right by $R{\hskip 0.5pt}(-{\hskip 0.5pt}u\,,-{\hskip 0.5pt}v)\otimes
1$ and then using (14). Inductively, this remark shows that the relations (65)
with $(i,j,r)\geqslant(k,l,s)$ imply all other relations (65). In the
induction argument, we use the $\mathbb{Z}{\hskip 0.5pt}$-filtration on
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ from Section
12.

If $(i,j)=(k,l)$ and $\,{\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}=0$ in
$\mathbb{Z}_{2}\,$, then the relation (37) simply means that the coefficients
of the series $T_{ij}(u)$ pairwise commute. Indeed, in this case the
expressions in the second and the third lines of (37) vanish. Then (37)
becomes

$T_{ij}(u)\,T_{ij}(v)-T_{ij}(v)\,T_{ij}(u)=\frac{T_{ij}(u)\,T_{ij}(v)-T_{ij}(v)\,T_{ij}(u)}{u-v}\,.$

Equivalently,

$T_{ij}(u)\,T_{ij}(v)=T_{ij}(v)\,T_{ij}(u)\,.\hskip 2.0pt$

In particular, we can eliminate from the definition of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ the relations
(65) with $(i,j,r)=(k,l,s)$ and $\,{\bar{\imath}\,}+{\hskip
0.5pt}{\bar{\jmath}\,}=0$ in $\mathbb{Z}_{2}$ as tautological relations. But
to define reductions via (65), we used (65) only when $(i,j,r)>(k,l,s)$ or
$(i,j,r)=(k,l,s)$ and $\,{\bar{\imath}\,}+{\hskip 0.5pt}{\bar{\jmath}\,}=1$ in
$\mathbb{Z}_{2}\,$. The above arguments show that the used relations imply all
other relations (65). Thus our reductions involve all defining relations of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

Let us now examine the _inclusion ambiguities_ of our set of reductions. They
correspond to misordered pairs of triples where one or both triples are not in
$\mathcal{J}{\hskip 0.5pt}$. Let us show that they are resolvable relative to
$\prec$ in the sense of (B, 4, Section 1).

Take any product $T^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}\,T^{{\hskip
0.5pt}(s+1)}_{{\hskip 0.5pt}kl}$ where the pair of triples $(i,j,r)$ and
$(k,l,s)$ is misordered. Suppose that $(i,j,r)\notin\mathcal{J}$ but
$(k,l,s)\in\mathcal{J}\,$. Then we can apply to the above product the
reduction via (64) in the first factor, or alternatively the reduction via
(65). The two results coincide by comparing (36) to the equality

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,{\hskip
0.5pt}S_{1}(u)\,T_{2}(v)=T_{2}(v)\,S_{1}(u)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,.$ (68)

Note that to prove the coincidence here we use equalities of linear
combinations of products from $\mathcal{S}$ which precede $T^{{\hskip
0.5pt}(r+1)}_{{\hskip 0.5pt}ij}\,T^{{\hskip 0.5pt}(s+1)}_{{\hskip 0.5pt}kl}$
relative to the partial ordering.

To get the equality (68) we apply the map
$\tau\,\pi\otimes{\mathrm{id}}\otimes{\mathrm{id}}$ to both sides of (36) and
then use (12). This yields the equality

$T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,({\hskip 0.5pt}R{\hskip 0.5pt}(u\,,-{\hskip
0.5pt}v)\otimes 1{\hskip 0.5pt})\,T_{2}(v)=T_{2}(v)\,({\hskip 0.5pt}R{\hskip
0.5pt}(u\,,-{\hskip 0.5pt}v)\otimes 1{\hskip
0.5pt})\,T^{\,\ast{\hskip-0.5pt}}_{1}(u)\,.$

By multiplying both sides of this equality by
$T^{\,\ast{\hskip-0.5pt}}_{1}(u)^{-1}$ on the left and on the right, changing
$u$ to $-{\hskip 0.5pt}u$ and then using (14) we indeed get (68).

The case when $(i,j,r)\in\mathcal{J}$ but $(k,l,s)\notin\mathcal{J}$ can be
treated in a similar way, but using instead of (68) the equality

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,{\hskip
0.5pt}T_{1}(u)\,S_{2}(v)=S_{2}(v)\,T_{1}(u)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,.$ (69)

To get this equality multiply each side of (36) by $R{\hskip 0.5pt}(-{\hskip
0.5pt}u\,,-{\hskip 0.5pt}v)\otimes 1$ on the left and on the right, and then
use (14). Apply the map ${\mathrm{id}}\otimes\tau\,\pi\otimes{\mathrm{id}}$ to
both sides of the resulting equality and then use (13). In this way we obtain
the equality

$T_{1}(u)\,({\hskip 0.5pt}R{\hskip 0.5pt}(u\,,-{\hskip 0.5pt}v)\otimes
1{\hskip
0.5pt})\,T^{\,\ast{\hskip-0.5pt}}_{2}(v)=T^{\,\ast{\hskip-0.5pt}}_{2}(v)\,({\hskip
0.5pt}R{\hskip 0.5pt}(u\,,-{\hskip 0.5pt}v)\otimes 1{\hskip
0.5pt})\,T_{1}(u)\,.$

Multiplying its both sides of this equality on the left and on the right by
$T^{\,\ast{\hskip-0.5pt}}_{2}(v)^{-1}$ and then changing $v$ to $-{\hskip
0.5pt}v$ yields (69).

The case when $(i,j,r)\notin\mathcal{J}$ and $(k,l,s)\notin\mathcal{J}$ can
also be treated in a similar way, but using instead of (68) the equality

$(R{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,{\hskip
0.5pt}S_{1}(u)\,S_{2}(v)=S_{2}(v)\,S_{1}(u)\,(R{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1)\,.$

To get this equality, multiply both sides of (44) on the left and on the right
first by $T^{\,\ast{\hskip-0.5pt}}_{1}(u)^{-1}$ and then by
$T^{\,\ast{\hskip-0.5pt}}_{2}(v)^{-1}\,$. Then change $u\,,v$ to $-{\hskip
0.5pt}u\,,{\hskip-0.5pt}-{\hskip 0.5pt}v$ respectively.

Finally, let us examine the _overlap ambiguities_ of our set of reductions.
They correspond to the sequences of triples of length three, such that the
pair of first two triples and the pair of the last two triples in the sequence
are misordered. Let us show that they are also resolvable relative to $\prec$
in the sense of (B, 4, Section 1).

Take any product $T^{{\hskip 0.5pt}(r+1)}_{{\hskip 0.5pt}ij}\,T^{{\hskip
0.5pt}(s+1)}_{{\hskip 0.5pt}kl}\,T^{{\hskip 0.5pt}(t+1)}_{{\hskip 0.5pt}gh}$
such that the pair of triples $(i,j,r)$ and $(k,l,s)$ is misordered, and so is
the pair of triples $(k,l,s)$ and $(g,h,t{\hskip 0.5pt})\,$. We can apply to
the product the reduction via (65) in the first two factors. Alternatively, in
the last two factors we can apply the reduction via the equality obtained by
replacing $(i,j,r)$ and $(k,l,s)$ in (65) respectively by $(k,l,s)$ and
$(g,h,t{\hskip 0.5pt})\,$.

To see that the results of two reductions coincide, we can continue using
(65). Thus for the first reduction we will use the equality

$\displaystyle({\hskip 0.5pt}R_{{\hskip
0.5pt}23}(v\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}12}(u\,,{\hskip-0.5pt}v)\otimes 1{\hskip 0.5pt})\,{\hskip
0.5pt}T_{1}(u)\,T_{2}(v)\,T_{3}(w)=$ $\displaystyle
T_{3}(w)\,T_{2}(v)\,T_{1}(u)\,{\hskip 0.5pt}({\hskip 0.5pt}R_{{\hskip
0.5pt}23}(v\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}12}(u\,,{\hskip-0.5pt}v)\otimes 1{\hskip 0.5pt})$

in the notation (35) with $n=3\,$. Here $w$ is another formal variable. In the
same notation, for the second reduction we will use the equality

$\displaystyle({\hskip 0.5pt}R_{{\hskip
0.5pt}12}(u\,,{\hskip-0.5pt}v)\,R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}23}(v\,,{\hskip-0.5pt}w)\otimes 1{\hskip 0.5pt})\,{\hskip
0.5pt}T_{1}(u)\,T_{2}(v)\,T_{3}(w)=$ $\displaystyle
T_{3}(w)\,T_{2}(v)\,T_{1}(u)\,{\hskip 0.5pt}({\hskip 0.5pt}R_{{\hskip
0.5pt}12}(u\,,{\hskip-0.5pt}v)\,R_{{\hskip
0.5pt}13}(u\,,{\hskip-0.5pt}w)\,R_{{\hskip
0.5pt}23}(v\,,{\hskip-0.5pt}w)\otimes 1{\hskip 0.5pt})\,.$

The two results will coincide due to (15). Note that to prove the coincidence
here we will use equalities of linear combinations of products from
$\mathcal{S}$ which precede $T^{{\hskip 0.5pt}(r+1)}_{{\hskip
0.5pt}ij}\,T^{{\hskip 0.5pt}(s+1)}_{{\hskip 0.5pt}kl}\,T^{{\hskip
0.5pt}(t+1)}_{{\hskip 0.5pt}gh}$ relative to the partial ordering.

So both the inclusion and the overlap ambiguities of our set of reductions are
resolvable. But the elements of the set $\mathcal{T}$ cannot be changed by the
reductions. The linear independence of $\mathcal{T}$ in
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is now
ensured by (B, 4, Theorem 1.2). ∎

###### Corollary 1

The elements of $\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ corresponding to $T^{{\hskip
0.5pt}(r+1)}_{{\hskip 0.5pt}ij}$ with the triples $(i,j,r)\in\mathcal{J}$ are
free generators of this supercommutative algebra.

Our proof of Theorem 2 also implies the Poincaré -Birkhoff -Witt Theorem for
the algebra $\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})\,$,
see (MM, 15, Theorem 5.15). Going back to our Section 10 we obtain

###### Corollary 2

The homomorphism $\operatorname{U}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\to\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ defined by the mapping (56) is an
embedding of $\mathbb{Z}_{2}$-graded Hopf algebras.

Let us relate the Yangian $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ to the Yangian
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip
0.5pt}N}{\hskip-0.5pt})$ of the general linear Lie algebra
$\mathfrak{gl}_{{\hskip 0.5pt}N}\,$. The latter Yangian is a deformation of
the universal enveloping algebra of the polynomial current Lie algebra
$\mathfrak{gl}_{{\hskip 0.5pt}N}[u]$ in the class of Hopf algebras. By
comparing (37) with the definition of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{gl}_{{\hskip 0.5pt}N}{\hskip-0.5pt})$ as given in (N4, 21,
Section 1) we immediately obtain another corollary to Theorem 2.

###### Corollary 3

The associative subalgebra of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ generated by elements $T^{{\hskip
0.5pt}(r)}_{{\hskip 0.5pt}ij}$ with $i\,,j>0$ is isomorphic to the algebra
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip
0.5pt}N}{\hskip-0.5pt})\,$.

Note that the associative subalgebra of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ appearing in Corollary 3 is not a Hopf
subalgebra, see the definition (54). Thus the isomorphism of Corollary 3 is
not an isomorphism of Hopf algebras.

### 15

In this section we will explicitly describe the centre of the Yangian
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. An element
of an $\mathbb{Z}_{2}{\hskip 0.5pt}$-graded associative algebra is central if
its supercommutator with every element of the algebra is zero.

First consider the subalgebra of the extended Yangian
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ generated by
the elements $T^{{\hskip 0.5pt}(r)}_{{\hskip 0.5pt}ij}$ with $i\,,j>0\,$. This
subalgebra of $\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$
is isomorphic to $\operatorname{Y}({\hskip 0.5pt}\mathfrak{gl}_{{\hskip
0.5pt}N}{\hskip-0.5pt})\,$. Similarly to Corollary 3 this isomorphism property
follows from Theorem 2 by (37). Consider the formal power series in $u^{-1}$
with coefficients in this subalgebra

$A{\hskip 0.5pt}(u)=\sum_{\sigma}\,(-1)^{{\hskip
0.5pt}\sigma}\,T_{\,1,\sigma(1)}(u)\,T_{\,2,\sigma(2)}(u+1)\,\ldots\,T_{\,N,\sigma(N)}({\hskip
0.5pt}u+N-1{\hskip 0.5pt})$ (70)

where $\sigma$ ranges over the symmetric group $\mathfrak{S}_{N}$ permuting
the indices $1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}N{\hskip 0.5pt}$. The power series in $(u+1)^{{\hskip 0.5pt}-1}{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}(u+N-1)^{{\hskip
0.5pt}-1}$ should be reexpanded in $u^{-1}\,$. The series $A{\hskip 0.5pt}(u)$
has the leading term $1\,$. The coefficients of $A{\hskip 0.5pt}(u)$ at
$u^{-1},u^{-2},{\hskip 0.5pt}\ldots$ are free generators of the centre of that
subalgebra of $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$, see (MNO, 16, Section 2). All these
coefficients have $\mathbb{Z}_{2}{\hskip 0.5pt}$-degree zero, hence the
supercommutator with any of them in the algebra $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is the usual commutator. By (MNO, 16,
Proposition 2.7)

$A{\hskip 0.5pt}(u)=\sum_{\sigma}\,(-1)^{{\hskip
0.5pt}\sigma}\,T_{\,\sigma(1),1}({\hskip 0.5pt}u+N-1{\hskip
0.5pt})\,\ldots\,T_{\,\sigma(N-1),N-1}({\hskip 0.5pt}u+1{\hskip
0.5pt})\,T_{\,\sigma(N),N}(u)\,.$

Now take the series in $u^{-1}$ with coefficients in $\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ and the leading term $1$

$A{\hskip 0.5pt}(u)\,A{\hskip 0.5pt}({\hskip 0.5pt}1-N-u{\hskip 0.5pt})\,.$
(71)

###### Theorem 3

The coefficients of the series (71) belong to the centre of
$\operatorname{X}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

We will prove this theorem in the following sections. Let us now employ it to
describe the centre of the algebra $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. We will denote by $B^{{\hskip
0.5pt}(1)},B^{{\hskip 0.5pt}(2)}\ldots$ the images in
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of the
coefficients of the series (71) at $u^{-1},u^{-2},{\hskip 0.5pt}\ldots$
respectively. These images are central in $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ by Theorem 3. They all are of
$\mathbb{Z}_{2}{\hskip 0.5pt}$-degree zero.

###### Corollary 4

The coefficients $B^{{\hskip 0.5pt}(2)},B^{{\hskip 0.5pt}(4)}\ldots$ freely
generate the centre of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

###### Proof

The elements $B^{{\hskip 0.5pt}(2)},B^{{\hskip 0.5pt}(4)}\ldots$ of
$\operatorname{Y}({\hskip 0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ are central
by Theorem 3. It suffices to prove now that the images of these elements in
the graded algebra $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ are free generators of its centre. For
$r\geqslant 1$ the image of $B^{{\hskip 0.5pt}({\hskip 0.5pt}2{\hskip
0.5pt}r)}$ in $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ equals

$2\,Y_{11}^{{\hskip 0.5pt}({\hskip 0.5pt}2{\hskip
0.5pt}r)}+\ldots+2\,Y_{NN}^{{\hskip 0.5pt}({\hskip 0.5pt}2{\hskip
0.5pt}r)}\,.$ (72)

By Theorem 2 the algebra $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ is isomorphic to the universal
enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{g}{\hskip-0.5pt})$ via (63). Under this isomorphism the
element (72) of $\operatorname{gr}^{\,\prime}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ with any $r\geqslant 1$ corresponds to
the element of $\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})$

$-\,2\,g_{{\hskip 0.5pt}11}^{{\hskip 0.5pt}({\hskip 0.5pt}2{\hskip
0.5pt}r-1)}-\ldots-2\,g_{{\hskip 0.5pt}NN}^{{\hskip 0.5pt}({\hskip
0.5pt}2{\hskip 0.5pt}r-1)}=-\,2\,({\hskip 0.5pt}e_{{\hskip
0.5pt}11}+e_{{\hskip 0.5pt}-1{\hskip 0.5pt},{\hskip
0.5pt}-1}+\ldots+e_{{\hskip 0.5pt}NN}+e_{{\hskip 0.5pt}-N{\hskip
0.5pt},{\hskip 0.5pt}-N}{\hskip 0.5pt})\,u^{{\hskip 0.5pt}2{\hskip 0.5pt}r-1}$

from our Section 14. The latter elements freely generate of the centre of
$\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})\,$.

Indeed, they are algebraically independent due to the Poincaré -Birkhoff-Witt
Theorem for $\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})\,$,
see (MM, 15, Theorem 5.15). To show that they generate the centre of
$\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})$ consider the
quotient of $\operatorname{U}({\hskip 0.5pt}\mathfrak{g}{\hskip-0.5pt})$ by
the ideal they generate. This quotient is isomorphic to the universal
enveloping algebra $\operatorname{U}({\hskip
0.5pt}\mathfrak{h}{\hskip-0.5pt})$ of the Lie superalgebra

$\mathfrak{h}=\\{\,h{\hskip 0.5pt}(u)\in\mathfrak{a}{\hskip
0.5pt}[u]:\omega{\hskip 0.5pt}({\hskip 0.5pt}h{\hskip 0.5pt}(u))=h{\hskip
0.5pt}(-{\hskip 0.5pt}u){\hskip 0.5pt}\\}$

where $\mathfrak{a}$ is the quotient of Lie superalgebra
$\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ by the span of a central element

$e_{{\hskip 0.5pt}11}+e_{{\hskip 0.5pt}-1{\hskip 0.5pt},{\hskip
0.5pt}-1}+\ldots+e_{{\hskip 0.5pt}NN}+e_{{\hskip 0.5pt}-N{\hskip
0.5pt},{\hskip 0.5pt}-N}\,.$

The automorphism $\omega$ of $\mathfrak{gl}_{{\hskip 0.5pt}N|N}$ maps this
element to its negative and therefore descends to the quotient
$\mathfrak{a}\,$. But the centre of the Lie superalgebra $\mathfrak{a}$ is
trivial. Hence the centre of $\operatorname{U}({\hskip
0.5pt}\mathfrak{h}{\hskip-0.5pt})$ is also trivial by (N3, 20, Proposition
3.6). ∎

To finish this section let us consider the images of the elements $B^{{\hskip
0.5pt}(1)},B^{{\hskip 0.5pt}(2)}\ldots$ of $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ in the supercommutative algebra
$\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$. In the notation of Section 12 let
$X(u)$ the $2{\hskip 0.5pt}N\times 2{\hskip 0.5pt}N$ matrix whose $ij$ entry
is the series

$X_{ij}(u)\,{(-1)}^{\,{\bar{\imath}\,}{\hskip 0.5pt}{\bar{\jmath}\,}{\hskip
0.5pt}+\,{\bar{\jmath}\,}}\,.$

We index the rows and columns of $X(u)$ by the numbers $1{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N,-{\hskip
0.5pt}1{\hskip 0.5pt},\ldots,-{\hskip 0.5pt}N{\hskip 0.5pt}$.

The matrix $X(u)$ is invertible. Let $X^{\,\prime}_{{\hskip 0.5pt}ij}(u)$ be
the $ij$ entry of the inverse matrix. Then the relations (57) imply that for
any indices $i$ and $j$

$X_{{\hskip 0.5pt}ij}(-{\hskip 0.5pt}u)=X^{\,\prime}_{{\hskip 0.5pt}-j{\hskip
0.5pt},{\hskip 0.5pt}-i{\hskip 0.5pt}}(u)\,.$

Therefore by using the definition (70) , the image in
$\operatorname{gr}\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt}){\hskip 0.5pt}[[u^{-1}]]$ of the series

$1+B^{{\hskip 0.5pt}(1)}{\hskip 0.5pt}u^{-1}+B^{{\hskip 0.5pt}(2)}{\hskip
0.5pt}u^{-2}+\ldots$ (73)

can be written as the product of the sum

$\sum_{\sigma}\,(-1)^{{\hskip 0.5pt}\sigma}\,X_{{\hskip
0.5pt}1,\sigma(1)}(u)\,\ldots\,X_{{\hskip 0.5pt}N,\sigma(N)}(u)\vspace{-8pt}$

by the sum

$\sum_{\sigma}\,(-1)^{{\hskip 0.5pt}\sigma}\,X^{\,\prime}_{{\hskip
0.5pt}-\sigma(1),-1}(u)\,\ldots\,X^{\,\prime}_{{\hskip
0.5pt}-\sigma(N),-N}(u)\,.$

This product is just the Berezinian or the superdeterminant of the matrix
$X(u)\,$. Hence the series (73) is an analogue for $\operatorname{Y}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})$ of the quantum Berezinian N1 (18) for
the Yangian of the general linear Lie superalgebra $\mathfrak{gl}_{{\hskip
0.5pt}M|N}\,$, see N5 (22) for details.

### 16

We will now begin our proof of Theorem 3. We will be use the same method as in
the proof of Proposition 3. However, the calculations will be more involved.
Fix any $n\geqslant 1$ and take the rational function of $u$ with values in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}n}$

$H(u)=\prod_{1\leqslant p<q\leqslant n}\,R_{\,pq{\hskip 0.5pt}}({\hskip
0.5pt}u+n-p\,,{\hskip-0.5pt}u+n-q{\hskip 0.5pt})$

where the factors are arranged from left to right by ordering their pairs of
indices $(p\,,q)$ lexicographically. The ordering of factors can be altered by
employing (15), without affecting the function. By using (36) repeatedly we
get the equality

$({\hskip 0.5pt}H(u)\otimes 1{\hskip 0.5pt}){\hskip
0.5pt}\,T_{1}(u+n-1)\,\ldots\,T_{n}(u)=T_{n}(u)\,\ldots\,T_{1}(u+n-1){\hskip
0.5pt}\,({\hskip 0.5pt}H(u)\otimes 1{\hskip 0.5pt})$

of formal power series in $u^{-1}$ with coefficients in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}n}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

###### Lemma 1

For any $1\leqslant p<q\leqslant n$ we have the equalities in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}n}{\hskip 0.5pt}(u)$

$P_{{\hskip 0.5pt}pq}{\hskip 0.5pt}H(u)=-\,H(u)\quad\text{and}\quad Q_{{\hskip
0.5pt}pq}{\hskip 0.5pt}H(u)=0\,.$

###### Proof

For $n=1$ we have nothing to prove. In particular, in this case $H(u)=1\,$.
Consider the case of $n=2\,$. Then $H(u)=R{\hskip
0.5pt}(u+1\,,{\hskip-0.5pt}u)$ while

$P\,R{\hskip 0.5pt}(u+1\,,{\hskip-0.5pt}u)=P\,\biggl{(}1-P+\frac{Q}{2{\hskip
0.5pt}u+1}\,\biggr{)}=P-1-\frac{Q}{2{\hskip 0.5pt}u+1}=-\,R{\hskip
0.5pt}(u+1\,,{\hskip-0.5pt}u)$

by the equality $P^{\,2}=1$ and by the first equality in (11). By the second
and third equalities in (11) we also have

$Q\,R{\hskip 0.5pt}(u+1\,,{\hskip-0.5pt}u)=Q\,\biggl{(}1-P+\frac{Q}{2{\hskip
0.5pt}u+1}\,\biggr{)}=Q-Q-\frac{0}{2{\hskip 0.5pt}u+1}=0\,.$

Now for any $n>2$ and for any given index $p<n$ let $q=p+1\,$. By (15) we can
rearrange the factors of $H(u)$ so that $R_{\,p,p+1{\hskip 0.5pt}}({\hskip
0.5pt}u+n-p\,,{\hskip-0.5pt}u+n-p-1{\hskip 0.5pt})$ is the leftmost factor.
Then the above calculations for $n=2$ imply the equalities of our lemma for
any $n{\hskip 0.5pt}$. Hence our lemma holds whenever $q=p+1\,$.

In turn, this implies the required equalities for every $q>p\,$. Indeed, the
set of all elements $P_{{\hskip 0.5pt}pq}$ and $Q_{{\hskip 0.5pt}pq}$ with
$p<q$ is generated in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}n}$
by those elements where $q=p+1\,$. ∎

It follows from Lemma 1 that every value of the normalized function
$H(u)/{\hskip 0.5pt}n{\hskip 0.5pt}!$ is idempotent in the algebra
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}n}\,$. Indeed, by using the lemma we get

$H(u)\,H(u)=\prod_{1\leqslant p<q\leqslant
n}\biggl{(}\,1+\frac{1}{p-q}\,\biggr{)}\cdot H(u)=n{\hskip 0.5pt}!\,H(u)\,.$

Elements of $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}n}$ act on
$({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}n}$ by
(1) and (2) . Take any vector $e_{{\hskip 0.5pt}i_{1}}\otimes\ldots\otimes
e_{{\hskip 0.5pt}i_{n}}$ with only positive indices $i_{{\hskip
0.5pt}1}{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}i_{{\hskip 0.5pt}n}\,$. Then all elements $Q_{{\hskip 0.5pt}pq}$ with
$1\leqslant p<q\leqslant n$ annihilate this vector. So the image $H(u){\hskip
0.5pt}({\hskip 0.5pt}e_{{\hskip 0.5pt}i_{1}}\otimes\ldots\otimes e_{{\hskip
0.5pt}i_{n}})$ equals

$\prod_{1\leqslant p<q\leqslant n}\,\biggl{(}\,1+\frac{P_{{\hskip
0.5pt}pq}}{p-q}\,\biggr{)}\cdot({\hskip 0.5pt}e_{{\hskip
0.5pt}i_{1}}\otimes\ldots\otimes e_{{\hskip
0.5pt}i_{n}})=\sum_{\sigma}\,(-1)^{{\hskip 0.5pt}\sigma}\,e_{{\hskip
0.5pt}i_{{\hskip 0.5pt}\sigma(1)}}\otimes\ldots\otimes e_{{\hskip
0.5pt}i_{{\hskip 0.5pt}\sigma(n)}}$

where $\sigma$ ranges over the group $\mathfrak{S}_{n}\,$. The last equality
is well known (MNO, 16, Section 2).

###### Lemma 2

For any $1\leqslant p<n$ we have an equality in $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}n}{\hskip 0.5pt}(u)$

$H(u)\,Q_{{\hskip 0.5pt}p,p+1}=({\hskip 0.5pt}2{\hskip 0.5pt}u+2{\hskip
0.5pt}n-2{\hskip 0.5pt}p-1{\hskip 0.5pt})\,H(u)\,(1+P_{{\hskip
0.5pt}p,p+1})\,.$

###### Proof

For $n=1$ we have nothing to prove. Let $n=2\,$. Then $H(u)=R{\hskip
0.5pt}(u+1\,,{\hskip-0.5pt}u)$ while by the relation (14) we have

$R{\hskip 0.5pt}(u+1\,,{\hskip-0.5pt}u)\,R{\hskip 0.5pt}(-\,u-1\,,-\,u)=0\,.$

Multiplying this equality by $2{\hskip 0.5pt}u+1$ and using the definition of
the second factor at the left hand side we get the equality

$R{\hskip 0.5pt}(u+1\,,{\hskip-0.5pt}u)\,({\hskip 0.5pt}({\hskip
0.5pt}2{\hskip 0.5pt}u+1){\hskip 0.5pt}({\hskip 0.5pt}1+P{\hskip
0.5pt})-Q{\hskip 0.5pt})=0\,.$

This implies our lemma for $n=2$ and $p=1\,$. For any $n>2$ and $1\leqslant
p<n$ by using (15) we can rearrange the factors of the product $H(u)$ so that
the factor

$R_{\,p,p+1{\hskip 0.5pt}}({\hskip 0.5pt}u+n-p\,,{\hskip-0.5pt}u+n-p-1{\hskip
0.5pt})$

is the rightmost. Then the above calculation with $u$ replaced by $u+n-p-1$
implies the lemma. ∎

### 17

Let us continue our proof of Theorem 3. For any $n\geqslant 1$ consider the
rational function of $u\,,v$ with values in the algebra
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}(n+1)}$

$F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)=R_{\,1,n+1{\hskip 0.5pt}}({\hskip
0.5pt}u+n-1\,,{\hskip-0.5pt}v{\hskip 0.5pt})\,\ldots\,R_{\,n,n+1{\hskip
0.5pt}}({\hskip 0.5pt}u\,,{\hskip-0.5pt}v{\hskip 0.5pt})\,.$

By using (36) repeatedly we get the equality of formal power series in
$u^{-1},v^{-1}$

$\displaystyle({\hskip 0.5pt}F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes
1{\hskip 0.5pt}){\hskip 0.5pt}\,T_{{\hskip 0.5pt}1}(u+n-1)\,\ldots\,T_{{\hskip
0.5pt}n}(u)\,T_{{\hskip 0.5pt}n+1}(v)=$ $\displaystyle T_{{\hskip
0.5pt}n+1}(v)\,T_{{\hskip 0.5pt}1}(u+n-1)\,\ldots\,T_{{\hskip
0.5pt}n}(u)\,{\hskip 0.5pt}({\hskip 0.5pt}F{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1{\hskip 0.5pt})$ (74)

which have their coefficients in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}(n+1)}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

###### Lemma 3

For any $n\geqslant 1$ the rational function $({\hskip 0.5pt}H(u)\otimes
1{\hskip 0.5pt})\,F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ of $u\,,v$ with values
in $(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}(n+1)}$ is equal to

$\biggl{(}\,1-\frac{P_{\,1,n+1}+\ldots+P_{\,n,n+1}}{u-v}+\frac{Q_{\,1,n+1}+\ldots+Q_{\,n,n+1}}{u+v+n-1}\,\biggr{)}\,({\hskip
0.5pt}H(u)\otimes 1{\hskip 0.5pt})$

plus terms divisible on the left by $Q_{{\hskip
0.5pt}pq}\in(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}(n+1)}$ with $1\leqslant
p<q\leqslant n\,$.

###### Proof

Using the relation (15) we can rewrite the product $({\hskip 0.5pt}H(u)\otimes
1{\hskip 0.5pt})\,F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ as

$R_{\,n,n+1{\hskip 0.5pt}}({\hskip 0.5pt}u\,,{\hskip-0.5pt}v{\hskip
0.5pt})\,\ldots\,R_{\,1,n+1{\hskip 0.5pt}}({\hskip
0.5pt}u+n-1\,,{\hskip-0.5pt}v{\hskip 0.5pt})\,({\hskip 0.5pt}H(u)\otimes
1{\hskip 0.5pt})\,.$ (75)

We will prove by induction on $n$ that the latter product is equal to the sum
in the lemma. If $n=1$ then $H(u)=1$ while
$F(u\,,{\hskip-0.5pt}v)=R(u\,,{\hskip-0.5pt}v)\,$. Then the required equality
is just the definition of $R(u\,,{\hskip-0.5pt}v)$ as given in Section 5.

Suppose that $n>1\,$. By using the induction assumption and the definition of
$H(u)$ for $n-1$ instead of $n\,$, the product (75) is equal to

$\displaystyle\biggl{(}\,1-\frac{P_{\,2,n+1}+\ldots+P_{\,n,n+1}}{u-v}+\frac{Q_{\,2,n+1}+\ldots+Q_{\,n,n+1}}{u+v+n-2}\,\biggr{)}\,\times$
$\displaystyle\biggl{(}\,1-\frac{P_{\,1,n+1}}{u-v+n-1}+\frac{Q_{\,1,n+1}}{u+v+n-1}\,\biggr{)}\,({\hskip
0.5pt}H(u)\otimes 1{\hskip 0.5pt})$

plus terms divisible on the left by $Q_{{\hskip
0.5pt}pq}\in(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}(n+1)}$ with $2\leqslant
p<q\leqslant n\,$. We also used the relations (15). The product of two sums in
the brackets equals

$\displaystyle 1-\,{\hskip
0.5pt}\frac{P_{\,1,n+1}}{u-v+n-1}+\frac{Q_{\,1,n+1}}{u+v+n-1}$
$\displaystyle-\,{\hskip
0.5pt}\frac{P_{\,2,n+1}+\ldots+P_{\,n,n+1}}{u-v}+\frac{Q_{\,2,n+1}+\ldots+Q_{\,n,n+1}}{u+v+n-2}\,{\hskip
0.5pt}+$ $\displaystyle\frac{({\hskip
0.5pt}P_{\,2,n+1}+\ldots+P_{\,n,n+1}{\hskip 0.5pt})\,P_{\,1,n+1}}{({\hskip
0.5pt}u-v{\hskip 0.5pt})\,({\hskip 0.5pt}u-v+n-1{\hskip
0.5pt})}-\frac{({\hskip 0.5pt}Q_{\,2,n+1}+\ldots+Q_{\,n,n+1}{\hskip
0.5pt})\,P_{\,1,n+1}}{({\hskip 0.5pt}u+v+n-2{\hskip 0.5pt})\,({\hskip
0.5pt}u-v+n-1{\hskip 0.5pt})}$ $\displaystyle-\,{\hskip 0.5pt}\frac{({\hskip
0.5pt}P_{\,2,n+1}+\ldots+P_{\,n,n+1}{\hskip 0.5pt})\,Q_{\,1,n+1}}{({\hskip
0.5pt}u-v{\hskip 0.5pt})\,({\hskip 0.5pt}u+v+n-1{\hskip
0.5pt})}+\frac{({\hskip 0.5pt}Q_{\,2,n+1}+\ldots+Q_{\,n,n+1}{\hskip
0.5pt})\,Q_{\,1,n+1}}{({\hskip 0.5pt}u+v+n-2{\hskip 0.5pt})\,({\hskip
0.5pt}u+v+n-1{\hskip 0.5pt})}\ .$

Multiplying the first fraction in the third line by $H(u)\otimes 1$ on the
right gives

$-\,{\hskip 0.5pt}\frac{(n-1)\,P_{\,1,n+1}}{({\hskip 0.5pt}u-v{\hskip
0.5pt})\,({\hskip 0.5pt}u-v+n-1{\hskip 0.5pt})}$ (76)

times $H(u)\otimes 1\,$. Here we used the first equality from Lemma 1 and the
relation

$P_{\,p,n+1}\,P_{\,1,n+1}=P_{\,1,n+1}\,P_{\,1{\hskip
0.5pt}p}\quad\text{where}\quad p=2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}n\,.$

Multiplying the second fraction in the third line by $H(u)\otimes 1$ on the
right gives just zero due to the second equality from Lemma 1 and to the
relations

$Q_{\,p,n+1}\,P_{\,1,n+1}=P_{\,1,n+1}\,Q_{\,p{\hskip 0.5pt}1}=-\,{\hskip
0.5pt}P_{\,1,n+1}\,Q_{\,1{\hskip 0.5pt}p}\quad\text{where}\quad p=2{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}n\,.$

The numerator of the first fraction in the fourth line can be written as the
sum of

$P_{\,p,n+1}\,Q_{\,1,n+1}=Q_{\,1{\hskip
0.5pt}p}\,P_{\,p,n+1}\quad\text{over}\quad p=2{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}n\,.$

These summands evidently have the divisibility property stated in the lemma.
Multiplying the second fraction in the fourth line by $H(u)\otimes 1$ on the
right gives

$-{\hskip 0.5pt}\,\frac{Q_{\,2,n+1}+\ldots+Q_{\,n,n+1}}{({\hskip
0.5pt}u+v+n-2{\hskip 0.5pt})\,({\hskip 0.5pt}u+v+n-1{\hskip 0.5pt})}$ (77)

times $H(u)\otimes 1\,$. Here we used the first equality from Lemma 1 and the
relation

$Q_{\,p,n+1}\,Q_{\,1,n+1}=Q_{\,p,n+1}\,P_{\,1p}\quad\text{where}\quad
p=2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}n\,.$

By adding to the first and the second lines the expressions (76) and (77)
instead of the third and the fourth lines and by collecting similar terms we
get the sum in brackets displayed in the lemma. Hence we make the induction
step. ∎

### 18

Set $n=N\,$. Thus the above introduced functions $H(u)$ and $F{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ will now take values in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}N}$ and $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}(N+1)}$ respectively. Elements of the latter algebra act on vectors in
$({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}(N+1)}$ by the conventions (1) and (2). From now on we will be denoting
$z=u-v$ and $w=u+v+N-1$ for short.

###### Lemma 4

Applying $({\hskip 0.5pt}H(u)\otimes 1{\hskip 0.5pt})\,F{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ to the vector $e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}1}$
has the same effect as applying $H(u)\otimes 1$ to the vector

$\frac{z-1}{z}\,\,e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}1}\,.$ (78)

###### Proof

For $1\leqslant p<q\leqslant N+1$ the elements $Q_{{\hskip
0.5pt}pq}\in(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip
0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip 0.5pt}(N+1)}$ annihilate the vector
$e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}1}\,$. Hence applying $({\hskip 0.5pt}H(u)\otimes 1{\hskip
0.5pt})\,F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to this vector has the same
effect as applying

$\displaystyle\prod_{1\leqslant p<q\leqslant N}\,\biggl{(}1+\frac{P_{{\hskip
0.5pt}pq}}{p-q}\,\biggr{)}\cdot\biggl{(}1-\frac{P_{\,1,N+1}}{u-v+N-1}\,\biggr{)}{\hskip
0.5pt}\ldots{\hskip 0.5pt}\biggl{(}1-\frac{P_{\,N,N+1}}{u-v}\,\biggr{)}=$
$\displaystyle\prod_{1\leqslant p<q\leqslant N}\,\biggl{(}1+\frac{P_{{\hskip
0.5pt}pq}}{p-q}\,\biggr{)}\cdot\biggl{(}\,1-\frac{P_{\,1,N+1}+\ldots+P_{\,N,N+1}}{u-v}\biggr{)}\,.$

The last equality is well known, see for instance (MNO, 16, Section 2.3). The
summands $P_{\,2,N+1}{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}P_{\,N,N+1}$ of the last denominator do not contribute to the image of
the vector $e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes
e_{{\hskip 0.5pt}1}$ by the remark made just before stating Lemma 2. The
summand $P_{\,1,N+1}$ does not change this vector. Using the above mentioned
remark once again we complete the proof of the lemma. ∎

###### Lemma 5

Applying $({\hskip 0.5pt}H(u)\otimes 1{\hskip 0.5pt})\,F{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ to the vector $e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}-1}$
has the same effect as applying $H(u)\otimes 1$ to the linear combination

$\displaystyle\frac{w+1}{w}\,\,e_{{\hskip 0.5pt}1}\otimes\ldots\otimes
e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}-1}$
$\displaystyle-\,\frac{1}{z}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip
0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}$
$\displaystyle-\,\frac{1}{w}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,.$

###### Proof

Applying $F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to $e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}-1}$
has the same effect as applying

$R_{\,1,N+1{\hskip 0.5pt}}({\hskip 0.5pt}u+N-1\,,{\hskip-0.5pt}v{\hskip
0.5pt})\,\biggl{(}\,1-\frac{P_{\,2,N+1}}{u-v+N-2}\,\biggr{)}{\hskip
0.5pt}\ldots{\hskip 0.5pt}\biggl{(}\,1-\frac{P_{\,N,N+1}}{u-v}\biggr{)}\,.$

The remark made before stating Lemma 2 remains valid if $n=N$ and one of the
indices $i_{{\hskip 0.5pt}1}{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}i_{{\hskip 0.5pt}N}$ is $-{\hskip 0.5pt}1$ while the
other $N-1$ indices are $2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}N$ in any order. Hence when applying $({\hskip
0.5pt}H(u)\otimes 1{\hskip 0.5pt})\,F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to
our vector we can replace $F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ by

$R_{\,1,N+1{\hskip 0.5pt}}({\hskip 0.5pt}u+N-1\,,{\hskip-0.5pt}v{\hskip
0.5pt})\,\biggl{(}\,1-\frac{P_{\,2,N+1}+\ldots+P_{\,N,N+1}}{u-v}\,\biggr{)}\,.$

Here we also use Lemma 2 and the above mentioned result from (MNO, 16, Section
2.3).

By simply opening the brackets the latter product equals

$\displaystyle
1-\frac{P_{\,1,N+1}}{u-v+N-1}+\frac{Q_{\,1,N+1}}{u+v+N-1}-\frac{\,P_{\,2,N+1}+\ldots+P_{\,N,N+1}}{u-v}\,{\hskip
0.5pt}+$
$\displaystyle\frac{\,P_{\,1,N+1}\,(\,P_{\,2,N+1}+\ldots+P_{\,N,N+1}{\hskip
0.5pt})}{({\hskip 0.5pt}u-v+N-1{\hskip 0.5pt})\,({\hskip 0.5pt}u-v{\hskip
0.5pt})}-\frac{\,Q_{\,1,N+1}\,(\,P_{\,2,N+1}+\ldots+P_{\,N,N+1}{\hskip
0.5pt})}{({\hskip 0.5pt}u+v+N-1{\hskip 0.5pt})\,({\hskip 0.5pt}u-v{\hskip
0.5pt})}\,.$

The last two denominators are respectively equal to the sums

$P_{\,12}\,P_{\,1,N+1}+\ldots+P_{\,1N}\,P_{\,1,N+1}\quad\text{and}\quad
P_{\,2,N+1}\,Q_{\,12}+\ldots+P_{\,N,N+1}\,Q_{\,1N}\,.$

The second of the two sums annihilates our vector. When applying the first sum
to our vector we can replace each of the factors $P_{\,12}{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}P_{\,1N}$ just by
$-{\hskip 0.5pt}1$. Here we again use Lemma 2. So when applying $({\hskip
0.5pt}H(u)\otimes 1{\hskip 0.5pt})\,F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to
our vector we can replace the function $F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$
by the sum

$\displaystyle
1-\frac{P_{\,1,N+1}}{u-v+N-1}+\frac{Q_{\,1,N+1}}{u+v+N-1}-\frac{P_{\,2,N+1}+\ldots+P_{\,N,N+1}}{u-v}$
$\displaystyle-\,\frac{(N-1)\,P_{\,1,N+1}}{({\hskip 0.5pt}u-v+N-1{\hskip
0.5pt})\,({\hskip 0.5pt}u-v{\hskip 0.5pt})}=$ $\displaystyle
1-\frac{P_{\,1,N+1}+P_{\,2,N+1}+\ldots+P_{\,N,N+1}}{z}+\frac{Q_{\,1,N+1}}{w}{\hskip
0.5pt}\,.$

Applying the expression in the last displayed line to $e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}-1}$
gives

$\displaystyle e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}-1}-\,\frac{1}{z}\,\,\sum_{k=1}^{N}\,{\hskip
0.5pt}e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes
e_{{\hskip 0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes
e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}k}$
$\displaystyle-\,\frac{1}{w}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}+\,\frac{1}{w}\,\,\sum_{k=1}^{N}\,{\hskip
0.5pt}e_{{\hskip 0.5pt}k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes
e_{{\hskip 0.5pt}N}\otimes e_{{\hskip 0.5pt}-{\hskip 0.5pt}k}\,.$

By Lemma 2 applying $H(u)\otimes 1{\hskip 0.5pt}$ to the last sum over
$k=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N$
annihilates all summands but one where $k=1\,$. This observation completes the
proof. ∎

### 19

Introduce the rational function of $u\,,v$ with values in
$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}(2{\hskip 0.5pt}N+1)}$

$G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)=F^{\,\prime}{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)\,F^{\,\prime\prime}{\hskip 0.5pt}({\hskip
0.5pt}1-u-N\,,{\hskip-0.5pt}v)$

where $F^{\,\prime}{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ and
$F^{\,\prime\prime}{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ are the images of
$F{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ by the two embeddings

$(\operatorname{End}\hskip 1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}(N+1)}\to(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}(2{\hskip 0.5pt}N+1)}$

defined as $\iota_{\,1}{\hskip-0.5pt}\otimes\ldots\otimes{\hskip
0.5pt}\iota_{\,N}\otimes{\hskip 0.5pt}\iota_{\,2{\hskip 0.5pt}N+1}$ and as
$\iota_{\,N+1}{\hskip-0.5pt}\otimes\ldots\otimes{\hskip
0.5pt}\iota_{\,2{\hskip 0.5pt}N}\otimes{\hskip 0.5pt}\iota_{\,2{\hskip
0.5pt}N+1}$ respectively. Using (74) twice we obtain that multiplying the
formal power series in $u^{-1},v^{-1}$

$T_{{\hskip 0.5pt}1}(u+N-1)\,\ldots\,T_{{\hskip 0.5pt}N}(u)\,T_{{\hskip
0.5pt}N+1}(-{\hskip 0.5pt}u)\,\ldots\,T_{\,2{\hskip 0.5pt}N}({\hskip
0.5pt}1-N-u{\hskip 0.5pt})\,T_{\,2{\hskip 0.5pt}N+1}(v)$

by $G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1$ on the left gives the same
result as as multuplying by $G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\otimes 1$ on
the right the series

$T_{\,2{\hskip 0.5pt}N+1}(v)\,T_{{\hskip 0.5pt}1}(u+N-1)\,\ldots\,T_{{\hskip
0.5pt}N}(u)\,T_{{\hskip 0.5pt}N+1}(-{\hskip 0.5pt}u)\,\ldots\,T_{\,2{\hskip
0.5pt}N}({\hskip 0.5pt}1-N-u{\hskip 0.5pt})\,.$

Both series have their coefficients in the algebra $(\operatorname{End}\hskip
1.0pt{\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip 0.5pt}\otimes{\hskip
0.5pt}(2{\hskip 0.5pt}N+1)}\otimes\operatorname{X}({\hskip
0.5pt}\mathfrak{p}_{N}{\hskip-0.5pt})\,$.

For any index $j=\pm\,1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}\\!\pm{\hskip-0.5pt}N$ denote

$f_{{\hskip 0.5pt}j}=e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}j}\,.$ (79)

Observe that replacing $u$ by $1-N-u$ maps $z=u-v$ to $1-N-u-v=-\,w\,$.
Therefore by using Lemma 4 with that replacement of the variable $u$ and then
in the original formulation we immeditaly get the following corollary to this
lemma.

###### Corollary 5

Applying $({\hskip 0.5pt}H(u)\otimes H({\hskip 0.5pt}1-N-u{\hskip
0.5pt})\otimes 1{\hskip 0.5pt})\,G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to the
vector $f_{{\hskip 0.5pt}1}$ has the same effect as applying $H(u)\otimes
H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})\otimes 1$ to the vector

$\frac{(z-1)\,(w+1)}{z\,w}\,f_{{\hskip 0.5pt}1}\,.$ (80)

Below is the counterpart of Corollary 5 for our Lemma 5 instead of Lemma 4.

###### Corollary 6

Applying $({\hskip 0.5pt}H(u)\otimes H({\hskip 0.5pt}1-N-u{\hskip
0.5pt})\otimes 1{\hskip 0.5pt})\,G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to
$f_{{\hskip 0.5pt}-1}$ has the same effect as applying $H(u)\otimes H({\hskip
0.5pt}1-N-u{\hskip 0.5pt})\otimes 1$ to the linear combination

$\displaystyle\frac{(z-1)\,(w+1)}{z\,w}\,f_{{\hskip 0.5pt}-1}$
$\displaystyle-\,{\hskip 0.5pt}\frac{z-1}{z^{{\hskip
0.5pt}2}}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip
0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}$ $\displaystyle\qquad\qquad-\,{\hskip
0.5pt}\frac{z-1}{z\,w}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,{\hskip 0.5pt}+$ $\displaystyle\hskip
10.0pt\frac{z-1}{z\,w}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip
0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,{\hskip 0.5pt}+$ $\displaystyle\hskip
36.0pt\frac{z-1}{z^{{\hskip 0.5pt}2}}\,\,\sum_{k=1}^{N}\,{\hskip
0.5pt}e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes
e_{{\hskip 0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,.$ (81)

###### Proof

By Lemma 5 with the variable $u$ replaced by $1-N-u$ we immediately obtain
that applying $({\hskip 0.5pt}1\otimes H({\hskip 0.5pt}1-N-u{\hskip
0.5pt})\otimes 1{\hskip 0.5pt})\,F^{\,\prime\prime}{\hskip 0.5pt}({\hskip
0.5pt}1-u-N\,,{\hskip-0.5pt}v)$ to the vector $f_{{\hskip 0.5pt}-1}$ has the
same effect as applying $1\otimes H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})\otimes
1$ to the linear combination

$\displaystyle\frac{z-1}{z}\,f_{{\hskip 0.5pt}-1}\,{\hskip 0.5pt}+$
$\displaystyle\frac{1}{w}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip
0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,{\hskip 0.5pt}+$
$\displaystyle\frac{1}{z}\,\,\sum_{k=1}^{N}\,{\hskip 0.5pt}e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,.$ (82)

Further, by the original Lemma 5 applying $({\hskip 0.5pt}H(u)\otimes 1\otimes
1{\hskip 0.5pt})\,F^{\,\prime}{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ to the
first of three lines (82) has the same effect as applying $H(u)\otimes
1\otimes 1$ to the sum of first three lines in (81). Further, applying
$({\hskip 0.5pt}H(u)\otimes 1\otimes 1{\hskip 0.5pt})\,F^{\,\prime}{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ to the second and to the third lines of (82) has
the same effect as applying the product

$({\hskip 0.5pt}H(u)\otimes 1\otimes 1{\hskip
0.5pt})\,\biggl{(}\,1-\frac{P_{\,1,2{\hskip
0.5pt}N+1}}{u-v+N-1}\,\biggr{)}{\hskip 0.5pt}\ldots{\hskip
0.5pt}\biggl{(}\,1-\frac{P_{\,N,2{\hskip 0.5pt}N+1}}{u-v}\biggr{)}\,.$

Applying the last displayed product to the second and third lines of (82)
gives the same result as applying the product

$({\hskip 0.5pt}H(u)\otimes 1\otimes 1{\hskip
0.5pt})\,\biggl{(}\,1-\frac{P_{\,1,2{\hskip 0.5pt}N+1}+\ldots+P_{\,N,2{\hskip
0.5pt}N+1}}{u-v}\biggr{)}\,.$ (83)

Here we used Lemma 2 like in the beginning of our proof of Lemma 5. Let us now
use the remark made just before stating Lemma 2. When applying (83) to the
summands with the index $k$ in the second and third lines of (82), all
summands of the numerator in (83) vanish except for $P_{\,k{\hskip
0.5pt},{\hskip 0.5pt}2{\hskip 0.5pt}N+1}\,$. The latter does not change the
summands with the index $k$ in the second and third lines of (82). So applying
$({\hskip 0.5pt}H(u)\otimes 1\otimes 1{\hskip 0.5pt})\,F^{\,\prime}{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ to the second and the third lines of (82) gives
the same as applying $H(u)\otimes 1\otimes 1$ respectively to the fourth and
the fifth lines of (81). ∎

### 20

So far we used the equality in Lemma 2 only in the cases when both sides of
the equality annihilate a given vector. In the next section we will use that
equality in more general cases. We will employ the following corollary to
Lemma 2. Denote

$d_{{\hskip 0.5pt}k}=e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip
0.5pt}k-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\quad\text{for}\quad k=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}N{\hskip 0.5pt}.$

###### Corollary 7

For any $k>1$ applying $H(u)$ to the vector $e_{{\hskip 0.5pt}-k}\otimes
e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}$ has the same
effect as applying $H(u)$ to the linear combination

$\displaystyle a(u)\,d_{{\hskip 0.5pt}k}\otimes e_{{\hskip
0.5pt}-1}\,(-1)^{{\hskip 0.5pt}N+k}-{\hskip 0.5pt}b{\hskip
0.5pt}(u)\,d_{{\hskip 0.5pt}1}\otimes e_{{\hskip 0.5pt}-k}\,(-1)^{{\hskip
0.5pt}N}$

where

$a(u)=\frac{2}{{\hskip 0.5pt}2u+N+1{\hskip 0.5pt}}\quad\text{and}\quad
b{\hskip 0.5pt}(u)=\frac{2u+N-1}{{\hskip 0.5pt}2u+N+1{\hskip 0.5pt}}\,\,.$

Similarly, for any index $k>1$ applying $H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})$
to the vector

$e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes
e_{{\hskip 0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes
e_{{\hskip 0.5pt}N}$

has the same effect as applying $H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})$ to the
linear combination

$\displaystyle a(u)\,e_{{\hskip 0.5pt}-k}\otimes d_{{\hskip 0.5pt}1}-b{\hskip
0.5pt}(u)\,e_{{\hskip 0.5pt}-1}\otimes d_{{\hskip 0.5pt}k}\,(-1)^{{\hskip
0.5pt}k}\,.$

###### Proof

For any index $i$ denote by $x_{{\hskip 0.5pt}i}(u)$ the result of applying
$H(u)$ to the vector

$e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes
e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes
e_{{\hskip 0.5pt}i}\otimes e_{{\hskip 0.5pt}-{\hskip 0.5pt}i}\,.$ (84)

By Lemma 2 applying $H(u)$ to the vector $e_{{\hskip 0.5pt}-k}\otimes
e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}$ gives the vector
$(-1)^{{\hskip 0.5pt}k}\,x_{{\hskip 0.5pt}-{\hskip 0.5pt}k}(u)\,$. Let us
apply both sides of the equality in Lemma 2 with $n=N$ and $p=N-1$ to the
vectors (84) with $i=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}N{\hskip 0.5pt}$. We obtain the equations

$x_{{\hskip 0.5pt}1}(u)-x_{{\hskip 0.5pt}-1}(u)+\ldots+x_{{\hskip
0.5pt}N}(u)-x_{{\hskip 0.5pt}-{\hskip 0.5pt}N}(u)=(2u+1)\,(\,x_{{\hskip
0.5pt}i}(u)+x_{{\hskip 0.5pt}-{\hskip 0.5pt}i}(u){\hskip 0.5pt})\,.$ (85)

But for $i=2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}k-1{\hskip 0.5pt},k+1{\hskip 0.5pt}{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N$ we have $x_{{\hskip 0.5pt}i}(u)=0$
by Lemma 2. So by (85) the vectors $x_{{\hskip 0.5pt}-{\hskip 0.5pt}i}(u)$
with $i=2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}k-1{\hskip 0.5pt},k+1{\hskip 0.5pt}{{\hskip 0.5pt},\hskip
1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N$ are equal to each other. Using
these two observations, it is easy to deduce from the equations (85) that

$x_{{\hskip 0.5pt}-k}(u)=a(u)\,x_{{\hskip 0.5pt}1}(u)-b{\hskip
0.5pt}(u)\,x_{{\hskip 0.5pt}k}(u)\,.$

By using Lemma 2 again we now obtain the first statement of our corollary.

Changing the notation, let $x_{{\hskip 0.5pt}i}(u)$ be the result of applying
$H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})$ to

$e_{{\hskip 0.5pt}-{\hskip 0.5pt}i}\otimes e_{{\hskip 0.5pt}i}\otimes
e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes
e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\,.$ (86)

By Lemma 2 applying $H({\hskip 0.5pt}1-N-u{\hskip 0.5pt})$ to $e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip
0.5pt}-1}\otimes e_{{\hskip 0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}$ gives the vector $(-1)^{{\hskip 0.5pt}k}\,x_{{\hskip 0.5pt}-{\hskip
0.5pt}1}(u)\,$. Let us replace $u$ by $1-N-u$ in the equality in Lemma 2 and
apply both sides of the resulting equality with $n=N$ and $p=1$ to the vectors
(86) with $i=1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip
0.5pt}}N\,$. In this way we obtain the same equations (85) although for the
changed vectors. For $i=2{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}k-1{\hskip 0.5pt},k+1{\hskip 0.5pt}{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N$ we still have
$x_{{\hskip 0.5pt}i}(u)=0$ by Lemma 2. So the equations (85) easily imply that

$x_{{\hskip 0.5pt}-1}(u)=a(u)\,x_{{\hskip 0.5pt}k}(u)-b{\hskip
0.5pt}(u)\,x_{{\hskip 0.5pt}1}(u)\,.$

By using Lemma 2 again we now get the second statement of the corollary. ∎

### 21

Let us relate two vectors in $({\mathbb{C}}^{{\hskip 0.5pt}N|N})^{{\hskip
0.5pt}\otimes{\hskip 0.5pt}(2{\hskip 0.5pt}N+1)}(u\,,v)$ by the symbol $\sim$
to each other if their difference vanishes under the action of $H(u)\otimes
H(1-N-u)\otimes 1\,$, or lies in the image of the action of $Q_{{\hskip
0.5pt}N,{\hskip 0.5pt}N+1}\,$. Extend the relation $\sim$ transitively. Recall
the notations $z=u-v$ and $w=u+v+N-1$ adopted in Section 18.

###### Proposition 8

For any index $j=\pm\,1{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}\\!\pm{\hskip-0.5pt}N$ we have the relation

$G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)\,f_{{\hskip
0.5pt}j}\,\sim\,\frac{(z-1)\,(w+1)}{z\,w}\,f_{{\hskip 0.5pt}j}\,.$

###### Proof

Proposition 8 follows from its particular cases of $j=1\,,-{\hskip 0.5pt}1\,$.
Indeed, the sums in (7) and (9) are invariant under any permutation of the
indices $1{\hskip 0.5pt}{{\hskip 0.5pt},\hskip 1.0pt\ldots{\hskip
0.5pt},{\hskip 0.5pt}}N$ with the corresponding permutation of $-{\hskip
0.5pt}1\,,\ldots,-{\hskip 0.5pt}N$ at the same time. So $H(u)$ and $G{\hskip
0.5pt}(u\,,{\hskip-0.5pt}v)$ are also invariant under any such permutation.
While the vector

$e_{{\hskip 0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}$ (87)

changes under the permutation, its image by the action of $H(u)\otimes
H(1-N-u)$ does not change due to Lemma 2. Finally, note that the product of
$G{\hskip 0.5pt}(u\,,{\hskip-0.5pt}v)$ by $H(u)\otimes H({\hskip
0.5pt}1-N-u{\hskip 0.5pt})\otimes 1$ on the left is also divisible by this
factor on the right. This remark follows by setting $n=N$ in the beginning of
the proof of Lemma 3.

For $j=1$ our proposition immediately follows from Corollary 5. For
$j=-{\hskip 0.5pt}1$ we will derive the proposition from Corollary 6. We will
show that the summands with the index $k$ in the second and fifth lines of
(81) cancel each other by $\sim$ and so do the summands with the index $k$ in
the third and fourth lines of (81).

Consider the second and the fifth lines of (81). For $k=1{{\hskip
0.5pt},\hskip 1.0pt\ldots{\hskip 0.5pt},{\hskip 0.5pt}}N$ by Lemma 2

$\displaystyle-{\hskip 0.5pt}\,e_{{\hskip 0.5pt}1}\otimes\ldots\otimes
e_{{\hskip 0.5pt}k-1}\otimes e_{{\hskip 0.5pt}-1}\otimes e_{{\hskip
0.5pt}k+1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}k}{\hskip 0.5pt}\,+$ $\displaystyle e_{{\hskip
0.5pt}1}\otimes\ldots\otimes e_{{\hskip 0.5pt}N}\otimes e_{{\hskip
0.5pt}-k}\otimes e_{{\hskip 0.5pt}2}\otimes\ldots\otimes e_{{\hskip
0.5pt}N}\otimes e_{{\hskip 0.5pt}k}\,\sim$ $\displaystyle-{\hskip
0.5pt}\,d_{{\hskip 0.5pt}k}\otimes e_{{\hskip 0.5pt}-1}\otimes e_{{\hskip
0.5pt}1}\otimes d_{{\hskip 0.5pt}1}\otimes e_{{\hskip 0.5pt}k}\,(-1)^{{\hskip
0.5pt}N+k}+d_{{\hskip 0.5pt}k}\otimes e_{{\hskip 0.5pt}k}\otimes e_{{\hskip
0.5pt}-k}\otimes d_{{\hskip 0.5pt}1}\otimes e_{{\hskip 0.5pt}k}\,(-1)^{{\hskip