1 Introduction

Let \(\mathfrak {g}\) be any finite-dimensional Lie superalgebra over a complex field \({\mathbb {C}}\,\). Let \(r{\hspace{0.5pt}}(u,v)\) be a meromorphic function of two complex variables u and v which takes values in \(\mathfrak {g}{\hspace{0.5pt}}\otimes \mathfrak {g}\,\). The classical Yang–Baxter equation for the function \(r{\hspace{0.5pt}}(u\,,v)\) is

$$\begin{aligned}{}[\,r_{{\hspace{0.5pt}}12}(u,v),r_{{\hspace{0.5pt}}13}(u,w)\,]+ [\,r_{{\hspace{0.5pt}}12}(u,v),r_{{\hspace{0.5pt}}23}(v,w)\,]+ [\,r_{{\hspace{0.5pt}}13}(u,w),r_{{\hspace{0.5pt}}23}(v,w)\,]=0 \end{aligned}$$

where w is another complex variable and the function of u, v, w at the left-hand side takes values in \(\mathfrak {g}{\hspace{0.5pt}}\otimes \mathfrak {g}{\hspace{0.5pt}}\otimes \mathfrak {g}\,\). Here we use the standard notation, for example \(r_{{\hspace{0.5pt}}23}(v,w)=1\otimes r{\hspace{0.5pt}}(v,w)\,\). For conventions on the tensor products see Sect. 2.

A solution of the above equation is called nondegenerate if not every value of the function \(r{\hspace{0.5pt}}(u,v)\) is degenerate as a quadratic tensor. For simple Lie algebras \(\mathfrak {g}\) the nondegenerate solutions were classified in [4, 5]. In particular, it was shown in [5] that for any nondegenerate solution \(r{\hspace{0.5pt}}(u,v)\) with a simple Lie algebra \(\mathfrak {g}\) one can find a domain \(D\subset {\mathbb {C}}\) and two holomorphic maps

$$\begin{aligned} \psi :D\rightarrow {\mathbb {C}}\quad \text {and}\quad \omega :D\rightarrow {\text {Aut}}\mathfrak {g}\end{aligned}$$

where \(\psi \) is not constant and the function \(({\hspace{0.5pt}}\omega (u)\otimes {\hspace{0.5pt}}\omega (v))\,r{\hspace{0.5pt}}({\hspace{0.5pt}}\psi (u),\psi (v))\) depends only on the difference \(u-v\,\).

This basic result of [5] does not hold for Lie superalgebras. Solutions \(r{\hspace{0.5pt}}(u,v)\) which depend not only on the difference \(u-v\) were constructed in [11]. There \(\mathfrak {g}\) is the general linear Lie superalgebra \(\mathfrak {gl}_{\,n|n}\) with any positive integer \(n\,\). These solutions are antisymmetric, that is

$$\begin{aligned} r_{{\hspace{0.5pt}}21}(v,u)=-\,r{\hspace{0.5pt}}(u,v). \end{aligned}$$
(1)

A general phenomenon underlying this construction was independently described in [1]. See [12] for further discussion of this remarkable phenomenon.

The solutions \(r{\hspace{0.5pt}}(u,v)\) constructed in [11] are rational functions of variables u and \(v\,\). Here we construct a solution \(r{\hspace{0.5pt}}(u,v)\) which is expressed in theta functions of \(u-v\) and \(u+v\,\). Our \(\mathfrak {g}\) is the quotient of the special linear Lie superalgebra \(\mathfrak {sl}_{\,n|n}\) by its one-dimensional centre. This quotient is denoted by \(\mathfrak {psl}_{\,n|n}\,\).

Elliptic solutions \(r{\hspace{0.5pt}}(u,v)\) for the Lie superalgebra \(\mathfrak {g}=\mathfrak {psl}_{\,n|n}\) were constructed in [9]. They have the form \(r{\hspace{0.5pt}}(u,v)=s{\hspace{0.5pt}}(u-v)\) where \(s{\hspace{0.5pt}}(u)\) is a function such that

$$\begin{aligned} s_{{\hspace{0.5pt}}21}(u)=-\,s{\hspace{0.5pt}}(-{\hspace{0.5pt}}u). \end{aligned}$$
(2)

Hence these solutions are antisymmetric.

Now let \(\eta \) be the involutive automorphism of \(\mathfrak {gl}_{\,n|n}\) defined in Sect. 2. The fixed point subalgebra of \(\mathfrak {gl}_{\,n|n}\) relative to \(\eta \) is the queer Lie superalgebra \(\mathfrak {q}_{{\hspace{0.5pt}}n}\,\). The automorphism \(\eta \) preserves the subalgebra \(\mathfrak {sl}_{\,n|n}\) and descends to \(\mathfrak {psl}_{\,n|n}\,\). In our Sect. 5 we show that the function \(s{\hspace{0.5pt}}(u)\) in [9] can be so chosen that

$$\begin{aligned} ({\hspace{0.5pt}}\eta \otimes \eta {\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u)=s{\hspace{0.5pt}}(-{\hspace{0.5pt}}u) \end{aligned}$$
(3)

and that

$$\begin{aligned} r{\hspace{0.5pt}}(u,v)=s{\hspace{0.5pt}}(u-v)+({\hspace{0.5pt}}{\textrm{id}}\otimes \eta {\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u+v) \end{aligned}$$
(4)

is another solution of classical Yang–Baxter equation for \(\mathfrak {g}=\mathfrak {psl}_{\,n|n}\,\).

It immediately follows from (3) and (4) that

$$\begin{aligned} r{\hspace{0.5pt}}(u,v)=s{\hspace{0.5pt}}(u-v)+({\hspace{0.5pt}}\eta {\hspace{0.5pt}}\otimes {\textrm{id}}{\hspace{0.5pt}})\,s{\hspace{0.5pt}}(-\,u-v). \end{aligned}$$
(5)

By comparing the definition (4) with (5) and by using (2) we see that our solution \(r{\hspace{0.5pt}}(u,v)\) is antisymmetric as well.

2 General Conventions

We shall use the following general conventions. Let \(\textrm{A}\) and \(\textrm{B}\) be any associative \(\mathbb {Z}_{{\hspace{0.5pt}}2}\)-graded algebras. Their tensor product \(\textrm{A}\otimes \textrm{B}\) is also an associative \(\mathbb {Z}_2\)-graded algebra such that for any homogeneous elements \(X,X^\prime \in \textrm{A}\) and \(Y,Y^\prime \in \textrm{B}\)

$$\begin{aligned} (X\otimes Y){\hspace{0.5pt}}(X^\prime \otimes Y^\prime )&=X{\hspace{0.5pt}}X^\prime \otimes Y{\hspace{0.5pt}}Y^\prime \,(-1)^{{\hspace{0.5pt}}\deg X^\prime \deg Y}, \\ \deg {\hspace{0.5pt}}(X\otimes Y)&=\deg X+\deg Y{\hspace{0.5pt}}. \end{aligned}$$

Furthermore, for any two \(\mathbb {Z}_2\)-graded modules U and V over \(\textrm{A}\) and \(\textrm{B}\), respectively, the vector space \(U\otimes V\) is a \(\mathbb {Z}_2\)-graded module over \(\textrm{A}\otimes \textrm{B}\) such that for any homogeneous elements \(x\in U\) and \(y\in V\)

$$\begin{aligned} (X\otimes Y){\hspace{0.5pt}}({\hspace{0.5pt}}x\otimes y{\hspace{0.5pt}})&=X{\hspace{0.5pt}}x\otimes Y{\hspace{0.5pt}}y \,(-1)^{{\hspace{0.5pt}}\deg x\,\deg Y}, \end{aligned}$$
(6)
$$\begin{aligned} \deg \hspace{0.5pt}({\hspace{0.5pt}}x\otimes y{\hspace{0.5pt}})&=\deg x+\deg y\,. \end{aligned}$$
(7)

Now let the indices i, j, k, l run through \({\hspace{0.5pt}}\pm 1{{\hspace{0.5pt}},\hspace{1.0pt}\ldots {\hspace{0.5pt}},{\hspace{0.5pt}}}{\hspace{-0.5pt}}\pm n\,\). Put \({\bar{\imath }\,}=0\) if \(i>0\) and \({\bar{\imath }\,}=1\) if \(i<0\,\). Consider the \(\mathbb {Z}_{{\hspace{0.5pt}}2}\)-graded vector space \({\mathbb {C}}^{{\hspace{0.5pt}}n|n}{\hspace{0.5pt}}\). Let \(e_i\in {\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) be an element of the standard basis. The \(\mathbb {Z}_{{\hspace{0.5pt}}2}\)-grading on \({\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) is defined by \(\deg e_i={\bar{\imath }\,}\,\).

Let \(E_{{\hspace{0.5pt}}ij}\in {\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) be the standard matrix unit, defined by \(E_{{\hspace{0.5pt}}ij}\,e_k=\delta _{{\hspace{0.5pt}}jk}\,e_i\,\). The associative algebra \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) is \(\mathbb {Z}_{{\hspace{0.5pt}}2}\)-graded by setting \(\deg E_{{\hspace{0.5pt}}ij}={\bar{\imath }\,}+{\hspace{0.5pt}}{\bar{\jmath }\,}\,\). Hence \({\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) is a \(\mathbb {Z}_2\)-graded module over \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\,\). For any positive integer m we can also identify the tensor product \(({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n})^{\otimes {\hspace{0.5pt}}m}\) with the algebra \({\text {End}}\hspace{1.0pt}(({\mathbb {C}}^{{\hspace{0.5pt}}n|n})^{\otimes {\hspace{0.5pt}}m})\) acting on the vector space \(({\mathbb {C}}^{{\hspace{0.5pt}}n|n})^{\otimes {\hspace{0.5pt}}m}\) by repeatedly using conventions (6) and (7).

The supertrace \(\textrm{str}\) is a linear function \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\rightarrow {\mathbb {C}}\) defined by setting

$$\begin{aligned} \textrm{str}\hspace{1.0pt}(E_{{\hspace{0.5pt}}ij})=\delta _{ij}\,(-1)^{\,{\bar{\imath }\,}}{\hspace{0.5pt}}. \end{aligned}$$

This definition implies that for any homogeneous elements \(X{\hspace{0.5pt}},Y\in {\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\)

$$\begin{aligned} \textrm{str}\hspace{1.0pt}({\hspace{0.5pt}}YX)=\textrm{str}\hspace{1.0pt}(X{\hspace{0.5pt}}Y)\,(-1)^{{\hspace{0.5pt}}\deg X\deg Y}. \end{aligned}$$

Further, we can define an involutive automorphism \(\eta \) of \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) by mapping

$$\begin{aligned} \eta :\,E_{{\hspace{0.5pt}}ij}\mapsto E_{\,-i{\hspace{0.5pt}},{\hspace{0.5pt}}-j}. \end{aligned}$$
(8)

This automorphism is the conjugation by the involutive odd element of \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\)

$$\begin{aligned} E_{{\hspace{0.5pt}}1,-1}+E_{{\hspace{0.5pt}}-1,1}+\cdots +E_{{\hspace{0.5pt}}n,-n}\,+E_{{\hspace{0.5pt}}-n,n}. \end{aligned}$$

We have

$$\begin{aligned}{}[\,E_{{\hspace{0.5pt}}ij},E_{{\hspace{0.5pt}}kl}\,]= \delta _{jk}\,E_{{\hspace{0.5pt}}il}- \delta _{{\hspace{0.5pt}}li}\,E_{{\hspace{0.5pt}}kj}\, {(-1)}^{{\hspace{0.5pt}}(\,{\bar{\imath }\,}+\,{\bar{\jmath }\,}\,)({\hspace{0.5pt}}{\bar{k}\,}+\,{\bar{l}\,}\,)} \end{aligned}$$
(9)

in \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\,\). Here the square brackets indicate the supercommutator. We will also consider each \(E_{{\hspace{0.5pt}}ij}\) as an element of the Lie superalgebra \(\mathfrak {gl}_{\,n|n}\,\). The special linear Lie superalgebra \(\mathfrak {sl}_{\,n|n}\) is the subalgebra of \(\mathfrak {gl}_{\,n|n}\) defined as the kernel of the function \(\textrm{str}\,\). The centre of \(\mathfrak {gl}_{\,n|n}\) is spanned by the element

$$\begin{aligned} E_{{\hspace{0.5pt}}11}+E_{{\hspace{0.5pt}}-1,-1}+\cdots +E_{{\hspace{0.5pt}}nn}\,+E_{{\hspace{0.5pt}}-n,-n}=1. \end{aligned}$$

The subalgebra \(\mathfrak {sl}_{\,n|n}\) contains this element. The quotient of the Lie superalgebra \(\mathfrak {sl}_{\,n|n}\) by the one-dimensional subspace spanned by this element is denoted by \(\mathfrak {psl}_{\,n|n}\,\). According to [8] the Lie superalgebra \(\mathfrak {psl}_{\,n|n}\) is simple if and only if \(n>1\,\). By using (9) we obtain that the Lie bracket on \(\mathfrak {psl}_{\,1|1}\) is just zero.

By (9) our \(\eta \) is also an involutive automorphism of the Lie superalgebra \(\mathfrak {gl}_{\,n|n}\,\). This automorphism preserves its subalgebra \(\mathfrak {sl}_{\,n|n}\,\). The queer Lie superalgebra \(\mathfrak {q}_{{\hspace{0.5pt}}n}\) is the fixed point subalgebra of \(\mathfrak {gl}_{\,n|n}\) by \(\eta \,\).

3 Theta Functions

Fix any complex number \(\tau \) with a positive imaginary part. For any real numbers a and b consider the Hermite theta function

$$\begin{aligned} \theta _{a,b}(u)=\sum _{m=-\infty }^\infty e^{\,\pi {\mathrm i}\tau {\hspace{0.5pt}}(a+m)^2+2\pi {\mathrm i}{\hspace{0.5pt}}(a+m){\hspace{0.5pt}}(b+u)}. \end{aligned}$$
(10)

The above series converges to a holomorphic function of the complex variable \(u\,\). All zeroes of this function are simple and form the subset

$$\begin{aligned} \textstyle (a+\frac{1}{2}+\mathbb {Z})\,\tau +(b+\frac{1}{2}+\mathbb {Z})\subset {\mathbb {C}}, \end{aligned}$$

see [7, pp. 196–199]. The numbers a and b here are called characteristics. For \(a=b=0\) the series (10) is the Jacobi theta function. It follows from (10) that

$$\begin{aligned} \theta _{a,b}(u+1)=e^{\,2\pi {\mathrm i}{\hspace{0.5pt}}a}\,\theta _{a,b}(u) \quad \text {and}\quad \theta _{a,b}(u+\tau )=e^{\,-2\pi {\mathrm i}{\hspace{0.5pt}}(u+b+\frac{\tau }{2})}\,\theta _{a,b}(u).\qquad \end{aligned}$$
(11)

By changing m to \(m+1\) in (10) and by using the first equation in (11) we obtain

$$\begin{aligned} \theta _{a+1,b}(u)=\theta _{a,b}(u) \quad \text {and}\quad \theta _{a,b+1}(u)=e^{\,2\pi {\mathrm i}{\hspace{0.5pt}}a}\,\theta _{a,b}(u) \end{aligned}$$
(12)

respectively. Further, by changing m to \(-{\hspace{0.5pt}}m\) in (10) we obtain the parity relation

$$\begin{aligned} \theta _{a,b}(-{\hspace{0.5pt}}u)=\theta _{-a,-b}{\hspace{0.5pt}}(u). \end{aligned}$$
(13)

Now let g and h be any integers not simultaneously divisible by \(2{\hspace{0.5pt}}n\) and by n, respectively. Consider the function of the complex variable u

$$\begin{aligned} \varphi _{{\hspace{0.5pt}}g,h}(u)= \frac{\,\theta _{\frac{g}{2n}+\frac{1}{2},\frac{1}{2}-\frac{h}{n}}(u)\, \theta ^{\,\prime }_{\frac{1}{2},\frac{1}{2}}(0)\,}{\,\theta _{\frac{g}{2n}+\frac{1}{2},\frac{1}{2}-\frac{h}{n}}(0)\, \theta _{\frac{1}{2},\frac{1}{2}}(u)\,}\,{\hspace{0.5pt}}. \end{aligned}$$

This function is meromorphic. It has simple poles at every point on the lattice

$$\begin{aligned} \mathcal {L}=\mathbb {Z}+\mathbb {Z}\,\tau \subset {\mathbb {C}}. \end{aligned}$$

Due to the chosen normalisation the residue of the pole of \(\varphi _{{\hspace{0.5pt}}g,h}(u)\) at \(u=0\) is \(1\,\).

It immediately follows from the relations (12) that

$$\begin{aligned} \varphi _{g+2n,h}(u)=\varphi _{g,h}(u) \quad \text {and}\quad \varphi _{g,h+n}(u)=\varphi _{g,h}(u). \end{aligned}$$

Thus \(\varphi _{{\hspace{0.5pt}}g,h}(u)\) depends on the integers g and h only modulo \(2{\hspace{0.5pt}}n\) and n, respectively. From now on g and h will run not through \(\mathbb {Z}\,\), but through the additive groups \(\mathbb {Z}_{{\hspace{0.5pt}}2{\hspace{0.5pt}}n}=\mathbb {Z}{\hspace{0.5pt}}/{\hspace{0.5pt}}2{\hspace{0.5pt}}n\,\mathbb {Z}\) and \(\mathbb {Z}_{{\hspace{0.5pt}}n}=\mathbb {Z}{\hspace{0.5pt}}/{\hspace{0.5pt}}n\,\mathbb {Z}\), respectively.

Let \(\varepsilon =e^{\,\pi {\hspace{0.5pt}}{\mathrm i}{\hspace{0.5pt}}/n}\) be a primitive root of unity of the \(2{\hspace{0.5pt}}n\,\). Direct calculation using (11) yields the periodicity properties

$$\begin{aligned} \varphi _{g,h}(u+1)=\varepsilon ^{\,g}\,\varphi _{g,h}(u) \quad \text {and}\quad \varphi _{g,h}(u+\tau )=\varepsilon ^{\,2{\hspace{0.5pt}}h}\,\varphi _{g,h}(u). \end{aligned}$$
(14)

Another direct calculation using (12) and (13) yields the parity relation

$$\begin{aligned} \varphi _{g,h}(-{\hspace{0.5pt}}u)=-\,\varphi _{-g,-h}{\hspace{0.5pt}}(u). \end{aligned}$$
(15)

4 Commuting Automorphisms

Consider the following two elements of the algebra \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\,\),

$$\begin{aligned} A= & {} E_{11}+\varepsilon ^{{\hspace{0.5pt}}2}{\hspace{0.5pt}}E_{{\hspace{0.5pt}}22}+\cdots +\varepsilon ^{\,2(n-1)}{\hspace{0.5pt}}E_{{\hspace{0.5pt}}nn} \\{} & {} +\,\,\varepsilon \, E_{{\hspace{0.5pt}}-1,-1}+\varepsilon ^{{\hspace{0.5pt}}-1}{\hspace{0.5pt}}E_{{\hspace{0.5pt}}-2,-2}+\cdots +\varepsilon ^{\,3-2n}{\hspace{0.5pt}}E_{{\hspace{0.5pt}}-n,-n} \end{aligned}$$

and

$$\begin{aligned} B= & {} E_{{\hspace{0.5pt}}12}+\cdots +E_{{\hspace{0.5pt}}n-1,n}+E_{{\hspace{0.5pt}}n1}\\{} & {} +\,\,E_{{\hspace{0.5pt}}-2,-1}+\cdots +E_{{\hspace{0.5pt}}-n,1-n}+\cdots +E_{{\hspace{0.5pt}}-1,-n}. \end{aligned}$$

These elements are invertible and of \(\,\mathbb {Z}_2{\hspace{0.5pt}}\)-degree zero. They satisfy the relations

$$\begin{aligned} A^{{\hspace{0.5pt}}2{\hspace{0.5pt}}n}=B^{{\hspace{0.5pt}}n}=1 \quad \text {and}\quad B{\hspace{0.5pt}}A=\varepsilon ^{{\hspace{0.5pt}}2} A{\hspace{0.5pt}}B. \end{aligned}$$
(16)

By (8) we get

$$\begin{aligned} \eta {\hspace{0.5pt}}(A)=\varepsilon \,A^{-1} \quad \text {and}\quad \eta {\hspace{0.5pt}}(B)=B^{{\hspace{0.5pt}}-1}. \end{aligned}$$
(17)

Let us define two automorphisms \(\alpha \) and \(\beta \) of the algebra \({\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) by setting

$$\begin{aligned} \alpha {\hspace{0.5pt}}(X)=A^{-1}{\hspace{-0.5pt}}X{\hspace{0.5pt}}A \quad \text {and}\quad \beta {\hspace{0.5pt}}(X)=B^{{\hspace{0.5pt}}-1}{\hspace{-0.5pt}}X{\hspace{0.5pt}}B \end{aligned}$$

for \(X\in {\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\,\). These automorphisms commute and are of degrees \(2{\hspace{0.5pt}}n\) and n, respectively. Here \(\alpha {\hspace{0.5pt}}(X)=\beta {\hspace{0.5pt}}(X)=X\) if and only if X is a linear combination of

$$\begin{aligned} E_{11}+\cdots +E_{nn} \quad \text {and}\quad E_{-1,-1}+\cdots +E_{-n,-n}. \end{aligned}$$

Since the elements A and B are of \(\mathbb {Z}_2{\hspace{0.5pt}}\)-degree \(0\,\), for each \(X\in {\text {End}}\hspace{1.0pt}{\mathbb {C}}^{{\hspace{0.5pt}}n|n}\) we have

$$\begin{aligned} \textrm{str}\hspace{1.0pt}(\alpha {\hspace{0.5pt}}(X))=\textrm{str}\hspace{1.0pt}(X) \quad \text {and}\quad \textrm{str}\hspace{1.0pt}(\beta {\hspace{0.5pt}}(X))=\textrm{str}\hspace{1.0pt}(X). \end{aligned}$$
(18)

Let us now regard \(\alpha \) and \(\beta \) as automorphisms of the Lie superalgebra \(\mathfrak {gl}_{\,n|n}\,\). It follows from the first equation in (16) that \(\alpha ^{\,2{\hspace{0.5pt}}n}=\beta ^{\,n}=1{\hspace{0.5pt}}\). The eigenvalues of \(\alpha \) and \(\beta \) are \(\varepsilon ^{\,g}\) and \(\varepsilon ^{\,2{\hspace{0.5pt}}h}\) where g and h range over \(\mathbb {Z}_{{\hspace{0.5pt}}2{\hspace{0.5pt}}n}\) and \(\mathbb {Z}_{{\hspace{0.5pt}}n}\), respectively. Let \(\mathfrak {gl}_{\,n|n}^{{\hspace{0.5pt}}\,g,h}\) be the joint eigenspace of \(\alpha \) and \(\beta \) corresponding to \(\varepsilon ^{\,g}\) and \(\varepsilon ^{\,2{\hspace{0.5pt}}h}\,\). By (18)

$$\begin{aligned} \mathfrak {gl}_{\,n|n}^{{\hspace{0.5pt}}\,g,h}\subset \mathfrak {sl}_{\,n|n}\quad \text {for}\quad (g,h)\ne (0,0). \end{aligned}$$
(19)

Consider the Casimir element of the tensor square \(\mathfrak {gl}_{\,n|n}\!\otimes \mathfrak {gl}_{\,n|n}\)

$$\begin{aligned} t\,=\sum _{i,j}\,E_{{\hspace{0.5pt}}ij}\otimes E_{{\hspace{0.5pt}}ji}\,(-1)^{\,{\bar{\jmath }\,}}{\hspace{0.5pt}}. \end{aligned}$$

This element is invariant by \(\mathfrak {gl}_{\,n|n}\) as for any indices k, l by using (9) we get

$$\begin{aligned}{}[{\hspace{0.5pt}}\,t,E_{{\hspace{0.5pt}}kl}\otimes 1+1\otimes E_{{\hspace{0.5pt}}kl}\,]=0. \end{aligned}$$
(20)

Note that

$$\begin{aligned} ({\hspace{0.5pt}}\eta {\hspace{0.5pt}}\otimes \eta {\hspace{0.5pt}}){\hspace{0.5pt}}\,t=-\,t. \end{aligned}$$
(21)

It follows from (20) that the Casimir element t is invariant by both \(\alpha {\hspace{0.5pt}}\otimes \alpha \) and \(\beta \otimes \beta \,\). Therefore t belongs to the direct sum of subspaces

$$\begin{aligned} \mathfrak {gl}_{\,n|n}^{{\hspace{0.5pt}}\,g,h}{\hspace{-0.5pt}}\otimes \mathfrak {gl}_{\,n|n}^{{\hspace{0.5pt}}\,-g,-h}\subset \mathfrak {gl}_{\,n|n}\!\otimes \mathfrak {gl}_{\,n|n}. \end{aligned}$$
(22)

Let \(t_{{\hspace{0.5pt}}g,h}\) be the projection of element t to the direct summand (22). By (17),(21)

$$\begin{aligned} ({\hspace{0.5pt}}\eta {\hspace{0.5pt}}\otimes \eta {\hspace{0.5pt}}){\hspace{0.5pt}}\,t_{{\hspace{0.5pt}}g,h}=-\,t_{{\hspace{0.5pt}}-g,-h}. \end{aligned}$$
(23)

Further, let \(\sigma \) be the linear transformation of \(\mathfrak {gl}_{\,n|n}\!\otimes \mathfrak {gl}_{\,n|n}\) defined by setting

$$\begin{aligned} \sigma \hspace{1.0pt}(X\otimes Y)=Y\otimes X\,(-1)^{{\hspace{0.5pt}}\deg X{\hspace{0.5pt}}\deg Y} \end{aligned}$$

for homogeneous elements X and \(Y\,\). By the definition of t we have \(\sigma (t)=t\,\). So

$$\begin{aligned} \sigma \hspace{1.0pt}(\,t_{{\hspace{0.5pt}}g,h}{\hspace{0.5pt}})={\hspace{0.5pt}}t_{{\hspace{0.5pt}}-g,-h}. \end{aligned}$$
(24)

We do not need explicit formulas for every projection \(t_{{\hspace{0.5pt}}g,h}\,\). We only note that

$$\begin{aligned} t_{\,0,0}=\frac{1}{{\hspace{0.5pt}}2{\hspace{0.5pt}}n{\hspace{0.5pt}}}\,({\hspace{0.5pt}}J\otimes 1+J\otimes 1{\hspace{0.5pt}}) \end{aligned}$$
(25)

where

$$\begin{aligned} J=E_{{\hspace{0.5pt}}11}-E_{{\hspace{0.5pt}}-1,-1}+\cdots +E_{{\hspace{0.5pt}}nn}\,-E_{{\hspace{0.5pt}}-n,-n}. \end{aligned}$$

5 Classical Yang–Baxter Equation

The central element 1 of the Lie superalgebra \(\mathfrak {gl}_{\,n|n}\) is contained in the eigenspace \(\mathfrak {gl}_{\,n|n}^{{\hspace{0.5pt}}\,0,0}\,\). Therefore for any \((g,h)\ne (0,0)\) the element \(t_{{\hspace{0.5pt}}g,h}\) can be identified with its image in \(\mathfrak {pgl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {pgl}_{\,n|n}\,\). By using this identification, introduce a function of u

$$\begin{aligned} s{\hspace{0.5pt}}(u)=\sum _{(g,h)\ne (0,0)}\varphi _{g,h}(u){\hspace{0.5pt}}\,t_{{\hspace{0.5pt}}g,h}. \end{aligned}$$

Observe that the function \(s{\hspace{0.5pt}}(u)\) takes all its values in the subspace

$$\begin{aligned} \mathfrak {psl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {psl}_{\,n|n}\subset \mathfrak {pgl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {pgl}_{\,n|n}. \end{aligned}$$

Changing the summation indices g, h to \(-g,-h\), respectively, and then using (15), (24) proves that \(s{\hspace{0.5pt}}(u)\) indeed satisfies the condition (2) for \(\mathfrak {g}=\mathfrak {psl}_{\,n|n}\,\). Using (15), (23) proves that \(s{\hspace{0.5pt}}(u)\) satisfies (3). Here we regard \(\eta \) as an automorphism of the Lie algebra \(\mathfrak {psl}_{\,n|n}\,\). The automorphisms \(\alpha \) and \(\beta \) of \(\mathfrak {gl}_{\,n|n}\) preserve \(\mathfrak {sl}_{\,n|n}\) and descend to \(\mathfrak {psl}_{\,n|n}\) too. By the periodicity properties (14) of the function \(\varphi _{g,h}(u)\)

$$\begin{aligned} s{\hspace{0.5pt}}(u+1)=({\hspace{0.5pt}}\alpha \otimes {\textrm{id}}{\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u)=({\hspace{0.5pt}}{\textrm{id}}\otimes \alpha ^{{\hspace{0.5pt}}-1}{\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u) \end{aligned}$$
(26)

and

$$\begin{aligned} \hspace{5pt} s{\hspace{0.5pt}}(u+\tau )=({\hspace{0.5pt}}\beta \otimes {\textrm{id}}{\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u)=({\hspace{0.5pt}}{\textrm{id}}\otimes \beta ^{{\hspace{0.5pt}}-1}{\hspace{0.5pt}})\,s{\hspace{0.5pt}}(u). \end{aligned}$$
(27)

By (25) the image in \(\mathfrak {pgl}_{\,n|n}\otimes \,\mathfrak {pgl}_{\,n|n}\) of the element \(t_{\,0,0}\) is zero. Hence the image of the element t is

$$\begin{aligned} \sum _{(g,h)\ne (0,0)}t_{{\hspace{0.5pt}}g,h}. \end{aligned}$$

Denote this image by \(p\,\). The residue of the function \(s{\hspace{0.5pt}}(u)\) at \(u=0\) equals \(p\,\).

Now consider the function \(r{\hspace{0.5pt}}(u,v)\) as defined by (4). We already observed in Sect. 1 that (2) and (3) imply the antisymmetry property (1). Let us show that \(r{\hspace{0.5pt}}(u,v)\) is indeed a solution of the classical Yang–Baxter equation for \(\mathfrak {g}=\mathfrak {psl}_{\,n|n}\,\). We will employ general arguments from [4, Sect. 5].

First observe that due to (4) and to the first equalities in (26) and (27) we get

$$\begin{aligned} r{\hspace{0.5pt}}(u+1,v)=({\hspace{0.5pt}}\alpha \otimes {\textrm{id}}{\hspace{0.5pt}})\,r{\hspace{0.5pt}}(u,v) \quad \text {and}\quad r{\hspace{0.5pt}}(u+\tau ,v)=({\hspace{0.5pt}}\beta \otimes {\textrm{id}}{\hspace{0.5pt}})\,r{\hspace{0.5pt}}(u,v). \end{aligned}$$
(28)

Similarly, due to (5) and to the second equalities in (26) and (27) we get

$$\begin{aligned} r{\hspace{0.5pt}}(u,v+1)=({\hspace{0.5pt}}{\textrm{id}}\otimes \alpha {\hspace{0.5pt}})\,r{\hspace{0.5pt}}(u,v) \quad \text {and}\quad r{\hspace{0.5pt}}(u,v+\tau )=({\hspace{0.5pt}}{\textrm{id}}\otimes \beta {\hspace{0.5pt}})\,r{\hspace{0.5pt}}(u,v). \end{aligned}$$

Let \(f{\hspace{0.5pt}}(u,v,w)\) be the left-hand side of the classical Yang–Baxter equation for our \(r{\hspace{0.5pt}}(u,v)\) with \(\mathfrak {g}=\mathfrak {psl}_{\,n|n}\,\). Since \(\alpha \) and \(\beta \) are Lie algebra automorphisms, we have

$$\begin{aligned} f{\hspace{0.5pt}}(u+1,v,w)=({\hspace{0.5pt}}\alpha \otimes {\textrm{id}}\otimes {\textrm{id}}{\hspace{0.5pt}})\,f{\hspace{0.5pt}}(u,v,w) \end{aligned}$$
(29)

and

$$\begin{aligned} f{\hspace{0.5pt}}(u+\tau ,v,w)=({\hspace{0.5pt}}\beta \otimes {\textrm{id}}\otimes {\textrm{id}}{\hspace{0.5pt}})\,f(u,v,w). \end{aligned}$$
(30)

Choose any values of v and w such that \(v-w,v+w\notin \mathcal {L}\,\). We will prove that then \(f{\hspace{0.5pt}}(u,v,w)\) is a holomorphic function of u in the whole \({\mathbb {C}}\,\). By (29) and (30) this function is bounded and hence a constant. This constant is then an element of \(\mathfrak {psl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {psl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {psl}_{\,n|n}\) invariant by \(\alpha \) and \(\beta \) applied in the first tensor factor. However the only element of \(\mathfrak {psl}_{\,n|n}\) invariant by both \(\alpha \) and \(\beta \) is zero. Therefore our function must be zero.

If \(u\pm v\in \mathcal {L}\), then \(u\pm w\not \in \mathcal {L}\) since \(v\pm w\not \in \mathcal {L}\,\). Then the function \(r_{{\hspace{0.5pt}}13}(u,w)\) has no pole. Consider the first two of the three summands of \(f{\hspace{0.5pt}}(u,v,w)\,\). By the definition (4) their sum can be written as

$$\begin{aligned}{} & {} [\,s_{12}(u-v),r_{13}(u,w)+r_{{\hspace{0.5pt}}23}(v,w)\,]\,+ \end{aligned}$$
(31)
$$\begin{aligned}{} & {} [\,({\hspace{0.5pt}}{\textrm{id}}\otimes \eta \otimes {\textrm{id}}{\hspace{0.5pt}})\,s_{{\hspace{0.5pt}}12}(u+v), r_{{\hspace{0.5pt}}13}(u,w)+r_{{\hspace{0.5pt}}23}(v,w)\,]. \end{aligned}$$
(32)

By multiplying (31) by \(u-v\) and then setting \(u=v\) we get

$$\begin{aligned}{}[{\hspace{0.5pt}}\,p,r_{13}(v,w)+r_{{\hspace{0.5pt}}23}(v,w)\,]=0. \end{aligned}$$

Here we employ (20) and the definition of p as the image of t in \(\mathfrak {pgl}_{\,n|n}\!\otimes {\hspace{0.5pt}}\mathfrak {pgl}_{\,n|n}\) . Therefore (31) has no pole at \(u-v=0\,\). It now follows from (26), (27), (28) that the summand (31) has no pole whenever \(u-v\in \mathcal {L}\,\).

By multiplying the summand (32) by \(u+v\) and then setting \(u=-\,v\) we get

$$\begin{aligned}{} & {} [\,({\hspace{0.5pt}}{\textrm{id}}\otimes \eta \otimes {\textrm{id}}{\hspace{0.5pt}})\,p, r_{{\hspace{0.5pt}}13}(-{\hspace{0.5pt}}v,w)+r_{{\hspace{0.5pt}}23}(v,w)\,]\\{} & {} \quad = ({\hspace{0.5pt}}{\textrm{id}}\otimes \eta \otimes {\textrm{id}}{\hspace{0.5pt}})\, [{\hspace{0.5pt}}\,p,r_{{\hspace{0.5pt}}13}(-{\hspace{0.5pt}}v,w)+ ({\hspace{0.5pt}}{\textrm{id}}\otimes \eta \otimes {\textrm{id}}{\hspace{0.5pt}})\,r_{{\hspace{0.5pt}}23}(v,w)\,] \\{} & {} \quad = ({\hspace{0.5pt}}{\textrm{id}}\otimes \eta \otimes {\textrm{id}}{\hspace{0.5pt}})\, [{\hspace{0.5pt}}\,p,r_{{\hspace{0.5pt}}13}(-{\hspace{0.5pt}}v,w)+r_{{\hspace{0.5pt}}23}(-{\hspace{0.5pt}}v,w)\,]=0. \end{aligned}$$

So (32) has no pole at \(u+v=0\,\). It follows from (26), (27), (28) that (32) has no pole whenever \(u+v\in \mathcal {L}\,\).

Thus the function \(f{\hspace{0.5pt}}(u,v,w)\) has no pole whenever \(u\pm v\in \mathcal {L}\,\). Further, by using the antisymmetry property (1) the function \(-{\hspace{0.5pt}}f{\hspace{0.5pt}}(u,v,w)\) can be written as

$$\begin{aligned}{}[\,r_{{\hspace{0.5pt}}13}(u,w),r_{{\hspace{0.5pt}}12}(u,v)\,]+ [\,r_{{\hspace{0.5pt}}13}(u,w),r_{{\hspace{0.5pt}}32}(w,v)\,]+ [\,r_{{\hspace{0.5pt}}12}(u,v),r_{{\hspace{0.5pt}}32}(w,v)\,]. \end{aligned}$$

Similarly to the above argument we can show that the latter function has no pole when \(u\pm w\in \mathcal {L}\,\). So \(r{\hspace{0.5pt}}(u,v)\) is a solution of the classical Yang–Baxter equation.

Following [2, 3, 6, 10] it would be interesting to find a solution of the quantum Yang–Baxter equation corresponding to our \(r{\hspace{0.5pt}}(u,v)\,\). For the rational solutions of the classical equation constructed in [11] this was already done in the same work.