Anticommutative extension of the Adler map

Let G be a ${{\mathbb{Z}}}_{2}$-graded algebra over ${\mathbb{C}}$ with a unity element e. As a linear space G is a direct sum $G={G}_{0}\oplus {G}_{1}$ (mod 2), such that ${G}_{i}{G}_{j}\subseteq {G}_{i+j}$. Those elements of G that belong either to G₀ or to G₁ are called homogeneous, the ones in G₁ are called odd (or fermionic), while those in G₀ are called even (or bosonic) and e ∈ G₀.

By definition, the parity $| a|$ of an even homogeneous element a is 0, and it is 1 for odd homogeneous elements. The parity of the product $| {ab}|$ of two homogeneous elements is a sum of their parities: $| {ab}| =| a| +| b|$. Grassmann commutativity means that ${ba}={(-1)}^{| a| | b| }{ab}$ for any homogeneous elements a and b. In particular, ${\alpha }^{2}=0$, for all odd elements $\alpha \in {G}_{1}$, and even elements commute with all the elements of G.

Lastly, for a square matrix, M, of the following block-form

$$\begin{eqnarray*}M=\left(\begin{array}{cc}P & {\rm{\Pi }}\\ {\rm{\Lambda }} & L\end{array}\right),\end{eqnarray*}$$

where P and L are matrices with even entries, while Π and Λ possess only odd entries (the block matrices are not necessarily square matrices), we can define the supertrace—and denote it by $\mathrm{str}(M)$—to be the following quantity

$$\begin{eqnarray*}&&\mathrm{str}(M)=\mathrm{tr}(P)-\mathrm{tr}(L),\end{eqnarray*}$$

where $\mathrm{tr}(.)$ is the usual trace of a matrix. We shall use the supertrace later on to generate invariants of the Grassmann extended Adler map.

Let ${V}_{G}=\{(a,\alpha )| \;a\in {G}_{0},\alpha \in {G}_{1}\}$. The definitions of Yang–Baxter maps in the Grassmann and commutative cases formally coincide; the only difference is the set of the objects of the maps. In particular, we say that a map $S\in \mathrm{End}({V}_{G}\times {V}_{G})$

$$\begin{eqnarray}&&((x,\chi ),(y,\psi ))\overset{S}{\mapsto }((u,\xi ),(v,\eta )),\end{eqnarray} \tag{ 1 }$$

is a Grassmann extended Yang–Baxter map if it satisfies the Yang–Baxter equation:

$$\begin{eqnarray*}&&{S}^{12}\;\circ \;{S}^{13}\;\circ \;{S}^{23}={S}^{23}\;\circ \;{S}^{13}\;\circ \;{S}^{12}.\end{eqnarray*}$$

Here ${S}^{{ij}}\in \mathrm{End}({V}_{G}\times {V}_{G}\times {V}_{G})$, $i,j=1,2,3$, $i\ne j$, are defined by the following relations

$$\begin{eqnarray*}&&{S}^{12}=S\times \mathrm{id},\quad {S}^{23}=\mathrm{id}\times S\quad \mathrm{and}\quad {S}^{13}={\pi }^{12}{S}^{23}{\pi }^{12},\end{eqnarray*}$$

where ${\pi }^{12}$ is the involution defined by ${\pi }^{12}((x,\chi ),(y,\psi ),(z,\zeta ))=((y,\psi ),(x,\chi ),(z,\zeta ))$. Moreover, map (1) is called reversible if ${S}^{21}\;\circ \;{S}^{11}=\mathrm{id}$, where S²¹ follows from the definition of S¹² if we change $(u,\xi )\leftrightarrow (v,\eta )$ and $(x,\chi )\leftrightarrow (y,\psi )$.

Furthermore, we shall be using the term Grassmann extended parametric Yang–Baxter map if two parameters $a,b\in {G}_{0}$ are involved in the definition of (1), namely we have a map

$$\begin{eqnarray}&&{S}_{a,b}\;:((x,\chi ),(y,\psi ))\mapsto ((u,\xi ),(v,\eta )),\end{eqnarray} \tag{ 2 }$$

satisfying the parametric Yang–Baxter equation

$$\begin{eqnarray}&&{S}_{a,b}^{12}\;\circ \;{S}_{a,c}^{13}\;\circ \;{S}_{b,c}^{23}={S}_{b,c}^{23}\;\circ \;{S}_{a,c}^{13}\;\circ \;{S}_{a,b}^{12}.\end{eqnarray} \tag{ 3 }$$

Similar to [11], we can consider a problem of refactorisation

$$\begin{eqnarray}&&{L}_{a}(u,\xi ;\lambda ){L}_{b}(v,\eta ;\lambda )={L}_{b}(y,\psi ;\lambda ){L}_{a}(x,\chi ;\lambda ),\end{eqnarray} \tag{ 4 }$$

where ${L}_{a}(x,\chi ;\lambda )$ is a matrix whose entries are Grassmann-valued, $x,a\in {G}_{0},\chi \in {G}_{1}$ and λ is a complex variable (a spectral parameter). The problem can be formulated as following: for a given set $\{a,b,x,y\in {G}_{0},\psi ,\chi \in {G}_{1}\}$ find such $\{u,v\in {G}_{0},\xi ,\eta \in {G}_{1}\}$ that equation (4) is satisfied identically in λ. If this problem of refactorisation has a unique solution, then it defines a map (2) which satisfies the YB equation (3) and it is reversible. The proof of the latter statement is exactly the same as in the commutative case [13].

The invariants of a Grassmann-valued Yang–Baxter map can be found using the supertrace. Applying the supertrace to equation (4), we obtain that $\mathrm{str}({L}_{b}(y,\psi ){L}_{a}(x,\chi ))$ is a generating function of the invariants associated Yang–Baxter map ${S}_{a,b}$.

According to the Grassmann extended Darboux transformation in (7), changing $({v}_{1}/2,\xi ,{\eta }_{[1]};{p}_{1})\to (x,{\chi }_{1},{\chi }_{2};a)$, we consider the following matrix

$$\begin{eqnarray}{M}_{a}(x,{\boldsymbol{\chi }})=\lambda {M}_{1}+{M}_{0}=\lambda \left(\begin{array}{ccc}0 & 0 & 0\\ e & 0 & 0\\ 0 & 0 & 0\end{array}\right)+\left(\begin{array}{ccc}x & e & 0\\ {x}^{2}-a+{\chi }_{1}{\chi }_{2} & x & {\chi }_{1}\\ {\chi }_{2} & 0 & e\end{array}\right),\end{eqnarray} \tag{ 8 }$$

where ${\boldsymbol{\chi }}=({\chi }_{1},{\chi }_{2})$, and we substitute it into the Lax equation. Then, we have the following.

(Noncommutative Adler map).

Proposition 3.1.1 Let $x+y$ be an invertible element of the Grassmann algebra. Then, the matrix refactorisation problem

$$\begin{eqnarray}&&{M}_{a}(u,{\boldsymbol{\xi }}){M}_{b}(v,{\boldsymbol{\eta }})={M}_{b}(y,{\boldsymbol{\psi }}){M}_{a}(x,{\boldsymbol{\chi }}),\end{eqnarray} \tag{ 9 }$$

where ${M}_{a}={M}_{a}(x,{\boldsymbol{\chi }})$ is given by (8) is equivalent to a map

$$\begin{eqnarray}&&((x,{\boldsymbol{\chi }}),(y,{\boldsymbol{\psi }}))\overset{{S}_{a,b}}{\;\longmapsto \;}((u,{\boldsymbol{\xi }}),(v,{\boldsymbol{\eta }})),\end{eqnarray} \tag{ 10 }$$

given by the following

$$\begin{eqnarray}&&x\mapsto u=y+\displaystyle \frac{a-b}{x+y-{\chi }_{1}{\psi }_{2}};\end{eqnarray} \tag{ 11a }$$

$$\begin{eqnarray}&&{\chi }_{1}\mapsto {\xi }_{1}={\psi }_{1}-\displaystyle \frac{a-b}{x+y}{\chi }_{1};\end{eqnarray} \tag{ 11b }$$

$$\begin{eqnarray}&&{\chi }_{2}\mapsto {\xi }_{2}={\psi }_{2};\end{eqnarray} \tag{ 11c }$$

$$\begin{eqnarray}&&y\mapsto v=x+\displaystyle \frac{b-a}{x+y-{\chi }_{1}{\psi }_{2}};\end{eqnarray} \tag{ 11d }$$

$$\begin{eqnarray}&&{\psi }_{1}\mapsto {\eta }_{1}={\chi }_{1};\end{eqnarray} \tag{ 11e }$$

$$\begin{eqnarray}&&{\psi }_{2}\mapsto {\eta }_{2}={\chi }_{2}+\displaystyle \frac{a-b}{x+y}{\psi }_{2},\end{eqnarray} \tag{ 11f }$$

which satisfies the Yang–Baxter equation (3), is a parametric, reversible, and birational map with invariants

$$\begin{eqnarray}&&{I}_{1}=x+y,\qquad {I}_{0}={\chi }_{1}{\chi }_{2}+{\psi }_{1}{\psi }_{2}.\end{eqnarray} \tag{ 12 }$$

Proof. Equation (9) implies that ${\xi }_{2}={\psi }_{2}$ and ${\eta }_{1}={\chi }_{1}$, and, using the latter, the following system of equations

$$\begin{eqnarray}&&u+v=x+y,\end{eqnarray} \tag{ 13a }$$

$$\begin{eqnarray}&&{uv}+{v}^{2}-b+{\chi }_{1}{\eta }_{2}={yx}+{x}^{2}-a+{\chi }_{1}{\chi }_{2},\end{eqnarray} \tag{ 13b }$$

$$\begin{eqnarray}&&{u}^{2}-a+{\xi }_{1}{\psi }_{2}+{uv}={y}^{2}-b+{\psi }_{1}{\psi }_{2}+{yx},\end{eqnarray} \tag{ 13c }$$

$$\begin{eqnarray}&&u{\chi }_{1}+{\xi }_{1}=y{\chi }_{1}+{\psi }_{1},\quad {\psi }_{2}v+{\eta }_{2}={\psi }_{2}x+{\chi }_{2},\end{eqnarray} \tag{ 13d }$$

$$\begin{eqnarray}&&({u}^{2}-a+{\xi }_{1}{\psi }_{2})v+u({v}^{2}-b+{\chi }_{1}{\eta }_{2})+{\xi }_{1}{\eta }_{2}\\ &&=({y}^{2}-b+{\psi }_{1}{\psi }_{2})x+y({x}^{2}-a+{\chi }_{1}{\chi }_{2})+{\psi }_{1}{\chi }_{2},\end{eqnarray} \tag{ 13e }$$

for variables u, v, ${\xi }_{1}$ and ${\eta }_{2}$. Now, from equations (13b) and (13c) we obtain

$$\begin{eqnarray}&&v(x+y)-b+{\chi }_{1}{\eta }_{2}=x(y+x)-a+{\chi }_{1}{\chi }_{2},\end{eqnarray} \tag{ 14a }$$

$$\begin{eqnarray}&&u(x+y)-a+{\xi }_{1}{\psi }_{2}=y(x+y)-b+{\psi }_{1}{\psi }_{2},\end{eqnarray} \tag{ 14b }$$

where we have made use of (13a). Moreover, expressing ${\xi }_{1}$ and ${\eta }_{2}$ in terms of u and v, using equations (13d), and substituting to (14) we find that u and v are given by (11a) and (11d), respectively. Using the latter, from (13d) follows that

$$\begin{eqnarray*}&&{\xi }_{1}={\psi }_{1}-\displaystyle \frac{a-b}{x+y-{\chi }_{1}{\psi }_{2}}{\chi }_{1},\qquad {\eta }_{2}={\chi }_{2}+\displaystyle \frac{a-b}{x+y-{\chi }_{1}{\psi }_{2}}{\psi }_{2}.\end{eqnarray*}$$

In the above expressions, we multiply both the numerators and the denominators with the conjugate expression of the latter, namely $x+y+{\chi }_{1}{\psi }_{2}$, and we use the fact that ${\chi }_{1}^{2}={\psi }_{2}^{2}=0$. Then, it follows that ${\xi }_{1}$ and ${\eta }_{2}$ are given by (11b) and (11f), respectively. Lastly, equation (13e) is satisfied in view of the rest ((13a)–(13d)).

The supertrace of the quantity ${M}_{b}(y,{\boldsymbol{\psi }}){M}_{a}(x,{\boldsymbol{\chi }})$ reads:

$$\begin{eqnarray*}&&\mathrm{str}{({M}_{b}(y,{\boldsymbol{\psi }}){M}_{a}(x,{\boldsymbol{\chi }}))=2\lambda +(x+y)}^{2}+{\chi }_{1}{\chi }_{2}+{\psi }_{1}{\psi }_{2}-a-b-e.\end{eqnarray*}$$

Thus, $I={(x+y)}^{2}+{\chi }_{1}{\chi }_{2}+{\psi }_{1}{\psi }_{2}={I}_{1}^{2}+{I}_{0}$ is invariant of the map (10)–(11), where I₁ and I₀ are given by (12). However, it can be readily verified that I₁ and I₀ are invariants themselves.

From the uniqueness of the refactorisation problem (9) it follows that the map (10)–(11) satisfies the Yang–Baxter equation and it is reversible. The birationality of the map follows from the fact that (9) admits the symmetry $(u,{\boldsymbol{\xi }},v,{\boldsymbol{\eta }},a,b)\leftrightarrow (y,{\boldsymbol{\psi }},x,{\boldsymbol{\chi }},b,a)$.

Setting all odd variables in (11) equal to zero and considering even variables and parameters $a,b$ to be ${\mathbb{C}}$-valued (the bosonic limit) we obtain the standard Adler map (15)

$$\begin{eqnarray}&&(x,y)\overset{{Y}_{a,b}}{\;\longmapsto \;}\left(y+\displaystyle \frac{a-b}{x+y},x+\displaystyle \frac{b-a}{x+y}\right).\end{eqnarray} \tag{ 15 }$$

The Adler map (15) occurs from the 3-D consistent discrete potential KdV equation [9, 10]. Its Lax representation [11, 13] occurs from the bosonic limit of (9); that is, it is given by

$$\begin{eqnarray*}&&{L}_{a}(u){L}_{b}(v)={L}_{b}(y){L}_{a}(x),\end{eqnarray*}$$

where ${L}_{a}={L}_{a}(x)$ is given by ${M}_{a}(x,{\boldsymbol{\chi }})$ in (8) if one sets the odd variables equal to zero, namely

$$\begin{eqnarray*}{L}_{a}(x)=\left(\begin{array}{cc}x & e\\ {x}^{2}+a-\lambda & x\end{array}\right).\end{eqnarray*}$$

It can be readily verified that the Adler map (15) is involutive; yet, for its Grassmann extension (10)–(11) we have the following.

Proposition 3.1.2. The map ${S}_{a,b}\in \mathrm{End}({V}_{G}\times {V}_{G})$ given by (11) is non-involutive.

Proof. If we denote

$$\begin{eqnarray*}&&(x,{\chi }_{1},{\chi }_{2},y,{\psi }_{1},{\psi }_{2})\overset{{S}_{a,b}}{\mapsto }(x^{\prime} ,{\chi }_{1}^{\prime },{\chi }_{2}^{\prime },y^{\prime} ,{\psi }_{1}^{\prime },{\psi }_{2}^{\prime })\overset{{S}_{a,b}}{\mapsto }(x^{\prime\prime} ,{\chi }_{1}^{\prime\prime },{\chi }_{2}^{\prime\prime },y^{\prime\prime} ,{\psi }_{1}^{\prime\prime },{\psi }_{2}^{\prime\prime }),\end{eqnarray*}$$

then it follows from (11c) and (11f) that

$$\begin{eqnarray*}&&{\chi }_{2}^{\prime\prime }={\psi }_{2}^{\prime }={\chi }_{2}+\displaystyle \frac{a-b}{x+y}{\psi }_{2}\ne {\chi }_{2}\end{eqnarray*}$$

which implies that

$$\begin{eqnarray*}&&{S}_{a,b}\;\circ \;{S}_{a,b}\ne \mathrm{id},\end{eqnarray*}$$

and this completes the proof.