On the classification of graded twisted planes

TL;DR

This paper classifies graded twisted tensor products of polynomial algebras using matrix representations, covering known quadratic cases, a new family with extension properties, and a partially classified family involving quasi-balanced sequences.

Contribution

It provides a nearly complete classification of graded twisted tensor products of polynomial algebras through matrix methods, including new families and extending previous classifications.

Findings

Classified quadratic algebras by Conner and Goetz.

Identified a new family $A(n,d,a)$ with extension properties.

Described a partially classified family $B(a,L)$ with quasi-balanced sequences.

Abstract

We use a representation of a graded twisted tensor product of $K[x]$ with $K[y]$ in $L(K^{\Bbb{N}_0})$ in order to obtain a nearly complete classification of these graded twisted tensor products via infinite matrices. There is one particular example and three main cases: quadratic algebras classified by Conner and Goetz, a family called $A(n,d,a)$ with the $n+1$-extension property for $n\ge 2$, and a third case, not fully classified, which contains a family $B(a,L)$ parameterized by quasi-balanced sequences.

Equations

$$y^{k}x=\sum _{i=0}^{k+1}{a}_i{x}^{k+1-i}{y}^i.$$

$$y^{k}x=x^{k+1}-x^{k}y+y^{k+1},\text{for all k<n},\text{and}{y}^{n}x\ne x^{n+1}-x^{n}y+y^{n+1}.$$

$$y^{n}x=d{x}^{n+1}-d{x}^{n}y-ax{y}^n+(a+1)y^{n+1},$$

$$y^{n}x=d{x}^{n+1}-x^{n}y+(a+1)y^{n+1},$$

$$\varphi :K[x]{\otimes }_{\sigma }K[y]\to L(K^{\text{mathds}{N}_0}).$$

$${\mu }_{\tau }=({\mu }_A\otimes {\mu }_B)\circ (A\otimes \tau \otimes B).$$

$$\tau (y^r\otimes a)=\sum _j{\gamma }_j^r(a)\otimes y^j.$$

$${\gamma }_j^r(ab)=\sum _{k=0}^{\infty }{\gamma }_k^r(a){\gamma }_j^k(b).$$

$${\gamma }_j^r=\sum _{l=0}^j{\gamma }_l^i\circ {\gamma }_{j-l}^{r-i}.$$

$$Y:=\left(\begin{array}{ccccc}0& 1& 0& 0& \dots \\ 0& 0& 1& 0& \\ 0& 0& 0& 1& \\ 0& 0& 0& 0& \\ ⋮& & & & \ddots \end{array}\right)\text{ and }Z:=\left(\begin{array}{ccccc}0& 0& 0& 0& \dots \\ 1& 0& 0& 0& \\ 0& 1& 0& 0& \\ 0& 0& 1& 0& \\ ⋮& & & & \ddots \end{array}\right)$$

$$\tau (y^i\otimes a)=\sum _j{\gamma }_j^i(a)\otimes y^j\in K[x]\otimes K[y],$$

$$\psi (1\otimes y^k)=Y^k\text{and}\psi (a\otimes y^k)=\psi ((a\otimes 1)(1\otimes y^k))=M(a)Y^k.$$

$$(1\otimes y^k)(a\otimes 1)=({\mu }_{K[x]}\otimes {\mu }_{K[y]})(1\otimes \tau (y^k\otimes a)\otimes 1)=\sum _u{\gamma }_u^k(a)\otimes y^u,$$

$$\psi (1\otimes y^k)\psi (a\otimes 1)=\sum _u\psi ({\gamma }_u^k(a)\otimes 1)\psi (1\otimes y^u)$$

$$(\psi (1\otimes y^k)\psi (a\otimes 1){)}_{ij}$$

$$K[x]{\otimes }_{\tau }K[y]\stackrel{\cong }{\to }\underset{i\in \text{mathds}{N}_0}{⨁}{K}_i,\text{where Ki≅K[x].}$$

$$K[x]{\otimes }_{\tau }K[y]\cong \sum _{n=0}^{\infty }{K}_i,\text{with Ki≅K[x],}$$

$$\left(\begin{array}{ccccc}0& 0& 0& 0& \dots \\ 1& 0& 0& 0& \\ 0& 1& 0& 0& \\ 0& 0& 1& 0& \\ ⋮& & & & \ddots \end{array}\right)\text{and}\left(\begin{array}{ccccc}x& {\gamma }_0^1(x)& {\gamma }_0^2(x)& {\gamma }_0^3(x)& \dots \\ 0& {\gamma }_1^1(x)& {\gamma }_1^2(x)& {\gamma }_1^3(x)& \\ 0& {\gamma }_2^1(x)& {\gamma }_2^2(x)& {\gamma }_2^3(x)& \\ 0& {\gamma }_3^1(x)& {\gamma }_3^2(x)& {\gamma }_3^3(x)& \\ ⋮& & & & \ddots \end{array}\right).$$

$$M_{kj}(M)=a_0\mathrm{Id}+a_{1}M+\cdots +a_n{M}^n\in L(K[x{]}^{\text{mathds}{N}_0}).$$

$$Y^{k}M=\sum _{j\ge 0}{M}_{kj}(M)Y^j.$$

$${\gamma }_j^r(x^{i+l})=(M^{i+l}{)}_{rj}=\sum _{k=0}^{\infty }(M^i{)}_{rk}(M^l{)}_{kj}=\sum _{k=0}^{\infty }{\gamma }_k^r(x^i){\gamma }_j^k(x^l),$$

$${\gamma }_j^{l+k}(x^i)=\sum _{u=0}^j{\gamma }_u^l({\gamma }_{j-u}^k(x^i)$$

$${\gamma }_j^{l+k}(x)$$

$$(M_{il}(M){)}_{rj}=\sum _s{a}_s(M^s{)}_{rj}=\sum _s{a}_s{\gamma }_j^r(x^s)={\gamma }_j^r\left(\sum _s{a}_s{x}^s\right)={\gamma }_j^r(M_{il}).$$

$${\gamma }_j^{l+k}(x^{i+1})$$

$$Y^{k}M=\sum _{j=0}^{k+1}{M}_{kj}{M}^{k+1-j}{Y}^j.$$

$${\gamma }_j^r(x^i):=(M^i{)}_{kj}{x}^{k+i-j}$$

$$(Y^k\tilde{M}{)}_{rs}$$

$$y^{2}x=y(bxy+c{y}^2)=b(yx)y+c{y}^3=b(bxy+c{y}^2)y+c{y}^3=b^{2}x{y}^2+c(b+1)y^3,$$

$$y^{k}x=b^{k}x{y}^k+c[k{]}_b{y}^{k+1},$$

Full text

Christian Valqui was supported by PUCP-DGI-2019-1-0015.

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

Corresponding author: Christian Valqui

1. 1 Preliminaries
2. 2 Construction of the matrices associated with a twisting map
3. 3 Roots of $Q_{n}$
4. 4 The case $yx=x^{2}-xy+y^{2}$
5. 5 The family $A(n,d,a)$
6. 6 Roots of $R_{k}$
7. 7 The case $y^{n}x=dx^{n+1}-x^{n}y+(a+1)y^{n+1}$
8. 8 The family $B(a,L)$ and quasi-balanced sequences
9. 9 Existence of twisting maps for the family $B(a,L)$

Introduction

In [CSV] the authors introduced the notion of twisted tensor product of unital $K$-algebras, where $K$ is a unital ring. We assume that $K$ is a field, and consider the basic problem of classifying all twisted tensor products of $A$ with $B$ for a given pair of algebras $A$ and $B$. In general this problem is out of reach, although some results have been obtained, mainly for finite dimensional algebras (see [A], [AGGV], [C], [GGV1], [JLNS] and [LN]). In particular, in [GGV1] some families of twisted tensor products of $K[x]$ with $K[y]$ were found. The full classification of these tensor products seems to be still out of reach, but in [CG2] (see also [CG]) the graded twisted tensor products of $K[x]$ with $K[y]$ which yield quadratic algebras were completely classified. On the other hand the twisted tensor product of $K[x]/\langle x^{n}\rangle$ with an algebra $A$ can be represented in $M_{n}(A)$ (see [GGV2]*Theorem 1.10). This representation can be generalized to finite dimensional algebras (see [A]).

In this article we start with a representation of a twisted tensor product of $K[x]$ with $K[y]$ in $L(K[x]^{\mathds{N}_{0}})$, which is very similar to the representation in [GGV2]*Theorem 1.10. In the graded case this representation can be simplified further to a representation of the graded twisted tensor product in the algebra $L(K^{\mathds{N}_{0}})$, embedded in the algebra of infinite matrices with entries in $K$. Thus we manage to translate the problem of classifying all graded twisted tensor products of $K[x]$ with $K[y]$ into the problem of classifying infinite matrices with entries in $K$ satisfying certain conditions (see Corollary 1.8). With this method we show that all graded twisted tensor products of $K[x]$ with $K[y]$ can be classified into three main cases, except one particular example (see Proposition 4.4). The first case is the case of quadratic algebras, already classified in [CG2], the second case yields a family called $A(n,d,a)$ and in the third case we have only some partial classification results, and obtain a family called $B(a,L)$.

One can describe a graded twisting tensor product of $K[x]$ with $K[y]$ by specifying how $y^{k}$ commutes with $x$, which means determining the coefficients $a_{i}$’s in

[TABLE]

For example, we obtain the quantum plane if the commutation relation is $yx=qxy$, and in Example 2.1 we explore the case $yx=bxy+cy^{2}$.

Definition 0.1**.**

A graded twisting tensor product of $K[x]$ with $K[y]$ has the $n$-extension property, if the multiplicative structure of the algebra is determined completely by the commutation relations (0.1) for $k=1,\dots,n-1$. For example, if the relation $yx=ax^{2}+bxy+cy^{2}$ determines the multiplicative structure of the algebra, then the algebra has the $2$-extension property, and is called quadratic.

In general, the relation $yx=ax^{2}+bxy+cy^{2}$ yields a quadratic algebra, provided that $ac\neq 1$ and that $(b,ac)$ is not a root of any member of a certain family of polynomials $Q_{n}(b,c)$. In this case our results match the results of [CG2], which were obtained with very different methods.

If $a\neq 0$, then such a tensor product is equivalent to one with $a=1$, and so we will focus on the case $yx=x^{2}+bxy+cy^{2}$. In the case $c=1$ one can show that necessarily $b=-1$ (see Lemma 2.6) and the resulting algebras are not quadratic, i.e., they do not have the 2-extension property. We obtain a particular algebra with $y^{k}x=x^{k+1}-x^{k}y+y^{k+1}$ for all $k\in\mathds{N}$ (see Proposition 4.4). This algebra does not have the $m$-extension property for any $m$.

For every graded twisted tensor product with $yx=x^{2}-xy+y^{2}$, which is not the particular case mentioned above, there exists $n\geq 2$ such that

[TABLE]

A central result is Proposition 4.6, which shows that we have exactly two possibilities for $y^{n}x$. In the first case

[TABLE]

for $a,d$ in $K$ satisfying certain conditions, namely, $(a,d)$ is not a root of any member of a certain family of polynomials $R_{j}(a,d)$. This yields a family $A(n,d,a)$ of twisted tensor products which have the $(n+1)$-extension property (see sections 5 and 6), which means that the multiplication is determined by the commuting relations up to degree $n+1$. The second case is treated in sections 7, 8 and 9, and the commutation relation at degree $n+1$ is

[TABLE]

where $(a+1)d=1$. Although the full classification is not achieved in this case, our methods show that one can achieve the classification of all possible twisting maps up to any degree, with increasing amount of computational work.

Moreover, we manage to find a family of twisted tensor products which we call $B(a,L)$, parameterized by $a\in K\setminus\{0,-1\}$ and $L\in\mathcal{L}$, where $\mathcal{L}$ is the set of quasi-balanced sequences of positive integers (see Definition 8.1). These sequences are interesting on their own, for example they show a surprising connection to Euler’s function $\varphi$. Every truncated quasi-balanced sequence can be continued in several ways, which implies that all members of the family $B(a,L)$ have not the $m$-extension property for any $m$.

The families that were found via the partial classification in the present article could be the smallest examples of non-quadratic graded algebras. On one hand we have a family of algebras $B(a,L)$, which have not the $m$-extension property for any $m$. On the other hand, for any chosen $n\geq 2$ there is an algebra with the $n+1$-extension property in $A(n,d,a)$. It would be very interesting to study the homological behaviour of these algebras.

The following table contains all possible graded twisting maps of $K[x]$ with $K[y]$. The only twisting maps that have not been fully classified are in the last row. For the families with $yx=x^{2}-xy+y^{2}$ we formulate the commuting relations with respect to the infinite matrices $Y=\phi(y)$, $M=\phi(x)$ and $\widetilde{M}=\phi(x-y)$, obtained from the faithful representation (see Remark 1.9)

[TABLE]

For example the relation $yx=x^{2}-xy+y^{2}$ corresponds to $Y\widetilde{M}=M\widetilde{M}$, and $Y^{k}\widetilde{M}=M^{k}\widetilde{M}$ stands for $y^{k}x=x^{k+1}-x^{k}y+y^{k+1}$.

1   Preliminaries

Let $K$ be a field and let $A$ and $B$ be unitary $K$-algebras. A twisted tensor product of $A$ with $B$ over $K$ is an associative algebra structure defined on $A\otimes B$, such that the canonical maps $i_{A}\colon A\longrightarrow A\otimes_{K}B$ and $i_{B}\colon B\longrightarrow A\otimes_{K}B$ are algebra maps satisfying $a\otimes b=i_{A}(a)i_{B}(b)$. We will classify the graded twisted tensor products of $A=K[x]$ with $B=K[y]$. As is known (see e.g. [CSV]) classifying the twisted tensor products is equivalent to classifying the twisting maps $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$, which are $K$-linear maps satisfying

1. (a)

$\tau(1\otimes a)=a\otimes 1$, 2. (b)

$\tau(y^{r}\otimes 1)=1\otimes y^{r}$, 3. (c)

$\tau(y^{r}\otimes ab)=(\mu_{A}\otimes C)\circ(A\otimes\tau)\circ(\tau\otimes A)(y^{r}\otimes a\otimes b)$, 4. (d)

$\tau(y^{r}y^{t}\otimes a)=(A\otimes\mu_{C})\circ(\tau\otimes C)\circ(C\otimes\tau)(y^{r}\otimes y^{t}\otimes a)$.

(See for example [CSV]*Remark 2.4). The multiplication on $A\otimes B$ is then defined by

[TABLE]

By definition two twisting maps $\tau$ and $\tau^{\prime}$ are isomorphic if and only if there are algebra automorphisms $g:A\to A$ and $h:B\to B$ such that $\tau^{\prime}=(g^{-1}\otimes h^{-1})\circ\tau\circ(h\otimes g)$.

Now, a linear map $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$ determines and is determined by linear maps $\gamma^{r}_{j}:K[x]\to K[x]$ for $r,j\in\mathds{N}_{0}$ such that $\gamma^{r}_{j}(a)=0$ for fixed $r,a$ and sufficiently big $j$; via the formula

[TABLE]

Proposition 1.1**.**

A linear map $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$ is a twisting map if and only if

1. (1)

$\gamma^{0}_{j}=\delta_{j0}\operatorname{Id}$. 2. (2)

$\gamma^{r}_{j}(1)=\delta_{jr}$. 3. (3)

For all $r,j$ and all $a,b\in K[x]$,

[TABLE]

Note that for fixed $a,b\in K[x]$, the sum is finite. 4. (4)

For all $r,j$ and $i<r$,

[TABLE]

Proof.

A straightforward computation shows that these four conditions correspond to the four conditions (a)–(d) that characterize a twisting map (see for example [GGV1]*Theorem 2.1). ∎

If $\tau$ is a twisting map, then we will define a representation of the twisted tensor product $K[x]\otimes_{\tau}K[y]$ on $K[x]^{\mathds{N}_{0}}$ along the lines of [GGV2]*Theorem 1.10. For this note that the elements of $L(K[x]^{\mathds{N}_{0}})$ are the infinite matrices with entries in $K[x]$ indexed by $\mathds{N}_{0}\times\mathds{N}_{0}$ such that each row has only a finite number of non zero entries.

Notation 1.2**.**

Throughout this paper we denote by $Y$ and $Z$ the infinite matrices

[TABLE]

in $L(K[x]^{\mathds{N}_{0}})$. Note that $YZ=\operatorname{Id}$ and $ZY=\operatorname{Id}-E_{00}$.

For any infinite matrix $B$ we have $(Y^{k}B)_{ij}=B_{i+k,j}$ and $(BY^{k})_{i,j}=\begin{cases}B_{i,j-k},&\mbox{if }j\geq k\\ 0,&\mbox{otherwise}.\end{cases}$

If now $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$ is a twisting map determined by the $K$-linear maps $\gamma^{i}_{j}$, for each $a\in K[x]$ we define the infinite matrix $M(a)\in L(K[x]^{\mathds{N}_{0}})$ by $M(a)_{ij}=\gamma_{j}^{i}(a)$. This matrix satisfies the finiteness condition, since

[TABLE]

so $\gamma_{j}^{i}(a)\neq 0$ only for a finite number of $j$’s.

Remark 1.3*.*

By conditions (2) and (3) we have $M(1)=\operatorname{Id}$ and $M(ab)=M(a)M(b)$ for all $a,b\in K[x]$.

Proposition 1.4**.**

Let $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$ be a twisting map. The formulas $\psi(a\otimes 1)=M(a)$ and $\psi(1\otimes y)=Y$ determine an injective algebra map (faithful representation) $\psi:K[x]\otimes_{\tau}K[y]\to L(K[x]^{\mathds{N}_{0}})$.

Proof.

By Remark 1.3 we have $\psi(1)=1$. For $\psi$ to be an algebra map, we need to define

[TABLE]

But then $\psi$ is compatible with the multiplication of elements of the form $(1\otimes y^{i})(1\otimes y^{k})$, and by Remark 1.3 it is also compatible with elements of the form $(a\otimes 1)(b\otimes 1)$. Since by (1.1) we know that

[TABLE]

we have to prove that

[TABLE]

for $a\in K[x]$. Using condition (4) of Proposition 1.1 we obtain

[TABLE]

which concludes the proof that $\psi$ is an algebra map. The injectivity follows from the fact that the composition of $\psi$ with the surjection onto the first row gives the canonical linear isomorphism

[TABLE]

∎

Remark 1.5*.*

The previous representation can be related to the right regular representation of the algebra $K[x]\otimes_{\tau}K[y]$. In fact, we can write $K[x]\otimes_{\tau}K[y]$ as a right module over itself, as

[TABLE]

so we are considering infinite column vectors with entries in $K[x]$. Then the multiplication by $y$ and $x$ from the right are represented by the multiplication from the left by the matrices

[TABLE]

This gives a representation of the opposite algebra $(K[x]\otimes_{\tau}K[y])^{Op}$ in the algebra of infinite matrices with only finitely many non-zero entries in each column. If we take the transposed matrices, then we obtain the representation in Proposition 1.4.

Notation 1.6**.**

Let $M\in L(K[x]^{\mathds{N}_{0}})$ and for fixed $k,j$ write $M_{kj}=a_{0}+a_{1}x+\dots+a_{n}x^{n}\in K[x]$. Then we will evaluate this polynomial at $M$ setting

[TABLE]

Proposition 1.7**.**

Let $M\in L(K[x]^{\mathds{N}_{0}})$ be such that $M_{0j}=x\delta_{0j}$ and

[TABLE]

(Note that the sum is finite). Then the maps $\gamma^{r}_{j}$ defined by $\gamma^{r}_{j}(x^{i})\coloneqq(M^{i})_{kj}\in K[x]$ determine a twisting map.

Proof.

We will prove that $\gamma^{r}_{j}$ satisfies (1) to (4) of Proposition 1.1.

(1): This is clear since $(M^{i})_{0j}=x^{i}\delta_{0j}$.

(2): Follows from $M^{0}=\operatorname{Id}$.

(3): Clearly

[TABLE]

as desired.

(4): For all $i,j,k,l$ we have to prove

[TABLE]

by induction in $i$. For $i=1$ we have

[TABLE]

where the fifth equality follows from the fact that for $M_{il}(x)=\sum_{s}a_{s}x^{s}$ we have

[TABLE]

Assume (1.3) is valid for $i$. Then

[TABLE]

as desired. ∎

Now we assume that the potential twisting map is graded, that means that $\tau(y^{r}\otimes x^{i})=\sum_{j=0}^{i+r}a_{j}x^{i+r-j}\otimes y^{j}$, and so the maps $\gamma^{r}_{j}$ are homogeneous of degree $r-j$.

Corollary 1.8**.**

Let $M\in L(K^{\mathds{N}_{0}})$ be such that $M_{0j}=\delta_{0j}$, $M_{kj}=0$ for $j>k+1$ and

[TABLE]

Then the maps $\gamma^{r}_{j}$ defined by

[TABLE]

determine a graded twisting map. Conversely, if $\tau:K[y]\otimes K[x]\to K[x]\otimes K[y]$ is a graded twisting map, then the corresponding $\gamma^{r}_{j}$’s determine via (1.5) a matrix $M$ satisfying (1.4), $M_{0j}=\delta_{0j}$ and $M_{kj}=0$ for $j>k+1$.

Proof.

The map $\tau$ determined by the $\gamma^{r}_{j}$’s is clearly graded, so we only need to show that the matrix $\widetilde{M}$ defined by $\widetilde{M}_{kj}=M_{kj}x^{k+1-j}$ satisfies (1.2), since $\gamma^{r}_{j}(x^{i})=(\widetilde{M}^{i})_{kj}$ and $\widetilde{M}_{0j}=x\delta_{0j}$. We obtain

[TABLE]

where we use that $\widetilde{M}_{kj}(\widetilde{M})=M_{kj}\widetilde{M}^{k+1-j}$. Now the result follows from Proposition 1.7, and the converse is straightforward. ∎

Remark 1.9*.*

In the graded case the formulas $\phi(x)=M$ and $\phi(y)=Y$ define an injective algebra map (faithful representation) $\phi:K[x]\otimes_{\tau}K[y]\to L(K^{\mathds{N}_{0}})$.

2   Construction of the matrices associated with a twisting map

In order to classify the graded twisting maps, we have to classify the matrices $M$ satisfying the conditions of Corollary 1.8. Note that $M=\psi(x)|_{x=1}=M(x)|_{x=1}$ in the notation of Proposition 1.4. We will write $M_{10}=a$, $M_{11}=b$ and $M_{12}=c$ for such a matrix. In some cases the values of $a$, $b$ and $c$ determine completely the matrix $M$ (and hence the twisting map).

Example 2.1**.**

If $a=0$, then the equality $yx=bxy+cy^{2}$ implies

[TABLE]

and a straightforward inductive argument shows that

[TABLE]

where $[k]_{b}$ denotes the $q$-number defined by $[k]_{b}=1+b+b^{2}+\dots+b^{k-1}$. So the only possible non zero entries of $M$ are $M_{nn}=b^{n}$ and $M_{n,n+1}=c[n]_{b}$ and the corresponding matrix is

[TABLE]

and the twisting map is given by

[TABLE]

By Corollary 1.8 the matrix equalities

[TABLE]

guarantee that $\tau$ is a twisting map. In order to prove these equalities, one notes first that the first equality implies all the others (use induction). Then we check directly that $YM=bMY+cY^{2}$:

[TABLE]

Lemma 2.2**.**

Let $A$ be an associative $K$-algebra, $k\geq 2$, $x,y\in A$, $M_{ij}\in K$ for $0\leq j<k$ and $0\leq i\leq j+1$ such that $M_{0j}=\delta_{j0}$, $(M_{10},M_{11},M_{12})=(1,b,c)$ and

[TABLE]

Then

[TABLE]

where

[TABLE]

for $s=1,\dots,k$.

Proof.

We have

[TABLE]

Since

[TABLE]

the result follows. ∎

Remark 2.3*.*

Let $M\in L(K^{\mathds{N}_{0}})$ be such that $M_{0j}=\delta_{0j}$, $M_{ji}=0$ for $i>j+1$ and for some $k>1$,

[TABLE]

Then $x=M$ and $y=Y$ satisfy the assumptions of the lemma. The equality (2.1) reads

[TABLE]

and if we take the entry $(0,i)$, then the left hand side gives

[TABLE]

and the right hand gives $\overline{M}_{ki}$, since

[TABLE]

Hence in a twisting map with $M_{k-1,k}\neq 1$, the coefficients $M_{ki}$ are determined uniquely by the coefficients $M_{ji}$ with $j<k$.

Remark 2.4*.*

Given a twisting map $\tau$ such that $a\neq 0$, we can replace $\tau$ by the isomorphic twisting map $\tau^{\prime}=(f^{-1}\otimes\operatorname{Id})\circ\tau\circ(\operatorname{Id}\otimes f)$ where $f(x)=ax$. Then for $\tau^{\prime}$ we have $a^{\prime}=1$, $b^{\prime}=b$ and $c^{\prime}=ca$. So we can and will assume that $a=1$.

Proposition 2.5**.**

Assume $M$ determines a graded twisting map. Assume $a=1$ and $M_{k,k+1}\neq 1$ for all $k\geq 1$. Then $b$ and $c$ determine uniquely the matrix $M$ (and hence the twisting map).

Proof.

By Remark 2.3 in that case the entries $M_{ji}$ with $j<k$ determine uniquely the entries $M_{ki}$, hence, by induction, the entries $M_{1j}$, i.e., $M_{10}=1$, $M_{1,1}=b$, and $M_{12}=c$, determine the whole matrix. ∎

However not every choice of $b$ and $c$ is valid, as we will see.

Lemma 2.6**.**

Let $M$ be a matrix of a graded twisting map and set $b_{n}=M_{nn}$ and $c_{n}=M_{n,n+1}$. Then

[TABLE]

Moreover, if $c\neq 1$, then $c_{n}\neq 1$ for all $n$, and if $c=1$, then $b=-1$ and $(1-c_{n})(1-c_{n+1})=0$ for all $n$.

Proof.

We have

[TABLE]

and

[TABLE]

since $M_{nk}=0$ for $k>n+1$ and $M_{k,n+2}=0$ for $k<n+1$. The entry at $(n,n+2)$ of the matrix equality

[TABLE]

which holds by Corollary 1.7, gives $c_{n+1}=c_{n}c_{n+1}+bc_{n}+c$, from which (2.3) follows.

Assume that $c\neq 1$ and assume by contradiction that $c_{n}=1$ for some $n>1$. Then from (2.3) for $n$ we obtain $b=-c$. Equation (2.3) again implies that $c_{2}=c$, $c_{3}=c_{2}$, and so on, contradicting $c_{n}\neq 1$.

Finally, assume that $c=1$, then from (2.3) for $n=1$ we obtain $b=-1$ and so (2.3) reads $c_{n+1}(1-c_{n})=-c_{n}+1$, hence $(1-c_{n})(1-c_{n+1})=0$, as desired. ∎

If $c\neq 1$, then not all values of $b$ and $c$ yield twisting maps. For example, if $c\neq 1$ and $1-c=c(b+1)$, then by (2.3) we would have $c_{2}=\frac{c(b+1)}{1-c}=1$, which is impossible by the previous lemma.

When $c\neq 1$, the first formulas for $c_{n}$ are

[TABLE]

and in general we have

[TABLE]

where $P_{n}$ and $Q_{n}$ are polynomials in $b$ and $c$. Moreover, the formula (2.3) yields the recursive rules

[TABLE]

Given $b$ and $c$, these values are defined even when some $c_{n}=1$, and that happens if and only if $Q_{n+1}=Q_{n+1}(b,c)=0$.

Corollary 2.7**.**

Let $K$ be a field and let $b,c\in K$ with $c\neq 1$. If $b$ and $c$ determine a (necessarily unique) twisting map via Proposition 2.5, then $Q_{n}(b,c)\neq 0$ for all $n\in\mathds{N}$, where the polynomials $P_{n},Q_{n}\in K[b,c]$ are defined by $P_{1}=1$, $Q_{1}=1$ and the recursive rules (2.4).

Proof.

By Proposition 2.5 and the previous discussion. ∎

In order to prove the converse of Corollary 2.7, we consider the valuation on the algebra $L(K[x]^{\mathds{N}_{0}})$ given by

[TABLE]

for $M=(m_{ij})_{i,j\in\mathds{N}_{0}}$, and we also set $w(0)=+\infty$. For example $w(E_{10})=1$ and $w(E_{01})=-1=w(E_{10}+E_{01})$. Note that for some $M$ we can have $w(M)=-\infty$.

Proposition 2.8**.**

Let $M,N\in L(K[x]^{\mathds{N}_{0}})$. Then

1. (1)

$w(M+N)\geq\min\{w(M),w(N)\}$. 2. (2)

If $w(M)\neq w(N)$, then $w(M+N)=\min\{w(M),w(N)\}$. 3. (3)

$w(MN)\geq w(M)+w(N)$.

Proof.

Straightforward. ∎

Definition 2.9**.**

We say that $M$ is homogeneous if $m_{ij}=0$ when $i-j\neq w(M)$, and we denote by $M^{(k)}$ the homogenous component of $M$ of weight $k$ given by $(M^{(k)})_{ij}=\delta_{i-j,k}M_{ij}$.

For example, consider the matrices $Y$ and $Z$ of Notation 1.2, given by $Y_{ij}=\delta_{i+1,j}$ and $Z_{ij}=\delta_{i,j+1}$. Then both are homogeneous with $w(Y)=-1$ and $w(Z)=1$.

Consider the subalgebra $\mathcal{R}\subset L(K[x]^{\mathds{N}_{0}})$ consisting of the homogeneous matrices of weight zero. If $M$ is homogeneous of weight $k>0$, then the matrix $N=(N_{ij})\in\mathcal{R}\subset L(K[x]^{\mathds{N}_{0}})$ given by $N_{ij}=\delta_{ij}M_{k+i,i}$, satisfies

[TABLE]

On the other hand, if $M$ is homogeneous of weight $k<0$, then the matrix $N=(N_{ij})\in\mathcal{R}\subset L(K[x]^{\mathds{N}_{0}})$ given by $N_{ij}=\delta_{ij}M_{i,i-k}$ satisfies

[TABLE]

It follows that for $M$ with $w(M)>-\infty$ we have a decomposition

[TABLE]

for some $N_{k}\in\mathcal{R}$, where the infinite sum converges in the $Z$-adic topology. Note that if $w(M)>0$, then the first sum is empty.

We define the shift operator and its left inverse on $\mathcal{R}$ by setting

[TABLE]

for $A=\operatorname{Diag}(a_{0},a_{1},a_{2},\dots)\in\mathcal{R}$. Note that for a matrix $A\in\mathcal{R}$ we have $ZAY=SA$ and $YAZ=TA$.

Proposition 2.10**.**

Let $K$ be a field and let $b,c\in K$ with $Q_{n}(b,c)\neq 0$ for all $n\in\mathds{N}$. Then there exists a unique matrix $M\in L(K[x]^{\mathds{N}_{0}})$ with $w(M)\geq-1$, such that

[TABLE]

Proof.

We will construct homogeneous components $M^{(j)}$ for $j\geq-1$, such that $M=\sum_{j\geq-1}M^{(j)}$ satisfies (2.8). Note that the equality (2.8) is true if and only if it holds for the homogeneous components of weight $k$ for all $k$, i.e. if

[TABLE]

for all $k\geq-2$. We will construct recursively $M^{(-1)}$, $M^{(0)}$, $M^{(1)}$, $M^{(2)}$,…, $M^{(j)}$, such that (2.9) holds for $k=-2,-1,0,1,\dots,j-1$. This yields an inductive construction of the unique $M$ such that (2.8) holds.

We write $M^{(-1)}=CY$ and $M^{(j)}=Z^{j}B_{j}$ for $j\geq 0$, for some diagonal matrices $C,B_{j}\in\mathcal{R}$, and so

[TABLE]

Note that

[TABLE]

for $j\geq 0$. Note also that

[TABLE]

for $j\geq 0$.

Finally,

[TABLE]

and

[TABLE]

for $j\geq 0$.

For $k=-2$ the equality (2.9) reads

[TABLE]

and since multiplying by $Y$ on the right is injective, we have

[TABLE]

Hence $(\mathds{1}-C)SC=bC+c\mathds{1}$, and the $n$th entry reads $(1-c_{n})c_{n+1}=bc_{n}+c$ which is equality (2.3). Thus we can construct recursively $c_{n+1}$, since $Q_{n}(b,c)\neq 0$ guarantees that $c_{n}\neq 1$ for all $n$. This proves that $b$ and $c$ determine uniquely $C=M_{-1}$ such that (2.9) holds for $k=-2$.

For $k=-1$ the equality (2.9) reads

[TABLE]

and since multiplying by $Y$ on the right is injective, we have

[TABLE]

So we have a recursive formula for $B_{0}$:

[TABLE]

Since $(1-c_{n})\neq 0$ and we already have $(B_{0})_{0}=1$ and $(B_{0})_{1}=b$, this formula determines a unique $B_{0}$ such that the equality (2.9) for $k=-1$ is satisfied.

For $j\geq 0$ the equality (2.9) reads

[TABLE]

and since multiplication by $Z^{j}$ at the left is injective we have

[TABLE]

Assume we have constructed inductively $C$ and $B_{i}$ for $i=0,\dots,j$ such that (2.9) is satisfied for $k=-2,\dots,j-1$. Then set $R:=\sum_{i=0}^{j}(S^{j-i}B_{i})B_{j-i}$, which depends only on $B_{i}$ for $i=0,\dots,j$, and we obtain a recursive formula

[TABLE]

which yields a unique $B_{j+1}$ such that (2.9) is satisfied for $k=-2,\dots,j$. Note that the formula is valid for $n=0$ setting $(B_{j+1})_{-1}=c_{-1}=0$.

This proves that there is a unique

[TABLE]

satisfying (2.8). ∎

Notation 2.11**.**

We define $E_{j}$ to be the infinite standard basis (row) vector, e.g., $E_{0}=(1,0,\dots)$, $E_{1}=(0,1,0,\dots)$.

Lemma 2.12**.**

Let the first row of $M\in L(K[x]^{\mathds{N}_{0}})$ be given by $M_{0*}=E_{0}$. If $M$ satisfies

[TABLE]

then $M_{kj}=a_{j}$ for $j=0,\dots,k+1$.

Proof.

Note that $(M^{r})_{0*}=E_{0}$ for all $r$, and so $(M^{r}Y^{i})_{0j}=M_{0,j-i}=\delta_{ij}$. Hence

[TABLE]

as desired. ∎

Theorem 2.13**.**

Let $K$ be a field and let $b,c\in K$ with $c\neq 1$. Assume that $Q_{n}(b,c)\neq 0$ for all $n\in\mathds{N}$. Then $b$ and $c$ determine a unique twisting map via Proposition 2.5.

Proof.

We will use Corollary 1.8. For this we first prove that the matrix $M$ constructed in Proposition 2.10 satisfies equality (1.4) for all $k$. For $k=0$ this is clear, and from Lemma 2.12 we obtain $M_{1*}=E_{0}+bE_{1}+cE_{2}$, hence, by (2.8), the equality (1.2) holds for $k=1$. Assume by induction hypothesis that (1.2) holds for $k<k_{0}$. Then Lemma 2.2 and the fact that $M_{k_{0},k_{0}+1}\neq 1$ yield

[TABLE]

But then, by Lemma (2.12) we have $M_{k_{0},s}=\frac{\overline{M}_{k_{0},s}}{1-M_{k_{0},k_{0}+1}}$ for $s=0,\dots,k_{0}+1$, which yields (1.2) for $k=k_{0}$ and completes the inductive step. Finally, Corollary 1.8 yields the desired twisting map, which is unique by Proposition 2.5. ∎

Remark 2.14*.*

Combining Corollary 2.7 and Theorem 2.13 we obtain that $b$ and $c\neq 1$ determine a (necessarily unique) twisting map via Proposition 2.5 if and only if $Q_{n}(b,c)\neq 0$ for all $n\in\mathds{N}$.

This condition is the same as the condition used in [CG2]*Theorem 3.4. In order to verify this, we first note that the polynomials $f_{n}(a,b)$ used by [CG2] satisfy $f_{n}(a,b)=Q_{n+1}(b,a)$. In fact, since $Q_{1}(b,a)=f_{0}(a,b)=1$, $P_{1}(b,a)=e_{0}(a,b)=1$, and the recursive relations are the same, i.e.,

[TABLE]

we conclude $e_{n}(a,b)=P_{n+1}(b,a)$ and $f_{n}(a,b)=Q_{n+1}(b,a)$, as desired.

Note that by Remark 2.4, when $a\neq 0$, the twisting map corresponding to

[TABLE]

is equivalent to a twisting map with

[TABLE]

and so our results match the results of [CG2].

3   Roots of $Q_{n}$

In view of Theorem 2.13, we want to analyze the polynomials $Q_{n}$ and their roots. In particular we are interested in the following question: Given a pair $(b,c)\in K^{2}$, does there exists an $n\in\mathds{N}$ such that $Q_{n}(b,c)=0$? If the answer is no, then $(b,c)$ defines a unique twisting map via the previous theorem. Else, if $(b,c)\neq(-1,1)$, there is no twisting map for that $(b,c)$.

For a fixed pair $(b,c)$, from the recursive relations (2.4) in matrix form, we obtain

[TABLE]

If the eigenvalues of $D:=\begin{pmatrix}b&1\\ -c&1\end{pmatrix}$ are different, then there exists an invertible matrix $T$ such that

[TABLE]

and so

[TABLE]

for some $r_{1},r_{2}\in K$. If $r_{1},\lambda_{2}\neq 0$ then $Q_{n}=0$ if and only if

[TABLE]

This condition is easier to verify than the infinite number of evaluations $Q_{n}(b,c)$. For example, if $K\subset\mathds{C}$, and $|\frac{r_{1}}{r_{2}}|\neq 1$, one can check the equality $Q_{n}=0$ using real logarithms on the modulus in order to find the (unique) possible $n$, and then verifying the equality (3.1) for that $n$.

The eigenvalues of $D$ are

[TABLE]

and from $Q_{0}=1=r_{1}+r_{2}$ and $Q_{1}=1=r_{1}\lambda_{1}+r_{2}\lambda_{2}$ we obtain

[TABLE]

If

[TABLE]

then $Q_{n}=0$ if and only if

[TABLE]

Note that if $b,c\in\mathds{R}$, then $4c<(b-1)^{2}$ implies $|\frac{r_{1}}{r_{2}}|\neq 1$, and so in this case it can be determined if $Q_{n}=0$ for some $n$.

Now we give a detailed account of each exceptional case in (3.2):

If the eigenvalues coincide ($\lambda_{1}-\lambda_{2}=0$) then $c=\frac{(b-1)^{2}}{4}$. In that case $\lambda=\lambda_{1}=\lambda_{2}=\frac{b+1}{2}\neq 0$, since $\lambda=0$ leads to $c=1$ and we also have

[TABLE]

Hence $Q_{n+1}=0$ if and only if $\lambda=\frac{n+1}{n}$ if and only if $b=1+\frac{2}{n}$.

If one of the values $\lambda_{1},\lambda_{2}$ is zero, then $\det(D)=b+c=0$. In that case (2.3) yields $c_{n}=c\neq 1$ for all $n$.

Note that $(1-\lambda_{1})(1-\lambda_{2})=c$, and so, if $1-\lambda_{1}=0$ or $1-\lambda_{2}=0$, then $c=0$, and in that case $c_{n}=0\neq 1$ for all $n$.

This covers all cases of (3.2). However there are some other interesting cases.

For example if we require $b=0$, then we recover the polynomials $S_{n}$ in [CG] via the equality $S_{n}(c)=Q_{n-1}(0,c)$.

Another exceptional case happens when $\lambda_{1}=-\lambda_{2}$. In that case $b=-1$, and then

[TABLE]

4   The case $yx=x^{2}-xy+y^{2}$

In this section we assume that $\sigma$ is a twisting map and that $Y$ and $M$ are as in Corollary 1.8. As before we write $M_{1,*}=(1,b,c,0,\dots)$ and assume that $(b,c)=(-1,1)$, which is the only case not covered by Theorem 2.13. This means that we are dealing with the commutation rule

[TABLE]

By Corollary 1.8 we have

[TABLE]

which implies $\widetilde{M}^{2}=0$, where $\widetilde{M}=M-Y$. The matrix $\widetilde{M}:=M-Y=\psi(x-y)\in L(K^{\mathds{N}_{0}})$ plays a central role in the classification of all the twisting maps with $(b,c)=(-1,1)$. Note that $\widetilde{M}_{0j}=\delta_{0j}-\delta_{1j}$ and that $\widetilde{M}_{1*}=\widetilde{M}_{0*}$.

Remark 4.1*.*

Let $\widetilde{M}\in L(K^{\mathds{N}_{0}})$ be such that $\widetilde{M}_{0j}=\delta_{0j}-\delta_{1j}$ and $\widetilde{M}_{kj}=0$ for $j>k+1$. Then a straightforward computation shows that $M:=\widetilde{M}-Y$ determines a twisting map via Corollary 1.8, if and only if for all $k\in\mathds{N}$ we have

[TABLE]

Lemma 4.2**.**

Let $d\neq 0$. Then

1. (1)

$\widetilde{M}_{k*}=\widetilde{M}_{k-1,*}$* if and only if $\widetilde{M}Y^{k-1}\widetilde{M}=0$.* 2. (2)

$\widetilde{M}_{k*}=d\widetilde{M}_{k-1,*}$* if and only if $\widetilde{M}Y^{k-1}\widetilde{M}=\frac{1-d}{d}Y^{k}\widetilde{M}$.* 3. (3)

$\widetilde{M}_{k*}=d\widetilde{M}_{k-2,*}$* if and only if $\widetilde{M}Y^{k-1}\widetilde{M}+Y\widetilde{M}Y^{k-2}\widetilde{M}=\frac{1-d}{d}Y^{k}\widetilde{M}$.*

Proof.

We only prove (2) and (3), since (1) follows from (2) with $d=1$. Assume $\widetilde{M}_{k*}=d\widetilde{M}_{k-1,*}$. Then

[TABLE]

where the first and the third equality follow from (4.1). Now $Y^{k}\widetilde{M}=d(\widetilde{M}+Y)Y^{k-1}\widetilde{M}$ implies $(1-d)Y^{k}\widetilde{M}=d\widetilde{M}Y^{k-1}\widetilde{M}$, and then $\widetilde{M}Y^{k-1}\widetilde{M}=\frac{1-d}{d}Y^{k}\widetilde{M}$ follows. On the other hand, $\widetilde{M}Y^{k-1}\widetilde{M}=\frac{1-d}{d}Y^{k}\widetilde{M}$ implies $Y^{k}\widetilde{M}=d(\widetilde{M}+Y)Y^{k-1}\widetilde{M}$ and then the first row of the matrix equality

[TABLE]

is

[TABLE]

which yields $\widetilde{M}_{k*}=d\widetilde{M}_{k-1,*}$.

Similarly $\widetilde{M}_{k*}=d\widetilde{M}_{k-2,*}$ if and only if $Y^{k}\widetilde{M}=dM^{2}Y^{k-2}\widetilde{M}=d(\widetilde{M}Y+Y\widetilde{M}+Y^{2})Y^{k-2}\widetilde{M}$ if and only if $(1-d)Y^{k}\widetilde{M}=d\widetilde{M}Y^{k-1}\widetilde{M}+dY\widetilde{M}Y^{k-2}\widetilde{M}$, as desired. ∎

Lemma 4.3**.**

Let $n\in\mathds{N}$ with $n\geq 2$. Then $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for $0<k<n$ if and only if $\widetilde{M}Y^{k}\widetilde{M}=0$ for $0\leq k\leq n-2$. Moreover, in this case

[TABLE]

Proof.

The first assertion follows directly from Lemma 4.2(1). In that case, expand $M^{k+1}=(\widetilde{M}+Y)^{k+1}$. If $k<n$, no term in the expansion can have two times the factor $\widetilde{M}$, and in the expansion of $M^{n+1}=(\widetilde{M}+Y)^{n+1}$, the only term with two times the factor $\widetilde{M}$ is $\widetilde{M}Y^{n-1}\widetilde{M}$. ∎

Proposition 4.4**.**

There exists a twisting map such that $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for all $k\in\mathds{N}$.

Proof.

Consider the matrix $\widetilde{M}$ such that for all $j$, $\widetilde{M}_{j0}=1$, $\widetilde{M}_{j1}=-1$ and $\widetilde{M}_{ji}=0$ for $i>1$. We will show that $\widetilde{M}$ satisfies the conditions of Remark 4.1. Clearly $\widetilde{M}_{0j}=\delta_{0j}-\delta_{1j}$ and $\widetilde{M}_{kj}=0$ for $j>k+1$. So we have to prove that for all $k$

[TABLE]

But clearly $Y^{k}\widetilde{M}=\widetilde{M}$ for all $k$, since the columns are constant. So we have to prove that $M^{k}\widetilde{M}=\widetilde{M}$. For $k=1$ we have

[TABLE]

since $\widetilde{M}^{2}=0$. This implies $M^{k}\widetilde{M}=\widetilde{M}$, which proves (4.2), finishing the proof of the proposition. ∎

Proposition 4.5**.**

Let $n\in\mathds{N}$ with $n\geq 2$ and assume that $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for $1<k<n$ and that $\widetilde{M}_{n*}\neq\widetilde{M}_{0*}$. Then $\widetilde{M}_{nj}=0$ for $1<j<n$. Moreover, if one sets $m_{i}:=\widetilde{M}_{ni}$ then

[TABLE]

and

[TABLE]

Proof.

Assume first that $m_{n+1}=\widetilde{M}_{n,n+1}=0$. We first prove that $\widetilde{M}_{nj}=0$ for $n\geq j>1$. Clearly $0=(\widetilde{M}^{2})_{nn}=(\widetilde{M}_{nn})^{2}$, so $\widetilde{M}_{nn}=0$. Now assume that $2<k\leq n$ and that $\widetilde{M}_{nj}=0$ for $j\geq k$. Then $0\leq n-k+1\leq n-2$ and so by Lemma 4.3 we have $\widetilde{M}Y^{n-k+1}\widetilde{M}=0$. But then $0=(\widetilde{M}Y^{n-k+1}\widetilde{M})_{k-1,k-1}=(\widetilde{M}_{n,k-1})^{2}$, hence $\widetilde{M}_{n,k-1}=0$. Thus inductively we obtain $\widetilde{M}_{nj}=0$ for $n\geq j>1$.

From $(\widetilde{M}^{2})_{n0}=0$ it follows that $m_{0}+m_{1}=0$, so (4.3) reads $(1-m_{0})Y^{n}\widetilde{M}=m_{0}\widetilde{M}Y^{n-1}\widetilde{M}$, which is satisfied by Lemma 4.2(2), and (4.4) is also trivially satisfied in this case.

Now we can assume that $m_{n+1}\neq 0$. Consider the equality (4.1) for $Y^{n}\widetilde{M}$ and expand the power of $M$ in each summand according to Lemma 4.3. We obtain

[TABLE]

We will evaluate the matrix equality (4.5) at the entries $(i,i+n+1)$ for $i=1,\dots,n-1$.

First we claim that $(\widetilde{M}Y^{n-1}\widetilde{M})_{i,i+n+1}=0$ for $i=1,\dots,n-1$. In fact,

[TABLE]

But, since $\widetilde{M}_{jk}=0$ if $k>j+1$, we also have $\widetilde{M}_{n-1,i+n+1}=0=\widetilde{M}_{n,i+n+1}$, which proves the claim.

Clearly $(Y^{n+1})_{i,i+n+1}=1$, so it remains to compute $(Y^{j}\widetilde{M}Y^{n-j})_{i,i+n+1}=\widetilde{M}_{i+j,i+j+1}$. Now we assert that

[TABLE]

In fact, since $0<i\leq n-1$ and $0\leq j\leq n$, we have $i+j>0$, and so, for $1\leq i+j\leq n$, we know that (4.6) holds. So it suffices to prove that

[TABLE]

But

[TABLE]

and so, since $m_{n+1}\neq 0$, we obtain (4.7), which proves (4.6).

Finally note that by (4.7) we have

[TABLE]

Gathering the entries at $(i,i+n+1)$ for all the terms of (4.5) we obtain

[TABLE]

Subtracting these equalities for consecutive values of $i$ yields $m_{n+1}m_{j}=0$ for $j=2,\dots,n-1$, hence $\widetilde{M}_{nj}=m_{j}=0$ for $1<j<n$. From the case $i=1$ we obtain $\sum_{k=0}^{n+1}m_{k}+m_{n+1}(m_{0}+m_{1})=0$, which gives (4.4). Finally, using $m_{j}=0$ for $1<j<n$ the equality (4.3) follows directly from (4.5). ∎

Proposition 4.6**.**

Let $n\in\mathds{N}$ with $n\geq 2$ and assume that $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for $1<k<n$ and that $\widetilde{M}_{n*}\neq\widetilde{M}_{0*}$. Rename the only possibly non zero entries in the $n$th row as

[TABLE]

Then either

1. (1)

$(d,c,b,a)=(d,-d,-a,a)$* with $(d,a)\neq(1,0)$ or* 2. (2)

$(d,c,b,a)=(d,-1,0,a)$* with $a\neq 0$ and $d(a+1)=1$.*

Proof.

We will prove the following four assertions:

1. i)

If $a=0$, then $b=0$ and $c=-d$. 2. ii)

If $a\neq 0$ and $d=1$, then $b=-a$ and $c=-1$. 3. iii)

If $a\neq 0$, $d\neq 1$ and $b\neq 0$, then $b=-a$ and $c=-d$. 4. iv)

If $a\neq 0$, $d\neq 1$ and $b=0$, then $c=-1$ and $d(a+1)=1$.

Note that in item i) we have $(a,d)\neq(0,1)$, since $\widetilde{M}_{n*}\neq\widetilde{M}_{0*}$. On one hand i), ii) and iii) imply condition (1) and on the other hand iv) implies condition (2). Since items i)–iv) cover all possible cases, it suffices to prove these items in order to show that one of the conditions (1) or (2) necessarily holds.

i): If $\widetilde{M}_{n,n+1}=0$, then $0=(\widetilde{M}^{2})_{nn}=(\widetilde{M}_{nn})^{2}$, so $b=\widetilde{M}_{nn}=0$. Now we obtain

[TABLE]

ii): If $a\neq 0$ and $d=1$, then the matrix equality (4.3) at the entry $(1,2)$ yields

[TABLE]

since for any matrix $C$ we have $(CY^{3})_{12}=0$. So

[TABLE]

If $n=2$, this gives $0=-b+(1+c)(1+c)+b\widetilde{M}_{10}=(1+c)^{2}$ and if $n>2$ this yields directly $0=(1+c)^{2}$.

So $c=-1$ and then (4.3) reads

[TABLE]

and multiplying by $\widetilde{M}$ from the left we obtain $0=(a+b)\widetilde{M}Y^{n+1}$ since $\widetilde{M}^{2}=0$. Hence $b=-a$, which concludes the proof of ii).

For the rest of the proof we assume $a\neq 0$ and $d=m_{0}\neq 1$ and we claim that

[TABLE]

For this we evaluate (4.3) at the entry $(1,0)$, noting that $(Y^{n}\widetilde{M})_{10}=\widetilde{M}_{n+1,0}$, $(\widetilde{M}Y^{n-1}\widetilde{M})_{10}=1-m_{0}$ and that $(Y^{j}\widetilde{M}Y^{n-j})_{10}=0$ if $j<n$. So we obtain

[TABLE]

and since $m_{0}\neq 1$, we have $\widetilde{M}_{n+1,0}=m_{0}=d$.

Now we compute $\widetilde{M}_{n+1,1}$. For this we evaluate (4.3) at the entry $(1,1)$, noting that $(Y^{n}\widetilde{M})_{11}=\widetilde{M}_{n+1,1}$, $(\widetilde{M}Y^{n-1}\widetilde{M})_{11}=-1-m_{1}$, $(Y^{n-1}\widetilde{M}Y)_{11}=\widetilde{M}_{n0}=m_{0}$ and that $(Y^{j}\widetilde{M}Y^{n-j})_{11}=0$ if $j<n-1$. So

[TABLE]

and since $m_{0}\neq 1$, we have $\widetilde{M}_{n+1,1}=-m_{0}$, concluding the proof of (4.8).

Now, the equalities $0=(\widetilde{M}^{2})_{n0}=(\widetilde{M}^{2})_{n1}$ give

[TABLE]

Adding these yields $(c+d)b=0$, and so, if $b\neq 0$, then $c=-d$. Moreover, (4.4) reads

[TABLE]

and so $b=-a$, proving iii).

Finally, if $b=0$, then (4.9) yields

[TABLE]

and from (4.10) it follows that $ac+a=0$, hence $c=-1$. Then (4.11) implies $d(a+1)=1$, which concludes the proof of iv). ∎

5   The family $A(n,d,a)$

In this section we will describe the case (1) of Proposition 4.6. We will prove that the resulting twisting map depends only on $n$, $d$ and $a$. We obtain a family of twisted tensor products $A(n,d,a)$, parameterized by $n\in\mathds{N}$, $n\geq 2$, and $(a,d)\in K^{2}$, such that for an infinite family of polynomials $R_{k}$ (see Definition 5.5) we have $R_{k}(a,d)\neq 0$.

Remark 5.1*.*

Let $\tau$ be a twisting map, assume that $Y$ and $M$ are as in Corollary 1.8 and set $\widetilde{M}=M-Y$. Let $n\in\mathds{N}$ with $n\geq 2$, take $a,d\in K$ with $(d,a)\neq(1,0)$, assume that $\widetilde{M}_{j,*}=\widetilde{M}_{0,*}$ for $j<n$, and that we are in the case (1) of Proposition 4.6, i.e.,

[TABLE]

which we write as

[TABLE]

Using Lemma 4.3(1) we obtain $M^{n}\widetilde{M}=\widetilde{M}Y^{n-1}\widetilde{M}+Y^{n}\widetilde{M}$, and so

[TABLE]

where $e:=1-d$.

Proposition 5.2**.**

Let $A$ be an associative $K$-algebra, $a,d\in K$ with $(d,a)\neq(1,0)$, $\widetilde{M},Y\in A$ satisfying (5.1). Then for all $k\geq 1$ we have

[TABLE]

where

[TABLE]

Proof.

If $d=0$, then $e=1$ and (5.1) reads $Y^{n}\widetilde{M}=-a\widetilde{M}Y^{n}$. A direct computation shows that then $Y^{kn}\widetilde{M}=(-a)^{k}\widetilde{M}Y^{kn}$, which is (5.2) in this case.

Now assume $d\neq 0$. Then a straightforward computation shows that

[TABLE]

Now we proceed by induction on $k$. For $k=1$, equality (5.2) is just (5.1). Assume that (5.2) holds for some $k$. Multiplying (5.2) by $\widetilde{M}Y^{n-1}$ from the left yields

[TABLE]

Replacing $\widetilde{M}Y^{n-1}\widetilde{M}$ using (5.1) and changing sides we obtain

[TABLE]

and by the inductive hypothesis we get

[TABLE]

From this and (5.4) it follows that

[TABLE]

and clearing denominators completes the induction step and concludes the proof. ∎

Theorem 5.3**.**

Let $\tau$ be a twisting map, assume that $Y$ and $M$ are as in Corollary 1.8 and set $\widetilde{M}=M-Y$. Let $n\in\mathds{N}$ with $n\geq 2$, take $a,d\in K$ with $(d,a)\neq(1,0)$, assume that $\widetilde{M}_{j,*}=\widetilde{M}_{0,*}$ for $j<n$, and that $a$, $d$, $Y$ and $\widetilde{M}$ satisfy (5.1). Let $d_{k}$ be defined by (5.3). Then $d_{k}+e^{k}\neq 0$ for all $k\in\mathds{N}$,

[TABLE]

and

[TABLE]

where $c_{k,0}=d\prod_{i=2}^{k}\frac{d_{i}}{d_{i}+e^{i}}$ and $c_{k,r}=\frac{(-a)^{r}}{d_{r}+e^{r}}\prod_{i=r+1}^{k}\frac{d_{i}}{d_{i}+e^{i}}$ for $1\leq r\leq k$.

Proof.

We first prove (5.5). Let $i\in\{0,\dots,n-2\}$. If $a\neq 0$, then multiplying the equality (5.2) by $\widetilde{M}Y^{i}$ from the left yields $\widetilde{M}Y^{kn+i}\widetilde{M}=0$ and similarly, if $e\neq 0$, then multiplying (5.2) by $Y^{i}\widetilde{M}$ from the right yields $\widetilde{M}Y^{kn+i}\widetilde{M}=0$. Since $(e,a)\neq(0,0)$ we obtain $\widetilde{M}Y^{kn+i}\widetilde{M}=0$ for $i=0,\dots,n-2$ and Lemma 4.2(1) implies (5.5).

Now we assume $d,a\neq 0$ and prove $d_{k}+e^{k}\neq 0$ and (5.6) by induction on $k$. For $k=1$ we use the equality $Y^{n-1}\widetilde{M}=M^{n-1}\widetilde{M}$ and (5.1) to obtain

[TABLE]

Since $e+d=1$, then clearing denominators and solving for $Y^{n}\widetilde{M}$ gives (5.6) for $k=1$ (note that $c_{1,0}=d$ and $c_{1,1}=-a$, as the empty product takes the value 1).

Now assume that (5.6) holds for $k-1\geq 1$ and that $e^{i}+d_{i}\neq 0$ for $i<k$. By (5.2) we have

[TABLE]

and so

[TABLE]

By the inductive hypothesis and (5.5) we have

[TABLE]

and so

[TABLE]

Evaluating the matrix equality (5.7) at the entry $(0,kn)$ yields $e^{k}+d_{k}\neq 0$. In fact, since $(M^{n(k-j)})_{0*}=(1,0,0,\dots,0,\dots)$ for $j<n$, we have $(M^{n(k-j)}\widetilde{M}Y^{nj})_{0,kn}=0$ for $j<n$ (note that $\widetilde{M}Y^{nj}$ is of degree $-nj$), which implies

[TABLE]

The equality (5.7) also implies that (5.6) holds for $k$ with

[TABLE]

This completes the induction step and concludes the proof in the case $a,d\neq 0$.

If $d=0$, then $d_{k}=0$ and $e^{k}=1$ for all $k$, $c_{k,j}=0$ for $j<k$, and from $Y^{n}\widetilde{M}=-a\widetilde{M}Y^{n}$ it follows that

[TABLE]

as desired.

Finally, if $a=0$, then $\frac{e_{k}}{d_{k}}=\frac{1-d}{d}$, hence by (5.2), Lemma 4.2(2) and (5.5) we obtain

[TABLE]

and so (5.6) holds with $c_{k,0}=d^{k}$ and $c_{k,j}=0$ for $j>0$, which concludes the proof, since then $e^{k}+d_{k}=e^{k-1}\neq 0$ (note that $e=0$ leads to the contradiction $(d,a)=(1,0)$). ∎

Remark 5.4*.*

Note that Theorem 5.3 yields explicit formulas for the entries of $\widetilde{M}$:

[TABLE]

[TABLE]

and all other entries of $\widetilde{M}$ are zero.

Definition 5.5**.**

For $a,d\in K$ and $k\in\mathds{N}$ we define the polynomial

[TABLE]

Note that $R_{k}(a,d)=e^{k}+d_{k}$, where $d_{k}$ is defined in (5.3). We have $R_{1}(a,d)=1$ and $R_{2}(a,d)=1-d-ad$, so $R_{2}(a,d)\neq 0$ implies $(a+1)d\neq 1$, in particular $(a,d)=(0,1)$ is not allowed if we require $R_{2}(a,d)\neq 0$.

Corollary 5.6**.**

Let $a,d\in K$ be such that for all $k$ we have $R_{k}(a,d)\neq 0$. Then the formulas in Remark 5.4 define a matrix $\widetilde{M}$ that determines a twisting map via Remark 4.1.

Proof.

Note that

[TABLE]

Then $M_{0*}=(1,0,0,\dots)$, since $\widetilde{M}_{0*}=(1,-1,0,0,\dots)$, and so by Corollary 1.8 we have to prove that $M$ satisfies (1.4) for all $k$.

For $k=0$ this is clear. For $k=1,\dots,n-1$ the equality (1.4) reads $Y^{k}M=M^{k+1}-M^{k}Y+Y^{k+1}$, which is equivalent to $Y^{k}\widetilde{M}=M^{k}\widetilde{M}$. A straightforward computation as in Lemma 4.2 shows that these equalities are satisfied if and only if

[TABLE]

We claim that

[TABLE]

A similar computation as above shows that then it suffices to prove (1.4) for all $k=rn$, because (5.8) implies (1.4) for all other $k$. Now we prove (5.8):

Since the only non zero entries in $\widetilde{M}_{k*}$ are of the form $\widetilde{M}_{k,rn}$ or $\widetilde{M}_{k,rn+1}$ for some $r\geq 0$, and $\widetilde{M}_{k,rn+1}=-\widetilde{M}_{k,rn}$, it suffices to verify that

[TABLE]

But this is equivalent to $\widetilde{M}_{(r+k)n+j,l}=\widetilde{M}_{(r+k)n+j+1,l}$ for $j=0,\dots,n-2$, which holds by the definition of $\widetilde{M}$ and so  (5.8) is true.

It only remains to prove (1.4) for $k=rn$ with $r\geq 1$. A straightforward computation using (5.8) shows that it suffices to prove (5.2) for all $k$, and by Proposition 5.2 we only have to prove that $\widetilde{M}$ satisfies (5.1).

We will prove the equality (5.1) in each entry $(l,nk+j)$. Since the columns $\widetilde{M}_{*,nk+j}$ vanish for $j=2,\dots,n-1$ and $\widetilde{M}_{*,nk}=-\widetilde{M}_{*,nk+1}$, it suffices to prove (5.1) at the entries $(l,nk)$. So we have to prove

[TABLE]

But

[TABLE]

and $\widetilde{M}_{l*}=\widetilde{M}_{rn,*}$ for $l=rn+j$ with $j=0,\dots,n-1$, hence it suffices to prove

[TABLE]

Note that $\widetilde{M}_{ij}=0$ if $i<0$ or $j<0$.

By definition

[TABLE]

where $E_{j}$ is the infinite vector with $(E_{j})_{i}=\delta_{ij}$,

[TABLE]

and $c_{k,r}=0$ for all other $(k,r)$.

Since $(Y^{n-1}\widetilde{M})_{*,nk}=(\widetilde{M}_{n-1,nk},\widetilde{M}_{n,nk},\widetilde{M}_{n+1,nk},\dots)$ we have

[TABLE]

Moreover,

[TABLE]

so (5.9) reads

[TABLE]

In order finish the proof it suffices to prove (5.12) for all $r,k$. For this we will use

[TABLE]

which follows directly from the definitions of $d_{r}$.

For $k>r+1$ the equality (5.12) is trivially true, since in that case both sides vanish. If $k=r+1$, then (5.12) reads

[TABLE]

Since $c_{r,r}(e^{r}+d_{r})=(-a)^{r}$ (note that $d_{0}=0$), this is equivalent to

[TABLE]

For $a=0$ this is true, and if $a\neq 0$, this is equivalent to $-d(-a)^{r}=e(e^{r}+d_{r})-(e^{r+1}+d_{r+1})$, which follows directly from (5.13), hence the case $k=r+1$ is proved.

Now we can assume that $k\leq r$ and we will use that for $i\leq r$ we have

[TABLE]

We prove (5.12) by induction on $r$ (assuming $k\leq r$ and using that (5.12) is true for $k=r+1$). For $r=0=k$ this means

[TABLE]

Using $c_{0,0}=1$ and $c_{1,0}=d=1-e$ we see that this equality is equivalent to $d(1-d)=ed$ which is true by definition of $e$.

Assume (5.12) is true for some $r-1\geq 0$. Multiplying (5.13) by $e^{r+1}$ we obtain

[TABLE]

and adding $ed_{r}d_{r+1}$ this reads

[TABLE]

Using that $(-a)^{r}=c_{r,r}(e_{r}+d_{r})$ we obtain

[TABLE]

which we can write as

[TABLE]

Next we multiply by $c_{r,k}$ and, since by (5.15) we know that $c_{r+1,k}=c_{r,k}\frac{d_{r+1}}{d_{r+1}+e^{r+1}}$, it follows that

[TABLE]

We claim that

[TABLE]

In fact, if $k<r$, this follows from the inductive hypothesis, and if $k=r$, then (5.14) gives the same equality. Now the equalities (5.17) and (5.15) imply

[TABLE]

So the equality (5.16) yields

[TABLE]

which is (5.12) for $r$. This completes the inductive step, proves (5.12) and concludes the proof. ∎

6   Roots of $R_{k}$

In view of Corollary 5.6, we want to analyze the polynomials $R_{k}$ and their roots. In particular we are interested in the following question: Given a pair $(a,d)$, does there exists $k\in\mathds{N}$ such that $R_{k}(a,d)=0$? If the answer is no, then for each $n\geq 2$ the pair $(a,d)$ defines a unique twisting map via Theorem 5.3. Else there is no twisting map satisfying item (1) of Proposition 4.6 for that $(a,d)$.

Fix $(a,d)$. If $-a=1-d=e$, then $d_{k}=kde^{k-1}$. So $R_{k}(a,d)=e^{k}+kde^{k-1}=e^{k-1}(kd+e)=0$ if and only if $0=kd+e=d(k-1)+1$. In that case $a=d-1=-\frac{k}{k-1}$.

Now assume $-a\neq 1-d=e$. Then

[TABLE]

so $R_{k}(a,d)=0$ if and only if

[TABLE]

In that case $e\neq 0$ and $e\neq 1$, since $e=0$ leads to $a=0$ and $e=1$ leads to $a=-1$, which contradicts $-a\neq e$. So $R_{k}(a,d)=0$ if and only if

[TABLE]

This condition is much easier to handle than the original condition. Assume $K\subset\mathds{C}$. If (6.1) is satisfied and $\left|\frac{a}{e}\right|\neq 1$, then

[TABLE]

Moreover, if (6.1) is satisfied and $\left|\frac{a}{e}\right|=1$, then necessarily $\left|\frac{1+a}{1-e}\right|=1$, and an elementary computation shows that then either $-a=e$ or $-a=\bar{e}$, where $\bar{e}$ is the complex conjugate of $e$. The first case is impossible by assumption, and another elementary computation shows that (6.1) is satisfied if and only if $r=\frac{u^{2k}-1}{u^{2k-1}-u}$, where $u=\frac{e}{|e|}$ is a unitary complex number and $r=|e|$. Hence, if $K\subset\mathds{C}$, we can describe a complete strategy in order to determine if for a given pair $(a,d)$ we have $R_{k}(a,d)\neq 0$ for all $k$.

1. (1)

If $-a=1-d=:e$ then $R_{k}(a,d)\neq 0$ for all $k$ if and only if $(a,d)\neq\left(-\frac{k+1}{k},-\frac{1}{k}\right)$ for all $k\in\mathds{N}$. 2. (2)

If $-a\neq 1-d=e$, then

1. a)

if $\left|\frac{a}{e}\right|\neq 1$ then

1. i)

if $\frac{\log\left|\frac{1+a}{1-e}\right|}{\log\left|\frac{a}{e}\right|}\notin\mathds{N}_{\geq 2}$, then $R_{k}(a,d)\neq 0$ for all $k$. 2. ii)

if $\frac{\log\left|\frac{1+a}{1-e}\right|}{\log\left|\frac{a}{e}\right|}=k_{0}\in\mathds{N}_{\geq 2}$, then $R_{k}(a,d)\neq 0$ for all $k$ if and only if $R_{k_{0}}(a,d)\neq 0$. 2. b)

if $\left|\frac{a}{e}\right|=1$ and $a\neq\bar{e}$, then $R_{k}(a,d)\neq 0$ for all $k$. 3. c)

if $\left|\frac{a}{e}\right|=1$ and $a=\bar{e}$, then $R_{k}(a,d)\neq 0$ for all $k$ if and only if $r\neq\frac{u^{2k}-1}{u^{2k-1}-u}$ for all $k\in\mathds{N}_{\geq 2}$, where $u=\frac{e}{|e|}$ and $r=|e|$.

Example 6.1**.**

This algorithm determines that $d=0$, $a=-1$ is a valid choice, since then $e=1$ and we are in the case (1), with $(a,d)=(-1,0)\neq\left(-\frac{k+1}{k},-\frac{1}{k}\right)$ for all $k\in\mathds{N}$. One verifies that this example corresponds to [CG]*Example 5.4.

7   The case $y^{n}x=dx^{n+1}-x^{n}y+(a+1)y^{n+1}$

Sections 7 and 8 are dedicated to the analysis of the case (2) of Proposition 4.6. So in this three sections $\sigma$ is a twisting map and $Y$ and $M$ are as in Corollary 1.8 and $\widetilde{M}=M-Y$. Moreover there exists $n\in\mathds{N}$ with $n\geq 2$, such that $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for $1<k<n$ and $\widetilde{M}_{n*}=dE_{0}-E_{1}+aE_{n+1}$ for some $a,d\in K^{\times}$ with $d(a+1)=1$. This implies that we are dealing with the case

[TABLE]

The corresponding matrix equality is

[TABLE]

We will use this equality and similar matrix equalities in order to compute the different possibilities for the resulting twisting maps. There is only one choice for the first $2n-1$ rows, but four choices for the $2n$’th row. In each of the four cases the rows are determined until the row $3n-1$. One can determine the rows $3n$, $3n+1$ and $3n+2$, in each of the four cases, and obtain again four cases in each of them (so we have 16 cases). As the number of possibilities grows, the systems of equations get more and more involved, so a full classification seems very difficult to achieve.

However, in section 8 we manage to describe a family of twisting maps, such that four of the above mentioned 16 cases coincide in the first $3n+2$ rows with members of this family.

In the present section we will establish some technical formulas. On one hand with these technical results one can carry out the computations mentioned above in order to determine the possibilities for the first rows of $\widetilde{M}$. On the other hand, in section 8, they will allow us to describe a certain family of twisting maps called $B(a,L)$. The following lemma is a result on lower infinite Hessenberg matrices, that should be well-known, but we couldn’t find any reference to it in the literature.

Lemma 7.1**.**

Let $A\in L(K^{\mathds{N}_{0}})$ be an infinite matrix such that $A_{ij}=0$ for $j>i+1$. Then

[TABLE]

Proof.

Since $w(A)=-1$, by Proposition 2.8 we have $w(A^{n+1})=-n-1$. A direct computation using Proposition 2.8, shows that then $(A^{n+1})^{(-n-1)}=(A^{(-1)})^{n+1}$, and so

[TABLE]

But

[TABLE]

hence the only term that survives in the sum is the term with $i_{k+1}=i_{k}+1$ for all $k$ and so

[TABLE]

as desired. ∎

Proposition 7.2**.**

Let $\widetilde{M}$ and $a$ be as above. We have

[TABLE]

and

[TABLE]

Moreover, set $m_{i}:=\widetilde{M}_{i,i+1}$ and let $L:=(L_{1}=n,L_{2},L_{3},\dots)$ be the increasing sequence of integers such that $m_{L_{i}}\neq 0$ and $m_{i}=0$ if $L_{k}<i<L_{k+1}$ for some $k$. Then

1. (1)

$\prod_{k=1}^{n+1}(m_{j+k}-a)=0$* for all $j\geq 0$.* 2. (2)

If $m_{j}\neq 0$, then $m_{j+k}=0$ for $k=1,\dots,n-1$. 3. (3)

We have $m_{L_{i}}=a$ and $(L_{i+1}-L_{i})\in\{n,n+1\}$ for all $i\geq 1$.

Proof.

From (7.1) and Lemma 4.3 we obtain

[TABLE]

So

[TABLE]

where the second equality follows from $(1-d)=ad$. Using again $(1-d)=ad$ and dividing by $d$ we obtain (7.2). Multiplying by $a^{n-1}$ we obtain

[TABLE]

But the right hand side is the expansion of $(aY-\widetilde{M})^{n+1}$, since in that expansion the only term with two times the factor $\widetilde{M}$ is $\widetilde{M}(aY)^{n-1}\widetilde{M}$. This proves (7.3). Item (1) follows directly from (7.3) and Lemma 7.1, since $(\widetilde{M}-aY)_{i,i+1}=m_{i}-a$.

By Lemma 4.3 we have $\widetilde{M}Y^{k}\widetilde{M}=0$ for $k=0,\dots,n-2$. But then

[TABLE]

for $k=1,\dots,n-1$, so item (2) is true.

Item (2) implies that $L_{i+1}-L_{i}\geq n$. Assume by contradiction that $L_{i+1}-L_{i}>n+1$. Then

[TABLE]

contradicting item (1), hence $(L_{i+1}-L_{i})\in\{n,n+1\}$ for all $i\geq 1$.

Finally, if $m_{L_{i}}\neq a$ for some $i>1$, then

[TABLE]

contradicting again item (1), hence $m_{L_{i}}=a$ for all $i$. ∎

8   The family $B(a,L)$ and quasi-balanced sequences

In this section we will describe a family of twisted tensor products that arise in the case (2) of Proposition 4.6. So $\sigma$ is a twisting map and $Y$ and $M$ are as in Corollary 1.8 and $\widetilde{M}=M-Y$. Moreover there exists $n\in\mathds{N}$ with $n\geq 2$, such that $\widetilde{M}_{k*}=\widetilde{M}_{0*}$ for $1<k<n$ and $\widetilde{M}_{n*}=dE_{0}-E_{1}+aE_{n+1}$ for some $a,d\in K^{\times}$ with $d(a+1)=1$.

By Proposition 7.2(3), the twisting map defines a sequence $L=(L_{1},L_{2},\dots)$ such that $\widetilde{M}_{L_{i},L_{i}+1}=a$ and such that $\widetilde{M}_{k,k+1}=0$ if $L_{i}<k<L_{i+1}$ for some $i$. Moreover $L_{j+1}-L_{j}\in\{n,n+1\}$ for all $j>1$, so $L$ belongs to the set of sequences

[TABLE]

We will construct an infinite matrix $M(L,a)$, where $L\in\Delta(n,n+1)$, $n\geq 2$ and $a\neq 0,1$; by setting $d:=\frac{1}{1+a}$, and defining each row of the infinite matrix in the following way

[TABLE]

Note that $M(L,a)_{ij}=0$ if $j\notin\{0,1\}\cup\{L_{k}\}_{k\in\mathds{N}}$. We will prove in Propositions 8.5 and 9.9 that the infinite matrix $\widetilde{M}:=M(L,a)$ defines a twisting map via Remark 4.1, if and only if $L$ is a quasi-balanced sequence, where the set of quasi-balanced sequences $\mathcal{L}$ is defined as follows:

[TABLE]

We first describe some properties of quasi-balanced sequences.

Definition 8.1**.**

We say that a sequence $L\in\Delta(n,n+1)$ is $r$-balanced, if $L_{r}=L_{j}+L_{r-j}$ for all $0<j<r$. We say that the sequence is $r$-quasi-balanced, if $L_{r}-1\leq L_{j}+L_{r-j}\leq L_{r}$ for all $0<j<r$. We also say that a finite sequence $(L_{1},\dots,L_{r_{0}})$ is quasi-balanced, if $L_{r}-1\leq L_{j}+L_{r-j}\leq L_{r}$ for all $0<j<r\leq r_{0}$.

Note that every sequence $L\in\Delta(n,n+1)$ is trivially $1$-balanced, and it is also easy to see that it is $2$-quasi-balanced and $3$-quasi-balanced. However the sequences beginning with $(n,2n,3n+1,4n+2,\dots)$ or $(n,2n+1,3n+1,4n+1,\dots)$ are not $4$-quasi-balanced.

Clearly a sequence $L\in\Delta(n,n+1)$ is quasi-balanced (i.e. belongs to $\mathcal{L}$), if and only if it is $r$-quasi-balanced for all $r$, if and only if all the sequences $L_{\leq r}:=(L_{1},\dots,L_{r})$ are quasi-balanced. For a given sequence $L$ and $0<j<r$ we define $\Delta_{r,j}:=L_{r}-L_{j}-L_{r-j}$. The fact that $(L_{1},\dots,L_{r_{0}})$ is quasi-balanced is equivalent to the fact that $\Delta_{r,j}\in\{0,1\}$ for all $0<j<r\leq r_{0}$.

Proposition 8.2**.**

Assume that the finite sequence $(L_{1},\dots,L_{r_{0}})$ is quasi-balanced. Then either $(L_{1},\dots,L_{r_{0}},L_{r_{0}}+n)$ is quasi-balanced or $(L_{1},\dots,L_{r_{0}},L_{r_{0}}+n+1)$ is quasi-balanced.

Proof.

We want to prove that either

[TABLE]

or that

[TABLE]

Note that

[TABLE]

hence $\Delta^{+}_{r_{0}+1,j}\in\{0,1,2\}$.

Now assume by contradiction that there exist $j_{0},j_{1}$ such that $\Delta^{-}_{r_{0}+1,j_{0}}=-1$ and such that $\Delta^{+}_{r_{0}+1,j_{1}}=2$, i.e., $\Delta^{-}_{r_{0}+1,j_{1}}=1$. We can assume that $j_{1}>j_{0}$ since $\Delta^{-}_{r_{0}+1,j}=\Delta^{-}_{r_{0}+1,r_{0}+1-j}$. Now we set $s:=j_{1}-j_{0}=(r_{0}+1-j_{0})-(r_{0}+1-j_{1})>0$ and obtain

[TABLE]

But $\Delta(r+1-j_{0},s),\Delta(j_{1},s)\in\{0,1\}$, so we obtain a contradiction, which proves that one of (8.1) or (8.2) is necessarily true. This concludes the proof. ∎

Corollary 8.3**.**

If $\tilde{L}=(L_{1},\dots,L_{r})$ is quasi-balanced, then there exists $L\in\mathcal{L}$ such that $L_{\leq r}=\tilde{L}$.

Proof.

Construct inductively $L_{r+1}$, $L_{r+2}$,…using Proposition 8.2. ∎

Lemma 8.4**.**

If $L\in\Delta(n,n+1)$, but $L\notin\mathcal{L}$, then there exists $k,r\in\mathds{N}$ such that either

1. (1)

$L_{k+r}=L_{k}+L_{r}-1$* and $L_{r-1}+2<L_{r}$ or* 2. (2)

$L_{k+r}=L_{k}+L_{r}+2$* and $L_{r}+2<L_{r+1}$.*

Proof.

Let $m:=\min\{i:\ L_{\leq i}\text{ is not quasi-balanced}\}$. For $0<j<m$ we have

[TABLE]

Since $\Delta_{m-1,j}\in\{0,1\}$, $(L_{m}-L_{m-1})\in\{n,n+1\}$ and $(L_{m-j}-L_{m-j-1})\in\{n,n+1\}$ necessarily $\Delta_{m,j}\in\{-1,0,1,2\}$. But $L_{\leq m}$ is not quasi-balanced, so there exists $j$ such that either $\Delta_{m,j}=-1$ or $\Delta_{m,j}=2$. In the case $\Delta_{m,j}=-1$ set $r:=\min\{j:\ \Delta_{m,j}=-1\}$ and $k:=m-r$. Then $\Delta_{m,r-1}=0$, since $|\Delta_{m,r}-\Delta_{m,r-1}|\leq 1$. But

[TABLE]

hence $0=-1+(L_{r}-L_{r-1})-(L_{m-r+1}-L_{m-r})$, and since $(L_{r}-L_{r-1}),(L_{m-r+1}-L_{m-r})\in\{n,n+1\}$, we have $L_{r}-L_{r-1}=n+1$ and $L_{m-r+1}-L_{m-r}=n$. So $L_{r}-L_{r-1}=n+1>2$ and together with $\Delta_{k+r,r}=L_{k+r}-L_{k}-L_{r}=-1$ this proves that we are in the case of (1).

On the other hand, in the case $\Delta_{m,j}=2$ set $r:=\max\{j:\ \Delta_{m,j}=2\}$ and $k:=m-r$. Then $\Delta_{m,r+1}=1$, since $|\Delta_{m,r}-\Delta_{m,r+1}|\leq 1$. But

[TABLE]

hence $1=2+(L_{m-r}-L_{m-r-1})-(L_{r+1}-L_{r})$, and since $(L_{m-r}-L_{m-r-1}),(L_{r+1}-L_{r})\in\{n,n+1\}$, we have $(L_{r+1}-L_{r})=n+1$ and $L_{m-r}-L_{m-r-1}=n$. So $L_{r+1}-L_{r}=n+1>2$ and together with $\Delta_{k+r,r}=L_{k+r}-L_{k}-L_{r}=2$ this yields (2) and concludes the proof. ∎

Proposition 8.5**.**

Let $L\in\Delta(n,n+1)$. If the infinite matrix $\widetilde{M}:=M(L,a)$ defines a twisting map via Remark 4.1, then $L\in\mathcal{L}$.

Proof.

Assume that $\widetilde{M}=M(L,a)$ defines a twisting map and assume by contradiction that $L\in\Delta(n,n+1)\setminus\mathcal{L}$. In the first case of Lemma 8.4 note that $L_{r-1}<L_{r}-2<L_{r}-1<L_{r}$, hence by definition $\widetilde{M}_{L_{r}-2,*}=\widetilde{M}_{L_{r}-1,*}=d^{r-1}(E_{0}-E_{1})$, and so, by Lemma 4.2(1) we have $\widetilde{M}Y^{L_{r}-2}\widetilde{M}=0$. But then, using that $L_{k+r}=L_{k}+L_{r}-1$ and that $a,d,1-d\neq 0$ we obtain

[TABLE]

a contradiction which discards this case. On the other hand, in the second case of Lemma 8.4 note that $L_{r}<L_{r}+1<L_{r}+2<L_{r+1}$, hence by definition $\widetilde{M}_{L_{r}+1,*}=\widetilde{M}_{L_{r}+2,*}=d^{r}(E_{0}-E_{1})$, and so, by Lemma 4.2(1) we have $\widetilde{M}Y^{L_{r}+1}\widetilde{M}=0$. But then

[TABLE]

But $\widetilde{M}_{L_{r}+1,L_{r+k}+1}=\widetilde{M}_{L_{r}+2,L_{r+k}+1}=0$ and $L_{k+r}=L_{k}+L_{r}+2$, and so we arrive at

[TABLE]

a contradiction that discards the second case of Lemma 8.4 and concludes the proof. ∎

9   Existence of twisting maps for the family $B(a,L)$

In Theorem 5.3 we determined the form of the matrices $\widetilde{M}$ corresponding to twisting maps of the family $A(n,d,a)$ and in Corollary 5.6 we proved conversely that each such matrix defines actually a twisting map. Similarly, in Proposition 8.5 we determined the form of the matrices $\widetilde{M}$ corresponding to twisting maps of the family $B(a,L)$, namely, that necessarily $L\in\mathcal{L}$. This last section is devoted to the proof that this condition is sufficient. So, along this last section, we assume that $L\in\Delta(n,n+1)$ and we set

[TABLE]

We will prove that if $L\in\mathcal{L}$, then the matrix $\widetilde{M}$ defines a twisting map via Remark 4.1.

Proposition 9.1**.**

Assume that the following conditions are satisfied

1. (1)

$\widetilde{M}Y^{k}\widetilde{M}=0$* for $k\in\mathds{N}_{0}$ with $k,k+1\neq L_{t}$ for all $t$,* 2. (2)

$\widetilde{M}Y^{L_{r}}\widetilde{M}+Y\widetilde{M}Y^{L_{r}-1}\widetilde{M}=aY^{L_{r}+1}\widetilde{M}$* for all $r\in\mathds{N}$,* 3. (3)

$\widetilde{M}Y^{L_{r}-1}\widetilde{M}=a\sum_{i=0}^{L_{r}}Y^{i}\widetilde{M}Y^{L_{r}-i}-ra^{2}Y^{L_{r}+1}$* for all $r\in\mathds{N}$.*

Set $L_{0}:=0$, then for all $r\in\mathds{N}_{0}$ and $L_{r}\leq j<L_{r+1}$, we have

[TABLE]

Proof.

We will prove (9.1) by induction on $j$. Note that it holds trivially for $j=L_{0}=0$. Now assume that (9.1) is true for $j$ and we will prove that it holds for $j+1$. We have to consider two cases for this inductive step: either $L_{r}\leq j<L_{r+1}-1$ for some $r\geq 0$, or $j=L_{r}-1$ for some $r\geq 1$.

First assume that $L_{r}\leq j<L_{r+1}-1$. Then we multiply (9.1) by $M=Y+\widetilde{M}$ and obtain

[TABLE]

We want to compute the last sum and note that for $i=0,\dots,j$ there are three possibilities:

1. $i=0$ or $L_{t}<i<L_{t+1}-1$ for some $0\leq t<j$, and then $Y^{j-i}\widetilde{M}Y^{i}\widetilde{M}=0$ by condition (1),

2. $i=L_{t}-1$ for some $1\leq t\leq r$, or

3. $i=L_{t}$ for some $1\leq t\leq r$.

Hence we have

[TABLE]

But by condition (2) we have

[TABLE]

and so, multiplying by $Y^{j-L_{t}}$, we obtain

[TABLE]

and then

[TABLE]

which implies (9.1) for $j+1$:

[TABLE]

Now, if $j=L_{r}-1$ for some $r\geq 1$, then $L_{r-1}<j<L_{r}$ and by the same argument as before, from (9.1) we obtain

[TABLE]

and we also obtain

[TABLE]

Consequently

[TABLE]

which gives

[TABLE]

Inserting into this equality the value of $\widetilde{M}Y^{L_{r}-1}\widetilde{M}$ according to condition (3) we obtain

[TABLE]

and using that $d(a+1)=1$ we obtain (9.1) for $j+1=L_{r}$. This completes the inductive step and concludes the proof. ∎

Proposition 9.2**.**

Assume the hypotheses of Proposition 9.1. Then (4.1) is valid for all $k\in\mathds{N}_{0}$.

Proof.

A straightforward computation shows that (4.1) holds for $k=0,1$. Moreover, for $k>0$ we have by definition

[TABLE]

Hence, if $L_{r}<k<L_{r+1}$ for some $r\geq 0$, then (4.1) reads

[TABLE]

and so we have to prove

[TABLE]

On the other hand, if $k=L_{r}$ for some $r>0$, then (4.1) reads

[TABLE]

and so we have to prove

[TABLE]

We first prove inductively (9.2). We know that it holds for $k=1$, and we will prove that if it holds for $k$ with $L_{r}<k<L_{r+1}-1$ for some $r\geq 0$, then it holds for $k+1$, and that if it holds for $k=L_{r}-1$ for some $r\geq 1$, then it holds for $k+2=L_{r}+1$.

Let $k$ satisfy $L_{r}<k<L_{r+1}-1$ for some $r\geq 0$. From condition (1) of Proposition 9.1, we obtain

[TABLE]

Hence, if $Y^{k}\widetilde{M}=d^{r}M^{k}\widetilde{M}$ for such $k$, then $Y^{k+1}\widetilde{M}=d^{r}M^{k+1}\widetilde{M}$, which is (9.2) for $k+1$.

Assume now that $k=L_{r}-1$ for some $r\geq 1$, and that (9.2) holds, i.e.,

[TABLE]

Multiplying this by $M^{2}=\widetilde{M}Y+Y\widetilde{M}+Y^{2}$ from the left we obtain

[TABLE]

and combined with $\widetilde{M}Y^{L_{r}}\widetilde{M}+Y\widetilde{M}Y^{L_{r}-1}\widetilde{M}=aY^{L_{r}+1}\widetilde{M}$, which holds by condition (2) of Proposition 9.1, this yields

[TABLE]

Using $(a+1)d=1$ this gives $Y^{L_{r}+1}\widetilde{M}=d^{r}M^{L_{r}+1}\widetilde{M}$, which is (9.2) for $k+2=L_{r}+1$.

Finally we prove (9.3), which is (4.1) for $k=L_{r}$. For this consider the equality

[TABLE]

which is the equality (9.1) for $j=L_{r}$, and the equality

[TABLE]

which is the equality (9.1) for $j=L_{r}-1$, multiplied by $Y$ from the right. Subtracting the second equality from the first, we obtain

[TABLE]

which is (9.3). Hence (4.1) holds for $k=L_{r}$, concluding the proof. ∎

In order to prove that for $L\in\mathcal{L}$ the matrix $\widetilde{M}:=M(L,a)$ defines a twisting map, we decompose the matrix $\widetilde{M}$ into three summands. Set

[TABLE]

and define the infinite matrix $M_{1}$ by $(M_{1})_{ij}\coloneqq\delta_{0j}m_{i}$. Now define the set

[TABLE]

and set

[TABLE]

We define the infinite matrix $B$ by $B_{ij}\coloneqq\delta_{ij}n_{i}$. Then $\widetilde{M}=BY+YM_{1}-M_{1}Y$ and

[TABLE]

Lemma 9.3**.**

The following equalities hold for all $k\in\mathds{N}_{0}$:

1. (1)

$M_{1}Y^{k}M_{1}=m_{k}M_{1}$, 2. (2)

$M_{1}Y^{k}B=n_{k}M_{1}Y^{k}$, 3. (3)

$(BY^{k}M_{1})_{ij}=\delta_{0j}n_{i}m_{k+i}$, 4. (4)

$(BY^{k}B)_{ij}=\delta_{i+k,j}(n_{i}n_{i+k})$.

Proof.

A straightforward computation. ∎

Remark 9.4*.*

Note that if $L\in\mathcal{L}$ and $k,k+1\notin|L|$, then $BY^{k+1}B=0$. In fact, by Lemma 9.3(4), it suffices to prove that if $n_{i}n_{i+k+1}\neq 0$ for some $i$, then either $k\in|L|$ or $k+1\in|L|$. But $n_{i}n_{i+k+1}\neq 0$ implies $i=L_{t}$ and $i+k+1=L_{r+t}$ for some $r,t\in\mathds{N}$. Since $L$ is quasi-balanced, then either $L_{r+t}=L_{t}+L_{r}$, which implies $k+1=L_{r}$, or $L_{r+t}=L_{t}+L_{r}+1$, which implies $k=L_{r}$.

Lemma 9.5**.**

Let $L\in\mathcal{L}$. If $i\in|L|$ and $k\notin|L|$, then $m_{i+k+1}=dm_{i}m_{k}$.

Proof.

We know that $i=L_{t}$ for some $t\in\mathds{N}$.

If $k<n=L_{1}$, then $L_{t}<i+k+1\leq L_{t}+n\leq L_{t+1}$, and so $m_{i+k+1}=d^{t}$. Since $m_{i}=d^{t-1}$ and $m_{k}=1$, this proves the result in this case. Else $L_{r}<k<L_{r+1}$ for some $r\in\mathds{N}$, and since $L$ is quasi-balanced, we have

[TABLE]

Hence

[TABLE]

and so $m_{i+k+1}=d^{r+t}$. Since $m_{i}=d^{t-1}$ and $m_{k}=d^{r}$, this concludes the proof. ∎

Lemma 9.6**.**

Assume $L\in\mathcal{L}$, and let $i,k\in\mathds{N}_{0}$ and $r\in\mathds{N}$. Then

[TABLE]

Proof.

We first prove (9.5) in each of the four possible cases. Note that for $k\in|L|$ we have $m_{k+1}=dm_{k}$, and for $k\notin|L|$ we have $m_{k+1}=m_{k}$.

Case $i,k\notin|L|$:  Here $n_{i}=n_{k}=0$, $m_{i+1}=m_{i}$, $m_{k+1}=m_{k}$, hence $m_{k}m_{i+1}-m_{k+1}m_{i}=0$ and both sides of (9.5) vanish.

Case $i\in|L|,k\notin|L|$: Here $n_{i}=a$, $n_{k}=0$, $m_{i+1}=dm_{i}$, $m_{k+1}=m_{k}$ and by Lemma 9.5 we have $m_{i+k+1}=dm_{i}m_{k}$. Hence

[TABLE]

and both sides of (9.5) vanish.

Case $i\notin|L|,k\in|L|$: Here $n_{i}=0$, $n_{k}=a$, $m_{i+1}=m_{i}$, $m_{k+1}=dm_{k}$ and by Lemma 9.5 we have $dm_{i}m_{k}=m_{i+k+1}$. Hence

[TABLE]

as desired.

Case $i,k\in|L|$: Here $n_{i}=a$, $n_{k}=a$, $m_{i+1}=dm_{i}$ and $m_{k+1}=dm_{k}$. Hence

[TABLE]

as desired, concluding the proof of (9.5).

Now we prove (9.6). If $L_{r}+i+1\notin|L|$, then both sides vanish. If $L_{r}+i+1\in|L|$, then we have to prove that $n_{i}+n_{i+1}=a$. There exists $t>0$ such that $L_{r}+i+1=L_{r+t}$. Since $L\in\mathcal{L}$, we have either $L_{r+t}=L_{r}+L_{t}$ or $L_{r+t}=L_{r}+L_{t}+1$.

In the first case $L_{t}=i+1$, $n_{i}=0$ and $n_{i+1}=a$; and in the second case $L_{t}=i$, $n_{i}=a$ and $n_{i+1}=0$. Hence in both cases $n_{i}+n_{i+1}=a$, which proves (9.6).

In order to prove (9.7) we consider three cases.

If $i<L_{1}=n$, then $L_{r}\leq L_{r}+i<L_{r+1}$, and so

[TABLE]

Since $n_{i}=0$, we obtain (9.7).

• -

If $L_{t}<i<L_{t+1}$, then $L_{r+t}\leq L_{r}+i<L_{r+t+1}$, since $L\in\mathcal{L}$ implies

[TABLE]

Hence

[TABLE]

and using $n_{i}=0$, we obtain (9.7) in this case.

• -

If $i=L_{t}$ for some $t$, then $an_{i+L_{r}}=n_{i}n_{L_{r}+i}$, and so it suffices to prove

[TABLE]

But $L\in\mathcal{L}$ implies

[TABLE]

and so

[TABLE]

as desired.

Thus (9.7) holds in all cases, concluding the proof. ∎

In order to verify the conditions of Proposition 9.1 we need to compute

[TABLE]

and so we have to compute $\widetilde{M}Y^{k}M_{1}$.

Lemma 9.7**.**

If $L\in\mathcal{L}$, then we have

[TABLE]

Proof.

We have

[TABLE]

and by Lemma 9.3 and equality (9.5), we also have

[TABLE]

as desired. ∎

Proposition 9.8**.**

If $L\in\mathcal{L}$, then

[TABLE]

Proof.

By Lemma 9.7 and equality (9.8) we have

[TABLE]

From Lemma 9.3(2) we obtain

[TABLE]

which concludes the proof. ∎

Proposition 9.9**.**

For each $a\in K\setminus\{0,-1\}$ and $L\in\mathcal{L}$, the matrix $\widetilde{M}=M(L,a)$ defines a twisting map via Remark 4.1.

Proof.

Since $\widetilde{M}_{0j}=\delta_{0j}-\delta_{1j}$ and $\widetilde{M}_{kj}=0$ for $j>k+1$, by Remark 4.1 and Propositions 9.1 and 9.2, it suffices to check the conditions (1)–(3) in Proposition 9.1.

If $k,k+1\notin|L|$, then by Proposition 9.8 we know that $\widetilde{M}Y^{k}\widetilde{M}=BY^{k+1}BY$. By Remark 9.4 we also know that $BY^{k+1}B=0$ for $k,k+1\notin|L|$, which proves item (1).

In order to prove item (2), we use Proposition 9.8 and compute

[TABLE]

So we have to prove

[TABLE]

We have

[TABLE]

and so (9.6) concludes the proof of item (2).

In order to prove item (3), we compute

[TABLE]

Since by Proposition 9.8 we know that

[TABLE]

we have to prove

[TABLE]

But

[TABLE]

hence it suffices to prove

[TABLE]

which holds by (9.7). This concludes the proof. ∎

Remark 9.10*.*

Our strategy contains two main components. On one hand the approach of equalities of infinite matrices yields conditions that reduce the possibilities to very few families. Even in the complicated case  (2) of Proposition 4.6 one can achieve the classification of all possible twisting maps up to any degree, with increasing amount of computational work. On the other hand proving that a given infinite matrix yields a twisting map requires to verify an infinite number of matrix equalities for infinite matrices. We are able to realize this difficult task in Corollary 5.6 and in Proposition 9.9. In the first case we only have to prove one of the equalities, since those twisting maps have the $n$-extension property, i.e., they are completely determined by the values of $\widetilde{M}_{k,*}$ for $k\leq n$. In the case of Proposition 9.9 we manage to decompose the infinite matrix into three simpler ones, and we prove the required matrix equalities using properties of these simpler matrices.

Notice that none of the twisting maps constructed in Proposition 9.9 has the $m$-extension property for any $m$. This is a direct consequence of the following property of quasi-balanced sequences:

Let $L_{\leq r}=(L_{1},\dots,L_{r})$ be a quasi-balanced partial sequence. Then there exists an extension of $L_{\leq r}$ of the form $(L_{1},\dots,L_{r},\dots,L_{r+k})$ such that both

[TABLE]

are quasi-balanced partial sequences.

In a forthcoming article this property will be proven, together with several other properties of these sequences. For example, the quasi-balanced sequences show a surprising connection to Euler’s totient function and so they are interesting on its own.

There are several open problems related to the results of this article, we want to highlight two of them:

1. (1)

In computations not shown in this paper we have found 16 different cases for the first $4n-1$ rows of the matrices $\widetilde{M}$ corresponding to the case (2) of Proposition 4.6, and 4 of these cases correspond to twisting maps of the family $B(a,L)$. Does there exist any twisting map corresponding to any of the other 12 cases? 2. (2)

Does any twisting map related to the case (2) of Proposition 4.6 has the $m$-extension property for any $m$?

Acknowledgement. We thank the anonymous referee for the thorough revision and numerous helpful suggestions.