A characterization of finite multipermutation solutions of the Yang-Baxter equation

TL;DR

This paper characterizes finite multipermutation solutions of the Yang-Baxter equation by linking their structure groups to left orderability and poly-infinite cyclic properties, providing a clear algebraic criterion.

Contribution

It establishes an equivalence between multipermutation solutions and the left orderability or poly-infinite cyclic nature of their structure groups.

Findings

Finite multipermutation solutions correspond to structure groups that are left orderable.

Such structure groups are exactly those that are poly-infinite cyclic.

Provides an algebraic characterization of multipermutation solutions.

Abstract

We prove that a finite non-degenerate involutive set-theoretic solution (X,r) of the Yang-Baxter equation is a multipermutation solution if and only if its structure group G(X,r) admits a left ordering or equivalently it is poly-(infinite cyclic).

Equations

$$(r\times {\mathrm{id}}_X)({\mathrm{id}}_X\times r)(r\times {\mathrm{id}}_X)=({\mathrm{id}}_X\times r)(r\times {\mathrm{id}}_X)({\mathrm{id}}_X\times r).$$

$$r(x,y)=({\sigma }_x(y),{\gamma }_y(x)),$$

$$r:X\times X\to X\times X,r(x,y)=(xy{x}^{-1},x),$$

$$a(b+c)+a=ab+ac,a,b,c\in A.$$

$$\mathrm{Soc}(A)=\{a\in A:{\lambda }_a=\mathrm{id}\}=\{a\in A:a+b=ab\text{ for all b∈A}\}.$$

$$r_A:A\times A\to A\times A,r_A(a,b)=({\lambda }_a(b),{\lambda }_{{\lambda }_a(b)}^{-1}(a)).$$

$$r_{G(X,r)}(\iota \times \iota )=(\iota \times \iota )r.$$

$$\overline{\pi }:G(X,r)/\mathrm{Soc}(G(X,r))\to \mathcal{G}(X,r)$$

$$a\ast b=ab-a-b=({\lambda }_a-\mathrm{id})(b),a,b\in B,$$

$$B=B^{(1)}\supseteq B^{(2)}\supseteq B^{(3)}\supseteq \cdots ,$$

$$\{({\lambda }_a-\mathrm{id})(b):a\in B^{(m)},b\in B\}$$

$$G^{(m+1)}$$

$$=⟨({\lambda }_a-\mathrm{id})(x):a\in G^{(m)},x\in X{⟩}_{+}$$

$$=⟨y-x:x,y\in X_i,1\le i\le r{⟩}_{+}.$$

$$\{1\}=G_0◃G_1◃\cdots ◃G_n=G$$

$${\lambda }_h(z)=hz-h=zh-h=z+h-h=z.$$

$$z_i=\sum _{t\in X_i}{n}_{t}t,$$

$$G^{(m)}=G^{(m+1)}+(\mathrm{Soc}(G)\cap G^{(m)})=H+(\mathrm{Soc}(G)\cap G^{(m)}).$$

$$y={\lambda }_g(x)={\lambda }_{g_2{g}_1}(x)={\lambda }_{g_2}{\lambda }_{g_1}(x)={\lambda }_{g_1}(x).$$

$${\sigma }_1=\mathrm{id},$$

$${\sigma }_3=(25)(3b4f)(7c8g)(9dea),$$

$${\sigma }_5=(38)(47)(bg)(cf),$$

$${\sigma }_7=(25)(3c7b)(4g8f)(9dea),$$

$${\sigma }_9=(38)(47)(9e)(ad),$$

$${\sigma }_b=(25)(3f4b)(7g8c)(9aed),$$

$${\sigma }_d=(9e)(ad)(bg)(cf),$$

$${\sigma }_f=(25)(3g7f)(4c8b)(9aed),$$

$$\mathcal{G}(X,r)=\{{\sigma }_x:x\in X\}.$$

$${\sigma }_1=\mathrm{id},$$

$${\sigma }_3=(34)(78)(9e)(ad)(bf)(cg),$$

$${\sigma }_5=(9a)(bc)(de)(fg),$$

$${\sigma }_7=(34)(78)(9d)(ae)(bg)(cf),$$

$${\sigma }_9=(56)(78)(de)(fg),$$

$${\sigma }_b=(34)(56)(9dae)(bgcf),$$

$${\sigma }_d=(56)(78)(9ead)(bgcf),$$

$${\sigma }_f=(34)(56)(9a)(fg),$$

$$\mathcal{G}(X,r)=\{{\sigma }_x:x\in X\}\simeq \mathbb{Z}/2\times {\mathbb{D}}_8,$$

Full text

The two first-named authors are partially supported by the grant MINECO MTM2014-53644-P. The third-named author is partially supported by PICT-2014-1376, MATH-AmSud 17MATH-01, ICTP, ERC advanced grant 320974 and the Alexander von Humboldt Foundation.

2010 MSC: Primary 16T25, 20F16, 20F60.

Introduction

According to Drinfeld [9], a set-theoretic solution of the Yang–Baxter equation is a pair $(X,r)$, where $X$ is a set and $r\colon X\times X\to X\times X$ is a bijective map such that

[TABLE]

The seminal papers [10] and [15] initiated the study of non-degenerate involutive set-theoretic solutions of the Yang–Baxter equation. Etingof, Schedler and Soloviev introduced the structure group $G(X,r)$ of a solution $(X,r)$ as the group presented with set of generators $X$ and with relations $xy=uv$ whenever $r(x,y)=(u,v)$. This group turned out to be very important to understand set-theoretic solutions. As proved by Gateva-Ivanova and Van den Bergh, the structure group $G(X,r)$ of a finite non-degenerate involutive set-theoretic solution of the Yang–Baxter equation is a Bieberbach group, i.e. a finitely generated torsion-free abelian-by-finite group.

In [10] multipermutation solutions were introduced. This is an important notion that was intensively studied [2, 3, 4, 5, 13, 14, 20]. In [17, Proposition 4.2] Jespers and Okniński proved that the structure group of a finite multipermutation solution is poly-$\mathbb{Z}$. The main result of this paper is to prove the converse: a finite solution $(X,r)$ such that $G(X,r)$ is poly-$\mathbb{Z}$ is a multipermutation solution.

To prove our result we use the language of braces introduced by Rump in [19]. Braces are algebraic structures that generalize radical rings. This fact allows us to use tools and techniques from ring theory to study set-theoretic solutions of the Yang–Baxter equation.

In [11, Theorem 23] Farkas proved that a Bieberbach group is poly-$\mathbb{Z}$ if and only if it admits a left ordering. Since Chouraqui proved in [6, Theorem 1] that the structure group of a finite non-degenerate involutive set-theoretic solution of the Yang–Baxter equation is a Garside group, our result in particular yields an infinite family of Garside groups that are not left orderable.

1. Preliminaries

A set-theoretic solution of the Yang–Baxter equation is a pair $(X,r)$, where $X$ is a set and $r\colon X\times X\to X\times X$ is a bijective map such that

[TABLE]

A solution $(X,r)$ is said to be involutive if $r^{2}=\mathrm{id}_{X^{2}}$ and it is said to be non-degenerate if

[TABLE]

where $\sigma_{x}$ and $\gamma_{x}$ are permutations of $X$ for all $x\in X$. The structure group $G(X,r)$ of a non-degenerate solution $(X,r)$ is defined as the group presented with set of generators $X$ and with relations $xy=uv$ whenever $r(x,y)=(u,v)$. In [6, Theorem 1] Chouraqui proved that the structure group of a non-degenerate involutive set-theoretic solution of the Yang–Baxter equation is a Garside group. A simpler proof of this result was recently given by Dehornoy in [8].

Example 1.1**.**

Let $X$ be a conjugacy class of a finite group $G$ such that the subgroup generated by $X$ is non-abelian. Then the map

[TABLE]

is a non-degenerate solution of the Yang–Baxter equation. This solution is not involutive. We claim that $G(X,r)$ is not a Garside group. Let $G(X,r)$ act on $X$ by conjugation. Then the center of $G(X,r)$ is the kernel of this action and hence it has finite index in $G(X,r)$. This implies that all conjugacy classes of $G(X,r)$ are finite. Thus the derived subgroup of $G(X,r)$ is a finite group by a theorem of Schur [18, Theorem 7.57]. In particular, $G(X,r)$ has torsion elements and hence $G(X,r)$ is not a Garside group.

Remark 1.2*.*

In [6] it is conjectured that structure groups of finite non-degenerate solutions are Garside groups. Example 1.1 shows that this conjecture is not true.

Convention 1.3**.**

A solution $(X,r)$ of the YBE will always be a non-degenerate involutive set-theoretic solution of the Yang–Baxter equation.

A left brace is an abelian group $(A,+)$ with another group structure with multiplication $A\times A\to A$, $(a,b)\mapsto ab$, such that

[TABLE]

It is known that in any left brace $A$ the neutral elements of the groups $(A,+)$ and $(A,\cdot)$ coincide. If $A$ is a left brace, then the map $\lambda\colon(A,\cdot)\to\mathrm{Aut}(A,+)$ given by $\lambda_{a}(b)=ab-a$ is a group homomorphism. It follows from the definition that $ab=a+\lambda_{a}(b)$ and $a+b=a\lambda^{-1}_{a}(b)$ for all $a,b\in A$.

An ideal $I$ of a left brace $A$ is a normal subgroup $I$ of the multiplicative group of $A$ such that $\lambda_{a}(y)\in I$ for all $a\in A$ and $y\in I$. The socle of a left brace $A$ is defined as the set

[TABLE]

The socle of $A$ is an ideal of $A$.

Rump proved that each left brace $A$ produces a solution of the YBE

[TABLE]

One of the main results of [10] is the following: If $(X,r)$ is a solution of the YBE, then there exists a bijective $1$-cocycle $G(X,r)\to\mathbb{Z}^{(X)}$, where $\mathbb{Z}^{(X)}$ is the free abelian group on $X$. From this it immediately follows that the canonical map $\iota\colon X\to G(X,r)$ is injective. Now using the language of braces the existence of a bijective $1$-cocycle can be written as follows: If $(X,r)$ is a solution, there exists a unique left brace structure over the structure group $G(X,r)$ such that the additive group of $G(X,r)$ is isomorphic to $\mathbb{Z}^{(X)}$, the multiplicative structure is that of $G(X,r)$ and such that

[TABLE]

The permutation group $\mathcal{G}(X,r)$ of a solution $(X,r)$ of the YBE is defined as the group generated by $\{\sigma_{x}:x\in X\}$, where $r(x,y)=(\sigma_{x}(y),\gamma_{y}(x))$. It is known that the map $x\mapsto\sigma_{x}$ extends to a homomorphism of groups $\pi\colon G(X,r)\to\mathcal{G}(X,R)$ such that $\mathrm{Ker}(\pi)=\mathrm{Soc}(G(X,r))$ and therefore $\mathcal{G}(X,r)$ has a unique structure of left brace such that the group isomorphism

[TABLE]

induced by $\pi$ is an isomorphism of left braces.

Remark 1.4*.*

Let $B$ be a left brace. Using the operation

[TABLE]

Rump introduced the series

[TABLE]

where $B^{(m+1)}=B^{(m)}*B$ is the additive group generated by

[TABLE]

for all $m\geq 1$. As a corollary of [19, Proposition 6] Rump proved that each $B^{(m)}$ is an ideal of $B$. Notice that this corollary refers to right braces.

For any left brace $A$ and a subset $X\subseteq A$ we will denote by $\langle X\rangle$ the subgroup of $(A,\cdot)$ generated by $X$. Similarly $\langle X\rangle_{+}$ will denote the subgroup of $(A,+)$ generated by $X$.

Remark 1.5*.*

Let $(X,r)$ be a finite solution of the YBE and let $G=G(X,r)$. Let $m$ be a positive integer and let $X_{1},\dots,X_{r}$ be the orbits of $X$ under the action of $G^{(m)}$. Then

[TABLE]

The second equality follows from the fact that $(G,+)$ is generated by $X$ and $\lambda_{a}$ is an automorphism of $(G,+)$. The third equality is obtained using that $\lambda_{a}(x)\in X$ for all $x\in X$ and all $a\in G$.

2. Multipermutation solutions

Let $(X,r)$ be a solution of the YBE. Consider the equivalence relation on $X$ given by $x\sim y$ if and only if $\sigma_{x}=\sigma_{y}$. The retraction of $(X,r)$ is defined as the solution $\mathrm{Ret}(X,r)$ induced by this equivalence relation. One defines recursively $\mathrm{Ret}^{m+1}(X,r)=\mathrm{Ret}(\mathrm{Ret}^{m}(X,r))$ for all $m$. A solution $(X,r)$ of the YBE is said to be a multipermutation solution of level $m$ if $m$ is the minimal positive integer such that $\mathrm{Ret}^{m}(X,r)$ has only one element. A solution $(X,r)$ of the YBE is said to be irretractable if $\mathrm{Ret}(X,r)=(X,r)$.

Recall that a group $G$ is said to be poly-$\mathbb{Z}$ if it has a subnormal series

[TABLE]

such that each quotient $G_{i}/G_{i-1}$ is isomorphic to $\mathbb{Z}$. A group $G$ is said to be left orderable if there is a total order $<$ on $G$ such that for any $x,y,z\in G$, $x<y$ implies $zx<zy$.

The main result of the paper is the following theorem.

Theorem 2.1**.**

Let $(X,r)$ be a finite non-degenerate involutive set-theoretic solution of the Yang–Baxter equation. Then the following statements are equivalent:

1. (1)

$(X,r)$* is a multipermutation solution.* 2. (2)

$G(X,r)$* is left orderable.* 3. (3)

$G(X,r)$* is poly-$\mathbb{Z}$.*

Proof.

Let $G=G(X,r)$ and $\mathcal{G}=\mathcal{G}(X,r)$. By [15, Theorem 1.6], $G$ is a Bieberbach group. Hence the equivalence between $(2)$ and $(3)$ follows from [11, Theorem 23]. The implication $(1)\implies(3)$ is [17, Proposition 4.2]; see also [7] for another proof of $(1)\Rightarrow(2)$. Let us prove $(2)\implies(1)$. For that purpose let us assume that $(X,r)$ is not a multipermutation solution. By [12, Theorem 5.15], the solution $(G,r_{G})$ is not a multipermutation solution. This implies that the solution $(\mathcal{G},r_{\mathcal{G}})$ is not a multipermutation solution. Using [3, Proposition 6] one obtains that $G^{(m)}\neq\{0\}$ and $\mathcal{G}^{(m)}\neq\{0\}$ for all $m$. Since $\mathcal{G}$ is finite, there exists $m$ such that $\mathcal{G}^{(m+1)}=\mathcal{G}^{(m)}\neq 0$. By [11, Theorem 23], to prove that $G$ is not left orderable it suffices to prove that the non-trivial subgroup $H=G^{(m+1)}$ of $(G,\cdot)$ has trivial center. Let $z\in Z(H)$. Since $\mathrm{Soc}(G)$ has finite index in $G$ and $G$ is torsion free, without loss of generality we may assume that $z\in\mathrm{Soc}(G)$. Notice that if $h\in H$, then

[TABLE]

Let $X_{1},\dots,X_{r}$ be the orbits of $X$ under the action of $\mathcal{G}^{(m)}$. These orbits are the orbits of $X$ under the action of $G^{(m)}$ through the map $\lambda$. Since $(G,+)$ is the free abelian group with basis $X$, the element $z$ can be uniquely written as $z=z_{1}+\cdots+z_{r}$, where each $z_{i}\in\langle X_{i}\rangle_{+}$. From the uniqueness of the decomposition of $z$ and (2.1) one obtains that $\lambda_{h}(z_{i})=z_{i}$ for all $i\in\{1,\dots,r\}$ and $h\in H$. Now write each $z_{i}$ as

[TABLE]

where each $n_{x}\in\mathbb{Z}$. Remark 1.5 implies that $\sum_{t\in X_{i}}n_{t}=0$. This decomposition is unique since $(G,+)$ is the free abelian group with basis $X$. Let $x,y\in X_{i}$ be such that $x\neq y$. Then there exists $g\in G^{(m)}$ such that $\lambda_{g}(x)=y$. From $\mathcal{G}^{(m+1)}=\mathcal{G}^{(m)}$ it follows

[TABLE]

Thus $g=g_{1}+g_{2}$, where $g_{1}\in H$ and $g_{2}\in\mathrm{Soc}(G)\cap G^{(m)}$. Since $g_{2}\in\mathrm{Soc}(G)$, $g=g_{2}g_{1}$. Therefore

[TABLE]

Since $z_{i}=\lambda_{g_{1}}(z_{i})=\sum_{t\in X_{i}}n_{t}\lambda_{g_{1}}(t)$, we conclude that $n_{x}=n_{y}$. Since $\sum_{t\in X_{i}}n_{t}=0$, it follows that $n_{t}=0$ for all $t\in X_{i}$ and all $i\in\{1,\dots,r\}$. Therefore $z=0=1$ and the result follows. ∎

Example 2.2**.**

Let $X=\{1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,g\}$ and let

[TABLE]

Then $r(x,y)=(\sigma_{x}(y),\sigma^{-1}_{\sigma_{x}(y)}(x))$ is an irretractable solution of the YBE. Hence its structure group $G(X,r)$ is not left orderable. In this case

[TABLE]

Example 2.3**.**

Let $X=\{1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,g\}$ and let

[TABLE]

Then $r(x,y)=(\sigma_{x}(y),\sigma^{-1}_{\sigma_{x}(y)}(x))$ is an irretractable solution of the YBE. Hence its structure group $G(X,r)$ is not left orderable. In this case

[TABLE]

where $\mathbb{D}_{8}$ denotes the dihedral group of size $8$.

Remark 2.4*.*

The solutions of Examples 2.2 and 2.3 correspond to the associated solutions to the only left braces of size $16$ with trivial socle. This was checked with GAP and the list of small braces of [16].

Example 2.5**.**

Let $X=\{1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,g,h,i,j,k,l,m,n,o\}$. Let

[TABLE]

Then $r(x,y)=(\sigma_{x}(y),\sigma^{-1}_{\sigma_{x}(y)}(x))$ is an irretractable solution of the YBE. Hence its structure group $G(X,r)$ is not left orderable. In this case

[TABLE]

The unique subgroup of index two of $\mathcal{G}(X,r)$ is an ideal of $\mathcal{G}(X,r)$. Hence $\mathcal{G}(X,r)$ is not a simple brace.

Remark 2.6*.*

There are two braces of size $24$ with trivial socle. One is the simple brace found in [1] and the other one is that of Example 2.5.