Then we say that an algebra $(X,f,g,\circ,1)$ is a Yang-Baxter monoid if it satisfies the following identities:

  1. $(X,\circ,1)$ is a monoid,

  2. $f(x,1)=1,f(1,x)=x,g(x,1)=x,g(1,x)=1$

  3. $x\circ y=f(x,y)\circ g(x,y)$

  4. $g(x,y\circ z)=g(g(x,y),z)$

  5. $f(x\circ y,z)=f(x,f(y,z))$

  6. $f(x,y\circ z)=f(x,y)\circ f(g(x,y),z)$

  7. $g(x\circ y,z)=g(x,f(y,z))\circ g(y,z)$,

  8. $g(f(x,y),f(g(x,y),z))=f(g(x,f(y,z)),g(y,z))$.

Suppose that $T:X^{2}\rightarrow X^{2}$ is a function. Then we say that $T$ satisfies the Yang-Baxter equations if $(T\times 1_{X})\circ(1_{X}\times T)\circ(T\times 1_{X})=(1_{X}\times T)\circ(T\times 1_{X})\circ(1_{X}\times T)$.

If $(X,f,g,\circ,1)$ is a Yang-Baxter monoid and $T:X^{2}\rightarrow X^{2},T(x,y)=(f(x,y),g(x,y))$, then $T$ automatically satisfies the Yang-Baxter equations.

An LD-monoid is an algebra $(X,*,\circ,1)$ that satisfies the identities

  1. $(X,\circ,1)$ is a monoid,

  2. $x\circ y=(x*y)\circ x$,

  3. $x*(y\circ z)=(x*y)\circ (x*z)$,

  4. $x*(y*z)=(x\circ y)*z$,

  5. $x*(y*z)=(x*y)*(x*z)$.

  6. $x*1=1,1*x=x$.

For example, let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$ and let $*$ be the operation on $\mathcal{E}_{\lambda}$ defined by $j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$. Then $(\mathcal{E}_{\lambda},*,\circ,1)$ is an LD-monoid. If $(G,\circ,1)$ is a group and $x*y=xyx^{-1}$, then $(G,*,\circ,1)$ is always an LD-monoid. Suppose that $(X,\circ,1)$ is a monoid, $*$ is a binary operation, and $f(x,y)=x*y,g(x,y)=x$. If $f(x,y)=x*y,g(x,y)=x$, then $(X,*,\circ,1)$ is an LD-monoid precisely when $(X,f,g,\circ,1)$ is a Yang-Baxter monoid.

The motivation for identities 1-8 comes from the notion of a permutative Yang-Baxter monoid which satisfies these identities.

What are some examples of Yang-Baxter monoids that do not trivially arise from LD-monoids? Are there any references for the notion of a Yang-Baxter monoid anywhere?

The closest thing that I found to the notion of a Yang-Baxter monoid is known as the structure group of a solution to the Yang-Baxter equation. The structure group of a function $T:X^{2}\rightarrow X^{2}$ that satisfies the Yang-Baxter equation is the group with a presentation consisting of the relations $xy=uv$ whenever $T(x,y)=(u,v)$ which is a version of property 3.

  • $\begingroup$ For me property 1 says that $(X,\circ,1)$ is an algebra (I guess you don't necessarily want $X$ to be an $R$-module for some commutative ring $R$). Are $f$ and $g$ in the first sentence also part of the data of the algebra? Should I think of them as alternative products with the same unit 1 as $\circ$, and if so, are they associative too? $\endgroup$ Jan 8, 2019 at 4:57
  • 1
    $\begingroup$ The word algebra is used in the universal algebraic sense and not the ring theoretic sense. By algebra, I only mean a set and a bunch of operations on that set. In the LD-monoid example, the operation $*$ is not associative nor commutative, so $f$ and $g$ do not have to satisfy any sort of associativity. By the axiom $x\circ y=f(x,y)\circ g(x,y)$, one should think of $f(x,y),g(x,y)$ as a sort of distinguished factorization of $x\circ y$. $\endgroup$ Jan 8, 2019 at 21:04

1 Answer 1


What about using the theory of braces? A skew brace is a triple $(A,+,\circ)$, where $(A,+)$ and $(A,\circ)$ are groups and $a\circ (b+c)=a\circ b-a+a\circ c$ holds for all $a,b,c\in A$. Notation: If $a\in A$, then $a'$ denotes the inverse of $a$ with respect to the circle operation $\circ$.

Facts: Let $A$ be a skew brace.

  1. The map $\lambda\colon (A,\circ)\to\mathrm{Aut}(A,+)$, $a\mapsto \lambda_a$, where $\lambda_a(b)=-a+a\circ b$, is a group homomorphism.
  2. The map $r\colon A\times A\to A\times A$, $r_A(a,b)=(\lambda_a(b),\lambda_a(b)'\circ a\circ b)$, is a solution. Moreover, $r^2=\mathrm{id}_{A\times A}$ if and only if $(A,+)$ is abelian.

Skew braces were the group $(A,+)$ is abelian were first considered by Rump to study involutive solutions to the Yang-Baxter equation:

  • Rump, Wolfgang. Braces, radical rings, and the quantum Yang-Baxter equation. J. Algebra 307 (2007), no. 1, 153-170. link


  1. Radical rings are examples of skew braces where the group $(A,+)$ is abelian. (A radical ring is a ring $A$ such that the operation $x\circ y=x+xy+y$ turns $A$ into a group. This, in particular, implies that radical rings produce solutions to the Yang-Baxter equation.)
  2. The structure group $G(X,r)$ of a non-degenerate solution $(X,r)$ is another example of a skew brace. (Non-degenerate means that you can write your solution $r$ as $r(x,y)=\sigma_x(y),\tau_y(x))$ for permutations $\sigma_x,\tau_x\colon X\to X$.) As you mentioned, the structure group is defined as the group $G(X,r)$ with generators $x\in X$ and relations $x\circ y=u\circ v$ whenever $r(x,y)=(u,v)$.

The following fact holds and it could be useful to address your question: If $(X,r)$ is non-degenerate solution, there exists a unique skew brace structure on the group $G(X,r)$ such that $r_{G(X,r)}(\iota\times\iota)=(\iota\times\iota)r$, where $\iota\colon X\to G(X,r)$ is the canonical map (which is general is not injective).


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