# Examples of Yang-Baxter monoids

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.

• 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? – Jules Lamers Jan 8 at 4:57
• 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 $x\circ y$. – Joseph Van Name Jan 8 at 21:04