5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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

Examples:

  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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.