Inverse semi-braces and the Yang-Baxter equation
Francesco Catino, Marzia Mazzotta, Paola Stefanelli
TL;DR
The paper addresses the problem of obtaining set-theoretical solutions to the Yang-Baxter equation that need not be bijective, introducing left inverse semi-braces as a framework linking inverse semigroups with semigroup-additive structures via $a(b+c)=ab+a(a^{-1}+c)$. It defines the associated solution map $r_S(a,b)=(\lambda_a(b),\rho_b(a))$ and provides sufficient conditions under which this map yields a Yang-Baxter solution, along with many explicit, non-bijective examples. It then develops three construction schemes—matched product, double semidirect product, and asymmetric product—to systematically generate broad families of non-bijective YBE solutions from given left inverse semi-braces, including detailed criteria and a host of concrete instances. The work situates these constructions within the brace/left-brace literature, connects to pentagon-type and cubic solutions, and discusses open questions about when these composite constructions preserve the solution property, offering a rich source of new examples and lower bounds (via finite inverse semigroups) for further study.
Abstract
The main aim of this paper is to provide set-theoretical solutions of the Yang-Baxter equation that are not necessarily bijective, among these new idempotent ones. In the specific, we draw on both to the classical theory of inverse semigroups and to that of the most recently studied braces, to give a new research perspective to the open problem of finding solutions. Namely, we have recourse to a new structure, the inverse semi-brace, that is a triple $(S,+, \cdot)$ with $(S,+)$ a semigroup and $(S, \cdot)$ an inverse semigroup satisfying the relation $a \left(b + c\right) = a b + a\left(a^{-1} + c\right)$, for all $a,b,c \in S$, where $a^{-1}$ is the inverse of $a$ in $(S, \cdot)$. In particular, we give several constructions of inverse semi-braces which allow for obtaining solutions that are different from those until known.