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Let $R\subset V\times V$ be a relation on a set $V$. For a subset $S\subset V$, define its orthogonal complement with respect to $R$ as $$S^l:=\{ x: \forall y\in S\ \ (x,y)\in R\},\ \ S^r:=\{y: \forall x\in S\ \ (x,y)\in R\},\ S^{lr}:=(S^l)^r,...$$

For two words $u,v$ in alphabet $\{l,r\}$, write $u =_{V,R,S} v$ iff $S^u=S^v$. Thus, $l=_{V,R,S} lrl$ and $r=_{V,R,S} rlr$.

This is the usual orthogonal complement when $V$ is a vector space and $R:=\{(u,v):\phi(x,y)=0\}$ for a bilinear form $\phi(-,-)$.

Fix $R$, and define a semi-group $O_{V,R}:=<l,r: u=v\ \text{ if }\ \forall S\subset V\ u=_{V,R,S}v >$ with two generators $l,r$ and relations $u=v$ whenever $\forall S\subset V\ u=_{R,S}v$. The relations $l=_{V,R,S} lrl$ and $r=_{V,R,S} rlr$ seem to hold only for the two generators, and in particular, it seems these semi-groups are not necessarily regular.

What is known about the regular semigroups of this form ? Have they been studied ? Is there a name ?

I am particularly interested in the case when $V$ is the set of morphisms of a category, and $R$ is the weak ortogonality relation, also known as the lifting property.

The simplest non-trivial case of interest is as follows: $X$ is the set of continuous maps in the category of topological spaces, and $R$ is the weak orthogonality relation, and $S:=\{ \emptyset\to \{pt\}\}$ is the class consisting of a single map from the empty set to a singleton.

Is something known when $V$ is a vector space and $R$ is a possibly non-symmetric bilinear form ? (I thank @BenjaminSteinberg for corrections).

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  • $\begingroup$ Are you sure the semigroup is regular? It isn't enough for the generators to be regular. $\endgroup$ Commented Jan 10, 2023 at 1:45
  • $\begingroup$ @BenjaminSteinberg: Yes, I think so, though it requires a little argument. Here is a (corrected) argument. $\endgroup$
    – user494312
    Commented Jan 10, 2023 at 9:27
  • $\begingroup$ @BenjaminSteinberg: Let $u^T$ denote the result of replacing each letter $l$ by $r$, and each $r$ by $l$, i.e. $l^T:=r$, $r^T:=l$, and $(xy)^T:=x^Ty^T$. Let $S^{-1}:=\{(y,x): (x,y)\in S\}$. Then $u=_{S,R}u u^T u$, as follows from the following. $S^r = ((S^{-1})^l)^{-1}$, and, more generally, $S^u=((S^{-1})^{u^T})^{-1}$. This implies that $l=_{S^u,V} lrl$ means that $ul=_{S,V} ul u^T r ul = (ul) (ul)^T (ul)$. $\endgroup$
    – user494312
    Commented Jan 10, 2023 at 9:28
  • $\begingroup$ What does S^{-1} mean? $\endgroup$ Commented Jan 10, 2023 at 12:09
  • $\begingroup$ The inverse of $S$: $S^{-1}:=\{ (y,x)\in V\times V: (x,y)\in S\}$. That is, $(x,y)\in S$ iff $(y,x)\in S^{-1}$. $\endgroup$
    – user494312
    Commented Jan 10, 2023 at 12:36

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