User talk:Dfeuer/Cone Condition Equivalent to Reflexivity

That's interesting.... If I'm not mistaken, this actually means that every transitive relation compatible with a group must be either reflexive or irreflexive.

Let's see... If $x \mathrel{R} x$, then $x \circ x^{-1} \mathrel{R} x \circ x^{-1}$, so $e \mathrel{R} e$, from which $y \mathrel{R} y$, so that holds. I wasn't expecting that. --Dfeuer (talk) 08:28, 31 January 2013 (UTC)


 * It appears that even transitivity is not necessary. You have only used compatibility. --Lord_Farin (talk) 08:54, 31 January 2013 (UTC)


 * Have you considered my question about cones in more general contexts (without a group)? I briefly pondered things like
 * $a \mathrel R b$ iff for some $c\in C$, $a \circ c = b$
 * but I don't know if anything of that general sort will actually produce compatible relations or not. --Dfeuer (talk) 09:09, 31 January 2013 (UTC)


 * Just some thoughts. Without associativity it can't be ensured that $R$ is transitive. Reflexivity amounts to a right-identity being in $C$. It is natural to investigate behaviour under the opposite operation ($a * b := b \circ a$). Currently this relation is only left-compatible; there is no immediate natural counterpart to make it properly compatible. --Lord_Farin (talk) 10:04, 31 January 2013 (UTC)