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)

I'm not seeing how to get compatibility without inverses and associativity. Unless you have some brilliant idea (as you did for the topological group question) I'm going to let it stand with the equivalence of "cone compatible with group" and "transitive relation compatible with group". In any case, it's already quite an improvement from the current mess spread through rings and integral domains, and it allows for a nice link to congruences as well (a congruence is induced by a symmetric cone containing the identity, where the cone is the equivalence class of the identity). Or that's how it looks to me, anyway. --Dfeuer (talk) 17:40, 31 January 2013 (UTC)


 * Associativity is really necessary. The only obvious thing coming to mind to ensure compatibility is $a \mathrel R b$ iff $\exists c,d \in C$, $a \circ c = b = d \circ a$ but this seems artificial to say the least. Maybe more later tonight - I have to go now. --Lord_Farin (talk) 18:31, 31 January 2013 (UTC)