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Timeline for Weirdos but algebraic

Current License: CC BY-SA 4.0

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May 20, 2020 at 23:21 comment added zeb With the edit, the problem seems to be roughly equivalent to classifying the directly indecomposable modules over the ring $\mathbb{Z}[x,y,x^{-1},y^{-1}]$.
May 20, 2020 at 10:42 history edited Wlod AA CC BY-SA 4.0
no (...)
May 20, 2020 at 10:03 history edited Wlod AA CC BY-SA 4.0
an EDIT :(
May 20, 2020 at 1:08 answer added Gerhard Paseman timeline score: 0
May 19, 2020 at 23:49 comment added YCor @zeb thanks for the terminology: medial quasigroup is axiom 3 for $(X,\sigma)$. However the magma should also be cancelative to ensure existence of $\lambda,\rho$ (which are then unique). For $X$ infinite, we can even assume that the whole magma law $X^2\to X$ is injective but not surjective. (Its injectivity obviously implies cancelative.)
May 19, 2020 at 23:46 answer added YCor timeline score: 1
May 19, 2020 at 23:43 comment added Wlod AA @zeb, the inverse operations give you surjectivity always, for all weirdos (infinite too). Or do you know of some extra-surjective notion?
May 19, 2020 at 23:40 comment added Wlod AA @YCor "I removed it" -- thus I removed my "semmi-reply". Thank you.
May 19, 2020 at 23:20 review Close votes
May 22, 2020 at 5:12
May 19, 2020 at 23:07 comment added YCor My "semilattice" remark was not obvious, but obviously false. I removed it.
May 19, 2020 at 23:04 comment added zeb In the finite case, weirdos are exactly the same as "medial quasigroups"/"entropic quasigroups", and the Bruck–Murdoch-Toyoda theorem describes the structure of these things. I don't know if that's enough to resolve the finite case completely. The infinite case seems harder, since there is no obvious guarantee that $\sigma$ is surjective.
May 19, 2020 at 22:44 history edited Wlod AA CC BY-SA 4.0
math-typos
May 19, 2020 at 22:41 comment added Wlod AA @Qfwfq, "commas" were useful before Chinese+Gutenberg, and not since them.
May 19, 2020 at 22:37 comment added Wlod AA @Qfwfq, "Also, could you put commas?" -- No.
May 19, 2020 at 22:36 comment added Qfwfq @YCor: Oh I see, thanks (I missed they were of arity 2)
May 19, 2020 at 22:35 comment added YCor @Qfwfq no, it's $\lambda(x,y)=xy^{-1}$ and $\rho(x,y)=x^{-1}y$. If you have an E-structure with $\sigma(x,y)=xy$ a group law, then $\lambda$ and $\rho$ automatically have these forms.
May 19, 2020 at 22:34 comment added Qfwfq Also, why don't you put the commas? :-)
May 19, 2020 at 22:34 comment added YCor No, I just started thinking about what it means for a W-structure to have $\sigma$ a group law. I ended up with the conclusion that is already apparent in your post, but when I read it it sounded like a weaker statement. Whence my comment. Besides being not fan of your terminology and in spite of the lack of focus, I'm curious about your structure, so please don't consider any participation as anything against you.
May 19, 2020 at 22:34 comment added Qfwfq "Abelian groups, where σ is the group operation, and λ ρ are the left and right inverse operations" - What are the left inverse and the right inverse operations in an abelian group? Are each of them just the inverse operation $x\mapsto x^{-1}$?
May 19, 2020 at 22:30 comment added Wlod AA @YCor and Haha (who is Haha?), about Abelian groups, you're repeating after my post, and it's trivial anyway. Why do you insist on destroying this thread? Why is it important to you? Simply say that you don't want to see me around MO. I may oblige, and if I do, I'll do it with a bit of pleasure and minor relief. ### I am not asking about your W-semigroups but about weirdos. Be polite and have the decency to leave my notation IN THIS THREAD alone. You may change my terminology in your own posts and publications.
May 19, 2020 at 22:24 comment added YCor Haha. About groups, if a E-structure has sigma a group law, then automatically $\lambda(x,y)=xy^{-1}$, $\rho(x,y)=x^{-1}y$, and being a W-structure is equivalent to being abelian. (I'm not sure whether this is what you said on groups, or whether you took as an additional assumption that $\lambda,\rho$ have these forms.
May 19, 2020 at 22:18 comment added Wlod AA @YCor, are you sure you want to stick to distracting from mathematics? #### YCor, are you sure you want to stick to imposing your notion of political correctness on me? #### YCor, are you sure you want to stick to ...? #### YCor, ...? #### ??...?
May 19, 2020 at 22:06 history edited Wlod AA CC BY-SA 4.0
extra observation and a typo
S May 19, 2020 at 21:52 history rollback Wlod AA
Rollback to Revision 2 - Edit approval overridden by post owner or moderator
May 19, 2020 at 20:21 history suggested JoshuaZ CC BY-SA 4.0
making clear that "challenge" is the question intended, and not a challenge in the sense of being already known to the author.
May 19, 2020 at 20:02 review Suggested edits
S May 19, 2020 at 21:52
May 19, 2020 at 11:56 comment added YCor @bof I guess you're joking... yes I know but all this don't specifically qualify people, and certainly not categories of people (also you're only quoting adjectives, while weirdo is a noun). The closest existing examples I'm aware of are "Polish spaces/groups", "Japanese rings", Chinese remainder theorem" although these are adjectives too. These are well-established, but it's safer not renew such neologisms.
May 19, 2020 at 10:17 comment added YCor Are you sure you want to stick to this terminology? I believe one should avoid to change words that usually qualify some category of people into mathematical words.
May 19, 2020 at 10:03 comment added Wlod AA I wouldn't ask about things which I already knew or at least I would make things 100% clear. ### As I have written at the end of my post, I formulated and proved a characterization only of some very special weirdos (in 1961/62).
May 19, 2020 at 10:02 history edited Wlod AA CC BY-SA 4.0
Open
May 19, 2020 at 9:51 comment added Todd Trimble What's the question? Is it the "challenge", and is this something you already know the answer to?
May 19, 2020 at 9:36 history asked Wlod AA CC BY-SA 4.0