Definition talk:Antiassociative Operation

The term "antiassociative" has not been coined for ProofWiki:

http://www.doc.ic.ac.uk/~sgc/html_papers/colton_ijcar01.html

--prime mover 05:19, 1 August 2012 (UTC)

Oh. I wonder why I couldn't find it. It was a pretty natural idea. Fun while it lasted though. Thinking it was my own. --Jshflynn 07:55, 1 August 2012 (UTC)

Now my idea of what antiassociativeness "ought" to mean would be: defined on a structure $\left({S, +, *}\right)$ with 2 operations, antiassociative would be:

$\forall a, b, c \in S: a * \left({b * c}\right) + \left({a * b}\right) * c = 0$

where $0$ is the identity element of $+$.

That is, one way round is specifically equal to the additive inverse of (i.e. "minus") the other way round.

But I haven't seen that defined like that, as such. Mind, I have a book on Lie Groups and Lie Algebras which I've never been able to penetrate in any depth because of the unconventionality of the language (physics rather than maths) which goes into this in some detail, discussing the "associator". --prime mover 11:47, 1 August 2012 (UTC)

So something like: $\left({S, +, \rightarrow}\right)$ where $\rightarrow$ denotes the right operation and all elements in $\left({S, +}\right)$ are self-inverses?

No, because the right operation is associative. --prime mover 16:47, 1 August 2012 (UTC)

What you said is true but I don't see what the associativity of the $*$ operation by itself has to do with your definition of antiassociativity. Observe:

\(\ds \left({a \rightarrow b}\right) \rightarrow c\)

\(=\)

\(\ds b \rightarrow c\)

\(\ds \)

\(=\)

\(\ds c\)

Then:

\(\ds a \rightarrow \left({b \rightarrow c}\right)\)

\(=\)

\(\ds a \rightarrow c\)

\(\ds \)

\(=\)

\(\ds c\)

And:

\(\ds c + c\)

\(=\)

\(\ds 0\)

By assumption

So:

$\forall a, b, c \in S: \left({a \rightarrow b}\right) \rightarrow c + a \rightarrow \left({b \rightarrow c}\right) = 0$

--Jshflynn 17:24, 1 August 2012 (UTC)

Regarding Lie algebras. I'm not quite there yet, still just a beginner. Someday hopefully, give it another 18 months. --Jshflynn 14:57, 1 August 2012 (UTC)