Definition talk:Anticommutative

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Just noticed this now: isn't anticommutative $a*b=-b*a$? wikipedia wolfram --Cynic (talk) 04:29, 19 April 2009 (UTC)

Good question. In a general algebraic structure (in particular semigroup) not every element has an inverse. So there is no concept of $-a*b$, and anticommutative can only be defined as $a*b=b*a\Longrightarrow a=b$ or its equivalent with $\ne$.

Trouble is, we can't logically call such a concept "non-commutative" because that just means $\exists a, b: a b \ne b a$, that is, there's nothing to stop $\exists c, d: c \ne d, c d = d c$ for a different pair of elements in the domain.

(Digression: When you're talking about a single-operation structure, in particular a group, there may be conditions under which $a \circ b \ne b \circ a \Longrightarrow a \circ b = \left({b \circ a}\right)^{-1}$, that is, the product inverting on commutation is a logical consequence of the anticommutativity as defined. Needs to be investigated.)

In the definition $a*b=-b*a$, we seem to be in the context of a ring (you've got two operations, $*$ and (by implication) $+$ which is where the $-$ comes from as the additive inverse).

It is clear that definining anticommutativity this more specific way (i.e. in a ring) is not quite compatible with the concept in its most general terms as I've defined it here, as if $a * b = b * a$ then that means $b * a = - (b * a)$ which can be satisfied with $b = -b$ which does not imply $a = b$ so more work needs to be done to tighten this up.

The sites you cited are deplorably lax in defining the context in which the concept holds. Wolfram doesn't specify it at all (wrong link you gave BTW, I've amended it) - the reader is left to assume it's a sort of general thingy which has two operations a-bit-ish which is not too useful. The wikipedia one (despite its abstruse wordiness) is not much better, unfortunately - it's as if someone has taken a chapter of something (oh I see, Bourbaki - haven't read it myself, I only have the first volume, Set Theory) and dumped it in without paying attention to the context in which it's set.

I'll put it on the back-burner.--Prime.mover 09:45, 19 April 2009 (UTC)

Occurs to me that an operation (the way I've defined it) can not be anticommutative in a group because then you'd have $a * e \ne e * a$ where $e$ is the identity.

So I think we'll be putting a split definition in here: one for a general algy struct / semigroup (as it is now) and one for a ring / skew field / possibly even module (which is the one found in wikipedia / wolfram.

I'll sleep on it and think about it during the day / week / month / whatever. --Prime.mover 21:56, 19 April 2009 (UTC)

I'm resurrecting this issue because it was never resolved and I'd like to have a name for $a \circ b = - b \circ a$ for putting up pages about cross products. I'm assuming it would be un-proofwiki-style to use "anticommutative" for two distinct meanings on the same website. I can have it as a disambiguation, but that's messy, and in any event the above objections to wolfram's/googlepedia's definition still stand. --GFauxPas 14:29, 27 March 2012 (EDT)

Perfectly appropriate for there to be two definitions for anticommutative:
a) On a structure with one operation: $a \ne b \implies a \circ b \ne b \circ a$
b) On a structure with one operations: $a \circ b = -\left({b \circ a}\right)$
But in order to define it properly, the nature of those operations needs to be defined. Williefram and twitipedia are both bletheringly illiterately incoherently on the topic, "assuming" that the elements in question are sort-of like numbers-a-bit-ish, as they take for granted the negative operator. But in order for this to make any sense, it needs to be a structure on which the negative is specifically defined. (If I were their schoolmaster I'd apply the cane to their arses.)
In the context of cross product, it's worth pointing out that as the cross product is not associative, a lot of the results proved about semigroups and rings don't apply - so care needs to be taken as to what results are accepted as true. --prime mover 15:05, 27 March 2012 (EDT)
Sorry, I thought I had made it clear that there's two different definitions for this thing. So it might be a good idea to put the second definition into at least a separate section. My bad, I have failed to communicate once again. ;-) --prime mover 17:01, 27 March 2012 (EDT)
... and it occurs to me that you haven't done "abstract algebra" yet so you might not have got your head round the subtlety of this. No matter. Someone will mop up. --prime mover 17:02, 27 March 2012 (EDT)
"Like dis?" No not really. The two definitions are completely different concepts which (apart from the fact that they are both defined in the field of abstract algebra) are not the same thing. Thinking more on it, it appears the two definitions may need to be on separate pages with a disambiguation. --prime mover 17:12, 27 March 2012 (EDT))