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March 9[edit]

Anti-commutativity[edit]

I am unsure about a canonical meaning of this notion. On one hand the WP-article about anticommutativity seems to refer to the algebraic notion of groups, involving an operation denoted by "∗", but without being explicit about the identitity and the notation of inverse elements for this assumed group. There is just the unexplained use of "-" and a sgn-operator (applied to permutations of indices).

On the other hand I have a vague image about an anticommutative binary Lie-bracket, defined on a k-module over some unitary ring k, involving the additive inverses of the group structure in the module, when commuting the operands of the bracket.

Does one need two operations (k-algebra + module group) or not (group alone)? In case of two, which operation necessarily supplies the relevant inverse? Are there several definitions of "anticommutativity", or are the above sketched situations in some way congruent? Purgy (talk) 17:05, 9 March 2018 (UTC)[reply]

Is this perhaps related to the notion of a group antihomomorphism?--Jasper Deng (talk) 18:51, 9 March 2018 (UTC)[reply]

Courtesy link: Anticommutativity. The article is a bit vague and could stand some clarification, but you really need both additive and multiplicative operations for the definition to make sense. It wouldn't necessarily have to be a ring or an algebra; I don't know what the most general structure would be called, but you\d probably want an Abelian group under + and the distributive law for *. Anticommutativity is one of the properties of a Lie algebra but it applies to quadratic forms as well. The definition is usually taken to be slightly different for characteristic 2 in order to make a distinction from commutativity. --RDBury (talk) 06:16, 10 March 2018 (UTC)[reply]

Thank you for the hints! I tried to improve on the Anticommutativity article, but without mentioning the quirks of characteristic 2. Apologies for not linking Anticommutativity beforehand, and thanks again. Purgy (talk) 15:20, 10 March 2018 (UTC)[reply]