Talk:Alternating algebra

Anticommutative

Is "anticommutative ring" = "graded-commutative ring"? The answer is yes if "anticommutative" = "skew-commutative". -- Taku (talk) 22:57, 21 December 2016 (UTC)[reply]

Can you give more context to why you're posting this here?

I think it is worth pointing out that anticommutative/skew-commutative seems to assume a Z₂-grading, whereas graded-commutative seems to assume a Z-grading, the latter being a more stringent criterion. (Every Z-grading induces/implies a Z₂-grading, but not the other way around.) Under this characterization, "anticommutative ring" ≠ "graded-commutative ring", because the latter will be a proper subset of the former. See my comment at Talk:Supercommutative_algebra#Supercommutative_vs._anticommutative. However, if "graded-commutative" assumes only that the grading monoid has Z₂ as a quotient monoid (and hence implies a Z₂-grading), then "anticommutative ring" = "graded-commutative ring". That article is not clear about it, however. —Quondum 23:44, 21 December 2016 (UTC)[reply]

I asked it because I think "anticommuattive ring" here should really be "graded-commutative algebra". I do agree we need to distinguish Z-graded algebra and Z/2-graded ones, basically because Z is not Z/2. Strictly speakingly, graded rings with different indexing sets need to be distinguished, especially the questions about the definitions. In practice, context makes it clear what is meant. I think, at least in mathematics, unqualified "graded ring" means either N or Z-graded rings and not Z/2-graded. While anticommutative algebra considers Z/2-algebra, graded-commutative ring considers a Z-graded algebra. So the latter seems like a better target. -- Taku (talk) 02:52, 22 December 2016 (UTC)[reply]

I agree with you (i.e. the suggested change to the link target), though there are so many subtleties. Notice though that a ring is not necessarily an algebra, and an alternating algebra is not necessarily a ring (not necessarily associative, but then, neither is a graded ring). I wish terminology was more consistent ... —Quondum 03:07, 22 December 2016 (UTC)[reply]

Simpler definition?

If someone has a reference that gives an intuitively simpler definition of an alternating algebra, this would be nice to add to the article. For example, something like "An alternating algebra is a $Z$ -graded algebra in which the homogeneous elements of degree 1 square to zero and which generate the entire algebra." I'm not sure that this is even correct (e.g. it may fall apart for non-associative algebras), but it should give the idea. —Quondum 22:03, 15 July 2017 (UTC)[reply]

This characterization should simplify by removing the reference to grading:

An alternating algebra A is an algebra for which there exists a generating set S ⊆ A such that x ∈ S implies x² = 0.
A skew-commutative algebra A is an algebra for which there exists a generating set S ⊆ A such that x,y ∈ S implies xy + yx = 0.

This should work generally, i.e. with nonunital nonassociative algebras. (Note to self: Bourbaki gives Proposition 13, which is very close to, but not identical my first characterization above; I'll add this at some point.) —Quondum 15:17, 30 July 2017 (UTC)[reply]

Note on possible merge

Serge Lang uses the name "alternating algebra" as a direct synonym for "exterior algebra". I have not seen anything to suggest that Bourbaki's definition of an alternating algebra is not equivalent to that of an exterior algebra. Unless there is a difference, it would seem appropriate to merge this article into Exterior algebra. —Quondum 15:32, 12 February 2021 (UTC)[reply]

The exterior algebra is a very specific construction that starts with vector spaces, and vector spaces remain central to the description and vocabulary. By contrast, here we start with a graded ring, where the grading is that of the abelian subgroups of the ring. Now, my gut sense intuition is that you can always organize these kinds of abelian subgroups into vector spaces, at least when the characteristic of the field is zero. But I don't know the name of that theorem or lemma. (It might not even be true.) If the characteristic of the field is two, it might not be true. At any rate, the "flavor" of graded rings and of vector spaces is very different to me. and I don't know enough about scheme (mathematics) to hand wave and say "oh yeah, these two, its all the same thing". It's not obviously so.

That said, we need a better compare-n-contrast of this and the supercommutative algebra. 67.198.37.16 (talk) 19:30, 24 May 2024 (UTC)[reply]