Talk:Superalgebra

Removed diagrams

I've removed the following material from the article as I'm not sure what to call these diagrams. I've heard them called tangle diagrams but they don't seem to have any relation to the notion discussed there. Also, I personally don't find these diagrams very illuminating.

Removed contet

An associative superalgebra (or Z₂-graded associative algebra) is one whose product is associative. Category theoretically, this means the commutative diagram expressing associativity commutes. Principal examples are Clifford algebras.

A supercommutative algebra is a superalgebra satisfying a graded version of commutivity. Category theoretically, $\nabla$ and $\nabla \circ \tau _{A,A}$ commute. The primary example being the exterior algebra on a vector space.

File:Commutative.png

A Lie superalgebra is nonassociative superalgebra which is the graded version of an ordinary Lie algebra. The product map is written as $[\cdot ,\cdot ]$ instead. Category theoretically, $[\cdot ,\cdot ]\circ (id+\tau _{A,A})=0$ and $[\cdot ,\cdot ]\circ ([\cdot ,\cdot ]\otimes id)\circ (id+\sigma +\sigma ^{2})=0$ where σ is the cyclic permutation braiding $(id\otimes \tau _{A,A})\circ (\tau _{A,A}\otimes id)$.

-- Fropuff 08:13, 7 February 2006 (UTC)

Superrings or superalgebras over a commutative ring

Is just started an article on superrings, but this strikes me as unnecessarily redundant—a superring can be defined simply as a superalgebra over the (purely even) ring of integers. I can't see that there is really anything to be gained by having a separated article. It would be nice to just merge the content here.

But as this article stands now, superalgebras are defined only over a (purely even) field K. This is, of course, an unnecessary restriction. It is just as easy to define a superalgebra over a (purely even) commutative ring R. One can go further, and define superalgebras over a commutative superring. Although, in order to avoid circular definitions, it is probably best to stick to the purely even case at the outset and mention generalizations later on. What do people think about relaxing the definition here and treating the superring case here as well? -- Fropuff (talk) 21:46, 8 February 2008 (UTC)

No comments, so I will use my best judgment and relax the definition here to include more general superalgebras. I'll then redirect superring here. Comments still welcome. -- Fropuff (talk) 04:34, 12 February 2008 (UTC)

Grade involution

Isn't grade involution an antiautomorphism as opposed to an automorphism? Molitorppd22 (talk) 17:01, 14 January 2010 (UTC)

Doesn't explain what's unusual about this algebra

There should be an example of the typical non-zero value assigned to the "supercommutator"; because if it's zero, that's just the classic way of commuting differential forms.

166.137.90.34 (talk) 10:26, 11 April 2015 (UTC)Collin237

Bingo! why, it is exactly that! Its just the classic way of commuting differential forms. Unless you are a physicist, in which case you say that it is exactly the classic way of commuting fermions. (follows from the Pauli exclusion principle). Here's the deal: a differential form assumes some underlying manifold, and some underlying derivative. But here, we've completely chucked that manifold, and the derivative out the window. There are no manifolds, no derivatives, no differential forms (and no fermions). Just the raw commutator remains, and nothing else.

Oh I just re-read what you wrote. Perhaps you wanted to be pointed at Clifford algebra or Jordan algebra or even Lie algebra. These have non-trivial (anti-) commutators. They can be used as ingredients to construct a superalgebra, much like you'd construct a supersalad at a lunch counter - a bit of this, a bit of that ... 67.198.37.16 (talk) 17:55, 28 September 2016 (UTC)