Talk:Supergroup (physics)

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Explain it!

This article should explain the claimed generalization of arbitrary groups, not just Lie groups. - 72.58.19.66 09:27, 8 May 2006 (UTC)[reply]

every group is a supergroup but not every supergroup is a group

You can not say so. What perhaps really is meant by this statement is that, a supergroup corresponds to each group in a definite way, and that not every supergroup can be obtained in this way. By the way, this does not logicaly exclude the converse. That is there may still exists some other way by which a group corresponds to each supergroup. Tamokk

I think the statement "every group is a supergroup but not every supergroup is a group" is fair enough. Rigorously speaking, ordinary groups are a full subcategory of supergroups (in a natural way). That is just as true as any other similar statement in mathematics e.g. any group is a semi-group but not vice versa, any field is a ring but not vice versa, etc. Top.Squark 20:58, 4 April 2007 (UTC)[reply]

Abstract groups have associated Hopf algebras over any ring. Reference is Group Hopf algebra.

More generally, an affine (say linear) group over a base (in algebraic geometry) is tautologically determined by the Hopf algebra structure on its algebras of regular functions over the base. For a finite (constant) groups one gets the dual of the algebra considered in Group Hopf algebra. For these objects one goes fully faithfully from groups to Hopf algebras

For infinite abstract groups (reduced 0-dimensional), and a chosen base ring, the algebra of regular functions looks like infinite product algebra. I did not check that taking the dual Hopf algebra one gets the one in Group Hopf algebra (with k[G] instead of k^G).

This accredits Hopf algebra generalises abstract groups. For continuous groups see Group algebra.

If one embeds fully faithfully vector spaces into supervector spaces as supervector spaces with even grading, it seems to embed algebra category into superalgebra and hopf algebra into hopf-superalgebras.

So group category is equivalent to full subcategory of hopf algebras and the same for hopf-superalgebras. Did not check everything, but commentary asking for reference and expert attention is irrelevant (though expert attention is still needed, as some parts, as written, looks like gibberish) http://fr.wikipedia.org/wiki/Utilisateur:RudeWolf

PS: I completely agree with Top.Squark on the Why physics issue. —Preceding unsigned comment added by 81.48.158.154 (talk) 08:20, 3 July 2008 (UTC)[reply]

Why physics?

I do not understand why this article is categorized as physics. Supergroups are a mathematical concept. They have great importance in physics, they originate from physics (like many other mathematical concepts, e.g. calculus, generalized functions, quantum groups etc.) but they are nevertheless an entirely abstract notion, independant of whatever physical model one can try to fit it in. It was perhaps true to classify it as "physics" when it was only invented and no rigorous mathematical definition was available, but not now. Top.Squark 20:58, 4 April 2007 (UTC)[reply]

Why not simply categorize it as Category\:mathematical physics? Jounce (talk) 16:37, 12 May 2017 (UTC)[reply]

Contradictory

This article states, confusingly, that supergroups are generalizations of groups yet that they also admit smooth functions which implies they are **topological** groups like Lie groups; this is contradictory. Could someone with more expertise than me correct this?

Jounce (talk) 16:37, 12 May 2017 (UTC)[reply]

The topology of a supergroup won't be locally Euclidean, so the supergroups are not manifolds. Manifolds are always locally Euclidean. The classical Lie groups are manifolds. I believe the correct mental picture is to think of the bosonic dimensions as manifolds, and the ferrmionic dimensions as having a product topology on the exterior algebra (so the product topology generates a Borel algebra of cylinder sets). Like any product topology, the sets are clopen. The get derivatives, start with the Berezin integral and go backwards. Derivatives exist to arbitrary order. So you can make things smooth. So, in some technical sense, I guess you could call it "topological" as long as you understand the topology is not Euclidean, but a weird cross-product of a cantor set and Eucliden space. (where you should think of the topology on the exterior algebra as being Cantor-like if you hold your head sideways, stand on one foot and hop up and down.) Someone should state this in the article. 67.198.37.16 (talk) 07:07, 17 November 2020 (UTC)[reply]