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Let $R$ be a (noncommutative) ring. (For me, the words "ring" and "algebra" are isomorphic, and all rings are associative with unit, and usually noncommutative.) Then I think I know what "linear algebra in characteristic $R$" should be: it should be the study of the category $R\text{-bimod}$ of $(R,R)$-bimodules. For example, an $R$-algebra on the one hand is a ring $S$ with a ring map $R \to S$. But this is the same as a ring object in the $R\text{-bimod}$. When $R$ is a field, we recover the usual linear algebra over $R$; in particular, when $R = \mathbb Z/p$, we recover linear algebra in characteristic $p$.

Suppose that $G$ is an algebraic group (or perhaps I mean "group scheme", and maybe I should say "over $\mathbb Z$"); then my understanding is that for any commutative ring $R$ we have a notion of $G(R)$, which is the group $G$ with coefficients in $R$. (Probably there are some subtleties and modifications to what I just said.)

My question: What is the right notion of an algebraic group "in characteristic $R$"?

It's certainly a bit funny. For example, it's reasonable to want $GL(1,R)$ to consist of all invertible elements in $R$. On the other hand, in $R\text{-bimod}$, the group $\text{Aut}(R,R)$ consists of invertible elements in the center $Z(R)$.

Incidentally, I'm much more interested in how the definitions must be modified to accommodate noncommutativity than in how they must be modified to accommodate non-invertibility. So I'm happy to set $R = \mathbb H$, the skew field of quaternions. Or $R = \mathbb K[[x,y]]$, where $\mathbb K$ is a field and $x,y$ are noncommuting formal variables.

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I'm curious: why do you use the word "characteristic" here? I would expect terminology to make sense when you pass to a special case (e.g., commutative rings), but here it does not. –  S. Carnahan Jan 17 '10 at 19:06

6 Answers 6

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It seems that you want some notions on noncommutative group scheme,right? In fact, A.Rosenberg has introduced noncommutative group scheme in his work with Kontsevich "noncommutative grassmannian and related constructions" (2008). Actually, this work gave a systematically treatment to the noncommutative grassmannian type space introduced in their early paper noncommutative smooth space and the work of Rosenberg himself on noncommutative spaces and schemes.

More comments: It seems that you want to know the linear algebra over noncommutative ring. I think you need to look at the paper by Gelfand and Retakh on Quasideterminants, I. And the main motivation for the "noncommutative grassmannian and related constructions" is to give a geometric explanation to the work of Gelfand and Retakh.

All of these work are based on functor of point of view.

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I was unable to find a general definition of noncommutative group scheme in the paper you linked. They only defined a noncommutative version of GL(V). –  S. Carnahan Jan 17 '10 at 21:10
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Sorry to response late. I just asked Kontsevich about this notion, he told me that in this paper I linked, they did not give the definition of general noncommutative group scheme. However, I remembered Rosenberg ever give a lecture course introducing general notions. I will reply when I got back to campus and found out what it is –  Shizhuo Zhang Jan 19 '10 at 1:49
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Maybe you can suggest to Kontsevich to check out mathoverflow :) –  Kevin H. Lin Jan 19 '10 at 8:36

Sorry for arriving here so late... I hope somebody will still notice my answer:

A possible definition would be to take a co-group in an appropriate category of noncommutative rings (or algebras), i.e. an object in such a category of noncommutative rings or algebras that represents a functor from this category into a category of groups (see also S. Carnahan's answer).

I know of several papers that follow this approach and study such co-groups, here is what I have been able to find:

Six papers and one book over a span of almost fifty years, citing each other rather sparsely (my list is certainly not complete, though). Compared to the vast literature on algebraic groups and group schemes, this does not seem to be a lot.

Question: Is there an explanation why noncommutative algebraic group theory (in the sense of the OP, maybe one should say noncocommutative?) is getting so little attention? E.g., lack of applications, technical difficulties, lack of interesting results?

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I'd like to add that there is an interesting paper http://arxiv.org/abs/math/0701399 that discusses Lie algebras and groups over noncommutative rings.

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The functoriality in this article is in the wrong direction; while it touches on some interesting algebra it is definitely not a canonical answer. –  Zoran Skoda Mar 8 '10 at 15:16
    
I agree it is not a canonical answer, and I am not sure there is a canonical answer at all, but I did not understand what you mean by functoriality being in the wrong direction. Can you please specify what statement in the article you are referring to? –  Pavel Etingof Mar 8 '10 at 22:21

I'd say an affine algebraic group over $R$ is an $R$-Hopf algebra, that is, a Hopf algebra object in the category of R-R bimodules. Further than that, it's hard for me to say.

[EDIT: This bit doesn't make any sense. Ignore it. I was up until 7am doing Mystery Hunt, so at least I have a good excuse.] I am pretty suspicious of a definition in terms of the functor of points; the whole problem with non-commutative geometry is that the points don't capture nearly enough information.

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But what is a Hopf algebra object in the category of R-R-bimodules? I thought Hopf algebra objects only make sense in symmetric (or at least braided) monoidal categories, as one needs to permute the factors in order to write down the axiom of compatibility between the multiplication and the comultiplication. –  Leonid Positselski Jan 17 '10 at 18:58
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@Ben Webster, the functor of points is the description by the Yoneda embedding, of course. It has all of the information available in the original representation plus more given to us by the noncommutative generalization of a grothendieck topology. –  Harry Gindi Jan 17 '10 at 19:23
    
@Leonid- My recollection is that the problem can be gotten around if one writes the axiom a bit differently, but we may need someone more expert than me in Hopfological algebra to be sure. –  Ben Webster Jan 17 '10 at 20:24
    
Given that Hopf algebras are Koszul dual to E_2 algebras, they shouldn't make sense in anything more than braided probably.. –  David Ben-Zvi Jan 18 '10 at 20:55

There is more than one category of noncommutative spaces of algebraic flavour, hence there is more than one notion of the algebraic group. In affine case, notice that the categorical product noncommutative affine schemes $NAff=Ass^{op}$ is opposite to the free product of the corresponding rings. There are extremely few such schemes, and they correspond to algebra which are very close to the free associative algebras (cf. I. Berstein, On cogroups in the category of graded algebras. Trans. Amer. Math. Soc. 115 (1965), 257–269)-- the example of $NGL_n$ like in Kontsevich-Rosenberg article mentioned by Zhang is just one of the few interesting examples. One can try not to work with categorical product, and work with tensor product like in some approaches to linear quantum groups (B. Parshall and J.Wang, Quantum linear groups. Mem. Amer. Math. Soc. 89 (1991), No. 439, vi+157 pp.), however then some categorical construction do not pass. However if we represent a space by the category of quasicoherent sheaves, then a group scheme is represented by a monoidal category, namely (up to various properness/finiteness conditions) the monoidal product is given by taking the external tensor product of sheaves on the group $G$ what gives a product on $G \times G$ and then one pushes doen this categorical product along the action to $G$. Similar pushdown along the action induces the action of this monoidal category of sheaves on the aprpopriate category of sheaves on the space the group acts on. Then in noncommutative case, we can replace Hopf algebra by its monoidal category of modules, and this category acts on the category of modules over any comodule algebra over that Hopf algebra in a canonical way. This way in the world of categories one indeed has actions of monoidal categories, which are in addition geometrically admissible in the sense explained in my article

Zoran Škoda, Some equivariant constructions in noncommutative algebraic geometry, Georgian Mathematical Journal 16 (2009), No. 1, 183–202, arXiv:0811.4770.

While Kontsevich-Rosenberg treatment of $NGL_n$ is nice functorialy (unfortunately the main part of the work from 1999 is still not a publically avaiulable article) and it was originallz motivated bz Gel'fand-Retakh quasideterminants this motivation is not fully justified by the results: namely various identities of quasideterminants were not explained as geometric statements about various maps of noncommutative schemes. There is another approach which I develop for a number of years and I hope to be able to finish and write down soon is by taking another version of $NGL_n$ namely the Manin's example of the Hopf envelope of the free matrix bialgebra on $n^2$-generators. This Hopf algebra has infinitely many generators and has interesting structure. There is a geometric quotient which I call a universal noncommutative flag variety. I succeeded to get some of the identities for quasideterminants as geometric statements on that variety. This variety is not a noncommutative scheme, but sort of noncommutative homotopy scheme as the descent is higher descent for *Cohn localization*s which do not have good flatness properties needed for the usual descent. On the other hand this variety is not represented by a group-valued functor on $NAff$ unlike the noncommutative flag variety of Kontsevich-Rosenberg (which is also glued using Cohn localizations as I was told).

Tomasz Maszczyk has his own approach to noncommutative group schemes (mainly unpublished) which emphasises on the categories of bimodules. But you should talk to him.

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You seem to be asking two different questions. The first is, "how do I define the notion of algebraic group over a noncommutative ring?" The second is, "given an algebraic group (viewed as a functor from commutative rings to groups), how do I evaluate it on noncommutative rings?" My answers are probably naive, but I don't understand noncommutative geometry.

First question: It should be a functor from rings to groups that preserves finite limits. You may need more conditions, but this is essentially what you get from the definition of formal groups by removing the "commutative Artinian" condition.

Second question: Evaluate the functor on the center of the ring. I can't think of a canonical alternative. Edit: Based on the helpful comments, I'd recommend evaluating the functor on the quotient by the two-sided ideal generated by commutators.

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Second answer: that sounds pretty terrible. Remember "center" is NOT a functor from noncommutative rings to commutative rings. –  Ben Webster Jan 17 '10 at 20:26
    
Are there any functors from rings to commutative rings that restrict to the identity on commutative rings? It seems unlikely, but I don't have a proof of nonexistence. –  S. Carnahan Jan 17 '10 at 21:12
    
What about associating to a ring R the new ring R/I, where I is the two-sided ideal generated by all elements of the form (ab-ba)? Since a ring homomorphism sends I of one ring to I of another, this gives a functor from rings to commutative rings, which is clearly the identity on commutative rings. Note that it sends most easy examples of non-commutative rings to the zero ring. –  Chris Schommer-Pries Jan 18 '10 at 2:32
    
Yeah, that's what I meant with the Hochschild homology statement. I probably should have deleted the previous comment. –  S. Carnahan Jan 18 '10 at 2:49
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I don't think Hochschild homology of a noncommutative ring is a ring in general.. it's a quotient by all commutators, not the two-sided ideal they generate –  David Ben-Zvi Jan 18 '10 at 20:53

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