MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone tell me what's the appalication of the Rosenberg' method on the definition of the noncommutative algebraic geometry (left spectrum)?
Now when talking about noncommutative algebraic geometry, we will follow the way of the category theory (abelian category, triangulated category and so on). What's the obstruction of the direct way from the algebraic geometry rather than the category theory? For example, have someone studied or founded the theorem like Riemann-Roch theorem on the noncommutative schemes?

share|cite|improve this question
Your question is very vague. Also, what have you read about Rosenberg's method? Even on mathoverflow there are various good answers which explain it. – Martin Brandenburg Oct 17 '11 at 10:38
At the risk of being accused of self-promotion, if you're interested in obstruction results for noncommutative spectra, you can find one here: – Manny Reyes Oct 17 '11 at 13:26
As noncommutative spectra have structure stacks (and by evidence of plethora of other phenomena like the importance of bimodule morphisms) one should not expect 1-categorical functoriality, but at least a generalization like 2-categorical (pseudo)functoriality... – Zoran Skoda Oct 17 '11 at 14:30
Your last sentence is discussed here:… or in brief, yes, in many settings. – B. Bischof Oct 17 '11 at 16:19
The answer to your first question can be found in Xin Tang's thesis, of which I do not have a link on hand(I'm typing on a phone). In short, the left spectrum was developed for purposes of application to representation theory. A natural place to look for a reference is the work of Rosenberg. In particular, his book from '95 uses the spectrum to consider representations of a certain class of algebras(those he refers to as hyperbolic algebras). Of course looking at the n-lab page on Rosenberg will provide you with many of his manuscripts of which many will answer you also. – B. Bischof Oct 17 '11 at 16:30

Quick answer (to the second part): noncommutative rings don't have enough ideals to make a decent spectrum. In words of Fred van Oystaeyen: "it doesn't matter how you try to define what is a point of a noncommutative space, you never have enough of them". There were some attempts of following more classical lines in the late seventies and early-mid eighties (see papers by Van Oystaeyen, Verschoren, and many others) but eventually everybody agreed that a more abstract approach was needed to get meaningful geometric information.

share|cite|improve this answer
what means "decent spectrum" ? – Alexander Chervov Oct 17 '11 at 11:23
@Alexander: "decent" is a synonyme for "nice", not a mathematical notion. – Johannes Hahn Oct 17 '11 at 11:54
He possibly meant faithful enough to reconstruct the scheme back and specializing to commutative one for commutative ring. These are usual minimal requirements. – Zoran Skoda Oct 17 '11 at 12:31
As Zoran says, by "decent" I mean nice enough to allow us to jump back and forth between the ring and the spectrum. In the classical situation if $R$ is a (commutative) ring and $X = Spec(R)$ then one can recover $R$ as the ring of global sections of the structure sheaf. In the noncommutative case, for instance if $R$ is the Weyl algebra $k<x,y>/(yx-xy-1)$ the spectrum is reduced to a single point and the global sections are constant functions. – javier Oct 17 '11 at 14:27
Weyl algebra is good example to make me doubt in "... you never have enough of them". Recall this famous result by Hollands (?) Berest Wilson - you classify projective modules over Weyl and you get exactly C^2 - the same amount as in the commutative limit of Weyl e.g. C[x,y]... Moreover you event get Hilbert scheme i.e. glued points... How to be with this ? – Alexander Chervov Oct 17 '11 at 17:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.