Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a complex Lie algebra $\mathfrak g$, we can form its universal enveloping algebra and interpret it as a noncommutative space.

Is this perspective useful? What does this space "look like"? How is it related to the space $\mathbb C[G]$ where $G$ is an algebraic group with Lie algebra $\mathfrak g$? How is it related to the flag variety (in the case of a semisimple Lie algebra)?

This question has also an incarnation which asks about group algebras.

share|improve this question
    
I've changed the tags to hopefully more useful ones (in my opinion). Feel free to roll back. –  Victor Protsak Oct 6 '10 at 6:34
    
You are right. The tags you suggest are better, thanks. –  Jan Weidner Oct 6 '10 at 7:02
1  
Off-topic and pedantic, but I feel compelled to insert my usual harrumph that algebras are not noncommutative spaces, but (can be regarded as) coordinate rings of noncommutative spaces. I notice this is what you said in your earlier question about group rings. –  Yemon Choi Oct 6 '10 at 7:17
add comment

4 Answers

You can think of $Ug$ as the algebra of distributions on $G$ supported at 1. Alternatively, you can think of $Ug$ as differential operators, for instance as global sections of monodromic differential operators on the flag variety.

"Useful" is very subjective. IMHO, differential operator interpretations are useful as it brings new insights into $Ug$ via Beilinson-Bernstein Localization. In other interpretations, info flows in the opposite direction as $Ug$ helps to understand Poisson geometry of $g*$ or geometry of $G$.

share|improve this answer
    
I really like your first description in terms of distributions. Do you by chance have a reference for where I might learn about this? –  Eric Finster Oct 5 '10 at 18:25
    
Try RAGS (Reps of Alg Groups by Jantzen). There is a chapter on distributions in the first half (sorry, don't remember the number). –  Bugs Bunny Oct 5 '10 at 19:55
    
Thanks, I also like this distributions picture, have to learn about it. –  Jan Weidner Oct 6 '10 at 7:04
add comment

Since the enveloping algebra has a filtration with corresponding associated graded algebra a symmetric algebra, you can look at $\mathcal{U}(\mathfrak g)$ as a non-commutative deformation of the ring of functions on an affine space, the dual space of $\mathfrak g$.

(Later) As observed by Ben in a comment, this is the deformation of that affine space in the direction of the Kostant-Kirillov Poisson bracket, so that the non-commutative geometry of this "space" is quite relevant to the representation theory and, in fact, already interesting by itself. In particular, this situation can be seen as an instance of what's called deformation quantization.

share|improve this answer
1  
Better yet, it's a deformation that corresponds to a well loved Poisson bracket on that space, the Kostant-Kirillov bracket. So the geometry of the universal enveloping algebra is really the geometry of coadjoint orbits. –  Ben Webster Oct 5 '10 at 15:23
add comment

This remark is just a caution for when one tries to think about what the Spec of $U\mathfrak g$ might mean.

Suppose that $\mathfrak g$ is semisimple, and let $I$ be the kernel of the natural augmentation $U\mathfrak g \to \mathbb C$. (Alternatively, $I$ is the kernel of the action of $U\mathfrak g$ on the trivial representation of $\mathfrak g$.) From the semisimplicity one finds that $I^2 = I$, and hence that $I^n = I$ for every $n \geq 1$.

One also checks (e.g. by a grading argument) that $U\mathfrak g$ is a non-commutative domain, in the sense that it has no zero divisors, and that $U\mathfrak g$ is Noetherian.

Now one easily checks that if $A$ is a commutative Noetherian domain, and if $I$ is a non-unit ideal such that $I^2 = I$, then necessarily $I = 0$. The point is that if $I = I^2$, the zero-locus of $I$ in Spec $A$ is "formally" isolated from its complement, and so morally provides a disconnection of Spec $A$ (and this moral argument is made rigorous via an application of the Krull intersection theorem). If $A$ is a domain, on the other hand, then Spec $A$ is irreducible, and so in particular connected, and so the only possibility is that $I = 0$.

So $U\mathfrak g$ has various properties that would be mutually incompatible in a commutative ring, and hence one has to be careful in importing geometric intuition naively in thinking about some kind of non-commutative Spec $U\mathfrak g$.

(Of course, the other answers here give a very non-naive discussion of various geometric points of view on $U\mathfrak g$, but I hope that the above remark will be useful to some people. Let me note that it is also somewhat related to this answer.)

share|improve this answer
    
Thanks for your nice answer Emerton! After interpreting the group algebra as $pt/G$ the "point like" properties of $U\mathfrak g$ don't come very surprising to me. –  Jan Weidner Oct 6 '10 at 7:01
2  
Another crucial difference is that primitive ideals need not be maximal, i.e. there are proper inclusions between primitive ideals, or "geometric" points need not be closed! Cf mathoverflow.net/questions/33161/… –  Victor Protsak Oct 6 '10 at 14:53
add comment

Orbit method As Mariano and Ben have already mentioned, $U(\mathfrak{g})$ quantizes coadjoint orbits of $\mathfrak{g}.$

In general, for a real algebraic group $G$ with Lie algebra $\mathfrak{g},$ the following three spaces are closely related:

  1. The space of primitive ideals of $U(\mathfrak{g}).$

  2. The set of isomorphism classes of irreducible unitary representations of $G.$

  3. The set of coadjoint orbits of $G.$

The original result is due to Kirillov and states that if $G$ is nilpotent and simply-connected then spaces 2 and 3 are in a natural bijection. Later this was extended in various directions by Auslander and Kostant, Gabriel, Borho and Rentschler, Duflo, and many others. For the case $\mathfrak{g}=\mathfrak{gl}_n,$ the Jacobson topology on the space of primitive ideals was determined by Oshima in the Advances Math paper. It is related to the topology on the space consisting of Zariski closures of the conjugacy classes of $n\times n$ matrices, but there are some subtle differences.


Relation to flag variety In the case when $G$ is a compact semisimple Lie group, the coadjoint orbit $\mathcal{O}_\lambda=G\cdot\lambda$ is equivariantly isomorphic to generalized flag variety $G_{\mathbb{C}}/P_\lambda;$ moreover, this isomorphism is compatible with symplectic structures and line bundles. Here $\lambda$ is an integral weight (after natural identification between $\mathfrak{g}$ and its dual) and when $\lambda$ is a regular integral dominant weight, $P_\lambda=B,$ so that the coadjoint orbit $\mathcal{O}_\lambda$ may be identified with the full flag variety $G_{\mathbb{C}}/B$ polarized by the line bundle $\mathcal{L}_\lambda.$

share|improve this answer
add comment

Your Answer

 
discard

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.