Is there a machinery describing all the irreducible representations ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:57:58Z http://mathoverflow.net/feeds/question/18559 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18559/is-there-a-machinery-describing-all-the-irreducible-representations Is there a machinery describing all the irreducible representations ? Peter Lee 2010-03-18T08:14:01Z 2010-03-18T18:47:12Z <p>Suppose we have a finite dimensional Lie algebra $g$， Is there a machinery to describe all the irreducible representation of $g$. </p> <p>Consider toy example: $sl_{2}$ or $sl_{3}$, how do we describe all the irreducible representations of them. </p> <p>Further, consider quantum case, Is there a machinery way(like algorithm)describing all the irreducible representations of $U_{q}(sl_{2})$</p> <p>EDIT: What I am looking for is an "mechanical" and canonical machinery describing <strong>all</strong> the irreducible representations(of course, not only finite dimensional representations,not only unitary representations)</p> <p>EDIT2: What I am looking for is some reference to describe them in explicitly(such as $sl_{3}$）</p> http://mathoverflow.net/questions/18559/is-there-a-machinery-describing-all-the-irreducible-representations/18564#18564 Answer by Bruce Westbury for Is there a machinery describing all the irreducible representations ? Bruce Westbury 2010-03-18T09:38:45Z 2010-03-18T13:25:13Z <p>The short answer is no. There is a classification of primitive ideals in the enveloping algebra (and quantised enveloping algebra). This reduces the problem to primitive rings. However the representation theory of primitive rings which are not Artinian is complicated. </p> <p>An example which I find easier is the Weyl algebra (or linear differential operators). This ring is primitive since the vector space of polynomials is an irreducible faithful representation. This ring is in fact simple (no proper ideal). However the representation theory encompasses the theory of linear differential equations with polynomial coefficients.</p> <p>So speaking heuristically, the representation theory of semisimple Lie algebras is at least as complicated as the representation theory of the Weyl algebra and it is unreasonable to expect an answer in this case.</p> <p>I don't know of a formal result that says this is an unreasonable request. For example: does this problem include the problem of classifying indecomposable representations of a wild algebra?</p> <p><strong>Edit</strong> I have just found this reference which solves the question for $sl(2)$.</p> <p>MR0605353 (83c:17010) Block, Richard E. The irreducible representations of the Lie algebra $sl(2)$ and of the Weyl algebra. Adv. in Math. 39 (1981), no. 1, 69--110.</p> http://mathoverflow.net/questions/18559/is-there-a-machinery-describing-all-the-irreducible-representations/18620#18620 Answer by Torsten Ekedahl for Is there a machinery describing all the irreducible representations ? Torsten Ekedahl 2010-03-18T16:20:02Z 2010-03-18T18:47:12Z <p>The problem of classifying irreducible $sl_2(\mathbb C)$-representations is essentially untractable as it contains a wild subproblem. Indeed, the action of the Casimir element $C$ on any irreducible representation is by a complex scalar (by a theorem of Quillen I believe). If we consider the case when $C$ acts by zero, by a result of Beilinson-Bernstein the category of $sl_2$-representations with $C=0$ is equivalent to the category of quasi-coherent $\mathcal D_{\mathbf P^1}$-modules. In this $1$-dimensional case every irreducible $\mathcal D_{\mathbf P^1}$-module is holonomic. If we restrict ourselves to irreducible regular holonomic modules we have two possibilities. One case is that they are supported at a single point and then the point is a complete invariant. In the other case they are classified by a finite collection of points of $\mathbf P^1$ and equivalence classes of irreducible representation of the fundamental group of the complement of the points which map the monodromy elements of the points non-trivially. In particular we can consider the case of three points in which case the fundamental group is free on two generators (they and the inverse of their product being the three monodromy elements). The irreducible representations where one of the monodromy elements act trivially correspond to removing the corresponding point and thinking of the representation as a reprentation of the fundamental group of that complement. </p> <p>Hence, we can embed the category of finite-dimensional representations of the free group on two elements as a full subcategory closed under kernels and cokernels of the category of $sl_2(\mathbb C)$-modules. This makes the latter category wild in the technical sense. However, the irreducible representations of the free group on two letters are also more or less unclassifiable.</p> <p>There is no contradiction between this and the result of Block. His result gives essentially a classification of irreducibles in terms of equivalence classes of irreducible polynomials in a twisted polynomial ring over $\mathbb C$. So the consequence is that such polynomials are essentially unclassifiable.</p> <p>[Added] Intractable depends on your point of view. As an algebraic geometer I agree with Mumford making (lighthearted) fun of representation theorists that think that wild problems are intractable. After all we have a perfectly sensible moduli space (in the case of irreducible representations) or moduli stack (in the general case). One should not try to "understand" the points of an algebraic variety but instead try to understand the variety geometrically. Today, I think that this view point has been absorbed to a large degree by representation theory.</p>