The finite-dimensional representations of a complex semisimple Lie algebra $\frak{g}$ are well known to be classifiable by their highest weight vectors, giving a convenient countable indexing set. I know that the situation for infinite dimensional representations is significantly more difficult, but am somewhat confused by the state of the art. Here I will ask a simpler question: For the special case of $\frak{sl}_2(\mathbb{C})$ are its infinite dimensional representations classified? If they are not is there some hope that they can be classified or is the problem too wild? If it is not too wild can the classifying set be expected to be countable or not? To be clear, I am interested in answers for the special cases of highest weight $\frak{sl}_2(\mathbb{C})$ modules, irreducible $\frak{sl}_2(\mathbb{C})$-modules, and arbitrary (anything goes) $\frak{sl}_2(\mathbb{C})$-modules.
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4$\begingroup$ One possible answer is the Beilinson-Bernstein localisation theorem. Others much more knowledgable than me will weigh in here but let me just point to arxiv.org/pdf/2002.01540.pdf which works out some very concrete examples for ${\frak sl}_2$ (both complex and real). $\endgroup$– BalazsCommented Mar 12, 2023 at 14:19
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5$\begingroup$ @ L Spice maybe I'm misunderstanding you but it is not the case that the irreps are finite dimensional. Indeed a Verma corresponding to non- (integral dominant) is irreducible isn't it? $\endgroup$– user108998Commented Mar 12, 2023 at 15:11
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3$\begingroup$ Yes. For a specific example, consider the representation on rational functions in two variables, as the Lie algebra generated by $x\partial/\partial y$ and $y\partial/\partial x$. The submodule generated by, say, $1/x$ is irreducible and infinitedimensional: it is spanned by $y^{k-1}/x^k$, $k=1,2,3,...$. $\endgroup$– მამუკა ჯიბლაძეCommented Mar 12, 2023 at 15:17
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2$\begingroup$ @EBz, hmm, as foretold, I embarrassed myself. I had in mind the analogies: representations of complex groups and Lie algebras might as well be algebraic; and representations of algebraic groups are locally finite, so irreducible ones are finite dimensional. I guess the (a?) wrong step in my reasoning is that the representations of the Lie algebra that you describe don't integrate to the group (and, in particular, their weights need not be preserved by the Weyl group), right? $\endgroup$– LSpiceCommented Mar 12, 2023 at 15:30
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1$\begingroup$ @მამუკაჯიბლაძე, re$\DeclareMathOperator\SL{SL}$, a cyclic representation need not be irreducible. In a representation of the algebraic group $\SL_2$, a la Jantzen, it will always be finite dimensional. I now see that there are representations of $\SL_2(\mathbb C)$ that are not algebraic in this sense; you have given one such. But your original representation was instead of $\mathfrak{sl}_2(\mathbb C)$. $\endgroup$– LSpiceCommented Mar 13, 2023 at 18:09
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1 Answer
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This is completely answered in Nonabelian Harmonic Analysis by Roger Howe and Eng-Chye Tan, Springer-Verlag 1992.
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1$\begingroup$ Thanks for the reference! I don't have access to the book. Might you be able to say you say in a few words what it says? Are there uncountably many irreps, for example? $\endgroup$ Commented Mar 14, 2023 at 10:35
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$\begingroup$ @LászlóSzabados Here's a shorter text maybe useful: www-users.cse.umn.edu/~garrett/m/v/intertwinings_SL2C.pdf or just google 'principal series'. In short there is a family of unitary irreps parameterized by a continuum. $\endgroup$– Peter WuCommented Mar 15, 2023 at 5:10
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$\begingroup$ You need some assumption on the representations: the standard one is that the diagonal Cartan subalgebra acts semisimply. Then an irreducible representation has weight spaces of dimension one. The representation can be finite dimensional; have weights bounded below; weights bounded above; or infinite in both directions. There is a continuous parameter. In the book they even classify the indecomposable representations (not a direct sum). The calculations are purely algebraic and not difficult. A key role is played by the Casimir operator (action of center of the universal enveloping algebra). $\endgroup$ Commented Mar 16, 2023 at 1:35
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$\begingroup$ @Jeffrey: What happens if the Cartan does not act semisimple. Does the problem then become "wild". What are examples of such representations? $\endgroup$ Commented Mar 16, 2023 at 11:27