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I would like to know a complete description of the indecomposable representations of the two dimensional non-abelian Lie algebra over the complex numbers. The finite dimensional representations would also be interesting for me.

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    $\begingroup$ Is there a reason why you are interested in this question? $\endgroup$ Oct 18, 2014 at 22:19
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    $\begingroup$ The way you handle the finite dimensional case is as follows. Let $V$ be an indecomposable representation, and $x,y\in\mathrm{End}(V)$ be the corresponding operators, $xy-yx=y$. Rewriting that as $xy=y(x+1)$, we immediately see that $y$ must be nilpotent. Indeed, we have $xy^k=y^k(x+k)$, so if $x(v)=\lambda v$ then $x(y^k v)=(\lambda+k)y^k v$, and a linear operator on a finite dimensional vector space cannot have infinitely many eigenvalues. Now, assume $y$ in Jordan normal form, and compute possible $x$. It seems not too hard to decide which of the obtained modules are indecomposable. $\endgroup$ Oct 19, 2014 at 2:15
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    $\begingroup$ I tried to follow Vladimir's recipe, since it does sound easy and I was curious to see where the wildness lies. Indeed, computing the normalizer algebra $\mathfrak{n}(y)$ of a nilpotent operator $y$ in canonical form is not difficult. But now we need to see which give equivalent representations, which means to separate the adjoint orbits of $N(y)$ in $\mathfrak{n}(y)$. I don't see a way to proceed uniformly here, so I guess this is it. $\endgroup$ Oct 20, 2014 at 15:42
  • $\begingroup$ An alternative argument for Vladimir's observation is that Lie's theorem implies that the only simple finite dimensional modules are one dimensional, together with the immediate fact that $y$ acts by zero on any one-dimensional module. $\endgroup$ Oct 20, 2014 at 17:37

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A further comment on Pasha's answer, in community-wiki mode:

The Russian article cited by Pasha (which doesn't yet have free access online in Russian) is referred to later that year in the same journal: E.A. Makedonskii, On wild and tame finite-dimensional Lie algebras. (Russian) Funktsional. Anal. i Prilozhen. 47 (2013), no. 4, 30–44; translation in Funct. Anal. Appl. 47 (2013), no. 4, 271–283.

This later paper suggests an answer to the question raised, but only in the limited finite dimensional setting. According to the MathSciNet review by V.V. Gorbatsevich, the only finite dimensional Lie algebras over an algebraically closed field of characteristic 0 for which the classification of finite dimensional representations is not a wild problem are the semisimple ones, or direct sums of semisimple ones and one-dimensional centers.

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I guess it follows as a very particular case, from the works about wild/tame dichotomy, representations of quivers, etc., that in characteristic zero the problem of classification of finite-dimensional indecomposable modules over a two-dimensional nonabelian Lie algebra, is wild. For example, it can be inferred from:

V.L. Ostrovskyi and Yu.S. Samoilenko, On pairs of quadratically related operators, Funct. Anal. Appl. 47 (2013), N1, 67-71,

where the problem of classification of pairs of linear operators A,B with [A,B] = polynomial (A) is reduced to some wild quiver problem.

On the other hand, the dimensions of restricted finite-dimensional indecmposable modules are bounded, as follows, for example, from:

R.D. Pollack, Restricted Lie algebras of bounded type, Bull. Amer. Math. Soc. 74 (1968), 326-331 DOI:10.1090/S0002-9904-1968-11943-3 ,

so they are probably, manageable. I do not know what the situation with arbitrary modules in positive characteristic is.

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    $\begingroup$ In the question, the Lie algebra is taken over $\mathbb{C}$. But it's natural to ask about any algebraically closed field of characteristic 0, maybe even about prime characteristic. However, Zassenhaus showed (1954) that all irreducible (not necessarily restricted) representations of a restricted f.d. Lie algebra in characteristic $p>0$ are of finite dimension bounded in terms of the given Lie algebra. As in the original question, one has to wonder why one would look for all indecomposables here. $\endgroup$ Oct 20, 2014 at 0:31
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Thank you very much to Pasha Zusmanovich for pointing out the reference Ostrovskyi, Samoilenko. (by the way I would like to vote his answer but I do not know how to do it, it is my first visit to mathoverflow). I gave a look to their paper and I can write down their answer.

They never write down the definition of wild but I think they use the following definition (for me the definition is new so I hope that what I write is correct)

A k algebra A is wild if the category of finite dimensional representation of A contains a subcategory equivalent to the category of representations of the free associative algebras in two generators over k.

If I understand correctly this implies that it contains a subcategory isomorphic to the category of the representations of any finitely generated k-algebra.

Also I do not understand if this definition of wild really implies that there is no hope for a classification. A priori it seems to me that a larger category could be simpler than the smaller one.

This is the argument of Ostrovskyi, Samoilenko:

Theorem: the category whose object are given by two vector spaces U,V and linear maps A:U-->U, B:U-->V, C:V-->V with AB=BC, and A^n=C^n=0 with n \geq 5 (and the natural definition of morphism) is wild.

This is the category of representations of the algebra: with five generators A,B,C,I,J with I^2=I, J^2=J, AI=IA=A, JC=CJ=C, BI=B=JB, AB=BC, IB=BJ=IJ=JI=IC=CI=JA=AJ=0 and B invertible

They do not prove this theorem, they refer to a paper of Hoshino and Miyachi (Tsukuba J. of Math. 1988) and to a paper of Han (J.of algebra 2002).

The claim that the representations of the algebra [x,y]=y is wild follows easily from the theorem: given a triple (A,B,C) consider representations of this algebra where x and y acts as 2x2 block matrices given by

x=(1+A 0 \ 0 C)

y=(0 B \ 0 0)

Then [x,y]=y is equivalent to AB=BC and if the representation defined by x',y' is isomorphic to that defined by x,y is equivalent to require that the representation of the algebra of the theorem above associated to (A,B,C) is equivalent to that defined by (A',B',C')

I gave a look to the proof of Theorem, however in these papers they have a general approach which for me is not easy to follow (you end up to prove that the affine quiver of type E_7 is wild). Maybe would be possible for these specific algebra to produce a more direct argument (notice however that according to their table if you require A^3=C^3=0, instead that A^5=C^5=0, this should not be wild)

thank you again

Andrea

PS: My interest in the problem does not come from any particular reason. I just tried to give a look to the smallest possible example. I thought it would have been simpler.

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    $\begingroup$ I'll let it stand this time, but please be aware that answer boxes should be reserved for precise answers to the posted question. You can always comment on anything you post yourself and on answers to your own posts, and once you have 50 points of MO reputation you can comment on any post. $\endgroup$
    – Todd Trimble
    Oct 28, 2014 at 15:28
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Quoth Dixmier, Enveloping algebras, p. xii: "But a deeper study reveals the existence of an enormous number of irreducible representations of [the 3-dimensional Heisenberg algebra]. It seems that these representations defy classification. A similar phenomenon exists for $\mathfrak g = \mathfrak{sl}(2)$, and most certainly for all non-commutative Lie algebras."

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    $\begingroup$ It is strange that he doesn't explicitly mention the smallest solvable non-abelian Lie algebra, though $\endgroup$
    – Yemon Choi
    Oct 19, 2014 at 2:13
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    $\begingroup$ Interestingly enough, this quotation appears in sciencedirect.com/science/article/pii/000187088190058X, an article which somehow shows that this question (classifying irreducible representations) is not as hopeless. The algebra that the OP is interested in is also featured in that article. $\endgroup$ Oct 19, 2014 at 2:33
  • $\begingroup$ (I heard a talk by an expert saying that very recently this has been done for sl2; I'll ask for a reference to check if I remember correctly) $\endgroup$ Oct 19, 2014 at 7:52
  • $\begingroup$ @MarianoSuárez-Alvarez: could the reference be books.google.ie/books?id=i6M2FQP3mIcC ? $\endgroup$ Oct 19, 2014 at 9:23
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    $\begingroup$ Note that the old paper by Richard Block deals mainly with the irreducible representations (mostly infinite dimensional) and not the problem of finding all indecomposables, while the notion of "classification" is a bit slippery. The article seems to have open access now: The irreducible representations of the Lie algebra $\mathfrak{sl}(2)$ and of the Weyl algebra, Adv. in Math. 39 (1981), no. 1, 69–110. $\endgroup$ Oct 19, 2014 at 15:07

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