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Well, the question is probably a rather basic one but I haven't been able to find the answer in literature or come up with it myself so here we go.

Do there exist irreducible representations of the Lie algebra $\mathfrak g$ which aren't a highest weight module with respect to some Borel subalgebra in the case when

a) $\mathfrak g$ is a complex finite dimensional semisimple algebra,

b) $\mathfrak g=\mathfrak{gl}_n(\mathbb C)$?

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    $\begingroup$ The derived representations of principal series give examples. There isn't really a classy algebraic construction, though. A post-hoc construction is available from wikipedia: en.wikipedia.org/wiki/… $\endgroup$
    – B R
    Commented Apr 23, 2012 at 14:50
  • $\begingroup$ Edit: "A post-hoc construction for ${\mathfrak sl}(2,\mathbb C)$". A similar construction can be done in general, but it seems like a pain. $\endgroup$
    – B R
    Commented Apr 23, 2012 at 14:54
  • $\begingroup$ 2BR. Thank you! This pretty much answers my question. I'd still appreciate a more detailed description if anyone's got the time. $\endgroup$
    – Igor Makhlin
    Commented Apr 23, 2012 at 15:06
  • $\begingroup$ Here's a simple construction: given a weight $\lambda$ of $\mathfrak h$, the principal series is the subrepresentation of $\mathfrak k$-finite vectors of ${\rm Hom}_{U({\mathfrak h})}\big(U({\mathfrak g}),{\mathbb C} _\lambda)\big)$. See Dixmier's "Enveloping Algebras", Section 9.3. $\endgroup$
    – B R
    Commented Apr 23, 2012 at 15:44
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    $\begingroup$ (I thought the agreement was to act as if all reps were highest weight modules: «the first rule of rep. theory is, you don't talk about...») $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 23, 2012 at 18:16

6 Answers 6

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Here's a more specific reference for the rank 1 simple Lie algebra; the "classification" by Block shows clearly how sparse the highest weight representations are among all irreducible representations here. Block was basically responding to the remarks of Dixmier and others to the effect that no "classification" would be possible, but of course his treatment is not very concrete.

Also, the close connection between irreducible representations of a general linear Lie algebra and its derived (special linear) algebra make it obvious how close the answers to (a) and (b) are.

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I think it's very far from true, and misleading to state too informally, that every representation is highest weight. For example if we think of representations (with a fixed infinitesimal character) geometrically on the flag manifold (on $P^1$ for $SL_2$), then highest weight modules with respect to a unipotent subgroup $N$ are locally constant (or rather have some central curvature) along the corresponding Schubert stratification (ie for SL2 they are attached geometrically to the stratification of $P^1$ by $A^1$ and a point), while a representation can have arbitrary "shape" - ie we can fix any stratification we want. (Note the finite dimensional representations, which as was stated are all highest weight, are attached to the stratification of the flag variety by just one piece - by Borel-Weil they arise from line bundles, which definitely satisfy this "shape requirement".) In particular the representations that people care most about tend to be Harish Chandra modules, which can be realized geometrically as stratified with respect to the (finitely many) orbits of a symmetric subgroup of $G$ - for instance, the admissible representations of $SL2R$ correspond to the stratification of $P^1$ by the two poles and $C^\times$, and most of these are evidently not highest weight.

What is true, and very useful and important (some might say the most fundamental statement about g-modules, but that's not a debate I want to start), is that representations DO live on the flag manifold - in other words, they are made out of "points" of the flag manifold, which correspond to representations that are highest weight (Verma modules). So the sense in which representations relate to highest weight modules is just that they can be realized geometrically on the flag manifold ($P^1$).

Technically this is captured by a version of Casselman's submodule theorem (reps of real groups live inside principal series) due to Beilinson-Bernstein: for every representation there IS an N so that the representation has nonvanishing N-coinvariants (ie not highest weight vectors but a form of highest weight covectors). Such N give the support of the corresponding picture of the representation on the flag manifold.

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  • $\begingroup$ This is a really attractive perspective. Do you have a resource or reference where I can learn more about this perspective on building (not necessarily highest weight) representations associated to more general stratifications of the flag variety? $\endgroup$
    – Marc Besson
    Commented Aug 5, 2020 at 19:55
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If you restrict yourself to finite-dimensional representations, the answer is NO. This is discussed in most thorough Lie theory textbooks, and is a standard step in some constructions of the algebraic groups associated to the semisimple Lie algebras. The textbook I have closest at hand is the draft Lie Groups book available on my website, based on notes from Mark Haiman's 2008 class; details are in Sections 5 and 6. The basic idea is that the raising operators have all eigenvalues equal to zero, and so on any finite-dimensional representation must act nilpotently, and so there is a highest weight.

In infinite dimensions, the story is much richer. In general, what people actually care about are things that can be considered representations of the group $G$ integrating $\mathfrak{g}$. The correct notion of such representations was found by Harish-Chandra. Details for $\mathrm{SL}(2,\mathbb R)$ are in Section 7.6 of loc. cit., among many places. If you just want representations of the Lie algebra $\mathfrak{g}$, then in general these won't "integrate" to any representation of the group, and are a bit easier to come by. For example, consider the case $\mathfrak{sl}(2)$ or $\mathfrak{gl}(2)$. These act on $\mathbb C^2$ in the usual way. On functions, we have $E = x\partial_y$, $F = y \partial_x$, and $H = x\partial_x - y\partial_y$. Consider the module of "functions" spanned by monomials of the form $(xy^{-1})^n x^{1/3}$. I think that this is an irrep of $\mathfrak{sl}(2)$, but I haven't checked.

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  • $\begingroup$ Seems to me it's not an irrep: $n=0$ provides vector which doesn't generate the whole module. It's actually a highest weight vector, which are not to be contained in the kind of irrep I'm looking for. $\endgroup$
    – Igor Makhlin
    Commented Apr 23, 2012 at 20:10
  • $\begingroup$ @Igor: Right, I wasn't thinking straight. Anyway, the idea is that you can build modules for $\mathfrak{sl}(2)$ as chains, and then tailor the various parameters. One parameter is the action of the Casimir, valued in $\mathbb C$. The other is valued in $\mathbb C / \mathbb Z$ and is the value of $H$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 24, 2012 at 0:22
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    $\begingroup$ Ok, second try. How about polynomials times $x^{1/3}y^{1/5}$? Note that if $f$ is a "function" and $Ef = 0$, then $f(x,y) = f(x)$ is independent of $x$. The game now is just to force that never to happen. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 24, 2012 at 0:25
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This is covered in Is there a machinery describing all the irreducible representations ?

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    $\begingroup$ @Bruce: Thanks for calling this to everyone's attention. Even though the question asked there was technically somewhat different from this one, the underlying theme is the same. It's a natural question about MO whether it will tend to recycle more and more such questions. To search past entries on MO often takes longer than answering the question from scratch. I wasn't aware of the 2010 question but was prepared to answer the question and had no strong incentive to waste time searching long lists of previous questions in case something might be relevant here. $\endgroup$
    – Jim Humphreys
    Commented Apr 24, 2012 at 0:32
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My understanding is that "formally" there are many irreps which are not highest weight (e.g. unitary principal series, Whitakker modules), however "informally" it is not true (highest weight vector is always nearby one need just to open eyes:).

The point is how do we think of representation space "V" - if we think it abstractly as some abstract vector space - okay - can be no highest weight vector. However if we think of $V$ as some subset of functions of some variables (x_1,...,x_n), which satisfy some condition (e.g. square integrable or whatever), then we can "open eyes" and find the highest weight vector - the subtlety is that it may not obey the "condition", so it does not belong $V$, but still it corresponds to some function $f(x_1,...,x_n)$.

Example.

Take $sl_2$ and its standard representation by differential operators (sorry for some misprints):

e = d/dx

h = xd/dx - l x

f = x^2d/dx - l^2x

1) Consider C[x] - then it is highest weight module with "1" as an highest weight vector.

2) Consider quasi-polynoms space {f(x) = p(x)exp(m x) } This is actually a Whittaker module wiht character "m", and function exp(m x) is Whittaker vector i.e. it is eigen for "e".

3) Consider square integrable functions with appropriate scalar product and condition on $l$. You can get unitary principal series representation in this way.

So to my taste all these representations are essentially equivalent and they have highest weight vector "1", but "formally" it is of course not true.

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a) No.
b) Consider determinant map. More generally every irrep of gln is a tensor product of an irrep of sln and (integer) power of the determinant.

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    $\begingroup$ Are you assuming the representations are finite dimensional? $\endgroup$
    – B R
    Commented Apr 23, 2012 at 14:51
  • $\begingroup$ a) Well, yeah. And why exactly? =) b) Since I'm considering Lie algebras, I assume you're talking about the trace representation: $$g(v)=k\mathrm{Tr}(g)v, k\in\mathbb C.$$ Then your statement (without the part about $k$ being an integer) is obviously true for the irreps which are highest-weight modules. Can't say why it holds for any other irrep though. $\endgroup$
    – Igor Makhlin
    Commented Apr 23, 2012 at 14:54
  • $\begingroup$ 2BR. Oh, yes, that would explain the misunderstanding. Just in case: in my question I'm not assuming the irreps to be finite dimensional. Otherwise the answer is "of course not", AFAIU. $\endgroup$
    – Igor Makhlin
    Commented Apr 23, 2012 at 14:58
  • $\begingroup$ Igor, that's good to know. By the way, you are correct that the answer is "of course not" for f.d. reps. $\endgroup$
    – B R
    Commented Apr 23, 2012 at 15:07
  • $\begingroup$ Bienvenue Monsieur Swiat! $\endgroup$
    – Alain Valette
    Commented Apr 23, 2012 at 15:38

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