Let $\lambda$ be a dominant integral weight of $\mathfrak g$, a finite-dimensional reductive Lie algebra over $\mathbb C$. Let $M(w\cdot \lambda)$ denote the Verma module with high weight $w\cdot \lambda := w(\lambda+\rho)-\rho$. Then there are known inclusions $M(w\cdot \lambda) \hookrightarrow M(\lambda)$, for each $w$, and for this question we identify them with their images.

Let $I_w := \{v \in W\ :\ v\geq w\}$ denote the principal upward order ideal in $W$. Does the assignment $I_w \to M(w\cdot \lambda)$ extend to a map of lattices, taking upward order ideals to submodules? EDIT: More concretely, is $$ M(w\cdot\lambda) \cap M(v\cdot\lambda) = \sum_{x\geq w,v} M(x\cdot\lambda),$$ or is the intersection strictly bigger? I believe this should follow from some geometric construction like "take $I$ to the $\mathcal D$-module of $B$-finite distributions on $G/B$ supported on $\bigcup_{i\in I} X_i$, twisted by $\mathcal L_\lambda$" but would rather quote a reference than hash out the geometric details.

Also, for each $w$ there is a quotient module $ M(w\cdot \lambda) \big/ \sum_{v > w} M(v\cdot \lambda)$, which further quotients to the irrep $L(w\cdot \lambda)$. I gather it was these that Verma erroneously thought were irreducible.

Is there a Kazhdan-Lusztig-polynomial way to compute the irreps in a composition series for this module? The standard use computes the multiplicity of $L(v\cdot \lambda)$ inside $M(\lambda)$ as a whole; my question is about how that multiplicity is distributed over the various $w\geq v$.

Of course references would be most appreciated.

  • $\begingroup$ I think that the lattice of submodules is isomorphic to $W$ ordered by the Bruhat order and that this is not the same in general as $\{I_w\}$ ordered by inclusion. $\endgroup$ Nov 14, 2014 at 7:40
  • $\begingroup$ $W$ isn't usually a lattice, so is not isomorphic to the lattice of submodules. $\endgroup$ Nov 14, 2014 at 13:42
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    $\begingroup$ Ah yes, $W/W_P$ is a lattice if $G/P$ is cominuscule, as I recall. $\endgroup$ Nov 14, 2014 at 18:25
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    $\begingroup$ @Allen: In this generality it's difficult to work out the module structures in much detail, as indicated already in Jantzen's LN 750 (1979), Chap. V, 5.16+. As early as 1972 Conze-Dixmier showed that a Verma module might have infinitely many submodules in spite of having finite length. When some composition factors have high multiplicity, your second question looks very hard. Maybe Ron Irving's papers in the late 1980s would be useful to consult, though some of his notes never got published. $\endgroup$ Nov 14, 2014 at 18:30
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    $\begingroup$ @Andre: These further maps (for example to a minimal submodule in the quotient) won't as a rule lift to maps of Verma modules. But nonzero maps between Verma modules are embeddings, and the Hom space has dimension 1 (see the early chapters of my textbook). $\endgroup$ Nov 16, 2014 at 19:02

1 Answer 1


Contrary to Verma's initial impression, the internal structure of a "Verma module" (Dixmier's terminology) tends to be extremely complicated. However, this complexity only shows up in ranks 3 or higher, by which time it's not easy to exhibit complete details in most examples. (Concerning Verma's subtle error in his thesis, see the overview in section 4.14 of my 2008 AMS Graduate Studies volume Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.)

Throughout the 1970s there was considerable activity by BGG, Jantzen, and others, who made serious progress in understanding the submodule structure. But the definitive conjecture, very soon proved, came from Kazhdan and Lusztig in their 1979 Inventiones paper. The "localization" approach of Beilinson-Bernstein to the proof was developed further by them in order to prove the earlier Jantzen conjecture on how composition factors are placed in layers of the Jantzen filtration of a Verma module.

All of this gets arbitrarily complicated combinatorially, even in type $A_n$. In particular, the estimate you give in your large shaded box for the behavior of the intersection of two Verma submodules is probably over-optimistic in most cases. The problem is that when composition factors appear with multiplicity greater than 1 (which typically happens), the extra factors occur "higher" in the structure than the corresponding Verma submodule. This is how the interpretation of coefficients of the Kazhdan-Lusztig polynomials fits with the Jantzen filtration.

Probably it's worth looking closely at the first difficult case $\mathfrak{sl}_4$, where the Weyl group $S_4$ has order 24 but there are typically 26 composition factors in a Verma module with dominant integral highest weight. Chapter 8 in my book outlines the general theory in detail, with plenty of references, and 8.6 discusses some of the data for type $C_3$ as an example. A "typical" Verma module in higher rank has a great many composition factors (with multiplicities which grow rapidly); the internal structure is almost impossible to work out explicitly when there are infinitely many submodules.

That said, your follow-up question in a shaded box might be more amenable to an answer involving the Kazhdan-Lusztig polynomial interpretation of the Jantzen filtration. This probably isn't straightforward to carry out in general, but a closer look at the rank 3 and 4 examples would be a good starting point. (The old tables of polynomials here computed by Mark Goresky might be helpful.)

UPDATED SUMMARY: The answer to the first question is almost certainly no, though it might take a large amount of computation to pin down a specific counterexample (probably in high rank). On the other hand, the second question seems to have a positive answer, but again the computations involved may be onerous.

To streamline notation, start with a weight $\lambda$ which is dominant regular (relative to the shifted dot-action of $W$ having origin $-\rho$). Take $\mu, \nu$ to be strongly linked subweights, with $\mu = w \cdot \lambda$ and $\nu = v \cdot \lambda$. So $M(\lambda)$ includes unique copies of $M(\mu)$ and $M(\nu)$, which intersect in a submodule $M$. It's clear that $M$ includes the sum of all Verma submodules $M(\pi)$ such that $\pi \leq \mu, \nu$ is strongly linked to both. The first question is whether $M$ must equal this sum.

Here the highest composition factors $\mu, \nu$ of those two Verma modules occur on certain layers of the Jantzen filtration of $M(\lambda)$, while the Jantzen layers of $M(\mu), M(\nu)$ embed in corresponding layers of $M(\lambda)$ determined by length differences in $W$ (Jantzen conjecture, proved by Beilinson-Bernstein). Say $M(\pi) \subset M$, with $\pi := u \cdot \lambda$. It may well happen that one or both of $M(\mu), M(\nu)$ has extra composition factors of type $L(\pi)$ (with some multiplicity). Necessarily such factors $L(\pi)$ occur in higher layers of the Jantzen filtration of $M(\mu)$ and/or $M(\nu)$. This happens just when a KL polynomial $P_{u,w}(q)$ or $P_{u,v}(q)$ has a nonzero coefficient of degree $>0$. Polo has shown that these polynomials (for type $A_n$, when $W = S_{n+1}$) can have arbitrarily high degrees if $n$ s large enough. In this situation $M$ will involve one or more composition factors isomorphic to $L(\pi)$ besides the single copy contained in the sum of Verma submodules.

At the same time, the ideas just sketched indicate that a detailed knowledge of coefficients of the relevant KL polynomials will be enough to determine the composition factor multiplicities in any quotient of a Verma module by the sum of its canonical Verma submodules. But again the required recursive computations may become quite lengthy, and no closed formula is likely.

  • $\begingroup$ Stroppel's 2003 paper "Category $\mathcal O$: quivers and endomorphism rings of projectives" math.uni-bonn.de/ag/stroppel/Quivers.pdf has the structure of regular $\mathfrak{sl}_4$ Vermas in the Appendix. She informs me that one can use it to verify that the intersections are indeed the sums, but like everyone else I've asked, is unwilling to believe that this will persist past type $A_9$ or so. $\endgroup$ Jan 21, 2015 at 21:45
  • $\begingroup$ @Allen: Yes, the lower rank type A examples are often misleadingly well-behaved. Best to be skeptical. $\endgroup$ Jan 21, 2015 at 22:43
  • $\begingroup$ @Allen: There are several fine points to add if you want a real algorithm, but my edited answer is already getting too long. $\endgroup$ Jan 23, 2015 at 13:56

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