2
$\begingroup$

Reposting my question from math.stackexchange: What is the relationship between Verma modules and delta functions? Here's the quote from Woit's notes on Lie theory (http://www.math.columbia.edu/~woit/LieGroups-2012/vermamodules.pdf):

The subject of "geometric representation theory" relates Verma modules to the geometry of the flag variety $G/B$. $B$ acts on the left, with a finite number $\lvert W \rvert$ of $B$-orbits (this is the Bruhat decomposition). For $G = SL(2;\mathbb C)$, $G/B = \mathbb CP_1$ and there are two $B$-orbits: the Riemann sphere minus the South pole, and the South pole, of dimension 2 and 0 respectively. Verma modules correspond to delta-function distributions on the $B$-orbits, dual Verma modules to holomorphic functions on the orbits, singular at the boundary.

Is there an analogue of meromorphic functions in this setting? Nevanlinna theory, perhaps? (I ask of Nevanlinna theory to, perhaps, make connections with Vojta's dictionary of some sort)

$\endgroup$
1
  • 3
    $\begingroup$ This seems to be a restatement of a theorem of Beilinson and Bernstein that gives an equivalence of categories between monodromic $D$-modules on the flag variety and representations of $G$. You can write generating elements of the $D$-modules as distributions corresponding to highest-weight vectors. $\endgroup$
    – S. Carnahan
    Dec 3, 2014 at 10:50

1 Answer 1

1
$\begingroup$

In general cases Verma modules are corresponding to D-modules on flag variety and then via Riemann-Hilbert correspondence to constructible sheaves on flag variety. Hence the suitable generalization of delta functions should be constructible sheaves.

As S. Carnathan pointed out, this is the standard topic of geometric representation theory. There are lots of references. For an out line you can see, for example, James E. Humphreys' book "Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$", Section 8.9, or the reference given in Woit's note.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.