Let $\mathfrak{g}$ be a semisimple Lie algebra and fix a root system. Let $\mathfrak{b}:=\mathfrak{h}\oplus\bigoplus_{\alpha\in R^+}\mathfrak{g}_\alpha$. The complex irreducible representation of $\mathfrak{g}$ with highest weight $\lambda \in P^+$ can be obtained via the following procedure:

  1. Form the Verma module $V_\lambda := U(\mathfrak{g})\otimes_{U(\mathfrak{b})} k_\lambda$, where $k_\lambda$ is a 1-dimensional representation of $\mathfrak{b}$. Here $h\in\mathfrak{h}$ acts by $\lambda(h)$ on $k_\lambda$ and the positive roots act trivially.
  2. Take the quotient of $V_\lambda$ by the maximal submodule $M_\lambda \subset V_\lambda$ to get the representation.

Now let $G$ be a connected semisimple algebraic group over $\mathbb{C}$. if I understood correctly, the Borel–Weil theorem states that irreducible representations of $G$ with highest weight $\lambda$ can be obtained as follows:

  1. Take the negative Borel subgroup $B^- \subset G$ and form the equivariant line bundle $L_\lambda := G\times_{B^-}\mathbb{C_\lambda}$, for $\mathbb{C}_\lambda$ the $B^-$-representation pulled back from $T = B^-/U^-$.
  2. Take $\Gamma_{\text{hol}}(L_\lambda) \subset \Gamma(L_\lambda)$ to get the representation.

It's hard to not notice the similarities between these constructions: both are "induced" from the corresponding representation of $\mathfrak{h}$, or $T$. However, the first construction is applied to the positive Borel subalgebra, while the second is applied to the negative Borel subgroup.

Is there a more precise relation between the two constructions, and is there an explanation for this discrepancy of choice of Borel subalgebra/subgroup?

  • $\begingroup$ It may be better to think of $G/B^-$ as the variety of all Borel subgroups. Any $T$ fixed point $B$ on this variety will do. If one has such a fixed point and an equivariant line bundle, it becomes a puzzle to compare representations of $G$ and $B$, where $B$ acts on the fiber. $\endgroup$ Sep 19, 2022 at 7:25

1 Answer 1


I don't think the $\pm$ issue is too deep, and I'm punting on it in favor of answering the other question.

You can get a hold of dual Verma modules by considering distributions on $G/B$ supported on a $B$-orbit. Note that $G$ will, naturally, not act on this space, but $B$ and $\mathfrak g$ will. If the $B$-orbit is the open dense one, then this space of distributions is simply functions, or, sections of a line bundle (trivialized) over the big cell. This $(\mathfrak g,B)$-module contains a minimal submodule (dual to the Verma quotient you asked about), which is the sections that extend over the whole $G/B$.

The big theorem relating these (and other) pictures is the Beilinson-Bernstein localization theorem.


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