MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Minimal relative Schubert modules

I am trying to better understand the definition of certain objects called minimal relative Schubert modules. My primary reference is Chapters 1 and 2 of Wilberd van der Kallen's Lectures on Frobenius splittings and B-modules.

Here is the setup: $G$ is a reductive algebraic group over an algebraically closed field $k$, $B$ is a Borel subgroup of $G$, $T$ is a maximal torus of $B$, $X(T)$ is the character group of $T$, and $W = N_G(T)/C_G(T)$ is the associated Weyl group. In van der Kallen's notes, the Borel subgroup is assumed to correspond to the set of positive roots of $T$ in $G$.

Given an element $w \in W$, one has the Bruhat cell $(BwB)/B \subset G/B$, and its closure in $G/B$ is the Schubert variety $X_w$. We also have the boundary $\partial X_w$.

Given a $B$-module $M$, one has an associated $G$-vector bundle $\mathcal{L}(M)$ over $G/B$. We can restrict the vector bundle to the Schubert varietiy $X_w$, and then consider the space $H^0(X_w,\mathcal{L}(M))$ of global sections. van der Kallen denotes this space by $H_w(M)$, and calls it a dual Joseph module.

It follows from Remark 2.2.3 and Corollary 2.2.7 in van der Kallen's notes that the functor $H_w(-)$ can be realized as a composition of induction (and restriction) functors $\text{ind}_B^{P_s}(-)$, where $P_s$ is a minimal parabolic subgroup of $G$. Specifically, if $w = s_{\alpha_1} s_{\alpha_2} \ldots s_{\alpha_t}$ is a reduced expression for $w$, then

$H_w(-) = \text{ind}_B^{P_1} \circ \text{ind}_B^{P_2} \circ \cdots \circ \text{ind}_B^{P_t}(-)$,

where $P_i = P_{\alpha_i}$ is the minimal parabolic corresponding to the simple root $\alpha_i$.

Now to the definitions I am grappling with: Let $\mu \in X(T)$, and let $w \in W$ such that $-w\mu \in X(T)_+$ (i.e., $w\mu$ is an antidominant weight). Then the dual Joseph module $P(\mu)$ is defined by $P(\mu) = H_{w^{-1}}(w\mu) = H^0(X_{w^{-1}},\mathcal{L}(w\mu))$. Assume now that $w$ was chosen to be of minimal length with the property that $-w\mu \in X(T)_+$. Then the minimal relative Schubert module $Q(\mu)$ is defined by

$Q(\mu) = \text{ker}(\text{res}: H^0(X_{w^{-1}},\mathcal{L}(w\mu)) \rightarrow H^0(\partial X_{w^{-1}},\mathcal{L}(w\mu)))$.

So $Q(\mu) \subset P(\mu)$.

Is there a way to understand $Q(\mu)$ in terms of the homological properties of the induction functors $\text{ind}_B^{P_i}(-)$ and their evaluation maps $\varepsilon:\text{ind}_B^{P_i}(M) \rightarrow M$ ?

I am not familiar or comfortable at this point with the geometric notions involved in the above definitions, but am much more comfortable with the homological properties of the induction functors, since these can be defined much more algebraically (say, as given in Chapter I.3 of Jantzen's Representations of Algebraic Groups).

-
Note that these induction functors are not the usual ones one might expect from discrete groups, or else the RHS would be infinite-dimensional. It's not an answer, but I'm pretty sure that this module was considered in one of Littelmann's path model papers. These $H^0$ for line bundles are basically Demazure modules, and the Lakshmibai-Seshadri-Littelmann model for them involve paths that move in directions $w\mu, w_1\mu, w_2\mu, ..., 1\mu$ for some chain $(w_i)$ in Bruhat order, possibly skipping some steps. The $Q(\mu)$ then have a path model where one does not skip the original $w$-step. – Allen Knutson Mar 21 2012 at 13:50
Maybe you'd like the following point of view for inductive study: $H_w \equiv \oplus_{v\leq w} Q_v(\mu)$ as $T$-representations. This is based on the fact that $\partial X_w$ is Frobenius split, so every intersection one meets (when trying to do Mobius inversion) is reduced. – Allen Knutson Mar 21 2012 at 13:53
@ Allen Knutson: Indeed in this theory induction stands for a right adjoint of restriction, not a left adjoint as one might expect from discrete groups. – Wilberd van der Kallen Mar 22 2012 at 9:32

This is not really an answer to your question -- more of a long comment, I suppose -- but there is a somewhat intuitive way to understand what these modules look like, representation-theoretically speaking. It's easier to describe what the duals to these modules look like, so let me do that. First, for any weight $\mu$, let $\mu^+$ denote the unique dominant element in the Weyl group orbit of $-\mu$ and let $V(\mu^+)$ denote the Weyl module for $G$ of highest weight $\mu^+$ (i.e., $V(\mu^+) = H^0(-w_0\mu^+)^*$). Then the Joseph module $P(\mu)^*$ is just the $B$-submodule of $V(\mu^+)$ generated by any nonzero weight vector of weight $-\mu$. (Dually, this now describes the surjection $H^0(-w_0\mu^+) \twoheadrightarrow P(\mu)$).

Now set $$\lbrace \mu_1, \ldots, \mu_r \rbrace := \lbrace s \mu : s \in W \textrm{ is a simple reflection and } s \mu < \mu \rbrace .$$ Let $I(\mu)$ be the $B$-submodule of $V(\mu^+)$ generated by $P(\mu_i)^*$, $1 \leq i \leq r$. Equivalently, $I(\mu)$ is the $B$-submodule of $V(\mu^+)$ generated by nonzero weight vectors of weights $-\mu_1, \ldots, -\mu_r$. Then $Q(\mu)^*$ fits into an exact sequence $$0 \to I(\mu) \to P(\mu)^* \to Q(\mu)^* \to 0 .$$ Remark that we can also describe $I(\mu)$ as the submodule of $V(\mu^+)$ generated by all Joseph modules properly contained in $P(\mu)^*$.

As for the evaluation map $\varepsilon$, it can be described as follows. Let $V \subseteq V(\mu^+)$ denote the highest weight subspace of weight $\mu^+$. Then $\varepsilon : P(\mu) \twoheadrightarrow k_{-\mu^+}$ is dual to the inclusion $V \hookrightarrow P(\mu)^*$ of the highest weight subspace. (Remark also that $V = P(-\mu^+)^*$).

-
Note also that in Chapter 7 Definition 7.2.1 the $Q(\lambda)$ are described in terms of iterated [induction and twist by a character], based on a reduced expression in the Weyl group. – Wilberd van der Kallen Mar 22 2012 at 9:37
@Wilberd van der Kallen: Thanks! I hadn't made it that far in the notes. That's exactly the kind of characterization I was looking for. – Christopher Drupieski Mar 22 2012 at 14:25
@George: Ah, good point. Thanks -- I've fixed that. – Chuck Hague Mar 22 2012 at 16:07