Talk on BGG resolution

characters

let \(\lambda\in \mathfrak{h}^*\)

\[ \operatorname{ch} M({\lambda})=\frac{e^{\lambda}}{\prod_{\alpha>0}(1-e^{-\alpha})} \]

let \(\lambda\in \Lambda^+\)

thm (Weyl character formula)

\[ \operatorname{ch}L({\lambda})=\frac{\sum_{w\in W} (-1)^{\ell(w)}e^{w\cdot \lambda}}{\prod_{\alpha>0}(1-e^{-\alpha})} \]

thus we have

\[ \operatorname{ch}L(\lambda)=\sum_{w \in W}(-1)^{\ell(w)}\operatorname{ch} M(w\cdot \lambda) \label{WCF} \]

prop

If \(0\to M' \to M \to M'' \to 0\) is a short exact sequence in \(\mathcal{O}\), we have \[ \operatorname{ch}M=\operatorname{ch}M'+\operatorname{ch}M'' \] or \[ \operatorname{ch}M'-\operatorname{ch}M+\operatorname{ch}M''=0 \]

if we have a long exact sequence, we still get a similar alternating sum = 0
- why? Euler-Poincare mapping : a long exact sequence can be decomposed into short exact sequences.
- then the Euler characteristic of a finite resolution makes sense
goal : realize the alternating sum \ref{WCF} as an Euler characteristic of a suitable resolution of \(L(\lambda)\)
The BGG resolution resolves a finite-dimensional simple \(\mathfrak{g}\)-module \(L(\lambda)\) by direct sums of Verma modules indexed by weights "of the same length" in the orbit \(W\cdot \lambda\)

thm (Bernstein-Gelfand-Gelfand Resolution)

Fix \(\lambda\in \Lambda^{+}\). There is an exact sequence of Verma modules \[ 0 \to M({w_0\cdot \lambda})\to \cdots \to \bigoplus_{w\in W, \ell(w)=k}M({w\cdot \lambda})\to \cdots \to M({\lambda})\to L({\lambda})\to 0 \] where \(\ell(w)\) is the length of the Weyl group element \(w\), \(w_0\) is the Weyl group element of maximal length. Here \(\rho\) is half the sum of the positive roots.

example of BGG resolution

\(\mathfrak{sl}_2\)

\(L({\lambda})\) : irreducible highest weight module
- weights \(\lambda ,-2+\lambda ,\cdots, -\lambda\)
\(M({\lambda})\) : Verma modules
- weights \(\lambda ,-2+\lambda ,\cdots, -\lambda, -\lambda-2,\cdots\)

thm

If \(\lambda\in \Lambda^+\), the maximal submodule \(N(\lambda)\) of \(M(\lambda)\) is the sum of submodules \(M(s_i\cdot \lambda)\) for \(1\le i \le l\), where \(l\) is the rank of \(\mathfrak{g}\).

\(s_{1}(\lambda+\rho)=-\lambda-\rho\), \(s_{1}\cdot \lambda=-\lambda-2\rho\)
if we identity \(\Lambda = \mathbb{Z} \omega_1\) with \(\mathbb{Z}\), then \(\rho=1,\alpha=2\)
we have

\[L({\lambda})=M({\lambda})/M({-\lambda-2})\] or \[0\to M({-\lambda-2})\to M({\lambda})\to L({\lambda})\to 0\]

this gives a BGG resolution
character of \(L({\lambda})\) = alternating sum of characters of Verma modules

\[\operatorname{ch}{L({\lambda})}=\operatorname{ch}{M({\lambda})}-\operatorname{ch}{M({-\lambda-2})}=\frac{e^{\lambda}}{1-e^{-2}}-\frac{e^{-\lambda-2}}{1-e^{-2}}\]

comparison with Weyl-Kac character formula

\[\operatorname{ch} L({\lambda})=\frac{\sum_{w\in W} (-1)^{\ell(w)}e^{w(\lambda+\rho)}}{e^{\rho}\prod_{\alpha>0}(1-e^{-\alpha})}=\frac{e^{\lambda+1}-e^{-\lambda-1}}{e^{1}(1-e^{-2})}\]

In general, there are more terms involved in a BGG resolution and choosing right homomorphisms is not easy
we take a detour

weak BGG resolution

def

We say that \(M \in O\) has a standard filtration (also called a Verma flag) if there is a sequence of submodules

\[0 = M_0 \subset M_1 \subset M_2 \subset \cdots \subset M_n = M\] for which each \(M^i := M_i/M_{i−1}\, (1 \le i \le n)\) is isomorphic to a Verma module.

thm (Weak BGG resolution)

There is an exact sequence \[ 0 \to M({w_0\cdot \lambda}) = D_m^{\lambda} \to D_{m-1}^{\lambda} \to \cdots \to D_1^{\lambda} \to D_0^{\lambda}=M(\lambda) \to L(\lambda) \to 0 \] where \(D_{k}^{\lambda}\) has a standard filtration involving exactly once each of the Verma modules \(M(w\cdot \lambda)\) with \(\ell(w)=k\)

strategy to construct a BGG resolution

construct a relative version of standard resoultion \(D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\Lambda^{k}(\mathfrak{g}/\mathfrak{b})\) for \(L(0)\)
construct a weak BGG resolution \(D_k^0:=D_k^{\chi_{0}}\) for \(L(0)\) by cutting down to the principal block component of each term
construct a weak BGG resolution \(D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}\) for \(L(\lambda)\) (we can also do this by applying the translation functor)
show that it is actually a BGG resolution by computing \(\operatorname{Ext}\) between Verma modules

standard resolution of trivial module

free \(U(\mathfrak{g})\)-modules \(U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k}(\mathfrak{g})\)
standard resolution of trivial module in Lie algebra cohomology

\[\cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{k}(\mathfrak{g})\to U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k-1}(\mathfrak{g})\to \cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{0}(\mathfrak{g})\to L(0)\]

the sequence of modules \(D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\wedge^{k}(\mathfrak{g}/\mathfrak{b})\) is a relative version of the standard resolution
we can describe \(D_0\) and \(D_m\) explicitly
define \(U(\mathfrak{g})\)-module homomorphism \(\partial_k : D_k \to D_{k-1}\) as

\[ \begin{align} \partial_k ( u\otimes \xi_1 \wedge \cdots \wedge \xi _k): &= \sum_{i=1}^k(-1)^{i+1}(uz_i\otimes \xi_1\wedge \cdots \wedge \hat{\xi_i}\wedge \cdots \wedge\xi_k)\\ &+\sum_{1\le i<j \le k} (-1)^{i+j}(u \otimes \overline{[z_iz_j]}\wedge \xi_1\wedge \cdots \wedge \hat{\xi_i}\wedge \cdots \wedge\hat{\xi_j} \wedge \cdots \xi_k) \end{align} \] where \(z_i\in \mathfrak{g}\) is a representative of \(\xi_i\in \mathfrak{g}/\mathfrak{b}\) and \(\overline{z}\) denotes the canonical surjection \(z\in\mathfrak{g}\) into \(\mathfrak{g}/\mathfrak{b}\).

need to show that \(\partial_k\) is well-defined and it is actually a complex and exact

exactness

exactness is tricky
- see Wallach, Real Reductive Groups I 6.A
- see Knapp, Lie Groups, Lie Algebras, and Cohomology IV.6
Let \(U_j(\mathfrak{g}):=U^j(\mathfrak{g})U(\mathfrak{b})\) where \(U^j(\mathfrak{g})\) is the span of the PBW basis whose degree is \(\le j\)
note that \(U_j(\mathfrak{g})=S_j(\mathfrak{n}^-)U(\mathfrak{b})\) where \(S_j(\mathfrak{n}^-)=\sum_{0\le k\le j}S^k(\mathfrak{n}^-)\).
\(S^k(\mathfrak{n}^-)\) denote the elements of \(\operatorname{Sym}(\mathfrak{n}^-)\) that are homogeneous of degree \(k\).
\(E_{j,k}:=U_j(\mathfrak{g})\otimes_{U(\mathfrak{b})}\wedge^{k}(\mathfrak{g}/\mathfrak{b})\)
\(\{E_{j,k}\}_{j\ge 0}\) gives a filtration of \(D_k\)
let \(\partial_0 : D_0\to L(0)\) be the canonical surjection
exercise \[\partial_k : E_{j,k}\to E_{j+1,k-1}, k\ge 1\]
\(\partial_k, k\ge 1\) induces \(\overline{\partial_k} : E_{j,k}/E_{j-1,k}\to E_{j+1,k-1}/E_{j,k-1}\)

prop

For each \(j\ge 1\), \(\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{-1\le k\le m+1}\) is an exact sequence. (here \(-1\) and \(m+1\) terms are zero)

proof

As a vector space, \(E_{j,k}/E_{j-1,k}\cong S^j(\mathfrak{n}^-)\otimes \wedge^{k}(\mathfrak{n}^-)\).

Now we can apply basic results on the Koszul complexes :

theorem

For each \(j\geq 1\), the following is exact \[ 0\to S^{j-m}(V)\otimes \wedge^m(V) \to S^{j-m+1}(V)\otimes \wedge^{m-1}(V) \to \cdots \to S^{j-1}(V)\otimes \wedge^{1}(V) \to S^{j}(V)\to 0 \] where \(S^j(V)=0\) for \(j<0\)

see Lang, Algebra ' Koszul complex'

examples

\[ 0\to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0 \] or \[ 0\to E_{0,2} \to E_{1,1}/E_{0,1} \to E_{2,0}/E_{1,0} \to 0 \]

\[ 0\to \wedge^1 \to S^1 \to 0 \] or \[ 0\to E_{0,1} \to E_{1,0}/E_{0,0} \to 0 \]

in fact, we also have

\[ 0\to E_{0,0}\to L(0)\to 0 \]

prop

For each \(j\ge 1\), \(\{(E_{j,k},\partial_k)\}_{-1\le k \le m+1}\) is exact. (here \(-1\) and \(m+1\) terms are zero)

proof

Suppose \(u\in E_{j,k},\, j,k\geq 1\) satisfies \(\partial(u)=0\) in \(E_{j+1,k-1}\).

show : there exists \(v\in E_{j-1,k+1}\) such that \(\partial(v)=u\) in \[ E_{j-1,k+1} \to E_{j,k}\to E_{j+1,k-1}. \]

We look at \[ E_{j-1,k+1}/E_{j-2,k+1} \to E_{j,k}/E_{j-1,k} \to E_{j+1,k-1}/E_{j,k}. \] As \(\overline{\partial}(\overline{u})=0\), we can find \(v_1\in E_{j-1,k+1}\) such that \(\overline{\partial}(\overline{v_1})=\overline{u}\).

As \(\overline{u}-\overline{\partial}(\overline{v_1})=0\) in \(E_{j,k,}/E_{j-1,k}\), there exists \(w\in E_{j-1,k}\) such that \(u-\partial(v_1)=w\).

Now \(\partial(w)=0\) in \(E_{j,k-1}\) and by the same argument applied to \[ E_{j-2,k+1} \to E_{j-1,k}\to E_{j,k-1}, \] there exists \(v_2\in E_{j-2,k+1}\) such that \(w-\partial (v_2)\in E_{j-2,k}\), i.e. \(u-\partial(v_1)-\partial(v_2)\ \in E_{j-2,k}\).

By repeating this, we can find \(v_1,\cdots, v_j\) such that each \(v_i\in E_{j-i,k}\) and \(u-\partial(v_1)-\cdots -\partial(v_j)\ \in E_{0,k}\).

As \(\overline{\partial} : E_{0,k}\to E_{1,k-1}/E_{0,k-1}\) is injective and \(\partial(u-\partial(v_1)-\cdots -\partial(v_j))=0 \in E_{1,k-1}\), we can conclude \(u-\partial(v_1)-\cdots -\partial(v_j)=0\) or, \[ u=\partial(v_1)+\cdots +\partial(v_j). \] Thus if we set \(v:=v_1+\cdots +v_j\in E_{j-1,k+1}\) and \(\partial(v)=u\). ■

thm

\(0\to D_m \to \cdots \to D_k\to \cdots \to D_1 \to D_0 \to L(0)\to 0\) is exact.

proof

The above proposition clearly proves the exactness at \(D_k,\, k\ge 1\).

Assume that \(u\in D_0\), hence \(u\in E_{j,0}\) for some \(j\). Let \(u=u_0+u_1\) where \(u_0\) is the degree zero piece and \(u_1\) other terms in PBW basis.

If \(\partial(u)=0\), then it implies \(u_0 = 0\) as only degree zero term survives under \(\partial\).

Therefore \(j\ge 1\) and the above proposition still applies to conclude that there exists \(v\in E_{j-1,1}\) such that \(\partial(v)=u\).

This proves the exactness at \(D_0\). ■

weak BGG resolution of \(L(0)\)

goal : find standard filtrations of \(D_k\) and \(D_k^0\) and their Verma subquotients

lemma

Let \(N\) be a finite-dimensional \(U(\mathfrak{b})\)-module having a basis of weight vectors. Then \(M=U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N\) has a standard filtration and each weight of \(N\) gives a corresponding Verma subquotient.

proof

Let \(\{v_1,\cdots, v_r \}\) be a basis of \(N\) consisting of weight vectors and let \(\mu_i\) be the weight of \(v_i\).

We order the basis so that \(i\le j\) whenever \(\mu_i\le \mu_j\)

Let \(N_k\) be a space spanned by \(\{v_k,\cdots, v_r \}\) for \(1\le k \le r\).

exercise. Check that each \(N_k\) is a \(U(\mathfrak{b})\)-submodule. (hint : weight cannot decrease under \(U(\mathfrak{b})\) action)

We have a flag of \(U(\mathfrak{b})\)-modules : \[ 0 \subset N_r \subset N_{r-1} \subset \cdots \subset N_1 = N \label{Nflag} \]

We get a standard filtration of \(M\) from \ref{Nflag} as the functor \(N\mapsto U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N\) is exact. (See Remark 1.3) ■

this lemma is true even if we drop the assumption about the existence of basis of weight vectors
but such module induces a module not necessarily on the BGG category

prop

\(D_k\) has a standard filtration with Verma subquotients associated to sums of \(k\) distinct negative roots.

proof

If we apply the above lemma, enough to answer :

Q. what are the weights of \(\wedge^k (\mathfrak{g}/\mathfrak{b})\) as \(\mathfrak{b}\)-module?

A : sum of \(k\) distinct negative roots ■

prop

\(D_k^0\) has a standard filtration with Verma subquotients \(M(w\cdot 0), w\in W^{(k)}\) where \(W^{(k)}:=\{w\in W|\ell(w)=k\}\)

proof

Taking a block preserves exactness :

If \(0\to M' \to M \to M(\mu)\to 0\), then \(0\to (M')^0 \to M^0 \to (M(\mu))^0 \to 0\).

\[ (M(\mu))^0 = \begin{cases} M(\mu) , & \text{if \]\mu\( is linked to \)0\( (\)\mu=w\cdot 0\( for some \)w\in W\()}\\ 0, & \text{otherwise} \\ \end{cases} \)

Thus we obtain a standard filtration of \(D_k^0\) from that of \(D_k\).

What are the Verma subquotients or when is \(\beta\) linked to \(0\) if \(\beta\) is given a sum of \(k\) distinct negative roots?

fact \[\ell(w)=|w \Phi^+ \cap \Phi^-|\].

exercise : Let \(\beta_w:=w\cdot 0\) for each \(w\in W\). Then \(\beta_w\) is a sum of elements in \(w \Phi^+ \cap \Phi^-\).

exercise : Let \(\Pi \subset \Phi^-\) be given. If the sum \(\beta\) of elements of \(\Pi\) is \(\beta_w\) for some \(w\in W\), then \(\Pi = w \Phi^+ \cap \Phi^-\). (we know the whole set by only looking at the sum of them)

Thus \(\beta\) is a sum of \(k\) distinct negative roots and linked to \(0\) iff there exists \(w\in W^{(k)}\).

Finally,

fact \[|W\cdot 0|=|W|\]. (this implies each \(\beta_w\) is distinct)

Therefore each \(M(w\cdot 0),\, w\in W^{(k)}\) appears only in once our standard filtration.

■

thus we have found a weak BGG resolution of \(L(0)\)

weak BGG resolution of \(L(\lambda)\)

based on Remark in 6.2
let \(\lambda\in \Lambda^+\)
goal : find a standard filtration of \(D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}\) and its Verma subquotients

prop

\(D_k^0\otimes L(\lambda)\) has a standard filtration with Verma subquotients \(M(w\cdot 0 + \mu)\) where \(w\in W^{(k)}\) and \(\mu\) is a weight of \(L(\lambda)\).

proof

Use the following :

thm (3.6)

Let \(M\) be a finite dimensional \(U(\mathfrak{g})\)-module. For any \(\lambda\in \mathfrak{h}^{*}\), \(T:=M(\lambda)\otimes M\) has a standard filtration with Verma subquotients \(M(\lambda+\mu)\). Here \(\mu\) ranges over the weights of \(M\), each occurring \(\dim M_{\mu}\) times in the filtration.

Tensoring with a finite-dimensional representation is an exact functor in BGG category (thm 1.1).

If \(0\to N\to M \to M(\lambda)\to 0\), then \(0\to N\otimes L(\lambda)\to M\otimes L(\lambda) \to M(\lambda)\otimes L(\lambda)\to 0\)

Use this to construct a standard filtration on \(D_k^0 \otimes L(\lambda)\) from that of \(D_k^0\). ■

prop

\(D_k^\lambda\) has a standard filtration with Verma subquotients \(M(w\cdot \lambda),\, w\in W^{k}\).

proof

Again taking the block component for \(\chi_{\lambda}\) is exact and it gives a standard filtration of \(D_k^\lambda\) from that of \(D_k^0\otimes L(\lambda)\).

We need to determine when \(w\cdot 0 + \mu\) is linked to \(\lambda\).

exercise \[w\cdot 0 + \mu\] is linked to \(\lambda\) iff \(\mu = w\lambda\).

As \(\lambda\) is the highest weight in \(L(\lambda)\), each \(w\lambda\) is with weight multiplicity 1. Thus each \(M(w\cdot \lambda),\, w\in W^{(k)}\) appears only once in our standard filtration. Note that each \(w\cdot \lambda\) is distinct as :

fact \[|W\cdot \lambda|=|W|\] for \(\lambda\in \Lambda^+\).■

extensions of Verma modules

we have constructed a weak BGG resolution of \(L(\lambda)\) involving \(D_k^{\lambda}\)
goal \[D_k^{\lambda}\] is a direct sum of Verma modules

def

Let \(\mu, \lambda\in \mathfrak{h}^{*}\). Write \(\mu \uparrow \lambda\) if \(\mu = \lambda\) or there exists \(\alpha\in \Phi^+\) such that \(\mu=s_{\alpha}\cdot \lambda < \lambda \) (\(\mathbb{Z}^+\)-linear combination of simple roots)

We say \(\mu\) is strongly linked to \(\lambda\) if \(\mu = \lambda\) or there exist \(\alpha_1,\cdots, \alpha_r\in \Phi^+\) such that \(\mu=(s_{\alpha_1}\cdots s_{\alpha_r})\cdot \lambda \uparrow (s_{\alpha_2}\cdots s_{\alpha_r})\cdot \lambda \uparrow \cdots \uparrow ( s_{\alpha_r})\cdot \lambda \uparrow \lambda \)

def

Let \(w,w'\in W\). Write \(w'\xrightarrow{t} w\) whenever \(w = t w' \) for some reflection \(t\in W\) and \(\ell(w') < \ell(w)\). Define \(w'<w\) if there is a sequence \(w'=w_0\to w_1\to \cdots \to w_n=w\). Extend this relation to a partial ordering of \(W\) and call it the Bruhat ordering.

example : http://groupprops.subwiki.org/wiki/File:Bruhatons3.png draw the edge of only the difference of length is 1.

exercise : Let \(w\in W, \alpha \in \Phi^+\) be given. The following are equivalent :

(i) There exists a \(\lambda\in \Lambda^{+}\) such that \(s_{\alpha}\cdot (w\cdot \lambda) \uparrow w\cdot \lambda\)

(ii) \(s_{\alpha}w > w\)

hint : use

fact \[w^{-1}\alpha>0\] iff \(\ell(s_{\alpha}w)> \ell(w)\).

thm

(a) Let \(\lambda,\mu\in \mathfrak{h}^{*}\). If \(\operatorname{Ext}_{\mathcal{O}}(M(\mu),M(\lambda))\neq 0\), then \(\mu\) is strongly linked to \(\lambda\) but \(\mu \neq \lambda\)

(b) Let \(\lambda\in \Lambda^{+}\) and \(w,w'\in W\). If \(\operatorname{Ext}_{\mathcal{O}}(M(w'\cdot\lambda),M(w\cdot\lambda))\neq 0\), then \(w<w'\) in the Bruhat ordering. In particular, \(\ell(w)<\ell(w')\).

proof of (a)

(a) uses projective cover, BGG reciprocity and BGG theorem from the previous chapters. (so we skip the proof) ■

proof of (b)

From (a), we see that \(w'\cdot\lambda\) is strongly linked to \(w\cdot\lambda\). Thus we can find \( w'\cdot\lambda=(s_{\alpha_1}\cdots s_{\alpha_r})\cdot (w\cdot\lambda) \uparrow (s_{\alpha_2}\cdots s_{\alpha_r})\cdot (w\cdot\lambda) \uparrow \cdots \uparrow ( s_{\alpha_r})\cdot (w\cdot\lambda) \uparrow (w\cdot\lambda) \) From exercise, \(w'=s_{\alpha_1}\cdots s_{\alpha_r}w> \cdots > s_{\alpha_r}w > w\). ■

finish

prop

A weak BGG resolution is a BGG resolution.

proof

Use induction on the length of standard filtration.

Let \(M\subset D_k^{\lambda}\) and \(D_k^{\lambda}/M \cong M(w'\cdot \lambda)\) for some \(w'\in W^{(k)}\).

By induction hypothesis, \(0\to M=\oplus_{w\in W^{(k)},w\neq w'} M(w\cdot \lambda)\to D_k^{\lambda} \to M(w'\cdot \lambda) \to 0\), then this splits as \(\operatorname{Ext}\left(M(w'\cdot \lambda ),\oplus M(w\cdot \lambda)\right)\) is zero (ext is additive).

■

memo

proof of Thm 3.6 uses the following (we don't need this for our goal)

thm Tensor Identity (56p)

Let \(M\) be a \(U(\mathfrak{g})\)-module and \(L\) a \(U(\mathfrak{b})\)-module. Then \[ (U(\mathfrak{g})\otimes_{U(\mathfrak{b})}L)\otimes M \cong U(\mathfrak{g})\otimes_{U(\mathfrak{b})}(L \otimes M) \]

prop (3.7)

Let \(M\in \mathcal{O}\) have a standard filtration. If \(\lambda\) is maximal among the weights of \(M\), then \(M\) has a submodule isomorphic to \(M(\lambda)\) and \(M/M(\lambda)\) has a standard filtration.

overview

principal block : filtering through central characters
- is a block a \(U(\mathfrak{g})\)-submodule? yes
- how to check that it preserves the exactness : any homomorphism between modules belonging to different blocks will be zero
- how to describe \(\chi_{\lambda}\)? use the twisted Harish-Chandra homomorphism \(\psi : Z(\mathfrak{g})\to S(\mathfrak{h})\). we have

\[ \chi_{\lambda}(z)=(\lambda+\rho)(\psi(z)),\quad z\in Z(\mathfrak{g}) \]

- see 26p for an example of \(\chi_{\lambda}\) in type \(A_1\)
combinatorial results
- longest elements satisfies \(w_0\cdot 0 = -2\rho\) (related to diagram automorphism)
- consider the set of sum of k distinct roots. Which elements are linked to \(0\)?
- Bruhat ordering
Bruhat ordering and strong linkage relation
- let \(\lambda \in \Lambda^+\) (which is regular for the dot-action of \(W\))
- \(w'\cdot \lambda< w \cdot \lambda \) translates into \(w < w'\) in the Bruhat ordering
strong linkage relation and extension of Verma modules
for exterior powers, see Lie Algebras of Finite and Affine Type by Carter