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 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

1. construct a relative version of standard resoultion $D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\Lambda^{k}(\mathfrak{g}/\mathfrak{b})$ for $\lambda=0$
2. 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
3. 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)
4. 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 $U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{k}(\mathfrak{g})$
• standard resolution of trivial module in Lie algebra cohomology

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

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

$$ \begin{align} \partial_k ( u\otimes \xi_1 \wedge \cdots \xi _k): &= \sum_{i=1}^k(-1)^{i+1}(uz_i\otimes \xi_1\wedge \cdots \hat{\xi_i}\wedge \cdots \xi_k)\\ &+\sum_{1\le i<j \le k} (-1)^{i+j}(u \otimes \overline{[z_iz_j]}\wedge \xi_1\wedge \cdots \hat{\xi_i}\wedge \cdots \hat{\xi_j} \cdots \xi_k) \end{align} $$

• need to show that it is actually a complex and exact
  • see Wallach, Real Reductive Groups I 6.A
  • see Knapp, Lie Groups, Lie Algebras, and Cohomology IV.6

weak BGG resolution of $L(0)$

• goal : find sstandard filtrations of $D_k$ and $D_k^0$

lemma

Let $N$ be a finite-dimensional $U(\mathfrak{b})$-module. 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 $\mu_i$ is a weight of $v_i$.

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

Let $N_k=\langle v_k,\cdots, v_r \rangle$ for $1\le k \le r$.

exercise. Check that each $N_k$ is a $U(\mathfrak{b})$-submodule.

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

Define $M=U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N$.

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) ■

prop

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

proof

Apply the above lemma.

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 $V(w\cdot 0), w\in W^{(k)}$ where $W^{(k)}:=\{w\in W|\ell(w)=k\}$

proof

Taking the principal block preserves exactness.

If we apply it to $0\to M' \to M \to M(\mu)\to 0$, then $0\to (M')^0 \to M^0 \to (M(\mu))^0 \to 0$.

$(M(\mu))^0$ is either 0 or $M(\mu)$ depending on whether $\mu$ is linked to $0$, i.e. $\mu=w\cdot 0$

We obtain a standard filtration of $D_k^0$ from that of $D_k$.

What are the Verma subquotients or which $\beta$ is linked to $0$ among $\beta$'s 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 $w \Phi^+ \cap \Phi^-$.

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

■

• thus we have found the principal block $D_k^0$ and Verma modules in its standard filtration

weak BGG resolution of $L(\lambda)$

• based on Remark in 6.2
• study $D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}$

claim

$D_k^0\otimes L(\lambda)$ has a standard filtration with Verma subquotients $V(w\cdot 0 + \mu)$ 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 $\lambda\in \mathfrak{h}^{*}$, the tensor product $T:=M(\lambda)\otimes M$ has a finite filtration with quotients isomorphic to Verma modules of the form $M(\lambda+\mu)$. 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).

In other words, $0\to N\to M \to M(\lambda)\to 0$ implies $0\to N\otimes L(\lambda)\to M\otimes L(\lambda) \to M(\lambda)\otimes L(\lambda)\to 0$

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

claim

$D_k^\lambda$ has a standard filtration.

proof

Taking the principal block does no harm in constructing standard filtration. ■

• first, find all Verma modules involved in a standard filtration of $D_k^0\otimes L(\lambda)$ : those associated to $w\cdot 0 + \mu$
• then determine which Verma modules are linked to $w\cdot \lambda$ : Ans. $\mu = w\lambda$
• as $\lambda$ is the highest weight in $L(\lambda)$ and thus with weight multiplicity 1, we also know that the Verma modules appear only once

extensions of Verma modules

• $\mu, \lambda\in \mathfrak{h}^{*}$
• $\mu \uparrow \lambda$ if $\mu = \lambda$ or there is a root $\alpha$ such that $\mu=s_{\alpha}\cdot \lambda < \lambda $
• we say $\mu$ is strongly linked to $\lambda$ if $\mu = \lambda$ or there exist root $\alpha_1,\cdots, \alpha_r\in \Phi^+$ such that $\mu=(s_{\alpha_1}\cdots s_{\alpha_r})\cdot \lambda \uparrow (s_{\alpha_2}\cdots s_{\alpha_r})\cdot \lambda \uparrow ( s_{\alpha_r})\cdot \lambda \uparrow \cdots \uparrow \lambda $

def (Bruhat ordering)

Define a partial order on the elements of $W$ as follows :

Write $w'\to w$ whenever $w = w' t$ for some reflection $t$ 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 $W$.

thm

Let $\lambda\in \mathfrak{h}^{*}$.

(a) If $\operatorname{Ext}_{\mathcal{O}}(M(\mu),M(\lambda))\neq 0$ for $\mu\in \mathfrak{h}^{*}$, then $\mu \uparrow \lambda$ 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

(a) uses BGG reciprocity and BGG theorem.

finish the construction of BGG resolution

• Now use induction on the length of standard filtration
• if we have $0\to \oplus M_{w\cdot \lambda}\to D_k^{\lambda} \to M(w'\cdot \lambda) \to 0$, then as $\operatorname{Ext}\left(M(w'\cdot \lambda ),\oplus M(w\cdot \lambda)\right)$ is zero since it is additive. Then we obtain a split extension.

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