Talk on BGG resolution

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

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
• 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
• \(M({\lambda})\) : Verma modules
  • note that the Verma modules are free modules of rank 1 over \(\mathbb{C}[F]\) where $F$ is the annihilation operator of $\mathfrak{sl}_2$
• \(\lambda ,-2+\lambda ,\cdots, -\lambda, -\lambda-2,\cdots\)
• \(L({\lambda})=M({\lambda})/M({-\lambda-2})\)
• BGG resolution

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

• number of modules = 2 (=order of Weyl group in general)
• character of $L({\lambda})$ = alternating sum of characters of Verma modules

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

• comparison with Weyl-Kac character formula

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

$\mathfrak{sl}_3$

• http://www.math.columbia.edu/~woit/LieGroups-2012/vermamodules.pdf

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 of the proof

• construct a relative version of standard resoultion $D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\Lambda^{k}(\mathfrak{g}/\mathfrak{b})$ for $\lambda=0$
• construct a weak BGG resolution $D_k^0:=D_k^{\chi_{0}}$ for $\lambda=0$ 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 $\lambda\in \Lambda^+$ 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 $U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{k}(\mathfrak{g})$
• standard resolution of trivial module

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

• 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 of the trivial module in Lie algebra cohomology
• we can describe $D_0$ and $D_m$ explicitly
• define $\partial_k : D_k \to D_{k-1}$ as

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

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

weights of exterior powers

• Q. what are the weights of $\wedge^k (\mathfrak{g}/\mathfrak{b})$ as $\mathfrak{b}$-module? A :
• let us find Verma modules in a standard filtration of $D_k$

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.

• note that $D_k^0$ now has a standard filtration as taking the principal block preserves exactness
  • if we apply it to $0\to M' \to M \to M(\mu)\to 0$, $0\to (M')^0 \to M^0 \to (M(\mu))^0 \to 0$.
  • the Verma quotient becomes either 0 or itself
• we want to find the relevant Verma modules
• let $\beta$ be a sum of $k$-distinct negative root. Can we find $w\in W$ such that $w\cdot 0=\beta$?
• note that $|W\cdot 0|=|W|$
• yes, if and only if $\beta$ is a sum of elements in $w \Phi^+ \cap \Phi^-$
• now we have found the principal block $D_k^0$ and Verma modules in its standard filtration

tensoring with simple module

• based on Remark in 6.2
• study $D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}$
• we want to know all Verma modules involved in a standard filtration of $D_k^\lambda$
• Q. Does it exist? Yes, because $D_k^0\otimes L(\lambda)$ has a standard filtration and taking the principal block does no harm in constructing standard filtration.

claim

$D_k^0\otimes L(\lambda)$ has a standard filtration.

proof

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 $0\to N\otimes L(\lambda)\to M\otimes L(\lambda) \to M(\lambda)\otimes L(\lambda)\to 0$

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

• 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 :

The vertex set is the set of elements of the Coxeter group and the edge set consists of directed edges $(u, v)$ whenever $u = t v$ for some reflection $t$ and $\ell(u) < \ell(v)$.

• example : http://groupprops.subwiki.org/wiki/File:Bruhatons3.png

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, $\ell(w)<\ell(w')$.

• 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.