"Talk on BGG resolution"의 두 판 사이의 차이

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imported>Pythagoras0
 
(다른 사용자 한 명의 중간 판 26개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==characters==
 
==characters==
* let $\lambda\in \mathfrak{h}^*$
+
* let <math>\lambda\in \mathfrak{h}^*</math>
$$
+
:<math>
 
\operatorname{ch} M({\lambda})=\frac{e^{\lambda}}{\prod_{\alpha>0}(1-e^{-\alpha})}
 
\operatorname{ch} M({\lambda})=\frac{e^{\lambda}}{\prod_{\alpha>0}(1-e^{-\alpha})}
$$
+
</math>
* let $\lambda\in \Lambda^+$
+
* let <math>\lambda\in \Lambda^+</math>
 
;thm ([[Weyl-Kac character formula|Weyl character formula]])
 
;thm ([[Weyl-Kac character formula|Weyl character formula]])
 
:<math>
 
:<math>
10번째 줄: 10번째 줄:
 
</math>
 
</math>
 
* thus we have
 
* thus we have
$$
+
:<math>
 
\operatorname{ch}L(\lambda)=\sum_{w \in W}(-1)^{\ell(w)}\operatorname{ch} M(w\cdot \lambda)  \label{WCF}
 
\operatorname{ch}L(\lambda)=\sum_{w \in W}(-1)^{\ell(w)}\operatorname{ch} M(w\cdot \lambda)  \label{WCF}
$$
+
</math>
  
 
;prop  
 
;prop  
If $0\to M' \to M \to M'' \to 0$ is a short exact sequence in $\mathcal{O}$, we have
+
If <math>0\to M' \to M \to M'' \to 0</math> is a short exact sequence in <math>\mathcal{O}</math>, we have
$$
+
:<math>
 
\operatorname{ch}M=\operatorname{ch}M'+\operatorname{ch}M''
 
\operatorname{ch}M=\operatorname{ch}M'+\operatorname{ch}M''
$$
+
</math>
 
or  
 
or  
$$
+
:<math>
 
\operatorname{ch}M'-\operatorname{ch}M+\operatorname{ch}M''=0
 
\operatorname{ch}M'-\operatorname{ch}M+\operatorname{ch}M''=0
$$
+
</math>
 
* if we have a long exact sequence, we still get a similar alternating sum = 0  
 
* if we have a long exact sequence, we still get a similar alternating sum = 0  
 
** why? [[Euler-Poincare principle|Euler-Poincare mapping]] : a long exact sequence can be decomposed into short exact sequences.  
 
** why? [[Euler-Poincare principle|Euler-Poincare mapping]] : a long exact sequence can be decomposed into short exact sequences.  
** then the Euler characteristic of a resolution makes sense
+
** 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 $L(\lambda)$
+
* goal : realize the alternating sum \ref{WCF} as an Euler characteristic of a suitable resolution of <math>L(\lambda)</math>
* 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 $W\cdot \lambda$
+
* The BGG resolution resolves a finite-dimensional simple <math>\mathfrak{g}</math>-module <math>L(\lambda)</math> by direct sums of Verma modules indexed by weights "of the same length" in the orbit <math>W\cdot \lambda</math>
 
;thm (Bernstein-Gelfand-Gelfand Resolution)
 
;thm (Bernstein-Gelfand-Gelfand Resolution)
Fix $\lambda\in \Lambda^{+}$. There is an exact sequence of Verma modules
+
Fix <math>\lambda\in \Lambda^{+}</math>. There is an exact sequence of Verma modules
$$
+
:<math>
 
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
 
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
$$
+
</math>
where $\ell(w)$ is the length of the Weyl group element $w$, $w_0$ is the Weyl group element
+
where <math>\ell(w)</math> is the length of the Weyl group element <math>w</math>, <math>w_0</math> is the Weyl group element
of maximal length. Here $\rho$ is half the sum of the positive roots.
+
of maximal length. Here <math>\rho</math> is half the sum of the positive roots.
  
 
==example of BGG resolution==
 
==example of BGG resolution==
===$\mathfrak{sl}_2$===
+
===<math>\mathfrak{sl}_2</math>===
 
* <math>L({\lambda})</math> : irreducible highest weight module
 
* <math>L({\lambda})</math> : irreducible highest weight module
 
** weights <math>\lambda ,-2+\lambda ,\cdots, -\lambda</math>
 
** weights <math>\lambda ,-2+\lambda ,\cdots, -\lambda</math>
43번째 줄: 43번째 줄:
 
** weights <math>\lambda ,-2+\lambda ,\cdots, -\lambda, -\lambda-2,\cdots</math>
 
** weights <math>\lambda ,-2+\lambda ,\cdots, -\lambda, -\lambda-2,\cdots</math>
 
;thm  
 
;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 $\mathfrak{g}$.
+
If <math>\lambda\in \Lambda^+</math>, the maximal submodule <math>N(\lambda)</math> of <math>M(\lambda)</math> is the sum of submodules <math>M(s_i\cdot \lambda)</math> for <math>1\le i \le l</math>, where <math>l</math> is the rank of <math>\mathfrak{g}</math>.
* $s_{1}(\lambda+\rho)=-\lambda-\rho$, $s_{1}\cdot \lambda=-\lambda-2\rho$
+
* <math>s_{1}(\lambda+\rho)=-\lambda-\rho</math>, <math>s_{1}\cdot \lambda=-\lambda-2\rho</math>
* if we identity $\Lambda = \mathbb{Z} \omega_1$ with $\mathbb{Z}$, then <math>\rho=1,\alpha=2</math>  
+
* if we identity <math>\Lambda = \mathbb{Z} \omega_1</math> with <math>\mathbb{Z}</math>, then <math>\rho=1,\alpha=2</math>  
 
* we have
 
* we have
$$L({\lambda})=M({\lambda})/M({-\lambda-2})$$ or
+
:<math>L({\lambda})=M({\lambda})/M({-\lambda-2})</math> or
 
:<math>0\to M({-\lambda-2})\to M({\lambda})\to L({\lambda})\to 0</math>
 
:<math>0\to M({-\lambda-2})\to M({\lambda})\to L({\lambda})\to 0</math>
 
* this gives a BGG resolution
 
* this gives a BGG resolution
* character of $L({\lambda})$ = alternating sum of characters of Verma modules
+
* character of <math>L({\lambda})</math> = alternating sum of characters of Verma modules
 
:<math>\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}}</math>
 
:<math>\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}}</math>
 
*  comparison with [[Weyl-Kac character formula]]
 
*  comparison with [[Weyl-Kac character formula]]
59번째 줄: 59번째 줄:
 
==weak BGG resolution==
 
==weak BGG resolution==
 
;def
 
;def
* We say that $M \in O$ has a standard filtration (also called a Verma flag) if there is a sequence of submodules
+
* We say that <math>M \in O</math> has a standard filtration (also called a Verma flag) if there is a sequence of submodules
$$0 = M_0 \subset M_1 \subset M_2 \subset \cdots \subset M_n = M$$
+
:<math>0 = M_0 \subset M_1 \subset M_2 \subset \cdots \subset M_n = M</math>
for which each $M^i := M_i/M_{i−1}\, (1 \le i \le n)$ is isomorphic to a Verma module.
+
for which each <math>M^i := M_i/M_{i−1}\, (1 \le i \le n)</math> is isomorphic to a Verma module.
 
;thm (Weak BGG resolution)
 
;thm (Weak BGG resolution)
 
There is an exact sequence
 
There is an exact sequence
$$
+
:<math>
 
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
 
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
$$
+
</math>
where $D_{k}^{\lambda}$ has a standard filtration involving exactly once each of the Verma modules $M(w\cdot \lambda)$ with $\ell(w)=k$
+
where <math>D_{k}^{\lambda}</math> has a standard filtration involving exactly once each of the Verma modules <math>M(w\cdot \lambda)</math> with <math>\ell(w)=k</math>
  
  
 
===strategy to construct a BGG resolution===
 
===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 relative version of standard resoultion <math>D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\Lambda^{k}(\mathfrak{g}/\mathfrak{b})</math> for <math>L(0)</math>
# 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 <math>D_k^0:=D_k^{\chi_{0}}</math> for <math>L(0)</math> 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)
+
# construct a weak BGG resolution <math>D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}</math> for <math>L(\lambda)</math> (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
+
# show that it is actually a BGG resolution by computing <math>\operatorname{Ext}</math> between Verma modules
  
 
==standard resolution of trivial module==
 
==standard resolution of trivial module==
* free $U(\mathfrak{g})$-modules $U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k}(\mathfrak{g})$
+
* free <math>U(\mathfrak{g})</math>-modules <math>U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k}(\mathfrak{g})</math>
 
* standard resolution of trivial module in [[Lie algebra cohomology]]
 
* standard resolution of trivial module in [[Lie algebra cohomology]]
$$\cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{k}(\mathfrak{g})\to U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k-1}(\mathfrak{g})\to \cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{0}(\mathfrak{g})\to L(0)$$
+
:<math>\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)</math>
* 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
+
* the sequence of modules <math>D_k:=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\wedge^{k}(\mathfrak{g}/\mathfrak{b})</math> is a relative version of the standard resolution
* we can describe $D_0$ and $D_m$ explicitly
+
* we can describe <math>D_0</math> and <math>D_m</math> explicitly
* define $U(\mathfrak{g})$-module homomorphism $\partial_k : D_k \to D_{k-1}$ as
+
* define <math>U(\mathfrak{g})</math>-module homomorphism <math>\partial_k : D_k \to D_{k-1}</math> as
$$
+
:<math>
 
\begin{align}
 
\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 \wedge \hat{\xi_i}\wedge \cdots \wedge\xi_k)\\
+
\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)
 
&+\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}
 
\end{align}
$$
+
</math>
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 $\mathfrak{g}/\mathfrak{b}$.
+
where <math>z_i\in \mathfrak{g}</math> is a representative of <math>\xi_i\in \mathfrak{g}/\mathfrak{b}</math> and <math>\overline{z}</math> denotes the canonical surjection <math>z\in\mathfrak{g}</math> into <math>\mathfrak{g}/\mathfrak{b}</math>.
* need to show that $\partial_k$ is well-defined and it is actually a complex and exact  
+
* need to show that <math>\partial_k</math> is well-defined and it is actually a complex and exact  
 
===exactness===
 
===exactness===
 
* exactness is tricky
 
* exactness is tricky
 
** see Wallach, Real Reductive Groups I 6.A
 
** see Wallach, Real Reductive Groups I 6.A
 
** see Knapp, Lie Groups, Lie Algebras, and Cohomology IV.6
 
** 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 $\le j$
+
* Let <math>U_j(\mathfrak{g}):=U^j(\mathfrak{g})U(\mathfrak{b})</math> where <math>U^j(\mathfrak{g})</math> is the span of the PBW basis whose degree is <math>\le j</math>
* note that $U_j(\mathfrak{g})=S_j(\mathfrak{n}^-)U(\mathfrak{b})$ where $S_j(\mathfrak{n}^-)=\sum_{0\le k\le j}S^k(\mathfrak{n}^-)$.
+
* note that <math>U_j(\mathfrak{g})=S_j(\mathfrak{n}^-)U(\mathfrak{b})</math> where <math>S_j(\mathfrak{n}^-)=\sum_{0\le k\le j}S^k(\mathfrak{n}^-)</math>.
* $S^k(\mathfrak{n}^-)$ denote the elements of $\operatorname{Sym}(\mathfrak{n}^-)$ that are homogeneous of degree $k$.
+
* <math>S^k(\mathfrak{n}^-)</math> denote the elements of <math>\operatorname{Sym}(\mathfrak{n}^-)</math> that are homogeneous of degree <math>k</math>.
* $E_{j,k}:=U_j(\mathfrak{g})\otimes_{U(\mathfrak{b})}\wedge^{k}(\mathfrak{g}/\mathfrak{b})$
+
* <math>E_{j,k}:=U_j(\mathfrak{g})\otimes_{U(\mathfrak{b})}\wedge^{k}(\mathfrak{g}/\mathfrak{b})</math>
* $\{E_{j,k}\}_{j\ge 0}$ gives a filtration of $D_k$
+
* <math>\{E_{j,k}\}_{j\ge 0}</math> gives a filtration of <math>D_k</math>
* '''exercise''' : $\partial_k : E_{j,k}\to E_{j+1,k-1}$
+
* let <math>\partial_0 : D_0\to L(0)</math> be the canonical surjection
* $\partial_k$ induces $\overline{\partial_k}  : E_{j,k}/E_{j-1,k}\to E_{j+1,k-1}/E_{j,k-1}$
+
* '''exercise''' : <math>\partial_k : E_{j,k}\to E_{j+1,k-1}, k\ge 1</math>
 +
* <math>\partial_k, k\ge 1</math> induces <math>\overline{\partial_k}  : E_{j,k}/E_{j-1,k}\to E_{j+1,k-1}/E_{j,k-1}</math>
  
 
;prop
 
;prop
For each $j\ge 1$, $\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{0\le k\le m}$ is an exact sequence
+
For each <math>j\ge 1</math>, <math>\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{-1\le k\le m+1}</math> is an exact sequence. (here <math>-1</math> and <math>m+1</math> terms are zero)
  
 
;proof
 
;proof
As a vector space, $E_{j,k}/E_{j-1,k}\cong S^j(\mathfrak{n}^-)\otimes \wedge^{k}(\mathfrak{n}^-)$.
+
As a vector space, <math>E_{j,k}/E_{j-1,k}\cong S^j(\mathfrak{n}^-)\otimes \wedge^{k}(\mathfrak{n}^-)</math>.
  
 
Now we can apply basic results on the Koszul complexes :  
 
Now we can apply basic results on the Koszul complexes :  
  
 
;theorem
 
;theorem
Let $j\geq 1$. The following sequences are exact :
+
For each <math>j\geq 1</math>, the following is exact
$$
+
:<math>
 
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
 
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
$$
+
</math>
where $S^j(V)=0$ for $j<0$
+
where <math>S^j(V)=0</math> for <math>j<0</math>
 
* see Lang, Algebra ' Koszul complex'
 
* see Lang, Algebra ' Koszul complex'
  
 +
* examples
 +
:<math>
 +
0\to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0
 +
</math>
 +
or
 +
:<math>
 +
0\to E_{0,2} \to E_{1,1}/E_{0,1} \to E_{2,0}/E_{1,0} \to 0
 +
</math>
  
 +
:<math>
 +
0\to \wedge^1 \to S^1 \to 0
 +
</math>
 +
or
 +
:<math>
 +
0\to E_{0,1} \to E_{1,0}/E_{0,0} \to 0
 +
</math>
 +
* in fact, we also have
 +
:<math>
 +
0\to E_{0,0}\to L(0)\to 0
 +
</math>
 
;prop
 
;prop
$\partial : E_{j,k}\to E_{j+1,k-1}$ gives an exact sequence
+
For each <math>j\ge 1</math>, <math>\{(E_{j,k},\partial_k)\}_{-1\le k \le m+1}</math> is exact. (here <math>-1</math> and <math>m+1</math> terms are zero)
  
 
;proof
 
;proof
Suppose $u\in E_{j,k}$ satisfies $\partial(u)=0$ in $E_{j+1,k-1}$.
+
Suppose <math>u\in E_{j,k},\, j,k\geq 1</math> satisfies <math>\partial(u)=0</math> in <math>E_{j+1,k-1}</math>.
  
show : there exists $v\in E_{j-1,k+1}$ such that $\partial(v)=u$ in
+
show : there exists <math>v\in E_{j-1,k+1}</math> such that <math>\partial(v)=u</math> in
$$
+
:<math>
 
E_{j-1,k+1} \to  E_{j,k}\to E_{j+1,k-1}.
 
E_{j-1,k+1} \to  E_{j,k}\to E_{j+1,k-1}.
$$
+
</math>
  
 
We look at
 
We look at
$$
+
:<math>
 
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}.
 
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}.
$$
+
</math>
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 <math>\overline{\partial}(\overline{u})=0</math>, we can find <math>v_1\in E_{j-1,k+1}</math> such that <math>\overline{\partial}(\overline{v_1})=\overline{u}</math>.  
  
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 $u-\partial(v_1)=w$.
+
As <math>\overline{u}-\overline{\partial}(\overline{v_1})=0</math> in <math>E_{j,k,}/E_{j-1,k}</math>, there exists <math>w\in E_{j-1,k}</math> such that <math>u-\partial(v_1)=w</math>.
  
Now $\partial(w)=0$ in $E_{j,k-1}$ and by the same argument applied to
+
Now <math>\partial(w)=0</math> in <math>E_{j,k-1}</math> and by the same argument applied to
$$
+
:<math>
 
E_{j-2,k+1} \to  E_{j-1,k}\to E_{j,k-1},
 
E_{j-2,k+1} \to  E_{j-1,k}\to E_{j,k-1},
$$
+
</math>
there exists $v_2\in E_{j-2,k+1}$ such that $w-\partial (v_2)\in E_{j-2,k}$, i.e. $u-\partial(v_1)-\partial(v_2)\ \in E_{j-2,k}$.
+
there exists <math>v_2\in E_{j-2,k+1}</math> such that <math>w-\partial (v_2)\in E_{j-2,k}</math>, i.e. <math>u-\partial(v_1)-\partial(v_2)\ \in E_{j-2,k}</math>.
  
By repeating this, we can find $v_1,\cdots, v_j$ such that $u-\partial(v_1)-\cdots \partial(v_j)\ \in E_{0,k}$.
+
By repeating this, we can find <math>v_1,\cdots, v_j</math> such that each <math>v_i\in E_{j-i,k}</math> and <math>u-\partial(v_1)-\cdots -\partial(v_j)\ \in E_{0,k}</math>.
  
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
+
As <math>\overline{\partial}  : E_{0,k}\to E_{1,k-1}/E_{0,k-1}</math> is injective and <math>\partial(u-\partial(v_1)-\cdots -\partial(v_j))=0 \in E_{1,k-1}</math>, we can conclude <math>u-\partial(v_1)-\cdots -\partial(v_j)=0</math> or,
$$
+
:<math>
 
u=\partial(v_1)+\cdots +\partial(v_j).
 
u=\partial(v_1)+\cdots +\partial(v_j).
$$
+
</math>
Thus if we set $v:=v_1+\cdots +v_j\in E_{j-1,k+1}$ and $\partial(v)=u$.
+
Thus if we set <math>v:=v_1+\cdots +v_j\in E_{j-1,k+1}</math> and <math>\partial(v)=u</math>. ■
 +
 
 +
 
 +
;thm
 +
<math>0\to D_m \to \cdots \to D_k\to \cdots \to D_1 \to D_0 \to L(0)\to 0</math> is exact.
 +
;proof
 +
The above proposition clearly proves the exactness at <math>D_k,\, k\ge 1</math>.
 +
 
 +
Assume that <math>u\in D_0</math>, hence <math>u\in E_{j,0}</math> for some <math>j</math>. Let <math>u=u_0+u_1</math> where <math>u_0</math> is the degree zero piece and <math>u_1</math> other terms in PBW basis.
 +
 
 +
If <math>\partial(u)=0</math>, then it implies <math>u_0 = 0</math> as only degree zero term survives under <math>\partial</math>.
 +
 
 +
Therefore <math>j\ge 1</math> and the above proposition still applies to conclude that there exists <math>v\in E_{j-1,1}</math> such that <math>\partial(v)=u</math>.
 +
 
 +
This proves the exactness at <math>D_0</math>. ■
  
==weak BGG resolution of $L(0)$==
+
==weak BGG resolution of <math>L(0)</math>==
* goal : find standard filtrations of $D_k$ and $D_k^0$ and their Verma subquotients
+
* goal : find standard filtrations of <math>D_k</math> and <math>D_k^0</math> and their Verma subquotients
 
;lemma
 
;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.
+
Let <math>N</math> be a finite-dimensional <math>U(\mathfrak{b})</math>-module having a basis of weight vectors. Then <math>M=U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N</math> has a standard filtration and each weight of <math>N</math> gives a corresponding Verma subquotient.
 
;proof
 
;proof
Let $\{v_1,\cdots, v_r \}$ be a basis of $N$ consisting of weight vectors and let $\mu_i$ be the weight of $v_i$.
+
Let <math>\{v_1,\cdots, v_r \}</math> be a basis of <math>N</math> consisting of weight vectors and let <math>\mu_i</math> be the weight of <math>v_i</math>.
  
We order the basis so that  $i\le j$ whenever $\mu_i\le \mu_j$
+
We order the basis so that  <math>i\le j</math> whenever <math>\mu_i\le \mu_j</math>
  
Let $N_k$ be a space spanned by $\{v_k,\cdots, v_r \}$ for $1\le k \le r$.   
+
Let <math>N_k</math> be a space spanned by <math>\{v_k,\cdots, v_r \}</math> for <math>1\le k \le r</math>.   
  
'''exercise'''. Check that each $N_k$ is a $U(\mathfrak{b})$-submodule. (hint : weight cannot decrease under $U(\mathfrak{b})$ action)
+
'''exercise'''. Check that each <math>N_k</math> is a <math>U(\mathfrak{b})</math>-submodule. (hint : weight cannot decrease under <math>U(\mathfrak{b})</math> action)
  
We have a flag of $U(\mathfrak{b})$-modules :
+
We have a flag of <math>U(\mathfrak{b})</math>-modules :
$$
+
:<math>
 
0 \subset N_r \subset N_{r-1} \subset \cdots \subset N_1 = N \label{Nflag}
 
0 \subset N_r \subset N_{r-1} \subset \cdots \subset N_1 = N \label{Nflag}
$$
+
</math>
 
 
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) ■
 
  
 +
We get a standard filtration of <math>M</math> from \ref{Nflag} as the functor <math>N\mapsto U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N</math> 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
 
;prop
$D_k$ has a standard filtration with Verma subquotients associated to sums of $k$ distinct negative roots.  
+
<math>D_k</math> has a standard filtration with Verma subquotients associated to sums of <math>k</math> distinct negative roots.  
 
;proof
 
;proof
 
If we apply the above lemma,  enough to answer :
 
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?  
+
Q. what are the weights of <math>\wedge^k (\mathfrak{g}/\mathfrak{b})</math> as <math>\mathfrak{b}</math>-module?  
  
A : sum of $k$ distinct negative roots
+
A : sum of <math>k</math> distinct negative roots
 
 
  
 
;prop
 
;prop
$D_k^0$ has a standard filtration with Verma subquotients $M(w\cdot 0), w\in W^{(k)}$ where $W^{(k)}:=\{w\in W|\ell(w)=k\}$
+
<math>D_k^0</math> has a standard filtration with Verma subquotients <math>M(w\cdot 0), w\in W^{(k)}</math> where <math>W^{(k)}:=\{w\in W|\ell(w)=k\}</math>
  
 
;proof
 
;proof
 
Taking a block preserves exactness :  
 
Taking a block preserves exactness :  
  
If $0\to M' \to M \to M(\mu)\to 0$, then $0\to (M')^0 \to M^0 \to (M(\mu))^0 \to 0$.
+
If <math>0\to M' \to M \to M(\mu)\to 0</math>, then <math>0\to (M')^0 \to M^0 \to (M(\mu))^0 \to 0</math>.
  
$$
+
:<math>
 
(M(\mu))^0 =
 
(M(\mu))^0 =
 
\begin{cases}  
 
\begin{cases}  
  M(\mu) , & \text{if $\mu$ is linked to $0$ ($\mu=w\cdot 0$ for some $w\in W$)}\\  
+
  M(\mu) , & \text{if </math>\mu<math> is linked to </math>0<math> (</math>\mu=w\cdot 0<math> for some </math>w\in W<math>)}\\  
 
  0, & \text{otherwise} \\  
 
  0, & \text{otherwise} \\  
 
\end{cases}
 
\end{cases}
$$
+
</math>
 +
 
 +
Thus we obtain a standard filtration of <math>D_k^0</math> from that of <math>D_k</math>.
  
Thus we obtain a standard filtration of $D_k^0$ from that of $D_k$.
+
What are the Verma subquotients or when is <math>\beta</math> linked to <math>0</math> if <math>\beta</math> is given a sum of <math>k</math> distinct negative roots?
  
What are the Verma subquotients or when is $\beta$ linked to $0$ if $\beta$ is given a sum of $k$ distinct negative roots?
+
'''fact''' : <math>\ell(w)=|w \Phi^+ \cap \Phi^-|</math>.
  
'''fact''' : $\ell(w)=|w \Phi^+ \cap \Phi^-|$.
+
'''exercise''' : Let  <math>\beta_w:=w\cdot 0</math> for each <math>w\in W</math>. Then <math>\beta_w</math> is a sum of elements in <math>w \Phi^+ \cap \Phi^-</math>.
  
'''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  <math>\Pi \subset \Phi^-</math> be given. If the sum <math>\beta</math> of elements of <math>\Pi</math> is <math>\beta_w</math> for some <math>w\in W</math>, then <math>\Pi = w \Phi^+ \cap \Phi^-</math>. (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 $w\in W^{(k)}$.
+
Thus <math>\beta</math> is a sum of <math>k</math> distinct negative roots and linked to <math>0</math> iff there exists <math>w\in W^{(k)}</math>.
  
 
Finally,
 
Finally,
  
'''fact''' : $|W\cdot 0|=|W|$. (this implies each $\beta_w$ is distinct)
+
'''fact''' : <math>|W\cdot 0|=|W|</math>. (this implies each <math>\beta_w</math> is distinct)
  
Therefore each $M(w\cdot 0),\, w\in W^{(k)}$ appears only in once our standard filtration.
+
Therefore each <math>M(w\cdot 0),\, w\in W^{(k)}</math> appears only in once our standard filtration.
  
 
 
* thus we have found a weak BGG resolution of $L(0)$
+
* thus we have found a weak BGG resolution of <math>L(0)</math>
  
==weak BGG resolution of $L(\lambda)$==
+
==weak BGG resolution of <math>L(\lambda)</math>==
 
* based on Remark in 6.2
 
* based on Remark in 6.2
* let $\lambda\in \Lambda^+$
+
* let <math>\lambda\in \Lambda^+</math>
* goal : find a standard filtration of $D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}$ and its Verma subquotients
+
* goal : find a standard filtration of <math>D_k^\lambda : = (D_k^0\otimes L(\lambda))^{\chi_{\lambda}}</math> and its Verma subquotients
 
;prop
 
;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 $L(\lambda)$.
+
<math>D_k^0\otimes L(\lambda)</math> has a standard filtration with Verma subquotients <math>M(w\cdot 0 + \mu)</math> where <math>w\in W^{(k)}</math> and <math>\mu</math> is a weight of <math>L(\lambda)</math>.
  
 
;proof
 
;proof
231번째 줄: 266번째 줄:
 
Use the following :  
 
Use the following :  
 
;thm (3.6)
 
;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.
+
Let <math>M</math> be a finite dimensional <math>U(\mathfrak{g})</math>-module. For any <math>\lambda\in \mathfrak{h}^{*}</math>, <math>T:=M(\lambda)\otimes M</math> has a standard filtration with Verma subquotients <math>M(\lambda+\mu)</math>. Here <math>\mu</math> ranges over the weights of <math>M</math>, each occurring <math>\dim M_{\mu}</math> times in the filtration.
  
 
Tensoring with a finite-dimensional representation is an exact functor in BGG category (thm 1.1).  
 
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 $0\to N\otimes L(\lambda)\to M\otimes L(\lambda) \to M(\lambda)\otimes L(\lambda)\to 0$
+
If <math>0\to N\to M \to M(\lambda)\to 0</math>, then <math>0\to N\otimes L(\lambda)\to M\otimes L(\lambda) \to M(\lambda)\otimes L(\lambda)\to 0</math>
  
Use this to construct a standard filtration on $D_k^0 \otimes L(\lambda)$ from that of $D_k^0$. ■
+
Use this to construct a standard filtration on <math>D_k^0 \otimes L(\lambda)</math> from that of <math>D_k^0</math>. ■
  
 
;prop
 
;prop
$D_k^\lambda$ has a standard filtration with Verma subquotients $M(w\cdot \lambda),\, w\in W^{k}$.
+
<math>D_k^\lambda</math> has a standard filtration with Verma subquotients <math>M(w\cdot \lambda),\, w\in W^{k}</math>.
  
 
;proof
 
;proof
Again taking the block component for $\chi_{\lambda}$ is exact and it gives a standard filtration of $D_k^\lambda$ from that of $D_k^0\otimes L(\lambda)$.
+
Again taking the block component for <math>\chi_{\lambda}</math> is exact and it gives a standard filtration of <math>D_k^\lambda</math> from that of <math>D_k^0\otimes L(\lambda)</math>.
  
We need to determine when $w\cdot 0 + \mu$ is linked to $\lambda$.  
+
We need to determine when <math>w\cdot 0 + \mu</math> is linked to <math>\lambda</math>.  
  
'''exercise''' : $w\cdot 0 + \mu$ is linked to $\lambda$ iff $\mu = w\lambda$.
+
'''exercise''' : <math>w\cdot 0 + \mu</math> is linked to <math>\lambda</math> iff <math>\mu = w\lambda</math>.
  
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 :
+
As <math>\lambda</math> is the highest weight in <math>L(\lambda)</math>, each <math>w\lambda</math> is with weight multiplicity 1. Thus each <math>M(w\cdot \lambda),\, w\in W^{(k)}</math> appears only once in our standard filtration. Note that each <math>w\cdot \lambda</math> is distinct as :
  
'''fact''' : $|W\cdot \lambda|=|W|$ for $\lambda\in \Lambda^+$.■
+
'''fact''' : <math>|W\cdot \lambda|=|W|</math> for <math>\lambda\in \Lambda^+</math>.■
  
 
==extensions of Verma modules==
 
==extensions of Verma modules==
* we have constructed a weak BGG resolution of $L(\lambda)$ involving $D_k^{\lambda}$
+
* we have constructed a weak BGG resolution of <math>L(\lambda)</math> involving <math>D_k^{\lambda}</math>
* goal : $D_k^{\lambda}$ is a direct sum of Verma modules
+
* goal : <math>D_k^{\lambda}</math> is a direct sum of Verma modules
  
 
;def
 
;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)
+
Let <math>\mu, \lambda\in \mathfrak{h}^{*}</math>. Write <math>\mu \uparrow \lambda</math> if <math>\mu = \lambda</math> or there exists <math>\alpha\in \Phi^+</math> such that <math>\mu=s_{\alpha}\cdot \lambda < \lambda </math> (<math>\mathbb{Z}^+</math>-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 $
+
We say <math>\mu</math> is ''strongly linked'' to <math>\lambda</math> if <math>\mu = \lambda</math> or there exist <math>\alpha_1,\cdots, \alpha_r\in \Phi^+</math> such that <math>\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 </math>
  
 
;def
 
;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]]''.
+
Let <math>w,w'\in W</math>. Write <math>w'\xrightarrow{t} w</math> whenever <math>w = t w' </math> for some reflection <math>t\in W</math> and <math>\ell(w') < \ell(w)</math>. Define <math>w'<w</math> if there is a sequence <math>w'=w_0\to w_1\to \cdots \to w_n=w</math>. Extend this relation to a partial ordering of <math>W</math> 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.
 
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 :
+
'''exercise''' : Let <math>w\in W, \alpha \in \Phi^+</math> be given. The following are equivalent :
  
(i) There exists a $\lambda\in \Lambda^{+}$ such that $s_{\alpha}\cdot (w\cdot \lambda) \uparrow w\cdot \lambda$
+
(i) There exists a <math>\lambda\in \Lambda^{+}</math> such that <math>s_{\alpha}\cdot (w\cdot \lambda) \uparrow w\cdot \lambda</math>
  
(ii) $s_{\alpha}w > w$
+
(ii) <math>s_{\alpha}w > w</math>
  
 
hint : use  
 
hint : use  
  
'''fact''' :  $w^{-1}\alpha>0$ iff $\ell(s_{\alpha}w)> \ell(w)$.
+
'''fact''' :  <math>w^{-1}\alpha>0</math> iff <math>\ell(s_{\alpha}w)> \ell(w)</math>.
  
 
;thm
 
;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 $\mu \neq \lambda$
+
(a) Let <math>\lambda,\mu\in \mathfrak{h}^{*}</math>. If <math>\operatorname{Ext}_{\mathcal{O}}(M(\mu),M(\lambda))\neq 0</math>, then <math>\mu</math> is strongly linked to <math>\lambda</math> but <math>\mu \neq \lambda</math>
  
(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, $\ell(w)<\ell(w')$.
+
(b) Let <math>\lambda\in \Lambda^{+}</math> and <math>w,w'\in W</math>. If <math>\operatorname{Ext}_{\mathcal{O}}(M(w'\cdot\lambda),M(w\cdot\lambda))\neq 0</math>, then <math>w<w'</math> in the [[Bruhat ordering]]. In particular, <math>\ell(w)<\ell(w')</math>.
  
 
;proof of (a)
 
;proof of (a)
286번째 줄: 321번째 줄:
  
 
;proof of (b)
 
;proof of (b)
From (a), we see that $w'\cdot\lambda$ is strongly linked to $w\cdot\lambda$. Thus we can find
+
From (a), we see that <math>w'\cdot\lambda</math> is strongly linked to <math>w\cdot\lambda</math>. Thus we can find
$
+
<math>
 
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)
 
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)
$
+
</math>
From exercise, $w'=s_{\alpha_1}\cdots s_{\alpha_r}w> \cdots > s_{\alpha_r}w > w$.
+
From exercise, <math>w'=s_{\alpha_1}\cdots s_{\alpha_r}w> \cdots > s_{\alpha_r}w > w</math>.
 
 
  
300번째 줄: 335번째 줄:
 
Use induction on the length of standard filtration.
 
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 $w'\in W^{(k)}$.  
+
Let <math>M\subset D_k^{\lambda}</math> and <math>D_k^{\lambda}/M \cong M(w'\cdot \lambda)</math> for some <math>w'\in W^{(k)}</math>.  
  
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).
+
By induction hypothesis, <math>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</math>, then this splits as <math>\operatorname{Ext}\left(M(w'\cdot \lambda ),\oplus M(w\cdot \lambda)\right)</math> is zero (ext is additive).
  
 
 
309번째 줄: 344번째 줄:
 
* proof of Thm 3.6 uses the following (we don't need this for our goal)
 
* proof of Thm 3.6 uses the following (we don't need this for our goal)
 
;thm Tensor Identity (56p)
 
;thm Tensor Identity (56p)
Let $M$ be a $U(\mathfrak{g})$-module and $L$ a $U(\mathfrak{b})$-module. Then
+
Let <math>M</math> be a <math>U(\mathfrak{g})</math>-module and <math>L</math> a <math>U(\mathfrak{b})</math>-module. Then
$$
+
:<math>
 
(U(\mathfrak{g})\otimes_{U(\mathfrak{b})}L)\otimes M \cong U(\mathfrak{g})\otimes_{U(\mathfrak{b})}(L \otimes M)
 
(U(\mathfrak{g})\otimes_{U(\mathfrak{b})}L)\otimes M \cong U(\mathfrak{g})\otimes_{U(\mathfrak{b})}(L \otimes M)
$$
+
</math>
  
  
 
;prop (3.7)
 
;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.
+
Let <math>M\in \mathcal{O}</math> have a standard filtration. If <math>\lambda</math> is maximal among the weights of <math>M</math>, then <math>M</math> has a submodule isomorphic to <math>M(\lambda)</math> and <math>M/M(\lambda)</math> has a standard filtration.
  
  
 
===overview===
 
===overview===
 
* principal block : filtering through central characters
 
* principal block : filtering through central characters
** is a block a $U(\mathfrak{g})$-submodule? yes
+
** is a block a <math>U(\mathfrak{g})</math>-submodule? yes
 
** how to check that it preserves the exactness : any homomorphism between modules belonging to different blocks will be zero
 
** 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
+
** how to describe <math>\chi_{\lambda}</math>? use the twisted Harish-Chandra homomorphism <math>\psi : Z(\mathfrak{g})\to S(\mathfrak{h})</math>. we have
$$
+
:<math>
 
\chi_{\lambda}(z)=(\lambda+\rho)(\psi(z)),\quad z\in Z(\mathfrak{g})
 
\chi_{\lambda}(z)=(\lambda+\rho)(\psi(z)),\quad z\in Z(\mathfrak{g})
$$
+
</math>
** see 26p for an example of $\chi_{\lambda}$ in type $A_1$
+
** see 26p for an example of <math>\chi_{\lambda}</math> in type <math>A_1</math>
 
* combinatorial results
 
* combinatorial results
** longest elements satisfies $w_0\cdot 0 = -2\rho$ (related to diagram automorphism)
+
** longest elements satisfies <math>w_0\cdot 0 = -2\rho</math> (related to diagram automorphism)
** consider the set of sum of k distinct roots. Which elements are linked to $0$?
+
** consider the set of sum of k distinct roots. Which elements are linked to <math>0</math>?
 
** Bruhat ordering
 
** Bruhat ordering
 
* Bruhat ordering and strong linkage relation
 
* Bruhat ordering and strong linkage relation
** let $\lambda \in \Lambda^+$ (which is regular for the dot-action of $W$)
+
** let <math>\lambda \in \Lambda^+</math> (which is regular for the dot-action of <math>W</math>)
**  $w'\cdot \lambda< w \cdot \lambda $ translates into $w < w'$ in the Bruhat ordering
+
**  <math>w'\cdot \lambda< w \cdot \lambda </math> translates into <math>w < w'</math> in the Bruhat ordering
 
* strong linkage relation and extension of Verma modules
 
* strong linkage relation and extension of Verma modules
 
* for exterior powers, see [[Lie Algebras of Finite and Affine Type by Carter]]
 
* for exterior powers, see [[Lie Algebras of Finite and Affine Type by Carter]]
342번째 줄: 377번째 줄:
 
[[분류:Talks and lecture notes]]
 
[[분류:Talks and lecture notes]]
 
[[분류:abstract concepts]]
 
[[분류:abstract concepts]]
 +
[[분류:migrate]]

2020년 11월 14일 (토) 02:14 기준 최신판

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

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

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

  • 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 \(L(0)\)
  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 \(U(\mathfrak{g})\otimes_{\mathbb{C}}\wedge^{k}(\mathfrak{g})\)
  • 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