"Talk on BGG resolution"의 두 판 사이의 차이
imported>Pythagoras0 (새 문서: ==overview== * Weyl character formula and characters of Verma modules * property of character map on short exact sequences * Euler-Poincare mapping * principal block : filtering throu...) |
imported>Pythagoras0 |
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22번째 줄: | 22번째 줄: | ||
;thm ([[Weyl-Kac character formula|Weyl character formula]]) | ;thm ([[Weyl-Kac character formula|Weyl character formula]]) | ||
:<math> | :<math> | ||
− | \operatorname{ch}L({\lambda})=\frac{\sum_{w\in W} (-1)^{\ell(w)}e^{w\cdot \lambda}}{\prod_{\alpha>0}(1-e^{-\alpha}) | + | \operatorname{ch}L({\lambda})=\frac{\sum_{w\in W} (-1)^{\ell(w)}e^{w\cdot \lambda}}{\prod_{\alpha>0}(1-e^{-\alpha})} |
</math> | </math> | ||
* thus we have | * thus we have | ||
$$ | $$ | ||
− | \operatorname{ch}L(\lambda)=\sum_{w \in W}(-1)^{\ell(w)}\operatorname{ch} M(w\cdot \lambda) | + | \operatorname{ch}L(\lambda)=\sum_{w \in W}(-1)^{\ell(w)}\operatorname{ch} M(w\cdot \lambda) \label{WCF} |
$$ | $$ | ||
2016년 4월 27일 (수) 00:46 판
overview
- Weyl character formula and characters of Verma modules
- property of character map on short exact sequences
- Euler-Poincare mapping
- 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
- combinatorial results
- 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
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
- goal : realize this formula \ref{WCF} as an Euler characteristic
- 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 $$ 0 \to M({w_0\cdot \lambda})\to \cdots \to \bigoplus_{w\in W, \ell(w)=k}M({w\cdot \lambda})\to \cdots M({\lambda})\to L({\lambda})\to 0 $$ 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
\[\chi_{L({\lambda})}=\chi_{M({\lambda})}-\chi_{M({-\lambda-2})}=\frac{q^{\lambda}}{1-q^{-2}}-\frac{q^{-\lambda-2}}{1-q^{-2}}\]
- comparison with Weyl-Kac character formula
\[\chi(L({\lambda}))=\frac{\sum_{w\in W} (-1)^{\ell(w)}e^{w(\lambda+\rho)}}{e^{\rho}\prod_{\alpha>0}(1-e^{-\alpha})}=\frac{q^{\lambda+1}-q^{-\lambda-1}}{q^{1}(1-q^{-2})}\] where I used \(\rho=1,\alpha=2\) and \(w(\lambda+\rho)=-\lambda-\rho\)
$\mathfrak{sl}_3$
Euler-Poincare characteristic
- $A$ : abelian category
- Euler-Poincare map $\varphi$ on an object in $A$
- $\varphi$ turns a short exact sequence into an alternating sum
- we can define Euler-Poincare characteristic $\chi_{\varphi}$ of a complex as the alternating sum of Euler-Poincare map on the homology
- the main result is that the Euler-Poincre characteristic can be computed from a different resolution and it is independent of the choice of it
Verma modules
maps between Verma modules
- 2 conditions to have non-zero homomorphisms \(V_{\lambda}\to V_{\mu}\) between two Verma modules
- \(\lambda+\rho, \mu+\rho\) are in the same orbit of Weyl group
- \(V_{\lambda}\leq V_{\mu}\), i.e. \(\lambda = \mu -\sum \alpha\), where the sum is over some positive roots.
example
- SL2
- \(\lambda = \mu -2n\), \(n=0,1,2,\cdots\)
- \((\lambda+1)^2 = (\mu+1)^2\)
composition series of Verma modules
- thm
The Verma module $M(\lambda)$ has a finite composition series $$ M(\lambda)=N_0\supset N_1\supset N_2\supset \cdots N_{r}=O $$ where each $N_i$ is a submodule of $M(\lambda)$ and $N_{i+1}$ is a maximal submodule of $N_i$. Moreover, $N_i/N_{i+1}$ is isomorphic to $L(w\cdot \lambda)$ for some $w\in W$.
action of center on Verma modules
- check
maximal submodule of Verma modules
- Maximal Submodule of $M(\lambda), \lambda \in \Lambda+$ (see 2.6)
weak BGG resolution
standard filtration
- We say that $M \in O$ has a standard filtration (also sometimes 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 $$0 = D_m^{\lambda} \to \subset D_{m-1}^{\lambda} \to \cdots \to D_2^{\lambda} \to D_1^{\lambda} \to L(0) \to 0$$ where $D_{k}^{\lambda}$ has a standard filtration involving exactly once each of the Verma modules $M(w\cdot \lambda)$ with $\ell(w)=k$
- we prove this for $\lambda=0$ and apply the translation function to extend it
strategy of the proof
- 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
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
$$\cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{p}(\mathfrak{g})\to U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{p-1}(\mathfrak{g})\to \cdots \to U(\mathfrak{g})\otimes_{\mathbb{C}}\Lambda^{0}(\mathfrak{g})\to L(0)$$
- $D_k$ are free only over $U(\mathfrak{n}^{−})$
weights of exterior powers
- what are the weights of $\wedge^k (\mathfrak{g}/\mathfrak{b})$ as $\mathfrak{b}$-module?
- 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 $w \Phi^+ \cap \Phi^-$
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 $
- 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, $\ell(w)<\ell(w')$.