# Wednesday February 12th ## Infinitesimal Blocks We'll break up category $\OO$ into smaller subcategories (blocks). Recall theorem 1.1 (e): $\mcz(\lieg)$ acts locally finitely on $M\in \OO$, and $M$ has a *finite* filtration with highest weight sections, so $M$ should involve only a finite number of central characters $\chi_\lambda$ (where $\lambda \in\lieh\dual$). > Note: an analog of Jordan decomposition works here because of this finiteness condition. > This discussion will parallel the JCF of a simple operator on a finite dimensional $\CC\dash$vector space. > However, this involves the *entire* center instead of just scalar matrices, so the analogy is diagonalizing a family of operators simultaneously. Let $\chi \in \hat \mcz(\lieg)$ and $M\in \OO$, and $$ M^\chi \definedas \theset{v\in M \suchthat ~\forall z\in \mcz(\lieg),~\exists n>0 \st (z- \chi(z))^n \cdot v = 0} $$ Idea: write $$ z = \chi(z) \cdot 1 + (z-\chi(z)\cdot 1), $$ where the first is a scalar operator and the second is (locally) nilpotent on $M^\chi$. Thus we can always arrange for $z$ to act by a sum of "Jordan blocks": ![Image](figures/2020-02-12-09:15.png)\ Some observations: - $M^\chi$ are $U(\lieg)\dash$submodules of $M$. - The subspaces $M^\chi$ are linearly independent - $\mcz(\lieg)$ stabilizes each $M_\mu$ since $\mcz(\lieg)$ and $U(\lieh)$ are a commuting family of operators on $M_\mu$. - We can write $$M_\mu = \bigoplus_{\chi \in \hat\mcz(\lieg)} (M_\mu \intersect M^\chi),$$ and since $M$ is generated by a finite sum of weight spaces, $M = \bigoplus_{\chi \in \hat\mcz(\lieg)} M^\chi$. - By Harish-Chandra's theorem, every $\chi$ is $\chi_\lambda$ for some $\lambda \in \lieh\dual$. Let $\OO_\chi$ be the full subcategory of modules $M$ such that $M = M^\chi$; we refer to this as a *block*. > Note: full subcategory means keep all of the hom sets. Proposition (O Factors into Blocks, Indecomposables/Highest Weight Modules Lie in a Single Block) : $\OO = \bigoplus_{\lambda \in \lieh\dual} \OO_{\chi_\lambda}$. Each indecomposable module in $\OO$ lies in a *unique* $\OO_\chi$. In particular, any highest weight module of highest weight $\lambda$ lies in $\OO_{\chi_\lambda}$. Thus we can reduce to studying $\OO_{\chi_\lambda}$. *Remark:* $\OO_{\chi_\lambda}$ has a finite number of simple modules $\theset{L(w\cdot \lambda) \suchthat w\in W}$ and a finite number of Verma modules $\theset{M(w\cdot \lambda) \suchthat w\in W}$. ## Blocks Let $\mcc$ be a category with is artinian and noetherian, with $L_1, L_2$ simple modules. We say $L_1 \sim L_2$ if there exists a non-split extension $$0 \to L_1 \to M \to L_2 \to 0,$$ i.e. $\ext^1_\OO(L_2, L_1) \neq 0$. In particular, $M$ equivalently needs to be indecomposable. We then extend $\sim$ to be reflexive/symmetric/transitive to obtain an equivalence relation. > $L_1$ ends up being the socle here. This partitions the simple modules in $\mcc$ into *blocks* $\mcb$. More generally, we say $M\in \mcc$ belongs to $\mcb$ iff all of the composition factors of $M$ belong to $\mcb$. Although not obvious, there are no nontrivial extensions between modules in different blocks. Thus each simple module (generally, just an object) $M\in \mcc$ decomposes as a direct sum of submodules (subobjects) with each belonging to a single block. *Question:* Is $\OO_\chi$ a block of $\OO$? The answer is not always. Because each indecomposable module in $\OO$ lives in a simple $\OO_\chi$ By the definition, it's clear that each block is contained in a single simple infinitesimal block $\OO_\chi$. > The block containing $L_1, L_2$ will be contained in the same infinitesimal block, and continuing the composition series puts all composition factors in a single block. Proposition (Integral Weights Yield Simple Blocks) : If $\lambda$ is an *integral* weight, so $\lambda \in \Lambda$, then $\OO_{\chi_\lambda}$ is a (simple) block of $\OO$. Proof : It suffices to show that all $L(w\cdot \lambda)$ for $w\in W$ lie in a single block. We'll induct on the length of $w$. Start with $2 = s_\alpha$ for some $\alpha\in \Delta$. Let $\mu = s_\alpha \cdot \lambda$. If $\mu = \lambda$, i.e. $\lambda$ is in the stabilizer, then we're done. Otherwise, assume WLOG $\mu < \lambda$ in the partial order, using the fact that $\lambda \in \Lambda$. (The difference between these is just an integer multiple of $\alpha$.) By proposition 1.4, we have the following maps: \begin{center} \begin{tikzcd} M(\mu) \arrow[rr, "\phi\neq 0"] & & N(\lambda) \arrow[rr, hook] & & M(\lambda) \\ & & & & \\ N(\mu) \arrow[rr, tail] \arrow[uu, hook] & & N = \phi(N(\mu)) \arrow[uu, hook] & & \end{tikzcd} \end{center} Then $\phi$ induces a map $L(\mu) \mapsvia{\bar \phi} M(\lambda)/N$, where the codomain here is a highest weight module with quotient $L(\lambda)$. Since highest weight modules are indecomposable and thus lie in a single bloc, $L(\mu)$ and $L(\lambda)$ are in the same block. > Note that if $v^+$ generates $M(\lambda)$, $v^+ + N$ generated the quotient. Now inducting on $\ell(w)$, iterating this argument yields all $L(w\cdot \lambda)$ (as $w$ varies) in the same block. Example : This isn't true for non-integral weights. Let $\lieg = \liesl(2, \CC)$ with $\lambda \in \RR \setminus \ZZ$ and $\lambda > -1$. Then \begin{align*} \mu &= s_\alpha \lambda \\ &= -\lambda - 2 \\ &<_{\RR} -1 \end{align*} with the usual ordering on $\RR$, but $\mu \not > \lambda$ in the ordering on $\lieh\dual$: we have $\lambda - \mu = 2\lambda + 2$, but $\alpha \equiv 2$ and thus these don't differ by an element of $2\ZZ$. Thus $\mu, \lambda$ are in different cosets of $\ZZ\Phi = \Lambda_r$ in $\lieh\dual$. However, $M(\lambda), M(\mu)$ are simple since $\lambda, \mu$ are not non-negative integers. By exercise 1.13, there can be no nontrivial extension, so they're in different homological blocks but in the same $\OO_{\chi_\lambda}$ since $\mu = s_\alpha \cdot \lambda$. So this infinitesimal block splits into multiple homological blocks. Friday: 1.14 and 1.15.