# Monday March 2nd ## Indecomposable Projectives :::{.remark} Recall that - $\OO$ has enough projectives - Every $M\in \OO$ has a projective cover, Let $\pi_\lambda: P(\lambda) \to L(\lambda)$ be the projective cover of $L(\lambda)$. ::: :::{.claim} $P(\lambda)$ is indecomposable. ::: :::{.proof title="of claim"} $\ker \pi_\lambda$ is a maximal submodule of $P(\lambda)$. Suppose $N$ were another submodule of $P(\lambda)$, then $\restrictionof{\pi_\lambda}{N} \neq 0$, so it maps to $N \to L(\lambda)$, which is a contradiction. So $P(\lambda)$ has a *unique* maximal submodule. ::: :::{.theorem title="?"} \envlist a. Every indecomposable projective in $\OO$ is isomorphic to $P(\lambda)$ for some $\lambda \in \lieh\dual$ b. Any projective $P = \bigoplus_{\lambda \in \lieh\dual} d_\lambda P(\lambda)$, where $d_\lambda$ counts the multiplicity in the direct sum. c. For all $M\in \OO$, $\dim \hom_\OO(P(\lambda), M) = [M: L(\lambda)]$. In particular, \[ \dim \Endo_\OO(P(\lambda)) = [P(\lambda): L(\lambda)] .\] ::: :::{.proof title="of a"} Suppose $P\in \OO$ is an indecomposable projective, then $P$ has at least one simple quotient $L(\lambda)$. [Diagram] Since $\pi_\lambda$ is essential, $\psi$ must be onto. [Diagram] This splits the sequence, so $P(\lambda) \divides P$. Since $P$ is indecomposable, $P = P(\lambda)$. ::: :::{.proof title="of b"} $P$ is a direct sum of indecomposable modules, but a direct summand of a projective module is projective. Thus $P = \bigoplus d_\lambda P(\lambda)$. So $P(\lambda)$ has $L(\lambda)$ as its unique simple quotient, and \[ \dim \hom_\OO(P(\lambda), L(\lambda)) = \dim_\OO \Endo(L(\lambda)) = 1 .\] Note that $\hom_\OO(P(\mu), L(\lambda)) = 0$ for $\mu \neq \lambda$, so \[ \dim \hom_\OO(\oplus d_\lambda P(\mu), L(\lambda)) = d_\lambda .\] ::: :::{.remark} Think about how to relate composition series in a SES. ::: :::{.proof title="of c"} Both sides are additive in $M$, i.e. for $0 \to M_1 \to M \to M_2 \to 0$, we have $[M: L(\lambda)] = [M_1: L(\lambda)] + [M_2: L(\lambda)]$. Since $P(\lambda)$ is projective, $\hom(P(\lambda), \wait)$ is exact, so the dimensions of the Hom spaces are additive, and induction can be used on the Jordan-Holder length of $M$. Start with $M = L(\lambda)$ a simple module, then \[ \dim \hom(P(\lambda), L(\mu)) = \delta_{\mu, \lambda} = [ L(\mu): L(\lambda)] .\] Note that the transpose dual functor $(\wait)\dual$ is an exact contravariant functor, taking - Projective to injectives (and vice versa) - Direct sums to direct sums Thus the indecomposable injectives are the duals of projectives, i.e. $P(\lambda)\dual \definedas Q(\lambda)$, the *injective hull* of $L(\lambda)$. This is characterized by an essential monomorphism $$ i_\lambda: L(\lambda) \injects Q(\lambda) ,$$ i.e. $\im i_\lambda$ intersects *every* nonzero submodule of $Q(\lambda)$. ::: ## Standard Filtrations of Projective :::{.remark} Every $M\in \OO$ admits a filtration whose sections are highest weight modules, but it's rare for the sections to be Vermas. ::: :::{.theorem title="?"} Each projective module in $\OO$ has a standard filtration. $(P(\lambda): M(\mu)) \neq 0 \implies \mu \geq \lambda$, and $(P(\lambda): M(\lambda)) = 1$. ::: :::{.proof title="?"} Recall the proof of theorem 3.8: there exists a dominant weight $\mu$ and a finite dimensional $L$ such that $P \definedas M(\mu) \tensor L \surjects M(\lambda) \surjects L(\lambda)$, and $P$ is projective by a previous theorem. Every projective is a direct sum of indecomposables, and $P(\lambda) \divides P$ is a direct summand. $P$ has a standard filtration involving $M(\lambda)$ once, and all other sections are of the form $M(\tau)$ for $\tau > \lambda$. By 3.7b, direct summands $P(\lambda)$ of$P$ inherit standard filtrations with the same properties. In particular, $P(\lambda) \surjects M(\lambda) \surjects L(\lambda)$, and the highest weights are the same as those for $P$. Since any projective is a finite direct sum of $P(\lambda)$s and any direct sum of modules with standard filtrations has a standard filtration, thus $P$ does as well. ::: :::{.corollary title="?"} Each projective $P\in \OO$ is determined up to isomorphism by its formal character. ::: :::{.remark} Note that a Verma and its dual may have the same formal character without being isomorphic. ::: :::{.proof title="?"} Write $P = \bigoplus d_\lambda P(\lambda)$, then it is sufficient to show that $\ch P$ determines the $d_\lambda$s. Use induction on the length $\ell(P)$. For $\ell(P) = 1$, we have $P = L(\lambda) = P(\lambda)$. For $\ell(P) > 1$, by 1.16 we can write $\ch P = \sum c_\mu \ch M(\mu)$ for unique $c_\mu \in \ZZ$, since the Vermas span $K(\OO)$. By the theorem, $c_\mu > 0$ for all $\mu$. Trick: choose $\lambda$ minimal for which $c_\lambda > 0$. By the theorem, $P(\lambda)$ must occur exactly $c_\lambda$ times as a summand of $P$. Thus $c_\lambda = d_\lambda$ by definition. Now mod out by $d_\lambda P(\lambda)$ and apply the induction hypothesis. ::: ## BGG Reciprocity :::{.theorem title="BGG Reciprocity"} For all $\lambda, \mu \in \lieh\dual$, \[ (P(\lambda): M(\mu)) = [M(\mu): L(\lambda)] = [M(\mu)\dual: L(\lambda)] ,\] where the last equality holds because $\ch M(\mu) = \ch M(\mu)\dual$. ::: :::{.remark title="Proof strategy"} We'll show that that first and last are both equal to $\dim \hom_\OO(P(\lambda), M(\mu)\dual)$. This lets us relate projectives, indecomposable projectives, and simple (?) modules. ::: ### Proof of BGG Reciprocity **Step 1:** Note that $\dim \hom_\OO(P(\lambda), M) = [M: L(\lambda)]$ for any $M\in \OO$ by theorem 3.9c. So take $M = M(\mu)\dual$. **Step 2:** For any $M\in \OO$ with a standard filtration, we know by theorem 3.7 that $(M: M(\mu)) = \dim \hom(M, M(\mu)\dual)$. So take $M = P(\lambda)$. :::{.example title="?"} Let $\lieg = \liesl(2, \CC)$. Let $\lambda \in \lieh\dual \cong \CC$. If $\lambda \not\in \Lambda^+$, then $\OO_{\chi_\lambda}$ is boring: it is semisimple with one simple module. For $\lambda \in \Lambda^+ \cong \ZZ^{>0}$, define $\mu = s_\alpha \cdot \lambda = -\lambda - 2$. Then \[ 0 \to M(\mu) \to M(\lambda) \to L(\lambda) \to 0 ,\] where $M(\mu) = L(\mu) = N(\lambda)$. ::: Note that $P(\lambda)$ has a filtration of length 1, since its quotients must be Vermas. If $\lambda$ is dominant, then $M(\lambda) = P(\lambda)$. $P(\mu)$ has a standard filtration with $M(\mu)$ occurring once at the top, and $(P(\mu) : M(\lambda)) = [M(\lambda): L(\mu)] = 1$. \begin{tikzpicture} \draw [decorate,decoration={brace,amplitude=10pt, mirror},xshift=-4pt,yshift=0pt] (0,0) -- (0,1.2) node [black,midway,xshift=0.8cm] {\footnotesize $M(\mu)$}; \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (0,-0.25) -- (0,-2) node [black,midway,xshift=0.8cm] {\footnotesize $M(\lambda)$}; \node at (-1,1) {$L(\mu)$}; \draw (-1.7, .25) -- (-.2, .25); \node at (-1,-.5) {$L(\lambda)$}; \draw (-1.7, -1) -- (-.2, -1); \node at (-1,-1.5) {$L(\mu)$}; \end{tikzpicture} This yields a SES \[ 0 \to M(\lambda) \to P(\mu) \to M(\mu) \to 0 ,\] which is a **nonsplit extension**. This gives an example of a self-dual projective, i.e. $P(\mu)\dual = Q(\mu) \cong P(\mu)$. ## Unknown Diagrams \todo[inline]{Find where these diagrams were supposed to go.} \begin{tikzcd} & & P \arrow[lldd, "\psi", dashed] \arrow[dd, "\varphi"] & & \\ & & & & \\ P(\lambda) \arrow[rr, "\pi_\lambda ?"'] & & L(\lambda) \arrow[rr] & & 0 \end{tikzcd} \begin{tikzcd} & & P \arrow[lldd, dashed] \arrow[dd, no head, Rightarrow] & & \\ & & & & \\ P \arrow[rr, "\psi"'] & & P(\lambda) \arrow[rr] & & 0 \end{tikzcd} \begin{align*} \frac{L(\lambda) }{L(\mu)} = P(\lambda) .\end{align*} \begin{tikzcd} M(\mu) \arrow[rr, "\varphi_2 \neq 0"', bend right] \arrow[rr, "\varphi_1 \neq 0", bend left] & & M(\lambda) & \\ & & & \\ & & & L_1 \arrow[luu, "\subset" description] \\ \soc M(\mu) = L \text{ simple} \arrow[rr, hook] \arrow[rrru, hook] \arrow[uuu, "\subset"] & & L_2 \arrow[uuu, "\subset"] & \end{tikzcd}