# Wednesday February 26th

We want to show the following identity:

\begin{align*}
\qty{ U(\lieg) \tensor_{U(\lieb)} L } \tensor_\CC M 
\cong
U(\lieg) \tensor_{U(\lieb)} \qty{ L \tensor_\CC M  }
.\end{align*}

Assume $L$ and $M$ are finite dimensional. Then for $N = L \tensor M$, there is a basis of weight vectors $v_1, \cdots, v_n, \nu_1, \cdots, \nu_m$ with $\nu_i \leq \nu_j \iff i\leq j$.
Moreover $$N_k = \CC \generators{v_k, \cdots, v_n} = U(\lieb)\generators{v_k, \cdots, v_n},$$ and we have a natural filtration

\begin{align*}
0 \subset N_n \subset \cdots \subset N_1 = N
\end{align*}
 with $N_i / N_{i+1} \cong \CC_{v_i}$ as $\lieb\dash$modules.

 We thus obtain $$\ind_\lieb^\lieg N_i / \ind_\lieb^\lieg N_{i+1} \cong \ind_\lieb^\lieg \CC_{v_i} = M(v_i)$$ by exactness of the Ind functor.
 Apply this to $L = \CC_\lambda$, then the LHS is $M(\lambda) \tensor_\CC M$, where $M$ is finite dimensional.
 On the RHS, $N = \CC_\lambda \tensor M$ has the same dimension as $M$ with weights $\lambda + \mu$ for $\mu \in \Pi(M)$.
 Thus $M(\lambda) \tensor M$ has filtration with quotients $M(\lambda + \mu)$ over $\mu \in \Pi(M)$, which was the theorem we had last time.

 Remark
 : The proof shows that $M(\lambda) \tensor M$ has a submodules $M(\lambda + \mu)$ for any maximal weight $\mu$ of $M$, and a quotient $M(\lambda + \nu)$ where $\nu$ is any minimal weight of $M$.
  We knew that every $M\in \OO$ has a finite filtration, but here the quotients are now Verma modules.
  This will help us study projectives later, which we need to study higher Exts.

## Standard Filtrations

There are several main players in the theory of highest-weight categories, of which $\OO_{\chi_\lambda}$ is one:

- Simple modules: $L(\lambda)$
- Standard modules $M(\lambda)$
- Costandard modules $M(\lambda)\dual$
- Indecomposable projectives $P(\lambda)$
- Tilting modules $T(\lambda)$.

Definition (Standard Filtration/Verma flag)
: A *standard filtration* of $M\in \OO$ is a filtration with subquotients isomorphic to Verma modules.

Note that when $M$ has a standard filtration, the submodules are *not* unique, but the length, subquotients, and multiplicities are unique.
We can thus use $K(\OO)$ or formal characters as an invariant, since the multiplicities $(M: M(\lambda))$ are well-defined.

If $M, N$ have standard filtration, then so does $M \oplus N$ by concatenation.
In this case, $(M\oplus N: M(\lambda)) = (M:M(\lambda)) + (N: M(\lambda))$.

Proposition (Submodules and Direct Summands Also Have Standard Filtrations)
:   Let $M\in \OO$ have a standard filtration. 
    Then

    a. If $\lambda$ is maximal in $\Pi(M)$, then $M$ has a submodule isomorphic to $M(\lambda)$ and $M/M(\lambda)$ has a standard filtration $$0 = M_0 \subset \cdots \subset M_n = M.$$
    b. If $M = M' \oplus M''$, then $M'$ and $M''$ have standard filtrations.
    c. $M$ is free as a $U(\lien^-)\dash$module.

Proof (of (a))
:   By assumption on $\lambda$, $M$ has a maximal vector of weight $\lambda$, and thus the universal property yields a nonzero morphism $\phi: M(\lambda) \to M$.
  
    The claim is that $\phi$ is injective, from which the proof follows.
    Proof of claim: choose a minimal index $i$ such that $\phi(M(\lambda)) \subset M_i$ in the filtration.
    Follow this with the subquotient map to yield $$\psi: M(\lambda) \to M^i \definedas M_i/M_{i-1} \cong M(\mu),$$ which is nonzero by minimality of $i$.

    Thus $\lambda \leq \mu$, and by our assumption, this implies $\lambda = \mu$.
    But then $\psi$ sends highest weight vectors to highest weight vectors and is free, so $\psi$ is an isomorphism.
    Thus $\phi$ is injective and $M(\lambda) \subset M$.

    We can now write $M(\lambda) \intersect M_{i-1} = \ker \psi = 0$, so we obtain a direct sum decomposition $M_i \cong M_{i-1} \oplus M(\lambda)$.
    We thus obtain a SES 
    $$0 \to M_{i-1} \to M/M(\lambda) \to M/M_i \to 0.$$

    We can easily construct standard filtrations for $M_{i-1}$ and $M/M_i$, so the middle term also has a standard filtration.
    Thus $M/M(\lambda)$ has a standard filtration of length one less than that of $M$.

Proof (of(b))
:   By induction of the filtration length $n$ of $M$. 
    If $n=0$, $M$ is a Verma module and thus indecomposable and there's nothing to show.
    
    For $n\geq 1$, let $\pi \in \Pi(M)$ be maximal (which we can always find for $M\in \OO)$) and WLOG $M_\lambda' \neq 0$.

    By the universal property, we have a nonzero composition
    
    \begin{center}
    \begin{tikzcd}
    M(\lambda) \arrow[rr] \arrow[rrrr, "\neq 0", dashed, bend right] &  & M' \arrow[rr, hook] &  & M
    \end{tikzcd}
    \end{center}

    Applying (a) to this composite map,

    1. It must be injective, so $M(\lambda) \injects M'$
    2. $M/M(\lambda) \cong M'/M(\lambda) \oplus M''$ has a standard filtration of length $n-1$.

    By induction, $M'/M(\lambda)$ and $M''$ have standard filtrations, and thus so does $M'$.

Proof (of (c))
: By induction on $n$: if $n=1$, then $M \cong M(\lambda)$ is $U(\lien^-)\dash$free.
  Otherwise, if $n > 1$, by (a) $M(\lambda) \subset M$ and $M/M(\lambda)$ has a standard filtration of length $n-1$.
  By induction, $M/M(\lambda)$ is $U(\lien^-)\dash$free, and hence so is $M$.


Theorem (Multiplicities of Vermas)
: If $M$ has a standard filtration, then $(M: M(\lambda)) = \dim \hom_\OO(M, M(\lambda)\dual)$.

Proof
: By induction on the filtration length $n$.
  If $n=1$, $M$ is a Verma module, and $(M(\mu) : M(\lambda)) = \delta_{\mu \lambda} = \dim \hom_\OO(M(\mu), M(\lambda)\dual)$ by Theorem 3.3c.
  
  For $n>1$, consider $$0 \to M_{n-1} \to M \to M(\mu) \to 0.$$
  Apply the left-exact contravariant functor $\hom_\OO(\wait, M(\lambda)\dual)$ to obtain
  
  \begin{center}
  \begin{wideeq}
    \begin{tikzcd}
0 \arrow[r]   & 
{\hom(M(\mu), M(\lambda)\dual)} \arrow[r] \arrow[dd, Rightarrow] &
{\hom(M, M(\lambda)\dual)} \arrow[r] \arrow[dd, Rightarrow]   &  
{\hom(M_{n-1}, M(\lambda)\dual)} \arrow[r] \arrow[dd, Rightarrow] &   
{\ext(M(\mu), M(\lambda)\dual)} \arrow[r]   &  
\cdots \\
& 
& 
& 
&  
&   
& \\
&  
\delta_{\mu\lambda} &  
\parbox{2cm}{\tiny $(M: M(\lambda)) = (M_{n-1}\text{:} M(\lambda)) + \delta_{\mu \lambda}$ } &  
\parbox{2cm}{\tiny $(M_{n-1}\text{:} M(\lambda)) \\ \text{ by induction }$ } &  
\text{ \tiny 0 by Thm 3.3d} \arrow[uu, Rightarrow] &  
\end{tikzcd}
\end{wideeq}
  \end{center}

## Projectives in $\OO$

We want to show that $\OO$ has *enough projectives*, i.e. every $M\in \OO$ is a quotient of a projective object.
We'll also want to show $\OO$ has *enough injectives*, i.e. every modules embeds into an injective object.

Definition (Projective Objects)
:   If $\mca$ is an abelian category, an object $P\in\mca$ is *projective* iff the left-exact functor $\hom_\mca(P, \wait)$ is exact, or equivalently

    \begin{center}
    \begin{tikzcd}
                &  & P \arrow[dd, "f"] \arrow[lldd, "\exists \tilde f"] &  &   \\
                &  &                                                    &  &   \\
    M \arrow[rr] &  & N \arrow[rr]                                       &  & 0
    \end{tikzcd}
    \end{center}

    In other words, there is a SES $$\hom(P, M) \to \hom(P, N) \to 0,$$ which precisely says that every $f$ in the latter has a lift $\tilde f$ in the former by surjectivity.

Definition (Injective Objects)
:   An object $Q\in\mca$ is *injective* iff $\hom_\mca(\wait, Q)$ is exact, i.e.

    \begin{center}
    \begin{tikzcd}
    0 \arrow[rr] &  & N \arrow[rr] \arrow[dd, "g"] &  & M \arrow[lldd, "\exists \tilde g"] \\
                &  &                              &  &                                    \\
                &  & Q                            &  &                                   
    \end{tikzcd}
    \end{center}

    i.e., $$\hom_\mca(M, Q) \to \hom_\mca(N, Q) \to 0$$ so every $g$ in the latter has a lift to $\tilde g$ in the former.

In $\OO$, having enough projectives is equivalent to having enough injectives because $(\wait)\dual$ is an exact contravariant endofunctor, which sends projectives to injectives and vice-versa.
Thus we'll focus on projectives.