# Friday February 28th
Recall that $\lambda \in \lieh\dual$ is *dominant* iff for all $\alpha \in \Phi^+$, we have $(\lambda + \rho, \alpha\dual) \not\in\ZZ^{<0}$.
Equivalently, as in Proposition 3.5c, $\lambda$ is maximal in its $W_{[\lambda]}\cdot$ orbit.
## Constructing Projectives
Proposition (Dominant Weights Yield Projective Vermas, Projective Tensor Finite-Dimensional is Projective)
: a. If $\lambda \in \lieh\dual$ is dominant, then $M(\lambda)$ is projective in $\OO$.
b. If $P\in \OO$ is projective and $\dim L < \infty$< then $P \tensor_\CC L$ is projective.
Proof
: a. We want to find a $\psi$ making this diagram commute:
\begin{center}
\begin{tikzcd}
& & v^+ \in M(\lambda) \arrow[dd, "\phi"] \arrow[lldd, "\psi", dashed] & & \\
& & & & \\
M \arrow[rr, "\pi"] & & p(v^+) \in N \arrow[rr] & & 0
\end{tikzcd}
\end{center}
Assume $\phi \neq 0$.
Since $M(\lambda) \in \OO_{\chi_\lambda}$, we have $\phi(M(\lambda)) \subset N^{\chi_\lambda}$.
WLOG, we can assume $M, N \in \OO_{\chi_\lambda}$, and if $v^+$ is maximal, $p(v^+)$ is maximal.
By surjectivity of $\pi$, there exists a $v\in M$ such that $v \mapsto p(v^+)$.
Then $M \supset U(\lien) v$ is finite dimensional, so it contains a maximal vector whose weight is linked to $\lambda$ since $M\in \OO_{\chi_\lambda}$.
But since $\lambda$ is dominant, there is no such weight greater than $\lambda$, so $v$ itself must be this maximal vector.
Then by the universal property of $M(\lambda)$, there is a map $\psi: M(\lambda) \to M$ where $v^+ \mapsto v$ making the diagram commute.
> Note nice property: Vermas are projective iff maximal in orbit.
b. We want to show $F = \hom_\OO(P\tensor L, \wait)$ is exact.
But this is isomorphic to
$$\hom_\OO(P, \hom_\CC(L, \wait)) \cong \hom_\OO(P, L\dual \tensor_\CC \wait).$$
Thus $F$ is the composition of two exact functors: first do $L\dual \tensor_\CC \wait$, which is exact since $\CC$ is a field, and $\hom_\OO(P,\wait)$ is exact since $P$ is projective.
Example
: Let $M(-\rho)$ be the Verma of highest weight $\rho$.
This is irreducible because $-\rho$ is antidominant, and projective since $-\rho$ is dominant.
In fact $W\cdot(-\rho) = \theset{-\rho}$ by a calculation.
Thus $$L(-\rho) = M(-\rho) = P(-\rho) = M(-\rho)\dual,$$ so all 4 members of the highest weight category here are equal.
> By convention, there is notation $M(-\rho) = \Delta(-\rho)$ and $M(-\rho)\dual = \nabla(-\rho)$.
Note that we always have $\ext_0(L(-\rho), L(-\rho)) = 0$, and every $\OO_{\chi_{-\rho}} \in M$ is equal to $\bigoplus L(-\rho)^{\oplus n}$.
> So this is referred to as a *semisimple category*.
Theorem (O has Enough Projectives and Injectives)
: $\OO$ has enough projectives and injectives.
Proof
: **Step 1**
For all $\lambda \in \lieh\dual$, there exists a projective mapping onto $L(\lambda)$.
Clearly $\mu \definedas \lambda + n\rho$ is dominant for $n\gg 0$, i.e. for $n$ large enough there are no negative integers resulting from inner products with coroots.
Thus $M(\mu)$ is projective, and since $n\rho \in \Lambda^+$, we have $\dim L(n\rho) < \infty$.
This implies $P \definedas M(\mu) \tensor L(n\rho)$ is projective by the previous proposition.
Apply $w_0$ reverses the weights, so $w_0(n\rho) = -n\rho$.
Note that this doesn't happen for all weights, so this property is somewhat special for $\rho$.
In particular, since $n\rho$ was a highest weight, $-n\rho$ is a lowest weight.
By remark 3.6, $P$ has a quotient isomorphic to $M(\mu - n\rho) = M(\lambda)$.
Thus $P \surjects M(\lambda) \to L(\lambda)$, and $L(\lambda)$ is a quotient of a projective.
This establishes the result for simple modules.
> Remark: By theorem 3.6, $P$ has a standard filtration with sections $M(\mu + \nu)$ for $\nu \in \Pi(L(n\rho))$.
> In particular $M(\lambda)$ occurs just once since $$\dim L(n\rho)_{-n\rho} = \dim L(n\rho)_{w_0(n\rho)} = \dim L(n\rho)_{n\rho} = 1,$$ with all $\mu + \nu > \lambda$.
**Step 2**
Use induction on Jordan-Hilder length to prove that any $0\neq M\in \OO$ is a quotient of a projective.
For $\ell = 1$, $M$ is simple, and by Step 1 this case holds.
Assume $\ell > 1$, then $M$ has a submodule $L(\lambda)$ obtained by taking the bottom of a Jordan-Holder series, so there is a SES
$$0 \to L(\lambda) \mapsvia{\alpha} M \mapsvia{\beta} N \to 0.$$
By induction, since $\ell(N) = \ell(M) - 1$, there exists a projective module $Q \mapsvia{\phi} N$ which extends to a map $\psi: Q \to M$.
If $\psi$ is surjective, we are done.
Otherwise, then the composition length forces $\psi(Q) \cong N$, and by commutativity there is a section $\gamma: N \to \psi(Q)$ splitting this SES.
Thus $M \cong L(\lambda) \oplus N$, and by 1, there are projectives $P \oplus Q$ projecting onto each factor, so $M$ is projective.
\begin{center}
\begin{tikzcd}
0 \arrow[rr] & & L(\lambda) \arrow[rr, "\alpha"] & & M \arrow[rr, "\beta"] & & N \arrow[rr] \arrow[ll, "\gamma", dotted, bend left] & & 0 \\
& & & & & & & & \\
& & & & & & P \arrow[uu, "\varphi"'] \arrow[lluu, "\psi", dashed] & &
\end{tikzcd}
\end{center}
## 3.9 Indecomposable Projectives
Definition (A Projective Cover)
: A *projective cover* of $M\in \OO$ is a map $\pi: P_M \to M$ where $P_M$ is projective and $\pi$ is an *essential epimorphism*, i.e. no proper submodule of $P_M$ is mapped surjectively onto $M$ by $\pi_M$.
It is an algebraic fact that in an Artinian (abelian) category with enough projectives, every module has a projective cover that is unique up to isomorphism.
> See Curtis and Reiner, Section 6c.
Definition (The Projective Cover for a Weight)
: For $\lambda \in \lieh\dual$, denote $\pi_\lambda: P(\lambda) \surjects L(\lambda)$ to be a fixed projective cover of $L(\lambda)$.