[proofplan]
We use the [PBW model of a Verma module](/theorems/9374) to isolate its one-dimensional highest weight space and its lower weight-space decomposition. First we prove that every proper submodule misses the highest weight line, because any nonzero vector on that line generates all of $M(\lambda)$. Then we prove that submodules split into their weight components, so the sum of all proper submodules still cannot acquire a highest weight vector. This sum is therefore the unique maximal proper submodule, and the corresponding quotient is simple and universal among simple quotients.
[/proofplan]
[step:Fix the highest weight generator and the PBW weight decomposition]
Let $v_\lambda:=1\otimes 1\in M(\lambda)$ denote the standard highest weight generator, where the first $1$ is the unit of $U(\mathfrak g)$ and the second $1$ is a nonzero basis vector of $\mathbb C_\lambda$. By the definition of $\mathbb C_\lambda$, for every $h\in\mathfrak h$ and every $x\in\mathfrak n^+$,
\begin{align*}
h\cdot v_\lambda=\lambda(h)v_\lambda.
\end{align*}
Also,
\begin{align*}
x\cdot v_\lambda=0.
\end{align*}
Let $Q_+$ denote the additive monoid generated by the positive roots determined by $\mathfrak b$. For each $\mu\in\mathfrak h^*$, define the $\mu$-weight space of $M(\lambda)$ by
\begin{align*}
M(\lambda)_\mu:=\{m\in M(\lambda):h\cdot m=\mu(h)m\text{ for every }h\in\mathfrak h\}.
\end{align*}
The PBW model of Verma modules, [citetheorem:9374], gives a weight-space [direct sum](/page/Direct%20Sum)
\begin{align*}
M(\lambda)=\bigoplus_{\beta\in Q_+}M(\lambda)_{\lambda-\beta}.
\end{align*}
Moreover,
\begin{align*}
M(\lambda)_\lambda=\mathbb C v_\lambda.
\end{align*}
Finally, $M(\lambda)$ is generated as a $U(\mathfrak g)$-module by $v_\lambda$, because $M(\lambda)=U(\mathfrak g)\cdot v_\lambda$ by construction of the induced module.
[/step]
[step:Show that a proper submodule cannot meet the highest weight line]
Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule. Suppose that
\begin{align*}
N\cap M(\lambda)_\lambda\neq \{0\}.
\end{align*}
Since $M(\lambda)_\lambda=\mathbb C v_\lambda$, there exists $c\in\mathbb C\setminus\{0\}$ such that $c v_\lambda\in N$. Because $N$ is a complex vector subspace, $v_\lambda=c^{-1}(c v_\lambda)\in N$. Since $N$ is stable under the $\mathfrak g$-action, it is stable under the induced $U(\mathfrak g)$-action, hence
\begin{align*}
U(\mathfrak g)\cdot v_\lambda\subseteq N.
\end{align*}
But $U(\mathfrak g)\cdot v_\lambda=M(\lambda)$, so $N=M(\lambda)$.
Therefore every proper $\mathfrak g$-submodule $N\subsetneq M(\lambda)$ satisfies
\begin{align*}
N\cap M(\lambda)_\lambda=\{0\}.
\end{align*}
[guided]
The key point is that the highest weight line is not just another weight space: it contains the generator of the whole Verma module. Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule and assume that $N$ contains a nonzero vector of highest weight $\lambda$. Since the highest weight space is one-dimensional,
\begin{align*}
M(\lambda)_\lambda=\mathbb C v_\lambda,
\end{align*}
there is a scalar $c\in\mathbb C\setminus\{0\}$ with $c v_\lambda\in N$. Because $N$ is a complex vector subspace, multiplying by $c^{-1}$ gives
\begin{align*}
v_\lambda\in N.
\end{align*}
Now use the defining generation property of a Verma module. The module $M(\lambda)$ is generated by $v_\lambda$, so
\begin{align*}
M(\lambda)=U(\mathfrak g)\cdot v_\lambda.
\end{align*}
Since $N$ is a $\mathfrak g$-submodule, it is stable under the action of every element of $U(\mathfrak g)$. Therefore
\begin{align*}
U(\mathfrak g)\cdot v_\lambda\subseteq N.
\end{align*}
Combining the two displayed equalities gives $M(\lambda)\subseteq N$, and the reverse inclusion $N\subseteq M(\lambda)$ is part of the definition of submodule. Hence $N=M(\lambda)$.
This proves the contrapositive form we need later: if $N$ is proper, then it cannot contain any nonzero vector in $M(\lambda)_\lambda$.
[/guided]
[/step]
[step:Project submodule elements onto their weight components]
Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule. We prove that
\begin{align*}
N=\bigoplus_{\mu\in\mathfrak h^*}\left(N\cap M(\lambda)_\mu\right),
\end{align*}
where only weights of $M(\lambda)$ contribute.
The inclusion from right to left follows because each $N\cap M(\lambda)_\mu$ is contained in $N$. For the reverse inclusion, let $n\in N$. Since $M(\lambda)$ is a direct sum of weight spaces, there are distinct weights $\mu_1,\dots,\mu_r\in\mathfrak h^*$ and nonzero vectors
\begin{align*}
n_i\in M(\lambda)_{\mu_i}
\end{align*}
such that
\begin{align*}
n=n_1+\cdots+n_r.
\end{align*}
Fix an index $i\in\{1,\dots,r\}$. Choose $H_i\in\mathfrak h$ such that the complex numbers $\mu_1(H_i),\dots,\mu_r(H_i)$ separate $\mu_i(H_i)$ from all the others. This is possible because the finitely many hyperplanes
\begin{align*}
\{H\in\mathfrak h:\mu_i(H)=\mu_j(H)\}
\end{align*}
with $j\neq i$ cannot cover $\mathfrak h$.
Define a polynomial $p_i\in\mathbb C[T]$ by
\begin{align*}
p_i(T):=\prod_{j\neq i}\frac{T-\mu_j(H_i)}{\mu_i(H_i)-\mu_j(H_i)}.
\end{align*}
Since $N$ is stable under $H_i$, it is stable under the operator $p_i(H_i)$. Therefore
\begin{align*}
p_i(H_i)\cdot n\in N.
\end{align*}
For each $j$, the vector $n_j$ is an eigenvector of $H_i$ with eigenvalue $\mu_j(H_i)$, so
\begin{align*}
p_i(H_i)\cdot n=n_i.
\end{align*}
Thus $n_i\in N$. Since this holds for every $i$, every weight component of $n$ belongs to $N$, and the asserted decomposition follows.
[/step]
[step:Form the sum of all proper submodules and prove it is still proper]
Let $\mathcal S$ denote the set of all proper $\mathfrak g$-submodules of $M(\lambda)$, and define
\begin{align*}
J(\lambda):=\sum_{N\in\mathcal S}N.
\end{align*}
This is a $\mathfrak g$-submodule, because it is a sum of $\mathfrak g$-submodules.
We prove that $J(\lambda)$ is proper. Suppose, toward contradiction, that
\begin{align*}
J(\lambda)\cap M(\lambda)_\lambda\neq \{0\}.
\end{align*}
Choose a nonzero vector $u\in J(\lambda)\cap M(\lambda)_\lambda$. By definition of a sum of subspaces, there exist finitely many proper submodules $N_1,\dots,N_s\in\mathcal S$ and vectors $u_k\in N_k$ such that
\begin{align*}
u=u_1+\cdots+u_s.
\end{align*}
By the weight-component decomposition for submodules, each highest weight component of $u_k$ lies in
\begin{align*}
N_k\cap M(\lambda)_\lambda.
\end{align*}
Each $N_k$ is proper, so the preceding step on the highest weight line gives
\begin{align*}
N_k\cap M(\lambda)_\lambda=\{0\}.
\end{align*}
Hence every $u_k$ has zero $\lambda$-weight component. Therefore the $\lambda$-weight component of $u_1+\cdots+u_s$ is zero, contradicting the fact that $u$ is a nonzero element of $M(\lambda)_\lambda$.
Thus
\begin{align*}
J(\lambda)\cap M(\lambda)_\lambda=\{0\}.
\end{align*}
By the previous highest-weight-line argument, any submodule meeting $M(\lambda)_\lambda$ nontrivially is all of $M(\lambda)$. Hence $J(\lambda)\neq M(\lambda)$, so $J(\lambda)$ is proper.
[guided]
The delicate point is that $J(\lambda)$ is a possibly infinite sum of proper submodules. We must rule out the possibility that infinitely many lower-weight pieces combine to produce a highest weight vector. The definition of a sum of subspaces prevents that: every element of the sum is a finite sum of elements from the summands.
Let $\mathcal S$ be the set of all proper $\mathfrak g$-submodules of $M(\lambda)$, and set
\begin{align*}
J(\lambda):=\sum_{N\in\mathcal S}N.
\end{align*}
The sum is a $\mathfrak g$-submodule because, if $x\in\mathfrak g$ and $u=u_1+\cdots+u_s$ with $u_k\in N_k$ for proper submodules $N_k$, then
\begin{align*}
x\cdot u=x\cdot u_1+\cdots+x\cdot u_s,
\end{align*}
and each $x\cdot u_k$ lies in $N_k$.
We now prove that $J(\lambda)$ is proper. Assume for contradiction that it contains a nonzero highest weight vector. Then there is
\begin{align*}
u\in J(\lambda)\cap M(\lambda)_\lambda
\end{align*}
with $u\neq 0$. Since $u$ lies in the sum defining $J(\lambda)$, there are finitely many proper submodules $N_1,\dots,N_s$ and vectors $u_k\in N_k$ such that
\begin{align*}
u=u_1+\cdots+u_s.
\end{align*}
Each $N_k$ is a submodule of the weight module $M(\lambda)$, and we proved that such a submodule decomposes into its weight components. Therefore the $\lambda$-weight component of $u_k$ lies in
\begin{align*}
N_k\cap M(\lambda)_\lambda.
\end{align*}
But $N_k$ is proper, and any submodule containing a nonzero vector from $M(\lambda)_\lambda$ equals the whole Verma module. Hence
\begin{align*}
N_k\cap M(\lambda)_\lambda=\{0\}
\end{align*}
for every $k$.
It follows that every $u_k$ has zero $\lambda$-weight component. Taking the $\lambda$-weight component of
\begin{align*}
u=u_1+\cdots+u_s
\end{align*}
therefore gives zero. This contradicts $u\neq 0$ and $u\in M(\lambda)_\lambda$. Thus $J(\lambda)$ does not meet the highest weight line nontrivially, so it cannot be all of $M(\lambda)$. Hence $J(\lambda)$ is a proper submodule.
[/guided]
[/step]
[step:Identify the sum as the unique maximal proper submodule]
By construction, every proper $\mathfrak g$-submodule $N\subsetneq M(\lambda)$ is contained in $J(\lambda)$. Since $J(\lambda)$ is itself proper, it is a maximal proper $\mathfrak g$-submodule.
It is also unique. If $K\subset M(\lambda)$ is any maximal proper $\mathfrak g$-submodule, then $K\in\mathcal S$, so
\begin{align*}
K\subseteq J(\lambda).
\end{align*}
Maximality of $K$ and properness of $J(\lambda)$ force
\begin{align*}
K=J(\lambda).
\end{align*}
Thus $J(\lambda)$ is the unique maximal proper $\mathfrak g$-submodule of $M(\lambda)$.
[/step]
[step:Prove that the quotient is simple and has highest weight $\lambda$]
Define the quotient $\mathfrak g$-module
\begin{align*}
L(\lambda):=M(\lambda)/J(\lambda),
\end{align*}
and let
\begin{align*}
\pi:M(\lambda)\to L(\lambda)
\end{align*}
be the quotient map. Since $J(\lambda)$ is proper, $\pi(v_\lambda)\neq 0$.
For every $h\in\mathfrak h$,
\begin{align*}
h\cdot \pi(v_\lambda)=\pi(h\cdot v_\lambda)=\lambda(h)\pi(v_\lambda).
\end{align*}
For every $x\in\mathfrak n^+$,
\begin{align*}
x\cdot \pi(v_\lambda)=\pi(x\cdot v_\lambda)=0.
\end{align*}
Also, since $M(\lambda)=U(\mathfrak g)\cdot v_\lambda$ and $\pi$ is surjective,
\begin{align*}
L(\lambda)=U(\mathfrak g)\cdot \pi(v_\lambda).
\end{align*}
Hence $L(\lambda)$ is a highest weight module of highest weight $\lambda$.
It remains to prove simplicity. Let $A\subset L(\lambda)$ be a nonzero $\mathfrak g$-submodule. Then $\pi^{-1}(A)$ is a $\mathfrak g$-submodule of $M(\lambda)$ containing $J(\lambda)$ properly, because $A\neq 0$. Since $J(\lambda)$ is maximal proper, the only such submodule is $M(\lambda)$. Therefore
\begin{align*}
\pi^{-1}(A)=M(\lambda).
\end{align*}
Applying $\pi$ gives $A=L(\lambda)$. Thus $L(\lambda)$ is simple.
[/step]
[step:Show that every simple quotient factors through the unique simple quotient]
Let $S$ be a simple $\mathfrak g$-module, and let
\begin{align*}
q:M(\lambda)\to S
\end{align*}
be a surjective $\mathfrak g$-[module homomorphism](/page/Module%20Homomorphism). Since $S\neq 0$, the kernel $\ker q$ is a proper $\mathfrak g$-submodule of $M(\lambda)$. Hence
\begin{align*}
\ker q\subseteq J(\lambda).
\end{align*}
The quotient $M(\lambda)/\ker q$ is isomorphic to $S$. Because $S$ is simple, $\ker q$ is a maximal proper submodule of $M(\lambda)$. By uniqueness of the maximal proper submodule,
\begin{align*}
\ker q=J(\lambda).
\end{align*}
Therefore $q$ factors through the quotient map $\pi:M(\lambda)\to L(\lambda)$, and the induced map
\begin{align*}
\overline q:L(\lambda)\to S
\end{align*}
is an isomorphism. Thus every simple quotient of $M(\lambda)$ is isomorphic to $L(\lambda)$ as a quotient generated by the image of $v_\lambda$.
This proves the [existence and uniqueness of the simple highest weight quotient](/theorems/9379) of $M(\lambda)$.
[/step]
