Virasoro Verma modules have a unique simple highest-weight quotient

The subspace $\mathfrak{b}$ is a Lie subalgebra: for $m,n\ge 0$, the bracket in [F1] is again in $\mathfrak{b}$; the central term can occur only when $m=n=0$, when its coefficient is zero. On ${\mathbb{C}}_{c,h}$, brackets of two positive modes and brackets of $L_0$ with a positive mode act by zero, as do all brackets with $C$. Thus the stated scalar action satisfies [F3] and defines a one-dimensional $\mathfrak{b}$-representation.

The inclusion $\mathfrak{b}\hookrightarrow \mathrm{V}\mathrm{i}\mathrm{r}$ induces $U(\mathfrak{b})\to U(\mathrm{V}\mathrm{i}\mathrm{r})$ by the quotient construction [F5]. Form $M(c,h)$ and $v$ as in the Statement. The balanced tensor relation gives $L_{n}v=0$ for $n>0$, $L_{0}v=hv$, and $Cv=cv$.

The modes $L_n$ are linearly independent: if a finite sum ${\sum }_n{a}_n{L}_n$ is zero, projection to $W$ gives ${\sum }_n{a}_n{d}_n=0$; applying this derivation to $z$ gives $-{\sum }_n{a}_n{z}^{n+1}=0$, so Laurent-monomial independence forces every $a_n=0$. The direct-sum definition of Vir also makes $C$ independent. Thus $\{L_n:n\in \mathbb{Z}\}\cup \{C\}$ is a supplied countable basis. Order it with all negative modes first, then $L_0,C$, then positive modes. This puts a basis of $\mathfrak{b}$ last. Applying [F6] to this ordered basis and to the basis of $\mathfrak{b}$ shows that $U(\mathfrak{b})\to U(\mathrm{V}\mathrm{i}\mathrm{r})$ is injective and that the PBW basis of $U(\mathrm{V}\mathrm{i}\mathrm{r})$ factors uniquely as a negative-mode monomial followed by a $\mathfrak{b}$-monomial. Tensoring over $U(\mathfrak{b})$ evaluates the latter monomial on ${\mathbb{C}}_{c,h}$, leaving exactly the ordered negative-mode monomials on $v$ as a basis of $M(c,h)$. The empty monomial $v$ is therefore nonzero, and the equalities in step 2.1 with [F4] show that it is a highest-weight vector. By [F1], $[L_0,L_{-n}]=n{L}_{-n}$; hence the monomial has weight $h+{\sum }_i{n}_i$. Only the empty monomial has weight $h$, so $M_h=\mathbb{C}v$ and every other basis vector has weight $h+N$ for an integer $N>0$.

Every submodule $S\subseteq M(c,h)$ is stable under $L_0$. Each vector is a finite sum of distinct $L_0$-eigencomponents by the basis in step 3.1; polynomial interpolation in $L_0$ projects any such vector onto each component, so every submodule is graded. If a proper submodule contained a vector with nonzero weight-$h$ component, the same projection would put a nonzero multiple of $v$ in $S$. Since $v$ generates $M(c,h)$, this would force $S=M(c,h)$, a contradiction. Thus every proper submodule lies in the sum of the positive-level spaces. Their sum $J$ also lies there and is proper because it misses $v$.

The quotient $M(c,h)\mathrm{/}J$ is nonzero because $v\mathrm{\notin }J$. If a submodule of this quotient is nonzero and proper, its preimage is a proper submodule of $M(c,h)$ strictly containing $J$, contradicting the definition of $J$ as the sum of all proper submodules. Hence the quotient is simple and highest weight. The kernel of any simple quotient is a maximal proper submodule, so it is contained in $J$ and, because $J$ is proper, maximality forces that kernel to equal $J$. Therefore this is the unique simple highest-weight quotient.