Universal property and pbw character of kac moody verma modules

Define $u\otimes c\mapsto cuv$. For $a\in U(\mathfrak{b})$ the defining highest-vector relations give $av$ equal to the action of $a$ on ${\mathbb{C}}_{\lambda }$ times $v$. Hence the tensor relation $ua\otimes c=u\otimes ac$ is respected, and the map is $\mathfrak{g}$-linear. The vector $1\otimes 1$ generates the induced module, so its image determines the map uniquely; its image is precisely $U(\mathfrak{g})v$. This includes $v=0$.

By F1 and F2, $M_A(\lambda )$ has basis all ordered monomials in a homogeneous basis of ${\mathfrak{n}}^{-}$ applied to $1\otimes 1$, including the empty monomial. A monomial of negative degree $-\beta$ has weight $\lambda -\beta$. The sign-changing involution in F5 sends ${\mathfrak{g}}_{\alpha }$ isomorphically onto ${\mathfrak{g}}_{-\alpha }$: it sends $h$ to $-h$, so applying it to $[h,x]=\alpha (h)x$ reverses the weight. Thus their dimensions agree.

Fix $\beta ={\sum }_i{b}_i{\alpha }_i\in Q^{+}$. Only negative basis vectors whose opposite root has coordinates between $0$ and the $b_i$ can occur in a monomial of degree $-\beta$. There are finitely many such integer tuples and finitely many basis vectors at each by F3. Each has positive height, so its exponent is at most ${\sum }_i{b}_i$. Consequently only finitely many monomials have that degree. For $\beta =0$ only the empty monomial occurs, giving top coefficient one. Weights outside $\lambda -Q^{+}$ have coefficient zero. This proves nonzero and $\mathcal{O}$ membership by F4.

For each individual negative basis vector of degree $-\alpha$, counting its possible exponent contributes ${\sum }_{m\ge 0}{e}^{-m\alpha }=(1-e^{-\alpha }{)}^{-1}$ as a formal geometric series. For any fixed $\beta$, step 2.1 reduces their product to finitely many factors and finitely many exponent choices. The ordered PBW basis makes each such choice exactly one basis monomial; multiplying by $e^{\lambda }$ therefore gives its actual weight multiplicity. There are $\dim {\mathfrak{g}}_{\alpha }$ factors at each root by 1.2, proving the displayed formula. Empty root sets give the empty product $1$; no choice is required beyond the fixed homogeneous-basis construction in F2.