The Jantzen sum formula for a Verma module

On a finite free weight block over $\mathbb{C}\mathrm{⟦}t\mathrm{⟧}$, the determinant formula and [F1] show that the deformed determinant is a nonzero power series: every factor is either a nonzero constant plus a linear term or a nonzero linear term. Smith normal form therefore has nonzero diagonal entries $t^{a_j}{u}_j(t)$ with $u_j(0)\mathrm{\ne }0$. Both the order of its determinant and ${\sum }_{i>0}\dim M^i(\lambda {)}_{\lambda -\beta }$ equal ${\sum }_j{a}_j$: each $a_j$ contributes once for each $1\le i\le a_j$.

Substitute the determinant formula along $\lambda +t\rho$. For each factor $⟨\lambda +t\rho +\rho ,{\alpha }^{\vee }⟩-n$, [F1] makes its $t$-coefficient nonzero, so it contributes valuation one exactly when $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=n$ and zero otherwise. Thus the determinant order in the $\lambda -\beta$ block is ${\sum }_{\alpha ,n:⟨\lambda +\rho ,{\alpha }^{\vee }⟩=n}K(\beta -n\alpha )$, exactly the coefficient of that weight in the right-hand character sum. Equality coefficientwise for every $\beta$ proves the formula.