Preliminary factorization of a Shapovalov determinant

Choose root vectors in opposite root spaces with nonzero pairing and PBW monomial bases in weight $\beta$. In the product of a positive and a negative PBW monomial, a factor of $\lambda$ arises only when commuting a positive root vector with its opposite and retaining their Cartan bracket. A maximal-degree contribution pairs every root factor this way. The PBW order makes those maximal-degree pairings diagonal: an unequal pair of exponent vectors leaves a root vector, or requires a non-Cartan commutator and loses at least one degree. For exponent vector $(k_{\alpha })$ the diagonal coefficient is a nonzero constant times ${\prod }_{\alpha }⟨\lambda ,{\alpha }^{\vee }{⟩}^{k_{\alpha }}$. Hence the determinant's top part is the product of these diagonal terms and is nonzero.

Across all PBW monomials of weight $\beta$, the total number of occurrences of $\alpha$ is ${\sum }_{n\ge 1}K(\beta -n\alpha )$: count a monomial with $k_{\alpha }$ copies once for each $1\le n\le k_{\alpha }$. This gives the displayed top part and its total degree.

The polynomial of step 2.1 is nonzero. Suppose $D_{\beta }(\lambda )=0$. A vector $v$ in its kernel belongs to the radical. The space $U({\mathfrak{n}}^{+})v$ is finite dimensional because only finitely many weights of $M(\lambda )$ lie above $\lambda -\beta$. Choose a nonzero vector in it of maximum weight height. It is singular, lies in the proper radical, and has weight $\lambda -\gamma$ with $0<\gamma \le \beta$. By [F2] and the Casimir scalar, the source and target have equal eigenvalues, giving $2(\lambda +\rho ,\gamma )=(\gamma ,\gamma )$. Thus the zero set of $D_{\beta }$ lies in the finite union of affine hyperplanes $H_{\gamma }$ defined by these equations, $0<\gamma \le \beta$.

Every irreducible polynomial factor $P$ of $D_{\beta }$ is proportional to one of the linear equations of the $H_{\gamma }$. Here is an elementary divisibility justification. If $P$ divided none of their product $F$, choose a variable $x$ in which $P$ has positive degree. Over the fraction field of the other variables, $P$ and $F$ are coprime; clearing a Bézout identity's denominators gives $AP+BF=g$, with $A,B$ polynomials and $g$ a nonzero polynomial in the other variables. Choose values of those variables where $g$ and the leading coefficient of $P$ in $x$ are nonzero. The specialized positive-degree polynomial $P$ has a complex root, at which $F\mathrm{\ne }0$ by the identity, contradicting step 3.1. If $P$ has no other variables, the same argument is the ordinary one-variable fact. Thus $P$ divides $F$, and irreducibility makes it proportional to one of its linear factors.

The highest homogeneous part of a factor $H_{\gamma }$ is the linear form $(\lambda ,\gamma )$. Since the product of the factors' highest parts is the top part in step 2.1, unique factorization forces this linear form to be proportional to $⟨\lambda ,{\alpha }^{\vee }⟩$ for some positive root $\alpha$. Hence $\gamma =c\alpha$ for a positive rational $c$ (both lie in the root lattice and positive cone). The equation of $H_{\gamma }$ becomes $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=c$.

By [F3], every root is primitive in the root lattice $Q$: an integral lattice automorphism carries it to a simple basis vector. Therefore $c$ in step 5.1 is a positive integer $n$. Substituting $\gamma =n\alpha$ into the equation of $H_{\gamma }$ gives the stated affine factor, while $\gamma \le \beta$ gives the support restriction. Distinct pairs $(\alpha ,n)$ give distinct affine hyperplanes; unique factorization supplies the exponents.