Verma Modules and Shapovalov Forms

Define $T(u\otimes c_{\lambda })=uv$. If $b=h+x\in \mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$, then $bv=\lambda (h)v$, exactly the scalar by which $b$ acts on $c_{\lambda }$; hence $T(ub\otimes c_{\lambda })=T(u\otimes b{c}_{\lambda })$ and $T$ descends to the induced module.

The descended map is $\mathfrak{g}$-linear and sends $v_{\lambda }$ to $v$. Conversely a homomorphism is determined by $v_{\lambda }$, since that vector generates $M(\lambda )$; its image necessarily has the two displayed properties.

Triangular PBW from Triangular decomposition from a chosen positive root system identifies $U(\mathfrak{g})$ with $U({\mathfrak{n}}^{-})\otimes U(\mathfrak{b})$. Tensoring over $U(\mathfrak{b})$ with ${\mathbb{C}}_{\lambda }$ therefore gives $U({\mathfrak{n}}^{-})\otimes {\mathbb{C}}_{\lambda }$.

Under that identification, $u\otimes c_{\lambda }$ maps to $u{v}_{\lambda }$, so the map is bijective. Applying the ordered PBW basis in the negative factor proves the asserted basis statement.

A monomial with negative roots $-\alpha$ occurring $k_{\alpha }$ times has $\mathfrak{h}$-weight $\lambda -\sum k_{\alpha }\alpha$, by commuting $h$ past its factors.

Such monomials span each displayed weight space. For fixed $\beta$, only finitely many nonnegative tuples $(k_{\alpha })$ have sum $\beta$, and the zero tuple is the only tuple for $\beta =0$.

For each positive root, its PBW exponent contributes the geometric series $1+e^{-\alpha }+e^{-2\alpha }+\cdots$.

Multiplying the finitely many root series and then multiplying by $e^{\lambda }$ counts precisely the PBW monomials of each weight. Each coefficient is finite by the fixed-$\beta$ finiteness in the given weight-space result, so the product belongs to the stated completion.

If $0\mathrm{\ne }c{v}_{\lambda }\in N$, scalar closure gives $v_{\lambda }\in N$.

The vector $v_{\lambda }$ generates $M(\lambda )$ by its induced construction, so $N=M(\lambda )$, contradicting properness.

A submodule is stable under $\mathfrak{h}$; projecting any finite weight decomposition by polynomials in elements of $\mathfrak{h}$ shows it is the direct sum of its weight intersections.

Every proper submodule has zero $\lambda$-weight intersection by the highest-line lemma. Hence their sum has zero $\lambda$-weight intersection, whereas $v_{\lambda }$ has that weight; therefore $J(\lambda )\mathrm{\ne }M(\lambda )$.

Every proper submodule is contained in $J(\lambda )$ by its definition, so $J(\lambda )$ is maximal and unique among proper maximal submodules.

A submodule of $M(\lambda )\mathrm{/}J(\lambda )$ lifts to a submodule containing $J(\lambda )$; it is either $J(\lambda )$ or all of $M(\lambda )$. Thus the quotient is simple, and the kernel of any simple quotient is a maximal submodule, necessarily $J(\lambda )$.

As in the weight-projection argument, a nonzero submodule has a nonzero weight vector. Choose one of weight $\lambda -\beta$ with $\beta \in Q^{+}$ of minimal height among its occurring weights.

If $x\in {\mathfrak{g}}_{\alpha }\subseteq {\mathfrak{n}}^{+}$, then $xv$ has weight $\lambda -(\beta -\alpha )$. If it were nonzero, $\beta -\alpha \in Q^{+}$ would have smaller height, contradicting the choice; hence ${\mathfrak{n}}^{+}v=0$.

Triangular PBW defines $p:U(\mathfrak{g})\to U(\mathfrak{h})$ by retaining the $U(\mathfrak{h})$ summand; put ${\phi }_{\lambda }={\mathrm{ev}}_{\lambda }\circ p$. For $u,w\in U({\mathfrak{n}}^{-})$ set $S_{\lambda }(u{v}_{\lambda },w{v}_{\lambda })={\phi }_{\lambda }(\tau (u)w)$.

The PBW model makes $u,w$ unique. Moreover $p\circ \tau =p$. If $a$ annihilates $v_{\lambda }$, then $za$ also annihilates it for every $z\in U(\mathfrak{g})$, so its $U(\mathfrak{h})$ PBW coefficient evaluates to $0$: ${\phi }_{\lambda }(za)=0$. Taking $z=\tau (w)$, and then using $p\circ \tau =p$, shows that either argument may be changed by an element annihilating $v_{\lambda }$; thus the displayed formula descends to $M(\lambda )$.

For $x\in U(\mathfrak{g})$, the anti-involution identity gives ${\phi }_{\lambda }(\tau (xu)w)={\phi }_{\lambda }(\tau (u)\tau (x)w)$, which is $S_{\lambda }(xu{v}_{\lambda },w{v}_{\lambda })=S_{\lambda }(u{v}_{\lambda },\tau (x)w{v}_{\lambda })$. Thus the descended form is contravariant, and $S_{\lambda }(v_{\lambda },v_{\lambda })={\phi }_{\lambda }(1)=1$.

For any contravariant form, repeatedly move the negative PBW monomial in its first argument to the second. Its value is therefore forced by its value on $(v_{\lambda },v_{\lambda })$, so the normalization proves uniqueness.

Choose $h\in \mathfrak{h}$ separating the two weights $\mu \mathrm{\ne }\nu$. Contravariance gives $\mu (h)S_{\lambda }(u,v)=S_{\lambda }(hu,v)=S_{\lambda }(u,hv)=\nu (h)S_{\lambda }(u,v)$.

Since $\mu (h)-\nu (h)\mathrm{\ne }0$, the displayed equality forces $S_{\lambda }(u,v)=0$.

Contravariance makes $R:=\mathrm{rad}{S}_{\lambda }$ a submodule. It is proper because $S_{\lambda }(v_{\lambda },v_{\lambda })=1$, hence $R\subseteq J(\lambda )$.

Let $N$ be proper. Its stability under $\mathfrak{h}$ lets one project every finite weight decomposition into $N$, so $N$ is the sum of its weight intersections; its $\lambda$-intersection is $0$. If $w\in N$ has weight different from $\lambda$, choose $h$ separating the weights and move $h$ across the form to get $S_{\lambda }(v_{\lambda },w)=0$, hence $S_{\lambda }(v_{\lambda },N)=0$.

For $u\in U(\mathfrak{g})$, $\tau (u)N\subseteq N$, so contravariance gives $S_{\lambda }(u{v}_{\lambda },N)=S_{\lambda }(v_{\lambda },\tau (u)N)=0$. The cyclicity of $v_{\lambda }$ therefore gives $N\subseteq R$. In particular $J(\lambda )\subseteq R$.

Hence $R=J(\lambda )$. Quotienting a bilinear form by its radical is well defined and nondegenerate, which yields the stated form on $L(\lambda )$.

In PBW bases, commute each positive-root factor past negative-root factors before evaluating on $v_{\lambda }$. The diagonal terms are the only terms of maximal total degree. Counting, for each occurrence of a root $\alpha$, the PBW monomials in which that occurrence can be removed gives the following leading term.

In particular, $D_{\beta }$ is nonzero and has the total degree displayed on the right.

Suppose $D_{\beta }(\lambda )=0$. The radical is then nonzero in weight $\lambda -\beta$. By The Shapovalov radical is the maximal submodule and Every nonzero Verma submodule contains a singular vector, it contains a singular vector of some weight $\lambda -\gamma$, with $\gamma \in Q^{+}\setminus \{0\}$. The universal property gives a nonzero map $M(\lambda -\gamma )\to M(\lambda )$. The Casimir has the same scalar on its source and image, giving the following identity.

Consequently every irreducible factor of $D_{\beta }$ is an affine linear form with normal direction $\gamma$. Comparing its leading direction with the product in step 1.1 shows that $\gamma =n\alpha$ for a positive root $\alpha$ and an integer $n\ge 1$. Substitution in step 3.1 then gives $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=n$. Thus, for some integers $m_{n\alpha }(\beta )\ge 0$, the following factorization holds.

This is the claimed preliminary factorization.

Fix $\alpha ,n$ and choose $\lambda$ generically on the hyperplane $H_{\alpha ,n}$. The generic-hyperplane result [L1] gives an injective map $M(\lambda -n\alpha )\to M(\lambda )$ whose image is exactly the Shapovalov radical. Under PBW, its part in weight $\lambda -\beta$ has dimension $K(\beta -n\alpha )$, including dimension $0$ when $\beta -n\alpha \mathrm{\notin }{Q}^{+}$.

Choose $\delta \in {\mathfrak{h}}^{\ast }$ with $⟨\delta ,{\alpha }^{\vee }⟩=1$ and restrict the form in weight $\lambda -\beta$ along $\lambda +t\delta$. Its kernel at $t=0$ is the space in step 6.1. By [L1], the first derivative of the form is nondegenerate on that kernel. Equivalently, a vector pairing to order $t^2$ would force the two relevant Casimir scalars to agree modulo $t^2$, although their difference is the following nonzero linear term.

This difference is nonzero modulo $t^2$, a contradiction. The elementary determinant lemma obtained by choosing bases adapted to the kernel now says that the transverse order of $D_{\beta }$ along $H_{\alpha ,n}$ is the kernel dimension. Therefore $m_{n\alpha }(\beta )=K(\beta -n\alpha )$.

Substitution in step 5.1 proves the formula up to the nonzero basis scalar. Finally $\beta -n\alpha \in Q^{+}$ bounds $n$ by the height of $\beta$, so only finitely many displayed exponents are nonzero.

If no displayed pairing is positive integral, every determinant block is nonzero by the formula. Orthogonality then makes the radical zero, so the maximal submodule is zero and $M(\lambda )$ is simple.

Conversely, if $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=n>0$, take $\beta =n\alpha$. The formula has the factor with exponent $K(0)=1$, so that block is singular; the radical and hence the maximal submodule is nonzero.