Verma Modules and Shapovalov Forms

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 $\beta$, since every positive root has nonnegative simple-root coordinates and each tuple coordinate is bounded by a coordinate of $\beta$. Conversely, write $\beta ={\sum }_i{n}_i{\alpha }_i$ with $n_i\ge 0$. The ordered PBW monomial formed from $n_i$ copies of each simple negative-root vector is nonzero in $M(\lambda )$ and has weight $\lambda -\beta$, so every claimed weight occurs. For $\beta =0$ the zero tuple is the only one, giving $M(\lambda {)}_{\lambda }=\mathbb{C}{v}_{\lambda }$.

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.

Filter $U({\mathfrak{n}}^{-})$ by PBW degree. PBW identifies its associated graded algebra with the symmetric algebra $S({\mathfrak{n}}^{-})$, a polynomial algebra and hence a domain.

If nonzero $x,y$ had $xy=0$, their nonzero leading symbols would have product zero in $S({\mathfrak{n}}^{-})$, impossible. Thus $U({\mathfrak{n}}^{-})$ is a domain.

In the PBW identifications, the image of the highest vector is $a{v}_{\lambda }$ for a nonzero $a\in U({\mathfrak{n}}^{-})$, and equivariance makes the map $u{v}_{\mu }\mapsto ua{v}_{\lambda }$.

If $u{v}_{\mu }$ is in the kernel, then $ua=0$ in $U({\mathfrak{n}}^{-})$; the domain property gives $u=0$. Hence the kernel is zero.

Suppose that $f,g:M(\mu )\to M(\lambda )$ are linearly independent. They are injective. Every endomorphism of $M(\mu )$ is scalar: it preserves the one-dimensional highest-weight space (visible in the character formula), and that highest vector generates the module. If the two simple images meet nontrivially, their intersection equals both images, so $f^{-1}g$ is such an endomorphism, contradicting independence. Thus $f(M(\mu ))\oplus g(M(\mu ))$ embeds in $M(\lambda )$.

Write $\delta =\lambda -\mu \in Q^{+}$, since a nonzero map sends the highest vector to a target weight vector, and put $d=\mathrm{ht}(\delta )$. List the positive roots as ${\alpha }_1,\dots ,{\alpha }_m$ and set $h_j=\mathrm{ht}({\alpha }_j)>0$. By the character formula, the sum of weight-space dimensions at heights at most $N$ below any Verma highest weight is $D(N)=\mathrm{\#}\{(k_1,\dots ,k_m)\in {\mathbb{Z}}_{\ge 0}^m:{\sum }_j{h}_j{k}_j\le N\}\mathrm{.}$ Both copies of the source's height-at-most-$N$ subspace map into the target's height-at-most-$N+d$ subspace, giving $2D(N)\le D(N+d)$. To compare these counts, let $S_t=\{x\in {\mathbb{R}}_{\ge 0}^m:{\sum }_j{h}_j{x}_j\le t\}$ and $H={\sum }_j{h}_j$. The union of unit cubes based at the integer points counted by $D(N)$ contains $S_N$ and is contained in $S_{N+H}$, up to boundaries of volume zero. Since $\mathrm{vol}(S_t)=t^m\mathrm{/}(m!{\prod }_j{h}_j)$, this yields $D(N)\sim N^m\mathrm{/}(m!{\prod }_j{h}_j)$ and hence $D(N+d)\mathrm{/}D(N)\to 1$. If $m=0$, both counts are instead exactly $1$. In either case $2D(N)\le D(N+d)$ is impossible for large $N$.

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 $L(\lambda )$.

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.

For $0<\gamma \in Q^{+}$ let $H_{\gamma }$ be the affine equation $2(\lambda +\rho ,\gamma )=(\gamma ,\gamma )$. When restricted to $H_{\alpha ,n}=H_{n\alpha }$, the equation $H_{\gamma }$ is identically satisfied only for $\gamma =n\alpha$: equality of its linear directions forces $\gamma =c\alpha$, and then its constant term forces $c=n$. For $M(\mu )$, a possible Casimir equality at lower weight $\mu -\gamma$ gives $2(\lambda -n\alpha +\rho ,\gamma )=(\gamma ,\gamma )$. This is never an identity on $H_{\alpha ,n}$: if $\gamma =c\alpha$ with $c>0$, the left side there is $-cn(\alpha ,\alpha )$ while the right side is $c^2(\alpha ,\alpha )>0$. Thus all undesired intersections are proper affine subspaces.

There are countably many $\gamma \in Q^{+}$, so omit the intersections in step 1.1. Their union cannot cover the complex affine space $H_{\alpha ,n}$: in dimension one it excludes only countably many points, and induction on dimension first chooses the preceding coordinates outside the countably many equations independent of the last coordinate, then chooses the last coordinate outside countably many points. For dimension zero step 1.1 says every omitted intersection is empty. We henceforth use such a $\lambda$. At this point the only possible non-highest singular weight in $M(\lambda )$ is $\mu$, and $M(\mu )$ has no non-highest singular weight.

If $M(\mu )$ had a nonzero proper submodule, [F1] would give a singular vector below $\mu$; its Casimir equality contradicts step 2.1. Thus $M(\mu )$ is simple. Since a determinant is divisible by the equation of $H_{\alpha ,n}$, it vanishes at $\lambda$, so $J(\lambda )$ is nonzero. Choose a weight in $J(\lambda )$ of minimum height below $\lambda$; its vector is singular, and step 2.1 forces its weight to be $\mu$. By [F2] it gives an injective map $M(\mu )\hookrightarrow J(\lambda )$.

The radical has no weight of height smaller than $\mathrm{ht}(n\alpha )$ below $\lambda$, since its minimum-height weight in that range would be singular and step 2.1 would force weight $\mu$. Hence every vector of $J(\lambda {)}_{\mu }$ is singular. The universal property and [F2] show $\dim J(\lambda {)}_{\mu }\le 1$; the embedded source supplies equality.

Suppose $J(\lambda )\mathrm{/}M(\mu )\mathrm{\ne }0$. It is a weight module with support below $\lambda$ and a weight of minimum height; a vector there is singular in the quotient. The Casimir still acts by the scalar of $M(\lambda )$, so [F3] and step 2.1 force this vector to have weight $\mu$. But step 4.1 says the quotient has zero $\mu$-space, a contradiction. Thus the embedded $M(\mu )$ equals $J(\lambda )$, and the quotient is the simple $L(\lambda )$.

The uniqueness of the map up to scalar follows from [F2] and its existence. PBW gives $\dim M(\mu {)}_{\mu -\eta }=K(\eta )$; the image at target weight $\lambda -\beta$ has $\eta =\beta -n\alpha$. Since the image is the radical, it is precisely the kernel of the restricted Shapovalov form.

Let $B_t$ be the restricted form in PBW coordinates, and let $R_{\beta }=\ker B_0$ in weight $\lambda -\beta$. For $x,y\in R_{\beta }$, the value $B_0^{\mathrm{\prime }}(x,y)$ is independent of their chosen lifts to first order, because $B_0$ vanishes whenever one argument is in $R_{\beta }$. Differentiating contravariance shows that these first derivatives assemble into a contravariant bilinear form on $R$: the extra terms from differentiating the module action are paired by $B_0$ with a radical vector and vanish. The form is symmetric.

Since $R\cong M(\mu )$, any contravariant form on it is determined by its value on the one-dimensional highest-weight line: move every negative-root operator across the form and use the highest-vector annihilation relations. Thus $B_0^{\mathrm{\prime }}{\mathrm{\mid }}_R$ is a scalar multiple of the Shapovalov form of $M(\mu )$. That form is nondegenerate because $M(\mu )$ is simple by the generic-radical lemma. It remains to prove the derivative's value on a nonzero highest vector $u_0\in R_{n\alpha }$ is nonzero.

Assume $B_0^{\mathrm{\prime }}(u_0,u_0)=0$. The matrix $B_0$ in weight $\mu$ is symmetric with kernel $\mathbb{C}{u}_0$, by the generic-radical lemma. The linear functional $-B_0^{\mathrm{\prime }}(u_0,-)$ annihilates that kernel, so it lies in the image of $B_0$. Choose a weight-$\mu$ vector $u_1$ with $B_0(u_1,-)=-B_0^{\mathrm{\prime }}(u_0,-)$ and set $u(t)=u_0+t{u}_1$. Then $B_t(u(t),w)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$ for every vector $w$ in that PBW weight space.

For each simple positive-root vector $e_i$ and any vector $w$ of the adjacent weight, contravariance gives $B_t(e_{i}u(t),w)=B_t(u(t),f_{i}w)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$. The specialized form is invertible at weight $\mu +{\alpha }_i$, since $R\cong M(\mu )$ has no weight above $\mu$. Therefore $e_{i}u(t)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$ for every $i$, and the same holds for all positive-root vectors because the $e_i$ generate ${\mathfrak{n}}^{+}$.

In the deformed module the Cartan weight of $u(t)$ is $\mu +t\delta$. Step 4.1 and the usual Casimir calculation on a highest vector give $Cu(t)=(\mu +t\delta ,\mu +t\delta +2\rho )u(t)(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$. But the same central element acts throughout $M(\lambda +t\delta )$ by $(\lambda +t\delta ,\lambda +t\delta +2\rho )$. Their constant terms agree because $\lambda \in H_{\alpha ,n}$; their linear coefficients differ by $2(\lambda -\mu ,\delta )=n(\alpha ,\alpha )\mathrm{\ne }0$. Since $u_0\mathrm{\ne }0$, these two congruences contradict each other. Thus $B_0^{\mathrm{\prime }}(u_0,u_0)\mathrm{\ne }0$, and step 2.1 proves perfectness on every radical weight space.

Choose bases adapted to the radical and a complementary subspace. The specialized matrix has a zero radical block and an invertible complementary block. The radical block of $B_t$ is $t{B}_0^{\mathrm{\prime }}{\mathrm{\mid }}_{R_{\beta }}+O(t^2)$, while its mixed blocks are $O(t)$. Since $B_0^{\mathrm{\prime }}{\mathrm{\mid }}_{R_{\beta }}$ and the complementary block are invertible, a block determinant expansion gives ${\mathrm{ord}}_{t=0}\det B_t=\dim R_{\beta }$. The generic-radical lemma identifies this dimension as $K(\beta -n\alpha )$, proving the asserted order.

The preliminary factorization gives nonnegative integers $m_{\alpha ,n}(\beta )$ supported where $n\alpha \le \beta$. Its top PBW term gives $\mathrm{deg}{D}_{\beta }={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}{\sum }_{n\ge 1}K(\beta -n\alpha )={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}{\sum }_{n\ge 1}{m}_{\alpha ,n}(\beta )$. Both sums are finite.

Fix a pair $(\alpha ,n)$ for which $m_{\alpha ,n}(\beta )>0$. The hyperplane $H_{\alpha ,n}$ is then a factor hyperplane of $D_{\beta }$, so the generic-radical lemma applies. At a generic point of this hyperplane the kernel in weight $\lambda -\beta$ has dimension $K(\beta -n\alpha )$, and the transverse-pairing lemma says the determinant's order along a transverse line is exactly this dimension. Other affine factors are nonzero at the chosen generic point. Thus $m_{\alpha ,n}(\beta )=K(\beta -n\alpha )$.

For every pair with $m_{\alpha ,n}(\beta )=0$ one has $m_{\alpha ,n}(\beta )\le K(\beta -n\alpha )$ trivially; step 2.1 gives equality for every positive multiplicity. The equality of finite total sums in step 1.1 therefore forces equality at every pair, including all pairs with vanishing preliminary multiplicity. Substituting these exponents into the factorization proves the formula.

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.