Homomorphisms Between Verma Modules and Linkage

A homomorphism $f$ sends the highest-weight vector $v_{\mu }$ to a vector of weight $\mu$ killed by ${\mathfrak{n}}^{+}$; evaluation is therefore a linear map into the displayed space.

Conversely, a vector $v$ in that space is a highest-weight vector of weight $\mu$, so the universal property supplies a unique homomorphism $M(\mu )\to M(\lambda )$ sending $v_{\mu }$ to $v$. The two constructions are inverse.

Suppose no embedded Verma submodule of $M(\lambda )$ is simple. Any proper nonzero submodule of an embedded Verma contains a singular vector; its generated submodule is another embedded Verma by injectivity. A proper inclusion strictly lowers the highest weight, so successive such submodules have weights $\lambda -{\beta }_j$ with strictly increasing heights of ${\beta }_j\in Q^{+}$.

Every such embedded Verma has the same Casimir scalar as $M(\lambda )$, so its nonzero ${\beta }_j$ satisfies $2(\lambda +\rho ,{\beta }_j)=({\beta }_j,{\beta }_j)$. Taking real parts in the real span of the roots places ${\beta }_j$ on a bounded sphere centred at the real part of $\lambda +\rho$. The root lattice is discrete, so the set $F$ of its solutions is finite.

Let $m=\mathrm{\mid }F\mathrm{\mid }$. Starting with $M(\lambda )$, apply step 1.1 only $m+1$ times. Finite induction and finite choice suffice to obtain $m+1$ proper, strictly nested embedded Vermas. Their distinct nonzero weights give $m+1$ distinct members of $F$, a contradiction. Hence some embedded Verma is simple.

Choose a simple Verma submodule $S\subset M(\mu )$. The restrictions of any two maps $f,g:M(\mu )\to M(\lambda )$ are proportional, say $f{\mathrm{\mid }}_S=cg{\mathrm{\mid }}_S$.

Then $(f-cg){\mathrm{\mid }}_S=0$. If $f-cg$ were nonzero, [L1] would make it injective, so it could not vanish on nonzero $S$. Hence $f=cg$, proving the bound.

The singular vector $f_i^m{v}_{\lambda }$ has weight $s_i\cdot \lambda$, so the universal property gives a homomorphism $M(s_i\cdot \lambda )\to M(\lambda )$ taking its highest vector to it.

By [F1], every source vector is uniquely $u{v}_{s_i\cdot \lambda }$ with $u\in U({\mathfrak{n}}^{-})$. Its image is $u{f}_i^m{v}_{\lambda }$, which vanishes only if $u{f}_i^m=0$ in $U({\mathfrak{n}}^{-})$, again by [F1]. If $u\mathrm{\ne }0$, its highest nonzero PBW symbol and the symbol of $f_i^m$ have nonzero product in the polynomial domain $S({\mathfrak{n}}^{-})$ of [F2]. Therefore $u{f}_i^m\mathrm{\ne }0$, and the map is injective.

Put $n=⟨\lambda +\rho ,{\alpha }^{\vee }⟩>0$ and $\mu =\lambda -n\alpha =s_{\alpha }\cdot \lambda$. At weight $\lambda -n\alpha$, the determinant formula has the factor $⟨\lambda +\rho ,{\alpha }^{\vee }⟩-n$ with exponent $K(0)=1$. Thus its hyperplane is a factor hyperplane, and at generic points $\eta$ on it the generic-radical lemma supplies a nonzero singular vector in $M(\eta {)}_{\eta -n\alpha }$.

In fixed PBW coordinates the maps $e_i:M(\eta {)}_{\eta -n\alpha }\to M(\eta {)}_{\eta -n\alpha +{\alpha }_i}$ have matrix entries polynomial in $\eta$. Stack them to form one matrix $E(\eta )$. Step 1.1 says it has a nonzero kernel at a dense set of points on the affine hyperplane $H_{\alpha ,n}$. Hence every maximal minor of $E$ vanishes identically on that hyperplane: a polynomial that vanishes outside a countable union of proper affine subspaces there is zero. In particular $E(\lambda )$ has a nonzero kernel. A vector in it is killed by every simple positive-root vector, hence by all of ${\mathfrak{n}}^{+}$, and has weight $\mu$.

By [F1] this vector defines a nonzero map $M(\mu )\to M(\lambda )$, which is injective. Since $\mu =s_{\alpha }\cdot \lambda$, this is the required embedding.

Every $z\in Z(U(\mathfrak{g}))$ acts on the cyclic highest-weight module $M(\lambda )$ by the scalar ${\chi }_{\lambda }(z)$.

The same scalar action descends to each subquotient; on the composition factor $L(\mu )$ it is by definition ${\chi }_{\mu }(z)$. Hence the two characters agree on every $z$.

Reducing the condition $S_t(v)\in t{M}_t(\lambda {)}^{\vee }$ modulo $t$ says exactly that the specialized Shapovalov form pairs $\bar{v}$ with every vector as zero. Thus $M^1(\lambda )$ is its radical.

The radical is $J(\lambda )$ by the radical theorem, so the two submodules agree.

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.

The hypothesis is precisely the absence of the positive-integral pairings in the criterion.

The criterion therefore says that $M(\lambda )$ is simple.

Suppose a proper nonzero submodule $N\subset M(\lambda )$ exists. By [L1], it contains a nonzero singular vector $v_{\mu }$ of weight $\mu =\lambda -\beta$ for some $\beta \in Q^{+}$. The highest-weight space has dimension one, so $\beta \mathrm{\ne }0$: otherwise $v_{\mu }$ would generate $M(\lambda )$, contrary to $N$ being proper. The cyclic submodule $U(\mathfrak{g})v_{\mu }$ is a highest-weight module of weight $\mu$.

By [L2], the Casimir acts on $M(\lambda )$ with scalar $(\lambda ,\lambda +2\rho )$ and on its cyclic submodule $U(\mathfrak{g})v_{\mu }$ with scalar $(\mu ,\mu +2\rho )$. These scalars agree on the nonzero vector $v_{\mu }$. Substituting $\mu =\lambda -\beta$ and expanding gives $2(\lambda +\rho ,\beta )=(\beta ,\beta )$.

Write $\beta ={\sum }_i{n}_i{\alpha }_i$ in the simple-root basis, where $n_i\in {\mathbb{Z}}_{\ge 0}$ and some $n_i>0$. By [L3], $2(\lambda +\rho ,{\alpha }_i)=({\alpha }_i,{\alpha }_i)⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩<0$ for every $i$. Therefore $(\lambda +\rho ,\beta )<0$, whereas positive definiteness gives $(\beta ,\beta )>0$, contradicting step 2.1. Thus $M(\lambda )$ has no proper nonzero submodule and is simple.