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.

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.

If no embedded Verma submodule were simple, repeatedly choose a nonzero proper submodule and then a singular vector in it; injectivity gives an infinite strictly descending chain of embedded Vermas $M(\lambda -{\beta }_j)\subset M(\lambda )$.

Their Casimir scalars equal that of $M(\lambda )$, so $2(\lambda +\rho ,{\beta }_j)=({\beta }_j,{\beta }_j)$. The ${\beta }_j$ lie in the positive lattice cone and strictly increase in height, while this positive-definite quadratic equation has only finitely many lattice solutions. This contradiction yields a simple embedded Verma module.

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 $N$.

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.

Choose nonzero $e_i\in {\mathfrak{g}}_{{\alpha }_i}$ and scale $f_i\in {\mathfrak{g}}_{-{\alpha }_i}$ so that $[e_i,f_i]={\alpha }_i^{\vee }$; the cited opposite-root bracket line permits this normalization. The resulting ${\mathfrak{s}\mathfrak{l}}_2$ relations give, by induction, $e_i{f}_i^r{v}_{\lambda }=r(⟨\lambda ,{\alpha }_i^{\vee }⟩-r+1)f_i^{r-1}{v}_{\lambda }\mathrm{.}$ At $r=m=⟨\lambda ,{\alpha }_i^{\vee }⟩+1$ this is zero, while [L1] shows $f_i^m{v}_{\lambda }\mathrm{\ne }0$.

For $j\mathrm{\ne }i$, $[e_j,f_i]=0$ because ${\alpha }_j-{\alpha }_i$ is not a root; hence every $e_j$ also kills $f_i^m{v}_{\lambda }$. Its weight is $\lambda -m{\alpha }_i=s_i(\lambda +\rho )-\rho$, so it is singular of the stated dot weight.

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.

In the PBW model the vector $f_i^m{v}_{\lambda }$ is nonzero. If $u{v}_{s_i\cdot \lambda }$ lay in the kernel, then $u{f}_i^m=0$ in $U({\mathfrak{n}}^{-})$. The PBW-degree associated graded is the domain $S({\mathfrak{n}}^{-})$, so $u=0$. Thus the nonzero homomorphism in step 1.1 is injective and is the asserted embedding.

Put $\nu =\lambda +\rho$ and $n=⟨\nu ,{\alpha }^{\vee }⟩$. Then $n\in {\mathbb{Z}}_{>0}$ and $s_{\alpha }\nu =\nu -n\alpha$, so $s_{\alpha }\nu$ is related to $\nu$ by one positive-integral root reflection.

The source theorem [L1] applies to this one-reflection relation and gives a unique embedding $M(s_{\alpha }\nu -\rho )\hookrightarrow M(\nu -\rho )$. Since $s_{\alpha }\nu -\rho =s_{\alpha }\cdot \lambda$, this is the asserted 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 $z$.

A factor $L(\mu )$ can occur only when $\mu$ is both a weight below $\lambda$ and in the dot orbit fixed by its central character. The Casimir equality restricts the possible lattice differences in each bounded weight cone to a finite set.

In particular, the $\mu$-weight space of $M(\lambda )$ is finite-dimensional and every copy of $L(\mu )$ contributes its one-dimensional highest-weight line there. Therefore the multiplicity is bounded by $\dim M(\lambda {)}_{\mu }<\mathrm{\infty }$.

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, Smith normal form has diagonal entries $t^{a_j}{u}_j(t)$. 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 $\lambda +t\rho$. Its order at $t=0$ 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.

Induct on the height of $\lambda -\mu$. Height zero gives $L(\lambda )$ and the empty linkage chain.

For positive height, the factor is in the maximal submodule $M^1(\lambda )$. The Jantzen sum and finite multiplicities place it in some $M(s_{\alpha }\cdot \lambda )$ with $⟨\lambda +\rho ,{\alpha }^{\vee }⟩\in {\mathbb{Z}}_{>0}$.

The new difference has smaller height, so induction gives $\mu \uparrow s_{\alpha }\cdot \lambda$; appending the indicated positive-integral reflection gives $\mu \uparrow \lambda$.

The image of the embedding is a submodule isomorphic to $M(\mu )$, so its simple quotient $L(\mu )$ is a subquotient, hence a composition factor, of $M(\lambda )$.

Applying the strong linkage principle to that factor gives $\mu \uparrow \lambda$.

If $\mu \uparrow \lambda$, each edge in a witnessing chain gives a positive-root embedding; composing them gives a nonzero homomorphism $M(\mu )\to M(\lambda )\mathrm{.}$

Conversely, write the image of the source highest vector as $a{v}_{\lambda }$ in the PBW model. For $a\mathrm{\ne }0$, a kernel vector $u{v}_{\mu }$ would give $ua=0$ in $U({\mathfrak{n}}^{-})$, impossible because its PBW-degree associated graded algebra is the domain $S({\mathfrak{n}}^{-})$. Thus the map is an embedding, and the necessity lemma gives $\mu \uparrow \lambda$.

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

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

If a proper nonzero submodule existed, it would contain a singular vector of some weight $\mu \mathrm{\ne }\lambda$; the universal property gives a nonzero map $M(\mu )\to M(\lambda )$.

By [L1] the map from step 1.1 is an embedding, so $\mu \uparrow \lambda$. A nonempty linkage chain begins with a positive-integral pairing for $\lambda$, contradicting the strictly negative antidominant inequalities. Thus no proper nonzero submodule exists.