Quantum Verma modules $M(\lambda )$, highest weight vectors, and the simple highest weight quotient $L(\lambda )$

${\chi }_{\lambda }$ is well defined: $U_q^{\ge 0}={⨁}_{\nu \in Q_{+}}{U}_q^{\ge 0}[\nu ]$ is graded with $U_q^{\ge 0}[0]=U_q^0$ by [F3], so every $b\in U_q^{\ge 0}$ has a unique decomposition $b=b_0+b_{+}$ with $b_0\in U_q^0$ and $b_{+}\in {⨁}_{\nu >0}{U}_q^{\ge 0}[\nu ]$. Define ${\chi }_{\lambda }(b):={\epsilon }_{\lambda }(b_0)$ with ${\epsilon }_{\lambda }$ as in [F2] and ${\chi }_{\lambda }(b_{+}):=0$; this is $\mathbb{Q}(q)$-linear and well defined by the uniqueness of the decomposition. It is multiplicative: for $\nu ,{\nu }^{\mathrm{\prime }}\in Q_{+}$, $U_q^{\ge 0}[\nu ]U_q^{\ge 0}[{\nu }^{\mathrm{\prime }}]\subseteq U_q^{\ge 0}[\nu +{\nu }^{\mathrm{\prime }}]$; if $\nu ={\nu }^{\mathrm{\prime }}=0$ both factors lie in $U_q^0$ and ${\epsilon }_{\lambda }$ is multiplicative; if $\nu >0$ then ${\chi }_{\lambda }(b{b}^{\mathrm{\prime }})=0={\chi }_{\lambda }(b){\chi }_{\lambda }(b^{\mathrm{\prime }})$, and symmetrically for ${\nu }^{\mathrm{\prime }}>0$. Also ${\chi }_{\lambda }(1)=1$. Hence $\mathbb{Q}(q{)}_{\lambda }$ is a well-defined left $U_q^{\ge 0}$-module.

By [F1] the map $U_q^{-}\otimes U_q^{\ge 0}\to U_q(\mathfrak{g})$, $u\otimes b\mapsto ub$, is a vector-space isomorphism compatible with right multiplication by $U_q^{\ge 0}$, so $U_q(\mathfrak{g})$ is a free right $U_q^{\ge 0}$-module on any basis of $U_q^{-}$; tensoring over $U_q^{\ge 0}$ with $\mathbb{Q}(q{)}_{\lambda }$ gives a linear isomorphism $U_q^{-}\to M(\lambda )$, $u\mapsto u\otimes 1=u{v}_{\lambda }$. Hence $M(\lambda )={⨁}_{\nu \in Q_{+}}{U}_q^{-}[-\nu ]v_{\lambda }$ with the component $M(\lambda {)}_{\lambda -\nu }:=U_q^{-}[-\nu ]v_{\lambda }$ and $M(\lambda {)}_{\mu }=0$ unless $\mu =\lambda -\nu$ with $\nu \in Q_{+}$; each component is finite dimensional and the top component is $\mathbb{Q}(q)v_{\lambda }$ by [F5]. The scalar conditions hold: $K_h{v}_{\lambda }=q^{⟨\lambda ,h⟩}{v}_{\lambda }$ because $K_h$ acts on $\mathbb{Q}(q{)}_{\lambda }$ by ${\chi }_{\lambda }(K_h)=q^{⟨\lambda ,h⟩}$ and $v_{\lambda }=1\otimes 1$; $E_i{v}_{\lambda }={\chi }_{\lambda }(E_i)v_{\lambda }=0$; and $M(\lambda )=U_q{v}_{\lambda }$. For the containment $E_{i}M(\lambda {)}_{\mu }\subseteq M(\lambda {)}_{\mu +{\alpha }_i}$ use [F4]: $E_i(u{v}_{\lambda })=u^{\mathrm{\prime }}{E}_i{v}_{\lambda }+(u^{\mathrm{\prime }\mathrm{\prime }}{b}^{\mathrm{\prime }\mathrm{\prime }})v_{\lambda }=u^{\mathrm{\prime }\mathrm{\prime }}{v}_{\lambda }$-terms with $u^{\mathrm{\prime }\mathrm{\prime }}\in U_q^{-}[-(\nu -{\alpha }_i)]$ when $u\in U_q^{-}[-\nu ]$, and $F_{i}u\in U_q^{-}[-(\nu +{\alpha }_i)]$; combined with the $K_h$-eigenvalue computation $K_h(u{v}_{\lambda })=q^{⟨\lambda -\nu ,h⟩}u{v}_{\lambda }$ this realizes $M(\lambda )$ as a $P$-graded type-one module with the stated grading.

Universal property: the $U_q^{\ge 0}$-module map $\mathbb{Q}(q{)}_{\lambda }\to V$, $1\mapsto v$, is well defined exactly because $K_{h}v=q^{⟨\lambda ,h⟩}v$ and $E_{i}v=0$, i.e. because ${\chi }_{\lambda }$ acts as prescribed; the map $u\otimes c\mapsto cuv$ is balanced because $bv={\chi }_{\lambda }(b)v$ for $b\in U_q^{\ge 0}$, and therefore defines a $U_q$-module map $M(\lambda )\to V$. Any such map is determined by $v_{\lambda }$, so it is unique, and it is surjective when $V=U_{q}v$. If $V$ is $P$-graded and $v\in V_{\lambda }$, the map sends $U_q^{-}[-\nu ]v_{\lambda }$ into $V_{\lambda -\nu }$ because $U_q^{-}[-\nu ]$ lowers labels by $\nu$, so it is grading-preserving.

Proper graded submodules have zero top component: if $N\subseteq M(\lambda )$ is graded and $N\cap M(\lambda {)}_{\lambda }\mathrm{\ne }0$, then $v_{\lambda }\in N$ (as $M(\lambda {)}_{\lambda }=\mathbb{Q}(q)v_{\lambda }$) and $M(\lambda )=U_q{v}_{\lambda }\subseteq N$, so $N=M(\lambda )$. Hence every proper graded submodule $N$ has $N_{\lambda }=0$, and then $({\sum }_j{N}_j{)}_{\lambda }={\sum }_j(N_j{)}_{\lambda }=0$ for any family of proper graded submodules, because the $\lambda$-component of a finite sum of vectors is the sum of components. Thus $N(\lambda )$, the sum of all proper graded submodules, is proper, graded, and contains every proper graded submodule.

The quotient $L(\lambda )=M(\lambda )\mathrm{/}N(\lambda )$ is nonzero and inherits the $P$-grading. If $\bar{P}\subseteq L(\lambda )$ is a graded submodule, its preimage $P\subseteq M(\lambda )$ is graded and contains $N(\lambda )$; if $P\mathrm{\ne }M(\lambda )$ then $P$ is proper graded, so $P\subseteq N(\lambda )$ and $\bar{P}=0$. Hence $L(\lambda )$ has no nonzero proper graded submodule. If $M(\lambda )\mathrm{/}Q$ is any simple graded quotient with $Q$ graded, then $Q$ is proper graded (else the quotient is $0$), so $Q\subseteq N(\lambda )$, and $N(\lambda )\mathrm{/}Q$ is a graded submodule of $M(\lambda )\mathrm{/}Q$; if $Q\mathrm{\ne }N(\lambda )$ it is nonzero, and it is proper because $N(\lambda )\mathrm{\ne }M(\lambda )$, contradicting simplicity. Hence $Q=N(\lambda )$ and the graded-simple quotient is unique. Finally assume the pairing separates $P$: then any submodule $N$ is graded, because the $K_h$ act semisimply on $M(\lambda )$ (by the label grading of step 2.1 and the eigenvalue prescription) and $N={⨁}_{\mu }(N\cap M(\lambda {)}_{\mu })$ — for a finite sum $x={\sum }_j{x}_{{\mu }_j}\in N$, fix $j$ and for each $l\mathrm{\ne }j$ choose $h_{jl}$ distinguishing their characters. The product ${\prod }_{l\mathrm{\ne }j}(K_{h_{jl}}-q^{⟨{\mu }_l,h_{jl}⟩})\mathrm{/}(q^{⟨{\mu }_j,h_{jl}⟩}-q^{⟨{\mu }_l,h_{jl}⟩})$ sends $x$ to $x_{{\mu }_j}$ and preserves $N$, so every component lies in $N$; here the label spaces coincide with the simultaneous eigenspaces by separation, so this is the usual weight decomposition. Therefore all proper submodules are proper graded, $N(\lambda )$ is the unique maximal proper submodule, and $L(\lambda )$ is the ordinary simple quotient of the Verma module. The identifications of $N(\lambda )$ with the radical of the contravariant form and the dominant-integral simplicity statements are the declared content of lem-cartan-projection-constructs-the-quantum-verma-contravariant-form and thm-integrable-quantum-highest-weight-module-exists-and-is-unique, both of which are proved on this page.