Verma module associated to a Lie algebra

Question

Let $L$ be a Lie algebra with Cartan decomposition $L=H \oplus(\sum_{\alpha \in \Phi}L_{\alpha})$. Let $B=\sum_{\alpha \in \Phi^+}L_{\alpha}$ and $N=\sum_{\alpha \in \Phi^-}L_{\alpha}$ so that $L=B \oplus N$. By definition Verma module is $\Delta(\lambda)=\mathcal{U}(L)\otimes_{\mathcal{U}(B)}\mathbb{C}_{\lambda}$ where $\mathcal{U}(L)$ is the universal enveloping algebra and $\mathbb{C}_{\lambda}$ is a one dimensional $\mathcal{U}(B)$ module. Now we can regard $\Delta(\lambda)$ as a $\mathcal{U}(N)$ module also and as $\mathcal{U}(N)$ modules I was able to prove that $\Delta(\lambda)$ is isomorphic to $\mathcal{U}(N) \otimes _{\mathbb{C}}\mathbb{C}_{\lambda}$.From this I want to conclude that $\Delta(\lambda)$ is a $\mathcal{U}(N)$ free module of rank 1. Also the map from $\mathcal{U}(N) \to \Delta(\lambda)$ mapping $n \to n.v_{\lambda }$ is an isomorphism where $v_{\lambda}$ is the canonical generator of $\Delta(\lambda)=\mathcal{U}(L)\otimes_{\mathcal{U}(B)}\mathbb{C}_{\lambda}$.

Answer 1

We need the following properties of tensor products:

1. Let $R$ be a ring and $M$ a left $R$-module. There exists a unique homomorpism of abelian groups $$ R ⊗_R M \to M \,, \quad r ⊗ m \mapsto rm \,, $$ and this homomorphism is an isomorphism of left $R$-modules.

2. Let $R$, $S$ and $T$ be three rings. Let $M$ be an $R$-$S$-bimodule, $N$ an $S$-$T$-bimodule, and $P$ a left $T$-module. There exists a unique homomorphism of abelian groups $$ (M ⊗_S N) ⊗_T P \to M ⊗_S (N ⊗_T P) \,, \quad (m ⊗ n) ⊗ p \mapsto m ⊗ (n ⊗ p) \,, $$ and this homomorphism is an isomorphism of left $R$-modules.

3. Let $S$ be a ring, $R$ a subring of $S$, and $F$ a free $R$-module. The left $S$-module $S ⊗_R F$ is again free. More precisely, if $(x_i)_{i ∈ I}$ is an $R$-basis of $F$, then $(1 ⊗ x_i)_{i ∈ I}$ is an $S$-basis of $S ⊗_R F$. This entails that $F$ and $S ⊗_R F$ have the same rank.

It follows from the decomposition $L = N ⊕ B$ and the PBW-theorem that the map $$ \mathcal{U}(N) ⊗_ℂ \mathcal{U}(B) \to \mathcal{U}(L) \,, \quad x ⊗ y \mapsto xy $$ is an isomorphism of vector spaces. It is also a homomorphism of $\mathcal{U}(N)$-$\mathcal{U}(B)$-bimodules, and thus an isomorphism of $\mathcal{U}(N)$-$\mathcal{U}(B)$-bimodules. We have therefore the isomorphism of left $\mathcal{U}(N)$-modules \begin{align*} Δ(λ) &= \mathcal{U}(L) ⊗_{\mathcal{U}(B)} ℂ_λ \\ &≅ \Bigl( \mathcal{U}(N) ⊗_ℂ \mathcal{U}(B) \Bigr) ⊗_{\mathcal{U}(B)} ℂ_λ \\ &≅ \mathcal{U}(N) ⊗_ℂ \Bigl( \mathcal{U}(B) ⊗_{\mathcal{U}(B)} ℂ_λ \Bigr) \\ &≅ \mathcal{U}(N) ⊗_ℂ ℂ_λ \end{align*} by properties (1) and (2) of tensor products.

The vector space $ℂ_λ$ is one-dimensional, and has thus a basis consisting of a single vector $v_λ$. It follows from property (3) of tensor products that $\mathcal{U}(N) ⊗_ℂ ℂ_λ$ is free as a left $\mathcal{U}(N)$-module, with a basis given by the single element $1 ⊗ v_λ$. This means that $\mathcal{U}(N) ⊗_ℂ ℂ_λ$ is free of rank $1$ as a left $\mathcal{U}(N)$-module, and that the map $$ \mathcal{U}(N) \to \mathcal{U}(N) ⊗_ℂ ℂ_λ \,, \quad x \mapsto x ⋅ (1 ⊗ v_λ) = x ⊗ v_λ $$ is an isomorphism of left $\mathcal{U}(N)$-modules. The overall isomorphism of left $\mathcal{U}(N)$-modules $$ \mathcal{U}(N) ≅ \mathcal{U}(N) ⊗_ℂ ℂ_λ ≅ \dotsb ≅ \mathcal{U}(L) ⊗_{\mathcal{U}(B)} ℂ_λ $$ is given by $$ x \mapsto x ⊗ v_λ \mapsto x ⊗ (1 ⊗ v_λ) \mapsto (x ⊗ 1) ⊗ v_λ \mapsto (x ⋅ 1) ⊗ v_λ = x ⊗ v_λ \,. $$