Proof Let $Z \subseteq Z(\mathfrak {g}_n)$ be the subalgebra generated by the elements (2.10). Since the inclusion $\operatorname {gr} Z \subseteq \operatorname {gr} Z(\mathfrak {g}_n)$ is an equality, it follows that the inclusion $Z \subseteq Z(\mathfrak {g}_n)$ is also an equality. If these elements admit a nontrivial algebraic relation, then taking $\operatorname {gr}$ and using (2.9), we see that the elements (2.6) would admit a nontrivial relation. By Theorem 2.1, we see that (2.10) are algebraically independent.

Proof Since $\mathfrak {h}_n$ preserves the weight spaces of M, which are finite-dimensional, it follows that each $M^\lambda $ decomposes into generalized eigenspaces for $\mathfrak {h}_n^{\ge 1}$ . Therefore, M admits a decomposition (3.5), and it suffices to show that each summand is a $\mathfrak {g}_n$ -module. This follows by a direct calculation, using the fact that $\mathfrak {g}$ admits an eigenbasis for $\mathfrak {h}$ (root space decomposition) and $\mathfrak {h}_n^{\ge 1}$ acts nilpotently on $\mathfrak {g}_n$ .

Proof Certainly, $p(L_\lambda )$ is a simple highest weight module of highest weight $\nu $ , and the proof follows.

Proof Since $\pm (\lambda - \nu )$ vanishes on $\mathfrak {g}_n^{\prime }$ , it defines a one-dimensional representation of $\mathfrak {g}_n$ , and the named functors are quasi-inverse autoequivalences of $\mathfrak {g}_n\operatorname {-mod}$ . To complete the proof, it suffices to observe that $(\bullet) \otimes _{U(\mathfrak {g}_n)} \mathbb {C}_{\lambda - \nu }$ sends ${\mathcal {O}}^{(\lambda _{\ge 1})}(\mathfrak {g}_n)$ to ${\mathcal {O}}^{(\nu _{\ge 1})}(\mathfrak {g}_n)$ , which follows directly from the definitions.

Proof Note that we have a Lie algebra embedding $\mathfrak {g}_1 \hookrightarrow \mathfrak {g}_n$ given by $x_0 \mapsto x_0$ and $x_1 \mapsto x_n$ for all $x\in \mathfrak {g}$ . We have $\xi _\mu (\nu ) - \eta _\mu (\nu ) \in S(\mathfrak {g}_1) \subseteq S(\mathfrak {g}_n)$ , and this reduces the claim to the case $n = 1$ . This was proven in [Reference ChaffeCh23, Lemma 4.6(b)].

Proof Since the center $Z(\mathfrak {g}_n)$ is generated by the elements $\omega (\partial ^{(k)} p_j)$ , it follows that $\chi _\lambda = \chi _{\lambda '}$ if and only if the characters coincide on these elements. Thus, if we can show that $\chi _\lambda (z_i^{(j)}) = \chi _{\lambda '}(z_i^{(j)})$ for all $j=0,...,n-1$ , then the proof will be complete.

We have $p_i \in S(\mathfrak {g}_n^n)$ , and it follows from (2.4) and (2.5) that $\partial ^{(j)}p_i \in S(\mathfrak {g}_n^{\ge n-j})$ . By the definition of $\omega $ (2.8), we have $z_i^{(j)} \in U(\mathfrak {g}_n^{\ge n-j})$ . Therefore, $\chi _\lambda (z_i^{(j)})$ only depends on $\lambda _{\ge n-j}$ . Since $\chi _\lambda $ is precisely the composition (4.2), we have reached the desired conclusion. This completes the proof.

Proof of Theorem 4.2

Let $\lambda , \lambda '$ satisfy the assumptions of the theorem. Thanks to Lemma 4.6, we must show that $\chi _\lambda (z_i^{(n)}) = \chi _{\lambda '}(z_i^{(n)})$ for all i is equivalent to the condition on $\lambda _0, \lambda _0^{\prime }$ .

Thanks to (2.7), we have $z_i^{(n)} = \omega (d^{(n)} p_j) + \omega (q_i^n)$ . Since $q_i^n \in S(\mathfrak {g}_n^{\ge 1})$ , it follows from (2.8) that $\chi _\lambda (z_i^{(n)}) = \chi _{\lambda '}(z_i^{(n)})$ if and only if $\chi _\lambda \circ \omega \circ d^{(n)}(p_j) = \chi _{\lambda '} \circ \omega \circ d^{(n)}(p_j)$ for all $i=1,...,r$ . We note that this second equality is well-defined because $d^{(n)}$ sends $\mathfrak {h}$ -invariants to $\mathfrak {h}$ -invariants (Lemma 2.2), and $\omega $ is $\mathfrak {h}_n$ -equivariant, and $\chi _\lambda $ coincides with the composition (4.2).

Since $\mu := \lambda _{\ge 1} = \lambda ^{\prime }_{\ge 1}$ , we can apply Lemma 4.5 to see that the central characters coincide if and only if $\eta _{\mu }(\lambda _0) = \eta _{\mu }(\lambda _0^{\prime })$ . Let $\pi : \mathfrak {h}^{*} \rightarrow \mathfrak {h}^{*}/W \cong \mathbb {C}^r$ be the quotient map. It follows from the Chevalley restriction theorem [Reference Chriss and GinzburgCGi10, Theorem 3.1.38] that we can write this in coordinates as $\pi (\nu ) = (p_1|_{\mathfrak {h}^{*}}(\nu ),...,p_r|_{\mathfrak {h}^{*}}(\nu ))$ . Now, it is easy to see by a direct comparison of the two definitions that $\eta _{\mu }(\lambda _0)$ coincides with the differential $d_{\lambda _n}\pi (\lambda _0)$ of the quotient map $\pi $ .

By [Reference ChaffeCh23, Lemma 4.6(e)], we see that $\ker \eta _\mu = \ker d_{\lambda _n} \pi = \mathbb {C}\Phi _{\lambda _n}$ . This implies that $\eta _{\mu }(\lambda _0) = \eta _{\mu }(\lambda _0^{\prime })$ if and only if $\lambda _0 - \lambda _0^{\prime } \in \mathbb {C} \Phi _{\mu _n}$ (this is where we use the fact that the centralizer in a standard Levi subalgebra). This completes the proof.

Proof Since M is indecomposable and admits a highest weight subquotient of weight $\lambda $ , we have $\lambda _n = \mu _n$ . Since $\mathfrak {g}^{\mu _n}$ is the Levi factor of a standard parabolic, we have $\mathfrak {r} = \operatorname {span}\{e_{\alpha }: \alpha \in \Phi ^+ \backslash \Phi _{\lambda _n}\}$ . By Lemma 3.3, we have a finite filtration $0 \subseteq M_1 \subseteq \dots \subseteq M_k = M$ such that $M_i / M_{i-1}$ has highest weight $\lambda ^{(i)} \in \mathfrak {h}_n^{*}$ . By (3.3), the weights of M lie in the set $\bigcup _{i=1}^k \{\lambda ^{(i)}_0 - \sum _{\beta \in \Phi ^+} k_\beta \beta \mid k_\beta \in \mathbb {Z}_{\ge 0}\}$ . Furthermore, by Corollary 4.3, there is an element $\Xi _M \in \mathfrak {h}^{*}/ \mathbb {C} \Phi _{\mu _n}$ such that $\lambda _0^{(i)} + \mathbb {C} \Phi _{\mu _n} = \Xi _M$ for all i. It follows that the weights of M actually lie in the set $\lambda ^{(i)} + \mathbb {C} \Phi _{\mu _n} - \sum _{\beta \in \Phi ^+} \mathbb {Z}_{\ge 0} \beta $ , for any choice of i.

In particular if $\nu \in \mathfrak {h}^{*}$ satisfies $\nu \in \lambda _0 + \Phi _{\lambda _n}$ , then $\nu + \alpha $ does not lie in $\lambda ^{(i)}_0 - \sum _{\beta \in \Phi ^+} \mathbb {Z}_{\ge 0} \beta $ for any $\alpha \in \Phi ^+ \setminus \Phi _{\lambda _n}$ and for any i. Therefore, ${\mathfrak {r}}_n \cdot M^{\nu } = 0$ .

Conversely, suppose $v\in M^{\mathfrak {r}_n}$ is of weight $\nu \in \mathfrak {h}^{*}$ . Since $\mathfrak {h}_n$ acts locally finitely and preserves weight spaces, we can find a common eigenvector for $\mathfrak {h}_n$ of weight $\nu $ in $U(\mathfrak {h}_n)\cdot v$ . Suppose the eigenvalue is $\lambda ' \in \mathfrak {h}_n^{*}$ (by assumption $\lambda ^{\prime }_0 = \nu $ ). Then a quotient of $M_{\lambda '}$ occurs as a submodule of M. All highest weight modules are indecomposable (they admit unique maximal submodules), and this forces the generalized central character of M to be $\chi _{\lambda '}$ . Now, Theorem 4.2 implies that $\nu \in \lambda _0^{\prime } + \mathbb {C}\Phi _{\mu _n}$ . By Corollary 4.3, we see that $\lambda _0^{\prime }$ and $\lambda _0$ lie in the same coset of $\mathfrak {h}^{*}$ modulo $\mathbb {C}\Phi _{\mu _n}$ , and so $\nu \in \lambda _0 + \mathbb {C}\Phi _{\mu _n}$ .

Proof It suffices to take a surjective morphism $M \to N$ in ${\mathcal {O}}^{(\mu )}(\mathfrak {g}_n)$ , and show that the restriction $M^{\mathfrak {r}_n} \rightarrow N^{\mathfrak {r}_n}$ is surjective. Without loss of generality, we can assume that $M, N$ are indecomposable. The existence of a nonzero map $M \to N$ forces $\Xi _M = \Xi _N$ . Since objects of ${\mathcal {O}}(\mathfrak {g}_n)$ are $\mathfrak {h}$ -semisimple, it follows that $M \to N$ is surjective on $\mathfrak {h}$ -weight spaces, and now the result follows immediately from Proposition 4.7.

Proof of Theorem 4.1

We let $\varphi _M$ and $\psi _N$ be the adjunction morphisms from (4.1).

Note that $\operatorname {Ind}$ is an exact functor because $U(\mathfrak {g}_n)$ is free over $U(\mathfrak {p}_n)$ , thanks to the PBW theorem. Furthermore, $(\bullet)^{{\mathfrak {r}}_n}$ is exact by Corollary 4.8. If we can check that $\varphi _M$ and $\psi _N$ are isomorphisms on highest weight modules, then a standard argument using the length of a highest weight filtration (Lemma 3.3) can be used to conclude that $\varphi _M$ and $\psi _N$ are isomorphisms for all $M \in {\mathcal {O}}^{(\mu )}(\mathfrak {g}_n)$ and all $N \in {\mathcal {O}}^{(\mu )}(\mathfrak {g}_n^{\mu _n})$

The map $\psi _M$ is an isomorphism for highest weight modules M, thanks to Proposition 4.7.

Now, suppose that $N \in {\mathcal {O}}^{(\mu )}(\mathfrak {g}_n)$ is a highest weight module with highest weight generator v. Since ${\mathfrak {r}}_n \cdot v = 0$ , it follows that $1\otimes v$ lies in the image of $\varphi _N$ and so $\varphi _N$ is surjective. To prove injectivity, let $K = \ker (\varphi _N)$ and consider the short exact sequence $K \to \operatorname {Ind}(N^{{\mathfrak {r}}_n}) \to N$ . By Corollary 4.8, we have another short exact sequence $K^{\mathfrak {r}_n} \to \operatorname {Ind}(N^{{\mathfrak {r}}_n})^{{\mathfrak {r}}_n} \to N^{{\mathfrak {r}}_n}$ .

Now, set $M = N^{{\mathfrak {r}}_n}$ , which is a highest weight $\mathfrak {g}^{\mu _n}_n$ -module generated by v. The map $\operatorname {Ind}(M)^{{\mathfrak {r}}_n} \to M$ is $U(\mathfrak {g}_n)$ -equivariant map uniquely determined by $1\otimes m \mapsto m$ . Therefore, it is the left inverse of $\psi _M$ , which we have already shown to be bijective. It follows that $K^{{\mathfrak {r}}_n} = 0$ , but since every nonzero object in ${\mathcal {O}}(\mathfrak {g}_n)$ admits a nonzero highest weight vector, it follows that $K = 0$ and so $\varphi _N$ is an isomorphism for highest weight N.

This concludes the proof.

Proof $U_{\alpha }$ satisfies the universal property of the localization by construction, and so it is isomorphic to both the left and right localization, thanks to [Reference McConnell and RobsonMR01, Corollary 2.1.4]. This proves (1), and also implies that U embeds inside $U_{\alpha }$ , and $U_{\alpha }$ is an integral domain.

Note that $U_{\alpha }$ is spanned by unordered monomials in $\mathfrak {g}_n$ and $F_{\alpha }^{-1}$ . Using the left Ore condition, we can rewrite any such monomial as a span of monomials of the form described in (2). Furthermore, if there is a linear dependence between the latter monomials, we can left multiply by appropriate elements of $F_{\alpha }$ to obtain a linear dependence between PBW monomials in U, which must be zero. This proves (2), and (3) follows by a symmetrical argument.

Proof These can be verified by multiplying by powers of $f_{\alpha , j}$ to obtain an expression which holds in U. We show the calculation for (5.3); the other relations are proved similarly. Using the relations in U, we have

$$ \begin{align*} f_{\alpha, j}^3 e_{\alpha, i} = f_{\alpha, j}^2 e_{\alpha, i} f_{\alpha, j} - f_{\alpha, j} h_{\alpha, i+j} f_{\alpha, j} - 2 f_{\alpha, i+2j} f_{\alpha, j}. \end{align*} $$

We then multiply on the left by $f_{\alpha , j}^{-3}$ and on the right by $f_{\alpha , j}^{-1}$ to obtain (5.3).

Proof For any module $M \in U\operatorname {-mod}$ ,

$$ \begin{align*} \phi_{\alpha}(S_{\alpha} \otimes_U M) \cong \phi_{\alpha}(S_{\alpha}) \otimes_U M, \end{align*} $$

and $(\bullet) \otimes _U M$ is right exact; hence, $T_{\alpha }$ is right exact.

Similarly, $G_{\alpha }$ is a composition of two left exact functors, $\operatorname {Hom}_U(S_{\alpha }, \phi ^{-1}_{\alpha }(\bullet))$ and $\mathcal {H}$ ; hence, it is also left exact.

Proof For $M \in \mathcal {O}^{(\mu )}(\mathfrak {g}_n)$ , we let $l(M)$ denote the minimal length of a filtration $0 = M_0 \subseteq M_1 \subseteq \dots \subseteq M_{k-1} \subseteq M_k = M$ such that the sections are highest weight modules (cf. Lemma 3.3). We have an exact sequence

$$\begin{align*}0 \rightarrow M_1 \rightarrow M \rightarrow M/M_1 \rightarrow 0,\end{align*}$$

and since $T_{\alpha }$ is right exact, we have an exact sequence

$$\begin{align*}T_{\alpha} M_1 \rightarrow T_{\alpha} M \rightarrow T_{\alpha}(M/M_1) \rightarrow 0.\end{align*}$$

Using Lemma 5.5 and the fact that $T_{\alpha } M_1$ is a highest weight module, we can reduce the claim that $T_{\alpha }(M) \in {\mathcal {O}}^{(\mu )}(\mathfrak {g}_n)$ to the case where $l(M) = 1$ , i.e., M is a highest weight module.

For the rest of the proof, we fix M highest weight of weight $\lambda \in \mathfrak {h}_n^{*}$ , and let $v \in M$ be a highest weight generator of M. We will show that $f_{\alpha , 0}^{-1} \dots f_{\alpha , n}^{-1} \otimes v$ is highest weight and generates $T_{\alpha } M$ , which will complete the proof of the lemma.

To see that the vector is maximal, use (5.11) to see that $e_{\alpha , i} f_{\alpha , 0}^{-1} \dots f_{\alpha , n}^{-1} \otimes v \in T_{\alpha } M$ lies in an $\mathfrak {h}$ -eigenspace whose weight is not a weight of $T_{\alpha } M$ (for any $i \ge 0$ and $\alpha \in \Phi ^+$ ). To see that it is a genuine highest weight vector, one can use (5.4) to show that $h_i$ acts via $s_{\alpha _i}(h_i)$ .

To see that $f_{\alpha , 0}^{-1} \dots f_{\alpha , n}^{-1} \otimes v$ generates $T_{\alpha } M$ , we use the fact that $Uv = M$ and that $S_{\alpha }$ has two bases, (5.7) and (5.8), to check that every element of $T_{\alpha } M$ lies in the submodule generated by the set $\{f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n} \otimes v \mid i_k> 0\}$ . Let L denote the span of this set. Note that it is an $(\mathfrak {sl}_2)_n$ -module, where $(\mathfrak {sl}_2)_n$ is the truncated current algebra on $\mathfrak {sl}_2 = \langle e_{\alpha }, h_{\alpha }, f_{\alpha }\rangle $ . To complete the current proof we show that L is a simple $(\mathfrak {sl}_2)_n$ -module.

Let $\mathfrak {t}_n \subseteq (\mathfrak {sl}_2)_n$ denote the span on $h_{\alpha ,0},...,h_{\alpha , n}$ , and let $\gamma = \lambda |_{\mathfrak {t}_n} \in \mathfrak {t}_n^{*}$ give the action on $f_{\alpha , 0}^{-1} \dots f_{\alpha , n}^{-1} \otimes v$ . If $M_\gamma $ denotes the Verma module of highest weight $\gamma $ , then there is a nonzero homomorphism $M_\gamma \to L$ and the dimensions of the weight spaces are the same. Therefore, it remains to show that the Verma module $M_\gamma $ is simple. By (5.4), we see that $\gamma (h_{\alpha , n}) = (s_{\alpha } \lambda )(h_{\alpha , n}) \ne 0$ and so we can apply Theorem 4.1 to see that ${\mathcal {O}}^{(\gamma _{\ge 1})}((\mathfrak {sl}_2)_n)$ is equivalent to ${\mathcal {O}}^{(\gamma _{\ge 1})}(\mathfrak {t}_n)$ . In the latter category, all highest weight modules are simple, and it follows that $M_\gamma $ is simple, as required. This completes the proof.

Proof We construct elements $m_{i_0, \dots , i_n}$ for $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1} \backslash \mathcal {I}$ inductively such that the $m_{i_0, \dots , i_n}$ satisfy conditions (i) and (ii) above for any $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1}$ . Then observe that (i) and (ii) ensure that setting $\varphi (f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n}) = m_{i_0, \dots , i_n}$ defines a $U(\mathfrak {a})$ -homomorphism $\varphi : A \rightarrow \phi _{\alpha }^{-1}(M)$ , since it is enough to check that $m_{i_0, \dots , i_k - 1, \dots i_n} = f_{\alpha , k} \cdot _{\alpha ^{-1}} \varphi (f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n}) = f_{\alpha , k} \cdot _{\alpha ^{-1}} m_{i_0, \dots , i_n} = e_{\alpha , k} \cdot m_{i_0, \dots , i_n}$ for any $0 \leq k \leq n$ and $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1}$ .

To construct such $m_{i_0, \dots , i_n}$ , let $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1} \backslash \mathcal {I}$ be such that $i_0 + \dots + i_n$ is minimal among elements of $\mathbb {Z}^{n+1} \backslash \mathcal {I}$ . Then, in particular, $(i_0, \dots , i_k - 1, \dots , i_n) \in \mathcal {I}$ for each $0 \leq k \leq n$ . Let $(\mathfrak {sl}_2)_n$ be the truncated current algebra on the copy of $\mathfrak {sl}_2$ spanned by $e_{\alpha }, h_{\alpha }, f_{\alpha }$ , and let N be the $(\mathfrak {sl}_2)_n$ -module generated by $\{m_{i_0, \dots , i_k - 1, \dots , i_n}: 0 \leq k \leq n\}$ , so $N \in \mathcal {O}^{(\mu )}((\mathfrak {sl}_2)_n)$ . Here, we make a slight abuse of notation, identifying $\mu $ with the restriction to $\mathfrak {h}_n \cap (\mathfrak {sl}_2)_n$ . Then we use the following claim to construct $m_{i_0, \dots , i_n} \in N \subseteq M$ such that $e_{\alpha , k} \cdot m_{i_0, \dots , i_n} = m_{i_0, \dots , i_k - 1, \dots , i_n}$ .

Claim (*) Consider the maps

$$\begin{align*}N \overset{\theta_1}{\longrightarrow} N^{n+1} \overset{\theta_2}{\longrightarrow} N^{\frac{1}{2}n(n+1)}\end{align*}$$

given by $\theta _1(x) = (e_{\alpha , 0} \cdot x, \dots , e_{\alpha , n} \cdot x)$ and $\theta _2(y_0, \dots , y_n) = (e_{\alpha , k} \cdot y_l - e_{\alpha , l} \cdot y_k)_{0 \leq k < l \leq n}$ . Then $\ker (\theta _2) = \operatorname {im}(\theta _1)$ . Furthermore, if $(y_0, \dots , y_n) \in \ker (\theta _2)$ and $y_k \in N \cap M^\lambda $ for all $0 \leq k \leq n$ , then there exists $x \in M^{\lambda - \alpha }$ such that $\theta _1(x) = (y_0, \dots , y_n)$ .

We now prove the claim, which will take several steps.

First, observe that $\theta _2 \circ \theta _1 = 0$ , so certainly $\ker (\theta _2) \supseteq \operatorname {im}(\theta _1)$ . Hence, we only need to show that $\operatorname {im}(\theta _1) \supseteq \ker (\theta _2)$ . Throughout the proof of the claim, we write $e_k$ for $e_{\alpha , k}$ .

We first deal with the case where $n = 1$ and $N = M_{\gamma }$ is a Verma module. In this case, we consider the restriction of these maps to certain weight spaces in the following way (for any $\lambda \in \mathfrak {h}^{*}$ ):

$$\begin{align*}N^\lambda \overset{\theta_1}{\longrightarrow} (N^{\lambda + \alpha})^2 \overset{\theta_2}{\longrightarrow} N^{\lambda + 2\alpha}.\end{align*}$$

Now, either $\dim (N^{\lambda + \alpha }) = 0$ , in which case $\operatorname {im}(\theta _1) = \ker (\theta _2)$ automatically, or the dimensions of these weight spaces satisfy $\dim (N^\lambda ) = n + 1$ , $\dim (N^{\lambda + \alpha }) = n$ , and $\dim (N^{\lambda + 2\alpha }) = n - 1$ . Hence, by considering dimensions and the fact that $\ker (\theta _2) \supseteq \operatorname {im}(\theta _1)$ , it is enough to show that $\theta _1$ is injective and $\theta _2$ is surjective. We can compute that $e_1$ acts on the basis vectors $f_1^i f_0^j \otimes 1$ by

(5.12) $$ \begin{align} e_1 \cdot (f_1^i f_0^j \otimes 1) = \mu j f_1^i f_0^{j-1} \otimes 1 - j(j-1)f_1^{i+1} f_0^{j-2} \otimes 1, \end{align} $$

so by considering $\theta _2(x, 0)$ , we see $\theta _2$ is surjective using (5.12) and an inductive argument. We also see that $e_1 \cdot v = 0$ if and only if $v = f_1^i \otimes 1$ . Since $\mu \neq 0$ , we can show that $e_0 \cdot f_1^i \otimes 1 \neq 0$ , so $\theta _1$ is injective as required.

We now deal with the case where $n \geq 1$ and $N = M_{\lambda }$ is a Verma in the $\mu $ -Jordan block, so $\lambda |_{\ge 1} = \mu $ . We use the following facts, which can be verified by computing the action of $e_n$ on the basis elements $f_0^{i_0} \dots f_n^{i_n} \otimes 1$ of $M_{\lambda }$ and recalling that $\mu (h_{\alpha }) \neq 0$ :

1. (a) $e_n \cdot N = N$ , which uses an inductive argument with (5.12).

2. (b) $e_n \cdot m = 0$ if and only if $m \in \operatorname {span}\{f_1^{i_1} \dots f_n^{i_n} \otimes 1: i_1, \dots , i_n \geq 0\}$ . The latter is isomorphic as a $U(\langle e_0, \dots , e_{n-1} \rangle )$ -module to the Verma module $M_{\gamma }$ over $\operatorname {span}\{ e_{\alpha , i}, h_{\alpha ,i}, f_{\alpha ,i} \mid i=0,...,n-1\} \cong ((\mathfrak {sl}_{2})_{\alpha })_{n-1}$ where $\gamma (h_{\alpha , i}) := \mu (h_{\alpha , i+1})$ .

3. (c) Applying (b) repeatedly, we see that for any $k \geq 1$ , we have that $e_l \cdot m = 0$ for all $k \leq l \leq n$ if and only if $m \in \operatorname {span}\{f_{n-k+1}^{i_{n-k+1}} \dots f_n^{i_n} \otimes 1: i_{n-k+1}, \dots , i_n \geq 0\}$ .

To ease notation slightly, we will write $\mathfrak {a}_{(i)} = \langle e_0,...,e_i\rangle $ , and write $(\mathfrak {sl}_2)_{(i)}$ for the Lie algebra $((\mathfrak {sl}_2)_{\alpha })_i$ and $\gamma _{(i)}$ for the character of $\operatorname {span}\{h_{\alpha , 0},..., h_{\alpha ,i}\}$ given by $\gamma _{(i)}(h_{\alpha ,j}) := \mu (h_{\alpha , n-i})$ .

Now, let $(y_0, \dots , y_n) \in \ker (\theta _2)$ , i.e., $e_k \cdot y_l - e_l \cdot y_k = 0$ for all $0 \leq k, l \leq n$ . By fact (a), there certainly exists an $x_n$ such that $e_n \cdot x_n = y_n$ . We then inductively construct $x_k$ for $2 \leq k \leq n$ such that $e_l \cdot x_k = y_l$ for all $k \leq l \leq n$ . Note that the cases $k = 0, 1$ will be dealt with by an additional argument immediately afterward, and we will then obtain $x_0,...,x_n$ such that $\theta (x_0,...,x_n) = (y_0,...,y_n)$ .

Suppose we have constructed $x_{k+1}$ such that $e_l \cdot x_{k+1} = y_l$ for all $k+1 \leq l \leq n$ . For any $k+1 \leq l \leq n$ , consider $e_l \cdot (e_k \cdot x_{k+1} - y_k) = e_k \cdot (e_l \cdot x_{k+1}) - e_l \cdot y_k = e_k \cdot y_l - e_l \cdot y_k = 0$ . Hence, by fact (c), we have $e_k \cdot x_{k+1} - y_k \in \operatorname {span}\{f_{n-k}^{i_{n-k}} \dots f_n^{i_n} \otimes 1: i_{n-k}, \dots , i_n \geq 0\}$ . But as a $U(\mathfrak {a}_{(k)})$ -module, this is isomorphic to $M_{\gamma _{(k)}}$ , so by fact (a), there exists $x_k^{\prime } \in \operatorname {span}\{f_{n-k}^{i_{n-k}} \dots f_n^{i_n} \otimes 1: i_{n-k}, \dots , i_n \geq 0\}$ such that $e_k \cdot x_k^{\prime } = e_k \cdot x_{k+1} - y_k$ , and by fact (c) $e_l \cdot x_k^{\prime } = 0$ for all $k+1 \leq l \leq n$ . Setting $x_k = x_{k+1} - x_k^{\prime }$ , we see that $e_k \cdot x_k = e_k \cdot x_{k+1} - e_k \cdot x_k^{\prime } = e_k \cdot x_{k+1} - (e_k \cdot x_{k+1} - y_k) = y_k$ , and for $k+1 \leq l \leq n$ , we have $e_l \cdot x_k = e_l \cdot x_{k+1} - e_l \cdot x_k^{\prime } = y_l - 0$ . Hence, we have constructed $x_k$ with the desired properties.

Now, consider $y_1^{\prime } = e_1 \cdot x_2 - y_1$ and $y_0^{\prime } = e_0 \cdot x_2 - y_0$ . For $2 \leq l \leq n$ , we have $e_l \cdot y_1^{\prime } = e_l \cdot (e_1 \cdot x_2 - y_1) = e_1 \cdot y_1 - e_l \cdot y_1 = 0$ , and similarly $e_l \cdot y_0^{\prime } = 0$ , so $y_0^{\prime }, y_1^{\prime } \in \operatorname {span}\{f_{n-1}^{i_{n-1}} f_n^{i_n} \otimes 1: i_{n-1}, i_n \geq 0\}$ , which is isomorphic to $M_{\gamma _{(1)}}$ (Verma module over $((\mathfrak {sl}_2)_{(1)}$ ) as a $U(\langle e_0, e_1 \rangle )$ -module. In addition, since $(y_0,...,y_n) \in \ker \theta $ , we have $e_0 \cdot y_1^{\prime } - e_1 \cdot y_0^{\prime } = 0$ , so we can apply the case where N is a Verma module and $n=1$ , proved earlier, to find $x'$ such that $e_0 \cdot x' = y_0^{\prime }$ , $e_1 \cdot x' = y_1^{\prime }$ , and $e_l \cdot x' = 0$ for $2 \leq l \leq n$ . Then setting $x = x_2 - x'$ , we have that $\theta _1(x) = (y_0, \dots , y_n)$ .

If N is not a Verma module, let $0 = N_0 \subseteq N_1 \subseteq \dots \subseteq N_{k-1} \subseteq N_k = N$ be a filtration of N such that each quotient is a highest weight module. Since $\mu (h_{\alpha }) \neq 0$ , all these highest weight modules must in fact be Verma modules by [Reference WilsonWi11, Theorem 7.1]. Then, given $(y_0, \dots , y_n) \in \ker (\theta _2)$ , in the quotient $N/N_{k-1}$ , we have that there exists $x \in N$ such that $e_l \cdot x + N_{k-1} {\kern-1pt}={\kern-1pt} y_l + N_{k-1}$ for all $0 {\kern-1pt}\leq{\kern-1pt} l {\kern-1pt}\leq{\kern-1pt} n$ . Hence, $e_l \cdot x - y_l {\kern-1pt}\in{\kern-1pt} N_{k-1}$ for all $0 \leq l \leq n$ . But $e_{l'} \cdot (e_l \cdot x - y_l) = e_l \cdot (e_{l'} \cdot x - y_{l'})$ for all $0 \leq l, l', \leq n$ , so by induction on k, there exists $x' \in N_{k-1}$ such that $e_l \cdot x' = e_l \cdot x - y_l$ for all $0 \leq l \leq n$ . Note that in this final argument, we have used the fact that $y_l^{\prime } := e_l \cdot x - y_l$ gives a collection of elements lying in the kernel of $\theta $ , which allows us to apply the inductive hypothesis. Hence, $e_l \cdot (x-x') = y_l$ for all $0 \leq l \leq n$ , so $\operatorname {im}(\theta _1) \supseteq \ker (\theta _2)$ as required.

Finally, if $y_0, \dots , y_n \in M^\lambda \cap N$ , then given $x \in N \subseteq M$ such that $\theta _1(x) = (y_0, \dots , y_n)$ , by considering weight spaces, we may replace x with its component $x^{\lambda - \alpha }$ in the $\lambda - \alpha $ weight space and $\theta _1(x) = \theta _1(x^{\lambda - \alpha })$ , proving the final part of Claim $(\ast )$ .

Since $e_{\alpha , k} \cdot m_{i_0, \dots , i_l -1, \dots , i_n} = m_{i_0, \dots , i_k - 1, \dots , i_l - 1, \dots i_n} = e_{\alpha , l} \cdot m_{i_0, \dots , i_k - 1, \dots i_n}$ for all $0 \leq k < l \leq n$ , this claim then allows us to pick $m_{i_0, \dots , i_n}$ such that $e_{\alpha , k} \cdot m_{i_0, \dots , i_n} = m_{i_0, \dots , i_k - 1, \dots i_n}$ for all $0 \leq k \leq n$ , and if $m_{i_0, \dots , i_k-1, \dots , i_n}$ has weight $\lambda $ for all $0 \leq k \leq n$ , then we can choose $m_{i_0, \dots , i_n}$ to have weight $\lambda - \alpha $ . Hence, applying this inductively, we can construct for all $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1}$ elements $m_{i_0, \dots , i_n} \in M$ satisfying conditions (i) and (ii).

Suppose there exists $\lambda \in \mathfrak {h}^{*}$ such that $\operatorname {wt}(m_{i_0, \dots , i_n}) = \lambda - (i_0 + \dots + i_n)\alpha $ for all $(i_0, \dots , i_n) \in \mathcal {I}$ . Then, by construction, $m_{i_0, \dots , i_n}$ is weight for all $(i_0, \dots , i_n) \in \mathbb {Z}^{n+1}$ , so by Corollary 5.7, the homomorphism we have constructed is a weight vector.

Proof First, we show $\epsilon _N$ is a homomorphism. Let $u \in U$ , let $s \in S_{\alpha }$ , and let $g \in G_{\alpha } N$ . Then

$$\begin{align*}u \cdot \epsilon_N(s \otimes g) &= u \cdot g(s) = \phi_{\alpha}^{-1} \phi_{\alpha} (u) \cdot g(s) = g(\phi_{\alpha}(u) \cdot s) = \epsilon_N((\phi_{\alpha}(u) \cdot s) \otimes g)\\ &= \epsilon_N(u \cdot (s \otimes g)),\end{align*}$$

so $\epsilon _N$ is certainly a U-homomorphism.

Now, let $m \in N$ be a weight vector, and choose $a_0, \dots , a_n$ such that $e_{\alpha , k}^{a_k} \cdot m = 0$ for each k. Let

$$ \begin{align*} \mathcal{I} &= \{(i_0, \dots, i_n) : i_k \leq 0 \mbox{ for some } k \mbox{ or } i_k \leq a_k \mbox{ for all } k\} \subseteq \mathbb{Z}^{n+1}\\ m_{i_0, \dots, i_n} &= e_{\alpha, 0}^{a_0 - i_0} \dots e_{\alpha, n}^{a_n - i_n} \cdot m \mbox{ if } i_k \leq a_k \mbox{ for all } k\\ m_{i_0, \dots, i_n} &= 0 \mbox{ otherwise.} \end{align*} $$

Then we can use Lemma 5.15 to construct $\varphi \in \operatorname {Hom}_U(S_{\alpha }, \phi _{\alpha }^{-1}(M))$ , which is a weight vector and therefore in $G_{\alpha } N$ such that $\varphi (f_{\alpha , 0}^{-a_0} \dots f_{\alpha , n}^{-a_n}) = m$ . Hence, $\epsilon _N(f_{\alpha , 0}^{-a_0} \dots f_{\alpha , n}^{-a_n} \otimes \varphi ) = m$ , so $\epsilon _N$ is surjective.

We now show $\epsilon _N$ is injective. By Lemma 5.12, any element of $T_{\alpha } G_{\alpha } N$ may be written as $f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n} \otimes g$ for some $g \in G_{\alpha } N$ . Suppose $\epsilon _N(f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n} \otimes g) = g(f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n}) = 0$ . We may assume g is weight (i.e., an $\mathfrak {h}$ -eigenvector). If not, we may write g as a sum of $g_\lambda \in (G_{\alpha } N)^\lambda $ , which, by considering weight spaces, must all satisfy $g_\lambda (f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n}) = 0$ , and then apply the following argument to each $g_\lambda $ .

Observe that

(5.13) $$ \begin{align} g(f_{\alpha, 0}^{-j_0} \dots f_{\alpha, n}^{-j_n}) = 0 \text{ if either } j_k \leq 0 \text{ for some } 0 \leq k \leq n \text{ or all } j_k \leq i_k. \end{align} $$

Now, choose

$$ \begin{align*} \mathcal{I} &= \Big\{(j_0, \dots, j_n) \in \mathbb{Z}^{n+1} : \begin{array}{c} j_k \leq i_k \mbox{ for all } 0 \leq k \leq n-1 \mbox{ or } j_n \leq i_n \\ \mbox{ or } j_k \leq 0 \mbox{ for some } 0 \leq k \leq n\end{array} \!\!\!\Big\} \subseteq \mathbb{Z}^{n+1} \\ m_{j_0, \dots, j_n} &= g(f_{\alpha, 0}^{-j_0} \dots f_{\alpha, n}^{-j_n}) \mbox{ if } j_k \leq i_k \mbox{ for all } 0 \leq k \leq n-1\\ m_{j_0, \dots, j_n} &= 0 \mbox{ if } j_n \leq i_n \mbox{ or } j_k \leq 0 \mbox{ for some } 0 \leq k \leq n. \end{align*} $$

Then, applying Lemma 5.15 to this, we construct $g_n \in \operatorname {Hom}_U(S_{\alpha }, \phi _{\alpha }^{-1}(M))$ , which is a weight vector (and hence in $G_{\alpha } N$ ) such that:

1. (a) $g_n(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = 0$ if $j_n \leq i_n$ ,

2. (b) $g_n(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = g(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n})$ if $j_k \leq i_k$ for all $0 \leq k \leq n-1$ .

Note that these two conditions are not mutually exclusive, but they are consistent by (5.13).

By (a), we can define $g_n^{\prime } \in G_{\alpha } N$ by setting $g_n^{\prime }(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n-1}^{-j_{n-1}} f_{\alpha , n}^{-j_n+i_n}) = g_n(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n})$ and so $g_n = f_{\alpha , n}^{i_n} \cdot g_n^{\prime }$ . By (b), we have $(g - g_n)(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = 0$ if either some $j_k \leq 0$ or if $j_k \leq i_k$ for all $0 \leq k \leq n-1$ .

We now construct $g_s \in G_{\alpha } N$ inductively by setting

$$ \begin{align*} \mathcal{I} &= \{(j_0, \dots, j_n) \in \mathbb{Z}^{n+1} : j_k \leq i_k \mbox{ for all } 0 \leq k \leq s-1 \mbox{ or } j_s \leq i_s \mbox{ or } j_k \leq 0\\ & \qquad\mbox{ for some } 0 \leq k \leq n\} \subseteq \mathbb{Z}^{n+1} \\ m_{j_0, \dots, j_n} &= (g - g_n - \dots - g_{s+1})(f_{\alpha, 0}^{-j_0} \dots f_{\alpha, n}^{-j_n}) \mbox{ if } j_k \leq i_k \mbox{ for all } 0 \leq k \leq s-1\\ m_{j_0, \dots, j_n} &= 0 \mbox{ if } j_s \leq i_s \mbox{ or } j_k \leq 0 \mbox{ for some } 0 \leq k \leq n. \end{align*} $$

The element $g - g_n - \dots - g_{s+1}$ satisfies a condition analogous to (5.13), which means that the definition of $m_{j_0,...,j_n}$ is consistent.

This $g_s$ satisfies:

1. (a) $g_s(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = 0$ if $j_s \leq i_s$ ,

2. (b) $g_s(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = (g - g_n - \dots - g_{s+1})(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n})$ if $j_k \leq i_k$ for all $0 \leq k \leq s-1$ .

By (a), we can write this $g_s$ in the form $g_s = f_{\alpha , s}^{i_s} \cdot g_s^{\prime }$ for some $g_s^{\prime } \in G_{\alpha } N$ . By (b), we have that $(g - g_n - \dots - g_s)(f_{\alpha , 0}^{-j_0} \dots f_{\alpha , n}^{-j_n}) = 0$ if $j_k \leq i_k$ for all $0 \leq k \leq s-1$ . Hence, we see that $g = f_{\alpha , 0}^{-i_0} \cdot g_0^{\prime } + \dots + f_{\alpha , n}^{-i_n} \cdot g_n^{\prime }$ for some $g_0^{\prime }, \dots , g_n^{\prime } \in G_{\alpha } N$ , so $f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n} \otimes g = \sum (f_{\alpha , 0}^{-i_0} \dots f_{\alpha , n}^{-i_n}) \otimes f_{\alpha , k}^{i_k} g_k^{\prime } = 0$ . Hence, $\epsilon _N$ is injective.

Proof First, observe that by Theorems 4.1 and 5.9, we may reduce to the case $\mu (\mathfrak {h}_n^n) = 0$ .

Recall the notation $\mathfrak {g}_n^n = \mathfrak {g} \otimes t^n \subseteq \mathfrak {g}_n$ .

First, suppose $n = 1$ . In this case, M has a filtration $M \supseteq \mathfrak {g}_1^1 M \supseteq (\mathfrak {g}_1^1)^2 M \supseteq \cdots $ . Each quotient $(\mathfrak {g}_1^1)^i M / (\mathfrak {g}_1^1)^{i+1} M$ lies in BGG category $\mathcal {O}$ for $\mathfrak {g}$ and hence has finite length, so this may be refined to a filtration satisfying (a) and (b).

For $n> 1$ , consider the filtration $M \supseteq \mathfrak {g}_{n}^n M \supseteq (\mathfrak {g}_{n}^n)^2 M \supseteq \cdots $ . Each quotient $(\mathfrak {g}_n^n)^i M / (\mathfrak {g}_n^n)^{i+1} M$ lies in the category $\mathcal {O}^{(\mu )}(\mathfrak {g}_{n-1})$ , where we identify $\mu $ with an element of $(\mathfrak {h}_{n-1}^{\ge 1})^{*}$ in the obvious manner, so we may set $M_{(i, 0, \dots , 0)} = (\mathfrak {g}_{n}^n)^i M$ and argue by induction that this may be refined to a filtration satisfying (a) and (b).

To show this filtration satisfies (c), it is enough to show that whenever $\mu (\mathfrak {h}_n^n) = 0$ , every $M \in \mathcal {O}^{(\mu )}(\mathfrak {g}_n)$ has the property that $\bigcap _{i \geq 0} (\mathfrak {g}_{n}^n)^i M = 0$ . By considering weight spaces, this property is preserved by taking quotients and extensions, so it is enough to verify this in the case where $M = M_{\lambda }$ is a Verma module. Since $\lambda _n = \mu _n = 0$ , this Verma module is graded: we place a grading on $U(\mathfrak {g}_n)$ by setting, for $x\in \mathfrak {g}$ ,

$$ \begin{align*} \deg x_{i} = \left\{ \begin{array}{rl} 0, & \text{ if } i=0,...,n-1,\\ 1, & \text{ if } i =n. \end{array} \right. \end{align*} $$

We transfer the induced grading on $U(\mathfrak {n}^-_n)$ to $M_\lambda $ via the isomorphism of $\mathfrak {n}^-_n$ -modules $U(\mathfrak {n}^-_n) \cong M_\lambda $ . Now, $M_\lambda $ is a positively graded $\mathfrak {g}_n$ -module and $\bigcap _{i \geq 0} (\mathfrak {g}_{n}^n)^i M_\lambda $ is contained in the intersection of all graded components, which is zero.

Finally, we observe that $\operatorname {ch}(M) = \sum _{(i_0, \dots , i_n) \in \mathcal {I}, i_n> 0} \operatorname {ch} (M_{i_0, \dots , i_n - 1}/ M_{i_0, \dots , i_n})$ , so by uniqueness in Lemma 6.1, the final part of the lemma holds.

Proof The proof is the same as [Reference ChaffeCh23, Corollary 6.3].

Proof Note that characters can be defined for any semisimple $\mathfrak {h}$ -module with finite-dimensional weight spaces. Let $\mathbb {C}_{\lambda _0}$ be the $\mathfrak {h}$ -module of weight $\lambda _0$ .

Since $M_\lambda \cong U(\mathfrak {n}^-_n) \otimes \mathbb {C}_{\lambda _0}$ as $\mathfrak {h}$ -modules, the lemma follows from the facts:

1. (i) The character of $U(\mathfrak {n}^-_n)$ is equal to the character of $S(\mathfrak {n}^-_n)$ .

2. (ii) $S(\mathfrak {n}^-_n)$ is a free module over $S(\mathfrak {n}^-_{n-1})$ and, for $\alpha \in \mathbb {Z}_{\ge 0} \Phi ^+$ , there are $p(\alpha )$ basis vectors of weight $-\alpha $ .

Proof By Lemma 3.6, we have $\operatorname {ch} L_{\nu } = \operatorname {ch} L_{\nu _{\le n-1}}$ for all $\nu \in \mathfrak {h}_n^{*}$ satisfying $\nu _n = 0$ , and so the claim follows from Lemma 6.4.

Proof This follows from Lemma 3.7, Theorem 4.1, and Corollaries 6.2 and 6.5. In order to apply these results, one should check that the functors appearing in the first two cited results send highest weight modules to highest weight modules of the corresponding weight, and we leave this verification to the reader.