#  Representations of the restricted Lie superalgebra $p(n)$

Chaowen Zhang
Department of Mathematics,
China University of Mining and Technology,
Xuzhou, 221116, Jiang Su, P. R. China

Mathematics Subject Classification 2010: 17B10; 17B45; 17B50

##  1 Introduction

Let $F$ be an algebraically closed field of characteristic $p>2$, and let
${\mathfrak{g}}$ be the restricted Lie superalgebra $p(n)$ over $F$. In this
paper, we first study the center of a quotient of the restricted enveloping
superalgebra of ${\mathfrak{g}}$, then apply the obtained results to
investigate the restricted representations of ${\mathfrak{g}}$.

For $n\geq 2$, the Lie superalgebra ${\mathfrak{g}}$ over $F$ consists of
block matrices of the form

$\begin{pmatrix}a&b\\\ c&-a^{t}\end{pmatrix}$

where $a$ is an arbitrary $n\times n$ matrix, $a^{t}$ denotes the transpose of
$a$, $b$ is a symmetric $n\times n$ matrix and $c$ is a skew-symmetric
$n\times n$ matrix ([14, 15, 16, 17]).

It is clear that ${\mathfrak{g}}_{\bar{0}}\cong\mathfrak{gl}(n)$ and hence
${\mathfrak{g}}_{\bar{0}}^{\prime}\cong\mathfrak{sl}(n)$ ([15, 17]). The Lie
superalgebra ${\mathfrak{g}}$ has the $\mathbb{Z}$-gradings
${\mathfrak{g}}={\mathfrak{g}}_{-1}\oplus{\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{1}$,
where ${\mathfrak{g}}_{-1}$ (resp. ${\mathfrak{g}}_{1}$) consists of above
matrices with $a=0$ and $b=0$ (resp. $a=0$ and $c=0$).

According to [14, 17], the universal enveloping superalgera of the Lie
superalgebra $p(n)$ over $\mathbb{C}$ has a nontrivial radical. Applying a
similar idea, we define a nontrivial nilpotent ideal $I$ for the restricted
enveloping superalgebra $u({\mathfrak{g}})$. Then the study of simple
$u({\mathfrak{g}})$-modules is reduced to that of the quotient superalgebra
$\bar{\mathfrak{u}}=u({\mathfrak{g}})/I$.

The paper is organized as follows. Section 2 gives the preliminaries. In
Section 3 we define the quotient superalgebra $\bar{\mathfrak{u}}$ and study
its center. Using the results from this section we study restricted
representations of ${\mathfrak{g}}$ in Section 4.

##  2 Preliminaries

A Lie superalgebra $L=L_{\bar{0}}\oplus L_{\bar{1}}$ is called restricted if
$L_{\bar{0}}$ is a restricted Lie algebra and if $L_{\bar{1}}$ is a restricted
$L_{\bar{0}}$-module under the adjoint action. Define the p-mapping $[p]$ on
${\mathfrak{g}}_{\bar{0}}$ to be the $p$-th power for each matrix in
${\mathfrak{g}}_{\bar{0}}$. Then ${\mathfrak{g}}_{\bar{0}}$ becomes a
restricted Lie algebra and ${\mathfrak{g}}$ is a restricted Lie superalgebra.

Let $L=L_{\bar{0}}\oplus L_{\bar{1}}$ be a restricted Lie superalgebra, let
$U(L)$ be its universal enveloping algebra and let $M=M_{\bar{0}}\oplus
M_{\bar{1}}$ be a $L$-module. We say that $M$ admits a $p$-character $\chi\in
L_{\bar{0}}^{*}$ if $(x^{p}-x^{[p]}-\chi^{p}(x)\cdot 1)M=0$ for all $x\in
L_{\bar{0}}$, where $x^{p}$ denotes the $p$-power of $x$ in $U(L)$. According
to [19, Lemma 2.1], each simple $L$-module admits a unique $p$-character.

Let $\chi\in{\mathfrak{g}}_{\bar{0}}^{*}$ and let $I_{\chi}$ be the (super)
ideal of $U({\mathfrak{g}})$ generated by elements
$x^{p}-x^{[p]}-\chi(x)^{p}$, $x\in{\mathfrak{g}}_{\bar{0}}$. The superalgebra
$u_{\chi}({\mathfrak{g}})=U({\mathfrak{g}})/I_{\chi}$ is called the
$\chi$-reduced enveloping algebra of ${\mathfrak{g}}$. If $\chi=0$, the
superalgebra $u_{\chi}({\mathfrak{g}})$, referred to as the reduced enveloping
algebra of $\mathfrak{g}$, is simply denoted $u({\mathfrak{g}})$. The
$\chi$-reduced enveloping algebra $u_{\chi}({\mathfrak{g}}_{\bar{0}})$ of the
Lie algebra ${\mathfrak{g}}_{\bar{0}}$ is defined similarly (see [18, 5.3]).
In particular, $u_{\chi}({\mathfrak{g}}_{\bar{0}})$ may be viewed as a
subalgebra of $u_{\chi}({\mathfrak{g}})$.

Let $\mathfrak{h}$ be the maximal torus of ${\mathfrak{g}}_{\bar{0}}$
consisting of diagonal matrices. Then $\mathfrak{h}$ has a basis

$h_{1}=:e_{11}-e_{n+1,n+1},\dots,h_{n}:=e_{nn}-e_{2n,2n}.$

Let $\epsilon_{1},\dots,\epsilon_{n}\in\mathfrak{h}^{*}$ be the basis defined
by $\epsilon_{i}(h_{j})=\delta_{ij}$. Take the natural basis of
${\mathfrak{g}}$ as follows. For $1\leq i<j\leq n$, let
$y_{ij}=:e_{i+n,j}-e_{j+n,i}\in{\mathfrak{g}}_{-1}$; for $1\leq i\leq j\leq
n$, let $z_{ij}=e_{i,j+n}+e_{i+n,j}\in{\mathfrak{g}}_{1}$; for $1\leq i,j\leq
n$, let $\tilde{e}_{ij}=e_{ij}-e_{j+n,i+n}\in{\mathfrak{g}}_{\bar{0}}$. These
are root vectors of ${\mathfrak{g}}$ and the roots are of the following form.

Roots of $\displaystyle{\mathfrak{g}}_{\bar{0}}:$
$\displaystyle\epsilon_{i}-\epsilon_{j},$ $\displaystyle 1\leq i\neq j\leq n;$
Roots of $\displaystyle{\mathfrak{g}}_{-1}:$
$\displaystyle-\epsilon_{i}-\epsilon_{j},$ $\displaystyle 1\leq i<j\leq n;$
Roots of $\displaystyle{\mathfrak{g}}_{1}:$
$\displaystyle\epsilon_{i}+\epsilon_{j},$ $\displaystyle 1\leq i\leq j\leq n.$

Note that $\text{dim}{\mathfrak{g}}_{-1}=d=:{1\over 2}n(n-1)$ and
$\text{dim}{\mathfrak{g}}_{1}=n^{2}$.

Denote by $G$ the linear algebraic group $GL_{n}(F)$. Let

$\rho:\ GL_{n}(F)\longrightarrow GL_{2n}(F)$

be defined by $\rho(a)=\text{diag}(a,a^{-1,t})$. Then $\rho$ is a monomorphism
of algebraic groups ([5, 7.4]). Under the action

$Ad\rho(a)x=\rho(a)x\rho(a)^{-1},\ a\in G,x\in{\mathfrak{g}},$

the Lie superalgebra ${\mathfrak{g}}$ becomes a (rational) $G$-module (see [5,
Corollary 10.3]). Then $U({\mathfrak{g}})$ is also a $G$-module.

Let $g\in G$ and let

$x=\begin{pmatrix}0&b\\\ 0&0\end{pmatrix}\in{\mathfrak{g}}_{1}.$

Then we get

$Ad\rho(g)x=\begin{pmatrix}0&gbg^{t}\\\
0&0\end{pmatrix}\in{\mathfrak{g}}_{1}.$

Hence ${\mathfrak{g}}_{1}$ is a $G$-submodule of ${\mathfrak{g}}$ by [10,
2.9(2), Part I]. Similarly ${\mathfrak{g}}_{-1}$ is also a $G$-submodule of
${\mathfrak{g}}$.

Let ${\mathfrak{g}}^{+}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{1}$.
According to [19, Lemma 2.4], each simple $U({\mathfrak{g}}^{+})$-module is
annihilated by ${\mathfrak{g}}_{1}$. Let
$\chi\in{\mathfrak{g}}^{*}_{\bar{0}}$, and let $M$ be a simple
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-module viewed as a
$u_{\chi}({\mathfrak{g}}^{+})$-module by letting ${\mathfrak{g}}_{1}\cdot
M=0$. Define the Kac module

$K_{\chi}(M)=u_{\chi}({\mathfrak{g}})\otimes_{u_{\chi}({\mathfrak{g}}^{+})}M.$

Since $M$ is finite dimensional ([18, Theorem 5.24]), so is $K_{\chi}(M)$. If
$\chi=0$, then we denote $K_{\chi}(M)$ simply by $K(M)$.

We have ${\mathfrak{g}}_{-1}$-module isomorphism
$K_{\chi}(M)\cong\wedge({\mathfrak{g}}_{-1})\otimes_{F}M$, where
$\wedge({\mathfrak{g}}_{-1})$ is the exterior algebra of
${\mathfrak{g}}_{-1}$. Let $v_{1},\dots,v_{d}$ be a basis of
${\mathfrak{g}}_{-1}$. Set

$\wedge^{k}({\mathfrak{g}}_{-1})=\langle v_{i_{1}}\cdots
v_{i_{k}}\in\wedge({\mathfrak{g}}_{-1})|1\leq i_{1}\leq\cdots\leq i_{k}\leq
d\rangle.$

Denote
$\wedge_{\bar{0}}({\mathfrak{g}}_{-1})=\sum_{k=2i}\wedge^{k}({\mathfrak{g}}_{-1})$
and
$\wedge_{\bar{1}}({\mathfrak{g}}_{-1})=\sum_{k=2i+1}\wedge^{k}({\mathfrak{g}}_{-1})$.
Then $\wedge({\mathfrak{g}}_{-1})$ is $\mathbb{Z}_{2}$-graded with
$\wedge({\mathfrak{g}}_{-1})=\wedge_{\bar{0}}({\mathfrak{g}}_{-1})\oplus\wedge_{\bar{1}}({\mathfrak{g}}_{-1})$.
It follows that $K_{\chi}(M)$ is $\mathbb{Z}_{2}$-graded with

$K_{\chi}(M)_{\bar{i}}=\wedge_{\bar{i}}({\mathfrak{g}}_{-1})\otimes_{F}M,\
i=0,1.$

######  Lemma 2.1.

Let $\chi\in{\mathfrak{g}}_{\bar{0}}^{*}$ and let
$\mathfrak{m}=\mathfrak{m}_{\bar{0}}\oplus\mathfrak{m}_{\bar{1}}$ be a simple
$u_{\chi}({\mathfrak{g}})$-module. Then there is an (even) epimorphism
$K_{\chi}(M)\longrightarrow\mathfrak{m}$ with
$M\subseteq\mathfrak{m}_{\bar{0}}$ a simple
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-submodule.

###### Proof.

Let $N\subseteq\mathfrak{m}_{\bar{0}}$ be a simple
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-submodule. Then
$\wedge^{k}({\mathfrak{g}}_{1})N\subseteq\mathfrak{m}_{\bar{0}}$ if $k$ is
even, and $\wedge^{k}({\mathfrak{g}}_{1})N\subseteq\mathfrak{m}_{\bar{1}}$ if
$k$ is odd. It is clear that each $\wedge^{k}({\mathfrak{g}}_{1})N$ is a
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-submodule of $\mathfrak{m}$. Let $k$ be
the largest such that $\wedge^{k}({\mathfrak{g}}_{1})N\neq 0$, and let
$M\subseteq\wedge^{k}({\mathfrak{g}}_{1})N$ be a simple
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-submodule. Then we have
${\mathfrak{g}}_{1}M=0$. We may renumber the $\mathbb{Z}_{2}$-gradings of
$\mathfrak{m}$ if necessary, so we always assume
$M\subseteq\mathfrak{m}_{\bar{0}}$. Then since
$\mathfrak{m}=\mathfrak{m}_{\bar{0}}\oplus\mathfrak{m}_{\bar{1}}$ is simple,
we obtain an (even) epimorphism from $K_{\chi}(M)$ to $\mathfrak{m}$. ∎

Let $\mathfrak{n}_{0}^{+}$ (resp. $\mathfrak{n}_{0}^{-}$) be the Lie
subalgebra of ${\mathfrak{g}}_{\bar{0}}$ spanned by positive root vectors

$\tilde{e}_{ij}=e_{ij}-e_{j+n,i+n},\ i<j\ (\text{resp.}\ i>j).$

######  Definition 2.2.

Let $\mathfrak{m}=\mathfrak{m}_{\bar{0}}\oplus\mathfrak{m}_{\bar{1}}$ be a
${\mathfrak{g}}$-module. A nonzero (homogeneous) vector $v\in\mathfrak{m}$ is
maximal of weight
$\mu=\mu_{1}\epsilon_{1}+\cdots+\mu_{n}\epsilon_{n}\in\mathfrak{h}^{*}$ if
$(\mathfrak{n}_{0}^{+}+{\mathfrak{g}}_{1})\cdot v=0$ and if $h_{i}\cdot
v=\mu_{i}v$, $i=1,\dots,n$.

If $\chi(\mathfrak{n}_{0}^{+})=0$, then each simple
$u_{\chi}({\mathfrak{g}})$-module has a maximal vector (see [19, Section 2]).
For $\mu=\sum^{n}_{i=1}\mu_{i}\epsilon_{i}\in\mathfrak{h}^{*}$, set
$\delta_{\mu}=:\Pi_{i<j}(\mu_{i}-\mu_{j}+j-i-1)\in F$.

In the following proposition, we assume $\chi(\mathfrak{n}_{0}^{+})=0$ and let
$M$ be a simple $u_{\chi}({\mathfrak{g}}_{\bar{0}})$-module generated by a
maximal vector of weight $\mu\in\mathfrak{h}^{*}$.

######  Proposition 2.3.

If $\delta_{\mu}\neq 0$, then $K_{\chi}(M)$ is simple.

###### Proof.

For an arbitrarily fixed order of the basis vectors $y_{ij}$’s of
${\mathfrak{g}}_{-1}$, set
$Y=\Pi_{i<j}y_{ij}\in\wedge^{d}({\mathfrak{g}}_{-1})$.

Let $\mathfrak{m}\subseteq K_{\chi}(M)$ be a simple
$u_{\chi}({\mathfrak{g}})$-submodule. We claim that $Y\otimes
M\subseteq\mathfrak{m}$. To see this, take a nonzero vector

$x=\sum_{i_{1},\dots,i_{k}}y_{i_{1},\dots,i_{k}}\otimes
m_{i_{1},\dots,i_{k}}\in\mathfrak{m},$

where each $y_{i_{1},\dots,i_{k}}$ is a monomial in
$\wedge^{k}({\mathfrak{g}}_{-1})$ and $m_{i_{1},\dots,i_{k}}\in M$. By
applying appropriate $y_{ij}$’s to $x$, we obtain $Y\otimes m\in\mathfrak{m}$
for some nonzero element $m\in M$. Since $FY$ is a $1$-dimensional
$\text{ad}_{{\mathfrak{g}}_{\bar{0}}}$-module, and since $M$ is a simple
$u_{\chi}({\mathfrak{g}}_{\bar{0}})$-module, we obtain $Y\otimes
M\subseteq\mathfrak{m}$.

Set $Z=\Pi_{i<j}z_{ij}$, where the $z_{ij}$’s are in an arbitrary fixed order.
Let $v_{\mu}\in M$ be the maximal vector of the weight $\mu$. Applying a
similar argument as that used in the proof of [15, Lemma 3.1], we have

$ZY\otimes v_{\mu}=\pm\delta_{\mu}\otimes v_{\mu}\in\mathfrak{m}.$

Since $\delta_{\mu}\neq 0$, we have $1\otimes v_{\mu}\in\mathfrak{m}$, and
hence $\mathfrak{m}=K_{\chi}(M)$, implying that $K_{\chi}(M)$ is simple. ∎

##  3 The quotient algebra $\bar{\mathfrak{u}}$

###  3.1  A nilpotent (super) ideal of $u({\mathfrak{g}})$

Since ${\mathfrak{g}}$ is a restricted Lie superalgebra, the element
$x^{p}-x^{[p]}$ is central in $U({\mathfrak{g}})$ for every
$x\in{\mathfrak{g}}_{\bar{0}}$. By [18, Theorem 5.1.2], these elements
generates a polynomial algebra $\mathcal{O}$ inside
$U({\mathfrak{g}}_{\bar{0}})\subseteq U({\mathfrak{g}})$. Let
$x\in{\mathfrak{g}}_{\bar{0}}$. Since $x^{[p]}$ is the $p$-th power of the
matrix $x$, we have $gx^{[p]}g^{-1}=(gxg^{-1})^{[p]}$ for every $g\in G$.
Therefore, $\mathcal{O}$ is $G$-invariant. Then the (super) ideal $I_{0}$
defined in Section 2 is a $G$-submodule of $U({\mathfrak{g}})$. It follows
that $u({\mathfrak{g}})$ is a $G$-module.

In the following we assume $p>3$. Let $x_{1},\dots,x_{t}$ be a basis of
${\mathfrak{g}}_{\bar{0}}$, $y_{1},\dots,y_{d}$ be a basis of
${\mathfrak{g}}_{-1}$ and $z_{1},\dots,z_{l}$ be a basis of
${\mathfrak{g}}_{1}$. By [1, Theorem 2.5] $u({\mathfrak{g}})$ has a basis of
the form

$y^{\delta}x^{s}z^{\delta^{\prime}}=:y_{1}^{\delta_{1}}\cdots
y_{d}^{\delta_{d}}x_{1}^{s_{1}}\cdots
x_{t}^{s_{t}}z_{1}^{\delta^{\prime}_{1}}\cdots z_{l}^{\delta^{\prime}_{l}},\
\delta_{i},\delta^{\prime}_{i}=0,1,\ 0\leq s_{j}\leq p-1.$

It follows that

$u({\mathfrak{g}})\cong\wedge({\mathfrak{g}}_{-1})\otimes
u({\mathfrak{g}}_{\bar{0}})\otimes\wedge({\mathfrak{g}}_{1}).$

If $y=y_{i_{1}}\cdots y_{i_{q}}\in\wedge^{q}({\mathfrak{g}}_{-1})$ or
$z=z_{i_{1}}\cdots z_{i_{q}}\in\wedge^{q}({\mathfrak{g}}_{1})$, we call $q$
the length of $y$ or $z$, and write $l(y)=q$ or $l(z)=q$. The following lemma
is proved in [14] for the Lie superalgebra $p(n)$ over $\mathbb{C}$. Applying
almost verbatim the proof there we obtain the following conclusion in
$u({\mathfrak{g}})$.

######  Lemma 3.1.

(cf. [14, Lemma 2.3(b)]) Let $y\in\wedge({\mathfrak{g}}_{-1})$ and
$z\in\wedge({\mathfrak{g}}_{1})$ are two monomials. Then
$zy=\sum_{\alpha}u_{\alpha}y_{\alpha}z_{\alpha}$, where $u_{\alpha}\in
u({\mathfrak{g}}_{\bar{0}})$, $y_{\alpha}\in\wedge({\mathfrak{g}}_{-1})$ and
$z_{\alpha}\in\wedge({\mathfrak{g}}_{1})$ such that
$l(y_{\alpha})-l(z_{\alpha})=l(y)-l(z)$.

Let $u({\mathfrak{g}})_{k}$ be spanned by above basis elements of
$u({\mathfrak{g}})$ with
$\sum_{j}{\delta_{j}^{\prime}}-\sum_{i}{\delta_{i}}=k$. Then we get

$u({\mathfrak{g}})=\oplus_{k=-d}^{l}u({\mathfrak{g}})_{k}.$

By Lemma 3.1, we obtain $u({\mathfrak{g}})_{i}u({\mathfrak{g}})_{j}\subseteq
u({\mathfrak{g}})_{i+j}$. So that $u({\mathfrak{g}})$ is a $\mathbb{Z}$-grade
algebra. It is easy to see that each $u({\mathfrak{g}})_{k}$ is a
$G$-submodule.

Let $I$ denote the (super) ideal of $u({\mathfrak{g}})$ generated by
$\wedge^{k}({\mathfrak{g}}_{1})$ with $k>d$. Then $I$ is spanned by basis
elements $y^{\delta}x^{s}z^{\delta^{\prime}}$ with
$\sum\delta^{\prime}_{j}\geq d+1$. It follows that $I$ is a $G$-submodule of
$u({\mathfrak{g}})$ such that

$I\subseteq\sum_{k\geq 1}u({\mathfrak{g}})_{k}.$

Applying Lemma 3.1 we have $I^{s}\subseteq\sum_{k\geq s}u({\mathfrak{g}})_{k}$
for any $s\geq 1$, implying that $I$ is nilpotent.

We say that $\mu\in\mathfrak{h}^{*}$ is compatible with
$\chi\in{\mathfrak{g}}_{\bar{0}}^{*}$ if $\mu(h)^{p}-\mu(h)=\chi(h)^{p}$ for
all $h\in\mathfrak{h}$. Set
$\mathfrak{b}_{0}=:\mathfrak{h}+\mathfrak{n}_{0}^{+}\subseteq{\mathfrak{g}}_{\bar{0}}$.

Let $\chi\in{\mathfrak{g}}_{\bar{0}}^{*}$ be such that
$\chi(\mathfrak{n}^{+}_{0})=0$ and let $\mu$ be compatible with $\chi$. Let
$Fv_{\mu}$ be the 1-dimensional $\mathfrak{b}_{0}+{\mathfrak{g}}_{1}$-module
of weight $\mu\in\mathfrak{h}^{*}$ and annihilated by
$\mathfrak{n}_{0}^{+}+{\mathfrak{g}}_{1}$. Define the baby Verma (super)
module

$Z_{\chi}(\mu)=u_{\chi}({\mathfrak{g}})\otimes_{u_{\chi}(\mathfrak{b}_{0}+{\mathfrak{g}}_{1})}Fv_{\mu}.$

If $\chi=0$, we denote it simply by $Z(\mu)$.

The following lemma follows immediately from Lemma 3.1.

######  Lemma 3.2.

(1) Let $K_{\chi}(M)$ be a Kac module. Then $IK_{\chi}(M)=0$.

(2) Let $Z_{\chi}(\mu)$ be a baby Verma module. Then $IZ_{\chi}(\mu)=0$.

###  3.2 The definition of $\bar{\mathfrak{u}}$

Define the quotient superalgebra $\bar{\mathfrak{u}}=u({\mathfrak{g}})/I$.
From Lemma 3.2 we see that each Kac module $K(M)$, each baby Verma module
$Z(\mu)$ or each simple module of $u({\mathfrak{g}})$ may be viewed as a
$\bar{\mathfrak{u}}$-module.

From above discussions we have
$\bar{\mathfrak{u}}=\oplus_{k=-d}^{d}\bar{\mathfrak{u}}_{k}$, where each
$\bar{\mathfrak{u}}_{k}$ has a basis

$y^{\delta}x^{s}z^{\delta^{\prime}},\
\sum_{j}\delta^{\prime}_{j}-\sum_{i}\delta_{i}=k.$

Set

$\mathfrak{a}=\langle y^{\delta}x^{s}z^{\delta^{\prime}}|\
\sum_{j}\delta^{\prime}_{j}=\sum_{i}\delta_{i}>0\rangle.$

Then we have
$\bar{\mathfrak{u}}_{0}=u({\mathfrak{g}}_{\bar{0}})\oplus\mathfrak{a}$. It is
clear that $\mathfrak{a}$, $u({\mathfrak{g}}_{\bar{0}})$ and each
$\bar{\mathfrak{u}}_{k}$ are all $G$-submodules of $\bar{\mathfrak{u}}$.

###  3.3 The center of $\bar{\mathfrak{u}}$

Recall the imbedding $\rho$ of $G$ into $GL_{2n}(F)$. We have

$\rho(G)=\\{\text{diag}(a,a^{-1,t})|a\in GL_{n}(F)\\}.$

The differential $\bar{\rho}$:
$\mathfrak{gl}(n,F)\longrightarrow\mathfrak{gl}(2n,F)$ is a Lie algebra
homomorphism. Using [5, 5.4] we obtain

$\bar{\rho}(a)=\begin{pmatrix}a&0\\\
0&-a^{t}\end{pmatrix}\in{\mathfrak{g}}_{\bar{0}},\ \ a\in\mathfrak{gl}(n,F).$

Define a bilinear form
$\theta:{\mathfrak{g}}_{\bar{0}}\times{\mathfrak{g}}_{\bar{0}}\longrightarrow
F$ by letting $\theta(x,y)=\text{tr}(ab)$

$\text{for}\ x=\text{diag}(a,-a^{t}),\
y=\text{diag}(b,-b^{t})\in{\mathfrak{g}}_{\bar{0}}.$

Then it is easy to see that $\theta$ is symmetric, non-degenerate and
$G$-invariant. Besides, the restriction of $\theta$ to $\mathfrak{h}$ is also
non-degenerate. For each $\lambda\in\mathfrak{h}^{*}$, let
$h_{\lambda}\in\mathfrak{h}$ be the unique element such that
$\theta(h_{\lambda},h)=\lambda(h)$ for all $h\in\mathfrak{h}$. It is easy to
show that $h_{\epsilon_{i}-\epsilon_{j}}=h_{i}-h_{j}$ for any $i\neq j$.

Note that $G$ satisfies the assumptions (H1)-(H3) in [9, 6.3]. Recall the
adjoint action $Ad\rho$ of $G$ on the Lie superalgebra ${\mathfrak{g}}$.

######  Lemma 3.3.

The differential of $Ad\rho$ is
$ad\bar{\rho}:\mathfrak{gl}(n,F)\longrightarrow\mathfrak{gl}({\mathfrak{g}})$
given by

$ad\bar{\rho}(x)(y)=[\bar{\rho}(x),y]=\bar{\rho}(x)y-y\bar{\rho}(x),\ \
x\in\mathfrak{gl}(n,F),\ y\in{\mathfrak{g}}.$

###### Proof.

Since $d(Ad\rho)=d(Ad)\bar{\rho}$ by [5, 5.4], the lemma follows from [5,
Theorem 10.4]. ∎

Define

$\mathcal{Z}=\\{u\in\bar{\mathfrak{u}}_{\bar{0}}|\ gu=u\ \text{for all}\ g\in
G\ \text{and}\ [u,{\mathfrak{g}}_{1}]=0\\}.$

Since $\bar{\mathfrak{u}}$ is a $G$-module, we obtain from [10, 7.11(5),I]
that $\mathcal{Z}\subseteq\\{u\in\bar{\mathfrak{u}}_{\bar{0}}|[x,u]=0\
\text{for all}\ x\in{\mathfrak{g}}\\}$. Thus, $\mathcal{Z}$ is a central
subalgebra of of $\bar{\mathfrak{u}}$.

Take the maximal torus $T=\\{diag(t_{1},\cdots t_{n})|t_{i}\in F^{*}\\}$ of
$G$. We also use the notation $\epsilon_{i}$, $1\leq i\leq n$, to denote the
natural basis of $X(T)$. Therefore the notation
$\epsilon_{i}\in\mathfrak{h}^{*}$ is the differential of $\epsilon_{i}\in
X(T)$ (see [13, 1.2]). Recall the root vectors of ${\mathfrak{g}}$ in Section
2. Then the $T$-weights of these root vectors (viewing $X(T)$ as a additive
group) are

$\text{wt}(y_{ij})=\epsilon_{i}+\epsilon_{j},\
\text{wt}(z_{ij})=-(\epsilon_{i}+\epsilon_{j}),\
\text{wt}(\tilde{e}_{ij})=\epsilon_{i}-\epsilon_{j}.$

Set $\bar{\mathfrak{u}}^{T}=\\{x\in\bar{\mathfrak{u}}|tx=x\ \text{for all}\
t\in T\\}$.

######  Lemma 3.4.

$\bar{\mathfrak{u}}^{T}\subseteq\bar{\mathfrak{u}}_{0}$.

###### Proof.

Recall that $\bar{\mathfrak{u}}$ has a basis consisting of elements
$y^{\delta}x^{s}z^{\delta^{\prime}}$. Let $u\in\bar{\mathfrak{u}}^{T}$. Since
each of these basis elements is a $T$-weight vector, it suffices to assume $u$
is such an element. Write

$u=\Pi_{i<j}y_{ij}^{\delta_{ij}}\Pi_{i\neq
j}\tilde{e}^{s_{ij}}_{ij}f(h)\Pi_{i\leq j}z_{ij}^{\delta^{\prime}_{ij}},\
\delta_{ij},\delta^{\prime}_{ij}=0,1,\ 0\leq s_{ij}\leq p-1,\ f(h)\in
u(\mathfrak{h}).$

Note that $\text{wt}(f(h))=0$.

Since $tu=t^{\text{wt}(u)}u=u$ for all $t\in T$, we have $\text{wt}(u)=0$.
Also, we have

$\displaystyle\text{wt}(\Pi_{i\neq j}\tilde{e}_{ij}^{s_{ij}})$
$\displaystyle=\sum_{i\neq j}s_{ij}(\epsilon_{i}-\epsilon_{j})$
$\displaystyle=\sum_{i<j}l_{ij}(\epsilon_{i}-\epsilon_{j}),$

where $l_{ij}=s_{ij}-s_{ji}$. It follows that

$\sum_{i<j}(\delta^{\prime}_{ij}-\delta_{ij})(\epsilon_{i}+\epsilon_{j})+\sum_{i=1}^{n}2\delta^{\prime}_{ii}\epsilon_{i}+\sum_{i<j}l_{ij}(\epsilon_{i}-\epsilon_{j})=0.$

Since $\epsilon_{1},\dots,\epsilon_{n}\in X(T)$ are linearly independent, the
coefficient of each $\epsilon_{i}$ is $0$. Then we have

$\sum_{i<j}(\delta^{\prime}_{ij}-\delta_{ij})+\sum_{k<i}(\delta^{\prime}_{ki}-\delta_{ki})+2\delta^{\prime}_{ii}+\sum_{i<j}l_{ij}-\sum_{k<i}l_{ki}=0.$

The sum of all these $n$ equations gives

$2\sum_{i<j}\delta^{\prime}_{ij}-2\sum_{i<j}\delta_{ij}+2\sum_{i=1}^{n}\delta^{\prime}_{ii}=0,$

and hence $\sum_{i\leq j}\delta^{\prime}_{ij}=\sum_{i<j}\delta_{ij}$, implying
that $u\in\bar{\mathfrak{u}}_{0}$. ∎

Immediately from the lemma, we have
$\mathcal{Z}\subseteq\bar{\mathfrak{u}}_{0}$.

###  3.4 The Harish-Chandra homomorphism on $\bar{\mathfrak{u}}$

Let $\Lambda_{0}=F_{p}\epsilon_{1}+\cdots+F_{p}\epsilon_{n}$ and let
$u(\mathfrak{h})$ be the restricted enveloping algebra of $\mathfrak{h}$ (see
[18, Section 5.3]).

######  Lemma 3.5.

Let $f(h)\in u(\mathfrak{h})$. If $f(h)(\mu)=0$ for all $\mu\in\Lambda_{0}$,
then $f(h)=0$.

###### Proof.

We proceed by induction on $n$. If $n=1$, then $f(h)$ is a polynomial of one
variable $h_{1}$ such that $\text{deg}f\leq p-1$. Since the set
$\\{h_{1}(\mu)|\ \mu\in\Lambda_{0}\\}\subseteq F$ contains $p$ elements, we
obtain $f(h)=0$.

If $n\geq 2$, then we may write $f(h)\in u(\mathfrak{h})$ as

$\displaystyle f(h_{1},\dots,h_{n})$
$\displaystyle=f_{0}(h_{1},\dots,h_{n-1})+f_{1}(h_{1},\dots,h_{n-1})h_{n}+\cdots+f_{p-1}(h_{1},\dots,h_{n-1})h_{n}^{p-1}.$

For each fixed $\mu\in\Lambda_{0}$, using the fact that
$\mu+F_{p}\epsilon_{n}\subseteq\Lambda_{0}$ together with the conclusion for
$n=1$, we obtain

$\displaystyle f_{0}(h_{1}(\mu),\dots,h_{n-1}(\mu))$
$\displaystyle=f_{1}(h_{1}(\mu),\dots,h_{n-1}(\mu))$ $\displaystyle=\cdots$
$\displaystyle=f_{p-1}(h_{1}(\mu),\dots,h_{n-1}(\mu))$ $\displaystyle=0.$

Then the induction hypotheses yields $f_{i}(h_{1},\dots,h_{p-1})=0$ for all
$i$, and hence $f(h)=0$. ∎

Denote by $\mathfrak{n}^{+}$(resp. $\mathfrak{n}^{-}$) the Lie subalgebra of
${\mathfrak{g}}$ spanned by positive (resp. negative) roots. We fix an order

$f_{\alpha_{1}},\dots,f_{\alpha_{l}},h_{1},\dots,h_{n},e_{\beta_{1}}\dots,e_{\beta_{m}},$

where $\alpha_{1},\dots,\alpha_{l}$ (resp. $\beta_{1},\dots\beta_{m}$) are all
the positive (resp. negative) roots of ${\mathfrak{g}}$. From Section 3.2,
$\bar{\mathfrak{u}}$ has a basis of the form

$f_{\alpha_{1}}^{d_{1}}\cdots f_{\alpha_{l}}^{d_{l}}h_{1}^{s_{1}}\cdots
h_{n}^{s_{n}}e_{\beta_{1}}^{d_{1}^{\prime}}\cdots
e_{\beta_{m}}^{d_{m}^{\prime}},\ 0\leq s_{k}\leq p-1,\sum_{\beta_{i}\ \text{is
odd}}d_{i}^{\prime}\leq d,$

where $d_{i}=0,1$ (resp. $d_{i}^{\prime}=0,1$) if $\alpha_{i}$ (resp.
$\beta_{i}$) is odd and $0\leq d_{i}\leq p-1$ (resp. $0\leq d_{i}^{\prime}\leq
p-1$) if $\alpha_{i}$ (resp. $\beta_{i}$) is even. Then
$\bar{\mathfrak{u}}^{T}$ is spanned by these vectors with

$\sum_{i=1}^{l}d_{i}\alpha_{i}=\sum_{j=1}^{m}d^{\prime}_{j}\beta_{j}.$

Denote by $\mathfrak{n}^{-}\bar{\mathfrak{u}}$ (resp.
$\bar{\mathfrak{u}}\mathfrak{n}^{+}$) the set spanned by these vectors with
$\sum_{i}d_{i}>0$ (resp. $\sum_{i}d_{i}^{\prime}>0$).

The following conclusion is analogous to [3, Lemma 7.4.2],

######  Lemma 3.6.

Set
$\mathfrak{l}=:\bar{\mathfrak{u}}\mathfrak{n}^{+}\cap\bar{\mathfrak{u}}^{T}$.

(1)
$\mathfrak{l}=\mathfrak{n}^{-}\bar{\mathfrak{u}}\cap\bar{\mathfrak{u}}^{T}$,
so that $\mathfrak{l}\triangleleft\bar{\mathfrak{u}}^{T}$;

(2) $\bar{\mathfrak{u}}^{T}=u(\mathfrak{h})\oplus\mathfrak{l}$.

###### Proof.

It suffices to prove (2).

It is clear that $\bar{\mathfrak{u}}^{T}=u(\mathfrak{h})+\mathfrak{l}$. Let
$x\in u(\mathfrak{h})\cap\mathfrak{l}$. Then we have $x=u_{0}(h)$ for some
$u_{0}(h)\in u(\mathfrak{h})$ and $x\in\bar{\mathfrak{u}}\mathfrak{n}^{+}$.
Applying $x$ to the maximal vector $v_{\mu}\in Z(\mu)$ for all
$\mu\in\Lambda_{0}$. Then we get $u_{0}(h)v_{\mu}=0$, and hence
$u_{0}(h)(\mu)=0$ for all $\mu\in\Lambda_{0}$. It follows that $x=u_{0}(h)=0$,
which completes the proof. ∎

Define the Harish-Chandra homomorphism
$h:\bar{\mathfrak{u}}^{T}\longrightarrow u(\mathfrak{h})$ to be the projection
with kernel $\mathfrak{l}$. Recall that
$\bar{\mathfrak{u}}_{0}=u({\mathfrak{g}}_{\bar{0}})\oplus\mathfrak{a}$. It is
easy to see that
$\bar{\mathfrak{u}}^{T}=\bar{\mathfrak{u}}_{0}^{T}=u({\mathfrak{g}}_{\bar{0}})^{T}\oplus{\mathfrak{a}}^{T}$.
Then the restriction of $h$ to $u({\mathfrak{g}}_{\bar{0}})^{T}$ and
$u({\mathfrak{g}}^{\prime}_{\bar{0}})^{T}$ define analogous homomorphisms on
$u({\mathfrak{g}}_{\bar{0}})$ and $u({\mathfrak{g}}^{\prime}_{\bar{0}})$.

Let

$\Phi_{0}=\\{\epsilon_{i}-\epsilon_{j}|1\leq i\neq j\leq n\\}\ (\text{resp.}\
\Phi^{+}_{0}=\\{\epsilon_{i}-\epsilon_{j}|1\leq i<j\leq n\\})$

denote the root system (resp. positive roots) of ${\mathfrak{g}}_{\bar{0}}$.
We denote the Weyl group of $G$ with respect to $T$ by $W$. This group is
generated by $s_{\alpha},\alpha\in\Phi^{+}_{0}\subseteq X(T)$. The action of
each $s_{\alpha}$ on $\mathfrak{h}^{*}$ is given by
$s_{\alpha}(\lambda)=\lambda-\lambda(h_{\alpha})\alpha$. Since $p>2$, each
$s_{\alpha}$ has order $2$. Since the elements

$h_{\epsilon_{1}-\epsilon_{2}}=h_{1}-h_{2},\dots,h_{\epsilon_{n-1}-\epsilon_{n}}=h_{n-1}-h_{n}$

are linearly independent, we can find $\rho\in\mathfrak{h}^{*}$ such that

$\rho(h_{\epsilon_{i}-\epsilon_{i+1}})=1\ \text{for}\ 1\leq i\leq n-1.$

The dot action on $\mathfrak{h}^{*}$ of any $w\in W$ is defined by
$w\cdot\lambda=w(\lambda+\rho)-\rho$. In particular, we have
$s_{\alpha}\cdot\lambda=s_{\alpha}\lambda-\alpha$ for all simple $\alpha$.
Since the $s_{\alpha}$ with $\alpha$ simple generate $W$, the dot action is
independent of the choice of $\rho$ (see [9, 9.2]).

Since the bilinear form $\theta_{|\mathfrak{h}}$ is $W$-invariant and since
$W$ permutes $\Phi_{0}$ ([4, Lemma 10.4 C], we obtain that $W$ permutes
$\\{h_{\alpha}|\alpha\in\Phi_{0}\\}$.

We may identify $U(\mathfrak{h})$ with the symmetric algebra
$S(\mathfrak{h})$, and hence with the algebra of polynomial functions on
$\mathfrak{h}^{*}$. Thus, the dot action on $\mathfrak{h}^{*}$ yields also a
dot action on $U(\mathfrak{h})$ defined by $(w\cdot
f)(\lambda)=f(w^{-1}\cdot\lambda)$ for $f\in U(\mathfrak{h})$. In particular,
we have $s_{\alpha}\cdot h=s_{\alpha}(h)-\alpha(h)1$ if $\alpha$ is a simple
root ([9, 9.2]).

It is easy to see that $W$ stabilizes $\Lambda_{0}\subseteq\mathfrak{h}^{*}$.
Then $W$ also acts on $\Lambda_{0}$ by the dot action. We have
$s_{\alpha}(h)=h-\alpha(h)h_{\alpha}$ for all simple roots $\alpha$, implying
that

$s_{\alpha}(h_{i})^{p}-s_{\alpha}(h_{i})\in\mathcal{O}\cap U(\mathfrak{h})$

for all $i$. Then $W$ acts also on the restricted enveloping algebra
$u(\mathfrak{h})$. Since

$(s_{\alpha}\cdot h_{i})^{p}-s_{\alpha}\cdot
h_{i}=s_{\alpha}(h_{i})^{p}-s_{\alpha}(h_{i})\in\mathcal{O}$

for all $i$, $W$ also acts on $u(\mathfrak{h})$ by the dot action.

Following [17], define an element $\Theta\in U(\mathfrak{h})$ by

$\displaystyle\Theta(\mu)$
$\displaystyle=\Pi_{\alpha\in\Phi_{0}}((\alpha,\mu+\rho)-1)$
$\displaystyle=\Pi_{\i\neq j}((h_{i}-h_{j})(\mu+\rho)-1)\ \text{for}\
\mu\in\mathfrak{h}^{*}.$

Since $W$ permute $\Phi_{0}$, we have $w\cdot\Theta=\Theta$ for all $w\in W$,
and hence $\Theta\in U(\mathfrak{h})^{W\cdot}$.

Note that

$\text{deg}\Theta=|\Phi_{0}|=2|\Phi_{0}^{+}|=2d.$

Assume $p>2d$ in the following. Denote also by $\Theta$ its image in
$u(\mathfrak{h})$. Then we have $\Theta\in u(\mathfrak{h})^{W\cdot}$.

Let $\alpha\in\Phi^{+}_{0}$ and let $x_{\alpha}$ be a root vector such that
$[x_{\alpha},x_{-\alpha}]=h_{\alpha}$. We may simply let
$x_{\alpha}={\tilde{e}}_{ij}$ and $x_{-\alpha}={\tilde{e}}_{ji}$ for
$\alpha=\epsilon_{i}-\epsilon_{j}\in\Phi^{+}_{0}$. Note that
$ad^{3}x_{\alpha}({\mathfrak{g}})=0$. With the assumption $p>3$, we may define
an automorphism of ${\mathfrak{g}}$ by letting

$\tilde{s}_{\alpha}=(\text{exp}\ adx_{\alpha})(\text{exp}\
adx_{-\alpha})(\text{exp}\ adx_{\alpha}).$

By [3, 1.10.19], we have ${\tilde{s}_{\alpha}}|_{\mathfrak{h}}=s_{\alpha}$.
Thus, each element of $W$ may be viewed as an automorphism of
${\mathfrak{g}}$. By essentially a similar argument as that for [4, 2.3 (*)],
we obtain $g\in G$ such that

$\displaystyle\tilde{s}_{\alpha}(x^{[p]})$
$\displaystyle=\rho(g)x^{[p]}\rho(g)^{-1}$
$\displaystyle=(\rho(g)x\rho(g)^{-1})^{[p]}$
$\displaystyle=(\tilde{s}_{\alpha}x)^{[p]}\ \text{for all}\
x\in{\mathfrak{g}}_{\bar{0}}.$

Then $\tilde{s}_{\alpha}$ can be extended to an automorphism of
$u({\mathfrak{g}})$. It is easy to see that
$\tilde{s}_{\alpha}({\mathfrak{g}}_{\pm 1})\subseteq{\mathfrak{g}}_{\pm 1}$.
Then each $w\in W$ can also be extended to an automorphism of
$\bar{\mathfrak{u}}$.

######  Lemma 3.7.

$h(\mathcal{Z})\subseteq u(\mathfrak{h})^{W\cdot}.$

###### Proof.

Let $\mu\in\Lambda_{0}\subseteq\mathfrak{h}^{*}$, and let $v_{\mu}$ be the
maximal vector in $Z(\mu)$. The Lie superalgebra ${\mathfrak{g}}$ has a
triangular decomposition

${\mathfrak{g}}=({\mathfrak{g}}_{-1}+\mathfrak{n}^{-})+\mathfrak{h}+(\mathfrak{n}^{+}+{\mathfrak{g}}_{1}).$

Let $x_{\alpha}$ be a root vector for $\alpha\in\Phi^{+}_{0}$. Note that
$x_{-\alpha}^{p-1}\otimes v_{\mu}\in Z(\mu)$ is a maximal vector for the new
triangular decomposition

$\displaystyle{\mathfrak{g}}=$
$\displaystyle\tilde{s}_{\alpha}({\mathfrak{g}}_{-1}+\mathfrak{n}^{-})+\mathfrak{h}+\tilde{s}_{\alpha}(\mathfrak{n}^{+}+{\mathfrak{g}}_{1})$
$\displaystyle=$
$\displaystyle({\mathfrak{g}}_{-1}+\tilde{s}_{\alpha}(\mathfrak{n}^{-}))+\mathfrak{h}+(\tilde{s}_{\alpha}(\mathfrak{n}^{+})+{\mathfrak{g}}_{1}).$

Therefore, $Z(\mu)$ is isomorphic to the baby Verma module
$Z(\mu-(p-1)\alpha)$ with respect to the new triangular decomposition.

Let $u\in\mathcal{Z}$. By Lemma 3.6, we may write $u=u_{1}+h(u)$ such that
$u_{1}\in\bar{\mathfrak{u}}\mathfrak{n}^{+}$. Since
$\mathcal{Z}\subseteq\bar{\mathfrak{u}}^{G},$ using the discussion preceding
the lemma we have

$u=\tilde{s}_{\alpha}(u)=\tilde{s}_{\alpha}(u_{1})+s_{\alpha}(h(u)).$

Since
$\tilde{s}_{\alpha}(u_{1})\in\bar{\mathfrak{u}}\tilde{s}_{\alpha}(\mathfrak{n}^{+})$,
the righthand side is a linear combination of a basis of $\bar{\mathfrak{u}}$
using the new triangular decomposition.

We now use Jantzen’s argument (see [9, 9.5]). Applying $u$ to
$x_{-\alpha}^{p-1}\otimes v_{\mu}$, we have

$u(x_{-\alpha}^{p-1}\otimes v_{\mu})=h(u)(\mu)x_{-\alpha}^{p-1}\otimes
v_{\mu}$

and

$u(x_{-\alpha}^{p-1}\otimes
v_{\mu})=\tilde{s}_{\alpha}(u)(x_{-\alpha}^{p-1}\otimes v_{\mu})$
$=s_{\alpha}(h(u))(\mu+\alpha)(x_{-\alpha}^{p-1}\otimes
v_{\mu})=h(u)(s_{\alpha}\cdot\mu)(x_{-\alpha}^{p-1}\otimes v_{\mu}),$

implying that $s_{\alpha}\cdot h(u)=h(u)$, and hence $h(u)\in
u(\mathfrak{h})^{W\cdot}$, since $\alpha$ is arbitrary. This completes the
proof. ∎

######  Lemma 3.8.

Let $\mu\in\Lambda_{0}$ be such that $\mu_{n-1}=\mu_{n}$ and let $h(z)\in
h(\mathcal{Z})$. Then

$h(z)(\mu+t(\epsilon_{n-1}+\epsilon_{n}))=h(z)(\mu)$

for any $t\in F_{p}$.

###### Proof.

Let $v_{\mu}\in Z(\mu)$ be the maximal vector of weight $\mu$. Recall the root
vector $\tilde{e}_{n.n-1}=e_{n,n-1}-e_{2n-1,2n}\in{\mathfrak{g}}_{\bar{0}}$.
Since $\mu_{n-1}=\mu_{n}$, $\tilde{e}_{n,n-1}v_{\mu}$ is a maximal vector of
weight $\mu-(\epsilon_{n-1}-\epsilon_{n})$. Then we obtain a homomorphism of
$\bar{\mathfrak{u}}$-modules
$\varphi:Z(\mu-(\epsilon_{n-1}-\epsilon_{n}))\longrightarrow Z(\mu)$. Since
$v_{\mu}\notin\text{Im}\varphi$, we have $\text{Im}\varphi\subsetneq Z(\mu)$.

Recall the root vector for $-\epsilon_{n}-\epsilon_{n-1}$ is
$y_{n-1,n}=e_{2n-1,n}-e_{2n,n-1}\in{\mathfrak{g}}_{-1}$. A straightforward
computation shows that, the image of the vector $y_{n-1,n}v_{\mu}$ in
$Z(\mu)/\text{Im}\varphi$ is a maximal vector of weight
$\mu-(\epsilon_{n-1}+\epsilon_{n})$. This implies that

$h(z)(\mu)=h(z)(\mu-(\epsilon_{n-1}+\epsilon_{n}))$

and hence $h(z)(\mu)=h(z)(\mu+t(\epsilon_{n-1}+\epsilon_{n}))$ for all $t\in
F_{p}$. ∎

######  Lemma 3.9.

Let $\mu\in\Lambda_{0}$ be such that $(\mu+\rho,\epsilon_{i}-\epsilon_{j})=1$
for $i\neq j$. Let $h(z)\in h(\mathcal{Z})$. Then
$h(z)(\mu+t(\epsilon_{i}+\epsilon_{j}))=h(z)(\mu)$ for any $t\in F_{p}$.

###### Proof.

Let $\alpha=\epsilon_{s}-\epsilon_{t},\ s<t$. Then

$s_{\alpha}(\epsilon_{k})=\begin{cases}\epsilon_{k},&\text{if $k\neq s,t$}\\\
\epsilon_{s},&\text{if $k=t$}\\\ \epsilon_{t},&\text{if $k=s$}.\end{cases}$

So we can find $w\in W$ satisfying
$w(\epsilon_{i}-\epsilon_{j})=\epsilon_{n-1}-\epsilon_{n}$ and
$w(\epsilon_{i}+\epsilon_{j})=\epsilon_{n-1}+\epsilon_{n}$. By assumption we
obtain $(w(\mu+\rho),\epsilon_{n-1}-\epsilon_{n})=1$ or, equivalently,

$(w\cdot\mu,\epsilon_{n-1}-\epsilon_{n})=0.$

Using the previous lemma we have

$h(z)(w\cdot\mu)=h(z)(w\cdot\mu+t(\epsilon_{n-1}+\epsilon_{n}))$

for all $t\in F_{p}$. Then we get from Lemma 3.7 that

$\displaystyle h(z)(\mu)$
$\displaystyle=h(z)(w\cdot\mu+t(\epsilon_{n-1}+\epsilon_{n}))$
$\displaystyle=h(z)(w\cdot(\mu+t(\epsilon_{i}+\epsilon_{j}))$
$\displaystyle=h(z)(\mu+t(\epsilon_{i}+\epsilon_{j}))$

for all $t\in F_{p}$, as required. ∎

For $i\neq j$, set
$s_{ij}=\\{\mu\in\Lambda_{0}|(\mu+\rho,\epsilon_{i}-\epsilon_{j})=1\\}$.

######  Lemma 3.10.

Each $h(z)\in h(\mathcal{Z})$ is constant on $\cup_{i\neq j}s_{ij}$.

###### Proof.

(cf. [17, Lemma 4.4]). By the proof of above lemma, we see that, for each
$s_{ij}$ and each $\mu\in s_{ij}$, there is $w\in W$ such that $w\cdot\mu\in
s_{n-1,n}$. Since $h(z)\in u(\mathfrak{h})^{W\cdot}$, it suffices to show that
$h(z)$ is constant on $s_{12}$. This is immediate from Lemma 3.8 when $n=2$.

Suppose $n>2$. Let $\mu=\sum_{i=1}^{n}\mu_{i}\epsilon_{i}\in s_{12}$. Set

$\lambda=\mu+(\mu_{3}-\mu_{1})(\epsilon_{1}+\epsilon_{2})\ \text{and}\
\nu=\lambda+(\mu_{1}-\mu_{3}+1)(\epsilon_{2}+\epsilon_{3}).$

Then we have $\lambda\in s_{23}$. By Lemma 3.9 we have

$h(z)(\mu)=h(z)(\lambda)=h(z)(\nu).$

Let $w=s_{\epsilon_{1}-\epsilon_{2}}s_{\epsilon_{1}-\epsilon_{3}}\in W$. By a
short computation we see that $w\cdot\nu=\mu+2\epsilon_{3}$. Then we have

$\displaystyle h(z)(\mu)$ $\displaystyle=h(z)(\nu)$
$\displaystyle=h(z)(w\cdot\nu)$ $\displaystyle=h(z)(\mu+2\epsilon_{3}),$

and hence $h(z)(\mu)=h(z)(\mu+2k\epsilon_{3})$ for all $k\in F_{p}$. Since
$p>2$, we get $h(z)(\mu)=h(z)(\mu+t\epsilon_{3})$ for all $t\in F_{p}$.
Similarly, we have

$h(z)(\mu)=h(z)(\mu+t\epsilon_{i})$

for all $i>2$ and all $t\in F_{p}$. Therefore, $h(z)$ is constant on $s_{12}$.
∎

######  Lemma 3.11.

Let $h(z)\in h(\mathcal{Z})$ be such that $h(z)|_{s_{n-1,n}}=0$. Then
$(h_{n-1}-h_{n})|h(z)$ in $u(\mathfrak{h})$.

###### Proof.

Write

$h(z)=\sum_{0\leq i,j\leq p-1}c_{ij}h_{n-1}^{i}h_{n}^{j},$

where each $c_{ij}\in u(\mathfrak{h})$ is a polynomial of the variables
$h_{1},\dots,h_{n-2}$. In particular, if $n=2$, each $c_{ij}$ is just a
scalar. By assumption, we have $h(z)(\mu)=0$ if $\mu\in\Lambda_{0}$ with
$\mu_{n-1}=\mu_{n}$.

Note that

$h(z)=\sum_{s=0}^{p-1}(\sum_{i+j\equiv s}c_{ij}h_{n-1}^{i}h_{n}^{j})$

and $k^{i+j}=k^{s}$ for all $k\in F_{p}$ if $i+j\equiv s$. Then we have for
any $\mu\in s_{n-1,n}$ that

$\displaystyle h(z)(\mu)$ $\displaystyle=\sum_{s=0}^{p-1}(\sum_{i+j\equiv
s}c_{ij}(\mu)\mu_{n}^{i+j})$ $\displaystyle=\sum_{s=0}^{p-1}(\sum_{i+j\equiv
s}c_{ij}(\mu))\mu_{n}^{s}$ $\displaystyle=0,$

implying that $\sum_{i+j\equiv s}c_{ij}(\mu_{1},\dots,\mu_{n-2})=0$ for all
$s\in F_{p}$. It follows from Lemma 3.5 that $\sum_{i+j\equiv s}c_{ij}=0\in
u(\mathfrak{h})$ for $s$. Since $h_{n}^{s}=h_{n}^{p+s}\in u(\mathfrak{h})$, we
obtain

$\displaystyle\sum_{i+j\equiv s}c_{ij}h_{n-1}^{i}h_{n}^{j}$
$\displaystyle=\sum_{i+j\equiv s}c_{ij}h_{n-1}^{i}h_{n}^{j}-\sum_{i+j\equiv
s}c_{ij}h_{n}^{s}$
$\displaystyle=\sum_{i+j=s}c_{ij}(h_{n-1}^{i}h_{n}^{j}-h_{n}^{s})+\sum_{i+j=p+s}c_{ij}(h_{n-1}^{i}h_{n}^{j}-h_{n}^{p+s})$
$\displaystyle=\sum_{i+j=s}c_{ij}(h_{n-1}^{i}-h_{n}^{i})h_{n}^{s-i}+\sum_{i+j=p+s}c_{ij}(h_{n-1}^{i}-h_{n}^{i})h_{n}^{p+s-i},$

which gives $(h_{n-1}-h_{n})|h(z)$, as required.

∎

For $i\neq j$, define $\Theta_{ij}\in u(\mathfrak{h})$ by

$\Theta_{ij}(\mu)=(h_{i}-h_{j})(\mu+\rho)-1=(\mu+\rho,\epsilon_{i}-\epsilon_{j})-1,\
\mu\in\Lambda_{0}.$

Then a short computation shows that $\Theta_{ij}=h_{i}-h_{j}+j-i-1$.

######  Corollary 3.12.

Let $i\neq j$ and let $h(z)\in h(\mathcal{Z})$ be such that
$h(z)|_{s_{ij}}=0$. Then $\Theta_{ij}|h(z)$ in $u(\mathfrak{h})$.

###### Proof.

There is $w\in W$ such that
$w(\epsilon_{i}-\epsilon_{j})=\epsilon_{n-1}-\epsilon_{n}$. Let $\mu\in
s_{ij}$. Since $(\mu+\rho,\epsilon_{i}-\epsilon_{j})=1$, so that
$(w(\mu+\rho),w(\epsilon_{i}-\epsilon_{j}))=1$, we have
$(w\cdot\mu+\rho,\epsilon_{n-1}-\epsilon_{n})=1$, and hence $w\cdot\mu\in
s_{n-1,n}$. Therefore, we get $w\cdot s_{ij}=s_{n-1,n}$.

Meanwhile, since

$\displaystyle\Theta_{ij}(\mu)$
$\displaystyle=(\mu+\rho,\epsilon_{i}-\epsilon_{j})-1$
$\displaystyle=(w\cdot\mu+\rho,\epsilon_{n-1}-\epsilon_{n})-1$
$\displaystyle=\Theta_{n-1,n}(w\cdot\mu)$
$\displaystyle=(w^{-1}\cdot\Theta_{n-1,n})(\mu)\ \text{for all}\
\mu\in\Lambda_{0},$

we have $\Theta_{ij}=w^{-1}\cdot\Theta_{n-1,n}$.

Since $w\cdot h(z)=h(z)$ and since $w^{-1}\cdot\lambda\in s_{ij}$ for
$\lambda\in s_{n-1,n}$, we have by assumption that

$h(z)(\lambda)=h(z)(w^{-1}\cdot\lambda)=0.$

Note that $\Theta_{n-1,n}=h_{n-1}-h_{n}$. Then Lemma 3.11 gives
$\Theta_{n-1,n}|h(z)$, and hence $w^{-1}\cdot\Theta_{n-1,n}|w^{-1}\cdot h(z)$,
implying that $\Theta_{ij}|h(z)$, as required. ∎

Let $z_{i}=h_{i}^{p}-h_{i}$ for $1\leq i\leq n$. By [18, Theorem 5.1.2],

$\mathcal{O}^{\prime}=:F[z_{1},\dots,z_{n}]\subseteq U(\mathfrak{h})$

is a polynomial ring. Clearly we see that
$\mathcal{O}^{\prime}=\mathcal{O}^{\prime}_{F_{p}}\otimes_{F_{p}}F$, where
$\mathcal{O}^{\prime}_{F_{p}}=F_{p}[z_{1},\dots,z_{n}]$. Then we have

$u(\mathfrak{h})=U(\mathfrak{h})/\mathcal{O}^{\prime}U(\mathfrak{h})=U(\mathfrak{h}_{F_{p}})\otimes_{F_{p}}F/(\mathcal{O}^{\prime}_{F_{p}}U(\mathfrak{h}_{F_{p}}))\otimes_{F_{p}}F\cong
u(\mathfrak{h}_{F_{p}})\otimes_{F_{p}}F.$

######  Lemma 3.13.

$u(\mathfrak{h})^{W\cdot}=u(\mathfrak{h}_{F_{p}})^{W\cdot}\otimes_{F_{p}}F$.

###### Proof.

Let $\\{a_{i}\\}_{i\in I}$ be a basis of the $F_{p}$-vector space $F$, and let

$x=\sum f_{i}(h)\otimes a_{i}\in(u(\mathfrak{h}_{F_{p}})\otimes F)^{W\cdot}.$

Since $w\cdot x=x$ for every $w\in W$, we get $\sum_{i}(w\cdot
f_{i}(h)-f_{i}(h))\otimes a_{i}=0$ for all $w$. Substituting $h$ by each
$\lambda\in\Lambda_{0}$, we get, for each $i$, $(w\cdot
f_{i}(h)-f_{i}(h))(\lambda)=0$ for all $\lambda\in\Lambda_{0}$. Then Lemma 3.5
yields $w\cdot f_{i}(h)-f_{i}(h)=0$ for all $w\in W$, and hence $f_{i}(h)\in
u(\mathfrak{h}_{F_{p}})^{W\cdot}$, as desired. ∎

######  Theorem 3.14.

Assume $p>2d$. Then $h(\mathcal{Z})\subseteq\Theta u(\mathfrak{h})^{W}+F$.

###### Proof.

Let $h(z)\in h(\mathcal{Z})$. By Lemma 3.10, there is $k\in F$ such that
$(h(z)-k)|_{s_{ij}}=0\ \text{for all}\ i\neq j.$ Then Corollary 3.12 gives
$\Theta_{ij}|(h(z)-k)$ in $u(\mathfrak{h})$ for all $i\neq j$. It follows that
$\Theta|(h(z)-k)^{2d}$ and hence $\Theta|(h(z)-k)^{p}$ in $u(\mathfrak{h})$.

Assume $(h(z)-k)^{p}=\Theta q(h)$ for some $q(h)\in u(\mathfrak{h})$. Since
$(h(z)-k)^{p},\Theta\in u(\mathfrak{h})^{W\cdot}$, we have
$(h(z)-k)^{p}=\Theta\ (w\cdot q(h))$ for every $w\in W$. Since
$p>2d=n(n-1)\geq n$, $|W|=n!\neq 0$. Let

$\tilde{q}(h)={1\over|W|}\sum_{w\in W}w\cdot q(h).$

Then we have $\tilde{q}(h)\in u(\mathfrak{h})^{W\cdot}$ and also
$(h(z)-k)^{p}=\Theta\tilde{q}(h)$. Thus, we may assume that $q(h)\in
u(\mathfrak{h})^{W\cdot}$.

Let $\\{a_{i}\\}_{i\in I}$ be a basis of the $F_{p}$-vector space $F$ as
above. Since $F$ is algebraically closed, the set $\\{a_{i}^{p}\\}_{i\in I}$
is also a basis. In fact, for each $a\in F$, we have
$a^{1/p}=\sum_{i}\lambda_{i}a_{i}$ for $\lambda_{i}\in F_{p}$, so that
$a=\sum\lambda_{i}a_{i}^{p}$. The linear independence of the set
$\\{a_{i}^{p}\\}$ is immediate.

By the lemma above, we have $h(z)-k=\sum_{i}h_{i}(z)\otimes a_{i}$, where
$h_{i}(z)\in u(\mathfrak{h}_{F_{p}})^{W\cdot}$. Then

$(h(z)-k)^{p}=\sum_{i}h_{i}(z)^{p}\otimes a_{i}^{p}=\sum_{i}h_{i}(z)\otimes
a_{i}^{p}.$

Assume $q(h)=\sum_{i}q_{i}(h)\otimes a_{i}^{p}$ with $q_{i}(h)\in
u(\mathfrak{h}_{F_{p}})^{W\cdot}$. Then we get

$\sum_{i}(h_{i}(z)-\Theta q_{i}(h))\otimes a_{i}^{p}=0.$

By a similar argument as that used in the proof of above lemma, we have
$h_{i}(z)=\Theta q_{i}(h)$ for all $i$, implying that $\Theta|(h(z)-k)$. Let
$h(z)-k=\Theta q^{\prime}(h)$. By the discussion above we may assume
$q^{\prime}(h)\in u(\mathfrak{h})^{W\cdot}$. This completes the proof.

∎

###  3.5 The image of $h$

In this subsection, we construct some central elements in
$\bar{\mathfrak{u}}$.

Set $\wedge^{\leq
d}({\mathfrak{g}}_{1})=:\sum_{i=0}^{d}\wedge^{i}({\mathfrak{g}}_{1})$. From
Section 3.2, we obtain an isomorphism $\bar{\mathfrak{u}}\cong\wedge^{\leq
d}({\mathfrak{g}}_{1})\otimes
u({\mathfrak{g}}_{\bar{0}})\otimes\wedge({\mathfrak{g}}_{-1})$ as vector
spaces. Then $\bar{\mathfrak{u}}$ has a basis of the form

$z^{\delta^{\prime}}x^{s}y^{\delta},\ z^{\delta^{\prime}}\in\wedge^{\leq
d}({\mathfrak{g}}_{1}),\ x^{s}\in u({\mathfrak{g}}_{\bar{0}}),\
y^{\delta}\in\wedge({\mathfrak{g}}_{-1}).$

Define a projection $(,)_{0}$: $\bar{\mathfrak{u}}\longrightarrow
u({\mathfrak{g}}_{\bar{0}})$ by

$(z^{\delta^{\prime}}x^{s}y^{\delta})_{0}=\begin{cases}x^{s},&\text{if
$z^{\delta^{\prime}}=y^{\delta}=1$}\\\ 0,&\text{otherwise.}\end{cases}$

Recall the notion $Y$ and $Z$ defined earlier.

######  Lemma 3.15.

Assume $p>2d$. Then $(Y\wedge^{d}({\mathfrak{g}}_{1}))_{0}\neq 0$.

###### Proof.

Let $L(\mu)$ be the restricted simple ${\mathfrak{g}}_{\bar{0}}$-module of
highest weight $\mu\in\Lambda_{0}$. Note that
$\delta_{\mu}=(\Pi_{i<j}\Theta_{ij})(\mu)$ for all $\mu\in\Lambda_{0}$. Since
$p>n(n-1)$, the degree of each $h_{i}$ in the polynomial
$\Pi_{i<j}\Theta_{ij}$ is less that $p$. By Lemma 3.5, there is
$\mu\in\Lambda_{0}$ such that $(\Pi_{i<j}\Theta_{ij})(\mu)\neq 0$. It then
follows from Proposition 2.3 that $K(L({\mu}))$ is simple as a
$u({\mathfrak{g}})$-module.

Let $v_{\mu}\in L(\mu)$ be a maximal vector. Write $YZ$ as a linear
combination of the above basis for $\bar{\mathfrak{u}}$. Then we have in
$K(L(\mu))$ that

$(YZ)_{0}Yv_{\mu}=YZYv_{\mu}=\delta_{\mu}Yv_{\mu}\neq 0,$

implying that $(YZ)_{0}\neq 0$. ∎

Since $\wedge({\mathfrak{g}}_{-1})$, $\wedge^{\leq d}({\mathfrak{g}}_{1})$ and
$u({\mathfrak{g}}_{0})$ are $G$-submodules of $u({\mathfrak{g}})$, it is easy
to see that $(,)_{0}$ is a homomorphism of $G$-modules.

Let $u^{d}({\mathfrak{g}}_{\bar{0}})=\langle x_{1}\cdots x_{k}\in
u({\mathfrak{g}}_{\bar{0}})|x_{1},\dots,x_{k}\in{\mathfrak{g}}_{\bar{0}},\
k\leq d\rangle$. Clearly $u^{d}({\mathfrak{g}}_{\bar{0}})$ is a $G$-submodule
of $u({\mathfrak{g}})$.

Define a linear map $\varphi:\wedge^{d}({\mathfrak{g}}_{1})\longrightarrow
u^{d}({\mathfrak{g}}_{\bar{0}})$ by letting $\varphi(x)=(Yx)_{0}$ for
$x\in\wedge^{d}({\mathfrak{g}}_{1})$.

######  Lemma 3.16.

$\varphi$ is a homomorphism of $G^{\prime}$-modules.

###### Proof.

Recall that $G$ acts on ${\mathfrak{g}}$ by the composition of the adjoint
action with $\rho$. A short computation shows that, for $c\in F$, $s<t$,

$Ad\rho(1+ce_{ij})y_{st}=\begin{cases}&y_{st}-cy_{jt},\ \text{if $i=s$}\\\
&y_{st},\ \ \text{otherwise.}\end{cases}$

Then we get $(1+ce_{ij})\cdot Y=Y$. Since $G^{\prime}=SL_{n}(F)$ is generated
by all $1+ce_{ij}$, $i\neq j$([8, Lemma 6.7.1]), we have $g\cdot Y=Y$ for all
$g\in G^{\prime}$.

For any $x\in\wedge^{d}({\mathfrak{g}}_{1})$ and $g\in G^{\prime}$, we then
have

$\displaystyle\varphi(g\cdot x)$ $\displaystyle=(Y\rho(g)x\rho(g)^{-1})_{0}$
$\displaystyle=(\rho(g)(Yx)\rho(g^{-1}))_{0}$ $\displaystyle=(g\cdot(Yx))_{0}$
$\displaystyle=g\cdot(Yx)_{0}$ $\displaystyle=g\cdot\varphi(x),$

as required. ∎

####  3.5.1 A simple ${\mathfrak{g}}_{\bar{0}}$-module of typical weight

In the following we study a simple ${\mathfrak{g}}_{\bar{0}}$-module used
later to construct central elements in $\bar{\mathfrak{u}}$.

######  Definition 3.17.

A weight $\lambda\in\Lambda_{0}$ is typical if $\Theta(\lambda)\neq 0$ and
atypical otherwise.

Example. Let
$\lambda=(n-1)\epsilon_{1}+(n-2)\epsilon_{2}+\cdots+\epsilon_{n-1}\in\Lambda_{0}$.
If $p>2n-1$, then the weight $\lambda$ is typical.

Note that if $\lambda$ is typical, then $\delta_{\lambda}\neq 0$.

Denote by ${\mathfrak{g}}_{\mathbb{C}}$ the Lie superalgebra $p(n)$ over
$\mathbb{C}$. Let $V(\tilde{\lambda})$ be the simple
${\mathfrak{g}}_{\bar{0},\mathbb{C}}$-module with highest weight
$\tilde{\lambda}=(n-1)\epsilon_{1}+(n-2)\epsilon_{2}+\cdots+\epsilon_{n-1}\in\mathfrak{h}^{*}$.

Since $\tilde{\lambda}$ is dominant integral, $V(\tilde{\lambda})$ is finite
dimensional (see [4, Theorem 21.2]). Assume
$\Phi_{0}^{+}=\\{\alpha_{1},\dots,\alpha_{d}\\}$, where
$\alpha_{i}=\epsilon_{i}-\epsilon_{i+1}$ for $1\leq i\leq n-1$. Then
$\beta=\alpha_{1}+\cdots+\alpha_{n-1}\in\Phi^{+}_{0}$ is the longest root. For
each $\alpha=\epsilon_{i}-\epsilon_{j}\in\Phi^{+}_{0}$, let
$e_{\alpha}=\tilde{e}_{ij},f_{\alpha}=\tilde{e}_{ji}$ and
$h_{\alpha}=h_{i}-h_{j}$.

Let $v^{+}\in V(\tilde{\lambda})$ be the maximal vector. Then
$V(\tilde{\lambda})$ is spanned by vectors of the form
$f_{\alpha_{1}}^{s_{1}}\cdots f_{\alpha_{d}}^{s_{d}}v^{+}$, $s_{i}\geq 0$. Let
$\mathfrak{sl}^{\beta}_{2}=\langle
e_{\beta},f_{\beta},h_{\beta}\rangle\subseteq{\mathfrak{g}}_{\bar{0},\mathbb{C}}$.
By [4, Theorem 6.3], $V(\tilde{\lambda})$ is completely reducible as a
$\mathfrak{sl}^{\beta}_{2}$-module. Moreover, each simple summand has maximal
weight $m\geq 0$ and has dimension $m+1$ ([4, Theorem 7.2]).

We claim that $\tilde{\lambda}(h_{\beta})$ is the largest of all these
weights. In fact, note that all weights (with respect to
$\mathfrak{h}_{\mathbb{C}}=\langle h_{1},\cdots,h_{n}\rangle$) in
$V(\tilde{\lambda})$ are of the form

$\tilde{\lambda}-\sum_{i=1}^{n-1}{k_{i}}\alpha_{i},\ k_{i}\geq 0.$

Since $\alpha_{i}=\epsilon_{i}-\epsilon_{i+1}$ for all $i$ and
$h_{\beta}=h_{1}-h_{n}$, we have
$\alpha_{i}(h_{\beta})=\delta_{i1}+\delta_{i+1,n}\geq 0$. Then the claim
follows.

Since $\tilde{\lambda}(h_{\beta})=n-1$, it follows from [4, Theorem 7.2] that
every simple summand of the $\mathfrak{sl}^{\beta}_{2}$-module
$V(\tilde{\lambda})$ has dimension $\leq n$. Thus, we obtain that
$f_{\beta}^{n}V(\tilde{\lambda})=0$, and similarly
$e_{\beta}^{n}V(\tilde{\lambda})=0$.

######  Lemma 3.18.

Let $\alpha\in\Phi_{0}^{+}$. Then
$e_{\alpha}^{n}V(\tilde{\lambda})=f_{\alpha}^{n}V(\tilde{\lambda})=0$.

###### Proof.

Note that $\Phi_{0}^{+}=\\{\epsilon_{i}-\epsilon_{j}|1\leq i<j\leq n\\}$. For
each $\alpha\in\Phi_{0}^{+}$, define

$x_{\alpha}(t)=\text{exp}(tade_{\alpha}),\ t\in\mathbb{C}.$

Then $x_{\alpha}(t)$ is an automorphism of
${\mathfrak{g}}_{\bar{0},\mathbb{C}}$. Set
$\tilde{\sigma}_{\alpha}=x_{\alpha}(1)x_{-\alpha}(-1)x_{\alpha}(1)$. According
to [2, Lemma 6.4.4] and [2, Proposition 6.4.2], we have
$\tilde{\sigma}_{\alpha}(h_{\beta})=h_{\sigma_{\alpha}(\beta)}$ and
$\tilde{\sigma}_{\alpha}(e_{\beta})=\pm e_{\sigma_{\alpha}(\beta)}$ for any
$\beta\in\Phi_{0}$, where $\sigma_{\alpha}$ is the refection defined with
$\alpha$ (see [4, Section 9.1]).

For each $\alpha\in\Phi_{0}^{+}$, we have a triangular decomposition of
${\mathfrak{g}}_{\bar{0},\mathbb{C}}$ with respect to $\alpha$:

${\mathfrak{g}}_{\bar{0},\mathbb{C}}=\tilde{\sigma}_{\alpha}(\mathfrak{n}_{0}^{+})+\mathfrak{h}+\tilde{\sigma}_{\alpha}(\mathfrak{n}_{0}^{-}).$

The corresponding positive root system(resp. bases) is
$\sigma_{\alpha}(\Phi_{0}^{+})$ (resp. $\sigma_{\alpha}(\Delta)$). Since
$e_{\alpha}$ acts nilpotently on $V(\tilde{\lambda})$, we may define an
automorphism of the vector space $V(\tilde{\lambda})$ by (see [11, Chapter 8]
or [4, 21.25])

$\sigma_{\alpha,V}=\text{exp}(e_{\alpha,V})\text{exp}(-e_{-\alpha,V})\text{exp}(e_{\alpha,V}),$

where $e_{\alpha,V}$ denotes the endomorphism of $V(\tilde{\lambda})$ given by
the action of $e_{\alpha}$. Then for $x\in{\mathfrak{g}}_{\bar{0}}$ and $v\in
V(\tilde{\lambda})$, we have

$\sigma_{\alpha,V}(xv)=\tilde{\sigma}_{\alpha}(x)\sigma_{\alpha,V}(v).$

This follows from [2, Lemma 4.5.2] (for the case $\alpha$ is simple, see also
[11, Chapter 8]).

Let $\mathfrak{b}=\mathfrak{h}+\mathfrak{n}_{0}^{+}$. We claim that
$\sigma_{\alpha,V}(v^{+})\in V(\tilde{\lambda})$ is the maximal vector with
respect to the Borel subalgebra $\tilde{\sigma}_{\alpha}(\mathfrak{b})$. In
fact, we have

$\tilde{\sigma}_{\alpha}(x)\sigma_{\alpha,V}(v^{+})=\sigma_{\alpha,V}(xv^{+})=0$

for all $x\in\mathfrak{n}_{0}^{+}$ and

$\displaystyle h\sigma_{\alpha,V}(v^{+})$
$\displaystyle=\sigma_{\alpha,V}(\tilde{\sigma}_{\alpha}^{-1}(h)v^{+})$
$\displaystyle=\tilde{\lambda}(\tilde{\sigma}_{\alpha}^{-1}(h))\sigma_{\alpha,V}(v^{+})$
$\displaystyle=(\sigma_{\alpha}\tilde{\lambda})(h)\sigma_{\alpha,V}(v^{+}),$

for $h\in\mathfrak{h}$. Then the claim follows. In addition, we see that the
highest weight of $V(\tilde{\lambda})$ is $\sigma_{\alpha}\tilde{\lambda}$.

For each $\alpha=\epsilon_{i}-\epsilon_{j}\in\Phi_{0}^{+}$, let
$\alpha_{1}=\epsilon_{1}-\epsilon_{i}$ and
$\alpha_{2}=\epsilon_{j}-\epsilon_{n}$. We let $\sigma_{\alpha_{2}}=1$ if
$j=n$ and $\sigma_{\alpha_{1}}=1$ if $i=1$. Then we have
$\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}(\beta)=\alpha$, where
$\beta=\epsilon_{1}-\epsilon_{n}$ is the longest root.

By above discussion, the highest weight of $V(\tilde{\lambda})$ with respect
to the Borel subalgebra
$\tilde{\sigma}_{\alpha_{2}}\tilde{\sigma}_{\alpha_{1}}(\mathfrak{b})$ is
$\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}\tilde{\lambda}$. Meanwhile, the
longest root in the positive system
$\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}(\Phi_{0}^{+})$ is $\alpha$. Decompose
$V(\tilde{\lambda})$ as a direct sum of simple
$\mathfrak{sl}^{\alpha}_{2}$-submodules. By a similar argument as above, we
obtain that the largest weight of these simple summands is

$\displaystyle(\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}\tilde{\lambda})(h_{\alpha})$
$\displaystyle=(\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}\tilde{\lambda})(h_{\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}(\epsilon_{1}-\epsilon_{n})})$
$\displaystyle=(\sigma_{\alpha_{2}}\sigma_{\alpha_{1}}\tilde{\lambda})(\tilde{\sigma}_{\alpha_{2}}\tilde{\sigma}_{\alpha_{1}}(h_{\epsilon_{1}-\epsilon_{n}}))$
$\displaystyle=\tilde{\lambda}(h_{\epsilon_{1}-\epsilon_{n}})$
$\displaystyle=n-1.$

This implies that $f_{\alpha}^{n}V(\tilde{\lambda})=0$ and similarly
$e_{\alpha}^{n}V(\tilde{\lambda})=0$. ∎

Set $d_{1}=n^{2}(n-1)$. Let

$U^{d_{1}}({\mathfrak{g}}_{\bar{0},\mathbb{C}})=\langle x_{1}\cdots x_{k}\in
U({\mathfrak{g}}_{\bar{0},\mathbb{C}})|x_{1},\dots,x_{k}\in{\mathfrak{g}}_{\bar{0}},k\leq
d_{1}\rangle.$

With the assumption $p>d_{1}$, the notation
$u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$($\subseteq u({\mathfrak{g}})$) can be
defined similarly.

######  Lemma 3.19.

Let
$\vartheta:U({\mathfrak{g}}_{\bar{0},\mathbb{C}})\longrightarrow\text{End}_{\mathbb{C}}V(\tilde{\lambda})$
be the representation afforded by the
${\mathfrak{g}}_{\bar{0},\mathbb{C}}$-module $V(\tilde{\lambda})$. Then
$\vartheta(U^{d_{1}}({\mathfrak{g}}_{\bar{0},\mathbb{C}}))=\text{End}_{\mathbb{C}}V(\tilde{\lambda})$.

###### Proof.

Set $\text{ann}(V(\tilde{\lambda}))=\\{u\in
U({\mathfrak{g}}_{\bar{0},\mathbb{C}})|uV(\tilde{\lambda})=0\\}$. Then
$\text{ann}V(\tilde{\lambda})$ is an ideal of $U({\mathfrak{g}}_{\bar{0}})$.
Moreover,
$\bar{U}({\mathfrak{g}}_{\bar{0},\mathbb{C}})=U({\mathfrak{g}}_{\bar{0},\mathbb{C}})/\text{ann}(V(\tilde{\lambda}))$
is primitive and $V(\tilde{\lambda})$ is a faithful
$\bar{U}({\mathfrak{g}}_{\bar{0},\mathbb{C}})$-module. By [7, Theorem 1.12, p.
420], $\vartheta(\bar{U}({\mathfrak{g}}_{\bar{0},\mathbb{C}}))$ is dense in
$\text{End}_{\mathbb{C}}V(\tilde{\lambda})$, implying that
$\vartheta(U({\mathfrak{g}}_{\bar{0},\mathbb{C}}))=\text{End}_{\mathbb{C}}V(\tilde{\lambda})$.

From the proof of above lemma, we see that, for $\alpha\in\Phi_{0}^{+}$, each
simple $\mathfrak{sl}^{\alpha}_{2}$-submodule of $V(\tilde{\lambda})$ has
dimension $\leq n$. Then $\vartheta(h_{\alpha})$ satisfying either

$(\vartheta(h_{\alpha})-(n-1))(\vartheta(h_{\alpha})-(n-3))\cdots(\vartheta(h_{\alpha})+(n-1))=0$

or

$(\vartheta(h_{\alpha})-(n-2))(\vartheta(h_{\alpha})-(n-4))\cdots(\vartheta(h_{\alpha})+(n-2))=0$

or both. Hence $\vartheta(U(\mathfrak{h}))$ is spanned by the images of
$h_{1}^{k_{1}}\cdots h_{n}^{k_{n}},0\leq k_{i}\leq n-1$. It follows that
$\text{End}_{\mathbb{C}}V(\tilde{\lambda})$ is spanned by the images under
$\vartheta$ of the elements

$e_{\alpha_{1}}^{s_{1}}\cdots
e_{\alpha_{d}}^{s_{d}}f_{\alpha_{1}}^{t_{1}}\cdots
f_{\alpha_{d}}^{t_{d}}h_{1}^{k_{1}}\cdots h_{n}^{k_{n}},\ 0\leq
s_{i},t_{i},k_{i}\leq n-1.$

All these elements are contained in $U^{d_{1}}({\mathfrak{g}}_{\bar{0}})$,
then the lemma follows. ∎

Let ${\mathfrak{g}}_{\bar{0},\mathbb{Z}}$ be the $\mathbb{Z}$-span of the
basis vectors $\tilde{e}_{ij}$ of ${\mathfrak{g}}_{\bar{0},\mathbb{C}}$. Let
$\mathcal{U}_{\mathbb{Z}}$ and $\mathcal{U}^{\prime}_{\mathbb{Z}}$ denote
respectively the Kostant $\mathbb{Z}$-form of
$U({\mathfrak{g}}_{\bar{0},\mathbb{C}})$ and
$U({\mathfrak{g}}^{\prime}_{\bar{0},\mathbb{C}})$ (see [4, 26]). Let
$\hbar=\sum_{i=1}^{n}h_{i}\in{\mathfrak{g}}_{\bar{0},\mathbb{Z}}$. Then
$[\hbar,\ {\mathfrak{g}}_{\bar{0},\mathbb{C}}]=0$. Let
$\mathcal{U}(\hbar)_{\mathbb{Z}}$ be the $\mathbb{Z}$-submodule of
$\mathcal{U}_{\mathbb{Z}}$ generated by all $\binom{\hbar}{a},a\geq 0$. Then
we have
$\mathcal{U}_{\mathbb{Z}}=\mathcal{U}^{\prime}_{\mathbb{Z}}\otimes_{\mathbb{Z}}\mathcal{U}(\hbar)_{\mathbb{Z}}$.

Let ${\bar{V}}(\lambda)=\mathcal{U}_{\mathbb{Z}}v^{+}\otimes F$. Then
${\bar{V}}(\lambda)$ is a rational $G$-module (hence restricted
${\mathfrak{g}}_{\bar{0}}$-module). Since $p>n$, the maximal vector
$v^{+}\otimes 1\in{\bar{V}}(\lambda)$ has the restricted weight
$\lambda=\tilde{\lambda}\otimes 1\in\Lambda_{0}$. Clearly we have
$\bar{V}(\lambda)=\mathcal{U}^{\prime}_{\mathbb{Z}}v^{+}\otimes F$. By [6,
Proposition 1.2], the ${\mathfrak{g}}_{\bar{0}}^{\prime}$-module
$\bar{V}(\lambda)$ has a unique maximal submodule. It follows that
$\bar{V}(\lambda)$, as a ${\mathfrak{g}}_{\bar{0}}$-module, also has a unique
maximal submodule. Therefore, $\bar{V}(\lambda)$ has a unique quotient
$L(\lambda)$ simple as a restricted ${\mathfrak{g}}_{\bar{0}}$-module (hence
as a $G$-module).

######  Theorem 3.20.

Let $\vartheta$ be the representation of $u({\mathfrak{g}}_{\bar{0}})$
afforded by ${\mathfrak{g}}_{\bar{0}}$-module $L(\lambda)$. Then
$\vartheta(u^{d_{1}}({\mathfrak{g}}_{\bar{0}}))=\text{End}_{F}L(\lambda)$.

###### Proof.

By Lemma 3.18 we have
$e_{\alpha}^{n}\bar{V}(\lambda)=f_{\alpha}^{n}\bar{V}(\lambda)=0$ and hence
$e_{\alpha}^{n}L(\lambda)=f_{\alpha}^{n}L(\lambda)=0$ for all
$\alpha\in\Phi_{0}^{+}$. From the proof of Lemma 3.19, we see that
$L(\lambda)$ is annihilated by

$\Pi_{k=-(n-1)}^{n-1}(\vartheta(h_{\alpha})-k)$

for each $\alpha\in\Phi_{0}^{+}$. Then the theorem follows from a similar
argument as that used in the proof of Lemma 3.19. ∎

####  3.5.2 An $\text{ad}_{{\mathfrak{g}}^{\prime}_{\bar{0}}}$-submodule of
$u^{d_{1}}({\mathfrak{g}})$

In the following, we continue making the preparation for constructing central
elements in $\bar{\mathfrak{u}}$.

Let $U({\mathfrak{g}}_{Q})$ be the universal enveloping superalgebra of the
Lie superalgebra $p(n)$ over $Q$. Set

$V_{Q}=:U^{d_{1}}({\mathfrak{g}}_{\bar{0},Q})+\wedge^{d}({\mathfrak{g}}_{1,Q}).$

Then $V_{Q}$ is an $\text{ad}_{{\mathfrak{g}}^{\prime}_{\bar{0},Q}}$-submodule
of $U^{d_{1}}({\mathfrak{g}}_{Q})$. Since
${\mathfrak{g}}_{\bar{0},Q}^{\prime}\cong\mathfrak{sl}(n,Q)$ is semisimple, we
have by [4, Theorem 6.3] that $V_{Q}=V_{1}\oplus V_{2}\oplus\cdots\oplus
V_{s}$, where each $V_{i}$ is a simple
${\mathfrak{g}}^{\prime}_{\bar{0},Q}$-module.

######  Lemma 3.21.

Each $V_{i}$ is generated by a maximal vector $v_{\lambda}\in V_{i}$ of
dominant integral weight $\lambda$.

###### Proof.

Since $\mathbb{C}$ is algebraically closed and since
${\mathfrak{g}}^{\prime}_{\bar{0}}$ is semisimple,
$V_{i}\otimes_{Q}\mathbb{C}$ is a direct sum of simple
${\mathfrak{g}}^{\prime}_{\bar{0},\mathbb{C}}$-modules each of the form
$V(\lambda)$ with $\lambda$ dominant integral (see [4, 21.2]).

Let $V(\lambda)$ be one of the summands, let $v_{\lambda}\in V(\lambda)$ be
the maximal vector and let $\\{\xi_{i}|i\in I\\}$ be a basis of $\mathbb{C}$
over $Q$. Write $v_{\lambda}=\sum_{j=1}^{t}\xi_{j}v_{j}$, $v_{j}\in V_{i}$.

If $t=1$, we let $v_{1}=v_{\lambda}\in V_{i}$. Then
$U({\mathfrak{g}}^{\prime}_{\bar{0},Q})v_{\lambda}\subseteq V_{i}$ is a
${\mathfrak{g}}^{\prime}_{\bar{0},Q}$-submodule of $V_{i}$, forcing
$V_{i}=U({\mathfrak{g}}^{\prime}_{\bar{0},Q})v_{\lambda}$, as required.

If $t>1$, then for all $\alpha\in\Phi_{0}^{+}$, we have
$e_{\alpha}v_{\lambda}=0$ and hence
$\sum^{t}_{j=1}\xi_{j}(e_{\alpha}v_{j})=0$. Applying every

$f\in\text{Hom}_{Q}(V_{i},Q)\subseteq\text{Hom}_{\mathbb{C}}(V_{i}\otimes_{Q}\mathbb{C},\mathbb{C})$

to this equation, we have $\sum_{j=1}^{t}\xi_{j}f(e_{\alpha}v_{j})=0$, and
hence $f(e_{\alpha}v_{j})=0$ for each $j$. Therefore, for each $j$, we have
$e_{\alpha}v_{j}=0$ for all $\alpha\in\Phi_{0}^{+}$. Similarly, we obtain
$h_{k}v_{j}=\lambda(h_{k})v_{j}$ for all $k=1,\dots,n$. It follows that each
$v_{j}$ is a maximal vector of weight $\lambda$. So that
$U({\mathfrak{g}}^{\prime}_{\bar{0},Q})v_{j}\subseteq V_{i}$ is a
${\mathfrak{g}}^{\prime}_{\bar{0},\mathbb{C}}$-submodule, implying that
$V_{i}=U({\mathfrak{g}}^{\prime}_{\bar{0},Q})v_{j}$, as required. ∎

Fix a PBW basis of $U({\mathfrak{g}}_{\mathbb{C}})$ consisting of monomials of
the natural basis of ${\mathfrak{g}}_{\mathbb{C}}$. Let
$U({\mathfrak{g}}_{\mathbb{Z}})$ denote the $\mathbb{Z}$-span of this basis,
and let $V_{\mathbb{Z}}$ be the $\mathbb{Z}$-span of the basis vectors
contained in $V_{Q}$.

By the lemma above, each $V_{i}$ may be written as $V(\lambda_{i})$ as in [4,
21.1]. In addition, we may assume $v_{\lambda_{i}}=\sum_{i=1}^{r}c_{i}v_{i}\in
V_{\mathbb{Z}}$ with each $v_{i}$ a PBW basis vector, each
$c_{i}\in\mathbb{Z}$ and $(c_{1},c_{2},\dots,c_{r})=1$. Then
$\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}$ is an admissible lattice in
$V(\lambda_{i})$. According to the proof of [4, Theorem 27.1],
$\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}$ is a free
$\mathbb{Z}$-module with $\mathbb{Z}$-rank $\text{dim}V(\lambda_{i})$.
Moreover, we have $\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}\subseteq
V_{\mathbb{Z}}$ by [4, Corollary 26.3].

Set
$\bar{V}(\lambda_{i})=\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}\otimes
F$. This is a rational $G^{\prime}$-module and hence a restricted
${\mathfrak{g}}^{\prime}_{\bar{0}}$-module (see [6]). Then the inclusion
$\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}\subseteq V_{\mathbb{Z}}$
induces a $G^{\prime}$-module homomorphism from $\bar{V}(\lambda_{i})$ to
$V=:V_{\mathbb{Z}}\otimes F$.

Remark: Since $p>2$, we have $U({\mathfrak{g}}_{\mathbb{Z}})\otimes F\cong
U({\mathfrak{g}})$, which gives $V\cong
U^{d_{1}}({\mathfrak{g}}_{\bar{0}})\oplus\wedge^{d}({\mathfrak{g}}_{1})$.

Now that
$\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{1}}\oplus\cdots\oplus\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{s}}\subseteq
V_{\mathbb{Z}}$ is a free $\mathbb{Z}$-module of $\mathbb{Z}$-rank

$\sum_{i=1}^{s}\text{dim}V(\lambda_{i})=\text{dim}V_{Q}.$

For each $V(\lambda_{i})$, there is a contravariant bilinear form whose
determinant on $\mathcal{U}^{\prime}_{\mathbb{Z}}v_{\lambda_{i}}$ is an
integer $D_{\lambda_{i}}$ ([11]). The induced bilinear form on
$\bar{V}(\lambda_{i})$ has as the radical the unique maximal
$G^{\prime}$-submodule.

Assume $(C1)$: $p\nmid\Pi^{s}_{i=1}D_{\lambda_{i}}$. Then the contravariant
form on each $\bar{V}(\lambda_{i})$ is nondegenerate, and hence each
$\bar{V}(\lambda_{i})$ is a simple ${\mathfrak{g}}_{\bar{0}}^{\prime}$-module
(and a simple $G^{\prime}$-module). Therefore, we obtain a completely
reducible $G^{\prime}$-submodule
$\bar{V}(\lambda_{1})+\cdots+\bar{V}(\lambda_{s})$ of $V$.

Let $v_{1},\dots,v_{l}$ denote the fixed PBW basis of $V_{\mathbb{Z}}$. Then
we have

$(*)\\\ \hskip 0.5pt\
(v_{\lambda_{1}},\dots,v_{\lambda_{s}})=(v_{1},\dots,v_{l})A_{l\times s},$

where $A_{l\times s}$ is an integral matrix. For each $v\in V_{\mathbb{Z}}$,
denote $v\otimes 1\in V$ by $\bar{v}$. Denote the image of $A_{l\times s}$
under the canonical map $\mathbb{Z}^{r\times s}\longrightarrow F_{p}^{r\times
s}$ by $\bar{A}_{l\times s}$. It follows that

$(\bar{v}_{\lambda_{1}},\dots,\bar{v}_{\lambda_{s}})=(\bar{v}_{1},\dots,\bar{v}_{l})\bar{A}_{l\times
s}.$

Since $v_{\lambda_{1}},\dots,v_{\lambda_{s}}$ are linearly independent, there
is a submatrix $A^{\prime}_{s\times s}$ in $A_{l\times s}$ with
$\text{det}A^{\prime}=m\neq 0$. Assume $(C1^{\prime}):p>|m|$. Then we have
$r(\bar{A}_{l\times s})=s$, and hence the elements
$\bar{v}_{\lambda_{1}},\dots,\bar{v}_{\lambda_{s}}\in V$ are linearly
independent.

######  Corollary 3.22.

With assumptions $(C1)$ and $(C1^{\prime})$, we have

$\bar{V}(\lambda_{1})\oplus\cdots\oplus\bar{V}(\lambda_{s})=V.$

###### Proof.

Since both sides have the same dimension, it suffices to show that
$\bar{V}(\lambda_{1})+\cdots+\bar{V}(\lambda_{s})\subseteq V$ is a direct sum.

Assume $x_{1}+\cdots+x_{s}=0$, $x_{i}\in\bar{V}(\lambda_{i})$, $1\leq i\leq
s$. Since each $\bar{V}(\lambda_{i})$ is a simple $G^{\prime}$-module, we may
assume that each $x_{i}$ is a weight vector with respect to the maximal torus
$T\cap G^{\prime}$.

Suppose that there is $i$, $1\leq i\leq s$, such that $x_{i}\neq 0$. If
$\text{wt}(x_{i})<\lambda_{i}$, then there is
$\alpha\in\\{\epsilon_{1}-\epsilon_{2},\dots,\epsilon_{n-1}-\epsilon_{n}\\}$
such that $e_{\alpha}x_{i}\neq 0$. Clearly we have
$\text{wt}(x_{i})<\text{wt}(e_{\alpha}x_{i})\leq\lambda_{i}$. Since each
$\bar{V}(\lambda_{i})$ has finite many nonzero weight spaces, by repeated
applications of appropriate $e_{\alpha}$’s we have

$x_{1}^{\prime}+\cdots+x_{s}^{\prime}=0,\
x^{\prime}_{i}\in\bar{V}(\lambda_{i}),$

where $x^{\prime}_{i}$’s are not all zero and each $x_{i}$ is a multiple of
$\bar{v}_{\lambda_{i}}$, contrary to the fact that
$\bar{v}_{\lambda_{1}},\dots,\bar{v}_{\lambda_{s}}$ are linearly independent.

Thus, we conclude that $x_{i}=0$ for all $1\leq i\leq s$, which completes the
proof. ∎

With the assumption $p>d_{1}$, we have an isomorphism of rational $G$-modules
(hence of restricted ${\mathfrak{g}}_{\bar{0}}$-modules)
$U^{d_{1}}({\mathfrak{g}}_{\bar{0}})\cong
u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$, and hence $V\cong
u^{d_{1}}({\mathfrak{g}}_{\bar{0}})\oplus\wedge^{d}({\mathfrak{g}}_{1})$.
Together with assumptions $(C1)$ and $(C1^{\prime})$, we obtain an isomorphism
of rational $G^{\prime}$-modules

$\bar{V}(\lambda_{1})\oplus\cdots\oplus\bar{V}(\lambda_{s})\cong
u^{d_{1}}({\mathfrak{g}}_{\bar{0}})\oplus\wedge^{d}({\mathfrak{g}}_{1}).$

Therefore, both $u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$ and
$\wedge^{d}({\mathfrak{g}}_{1})$ are completely reducible as restricted
$\text{ad}_{{\mathfrak{g}}^{\prime}_{\bar{0}}}$-modules.

In the remainder of the paper we assume the following conditions:

(H1) $(C1)$ and $(C1^{\prime})$;

(H2) $p>d_{1}$;

(H3) $p>\text{dim}\wedge^{d}({\mathfrak{g}}_{1})=C^{d}_{n^{2}}$.

Under these assumptions, since $d_{1}>2d$, the condition on $p$ in Theorem
3.14 is satisfied.

####  3.5.3 Central elements in $\bar{\mathfrak{u}}$

Recall the weight
$\lambda=(n-1)\epsilon_{1}+\cdots+\epsilon_{n-1}\in\Lambda_{0}$. From 3.5.1 we
see that $\lambda$ is typical, since $p>d_{1}=(n-1)n^{2}\geq n^{2}\geq 2n-1$.

Let $\vartheta$ be the representation of $u({\mathfrak{g}}_{\bar{0}})$
afforded by the simple G-module $L(\lambda)$ in 3.5.1. According to Theorem
3.20, $\vartheta$ is an epimorphism from $u({\mathfrak{g}}_{\bar{0}})$ to
$\text{End}_{F}L(\lambda)$. This induces an isomorphism of algebras

$u({\mathfrak{g}}_{\bar{0}})/\text{ker}\vartheta\cong\text{End}_{F}L(\lambda).$

Denote the restriction of $\vartheta$ to $u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$
by $\vartheta_{1}$. Then $\text{ker}\vartheta_{1}=\text{ker}\vartheta\cap
u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$, and which is easily seen to be an
$\text{Ad}G$-submodule of $u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$. By Theorem
3.20 we obtain an isomorphism of $G$-modules

$u^{d_{1}}({\mathfrak{g}}_{\bar{0}})/\text{ker}\vartheta_{1}\cong\text{End}_{F}L(\lambda).$

Recall the mapping $\varphi$. Let $v_{\lambda}\in L(\lambda)$ be the maximal
vector. Since $\lambda$ is typical, we have

$(YZ)v_{\lambda}=(YZ)_{0}v_{\lambda}=\delta_{\lambda}v_{\lambda}\neq 0,$

implying that $\text{Im}\varphi\varsubsetneq\text{ker}\vartheta$. Since
$\text{Im}\varphi\subseteq u^{d}({\mathfrak{g}}_{\bar{0}})$ and $d<d_{1}$, we
get $\text{Im}\varphi\varsubsetneq\text{ker}\vartheta_{1}$. Since
$u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$ is a completely reducible
${\mathfrak{g}}_{\bar{0}}^{\prime}$-module, so is $\text{Im}\varphi$. It
follows that $L(\mu)\varsubsetneq\text{ker}\vartheta_{1}$ for some simple
summand $L(\mu)$ of $\text{Im}\varphi$.

Since $L(\mu)$ is simple, we have $L(\mu)\cap\text{ker}\vartheta_{1}=\\{0\\}$.
Since $u^{d_{1}}({\mathfrak{g}}_{\bar{0}})$ is completely reducible, we have

$u^{d_{1}}({\mathfrak{g}}_{\bar{0}})=(L(\mu)+\text{ker}\vartheta_{1})\oplus W$

for some $G^{\prime}$-submodule $W$.

Let $T_{1}$ be the toral subgroup of $G$ defined by

$T_{1}=\\{diag(t,\dots,t)|t\in F^{*}\\}.$

Then we have $G=G^{\prime}\times T_{1}$. Since $T_{1}$ acts trivially on
$u({\mathfrak{g}}_{\bar{0}})$, every $G^{\prime}$-submodule of
$u({\mathfrak{g}}_{\bar{0}})$ is naturally a $G$-submodule. Then we have a
$G$-module isomorphism

$u^{d_{1}}({\mathfrak{g}}_{\bar{0}})/\text{ker}\vartheta_{1}\cong
L(\mu)+W\cong\text{End}_{F}L(\lambda).$

Since the trace form on $\text{End}_{F}L(\lambda)$ is $G$-invariant,
nondegenerate and symmetric, so is the induced form on $L(\mu)+W$.

Since $\wedge^{d}({\mathfrak{g}}_{1})$ is completely reducible, we have an
isomorphism of $u({\mathfrak{g}}_{\bar{0}}^{\prime})$-modules (also
$G^{\prime}$-modules)

$\text{ker}\varphi\oplus\text{Im}\varphi\cong\wedge^{d}({\mathfrak{g}}_{1}).$

Let $Z_{1},\dots,Z_{r}$ be a basis of the simple
$u({\mathfrak{g}}^{\prime}_{\bar{0}})$-submodule of
$\wedge^{d}({\mathfrak{g}}_{1})$ whose image under $\varphi$ is $L(\mu)$. Then
$(YZ_{1})_{0},\dots,(YZ_{r})_{0}$ is a basis of $L(\mu)$, from which we extend
to a basis of $L(\mu)+W$:

$(YZ_{1})_{0},\dots,(YZ_{r})_{0},w_{1},\dots,w_{t}.$

Let
$(YZ_{1})_{0}^{\lor},\dots,(YZ_{r})_{0}^{\lor},w_{1}^{\lor},\dots,w_{t}^{\lor}$
be the dual basis. Set

$W^{\perp}=\\{x\in L(\mu)+W|(x,W)=0\\}.$

Since $W$ is a $G$-submodule of $u({\mathfrak{g}}_{\bar{0}})$, so is
$W^{\perp}$. By [8, Section 6.1], we have

$\text{dim}W^{\perp}=\text{dim}(L(\mu)+W)-\text{dim}W=\text{dim}L(\mu)=r,$

implying that
$W^{\perp}=\langle(YZ_{1})_{0}^{\lor},\dots,(YZ_{r})_{0}^{\lor}\rangle$.

For $g\in G^{\prime}$, assume

$g(Z_{1},\dots,Z_{r})^{t}=(b^{g}_{ij})_{r\times r}(Z_{1},\dots,Z_{r})^{t}$

and

$g((YZ_{1})_{0}^{\lor},\dots,(YZ_{r})_{0}^{\lor})^{t}=(c_{ij}^{g})_{r\times
r}((YZ_{1})_{0}^{\lor},\dots,(YZ_{r})_{0}^{\lor})^{t}.$

Then since $Y$ is a $G^{\prime}$-invariant, we have

$g(YZ_{1},\dots,YZ_{r})^{t}=(b^{g}_{ij})_{r\times
r}(YZ_{1},\dots,YZ_{r})^{t}.$

Since the mapping $(,)_{0}$ is a $G$-module homomorphism, we obtain

$g((YZ_{1})_{0},\dots,(YZ_{r})_{0})^{t}=(b_{ij}^{g})_{r\times
r}((YZ_{1})_{0},\dots,(YZ_{r})_{0})^{t}.$

Then the $G$-invariance of the bilinear form implies that
$(b_{ij}^{g})_{r\times r}^{t}=(c_{ij}^{g})_{r\times r}^{-1}.$

Set $\omega=\sum_{i=1}^{r}Z_{i}(YZ_{i})_{0}^{\lor}\in u({\mathfrak{g}})$. The
following conclusion follows from a standard argument in linear algebra.

######  Lemma 3.23.

For each $g\in G^{\prime}$, we have $g\omega=\omega$.

Then by [10, 7.11(5), Part I], we have $[x,\omega]=0$ for all
$x\in{\mathfrak{g}}_{\bar{0}}^{\prime}$.

######  Lemma 3.24.

$(Y\omega)_{0}\neq 0$.

###### Proof.

Since $(Y\omega)_{0}=\sum_{i=1}^{r}(YZ_{i})_{0}(YZ_{i})_{0}^{\lor}$, we have

$tr\vartheta(Y\omega)_{0}=\sum_{i=1}^{r}tr\vartheta((YZ_{i})_{0})\vartheta((YZ_{i})_{0}^{\lor})=\sum_{i=1}^{r}((fZ_{i})_{0},(fZ)_{0}^{\lor})=r\neq
0,$

where the last inequality follows from the assumption
$p>\text{dim}\wedge^{d}({\mathfrak{g}}_{1})\geq r$. It follows that
$(Y\omega)_{0}\neq 0$. ∎

Let $M=M_{\bar{0}}\oplus M_{\bar{1}}$ be a restricted ${\mathfrak{g}}$-module.
If both $M_{\bar{0}}$ and $M_{\bar{1}}$ are rational $T$-modules such that the
$\mathfrak{h}$-action is induced from the differential of the $T$-action. In
addition, for all $t\in T$, $x\in{\mathfrak{g}}_{\bar{0}}$ and $m\in M$, we
have $t(xm)=\text{Ad}(t)(x)(tm)$. Then we call $M$ a
$u({\mathfrak{g}})-T$-module ([13, 1.8]). For example, $u({\mathfrak{g}})$ and
$\bar{\mathfrak{u}}$ are both $u({\mathfrak{g}})-T$-modules.

Fix an order of the root vectors $y_{ij}$ of ${\mathfrak{g}}_{-1}$ in the
following lemma.

######  Lemma 3.25.

(1) Let $S=\Pi_{i<j}\text{ad}y_{ij}(\omega)\in u({\mathfrak{g}})$ and let
$\bar{S}$ be its image in $\bar{\mathfrak{u}}$. Then $\bar{S}\in\mathcal{Z}$
and $h(\bar{S})=k\Theta$ for some $k\in F^{*}$.

(2) For $z\in u({\mathfrak{g}}_{\bar{0}})^{G}$, let
$S_{z}=\Pi_{i<j}\text{ad}y_{ij}(z\omega)\in u({\mathfrak{g}})$ and let
$\bar{S}_{z}$ be its image in $\bar{\mathfrak{u}}$. Then
$\bar{S}_{z}\in\mathcal{Z}$ and $h(\bar{S}_{z})=h(\bar{S})h(z)$.

###### Proof.

(1) Since $FY$ is a 1-dimensional
$\text{ad}_{{\mathfrak{g}}_{\bar{0}}}$-module, $Y\otimes_{F}M$ is a restricted
simple ${\mathfrak{g}}_{\bar{0}}$-module, for each restricted simple
${\mathfrak{g}}_{\bar{0}}$-module $M$. Then the mapping $Y\otimes_{F}-$
defines a bijection on the set of all restricted simple
${\mathfrak{g}}_{\bar{0}}$-modules.

Now let $L(\nu)$ be a restricted simple ${\mathfrak{g}}_{\bar{0}}$-module such
that $Y\otimes_{F}L(\nu)$ is isomorphic to $L(\lambda)$ above. By the proof of
Lemma 3.15, we have $(Y\omega)_{0}Y\otimes m\neq 0$ for some $m\in L(\nu)$.
Applying Proposition 2.3 to $K(L(\nu))$, we have

$(Y\omega)_{0}Y\otimes m=Y\omega Y\otimes m=SY\otimes m=\bar{S}Y\otimes m,$

and hence $\bar{S}\neq 0$.

Since $\bar{\omega}\in\bar{\mathfrak{u}}_{d}$,
${\mathfrak{g}}_{1}\bar{\omega}=\bar{\omega}{\mathfrak{g}}_{1}=0$. We also
have $g\bar{\omega}=\bar{\omega}$ for all $g\in G^{\prime}$. Since
$(Y\omega)_{0}$ is a $G$-invariant, it follows that $\bar{\omega}$ has
$T$-weight $-\sigma$, where $\sigma$ is the $T$-weight of $Y$. Let
$V\subseteq\bar{\mathfrak{u}}$ be an $\text{ad}_{{\mathfrak{g}}}$-submodule
generated by $\bar{\omega}$, which is also a $G$-submodule. Since $\bar{S}\neq
0$, we have

$V\cong
u({\mathfrak{g}})\otimes_{u({\mathfrak{g}}_{\bar{0}}+{\mathfrak{g}}_{1})}F\bar{\omega}=K(F\bar{\omega}).$

Clearly $V$ is a $u({\mathfrak{g}})-T$-module. Since
$V\cong\wedge({\mathfrak{g}}_{-1})\otimes F\bar{\omega}$, each element of $V$
may be written as $\sum_{i}u_{i}\bar{\omega}$ with
$u_{i}\in\wedge^{i}({\mathfrak{g}}_{-1})$.

Let $M\subseteq V$ be a simple $u({\mathfrak{g}})-T$-submodule and let
$\sum_{i}u_{i}\bar{\omega}\in M$ be a nonzero $T$-weight vector. By applying
appropriate $y_{ij}$’s to this vector, we obtain $Y\bar{\omega}=\bar{S}\in M$.
Since $M$ is simple, we obtain

$M=\bar{\mathfrak{u}}\cdot\bar{S}=\wedge^{\leq
d}({\mathfrak{g}}_{1})\cdot\bar{S}.$

Let $\wedge({\mathfrak{g}}_{1})^{+}=:\sum_{i\geq
1}\wedge^{i}({\mathfrak{g}}_{1})$. Then a short computation shows that
$\wedge({\mathfrak{g}}_{1})^{+}\bar{S}$ is a $u({\mathfrak{g}})-T$-submodule
of $M$. By comparing the $T$-weights we see that
$\bar{S}\notin\wedge({\mathfrak{g}}_{1})^{+}\bar{S}$, implying that
$M=F\bar{S}$, and hence $[\bar{S},{\mathfrak{g}}_{\bar{1}}]=0$. From the proof
of Lemma 3.16, we see that $Y$ is a $G^{\prime}$-invariant. It follows that
$\bar{S}$ is also a $G^{\prime}$-invariant. This gives
$\bar{S}\in\mathcal{Z}$, since the $T$-weight of $\bar{S}$ is 0.

Note that $\bar{S}\in u^{2d}({\mathfrak{g}})$. Then we have by Theorem 3.14
that $h(\bar{S})=k\Theta+k^{\prime}$ for some $k,k^{\prime}\in F$. Since
$\bar{S}$ annihilates the trivial ${\mathfrak{g}}$-module, in view of the fact
that $\Theta(0)=0$, we have $k^{\prime}=h(\bar{S})(0)=0$. From above we have
$\bar{S}K(L(\nu))\neq 0$, implying that $h(\bar{S})(\nu)=k\Theta(\nu)\neq 0$
and hence $k\neq 0$.

(2) By a similar argument as above we have $\bar{S}_{z}\in\mathcal{Z}$, for
$z\in u({\mathfrak{g}}_{\bar{0}})^{G}$. Let $\mu\in\Lambda_{0}$ and let
$v_{\mu}\in K(L(\mu))$ be the maximal vector of weight $\mu$. Since $FY$ is an
$\text{ad}_{{\mathfrak{g}}_{\bar{0}}}$-module, implying that $Yz=(z+c(z))Y$
for some $c(z)\in F$, we have

$\displaystyle h(\bar{S}_{z})(\mu)Y\otimes v_{\mu}$
$\displaystyle=S_{z}Y\otimes v_{\mu}$ $\displaystyle=Yz\omega Y\otimes
v_{\mu}$ $\displaystyle=(z+c(z))Y\omega Y\otimes v_{\mu}$
$\displaystyle=(z+c(z))\bar{S}Y\otimes v_{\mu}$
$\displaystyle=\bar{S}(z+c(z))Y\otimes v_{\mu}$ $\displaystyle=\bar{S}Y\otimes
zv_{\mu}$ $\displaystyle=h(z)(\mu)\bar{S}Y\otimes v_{\mu}$
$\displaystyle=h(z)(\mu)h(\bar{S})(\mu)Y\otimes v_{\mu},$

implying that $h(\bar{S}_{z})(\mu)=h(z)(\mu)h(\bar{S})(\mu)$ for all
$\mu\in\Lambda_{0}$, and hence $h(\bar{S}_{z})=h(z)h(\bar{S})$. ∎

Immediately, we have the following conclusion.

######  Corollary 3.26.

$h(\mathcal{Z})=F+\Theta u(\mathfrak{h})^{W\cdot}.$

##  4 Representations of $\bar{\mathfrak{u}}$

Fir each $\mu\in\Lambda_{0}$, we define an algebra homomorphism
$\chi_{\mu}:\mathcal{Z}\longrightarrow F$ by

$\chi_{\mu}(z)=h(z)(\mu),\ z\in\mathcal{Z}.$

Two weights $\mu,\nu\in\Lambda_{0}$ are referred to be linked and denoted
$\mu\sim\nu$, if there is $\sigma\in W$ such that $\mu+\rho=\sigma(\nu+\rho)$
([6]).

Let $U({\mathfrak{g}}^{\prime}_{\bar{0}})$ be the universal enveloping algebra
of ${\mathfrak{g}}_{\bar{0}}^{\prime}$ over $F$, and let

$U({\mathfrak{g}}^{\prime}_{\bar{0}})^{G^{\prime}}=\\{u\in
U({\mathfrak{g}}_{\bar{0}}^{\prime})|g\cdot u=u\ \text{for all}\ g\in
G^{\prime}\\}.$

Set $\mathfrak{h}_{1}=\mathfrak{h}\cap{\mathfrak{g}}^{\prime}_{\bar{0}}$.
According to [9, 9.1-9.4], there is an algebra isomorphism

$\pi:U({\mathfrak{g}}^{\prime}_{\bar{0}})^{G^{\prime}}\longrightarrow
U(\mathfrak{h}_{1})^{W\cdot}$

with kernel
$\mathfrak{n}_{0}^{-}U({\mathfrak{g}}_{\bar{0}}^{\prime})+U({\mathfrak{g}}_{\bar{0}}^{\prime})\mathfrak{n}_{0}^{+}$.

For each $\lambda\in\mathfrak{h}_{1}^{*}$, define an algebra homomorphism
$\text{cen}_{\lambda}:U({\mathfrak{g}}_{\bar{0}}^{\prime})^{G^{\prime}}\longrightarrow
F$ by $\text{cen}_{\lambda}(u)=\pi(u)(\lambda)$, $u\in
U({\mathfrak{g}}_{\bar{0}}^{\prime})^{G^{\prime}}$ ([9, 9.4]).

Let $\xi:U(\mathfrak{h}_{1})\longrightarrow u(\mathfrak{h}_{1})$ be the
canonical homomorphism. By [6, Lemma 3.3], the restriction
$\xi:U(\mathfrak{h}_{1})^{W\cdot}\longrightarrow u(\mathfrak{h}_{1})^{W\cdot}$
is an epimorphism. It follows that
$\xi\pi:U({\mathfrak{g}}^{\prime}_{\bar{0}})^{G^{\prime}}\longrightarrow
u(\mathfrak{h}_{1})^{W\cdot}$ is also an epimorphism.

Remark: The assumption on $p$ in [6, Lemma 3.3] is satisfied in our case here,
since $|W|=n!$.

For each
$\lambda=\lambda_{1}\epsilon_{1}+\cdots+\lambda_{n}\epsilon_{n}\in\Lambda_{0}$.
we have

$h_{i}^{p}(\lambda)-h_{i}(\lambda)=\lambda_{i}^{p}-\lambda_{i}=0\ \text{for}\
1\leq i\leq n.$

Then we get $f(h)(\lambda)=\xi(f(h))(\lambda)$ for all $f(h)\in
U(\mathfrak{h})$. In particular, if
$\lambda\in\Lambda_{0}\cap\mathfrak{h}_{1}^{*}$, then we get

$\text{cen}_{\lambda}(u)=\pi(u)(\lambda)=\xi\pi(u)(\lambda)$

for all $u\in U({\mathfrak{g}}_{\bar{0}}^{\prime})^{G^{\prime}}$.

######  Theorem 4.1.

Let $\mu,\nu\in\Lambda_{0}$. Then $\chi_{\mu}=\chi_{\nu}$ if and only if both
$\mu$ and $\nu$ are atypical or both are typical with $\mu\sim\nu$.

###### Proof.

Suppose $\chi_{\mu}=\chi_{\nu}$. Then we have $h(z)(\mu)=h(z)(\nu)$ for all
$z\in\mathcal{Z}$. Write

$h(z)=\Theta f(h)+c,\ f(h)\in u(\mathfrak{h})^{W\cdot},\ c\in F.$

Then we have $\Theta(\mu)f(h)(\mu)=\Theta(\nu)f(h)(\nu)$ for all $f(h)\in
u(\mathfrak{h})^{W\cdot}$.

If $\mu$ is atypical, then we obtain $\Theta(\nu)f(h)(\nu)=0$ for all $f(h)\in
u(\mathfrak{h})^{W\cdot}$. Let $f(h)=1$. Then we get $\Theta(\nu)=0$, and
hence $\nu$ is atypical.

Suppose both $\mu$ and $\nu$ are typical. Let $h(z)=\Theta$. Then we get
$\Theta(\mu)=\Theta(\nu)$ and hence $f(h)(\mu)=f(h)(\nu)$ for all $f(h)\in
u(\mathfrak{h})^{W\cdot}$. Since $u(\mathfrak{h}_{1})^{W\cdot}\subseteq
u(\mathfrak{h})^{W\cdot}$, we obtain $f(h)(\mu)=f(h)(\nu)$ for all $f(h)\in
u(\mathfrak{h}_{1})^{W\cdot}$. It follows that

$\text{cen}_{\mu}(u)=\xi\pi(u)(\mu)=\xi\pi(u)(\nu)=\text{cen}_{\nu}(u)$

for all $u\in U({\mathfrak{g}}^{\prime}_{\bar{0}})^{G^{\prime}}$, and hence
$\text{cen}_{\mu}=\text{cen}_{\nu}$, and whence
$\mu_{|\mathfrak{h}_{1}}=(w\cdot\nu)_{|\mathfrak{h}_{1}}$ for some $w\in W$,
by [9, Corollary 9.4].

Since $\hbar=\sum_{i=1}^{n}h_{i}\in u(\mathfrak{h})^{W\cdot}$, we also have
$\hbar(\mu)=\hbar(\nu)$, implying that $\mu(\hbar)=(w\cdot\nu)(\hbar)$, and
hence $\mu=w\cdot\nu$.

If both $\mu$ and $\nu$ are atypical, then Corollary 3.26 gives
$\chi_{\mu}=\chi_{\nu}$. Assume both $\mu$ and $\nu$ are typical and
$\mu\sim\nu$. For each $z\in\mathcal{Z}$, we have by Lemma 3.7 that $h(z)\in
u(\mathfrak{h})^{W\cdot}$, implying that $\chi_{\mu}(z)=\chi_{\nu}(z)$.

∎

For $\mu\in\Lambda_{0}$, let $\bar{\mathfrak{u}}_{\mu}$ denote the
superalgebra $\bar{\mathfrak{u}}/\bar{\mathfrak{u}}\text{ker}\chi_{\mu}$. Let

$\chi^{0}_{\mu}:u({\mathfrak{g}}_{\bar{0}})^{G}\longrightarrow F$

be the homomorphism defined by $\chi^{0}_{\mu}(z)=h(z)(\mu),\ z\in
u({\mathfrak{g}}_{\bar{0}})^{G}$. Define the quotient algebra
$u^{0}_{\mu}=:u({\mathfrak{g}}_{\bar{0}})/u({\mathfrak{g}}_{\bar{0}})\text{ker}\chi^{0}_{\mu}$.

Note: It is easy to see that $\hbar$ is an element of
$u({\mathfrak{g}}_{\bar{0}})^{G}$, but not necessarily in $\mathcal{Z}$.

######  Definition 4.2.

If $M=M_{\bar{0}}\oplus M_{\bar{1}}$ is a $\bar{\mathfrak{u}}_{\mu}$-module
and also a $u({\mathfrak{g}})-T$-module, then we call $M$ a
$\bar{\mathfrak{u}}_{\mu}-T$-module.

Example. For $\mu\in\Lambda_{0}$, the baby Verma module $Z(\mu)$ is a
$\bar{\mathfrak{u}}_{\mu}-T$-module. A $u^{0}_{\mu}-T$-module can be defined
similarly.

######  Theorem 4.3.

Assume $\mu$ is typical.

(1) If $N$ is a simple $u^{0}_{\mu}-T$-module, then $K(N)$ is a simple
$\bar{\mathfrak{u}}_{\mu}-T$-module.

(2) If $M$ is a simple $\bar{\mathfrak{u}}_{\mu}-T$-module, then
$M^{{\mathfrak{g}}_{1}}=\\{m\in M|xm=0\ \text{for all}\
x\in{\mathfrak{g}}_{1}\\}$ is a simple $u^{0}_{\mu}-T$-module.

###### Proof.

(1) By definition $N$ is a restricted simple ${\mathfrak{g}}_{\bar{0}}$-module
annihilated by $\text{ker}\chi^{0}_{\mu}$. Let $v_{\lambda}\in N$ be the
maximal vector of weight $\lambda$. For each $z\in
u({\mathfrak{g}}_{\bar{0}})^{G}$, we have

$zv_{\lambda}=h(z)v_{\lambda}=h(z)(\lambda)v_{\lambda}=\chi^{0}_{\lambda}(z)v_{\lambda}$

and

$zv_{\lambda}=(z-\chi^{0}_{\mu}(z))v_{\lambda}+\chi^{0}_{\mu}(z)v_{\lambda}=\chi^{0}_{\mu}(z)v_{\lambda},$

since $z-\chi^{0}_{\mu}(z)\in\text{ker}\chi^{0}_{\mu}$. Therefore, we get
$\chi^{0}_{\lambda}=\chi^{0}_{\mu}$. Applying [6, Theorem 3.1], we get
$\lambda_{|\mathfrak{h}_{1}}\sim\mu_{|\mathfrak{h}_{1}}$, and hence
$\lambda_{|\mathfrak{h}_{1}}=(\sigma\cdot\mu)_{|\mathfrak{h}_{1}}$ for some
$\sigma\in W$.

By [6, Lemma 3.2, 3.3], the Harish-Chandra morphism
$h:u({\mathfrak{g}}_{\bar{0}}^{\prime})^{G^{\prime}}\longrightarrow
u(\mathfrak{h}_{1})^{W\cdot}$ is an epimorphism. Note that

$u({\mathfrak{g}}_{\bar{0}})^{G}\cong
u({\mathfrak{g}}_{\bar{0}}^{\prime})^{G^{\prime}}\otimes u(F\hbar)$

and $h(\hbar)=\hbar\in u(\mathfrak{h})^{W\cdot}$. Then
$h:u({\mathfrak{g}}_{\bar{0}})^{G}\longrightarrow u(\mathfrak{h})^{W\cdot}$ is
also an epimorphism. From $\chi^{0}_{\lambda}(\hbar)=\chi^{0}_{\mu}(\hbar)$,
we obtain $\hbar(\lambda)=\hbar(\mu)$, which gives
$\lambda(\hbar)=(\sigma\cdot\mu)(\hbar)$, since $\hbar\in
u(\mathfrak{h})^{W\cdot}$. Thus, we obtain $\lambda\sim\sigma\cdot\mu$. Then
$\lambda$ is also typical by Theorem 4.1. Hence $K(N)$ is simple as
$u({\mathfrak{g}})$-module by Proposition 2.3, and also as
$\bar{\mathfrak{u}}$-module by Lemma 3.2.

Let the $T$-action on $K(N)$ be induced from that of $N$. It remains to show
that $K(N)$ is annihilated by $\text{ker}\chi_{\mu}$. To see this, let
$z\in\text{ker}\chi_{\mu}$. Then we have

$\displaystyle zv_{\lambda}$ $\displaystyle=h(z)v_{\lambda}$
$\displaystyle=h(z)(\lambda)v_{\lambda}$ $\displaystyle=h(z)(\mu)v_{\lambda}$
$\displaystyle=\chi_{\mu}(z)v_{\lambda}$ $\displaystyle=0,$

and hence $zK(N)=0$.

(2) Let $M=M_{\bar{0}}\oplus M_{\bar{1}}$ and let $m\in M_{\bar{0}}\cup
M_{\bar{1}}$ be a nonzero $T$-weight vector. Set
$M^{\prime}=\wedge({\mathfrak{g}}_{1})m\subseteq M$. Since
$\wedge^{i}({\mathfrak{g}}_{1})=0$ in $\bar{\mathfrak{u}}$ is for $i>d$, we
have

$M^{\prime}=\wedge^{\leq d}({\mathfrak{g}}_{1})m.$

Let $i_{0}$ be the largest integer such that
$\wedge^{i_{0}}({\mathfrak{g}}_{1})m\neq 0$. Then we have $0\leq i_{0}\leq d$
and ${\mathfrak{g}}_{1}\wedge^{i_{0}}({\mathfrak{g}}_{1})m=0$. Therefore we
have $M^{{\mathfrak{g}}_{1}}\neq 0$. Clearly $M^{{\mathfrak{g}}_{1}}$ is
$\mathbb{Z}_{2}$-graded.

It is easy to see that $M^{{\mathfrak{g}}_{1}}$ is a $T$-submodule and a
${\mathfrak{g}}_{\bar{0}}$-submodule of $M$. Since $M$ is a restricted
${\mathfrak{g}}_{\bar{0}}$-module, so is $M^{{\mathfrak{g}}_{1}}$. It is no
loss of generality to assume that $M^{{\mathfrak{g}}_{1}}_{\bar{0}}\neq 0$.

Let $L(\lambda)\subseteq M^{{\mathfrak{g}}_{1}}_{\bar{0}}$ be a simple
$u({\mathfrak{g}}_{\bar{0}})$-submodule of highest weight $\lambda$. Then
$L(\lambda)$ is a $u({\mathfrak{g}}_{\bar{0}})-T$-submodule. The inclusion
$L(\lambda)\subseteq M_{\bar{0}}$ induces a $u({\mathfrak{g}})$-module
homomorphism from $K(L(\lambda))$ onto $M$. Since $M$ is a
$\bar{\mathfrak{u}}_{\mu}$-module, by applying $z\in\mathcal{Z}$ to the
maximal vector $v_{\lambda}\in L(\lambda)$ we obtain
$\chi_{\lambda}=\chi_{\mu}$, implying that $\lambda\sim\mu$ by Theorem 4.1. It
follows that $\lambda$ is also typical, and hence $K(L(\lambda))$ is simple,
and whence $M\cong K(L(\lambda))$. We simply write $M=K(L(\lambda))$. Then
$L(\lambda)\subseteq M^{{\mathfrak{g}}_{1}}$.

We claim that $M^{{\mathfrak{g}}_{1}}=L(\lambda)$. To see this, note that

$M=L(\lambda)\oplus{\mathfrak{g}}_{-1}M.$

Let $N=M^{{\mathfrak{g}}_{1}}\cap{\mathfrak{g}}_{-1}M$. Then we have

$M^{{\mathfrak{g}}_{1}}=L(\lambda)+N.$

Let $\tilde{N}\subseteq M$ be the ${\mathfrak{g}}$-submodule generated by $N$.
By the definition of $K(L(\lambda))$ we have
$\tilde{N}\subseteq{\mathfrak{g}}_{-1}M$, and hence $\tilde{N}\cap
L(\lambda)=0$. Since $M$ is simple, we have $\tilde{N}=0$ and hence $N=0$. So
the claim follows.

It remains to show that $L(\lambda)$ is annihilated by
$\text{ker}\chi^{0}_{\mu}$. Let $z\in u({\mathfrak{g}}_{\bar{0}})^{G}$. Since
$\lambda\sim\mu$, we have

$\displaystyle zv_{\lambda}$ $\displaystyle=h(z)(\lambda)v_{\lambda}$
$\displaystyle=h(z)(\mu)v_{\lambda}$
$\displaystyle=\chi^{0}_{\mu}(z)v_{\lambda}.$

Then $z$ acts on $L(\lambda)$ as a multiplication by $\chi^{0}_{\mu}(z)$ or,
equivalently, $L(\lambda)$ is annihilated by $\text{ker}\chi^{0}_{\mu}$. It
follows that $L(\lambda)$ is a simple $u^{0}_{\mu}-T$-module. ∎

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Q_{k}{f_{2}}(H_{a}){R_{a}(z)}\rVert_{{\mathcal{L}}({\mathcal{B}},\widetilde{{\mathcal{H}}})}<\infty;\quad(\widetilde{{\mathcal{H}}}={\mathcal{H}}\text{
or }\widetilde{{\mathcal{H}}}={\mathcal{H}}^{d}).\end{split}$ (A.7)

These bounds allow us to take the following weak limit, cf. Remark 2.3 i,

$Q_{k}f_{2}(H)R(\lambda+\mathrm{i}0)\psi:=\operatorname*{{w-\widetilde{\mathcal{H}}}-lim}2lim_{\epsilon\to
0_{+}}\,\,Q_{k}f_{2}(H)R(\lambda+\mathrm{i}\epsilon)\psi.$ (A.8)

Using in turn that $\langle x\rangle^{s}f_{2}(H_{a}){Q_{k}}^{*}B_{k}$ is
bounded, (A.6) and (A.8) we can compute the above third term as follows.
Taking limits in the weak sense in $L^{2}_{-1}$

$\displaystyle-\mathrm{i}$ $\displaystyle\lim_{\epsilon\to
0_{+}}R_{a}(\lambda+\mathrm{i}\epsilon)\chi^{2}_{+}(2B/\epsilon_{0})T_{a}R(\lambda+\mathrm{i}\epsilon)\psi$
$\displaystyle=-\mathrm{i}\lim_{\epsilon\to
0_{+}}R_{a}(\lambda+\mathrm{i}0)\chi^{2}_{+}(2B/\epsilon_{0})T_{a}R(\lambda+\mathrm{i}\epsilon)\psi\quad\text{
by \eqref{eq:strongED_+SDY}}$
$\displaystyle=-\mathrm{i}R_{a}(\lambda+\mathrm{i}0)\chi^{2}_{+}(2B/\epsilon_{0})T_{a}R(\lambda+\mathrm{i}0)\psi\quad\text{
by \eqref{eq:weak} }$ $\displaystyle=-\mathrm{i}\lim_{\epsilon\to
0_{+}}R_{a}(\lambda+\mathrm{i}\epsilon)\chi^{2}_{+}(2B/\epsilon_{0})T_{a}R(\lambda+\mathrm{i}0)\psi\quad\text{
by \eqref{eq:strongED_+SDY} }.$

We may summerize our interpretation of (A.5) as

$\psi_{a}\in(L^{2}_{s}\subseteq)\,{\mathcal{H}}\text{ \,and\,
}\phi_{a}=R_{a}(\lambda+\mathrm{i}0)\psi_{a}=\lim_{\epsilon\to
0_{+}}\,R_{a}(\lambda+\mathrm{i}\epsilon)\psi_{a}\text{ weakly in
}L^{2}_{-1}.$

Moreover it follows from (A.7), (A.8) and a commutation that the vector
$\phi_{a}$ defined this way is actually an element of ${\mathcal{B}}^{*}$,
although this is of course obvious from the representation

$\phi_{a}=\lim_{\epsilon\to 0_{+}}S_{a}R(\lambda+\mathrm{i}\epsilon)\psi.$

Anyway the above considerations combined with those in the beginning of
Section 5 as well as with Theorem A.1 finish our justification of (5.6).

###### Proof of Lemma 5.2.

We need to construct a sequence of vectors $\psi_{a,n}\in L^{2}_{\infty}$ such
that $\psi_{a,n}\to\psi_{a}\in L^{2}_{1/2}$ as well as

$R_{a}(\lambda+\mathrm{i}0)\psi_{a,n}\to
R_{a}(\lambda+\mathrm{i}0)\psi_{a}\text{ in }{\mathcal{B}}^{*}\text{ for
}n\to\infty.$ (A.9)

First note that the above discussion allows us to represent, in terms of
‘$Q$-operators’ $Q_{k}$ (and bounded $B_{k}$),

$\psi_{a}=\sum_{k}\,f_{2}(H_{a}){Q_{k}}^{*}B_{k}Q_{k}f_{2}(H)R(\lambda+\mathrm{i}0)\psi+{\breve{\psi}};\quad{\breve{\psi}}\in{\mathcal{B}}.$
(A.10a) We are let to define, cf. the proof of [Sk1, Lemma 9.12],
$\psi_{a,n}=\sum_{k}\,f_{2}(H_{a}){Q_{k}}^{*}\chi_{n}B_{k}Q_{k}f_{2}(H)R(\lambda+\mathrm{i}0)\psi+\chi_{n}{\breve{\psi}};\quad\chi_{n}=\chi_{-}(\langle
x\rangle/n).$ (A.10b)

Obviously $\chi_{n}{\breve{\psi}}\to{\breve{\psi}}$ in
${\mathcal{B}}\,(\subseteq L^{2}_{1/2})$, and due to the fact that
$\chi_{n}B_{k}\to B_{k}$ strongly on $\widetilde{{\mathcal{H}}}$ it follows
from (A.8) that $L^{2}_{\infty}\ni\psi_{a,n}\to\psi_{a}\in L^{2}_{1/2}$.
Moreover it follows from (A.7), (2.4b) and the mentioned strong convergence
that

$\sup_{\epsilon>0}\,\big{\lVert}R_{a}(\lambda+\mathrm{i}\epsilon)\big{(}\psi_{a}-\psi_{a,n}\big{)}\big{\rVert}_{{\mathcal{B}}^{*}}\to
0\text{ for }n\to\infty.$

This uniform convergence and (2.4b) yield (A.9). ∎

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