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\begin{document}
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\title{\bf  Generalized  Projective  Representations
 for sl(n+1)\footnote{Research
supported by NSFC Grant 10701002}}

\author{ $\mbox{Yufeng Zhao}^1$  and  $\mbox{Xiaoping Xu}^2$\\\\\\ 1 LMAM, School of
Mathematical Sciences \\Peking University,
  Beijing, 100871, P. R. China\\\\ 2 Hua Loo-Keng Key Mathematical Laboratory\\Institute of
Mathematics, Academy of Mathematics and Systems Sciences\\ Chinese
Academy of  Sciences, Beijing, 100190, P. R. China.}
\date{ }
\maketitle
\begin {abstract}
\quad

It is well known that $n$-dimensional projective group gives rise to
a non-homogenous representation of the Lie algebra $sl(n+1)$ on the
polynomial functions of the projective space. Using Shen's mixed
product for Witt algebras (also known as Larsson functor), we
generalize the above representation of $sl(n+1)$ to a non-homogenous
representation on the tensor space of any finite-dimensional
irreducible $gl(n)$-module with the polynomial space. Moreover, the
structure of such a representation is completely determined by
employing projection operator techniques and  well-known Kostant's
characteristic identities for certain matrices with entries in
 the universal enveloping algebra. In particular, we
obtain a new one parameter family of infinite-dimensional
irreducible
 $sl(n+1)$-modules, which are in general not highest-weight type, for any given finite-dimensional irreducible
 $sl(n)$-module.
 The results could also be used to study the quantum field theory with the projective group as
the symmetry.

\end{abstract}

\section {Introduction}

A projective transformation  on $\mathbb{F}^n$ for a field
$\mathbb{F}$ is given by
$$u\mapsto \frac{Au+\vec b}{\vec c\:^t u+d}\qquad\mbox{for}\;\;u\in
\mathbb{F}^n,\eqno(1.1)$$ where all the vectors in $\mathbb{F}^n$
are in column form and
$$\left(\begin{array}{cc}A&\vec b\\ \vec c\:^t&
d\end{array}\right)\in GL(n).\eqno(1.2)$$ It is well-known that a
transformation of mapping straight lines to lines must be a
projective transformation. The group of projective transformations
is the fundamental symmetry of $n$-dimensional projective geometry.
Physically, the group  with $n=4$ and $\mathbb{F}=\mathbb{R}$
consists of all the transformations of keeping free particles
including light signals moving with constant velocities along
straight lines (e.g., cf. [GWZ1-2]). Based on the embeddings of the
poincar\'{e} group and De Sitter group into the projective group
with $n=4$ and $\mathbb{F}=\mathbb{R}$, Guo, Wu and Zhou [GWZ1-2]
proposed three kinds of special relativity.

In this paper, we give a representation-theoretic exploration on the
impact of projective transformations. Note that the Lie algebra of
$n$-dimensional projective group is spanned by the following
differential operations
$$\{\partial_{x_j},x_i\partial_{x_j},x_i\sum_{r=1}^nx_r\partial_{x_r}\mid
i,j=1,2,...,n\},\eqno(1.3)$$ which is isomorphic to the special
linear Lie algebra $sl(n+1)$. Through the above operators, we obtain
a representation of $sl(n+1)$ on the polynomial algebra
$\mathbb{F}[x_1,...,x_n]$. The non-homogeneity of (1.3) motivates us
to generalize the above representation of $sl(n+1)$ to a
non-homogenous representation on the tensor space of any
finite-dimensional irreducible $gl(n)$-module with
$\mathbb{F}[x_1,...,x_n]$ via Shen's mixed product for Witt algebras
(cf. [Sg1-3]) (also known as Larsson functor (cf. [La])). It turns
out that the structure of such generalized projective
representations can be completely determined by employing projection
operator techniques (cf. [Gm1]) and  well-known Kostant's
characteristic identities for certain matrices with entries in  the
universal enveloping algebra (cf. [K]). In particular, we obtain a
new one parameter family of infinite-dimensional irreducible
 $sl(n+1)$-modules for any given finite-dimensional irreducible
 $sl(n)$-module.

  Denote by  $\mathbb Z $ the ring of integers and by $
\mathbb N $ the additive semigroup of nonnegative integers. For any
two integers $m$ and $n$, we denote

$$\ol{m,n}=\left\{\begin{array}{ll}\{m,m+1,\cdots,n\}&\mbox{if}\;m\leq n,\\
\emptyset&\mbox{otherwise}.\end{array}\right.\eqno(1.4)$$
 Let
${\cal A}$ be a commutative associative algebra over $\mathbb{F}$.
If $\{D_{i} \ | \ i \in \overline{1,n} \}$ is a set of commuting
derivations of ${\cal A}$, the set of derivations ${\cal
W}(n)=\{\sum\limits_{i=1}^{n}a_{i}D_{i} | \ a_{i} \in {\cal A}  \}$
forms a Lie algebra via the following Lie brackets:
$$[\sum\limits_{i=1}^{n}a_{i}D_{i},\sum\limits_{i=1}^{n}b_{i}D_{i}]=
\sum\limits_{i,j=1}^{n}(a_{j}D_{j}
(b_{i})-b_{j}D_{j}(a_{i}))D_{i}.\eqno(1.5)$$

Let $E_{r,s}$ be the square matrix with 1 as its $(r,s)$-entry and 0
as the others. The general linear Lie algebra $gl(n)$ is the Lie algebra of $n\times n$
matrices over ${\mathbb F}$ with a vector space basis $\{E_{i,j}\mid
i, j \in \overline{1,n}\}$.  For any $gl(n)$-module $V$, we define
an action $\pi$ of the Lie algebra ${\cal W}(n)$ on ${\cal
A}\otimes_{\mathbb{F}}V$ by
$$\pi(\sum\limits_{i=1}^{n}a_{i}D_{i})=\sum_{i,j=1}^nD_i(a_j)\otimes E_{i,j}
+\sum\limits_{i=1}^{n} a_{i}D_{i}\otimes \mbox{Id}_V.\eqno(1.6)$$
Then $\pi$ gives a representation of ${\cal W}(n)$ (cf. [Sg1-3]) and
the functor from $gl(n)$-modules to ${\cal W}(n)$-modules was also
later known as Larsson functor (cf. [L]). The structure of the
module ${\cal A}\otimes_{\mathbb{F}}V$ was determined by Rao [R]
when ${\cal A}=\mathbb{F}[x_1^{\pm 1},...,x_n^{\pm 1}]$,
$D_i=\partial_{x_i}$ and $V$ is a finite-dimensional irreducible
$gl(n)$-module. Lin and Tan [LT] did the similar thing when ${\cal
A}$ is the algebra of quantum torus. The first author of this paper
[Z] determined the ${\cal W}(n)$-module structure of ${\cal
A}\otimes_{\mathbb{F}}V$ in the case that ${\cal A}$ is a certain
semi-group algebra and $D_i$ are locally-finite derivations in [X].
Throughout this paper, we always assume $\mbox{char}\:\mathbb{F}=0$.

 Take ${\cal
H}=\sum\limits_{i=1}^n\mathbb{F}E_{i,i}$ as a Cartan subalgebra of
$gl(n)$. Assume ${\cal A}=\mathbb{F}[x_1,...,x_n]$ and let
$D_i=\partial_{x_i}$. Embed $sl(n+1)$ into ${\cal W}(n)$ via (1.3).
Then the space ${\cal A}\otimes_{\mathbb{F}}V$ forms an
$sl(n+1)$-module with the representation $\pi|_{sl(n+1)}$, which we
call a {\it generalized projective representation}.

Characteristic identities have a long history. The first person to
exploit them was Dirac [D], who wrote down what amounts to the
characteristic identity for the Lie algebra $so(1,3)$. This
particular example is intimately connected with the problem of
describing the structure of  relativistically invariant wave
equations.
 Such identities have been shown to be  powerful tools for the analysis
of finite dimensional representations of Lie groups (cf.  [BB],
[BG], [F]). It has been shown by Kostant [K] (also cf. [Gm4]) that
the characteristic identities for semi-simple Lie algebras also hold
for infinite dimensional representations. Moreover, one may
construct projection operators analogous to the projection operators
of Green [G], and Bracken and Green [BG] in finite dimensions. We
refer
  [Gh], [Gm1-Gm3], [Ha],  [L], [LG], [M], [OCC] and [O] for the
 other works on the identities. Using
projection operator techniques and Kostant's characteristic
identities, we prove:

\vspace{0.2cm}
 {\bf
Main Theorem}. {\it Let $V$ be a finite-dimensional irreducible
$gl(n)$-module with highest weight  $\mu$. We have the following
conclusions:

(i) The space ${\cal A}\otimes_{\mathbb{F}}V$ is an irreducible
$sl(n+1)$-module if and only if
$$\mu(E_{1,1})+\sum\limits_{j=1}^{n}\mu(E_{j,j})\not\in
-\mathbb{N}\bigcup\ol{2,1+\mu(E_{1,1}-E_{2,2})}\eqno(1.7)$$ and
$$\mu(E_{i,i})+\sum\limits_{j=1}^{n}\mu(E_{j,j})-i\not\in\ol{1,
\mu(E_{i,i}-E_{i+1,i+1})}\qquad\mbox{for}\;i\in\ol{2,n-1}.\eqno(1.8)$$

(ii) If one of the conditions in (1.7) and (1.8) fails, then both
$U(sl(n+1))(1\otimes V)$ and $({\cal
A}\otimes_{\mathbb{F}}V)/(U(sl(n+1))(1\otimes V))$ are irreducible $sl(n+1)$-modules .}\vspace{0.2cm}

Note that all $\mu(E_{i,i}-E_{i+1,i+1})$ are nonnegative integers,
which determine the corresponding $sl(n)$-module $V$ uniquely.
Moreover, the identity matrix in $gl(n)$ are allowed to be any
constant map. The above theorem says that given a finite-dimensional
irreducible $sl(n)$-module, we can construct a new one-parameter
family of explicit irreducible $sl(n+1)$-modules via its projective
representation and Shen's mixed product.

 A quantum field is an operator value function on a certain Hilbert space,
which is often a direct sum of  infinite-dimensional irreducible
modules of a certain Lie algebra (group). The Lie algebra of
two-dimensional conformal group is exactly the Virasoro algebra,
which is infinite-dimensional. The minimal models of two-dimensional
conformal field theory were constructed from direct sums of certain
infinite-dimensional irreducible modules of the Virasoro algebra,
where a distinguished module gives rise to a vertex operator
algebra. When $n>2$, the $n$-dimension conformal group is
finite-dimensional, whose Lie algebra is exactly isomorphic to
$so(n,2)$. It is still unknown what should a higher-dimensional
conformal field theory be. Part of reason is that we lack of enough
knowledge on the infinite-dimensional irreducible $so(n,2)$-modules
that are compatible to the natural conformal representation of
$so(n,2)$. This motivates us to study explicit infinite-dimensional
irreducible modules of finite-dimensional simple Lie algebras by
using non-homogeneous polynomial representations and Shen's mixed
product for Witt algebras. This paper is the first work in this
direction.

As we mentioned earlier, projective groups are important groups in
physics. In comparison with the minimal models of two-dimensional
conformal field theory, the underlying module of the projective
representation of $sl(n+1)$ should be the distinguished  module in
the possible quantum field theory with the projective group as the
symmetry. The other modules would make the theory more substantial.

 The paper is organized as follows. In Section 2, we
slightly generalize Kostant's characteristic identities and recall
some facts about projection operators based on Kostant's work [K]
and Gould's works [Gm1, Gm4]. In Section 3, we prove (i) and (ii) in
the theorem. \vspace{0.2cm}

{\bf Acknowledgement:} We would like to thank Professor Han-Ying Guo
for his interesting talk that motivates this work.

\section {Characteristic  Identities and Projection Operators}

\quad \quad In order to keep the paper self-contained, we will first prove
certain characteristic identities
  for $gl(n)$, which will be used to study  the irreducibility of the generalized projective representations for $sl(n+1)$ . Then we will recall some facts about the projection operators for $gl(n)$, although part of
 the results in this section has appeared in  [Gm1, Gm4], [OCC] and [K].

\subsection{Some Standard Facts for $gl(n)$ and $sl(n)$ }

\quad \quad Recall that $gl(n)$ is the Lie algebra of $n\times n$
matrices over ${\mathbb F}$ with a  basis $\{E_{i,j}\mid i, j \in
\overline{1,n}\}$ and the Lie bracket:
$$[E_{i,j},E_{k,l}]=\delta_{k,j}E_{i,l}-\delta_{i,l}E_{k,j}\qquad\mbox{for}\;\;
i,j,k,l \in \overline{1,n}.\eqno(2.1)$$  Note that $gl(n)$ is
reductive  with the following decomposition of ideals
$gl(n)=sl(n)\oplus {\mathbb F}\mbox{I}$, where $sl(n)$ is the
special linear Lie algebra of  matrices with zero trace and
$\mbox{I}=\sum\limits_{i=1}^n E_{i,i}$ is the identity matrix, which
is central.

 Take ${\cal
H}=\sum\limits_{i=1}^n\mathbb{F}E_{i,i}$ as  a Cartan subalgebra of
$gl(n)$. If $\lambda \in {\cal{H}}^{*}$ is a weight of $gl(n)$, we
identify $\lambda$ with the $n$-tuple
$\lambda=(\lambda_{1},\cdots,\lambda_{n})$, where
$\lambda_{i}=\lambda(E_{i,i})$. For $\lambda, \mu  \in
{\cal{H}}^{*}$, we define $$(\lambda,
\mu)=\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}.\eqno(2.2)$$

 Denote by $\varepsilon_i$ the weight with 1 as its $i$th coordinate
 and 0 as the others, i.e. $$\varepsilon_{i}=(0,\cdots, 0, \stackrel{i}{1}, 0, \cdots, 0).\eqno(2.3)$$
 The set $\Phi^+=\{\varepsilon_{i}-\varepsilon_{j}\mid 1\leq i <
 j\leq n\}$ forms a set of positive roots of $gl(n)$. In this case, the
half-sum of the positive roots is given
by$$\delta=\frac{1}{2}\sum\limits_{i <
j}(\varepsilon_{i}-\varepsilon_{j})=\frac{1}{2}\sum\limits_{i=1}^{n}
(n+1-2i)\varepsilon_{i}.\eqno(2.4)$$ Moreover, there exists a
one-to-one correspondence  between the set of finite-dimensional
irreducible $gl(n)$-modules and the set of $n$ tuples
$\lambda=(\lambda_{1},\cdots,\lambda_{n})$ such that
$\lambda_{i}-\lambda_{i+1} \in {\mathbb{N} } \  \mbox{for} \ i \in
\overline{1,n-1}$. Such an $n$ tuple $\lambda$ is called the {\it
highest weight} of the corresponding module which we  denote as
$V(\lambda)$.\vspace{0.2cm}

 Let $U$ denote the universal enveloping
algebra of $gl(n)$ and let $Z$ be the center of $U$. Set
$$\sigma_{1}=\mbox{I}, \ \sigma_r=\sum\limits_{i_1,..,i_r=1}^n
E_{i_1,i_2}E_{i_2,i_3}\cdots
E_{i_{r-1},i_r}E_{i_r,i_1}\qquad\mbox{for}\;r\in\ol{2,n}.\eqno(2.5)$$
Then the center
$$Z=\mathbb{F}[\sigma_{1},\cdots,\sigma_{n}].\eqno(2.6)$$
The subspace
${\cal H}_0={\cal H}\bigcap sl(n)$ is a Cartan
subalgebra of $sl(n)$ with the standard basis
$\{\alpha_{i}^\vee=E_{i,i}-E_{i+1,i+1}\mid i \in
\overline{1,n-1}\}.$ Denote the dual vector space of ${\cal
H}_{0}$ by ${\cal H}_{0}^*$ and let
$\omega_{1},\omega_{2},\cdots,\omega_{n-1}$ be the fundamental
integral dominant weights in ${\cal H}_{0}^*$ defined by
$\omega_{i}(\alpha_{j}^\vee)=\delta_{i,j}.$\vspace{0.3cm}

Let $V(\psi)$ be the  finite dimensional irreducible $sl(n)$-module
with the highest weight $\psi$. We can make $V(\psi)$ as a
$gl(n)$-module $V(\psi,b)$ by letting the central element $\mbox{I}$ act
as the scalar map $b\mbox{Id}_{V(\psi)}$.

For $\vec a=(a_1,...,a_{n-1})\in\mathbb{N}^{n-1}$ and
$0<k\in\mathbb{Z}$, we denote \begin{eqnarray*}\hspace{1.4cm}I(\vec
a,k)&=&\{(a_1+c_1-c_2,a_2+c_2-c_3,...,a_{n-1}+c_{n-1}-c_n)\mid
c_i\in\mathbb{N}\\ & &\mbox{such
that}\;\sum_{i=1}^nc_i=k\;\mbox{and}\;c_{s+1}\leq
a_s\;\mbox{for}\;s\in\ol{1,n-1}\}.\hspace{3cm}(2.7)\end{eqnarray*}
Moreover, we set
$$\omega_{\vec a}=\sum_{i=1}^{n-1}a_i\omega_i\qquad\mbox{for}\;
\vec a\in\mathbb{N}^{n-1}.\eqno(2.8)$$\vspace{0.1cm}

{\bf Lemma 2.1.1} (e.g., cf. Proposition 15.25 in  [FH]) {\it \quad
For any $\vec a\in\mathbb{N}^{n-1}$, the tensor product of
$sl(n)$-module $V(\omega_{\vec a})$ with
 $V(k\omega_{1})$ decomposes into a direct sum:
$$V(\omega_{\vec a})\otimes_{\mathbb{F}}
V(k\omega_{1})=\bigoplus_{\vec b\in I(\vec a,k)} V(\omega_{\vec
b}).\eqno(2.9)$$}
\vspace{0.1cm}

{\bf Lemma 2.1.2} {\it \quad  Let  $\Pi$ be the weight set of $gl(n)$-module $V(\psi,b)$. Assume $\psi=\sum\limits_{i=1}^{n-1}a_i\omega_i$, $\nu=(\nu_{1},\cdots,\nu_{n})\in \Pi$ and $(\nu_{1}-\nu_{2},\cdots,\nu_{n-1}-\nu_{n})=\psi-\sum\limits_{i=1}^{n-1}k_i\alpha_i$. Then
$$\nu_{1}=\sum\limits_{i=1}^{n-1}a_i+\frac{b-\sum\limits_{i=1}^{n-1}ia_{i}}{n}-k_{1},
 \ \nu_{n}=\frac{b-\sum\limits_{i=1}^{n-1}ia_{i}}{n}+k_{n-1}, $$
 $$ \nu_{j}=\sum\limits_{i=j}^{n-1}a_i+\frac{b-\sum\limits_{i=1}^{n-1}ia_{i}}{n}+k_{j-1}-k_{j},\;\;j \in \overline{2,n-1}.\eqno(2.10) $$  }\vspace{0.1cm}

Denote
$$\underline{k}=(k_1,k_2,...,k_n)\in \mathbb{N}^n,\quad
|\underline{k}|=\sum_{i=1}^nk_i, $$$$I(\mu,j)=\{\underline{c}=(c_{1},\cdots,c_{n}) \  | \  c_i \in\mathbb{N} \ , \ |\underline{c}|=j , \ c_{s+1}\leq
\mu_{s}-\mu_{s+1} \;\mbox{for}\;s\in\ol{1,n-1}\}. \eqno(2.11)$$

It is easy to deduce the following lemma from  the above two
lemmas.\vspace{0.3cm}

{\bf Lemma 2.1.3} {\it \quad  The tensor product of $gl(n)$-module $V(\mu)$ with
 $V(k\varepsilon_{1})$ decomposes into a direct sum:
$$V(\mu)\otimes_{\mathbb{F}}
V(k\varepsilon_{1})=\bigoplus_{\underline{c}\in I(\mu,k)} V(\mu+\underline{c}).\eqno(2.12)$$}\vspace{0.3cm}

\subsection{Characteristic Identities and Projection Operators for $gl(n)$}

\quad Now we will introduce the characteristic identities for $gl(n)$.
We know that the universal enveloping algebra $U$ of $gl(n)$ can be imbedded
into $U\otimes U$ by the associative algebra homomorphism $d: U
\rightarrow U\otimes U$ determined  by $$d(u)=u\otimes 1 +1 \otimes
u \qquad \mbox{ for} \  u\in gl(n).\eqno(2.13)$$
 Let $V(\lambda)$ be a fixed finite-dimensional $gl(n)$-module  with highest weight $\lambda$
  and let $\pi_{\lambda}$ be the corresponding
representation. Kostant [K] considered the map
$$\partial: U \rightarrow (\mbox{End}\:V(\lambda))\otimes_{\mathbb{F}} U; $$$$\ u
\mapsto 1\otimes u+\pi_{\lambda}(u)\otimes 1\eqno(2.14)$$ for $u \in
gl(n)$ and extended $\partial$ to an associative algebra
homomorphism from $U$ to
$(\mbox{End}\:V(\lambda))\otimes_{\mathbb{F}} U$. More generally, if
$d(u)=\sum\limits_{r}u_{r}\otimes v_{r}$, we have $\partial(u)=
\sum\limits_{r}\pi_{\lambda}(u_{r})\otimes v_{r}$. For $z \in Z$, we
denote  $$
\tilde{z}=-\frac{1}{2}[\partial(z)-\pi_{\lambda}(z)\otimes
1-1\otimes z],\eqno(2.15)$$which may be viewed as an $m \times m$
($m=\mbox{dim} V(\lambda)$)  matrix with entries in $U$.

Denote by $\chi_\zeta$ the central character of a highest weight
$gl(n)$-module with highest weight $\zeta$. Suppose now that $W$ is
another $gl(n)$-module  admitting the central character $\chi_{\mu}$
and $\pi_{\mu}$ is the corresponding representation. We extend
$\pi_{\mu}$ to an algebra homomorphism
$$ \tilde{\pi_{\mu}}: (\mbox{End}\:V(\lambda)) \otimes_{\mathbb{F}} U \rightarrow
(\mbox{End}\:V(\lambda)) \otimes_{\mathbb{F}} (\mbox{End}\:W); $$$$\
\sum\limits_{i}\rho_{i}\otimes u_{i} \mapsto
\sum\limits_{i}\rho_{i}\otimes \pi_{\mu}(u_{i}),\eqno(2.16)
$$where $\rho_{i} \in \mbox{End}\:V(\lambda)$ and $u_{i} \in U$. In particular, we have
$$\tilde{\pi_{\mu}}(\tilde{z})=
-\frac{1}{2}[(\pi_{\lambda}\otimes \pi_{\mu})
(z)-\pi_{\lambda}(z)\otimes 1-1\otimes \pi_{\mu}(z)]\eqno(2.17)$$
for $z \in Z$. Clearly, $\tilde{\pi_{\mu}}(\tilde{z})$ is a linear
operator on $V(\lambda) \otimes_{\mathbb{F}} W$ which may be viewed
as an $m \times m$  matrix with entries from $\mbox{End}\:W$ under a
 basis of $V(\lambda)$. For $\nu\in {\cal H}^\ast$, we define
$$f_\nu=-\frac{1}{2}(\chi_{\mu+\nu}-\chi_\lambda-\chi_\mu).\eqno(2.18)$$
Denote by $\Pi_\lambda$ the weight set of $V(\lambda)$.
\vspace{0.3cm}

{\bf Lemma 2.2.1} (cf.  [K], [G], [OCC] ) \quad {\it On the space
$W$, the matrix $\tilde{z}$ satisfies the following characteristic
identity:
$$\prod\limits_{\nu\in\Pi_\lambda}(\tilde{z}-f_\nu(z))=0
\qquad\mbox{for}\;\;z\in Z.\eqno(2.19)$$ }

By varying the module $V(\lambda)$ and the central element $z$, we
obtain a series of characteristic  identities. In particular, the
following characteristic  identity will be used in the proof of the
main theorem:\vspace{0.3cm}

 {\bf Corollary  2.2.2}\quad {\it
 Take $V(\lambda)$ to be the dual module of
 $gl(n)$-module $V(2,1,\cdots,1)$.  Then the matrix $\tilde{\sigma_{2}}$ satisfies the
following characteristic identity on $W$:
$$\prod\limits_{i=1}^{n}(\tilde{\sigma_{2}}-m_i)=0,
\ \
m_i=\frac{1}{2}(\lambda,\lambda+2\delta)-\frac{1}{2}(\lambda_{i},
\lambda_{i}+2(\mu+\delta)),\eqno(2.20)$$
where $\lambda_{i}=(-1,\cdots, \stackrel{i}{-2},\cdots, -1), \ i \in
\overline{1,n}$.

}\vspace{0.3cm}

{\it Proof}\quad Note that the module $V(2,1,\cdots,1)$ has a basis
$\{e_{1},\cdots,e_{n}\}$ such that
$$E_{i,i}(e_{k})=(1+\delta_{i,k})e_{k}, \
E_{i,j}(e_{k})=\delta_{j,k}e_{i} \ (i \neq j).\eqno(2.21)$$
 Let $\pi^{*}$ be
the dual module of $gl(n,{\mathbb F})$-module $V(2,1,\cdots,1)$.
Then $$\pi^{*}(E_{i,i})=-(E_{i,i}+\mbox{I}), \
\pi^{*}(E_{i,j})=-E_{j,i} \ (i \neq j).\eqno(2.22)$$ Obviously, the
representation $\pi^{*}$ is $n$-dimensional and all its weights
$$\{\lambda_{1}=(-2,-1,\cdots,-1),..., \lambda_{i}=(-1,\cdots,
\stackrel{i}{-2},\cdots, -1), \cdots,
\lambda_{n}=(-1,\cdots,-1,-2)\} \eqno(2.23)$$occur with multiplicity
one. The matrix $\tilde{\sigma_{2}}$ in this case is given by
$$\tilde{\sigma_{2}}=-\sum\limits_{i,j}^{n}\pi^{*}(E_{j,i})E_{i,j}.\eqno(2.24)$$
This is the matrix
$$\begin{array}{rcl}
\tilde{\sigma_{2}}&=& \left[\begin{array}{cccc}
\mbox{I}+E_{1,1}&E_{1,2}&\cdots&E_{1,n}\\
E_{2,1}&\mbox{I}+E_{2,2}&\cdots&E_{2,n}\\
\vdots&\vdots&\cdots&\vdots\\
E_{n,1}&E_{n,2}&\cdots&\mbox{I}+E_{n,n}
\end{array}
\right].
\end{array} \eqno(2.25)$$
By Lemma 2.2.1, the operator $\tilde{\sigma_{2}}$ satisfies the
characteristic identity on $W$:
$$\prod\limits_{i=1}^{n}(\tilde{\sigma_{2}}-m_i)=0,
\ \
m_i=\frac{1}{2}(\lambda,\lambda+2\delta)-\frac{1}{2}(\lambda_{i},
\lambda_{i}+2(\mu+\delta)),
 \ i \in
\overline{1,n}.\eqno(2.26)$$\hspace{1cm} $\Box$\vspace{0.3cm}

In the rest of this section, we will recall some facts about
projection operators appeared in [Gm2].

 Take $gl(n)$-module $V(\lambda)=V(\varepsilon_{1})^{*}$ (resp. $V(\varepsilon_{1})$). Then the corresponding
 matrices
 $\tilde{\sigma_{2}}$ are
 $$M=
(E_{i,j})_{i,j=1}^{n}, \ \tilde{M}=-M^{T} \in U(gl(n)),\eqno(2.27)$$
respectively. On the space $W$, the matrices $M$ and $\tilde{M}$
satisfy the following characteristic identities:
$$\prod\limits_{i=1}^{n}(M-d_i)=0, \ d_{i}=\mu_{i}+n-i;\eqno(2.28)$$
$$\prod\limits_{i=1}^{n}(\tilde{M}-\tilde{d}_i)=0, \ \tilde{d}_{i}=n-1-d_{i}, \ i \in \overline{1,n}.\eqno(2.29) $$
For $r \in \overline{1,n}$,
$$P_{r}=\prod\limits_{r\neq l \in \overline{1,n}}(\frac{M-d_{l}}{d_{r}-d_{l}}),\qquad \tilde{P}_{r}=\prod\limits_{r\neq l \in \overline{1,n}}(\frac{\tilde{M}-\tilde{d}_{l}}{\tilde{d}_{r}-\tilde{d}_{l}})\eqno(2.30)$$are called {\it  projection  operators}, which  project the tensor product space
$V(\varepsilon_{1})^{*}{\otimes}_{\mathbb{F}} W$ (resp. $V(\varepsilon_{1})\otimes_{\mathbb{F}} W$ ) onto the irreducible
 module $W_{r}=P_{r}(V(\varepsilon_{1})^{*}\otimes_{\mathbb{F}} W)$ (resp. $\tilde{W}_{r}=\tilde{P}_{r}(V(\varepsilon_{1})\otimes_{\mathbb{F}} W)$)
 with central character $\chi_{\nu-\varepsilon_{r}}$ (resp. $\chi_{\nu+\varepsilon_{r}}$).

\section{Generalized Projective Representations}

\quad \quad In this section, we will   give the
  detailed
  construction of generalized projective representations
 for  the special linear Lie
 algebra $sl(n+1)$ and study their irreducibility.

  \subsection{Construction of the Representations}

\qquad Let ${\cal A}=\mathbb{F}[x_{1},\cdots,x_{n}]$. The operator
$p_{i}=x_{i}\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}$ is called
{\it pseudo-translation operator} on ${\cal A}$ in physics. Note
that the Lie algebra of $n$-dimensional projective group
$$L_{n+1}=\sum\limits_{i,j=1}^{n}\mathbb{F}x_{i}\partial_{x_{j}}+
\sum\limits_{i=1}^{n}\mathbb{F}\partial_{
x_{i}}+\sum\limits_{i=1}^{n}\mathbb{F}p_{i},\eqno(3.1)$$ forms a Lie
subalgebra of Witt algebra
$${\cal W}(n)=\{\sum\limits_{i=1}^{n}f_{i}\partial_{ x_{i}} \ | \
f_{i} \in {\cal A} \} .\eqno(3.2) $$
 Moreover, we have
the following Lie brackets:
$$[\partial_{x_{j}},p_{i}]=\left\{\begin{array}{llll}
x_{i}\partial_{x_{j}},  & i \neq j,\\
\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}+x_{i}\partial_{x_{i}}, &
i
=j,\end{array}\right.\eqno(3.3)$$$$[x_{i}\partial_{x_{j}},p_{k}]=\delta_{j,k}{p}_{i},
\
[x_{i}\partial_{x_{j}},\partial_{x_{k}}]=-\delta_{i,k}\partial_{x_{j}},
\
[x_{i}\partial_{x_{j}},x_{k}\partial_{x_{l}}]=\delta_{j,k}x_{i}\partial_{x_{l}}-
\delta_{i,l}x_{k}\partial_{x_{j}}.\eqno(3.4)$$To abbreviate,  we
denote
$$P=\sum\limits_{i=1}^{n}\mathbb{F}p_{i}, \
S=\sum\limits_{i=1}^{n}\mathbb{F}\partial_{ x_{i}}, \
\overline{L}_{n}=\sum\limits_{i,j=1}^{n}\mathbb{F}x_{i}\partial_{x_{j}}, \ \overline{L}_{n}'=[\overline{L}_{n},\overline{L}_{n}].\eqno(3.5)$$
Then
$$L_{n+1}=P \oplus S \oplus \overline{L}_{n}, \  [P,P]=\{0\}, \
[S,S]=\{0\}.
 \eqno(3.6)$$
Moreover, $\overline{L}_{n}$ (resp. $\overline{L}_{n}'$) is
isomorphic to $gl(n)$ (resp. $sl(n)$).  It is easy to
verify:\vspace{0.3cm}

 {\bf Lemma 3.1.1 } {\it The special linear Lie algebra $sl(n+1)$ is isomorphic
to $L_{n+1}$ with the following identification of Chevalley
generators:
$$h_{i}=x_{i}\partial_{x_{i}}-x_{i+1}\partial_{x_{i+1}},  \qquad
h_{n}=\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}+x_{n}\partial_{x_{n}},\eqno(3.7)$$
$$e_{i}=x_{i}\partial_{x_{i+1}}, \;\; f_{i}=x_{i+1}\partial_{x_{i}},
\qquad e_{n}=p_{n}, \;\; f_{n}=-\partial_{x_{n}},\eqno(3.8)$$ for $i
\in \overline{1,n-1}$. }\vspace{0.1cm}

For any finite dimensional $gl(n)$-module $V$ with highest weight
$\mu=(\mu_{1},\cdots,\mu_{n}) $, we define an action $\pi$ of Witt
Lie algebra ${\cal W}(n)$ on ${\cal A}\otimes_{\mathbb{F}}V$ by
$$\pi(\sum\limits_{i=1}^{n}a_{i}D_{i})=\sum_{i,j=1}^nD_i(a_j)\otimes E_{i,j}
+\sum\limits_{i=1}^{n} a_{i}D_{i}\otimes \mbox{Id}_V.\eqno(3.9)$$
Embed $sl(n+1)$ into ${\cal W}(n)$ by (3.7) and (3.8). The space
${\cal A}\otimes_{\mathbb{F}}V$ forms an $sl(n+1)$-module with the
representation $\pi|_{sl(n+1)}$, which we call a {\it generalized
projective representation} of $sl(n+1)$.\vspace{0.3cm}

It follows from (3.7)-(3.9) that the explicit $L_{n+1} $-module ${\cal A}\otimes_{\mathbb{F}}V$
structure is given by:

$$(x_{i}\partial_{x_{j}}).(f\otimes v)=(x_{i}\partial_{x_{j}})f\otimes
v +f\otimes E_{i,j}.v,
 \eqno(3.10)$$$$\partial_{x_{i}}.(f\otimes v)=\partial_{x_{i}}(f)\otimes v,\eqno(3.11)$$$$
p_{i}.(f\otimes v)=p_{i}(f)\otimes v+x_{i}f\otimes
\sum\limits_{i=1}^n E_{i,i}.v+\sum\limits_{j=1}^n x_{j}f\otimes
E_{i,j}.v \eqno(3.12)$$ for $i,j\in\ol{1,n}$, where $f \in {\cal A}$
and  $v \in V$.

\subsection{Irreducibility Criteria for Generalized Projective Representations}

 \qquad In this section,  we will give two irreducibility criteria for  $L_{n+1}$-module  ${\cal A}\otimes_{\mathbb{F}}V$ (cf. Proposition 3.2.3 and Proposition 3.2.5).\vspace{0.3cm}

{\bf Lemma 3.2.1} \quad {\it The vector space $U(P)(1\otimes_{\mathbb{F}} V)$ is an
 irreducible $L_{n+1}$-submodule of ${\cal A}\otimes_{\mathbb{F}}V$,
where $U(P)$ denote the universal enveloping algebra of abelian Lie
algebra $P$.}\vspace{0.3cm}

 {\it Proof} \quad Denote
$$x^{\underline{c}}=x_{1}^{c_1}x_{2}^{c_2}\cdots x_{n}^{c_n}  \qquad \mbox{for} \ \underline{c}=(c_{1},\cdots,c_{n}) \in \mathbb{N}^{n}.\eqno(3.13)$$
Recall that $\Pi_\mu$ denotes the weight set of $V$. By (3.10), we
have
$$(x_{i}\partial_{x_{i}}).(x^{\underline{c}}\otimes v_{\nu})=(c_{i}+\nu_{i})
x^{\underline{c}}\otimes v_{\nu}, \eqno(3.14)$$
$$(\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}).(x^{\underline{c}}\otimes
v_{\nu})=(|\underline{c}|+\sum\limits_{i=1}^{n}\nu_{i}) x^{\underline{c}}\otimes
v_{\nu},
 \eqno(3.15)$$ for any $
x^{\underline{c}} \in {\cal A}$ and $\nu \in \Pi_\mu$.

Assume that
$\{v_{1},\cdots,v_{\ell}\}$ is a basis of $V$. For $1 \leq k \in
\mathbb{N}$, we set
$$ U(P)(1\otimes_{\mathbb{F}}
V)_{{\la}k\ra}=\mbox{Span}_{\mathbb{F}}\{ p_{i_{1}}p_{i_{2}}\cdots
p_{i_{k}}(1\otimes v_{j}) \ | \ 1 \leq i_{1}\leq i_{2}\leq \cdots
\leq i_{k}\leq n, j \in \overline{1,d} \}, \eqno(3.16)$$
$$({\cal A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}=\mbox{Span}_{\mathbb{F}}\{x^{\underline{c}}\otimes v_{j} \ | \
x^{\underline{c}} \in {\cal A}, \ |\underline{c}|=k, \ j \in \overline{1,l}\}.
\eqno(3.17)$$
 Then (3.14) implies
that
 $$U(P)(1\otimes_{\mathbb{F}} V)=\bigoplus\limits_{k \in \mathbb{N}}(U(P)(1\otimes
V))_{{\la}k\ra}.\eqno(3.18)$$Denote
$$\triangle_{i,j}^{k}=\left\{\begin{array}{llll}
x_{j}\partial_{x_{i}}  & \mbox{if}\;i \neq j,\\
\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}+x_{i}\partial_{x_{i}}+k-1
& \mbox{if}\;i =j,\end{array}\right.\eqno(3.19)$$By induction, we can easily verify the following fomula:
$$\partial_{x_{i}}.p_{i_{1}}p_{i_{2}}\cdots p_{i_{k}}(1\otimes
v_{j})=\sum\limits_{s=1}^{n}p_{i_{1}}p_{i_{2}}\cdots\hat{p}_{i_{s}}\cdots
p_{i_{k}}\triangle_{i,i_{s}}^{k}(1\otimes v_{j}),\eqno(3.20)$$
\begin{eqnarray*}\hspace{1cm}& &
(x_{i}\partial_{x_{l}}).p_{i_{1}}p_{i_{2}}\cdots p_{i_{k}}(1\otimes
v_{j})\\
&=&\sum\limits_{s=1}^{n}\delta_{l,i_{s}}p_{i_{1}}p_{i_{2}}\cdots\hat{p}_{i_{s}}\cdots
p_{i_{k}}p_{i}(1\otimes
v_{j})+p_{i_{1}}p_{i_{2}}\cdots{\cal}{p}_{i_{k}}(1\otimes
E_{i,l}.v_{j}),\hspace{2.6cm}(3.21)\end{eqnarray*}
$${p}_{i}.{\cal}{p}_{i_{1}}{p}_{i_{2}}\cdots p_{i_{k}}(1\otimes
v_{j})={p}_{i}p_{i_{1}}p_{i_{2}}\cdots {p}_{i_{k}}(1\otimes
v_{j}).\eqno(3.22)$$  So (3.20)-(3.22) imply that
$U(P)(1\otimes_{\mathbb{F}} V)$ is an $L_{n+1}$-submodule of  ${\cal
A}\otimes_{\mathbb{F}}V$. Furthermore, it is easy to verify that
$$\bigcap\limits_{i=1}^{n} \mbox{Ker} \ \partial_{x_{i}}|_{({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}}=\{0\} \  \mbox{for \ any } \ 1
\leq k \in \mathbb{N}. \eqno(3.23)
$$Thus any non-trivial submodule of $U(P)(1\otimes_{\mathbb{F}} V)$ must contain $1\otimes V$.
The irreducibility of  $U(P)(1\otimes_{\mathbb{F}} V)$ follows.
\hspace{1cm}
$\Box$\vspace{0.3cm}

In the rest of this section,  we will investigate the condition for
${\cal A}\otimes_{\mathbb{F}}V=U(P)(1\otimes_{\mathbb{F}}
V)$.\vspace{0.3cm}

 It is obvious that
${\cal A}\otimes_{\mathbb{F}}V=U(P)(1\otimes_{\mathbb{F}} V)$ if and
only if $({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}=(U(P)(1\otimes_{\mathbb{F}}
V))_{{\la}k\ra}$ holds for any $   k \in \mathbb{N}$.\vspace{0.3cm}

 Suppose that $B_{i}$ is an ordered basis for $({\cal
   A}\otimes_{\mathbb{F}}V)_{{\la}i\ra}$. Denote by
${P}_{i+1,i}^{j}$ the matrix of the linear map ${p}_{j}|_{({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}i\ra}}$ (cf. (3.12))
  with respect to the bases $B_{i}$ and $B_{i+1}$. For any
  $0<j\in\mathbb{Z}$, we denote
  $$\Gamma_j=\{\hat i=(i_1,i_2,...,i_j)\mid i_s\in\ol{1,n};i_1\leq
  i_2\leq\cdots\leq i_j\}.\eqno(3.24)$$
Set
$$P_{\hat
i}={P}_{j,j-1}^{i_{1}}{P}_{j-1,j-2}^{i_{2}}\cdots{P}_{1,0}^{i_{j}}.\eqno(3.25)$$
We order
$$\Gamma_j=\{\hat k^1,\hat k^2,...,\hat k^{\ell_j}\}\eqno(3.26)$$
lexically. In particular,
$$\hat k^1=(1,1,...,1),\qquad \hat
k^{\ell_j}=(n,n,...,n).\eqno(3.27)$$
  Then $(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}$ is, as a vector
  space,
isomorphic to the column space of the $\ell\ell_j\times \ell\ell_j$
matrix
$$
M_{j}= [P_{\hat k^1},P_{\hat k^2},...,P_{\hat k^{\ell_j}}]
\eqno(3.28)$$ (recall that $\ell=\dim V$).
 Denote
$$I_{1}=\{1\}\bigcup\{i \in \overline{2,n} \ | \ \mu_{i-1}-\mu_{i}\geq 1\}.
\eqno(3.29)$$ By means of the characteristic identity  in Corollary
2.2.2, we get the following necessary but not sufficient condition
for the irreducibility of  $L_{n+1}$-module  ${\cal
A}\otimes_{\mathbb{F}}V$:\vspace{0.3cm}

 {\bf Lemma 3.2.2 } {\it   \quad For any $1 \leq k \in \mathbb{N}$. If $\mu_{i}+|\mu|-i+s \neq 0$ for any $ s \in
\overline{1,k}, \ i \in I_{1}$, then  $L_{n+1}$-module  ${\cal A}\otimes_{\mathbb{F}}V$ is irreducible.
 }\vspace{0.3cm}

 {\it Proof} \quad   The Lemma is based on the following result:\vspace{0.3cm}

 {\it Claim. For any $1 \leq k \in \mathbb{N}$. If $\mu_{i}+|\mu|-i+s \neq 0,  \ \forall \ s \in
\overline{1,k}, \ i \in I_{1}$, then
$({\cal A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}=(U(P)(1\otimes_{\mathbb{F}}
V))_{{\la}k\ra}$ holds. }\vspace{0.3cm}

 We will prove this claim by induction on $k$.
For $k=1$, $(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}1\ra}$ is
isomorphic as vector space to the column space of $\ell n\times \ell
n$ matrix
$$\begin{array}{rcl}
&& [{P}_{1,0}^{1},{P}_{1,0}^{2},\cdots,{P}_{1,0}^{n} ]
\end{array} \mbox{with} \ \begin{array}{rcl}
{P}_{1,0}^{k}&=& \left[\begin{array}{c} E_{k,1}|_V\\E_{k,2}|_V\\\vdots\\
(\mbox{I}+E_{k,k})|_V\\\vdots\\E_{k,n}|_V
\end{array}
\right],
\end{array}\eqno(3.30)$$
which is exactly $\tilde{\sigma_{2}}|_V$ (cf. (2.25)).
By Corollary
2.2.2, the matrix $\tilde{\sigma_{2}}|_V$ is diagonalizable and it has
full rank if and only if all its eigenvalues are not zero, i.e.
$$m_{i}=\frac{1}{2}(\lambda,\lambda+2\delta)-\frac{1}{2}
 (\lambda_{i},\lambda_{i}+2(\mu+\delta))
 =\mu_{i}+ |\mu|-i+1\neq
 0\eqno(3.31)$$ for any $i \in I_{1}$.

Now suppose that the lemma holds for $k=\iota-1$. Assume $k=\iota$.
Note that the eigenvalues of the matrix
$$ \left[\begin{array}{cccc}
(\mbox{I}+E_{1,1}+s-1)|_V&E_{1,2}|_V&\cdots&E_{1,n}|_V\\
E_{2,1}|_V&(\mbox{I}+E_{2,2}+s-1)|_V&\cdots&E_{2,n}|_V\\
\vdots&\vdots&\cdots&\vdots\\
E_{n,1}|_V&E_{n,2}|_V&\cdots&(\mbox{I}+E_{n,n}+s-1)|_V
\end{array}
\right]  \eqno(3.32)$$  are  $\mu_{i}+\sum\limits_{ j
=1}^{n}\mu_{j}-i+s  \ ( i \in I_{1})$ by (3.31). Thus it is
invertible if and only if $\mu_{i}+|\mu|-i+s \neq 0,  \ \forall \ i
\in I_{1}$.\vspace{0.3cm}

 Observe that
$({\cal A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}=(U(P)(1\otimes_{\mathbb{F}}
V))_{{\la}k\ra}$
 if and only if  the vectors
 $$\{
 p_{i_{1}}\cdots
p_{i_{k}}(1\otimes v_{j}) \ | \ 1 \leq i_{1} \leq \cdots \leq i_{k}
\leq n, j \in \overline{1,\ell} \}\eqno(3.33)$$ are linearly
independent. Equivalently, the corresponding  $\ell\ell_{k}\times
\ell \ell_{k}$ system of homogeneous linear equations
$$M_{k}X=0\eqno(3.34)$$ has only zero solution by (3.31).

Suppose $$\sum\limits_{1 \leq i_{1}\leq i_{2}\leq \cdots \leq
i_{k}\leq n, j \in \overline{1,d}}
a_{i_{1},i_{2},\cdots,i_{k}}^{j}p_{i_{1}}p_{i_{2}} \cdots p_{i_{k}}
(1\otimes v_{j})=0, \ a_{i_{1}, i_{2}, \cdots , i_{k}}^{j} \in
\mathbb{F}.\eqno(3.35)$$ It can be written as
$$\sum\limits_{1 \leq i_{1}\leq i_{2}\leq \cdots \leq i_{k}\leq n}
p_{i_{1}}p_{i_{2}}\cdots p_{i_{k}}(1\otimes w_{i_{1}, i_{2}, \cdots
, i_{k}})=0, \ w_{i_{1}, i_{2}, \cdots , i_{k}} \in V.\eqno(3.36)$$
 Then $$\partial_{x_{l}}.\sum\limits_{1 \leq i_{1}\leq
i_{2}\leq \cdots \leq i_{k}\leq n} p_{i_{1}}p_{i_{2}}\cdots
p_{i_{k}}(1\otimes w_{i_{1}, i_{2}, \cdots , i_{k}})=0, \  \forall \
l \in \overline{1,n}.\eqno(3.37)$$ Equivalently,
$$
\sum\limits_{1 \leq i_{1}\leq i_{2}\leq \cdots \leq i_{k}\leq
n}\sum\limits_{s=1}^{n}p_{i_{1}}p_{i_{2}}\cdots\hat{p}_{i_{s}}\cdots
p_{i_{k}}\triangle_{l,i_{s}}^{k}(1\otimes w_{i_{1}, i_{2}, \cdots ,
i_{k}})=0\eqno(3.38)$$ by (3.20). Moreover, it can be written as the
form
$$ \sum\limits_{1 \leq j_{1}\leq j_{2}\leq \cdots \leq j_{k-1}\leq
n} p_{j_{1}}p_{j_{2}}\cdots p_{j_{k-1}}(1\otimes u_{j_{1},
j_{2}, \cdots , j_{k-1}})=0,\eqno(3.39)$$where
\begin{eqnarray*}\hspace{2cm}& &
1\otimes u_{j_{1}, j_{2}, \cdots ,
j_{k-1}}\\
&=&\sum\limits_{i=1}^{j_{1}}\triangle_{l,i}^{k}(1\otimes w_{i,j_{1},
j_{2}, \cdots ,
j_{k-1}})+\sum\limits_{i=j_{1}}^{j_{2}}\triangle_{l,i}^{k}(1\otimes
w_{j_{1},i,j_{2}, \cdots , j_{k-1}})+\cdots\\&&+
\sum\limits_{i=j_{s-1}}^{j_{s}}\triangle_{l,i}^{k}(1\otimes
w_{j_{1},j_{2}, \cdots ,j_{s-1},i,j_{s},\cdots j_{k-1}})+\cdots\\
& &+\sum\limits_{i=j_{k-1}}^{n}\triangle_{l,i}^{k}(1\otimes
w_{j_{1},j_{2}, \cdots ,
j_{k-1},i})=N_{k-1}\varpi_{n}\hspace{5cm}(3.40)
\end{eqnarray*}
 with
$$ N_{k-1}={\small \left[\begin{array}{cccc}
\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}+x_{1}\partial_{x_{1}}+k-1&x_{2}\partial_{x_{1}}
&\cdots&x_{n}\partial_{x_{1}}\\
x_{1}\partial_{x_{2}}&\sum\limits_{i=1}^{n}x_{i}\partial_{x_{i}}+x_{2}
\partial_{x_{2}}+k-1&
\cdots&x_{n}\partial_{x_{2}}\\
\vdots&\vdots&\cdots&\vdots\\
x_{1}\partial_{x_{n}}&x_{2}\partial_{x_{n}}&\cdots&\sum\limits_{i=1}^{n}
x_{i}\partial_{x_{i}}+x_{n}
\partial_{x_{n}}+k-1
\end{array}
\right]}\eqno(3.41)$$ and
$$
\varpi_{n}= \left[\begin{array}{c} X_{1,j_{1}-1}\\
X_{j_{1},j_{2}-1}\\X_{j_{2},j_{3-1}}\\\vdots\\
X_{j_{k-1},n}\end{array} \right], \eqno(3.42)$$ in which
$$
X_{1,j_{1}-1}= \left[\begin{array}{c} 1 \otimes w_{1,j_{1},j_{2},
\cdots , j_{k-1}} \\ 1\otimes w_{2,j_{1},j_{2}, \cdots ,
j_{k-1}}\\\vdots \\1\otimes w_{j_{1}-1,j_{1}, j_{2},\cdots ,
j_{k-1}}\end{array} \right], \;\;X_{j_{1},j_{2}-1}=
\left[\begin{array}{c} 2\otimes w_{j_{1},j_{1},j_{2}, \cdots ,
j_{k-1}}\\1\otimes w_{j_{1},j_{1}+1,j_{2}, \cdots , j_{k-1}}
\\\vdots \\1\otimes w_{j_{1},j_{2}-1,j_{2}, \cdots , j_{k-1}}
\end{array} \right],\eqno(3.43) $$
$$X_{j_{2},j_{3-1}}=\left[\begin{array}{c} 2\otimes
w_{j_{1},j_{2},j_{2}, \cdots , j_{k-1}}
 \\1\otimes w_{j_{1},j_{2},j_{2}+1, \cdots , j_{k-1}}
\\\vdots \\1\otimes w_{j_{1},j_{2},j_{3}-1, \cdots , j_{k-1}}
\end{array} \right],\cdots,\;\;
 X_{j_{k-1},n}=\left[\begin{array}{c}
2\otimes w_{j_{1},j_{2}, \cdots , j_{k-1},j_{k-1}}\\1\otimes
w_{j_{1},j_{2}, \cdots ,j_{k-1}, j_{k-1}+1}\\\vdots \\1\otimes
w_{j_{1},j_{2}, \cdots , j_{k-1},n}
\end{array}
\right]. \eqno(3.44)$$ We know that $N_{k-1}\varpi_{n}=0$ has only
zero solution if and only if $\mu_{i}+|\mu|-i+k \neq 0,  \ \forall \
i \in I_{1}$. By Lemma 3.2.1 and the claim, the Lemma is followed.
\hspace{1cm} $\Box$\vspace{0.3cm}

Next, we study the structure $(U(P)(1\otimes_{\mathbb{F}}
V))_{{\la}j\ra}$ as an $\bar L_n$-module and finally get a
necessary and sufficient condition for the irreducibility of  $L_{n+1}$-module  ${\cal A}\otimes_{\mathbb{F}}V$ (cf. Proposition 3.2.5 ).\vspace{0.3cm}

 Note that $\bar L_n \simeq gl(n)$  ( resp. $\bar L_n'\simeq sl(n)$ ) according to (3.5). Obviously, (3.12) implies that
 as $\overline{L}_{n}$-module
$({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra} $ is isomorphic  to
tensor product module $V(j\epsilon_{1})\otimes_{\mathbb{F}}V$
$$({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra} \cong
V(j\varepsilon_{1})\otimes_{\mathbb{F}}V(\mu)=\bigoplus_{\underline{c}\in
I(\mu,j)} V(\mu+\underline{c})\eqno(3.45)$$ (cf. Lemma 2.1.3).
Denote
$$ p^{\underline{c}}=p_{1}^{c_1}p_{2}^{c_2}\cdots p_{n}^{c_n}\qquad \mbox{for} \
\underline{c}=(c_{1},\cdots,c_{n}) \in \mathbb{N}^{n}.\eqno(3.46)$$
Define the linear map
$$\varphi_{j}:({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}\rightarrow (U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}$$$$\sum\limits_{|\underline{l}|=j, \  q \in
\overline{1,l}}a_{\underline{l},q}x^{\underline{l}}\otimes v_{q} \mapsto \sum\limits_{|\underline{l}|=j, \  q \in
\overline{1,l}}a_{\underline{l},q}p^{\underline{l}}.(1\otimes  v_{q}). \eqno(3.47)$$It is easy to verify that
$\varphi_{j}$ is an $\overline{L}_{n} $-module homomorphism by (3.21).\vspace{0.3cm}

For any $\underline{c}\in I(\mu,j)$ (cf. (2.11)), let
$\xi_{\underline{c}}$ be a maximal vector
 for  highest weight module $V(\mu+\underline{c})\subseteq ({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}$.
 Then
$$\hat{\xi}_{\underline{c}}=\varphi_{j}(\xi_{\underline{c}})\eqno(3.48)$$
is also a maximal vector of $V(\mu+\underline{c}) \subseteq
(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}$.
 Write
$$\hat{\xi}_{\underline{c}}=q_{\underline{c}}\xi_{\underline{c}},\; \ \ q_{\underline{c}}\in \mathbb{F}.\eqno(3.49)$$We know that
$$(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}\bigcap
V(\mu+\underline{c}) \neq \{0\}\Rightarrow
V(\mu+\underline{c})\subseteq (U(P)(1\otimes_{\mathbb{F}}
V))_{{\la}j\ra}.\eqno(3.50)$$
So
$$(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}\bigcap
V(\mu+\underline{c}) \neq \{0\} \qquad \mbox{iff} \qquad  q_{\underline{c}} \neq 0.\eqno(3.51)$$In the following Lemma, we will calculate $q_{\underline{c}} $ explicitly:
 \vspace{0.3cm}

{\bf Lemma 3.2.3 } {\it  For any $\underline{c}\in I(\mu,j)$, we have $$q_{\underline{c}}=\prod\limits_{s=1}^{n}\prod\limits_{i=1}^{c_{s}}(\mu_{s}+|\mu|-s+i),
\eqno(3.52)$$where we treat
$\prod\limits_{i=1}^{c_{s}}(\mu_{s}+|\mu|-s+i)=1$ if $c_{s}=0$. }
\vspace{0.3cm}

{\it Proof}\quad  Let $v_{\mu}$ be a highest weight vector of the
given $gl(n)$-module $V$.

For any $\underline{c}\in I(\mu,j)$,  we write a maximal vector
$\xi_{\underline{c}}$
 for  highest weight module $V(\mu+\underline{c})$ as
$$\xi_{\underline{c}}=a_{\underline{c}}x^{{\underline{c}}}\otimes v_{\mu}+
\sum\limits_{(\nu,\underline{l})\neq (\mu, \underline{c} ),
{\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}x^{\underline{l}}\otimes
v_{\nu}^{i},\qquad 0 \neq a_{\underline{c}},  \ a_{\nu,\underline{l}}^{i}
\in \mathbb{F},\eqno(3.53)$$where  $m({\nu})$ denotes the
multiplicity of weight $\nu \in \Pi_\mu$.\vspace{0.3cm}

{\it Case 1.  $
c_{n}\neq 0$.}\vspace{0.3cm}

Note that $[x_{s}\partial_{x_{t}},\partial_{x_{n}}]=0, \ \forall \ 1 \leq s
< t \leq n$. We know
$$0 \neq
\partial_{x_{n}}. \xi_{\underline{c}}=
a_{\underline{c}}c_{n}x^{{\underline{c}}-\varepsilon_{n}}\otimes v_{\mu}+
\sum\limits_{(\nu,\underline{l})\neq (\mu, \underline{c} ),
{\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}l_{n}x^{\underline{l}
-\varepsilon_{n}}\otimes v_{\nu}^{i}\eqno(3.54)$$ is a maximal
vector for the  $\overline{L}_{n}$- highest weight module
$V(\mu+\underline{c}-\varepsilon_{n})$. Set $$\xi_{\underline{c}-\varepsilon_{n}}=\partial_{x_{n}}. \xi_{\underline{c}}.\eqno(3.55)$$Obviously,
$$\hat{\xi}_{\underline{c}}=a_{\underline{c}}p^{{\underline{c}}}.(1\otimes v_{\mu})+
\sum\limits_{(\nu,\underline{l})\neq (\mu, \underline{c} ),
{\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}p^{\underline{l}}.(1\otimes
v_{\nu}^{i}),\eqno(3.56)$$
$$
\hat{\xi}_{\underline{c}-\varepsilon_{n}}=\varphi_{j}(\partial_{x_{n}}. \xi_{\underline{c}})
=a_{\underline{c}}c_{n}p^{
{\underline{c}}-\varepsilon_{n}}.(1\otimes v_{\mu})+
\sum\limits_{(\nu,\underline{l})\neq (\mu, \underline{c} ),
{\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}l_{n}p^{
\underline{l}-\varepsilon_{n}}.(1\otimes v_{\nu}^{i}).\eqno(3.57)
$$
Write $$ \partial_{x_{n}}.
\hat{\xi}_{\underline{c}}=b_{\underline{c}}(n)
\hat{\xi}_{\underline{c}-\varepsilon_{n}}, \qquad b_{\underline{c}}(n) \in \mathbb{F}.\eqno(3.58)$$By
(3.21) and (3.54), we have
\begin{eqnarray*}& &
\partial_{x_{n}}.
\hat{\xi}_{\underline{c}}=\partial_{x_{n}}.\varphi_{j}( \xi_{\underline{c}})\\
&=&c_{n}
a_{\underline{c}}p^{{\underline{c}}-\varepsilon_{n}}\Delta_{n,n}^{j}(
1 \otimes v_{\mu})\\&&+ \sum\limits_{(\nu,\underline{l})\neq (\mu,
\underline{c} ), {\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}[\sum\limits_{s=1}
^{n-1}l_{s}p^{\underline{l}-\varepsilon_{s}}(1\otimes
E_{s,n}v_{\nu}^{i})+l_{n}p^{\underline{l}-\varepsilon_{n}}\Delta_{n,n}^{j}(1\otimes
v_{\nu}^{i})] .\hspace{1cm}(3.59)
\end{eqnarray*}
 Denote
$$E_{s,n}.v_{\mu+\varepsilon_{n}-\varepsilon_{s}}^{i}=\Im_{i}v_{\mu}, \qquad \Im_{i} \in \mathbb{F}.\eqno(3.60)$$
Then, we have
$$b_{\underline{c}}(n)=\frac{c_{n}a_{\underline{c}}(j-1+\mu_{n}+|\mu|)
+\sum\limits_{s=1}^{n-1}(1+c_{s})\sum\limits_{i=1}^{m(\mu-\varepsilon_{s}
+\varepsilon_{n})}
a_{\mu-\varepsilon_{s}+\varepsilon_{n},\underline{c}+\varepsilon_{s}-\varepsilon_{n}}^{i}
\Im_{i}}{c_{n}a_{\underline{c}}}\eqno(3.61)$$ by (3.58)-(3.61). For any $s
\in \overline{1,n-1}$, we get
\begin{eqnarray*}& &
0=x_{s}\partial_{x_{n}}.
\xi_{\underline{c}}\\
&=&c_{n}a_{\underline{c}}x^{\underline{c}+\varepsilon_{s}-\varepsilon_{n}}\otimes
v_{\mu}+ \sum\limits_{(\nu,\underline{l})\neq (\mu, \underline{c} ),
{\underline{c}}+\mu={\underline{l}}+\nu, \ i \in
\overline{1,m({\nu})}}a_{\nu,\underline{l}}^{i}[l_{n}x^
{\underline{l}+\varepsilon_{s}-\varepsilon_{n}}\otimes
v_{\nu}^{i}+x^{\underline{l}}\otimes
E_{s,n}.v_{\nu}^{i}].\hspace{0.2cm}(3.62)
\end{eqnarray*}
Therefore,
$$c_{n}a_{\underline{c}}+\sum\limits_{i=1}^{m(\mu-\varepsilon_{s}+\varepsilon_{n})}
a_{\mu-\varepsilon_{s}+\varepsilon_{n},\underline{c}+\varepsilon_{s}-
\varepsilon_{n}}^{i}\Im_{i}=0\eqno(3.63)$$ Hence,  we have
$$b_{\underline{c}}(n)=c_{n}-n+\mu_{n}+|\mu|.\eqno(3.64)$$Furthermore,
by (3.55), we obtain
$$  \partial_{x_{n}}.
\hat{\xi}_{\underline{c}}=q_{\underline{c}}\partial_{x_{n}}.\xi_{\underline{c}}
=q_{\underline{c}}\xi_{\underline{c}-\varepsilon_{n}}=
b_{\underline{k}}(n)
\hat{\xi}_{\underline{c}-\varepsilon_{n}}
=b_{\underline{k}}(n)q_{\underline{c}-\varepsilon_{n}}\xi_{\underline{c}-\varepsilon_{n}}
\eqno(3.65)$$which implies that
$$q_{\underline{c}}=b_{\underline{c}}(n)q_{\underline{c}-\varepsilon_{n}}\eqno(3.66)$$

{\it Case 2. $c_{n}=0$.}

Suppose $c_{n-1}\neq 0$. We claim that  $l_{n}=0$ for any
$\underline{l}$ of $ a_{\nu,\underline{l}}^{i} \neq 0$ in (3.54).
Indeed, we can write $$ (\nu_{1}-\nu_{2},\cdots,\nu_{n-1}-\nu_{n})
=\sum\limits_{i=1}^{n-1}(\mu_{i}-\mu_{i+1})\varpi_{i}
-\sum\limits_{i=1}^{n-1}k_i\alpha_i.\eqno(3.67)$$ Since $
\nu_{n}+l_{n}=\mu_{n}+c_{n}$, we have
$l_{n}=\mu_{n}-\nu_{n}=\mu_{n}-(\mu_{n}+k_{n-1})$ by (2.9). Thus we
get $l_{n}+k_{n-1}=0$. So $l_{n}=0$.

Hence,
$$x_{n-1}\partial_{x_{n}}.\partial_{x_{n-1}}.\xi_{\underline{c}}
=-\partial_{x_{n}}.\xi_{\underline{c}}+\partial_{x_{n-1}}.
x_{n-1}\partial_{x_{n}}.\xi_{\underline{c}}=0.\eqno(3.68)$$This
implies that  $\partial_{x_{n-1}}.\xi_{\underline{c}}$ is  a maximal
vector for the highest weight $\overline{L}_{n}$-module
$V(\mu+\underline{c}-\varepsilon_{n-1})$.  Repeating the process
from (3.56) to (3.64), we get
$$q_{\underline{c}}=(\mu_{n-1}+|\mu|-(n-1)+c_{n-1})
q_{\underline{c}-\varepsilon_{n-1}}.\eqno(3.69)$$ Then induction
implies that $$ q_{\underline{c}}=
\prod\limits_{s=1}^{n}\prod\limits_{i=1}^{c_{s}}(\mu_{s}+|\mu|-s+i).
\eqno(3.70)$$Thus the lemma is proved. \hspace{1cm}
$\Box$\vspace{0.3cm}

 By Lemma 3.2.3, we know that $(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}= ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}$ iff  $q_{\underline{c}}\neq 0$ for any $\underline{c}\in I(\mu,j), \ 0 < j \in \mathbb{N}$. Thus we  get the following sufficient and necessary condition for the irreducibility of $L_{n+1}$-module  ${\cal A}\otimes_{\mathbb{F}}V$:
 \vspace{0.3cm}

{\bf Proposition 3.2.4 } {\it The $L_{n+1}$-module  ${\cal A}\otimes_{\mathbb{F}}V$ is irreducible iff  $\ q_{\underline{c}}\neq 0$ for any $\underline{c}\in I(\mu,j), \ 0 < j \in \mathbb{N}$.
}\vspace{0.3cm}

\subsection{Jordan-Holder Series for
  $L_{n+1}$-module ${\cal A}\otimes_{\mathbb{F}}V$}

\qquad In this section, we will assume that ${\cal A}\otimes_{\mathbb{F}}V \neq U(P)(1\otimes_{\mathbb{F}}
V)
 $ and prove the irreducibility of the quotient module $({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}} V)$.\vspace{0.3cm}

First, we study the structure of $({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}} V)$ as an $\overline{L}_{n} $-module (cf.
Lemma 3.3.1 and Lemma 3.3.2). Recall the $\overline{L}_{n} $-module
homomorphism $\varphi_{j}$ defined by (3.47). Denote
$\mbox{Ker}(\varphi_{j})$ by ${\cal{R}}_{{\la}j\ra}$ and set
 $$I(\mu,j)'=\{\underline{c}\in I(\mu,j) \ | \ \ q_{\underline{c}} =0\}.\eqno(3.71)$$According to Lemma 3.2.3, we have the following result:\vspace{0.3cm}

{\bf Lemma 3.3.1} {\it \quad For  any $  1 \leq j \in
\mathbb{N}$, we have ${\cal{R}}_{{\la}j\ra}=\bigoplus\limits_{\underline{c}\in I(\mu,j)'}V(\mu+\underline{c})$ and the following direct sum of submodules for
$\overline{L}_{n} $:
$ ({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}= (U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j\ra}\oplus
{\cal{R}}_{{\la}j\ra}.$
}
\vspace{0.3cm}

Let $1 \leq k \in \mathbb{N}$. Suppose that
 $$({\cal A}\otimes_{\mathbb{F}}V)_{{\la}i\ra}=(U(P)(1\otimes V))_{{\la}i\ra} \;\;  \mbox{for any
} \ \;i \in \overline{0,k},\eqno(3.72)$$ but $$({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}\neq (U(P)(1\otimes
V))_{{\la}j\ra} \;\;\mbox{when}\ j \geq k+1. \eqno(3.73)$$ Then by
Lemma 3.3.1, as $\overline{L}_{n} $-modules,
$$({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}} V) \cong\bigoplus\limits_{j=
k+1}^\infty{\cal{R}}_{{\la}j\ra}.\eqno(3.74)$$

{\bf  Lemma 3.3.2} {\it \quad Let $1 \leq k \in \mathbb{N}$ such
that (3.72) and (3.73) hold. Then $\bar{L}_n$-module
${\cal{R}}_{{\la}k+1\ra}$ is irreducible.}\vspace{0.3cm}

{\it Proof}\quad Since $({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}=(U(P)(1\otimes V))_{{\la}k\ra}
$, Lemma 3.2.3 implies
$$ q_{\underline{c}}\neq 0 \qquad  \forall \  \underline{c}\in I(\mu,k)
.\eqno(3.75)$$
For convenience, we denote
$$q_{s}(\underline{c})=\prod\limits_{i=1}^
{c_{s}}(\mu_{s}+|\mu|-s+i).\eqno(3.76)$$So
$$q_{\underline{c}}=\prod\limits_{s=1}^ {n}q_{s}(\underline{c}).\eqno(3.77)$$
Assume $\underline{m}\in I(\mu,k+1)'$, i.e.
$q_{\underline{m}}=0$. Since the $gl(n)$-modules
$$
V((k+1)\varepsilon_{1})\otimes_{\mathbb{F}}V\simeq ({\cal A}\otimes_{\mathbb{F}}V)_{{\la}k+1\ra}\subset
V(\varepsilon_1)\otimes_{\mathbb{F}} ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}k\ra}\simeq V(\varepsilon_{1})\otimes_{\mathbb{F}}(V
(k\varepsilon_{1})
\otimes_{\mathbb{F}}V) \eqno(3.78) $$ in the sense of
monomorphism,
 we know that $$\exists \quad \underline{t}\in {\mathbb{N}}^{n} \
\mbox{and} \ r \in \overline{1,n}, \  \mbox{such \ that}  \
\underline{t}\in I(\mu,k) \ \mbox{and} \
\underline{m}=\underline{t}+\varepsilon_{r}.\eqno(3.79)$$ Therefore,
$m_{r}=t_{r}+1$. The fact $q_{\underline{m}}=0$ implies that
$q_{r}(\underline{m})=0$ because
$q_{s}(\underline{m})=q_{s}(\underline{t})\neq 0$ for $s \neq r$.
Then $q_{r}(\underline{t})\neq 0$ and
$q_{r}(\underline{m})=0=q_{r}(\underline{t})(\mu_{r}+|\mu|-r+t_{r}+1)$
imply that
$$\mu_{r}+|\mu|-r+t_{r}+1=0=\mu_{r}+|\mu|-r+m_{r}.\eqno(3.80)$$
Obviously, $1 \leq m_{r}=t_{r}+1 \leq |\underline{m}|=k+1$ and $\underline{m}\in I(\mu,k+1)'\subset I(\mu,k+1) $ imply that $$ m_{r}\leq
\mu_{r-1}-\mu_{r}\eqno(3.81)$$ by (2.11).
Assume $1 \leq m_{r}< k$. Then by (2.11) and (3.81), we know
there exists some $\underline{l}\in {\mathbb{N}}^{n}$ satisfying
$l_{r}=m_{r}$ and $\underline{l}\in I(\mu,k) $. So
$q_{r}(\underline{l})=q_{r}(\underline{m})=0$. Furthermore,
$q_{\underline{l}}=0$, which contradicts (3.75). Therefore,
$m_{r}=k+1$, i.e. $\underline{m}=(k+1)\epsilon_{r}$.

Suppose there exists another $ (k+1)\varepsilon_{s}\in I(\mu,k+1)'$ but $s \neq r$. Then
$0=\mu_{r}+|\mu|-r+k+1=\mu_{s}+|\mu|-s+k+1$. This is impossible, since $\mu_{s}\geq \mu_{r}$ whenever $s< r$. Thus we prove that $|I(\mu,k+1)'|=1$, i.e.
$\bar{L}_n$-module ${\cal{R}}_{{\la}k+1\ra}$ is irreducible.\hspace{1cm} $\Box$\vspace{0.3cm}

Set$$I_{s}=\{1\}\bigcup\{j \in \overline{2,n} \ | \
\mu_{j-1}-\mu_{j} \geq s\}.\eqno(3.82)$$
By the above two lemmas, we can  give the proof of (i) in the Main Theorem:\vspace{0.3cm}

{\bf Proposition 3.3.3} \quad {\it The vector space ${\cal
A}\otimes_{\mathbb{F}}V$ is an irreducible $sl(n+1)$-module if and
only if
$$\forall \  1 \leq s \in \mathbb{N}, \ \mu_{i}+|\mu|-i+s \neq 0
\qquad\mbox{for\ any}\;\;  i\in I_{s}.\eqno(3.83)$$} \vspace{0.1cm}

{\it Proof }
Assume that there exist $1 \leq s \in \mathbb{N} $ and $i\in I_{s}$
satisfying
 $\mu_{i}+|\mu|-i+s  = 0 $. Then $V(\mu+s\varepsilon_{i})\subseteq {\cal{R}}_{{\la}s\ra}$ by (2.11), (3.71), (3.82) and
 Lemma 3.3.1. Hence, $({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}s\ra}\neq (U(P)(1\otimes
V))_{{\la}s\ra} $, i.e. $sl(n+1)$-module ${\cal
A}\otimes_{\mathbb{F}}V$ is reducible.

Suppose that $sl(n+1)$-module ${\cal A}\otimes_{\mathbb{F}}V$ is
reducible. Let $k$ satisfying (3.72) and (3.73). By Lemma 3.3.2, we
have $V(\mu+(k+1)\epsilon_{s})={\cal{R}}_{{\la}k+1\ra}$ for some $s
\in I_{k+1}$. So Lemma 3.2.3 implies $\mu_{s}+|\mu|-s+k+1 = 0
$.\hspace{1cm} $\Box$\vspace{0.3cm}

{\bf Remark 3.3.4}\quad The condition (3.83) is equivalent  to (1.7) and
(1.8) given in the Main Theorem.\vspace{0.3cm}

In the rest of this section, we study the relationship between any
$\bar{L}_n$-module ${\cal{R}}_{{\la}j\ra}$ and  $\bar{L}_n$-module
${\cal{R}}_{{\la}j+1\ra}$  for any $j\geq k+1$ based on the
decomposition of tensor  module and the projection operator
techniques for $gl(n)$ (Recall Lemma 2.1.3 and projection operators
appeared in Section 2.2). And we finally prove the irreducibility of
quotient module $({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}} V)$ (cf. Proposition 3.3.8)
.\vspace{0.3cm}

{\bf  Lemma 3.3.5} {\it \quad Let $k+1\leq s\in\mathbb{N}$. For
$\nu=\mu+\underline{l}$ with $\underline{l} \in I(\mu,s+1)'$, there
exist $ \nu'=\mu+\underline{m}$ with  $\underline{m} \in I(\mu,s)'$
and $r \in \overline{1,n}$ such that $\nu=\nu'+\varepsilon_{r}$ and
$\mu_{r}+|\mu|-r+m_{r}+1\neq 0$. }\vspace{0.3cm}

{\it Proof}\quad Suppose $\nu=\mu+\underline{l}$ with
$\underline{l}\in I(\mu,s+1)'$. Then
$q_{\underline{l}}=0$.\vspace{0.1cm}

 {\it Claim. There exists $r \in  \overline{1,n}$ such that $\mu_{r}+|\mu|-r+l_{r}\neq 0$ and $l_{r}\geq 1$.}\vspace{0.1cm}

Set $$I_{\underline{l}}=\{t \in \overline{1,n} \ | \
\mu_{t}+|\mu|-t+l_{t}=0\}.\eqno(3.84)$$ It follows that
$|I_{\underline{l}}|=0,1$. Otherwise,
$\mu_{t}+|\mu|-t+l_{t}=0=\mu_{q}+|\mu|-q+l_{q}$ for some $t< q$.
This is impossible because $\mu+\underline{l}$ is a highest weight
implies that $l_{t}+\mu_{t}\geq l_{q}+\mu_{q}$ whenever $t< q$. It
is obvious that the claim holds when $|I_{\underline{l}}|=0$.

Now assume $|I_{\underline{l}}|=1$ and $r_{0}\in I_{\underline{l}}$. If the claim does not hold, then
$\underline{l}=(s+1)\varepsilon_{r_{0}}$ and $\mu_{t}+|\mu|-t \neq 0$ for any $t \neq r_{0}$.
 From Lemma 3.3.2, we know ${\cal{R}}_{{\la}k+1\ra}=V(\mu+(k+1)\varepsilon_{t})$ for some $t \in I_{k+1}$.
 Thus, Lemma 3.2.3 implies $q_{(k+1)\epsilon_{t}}=\prod\limits_{i=1}^{k+1}(\mu_{t}+|\mu|-t+i)=0$. Assume
  $\mu_{t}+|\mu|-t+r=0$ for some $r \in \overline{1,k+1}$. On the other hand,
  $\mu_{r_{0}}+|\mu|-r_{0}+s+1=0$ by (3.84) due to $r_{0}\in I_{\underline{l}}$. Hence, we have $\mu_{r_{0}}-\mu_{t}=r_{0}+r-(t+s+1)$. If $r_{0} \leq t$,
  then $\mu_{r_{0}}-\mu_{t} \geq 0$; which contradicts $r_{0}+r-(t+s+1)= r_{0}-t+r-(s+1)< 0$. If $r_{0} > t$, then
  $\mu_{t}+l_{t}=\mu_{t} \geq \mu_{r_{0}}+l_{r_{0}}=\mu_{r_{0}}+s+1$ because $\mu+\underline{l}=\mu+(s+1)\varepsilon_{r_{0}}$ is
  a highest weight. Therefore, $\mu_{t}-\mu_{r_{0}}\geq s+1$; i.e. $t+s+1-(r_{0}+r)\geq s+1$. Hence, $r_{0}-t+r \leq 0$. A contradiction arises. Thus  the claim holds.

Suppose that $r$ satisfies the claim. Take $\nu'=\nu-\varepsilon_{r}, \
\underline{m}=\underline{l}-\varepsilon_{r}$.
 We claim that $\underline{m}\in I(\mu,s)'$. In fact,  $\underline{m}\in I(\mu,s)$ by (2.11)
 because $l_{r}-1\leq s$ and $l_{r}\leq \mu_{r-1}-\mu_{r}$ implies $ l_{r}-1 < \mu_{r-1}-\mu_{r}$. Furthermore,
 $q_{\underline{l}}=(\mu_{r}+|\mu|-r+l_{r})q_{\underline{m}}=0$ implies that $q_{\underline{m}}=0$. Therefore, $\underline{m}\in I(\mu,s)'$. Thus the Lemma follows.
\hspace{1cm} $\Box$\vspace{0.3cm}

{\bf Lemma 3.3.6} {\it \quad We have
$\partial_{l}({\cal{R}}_{{\la}j\ra})\not\subseteq (U(P)(1\otimes
V))_{{\la}j-1\ra}$ for any $l \in \overline{1,n}$ and $j >
k+1.$}\vspace{0.3cm}

{\it Proof}\quad Assume
 $$\partial_{l}({\cal{R}}_{{\la}j\ra})\subseteq (U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j-1\ra}\ \mbox{for \ some}   \ l \in
 \overline{1,n}\;\mbox{and}\;
  j > k+1.\eqno(3.85)$$
By (3.11), we know that
$$\partial_{l}(U(P)(1\otimes_{\mathbb{F}} V)_{{\la}j\ra})\subseteq
(U(P)(1\otimes_{\mathbb{F}} V))_{{\la}j-1\ra}.\eqno(3.86)$$ Then (3.85), (3.86) and Lemma 3.3.1 imply that
 $$\partial_{l}(({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra})\subseteq (U(P)(1\otimes
 V))_{{\la}j-1\ra},\eqno(3.87)$$ that is,
$$\partial_{l}(({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra})
\varsubsetneq ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j-1\ra}.\eqno(3.88)$$
This contradicts the fact
$$\partial_{l}(({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}) = ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j-1\ra}\;\; \ \mbox{for any} \ l \in
\overline{1,n}. \eqno(3.89)$$Thus the lemma follows.\hspace{1cm}
$\Box$\vspace{0.3cm}

Let $\{e_{1}',\cdots,e_{n}'\}$ be a basis for
$\overline{L}_{n}$-module $V(\varepsilon_{1})$. For any $1 \leq j
\in \mathbb{N}$, we define the following linear map:
$$T_{j} :V(\varepsilon_{1})\otimes_{\mathbb{F}}({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}\rightarrow({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j+1\ra}$$$$T_{j}(e_{i}'\otimes
v)=p_{i}.v, \ \forall \  v \in ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}.\eqno(3.90)$$

{\bf Lemma 3.3.7 } \quad {\it The linear map  $\  T_{j} \ $  is an intertwining  operator from the tensor module $V(\varepsilon_{1})\otimes_{\mathbb{F}}({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}$ to  $({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}j+1\ra}$ for $\overline{L}_{n}$. }\vspace{0.3cm}

{\it Proof}\quad For any $s,t,i \in \overline{1,n}$ and $v \in
({\cal A}\otimes_{\mathbb{F}}V)_{{\la}j\ra}$, we have
\begin{eqnarray*}& & T_{j} (x_{s}\partial_{x_{t}}.(e_{i}'\otimes v))\\&=&T_{j} (x_{s}\partial_{x_{t}}(e_{i}')\otimes v+e_{i}'\otimes x_{s}\partial_{x_{t}}.v)=T_{j}(\delta_{t,i}e_{s}'\otimes v+e_{i}'\otimes x_{s}\partial_{x_{t}}.v)\\&=&\delta_{t,i}p_{s}.v+p_{i}.x_{s}\partial_{x_{t}}.v
=x_{s}\partial_{x_{t}}.p_{i}.v=x_{s}\partial_{x_{t}}.T_{j}(e_{i}'\otimes
v)\hspace{5.1cm}(3.91)
\end{eqnarray*}
by (3.21). Thus the lemma follows.\hspace{1cm} $\Box$\vspace{0.3cm}

Based on the Lemma 3.3.5, Lemma 3.3.6 and Lemma 3.3.7, we can prove (ii) of the Main Theorem in the following:\vspace{0.3cm}

{\bf Proposition 3.3.8 } {\it If ${\cal A}\otimes_{\mathbb{F}}V \neq
U(P)(1\otimes_{\mathbb{F} } V)
 $, then
 $\{0\} \subset U(P)(1\otimes_{\mathbb{F}} V) \subset {\cal A}\otimes_{\mathbb{F}}V$ is a Jordan-Holder Series for
  $L_{n+1}$-module ${\cal A}\otimes_{\mathbb{F}}V$.}\vspace{0.3cm}

 {\it Proof} \quad
Suppose  that $W+U(P)(1\otimes_{\mathbb{F}} V)$ is any nonzero
submodule of quotient module $({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}} V)$, where $W=\bigoplus\limits_{j\geq
k+1}W\bigcap {\cal{R}}_{{\la}j\ra}$ is a weighted subspace of ${\cal
A}\otimes_{\mathbb{F}}V$. By Lemma 3.3.6, we get $W\bigcap
{\cal{R}}_{{\la}k+1\ra}\neq \{0\}$.

By Lemma 3.3.5, we know that for any $s \geq k+1$ and
$\nu=\mu+\underline{l}$ with $\underline{l} \in I(\mu,s+1)'$, there
exist $ \nu'=\mu+\underline{m}$ with $\underline{m} \in I(\mu,s)'$
and $r \in \overline{1,n}$ such that $\nu=\nu'+\varepsilon_{r}$ and
$\mu_{r}+|\mu|-r+m_{r}+1\neq 0$. Therefore, the highest weight
module $V(\nu)\; (\subseteq {\cal{R}}_{{\la}s+1\ra} )$ of highest
weight $\nu$ appears in the decomposition of
$\overline{L}_{n}$-tensor module
$V(\varepsilon_{1})\otimes_{\mathbb{F}}V(\nu')$ ($ \subseteq
V(\varepsilon_{1})\otimes_{\mathbb{F}}{\cal{R}}_{{\la}s\ra}
\subseteq V(\varepsilon_{1})\otimes_{\mathbb{F}}({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}s\ra} $). \vspace{0.3cm}

{\it Claim. \quad  There exists some maximal vector $v_{\nu}$ (resp.
$\xi_{\underline{m}+\varepsilon_{r}}$ ) of highest weight module
$V(\nu)\subseteq V(\varepsilon_{1})\otimes_{\mathbb{F}}V(\nu') $
(resp. $V(\nu) \subseteq {\cal{R}}_{{\la}s+1\ra} $ ) satisfying
$$T_{s}(v_{\nu})=(\mu_{r}+|\mu|-r+m_{r}+1)\xi_{\underline{m}+\varepsilon_{r}}\neq
0.\eqno(3.92)$$} Assume that $w_{\nu}$ is a maximal vector of
irreducible module $V(\nu)\subseteq
V(\varepsilon_{1})\otimes_{\mathbb{F}}V(\nu')$. Since
$T_{s}:V(\varepsilon_{1})\otimes_{\mathbb{F}}({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}s\ra}\rightarrow({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}s+1\ra}$ is an intertwining  operator
for $\overline{L}_{n}$, we know  $T_{s}(w_{\nu})$ is also a maximal
vector of $\overline{L}_{n}$-module $V(\nu)\subseteq
{\cal{R}}_{{\la}s+1\ra} \;( \subseteq ({\cal
A}\otimes_{\mathbb{F}}V)_{{\la}s+1\ra})$. Since the  maximal vector
$w_{\nu}$ must take the following form:
$$w_{\nu}=a_{r}e_{r}'\otimes \varpi_{\nu'}+
\sum_{j=1}^{r-1}\sum_{i=1}^{m(\nu'-\varepsilon_{j}+\varepsilon_{r})}
a_{j}^{i}e_{j}'\otimes
v_{\nu'-\varepsilon_{j}+\varepsilon_{r}}^{i},\qquad 0 \neq a_{r}, \
a_{j}^{i} \in \mathbb{F},\eqno(3.93)$$ where $\varpi_{\nu'}$ is a
maximal vector of the highest weight module $V(\nu')$ and the set
$\{v_{\nu'-\varepsilon_{j}+\varepsilon_{r}}^{i}\mid
i\in\ol{1,m(\nu'-\varepsilon_{j}+\varepsilon_{r}))}\}$ is a basis
 of the weight subspace
 $[V(\nu')]_{\nu'-\varepsilon_{j}+\varepsilon_{r}}$.
 For
convenience, we write
$$v_{\nu'-\varepsilon_{j}+\varepsilon_{r}}^{i}=\sum\limits_{ \varepsilon_{j}-\varepsilon_{r}=\sum\limits_{k=1}^{m}i_{k}
(\varepsilon_{s_{k}}-\varepsilon_{t_{k}});\: s_{i}> t_{i}
}a_{\underline{s},\underline{t}}^{\underline{i}}(x_{s_{1}}
\partial_{t_{1}})^{i_{1}}.\cdots.(x_{s_{m}}\partial_{t_{m}})^{i_{m}}.
\varpi_{\nu'}.\eqno(3.94)
$$
Therefore,$$T_{s}(w_{\nu})=a_{r}p_{r}.\varpi_{\nu'}+
\sum_{j=1}^{r-1}\sum_{i=1}^{m(\nu'-\varepsilon_{j}+\varepsilon_{r})}a_{j}^{i}p_{j}.
v_{\nu'-\varepsilon_{j}+\varepsilon_{r}}^{i}.\eqno(3.95)$$So (3.94) and (3.95) imply that
$$T_{s}(w_{\nu})=n_{\varepsilon_{r}}.\varpi_{\nu'} \qquad  \mbox{for \ some} \  n_{\varepsilon_{r}} \in U(P\oplus \bar{L}_{n}).\eqno(3.96)$$
Let $\varpi_{\nu'}={\xi}_{\underline{m}}$ (resp.
$\varpi_{\nu'}=\hat{\xi}_{\underline{m}}$ ) in (3.96), where
$$\xi_{\underline{m}}=a_{\underline{m}}x^{{\underline{m}}}\otimes
v_{\mu}+ \sum\limits_{(\eta,\underline{l})\neq (\mu, \underline{m}
), {\underline{m}}+\mu={\underline{l}}+\eta, \ i \in
\overline{1,m({\eta})}}a_{\eta,\underline{l}}^{i}x^{\underline{l}}\otimes
v_{\eta}^{i},\qquad 0 \neq a_{\underline{m}},  \
a_{\eta,\underline{l}}^{i} \in \mathbb{F};\eqno(3.97)$$
$$\hat{\xi}_{\underline{m}}=a_{\underline{m}}p^{{\underline{m}}}.(1\otimes v_{\mu})+
\sum\limits_{(\eta,\underline{l})\neq (\mu, \underline{m} ),
{\underline{m}}+\mu={\underline{l}}+\eta, \ i \in
\overline{1,m({\eta})}}a_{\eta,\underline{l}}^{i}p^{\underline{l}}.(1\otimes
v_{\eta}^{i}),\qquad 0 \neq a_{\underline{m}},  \ a_{\eta,\underline{l}}^{i}
\in \mathbb{F}.\eqno(3.98)$$
Take $$\hat{\xi}_{\underline{m}+\varepsilon_{r}}=n_{\varepsilon_{r}}.
\hat{\xi}_{\underline{m}}.\eqno(3.99)$$
Then by Lemma 3.2.3, we have$$\hat{\xi}_{\underline{m}+\varepsilon_{r}}=q_{\underline{m}+\varepsilon_{r}}
{\xi}_{\underline{m}+\varepsilon_{r}}=
n_{\varepsilon_{r}}.
\hat{\xi}_{\underline{m}}=q_{\underline{m}}n_{\varepsilon_{r}}.
{\xi}_{\underline{m}}.\eqno(3.100)$$Hence, we have
$$(\mu_{r}+|\mu|-r+m_{r}+1){\xi}_{\underline{m}+\varepsilon_{r}}=
n_{\varepsilon_{r}}. {\xi}_{\underline{m}}.\eqno(3.101)$$ Now we
take
\begin{eqnarray*}v_{\nu}&=&
\sum_{j=1}^{r-1}\sum_{i=1}^{m(\nu'-\varepsilon_{j}+\varepsilon_{r})}\sum\limits_{
\varepsilon_{j}-\varepsilon_{r}=\sum\limits_{k=1}^{m}i_{k}
(\varepsilon_{s_{k}}-\varepsilon_{t_{k}});\: s_{i}> t_{i}
}a_{j}^{i}a_{\underline{s},\underline{t}}^{\underline{i}}e_{j}'\otimes(x_{s_{1}}
\partial_{t_{1}})^{i_{1}}.\cdots.(x_{s_{m}}\partial_{t_{m}})^{i_{m}}.
{\xi}_{\underline{m}}\\ &
&+a_{r}e_{r}'\otimes
{\xi}_{\underline{m}}.\hspace{11.2cm}(3.102)\end{eqnarray*}
Then $ T_{s}(v_{\nu})=(\mu_{r}
+|\mu|-r+m_{r}+1)\xi_{\underline{m}+\varepsilon_{r}}\neq 0$ by
(3.101) and (3.102). Therefore, (2.30) and (3.92) imply that
$$\{0\}\neq
(T_{s}|_{V(\varepsilon_{1})\bigotimes_{\mathbb{F}}V(\nu')}\circ
\tilde{P}_{r})(V(\varepsilon_{1})\otimes_{\mathbb{F}}V(\nu'))=V(\nu),\eqno(3.103)$$
where $$\tilde{P}_{r}=\prod\limits_{l\neq
r}(\frac{\tilde{M}-\tilde{d}_{l}}{\tilde{d}_{r}-\tilde{d}_{l}}), \
\tilde{d}_{i}=i-1-m_{i}-\mu_{i},\eqno(3.104)$$ and  $\tilde{M}$ is
the matrix in (2.27). Hence, ${\cal{R}}_{{\la}s\ra} \subseteq W$ for
any $s \geq k+1$.
 Thus we prove  $({\cal A}\otimes_{\mathbb{F}}V) / U(P)(1\otimes_{\mathbb{F}} V)$ is
irreducible.\hspace{1cm} $\Box$ \psp

Suppose ${\cal{R}}_{{\la}k+1\ra}=V(\mu+(k+1)\varepsilon_{r})$ for
some $r \in I_{k+1}$. From the proof of Proposition 3.3.8, we know $({\cal
A}\otimes_{\mathbb{F}}V) / U(P)(1\otimes_{\mathbb{F}} V)$ is
generated by
$\bar{\xi}_{\mu+(k+1)\varepsilon_{r}}={\xi}_{\mu+(k+1)
\varepsilon_{r}}+U(P)(1\otimes_{\mathbb{F}}
V)$, i.e. $({\cal A}\otimes_{\mathbb{F}}V) /
U(P)(1\otimes_{\mathbb{F}}
V)=U(L_{n+1}).\bar{\xi}_{\mu+(k+1)\varepsilon_{r}}$. Hence, we get
the following result:\vspace{0.3cm}

{\bf Proposition 3.3.9}\quad {\it The  $L_{n+1}$-module $({\cal
A}\otimes_{\mathbb{F}}V) / U(P)(1\otimes_{\mathbb{F}} V)$  is
isomorphic to the irreducible module $ U(P)(1\otimes_{\mathbb{F}}
M)$, where $M$ is a $gl(n)$-irreducible module admitting the
character $\chi_{\mu+(k+1)\varepsilon_{r}}$.}\vspace{0.3cm}

{\it Proof} \quad Since $\partial_{i}.
{\xi}_{\mu+(k+1)\varepsilon_{r}} \in (U(P)(1\otimes
V))_{{\la}k\ra}=({\cal A}\otimes_{\mathbb{F}}V)_{{\la}k\ra} $, we
get $\partial_{i}. \bar{\xi}_{\mu+(k+1)\varepsilon_{r}}=0$ for any
$i \in \overline{1,n}$. It follows from Proposition 3.3.8 that both
$({\cal A}\otimes_{\mathbb{F}}V) / U(P)(1\otimes_{\mathbb{F}} V)$
and $ U(P)(1\otimes_{\mathbb{F}} M)$ are irreducible
$L_{n+1}$-modules. So it is easy to verify that
$$\vartheta:({\cal A}\otimes_{\mathbb{F}}V) / U(P)(1\otimes_{\mathbb{F}} V)\rightarrow U(P)(1\otimes_{\mathbb{F}} M);
x.\bar{\xi}_{\mu+(k+1)\varepsilon_{r}}\mapsto x.(1 \otimes
v_{\mu+(k+1)\varepsilon_{r}})\eqno(3.105)$$ is an $L_{n+1}$-module
isomorphism, where $x \in U(P\oplus\bar{L}_{n})$ and
$v_{\mu+(k+1)\varepsilon_{r}}$ is a maximal vector of
$M$.\hspace{1cm} $\Box$ \psp

From Proposition 3.3.3, Proposition 3.3.8 and Proposition 3.3.9, we get the Main Theorem.\vspace{0.3cm}

{\bf Remark 3.3.10}\quad The irreducible module $U(P)(1\otimes_{\mathbb{F}} V)$ is cyclic, i.e. it is generated by one vector. From Lemma 3.1.1 and (3.12), we know that $U(P)(1\otimes_{\mathbb{F}} V)$ is in general not a highest weight module. We can easily verify the following result from the Main Theorem:\vspace{0.3cm}

{\bf Corollary 3.3.11}\quad {\it Assume one of the conditions in (1.7) and (1.8) fails. Denote $$i_{0}=\mbox{min}\{i\in \overline{1,n} \ | \ \mu_{i}+|\mu|-i+1 \in -\mathbb{N}\}.\eqno(3.106)$$ We have:

(i) The integer $k=-\mu_{i_{0}}-|\mu|+ i_{0}-1$ satisfies  (3.72) and (3.73).

(ii) The irreducible module $U(P)(1\otimes_{\mathbb{F}} V)$ is finite dimensional highest weight module with highest weight $k\omega_{1}+\sum\limits_{i=2}^{n}m_{i-1}\omega_{i}$ iff  $ \ i_{0}=1$, where $m_{i}=\mu_{i}-\mu_{i+1}$ for $i \in \overline{1,n-1}$.

}\vspace{0.3cm}


TITLE: Why does casting out $9$ and $11$ work to compute remainders?
QUESTION [2 upvotes]: I found two examples that I'd like to understand:
Example 1: Find $112358132134 \mod 9$.
Solution is:
$1+1+2+3+5+8+1+3+2+1+3+4 \mod 9 = 7 \mod 9 = 7$
Example 2: Find $62831853 \mod 11$.
Solution is:
$(-6)+2+(-8)+3+(-1)+8+(-5)+3 \mod 11 = -4 \mod 11 = 7$
My question is: Based on what property we sum the digits of the number ( as Example 1) or subtract (as Example 2)?

REPLY [0 votes]: Remainder after division by $11$
The OP's question asked for the properties that 'force' the technique. They should be aware that there are often several ways to tackle math problems.
Working in $\Bbb N = \{0,1,2,3,\dots\}$ with numbers presented in their base ten representation.
Problem: For any $m$ find the remainder $r$ after division by $11$: $\;m = 11q + r$.
Without using modular arithmetic and negative numbers, you can employ symbolic homotopy rules to transform $m$ down to $r$. Since this is recreational math, we leave it to the reader to formalize these rules; here we only give some examples.
This method is useful for an organized approach where you can carefully check each step for errors.
Example 1: If $m \lt 11$ then $m \sim [m]$
Example 2: $20]. \sim [9]$
Example 3: $30]. \sim [8]$
Example 4: $90]. \sim [2]$
Example 5: $55] \sim [0]$
Example 6: $58] \sim [3]$
Example 7: $85] \sim 30]. \sim [8]$
Example 8: Find the remainder $r$ for $\frac{62831853}{11}$:
$\quad 62]831853 \sim 408]31853 \sim 12]31853 \sim 13]1853 \sim 21]853 \sim 108]53 \sim 95]3 \sim 403] \sim [7]$
Example 9: $29]0 \sim 70]. \sim [4]$
Example 10: $100]0000 \sim 100]00] \sim 100] \sim [1]$
Example 11: $11]11111 \sim 11]111 \sim 11]1 \sim [1]$


TITLE: Does the distance map send closed sets to closed sets?
QUESTION [1 upvotes]: Let $(X,d)$ be a metric space and $x_0$ be an element in $X$. Define $f:X \to \mathbb R$ by $f(x)=d(x_0,x)$. Does this map send closed sets to closed sets? Further, let $C$ be a closed subset of $X$. Is $d(x_0,C) \in f(C)$?

REPLY [4 votes]: No.  Let X = $\{-1\} \cup \{1/n : n \text{ a positive integer} \}$, let $C = \{ 1/n : n \text{ a positive integer } \}$,  $x_0 = -1$.
$X$ with the inherited metric is a discrete space. $C$ is a closed subset of $X$, but $f(C) = \{ 1 + 1/n : n \text{ a positive integer } \}$ isn't a closed subset of $\mathbb R$. Further, $d(-1, C) = 1 \notin f(C)$.


TITLE: Generating multivariate Gaussian samples--Why does it work?
QUESTION [2 upvotes]: I came across the method for generating multivariate normal samples on wikipedia:
https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Drawing_values_from_the_distribution

A widely used method for drawing (sampling) a random vector $X$ from the $N$-dimensional multivariate normal distribution with mean vector $μ$ and covariance matrix $Σ$ works as follows:[28]

Find any real matrix $A$ such that $AA^T = Σ$. When $Σ$ is positive-definite, the Cholesky decomposition is typically used, and the extended form of this decomposition can always be used (as the covariance matrix may be only positive semi-definite) in both cases a suitable matrix $A$ is obtained. An alternative is to use the matrix $A = UΛ^½$ obtained from a spectral decomposition $Σ = UΛU^T$ of $Σ$. The former approach is more computationally straightforward but the matrices $A$ change for different orderings of the elements of the random vector, while the latter approach gives matrices that are related by simple re-orderings. In theory both approaches give equally good ways of determining a suitable matrix $A$, but there are differences in computation time.

Let $Z = (z_1, …, z_N)^T$ be a vector whose components are $N$ independent standard normal variates (which can be generated, for example, by using the Box–Muller transform).

Let $X$ be $μ + AZ$. This has the desired distribution due to the affine transformation property.

Why does the cholesky decomposition matrix '$A$' multiplied by the vector of samples chosen from the standard normal distribution '$Z$' plus '$μ$' give us our result (ie $X = μ + AZ$)?
Why does this work? What is the proof?

REPLY [3 votes]: Simply take the vector you have generated, $\boldsymbol{x}$, and compute its covariance: $$\mathbb E[(\boldsymbol{x}-\boldsymbol{\mu})(\boldsymbol{x}-\boldsymbol{\mu})^T] = \mathbb E[\boldsymbol A\boldsymbol z\boldsymbol z^t\boldsymbol A^t] =  \boldsymbol A\mathbb E[\boldsymbol z\boldsymbol z^t]\boldsymbol A^t = \boldsymbol A \boldsymbol I \boldsymbol A = \boldsymbol A \boldsymbol A^T = \boldsymbol\Sigma,$$
as desired.


TITLE: Modular Arithmetic: Congruence System
QUESTION [0 upvotes]: How to solve this congruence system:
        \begin{cases}
x≡2 &\bmod3 \\
     2x≡1& \bmod5 \\
     3x≡3 &\bmod6
\end{cases}

REPLY [1 votes]: $x\equiv 2 \pmod{3}$ is equivalent to say $x=2+3k$ for some $k\in \mathbb{Z}$. Substitute this into the second congruence and you have
$$2(2+3k)\equiv 1 \pmod{5}$$
or
$$k\equiv 2 \pmod{5}$$
Which means $k=2+5\ell$ and that $x=2+3(2+5\ell)=8+15\ell$ for $\ell\in \mathbb{Z}$. Finally, put this into the last congruence and then
$$3(8+15\ell)\equiv 3 \pmod{6}$$
which is simplified as
$$\ell\equiv1 \pmod{2}$$
or well $\ell=1+2t$ for some integer $t$. Therefore $x=8+15(1+2t)=23+30t$ or equivalently
$$x\equiv 23 \pmod{30}$$
and this is the solution.


TITLE: For what $n$ is $\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ an integer?
QUESTION [0 upvotes]: For what $n$ is $\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ an integer?

I tried setting $a=\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}$ , $b=\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ and playing with their squares/fourth powers, but couldn't find anything.

REPLY [5 votes]: If you set the expression equal to $x$ and square the equation, you get $$x^2=25+2\sqrt{n}.$$ This is equivalent to $$n=\left(\frac{x^2-5^2}{2}\right)^2.$$ So just pick $x$ equal to an odd integer and then this above expression is $n.$
As Crostul points out, in order for the original expression to be well-defined, we need $n$ to be less than or equal to $\frac{625}{4}=156.25.$ And the fact that $n$ is a square means that it must be non-negative.


text:
\begin{document}

\begin{frontmatter}

\dochead{}

\title{Severi's theorem for $d$-uple Veronese varieties\\Le th\'eor\`eme de Severi pour les vari\'et\'es $d$-uple de Veronese }

\author[JS]{Jeroen Schillewaert \fnref{JJ}}
\ead{jschillewaert@gmail.com}
\ead[JS]{www.jeroenschillewaert.com}
\fntext[JJ]{Research of the first author supported by Marie Curie IEF grant GELATI (EC grant nr 328178)}

\author[KS]{Koen Struyve \fnref{KK}}
\ead{kstruy@gmail.com}
\ead[KS]{http://cage.ugent.be/~kstruyve/}
\fntext[KK]{Research of the second author supported by the Fund for Scientific Research -- Flanders (FWO - Vlaanderen)}

\address[JS]{Department of Mathematics, Imperial College, London, United Kingdom}
\address[KS]{Department of Mathematics, Ghent University, Ghent, Belgium}

\begin{abstract}
We characterize $d$-uple Veronese embeddings of finite-dimensional projective spaces. The easiest non-trivial instance of our theorem is the embedding of the projective plane in 5-dimensional projective space, a result obtained in 1901 by Severi when the underlying field is complex.\vspace{1mm}

\flushleft{\normalsize \bf Resum\'e}\\
Le plongement de Veronese $d$-uple d'un espace projectif de dimension finie est caracteris\'e. L'exemple le plus simple et non-trivial de notre th\'eor\`eme est le plongement
d'un plan projectif dans un espace projectif de dimension 5, un r\'esultat obtenu par Severi en 1901 si le corps sous-jacent est le corps des nombres complexes.
\end{abstract}

\begin{keyword}

Embedding \sep Veronese variety \sep Rational Normal Curve

\MSC 51M35 \sep 14N05 \sep14M17
\end{keyword}

\end{frontmatter}

\section{Introduction}
In 1901 Severi \cite{Severi} characterized the Veronese surface as the only smooth irreducible and non-degenerate projective surface in a 5-dimensional complex projective space which can be projected isomorphically into a 4-dimensional complex projective space.

In 1984, Mazzocca and Melone \cite{Maz-Mel:84} used three simple axioms (the MM-axioms) to characterize the ordinary quadric Veronese variety over finite fields of odd order. This was extended to all finite fields by Hirschfeld and Thas \cite{Hir-Tha:91} and to arbitrary fields by Schillewaert and Van Maldeghem \cite{SVM2}. Our main result extends Severi's original characterization \cite{Severi} to arbitrary finite degree $d$ and arbitrary fields of cardinality at least $d^{\frac{3}{2}}$ using a generalization of the MM-axioms, as Severi's conditions imply the MM-axioms \cite{LVDV,Zak}.

\subsection{Definitions}\label{section:defs}

The \emph{Veronese embedding} $\phi_{n,d}:\mathbb{P}^n(\K)\to \mathbb{P}^{{n+d \choose d}-1}(\K)$ maps $(x_0,\cdots,x_n)$ to $(x_0^d,x_0^{d-1}x_1,\cdots,x_n^d)$. The \emph{Veronese variety} $\mathcal{V}_n^{d^n}$ is the image of this map. A \emph{rational normal curve} $C$ in a $d$-dimensional projective space $\Sigma$ is a set of points of $\Sigma$ which is projectively equivalent to $\mathcal{V}_1^{d}$.  For any $x \in C$ one can define a unique {\em tangent line} $T_x(C)$ which is determined by the points of the curve if either $d=2$ or $d \geq | \K |+2$.

Let $V$ be a possibly infinite-dimensional, non-trivial vector space over some field $\K$, and let $\mathbb{P}(V)$ be the corresponding projective space.  Let $X$ be a spanning point set of $\mathbb{P}(V)$ and let $\Xi$ be a collection of  $d$-dimensional subspaces of $\mathbb{P}(V)$, the {\em rational spaces} of $X$, such that, for any $\xi\in\Xi$, the intersection $\xi\cap X$ is a rational normal curve $X(\xi)$ in $\xi$ of dimension $d$  ($d \geq 2$) and then, for $x\in X(\xi)$, we sometimes denote $T_x(X(\xi))$ simply by $T_x(\xi)$. We call $\mathcal{V}=(X,\Xi)$, or briefly $X$, a {\em Veronesean cap of degree $d$} if the following properties (V1), (V2) and (V3) hold.
\begin{itemize}\addtolength{\itemsep}{-0.4\baselineskip}

\item[(V1)] Any two points $x$ and $y$ of $X$ lie in a unique element of $\Xi$, denoted by $[x,y]$.

\item[(V2)] If $\xi_1,\xi_2\in \Xi$, with $\xi_1\neq \xi_2$, then $\xi_1\cap\xi_2\subset X$.

\item[(V3)] For every $x\in X$, every $\xi\in\Xi$, with $x\notin \xi$, $\cup_{y\in X(\xi)} T_x([x,y])$ is a plane $T(x,\xi)$  depending on $x$ and $\xi$.
\end{itemize}

In the complex case, every (possibly singular) non-degenerate curve in $\mathbb{P}^{d}$ of degree $d$ is a rational normal curve \cite[p.121]{Shafarevich}. In the finite case one can sometimes relax the conditions \cite{Casse-Glynn}, but not always \cite{Glynn,Segre}.

\subsection{A Veronese variety is a Veronesean cap}

By Lemma 2.3 of \cite{KantorShult} the images of lines are rational normal curves. Property (V1) is immediate. To verify (V2), by homogeneity, see \cite{KantorShult}, we only need to check two situations, namely $\psi_{1,2}:=\<\phi_{n,d}(L_1)\>\cap \<\phi_{n,d}(L_2)\>$ and if $n\geq 3$ we need to compute also $\psi_{1,3}:=\<\phi_{n,d}(L_1)\>\cap\<\phi_{n,d}(L_3)\>$ where $L_1$ is given by $X_i=0,2\leq i\leq N$, $L_2$ by $X_i=0, i\neq 2,1\leq i\leq N $ and $L_3$ by $X_i=0;0\leq i \leq N-2$. Then a computation yields $\psi_{1,2}:=\phi_{n,d}(L_1\cap L_2)$ and $\psi_{1,3}=\emptyset$.
For (V3), again by homogeneity, consider the point $r:=(0,\dots,0,1)$ and the points $p_1:=(1,0,\dots,0)$ and $p_2:=(0,1,0,\dots,0)$ on $L_1$.
Then a computation yields that the plane spanned by the tangent lines at $\phi_{n,d}(r)$ to $\phi_{n,d}(\langle r,p_1 \rangle)$ and $\phi_{n,d}(\langle r,p_2 \rangle)$
is the requested one.

\section{Statement of the main results}

\begin{prop}\label{tangentspace}
If $X$ is a Veronesean cap, then for every $x\in X$ the set $\cup_{\xi \in\Xi, x \in \xi } T_x(\xi)$ is a subspace of constant dimension.
\end{prop}

We call the dimension of these subspaces the \emph{index}. The next result classifies Veronesean caps of sufficiently large dimension with respect to its degree and index over fields of sufficiently large cardinality with respect to the degree.

\begin{Theorem}\label{Thm:main}
If $X$ is a Veronesean cap of degree $d$ and finite index $n\geq 2$ in a projective space $\mP(V)$ over a field $\K$ such that $\dim V\geq M:= {n+d \choose d}$ and such that $|\K | \geq d^{3/2}$, then $\dim V=M$, and $X$ is projectively equivalent to the Veronese variety $\mathcal{V}_n^{d^n}$ over $\K$.
\end{Theorem}

\begin{remark}\label{rem:ineq}
One can strengthen the bound on the field size in the Main Result to the combination of the bounds $$|\K | \geq d + 2   \mbox{ when $d \neq 2$, and } \frac{|\K|^{n} - 1}{|\K| -1} \geq 2 + \sum_{i=2}^{\min(n-1,d) } (-1)^{i} {d+1 \choose i+1} \frac{|\K|^{n-i} -1}{|\K|-1}.$$ The first bound is necessary as otherwise the tangent lines are not determined by the points of the curve. The second bound becomes irrelevant in view of the first one for $n \leq 2$ or $d \leq 10$.
\end{remark}

\section{Basic properties of rational normal curves}\label{rncprop}
Let $C$ be a rational normal curve in a $d$-dimensional projective space $\Sigma$ as defined in Section~\ref{section:defs} (with $d \geq 2$).

\begin{lemma}
Every $d+1$ points are linearly independent. \cite[Theorem 1.1]{KantorShult}
\end{lemma}

\begin{observ}\label{obs:1}
The rational normal curve $C$ comes equipped with a notion of crossratio on quadruples of points, such that the group $\PGL(2,\K)$ acts three-transitively on $C$ while preserving the crossratio. (See~\cite[Ex. 1.19]{Har:92})
\end{observ}

\begin{lemma}\label{lem:proj1}
Let $x$ be a point of the curve $C$, then a projection from $x$ to a hyperplane not incident with it maps $C \setminus \{x\}$ and the tangent line through $x$ bijectively to the points of a rational normal curve in a $(d-1)$-dimensional projective space. This projection preserves the crossratio on $C$.
\end{lemma}
\proof
By Observation~\ref{obs:1} we may fix $x=(1,0,\dots,0)$ in $\Sigma$. This reduces the problem to an easy calculation.\qed

Repeated application of Lemma~\ref{lem:proj1} shows
\begin{cor}\label{cor:proj1}
Let $x_1, \dots, x_i$ be $i$ pairwise different points of $C$ such that $1 \leq i \leq d-1$. Then a projection from the span of $x_1, \dots, x_i$ to a complementary subspace maps $C \setminus \{x\}$ and the tangent lines through $x_1, \dots, x_i$ bijectively to the points of a rational normal curve in a $(d-i)$-dimensional projective space. This projection preserves the crossratios $C$.
\end{cor}

\begin{observ}\label{obs:2}
By considering 4 points of $C$ and applying Corollary~\ref{cor:proj1} to $d-1$  other points on the curve (which always exist when $| \K | \geq 4$ by the first bound of Remark~\ref{rem:ineq}), one obtains that the crossratio mentioned in Observation~\ref{obs:1} solely depends on the point set of $C$. For $|\K| = 2$ or $3$ this is trivial as there is only one way to define a valid crossratio on 3 or 4 points.
\end{observ}

Using Corollary~\ref{cor:proj1} considering a projection from $\langle C\cap L_1,\cdots,C\cap L_i \rangle$ yields

\begin{cor}\label{cor:proj2}
Let $L_1, \dots, L_i$ and $x_1, \dots x_j$ be respectively $i$ tangent lines on $C$ and $j$ points of $C$, none of which are equal or incident with each other. If $2i +j \leq d+1$, then the span of these lines and points is of dimension $2i +j -1$.
\end{cor}

\section{Proof of the Main Result}

\subsection{The projective space associated with a Veronesean cap}\label{projspace}
We use the same notation as in Section~\ref{section:defs}. Associated with $\cV$ we can consider the geometry $\cP$ having point set $X$ and line set $\Xi$, endowed with the natural incidence.
Then a proof similar to~\cite[Prop 2.2]{ThasHVM} yields

\begin{prop}\label{projplane}
$\cP$ is  a projective space of dimension $n \geq 2$.
\end{prop}

Also the proof of Proposition~\ref{tangentspace} is essentially the same as the proof of~\cite[Prop. 2.4]{ThasHVM}. The index obtained is the same $n$ as in Proposition~\ref{projplane}.

From now on, we denote a point of $\cP$ with a $\bar{.}$, e.g. $\bar{x}$, and the corresponding point in $\mP(V)$ without a $\bar{.}$, e.g. $x$. Similarly we denote subspaces of $\cP$ with a $\bar{.}$, and if $\bar\pi$ is a subspace of $\cP$, then $\pi:=\langle  x \in X \vert \bar{x} \in \bar\pi  \rangle$. The following easy observation will be crucial for induction arguments.

\begin{lemma}\label{lem:ind}
Let $\bar\pi$ be some non-empty subspace of $\cP$, then $(X_\pi, \Xi_\pi)$ with $X_\pi := \{x \vert \bar{x} \in \bar{\pi}\}$, and $\Xi_\pi :=\{\xi \in \Xi \vert  X(\xi) \subset X_\pi\}$, is a Veronesean cap of degree $d$ inside $\pi$, which we call a {\em subcap}.
\end{lemma}

\subsection{Dimensional analysis}

In the remainder of the proof let $(X,\Xi)$ be a Veronesean cap of degree $d$ and index $n$, satisfying the restrictions listed in the Main Result and let $M={n+d \choose d}$.
We assume as induction hypothesis that the Main Result holds for Veronesean caps of index up to $n-1$,  note that the cases of index 0 and 1 are trivial.

\begin{lemma}\label{lem:secants}
Let $\bar\pi_1, \bar\pi_2, \dots \bar\pi_{d+1}$ be a set of hyperplanes of $\cP$ in general position. If $\bar{x}$ is a point of $\cP$ not contained in any of these hyperplanes, then there exist at least two different lines through $x$ such that each point on these lines is contained in at most two of the hyperplanes  $\bar\pi_1, \bar\pi_2, \dots \bar\pi_{d+1}$.
\end{lemma}
\proof
This holds as the right hand side of the second inequality of Remark~\ref{rem:ineq} counts the number of intersection points of at least three of these subspaces (via the inclusion-exclusion principle) plus two, while the left hand side of this inequality counts the number of lines through $x$.
\qed

For the following proposition, note that whenever $| \K | \geq d$ there do exist hyperplanes $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+1}$ in general position, for example the dual of the point set of a rational normal curve in $\cP$.

\begin{prop}\label{prop:bound}
Let $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_i$ ($0 \leq i \leq d+1$) be a number of hyperplanes of $\cP$ in general position which is extendable to a set of $d+1$ hyperplanes in general position. Then
\begin{itemize}\addtolength{\itemsep}{-0.4\baselineskip}
\item[(i)] $\mathrm{codim} \<\pi_1, \pi_2 , \dots, \pi_i \> ={n + d  -i\choose d -i}$.
\item[(ii)]$\dim V=M$.
\item[(iii)] $\pi_i \cap \<\pi_1, \dots,\pi_{i-1}  \>=\< \pi_i \cap \pi_1, \dots, \pi_i \cap \pi_{i-1}  \>$.
\item[(iv)]  If the geometric dimension of $\bar\pi$ is $m$, then the geometric dimension of  $\pi$ is ${m+d \choose d} -1$. In particular  $(X_\pi, \Xi_\pi)$ is a Veronesean cap of degree $d$ and index $m$ satisfying the restrictions listed in the Main Result.
\end{itemize}
\end{prop}
\proof
First we prove $\mathrm{codim} \<\pi_1, \pi_2 , \dots, \pi_i \>  \leq {n + d  -i\choose d -i}$, not using $\dim V \geq M$, by induction on both $n$ and $d-i$, making use of Lemma~\ref{lem:ind}.  The case $n = 1$ follows by Section \ref{rncprop}, so assume $n\geq 2$.

First assume that $d- i = -1$. Let $x$ be any point in $X \setminus (\cup_{j=1}^{d+1} \pi_j  )$, and let $\xi \in \Xi$ be a rational normal space through $x$ such that each $y \in X(\xi)$ is contained in at most two of the subspaces $\pi_i$ (which exists by Lemma~\ref{lem:secants}).
For such a $\xi$, we have that if $y$ is contained in some subspace $\pi_i$, then it is contained in $\<\pi_1, \pi_2 , \dots \pi_i \>$. If it is contained in two of these subspaces, then also its tangent line is contained in $\<\pi_1, \pi_2 , \dots \pi_{d+1} \>$. Hence $\<\pi_1, \pi_2 , \dots \pi_{d+1} \>$ contains $\xi$, and by varying $x$ all of $X$.

Secondly assume that $d-i \geq 0$. In this case we pick a hyperplane $\bar\pi_{i+1}$ of $\cP$ such that $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{i+1}$ are in general position. The codimension of the span $ \<\pi_1, \pi_2 , \dots ,\pi_i \> $ in $\mP(V)$ equals the sum of the codimension $c_1$ of $ \<\pi_1, \pi_2 , \dots, \pi_i \> $ in $ \<\pi_1, \pi_2 , \dots, \pi_{i+1} \> $ and the codimension $c_2$ of $ \<\pi_1, \pi_2 , \dots, \pi_{i+1} \> $ in $\mP(V)$. Then $c_1$ is at most the codimension of $ \<\pi_1\cap \pi_{i+1} , \dots , \pi_i \cap \pi_{i+1}\> $ in $\pi_{i+1}$, which is bounded from above by ${(n-1)+d - i  \choose  d-i }$ by induction on $n$, and $c_2$ is bounded from above by ${n+d - (i + 1) \choose  d-(i+1) }$ by induction on $d-i$. The sum of both is at most ${ n+d - i \choose d-i }$.

Combining the bound for $i=0$ with the restriction $\dim V \geq M$ yields $\dim V = M$ (proving (ii)), and that all bounds above have to be sharp, which proves (i) and (iii). Repeated application of (i) with $i=1$ proves (iv).\qed

\subsection{The projective space $\cP$ is isomorphic to $\mathbb{P}^{n}(\K)$} \label{section:projspace}

As we know the dimension of $\cP$  from Proposition~\ref{projplane} one can use part (iv) of Proposition~\ref{prop:bound} to restrict to the case $n=2$ for the remainder of Section~\ref{section:projspace}.
Remember from Observations~\ref{obs:1} and~\ref{obs:2} that the point sets of the rational normal curves $\xi \cap X$ with $\xi \in \Xi$ come equipped with a natural crossratio on quadruples. By construction of $\cP$ this yields a crossratio on the point sets of lines of $\cP$.  The next step is now to define a crossratio on line pencils of $\cP$.

\begin{observ}
Let  $\bar{x}$ be an arbitrary point of $\cP$ and identify each line $\bar\zeta$ through $\bar{x}$ with the tangent line $T_x(\zeta)$. Hence, by (V3), we obtain a correspondence of lines through $\bar{x}$ with lines in the tangent plane on $x$. The latter, being a line pencil in the projective space $\mP(V)$, comes with a natural notion of a crossratio, which via identification provides a crossratio on the line pencil through $\bar{x}$ in $\cP$.
\end{observ}

The next lemma links these crossratios together.

\begin{lemma}\label{lem:cros} The perspectivity from the line pencil through a point $\bar{x}$ of $\cP$ to the points on a line $\bar\xi$ of $\cP$ not containing $\bar{x}$ preserves the crossratios defined on these.
\end{lemma}
\proof
Extend the line $\bar\xi$ to a set of lines $\{\bar\xi, \bar\xi_2 , \dots \bar\xi_{d+1}  \}$ in general position, such that only $\bar\xi_{d+1}$ contains $\bar{x}$.

Parts (i) and (ii) of Proposition~\ref{prop:bound} imply that the codimension of $\langle  \xi_2, \dots, \xi_{d} \rangle$ is 3. Hence we may consider a projection $\tau$ of $\mP(V)$ from $\langle \xi_2, \dots, \xi_{d} \rangle$ on some two-dimensional subspace disjoint from it. Proposition~\ref{prop:bound} also implies that $\tau$ maps $\xi$ on some line $L$ (as $\langle \xi, \xi_2, \dots, \xi_{d} \rangle$ is of codimension one)  and that  $x$ is mapped to a point disjoint from $L$ (by part (iii) applied to $\xi_{d+1}$ and $\langle \xi, \xi_2, \dots, \xi_{d} \rangle$).

The points of the curve $X(\xi)$ not in $ \xi_2 \cup \dots \cup \xi_{d}$ and the tangent lines on $\xi$ through the remaining points on this curve project to pairwise distinct points of $L$ while preserving the crossratios,  by Corollary~\ref{cor:proj1} and part (iii) of Proposition~\ref{prop:bound}. This Corollary~\ref{cor:proj1} also states that this projection preserves crossratios, hence we have connected the crossratio on $\xi \cap X$ and the crossratio on $L$.

Let $\xi'$ be some rational normal space through $x$. From Corollary~\ref{cor:proj2} and counting the number of  points of $X \cap \xi'$ contained in any of the $\xi, \xi_2 , \dots \xi_{d-1}$, while accounting for the tangent lines of the points that are contained in two of these rational normal spaces (applying (V3)), it follows that the image of $\xi'$ under $\tau$ is at most one-dimensional. As this image contains $\tau(x)$ and a point of $L$ it has to be a line.

In the image of the projection we hence obtain a perspectivity between the line pencil through the point $\bar{x}$ of $\cP$ and the points on $\bar\xi$, preserving the crossratios on both. \qed

This has the following immediate consequence.

\begin{cor}
The induced action on the rational normal curve $X(\xi)$ of the projectivity group of $\cP$ w.r.t. to a line $\bar\xi$ of $\cP$ preserves the natural crossratio on it.
\end{cor}

Let $G$ be the projectivity group of $\cP$ w.r.t. to some line $\bar\xi$ of $\cP$.  Then $G \leq \mathrm{PGL}(2,\K)$ as the latter consists of those permutations preserving the crossratio. As $\mathrm{PGL}(2,\K)$ acts sharply 3-transitive while $G$ is at least 3-transitive, one has $G=\mathrm{PGL}(2,\K)$. Moreover, by von Staudt's theorem \cite{VonStaudt}, this implies that the projective plane $\cP$ is Pappian, hence $\K$ is a field and $\cP$ is isomorphic to $\mathbb{P}^{2}(\K)$.

Let us mention another corollary.

\begin{cor}\label{cor:findtangents}
Let $\bar{x}$ and $\bar\xi$ be a point and line of $\cP$ such that $\bar{x} \notin \bar\xi$. Then the map
\begin{align*}
X(\xi)  \to T(x,\xi) : y  \mapsto  T_x([x,y])
\end{align*}
is completely determined by the curve $X(\xi)$, the point $x$, and the images of at least three points in $X(\xi)$ under this map.
\end{cor}
\proof
By Lemma~\ref{lem:cros} this map preserves crossratios, so  knowing the image of three points suffices to completely determine the map.
\qed

\subsection{Mapping $X$ onto a Veronese variety}

Let $Y$ represent the point set of the projective space $\cP$. By construction of $\cP$ we have a bijective map $\iota: Y \to X: \bar{x} \mapsto x$. If we set $X'$ and $\Xi'$ to be respectively the point set and the set of rational spaces of a Veronese variety $\mathcal{V}^{d^n}_n$ over the field $\K$ with ambient vector space $\mP(V')$, then, by construction of the variety, we also have a bijective map $\iota' : Y \to X'$, i.e. the Veronese embedding.

Our goal is to show that the bijective map $\varphi := \iota' \circ \iota^{-1}$ from $X$ to $X'$ lifts uniquely to a collineation between the ambient projective spaces. We do this by induction on subcaps $Z$ of increasing index $m$.

\subsubsection{The case $m=1$}
Here we have to consider a map $\varphi$ defined on a subcap $Z$ of index 1, so on the points of a certain rational normal curve $X(\xi)$ (with $\xi \in \Xi$). We want to show that $\varphi$ extends uniquely to a collineation defined on the subspace $\xi$ of $\mP(V)$. This is the case if and only if $\varphi$ preserves the crossratios on the rational normal curve up to a possible field automorphism (see for example~\cite[Ex. 1.19]{Har:92}).

We may assume that the index of $X$ is 2 (by using (iv) of Proposition~\ref{prop:bound}), so  that $\cP$ is a projective plane.
As the projectivity group determines crossratios up to a possible field automorphism $\sigma$, and by the fact that this group is a feature of $\cP$ and does not depend on the actual structure of $X$ and $X'$, it is implied that $\varphi$ preserves the crossratio on $X(\xi)$ up to a field automorphism $\sigma$.

\begin{remark} \label{rem:indep}
Moreover, by Lemma~\ref{lem:cros}, the field automorphism $\sigma$ does not depend on the choice of the subcap $Z$.
\end{remark}

\subsubsection{The case $m \geq 2$}
We may suppose without loss of generality that $Z=X$, so $m=n$. Fix $d+1$ hyperplanes $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+1}$ of $\cP$ in general position, obtained by considering the dual of a rational normal curve in $\mP^n(\K)$. The spaces  $\pi_1, \dots, \pi_{d+1}$ span $\mP(V)$ by Proposition~\ref{prop:bound}.

The induction hypothesis allows us to define a collineation $\rho$ mapping the subspace $\pi_1$ to the subspace $\varphi(\pi_1)$, in such a way that $\rho$ extends $\varphi$ on $X \cap \pi_1$. We now extend $\rho$ recursively to a collineation agreeing with $\varphi$ on $X \cap (\cup_{j=1}^{d+1} \pi_j)$. Suppose that we already have defined $\rho$ on the span of $\pi_1$ up to $\pi_i$, agreeing with $\varphi$ on $X \cap (\cup_{j=1}^{i} \pi_j)$ ($1 \leq i \leq d$). Then, in order to extend $\rho$ to a collineation defined on the span up to $\pi_1$ up to $\pi_{i+1}$, it suffices to extend the restriction of $\rho$ on $\pi_{i+1} \cap \<\pi_1, \dots , \pi_i \>$ to $\pi_{i+1}$. This is possible because of the induction hypothesis and (iii) and (iv) of Proposition~\ref{prop:bound} (and using Remark~\ref{rem:indep} in the case $m=2$ to avoid having conflicting field automorphisms). Eventually we obtain a collineation $\rho$ defined on the entirety of $\mP(V)$, agreeing with $\varphi$ on $X \cap (\cup_{j=1}^{d+1} \pi_j)$.

\begin{prop}
The collineation $\rho$ agrees with $\varphi$ on the entirety of  $X$.
\end{prop}
\proof
Let $x$ be any point in $X \setminus \cup_{j=1}^{d+1} \pi_j$. It suffices to show that $x$ is uniquely determined by the points  $X \cap (\cup_{j=1}^{d+1} \pi_j)$.

By Lemma~\ref{lem:secants} we can find two lines $\bar\xi_1$ and $\bar\xi_2$ through $\bar{x}$ such that each point on these lines is contained in at most two of the hyperplanes $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+1}$. If $d\leq 3$ we can pick these such that each point on these lines is contained in at most one such hyperplane.

Assume $d > 3$, and let $y$ be a point on $\xi_1$ contained in two of the hyperplanes, we may suppose these are $\pi_1$ and $\pi_{d+1}$. By the bounds in Remark~\ref{rem:ineq} we have $| \K | \geq d + 2$. Next we reconstruct the tangent line $L:= T_y(\xi_1)$ from the points in  $X \cap (\cup_{j=1}^{d+1} \pi_j)$. We already know the position of $y$ and some plane in which $L$ is contained by property (V3). If $n >2$ we know at least two of these planes which determines $L$ uniquely, so we are left with the case that $n=2$.

As  $| \K | \geq d+ 2$ and by the construction of the lines $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+1}$, there exists some lines $\bar\pi_{d+2}$ and $\bar\pi_{d+3}$ such that $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+3}$ are in general position. As $\bar\pi_{d+2}$ and $\bar\pi_{d+3}$ intersect the $\bar\pi_1, \bar\pi_2, \dots, \bar\pi_{d+1}$ in $d+1$ pairwise different points, the position of the subspaces $\pi_{d+2}$ and $\pi_{d+3}$ is uniquely determined. Also the unique intersection point of both, which we denote by $z$, is  determined.

Applying part (iii) of Proposition~\ref{prop:bound} to  $\pi_1$ and $\pi_{d+2}$ yields that $y,z\notin \langle \pi_{i}\vert 2 \leq i\leq d \rangle$. Part (iii) of Proposition~\ref{prop:bound} applied to $\pi_1$ and $\pi_2, \dots, \pi_{d+1}$ also yields $T_y(\pi_1)\not\subset \langle \pi_{i} \vert 2 \leq i\leq d \rangle$, in particular the projections of the tangent lines $T_y(\pi_1)$ and $T_y(\pi_{d+1})$, and hence all tangent lines in $T(x,\pi_2)$, are pairwise different. Using this projection we can also determine the tangent line $T_y([y,z])$ (this is the unique line in $T(x,\pi_2)$ mapped to the line containing the projections of $y$ and $z$). Corollary~\ref{cor:findtangents} now allows us to determine all tangent lines through $y$, in particular $L$.

By Corollary~\ref{cor:proj2} the points in $X \cap (\cup_{j=1}^{d+1} \pi_j)$ uniquely determine $\xi_1$ and analogously $\xi_2$. The point $x \in \xi_1\cap\xi_2$ is hence also uniquely determined.
\qed

This concludes the proof of the Main Result.

\small

config: arxiv


TITLE: Is the square root of -1 rational?
QUESTION [26 upvotes]: This is not a deep question, but if there is a definite answer then here is the place where I will find it.
Is it justified to say that $i =\sqrt{-1}$  is rational?
The origin of this question lies in a regular discussion I have over this t-shirt of mine:

While obvious $\pi$'s comment is completely legit, $\sqrt{-1}$'s might be hypocritical, if the rationality of $\sqrt{-1}$ is questionable.

REPLY [3 votes]: Echoing what other people have already said: no $i= \sqrt{-1}$ is not a rational number.
You have
$$
\begin{align}
&\mathbb{C} \;\text{ the complex numbers}\\
&\cup \\
&\mathbb{R} \;\text{ the real numbers}\\
&\cup \\
&\mathbb{Q} \;\text{ the rational numbers}\\
&\cup \\
&\mathbb{Z} \;\text{ the integers}\\
&\cup \\
&\mathbb{N} \;\text{ the natural numbers}
\end{align}
$$
Here the $\cup$ denotes that the lower is contained in the upper. So for example all real numbers are complex numbers. And: an integer is a real number. Note that for example not all complex numbers are real numbers. Not all complex numbers are rational numbers. Not all integers are natural numbers.
So the question now is whether $i = \sqrt{-1}$ (which is a complex number) is a rational number. And here we first note that the rational numbers are those numbers that can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers (so belongs to $\mathbb{Z}$) ($b\neq 0$). So is $i = \frac{a}{b}$ for any integers $a$, and $b$? As provided in Austin's fine answer, one can show that indeed the answer is no.
To have something to think about, maybe you can answer the question: Is $\sqrt{2}$ a rational number? (You can find the answer here on M.SE).


TITLE: How to evaluate one improper integral
QUESTION [2 upvotes]: Please show me the detailed solution to the question:
Compute the value of
$$\int_{0}^{\infty }\frac{\left( \ln x\right) ^{40021}}{x}dx$$
Thank you a million!

REPLY [0 votes]: Since this is an exercise on improper integrals, it is natural to replace the
upper and lower limits by $R$, $\frac{1}{R}$ respectively and define the integral to be the limit as $R \rightarrow \infty$ . Then write the integral as the sum of the integral from $\frac{1}{R}$ to $1$ and from $1$ to $R$. In the second integral make the usual transformation replacing $x$ by $\frac{1}{x}$. The two integrals cancel.