%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% LaTeX file. Please LaTeX twice. %%%%% % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[12pt]{article} \usepackage{amssymb,latexsym} \usepackage{amsbsy} %%for italic bold: \pmb % % definitions concerning automatic numbering of theorems, lemmas, etc. % \newtheorem{theo}{Theorem}%[section] \newtheorem{defi}[theo]{Definition} \newtheorem{lemm}[theo]{Lemma} \newtheorem{coro}[theo]{Corollary} \newtheorem{rema}[theo]{Remark} \newtheorem{prop}[theo]{Proposition} \newtheorem{conj}[theo]{Conjecture} \newtheorem{exam}[theo]{Example} % \setlength{\topmargin}{0cm} \setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm} \setlength{\textheight}{22.5cm} \setlength{\textwidth}{16cm} % \addtolength{\parskip}{2mm} % % special definitions or abbreviations for the paper % \def\lb{[\![} \def\rb{]\!]} \def\gl{\mathfrak{gl}} \def\g{\mathfrak{g}} \def\h{\mathfrak{h}} \def\n{\mathfrak{n}} \def\ch{\mathop{\rm ch}\nolimits} \def\mult{\mathop{\rm mult}\nolimits} \def\rank{\mathop{\rm rank}\nolimits} \def\tr{\mathop{\rm tr}\nolimits} \def\al{\alpha} \def\be{\beta} \def\ga{\gamma} \def\de{\delta} \def\ep{\epsilon} \def\varep{\varepsilon} \def\et{\eta} \def\th{\theta} \def\la{\lambda} \def\rh{\rho} \def\si{\sigma} \def\ta{\tau} \def\ph{\phi} \def\vp{\varphi} \def\ps{\psi} \def\Ga{\Gamma} \def\De{\Delta} \def\La{\Lambda} \def\Up{\Upsilon} \def\C{\mathbb{C}} \def\Z{\mathbb{Z}} % % LaTeX begin and end of numbered equation % \def\beq{\begin{equation}} \def\eeq{\end{equation}} \def\bea{\begin{eqnarray}} \def\eea{\end{eqnarray}} \def\beas{\begin{eqnarray*}} \def\eeas{\end{eqnarray*}} \def\nn{\nonumber} \def\ds{\displaystyle} % the end-of-proof box : \def\mybox{\hfill\llap{$\Box$}} % % numbering of equations per section: %\renewcommand{\theequation}{\arabic{section}.\arabic{equation}} % \begin{document} \addtolength{\baselineskip}{1mm} \begin{center} {\large \bf Fock representations of the Lie superalgebra $q(n+1)$}\\[1cm] T Palev\footnote{Permanent address~: Institute for Nuclear Research and Nuclear Energy, 1784 Sofia, Bulgaria (e-mail~: tpalev@inrne.bas.bg).} and J Van der Jeugt\footnote{Research Associate of the Fund for Scientific Research -- Flanders (Belgium);\\ (e-mail~: Joris.VanderJeugt@rug.ac.be).} \\ Department of Applied Mathematics and Computer Science,\\ University of Ghent, Krijgslaan 281-S9, B-9000 Gent, Belgium \end{center} \vskip 1cm \begin{abstract} For the Lie superalgebra $q(n+1)$ a description is given in terms of creation and annihilation operators, in such a way that the defining relations of $q(n+1)$ are determined by quadratic and triple supercommutation relations of these operators. Fock space representations $V_p$ of $q(n+1)$ are defined by means of these creation and annihilation operators. These new representations are introduced as quotient modules of some induced module of $q(n+1)$. The representations $V_p$ are not graded, but they possess a number of properties that are of importance for physical applications. For $p$ a positive integer, these representations $V_p$ are finite-dimensional, with a unique highest weight (of multiplicity 1). The Hermitian form that is consistent with the natural adjoint operation on $q(n+1)$ is shown to be positive definite on $V_p$. For $q(2)$ these representations are ``dispin''. For the general case of $q(n+1)$, many structural properties of $V_p$ are derived. \end{abstract} \vskip 2cm \noindent Short title~: The Lie superalgebra $q(n+1)$\\[2mm] \noindent PACS Classification~: 02.10w, 02.20b, 03.65w \newpage % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction} Lie superalgebras and their irreducible representations (simple modules) have been the subject of much attention in both the mathematical~\cite{Kac1,Kac2,Scheunert} and the physics~\cite{Corwin,Balantekin,Hurni} literature. However, even for the simplest family of basic classical Lie superalgebras, namely $sl(m/n)$ or $gl(m/n)$, the understanding of all finite-dimensional simple modules has been a very difficult problem. The main reason for this difficulty has been the existence of so-called {\em atypical} modules~\cite{Kac2}. Although partial progress was made in determining a character formula for atypical modules~\cite{VHKT1,VHKT2,KW}, the problem of determining the character for $gl(m/n)$ modules was solved only recently~\cite{Serganova1,Serganova2} (see also~\cite{VZ} for a simpler algorithm). The general linear Lie algebra $gl(n)$ has two super-analogues. The first is the Lie superalgebra $gl(n/1)$, for which the representations (even the atypical ones) are now well known~: e.g., a Gelfand-Zetlin basis has been introduced and its transformations have been determined~\cite{pal78,pal82}, and representations of $sl(n/1)$ have been studied~\cite{pal79,pal81}. The second super-analogue is the strange Lie superalgebra $q(n)$. Also this Lie superalgebra has received attention recently. In particular, characters of finite-dimensional irreducible graded representations of $q(n)$ have been determined~\cite{penkov0,penkov1,penkov2}, both in the typical and atypical case. In a different context, oscillator realizations have been given~\cite{frappat}. The finite-dimensional irreducible graded representations of $q(n)$ have the strange property that the multiplicity of the highest (or lowest) weight is in general greater than 1~\cite{penkov1}. From the physical point of view, where one wishes to interpret the representation space as a Hilbert space with a unique vacuum, this situation is not very favourable. In the present paper, the purpose is to study a new class of irreducible finite-dimensional representations of the Lie superalgebra $q(n)$, which have certain properties that are required in a physical context. In particular, the highest weight has multiplicity 1 (so there is a unique highest weight vector, up to a factor), and the representation space can be naturally equipped with a symmetric and nondegenerate positive definite Hermitian form (inner product). Moreover, creation and annihilation generators (or operators) are introduced for $q(n)$ such that the representation space is a Fock space. The only property that has to be abandoned is the grading of the representation space (but from the physical point of view, this grading is no requirement). The structure of the paper is as follows. In section~2 the main definitions are given concerning the Lie superalgebra $q(n+1)$. In section~3 a new basis for $q(n+1)$ is given in terms of so-called creation and annihilation operators. The new class of representations of $q(n+1)$ is introduced in section~4. These representations are defined by means of an induced module $\bar V_p$; the actual irreducible representation $V_p$ which is of interest is then a quotient module of $\bar V_p$. To have some idea about the structure of $V_p$, we first consider the low rank case of $q(2)$ in section~5. Here, the representation has a ``dispin'' structure. Section~6 goes back to the general case $q(n+1)$, and includes several (technical) properties concerning the structure of $\bar V_p$, paving the way to determining the structure of the simple modules $V_p$. This is performed in section~7, where in particular we give the dimension and character of $V_p$, and show that the Hermitian form is positive definite. \section{The Lie superalgebra $q(n+1)$} %\label{sec-intro} %\setcounter{equation}{0} The Lie superalgebra $q(n+1)$ can be determined through its defining representation, i.e.\ \beq q(n+1) = \left\{ \left( \begin{array}{cc} A&B\\B&A \end{array} \right) \; | \; A,B\in gl(n+1) \right\}, \eeq where the matrices with $B=0$ are even, or elements of $q(n+1)_{\bar 0}$, and those with $A=0$ odd, or elements of $q(n+1)_{\bar 1}$. The subalgebra $sq(n+1)$ consists of those elements with $\tr(B)=0$. The Lie superalgebras $q(n+1)$ and $sq(n+1)$ still contain a one-dimensional center $\C I$, where $I$ is the identity matrix. Hence one defines the quotient Lie superalgebra $psq(n+1)$ as $sq(n+1)/ \C I$. The notation $q(n+1)$, $sq(n+1)$ and $psq(n+1)$ is due to Penkov and others~\cite{penkov0}; in the notation of Kac~\cite{Kac1,Kac2} we have $\bar Q (n)=sq(n+1)$ and $Q(n)=psq(n+1)$. Recall that $Q(n)$ is a simple Lie superalgebra for $n\geq 2$, and that it is one of the series of classical (but ``strange'') Lie superalgebras in the classification of Kac~\cite{Kac1}. For the development of representation theory, we shall be working mainly with $q(n+1)$. Thus $q(n+1)$ can be defined as the Lie superalgebra with $(n+1)^2$ even basis elements $e_{ij}^{\bar 0}$ ($i,j=0,1,\ldots,n$) and $(n+1)^2$ odd basis elements $e_{ij}^{\bar 1}$ ($i,j=0,1,\ldots,n$), satisfying the bracket relation \beq \lb e_{ij}^\si, e_{kl}^\theta \rb = \de_{jk} e_{il}^{\si+\theta} - (-1)^{\si\theta} \de_{il} e_{kj}^{\si+\theta}, \eeq where $\si,\theta\in\Z_2=\{\bar 0, \bar 1\}$, and $i,j,k,l\in\{0,1, \ldots,n\}$. In this paper, we shall use $\lb\,,\,\rb$ for the Lie superalgebra bracket, and write explicitly $[\,,\,]$ ( resp.\ $\{\,,\,\}$) if this stands for a commutator (resp.\ anti-commutator). Let $E_{ij}$ denote the $(n+1)\times(n+1)$ matrix with 1 in position $(i,j)$ and 0 elsewhere (indices running from 0 to $n$), then the defining representation of $q(n+1)$ is given by the map \beq e_{ij}^{\bar 0} \rightarrow \left( \begin{array}{cc} E_{ij}&0\\0&E_{ij} \end{array} \right),\qquad e_{ij}^{\bar 1} \rightarrow \left( \begin{array}{cc} 0&E_{ij}\\E_{ij}&0 \end{array} \right) \eeq onto matrices of order $2(n+1)$. Following the definition of Cartan subalgebra as a maximal nilpotent subalgebra coinciding with its own normalizer, a Cartan subalgebra $H$ of $G=q(n+1)$ is given by $H=H_{\bar 0} \oplus H_{\bar 1}$, where $H_{\bar 0}={\rm span}\{e_{ii}^{\bar 0}\;|\; i=0,1,\ldots n\}$ and $H_{\bar 1}={\rm span}\{e_{ii}^{\bar 1}\;|\; i=0,1,\ldots n\}$ \cite{penkov0,dict}. This subalgebra is not abelian, and since the elements of $H_{\bar 1}$ are odd, the root generators $e_{ij}^{\bar 1}$ are not eigenvectors of $H_{\bar 1}$. Therefore, to give a root decomposition of $G$ it is more convenient~\cite{dict} to work with the abelian subalgebra $H_{\bar 0}$. The dual space $H_{\bar 0}^*$ has the standard basis $\{ \ep_0, \ep_1, \ldots, \ep_n\}$, in terms of which the roots of $G$ can be described. The elements $e_{ij}^{\bar 0}$ ($i\ne j$) are even root vectors corresponding to the root $\ep_i-\ep_j$; the elements $e_{ij}^{\bar 1}$ ($i\ne j$) are odd root vectors also corresponding to the root $\ep_i-\ep_j$; the elements $e_{ii}^{\bar 1}$ from $H_{\bar 1}$ can then be interpreted as odd root vectors corresponding to the root 0. Note that every root $\ep_i-\ep_j$ ($i\ne j$) has multiplicity~2 (counting once as even and once as odd root). Let, as usual, $\De$ be the set of all roots, $\De^{\bar 0}$ (resp.\ $\De^{\bar 1}$) be the set of even (resp.\ odd) roots~: \beq \De^{\bar 0} = \{ \ep_i -\ep_j \;|\; 0\leq i\ne j \leq n \},\qquad \De^{\bar 1} = \De^{\bar 0} \cup \{0\}. \eeq The positive roots are \beq \De_+=\De^{\bar 0}_+ = \De^{\bar 1}_+ = \{ \ep_i -\ep_j \;|\; 0\leq i1). \eeq So $V_p$ splits into two irreducible $gl(2)$ modules, both of which have been labelled by their highest weight (in the $(\ep_0,\ep_1)$-basis). So for $p>1$ these $q(2)$ representations could be referred to as ``dispin'' representations, similar to the known dispin representations of $osp(1,2)$~\cite{SNR,Hughes}. For $p=1$, the decomposition is simply $V_p \rightarrow (p,0)$. The dimension follows easily~: \beq \dim V_p = 2p. \eeq It is possible to give an orthonormal basis for $V_p$, in terms of the above basis vectors $v_k$, $w_k$. Since \bea &&\langle v_k\pm \sqrt{k} w_k | v_k\pm \sqrt{k} w_k \rangle = 2(1\pm\sqrt{k/p}) k!p!/(p-k)!,\\ &&\langle v_k+ \sqrt{k} w_k | v_k- \sqrt{k} w_k \rangle =0, \eea one can define \bea &&\phi_k = \left( (p-k)!\over 2k!p!(1+\sqrt{k/p}) \right)^{1/2} (v_k+ \sqrt{k} w_k),\;(k=1,\ldots,p)\\ &&\psi_k = \left( (p-k)!\over 2k!p!(1-\sqrt{k/p}) \right)^{1/2} (v_k- \sqrt{k} w_k),\;(k=0,\ldots,p-1). \eea These vectors are orthonormal~: \beq \langle \phi_k | \phi_l \rangle = \langle \psi_k | \psi_l \rangle = \de_{kl},\qquad \langle \phi_k | \psi_l \rangle = 0. \eeq The action of the creation and annihilation operators on this basis can be computed. We have~: \bea f^+ \phi_k &=& {1\over 2}\left[ (\sqrt{p}-\sqrt{k})(\sqrt{p}+\sqrt{k+1}) \right]^{1/2} \phi_{k+1} \nn\\ &&-{1\over 2}\left[ (\sqrt{p}-\sqrt{k})(\sqrt{p}-\sqrt{k+1}) \right]^{1/2} \psi_{k+1}\\ f^+ \psi_k &=& {1\over 2}\left[ (\sqrt{p}+\sqrt{k})(\sqrt{p}+\sqrt{k+1}) \right]^{1/2} \phi_{k+1} \nn\\ &&-{1\over 2}\left[ (\sqrt{p}+\sqrt{k})(\sqrt{p}-\sqrt{k+1}) \right]^{1/2} \psi_{k+1}\\ b^+ \phi_k &=& {1\over 2}(\sqrt{k+1}+\sqrt{k}) \left[ (\sqrt{p}-\sqrt{k})(\sqrt{p}+\sqrt{k+1}) \right]^{1/2} \phi_{k+1} \nn\\ &&+{1\over 2}(\sqrt{k+1}-\sqrt{k}) \left[ (\sqrt{p}-\sqrt{k})(\sqrt{p}-\sqrt{k+1}) \right]^{1/2} \psi_{k+1}\\ b^+ \psi_k &=& {1\over 2}(\sqrt{k+1}-\sqrt{k}) \left[ (\sqrt{p}+\sqrt{k})(\sqrt{p}+\sqrt{k+1}) \right]^{1/2} \phi_{k+1} \nn\\ &&+{1\over 2}(\sqrt{k+1}+\sqrt{k}) \left[ (\sqrt{p}+\sqrt{k})(\sqrt{p}-\sqrt{k+1}) \right]^{1/2} \psi_{k+1}. \eea The action of the annihilation operators follows immediately from $b^-=(b^+)^\dagger$, $f^-=(f^+)^\dagger$. For example, \beq f^- \phi_k = {1\over 2}\left[ (\sqrt{p}-\sqrt{k-1})(\sqrt{p}+\sqrt{k}) \right]^{1/2} \phi_{k-1} + {1\over 2}\left[ (\sqrt{p}+\sqrt{k-1})(\sqrt{p}+\sqrt{k}) \right]^{1/2} \psi_{k-1}. \eeq \section{Structure of the module $\bar V_p$} In this section we return to the general case $q(n+1)$. By calculating the action of creation and annihilation operators on basis vectors of the induced module $\bar V_p$, the way is prepared to determine the structure of the irreducible quotient module $V_p$ of $\bar V_p$. Recall that a basis for $\bar V_p$ is given by the vectors \bea |p;{\pmb k}, {\pmb l } \rangle &=&|p;k_1,l_1,k_2,l_2,\ldots,k_n,l_n\rangle = (b_1^+)^{k_1}(f_1^+)^{l_1}(b_2^+)^{k_2}(f_2^+)^{l_2}\cdots (b_n^+)^{k_n}(f_n^+)^{l_n} v_0,\nn\\ &&\qquad l_i\in\{0,1\},\; k_i=0,1,2,\ldots . \label{pkkll} \eea Since all creation operators $b_i^+$ and $f_i^+$ supercommute, a different order of the creation operators in (\ref{pkkll}) can produce only a sign change. In the standard basis, the weight of the vector $|p;{\pmb k}, {\pmb l } \rangle$ is given by \beq {\rm weight}(|p;{\pmb k}, {\pmb l } \rangle) = p\ep_0+\sum_{i=1}^n (k_i+l_i)(\ep_i-\ep_0)= \left( p-\sum_{i=1}^n(k_i+l_i), k_1+l_1,\ldots, k_n+l_n\right). \eeq Thus every weight of $\bar V_p$ is of the form \beq \la_m = (p-\sum_{i=1}^n m_i, m_1,\ldots, m_n),\qquad m_i=0,1,2,\ldots ; \label{la_m} \eeq conversely, every weight of the form (\ref{la_m}) is a weight of $\bar V_p$. Since the $l_i$ in (\ref{pkkll}) are either 0 or 1, it follows that the multiplicity of the weight $\la_m$ in $\bar V_p$ is given by \beq \mult_{\bar V_p} (p-\sum_{i=1}^n m_i, m_1,\ldots, m_n)= 2^{\ga(m_1)+\cdots +\ga(m_n)}, \eeq where $\ga(m_i)=0$ if $m_i=0$ and $\ga(m_i)=1$ if $m_i\ne 0$. The action of the creation operators on the basis of ${\bar V_p}$ is very simple. It will be convenient to denote the basis vectors in the right hand side of such actions only by means of the labels that are effectively changed by the action. So, instead of writing \beas && b^+_j |p;{\pmb k}, {\pmb l } \rangle = |p;k_1,l_1,\ldots,k_j+1,l_j,\ldots,k_n,l_n\rangle,\\ && f^+_j |p;{\pmb k}, {\pmb l } \rangle = \de_{l_j,0} (-1)^{l_1+\cdots+l_{j-1}} |p;k_1,l_1,\ldots,k_j,l_j+1,\ldots,k_n,l_n\rangle, \eeas we abbreviate this to~: \bea && b^+_j |p;{\pmb k}, {\pmb l } \rangle = |k_j+1 \rangle,\label{bj+}\\ && f^+_j |p;{\pmb k}, {\pmb l } \rangle = \de_{l_j,0} (-1)^{l_1+\cdots+l_{j-1}} |l_j+1 \rangle. \label{fj+} \eea The action of the annihilation operators is more complicated, and here the notational convention just introduced will be very useful. \begin{prop} The action of the annihilation operators in the module $\bar V_p$ is given by~: \bea f_j^- |p;{\pmb k}, {\pmb l } \rangle &=& (-1)^{l_1+\cdots+l_{j-1}} l_j \bigl(p+1+k_j-\sum_{i=1}^n(k_i+l_i)\bigr) |l_j-1\rangle \nn\\ && + (-1)^{l_1+\cdots+l_n} k_j \sqrt{p} |k_j-1\rangle - (-1)^{l_1+\cdots+l_{j-1}} \de_{l_j,0} k_j(k_j-1) |k_j-2,l_j+1\rangle\nn\\ && - \sum_{i=1 \atop i\ne j}^n (-1)^{l_1+\cdots+l_{i-1}} \de_{l_i,0} k_i k_j |k_j-1, k_i-1,l_i+1 \rangle \nn\\ && + \sum_{i=1 \atop i\ne j}^n (-1)^{l_1+\cdots+l_{i-1}} l_i k_j |k_j-1, k_i+1,l_i-1 \rangle; \label{fj-} \eea \bea (-1)^{l_{j+1}+\cdots+l_n}b_j^- |p;{\pmb k}, {\pmb l } \rangle &=& (-1)^{l_{j+1}+\cdots+l_n} k_j \bigl(p+1-l_j-\sum_{i=1}^n(k_i+l_i)\bigr) |l_j-1\rangle \nn\\ &&+ l_j \sqrt{p} |l_j-1\rangle + \sum_{i=1 \atop i\ne j}^n (-1)^{l_i+\cdots+l_n}\th_{ij}\de_{l_i,0} k_i l_j |l_j-1, k_i-1,l_i+1 \rangle \nn\\ && - \sum_{i=1 \atop i\ne j}^n (-1)^{l_i+\cdots+l_n} \th_{ij} l_i l_j |l_j-1, k_i+1,l_i-1 \rangle, \label{bj-} \eea where $\th_{ij}=+1$ if $ij$. \end{prop} \noindent {\bf Proof.} We shall sketch the proof for the action of $f_j^-$; that for $b_j^-$ is similar. The proof uses induction on $n$. As a first step, the action of $f_1^+$ will be determined. Denote $y_i=(b_i^+)^{k_i}(f_i^+)^{l_i}$. Then, \beq f_1^- |p;{\pmb k}, {\pmb l } \rangle = f_1^- y_1 y_2\cdots y_n v_0 = \lb f_1^-,y_1\rb y_2\cdots y_n v_0 + (-1)^{l_1} y_1 f_1^- y_2 \cdots y_n v_0. \label{54} \eeq Since the weight of $f_1^- y_2 \cdots y_n v_0$, which is $\bigl(p+1-\sum_{i=2}^n(k_i+l_i),-1,k_2+l_2,\ldots, k_n+l_n\bigr)$, is not of the form (\ref{la_m}), the vector cannot belong to $\bar V_p$, so the second term in (\ref{54}) has to be zero. Using $\lb f_1^-,b_1^+\rb=e_{00}^{\bar 1}- e_{11}^{\bar 1}$ and $\lb f_1^-,f_1^+\rb=e_{00}^{\bar 0}+ e_{11}^{\bar 0}$, one finds \beq \lb f_1,y_1\rb = \sum_{r=1}^{k_1-1} (b_1^+)^r (e_{00}^{\bar 1}- e_{11}^{\bar 1}) (b_1^+)^{k_1-r-1}(f_1^+)^{l_1} + l_1 (b_1^+)^{k_1}(e_{00}^{\bar 0}+ e_{11}^{\bar 0}). \eeq {}From the weight of $y_i$ and $v_0$ one obtains \beq (e_{00}^{\bar 0}+ e_{11}^{\bar 0})y_2\cdots y_n v_0 = (p-k_2-l_2-\cdots-k_n-l_n) y_2\cdots y_n v_0. \eeq Next we need to determine the action of $e_{00}^{\bar 1}$ and $e_{11}^{\bar 1}$ on vectors of the form $y'_1 y_2\cdots y_n v_0$, where $y'_1=(b_1^+)^{k'_1}(f_1^+)^{l_1}$ with $k'_1=k_1-r-1$~: \beas e_{00}^{\bar 1} y'_1 y_2\cdots y_n v_0 &=& \lb e_{00}^{\bar 1}, y'_1 y_2\cdots y_n\rb v_0 + (-1)^{l_1+\cdots l_n} y'_1 y_2\cdots y_n \sqrt{p}\; v_0\\ \nn &=& \lb e_{00}^{\bar 1}, y'_1\rb y_2\cdots y_n v_0 + \sum_{i=2}^n y'_1\cdots y_{i-1}\lb e_{00}^{\bar 1}, y_i\rb y_{i+1}\cdots y_n v_0 \\ \nn && \qquad\qquad + (-1)^{l_1+\cdots l_n} \sqrt{p}\; y'_1 y_2\cdots y_n v_0. \eeas Every term in this expression can be determined explicitly using \beq \lb e_{00}^{\bar 1}, y_i\rb = -\de_{l_i,0} k_i (b_i^+)^{k_i-1} f_i^+ + l_i (b_i^+)^{k_i+1}, \eeq which follows from the supercommutator of $e_{00}^{\bar 1}$ with $b_i^+$ and $f_i^+$. Similarly, for the action of $e_{11}^{\bar 1}$ one finds \beq e_{11}^{\bar 1} y'_1 y_2\cdots y_n v_0 = \lb e_{11}^{\bar 1}, y'_1 \rb y_2\cdots y_n v_0 , \eeq since $\lb e_{11}^{\bar 1}, y_i \rb =0$ for $i>1$ and $e_{11}^{\bar 1} v_0=0$. Using the supercommutator of $e_{11}^{\bar 1}$ with $b_1^+$ and $f_1^+$, there comes \beq \lb e_{11}^{\bar 1}, y'_1 \rb = \de_{l_1,0} k'_1 (b_1^+)^{k'_1-1} f_1^+ + l_1 (b_1^+)^{k'_1+1}. \eeq Collecting now all contributions yields the action of $f_1^+$ on $|p;{\pmb k}, {\pmb l } \rangle$, as given in the proposition. In the second step, we use induction in the following way. First observe that for $j=2$, \beq f_2^- y_1 y_2\cdots y_n v_0 = \lb f_2^-, y_1\rb y_2\cdots y_n v_0 + (-1)^{l_1} y_1 f_2^- y_2\cdots y_n v_0. \eeq But the action $f_2^- y_2\cdots y_n v_0$ is formally the same as the (known) action $f_1^- y_1 y_2\cdots y_n v_0$, by relabelling of indices. More generally, for $j\geq 2$, \beq f_j^- y_1 y_2\cdots y_n v_0 = \lb f_j^-, y_1\rb y_2\cdots y_n v_0 + (-1)^{l_1} y_1 f_j^- y_2\cdots y_n v_0.\label{61} \eeq Once again, $f_j^- y_2\cdots y_n v_0$ can formally be reduced to $f_{j-1}^- y_1\cdots y_{n-1} v_0$, on which one uses the induction hypothesis. So what remains to be determined is the first term in (\ref{61}). Since $[f_j^-,b_1^+]$ commutes with $b_1^+$ (see (\ref{Q2})), one finds \beq \lb f_j^-,(b_1^+)^{k_1}(f_1^+)^{l_1}\rb = (-1)^{l_1} k_1 (b_1^+)^{k_1-1}(f_1^+)^{l_1} [ f_j^-,b_1^+ ] + l_1 (b_1^+)^{k_1} \{ f_j^-, f_1^+ \}. \eeq To calculate $[ f_j^-,b_1^+ ] y_2\cdots y_n v_0$, one uses again the triple relations (\ref{Q2}) and (\ref{Q3})~: \beq [ f_j^-,b_1^+ ] y_2\cdots y_n v_0 = (-1)^{l_2+\cdots+l_{j-1}} y_2\cdots y_{j-1} \lb [ f_j^-,b_1^+ ], y_j \rb y_{j+1}\cdots y_n v_0; \eeq furthermore \beq \lb [ f_j^-,b_1^+ ], y_j \rb = -k_j f_1^+ (b_j^+)^{k_j-1} (f_j^+)^{l_j} -l_j (b_j^+)^{k_j} b_1^+, \eeq where again the triple relations have been used. In a similar way, the action $\{ f_j^-,f_1^+ \} y_2\cdots y_n v_0$ is determined. Together they yield the first term in the right hand side of (\ref{61}). Combining then the coefficients of all identical vectors in the right hand side of (\ref{61}) proves the proposition. \mybox Now we wish to determine the vectors in $\bar V_p$ that are annihilated by all annihilation operators, i.e. by \beq P={\rm span} \{b_1^-,\ldots,b_n^-, f_1^-,\ldots,f_n^-\}. \eeq For this purpose, and inspired by the right hand sides in (\ref{bj-}) and (\ref{fj-}), we introduce the following weight vectors (still using the notational convention introduced in (\ref{bj+})-(\ref{fj+}))~: \bea X(p;{\pmb k},{\pmb l}) &=& (-1)^{l_1+\cdots+l_n} \sqrt{p} |p;{\pmb k}, {\pmb l } \rangle - \sum_{i=1}^n (-1)^{l_1+\cdots+l_i} l_i |k_i+1,l_i-1 \rangle \nn \\ && - \sum_{i=1}^n (-1)^{l_1+\cdots+l_i} \de_{l_i,0} k_i |k_i-1,l_i+1 \rangle . \eea Then we have~: \begin{prop} For the weight vectors $|p;{\pmb k},{\pmb l} \rangle$ and $X(p;{\pmb k},{\pmb l})$ the following equalities hold (again we use the convention that in the right hand side only the labels $k_i$ and $l_i$ that are effectively changed are withheld)~: \bea b_j^- |p;{\pmb k},{\pmb l} \rangle &=& k_j(p+1-\sum_{i=1}^n(k_i+l_i)) |k_j-1\rangle + (-1)^{l_1+\cdots+l_{j-1}}l_j X(p;l_j-1),\label{bj-p}\\ f_j^- |p;{\pmb k},{\pmb l} \rangle &=& (-1)^{l_1+\cdots+l_{j-1}}l_j(p+1-\sum_{i=1}^n(k_i+l_i)) |l_j-1\rangle + k_j X(p;k_j-1),\label{fj-p}\\ b_j^+ X(p;{\pmb k},{\pmb l}) &=& X(p;k_j+1)+(-1)^{l_1+\cdots+l_j} \de_{l_j,0} |l_j+1\rangle,\label{bj+x}\\ f_j^+ X(p;{\pmb k},{\pmb l}) &=& -\de_{l_j,0} (-1)^{l_1+\cdots+l_{j-1}} X(p;l_j+1)+ |k_j+1\rangle,\label{fj+x}\\ b_j^- X(p;{\pmb k},{\pmb l}) &=& k_j(p-\sum_{i=1}^n(k_i+l_i)) X(p;k_j-1),\label{bj-x}\\ f_j^- X(p;{\pmb k},{\pmb l}) &=& (-1)^{l_1+\cdots+l_{j-1}} l_j(p-\sum_{i=1}^n(k_i+l_i)) X(p;l_j-1). \label{fj-x} \eea \end{prop} The proof is by direct computation, using proposition~1. The first two relations are just a reformulation of the equalities in proposition~1. The next two relations follow immediately from the definition of $X(p;{\pmb k},{\pmb l})$ and the actions (\ref{bj+})-(\ref{fj+}). The last two relations are the most difficult to verify. They follow from the action of the annihilation operators on each part of $X(p;{\pmb k},{\pmb l})$, using proposition~1, and then collecting the terms according to equal weight vectors $|p;{\pmb k},{\pmb l} \rangle$. This computation is long but straightforward, and will not be given in detail here. \mybox Note that all vectors $X(p;{\pmb k},{\pmb l})$ with $\sum_i (k_i+l_i)=p$ are annihilated by $P$. In order to understand the linear (in)dependance of the newly introduced weight vectors $X(p;{\pmb k},{\pmb l})$, it is convenient to prove first an interesting lemma about the determinant and rank of a matrix. Let $r$ be a positive integer, and consider the $2^r\times 2^r$ matrix $A$, where the rows and columns of $A$ are labelled by the binary sequences ${\pmb l}=(l_1,\ldots,l_r)$ ($l_i\in\{0,1\}$) of length $r$ in reverse binary order. The elements $a_{{\pmb l},{\pmb l'}}$ of $A\equiv A(s;t_1,\ldots,t_r)$ are as follows \bea && {\rm if}\ {\pmb l}={\pmb l'}\ {\rm then}\ a_{{\pmb l},{\pmb l'}} = s,\nn \\ \nn && {\rm if}\ {\pmb l}\ {\rm and}\ {\pmb l'}\ {\rm differ\ in\ only\ position}\ i \ {\rm then} \\ \label{defA} && \qquad\qquad {\rm if}\ l_i=0\ {\rm then}\ a_{{\pmb l},{\pmb l'}} = -(-1)^{l_{i+1}+\cdots+l_r} t_i,\\ \nn && \qquad\qquad {\rm if}\ l_i=1\ {\rm then}\ a_{{\pmb l},{\pmb l'}} = (-1)^{l_i+\cdots+l_r} t_i,\\ \nn && {\rm otherwise}\ a_{{\pmb l},{\pmb l'}} = 0. \eea Herein, $s$ and $t_1,\ldots,t_r$ are arbitrary real numbers or variables. It is constructive to consider an example. For $r=2$, the binary sequences labelling the rows and columns of $A$ are $(0,0), (1,0), (0,1), (1,1)$ (in this order); for $r=3$ the sequences are $(0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1), (0,1,1), (1,1,1)$. The matrices take the following form~: \beq A(s;t_1,t_2) = \left( \begin{array}{cccc} s & -t_1 & -t_2 & 0 \\ -1 & s & 0 & -t_2 \\ -1 & 0 & s & t_1 \\ 0 & -1 & 1 & s \end{array} \right), \eeq \beq A(s;t_1,t_2,t_3) = \left( \begin{array}{cccccccc} s & -t_1 & -t_2 & 0 & -t_3 &0&0&0\\ -1 & s & 0 & -t_2 &0&-t_3&0&0\\ -1 & 0 & s & t_1 &0&0&-t_3&0\\ 0 & -1 & 1 & s &0&0&0&-t_3\\ -1&0&0&0&s & t_1 & t_2 & 0 \\ 0&-1&0&0&1 & s & 0 & t_2 \\ 0&0&-1&0&1 & 0 & s & -t_1 \\ 0&0&0&-1&0 & 1 & -1 & s \end{array} \right) \eeq \begin{lemm} The matrix $A(s;t_1,\ldots,t_r)$ ($r>1$) defined above satisfies~: \begin{itemize} \item[(a)] the determinant is given by \beq \det A(s;t_1,\ldots,t_r) = (s^2-\sum_{i=1}^r t_i)^{2^{r-1}}; \label{detA} \eeq \item[(b)] if the elements $t_i$ are positive real numbers such that $\sum_{i=1}^r t_i = s^2$ then $\rank(A(s;t_1,\ldots,t_r))=2^{r-1}$. \item[(c)] \beq A(s;t_1,\ldots,t_r)\cdot A(-s;t_1,\ldots,t_r)= (\sum_{i=1}^r t_i-s^2)I, \label{invA} \eeq where $I$ is the identity matrix of order $2^r$. \end{itemize} \end{lemm} \noindent {\bf Proof.} The proof of (a) is by induction on $r$. Clearly it holds for $r=2$. By definition, the matrix $A(s;t_1,\ldots,t_r)$ can be written in block form as \beq A(s;t_1,\ldots,t_r) = \left( \begin{array}{cc} A(s;t_1,\ldots,t_{r-1}) & -t_r I \\ -I & -A(-s;t_1,\ldots,t_{r-1}) \end{array} \right), \label{block} \eeq where $I$ is the identity matrix of order $2^{r-1}$. So by induction we have $\det(A(s;t_1,\ldots,t_{r-1}))= (s^2-\sum_{i=1}^{r-1} t_i)^{2^{r-2}}$ and $\det(-A(-s;t_1,\ldots,t_{r-1}))= ((-s)^2-\sum_{i=1}^{r-1} t_i)^{2^{r-2}}$. Then it follows from (\ref{block}) that \bea &&\det(A(s;t_1,\ldots,t_{r-1},0))=\nn\\ &&\det(A(s;t_1,\ldots,t_{r-1})) \det(-A(-s;t_1,\ldots,t_{r-1})) = (s^2-\sum_{i=1}^{r-1} t_i)^{2^{r-1}}. \label{78} \eea On the other hand, exchanging $t_i$ and $t_j$ in $A(s;t_1,\ldots,t_r)$ corresponds to a permutation of the rows and corresponding columns of $A(s;t_1,\ldots,t_r)$; a closer examination shows that the signature of such a permutation is positive. Thus $A(s;t_1,\ldots,t_r)$ is invariant for transpositions of the form $t_i \leftrightarrow t_j$. Therefore, $\det A(s;t_1,\ldots,t_r)$ is a symmetric polynomial in the elements $t_i$. Since the power sum symmetric functions form a basis of the ring of symmetric polynomials~\cite[p.~24]{Macdonald}, it follows that \beq \det A(s;t_1,\ldots,t_{r-1},t_r)= \sum_{k=0}^N \sum_{\kappa\vdash k} c_\kappa(s) p_\kappa(t_1,\ldots,t_{r-1},t_r), \label{refnew1} \eeq where $N=2^{r-1}$, $\kappa$ is summed over all partitions of $k$, and $p_\kappa$ is the multiplicative power sum function~\cite[p.~24]{Macdonald}. Combining (\ref{refnew1}) with (\ref{78}) gives \beq \det A(s;t_1,\ldots,t_{r-1},0)= \sum_{k=0}^N \sum_{\kappa\vdash k} c_\kappa(s) p_\kappa(t_1,\ldots,t_{r-1}) = (s^2-p_1(t_1,\ldots,t_{r-1}))^N, \label{refnew2} \eeq since $p_1(t_1,\ldots,t_{r-1})=\sum_{i=1}^{r-1} t_i$. Thanks to the linear independence of the $p_\kappa$, the expansion of the factor to the $N$th power in (\ref{refnew2}) fixes all the coefficients $c_\kappa(s)$. Substituting these back into (\ref{refnew1}), it follows that we must have \beq \det A(s;t_1,\ldots,t_{r-1},t_r)= \sum_{k=0}^N \sum_{\kappa\vdash k} c_\kappa(s) p_\kappa(t_1,\ldots,t_{r-1},t_r) = (s^2-p_1(t_1,\ldots,t_{r-1},t_r))^N, \label{refnew3} \eeq which proves (a). To prove (b), write $A(s;t_1,\ldots,t_r)$ as $A(s;t_1,\ldots,t_r)=sI-B$, where $I$ is the identity matrix of order $2^r$. Note that $B$ is a matrix with elements similar to those of $A$ but with zeros on the diagonal. Introducing a diagonal matrix $D$ of order $2^r$ by \beq d_{{\pmb l},{\pmb l}}= \left( t_1^{l_1} t_2^{l_2}\cdots t_r^{l_r} \right)^{1/2}, \eeq it is easy to see that $DBD^{-1}$ is a real and symmetric matrix. For such matrices, all eigenvalues are real, with geometric multiplicity equal to the algebraic multiplicity. Thus this also holds for the eigenvalues of $B$. The characteristic equation of $B$ is \beq \det(B-\la I)=\det(A(\la;t_1,\ldots,t_r)) = (\la^2-\sum_{i=1}^r t_i)^{2^{r-1}} = (\la-\mu)^{2^{r-1}}(\la+\mu)^{2^{r-1}}, \label{detB} \eeq where $\mu=+\sqrt{\sum_{i=1}^r t_i}$. So for the eigenvalue $\mu$ the geometric multiplicity is $2^{r-1}$. Since the geometric multiplicity is also equal to $2^r - \rank(B-\mu I)$, it follows that $\rank(B-\mu I)=2^r-2^{r-1}=2^{r-1}$, or $\rank(A(\mu;t_1\ldots,t_r)) =2^{r-1}$, implying the statement (b). To prove (c), consider the element $c_{{\pmb l},{\pmb l'}}$ in the multiplication of $A(s;t_1,\ldots,t_r)$ with $A(-s;t_1,\ldots,t_r)$. From the definition (\ref{defA}) it follows immediately that $c_{{\pmb l},{\pmb l'}}=0$ if ${\pmb l}\ne{\pmb l'}$, and that $c_{{\pmb l},{\pmb l}}=\sum_i t_i-s^2$. \mybox This lemma can now be used to determine the linear (in)dependence of the weight vectors $X(p;{\pmb k},{\pmb l})$. Consider a weight $\la_m=(p-\sum_i m_i,m_1, \ldots, m_n)$, with all $m_i\geq 0$. Then the multiplicity of $\la_m$ in $\bar V_p$, or equivalently the dimension of the weight space $\bar V_p(\la_m)$, is given by \beq d_m=\dim \bar V_p(\la_m) = 2^{\ga(m_1)+\cdots+\ga(m_n)}=2^r, \eeq where $r$ is the number of nonzero $m_i$'s. A basis of $\bar V_p(\la_m)$ is given by the set of vectors $|p;{\pmb k},{\pmb l}\rangle$ with every $k_i+l_i=m_i$ (or ${\pmb k}+ {\pmb l}={\pmb m}$). One can consider another set of $d_m$ vectors $X(p;{\pmb k},{\pmb l})$ with ${\pmb k}+{\pmb l}={\pmb m}$. The coefficient matrix of the vectors $(-1)^{l_1+\cdots+l_n} X(p;{\pmb k},{\pmb l})$ expressed in terms of the vectors $|p;{\pmb k},{\pmb l}\rangle$ coincides with the matrix $A(\sqrt{p};t_1,\ldots,t_r)$ defined in Lemma~3, with $t_i$ corresponding to the nonzero $m_i$'s. Thus the determinant of this matrix is \beq (p-\sum_{i=1}^n m_i)^{d_m/2}. \eeq In other words, if $\sum_{i=1}^n m_i\ne p$, then the coefficient matrix is nonsingular, and the $d_m$ vectors $X(p;{\pmb k},{\pmb l})$ with ${\pmb k}+{\pmb l}={\pmb m}$ form a basis for $\bar V_p(\la_m)$. When $\sum_{i=1}^n m_i=p$, it follows from Lemma~3(b) that the span of the $d_m$ vectors $X(p;{\pmb k},{\pmb l})$ with ${\pmb k}+{\pmb l}={\pmb m}$ is a subspace of $\bar V_p(\la_m)$ of dimension $d_m/2$. \section{The simple module $V_p$} Denote by $M_p$ the maximal $G=q(n+1)$ submodule of $\bar V_p$ (different {}from $\bar V_p$ itself). Then the quotient module $V_p=\bar V_p / M_p$ is an irreducible (or simple) $q(n+1)$ module. In this section we shall show that $V_p$ is finite dimensional, and give its weight structure, character, and dimension. By definition, $V_p$ and $M_p$ are weight modules, and \beq v\in M_p \Leftrightarrow v_0 \not\in U(G) v. \label{Mp} \eeq For the weight vectors $|p;{\pmb k},{\pmb l}\rangle$ or $X(p;{\pmb k},{\pmb l})$ it will be useful to refer to the quantity $\sum_{i=1}^n (k_i+l_i)=\sum_i m_i$ as the {\em level} of the vector (or of the corresponding weight). \begin{prop} The weight vectors $v$ of $\bar V_p$ satisfy the following~: \begin{itemize} \item[(a)] if the level of $v$ is greater than $p$ then $v\in M_p$; \item[(b)] if the level of $v$ is less than $p$ then $v\not\in M_p$, and denoting the vectors of the quotient module by their representatives in $\bar V_p$ we can write $v\in V_p$; \item[(c)] if the level of $v$ is equal to $p$, consider its weight $\la_m=(0,m_1,\ldots,m_n)$. With $d_m=\dim(\bar V_p(\la_m))$, we have that $\dim M_p(\la_m)=\dim(V_p(\la_m))=d_m/2$. Moreover, the vectors $X(p;{\pmb k},{\pmb l})$ of level $p$ with $k_i+l_i=m_i$ span $M_p(\la_m)$. \end{itemize} \end{prop} \noindent {\bf Proof.} Consider a fixed weight $\la_m$, and the corresponding weight vectors $|p;{\pmb k},{\pmb l}\rangle$ and $X(p;{\pmb k},{\pmb l})$. If the level of $\la_m$ is less than $p$, then the vectors $X(p;{\pmb k},{\pmb l})$ form a basis for $\bar V_p(\la_m)$ (see end of section 6). By (\ref{bj-x}) and (\ref{fj-x}), \[ (b_1^-)^{k_1}(f_1^-)^{l_1}\cdots (b_n^-)^{k_n} (f_n^-)^{l_n} X(p;{\pmb k},{\pmb l}) \] is equal to a nonzero constant times $v_0$. Thus $v_0\in U(G) X(p;{\pmb k},{\pmb l})$, in other words all vectors of weight $\la_m$ are not in $M_p$. Let the level of $v$ be greater than $p$. Applying $b_j^-$ or $f_j^-$ reduces the level by one. But for the vectors at level $p+1$ it follows from (\ref{bj-p}) and (\ref{fj-p}) that the action of $b_j^-$ or $f_j^-$ yields only vectors of the form $X(p;{\pmb k},{\pmb l})$ at level $p$, and all these vectors themselves are annihilated by $b_j^-$ and $f_j^-$. Thus one deduces that $v_0$ cannot belong to $U(G)v$ if $v$ has level greater than $p$. Finally, consider a weight $\la_m=(0,m_1,\ldots,m_n)$ of level $p$. The vectors $X(p;{\pmb k},{\pmb l})$ with $k_i+l_i=m_i$ are all annihilated by $b_j^-$ and $f_j^-$, so it follows that they belong to $M_p$. In this case, we know (see end Section 6) that these vectors span a subspace of dimension $d_m/2$. Thus $\dim M_p(\la_m) \geq d_m/2$, and we still need to show that the dimension of $M_p(\la_m)$ does not exceed $d_m/2$. The space $\bar V_p(\la_m)$, of dimension $d_m$, is spanned by the $d_m$ vectors $|p;{\pmb k},{\pmb l}\rangle$ with $k_i+l_i=m_i$. Assume that $m_n\ne 0$ (the same argument works for another $m_i\ne 0$). Consider the sets \bea S_0 &=& \{ X(p;k_1,l_1,\ldots,k_{n-1},l_{n-1},m_n,0) | k_i+l_i=m_i \},\\ S_1 &=& \{ |p;k_1,l_1,\ldots,k_{n-1},l_{n-1},m_n-1,1\rangle | k_i+l_i=m_i \}, \\ S&=& S_0 \cup S_1. \eea Clearly, $\# S_0= \# S_1 = d_m/2$. The vectors in $S_1$ are obviously linearly independent. By considering the coefficient matrix of the vectors of $S_0$ in terms of the $d_m$ vectors $|p;{\pmb k},{\pmb l}\rangle$, and using Lemma~3, it follows that the vectors of $S_0$ are also linearly independent, and furthermore that the vectors of $S$ are linearly independent. Thus $S$ constitutes a basis for $\bar V_p(\la_m)$. The elements of $S_0$ all belong to $M_p(\la_m)$. On the other hand, ${\rm span}(S_1)$ contains no vectors that are annihilated by $P$. Indeed, consider a linear combination $c$ of the vectors in $S_1$, and express that $c$ is annihilated by $b_n^-$. Using (\ref{bj-p}), and linear independence of the vectors appearing in $b_n^- c$, it follows that $b_n^- c=0$ only if all the coefficients in the linear combination $c$ are zero. One can now deduce that no linear combination of $S_1$ can yield a vector of $M_p(\la_m)$. This shows that $\dim M_p(\la_m)=d_m/2$, hence $\dim V_p(\la_m)=d_m/2$. \mybox It can be verified that $M_p$, which is a $q(n+1)$ module, is generated by one vector $w$ as a $q(n+1)$ module (otherwise said~: $\bar V_p$ containes one $q(n+1)$ highest weight singular vector $w$). In our notation, this vector $w$ is equal to $w=X(p;p,0,\ldots,0, 0,\ldots 0)$. This proposition gives us explicitly the vectors of $M_p$. Hence it also gives the (representatives of) the vectors of $V_p= \bar V_p/M_p$. In particular, $V_p$ is finite dimensional, and the weight structure of $V_p$ can be deduced. For a weight $\la_m=(p-\sum_i m_i,m_1,\ldots,m_n)$ (with all $m_i\geq 0$), let $r$ be the number of nonzero $m_i$'s. Then the multiplicity of $\la_m$ is $2^r$ if the level of $\la_m$ is less than $p$, and $2^r/2=2^{r-1}$ if the level of $\la_m$ is equal to $p$ (and, of course, zero of the level is larger than $p$). Once the weight structure is known, it is possible to write down the character and dimension of $V_p$. To do this, it will be useful to first determine the decomposition of $V_p$ with respect to the subalgebra $gl(n+1)\subset G$. Since $V_p$ is finite dimensional, it will decompose into a direct sum of simple finite dimensional $gl(n+1)$ modules. This decomposition can be derived from the weight structure. The highest weight is $(p,0,\ldots,0)$, with multiplicity 1. So the $gl(n+1)$ module with highest weight $(p,0,\ldots,0)$ is a component of the decomposition. Subtracting the (known) weights of this $gl(n+1)$ module from the set of weights of $V_p$, leaves $(p-1,1,0,\ldots,0)$ as the next highest weight, also with multiplicity one. Then we go on~: first subtract all weights (including multiplicities) of the $gl(n+1)$ module labelled by $(p-1,1,0,\ldots,0)$; then determine the highest weight of the remaining ones, etc. Finally, one obtains~: \begin{prop} The decomposition of the $q(n+1)$ module $V_p$ into $gl(n+1)$ modules (with each $gl(n+1)$ module characterized by its highest weight) is as follows~: \beq V_p \rightarrow (p,0,\ldots,0)\oplus (p-1,1,0,\ldots,0) \oplus (p-2,1,1,0,\ldots,0)\oplus \cdots \oplus (p-n,1,1,\ldots,1). \label{decomp} \eeq The dimension of $V_p$ is given by~: \beq \dim V_p = \sum_{i=0}^n {p-1 \choose i}{p+n-i\choose n-i}. \label{dimen} \eeq \end{prop} \noindent {\bf Proof.} The decomposition follows from the known weight structure determined in Proposition~4. From the known dimension formula (e.g.\ \cite[\S 4]{Wybourne}) of simple $gl(n+1)$ modules, (\ref{dimen}) follows. \mybox The (formal) character of a $G$ module $V$ is defined as usual~: \beq \ch V = \sum_{\la} \dim V(\la) x_0^{\la_0}\cdots x_n^{\la_n}, \eeq where $\dim V(\la)$ is the multiplicity of a weight $\la$ in $V$, and the $x_i$ can be considered as formal variables. The character of $V_p$ follows from (\ref{decomp}), using the (known) characters of $gl(n+1)$ modules. For a $gl(n+1)$ module with highest weight $\la$, the character is equal to the Schur function $s_\la(x_0,\ldots,x_n)$. In this case, the highest weights appearing in the decomposition are of a special form; in fact, they are of Frobenius form $(p-1-i|i)$~\cite[p.\ 3]{Macdonald}. Since in such a case the character is given by~\cite[p.\ 47]{Macdonald}~: \beq s_{(a|b)}= h_{a+1}e_b- h_{a+2}e_{b-1}+ \cdots +(-1)^b h_{a+b+1}, \eeq where $e_r$ (resp.\ $h_r$) is the $r$th elementary (resp.\ complete) symmetric function in the $x_i$, it follows that \beq \ch V_p = h_{p-n}e_n + h_{p-n-2}e_{n-2}+ \cdots, \eeq ending with $h_p$ if $n$ is even and with $h_{p-1}e_1$ if $n$ is odd. To have the interpretation of $V_p$ as a Fock space of $q(n+1)$, we still need to show that the Hermitian form is positive definite. \begin{prop} The Hermitian form on $V_p$, induced by (\ref{form}), is positive definite. \end{prop} \noindent {\bf Proof.} It is clear from (\ref{form}) and (\ref{Mp}) that the Hermitian form is zero on $M_p$, so (\ref{form}) induces indeed a Hermitian form on $V_p= \bar V_p/M_p$. It also follows from (\ref{form}) that $\langle v | w\rangle =0$ if the weight of $v$ and $w$ is different. So it is sufficient to study the behaviour of the Hermitian form on a weight space $V_p(\la_m)$ only. Let $\la_m$ be fixed, and assume that the level of $\la_m$ is less than $p$ ($\sum m_i