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% On the Lie superalgebra embedding $C(n+1)\supset B(0,n)$
% and dimension formulas
% J. Van der Jeugt
% J. Math. Phys. 37 (1996) 4176-4186.
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\begin{document}
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\begin{center}
{\Large \bf On the Lie superalgebra embedding $C(n+1)\supset B(0,n)$\\[3mm] 
and dimension formulas}\\[2cm]
J.~Van der Jeugt\footnote{Senior Research Associate N.F.W.O. (National Fund
for Scientific Research of Belgium)} \\[8mm]
Department of Applied Mathematics and Computer Science,
University of Ghent,\\
Krijgslaan 281--S9, B9000 Gent, Belgium\\
E-mail~: Joris.VanderJeugt@rug.ac.be
\end{center}

\vskip 3cm
\begin{abstract}
It is shown that the orthosymplectic Lie superalgebras $osp(m/2n)$ have
a non-regular subalgebra $osp(1/2n)$. This implies that paraboson
operators can be realized as elements of $osp(m/2n)$. The embedding
$osp(2/2n)\supset osp(1/2n)$, or, in a different notation,
$C(n+1)\supset B(0,n)$, is studied in more detail. In particular,
branching rules are determined for all typical and atypical irreducible
representations of $osp(2/2n)$ with respect to the subalgebra
$osp(1/2n)$. Finally, dimension and superdimension formulas are given
for the Lie superalgebras under consideration.
\end{abstract}
\vskip 5mm
\noindent PACS~: 02.20b
\vskip 5mm
\noindent Short title~: $C(n+1)\supset B(0,n)$.
\newpage
%
\section{Introduction}  \label{sec-intro}

Lie superalgebras and their irreducible representations (simple modules)
have been the subject of much attention in both the
mathematical~\cite{k75,k77a,k77b,sch} and the
physics~\cite{bal,cor,don,hur,snr} literature.
However, even for the simplest family of  basic
classical Lie superalgebras, namely $sl(m/n)$,
a number of open problems remain with respect to understanding all
finite-dimensional simple modules. The main reason is the existence of
so-called {\em atypical} modules~\cite{k77a,k77b}, besides the {\em
typical} modules which are well understood. For singly atypical modules
(for which the highest weight $\La$ is atypical with respect to only
one single odd root) of $sl(m/n)$, a character formula has been
constructed~\cite{van2} and also the dimensions can be
obtained~\cite{vandim}. For the remaining atypical modules, a number of
conjectures have been formulated~\cite{van1,pen,ser}. 

The Lie superalgebras considered in this paper are the orthosymplectic
superalgebras $osp(m/2n)$, and in particular those with $m=1$ and
$m=2$. For $osp(1/2n)$, or $B(0,n)$ in Kac's notation, all
representations are typical and thus a character formula is known for
all of its finite-dimensional simple modules. An interesting feature of
$osp(1/2n)$ is its relation with the so-called paraboson algebra,
established by Palev~\cite{pal}. 
We shall use this relation here in order to
identify a non-regular $osp(1/2n)$ subalgebra in $osp(m/2n)$. 
The Lie superalgebra $osp(2/2n)$, or $C(n+1)$ in Kac's notation, does
have both typical and atypical representations. However, the atypical
modules are only singly atypical, and thus the techniques
of~\cite{van1} could be used to construct their characters~\cite{vanc}.
Thus, also here a character formula is available for all
finite-dimensional simple modules. 

Little work has been done on branching rules for Lie superalgebras. The
reason is quite obvious~: all simple Lie superalgebras except for
$B(0,n)$ have finite-dimensional indecomposable representations. Thus
for a subalgebra pair of Lie superalgebras $G_1\supset G_2$, the
decomposition of a simple $G_1$ module with respect to $G_2$ is in
general not completely reducible. The branching considered in this
paper, $C(n+1)\supset B(0,n)$, does not have this disadvantage.
Moreover, since the characters of all simple modules (both typical and
atypical) of $C(n+1)$ are known, we can give a complete set of
branching rules (proposition~2). In particular, the branching rules
shed some further light on the structure of atypical representations of
$C(n+1)$. 

The structure of the paper is as follows. Section~II gives the main
definitions for 
$osp(m/2n)$. Section~III considers the non-regular embedding
$osp(m/2n)\supset osp(1/2n)$, and as a byproduct gives a realization of
$n$ paraboson operators in $osp(m/2n)$. In Section~IV, the
representation theory of $osp(2/2n)=C(n+1)$ is recalled, and this is
used in Section~V to find the decomposition of simple $osp(2/2n)$
modules with respect to its subalgebra $osp(1/2n)$. Examples are given,
and can be checked by counting dimensions and superdimensions. In a
final section, (super)dimension formulas are given for $sp(2n)$,
$osp(1/2n)$ and $osp(2/2n)$. Some of these formulas are not new, but were
never given in the partition notation for the highest weight. The
dimension formula for atypical representations of $osp(2/2n)$ however,
is new; an alternative expression for this formula is constructed in the
appendix. 

\section{The Lie superalgebras $osp(m/2n)$ } \label{sec-osp}

The general linear Lie superalgebra
$G=gl(m/n)$ with $m,n\in\Nat$ is defined~\cite{k77a,sch} by~:
\beq
\begin{array}{rl}
gl(m/n)=\Bigl\lbrace x=\left(\matrix{A&B\cr C&D\cr}\right) \mid &
 A\in M_{m\times m}, B\in M_{m\times n}, \\
 & C\in M_{n\times m}, D\in M_{n\times n} \Bigr\rbrace ,
\end{array}
\label{defgl}
\eeq
where $M_{p\times q}$ is the space of all $p\times q$ complex
matrices. The even subspace $gl(m/n)_{\bar 0}$ has $B=0$ and
$C=0$; the odd subspace $gl(m/n)_{\bar 1}$ has $A=0$ and
$D=0$. The Lie superalgebra bracket is determined by
\beq
[a,b]=ab-(-1)^{\al\be}ba,\qquad\forall a\in G_\al\hbox{ and }
\forall b\in G_\be \quad (\al,\be\in\{ {\bar 0},{\bar 1} \}).
\label{comm-rel}
\eeq
One defines the {\it supertrace} $\hbox{str}(x)$
of $x=\bigl({A\atop C}{B\atop D}\bigr)$ as~\cite{k77a,sch}
$\hbox{str}(x)=\hbox{tr}(A)-\hbox{tr}(D)$,
where {\sl tr} is the ordinary trace.
The {\it special linear Lie superalgebra} $sl(m/n)$ is:
\beq
sl(m/n)=\lbrace x\in gl(m/n)\mid \hbox{str}(x)=0\rbrace.
\eeq
Using the definition of {\it supertranspose} of $x=\bigl({A\atop
C}{B\atop D}\bigr)$,
\beq
x^T=\left(\matrix{A^t& -C^t\cr B^t&D^t\cr}\right) ,
\eeq
with $A^t$ the ordinary transpose of $A$, the orthosymplectic
subalgebra $osp(m/2n)$ of $sl(m/2n)$ is defined~\cite{k77a,sch} as 
\bea
osp(m/2n)&=&\bigl\{ x=\left(\matrix{A&B\cr C&D\cr}\right) \in sl(m/2n) |
 x^T_\al J + (-1)^\al J x_\al =0, \nn\\
& & \qquad \forall x_\al \in sl(m/2n)_\al \; (\al={\bar 0}, {\bar 1})
\bigr\}, 
 \label{osp}
\eea
where
\beq
J=\left( \matrix{ I_m& & \cr  & 0 & I_n \cr & -I_n&0 \cr} \right)
\eeq
and $I_q$ is the $q\times q$ identity matrix. In the notation of
Kac~\cite{k77a}, $B(m,n)=osp(2m+1/2n)$, $D(m,n)=osp(2m/2n)$ for $m\ne
1$ and $C(n+1)=osp(2/2n)$.

\section{The non-regular embedding $osp(m/2n)\supset osp(1/2n)$ } 
 \label{sec-emb}

In this section the embedding $osp(m/2n)\supset osp(1/2n)$ ($m>1$) will be
established. This is a non-regular embedding, in the sense that the
root vectors of $osp(1/2n)$ are not root vectors of $osp(m/2n)$
(see Ref.~\cite{k77a}, or the following section, for the set of roots).

Let $E_{i,j} \in gl(m/2n)$ denote the matrix with entry 1 at position
$(i,j)$ and 0 elsewhere. The following matrices are odd elements of
$osp(m/2n)$~: 
\bea
b_k^+ & = & \sqrt{2\over m} \sum_{i=1}^m (E_{i,m+n+k}+E_{m+k,i}),\\
b_k^- & = & \sqrt{2\over m} \sum_{i=1}^m (E_{i,m+k}-E_{m+n+k,i}),
\qquad k=1,2,\ldots,n.
\label{bees}
\eea

\begin{prop}
The elements $b_k^\xi$ satisfy the following relation~:
\bea
[[b_i^\xi,b_j^\et],b_k^\th]& =& 2\th\de_{j,k}\de_{\th,-\et} b_i^\xi +
 2\th\de_{i,k}\de_{\th,-\xi} b_j^\et,
\label{pabos} \\[2mm]
 && \;(i,j,k=1,2,\ldots,n;\; \xi,\et,\th=\pm). \nn
\eea
\end{prop}

Note that in~(\ref{pabos}) the inner bracket stands for an
anticommutator, and the outer bracket for a commutator.
Relation~(\ref{pabos}) is the defining equation for paraboson
operators~\cite{pal,green,ohnuki}. Thus the proposition claims that a
set of $n$ paraboson operators have a realization in $osp(m/2n)$. 

The proof of~(\ref{pabos}) is by straightforward computation~: for
$(\xi,\et,\th)=(+,+,+)$ or $(-,-,-)$ the calculation is trivial; due to
symmetry of the anticommutator the remaining cases to be checked are
$(+,+,-)$, $(+,-,+)$, $(+,-,-)$ and $(-,-,+)$. 

Next, a theorem of Palev~\cite{pal} is used~: the Lie superalgebra
generated by $2n$ odd elements $b^\xi_k$ satisfying the
relations~(\ref{pabos}) is $osp(1/2n)$. Therefore it follows that
$osp(m/2n)$ contains $osp(1/2n)$ as a (non-regular) subalgebra
(algebra--subalgebra in the sense of Lie superalgebras).

\section{The Lie superalgebra $C(n+1)=osp(2/2n)$ and its representations } 
 \label{sec-cn}

In this section, $G=C(n+1)=osp(2/2n)$. The Lie superalgebra $G$ has a
Cartan subalgebra $H$ spanned by the diagonal matrices in~(\ref{osp}) with
$m=2$. 
The dual space $H^*$ is described in
the basis of forms $\ep$ and $\de_j$ ($j=1,\ldots,n$), where
$\ep:x\rightarrow A_{1,1}$ and $\de_j:x\rightarrow D_{jj}$ for
$x=\bigl({A\atop C}{B\atop D}\bigr) $.
Denote by $\De$ the set of all roots of $G$, by $\De_0$ the set of even
roots, and by $\De_1$ the set of odd roots. We have~\cite{k77a,vanc}~:
\bea
\De_0&=&\{\de_j-\de_k\;(j\ne k),\; \pm(\de_j+\de_k),\; (1\leq j,k \leq
n)\},\\ 
\De_1&=&\{\ep\pm\de_k, -\ep\pm\de_k,\;(1\leq k\leq n)\},\\
\De&=&\De_0\cup\De_1.
\eea
A set of simple roots $\{\al_0,\al_1,\ldots,\al_n\}$
of $\De$ may be chosen as follows~\cite{k77a}: $\al_0=\ep-\de_1$,
$\al_j=\de_j-\de_{j+1}$ ($1\leq j\leq n-1$), and $\al_n=2\de_n$. This
choice corresponds to the ``distinguished basis''~\cite{k77b}. 
The even and odd {\it positive} roots of $G$ are then given by
\begin{eqnarray}
\De_0^+ & = & \{\de_j-\de_k\;(j<k), \; \de_j+\de_k\},\nonumber\\
\De_1^+ & = & \{\ep\pm\de_k \}.
\label{deltas}
\end{eqnarray}

The non-degenerate bilinear form $\langle\;|\;\rangle$ on $H^*$ is determined
by~\cite{vanc} 
\beq
\langle\ep | \ep\rangle=1, \;
\langle\ep | \de_j\rangle=0, \;\hbox{ and }
\langle\de_j | \de_k\rangle=-\de_{jk}.
\eeq
An element $\La\in H^*$
with $\La=\la_0\ep+\sum_{j=1}^n\la_j\de_j$ can be written
as $\La=(\la_0;\la_1$, $\ldots$, $\la_n)$
in terms of its {\it components} in the $\ep\de$-basis,
or in terms of its {\it Dynkin
labels} $\La=[a_0;a_1,\ldots,a_n]$, where
\beq
a_0=\la_0+\la_1,\;\; a_k=\la_k-\la_{k+1}\; (1\leq k\leq n-1),
\hbox{ and } a_n=\la_n. 
\eeq
A weight $\La\in H^*$ is {\it integral dominant} if all
$a_i\in\Nat$ for $i=1,\ldots,n$. This implies that $\la_1\geq \la_2\geq
\cdots$. Thus in terms of its components, an integral
dominant weight is characterized by a complex number $\la_0$ and a
{\it partition} $\la=(\la_1,\ldots,\la_n)$ consisting of at most $n$
parts. 

Kac~\cite{k77b} showed that every integral dominant weight $\La$ uniquely
determines a simple finite-dimensional highest weight module $V(\La)$,
and vice versa that every simple finite-dimensional $G$-module is
uniquely characterized by its integral dominant highest weight $\La$.
In order to describe the structure of such a simple module $V(\La)$ by
means of its character, we need to introduce a few more objects.
The {\em Weyl group $W$} of $G$ is defined to be the Weyl group of
$G_{\bar 0}$, so in this case it is just the Weyl group of the
symplectic Lie algebra $sp(2n)$. As usual, $\vep(w)$ denotes the
signature of $w\in W$. Further on, we denote $\rh=\rh_0-\rh_1$ with
\beq
\rh_0={1\over 2} \sum_{\al\in\De_0^+} \al,\qquad
\rh_1={1\over 2} \sum_{\be\in\De_1^+} \be,
\eeq
or explicitly, $\rh_0=\sum_{k=1}^n (n+1-k)\de_k$ and $\rh_1=n\ep$. 

Kac~\cite{k77b} distinguished between {\it typical} and {\it atypical}
modules~: $V(\La)$ and $\La$ are said to be typical if and only if
$\langle \La+\rh|\be \rangle \ne 0$ for all $\be\in\De_1^+$. Otherwise,
$V(\La)$ and $\La$ are said to be atypical. Explicitly, an integral
dominant weight $\La=(\la_0;\la)$ is typical if 
\beq
\la_0\ne n\pm(\la_k+n+1-k),\qquad (k=1,2,\ldots,n).
\eeq
For a typical module $V(\La)$, Kac proved the following character
formula~\cite{k77b}~: 
\beq
\cha V(\La)=L_0^{-1}\sum_{w\in W}\vep(w) w\bigl( e^{\La+\rh_0}
\prod_{\be\in\De_1^+} (1+e^{-\be}) \bigr),
\label{typchar}
\eeq
where $L_0=\prod_{\al\in\De_0^+}(e^{\al/2}-e^{-\al/2})$ and 
$e^\la$ ($\la\in H^*$) is the {\em formal exponential}.

For atypical modules of $C(n+1)$, the characters were determined
in~\cite{vanc}, where two alternative expressions were given. The one we
shall use here is~:
\beq
\cha V(\La)=L_0^{-1}\sum_{w\in W}\vep(w) w\bigl( e^{\La+\rh_0}
\prod_{\ss\be\in\De_1^+ \atop \ss \langle\La+\rh|\be\rangle\ne 0}
(1+e^{-\be}) \bigr). 
\label{atypchar}
\eeq
These character formulas contain all the information on the weight
structure of the simple modules. Hence, to determine the decomposition
of a $G$ module with respect to a subalgebra of $G$, the character is
in priniciple all that is needed.

\section{The branching $C(n+1)=osp(2/2n)\supset osp(1/2n)=B(0,n)$ } 
 \label{sec-bra}

The embedding $osp(2/2n)\supset osp(1/2n)$ was described in
Section~\ref{sec-emb}.
The two Lie superalgebras have almost the same even subalgebra~:
$osp(2/2n)_{\bar 0}=\C\oplus sp(2n)$ and $osp(1/2n)_{\bar 0}=sp(2n)$.
They have the same Weyl group $W$, namely the Weyl group of $sp(2n)$.
And the weight space of $osp(1/2n)$ can be identified with the subspace
of $H^*$ spanned by the $\de_k$'s only. Thus for an
$osp(2/2n)$ module $V(\La)$ reduced to $osp(1/2n)$, the
$osp(1/2n)$ weights are determined by projecting the $osp(2/2n)$
weights on this subspace, i.e.\ by putting the $\ep$-component equal 
to~0. Since all finite-dimensional modules of $osp(1/2n)$ are
completely reducible (cfr.\ Kac~\cite{k77b}; note that $osp(1/2n)$ is
the only simple Lie superalgebra with this property),
one simply has to decompose the character of this $osp(1/2n)$ module
into simple $osp(1/2n)$ characters in order to obtain the decomposition
of $V(\La)$ into simple $osp(1/2n)$ modules.

A simple $osp(1/2n)$ module $V=V(\mu)$ is uniquely characterized by its
highest weight $\mu=\sum_{j=1}^n \mu_j\de_j =(\mu_1,\ldots,\mu_n)$
which is integral dominant, i.e.\ $\mu$ forms a partition. All $V(\mu)$
are typical, with their character given by
\beq
\cha V(\mu)=L_0^{-1}\sum_{w\in W}\vep(w) w\bigl( e^{\mu+\rh_0}
\prod_{j=1}^n (1+e^{-\de_j}) \bigr); \label{chmu}
\eeq
herein, $L_0$, $W$ and $\rh_0$ are the same as in (\ref{typchar}).
The procedure to determine the decomposition of a simple $osp(2/2n)$
module $V(\La)$ into simple $osp(1/2n)$ modules $V(\mu)$ is thus as
follows~: consider the character $\cha V(\La)$, substitute herein
$\ep=0$, and rewrite the result as a sum of characters $\cha V(\mu)$ of
the type~(\ref{chmu}). 
This procedure can formally be executed for both typical and atypical
representations of $osp(2/2n)$. Without going into the details of the
computation, we give the main result in Proposition~2. Before
formulating this result, recall that a partition
$\la=(\la_1,\la_2,\cdots)$ is conveniently represented~\cite{mac} by its Young
diagram $F^\la$ consisting of left-adjusted rows of boxes of length
$\la_i$. A diagram $F^\si$ obtained from $F^\la$ by adding or removing
certain boxes is called {\it standard} if it still corresponds to a
partition, i.e.\ if $\si_1\geq \si_2\geq \cdots$. For example, let
\beq
F^{(3,2,2)} = \vcenter
 {\offinterlineskip
 \halign{&\mystrut\vrule#&\hbox to 11.6pt{\hss$#$\hss}\cr
  \multispan{7}\hrulefill\cr
  & && && &\cr
  \multispan{7}\hrulefill\cr
  & && &\cr
  \multispan{5}\hrulefill\cr
  & && &\cr
  \multispan{5}\hrulefill\cr
  }} \; .
\eeq
Adding one box in the second row yields a standard diagram; adding one
box in the third row yields a non-standard diagram.

\begin{prop}
Let $\La=(\la_0;\la)=(\la_0;\la_1,\la_2,\ldots,\la_n)\in H^*$ be
integral dominant and $V(\La)$ the simple $osp(2/2n)$ module with
highest weight $\La$. Then the set of $\mu$'s determining the
decomposition of $V(\La)$ into simple $osp(1/2n)$
modules $V(\mu)$ is found as follows.
\begin{itemize}
\item[(a)] If $\La$ is typical, construct all standard diagrams $F^\mu$
obtained from $F^\la$ by adding either no or else 1 box into each row
of $F^\la$.
\item[(b)] If $\La$ is atypical with respect to $\ep-\de_k$, i.e.\
$\la_0= -\la_k+k-1$, then construct all standard diagrams $F^\mu$
obtained from $F^\la$ by adding no box in row $k$ and by adding 
either no or else 1 box into each row different from $k$.
\item[(c)] If $\La$ is atypical with respect to $\ep+\de_k$, i.e.\
$\la_0= \la_k-k+2n+1$, then construct all standard diagrams $F^\mu$
obtained from $F^\la$ by adding one box in row $k$ and by adding 
either no or else 1 box into each row different from $k$.
\end{itemize}
\end{prop}

The proposition makes the distinction between typical and atypical
modules very clear.
To illustrate this, consider $osp(2/6)\supset osp(1/6)$ for $V(\La)$
with $\La=(\la_0;3,2,2)$. For $\la_0\not\in\{-3,-1,0,6,7,9\}$, $V(\La)$
is typical and the decomposition is given by
$$
\begin{array}{ccccccccccccc}
(\la_0;3,2,2)&\rightarrow&(3,2,2)&\oplus&(4,2,2)&\oplus& (3,3,2)& \oplus&
(4,3,2)& \oplus& (3,3,3)& \oplus& (4,3,3).\\
24192 & = &1386&+&4095&+&2310&+&9009&+&1386&+&6006\\
0 & = &54&-&135&-&70&+&231&+&30&-&110
\end{array}
$$
If $\la_0=-1$, $\La$ is atypical with respect to $\ep-\de_2$ and in this
case the decomposition is~:
\beq
\begin{array}{ccccc}
(-1;3,2,2)&\rightarrow&(3,2,2)&\oplus&(4,2,2).\\
 5481&=&1386&+&4095\\
 81&=&-54&+&135
\end{array}
\eeq
If $\la_0=6$, $\La$ is atypical with respect to $\ep+\de_3$ and then
the decomposition is given by~:
\beq
\begin{array}{ccccc}
(6;3,2,2)&\rightarrow& (3,3,3)& \oplus& (4,3,3).\\
 7392&=&1386&+&6006\\
 80&=&-30&+&110
\end{array} \label{23}
\eeq

Below the representation labels are given the dimensions ($\dim V= \dim
V_{\bar 0}+\dim V_{\bar 1}$) and so-called superdimensions
($\hbox{sdim} V= \dim V_{\bar 0}-\dim V_{\bar 1}$). Some but not all of
these dimensions can be found in the tables of Thierry-Mieg~\cite{t-m}.
In the next section, we shall list the known dimension formulas for
$osp(m/2n)$ ($m=0,1,2$) and give some new expressions.

\section{Dimension formulas}
 \label{sec-dim}

Below we summarize the dimension and superdimension formulas for the
simple modules of the Lie superalgebras treated in this paper, in an
obvious notation. First we recall the dimension formula of the simple
module with highest weight $\la=\sum_{i=1}^n \la_i\de_i$ for the Lie
algebra $sp(2n)$, taken from Ref.~\cite{weyl}. Next are given the
superdimension and dimension formula for $osp(1/2n)$; these can be
deduced from~\cite{k77b} (up to some printing errors). From the same
reference follow the dimension and superdimension of {\em typical}
representations of $osp(2/2n)$ (here, the highest weight of the module
is $\La=(\la_0;\la)$, but $\la_0$ does not appear in the dimension
formulas). 

\bea
&&\dim_{sp(2n)}(\la)=
 \prod_{i=1}^n {\la_i+n+1-i\over n+1-i} 
 \prod_{1\leq i<j\leq n} {(\la_i-\la_j+j-i)(\la_i+\la_j+2n+2-i-j) \over
 (j-i)(2n+2-i-j)} ;\nn\\
&& \\
&&\hbox{sdim}_{osp(1/2n)}(\la)=
 \prod_{1\leq i<j\leq n} {(\la_i-\la_j+j-i)(\la_i+\la_j+2n+1-i-j) \over
 (j-i)(2n+1-i-j)} ;\\[2mm]
&&\dim_{osp(1/2n)}(\la)=
\prod_{i=1}^n {\la_i+n-i+{1\over 2}\over n-i+{1\over2}} \quad
 \hbox{sdim}_{osp(1/2n)}(\la)
; \\[2mm]
&&\hbox{sdim}_{osp(2/2n)}^{typ}(\la_0;\la)= 0 ;\\[2mm]
&&\dim_{osp(2/2n)}^{typ}(\la_0;\la)= 2^{2n} \dim_{sp(2n)}(\la).
\eea

What remains to be determined are formulas for the dimensions and
superdimensions of {\em atypical} irreducible representations of
$osp(2/2n)$. For the superdimension, a formula can be derived from
equation~(6.21) of Ref.~\cite{vanc}~:
\beq
\hbox{sdim}_{osp(2/2n)}^{atyp : \ep\pm\de_k}(\la_0;\la) =
 \dim_{sp(2n-2)} (\la_1+1,\la_2+1,\ldots,\la_{k-1}+1,
  \la_{k+1},\la_{k+2},\ldots,\la_n).
\eeq
Again, the notation should be obvious~:
$\hbox{sdim}_{osp(2/2n)}^{atyp : \be}(\la_0;\la) $ stands for the
superdimension of a simple module with highest weight $\La=(\la_0;\la)$
atypical with respect to a positive odd root $\be$.
The dimension formula for an atypical $osp(2/2n)$ module is more
difficult to obtain. In fact, contrary to the dimension formulas given
so far, it is not completely but only partially factorizable into
monomials in the $\la_i$.
A technique to obtain a dimension formula from a character formula of
the type~(\ref{atypchar}) was given by Kac and Wakimoto~\cite{KW}.
Following this technique leads in the present case to the following
formulae~: 
\bea
&&\dim_{osp(2/2n)}^{atyp : \ep+\de_k} (\la_0;\la) =
{2^{2n-1}\over n!} \left(\prod_{i=1}^n {(2i)!\over (n-i)!(n+i)!}\right)
\prod_{\ss i=1 \atop \ss i\ne k}^n x_i
\prod_{\ss 1\leq i<j\leq n \atop \ss i\& j \ne k}(x_i-x_j)(x_i+x_j) \nn\\
&&\times (-1)^{k-1}\sum_{j=0}^{2n-1}\sum_{l=0}^{j}(-1)^l 2^{-j}
\left(j \atop l\right) (x_k-l)\prod_{\ss i=1 \atop \ss i\ne k}^n
(x_k-l-x_i)(x_k-l+x_i),  \label{atypplus}\\[2mm]
&&\qquad\hbox{ where } x_i=\la_i+n+1-i; \nn
\eea
\bea
&&\dim_{osp(2/2n)}^{atyp : \ep-\de_k} (\la_0;\la) =
{2^{2n}\over n!} \left(\prod_{i=1}^n {(2i)!\over (n-i)!(n+i)!}\right)
\prod_{\ss i=1 \atop \ss i\ne k}^n x_i
\prod_{\ss 1\leq i<j\leq n \atop \ss i\& j \ne k}(x_i-x_j)(x_i+x_j) \nn\\
&&\times (-1)^{k-1}\sum_{j=0}^{2n-1}\sum_{l=0}^{j}(-1)^l 2^{-j}
\left(j \atop l\right) (x_k-l)\prod_{\ss i=1 \atop \ss i\ne k}^n
(x_k-l-x_i)(x_k-l+x_i),  \label{atypmin}\\[2mm]
&&\qquad\hbox{ where } x_i=\la_i+n+1-i\;(i\ne k)\hbox{ and } 
x_k=\la_k+n-k. \nn
\eea
For example, consider $osp(2/6)$ and representations that are atypical
with respect to $\ep+\de_3$; then~(\ref{atypplus}) yields
\bea
&&\dim_{osp(2/6)}^{atyp : \ep+\de_3} (\la_0=\la_3+4;\la_1,\la_2,\la_3)
=\nn\\ 
&&\quad{1\over 90}(\la_1+3)(\la_2+2)(\la_1-\la_2+1)(\la_1+\la_2+5)
(2\la_3+3)(2\la_1^2\la_2^2-2\la_1^2\la_3^2\nn\\
&&\quad-2\la_2^2\la_3^2+2\la_3^4+8\la_1^2\la_2+12\la_1\la_2^2
-6\la_1^2\la_3-6\la_2^2\la_3-12\la_1\la_3^2-8\la_2\la_3^2+
12\la_3^3+5\la_1^2\nn\\
&&\quad+48\la_1\la_2+15\la_2^2
-36\la_1\la_3-24\la_2\la_3-4\la_3^2+30\la_1+60\la_2-66\la_3+35).
\label{expl}
\eea
Using this formula, the dimension in (\ref{23}) can be calculated.
Note in~(\ref{expl}) that the polynomial obtained from the sum over $j$
and $l$ part in~(\ref{atypplus}), with the proper
arguments, gives rise to one more factor, namely $(2\la_3+3)=(2x_3+1)$.
All other examples we have worked through confirm this extra factor. 
This comes as no surprise. Indeed, the branching rule for an atypical
representation (of type $\ep+\de_k$, for example) determined in the
previous section implies that all $osp(1/2n)$ components of this
representation have $(x_k+{1\over 2})$ as a factor for their
dimensions, thus also the total $osp(2/2n)$ dimension should have this
factor. 
This means that the polynomial part (i.e.\ sum over $j$ and
$l$) in (\ref{atypplus}) or~(\ref{atypmin}) can be rewritten in another
form, giving this extra factor explicitly. 
Identifying this polynomial in~(\ref{atypplus}) as
$P_{n-1}(x_k;x_1,x_2, \ldots, x_{k-1},x_{k+1}, \ldots, x_n)$, it is of
the type
\beq
P_N(z;z_1,\ldots,z_N)=\sum_{j=0}^{2N+1}\sum_{l=0}^j (-1)^l 
\left( j \atop l\right) 2^{-j}
 (z-l)\prod_{i=1}^N (z-l-z_i)(z-l+z_i),
\label{PN1}
\eeq
and the presently obtained branching rule implies that this polynomial
has a factor $(2z+1)$.
After various combinatorial manipulations, the details of which are
presented in the Appendix, we arrived indeed at the following expression~:
\beq
P_N(z;z_1,\ldots,z_N)=(2z+1) \sum_{p=0}^N \left(\sum_{m=0}^p \al_{p,m}
\prod_{q=2}^{m+1} (2z+q) \right) e_{N-p}.
\label{PN2}
\eeq
Herein, $e_r$ is the $r^{\rm th}$ elementary symmetric
function~\cite{mac} in the variables $z^2-z_1^2,\ldots ,z^2-z_N^2$, and
an expression for the coefficient $\al_{p,m}$ is given by
\beq
\al_{p,m}=\sum_{j=0}^{2p+1} \sum_{a=0}^j \sum_{b=0}^m 2^{-j-1}
(-1)^{a+b} {1\over(m+1)!} \left(j\atop a\right) \left(m+1\atop
b+1\right) a^p(a+b+2)^p(2a+b+2).
\label{coef}
\eeq
In the last formula, $0^0$ should be interpreted as 1. 

\section*{Appendix}
\renewcommand{\theequation}{A.{\arabic{equation}}}
\setcounter{equation}{0}
\setcounter{theo}{0}

First, we prove two short lemmas.

\begin{lemm} Let $f(z)$ and $g(z)$ be real polynomials in $z$ with
$g(z-{1\over 2})+g(z+{1\over 2})=f(z)$. If $f(-z)=-f(z)$, then also
$g(-z)=-g(z)$. 
\end{lemm}

\noindent{\bf Proof.}
Putting $z=0$, and using $f(0)=0$, yields $g(-1/2)=-g(1/2)$. Assume now
$g(-k/2)=-g(k/2)$ for $k$ an odd positive integer. Then,
\[
g({k\over 2})+g({k\over 2}+1)=f({k+1\over 2})=
-f(-{k+1\over 2})=-g(-{k\over 2})-g(-{k\over 2}-1),
\]
thus also $g(-{k\over 2}-1)=-g({k\over 2}+1)$. By induction
$g(-k/2)=-g(k/2)$ for all odd positive integers $k$. Since $g(z)$ is a
polynomial, this can only happen when $g(-z)=-g(z)$.

The second lemma follows from numerical analysis results, and uses the
translation operator $Ef(z)=f(z+1)$ and the forward difference operator
$\Delta f(z)=f(z+1)-f(z)$; thus $\Delta=E-1$.

\begin{lemm} Let $f(z)$ be a polynomial in $z$ of degree $n$. Then
\[
\sum_{k=0}^j (-1)^k \left(j\atop k\right) f(z+k)=0\qquad
\hbox{for } j>n.
\]
\end{lemm}

\noindent {\bf Proof.} The sum can be written as
\[
\sum_{k=0}^j (-1)^k \left(j\atop k\right) f(z+k)=
\sum_{k=0}^j (-1)^k \left(j\atop k\right) E^k f(z)=
(1-E)^j f(z) = (-\Delta)^j f(z),
\]
and this is zero, since it is a classical result that $\Delta^j f(z)=0$
for $f(z)$ a polynomial with degree $n<j$.

\def\com#1#2{\left(#1\atop #2\right)}

Consider now (\ref{PN1}); using 
\[
\prod_{i=1}^N(z-l-z_i)(z-l+z_i)=\prod_{i=1}^n(l^2-2zl+(z^2-z_i^2)),
\]
and
\[ 
 \prod_{i=1}^N (x+x_i)=\sum_{n=0}^N x^n e_{N-n}(x_1,\ldots,x_N),
\]
with $e_r$ the $r^{\rm th}$ elementary symmetric function, we see that
\[
P_N(z;z_1,\ldots,z_N)=\sum_{n=0}^N p_n(z)
e_{N-n}(z^2-z_1^2,\ldots,z^2-z_N^2), \]
with
\[
p_n(z)=\sum_j\sum_l (-1)^l \com{j}{l} 2^{-j} (z-l)l^n(l-2z)^n.
\]
By Lemma~2, the sum over $j$ runs from 0 to $j>2n+1$; the sum over $l$ is
restricted by the binomial coefficient. Next, we use
\beas
(z-l)l^n(l-2z)^n&=&\sum_i \com{n}{i}l^{2n-i} (-2z)^i (z-l)\\
&=& -{1\over 2}\sum_i \com{n}{i} l^{2n-i}(-2z)^{i+1}-
 \sum_i \com{n}{i} l^{2n+1-i}(-2z)^{i}\\
&=& -{1\over 2}\sum_i \com{n}{i-1} l^{2n+1-i}(-2z)^{i}-
 \sum_i \com{n}{i} l^{2n+1-i}(-2z)^{i}\\
&=&-{1\over 2} \sum_i l^{2n+1-i}(-2z)^i \left\{
\com{n}{i-1}+2\com{n}{i} \right\}\\
&=&-{1\over 2} \sum_i l^{2n+1-i}(-2z)^i \left\{
\com{n}{i}+\com{n+1}{i} \right\},
\eeas
so
\beq
p_n(z)=\sum_{i,j,l}(-1)^{l+1} \com{j}{l} 2^{-j-1} \left\{
\com{n}{i}+\com{n+1}{i} \right\} l^{2n+1-i} (-2z)^i.
\label{pn}
\eeq
In order to see that $p_n(z)$ has $2z+1$ as a divisor, and to rewrite
the quotient in an appropriate form, we shall express all quantities in
terms of the functions $w_m(z)$, where $w_0(z)=1$ and
\[
 w_m(z)=\prod_{q=1}^m (-2z-q).
\]
Using the following formula from combinatorial analysis,
\[ 
x^r=\sum_{m,k} {(-1)^{m-k}\over m!} \com{m}{k} k^r x(x-1)\cdots
(x-m+1),
\] 
on $(-2z)^i=\sum_r\com{i}{r} (-2z-1)^r$, yields
\beas
(-2z)^i&=& \sum_r \com{i}{r} \sum_{m,k} {(-1)^{m-k}\over m!} \com{m}{k}
k^r w_m(z) \\
&=& \sum_{m,k} {(-1)^{m-k}\over m!} \com{m}{k} (k+1)^i w_m(z) .
\eeas
Putting this back in (\ref{pn}) gives
\[
p_n(z)=\sum_{m=0}^{2n+1} \beta_{n,m} w_m(z),
\]
with
\beas
\beta_{n,m}&=&\sum_{i,j,l,k} (-1)^{l+1} \com{j}{l} 2^{-j-1} \left\{
\com{n}{i}+\com{n+1}{i} \right\} l^{2n+1-i} {(-1)^{m-k}\over m!}
\com{m}{k} (k+1)^i \\
&=&\sum_{j,l,k} {(-1)^{l+m+k+1}\over m!} \com{j}{l} \com{m}{k} 2^{-j-1}
\sum_i\left\{\com{n}{i}+\com{n+1}{i} \right\} l^{2n+1-i} (k+1)^i \\
&=&\sum_{j,l,k} {(-1)^{l+m+k+1}\over m!} \com{j}{l} \com{m}{k} 2^{-j-1}
\left\{ l^n(l+k+1)^{n+1} + l^{n+1}(l+k+1)^n \right\}\\
&=&\sum_{j,l,k} {(-1)^{l+m+k+1}\over m!} \com{j}{l} \com{m}{k} 2^{-j-1}
l^n(l+k+1)^{n}(2l+k+1).
\eeas

This gives a proper form for the coefficients of $p_n(z)$ in terms of
$w_m(z)$. It remains to be shown that the coefficient of $w_0(z)$
vanishes, or $\beta_{n,0}=0$ for all $n$, where
\[
\beta_{n,0}=\sum_{j,l} (-1)^{l+1} \com{j}{l}
2^{-j-1}l^n(l+1)^{n}(2l+1). 
\]
Consider
\[
t_n=\sum_{k=0}^n \com{n}{k} 4^{n-k} \beta_{n-k,0};
\]
since this is a traingular system of the $\beta_{n-k,0}$ in terms of
the $t_n$, all $\beta_{j,0}$ are zero if we prove that all $t_n$ are
zero for $n=0,1,2,\ldots$. For $t_n$, there comes the expression
\beas
t_n&=& \sum_{k,j,l} \com{n}{k} 4^{n-k} (-1)^{l+1} \com{j}{l}
2^{-j-1}(l^2+l)^{n-k}(2l+1)\\
&=& \sum_{j,l} (-1)^{l+1} \com{j}{l} 2^{-j-1}(2l+1)^{2n+1}.
\eeas
Consider
\[
T_n(z)=\sum_{j} 2^{-j-1} \sum_l (-1)^{l+1} \com{j}{l} (2z+2l+1)^{2n+1};
\]
then $t_n=T_n(0)$, and with $f(z)=(2z+1)^{2n+1}$ we have (using the
notation of Lemma~2)
\beas
T_n(z)&=& \sum_{j} 2^{-j-1} \sum_l (-1)^{l+1} \com{j}{l} f(z+l)
= -\sum_j 2^{-j-1} (1)^j \Delta^j f(z)\\
&=& -{1\over 2} \sum_j (-{\Delta\over 2})^j f(z)
=  -{1\over 2} {1\over 1+\Delta/2} f(z) = {-1 \over 1+E} f(z).
\eeas
Hence $T_n(z)$ is the polynomial solution of
$T_n(z+1)+T_n(z)=-(2z+1)^{2n+1}$.  Replacing herein $z$ by $z-1/2$
leads to $T_n(z+1/2)+T_n(z-1/2)=-(2z)^{2n+1}$; using Lemma~1 implies
that $T_n(z)$ is an odd function of $z$, thus $t_n=T_n(0)=0$.
Combining all results now leads to (\ref{PN2}) and (\ref{coef}).

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\end{document}
