\section{Introduction}
There are a number of results which connect classical affine Lie algebras with interchanged level and rank; such phenomena are generally called {\em level-rank} duality. One manifestation of this is that the braided fusion categories associated with such Lie algebras are closely related.  
In type $A$ a result of this type was proved in \cite{OS}. It states that there is a braid-reversing tensor equivalence between $\Cc(\ssl_n)_k^0$ and $\Cc(\ssl_k)_n^0$ where $\Cc({\ssl}_n)_k^0$
is the adjoint subcategory of the modular tensor category $\Cc(\ssl_n)_k$ obtained from the affine Lie algebra $\widehat\ssl_n$ at level $k$. Recall that an alternative construction of $\Cc(\ssl_n)_k$ is via the semisimplification of the category of tilting modules of the quantum group $U_q\ssl_n$ specialized at $q=e^{\pi i/(n+k)}$, typically denoted $\Cc(\ssl_n,n+k)$.


In this paper we examine the analogous situation for the type $C$ case associated with symplectic Lie algebras $\ssp_{2n}$.  Just as in the type $A$ case, the unitary modular category $\Cc(\ssp_{2n})_k$  may be constructed in (at least) two ways: 1) as the semisimplification $\Cc(\ssp_{2n},2k+2n+2)$ of the subcategory of tilting modules in $Rep(U_q \ssp_{2n})$ for the choice $q=e^{\pi i/(2k+2n+2)}$ and 2) as the category $\Cc(\ssp_{2n})_k$ of level $k$ representations of the affine Lie algebra $\widehat{\ssp}_{2n}$ under the level preserving tensor product (see \cite{BK}). The braided monoidal equivalence of the categories constructed from these two approaches can be found in \cite{finkelberg,Huang}.  Despite this equivalence we will use both notations to distinguish the approaches.

From any braided fusion category $\Cc$ we may obtain a new braided fusion category $\Cc^{rev}$ with the same underlying fusion category by replacing the braiding isomorphisms $c_{X,Y}$ with $c_{Y,X}^{-1}$.
For any $\Z/2$-graded braided fusion category $\Cc=\Cc_0\oplus\Cc_1$ we may also obtain a new braided structure by replacing the braiding $c_{X,Y}$ for $X,Y\in\Cc_1$  by $-c_{X,Y}$, leaving the braiding unchanged if either of $X$ or $Y$ are in $\Cc_0$.  One way to construct this new braided category directly is to take the diagonal of the $\Z/2\times\Z/2$-graded category $\Cc\boxtimes sVec$.   We will denote by $\Cc^{-}$ the braided fusion category obtained in this way.
We prove the following:
\begin{theorem} There is a braid-reversing equivalence between
$\Cc(\ssp_{2n})_k$ and $\Cc(\ssp_{2k})_n^{-}$.
\end{theorem}
We provide two proofs of this theorem.  The first, direct, proof follows the strategy of \cite{OS} using the conformal embedding $(\widehat{\mathfrak{sp}}_{2n})_k\oplus (\widehat{\mathfrak{sp}}_{2k})_n\subset (\widehat{\mathfrak{so}}_{4nk})_1.$  The second employs reconstruction techniques for braided fusion categories with fusion rules of type $C$, found in \cite{TW}.  While the latter proof is shorter, it invokes some heavy categorical machinery.  On the other hand, this categorical approach allows us to prove a related result for $\Cc(\ssp_{2n},\ell)$ for \emph{odd} $\ell$ where now the category $\Cc(\sso_{2k},\ell)$ with $\ell=2k+2n+1$ plays the role of the dual.  These are typically non-unitarizable categories, and hence cannot be constructed from affine Lie algebras in the standard way.  

It would be interesting to extend our results to the orthogonal case where the level-rank duality
connects affine Lie algebras $(\widehat{\mathfrak{so}}_{n})_k$ and $(\widehat{\mathfrak{so}}_{k})_n$. However this case seems to be technically more involved than the symplectic case 
and it is not considered in this paper. We refer the reader to \cite{Has,JL,Muk} for some interesting results in this direction.

This article is organized as follows.  In Section \ref{prelim} we lay the basic Lie theoretic and combinatorial groundwork and in Section \ref{branching} we describe the key conformal embedding.  Section \ref{main results} contains two proofs of our level-rank duality theorem.  The appendix contains a detailed proof of the Kac-Peterson formula in the symplectic case.


\section{Preliminaries}\label{prelim}
\subsection{Combinatorics}\label{combinatorics}
Let $\lambda = (\lambda_1\geq\dots\geq\lambda_n\ge 0)$ be a \emph{partition} of $|\lambda | := \lambda_1 + \dots + \lambda_n$ with $n$ parts (see e.g. \cite{Mac}). We will identify $\lambda$ with its corresponding \emph{Young diagram}. Let $I_{n,k}$ be the set of all partitions $\lambda$ whose Young diagram fits into an $n\times k$ rectangle; in other words $I_{n,k}$ consists of partitions with $n$ parts and $\lambda_1\le k$. We will denote by $I_{n,k}^0$ (respectively 
$I_{n,k}^1$) the subset of $I_{n,k}$ consisting of partitions $\lambda$ such that $|\lambda|$ is
even (respectively odd).

Denote by $\lambda^t$ the transposed partition of $\lambda$. Clearly, $\lambda\in I_{n,k}$ implies $\lambda^t\in I_{k,n}$. Denote by $\lambda^c$ the transpose of the complement of $\lambda$ in an $n\times k$ rectangle. Again,  $\lambda\in I_{n,k}$ implies $\lambda^c\in I_{k,n}$. We will also
consider a composition $(\lambda^t)^c=(\lambda^c)^t=:\lambda^{tc}$ which preserves the set
$I_{n,k}$.

\begin{eg} 
Let $\lambda =(2,1,1)\in I_{3,2}$. Then $\lambda^t=(3,1)\in I_{2,3}$, $\lambda^c=(2,0)\in I_{2,3}$,
and $\lambda^{tc}=(1,1,0)\in I_{3,2}$.
\end{eg}

Let $C_{n,k} :=\{(k_0,k_1,..., k_n) \in \N^{n+1} | \sum_i k_i = k\}$ be the set of  dominant weights for $\widehat{\mathfrak{sp}}_{2n}$
of level $k$. We identify the sets $C_{n,k}$ and $I_{n,k}$ via the mutually inverse bijections
$c_{n,k}: I_{n,k}\to C_{n,k}$, 
$c_{n,k}(\lambda) := (k-\lambda_1,\lambda_1- \lambda_2,\lambda_2-\lambda_3,\dots,\lambda_{n-1}-\lambda_n, \lambda_n)$ and $d_{n,k}: C_{n,k}\to I_{n,k}$, $d_{n,k}(k_0, k_1, \dots, k_{n})=(k_1+\dots+k_{n},k_2+\cdots+k_{n},\dots, k_{n})$. With these identifications the bijection $C_{n,k}\to C_{k,n}$
corresponding to $\lambda \mapsto \lambda^c$ can be described as follows (see \cite{KP}):
if we consider $n+k$ dots, first represent $(k_0,\dots, k_n)\in C_{n,k}$ by picking $n$ dots to be black (leaving the other $k$ dots white) so that there are $k_0$ white dots before the first black dot, $k_1$ white dots between the first and second black dot, and continue this until the last $k_n$ dots are after the last black dot. In this way the black dots partition the white dots according to an element of $C_{n,k}$. Its corresponding element in $C_{k,n}$ is how the white dots partition the black dots (see example below). 
\begin{eg}\label{bwdots} Let $n=7$, $k=6$ consider $n+k=13$ dots
$$\bullet\bullet\circ\bullet\bullet\bullet\bullet\circ\circ\circ\bullet\circ \circ$$
$$C_{7,6}\ni (0,0,1,0,0,0,3,2) \mapsto(2,4,0,0,1,0,0)\in C_{6,7}.$$
The corresponding partitions are $(6,6,5,5,5,5,2)\in I_{7,6}$ and $(5,1,1,1)\in I_{6,7}$.
\end{eg}

The bijection $\lambda \mapsto \lambda^{tc}$ also has a simple interpretation in this language:
one has to read the diagram of black and white dots representing $\lambda$ backwards.
\subsection{Symplectic Lie algebra} \label{symplectic}

Let $\mathfrak{sp}_{2N}$ be the Lie algebra of $2N\times 2N$ symplectic matrices over $\C$,
see e.g. \cite{Hum}.
\subsubsection{The root system of $\mathfrak{sp}_{2N}$}
Following \cite{Hum}, let $e_1,e_2,\dots,e_N$ be an orthonormal basis for $\mathfrak{h}^*$ such that the simple roots are given by
$$e_1-e_2, e_2-e_3,\dots, e_{N-2}-e_{N-1}, e_{N-1}-e_N,2e_N.$$
The positive long roots are therefore given by
$$2e_1,2e_2,\dots, 2e_N$$
Let $W$ be the Weyl group of $\mathfrak{sp}_{2N}$. We recall that the group $W$ identifies
with the group of signed permutations of the basis $e_1,e_2,\dots,e_N$, see \cite{Hum}.

\subsubsection{Affinization} Let $\widehat{\mathfrak{sp}}_{2N}$ be the affinization of $\mathfrak{sp}_{2N}$ (see e.g. \cite[Chapter 7]{Kac}). 
Thus $\widehat{\mathfrak{sp}}_{2N} = (\mathfrak{sp}_{2N}\otimes\C[t,t^{-1}])\oplus\C K$ where the element $K$ is central and (for $\kappa$ the Killing form)
$$[x\otimes t^a,y\otimes t^b]=[x,y]\otimes t^{a+b}+a\kappa(x,y)\delta_{a,-b}K.$$
\subsubsection{The root system of $\widehat{\mathfrak{sp}}_{2N}$}\label{affroots}
Let $\hat{W}$ be the affine Weyl group. Let $T\subset \hat{W}$ be the subgroup of translations, so $\hat{W}=W\rtimes T$ (\cite[Proposition 6.5]{Kac}). Write $\hat{\Delta}$ and $\hat{\Delta}^+$ for the affine root system and its positive roots respectively. Let $\delta$ be the indivisible positive imaginary root. Then $w(\delta)=\delta$ for all $w\in\hat{W}$ and the imaginary roots are nonzero multiples of $\delta$. The real roots of $\hat{\Delta}$ are $\Delta+\Z\delta$, with $\hat{\Delta}^+=\Delta^+\cup (\Delta+\Z_{>0}\delta)$. 
\subsubsection{Representations}Let $\Cc(\mathfrak{sp}_{2n})_k$ be the category of highest weight integrable $\widehat{\mathfrak{sp}}_{2n}$ modules of level $k$, see e.g. \cite[Chapter 10]{Kac}.
 The simple objects of $\Cc(\mathfrak{sp}_{2n})_k$ are labeled by partitions $\lambda \in I_{n,k}$. Write $\hat{\lambda}$ for the simple $\Cc(\mathfrak{sp}_{2n})_k$ module of highest weight $c_{n,k}(\lambda)$. 
We will also denote by $\underline{\lambda}$ the corresponding irreducible finite dimensional
module over $\mathfrak{sp}_{2n}$; thus the highest weight of $\underline{\lambda}$ is
$(k_1,\ldots,k_n)$ if $c_{n,k}(\lambda)=(k_0,k_1,\ldots,k_n)$.

\subsection{Orthogonal Lie algebra} \label{ortho}
Let $\mathfrak{so}_{N}$ be the Lie algebra of form preserving endomorphisms on an $N$-dimensional vector space equipped with a symmetric non-degenerate bilinear form and let $\widehat{\mathfrak{so}}_{N}$ be its affinization. We will use the same
notations as in Section \ref{symplectic} to discuss the  highest weight integrable representations
of $\widehat{\mathfrak{so}}_{N}$.
Recall that in the case when $N$ is divisible by 4 the simple objects of $\Cc(\mathfrak{so}_{N})_1$ are $\aLam_0, \aLam_1, \aLam_+, \aLam_-$
where $\fLam_0$ is the trivial $\mathfrak{so}_{N}-$module, $\fLam_1$ is the natural $\mathfrak{so}_{4nk}-$module
of dimension $N$, and $\fLam_\pm$ are two half-spinor $\mathfrak{so}_{N}-$modules. 

Now assume that $N=4nk$ and that the symmetric bilinear form on $\C^N$ is the tensor product
of the symplectic forms on the spaces $\C^{2n}$ and $\C^{2k}$. Thus we have an embedding 
$\mathfrak{sp}_{2n}\oplus \mathfrak{sp}_{2k}\subset \mathfrak{so}_{4nk}$ and we can
distinguish
between $\fLam_+$ and $\fLam_-$ in the following way: $\fLam_+$ has some nonzero
$\mathfrak{sp}_{2n}-$invariant vector while $\fLam_-$ has no nonzero $\mathfrak{sp}_{2n}-$invariant vectors, see
Remark \ref{ccfinite} below.

{\bf Warning:} Notation convention for $\aLam_\pm$ may change if we interchange the roles
of $\mathfrak{sp}_{2n}$ and $\mathfrak{sp}_{2k}$, see Remark \ref{ccfinite}.


\subsection{Conformal embeddings and Kac-Peterson formula}
It is a well known fact and a consequence of general results in \cite{KP} that the embedding
$\mathfrak{sp}_{2n}\oplus \mathfrak{sp}_{2k}\subset \mathfrak{so}_{4nk}$ induces 
a {\em conformal embedding}
\begin{equation}\label{tg}
(\widehat{\mathfrak{sp}}_{2n})_k\oplus (\widehat{\mathfrak{sp}}_{2k})_n\subset (\widehat{\mathfrak{so}}_{4nk})_1.
\end{equation}
This means that a restriction of a highest weight integrable $\widehat{\mathfrak{so}}_{4nk}-$module 
of level 1 to $\widehat{\mathfrak{sp}}_{2n}\oplus \widehat{\mathfrak{sp}}_{2k}$ is again a highest weight integrable module on which the central elements of $\widehat{\mathfrak{sp}}_{2n}$ and $\widehat{\mathfrak{sp}}_{2k}$ act as $k\mbox{Id}$ and $n\mbox{Id}$ respectively.

The following result is \cite[Proposition 2]{KP}. Since the proof is omitted in {\em loc. cit.} 
we provide its derivation from \cite[Proposition 1]{KP} in  the appendix (Section \ref{KPproof}).


\begin{prop}\label{KPcc}
There is an isomorphism of 
$(\widehat{\mathfrak{sp}}_{2n})_k\oplus (\widehat{\mathfrak{sp}}_{2k})_n-$modules:
$$\aLam_+\oplus \aLam_-\cong \bigoplus_{\lambda \in I_{n,k}}
\alam \boxtimes \widehat{\lambda^c}.$$
\end{prop}

\subsection{Tensor categories}\label{tcat}
Recall that 
a monoidal category $(\Cc,\otimes)$ is \emph{braided} if there is a natural bifunctor
isomorphism $c_{X,Y} : X \otimes Y \to Y \otimes X$ called a braiding subject to the
hexagon axioms (see e.g \cite[Definition 8.1.1]{egnobook}). For every braided tensor category $(\Cc, \otimes, c)$, there is a
reversed braiding on $\Cc$ given by $c_{X,Y}^{rev} = c_{Y,X}^{-1}$. A braided tensor category
$\Cc$ endowed with the reversed braiding will be denoted $\Cc^{rev}$. A monoidal functor (a functor $T : \Cc \to \Dd$, together with a natural isormorphism $J:T(-\otimes-)\to T(-)\otimes T(-)$) is said to be a braided equivalence of categories if $c_{T(X),T(Y)}J_{X,Y}=J_{Y,X}T(c_{X,Y})$ and $T$ is an equivalence of the underlying categories (see e.g. \cite[Definition 8.1.7]{egnobook}). A \emph{braid--reversing} equivalence of $\Cc$ and $\Dd$ is a braided equivalence $\Cc \simeq \Dd^{rev}$. Two objects $X,Y$ in a braided tensor category $\Cc$ are said to mutually centralize each other, in the sense of \cite{Muger}, if $c_{X,Y}c_{Y,X}=\mbox{id}_{X\otimes Y}$. 

It is a deep and important fact (see e.g. \cite[Chapter 7]{BK}) that the categories
$\Cc(\mathfrak{sp}_{2n})_k$ and $\Cc(\mathfrak{so}_{4nk})_1$ have a natural structure 
of {\em modular tensor categories}. In particular they are braided rigid monoidal categories
which are {\em non-degenerate} in the sense of \cite[8.19]{egnobook}. We will often use
the following special property: in the categories $\Cc(\mathfrak{sp}_{2n})_k$ and $\Cc(\mathfrak{so}_{4nk})_1$ every object is self-dual.

It is well known that the category $\Cc(\mathfrak{sp}_{2n})_k$ is $\Z/2-$graded (see \cite[4.14]{egnobook});
the grading is given by the weights of an object modulo the root lattice. Thus the parity of the object $\alam$ is $|\lambda|\pmod{2}$.

Similarly the category $\Cc(\mathfrak{so}_{4nk})_1$ is graded by the quotient of the weight lattice
of $\mathfrak{so}_{4nk}$ modulo the root lattice; this group is known to be $\Z/2\oplus \Z/2$.
This implies that the category $\Cc(\mathfrak{so}_{4nk})_1$ is {\em pointed} (i.e. all simple objects are invertible) with group of simple objects isomorphic to
$\Z/2\oplus \Z/2$. The unit object of the category $\Cc(\mathfrak{so}_{4nk})_1$ is $\aLam_0$. It is a well known and easy to check fact that the object $\aLam_1\in \Cc(\mathfrak{so}_{4nk})_1$ is a {\em fermion}. This means that the braiding $c_{\aLam_1,\aLam_1}$ equals $-\mbox{id}_{\aLam_1,\aLam_1}$ or, equivalently, that the
subcategory of $\Cc(\mathfrak{so}_{4nk})_1$ generated by $\aLam_1$ is equivalent to the category of super vector spaces as a braided fusion category. This gives a simple construction of 
the category $\Cc^-$ which appears in the Introduction: let $\Cc=\Cc^0\oplus \Cc^1$ be a 
$\Z/2-$graded braided fusion category. Consider the subcategory $\Cc_0=\Cc^0\boxtimes \aLam_0\oplus \Cc^1\boxtimes \aLam_1$ of $\Cc \boxtimes \Cc(\mathfrak{so}_{4nk})_1$. It
is clear that $\Cc_0$ is closed under the tensor product; moreover we have a braided equivalence
$\Cc^-\simeq \Cc_0$ which sends $X\in \Cc_0$ to $X\boxtimes \aLam_0$ and $X\in \Cc_1$ to
$X\boxtimes \aLam_1$.


\subsection{\'Etale algebras from conformal embeddings}
An \'etale algebra in a semisimple braided tensor category $\Cc$ is defined to be an object
$A \in \Cc$ endowed with an associative commutative unital multiplication
and that the category $\Cc_A$ of right $A$-modules is semisimple, see \cite[
Definition 3.1]{KO2}. An \'etale algebra $A$ is called connected if the unit object
appears in $A$ with multiplicity 1. For a connected \'etale algebra $A \in\Cc$
the category $\Cc_A$ with operation $\otimes_A$ of tensor product over $A$ is naturally
a fusion category, see e.g. \cite[Section 3.3]{KO2}. The category $\Cc_A$ contains
a full tensor Serre subcategory $\Cc_A^{dys}$ of dyslectic modules, which is also
naturally braided. See for example \cite[Section 3.5]{KO2}. 

A general result observed in \cite[Theorem 5.2]{KO} states that for any conformal embedding the pullback of the vacuum module is an \'etale algebra (see \cite{HKL,CKM2} for a proof); moreover taking pullbacks is a braided equivalence with the category of dyslectic modules over this algebra, see \cite{CKM} for a proof. Specializing this to the conformal embedding (\ref{tg}) we get
 
 \begin{thm}\label{etale}
 Let $A$ be the restriction of $\hat{\Lambda}_0$ under the embedding (\ref{tg}).
Then A is a connected \'etale algebra in $ \Cc(\mathfrak{sp}_{2n})_k\boxtimes \Cc(\mathfrak{sp}_{2k})_n$. Moreover,
the restriction functor is a braided equivalence $\Cc(\mathfrak{so}_{4nk})_1\simeq  (\Cc(\mathfrak{sp}_{2n})_k\boxtimes \Cc(\mathfrak{sp}_{2k})_n)_A^{dys}$. Hence the pullbacks of $\aLam_1$, $\aLam_+$ and $\aLam_-$ are precisely the simple dyslectic $A$-modules in $\Cc(\mathfrak{sp}_{2n})_k\boxtimes \Cc(\mathfrak{sp}_{2k})_n$. 
\end{thm}

\subsection{Lagrangian algebras in products} We recall that a connected \'etale algebra $A$ in a non-degenerate braided fusion category $\Cc$ is {\em Lagrangian} if the category $\Cc_A^{dys}$ is
trivial or, equivalently, $\mbox{FPdim}(A)^2=\mbox{FPdim}(\Cc)$, see \cite[Definition 4.6]{KO2}. 

The following result is a version of \cite[Theorem 3.6]{DNO}.

\begin{thm} \label{equivalences}
Let $\Cc$ and $\Dd$ be two non-degenerate braided fusion categories and let $A\in \Cc \boxtimes \Dd$ be a Lagrangian algebra. Let $A_1=A\cap (\Cc \boxtimes {\bf 1})$ and let
$A_2=A\cap ({\bf 1}\boxtimes \Dd)$. Then $A_1\in \Cc$ and $A_2\in \Dd$ are connected \'etale algebras
and there exists a braid reversing equivalence $\phi: \Cc_{A_1}^{dys}\simeq \Dd_{A_2}^{dys}$
such that $A=\bigoplus_{M\in {\mbox{Irr}(\Cc_{A_1}^{dys})}}M^*\boxtimes \phi(M)$ as an object of
$\Cc \boxtimes \Dd$ (the summation is over the isomorphism classes of simple objects of
$\Cc_{A_1}^{dys}$).

Conversely, given two connected \'etale algebras $A_1\in \Cc$ and $A_2\in \Dd$ and 
a braid reversing equivalence $\phi: \Cc_{A_1}^{dys}\simeq \Dd_{A_2}^{dys}$, the object
$A=\bigoplus_{M\in {\mbox{Irr}(\Cc_{A_1}^{dys})}}M^*\boxtimes \phi(M)\in \Cc \boxtimes \Dd$
has the structure of a Lagrangian algebra.
\end{thm}

\begin{eg}\label{good equiv} 
Assume that $A\in \Cc \boxtimes \Dd$ is a Lagrangian algebra such that 
$A_1=A\cap (\Cc \boxtimes {\bf 1})={\bf 1}\boxtimes {\bf 1}$. Then Theorem \ref{equivalences}
says that there exists a braid reversing equivalence $\phi :\Cc \simeq \Dd_{A_2}^{dys}$.
If we write $A=\bigoplus_{X_i\in {\mbox{Irr}(\Cc)}}X_i^*\boxtimes Y_i$ then the functor
$\phi$ sends $X_i$ to $A_2-$module which is $Y_i$ as an object of $\Dd$.
\end{eg} 

\section{Branching rules}\label{branching}


In this Section we will present the branching rules for conformal embedding \eqref{tg}.





The unit object $\mathbf{1}_{n,k}$ of $\Cc(\mathfrak{sp}_{2n})_k$ corresponds to the empty Young diagram in $I_{n,k}$ and $\mathbf{1}_{n,k}^c$ is a unique nontrivial invertible object corresponding in $I_{k,n}$ to the unique diagram with $kn$ boxes. For $\lambda\in I_{n,k}$, tensoring in $\Cc(\mathfrak{sp}_{2k})_n$ is given by
$$\mathbf{1}_{n,k}^c \otimes \alam \cong \widehat{\lambda}^{tc},$$
see \cite{Fuchs}. We will just write $\mathbf1$ and $\mathbf1^c$ for convenience.



\begin{lem} \label{invcc}
The $A-$modules $(\mathbf{1}\boxtimes\mathbf{1}^c)\otimes A$ and $(\mathbf{1}^c \boxtimes\mathbf1)\otimes A$
are simple and dyslectic. Moreover there exists a sign $s=\pm$ such that
$(\mathbf{1}\boxtimes\mathbf1^c)\otimes A\cong \aLam_s$.
\end{lem} 
\begin{proof}
By Proposition \ref{KPcc},
$\mathbf{1}\boxtimes \mathbf{1}^c$ and $\mathbf{1}^c \boxtimes \be$ are direct summands of
$\aLam_+\oplus \aLam_-$.  Hence the result follows from Theorem \ref{etale}. 
\end{proof}

{\bf Warning:} We do not claim that these two modules are (or are not) isomorphic. We will
see that this depends on the values of $n$ and $k$.

Recall that the category $\Cc(\mathfrak{sp}_{2n})_k$ is $\Z/2-$graded. We denote
by $\Cc(\mathfrak{sp}_{2n})_k^0$ and $\Cc(\mathfrak{sp}_{2n})_k^1$ its trivial and nontrivial 
components. Recall that $\Cc(\mathfrak{sp}_{2n})_k^0$ coincides with the centralizer of the object
$\mathbf1^c$.

\begin{cor}\label{paircc} (i) The object $\aLam_0=A$ is contained in $\Cc(\mathfrak{sp}_{2n})_k^0\boxtimes \Cc(\mathfrak{sp}_{2k})_n^0$.

(ii) The object $\aLam_1$ is contained in $\Cc(\mathfrak{sp}_{2n})_k^1\boxtimes \Cc(\mathfrak{sp}_{2k})_n^1$.
\end{cor}

\begin{proof} By Lemma \ref{invcc} and \cite[Lemma 3.15]{DGNO} $A$ centralizes $\mathbf{1}\boxtimes\be^c$ and $\be^c \boxtimes \be$. This implies (i). It is clear that
$\aLam_1$ contains $\alam_1\boxtimes\alam_1$ where $\alam_1$ is the one box Young diagram;
hence $\aLam_1$ is a direct summand of $(\alam_1\boxtimes\alam_1)\otimes A$. Thus (i) implies (ii).
\end{proof}


We arrive at the main result of this section:
\begin{thm}[cf \cite{Has}]\label{maincc}
There are isomorphisms of 
$(\widehat{\mathfrak{sp}}_{2n})_k\oplus (\widehat{\mathfrak{sp}}_{2k})_n-$modules:
$$\aLam_0\cong \bigoplus_{\lambda \in I_{n,k}^0}\alam \boxtimes \alamt ,$$
$$\aLam_1\cong \bigoplus_{\lambda \in I_{n,k}^1}\alam \boxtimes \alamt ,$$
$$\aLam_+\cong \bigoplus_{\lambda \in I_{n,k}^0}\alam \boxtimes \widehat{\lambda^c},$$
$$\aLam_-\cong \bigoplus_{\lambda \in I_{n,k}^1}\alam \boxtimes \widehat{\lambda^c}.$$

\end{thm}
\begin{proof}
Let $s=\pm$ be as in Lemma \ref{invcc}.
The multiplication rules in the group $\Z/2\oplus \Z/2$ imply that 
$$\aLam_s\otimes (\aLam_+\oplus \aLam_-)\cong \aLam_0\oplus \aLam_1,$$
where the tensor product is taken in the category $\Cc(\mathfrak{so}_{4nk})_1$. Using Theorem \ref{etale},
the definition of $s$, and Proposition \ref{KPcc} we have equivalently
$$\begin{aligned}\aLam_0\oplus \aLam_1&\cong \aLam_s\otimes_A (\aLam_+\oplus \aLam_-)\\
&\cong((\mathbf{1}\boxtimes{\mathbf1}^c)\otimes A)\otimes_A\bigoplus_{\lambda \in I_{n,k}}(\alam \boxtimes\alam^c)\\
&\cong(\mathbf{1}\boxtimes{\mathbf1}^c)\otimes\bigoplus_{\lambda \in I_{n,k}}(\alam \boxtimes\alam^c)\\
&\cong\bigoplus_{\lambda \in I_{n,k}}\alam \boxtimes(\mathbf{1}^c \otimes \alam^c)\\
&\cong\bigoplus_{\lambda \in I_{n,k}}\alam \boxtimes\alamt,\end{aligned}$$
where the tensor product in the third and fourth line is taken in $\Cc(\mathfrak{sp}_{2n})_k\boxtimes\Cc(\mathfrak{sp}_{2k})_n$ and in 
$\Cc(\mathfrak{sp}_{2k})_n$ respectively. Using Corollary \ref{paircc} we get the required decompositions of $\aLam_0$ and
$\aLam_1$. Now by definition of $s$ we get
$\aLam_s=(\mathbf{1}\boxtimes\mathbf{1}^c)\otimes A\cong \bigoplus_{\lambda \in I_{n,k}^0}\alam \boxtimes \widehat{\lambda^c}$ and therefore $\aLam_{-s}=\bigoplus_{\lambda \in I_{n,k}^1}\alam \boxtimes \widehat{\lambda^c}$. Then by Remark \ref{ccfinite} we see that $s=+$
and the result follows.
\end{proof}
\begin{rem} \label{ccfinite} 
(i) We see that $\mathbf{1}^c \boxtimes \be$ appears
in the decomposition of the same $\aLam_+$ if and only if $nk$ is even. Thus the modules
$(\mathbf{1}\boxtimes\mathbf{1}^c)\otimes A$ and $(\mathbf{1}^c \boxtimes\mathbf1)\otimes A$
from Lemma \ref{invcc} are isomorphic if and only if $nk$ is even.

(ii) It was observed by Hasegawa in \cite{Has} that all summands in the decompositions of $\aLam_\pm$ have the same {\em conformal dimension}. It follows that we have similar decompositions
$$\fLam_{\, +}\cong \bigoplus_{\lambda \in I_{n,k}^0}\flam \boxtimes \underline{\lambda^c},\;\;\;\;
\fLam_{\, -}\cong \bigoplus_{\lambda \in I_{n,k}^1}\flam \boxtimes \underline{\lambda^c}$$
for the branching under the finite dimensional algebras embedding $\mathfrak{sp}_{2n}\oplus \mathfrak{sp}_{2k}\subset \mathfrak{so}_{4nk}$ (see Proposition \ref{affine} for a closely related statement). In particular the summand of $\fLam_{\, +}$
corresponding to the empty Young diagram contains vectors invariant under the action of
$\mathfrak{sp}_{2n}$ which justifies our choice for labeling of $\fLam_\pm$ in Section \ref{ortho}.
Note that the conformal dimensions of the summands of $\aLam_0$ and $\aLam_1$ are not constant.
\end{rem}





\section{Level rank duality}\label{main results}
In this section we prove the main result of this paper. Recall the construction of the category
$\mathcal{C}(\mathfrak{sp}_{2n})_k^-$ from the introduction.

\begin{thm}\label{main theorem} There is a braid-reversing monoidal equivalence $$\mathcal{C}(\mathfrak{sp}_{2n})_k\simeq\mathcal{C}(\mathfrak{sp}_{2k})_n^{-}$$
sending
$\widehat{\lambda}\mapsto\widehat{\lambda^t}$.
In particular, $\mathcal{C}(\mathfrak{sp}_{2n})_k^0$ and $\mathcal{C}(\mathfrak{sp}_{2k})_n^0$ are braid reversing equivalent.
\end{thm}
\subsection{First proof}
\begin{proof} By Theorems \ref{etale} and \ref{maincc} we see that an object $$A=\bigoplus_{\lambda \in I_{n,k}^0}\alam \boxtimes \alamt \in \Cc(\mathfrak{sp}_{2n})_k\boxtimes \Cc(\mathfrak{sp}_{2k})_n$$ has the structure of a connected \'etale algebra 
such that we have a braided equivalence $(\Cc(\mathfrak{sp}_{2n})_k\boxtimes \Cc(\mathfrak{sp}_{2k})_n)_A^{dys}\simeq \Cc(\mathfrak{so}_{4nk})_1$. Thus
by Theorem \ref{equivalences} we get a Lagrangian algebra
 $$\tilde{A}\in \mathcal{C}(\mathfrak{sp}_{2n})_k\boxtimes\mathcal{C}(\mathfrak{sp}_{2k})_n\boxtimes\mathcal{C}(\mathfrak{so}_{4nk})_1^{rev}$$
  given by
$$\left(\bigoplus_{\lambda\in I_{n,k}^0}\hat{\lambda}\boxtimes\hat{\lambda^t}\boxtimes\hat{\Lambda}_0\right)\oplus\left(\bigoplus_{\lambda\in I_{n,k}^1}\hat{\lambda}\boxtimes\hat{\lambda^t}\boxtimes\hat{\Lambda}_1\right)$$
$$\oplus\left(\bigoplus_{\lambda\in I_{n,k}^0}\hat{\lambda}\boxtimes\hat{\lambda^c}\boxtimes\hat{\Lambda}_+\right)\oplus\left(\bigoplus_{\lambda\in I_{n,k}^1}\hat{\lambda}\boxtimes\hat{\lambda^c}\boxtimes\hat{\Lambda}_{-}\right).$$

Let us use Theorem \ref{equivalences} again with $\Cc =\mathcal{C}(\mathfrak{sp}_{2n})_k$
and $\Dd =\mathcal{C}(\mathfrak{sp}_{2k})_n\boxtimes\mathcal{C}(\mathfrak{so}_{4nk})_1^{rev}$.
We have $\tilde A\cap (\Cc \boxtimes {\bf 1})={\bf 1}\boxtimes {\bf 1}\boxtimes \aLam_0$ and
$\tilde A\cap ({\bf 1}\boxtimes \Dd)={\bf 1}\boxtimes {\bf 1}\boxtimes \aLam_0\oplus {\bf 1}\boxtimes {\bf 1}^c\boxtimes \aLam_+=:B$ (note that the second summand is invertible of order 2). Thus by Example \ref{good equiv} we have a braid reversing equivalence
$\Cc \simeq \Dd_B^{dys}$ sending $\alam$ to the $B-$module $\alam^t\boxtimes \aLam_0\oplus
\alam^c\boxtimes \aLam_+=(\alam^t\boxtimes \aLam_0)\otimes B$ if $|\lambda|$ is even and to $\alam^t\boxtimes \aLam_1\oplus
\alam^c\boxtimes \aLam_-=(\alam^t\boxtimes \aLam_1)\otimes B$ if $|\lambda|$ is odd. 

Now consider the additive subcategory $\Dd_0$ of $\Dd =\mathcal{C}(\mathfrak{sp}_{2k})_n\boxtimes\mathcal{C}(\mathfrak{so}_{4nk})_1^{rev}$ generated by the objects
$\alam \boxtimes \aLam_0$ with $\lambda \in I_{k,n}^0$ and $\alam \boxtimes \aLam_1$ with $\lambda \in I_{k,n}^1$. It is clear that $\Dd_0$ is fusion subcategory of $\Dd$; moreover since 
the object $\aLam_1$ is a fermion (see Section \ref{tcat}), the category $\Dd_0$ identifies with
the category $\mathcal{C}(\mathfrak{sp}_{2k})_n^{-}$. On the other hand, the free module functor
$X\mapsto X\otimes B$ gives a braided equivalence $\Dd_0\simeq \Dd_B^{dys}$. Thus we get
a chain of equivalences 
$$\mathcal{C}(\mathfrak{sp}_{2n})_k=\Cc \simeq \Dd_B^{dys}\simeq \Dd_0=\mathcal{C}(\mathfrak{sp}_{2k})_n^{-}$$
sending $\alam$ to $\alam^t$. This completes the proof.

\end{proof}

\begin{rem} The proof above gives no information on uniqueness of the structure of braided tensor functor on the functor sending $\alam$ to $\alam^t$.
\end{rem}

\subsection{Second Proof}
In \cite{TW} braided fusion categories with the same fusion rules as $\mathcal{C}(\ssp_{2n},2n+2k+2)$ and $\Cc(\ssp_{2n},2n+2k+1)$ are reconstructed from their Grothendieck (semi-)rings and the eigenvalues of the braiding $c_{X,X}$ on the generating object $X$ analogous to the $2n$-dimensional representation of $\ssp_{2n}$, i.e. with highest weight $(1,0,\ldots,0)$. In particular they prove the following (cf. \cite[Section 9.3]{TW}):
\begin{thm}\label{twtheorem}
Let $\mathcal{C}$ and $\widetilde{\mathcal{C}}$ be braided fusion categories with  Grothendieck rings isomorphic to that of $\Cc(\ssp_{2n},\ell)$. Denote by $X$, $\tilde{X}$ the generating objects of $\Cc,\tilde{\Cc}$ corresponding to $X_{(1)}\in\Cc(\ssp_{2n},\ell)$, and $\be,\tilde{\be}$, $Y,\tilde{Y}$ and $W,\tilde{W}$ the objects corresponding to $X_{(0)},X_{(2)}$ and $X_{(1,1)}$ in $\Cc(\ssp_{2n},\ell)$.  If the eigenvalues of $c_{X,X}$ and $c_{\tilde{X},\tilde{X}}$ coincide on $[\be,Y,W]$ and $[\tilde{\be},\tilde{Y},\tilde{W}]$ then $\Cc$ and $\widetilde{\Cc}$ are equivalent as braided fusion categories.
\end{thm}

We remark that if $\ell$ is even we are in the symplectic case, while for $\ell$ odd this is called the ortho-symplectic case.  Notice also that in this subsection we denote the weights by the corresponding Young diagrams, i.e. $\lambda$ etc. rather than as $\widehat{\lambda}$ as is customary in the quantum group approach.


We now provide an alternate proof to Theorem \ref{main theorem} using Theorem \ref{twtheorem}.  

\begin{proof} 
Let $\ell=2k+2n+2$, and denote by $X_\lambda$ the simple objects in $\Cc(\ssp_{2n},\ell)$ and by $\tilde{X}_\mu$ the simple objects in $\Cc(\ssp_{2k},\ell)$, where $\lambda$ is a Young diagram with at most $n$ rows and at most $k$ columns and $\mu$ is a Young diagram with at most $k$ rows and at most $n$ columns.  Notice that $\Cc(\ssp_{2k},\ell)^{-rev}$ and $\Cc(\ssp_{2k},\ell)$ are identical as fusion categories.

We first must verify that $\Cc(\ssp_{2k},\ell)$ has the same fusion rules as $\Cc(\ssp_{2n},\ell)$.  Clearly there is a bijection between these label sets given by transpose.  To see that transpose provides an isomorphism of Grothendieck rings we note by \cite[Proposition 8.6]{TW} that the fusion rules for $\Cc(\ssp_{2n},\ell)$ and $\Cc(\ssp_{2k},\ell)$ are determined by the rule for tensoring with $X:=X_{(1)}$ (resp. $\tilde{X}:=\tilde{X}_{(1)}$.  These are as follows: $X_{(1)}\otimes X_{\lambda}$ is the direct sum of all $X_{\mu}$ whose Young diagram has one box more or one box less than $\hat{\lambda}$ with the same rule for $\tilde{X}$.  Clearly this rule is preserved when taking the transpose. Notice that under this isomorphism the object $X_{(2)}$ is mapped to $\tilde{X}_{(1,1)}$ and $X_{(1,1)}$ is mapped to $\tilde{X}_{(2)}$.


Next we must compare the eigenvalues of the braiding isomorphisms $c_{X,X}$ and $c_{\tilde{X},\tilde{X}}$.


 Observe that $X\otimes X\cong \mathbf{1}\oplus X_{(1,1)}\oplus X_{(2)}$.  The eigenvalues of $c_{X,X}$ on $X_\lambda\subset X^{\otimes 2}$ are computed as in \cite{Resh}: $\pm q^{\frac{1}{2}c_\lambda-c_{(1)}}$ where $c_\lambda:=\langle \lambda+2\rho,\lambda\rangle$ and where we take $+$ if $X_{\lambda}$ appears in $S^2X$ as representations of $U_q\ssp_{2n}$ and $-$ otherwise.  In this case $X_{(2)}\subset S^2 X$ while $\be=X_{(0)},X_{(1,1)}\subset \wedge^2 X$.  It follows that the eigenvalues for $c_{X,X}$ on $[\be,X_{(2)},X_{(1,1)}]$ are $[-q^{-2n-1},q,-q^{-1}]$, whereas for  $c_{\tilde{X},\tilde{X}}$ on $[\be,\tilde{X}_{(2)},\tilde{X}_{(1,1)}]$ are $[-q^{-2k-1},q,-q^{-1}]$.  The effect of transposing diagrams, followed by reversing the braiding and then an overall sign change takes $[-q^{-2k-1},q,-q^{-1}]$ to $[q^{2k+1},q,-q^{-1}]$, and since $q^{2k+1}=-q^{-2n-1}$ we have verified the eigenvalues match.

\end{proof}

In \cite{RowellMathZ} it is shown that the non-unitary ribbon category $\Cc(\ssp_{2n},2n+2k+1)$ has the same fusion rules as $\Cc(\sso_{2k+1},2n+2k+1)$.  We will prove a more precise result, but first we need some additional notation.  One may construct ribbon fusion categories from $U_q \sso_{2k+1}$ and $U_q \ssp_{2n}$ for any choice of $q$ with $q^2$ a primitive $\ell$th root of unity.  Since $\ell=2k+2n+1$ is odd there are, up to Galois conjugation, two choices of this parameter: $e^{\pi i/\ell}$ or $e^{\pi i/\ell}$. For ease of notation we will define $q:=e^{\pi i/\ell}$, and emphasize this parameter dependence by denoting the categories by $\Cc(\ssp_{2n},Q,\ell)$ and $\Cc(\sso_{2k+1},Q,\ell)$, where $Q^2$ is a primitive $\ell$th root of unity.

We first establish the following:
\begin{lem}\label{bclemma} $\Cc(\ssp_{2n},Q,\ell)$ and $\Cc(\sso_{2k+1},Q,\ell)$ are $\Z/2$-graded, with modular trivially graded component, for any choice of $Q$.  In fact we have $\Cc(\ssp_{2n},Q,\ell)\cong\Cc(\ssp_{2n},Q,\ell)_0\boxtimes \Cc(\Z/2,P_1)$ and $\Cc(\sso_{2n},Q,\ell)\cong\Cc(\sso_{2n},Q,\ell)_0\boxtimes \Cc(\Z/2,P_2)$ where $\Cc(\Z/2,P_i)$ are pointed ribbon categories associated with the pre-metric group $(\Z/2,P_i)$.
\end{lem}

\begin{proof} $\Cc(\ssp_{2n},Q,\ell)$ and $\Cc(\sso_{2k+1},Q,\ell)$ each have one non-trivial invertible object (see \cite{RowellMathZ}), labeled by the weight $(\ell-2n,0,\ldots,0)$ for type $C$ and $\frac{1}{2}(\ell-2k,\ldots,\ell-2k)$ for type $B$.  Let us denote these by $\eta$ for type $C$ and $\gamma$ for type $B$.  

Computing the twists and dimension we find that $\eta$ is a non-unitary fermion ($\dim(\eta)=-1$, $\theta_\eta=1)$ if $Q^\ell=1$ and a non-unitary boson $(\dim(\eta)=-1$, $\theta_\eta=-1)$ if $Q^\ell=-1$.  Moreover, we compute $$c_{\eta,X_{(1)}}c_{X_{(1)},\eta}=\frac{\theta_\eta\theta_{X_{(1)}}}{\theta_{\eta\otimes X_{(1)}}}Id_{X_{(1)}\otimes\eta}=Id_{X_{(1)}\otimes\eta}$$ so that $\eta$ is transparent regardless of the value of $Q^\ell$, and in fact generates the M\"uger center of $\Cc(\ssp_{2n},Q,\ell)$.  In particular the subcategory with simple objects labeled by Young diagrams with an even number of boxes $\Cc(\ssp_{2n},Q,\ell)_0$ does not contain $\eta$ but has centralizer $\langle\eta\rangle$, giving the desired factorization and modularity $\Cc(\ssp_{2n},Q,\ell)_0$.

Similarly we compute that $\gamma$ is a transparent boson ($\dim(\gamma)=\theta_\gamma=\pm 1)$ or fermion ($\dim(\gamma)=-\theta_\gamma=\pm 1$) if $k$ is even or if $k$ is odd
 and $Q^\ell=1$ and a semion ($\dim(\gamma)=\pm 1$ and $\theta_\gamma=\pm i$) otherwise. In either case this shows that the subcategory $\Cc(\sso_{2k+1},Q,\ell)_0$ with simple objects labeled by integer weights forms a modular subcategory, and we have the desired factorization.


\end{proof}

We can now prove the following:
\begin{thm} \label{nonunitary}
Let $\ell=2k+2n+1$ and $q=e^{\pi i/\ell}$. Then there are braid-reversing equivalences between
\begin{enumerate}
\item $\Cc(\ssp_{2n},q,\ell)_0$ and $\Cc(\sso_{2k+1},q^{\frac{\ell+1}{2}},\ell)_0$  and
\item $\Cc(\ssp_{2n},q^2,\ell)_0$ and $\Cc(\sso_{2k+1},q,\ell)_0$.
\end{enumerate}
\end{thm}

\begin{proof}
We again employ Theorem \ref{twtheorem}.  The Grothendieck rings of $\Cc(\sso_{2k+1},\ell)$ and $\Cc(\ssp_{2n},\ell)$ were already established to be isomorphic in \cite{RowellMathZ}, denote that isomorphism by $\Phi$. To avoid confusion we will denote simple objects in $\Cc(\sso_{2k+1},Q,\ell)$ by $Y_{\lambda}$ and those in $\Cc(\ssp_{2n},P,\ell)$ by $X_{\mu}$. Under the isomorphism $\Phi$ we have that $\Phi(Y_{(1)})=X_{(1)}\otimes X_{\eta}=X_{(\ell-2n-1,0,\ldots,0)}=X^\prime$, and these objects generate the respective modular subcategories $\Cc(\sso_{2k+1},Q,\ell)_0$ and $\Cc(\ssp_{2n},P,\ell)_0$ as in Lemma \ref{bclemma}.  Notice that it is enough to identify the eigenvalues of the braidings on $Y_{(1)}^{\otimes 2}$ and $(X^\prime)^{\otimes 2}$: we may lift this identification to $\Cc(\sso_{2k+1},Q,\ell)$ and $\Cc(\ssp_{2n},P,\ell)$ by tensoring with pointed categories $\Cc(\Z/2,P_i)$ and then apply Theorem \ref{twtheorem}.

The eigenvalues of $c_{Y_{(1)},Y_{(1)}}$ on $[\be,Y_{(2)},Y_{(1,1)}]$ for $\Cc(\sso_{2k+1},Q,\ell)$ are found in \cite[Table 6.1]{LRW}, the are $[Q^{-4k},Q^2,-Q^{-2}]$.  We have $\Phi(Y_{(2)})=X_{(1,1)}$ and vice versa, so transpose followed by braid-reversing gives us the eigenvalues $[Q^{4k},-Q^2,Q^{-2}]$.

For $\Cc(\ssp_{2n},P,\ell)$ the eigenvalues of $c_{X_{(1)},X_{(1)}}$ on $[\be,X_{(2)},X_{(1,1)}]$ are $[-P^{-2n-1},P,-P^{-1}]$ \cite[Table 6.1]{LRW}.  If $P^\ell=-1$ then $\eta$ is a boson so the braiding eigenvalues on $X^\prime$ and $X_{(1)}$ are identical.  Thus if $P=q=e^{\pi i/\ell}$ we must take $Q=q^{\frac{\ell+1}{2}}$ so that $Q^{4k}=q^{2k(\ell+1)}=q^{2k}=-q^{-2n-1}$, and $-Q^2=-(q^{\ell+1})=q$ and $Q^{-2}=-q^{-1}$ so that the eigenvalues match.  If $P^\ell=1$ then $\eta$ is a fermion so that the braiding eigenvalues on $X^\prime$ differ from those on $X_{(1)}$ by an overall sign, giving: $[P^{-2n-1},-P,P^{-1}]$.  Thus if we take $P=q^2=e^{2\pi i/\ell}$ we choose $Q=q$ so that $Q^{4k}=P^{2k}=P^{-2n-1}$, $-Q^2=-P$ and $Q^{-2}=P^{-1}$.  

\end{proof}


We close this section with some remarks on the advantages of this categorical approach.
\begin{rem}
\begin{enumerate}
\item The non-unitary categories $\Cc(\ssp_{2n},2n+2k+1)$ and $\Cc(\sso_{2k+1},2n+2k+1)$ cannot be constructed from affine Lie algebras, to our knowledge.  Twisted affine Lie algebras at fractional levels (see e.g. \cite{Kac}) provide similar combinatorics, but there is no level-preserving fusion product.
\item The results of \cite{TW} provide a description of the categories $\Cc(\ssp_{2n},l)$ via generators and relations. One deduces that the functor between $\Cc(\ssp_{2k},2k+2n+2)$ and $\Cc(\ssp_{2n},2k+2n+2)$ sending $X$ to $\tilde X$ has a \emph{unique} braided tensor structure up to
isomorphism of tensor functors, see \cite{EM}. Similar uniqueness holds for functors from Theorem \ref{nonunitary} provided that the functor sends $X$ to $X^\prime$.
\end{enumerate}
\end{rem}



\section{Appendix: Kac-Peterson formula in the symplectic case} \label{KPproof}
\subsection{}
The main goal of this Section is to give a proof of Proposition \ref{KPcc} based on
\cite[Proposition 1]{KP}. 

Let us recall the setup. Let $\mathfrak{g}$ be a simple finite dimensional Lie algebra and
let $\theta$ be an automorphism of $\mathfrak{g}$ of order 2. Let $\mathfrak{t}=\{ x\in \mathfrak{g}|
\theta(x)=x\}$ be the subspace of $\theta-$invariant vectors and let $\mathfrak{p}=\{ x\in \mathfrak{g}| \theta(x)=-x\}$. The following result is standard:

\begin{lem}[Cartan decomposition]\label{cartandecomp} We have a decomposition $\mathfrak{g} = \mathfrak{t}\oplus\mathfrak{p}$ satisfying the following 
\begin{itemize}
\item[(i)] $\mathfrak{t}$ is a reductive Lie subalgebra of $\mathfrak{g}$ and $[\mathfrak{t}, \mathfrak{p}] \subset \mathfrak{p}$.
\item[(ii)] The restricted Killing form $\kappa|_{\mathfrak{p}\times\mathfrak{p}}$ is a non-degenerate symmetric bilinear form preserved by the action of $\mathfrak{t}$.
\item[(iii)] The action of $\mathfrak{t}$ on $\mathfrak{p}$ gives an embedding $\mathfrak{t}\hookrightarrow\mathfrak{so}(\mathfrak{p}, \kappa|_{\mathfrak{p}\times\mathfrak{p}})\cong \mathfrak{so}_{\dim\mathfrak{p}}$.
\end{itemize}
\end{lem}


We will be interested in the following special case.

\begin{eg}\label{tpg} Let $\C^{2n+2k}=\C^{2n}\oplus \C^{2k}$ be a direct sum of two symplectic 
spaces. Let $\theta \in GL(\C^{2n+2k})$ be the linear operator acting by -1 on $\C^{2n}$ and
by 1 on $\C^{2k}$. It is clear that $\theta$ preserves the symplectic form, so it acts by
conjugations on $\mathfrak{g}= \mathfrak{sp}(\C^{2n+2k})=\mathfrak{sp}_{2n+2k}$.

Then $\mathfrak{t}\cong \mathfrak{sp}_{2n}\oplus\mathfrak{sp}_{2k}$ and $\mathfrak{p}$ must have dimension $4nk$ so we get an embedding $\mathfrak{sp}_{2n}\oplus\mathfrak{sp}_{2k}\subset \mathfrak{so}_{4nk}$ from Section \ref{ortho}.
\end{eg}

This example enjoys the following extra property: there exists a Cartan subalgebra $\mathfrak{h}$
of $\mathfrak{g}$ which is contained in $\mathfrak{t}$. We choose and fix such a subalgebra.
Let $\Delta$ and $\Delta_{\mathfrak{t}}$ denote the root systems of $\mathfrak{g}$ and $\mathfrak{t}$ with respect to $\mathfrak{h}$. Let $W$ and $W_{\mathfrak{t}}$ be the corresponding
Weyl groups. We choose a set $\Delta^+$ of positive roots for $\Delta$; then $\Delta^+_{\mathfrak{t}}=\Delta_{\mathfrak{t}}\cap \Delta$ is a set of positive roots for $\Delta_{\mathfrak{t}}$. Let $\rho$ and $\rho_{\mathfrak{t}}$ be the sums of fundamental weights.
Finally let $\hat W, \hat W_{\mathfrak{t}}, \hat \Delta, \hat \Delta_{\mathfrak{t}}$ etc denote the
affine versions of the notions above.

The set
$$W_{\mathfrak{t}}^1=\{w\in W\,|\, \Delta_{\mathfrak{t}}^+\subset w\Delta^+\}=\{w\in W\,|\, w^{-1}\Delta_{\mathfrak{t}}^+\subset \Delta^+\}$$
and its affine counterpart
$$\hat{W}_{\mathfrak{t}}^1=\{w\in \hat{W}\,|\,\hat{\Delta}_{t}^+\subset w\hat{\Delta}^+\}=\{w\in \hat{W}\,|\,w^{-1}\hat{\Delta}_{t}^+\subset \hat{\Delta}^+\}$$
will play a significant role in what follows in view of the following result:


\begin{thm}\label{KP} Let $\mathfrak{g}$, $\mathfrak{t}$ and $\mathfrak{h}$ be as above. 

{\em (i) (Lemma 2.2 of \cite{Part})} Let $S$ be the spinor representation of $\mathfrak{so}_{\dim\mathfrak{p}}$ restricted to $\mathfrak{t}$ and let $L(\mu)$ be the irreducible $\mathfrak{t}-$module of highest weight $\mu$.. Then there is an isomorphism of $\mathfrak{t}-$modules
$$S\cong\bigoplus_{w\in {W}^1} L(w({\rho})-{\rho_{\mathfrak{t}}}).$$

{\em (ii) (Proposition 1 of \cite{KP})}
The affinization of the embedding $\mathfrak{t}\subset \mathfrak{so}_{\dim\mathfrak{p}}$ is a conformal embedding $\widehat{\mathfrak{t}}\subset (\widehat{\mathfrak{so}}_{\dim\mathfrak{p}})_1$. Let $\hat S$ be the spinor representation
of $\widehat{\mathfrak{so}}_{\dim\mathfrak{p}}$ and let $\hat L(\mu)$ be the irreducible $\widehat{\mathfrak{t}}-$module of highest weight $\mu$.
Then there is an isomorphism of $\widehat{\mathfrak{t}}$-modules
$$\hat S\cong\bigoplus_{w\in \hat{W}^1} \hat L(w(\hat{\rho})-\hat{\rho_{\mathfrak{t}}}).$$
\end{thm}

\begin{rem} 
(i) The levels of affine Lie algebras appearing in Theorem \ref{KP} (ii) can be computed as follows:
Let $\mathfrak{t}_1$ be a direct summand of the Lie algebra $\mathfrak{t}$. Then the level
of $\widehat{\mathfrak{t}}_1$ is the ratio of the Killing form of $\mathfrak{g}$ restricted to
$\mathfrak{t}_1$ and the Killing form of $\mathfrak{t}_1$. In particular in the setup of Example
\ref{tpg} we get the conformal embedding \eqref{tg}.

(ii) Note that the dimension of space $\mathfrak{p}$ is even (this is twice the cardinality of
$\Delta^+ \setminus \Delta^+_{\mathfrak{t}}$). Thus both $S$ and $\hat S$ are the sums of two half-spinor modules. In particular in the setup of Example \ref{tpg} we have $\hat S=\aLam_+\oplus \aLam_-$.
\end{rem}
\subsection{The symplectic case}
We will make Theorem \ref{KP} explicit in the setup of Example \ref{tpg}.
Let $$\{e_1-e_2,e_2-e_3,\dots,e_{n+k-1}-e_{n+k},2e_{n+k}\}$$ be the simple roots of $\mathfrak{sp}_{2n+2k}$. Then the simple roots of $\mathfrak{sp}_{2n}$ are
$$\{e_1-e_2,e_2-e_3,\dots,e_{n-1}-e_n,2e_n\}$$and the simple roots of $\mathfrak{sp}_{2k}$ are
$$\{e_{n+1}-e_{n+2},e_{n+2}-e_{n+3},\dots,e_{n+k-1}-e_{n+k},2e_{n+k}\}.$$

 Let $\Delta_{\mathrm{long}}^+=\{ 2e_1, \ldots , 2e_{n+k}\}$ be the set of long positive roots of $\mathfrak{sp}_{2n+2k}$. 
 Since the long positive roots of $\mathfrak{sp}_{2n}$ are $2e_{1},\dots , 2e_{n}$ and of $\mathfrak{sp}_{2k}$ are $2e_{n+1},\dots,2e_{n+k}$, we have $\Delta_{\mathrm{long}}^+\subset \Delta^+_{\mathfrak{t}}$.



\begin{prop}\label{affine}  In the setup of Example \ref{tpg}, $\hat{W}_{\mathfrak{t}}^1=W_{\mathfrak{t}}^1$. 
\end{prop}
\begin{proof} Let $w'\in\hat{W}_{\mathfrak{t}}^1$. Recall $\hat{W}=W\rtimes T$ (\cite[Proposition 6.5]{Kac}) where
$T$ is the group of translations. So $w'=wt$ for some $w\in W$ and $t^{-1}=t_{m_1e_1+\dots +m_{n+k}e_{n+k}}\in T$ for $m_i\in\Z$. Let $\delta\in \hat{\Delta}_{t}^+$ be the indivisible imaginary root. Consider

$$\begin{aligned}t^{-1}w^{-1}(w(\delta-2e_i))&=t^{-1}(\delta)-t^{-1}(2e_i)\\
                        &=\delta-2e_i+2m_i\delta.\end{aligned}$$
                       
Since $w(\delta-2e_i)=\delta-w(2e_i)\in\Delta^+_{\mathfrak{t}}$ and $w'\in\hat{W}_{\mathfrak{t}}^1$, it follows $m_i\geq0$, see Section \ref{affroots}.

Similarly $w(\delta+2e_i)\in\Delta^+_{\mathfrak{t}}$ and $(w')^{-1}w(\delta+2e_i)=\delta+2e_i-2m_i\delta$. Thus $w'\in\hat{W}_{\mathfrak{t}}^1$ implies $m_i\leq0$. Hence $m_i=0$.
\end{proof}

\begin{rem} In view of Theorem \ref{KP} the Proposition \ref{affine} says that the branching rules
for the finite dimensional and affine cases are ``the same'', cf Remark \ref{ccfinite} (ii).
\end{rem}

Let us describe $W_{\mathfrak{t}}^1$. Recall that the group $W$ is the group of signed permutations; let $S_{n+k}\subset W$ be the subgroup of permutations without signs.
Recall that $\Delta_{\mathrm{long}}^+\subset \Delta^+_{\mathfrak{t}}$. Thus we have
$$W^1=\{w\in W \,|\, w^{-1}\Delta_{\mathrm{long}}^+=\Delta_{\mathrm{long}}^+, w^{-1}(e_i-e_{i+1}) \in \Delta^+, \text{for all }i\neq n\}.$$
Observe that the the first condition implies that $w^{-1}\in S_{n+k}$. Thus we have

\begin{lem}\label{permute} The set $W_{\mathfrak{t}}^1$ is contained in $S_{n+k}\subset W$. 
A permutation $s\in S_{n+k}$ is in $W_{\mathfrak{t}}^1$ if and only if
$$s^{-1}(1)<s^{-1}(2)<\dots <s^{-1}(n), s^{-1}(n+1)<s^{-1}(n+2)<\dots<s^{-1}(n+k).\;\; \square$$
\end{lem}

Now if we consider $n+k$ dots on a straight line and paint the dots numbered $s^{-1}(1), s^{-1}(2), \dots ,s^{-1}(n)$ in black and the dots numbered $s^{-1}(n+1), s^{-1}(n+2), \dots ,s^{-1}(n+k)$ in white we get precisely ``black and white dots diagram'' as in Example \ref{bwdots}.
Conversely from such a diagram we get a unique permutation as in Lemma \ref{permute}.
Thus we constructed a bijection between $W_{\mathfrak{t}}^1$ and the set $C_{n,k}=I_{n,k}$ from
Section \ref{combinatorics}. 

\begin{eg} The permutation $s^{-1}$ corresponding to the diagram
$$\bullet\bullet\circ\bullet\bullet\bullet\bullet\circ\circ\circ\bullet\circ \circ$$
from example \ref{bwdots} is 
$$\left( \begin{array}{ccccccccccccc}1&2&3&4&5&6&7&8&9&10&11&12&13\\
1&2&4&5&6&7&11&3&8&9&10&12&13\end{array}\right) .$$
\end{eg}

It remains to compute the weights $w({\rho})-{\rho_{\mathfrak{t}}}$ for $w\in W_{\mathfrak{t}}^1$
(we can restrict ourselves to the finite case in view of Proposition \ref{affine}). We have
$$\rho =(n+k)e_1+(n+k-1)e_2+\ldots +2e_{n+k-1}+e_{n+k}=\sum_{i=1}^{n+k}(n+k+1-i)e_i,$$
and
$$\rho_{\mathfrak{t}}=ne_1+(n-1)e_2+\ldots +e_n+ke_{n+1}+(k-1)e_{n+2}\ldots +e_{n+k}.$$
For a permutation $s\in S_{n+k}$ we have
$$s\rho =\sum_{i=1}^{n+k}(n+k+1-i)e_{s(i)}=\sum_{i=1}^{n+k}(n+k+1-s^{-1}(i))e_i.$$

Thus we have
$$s\rho -\rho_{\mathfrak{t}}=\sum_{i=1}^{n}(k+i-s^{-1}(i))e_i+\sum_{i=1}^{k}(n+i-s^{-1}(i))e_{n+i}.$$
Here the first summand represents the weight of $\mathfrak{sp}_{2n}\subset \mathfrak{sp}_{2n}\oplus\mathfrak{sp}_{2k}=\mathfrak{t}$ and the second summand represents the weight of $\mathfrak{sp}_{2k}\subset \mathfrak{sp}_{2n}\oplus\mathfrak{sp}_{2k}=\mathfrak{t}$. It is clear
that the second weight is computed similarly to the first one with replacement of the sequence
$s^{-1}(1)<s^{-1}(2)<\dots <s^{-1}(n)$ by the sequence $s^{-1}(n+1)<s^{-1}(n+2)<\dots<s^{-1}(n+k)$ or, equivalently, by replacing all the black dots by the white ones and vice versa. This is precisely the description of the bijection $\lambda \mapsto \lambda^c$ in the language of diagrams. 
Thus Proposition \ref{KPcc} is proved.




