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Foata ed.} \newcommand{\caup}{Cambridge University Press} \newcommand{\oup}{Oxford University Press} \newcommand{\pr}{preprint} \newcommand{\ip}{in preparation} \newcommand{\ds}{\displaystyle} \newcommand{\U}{\mathop{U}} \newcommand{\Z}{\mathop{Z}} \newcommand{\C}{\mathbb{C}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\PD}{\Cal{PD}} \newcommand{\gl}{\mathfrak{gl}} \newcommand{\dime}{\mathop{\rm dim}} \newcommand{\di}{\partial} \newcommand{\ot}{\otimes} \newcommand{\ts}{\,} \newcommand{\Proof}{\noindent{\bf Proof.}\ \ } \newtheorem{thm}{Theorem}[section] \newtheorem{prop}[thm]{Proposition} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{conj}[thm]{Conjecture} \newtheorem{exa}[thm]{Example} \newtheorem{defn}[thm]{Definition} \begin{document} \pagestyle{empty} \title{A Littlewood-Richardson Rule for\\ factorial Schur functions } \author{Alexander I. Molev\\ Centre for Mathematics and its Applications\\ Australian National University\\ Canberra, ACT 0200\\ AUSTRALIA\\ molev@pell.anu.edu.au\\[5pt] Bruce E. Sagan\\ Department of Mathematics\\ Michigan State University\\ East Lansing, MI 48824-1027\\ U.S.A.\\ sagan@math.msu.edu} \date{\today \\[1in] \begin{flushleft} Key Words: Capelli operator, factorial Schur function, Littlewood-Richardson Rule, quantum immanant, Young tableau\\[1em] AMS subject classification (1991): Primary 05E05; Secondary 05E10, 17B10, 17B35, 20C30. \end{flushleft} } \maketitle \begin{flushleft} Proposed running head: \end{flushleft} \begin{center} A factorial L-R Rule \end{center} Send proofs to: \begin{center} Bruce E. Sagan \\ Department of Mathematics \\Michigan State University \\ East Lansing, MI 48824-1027\\[5pt] Tel.: 517-355-8329\\ FAX: 517-432-1562\\ Email: sagan@math.msu.edu \end{center} \begin{abstract} We give a combinatorial rule for calculating the coefficients in the expansion of a product of two factorial Schur functions. It is a special case of a more general rule which also gives the coefficients in the expansion of a skew factorial Schur function. Applications to Capelli operators and quantum immanants are also given. \end{abstract} \pagestyle{plain} \section{Introduction} \label{i} As $\lambda$ runs over all partitions with length $l(\lambda)\leq n$, the Schur polynomials $s_{\lambda}(x)$ form a distinguished basis in the algebra of symmetric polynomials in the independent variables $x=(x_1,\dots,x_n)$. By definition, $$ s_{\lambda}(x)= \frac{\det (x_j^{\lambda_i+n-i})_{1\leq i,j\leq n}} {\prod_{i|\lambda|+|\mu|$.} \eeq Contrary to the classical case, the coefficients $c_{\lambda\mu}^{\nu}(a)$ turn out to be nonzero if $|\nu|<|\lambda|+|\mu|$ and $\lambda,\mu\subseteq\nu$. This makes it possible to compute them using induction on $|\nu/\mu|$ while keeping $\lambda$ fixed. The starting point of our calculation is the fact that the polynomials $s_{\lambda}(x|a)$ possess some (characteristic) vanishing properties; see~\cite{sah:sci,oko:qih}. We use the following result from~\cite{oko:qih}. For a partition $\rho$ with $l(\rho)\leq n$ define an $n$-tuple $a_{\rho}=(a_{\rho_1+n},\dots,a_{\rho_n+1})$. \bth[Vanishing Theorem] \label{VT} Given partitions $\la,\rho$ with $l(\la),l(\rho)\leq n$ $$ s_{\lambda}(a_{\rho}|a)= \case{0}{if $\lambda\not\subseteq\rho,$} {\prod_{(i,j)\in\lambda}(a_{\lambda_i+n-i+1}-a_{n-\lambda^t_j+j})} {if $\la=\rho$.} $$ where $\lambda^t$ is the diagram conjugate to $\lambda$. \eth In particular, $s_{\lambda}(a_{\lambda}|a)\ne 0$ for any specialization of the sequence $a$ such that $a_i\ne a_j$ if $i\ne j$. We reproduce the proof of the Vanishing Theorem from~\cite{oko:qih} for completeness. \bigskip \Proof The $ij$-th entry of the determinant in the numerator of the right hand side of~(\ref{det}) for $x=a_{\rho}$ is \beq \label{arho} (a_{\rho_j+n-j+1}-a_1)\cdots (a_{\rho_j+n-j+1}-a_{\lambda_i+n-i}). \eeq The condition $\lambda\not\subseteq\rho$ implies that there exists an index $k$ such that $\rho_k<\lambda_k$. Then for $i \leq k\leq j$ we have $$ 1\leq \rho_j+n-j+1\leq \rho_k+n-k+1\leq \lambda_k+n-k\leq \lambda_i+n-i, $$ and so all the entries~(\ref{arho}) with $i \leq k\leq j$ are zero. Hence, the determinant is zero which proves the first part of the theorem. Let us set now $x=a_{\lambda}$ in~(\ref{det}). Then the $ij$-th entry of the determinant is $$ (a_{\lambda_j+n-j+1}-a_1)\cdots (a_{\lambda_j+n-j+1}-a_{\lambda_i+n-i}), $$ which equals zero for $i1$. (In the denominator an empty product is, as usual, equal to 1.)\hfill\Qed Note that a different expression for the $c_{\theta\mu}^{\nu}(a,b)$ in terms of divided differences can be deduced from the Newton interpolation formula in several variables~\cite{ls:in}. \bpr We have the recurrence relation \beq \label{rr} c_{\theta\mu}^{\nu}(a,b)=\frac{1}{|a_{\nu}|-|a_{\mu}|}\left( \sum_{\mu\to\mu'}c_{\theta\mu'}^{\nu}(a,b) -\sum_{\nu'\to\nu}c_{\theta\mu}^{\nu'}(a,b)\right). \eeq \epr \Proof By Proposition~\ref{HH'} it suffices to check that $$ H(\mu,\rho) H'(\rho,\nu)=\frac{1}{|a_{\nu}|-|a_{\mu}|}\left( \sum_{\mu\to\mu'}H(\mu',\rho) H'(\rho,\nu) -\sum_{\nu'\to\nu}H(\mu,\rho) H'(\rho,\nu')\right). $$ This follows from the relations $$ \sum_{\mu\to\mu'}H(\mu',\rho)=(|a_{\rho}|-|a_{\mu}|)H(\mu,\rho) $$ and $$ \sum_{\nu'\to\nu}H'(\rho,\nu')=(|a_{\rho}|-|a_{\nu}|)H'(\rho,\nu).\qquad\Qed $$ Given a sequence $$ R: \mu=\rho^{(0)}\to\rho^{(1)}\to \ldots\to\rho^{(l-1)}\to\rho^{(l)}=\nu, $$ and an index $k\in\{1,\dots,l\}$ introduce a set of $\theta$-tableaux $\cT_k(\theta,R)$ having entries from the set $\{1,\ldots,n\}$ as follows. Each tableau $T\in \cT_k(\theta,R)$ contains cells $\al_1,\dots,\al_{k-1},\al_{k+1},\dots,\al_l$ such that $$ \al_1<\ldots<\al_{k-1}<\al_{k+1}<\ldots<\al_l\qmq{and}T(\al_i)=r_i,\ 1\le i\le l,i\neq k. $$ As usual, we distinguish the entries in the $\al_i, i\neq k$, by barring them. Now modify the $\rho(\al)$ for $R$ by defining, for cells with unbarred entries, $$ \rho^{+}(\alpha)= \case{\rho^{(k)}}{if $\al_{k-1}<\al<\al_{k+1}$,}{\rho(\al)}{otherwise,} $$ and $$ \rho^{-}(\alpha)= \case{\rho^{(k-1)}}{if $\al_{k-1}<\al<\al_{k+1}$,}{\rho(\al)}{otherwise.} $$ Also define corresponding weights \bea \cS(R)&=& \sum_{T\in\cT(\th,R)}\prod_{\scl\al\in\th\atop\scl T(\al)\mbox{\scz unbarred}} \left((a_{\rho(\alpha)})_{T(\alpha)}-b_{T(\alpha)+c(\alpha)}\right),\\ \cS_k^{+}(R)&=& \sum_{T\in\cT_k(\th,R)}\prod_{\scl\al\in\th\atop\scl T(\al)\mbox{\scz unbarred}} \left((a_{\rho^{+}(\alpha)})_{T(\alpha)}-b_{T(\alpha)+c(\alpha)}\right), \eea and similarly for $\cS_k^-(R)$. So Theorem~\ref{lrab} is equivalent to \beq \label{SR} c_{\th\mu}^\nu(a,b)=\sum_R \cS(R). \eeq \bpr Given a sequence $R$ we have \beq \label{S+-} \cS(R)=\frac{1}{|a_{\nu}|-|a_{\mu}|} \sum_{k=1}^l \left(\cS_k^{+}(R)-\cS_k^{-}(R)\right). \eeq \epr \Proof It suffices to show that for each $k$ we have $$ \cS_k^{+}(R)-\cS_k^{-}(R)=(|a_{\rho^{(k)}}|-|a_{\rho^{(k-1)}}|)\cS(R). $$ Formula (\ref{S+-}) will then follow from the relation $$ \sum_{k=1}^l(|a_{\rho^{(k)}}|-|a_{\rho^{(k-1)}}|)=|a_{\nu}|-|a_{\mu}|. $$ For a given $T\in\cT_k(\th,R)$ the factors in the formulas for $\cS_k^{+}(R)$ and $\cS_k^{-}(R)$ are identical except for the case where $\al_{k-1}<\al<\al_{k+1}$ and $T(\alpha)=r_k$. To see what happens when we divide $\cS_k^{+}(R)-\cS_k^{-}(R)$ by $$ |a_{\rho^{(k)}}|-|a_{\rho^{(k-1)}}|=(a_{\rho^{(k)}})_{r_k}-(a_{\rho^{(k-1)}} )_{r_k}, $$ fix $T$ and consider its contribution to the quotient. We need the following easily proved formula, where we are thinking of $u=(a_{\rho^{(k)}})_{r_k}$, $v=(a_{\rho^{(k-1)}})_{r_k}$ and $m_i=b_{T(\al)+c(\al)}$ as $\al$ runs over all cells of $T$ with $\al_{k-1}<\al<\al_{k+1}$ and $T(\al)=r_k$: $$ \frac{\prod_{i=1}^s (u-m_i)-\prod_{i=1}^s (v-m_i)}{u-v}= \sum_{j=1}^s (u-m_1)\cdots\widehat{(u-m_j)}(v-m_{j+1})\cdots(v-m_s) $$ (a hat indicates the factor is to be omitted). The right-hand side of this expression can now be interpreted as the contribution to $\cS(R)$ of all tableaux gotten from $T$ by barring one of the $r_k$ between $\al_{k-1}$ and $\al_{k+1}$ in column order.\hfill \Qed We now prove~(\ref{SR}) by induction on $|\nu/\mu|$. Equation~(\ref{low}) takes care of the case $|\nu/\mu|=0$. By the induction hypothesis, $$ \sum_{\mu\to\mu'}c_{\theta\mu'}^{\nu}(a,b)=\sum_R \cS_1^{+}(R) \qmq{and} \sum_{\nu'\to\nu}c_{\theta\mu}^{\nu'}(a,b)=\sum_R \cS_l^{-}(R). $$ So formulas~(\ref{rr}), (\ref{S+-}) and the following proposition complete the proof of~(\ref{SR}) and hence Theorem~\ref{lrab}. \bpr We have $$ \sum_R \sum_{k=1}^{l-1}\cS_k^{-}(R)=\sum_R \sum_{k=2}^l\cS_k^{+}(R). $$ \epr \Proof We can rewrite this formula as follows: \beq \label{wt} \sum_{R,k,T}\wt^{-}(R,k,T)=\sum_{R',k',T'}\wt^{+}(R',k',T'), \eeq where $T\in\cT_k(\theta,R)$, $k=1,\dots,l-1$ and $T'\in\cT_{k'}(\theta,R')$, $k'=2,\dots,l$ with weights defined by $$ \wt^-(R,k,T)= \prod_{\alpha\in\theta\atop T(\alpha)\ \mbox{\scriptsize unbarred}} \left((a_{\rho^{-}(\alpha)})_{T(\alpha)}-b_{T(\alpha)+c(\alpha)}\right) $$ and similarly define $\wt^+(R',k',T')$. To prove~(\ref{wt}) we will construct a bijection $(R,k,T)\longleftrightarrow (R',k',T')$ preserving the weights in the sense that $\wt^{-}(R,k,T)=\wt^{+}(R',k',T')$. There are three cases. \medskip {\sl Case 1.} Suppose that the skew diagram $\rho^{(k+1)}/\rho^{(k-1)}$ consists of two cells in different rows and columns. Then $R'$ is the sequence obtained from $R$ by replacing $\rho^{(k)}$ by the other diagram ${\rho'}^{(k)}$ such that $\rho^{(k-1)}\to {\rho'}^{(k)}\to \rho^{(k+1)}$ while $k'=k+1$ and $T'=T$. \medskip {\sl Case 2.} Let $\rho^{(k+1)}/\rho^{(k-1)}$ have two cells in the same row. Then $R'=R$, $k'=k+1$ and $T'=T$. \medskip {\sl Case 3.} Let $\rho^{(k+1)}/\rho^{(k-1)}$ have two cells in the same column, say in rows $r$ and $r+1$. Let $(i+1,j)=(i_1,j_1)$ be the cell of $T$ containing the correponding $\ol{r+1}$. If there is an $r$ in cell $(i,j)$ then it must be unbarred. In this case let $T'$ be $T$ with the bar moved from the $r+1$ to the $r$, $R'=R$ and $k'=k+1$. Weights are preserved since $(a_{\rho^{(k-1)}})_r=(a_{\rho^{(k+1)}})_{r+1}$ and $T(\al)+c(\al)$ is invariant under the change. Now suppose cell $(i,j)$ of $T$ contains an entry less than $r$ or $(i,j)\not\in\th$ and let $j'=j'_1$ be the column of the left-most $r+1$ in row $i+1$. Since this subcase is more complicated than the others, the reader may wish to follow along with the example given after the end of this proof. Let $s$ be the maximum integer such that for $1\le t\le s$ we have \ben \item there is an $\ol{r+t}$ in cell $(i+t,j_t)$ for some $j_t$ corresponding to a cell in the same column as those of $\rho^{(k+1)}/\rho^{(k-1)}$, \item if $(i+t,j'_t)$ contains the left-most $r+t$ in row $i+t$ then $(i+t,j_t)$ is between $(i+t-1,j_{t-1})$ and $(i+t-1,j'_{t-1})$ in column order. (Assume this is true vacuously when $t=1$.) \een Note that the condition on cell $(i,j)$ implies that none of the $r+t$'s to the left of the one in $(i+t,j_t)$ can be barred. We now form $T'$ by moving the bar in cell $(i+t,j_t)$ to cell $(i+t,j'_t)$ and replacing the $r+t$'s in cells $(i+t,j'_t),(i+t,j'_t+1),\ldots,(i+t,j_t)$ by $r+t-1$'s. Note that the result will still be a semistandard tableau because of the assumption about $(i,j)$ and the choice of elements to decrease. Since the elements from the given column of $\rho^{(k+1)}/\rho^{(k-1)}$ are still added in the correct order in $T'$ it determines a valid $R'$, complete except for the step where a cell is added in row $r+s$ of that column which should be done immediately following the addition of $r+s-1$. Then $k'$ is the position of this $r+s$. Invariance of weights follows from considerations like those in the first subcase, noting that the contribution to $\wt^-$ of each entry decreased in $T$ is the same as that of the element on its right to $\wt^+$ in $T'$. The inverse of this construction is similar and left to the reader. This completes the proof of the Theorem~\ref{lrab}.\hfill\Qed As an example of the last subcase, suppose we have the $R$ sequence $$ (3,2,2,2)\to(3,3,2,2)\to(3,3,3,2)\to(4,3,3,2)\to(4,3,3,2,1)\to(4,3,3,3,1) $$ with Yamanouchi symbol $r_1\ldots r_5=23154$. Let $k=1$ so $r=2$ and consider $$ T=\barr{cccccc} 1 &1 &1 &1 &\ol{1} &1\\ 2 &3 &3 &3 &3 &\ol{3}\\ 4 &4 &\ol{4} &4 &\ol{5} \earr. $$ Then $(i+1,j)=(2,6)$ and $s=2$ with $r+1,r+2=3,4$ so $$ T'=\barr{cccccc} 1 &1 &1 &1 &\ol{1} &1\\ 2 &\ol{2} &2 &2 &2 &2\\ \ol{3} &3 &3 &4 &\ol{5} \earr. $$ Column reading the barred elements of $T'$ and inserting $r+2=4$ after $r+1=3$ gives the Yamanouchi symbol $15234$ corresponding to the $R'$ sequence $$ (3,2,2,2)\!\to\!(4,2,2,2)\!\to\!(4,2,2,2,1)\!\to\!(4,3,2,2,1)\!\to\!(4,3,3,2 ,1) \!\to\!(4,3,3,3,1) $$ and $k'=5$. \section{Multiplication rules for Capelli operators and quantum immanants} \label{mrc} Let $E_{ij}$, $i,j=1,\dots,n$ denote the standard basis of the general linear Lie algebra $\gl(n)$. Denote by $\Z(\gl(n))$ the center of the universal enveloping algebra $\U(\gl(n))$. Given $\kappa=(\kappa_1,\dots,\kappa_n)\in\C^n$ consider a $\gl(n)$-module $L(\kappa)$ of highest weight $\kappa$. That is, $L(\kappa)$ is generated by a nonzero vector $v$ such that \bea E_{ii}\cdot v&=&\kappa_i\ts v,\qmq{for}i=1,\dots,n;\\ E_{ij}\cdot v&=&0 \qmq{for}1\leq i