%  Jacobians of symmetric functions    Feb 20, 2002   Correct Juin02
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\title{\large  JACOBIANS OF SYMMETRIC POLYNOMIALS}

\author{ Alain Lascoux \  and \  Piotr Pragacz}

\newtheorem{prop}{Proposition}
\newtheorem{rem}{Remark}

\begin{document}
\date{}
\maketitle

\begin{abstract}
We give the Jacobian of any family of complete symmetric functions,
or of power sums, in a finite number of variables. 
\end{abstract}

Changes of bases of the ring of symmetric functions are usually performed 
using a canonical scalar product on this ring
(\cite{Macdonald}, Ch.I). On the other hand,
Cayley \cite{Cayley}, Sylvester \cite{Sylvester} and Mac Mahon \cite{McMahon}
were rather led to characterize 
invariants like, e.g., the discriminant, by 
using differential calculus on symmetric polynomials.  
The aim of the present note is to write explicitly the Jacobians
of different families of symmetric polynomials 
in $n$ indeterminates $\X=\{x_1,\ldots,\, x_n\}$, with respect to the 
indeterminates, or to the elementary symmetric functions.

Define 
$$ \lambda_z(\X):= \prod_{i=1}^{n}(1+zx_i) = \sum_{j=0}^n z^j\, e_j(\X)\ \ 
, \ \ \sigma_z(\X): = 1/\lambda_{-z}(\X)= 
\sum_{j=0}^\infty z^j h_j(\X)\,,$$
and
$$
\sum_{j= 1}^\infty  z^j\, p_j(\X)/j:=\log(\sigma_z(\X))\,,
$$
the $e_j=e_j(\X)$ being the \emph{elementary symmetric functions},
the $h_j=h_j(\X)$ being the \emph{complete functions}, the
$p_j=p_j(\X)$ being the \emph{power sums}.

More generally, for any $k\in\C$, let
$$\sigma_z(k\,\X)= \bigl( \sigma_z(\X) \bigr)^k = 
  \sum_{j=0}^\infty z^j h_j(k\, \X)=
\exp\bigl(\sum_{j=1}^{\infty} z^j \,p_j(k\X)/j \bigr)\,.$$

For any $n$-tuple $f_1,\ldots,\, f_n$ in the ring of 
symmetric polynomials in $\X$, one has two \emph{Jacobians}\,, expressed
by $(n\times n)$-determinants :
$$ J(f_1,\ldots,\, f_n):= 
     \Bigl| \tfrac{\d}{\d_{x_i}}(f_j) \Bigr|_{1\leq i,j\leq n}  
   \qquad {\rm and} \qquad 
  J_e(f_1,\ldots,\, f_n) := 
       \Bigl| \tfrac{\d}{\d_{e_i}}(f_j) \Bigr|_{1\leq i,j\leq n} \, .$$

Since 
$$\tfrac{\d}{\d_{x_i}}(e_j) =\sum_{r=0}^{j-1}(-1)^r x_i^r \, e_{j-1-r} \,,
$$
by subtracting suitable combinations of columns in 
$J(e_1,\ldots,e_n)$, we get
$$ 
J(e_1,\ldots, e_n) 
=\Bigl|\sum_{r=0}^{j-1}(-1)^r x_i^r \, e_{j-1-r}
\Bigr|_{1\leq i,j\leq n} 
$$
$$
= \Bigl|(-1)^{j-1} x_i^{j-1} \Bigr|_{1\leq i,j\leq n} = 
\prod_{1\leq i<j\leq n} (x_i-x_j) = :
 \Delta(\X)= \Delta\,, $$
the Vandermonde determinant.
Hence, for any $n$-tuple of symmetric polynomials, we have 
the factorization
\begin{equation}
 J(f_1,\ldots,\, f_n) =J_e(f_1,\ldots,\, f_n) \cdot \Delta \ .
\end{equation}

The following proposition gives the expressions of some $J_e$ in terms
of Schur functions $s_{\nu}(k\X)$ which are defined as the 
determinants 
$$
\Bigl|h_{\nu_i-i+j}(k\X)\Bigr|_{1\leq i,j\leq n}
$$
(cf. (\cite{Macdonald}, I.3.4). 
This definition remains valid when $\nu=(\nu_1,\ldots,\nu_n)$ is not a
partition.
If $k=1$ and $\nu_i\ge -(n-i)$ for $i=1,\ldots,n$, then 
the expression 
$$
\Bigl|x_i^{\nu_j+n-j}\Bigr|_{1\leq i,j\leq n} / \Delta
$$
(cf. (\cite{Macdonald}, I.3.1)) in terms of the powers of 
the $x_i$'s, gives $s_{\nu}(\X)$. 

\begin{prop}\label{} \ For $\mu=(\mu_1,\ldots,\mu_n) \in (\N^*)^n$, 
we have

\begin{equation}
  J_e\bigl(p_{\mu_1}(k\X),\ldots,\, p_{\mu_n}(k\X)  \bigr) =
  k^n \mu_1\cdots \mu_n\, s_{(\mu_1-n,\ldots,\, \mu_n-1)}(\X)   
\end{equation}
and for $\mu \in \N^n$\,, 
\begin{equation}
   J_e\bigl(h_{\mu_1}(k\X),\ldots,\, h_{\mu_n}(k\X)  \bigr) =
  k^n  s_{(\mu_1-n,\ldots,\, \mu_n-1)}\bigl((k\!+\!1)\, \X\bigr)  \ .
\end{equation} 
\end{prop}
\noindent
(If $\mu_1>\cdots >\mu_n\ge 1$, then $\mu_1-n\geq \cdots \geq \mu_n-1$ is
a partition.)

\medskip

Firstly, since $p_j(k\X)=kp_j(\X)$, we have
$$ 
\tfrac{\d}{\d_{x_i}}\bigl(p_j(k\X)\bigr) = 
 k\tfrac{\d}{\d_{x_i}}\big(p_j(\X)\bigr)= kj\, x_i^{j-1} \,,$$
and
$$
J\bigl(p_{\mu_1}(k\X),\ldots,\, p_{\mu_n}(k\X)  \bigr) =
  k^n \mu_1\cdots \mu_n\,\cdot \Bigl| x_i^{\mu_j -1}\Bigr|_{1\leq i,j\leq n} \ .
$$
Using the expression of a Schur function in terms of the powers of the 
$x_i$'s, one gets equation (2).

Secondly, using the Leibniz rule for differentiation
of the product, we have
\begin{equation}
\tfrac{\d}{\d_{e_i}}\big( \sigma_z(k\, \X)\bigr)=
\tfrac{\d}{\d_{e_i}}\big( 1/\lambda_{-z}(\X)^k\bigr)=
   -k(-z)^i/ \bigl(\lambda_{-z}(\X) \bigr)^{k+1}= 
       -k(-z)^i \sigma_z\bigl((k\!+\!1)\, \X\bigr)\,.
\end{equation}
Thus the expression of a Schur function in terms of complete functions 
entails equation (3). The proposition has been proved.

\bigskip

For example, we have
$$
(n!)^{-1}\, J(p_1,p_2,\ldots,p_n)=J(h_1,h_2,\ldots,h_n)=(-1)^{n(n-1)/2}\Delta \,,
$$
and for $n=3$, 
$$
J_e(h_5(\X),h_3(\X),h_2(\X)) = s_{2,1,1}(2\X) \,.
$$


\begin{rem}\label{2r.1}
In analogy to (4), we record 
\begin{equation}
\tfrac{\d}{\d_{e_i}}\big( \lambda_z(k\, \X)\bigr)=
       k z^i \lambda_z\bigl((k\!-\!1)\, \X\bigr)\,.
\end{equation}
\end{rem}


\begin{rem}\label{2r.2}
For any three partitions $\mu,\nu,\eta$ of the same weight $d$, let
$(\mu,\nu,\eta)$ be the multiplicity of the trivial representation
of the symmetric group $\S_d$ in the tensor product of the three
irreducible representations of $\S_d$ labeled by the partitions 
$\mu,\nu,\eta$ (\cite{Macdonald}, I.7). Then, for any $k\in\C$,  
any partition $\mu$ of $d$, one has the expansion 
$$ s_\mu(k\X)= \sum_{\nu,\eta} (\mu,\nu,\eta) s_\eta(k)  s_\nu(\X) \ ,$$
sum over all pair of partitions of weight $d$, $s_\eta(k)$
meaning the specialization  of the Schur function $s_\eta$
under $h_j \to \binom{k+j-1}{j}$, $j\geq 0$. 
\end{rem}


For example, for $n=3$, $d=4$, one has the following expression for
$s_{2,1,1}(2\X)$:
$$
\bigl( s_4(2)+s_{3,1}(2)+s_{2,2}(2) \bigr)s_{2,1,1}(\X) +
     \bigl( s_{3,1}(2)\bigr)s_{2,2}(\X)
 + \bigl( s_{3,1}(2)+s_{2,2}(2) \bigr)s_{3,1}(\X) 
$$
$$
= 9 s_{2,1,1}(\X) + 3 s_{2,2}(\X) + 4 s_{3,1}(\X)
$$ 
(in that case, all coefficients $((2,1,1), \nu,\eta)$ are $0$ or $1$).


\begin{rem}\label{2r.3}
Scott \cite{Scott}, followed by Muir \cite{Muir}, p.148,
wrongly asserted that the
Jacobians of $p_{\mu_1}(\X),\ldots,\, p_{\mu_n}(\X)$ and
$h_{\mu_1}(\X),\ldots,\, h_{\mu_n}(\X)$ differ by a scalar factor. 
The error was to deduce $\tfrac{\d}{\d_{p_j}}(h_j)=1/j $ from
the Newton relation: 
$$ jh_j = \sum_{r=0}^{j-1} p_{j-r}\, h_r \,, $$
the power sums appearing in this formula not being algebraically 
independent when $j>n$. 
One ``raison d'\^etre" of the present note was to correct this error.
\end{rem}
 

\begin{thebibliography}{1}

\bibitem{Cayley} A. Cayley, {\em On symmetric functions and seminvariants },
Amer. J. of Math. {\bf 15} (1893) 1--74.

\bibitem{Macdonald}
I.~G. Macdonald,
\newblock {\em Symmetric functions and {Hall} polynomials},
\newblock Oxford University Press, Oxford, 2nd edition (1995).

\bibitem{McMahon} Capt. McMahon, 
{\em Operators in the theory of seminvariants},
{Quart. J. Math.} {\bf 20} (1885) 362--365.

\bibitem{Muir}  T. Muir, \emph{History of the Theory of Determinants}, Vol.
III, Dover reprint (1960).
 
\bibitem{Scott} R.~F. Scott, 
{\em On some alternating functions of $n$ variables}, Messenger of Math.
{\bf 11} (1882), 98-103. 

\bibitem{Sylvester} J.J. Sylvester,
{\em Note on Capt. Mac Mahon's transformationn of the theory of invariants},
{Messenger of Math.} {\bf 14} (1884) 163--165.

\end{thebibliography}


\vskip30pt
\noindent
{\small A.L.: CNRS, Institut Gaspard Monge, Universit\'e de 
Marne-la-Vall\'ee, F-77454 Marne-la-Vall\'ee Cedex, France

\noindent
Alain.Lascoux@univ-mlv.fr}

\bigskip
\noindent
{\small P.P.: Institute of Mathematics, Polish Academy of Sciences,
\'Sniadeckich 8, PL-00950 Warsaw, Poland

\noindent
pragacz@impan.gov.pl}


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