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Let $f_i=f_i(x_1,x_2,\ldots, x_n),i=0,1,2, \ldots $ be a family of symmetric polynomials. For the partition $\lambda=(\lambda_1,\lambda_2, \ldots, \lambda_n)$ consider the determinant $$ D_\lambda(f)=\left | \begin{array}{lllll} f_{\lambda_1} & f_{\lambda_1+1} & f_{\lambda_1+2} & \ldots & f_{\lambda_1+n-1}\\ f_{\lambda_2-1} & f_{\lambda_2} & f_{\lambda_2+1} & \ldots & f_{\lambda_2+n-2}\\ \vdots & \vdots & \vdots & \ldots & \vdots \\ f_{\lambda_n-(n-1)} & f_{\lambda_n-(n-2)} & f_{\lambda_n-(n-3)} & \ldots & f_{\lambda_n} \end{array} \right|. $$ It is well known (Jacobi−Trudi formulas) that for the elementary symmetric polynomials $e_i=e_i(x_1,x_2,\ldots, x_n)$ and for the complete homogeneous symmetric polynomials $h_i=h_i(x_1,x_2,\ldots, x_n)$ we have $$ D_\lambda(h)=s_{\lambda}(x_1,x_2,\ldots, x_n) \text{ and } D_\lambda(e)=s_{\lambda'}(x_1,x_2,\ldots, x_n), $$ where $s_{\lambda}(x_1,x_2,\ldots, x_n)$ is the Schur polynomial and $\lambda'$ is the conjugate partition.

Question. Is there a similar expression for $D_\lambda(p)$ where $p_i=p_i(x_1,x_2,\ldots, x_n)$ is the power sum symmetric polynomials?

By direct calculation for $n=2, \lambda_2\geq 1$ I got that

$$D_{(\lambda_1,\lambda_2)}(p)=-s_{(\lambda_1-1,\lambda_2-1)} V(x_1,x_2)^2$$ and for $n=3,\lambda_3\geq 2$

$$D_{(\lambda_1,\lambda_2,\lambda_3)}(p)=-\frac{s_{(\lambda_1-1,\lambda_2-1,\lambda_3-1)}}{s_{(1,1,1)}} V(x_1,x_2,x_3)^2$$ and so on.

Here $V$ is the Vandermonde determinant.

I hope that must be a nice formula and for any $n$.

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1 Answer 1

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I don't know a fully general result, but your pattern for partitions $\lambda$ of length $\leq n$ with $n$-th entry $\lambda_{n}\geq n-1$ and with $n$ indeterminates persists:

Theorem 1. Let $n$ be a positive integer. Let $\lambda=\left( \lambda _{1},\lambda_{2},\ldots,\lambda_{n}\right) $ be an integer partition with at most $n$ parts. Assume that $\lambda_{n}\geq n-1$. Consider polynomials in $n$ indeterminates $x_{1},x_{2},\ldots,x_{n}$. For each nonnegative integer $k$, we set \begin{align*} p_{k}:=x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}. \end{align*} (This is the $k$-th power-sum symmetric polynomial in $x_{1},x_{2} ,\ldots,x_{n}$ when $k>0$. We have $p_{0}=n$.) Define the $n\times n$-matrix \begin{align*} P:=\left( p_{\lambda_{i}-i+j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}=\left( \begin{array} [c]{cccc} p_{\lambda_{1}} & p_{\lambda_{1}+1} & \cdots & p_{\lambda_{1}+n-1}\\ p_{\lambda_{2}-1} & p_{\lambda_{2}} & \cdots & p_{\lambda_{2}+n-2}\\ \vdots & \vdots & \ddots & \vdots\\ p_{\lambda_{n}-n+1} & p_{\lambda_{n}-n+2} & \cdots & p_{\lambda_{n}} \end{array} \right) . \end{align*} Let $\mu=\left( \mu_{1},\mu_{2},\ldots,\mu_{n}\right) $ be the partition defined by \begin{align*} \mu_{i}=\lambda_{i}-\left( n-1\right) \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{ 1,2,\ldots,n\right\} . \end{align*} (This is indeed a partition, since $\mu_{n}=\underbrace{\lambda_{n}}_{\geq n-1}-\left( n-1\right) \geq0$.) Let $s_{\mu}$ be the corresponding Schur polynomial in the $n$ indeterminates $x_{1},x_{2},\ldots,x_{n}$. Furthermore, let \begin{align*} V_{n}:=\prod_{1\leq i<j\leq n}\left( x_{i}-x_{j}\right) \end{align*} be the Vandermonde determinant. Then, \begin{align*} \det P=\left( -1\right) ^{n\left( n-1\right) /2}s_{\mu}\cdot V_{n}^{2}. \end{align*}

Proof. Let $A_{\mu}$ be the $n\times n$-matrix \begin{align*} \left( x_{j}^{\mu_{i}+n-i}\right) _{1\leq i\leq n,\ 1\leq j\leq n}=\left( \begin{array} [c]{cccc} x_{1}^{\mu_{1}+n-1} & x_{2}^{\mu_{1}+n-1} & \cdots & x_{n}^{\mu_{1}+n-1}\\ x_{1}^{\mu_{2}+n-2} & x_{2}^{\mu_{2}+n-2} & \cdots & x_{n}^{\mu_{2}+n-2}\\ \vdots & \vdots & \ddots & \vdots\\ x_{1}^{\mu_{n}+n-n} & x_{2}^{\mu_{n}+n-n} & \cdots & x_{n}^{\mu_{n}+n-n} \end{array} \right) . \end{align*} It is then well-known that \begin{equation} s_{\mu}=\dfrac{\det\left( A_{\mu}\right) }{V_{n}} . \label{darij1.eq.slam=frac} \tag{1} \end{equation} Indeed, this is the alternant formula for Schur polynomials. For a proof, see, e.g., Corollary 2.6.7 in the lecture notes Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, arXiv:1409.8356v7. (The notations in those notes are not quite ours. Namely, our matrix $A_{\mu}$ is the transpose of the matrix whose determinant is $a_{\mu+\rho}$ in the notes, whereas our $V_{n}$ is $a_{\rho}$ in these notes. Corollary 2.6.7 has to be applied to $\mu$ instead of $\lambda$.)

Let $B$ be the $n\times n$-matrix \begin{align*} \left( x_{i}^{j-1}\right) _{1\leq i\leq n,\ 1\leq j\leq n}=\left( \begin{array} [c]{cccc} 1 & x_{1} & \cdots & x_{1}^{n-1}\\ 1 & x_{2} & \cdots & x_{2}^{n-1}\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_{n} & \cdots & x_{n}^{n-1} \end{array} \right) . \end{align*} The Vandermonde determinant formula yields \begin{align*} \det B & =\prod_{1\leq i<j\leq n}\underbrace{\left( x_{j}-x_{i}\right) }_{=-\left( x_{i}-x_{j}\right) }=\prod_{1\leq i<j\leq n}\left( -\left( x_{i}-x_{j}\right) \right) \\ & =\left( -1\right) ^{n\left( n-1\right) /2}\underbrace{\prod_{1\leq i<j\leq n}\left( x_{i}-x_{j}\right) }_{=V_{n}}=\left( -1\right) ^{n\left( n-1\right) /2}V_{n}. \end{align*}

However, we have \begin{equation} A_{\mu}B=P. \label{darij1.eq.AB=P} \tag{2} \end{equation} (Indeed, for any $i,j\in\left\{ 1,2,\ldots,n\right\} $, the $\left( i,j\right) $-th entry of the matrix $A_{\mu}B$ is \begin{align*} \sum_{k=1}^{n}\underbrace{x_{k}^{\mu_{i}+n-i}x_{k}^{j-1}}_{\substack{=x_{k} ^{\mu_{i}+n-i+j-1}=x_{k}^{\lambda_{i}-i+j}\\\text{(since }\mu_{i}=\lambda _{i}-\left( n-1\right) \text{ and}\\\text{thus }\mu_{i}+n-i+j-1=\lambda _{i}-\left( n-1\right) +n-i+j-1=\lambda_{i}-i+j\text{)}}} & =\sum_{k=1} ^{n}x_{k}^{\lambda_{i}-i+j}\\ & =x_{1}^{\lambda_{i}-i+j}+x_{2}^{\lambda_{i}-i+j}+\cdots+x_{n}^{\lambda _{i}-i+j}=p_{\lambda_{i}-i+j}, \end{align*} which happens to be precisely the $\left( i,j\right) $-th entry of the matrix $P$. Thus, \eqref{darij1.eq.AB=P} follows.)

Now, the two matrices $A_{\mu}$ and $B$ are square matrices. Hence, \begin{align*} \det\left( A_{\mu}B\right) & =\underbrace{\det\left( A_{\mu}\right) }_{\substack{=s_{\mu}V_{n}\\\text{(by \eqref{darij1.eq.slam=frac})}} }\cdot\underbrace{\det B}_{=\left( -1\right) ^{n\left( n-1\right) /2} V_{n}}\\ & =s_{\mu}V_{n}\cdot\left( -1\right) ^{n\left( n-1\right) /2}V_{n}=\left( -1\right) ^{n\left( n-1\right) /2}s_{\mu}\cdot V_{n}^{2}. \end{align*} In view of \eqref{darij1.eq.AB=P}, we can rewrite this as \begin{align*} \det P=\left( -1\right) ^{n\left( n-1\right) /2}s_{\mu}\cdot V_{n}^{2}. \end{align*} This proves Theorem 1. $\blacksquare$

The claim of Theorem 1 can further be rewritten by observing that (in $n$ indeterminates $x_{1},x_{2},\ldots,x_{n}$) we have \begin{align*} s_{\lambda}=s_{\mu}\cdot\left( x_{1}x_{2}\cdots x_{n}\right) ^{n-1} \end{align*} (because the entries of $\lambda$ are the respective entries of $\mu$ plus $n-1$). The product $x_{1}x_{2}\cdots x_{n}$ can also be rewritten as $s_{\left( 1^{n}\right) }$, where $\left( 1^{n}\right) $ is the partition $\left( 1,1,\ldots,1\right) $ with $n$ entries.

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  • $\begingroup$ Thank you. The (2) is very nice $\endgroup$
    – Leox
    Jan 21, 2022 at 18:20
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    $\begingroup$ Is there a skew version of this? $\endgroup$ Mar 18, 2022 at 19:03

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