Let's say I only know the following about Schur functions: you give me a partition $\lambda$ of $d$ such that $\lambda$ has $n$ parts $\lambda_1,\ldots,\lambda_n$, and I can compute the Schur function $s_\lambda$. This function $s_\lambda$ is a function of $n$ variables $x_1,x_2,\ldots,x_n$ and I compute it by the formula I find on Wikipedia:

$$s_\lambda(x_1,\ldots,x_n) = \frac{\det\begin{pmatrix} x_1^{\lambda_1+n-1} & x_2^{\lambda_1+n-1} & \cdots & x_n^{\lambda_1+n-1} \\ x_1^{\lambda_2+n-2} & x_2^{\lambda_2+n-2} & \cdots & x_n^{\lambda_2+n-2} \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{\lambda_n} & x_2^{\lambda_n} & \cdots & x_n^{\lambda_n} \end{pmatrix}}{\det\begin{pmatrix} x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}}$$ For example, if $\lambda = (2,2)$, (so above $n=2$, $d = 4$, and $\lambda_1=\lambda_2=2$), I can calculate $$s_{(2,2)}(x_1,x_2) = \frac{\det\begin{pmatrix} x_1^3 & x_2^3 \\ x_1^2 & x_2^2\end{pmatrix}}{\det\begin{pmatrix} x_1 & x_2 \\ 1 & 1\end{pmatrix}} = x_1^2x_2^2$$

Now someone comes along and teaches me the Jacobi-Trudi formula: if $\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_n)$ and $e_i$ is the $i$th elementary symmetric polynomial (in some variables which are part of my issue), I can compute $s_\lambda$ as $$s_\lambda = \det(e_{\lambda_i'+j-i})$$ where $\lambda_i'$ is the $i$th part of the conjugate partition. If I apply this formula to my example above, I get (again, note that I'm balking on variables) $$s_{(2,2)} = \det\begin{pmatrix}e_2 & e_3 \\ e_1 & e_2\end{pmatrix} = e_2^2-e_1e_3$$ Here is my problem: This last expression requires at least three variables. What is the convention I am missing? How do I correctly apply the Jacobi-Trudi formula? In general, I think I should be able to evaluate $s_\lambda(x_1,\ldots,x_n)$ for $n$ greater than or equal to the number of parts of $\lambda$, and I don't see how this is possible.

Note: If we use three variables, $x_1,x_2,x_3$ and use the Jacobi-Trudi formula for the partition (2,2,0), everything makes sense and $$s_{(2,2,0)}(x_1,x_2,0)=x_1^2x_2^2$$


1 Answer 1


In order to see the "correct" behavior of the Schur polynomials and to be able to compare different formulas for the Schur polynomials, the number of variables you need is equal to the total number of boxes in the partition (the value $d$, in the Wikipedia notation), not just the number of (nonzero) parts in the partition. Try taking $n=d=4$ in both of your formulas above (with the partition $\lambda$ extended with zeroes); you should find that they agree.

The Wikipedia article alludes to this with this definition of Schur polynomials: $$s_\lambda = \sum_T x^T$$ where $T$ is a semistandard Young tableau. A semistandard Young tableau fills the boxes of the partition with positive integers from 1 through $n$ such that rows are nondecreasing and columns are (strictly) increasing. If $t_i$ is the number of occurrences of the value $i$ in the tableau, then the term $x^T = \prod x_i^{t_i}$. Since each box could contain a different integer, you need $n$ to be at least the number of boxes in the partition; otherwise, you will miss some of the terms in the Schur polynomial.

Even better is to take infinitely many variables $x_1, x_2, \ldots$; then the Schur polynomial becomes a power series (each term is still finite though). This allows you to write formulas comparing Schur polynomials of different partitions without worrying about the number of variables. The same holds for other symmetric polynomials.

(Note: Of course, the formulas are not nonsense if you take fewer than $d$ variables. You just have to be careful when comparing two different formulas by doing exactly what you did above: take as many variables as you need, then setting equal to 0 the extra variables in whichever formula has more variables.)


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