# Generalization of symmetric functions

A $$n$$-variable function $$f$$ is a symmetric function if $$f(x_1,x_2, \ldots, x_n) = f(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n)})$$ for every permutation $$\sigma \in S_n$$. In particular, if $$f$$ is a polynomial, then $$f$$ is a symmetric polynomial. These objects have been studied extensively.

I wonder if the following generalization has been studied. A $$n^2$$-variable function $$f$$ is an $$S_n$$-symmetric function if $$f(x_{11}, x_{12}, \ldots, x_{1n}, \ldots, x_{nn}) = f(x_{\sigma(1)\sigma(1)}, x_{\sigma(1)\sigma(2)}, \ldots, x_{\sigma(1)\sigma(n)}, \ldots, x_{\sigma(n)\sigma(n)})$$ for every permutation $$\sigma \in S_n$$.

I think such objects must have been studied as they are so natural. But I don't know the keywords and couldn't find the literature.

I'm very grateful if anyone could provide information on them. Thanks in advance.

• if you only permute the first index you would have what is called a multisymmetric function Oct 1 '20 at 21:07

Let $$w\in S_n$$ (the symmetric group) have cycle type $$\lambda =(\lambda_1,\dots, \lambda_\ell)\vdash n$$, where $$\ell=\ell(\lambda)$$ is the length (number of nonzero parts) of $$\lambda$$. Then the induced action of $$w$$ on $$[n]\times [n]$$ (where $$[n]=\{1,2,\dots,n\}$$) has cycle enumerator $$\prod_{i=1}^{\ell(\lambda)} \prod_{j=1}^{\ell(\lambda)} z_{\mathrm{lcm}(\lambda_i,\lambda_j)}^{\mathrm{gcd}(\lambda_i,\lambda_j)}.$$ Let $$f_n(d)$$ be the dimension of the space of complex polynomials in the variables $$x_{ij}$$, $$1\leq i,j\leq n$$, that are homogeneous of degree $$d$$ and $$G$$-invariant. Then by Molien's theorem, $$F_n(x):=\sum_{d\geq 0} f_n(d)x^d$$ $$\ \ = \frac{1}{n!}\sum_{\lambda\vdash n} \frac{n!}{z_\lambda}\prod_{i=1}^{\ell(\lambda)} \prod_{j=1}^{\ell(\lambda)} \frac{1}{\left(1-x^{\mathrm{lcm}(\lambda_i,\lambda_j)}\right)^ {\mathrm{gcd}(\lambda_i,\lambda_j)}}.$$ I am using standard symmetric function notation, so $$n!/z_\lambda$$ is the number of permutations in $$S_n$$ of cycle type $$\lambda$$. For instance, $$F_1(x) = \frac{1}{1-x}$$ $$F_2(x) = \frac{1+x^2}{(1-x)^4(1+x)^2}$$ $$F_3(x) = \frac{1+3x^2+10x^3+16x^4+12x^5+16x^6+10x^7+3x^8+x^{10}} {(1-x)^9(1+x)^4(1+x+x^2)^3}.$$ Addendum. The invariant theory of finite groups, such as can be found here, can be used to obtain some further information about the ring $$R$$ of invariant polynomials. For instance, if $$S$$ is the subring of all symmetric functions in the $$x_{ij}$$'s, then $$R$$ is a finitely-generated free $$S$$-module of rank $$n^2!/n!$$.

• I am confused. I think $f_n(1)$ should be $2$. The sum of diagonal elements and the sum of off-diagonal elements form a basis. But if $F_1(x) = \frac{1}{1-x}$, then $f_n(1) = 1$. I am sorry that the origonal notation "$S_n \times S_n$-invariant" is misleading.
– MMM
Oct 16 '20 at 6:13
• @MMM $f_n(1)$ is not the coefficient of $x^n$ in $F_1(x)$, it is the coefficient of $x$ in $F_n(x)$. It is easy to see that for $F_2(x)$ and $F_3(x)$ as above, the coefficient of $x$ is indeed equal to $2$. Oct 16 '20 at 6:46