Separating Invariants for Two Copies of the Natural $S_n$-Action

TL;DR

This paper identifies a small, efficient set of invariants that distinguish two copies of the natural $S_n$-representation, providing minimal separating sets for small n, advancing understanding of invariant rings.

Contribution

It introduces a significantly smaller set of separating invariants for the ring of vector invariants of two copies of the natural $S_n$-representation, with minimality proven for small n.

Findings

Set of separating invariants is much smaller than generating sets.

For n ≤ 4, the set is minimal among all separating sets.

Provides explicit construction of separating invariants.

Abstract

This note provides a set of separating invariants for the ring of vector invariants $K[V^2]^{S_n}$ of two copies of the natural $S_n$-representation $V = K^n$ over a field of characteristic 0. This set is much smaller than generating sets of $K[V^2]^{S_n}$. For $n \leq 4$ we show that this set is minimal with respect to inclusion among all separating sets.

Equations

$$x_1+\dots +x_n,x_1^2+\dots +x_n^2,\dots ,x_1^n+\dots +x_n^n.$$

$$f_{k_1,\dots ,k_m}:=\sum _{i=1}^n{x}_{i1}^{k_1}\cdot \dots \cdot x_{im}^{k_m}\text{ with ki∈N such that k1+…+km≤n},$$

$$M=\{\underset{=:f_{j,k}}{\underbrace{\sum _{i=1}^n{x}_i^j{y}_i^k}}\mid j,k\in \mathbb{N}\text{ such that }j+k\le n\text{ and }jk\ne 0\}.$$

$$|M|=\left(\sum _{j=0}^n(n-j+1)\right)-1=\frac{1}{2}(n^2+3n).$$

$$I=\{(j,k)\in {\mathbb{N}}^2\mid 0\le j\le n,0\le k\le \lfloor \frac{n}{j+1}\rfloor ,jk\ne 0\}.$$

$$|S|=\left(\sum _{j=0}^n(\lfloor \frac{n}{j+1}\rfloor +1)\right)-1=n+\sum _{j=1}^n\lfloor \frac{n}{j}\rfloor .$$

$$|S|=\mathcal{O}(n\log (n))\text{vspace}1mm$$

$$p=(a_1,\dots ,a_n,b_1,\dots ,b_n)\text{and}q=(a_1,\dots ,a_n,c_1,\dots ,c_n).$$

$$\sum _{i=1}^n{a}_i^j{b}_i=f_{j,1}(p)=f_{j,1}(q)=\sum _{i=1}^n{a}_i^j{c}_i.$$

$$\left(\begin{array}{ccc}1& \dots & 1\\ a_1& \dots & a_n\\ ⋮& & \\ a_1^{n-1}& \dots & a_n^{n-1}\end{array}\right)\cdot \left(\begin{array}{c}{b}_1\\ ⋮\\ b_n\end{array}\right)=\left(\begin{array}{ccc}1& \dots & 1\\ a_1& \dots & a_n\\ ⋮& & \\ a_1^{n-1}& \dots & a_n^{n-1}\end{array}\right)\cdot \left(\begin{array}{c}{c}_1\\ ⋮\\ c_n\end{array}\right).$$

$$\{{\lambda }_1,\dots ,{\lambda }_r\}=\{a_1,\dots ,a_n\}.$$

$$\tau (a_1,\dots ,a_n)=(\underset{n_1}{\underbrace{{\lambda }_1,\dots ,{\lambda }_1}},\underset{n_2}{\underbrace{{\lambda }_2,\dots ,{\lambda }_2}},\dots ,\underset{n_r}{\underbrace{{\lambda }_r,\dots ,{\lambda }_r}}),$$

$$p$$

$$q$$

$$j\le r-1\le n-1,$$

$$k\le n_r\le \frac{n}{r}\le \frac{n}{j+1}.$$

$$f_{j,k}(p)=\sum _{i=1}^n{a}_i^j{b}_i^k$$

$$=\sum _{i=1}^r{\lambda }_i^j\left(\sum _{l=1}^{n_i}{b}_{n_1+\dots +n_{i-1}+l}^k\right),$$

$$f_{j,k}(q)=\sum _{i=1}^r{\lambda }_i^j\left(\sum _{l=1}^{n_i}{c}_{n_1+\dots +n_{i-1}+l}^k\right).$$

$$\left(\begin{array}{c}{b}_1^k+\dots +b_{n_1}^k\\ b_{n_1+1}^k+\dots +b_{n_1+n_2}^k\\ ⋮\\ b_{n-n_r+1}^k+\dots +b_n^k\end{array}\right)\text{and}\left(\begin{array}{c}{c}_1^k+\dots +c_{n_1}^k\\ c_{n_1+1}^k+\dots +c_{n_1+n_2}^k\\ ⋮\\ c_{n-n_r+1}^k+\dots +c_n^k\end{array}\right)\in K^r$$

$$\left(\begin{array}{ccc}1& \dots & 1\\ {\lambda }_1& \dots & {\lambda }_r\\ ⋮& & ⋮\\ {\lambda }_1^{r-1}& \dots & {\lambda }_r^{r-1}\end{array}\right).$$

$$b_{n-n_r+1}^k+\dots +b_n^k=c_{n-n_r+1}^k+\dots +c_n^k.$$

$$p^{\prime }=(a_1,\dots ,a_{n-n_r},b_1,\dots ,b_{n-n_r})$$

$$q^{\prime }=(a_1,\dots ,a_{n-n_r},c_1,\dots ,c_{n-n_r}),$$

$$I^{\prime }=\{(j,k)\in {\mathbb{N}}^2\mid 0\le j\le n-n_r,0\le k\le \lfloor \frac{n-n_r}{j+1}\rfloor ,jk\ne 0\}.$$

$$f_{j,k}^{\prime }:=\sum _{i=1}^{n-n_r}{x}_i^j{y}_i^k$$

$$K[x_1,\dots ,x_{n-n_r},y_1,\dots ,y_{n-n_r}{]}^{S_{n-n_r}}.$$

$$f_{j,k}^{\prime }(p^{\prime })=\sum _{i=1}^{n-n_r}{a}_i^j{b}_i^k$$

$$=f_{j,k}(q)-\sum _{i=n-n_r+1}^n{a}_i^j{c}_i^k=f_{j,k}^{\prime }(q^{\prime }),$$

$$\{\sum _{i=1}^r{x}_i^k\mid k=1,\dots ,r-1\}$$

$$\sum _{i=1}^r{b}_i^k=\sum _{i=1}^r{c}_i^k\text{for }k=1,\dots ,r-1.$$

$$p=(\underset{r}{\underbrace{1,\dots ,1}},\underset{r}{\underbrace{2,\dots ,2}},b_1,\dots ,b_r,c_1,\dots ,c_r)\in K^n\times K^n$$

Full text

For $n\leq 4$ we show that this set is minimal with respect to inclusion among all separating sets.

Introduction

Let $K$ be a field of characteristic 0. Let $G=S_{n}$ be the symmetric group and $V=K^{n}$ its natural representation over $K$. The $K$-algebra of symmetric polynomials $K[V]^{S_{n}}=K[x_{1},\ldots,x_{n}]^{S_{n}}$ is generated by the first $n$ power sums

[TABLE]

The action of $S_{n}$ on $V$ extends to a diagonal action of $S_{n}$ on $V^{m}$ for $m\geq 2$. The ring of vector invariants $K[V^{m}]^{S_{n}}$ is not a polynomial ring anymore. Its elements are called multisymmetric polynomials and they have been classically studied (see Weyl [5]). If we write the variables of $K[V^{m}]$ as $x_{ij}$ with $i=1,\ldots,n$, $j=1,\ldots,m$, then $K[V^{m}]^{S_{n}}$ is minimally generated by the following polarizations of the above power sums:

[TABLE]

see Domokos [2, Theorem 2.5 & Remark 2.6]. Asymptotically, this minimal generating set has size $\mathcal{O}(n^{m})$. But by the general upper bound for separating invariants (see Derksen and Kemper [1, Chapter 2.4]), a subset of $K[V^{m}]^{S_{n}}$ of size $2nm+1=\mathcal{O}(nm)$ which separates all $S_{n}$-orbits in $V^{m}$ exists. A lower bound of $(n+1)m-1$ for the size of a separating set was achieved by Dufresne and Jeffries [4, Theorem 3.4].

However, to the best of my knowledge no explicit small separating sets for multisymmetric polynomials are known. Using the concept of “cheap polarizations” Draisma et al. [3, Corollary 2.12] give a separating set of size $\mathcal{O}(n^{2}m)$.

This note considers the case $m=2$. Here the two sets of variables are denoted by $x_{1},\ldots,x_{n}$ and $y_{1},\ldots,y_{n}$. Then $K[V^{2}]^{S_{n}}$ is minimally generated by the set of power sums

[TABLE]

We use the notation $f_{j,k}=\sum_{i=1}^{n}x_{i}^{j}y_{i}^{k}$ and $M$ as in (0.1) throughout this paper. Observe that

[TABLE]

The main result of this paper is the separating set $S$ given in the following theorem, which is much smaller than $M$.

Theorem 1**.**

Let $I$ be the set of double indices

[TABLE]

*Then $S:=\{f_{j,k}\mid(j,k)\in I\}$ is a separating subset of the ring of bisymmetric poly-

nomials $K[V^{2}]^{S_{n}}$.*

Remark**.**

(a) Note that

[TABLE]

The sum $D(n):=\sum_{j=1}^{n}\lfloor\frac{n}{j}\rfloor$ is the so-called Divisor summatory function. Asymptotically, we have

[TABLE]

in contrast to $|M|=\mathcal{O}(n^{2})$. For example, for $n=100$ we have $|M|=5150$ and $|S|=582$.

(b) The invariants from $S$ are homogeneous, while the general upper bound of $2nm+1$ for separating invariants can typically only be achieved with inhomogeneous invariants.

(c) The set $I$ is not symmetric in $(j,k)$. For example, for $n=3$ we have $f_{2,1}\in S$, but $f_{1,2}\notin S$.

We give the proof of this theorem in Section 1. In Section 2 we analyze the cases $n=2$, $3$, $4$ more closely and show that $S$ is indeed a minimal separating set.

1. Proof of the main theorem

We use induction on $n$ to proof Theorem 1. For $n=1$ we have $S=M=\{x_{1},y_{1}\}$ which is clearly separating (and generating) for the trivial action on $K[V^{2}]=K[x_{1},y_{1}]$.

Now let $n\geq 2$ and take two points $p$, $q\in V^{2}$ such that $f(p)=f(q)$ for all $f\in S$. We need to show that $p$ and $q$ lie in the same $S_{n}$-orbit.

For this let $p=(a_{1},\ldots,a_{n},b_{1},\ldots,b_{n})$. The „original“ power sums $f_{j,0}=\sum_{i=1}^{n}x_{i}^{j}$ with $j=1,\ldots,n$ are elements of $S$. From $f_{j,0}(p)=f_{j,0}(q)$ for all $j$ we get that the $x_{i}$-coordinates of $p$ and $q$ are the same upto a permutation. So there exists an element $\sigma\in S_{n}$ with $\sigma q=(a_{1},\ldots,a_{n},c_{1},\ldots,c_{n})$. Since we only want to show that $p$ and $q$ are equivalent under the $S_{n}$-action, we may replace $q$ by $\sigma q$. So now we may assume that

[TABLE]

To exhibit the main idea of the following proof let us first do the much easier special case that all $a_{i}$ are distinct. The elements $f_{j,1}$ with $j=0,\ldots,n-1$ are in $S$, hence for all these $j$ we have

[TABLE]

Using a Vandermonde matrix these equations can be written as

[TABLE]

Now if we assume that the $a_{i}$ are pairwise distinct, we can conclude $b_{i}=c_{i}$ for all $i$ and hence $p=q$. So indeed they lie in the same $S_{n}$-orbit.

For the general case let $r:=|\{a_{1},\ldots,a_{n}\}|$ be the number of distinct $a_{i}$’s and let

[TABLE]

We can use a permutation $\tau\in S_{n}$ to rearrange the $a_{i}$’s such that

[TABLE]

where $(n_{1},\ldots,n_{r})$ is a partition of $n$, i.e., $n=n_{1}+\ldots+n_{r}$ and $n_{1}\geq n_{2}\geq\ldots\geq n_{r}$.

By replacing $p$ and $q$ with $\tau p$ and $\tau q$, respectively, we may now assume that

[TABLE]

where the $\lambda_{i}$ are pairwise distinct and $n_{1}\geq n_{2}\geq\ldots\geq n_{r}$ holds.

Observe that $rn_{r}\leq n$, i.e., $n_{r}\leq\frac{n}{r}$. For every $j=0,\ldots,r-1$ and $k=1,\ldots,n_{r}$ we have $(j,k)\in I$, since:

[TABLE]

So $f_{j,k}\in S$ for $j=0,\ldots,r-1$ and $k=1,\ldots,n_{r}$. Hence for these $j$ and $k$ we have $f_{j,k}(p)=f_{j,k}(q)$ by assumption. It is

[TABLE]

and similarly

[TABLE]

For a fixed $k\in\{1,\ldots,n_{r}\}$ this holds for every $j=0,\ldots,r-1$. So we conclude that the vectors

[TABLE]

have the same image when multiplied from the right to the Vandermonde matrix

[TABLE]

Since the $\lambda_{i}$ are pairwise distinct, the vectors in (1.2) must be equal. In the last component this means

[TABLE]

But this holds or every $k=1,\ldots,n_{r}$, so we can conclude that the $n_{r}$ elements $b_{n-n_{r}+1},\,b_{n-n_{r}+2},\ldots,b_{n}$ are equal to the $n_{r}$ elements $c_{n-n_{r}+1},\,c_{n-n_{r}+2},\ldots,c_{n}$ upto a permutation $\pi\in S_{n_{r}}$.

We read $\pi$ as an element of $S_{n}$ which permutes only the last $n_{r}$ positions. So applying $\pi$ to a point in $V^{2}=K^{n}\times K^{n}$ can only change the last $n_{r}$ of the $x_{i}$- and the last $n_{r}$ of the $y_{i}$-coordinates of this point. But observe that because of (1) applying $\pi$ to $q$ has no affect on the last $n_{r}$ of the $x_{i}$-coordinates of $q$. So after replacing $q$ with $\pi q$ we may now assume that the last $n_{r}$ of the $y_{i}$-coordinates of $p$ and $q$ are equal as well, while $p$ and $q$ are still in the form of (1).

Now we project onto the first $n-n_{r}$ of both the $x_{i}$- and $y_{i}$-coordinates. This projection $K^{n}\times K^{n}\to K^{n-n_{r}}\times K^{n-n_{r}}$ maps $p$ and $q$ to

[TABLE]

and

[TABLE]

respectively. For $n-n_{r}$ instead of $n$ we have a corresponding set

[TABLE]

of double indices, which leads to a corresponding set $S^{\prime}=\{f_{j,k}^{\prime}\mid(j,k)\in I^{\prime}\}$ of polynomials

[TABLE]

in the ring of bisymmetric polynomials

[TABLE]

By induction assumption, $S^{\prime}$ is separating.

For every double index $(j,k)\in I^{\prime}$ we have $(j,k)\in I$. So for every $(j,k)\in I^{\prime}$ we get

[TABLE]

where we have used that the last $n_{r}$ coordinates of $p$ and $q$ are equal. Since $S^{\prime}$ is separating, there exists a permutation $\varphi\in S_{n-n_{r}}$ such that $p^{\prime}=\varphi q^{\prime}$. We read $\varphi$ naturally as an element of $S_{n}$ fixing the last $n_{r}$ positions. Since $p$ and $q$ were already equal at the last $n_{r}$ of the $x_{i}$- and the last $n_{r}$ of the $y_{i}$-coordinates, we conclude that $p=\varphi q$. ∎

2. Minimality in low-dimensional cases

In this section we consider the question of minimality (w. r. t.  inclusion) of the set $S$ from Theorem 1 among all separating subsets of $K[V\times V]^{S_{n}}$. For $n\leq 4$ we show that $S$ is indeed a minimal separating set.

First we note in the following lemma that the power sums in $x_{1},\ldots,x_{n}$ (and similarly the power sums in $y_{1},\ldots,y_{n}$) cannot be left out from $S$.

Lemma 1**.**

Let $S$ be the set from Theorem 1. For all $j\in\{1,\ldots,n\}$ the set $S\setminus\{f_{j,0}\}$ is not separating. Similarly, for all $k\in\{1,\ldots,n\}$ the set $S\setminus\{f_{0,k}\}$ is not separating.

Proof.

We can view $T=\{f_{j,0}\mid j=1,\ldots,n\}$ as a subset of the ring of symmetric polynomials $K[x_{1},\ldots,x_{n}]^{S_{n}}$. Here a minimal separating set has size $n$, so for all $j\in\{1,\ldots,n\}$ the set $T\setminus\{f_{j,0}\}$ is not separating. Hence there exist $a=(a_{1},\ldots,a_{n})$ and $a^{\prime}=(a_{1}^{\prime},\ldots,a_{n}^{\prime})\in K^{n}$ such that $g(a)=g(a^{\prime})$ for all $g\in T\setminus\{f_{j,0}\}$. But then $(a,0)$, $(a^{\prime},0)$ in $K^{2n}$ cannot be separated by $S\setminus\{f_{j,0}\}$. The second statement is analogous. ∎

Next we note in the following lemma another polynomial that cannot be left out from $S$.

Lemma 2**.**

Let $S$ be the set from Theorem 1. Assume that $n$ is even and set $r:=\frac{n}{2}$. Then the set $S\setminus\{f_{1,r}\}$ is not separating.

Proof.

In $K[x_{1},\ldots,x_{r}]^{S_{r}}$ the set

[TABLE]

is not separating (since the last power sum is missing). So there exist $b=(b_{1},\ldots,b_{r})$ and $c=(c_{1},\ldots,c_{r})$ which are not in the same $S_{r}$-orbit, but satisfy

[TABLE]

Consider

[TABLE]

and

[TABLE]

We have

[TABLE]

hence

[TABLE]

So if $j=0$ or if $k\leq r-1$ we have $f_{j,k}(p)-f_{j,k}(q)=0$. Looking at the definition of $S$ in Theorem 1 we see that for all $f_{j,k}\in S$ with $(j,k)\neq(1,r)$ we have $j=0$ or $k\leq r-1$, hence $f_{j,k}(p)=f_{j,k}(q)$. But $p$ and $q$ are not in the same $S_{n}$-orbit, so $S\setminus\{f_{1,r}\}$ is not separating. ∎

Theorem 2**.**

For $n\leq 4$ the set $S$ from Theorem 1 is a minimal separating set with respect to inclusion.

Proof.

Case $n=2$: Here we have

[TABLE]

By Lemma 1, $S\setminus\{f_{j,k}\}$ with $j=0$ or $k=0$ is not separating. By Lemma 2, $S\setminus\{f_{1,1}\}$ is not separating.

Case $n=3$: Here

[TABLE]

By Lemma 1 there are only two cases left to check.

For example

[TABLE]

show that $S\setminus\{f_{2,1}\}$ is not separating, while

[TABLE]

show that $S\setminus\{f_{1,1}\}$ is not separating.

Case $n=4$: Here

[TABLE]

Again by Lemma 1 there are only four cases left to check. Lemma 2 shows that $S\setminus\{f_{1,2}\}$ is not separating.

For example

[TABLE]

show that $S\setminus\{f_{3,1}\}$ is not separating.

For example

[TABLE]

show that $S\setminus\{f_{2,1}\}$ is not separating.

For example

[TABLE]

show that $S\setminus\{f_{1,1}\}$ is not separating. ∎