Let $W=S_n$ (the symmetric group) acting on $V=k^n$ via permutation of the indices, where $k$ is an algebraically closed field. Closely related to this are the reflection (sub-)representation $V_0=\{ (v_1,\ldots ,v_n)\in k^n : \sum_{i=1}^n v_i=0\}$ and its dual $V/{\mathfrak z}$, where ${\mathfrak z}$ is the subspace spanned by $(1,1,\ldots ,1)$. Of course, if ${\rm char}\, k=0$ or is a prime not dividing $n$, then these representations are isomorphic. But this is no longer true if ${\rm char}\, k$ divides $p$.

It's well known that the $W$-invariant polynomials in $k[V]=S(V^*)$ are freely generated by the elementary symmetric polynomials in the coefficients. Since $V_0$ is the zero set of the first elementary symmetric polynomial $(v_1,\ldots ,v_n)\mapsto v_1+\ldots +v_n$, it follows that $k[V_0]^W$ is freely generated by homogeneous polynomials of degree $2,3,\ldots ,n$.

What (if anything) is known about the invariants $k[V/{\mathfrak z}]^W\cong S(V_0)^W$ when $p|n$?

The motivation for the question is: by the Chevalley restriction theorem (proved in maximal generality by Chaput and Romagny https://arxiv.org/abs/0805.2140), $k[V/{\mathfrak z}]^W$ is isomorphic to $k[\mathfrak{pgl}_n]^{{\rm GL}_n}$, and the generators of the ideal $\{ f\in k[V/{\mathfrak z}]^W: f(0)=0\}$ are mapped to generators of the ideal determining the nilpotent cone ${\mathcal N}(\mathfrak{pgl}_n)$. The geometry here may be somewhat different to the geometry of ${\mathcal N}(\mathfrak{sl}_n)$ (e.g. ${\mathcal N}(\mathfrak{pgl}_2)$ is smooth in characteristic 2).

There is always a linear generator, namely $(1,1,\ldots ,1)$. To see how different this can be from $k[V_0]$: for $n=p=3$ I think $S(V_0)^W$ is freely generated by elements of degree $1$ and $6$. For $n=p=5$ there are generators of the invariant ring of degree $1, 4, 6, 8$, but I don't know whether they generate all of $S(V_0)^W$. (I suspect not.)

  • 2
    $\begingroup$ (1) In other words, we're looking at the fixed points in the ring of elementary symmetric polynomials $k[e_1,\dots, e_n]$ of the one-parameter family of automorphisms given by translation, which if I calculated right is $e_i \mapsto \sum_{j=0}^i e_j \binom{n -j}{i-j} \lambda^{i-j}$. I don't know if this description helps much. (2) The degree 6 invariant can be interpreted as the discriminant of the cubic, i.e. the product of the squares of the differences of the $v_i$. I wonder if the discriminant of the quintic can be expressed in terms of the generators you found. $\endgroup$
    – Will Sawin
    Oct 20, 2021 at 21:09
  • $\begingroup$ Good idea, and in fact I think the discriminant shows that my generators of degree $1,4,6,8$ aren't a complete set for the $n=p=5$ case. I agree with your statement about the translations of $k[e_1,\ldots , e_n]$ - in concrete cases this seems to help with some calculations, but I don't know if it is any use in general. $\endgroup$
    – Paul Levy
    Oct 20, 2021 at 22:54
  • 2
    $\begingroup$ The invariant ring $S(V_0)^W$ tends to be very nasty. By Corollary 2.8 in my article "On the Cohen-Macaulay Property of Modular Invariant Rings" in J. of Algebra 215, it's not Cohen-Macauly if $p \ge 5$. So your hunch that trouble starts there is exactly right. In fact, for n = p = 5, minimal generators of the invariant ring turn out to be of degrees 1,4,6,8,10,12,14,15,17,18,20,21. Really nasty! By another paper of mine in J. of Algebra 245, if $n < 2 p$, then the invariant ring has depth $n/p + 2$, so it's arbitrarily far away from being Cohen-Macaulay. $\endgroup$ Oct 21, 2021 at 9:39
  • $\begingroup$ I actually laughed at how arbitrary those degrees are :D Excellent answer. I'm reassured I wasn't missing something obvious! $\endgroup$
    – Paul Levy
    Oct 21, 2021 at 13:13


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