6
$\begingroup$

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.)

Improve this question
$\endgroup$
4
  • 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. $\endgroup$
    – Gregor Kemper
    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

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.