# $S_n$-invariant polynomials on the dual of reflection representation

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.)

• (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. Oct 20, 2021 at 21:09
• 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. Oct 20, 2021 at 22:54
• 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. Oct 21, 2021 at 9:39
• I actually laughed at how arbitrary those degrees are :D Excellent answer. I'm reassured I wasn't missing something obvious! Oct 21, 2021 at 13:13