Tail Positive Words and Generalized Coinvariant Algebras

NEW

Problem: For $n,k,r\in\mathbb Z_{\ge0}$ with $r\le n$, let

$$P_n=\mathbb Q[x_1,\dots,x_n],$$

with $S_n$ acting by permutation of variables. Let $h_d,e_d,p_d$ be the complete, elementary, and power-sum symmetric polynomials in $x_1,\dots,x_n$. Define

$$I_{n,k,r}:=\langle h_{k+1},\dots,h_{k+n},e_n,e_{n-1},\dots,e_{n-r+1}\rangle$$

and

$$I'_{n,k,r}:=\langle p_{k+1},\dots,p_{k+n},e_n,e_{n-1},\dots,e_{n-r+1}\rangle .$$

Let

$$R_{n,k,r}=P_n/I_{n,k,r},\qquad R'_{n,k,r}=P_n/I'_{n,k,r}.$$

The reconstructed conjecture is: there is an isomorphism of graded $S_n$-modules

$$R_{n,k,r}\cong R'_{n,k,r}.$$

This is the intended formalization of Conjecture 5.1; the paper has a minor typographical slip where $R'_{n,k,r}$ is momentarily written using $I_{n,k,r}$, but the surrounding text says $I'_{n,k,r}$ is obtained by replacing $h_d$ by $p_d$.

Result: The conjecture is true.

Let $\Lambda_n=P_n^{S_n}=\mathbb Q[e_1,\dots,e_n]$. Since $I_{n,k,r}$ and $I'_{n,k,r}$ are generated by symmetric polynomials, they are extensions of homogeneous ideals $J,J'\subseteq \Lambda_n$. By the Chevalley decomposition,

$$P_n\cong \Lambda_n\otimes H_n$$

as graded $S_n$-modules, where $S_n$ acts trivially on $\Lambda_n$. Hence

$$P_n/JP_n\cong (\Lambda_n/J)\otimes H_n,$$

and similarly for $J'$. Therefore it suffices to show that $\Lambda_n/J$ and $\Lambda_n/J'$ have the same Hilbert series.

Put $m=n-r$. Modding out by

$$e_n,e_{n-1},\dots,e_{m+1}$$

identifies $\Lambda_n/(e_{m+1},\dots,e_n)$ with $\Lambda_m=\mathbb Q[e_1,\dots,e_m]$. Under this identification, $h_d(x_1,\dots,x_n)$ and $p_d(x_1,\dots,x_n)$ become the corresponding $h_d,p_d$ in $m$ variables. Thus

$$\Lambda_n/J\cong \Lambda_m/\langle h_{k+1},\dots,h_{k+n}\rangle,$$ $$\Lambda_n/J'\cong \Lambda_m/\langle p_{k+1},\dots,p_{k+n}\rangle .$$

In $\Lambda_m$, the recurrences for complete functions and power sums imply that the later generators are redundant:

$$\langle h_{k+1},\dots,h_{k+n}\rangle=\langle h_{k+1},\dots,h_{k+m}\rangle,$$ $$\langle p_{k+1},\dots,p_{k+n}\rangle=\langle p_{k+1},\dots,p_{k+m}\rangle .$$

Both sequences

$$h_{k+1},\dots,h_{k+m} \quad\text{and}\quad p_{k+1},\dots,p_{k+m}$$

are homogeneous systems of parameters in $\Lambda_m$. Indeed, over $\mathbb C$, if $h_{k+1}=\cdots=h_{k+m}=0$, then the sequence $h_d$ has $m$ consecutive zero terms. Removing zero roots, it satisfies an invertible linear recurrence whose constant coefficient is nonzero; back-substitution forces $h_0=0$, contradiction. Hence all roots are zero. The same argument applies to $p_d=\sum_i x_i^d$: after grouping distinct nonzero roots, the power sums satisfy an invertible recurrence, and $m$ consecutive zero terms force $p_0=0$, contradiction unless all roots are zero.

Thus both are regular sequences in the polynomial ring $\Lambda_m$, with degrees

$$k+1,k+2,\dots,k+m.$$

Therefore both quotient Hilbert series equal

$$\prod_{i=1}^{m}\frac{1-q^{k+i}}{1-q^i}.$$

So $\Lambda_n/J$ and $\Lambda_n/J'$ are isomorphic as graded vector spaces. Tensoring such a graded vector-space isomorphism with $H_n$ gives a graded $S_n$-module isomorphism

$$R_{n,k,r}\cong R'_{n,k,r}.$$

The boundary cases are included: if $r=n$, then $m=0$ and both invariant quotients are $\mathbb Q$; if $k=0$, both ideals give the ordinary coinvariant algebra.

Citation: No exact prior resolution is cited here. The proof uses the standard Chevalley freeness/decomposition theorem for reflection groups: C. Chevalley, “Invariants of finite groups generated by reflections,” Amer. J. Math. 77 (1955), 778–782.