In last week’s post, we made use of the coinvariant ring $$\mathbb{C}[x_1,\ldots,x_n]/I$$ where $I=(p_1,\ldots,p_n)$ is the ideal generated by the positive-degree homogeneous $S_n$-invariants (symmetric polynomials). We saw that this was an $S_n$-module with Hilbert series $(n)_q!$, and claimed that it was the regular representation.

Let’s see why that is, and see if we can understand where the irreducible components occur.

More precisely, our goal is to understand the series $$\sum_{d} H_{\chi^\mu}(d)q^d$$ where $H_{\chi^\mu}(d)$ is the number of copies of the $\mu$th irreducible representation of $S_n$ occurring in the $d$th degree component of $\mathbb{C}[x_1,\ldots,x_n]/I$. In Stanley’s paper on invariants of finite groups, he states without proof the answer as the following “unpublished result of Lusztig”:

Let $G$ be the group of all $m \times m$ permutation matrices, and let $\chi$ be the irreducible character of $G$ corresponding to the partition $\mu$ of $m$. Then $H_{\chi}(n)$ is equal to the number of standard Young tableaux $T$ of shape $\mu$ such that $n$ is equal to the sum of those entries $i$ of $T$ for which $i$ appears in a column to the left of $i+1$.

To prove this, let’s start with the identity we showed last time, using boldface $\mathbf{x}$ to denote $x_1,\ldots,x_n$ as a shorthand:

$$\mathbb{C}[\mathbf{x}]=\Lambda_{\mathbb{C}}(\mathbf{x})\otimes_{\mathbb{C}[S_n]}\mathbb{C}[\mathbf{x}]/I$$

Since $\Lambda_{\mathbb{C}}(\mathbf{x})$, the ring of symmetric functions, consists entirely of copies of the trivial representation of $S_n$, we see that the irreducible components of type $\chi^\mu$ in degree $d$ on the right hand side come from those of that type in $\mathbb{C}[\mathbf{x}]/I$ of degree $d-k$, tensored with the trivial representations in degree $k$ in $\Lambda$, for some $k$. Moreover, there are $p_n(d)$ copies of the trivial representation in the $d$th degree in $\Lambda$ for all $d$, where $p_n(d)$ is the number of partitions of $d$ into parts of size at most $n$. (One can use the elementary or power sum symmetric function bases to see this.) From this we obtain the following series identity:

$$\left(\sum \left\langle (\mathbb{C}[\mathbf{x}])_d,\chi^\mu \right\rangle q^d\right)= \left(\sum p_n(d)q^d\right)\cdot \left(\sum H_{\chi^\mu}(d) q^d\right)$$

To simplify the left hand side, we can use the generalized version of Molien’s theorem for isotypic components (see here.) This gives us

$$\sum \left\langle (\mathbb{C}[\mathbf{x}])_d,\chi^\mu \right\rangle q^d=\frac{1}{n!}\sum_{\pi\in S_n}\frac{\overline{\chi^\mu}(\pi)}{\prod (1-q^{c_i(\pi)})}$$ where the $c_i(\pi)$’s are the cycle lengths of $\pi$.

(See this post for details on the above simplification in the case of the trivial character. The case of a general $\chi^\mu$ is analogous.)

If we group the permutations $\pi$ in the sum above according to cycle type (i.e. by conjugacy class), and use the fact that characters of $S_n$ are integers and hence $\overline{\chi^\mu}=\chi^\mu$, we have $$\sum \left\langle (\mathbb{C}[\mathbf{x}])_d,\chi^\mu \right\rangle q^d=\frac{1}{n!}\sum_{\lambda\vdash n} \frac{n!}{z_\lambda}\frac{\chi^\mu(\lambda)}{\prod_i (1-q^{\lambda_i})}.$$ Here $z_\lambda$ are the numbers such that $n!/z_\lambda$ is the size of the conjugacy class corresponding to the partition $\lambda$. It is not hard to see that this simplifies to a specialization of the power sum symmetric functions: $$\sum \frac{\chi^\mu(\lambda)}{z_\lambda} p_\lambda(1,q,q^2,\ldots)$$

Finally, by the representation-theoretic definition of the Schur functions, we see that this is simply $$s_\mu(1,q,q^2,\ldots).$$

Substituting for the left hand side of our original equation, we now have $$s_\lambda(1,q,q^2,\ldots)=\left(\sum p_n(d) q^d\right)\cdot \left(\sum H_{\chi^\mu}(d) q^d\right).$$ We can simplify this further by using the series identity $$\sum p_n(d) q^d=\frac{1}{(1-q)(1-q^2)\cdots(1-q^n)}.$$ In addition, there is a well-known identity (see also Enumerative Combinatorics vol. 2, Proposition 7.19.11) $$s_\mu(1,q,q^2,\ldots)=\frac{\sum_T q^{\operatorname{maj} T}}{(1-q)(1-q^2)\cdots (1-q^n)}$$ where the sum ranges over all standard Young tableaux $T$ of shape $\mu$, and where $\operatorname{maj} T$ denotes the sum of all entries $i$ of $T$ that occur in a higher row than $i+1$ (written in English notation).

This does it: putting everything together and solving for $\sum H_{\chi^\mu}(d) q^d$, we obtain $$\sum H_{\chi^\mu}(d) q^d=\sum_{T}q^{\operatorname{maj} T},$$ which is just about equivalent to Lusztig’s claim. (The only difference is whether we are looking at the rows or the columns that $i$ and $i+1$ are in. There must have been a typo, because the two are not the same $q$-series for the shape $(3,1)$. Replacing “column to the left of” by “row above” or replacing $\mu$ by its conjugate would fix the theorem statement above.)

One final consequence of the formulas above is that it is easy to deduce that the ring of coinvariants, $\mathbb{C}[\mathbf{x}]/I$, is isomorphic to the regular representation of $S_n$. Indeed, setting $q=1$ we see that the total number of copies of the irreducible representation corresponding to $\mu$ is equal to the number of standard Young tableaux of shape $\mu$, giving us the regular representation.

*Acknowledgments: The above techniques were shown to me by Vic Reiner at a recent conference. Thanks also to Federico Castillo for many discussions about the ring of coinvariants.*

Thanks Maria, this is really cool! Are you interested in diagonal harmonics too?

Yes! I’m looking at this stuff right now because I realized I didn’t even understand the one-variable case all that well. It’s making diagonal harmonics make a lot more sense. ðŸ™‚

Are you interested in DH? What aspects are you interested in?