A result of Newell on plethysms of symmetric functions

It’s often said that part of the unique character of mathematics is that it builds on itself. Old results may be forgotten, but are only very rarely found to be incorrect. But changes in the ‘expected general background’ make many old papers impenetrable. For example, a long standing conjecture of Foulkes was introduced in a paper with the title ‘On the concomitants of the quintic and sextic up to degree four in the coefficients of the ground form‘. How many algebraists working today would immediately know even roughly what it is about? Of them, I’m sure still fewer would expect to read his paper with any pleasure. Certainly I cannot.

The purpose of this post is to do a small, but detailed, translation exercise on a paper of Newell from 1951, with the title ‘A theorem on the plethysm of $S$-functions‘. This is a relatively late publication from the stable of British algebraists working on invariant theory, so one might hope it would be fairly accessible.

Overview

Fix $m, n \in \mathbb{N}$ and let $\lambda$ be a partition of $mn$. Newell’s paper is cited in an important paper of Weintraub for the pair of results stated verbatim below:

\begin{aligned} ([m+1] \odot [1^n],[\lambda_1+1,\ldots,\lambda_n+1]) &= ([m] \odot [n], [\lambda]) \\ ([m+1] \odot [n],[\lambda_1+1,\ldots,\lambda_n+1]) &= ([m] \odot [1^n], [\lambda]) \end{aligned}.

Weintraub defines all his notation (which was standard for the time) clearly. For example, $[\lambda]$ denotes the irreducible representation of the symmetric group labelled by the partition $\lambda$, and $\odot$ is the analogue of the plethysm product $\circ$ for the symmetric group, with the variables swapped. Writing $s_\lambda$ for the Schur function labelled by the partition $\lambda$, an equivalent statement in the language of symmetric functions is:

\begin{aligned} \langle s_{(1^n)} \circ s_{(m+1)}, s_{\lambda + (1^n)} \rangle &= \langle s_{(n)} \circ s_{(m)}, s_\lambda \rangle \\ \langle s_{(n)} \circ s_{(m+1)}, s_{\lambda + (1^n)} \rangle &= \langle s_{(1^n)} \circ s_{(m)}, s_\lambda \rangle \end{aligned}.

Even a careful visual inspection of Newell’s paper reveals nothing that looks remotely like Weintraub’s statement (or my restatement). But in fact, the upper displayed equations are a special case of Newell’s Theorem 1, stated verbatim below:

Theorem 1.

If $\{m\} \otimes \{n\} = \sum \{\nu\}$ where $m$ and $n$ are integers then for any integer $k \le n$

$\sum g_{(1^k)\xi\nu} \{\nu\} = [(m-1)\otimes (1^k)][\{m\} \otimes \{n-k\}]$.

To be fair to Newell, $g$ is defined earlier: one reads ‘… where $g_{rst}$ is defined from the multiplication of $S$-functions by means of $\{r\}\{s\} = g_{rst}\{t\}$‘. So alles klar?

Translation up to Theorem 1

Much of Newell’s notation was standard at its time: $\{\nu\}$ is the Schur function $s_\nu$ and $\otimes$ is the plethystic product corresonding to Weintraub’s $\odot$. (So again, the order is reversed compared to $\circ$.) The hypothesis $\{m\} \otimes \{n\} = \sum \{\nu\}$ may still seem a little mysterious to modern eyes, since a general plethysm is certainly not multiplicity-free: nowadays we might write ‘$s_{(n)} \circ s_{(m)} = \sum c_\nu s_{\nu}$ where $c_\nu \in \mathbb{N}_0$ for each partition $\nu$ of $mn$‘. Only the $g$ remains to be understood: Newell defines $g_{rst}$ for natural numbers $r, s, t$, but then uses general partitions as the coefficients. Even in the special case, and with the benefit of knowing what is meant, his definition seems highly unclear to me.

A clue to the correct interpretation is given by the start of the sentence defining $g$ quoted above: ‘It is known [(2) 349] that if $\{\lambda\} \otimes \{n\} = \sum \{\nu\}$ then $\sum g_{1\xi\nu} \{\xi\} = \{\lambda\} \otimes \{n-1\}[g_{1\mu \lambda} \{\mu\}]$ …’. This must be parsed bearing in mind the summation convention that the repeated letters $\xi$ and $\mu$ are summed over. So, in modern language, it says: ‘if $s_n \circ s_\lambda = \sum c_\nu s_\nu$ then

$\sum_\nu c_\nu \sum_\xi g_{1\xi\nu} s_\xi = (s_{(n-1)} \circ s_\lambda)\sum_\mu g_{1\mu\lambda} s_\mu.'$

(I have included the parentheses around $s_{(n-1)} \circ s_\lambda$ as a small editorial mercy.) After staring at this for a while, I decided it must express the following symmetric function identity:

$(s_{(n)} \circ s_\lambda)\!\downarrow = (s_{(n-1)} \circ s_\lambda)(s_\lambda\!\downarrow)$,

where $s_\lambda\!\downarrow$ is the sum of all Schur functions $s_\mu$ labelled by partitions $\mu$ obtained from $\lambda$ by removing a box. (This identity is proved below, using the symmetric group.) Reciprocally, we have

$\langle s_\lambda\!\downarrow, s_\mu\rangle = \langle s_\lambda, s_{(1)}s_\mu \rangle.$

Thus $\downarrow$ is the adjoint to multiplication by $s_{(1)}$, and Newell’s $g_{1\mu \nu}$ are the corresponding coefficients. Generalizing freely, we can guess the correct definition of $g_{\theta\xi \nu}$.

Definition. Given partitions $\theta, \mu, \nu$ we define

$g_{\theta \mu\nu} = \langle s_\nu, s_\theta s_\mu \rangle.$

If we believe this is correct, then after one more piece of guesswork where we amend $g_{(1^k)\xi\nu}\{\nu\}$ in the statement of Theorem 1 to $g_{(1^k)\xi\nu}\{\xi\}$ (making it consistent with the above, and also with the analogous Theorem 1A), the conclusion of Theorem 1 becomes

$\sum_\nu c_\nu \sum_\xi \langle s_\xi s_{(1^k)}, s_\nu \rangle s_\xi = (s_{(1^k)}\circ s_{(m-1)})(s_{(n-k)} \circ s_{(m)}).$

The left-hand side is

$\sum_\xi \langle s_\xi s_{(1^k)}, \sum_\nu c_\nu s_\nu \rangle s_\xi = \sum_\xi \langle s_\xi s_{(1^k)}, s_{(n)} \circ s_{(m)} \rangle s_\xi.$

Therefore Newell’s Theorem 1 can be stated as follows:

$\sum_{\xi} \langle s_{\xi} s_{(1^k)}, s_{(n)} \circ s_{(m)} \rangle s_\xi = (s_{(1^k)}\circ s_{(m-1)})(s_{(n-k)} \circ s_{(m)}).$

If $s_\nu$ appears in $s_{(n)} \circ s_{(m)}$ then $\nu$ has at most $n$ parts. On the other hand, by Pieri’s rule, $s_{\xi}s_{(1^k)}$ is the sum of all partitions $\nu$ obtained from $\xi$ by adding $k$ boxes, no two in the same column. Therefore, in the special case when $k=n$, we have

$\langle s_\xi s_{(1^n)}, s_{(n)} \circ s_{(m)} \rangle = \langle s_{\xi + (1^n)}, s_{(n)} \circ s_{(m)} \rangle.$

Since $s_{(0)} \circ s_{(m)} = s_{(0)} = 1$, the unit symmetric function, we obtain

$\langle s_{\xi + (1^n)}, s_{(n)} \circ s_{(m)} \rangle = \langle s_\xi, s_{(1^n)} \circ s_{(m-1)}\rangle.$

This is equivalent to the first displayed equation in my restatement of Weintraub’s version of Newell’s result. The second displayed equation can be translated similarly.

Proof of Theorem 1

We prove Newell’s theorem in the symmetric group, replacing $m$ with $m+1$ for consistency with Weintraub’s statement. Let $\Omega$ be the collection of all set partitions of $\{1,\ldots,(m+1)n\}$ into $n$ sets each of size $m+1$, and let $M$ denote the corresponding permutation module for $\mathbb{C}S_{(m+1)n}$ with basis $\Omega$. In Weintraub’s notation, the permutation character of $M$ is $[m+1] \odot [n]$.

Consider the restricted module $M\!\downarrow_{S_k \times S_{(m+1)n-k}}$, where $S_k$ permutes $\{1,\ldots, k\}$ and $S_{(m+1)n-k}$ permutes $\{k+1,\ldots, (m+1)n\}$. The maximal summand of $M$ on which $S_k$ acts as the sign representation is spanned by the symmetrized set partitions $v_\mathcal{P}$ for $\mathcal{P} \in \Omega$, where

$v_\mathcal{P} = \mathcal{P} \sum_{\sigma \in S_k} \sigma \mathrm{sgn}(\sigma).$

Observe that $v_\mathcal{P} = 0$ unless $\mathcal{P}$ has at most one entry from $\{1,\ldots, k\}$ in each of its $n$ constituent $(m+1)$-sets. If $X$ and $Y$ are two such subsets, with entries $x_1, \ldots, x_{m}$ and $y_1, \ldots, y_{m}$ in $\{k+1,\ldots,(m+1)n\}$ respectively, then

$v_\mathcal{P} (x_1,y_1)\ldots (x_{m},y_{m}) = -v_\mathcal{P}.$

It follows that

\begin{aligned}\quad [m+1] \odot [n] & \!\downarrow_{S_k \times S_{(m+1)n-k}} \\ & \hskip0.1in = \mathrm{sgn}_{S_k} \times ([m] \odot [(1^k)])([m+1] \odot [n-k]) + \pi \end{aligned}

where $\pi$ is a sum of irreducible characters of $S_k \times S_{(m+1)n-k}$ not of the form $\mathrm{sgn}_{S_k} \times \chi^\tau$. By Frobenius reciprocity,

\begin{aligned} \langle [m+1] \odot [n], & \mathrm{sgn}_{S_k} \times \chi^\xi \uparrow^{S_{(m+1)n}} \rangle \\ &\hskip0.2in = \langle ([m] \odot [(1^k)])([m+1] \odot [n-k]), \chi^\xi \rangle \end{aligned}

for each partition $\xi$ of $(m+1)n -k$. This is equivalent to

$\langle s_{(n)} \circ s_{(m+1)}, s_{(1^k)}s_\xi \rangle = \langle (s_{(1^k)} \circ s_m)(s_{(n-k)} \circ s_{(m+1)}, s_\xi \rangle$

which is obtained from my restatement of Newell’s theorem by taking the inner product with $s_\xi$. The companion result follows by a sign twist. $\Box$

No wonder old results are frequently reproved: its invariably less work than reading the original.