# The Frobenius transform of a symmetric function

Mitchell Lee

###### Abstract.

We define an abelian group homomorphism $\mathscr{F}$, which we call the
_Frobenius transform_ , from the ring of symmetric functions to the ring of
the symmetric power series. The matrix entries of $\mathscr{F}$ in the Schur
basis are the _restriction coefficients_
$r_{\lambda}^{\mu}=\dim\operatorname{Hom}_{\mathfrak{S}_{n}}(V_{\mu},\mathbb{S}^{\lambda}\mathbb{C}^{n})$,
which are known to be nonnegative integers but have no known combinatorial
interpretation. The Frobenius transform satisfies the identity
$\mathscr{F}\\{fg\\}=\mathscr{F}\\{f\\}\ast\mathscr{F}\\{g\\}$, where $\ast$
is the Kronecker product.

We prove for all symmetric functions $f$ that
$\mathscr{F}\\{f\\}=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}\cdot(1+h_{1}+h_{2}+\cdots)$,
where
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}$
is a symmetric function with the same degree and leading term as $f$. Then, we
compute the matrix entries of
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
and
$\mathscr{F}^{-1}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
in the complete homogeneous, elementary, and power sum symmetric function
bases, giving combinatorial interpretations of the coefficients where
possible. In particular, the matrix entries of
$\mathscr{F}^{-1}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
in the elementary basis count words with a constraint on their Lyndon
factorization.

As an example application of our main results, we prove that
$r_{\lambda}^{\mu}=0$ if the Young diagram of $\mu$ contains a square of side
length greater than $2^{\lambda_{1}-1}$, and this inequality is tight.

##  1\. Introduction

Let $n\geq 0$ and let $\lambda$ be a partition with at most $n$ parts. There
is a corresponding irreducible $GL_{n}(\mathbb{C})$-module: the Schur module
$\mathbb{S}^{\lambda}\mathbb{C}^{n}$. Because the symmetric group
$\mathfrak{S}_{n}$ embeds in $GL_{n}(\mathbb{C})$ (by permutation matrices),
one may ask: how does the restriction of $\mathbb{S}^{\lambda}\mathbb{C}^{n}$
to $\mathfrak{S}_{n}$ decompose into irreducible $\mathfrak{S}_{n}$-modules?

In other words, let $\lambda$ and $\mu$ be partitions and let $n=|\mu|$. What
is the value of the restriction coefficient

$r_{\lambda}^{\mu}=\dim\operatorname{Hom}_{\mathfrak{S}_{n}}(V_{\mu},\mathbb{S}^{\lambda}\mathbb{C}^{n}),$

where $V_{\mu}$ is the Specht module corresponding to the partition $\mu$?
This problem, called the _restriction problem_ , has held considerable recent
interest [2, 7, 15, 16, 19, 21]. However, there remains no known combinatorial
interpretation for $r_{\lambda}^{\mu}$.

Let $\Lambda$ be the ring of symmetric functions in the variables
$x_{1},x_{2},x_{3},\ldots$ and let $\overline{\Lambda}$ be the ring of
symmetric power series in $x_{1},x_{2},x_{3},\ldots$. In this paper, we will
consider the abelian group homomorphism
$\mathscr{F}\colon\Lambda\to\overline{\Lambda}$ defined on the basis
$\\{s_{\lambda}\\}$ of Schur functions by

$\mathscr{F}\\{s_{\lambda}\\}=\sum_{\mu}r_{\lambda}^{\mu}s_{\mu}.$

Equivalently, $\mathscr{F}\\{s_{\lambda}\\}$ is the result of applying the
Frobenius character map to the representation
$\bigoplus_{n}\mathbb{S}^{\lambda}\mathbb{C}^{n}$ of
$\bigoplus_{n}\mathbb{C}[\mathfrak{S}_{n}]$. For this reason, we call
$\mathscr{F}$ the _Frobenius transform_. It encodes all information about all
of the restriction coefficients.

Section 3 will cover the basic properties of the Frobenius transform, many of
which are implicit in the work of Orellana and Zabrocki. For example, for any
symmetric functions $f,g$, we have
$\mathscr{F}\\{fg\\}=\mathscr{F}\\{f\\}\ast\mathscr{F}\\{g\\}$, where $\ast$
is the Kronecker product of symmetric functions. Moreover, for any
$f\in\Lambda$, there exists
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}\in\Lambda$
with the same degree and leading term as $f$ such that
$\mathscr{F}\\{f\\}=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}\cdot(1+h_{1}+h_{2}+\cdots)$.
We refer to the map
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\colon\Lambda\to\Lambda$
as the _surjective Frobenius transform_. Because the surjective Frobenius
transform preserves degree and leading term, it has an inverse, which we
denote by
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}$.

The _induced trivial character basis_ $\\{\tilde{h}_{\lambda}\\}_{\lambda}$
and the _irreducible character basis_ $\\{\tilde{s}_{\lambda}\\}_{\lambda}$,
which were introduced by Orellana and Zabrocki in 2021 [19], can be defined in
terms of the inverse surjective Frobenius transform. Namely,
$\tilde{h}_{\lambda}=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{h_{\lambda}\right\\}$
and
$\tilde{s}_{\lambda}=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{(H^{\perp})^{-1}s_{\lambda}\right\\}$,
where $H^{\perp}\colon\Lambda\to\Lambda$ is the operator adjoint (under the
Hall inner product) to multiplication by $H=1+h_{1}+h_{2}+\cdots$.

In Section 4, we will use the Frobenius transform to study _stable restriction
coefficients_ ; that is, the limits

$a_{\lambda}^{\mu}=\lim_{n\to\infty}r_{\lambda}^{(n-|\mu|,\mu_{1},\ldots,\mu_{\ell(\mu)})},$

which exist for all $\lambda,\mu$ by a classical result of Littlewood [9]. In
2019, Assaf and Speyer found a formula for $a_{\lambda}^{\mu}$ and for the
entries $b_{\lambda}^{\mu}$ of the inverse matrix
$[b_{\lambda}^{\mu}]=[a_{\lambda}^{\mu}]^{-1}$ [2]. We will provide an
alternative proof of these formulas using the Frobenius transform.

In Section 5, we will compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}$
for various symmetric functions $f$. We will show that
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{e_{n}\right\\}=e_{n}$
and
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{p_{n}\right\\}=\sum_{d\mid
n}p_{d}$ for all $n$. More generally, with $\mathbb{N}=\\{0,1,2,\ldots\\}$:

theoremcomputation Let $\lambda$ be a partition and let $\ell=\ell(\lambda)$
be its length.

  1. (a)

Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{h_{\lambda}\right\\}=\sum_{M}\prod_{j\in\mathbb{N}^{\ell}}h_{M(j)}$

where the sum is over all functions $M\colon\mathbb{N}^{\ell}\to\mathbb{N}$
such that $M(0,\ldots,0)=0$ and
$\sum_{j\in\mathbb{N}^{\ell}}j_{i}M(j)=\lambda_{i}$ for $i=1,\ldots,\ell$.

  2. (b)

Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{e_{\lambda}\right\\}=\sum_{M}\prod_{j\in\\{0,1\\}^{\ell}}\begin{cases}h_{M(j)}&\mbox{if
$j_{1}+\cdots+j_{\ell}$ is even}\\\ e_{M(j)}&\mbox{if $j_{1}+\cdots+j_{\ell}$
is odd}\end{cases}$

where the sum is over all functions $M\colon\\{0,1\\}^{\ell}\to\mathbb{N}$
such that $M(0,\ldots,0)=0$ and
$\sum_{j\in\\{0,1\\}^{\ell}}j_{i}M(j)=\lambda_{i}$ for $i=1,\ldots,\ell$.

  3. (c)

Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{p_{\lambda}\right\\}=\sum_{\pi}\prod_{U\in\pi}\left(\sum_{d\mid\gcd\\{\lambda_{i}\colon
i\in U\\}}d^{|U|-1}p_{d}\right)$

where the outer sum is over all partitions $\pi$ of $\\{1,\ldots,\ell\\}$ into
nonempty sets.

A statement equivalent to parts (a) and (b) of this theorem has appeared
previously in the work of Orellana and Zabrocki [17, Equation (6)].

Section 1 has the following interesting consequence. For any partition $\mu$,
denote by $D(\mu)$ the size of the Durfee square of $\mu$; that is, $D(\mu)$
is the largest integer $d$ such that $\mu_{d}\geq d$ [1, Chapter 8].

theoremdurfeebound Let $\mu$ be a partition and let $k\geq 1$ be an integer.
The following are equivalent:

  1. (A)

There exists a partition $\lambda$ such that $\lambda_{1}\leq k$ and
$r_{\lambda}^{\mu}>0$.

  2. (B)

$D(\mu)\leq 2^{k-1}$.

In particular, $r_{\lambda}^{\mu}=0$ if $D(\mu)>2^{\lambda_{1}-1}$.

Finally, in Section 6, we will compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}$
and
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{h_{\lambda}\right\\}$
(Theorem 6.7). In particular, we will prove that

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}=\sum_{\mu}(-1)^{|\lambda|-|\mu|}L_{\lambda}^{\mu}e_{\mu},$

where $L_{\lambda}^{\mu}$ is a nonnegative integer with an explicit
combinatorial interpretation involving Lyndon words.

## Acknowledgements

The author thanks Rosa Orellana and Mike Zabrocki for helpful correspondence.

##  2\. Preliminaries

See [5], [13, Chapter I], or [22, Chapter 7] for a complete introduction to
the theory of symmetric functions.

Let $\Lambda=\Lambda_{\mathbb{Z}}$ denote the ring of symmetric functions over
$\mathbb{Z}$ in the variables $x_{1},x_{2},x_{3},\ldots$. For $n\geq 0$, let
$\Lambda_{n}$ denote the subgroup of $\Lambda$ consisting of all symmetric
functions that are homogeneous of degree $n$. Let $\overline{\Lambda}$ denote
the ring of symmetric formal power series (that is, formal sums
$\sum_{n}f_{n}$, where each $f_{n}\in\Lambda_{n}$). Let
$\langle\cdot,\cdot\rangle\colon\Lambda\times\overline{\Lambda}\to\mathbb{Z}$
denote the Hall inner product.

For any partition $\lambda=(\lambda_{1},\ldots,\lambda_{\ell})$, define the
_length_ $\ell(\lambda)=\ell$ and the _size_
$|\lambda|=\lambda_{1}+\cdots+\lambda_{\ell}$. Let $m_{i}(\lambda)$ be the
number of times the part $i$ appears in $\lambda$, and let
$z_{\lambda}=\prod_{i}i^{m_{i}(\lambda)}(m_{i}(\lambda))!$. Let $\lambda^{T}$
denote the dual of $\lambda$. Let
$m_{\lambda},e_{\lambda},h_{\lambda},p_{\lambda},s_{\lambda}\in\Lambda$ denote
the monomial, elementary, homogeneous, power sum, and Schur symmetric
functions respectively. Let $V_{\lambda}$ denote the corresponding _Specht
module_ , which is an irreducible $\mathfrak{S}_{|\lambda|}$-module. Let
$\chi_{\lambda}$ denote the character of $V_{\lambda}$. Let
$\mathbb{S}^{\lambda}$ denote the corresponding _Schur functor_ , which is an
endofunctor of the category of vector spaces over $\mathbb{C}$.

Let $\omega\colon\Lambda\to\Lambda$ be the ring homomorphism given by
$\omega(p_{k})=(-1)^{k-1}p_{k}$. Recall that $\omega$ is an involution and
that for all partitions $\lambda$, we have $\omega(h_{\lambda})=e_{\lambda}$
and $\omega(s_{\lambda})=s_{\lambda^{T}}$.

Define the Lyndon symmetric function $L_{n}$ for $n\geq 1$ by

$L_{n}=\frac{1}{n}\sum_{d\mid n}\mu(d)p_{d}^{n/d},$

where $\mu\colon\\{1,2,3,\ldots\\}\to\\{-1,0,1\\}$ is the Möbius function.

Let $C_{n}$ denote the space of all $\mathbb{C}$-valued class functions on
$\mathfrak{S}_{n}$. Let $R_{n}$ denote the additive group of all virtual
characters on $\mathfrak{S}_{n}$. In other words, $R_{n}$ is the subgroup of
$C_{n}$ generated by the irreducible characters $\chi_{\lambda}$.

######  Definition 2.1.

The $n$th _Frobenius character map_ is the map $\operatorname{ch}_{n}\colon
R_{n}\to\Lambda_{n}$ defined by

$\operatorname{ch}_{n}(\chi)=\frac{1}{n!}\sum_{w\in\mathfrak{S}_{n}}\chi(w)p_{c(w)},$

where $c(w)$ denotes the cycle type of $w$.

It is well-known that $\operatorname{ch}_{n}$ is an isomorphism and that
$\operatorname{ch}_{n}(\chi_{\lambda})=s_{\lambda}$ for all $\lambda$ with
$|\lambda|=n$.

######  Definition 2.2.

The _Kronecker product_ is the unique bilinear operator
$\ast\colon\Lambda\times\Lambda\to\Lambda$ satisfying $p_{\lambda}\ast
p_{\mu}=\delta_{\lambda\mu}z_{\lambda}p_{\lambda}$ for all $\lambda,\mu$. It
extends to a bilinear operator
$\ast\colon\overline{\Lambda}\times\overline{\Lambda}\to\overline{\Lambda}$ by
continuity.

The Frobenius character map and the Kronecker product are related in the
following way: for any $\chi_{1},\chi_{2}\in R_{n}$, we have
$\operatorname{ch}(\chi_{1}\chi_{2})=\operatorname{ch}(\chi_{1})\ast\operatorname{ch}(\chi_{2})$.

For any $f,g\in\overline{\Lambda}$, let $f[g]$ denote the plethysm of $f$ by
$g$. This is well-defined as long as $f\in\Lambda$ or $g$ has no constant
term.

######  Definition 2.3.

Let $t$ be a variable. Let

$\displaystyle H(t)$ $\displaystyle=\sum_{n\geq 0}h_{n}t^{n}$
$\displaystyle=\prod_{n\geq 1}\frac{1}{1-x_{i}t}$
$\displaystyle\in\Lambda\llbracket t\rrbracket$ $\displaystyle E(t)$
$\displaystyle=\sum_{n\geq 0}e_{n}t^{n}$ $\displaystyle=\prod_{n\geq
1}(1+x_{i}t)$ $\displaystyle\in\Lambda\llbracket t\rrbracket.$

Let $H=H(1)=1+h_{1}+h_{2}+\cdots$ and let
$H_{+}=h_{1}+h_{2}+h_{3}+\cdots=H-1$.

It is clear that $E(t)=\frac{1}{H(-t)}$.

######  Definition 2.4.

Let $\lambda,\mu,\nu$ be partitions. The _Littlewood-Richardson coefficient_
$c^{\nu}_{\lambda\mu}$ is the Hall inner product $\langle
s_{\nu},s_{\lambda}s_{\mu}\rangle$.

######  Definition 2.5.

Let $\lambda,\mu,\nu$ be partitions. Define $g_{\lambda\mu\nu}=\langle
s_{\nu},s_{\lambda}\ast s_{\mu}\rangle$.

######  Definition 2.6.

Let $\lambda,\mu$ be partitions, and let $n=|\mu|$. Then, the Schur module
$\mathbb{S}^{\lambda}\mathbb{C}^{n}$ can be considered as an
$\mathfrak{S}_{n}$-module, with $\mathfrak{S}_{n}$ acting on $\mathbb{C}^{n}$
by permutation matrices. The _restriction coefficient_

$r_{\lambda}^{\mu}=\dim\operatorname{Hom}_{\mathfrak{S}_{n}}(V_{\mu},\mathbb{S}^{\lambda}\mathbb{C}^{n})$

is the multiplicity of the Specht module $V_{\mu}$ in
$\mathbb{S}^{\lambda}\mathbb{C}^{n}$.

##  3\. The Frobenius transform: definition and basic properties

Recall from the introduction that it is a long-standing open problem to find a
combinatorial interpretation of $r_{\lambda}^{\mu}$. As a potential way to
approach this problem, we now define the Frobenius transform, which is the
primary object of study in this paper.

######  Definition 3.1.

The _Frobenius transform_ is the abelian group homomorphism
$\mathscr{F}\colon\Lambda\to\overline{\Lambda}$ defined on the basis
$\\{s_{\lambda}\\}$ by

$\mathscr{F}\\{s_{\lambda}\\}=\sum_{\mu}r_{\lambda}^{\mu}s_{\mu},$

where the sum is over all partitions $\mu$.

######  Remark 3.2.

By a classical result of Littlewood [9], we have

$r_{\lambda}^{\mu}=\langle s_{\lambda},s_{\mu}[H]\rangle.$

Hence, $\mathscr{F}$ is adjoint to plethysm by $H$ under the Hall inner
product.

Here is the reason for calling $\mathscr{F}$ the Frobenius transform. Let
$n\geq 0$ and let $\lambda$ be any partition. Then
$\mathbb{S}^{\lambda}\mathbb{C}^{n}$, considered as an
$\mathfrak{S}_{n}$-module, can be expressed as a direct sum of Specht modules:

$\bigoplus_{|\mu|=n}r_{\lambda}^{\mu}V_{\mu}=\mathbb{S}^{\lambda}\mathbb{C}^{n}.$

Taking the character of both sides and applying the Frobenius character map,
we obtain

(1)
$\sum_{|\mu|=n}r_{\lambda}^{\mu}s_{\mu}=\operatorname{ch}_{n}(\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}).$

In other words, the degree $n$ part of $\mathscr{F}\\{s_{\lambda}\\}$ is equal
to the Frobenius character of $\mathbb{S}^{\lambda}\mathbb{C}^{n}$.

######  Example 3.3.

Let $r\geq 0$. We will compute $\mathscr{F}\\{e_{r}\\}$. First,
$e_{r}=s_{\lambda}$, where $\lambda=(1^{r})$. Hence, for any $n$, the degree
$n$ part of $\mathscr{F}\\{e_{r}\\}$ is the Frobenius character of

$\mathbb{S}^{\lambda}\mathbb{C}^{n}=\wedge^{r}\mathbb{C}^{n}=\operatorname{Ind}_{\mathfrak{S}_{r}\times\mathfrak{S}_{n-r}}^{\mathfrak{S}_{n}}V_{(1^{r})}\otimes
V_{(n-r)},$

considered as an $\mathfrak{S}_{n}$-module. Thus, it is equal to
$e_{r}h_{n-r}$ [22, Proposition 7.18.2]. Taking the sum over all $n$ yields
$\mathscr{F}\\{e_{r}\\}=e_{r}\cdot H$.

###  3.1. The Frobenius transform and representations of combinatorial
categories

The purpose of this subsection is to provide an alternate perspective on the
Frobenius transform. This subsection is not essential to the proofs of our
main results, so the reader may skip it. For a more complete introduction to
the representation theory of categories, see Wiltshire-Gordon’s 2016 PhD
thesis [23].

In what follows, let $\text{Vect}_{\mathbb{C}}$ be the category whose objects
are vector spaces over $\mathbb{C}$ and whose morphisms are linear
transformations. The ground field $\mathbb{C}$ can be replaced by any
algebraically closed field of characteristic $0$.

######  Definition 3.4.

Let $\mathcal{C}$ be a category. A _$\mathcal{C}$ -module_ (over the ground
field $\mathbb{C}$) is a functor
$M[\bullet]\colon\mathcal{C}\to\text{Vect}_{\mathbb{C}}$. The _category of
$\mathcal{C}$-modules_ is the functor category
$\text{Mod}^{\mathcal{C}}=(\text{Vect}_{\mathbb{C}})^{\mathcal{C}}$.

We say that a $\mathcal{C}$-module $M$ is _finite-dimensional_ if $M[U]$ is
finite-dimensional for all $U\in\operatorname{Ob}(\mathcal{C})$.

If $M$ and $N$ are $\mathcal{C}$-modules, we may form the _direct sum_
$M\oplus N$ by $(M\oplus N)[U]=M[U]\oplus N[U]$.

Let $\mathrm{Bij}$ be the category whose objects are finite sets and whose
morphisms are bijections and let $M$ be a $\mathrm{Bij}$-module. (Previous
authors have referred to $\mathrm{Bij}$-modules as _linear species_ or _tensor
species_ [8, 14].) Denote by $M[n]$ the vector space
$M[[n]]=M[\\{1,\ldots,n\\}]$, which is an $\mathfrak{S}_{n}$-module. Then $M$
is uniquely determined, up to natural isomorphism, by the sequence
$\\{M[n]\\}_{n}$; namely,

(2)
$M[U]=\mathbb{C}\mathchoice{\operatorname{Bij}}{\operatorname{Bij}}{\mathrm{Bij}}{\mathrm{Bij}}([n],U)\otimes_{\mathfrak{S}_{n}}M[n]$

for all finite sets $U$ with $|U|=n$. If $M$ is finite-dimensional, then we
may form the series

$\operatorname{ch}(M)=\sum_{n}\operatorname{ch}_{n}(\chi_{M[n]})\in\overline{\Lambda}$

which we call the _Frobenius character of $M$_. It also uniquely determines
$M$ up to natural isomorphism.

Not every symmetric power series can be written in the form
$\operatorname{ch}(M)$ where $M$ is a finite-dimensional
$\mathrm{Bij}$-module. However, it is still helpful to think of
$\operatorname{ch}$ as a sort of correspondence between (isomorphism classes
of) finite-dimensional $\mathrm{Bij}$-modules and symmetric power series. Many
concepts from the theory of symmetric power series have analogous concepts in
the theory of $\mathrm{Bij}$-modules. For example, in Definition 3.10 below,
we will define the _product_ $M\cdot N$ of two $\mathrm{Bij}$-modules $M,N$.
Via the Frobenius character, this is analogous to the product of symmetric
power series, in the sense that $\operatorname{ch}(M\cdot
N)=\operatorname{ch}(M)\cdot\operatorname{ch}(N)$ (Proposition 3.11).

In what follows, we will find the construction in the theory of
$\mathrm{Bij}$-modules which is analogous to the Frobenius transform. More
precisely, suppose that $M$ is a $\mathrm{Bij}$-module such that $M_{n}=0$ for
all but finitely many $n$. Then $\operatorname{ch}(M)$ is in fact a symmetric
function, so its Frobenius transform $\mathscr{F}\\{\operatorname{ch}(M)\\}$
is well-defined. We will define a $\mathrm{Bij}$-module whose Frobenius
character is $\mathscr{F}\\{\operatorname{ch}(M)\\}$. Before we do, we need
the following definitions.

######  Definition 3.5 ([23, Definition 2.6.1]).

Let $F\colon\mathcal{C}\to\mathcal{D}$ be a functor. The _pullback functor_
$F^{*}\colon\text{Mod}^{\mathcal{D}}\to\text{Mod}^{\mathcal{C}}$ is given by
$F^{*}M=M\circ F$.

######  Definition 3.6 ([23, Proposition 2.6.2]).

Let $F\colon\mathcal{C}\to\mathcal{D}$ be a functor. The _left Kan extension
functor_ $F_{!}\colon\text{Mod}^{\mathcal{C}}\to\text{Mod}^{\mathcal{D}}$ is
the left adjoint to the pullback $F^{*}$:

$\operatorname{Hom}_{\text{Mod}^{\mathcal{D}}}(F_{!}M,N)=\operatorname{Hom}_{\text{Mod}^{\mathcal{C}}}(M,F^{*}N).$

Let $\mathrm{Fun}$ be the category whose objects are finite sets and whose
morphisms are functions. Clearly, $\mathrm{Bij}$ is a subcategory of
$\mathrm{Fun}$; let $\iota\colon\mathrm{Bij}\to\mathrm{Fun}$ be the inclusion
functor.

######  Proposition 3.7.

Let $M$ be a $\mathrm{Bij}$-module. Then the left Kan extension $\iota_{!}M$
is given on finite sets $U$ by

$(\iota_{!}M)[U]=\bigoplus_{n}\mathbb{C}U^{n}\otimes_{\mathfrak{S}_{n}}M[n].$

Here $\mathbb{C}U^{n}\otimes_{\mathfrak{S}_{n}}M[n]$ denotes the quotient of
the tensor product $\mathbb{C}U^{n}\otimes M[n]$ by the relation that
$a\otimes wb\sim aw\otimes b$ for all $a\in\mathbb{C}U^{n}$, $b\in M_{n}$, and
$w\in\mathfrak{S}_{n}$.

###### Proof.

This follows from [23, Proposition 2.6.7]. ∎

######  Remark 3.8.

Joyal refers to $\iota_{!}M\colon\mathrm{Fun}\to\text{Vect}_{\mathbb{C}}$ as
the _analytic functor_ corresponding to the tensor species $M$ [8, Definition
4.2].

We are now ready to show that $\iota^{*}\iota_{!}$ is the construction
analogous to the Frobenius transform.

######  Proposition 3.9.

Let $M$ be a finite-dimensional $\mathrm{Bij}$-module such that $M[n]=0$ for
all but finitely many $n$. Then $\iota^{*}\iota_{!}M$ is finite-dimensional
and
$\mathscr{F}\\{\operatorname{ch}(M)\\}=\operatorname{ch}(\iota^{*}\iota_{!}M)$.

###### Proof.

For any partition $\lambda$, define the $\mathrm{Bij}$-module $M_{\lambda}$ by

$(M_{\lambda})[n]=\begin{cases}V_{\lambda}&\mbox{if $n=|\lambda|$}\\\
0&\mbox{otherwise}\end{cases},$

extending to all of $\mathrm{Bij}$ using (2).

We have that $M$ can be written as a direct sum of the $M_{\lambda}$. Since
$\iota^{*}$ and $\iota_{!}$ preserve direct sums, it is enough to prove the
proposition for $M=M_{\lambda}$. Then
$\operatorname{ch}(M)=\operatorname{ch}_{n}(\chi_{\lambda})=s_{\lambda}$.

On the other hand, by Proposition 3.7, we have for $m\geq 0$ that

$\displaystyle(\iota^{*}\iota_{!}M)[m]$
$\displaystyle=\bigoplus_{n}\mathbb{C}[m]^{n}\otimes_{\mathfrak{S}_{n}}M[n]$
$\displaystyle=\bigoplus_{n}(\mathbb{C}^{m})^{\otimes
n}\otimes_{\mathfrak{S}_{n}}M[n]$
$\displaystyle=(\mathbb{C}^{m})^{\otimes|\lambda|}\otimes_{\mathfrak{S}_{|\lambda|}}V_{\lambda}.$

By Schur-Weyl duality, this is isomorphic as a $GL_{m}$-module to
$\mathbb{S}^{\lambda}\mathbb{C}^{m}$. Hence, as a $\mathfrak{S}_{m}$-module,
it decomposes into irreducibles as

$(\iota^{*}\iota_{!}M)[m]=\bigoplus_{|\mu|=m}r^{\mu}_{\lambda}V_{\mu}.$

It follows that

$\operatorname{ch}(\iota^{*}\iota_{!}M)=\sum_{\mu}r^{\mu}_{\lambda}s_{\mu}=\mathscr{F}\\{s_{\lambda}\\}=\mathscr{F}\\{\operatorname{ch}(M)\\}$

as desired. ∎

We will occasionally make use of the following definition in later remarks.

######  Definition 3.10 ([8, Section 4.1]).

Let $M,N$ be $\mathrm{Bij}$-modules. Define the _product_ $M\cdot N$ to be the
$\mathrm{Bij}$-module given by

$(M\cdot N)[U]=\bigoplus_{\begin{subarray}{c}U_{1}\cup U_{2}=U\\\ U_{1}\cap
U_{2}=\emptyset\end{subarray}}M[U_{1}]\otimes N[U_{2}].$

######  Proposition 3.11 ([14, Proposition 2.1]).

Let $M$ and $N$ be finite-dimensional $\mathrm{Bij}$-modules. Then

$\operatorname{ch}(M\cdot N)=\operatorname{ch}(M)\operatorname{ch}(N).$

###  3.2. The Frobenius transform and evaluation at roots of unity

Let us now describe the expansion of $\mathscr{F}\\{s_{\lambda}\\}$ in the
power sum basis. In order to simplify the description, we will do it one
degree at a time. In this subsection, we will write $f_{n}$ to mean the degree
$n$ part of $f$.

By (1) and Definition 2.2, we have

(3)
$(\mathscr{F}\\{s_{\lambda}\\})_{n}=\operatorname{ch}_{n}(\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}})=\frac{1}{n!}\sum_{w\in\mathfrak{S}_{n}}\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(w)p_{c(w)}.$

It is well-known [5, Section 8.3] that the character of
$\mathbb{S}^{\lambda}\mathbb{C}^{n}$ as a $GL_{n}$-module is the Schur
function $s_{\lambda}$ itself. In other words, if $g\in GL_{n}(\mathbb{C})$
has eigenvalues $x_{1},\ldots,x_{n}$, then
$\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(g)$ is equal to the evaluation
$s_{\lambda}(x_{1},\ldots,x_{n})$.

For $w\in\mathfrak{S}_{n}$, we have
$\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(w)=s_{\lambda}(x_{1},\ldots,x_{n})$,
where $x_{1},\ldots,x_{n}$ are the eigenvalues of the permutation matrix
$P_{w}$. Let $\mu=(\mu_{1},\ldots,\mu_{\ell})$ be the cycle type of $w$. Then
the eigenvalues of $P_{w}$ are the roots of unity

$\displaystyle 1,\exp\left(\frac{2\pi i}{\mu_{1}}\right),\exp\left(\frac{4\pi
i}{\mu_{1}}\right),\ldots,\exp\left(\frac{2(\mu_{1}-1)\pi i}{\mu_{1}}\right),$
$\displaystyle\vdots$ $\displaystyle 1,\exp\left(\frac{2\pi
i}{\mu_{\ell}}\right),\exp\left(\frac{4\pi
i}{\mu_{\ell}}\right),\ldots,\exp\left(\frac{2(\mu_{\ell}-1)\pi
i}{\mu_{\ell}}\right).$

Following Orellana and Zabrocki [18, 19], let $\Xi_{\mu}\in\mathbb{C}^{n}$
denote this sequence. Then $\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(w)$ is
the result of evaluating $s_{\lambda}$ at these roots of unity:

$\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(w)=s_{\lambda}\left(\Xi_{\mu}\right).$

Now, let us group the terms on the right-hand side of (3) according to the
cycle type $\mu=c(w)$. The number of permutations $w\in\mathfrak{S}_{n}$ with
cycle type $\mu$ is $\frac{n!}{z_{\mu}}$, so

$\displaystyle(\mathscr{F}\\{s_{\lambda}\\})_{n}$
$\displaystyle=\frac{1}{n!}\sum_{w\in\mathfrak{S}_{n}}\chi_{\mathbb{S}^{\lambda}\mathbb{C}^{n}}(w)p_{c(w)}$
$\displaystyle=\frac{1}{n!}\sum_{|\mu|=n}\frac{n!}{z_{\mu}}s_{\lambda}(\Xi_{\mu})p_{\mu}$
$\displaystyle=\sum_{|\mu|=n}s_{\lambda}(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}.$

Summing over all $n$, we obtain

$\mathscr{F}\\{s_{\lambda}\\}=\sum_{\mu}s_{\lambda}(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}.$

Finally, by linearity, we may extend this result to any $f\in\Lambda$. We have
proved the following.

######  Proposition 3.12.

Let $f\in\Lambda$. Then

$\mathscr{F}\\{f\\}=\sum_{\mu}f(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}.$

In the notation of Orellana and Zabrocki [18], we may write

$\mathscr{F}\\{f\\}=\phi_{0}(f)+\phi_{1}(f)+\phi_{2}(f)+\cdots.$

###  3.3. The Frobenius transform and the Kronecker product

The Frobenius transform relates the ordinary product of symmetric functions to
the Kronecker product in the following way.

######  Proposition 3.13 (cf. [18, Section 2.3]).

Let $f,g\in\Lambda$. Then
$\mathscr{F}\\{fg\\}=\mathscr{F}\\{f\\}\ast\mathscr{F}\\{g\\}$.

We provide two different proofs of Proposition 3.13: a category-theoretic
proof using Proposition 3.9 and a direct computational proof using Proposition
3.12.

###### First proof.

Because both sides of the desired equation are bilinear, we may assume that
there exist $\mathrm{Bij}$-modules $M,N$ such that $\operatorname{ch}(M)=f$
and $\operatorname{ch}(N)=g$. By [14, Proposition 2.1], we have
$\operatorname{ch}(M\cdot N)=\operatorname{ch}(M)\operatorname{ch}(N)=fg$,
where $M\cdot N$ is the product as defined in Definition 3.10.

By [8, Equation 2.1(ii)], we have $\iota^{*}\iota_{!}(M\cdot
N)=(\iota^{*}\iota_{!}M)\otimes_{\mathbb{C}}(\iota^{*}\iota_{!}N)$, where
$\otimes_{\mathbb{C}}$ denotes the object-wise tensor product of
$\mathrm{Bij}$-modules. Applying $\operatorname{ch}$ to both sides, we obtain
$\mathscr{F}\\{fg\\}=\mathscr{F}\\{f\\}\ast\mathscr{F}\\{g\\}$, as desired. ∎

###### Second proof.

By Proposition 3.12, we have

$\displaystyle\mathscr{F}\\{f\\}\ast\mathscr{F}\\{g\\}$
$\displaystyle=\left(\sum_{\mu}f(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}\right)\ast\left(\sum_{\mu}g(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}\right)$
$\displaystyle=\sum_{\mu}f(\Xi_{\mu})g(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}$
$\displaystyle=\mathscr{F}\\{fg\\}$

as desired. ∎

######  Corollary 3.14.

Let $\lambda,\mu,\nu$ be partitions. Then

$\sum_{\nu^{\prime}}r^{\nu}_{\nu^{\prime}}c^{\nu^{\prime}}_{\lambda\mu}=\sum_{\lambda^{\prime},\mu^{\prime}}r^{\lambda^{\prime}}_{\lambda}r^{\mu^{\prime}}_{\mu}g_{\lambda^{\prime}\mu^{\prime}\nu}.$

###### Proof.

In Proposition 3.13, take $f=s_{\lambda}$ and $g=s_{\mu}$. Then take the Hall
inner product of both sides with $s_{\nu}$. ∎

###  3.4. The surjective Frobenius transform

######  Proposition 3.15.

Let $f\in\Lambda$. Then there exists a symmetric function
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}\in\Lambda$
such that
$\mathscr{F}\\{f\\}=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}\cdot
H$. Moreover,
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}$
has the same degree and leading term as $f$.

For example, in Example 3.3 we showed that $\mathscr{F}\\{e_{r}\\}=e_{r}\cdot
H$. Thus,
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{e_{r}\right\\}=e_{r}$.
Note, however, that in general,
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
does not preserve the property of being homogeneous. Again, we provide two
separate proofs of this proposition: a category-theoretic proof and a direct
computational proof.

###### First proof.

Because both sides of the desired equation are linear in $f$, we may assume
that there exists a $\mathrm{Bij}$-module $M$ such that
$\operatorname{ch}(M)=f$.

Let $\mathrm{Sur}$ be the category whose objects are finite sets and whose
morphisms are surjections, and let $\kappa\colon\mathrm{Bij}\to\mathrm{Sur}$
be the inclusion functor. We claim that the choice
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}=\operatorname{ch}(\kappa^{*}\kappa_{!}M)$
satisfies the desired properties. For this, we will show that there is a
natural isomorphism

(4) $\iota^{*}\iota_{!}M=\kappa^{*}\kappa_{!}M\cdot\mathbb{C}E,$

where $\iota$ and $\cdot$ are defined as in Section 3.1 and $\mathbb{C}E$ is
the $\mathrm{Bij}$-module given by $(\mathbb{C}E)[U]=\mathbb{C}$ for all
finite sets $U$.

By Proposition 3.7, we have

(5)
$(\iota^{*}\iota_{!}M)[U]=\bigoplus_{n}\mathbb{C}U^{n}\otimes_{\mathfrak{S}_{n}}M[n].$

Think of $U^{n}$ as the set of all functions $[n]\to U$. By grouping those
functions by their image $V$, we may write $U^{n}$ as a disjoint union:

$U^{n}=\sum_{V\subset
U}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],V).$

Substituting into (5), we obtain

$\displaystyle(\iota^{*}\iota_{!}M)[U]$
$\displaystyle=\bigoplus_{n}\mathbb{C}\left(\sum_{V\subset
U}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],V)\right)\otimes_{\mathfrak{S}_{n}}M[n]$
$\displaystyle=\bigoplus_{n}\left(\bigoplus_{V\subset
U}\mathbb{C}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],V)\right)\otimes_{\mathfrak{S}_{n}}M[n]$
$\displaystyle=\bigoplus_{V\subset
U}\bigoplus_{n}\mathbb{C}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],V)\otimes_{\mathfrak{S}_{n}}M[n]$
$\displaystyle=\bigoplus_{V\subset U}(\kappa^{*}\kappa_{!}M)[V]$

where the last step follows from [23, Proposition 2.6.7]. Recognizing the
latter as $(\kappa^{*}\kappa_{!}M\cdot\mathbb{C}E)[U]$, we have shown (4).
Applying $\operatorname{ch}$ to both sides and using Propositions 3.9 and 3.11
yields $\mathscr{F}\\{f\\}=\operatorname{ch}(\kappa^{*}\kappa_{!}M)\cdot H$.

It remains to show that $\operatorname{ch}(\kappa^{*}\kappa_{!}M)$ has the
same degree and leading term as $f$. For this, we will prove that
$(\kappa^{*}\kappa_{!}M)[m]$ and $M[m]$ are isomorphic as
$\mathfrak{S}_{m}$-modules for all $m\geq\deg f$. Consider the natural
isomorphism

$(\kappa^{*}\kappa_{!}M)[U]=\bigoplus_{n}\mathbb{C}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],U)\otimes_{\mathfrak{S}_{n}}M[n]$

with $U=[m]$. In each summand, the factor
$\mathbb{C}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([n],[m])$
vanishes when $n<m$ and the factor $M[n]$ vanishes when $n>\deg f$. If
$m\geq\deg f$, this means that every term vanishes except possibly the term
$n=m$. Hence,

$\displaystyle(\kappa^{*}\kappa_{!}M)[m]$
$\displaystyle=\mathbb{C}\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}([m],[m])\otimes_{\mathfrak{S}_{m}}M[m]$
$\displaystyle=\mathbb{C}\mathfrak{S}_{m}\otimes_{\mathfrak{S}_{m}}M[m]$
$\displaystyle=M[m]$

as desired. ∎

###### Second proof.

We claim that the choice

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}=\sum_{\mu}\langle
f,s_{\mu}[H_{+}]\rangle s_{\mu}$

satisfies the desired properties. In other words, we may take
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
to be the adjoint to plethysm by $H_{+}$ under the Hall inner product.

By Remark 3.2, we have

$\displaystyle\mathscr{F}\\{f\\}$ $\displaystyle=\sum_{\lambda}\langle
f,s_{\lambda}[H]\rangle s_{\lambda}$ $\displaystyle=\sum_{\lambda}\langle
f,s_{\lambda}[H_{+}+1]\rangle s_{\lambda}.$

By the plethystic addition formula [10, Section 3.2], we have

$\displaystyle\mathscr{F}\\{f\\}$ $\displaystyle=\sum_{\lambda,\mu}\langle
f,s_{\mu}[H_{+}]s_{\lambda/\mu}[1]\rangle s_{\lambda}.$

We have

$s_{\lambda/\mu}[1]=s_{\lambda/\mu}(1,0,0,\ldots)=\begin{cases}1&\mbox{if
$\lambda/\mu$ is a horizontal strip}\\\ 0&\mbox{otherwise}\end{cases},$

so

$\displaystyle\mathscr{F}\\{f\\}$
$\displaystyle=\sum_{\begin{subarray}{c}\lambda,\mu\\\ \text{$\lambda/\mu$ h.
strip}\end{subarray}}\langle f,s_{\mu}[H_{+}]\rangle s_{\lambda}$
$\displaystyle=\sum_{\mu}\langle
f,s_{\mu}[H_{+}]\rangle\left(\sum_{\begin{subarray}{c}\lambda\\\
\text{$\lambda/\mu$ h. strip}\end{subarray}}s_{\lambda}\right)$
$\displaystyle=\sum_{\mu}\langle f,s_{\mu}[H_{+}]\rangle s_{\mu}\cdot H,$

where the last equality follows from the Pieri rule.

Now, the only thing left to show is that $f$ has the same degree and leading
term as

$\sum_{\mu}\langle f,s_{\mu}[H_{+}]\rangle s_{\mu}.$

This follows directly from the observation that if $|\mu|\geq\deg f$, then
$\langle f,s_{\mu}[H_{+}]\rangle=\langle f,s_{\mu}\rangle$. ∎

We refer to
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\colon\Lambda\to\Lambda$
as the _surjective Frobenius transform_. Clearly, it is invertible:

######  Corollary 3.16.

There exists a two-sided inverse
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\colon\Lambda\to\Lambda$
of
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$.

###### Proof.

Define $\mathcal{M}\colon\Lambda\to\Lambda$ by
$\mathcal{M}\\{f\\}=f-\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}$.
By Proposition 3.15, we have

$\deg(\mathcal{M}\\{f\\})<\deg f$

for any $f\in\Lambda\setminus\\{0\\}$. Hence, $\mathcal{M}^{k}\\{f\\}=0$ for
any $k>\deg f$.

Define
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\colon\Lambda\to\Lambda$
by

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{f\right\\}=f+\mathcal{M}\\{f\\}+\mathcal{M}^{2}\\{f\\}+\cdots.$

This is well-defined by the above, and it is easy to check that it is the two-
sided inverse of
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$.
∎

Like the ordinary Frobenius transform, the surjective Frobenius transform can
be described in terms of its matrix entries in the Schur basis.

######  Definition 3.17.

Let $\lambda,\mu$ be partitions. Define the _surjective restriction
coefficient_

$t_{\lambda}^{\mu}=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\},s_{\mu}\rangle$

and define the _inverse surjective restriction coefficient_

$u_{\lambda}^{\mu}=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{s_{\lambda}\right\\},s_{\mu}\rangle.$

By the above, we have
$t_{\lambda}^{\mu}=u_{\lambda}^{\mu}=\delta_{\lambda\mu}$ for
$|\mu|\geq|\lambda|$.

##  4\. Stable restriction

Define the _stable restriction coefficients_ $a_{\lambda}^{\mu}$ as follows.
For any partition $\mu=(\mu_{1},\ldots,\mu_{\ell})$ and any
$n\geq\mu_{1}+|\mu|$, define $\mu^{(n)}=(n-|\mu|,\mu_{1},\ldots,\mu_{\ell})$.
Then, for any partitions $\lambda,\mu$, the stable restriction coefficient
$a_{\lambda}^{\mu}$ is defined by the following limit, which exists by a
classical result of Littlewood [9]:

$a_{\lambda}^{\mu}=\lim_{n\to\infty}r_{\lambda}^{\mu(n)}.$

Littlewood also showed that $a_{\lambda}^{\mu}=\delta_{\lambda\mu}$ if
$|\lambda|\leq|\mu|$, so the infinite matrix $[a_{\lambda}^{\mu}]$, with rows
and columns indexed by partitions in increasing order of size, is upper
unitriangular. In 2019, Assaf and Speyer [2] found the following formula for
the entries $b_{\lambda}^{\mu}$ of the inverse matrix
$[b_{\lambda}^{\mu}]=[a_{\lambda}^{\mu}]^{-1}$: [[2, Theorem
2]]theoremassafspeyer Let $\lambda,\mu$ be partitions. Then

$b_{\lambda}^{\mu}=(-1)^{|\lambda|-|\mu|}\langle
s_{\lambda^{T}},s_{\mu^{T}}[L_{1}+L_{2}+L_{3}+\cdots]\cdot H\rangle.$

In particular, $(-1)^{|\lambda|-|\mu|}b_{\lambda}^{\mu}$ is a nonnegative
integer. In this section, we will provide an alternative proof of Section 4.
In fact, we will prove similar plethystic formulas for all five kinds of
restriction coefficients defined so far. These formulas have all been
collected into Theorem 4.1 below.

######  Theorem 4.1.

Let $\lambda,\mu$ be partitions with $|\lambda|=m$ and $|\mu|=n$.

  1. (a)

The restriction coefficient $r_{\lambda}^{\mu}$ is given by

$\displaystyle r_{\lambda}^{\mu}$
$\displaystyle=\langle\mathscr{F}\\{s_{\lambda}\\},s_{\mu}\rangle$
$\displaystyle=\langle s_{\lambda},s_{\mu}[H]\rangle$

  2. (b)

The surjective restriction coefficient $t_{\lambda}^{\mu}$ is given by

$\displaystyle t_{\lambda}^{\mu}$
$\displaystyle=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\},s_{\mu}\rangle$
$\displaystyle=\langle s_{\lambda},s_{\mu}[H_{+}]\rangle.$

  3. (c)

The inverse surjective restriction coefficient $u_{\lambda}^{\mu}$ is given by

$\displaystyle u_{\lambda}^{\mu}$
$\displaystyle=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{s_{\lambda}\right\\},s_{\mu}\rangle$
$\displaystyle=\langle
s_{\lambda},s_{\mu}[\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots]\rangle$
$\displaystyle=(-1)^{m-n}\langle
s_{\lambda^{T}},s_{\mu^{T}}[L_{1}+L_{2}+L_{3}+\cdots]\rangle.$

  4. (d)

The stable restriction coefficient $a_{\lambda}^{\mu}$ is given by

$\displaystyle a_{\lambda}^{\mu}$ $\displaystyle=\langle
H^{\perp}\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\},s_{\mu}\rangle$
$\displaystyle=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\},s_{\mu}\cdot
H\rangle$ $\displaystyle=\langle s_{\lambda},(s_{\mu}\cdot H)[H_{+}]\rangle,$

where $H^{\perp}\colon\Lambda\to\Lambda$ is the operator adjoint (under the
Hall inner product) to multiplication by $H$.

  5. (e)

The inverse stable restriction coefficient $b_{\lambda}^{\mu}$ is given by

$\displaystyle b_{\lambda}^{\mu}$
$\displaystyle=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{(H^{\perp})^{-1}s_{\lambda}\right\\},s_{\mu}\rangle$
$\displaystyle=\langle
s_{\lambda},s_{\mu}[\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots]\cdot(1-e_{1}+e_{2}-e_{3}+\cdots)\rangle$
$\displaystyle=(-1)^{m-n}\langle
s_{\lambda^{T}},s_{\mu^{T}}[L_{1}+L_{2}+L_{3}+\cdots]\cdot H\rangle.$

Before proving Theorem 4.1, we restate a basic result of plethystic calculus.

######  Lemma 4.2 (Negation Rule, [10, Theorem 6]).

Let $f\in\Lambda$ and $g\in\overline{\Lambda}$. If $f$ is homogeneous, then

$f[-g]=(-1)^{\deg f}(\omega(f))[g].$

###### Proof of Theorem 4.1.

  1. (a)

The first equality is true by definition. The second follows from Remark 3.2.

  2. (b)

The first equality is true by definition. The second is demonstrated in the
second proof of Proposition 3.15.

  3. (c)

The first equality is true by definition.

For the second, Cadogan showed in 1971 that
$\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots$ is the plethystic inverse
of $H_{+}$ [3]. In part (b), we showed that
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
is adjoint to plethysm by $H_{+}$. Hence,
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}$
is adjoint to plethysm by $\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots$,
as desired.

For the third, by Lemma 4.2 and the associativity of plethysm, we have

$\displaystyle\langle
s_{\lambda},s_{\mu}[\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots]\rangle$
$\displaystyle=$ $\displaystyle\langle
s_{\lambda},s_{\mu}[-(L_{1}+L_{2}+L_{3}+\cdots)[-p_{1}]]\rangle$
$\displaystyle=$ $\displaystyle\langle
s_{\lambda},(s_{\mu}[-(L_{1}+L_{2}+L_{3}+\cdots)])[-p_{1}]\rangle$
$\displaystyle=$ $\displaystyle\langle
s_{\lambda}[-p_{1}],s_{\mu}[-(L_{1}+L_{2}+L_{3}+\cdots)]\rangle$
$\displaystyle=$
$\displaystyle\langle(-1)^{m}\omega(s_{\lambda})[p_{1}],(-1)^{n}\omega(s_{\mu})[-(L_{1}+L_{2}+L_{3}+\cdots)]\rangle$
$\displaystyle=$ $\displaystyle(-1)^{m-n}\langle
s_{\lambda^{T}},s_{\mu^{T}}[L_{1}+L_{2}+L_{3}+\cdots]\rangle$

as desired.

  4. (d)

By part (a), we have

(6)
$a^{\mu}_{\lambda}=\lim_{n\to\infty}\langle\mathscr{F}\\{s_{\lambda}\\},s_{\mu(n)}\rangle=\lim_{n\to\infty}\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\}\cdot
H,s_{\mu(n)}\rangle.$

Now, we claim that for any $f\in\Lambda$, we have

(7) $\lim_{n\to\infty}\langle f\cdot H,s_{\mu(n)}\rangle=\langle
f,s_{\mu}\cdot H\rangle.$

By linearity, it is enough to show (7) when $f=s_{\nu}$ for some partition
$\nu$. In that case, by the Pieri rule,

$\lim_{n\to\infty}\langle f\cdot
H,s_{\mu(n)}\rangle=\lim_{n\to\infty}\begin{cases}1&\mbox{if $\mu(n)/\nu$ is a
horizontal strip}\\\ 0&\mbox{otherwise}\end{cases}.$

For $n$ sufficiently large, it is easy to see that $\mu(n)/\nu$ is a
horizontal strip if and only if $\nu/\mu$ is a horizontal strip. So the above
limit is equal to

$\begin{cases}1&\mbox{if $\nu/\mu$ is a horizontal strip}\\\
0&\mbox{otherwise}\end{cases},$

which is just $\langle f,s_{\mu}\cdot H\rangle$. Hence, (7) indeed holds.
Substituting (7) into (6) with
$f=\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\}$,
we obtain

$a^{\mu}_{\lambda}=\langle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{s_{\lambda}\right\\},s_{\mu}\cdot
H\rangle.$

The result now follows from the fact that
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
is adjoint to plethysm by $H_{+}$.

  5. (e)

The first and second equalities follow from the definition of
$b^{\mu}_{\lambda}$ and from (d).

For the third equality, we again use Lemma 4.2, together with the fact that
plethysm by $-p_{1}$ is an isometry and a ring automorphism:

$\displaystyle\langle
s_{\lambda},s_{\mu}[\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots]\cdot(1-e_{1}+e_{2}-e_{3}+\cdots)\rangle$
$\displaystyle=$ $\displaystyle\langle
s_{\lambda},s_{\mu}[-(L_{1}+L_{2}+L_{3}+\cdots)][-p_{1}]\cdot(1-e_{1}+e_{2}-e_{3}+\cdots)\rangle$
$\displaystyle=$ $\displaystyle\langle
s_{\lambda}[-p_{1}],s_{\mu}[-(L_{1}+L_{2}+L_{3}+\cdots)]\cdot(1-e_{1}+e_{2}-e_{3}+\cdots)[-p_{1}]\rangle$
$\displaystyle=$
$\displaystyle\langle(-1)^{m}\omega(s_{\lambda})[p_{1}],(-1)^{n}\omega(s_{\mu})[L_{1}+L_{2}+L_{3}+\cdots]\cdot
H\rangle$ $\displaystyle=$ $\displaystyle(-1)^{m-n}\langle
s_{\lambda^{T}},s_{\mu^{T}}[L_{1}+L_{2}+L_{3}+\cdots]\cdot H\rangle,$

as desired. ∎

##  5\. Computations of the surjective Frobenius transform

In this section, we will compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}$
for various symmetric functions $f$. Recall from the introduction:

*

######  Example 5.1.

Let us use Section 1(a) to compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{h_{2,2}\right\\}$.
First, we list all the functions $M\colon\mathbb{N}^{2}\to\mathbb{N}$ such
that $M(0,0)=0$ and $\sum_{j\in\mathbb{N}^{2}}jM(j)=(2,2)$. There are nine
such functions $M_{1},\ldots,M_{9}$. Here are all of their nonzero
values.111For readers who are familiar with the language of multisets and
multiset partitions [19], it can be helpful to remember that such functions
$M$ are in bijection with multiset partitions of
$\\{\mskip-5.0mu\\{1,1,2,2\\}\mskip-5.0mu\\}$. The multiset partition
corresponding to the function $M$ contains $M(j)$ copies of
$\\{\mskip-5.0mu\\{1^{j_{1}},2^{j_{2}}\\}\mskip-5.0mu\\}$ for all
$j\in\mathbb{N}^{2}$. For example, the function $M_{6}$ corresponds to the
multiset partition
$\\{\mskip-5.0mu\\{\\{\mskip-5.0mu\\{1,1\\}\mskip-5.0mu\\},\\{\mskip-5.0mu\\{2\\}\mskip-5.0mu\\},\\{\mskip-5.0mu\\{2\\}\mskip-5.0mu\\}\\}\mskip-5.0mu\\}\Vdash\\{\mskip-5.0mu\\{1,1,2,2\\}\mskip-5.0mu\\}$.

$\begin{gathered}M_{1}(2,2)=1\\\ M_{2}(1,1)=2\\\ M_{3}(2,1)=1\qquad
M_{3}(0,1)=1\\\ M_{4}(1,2)=1\qquad M_{4}(1,0)=1\\\ M_{5}(2,0)=1\qquad
M_{5}(0,2)=1\\\ M_{6}(2,0)=1\qquad M_{6}(0,1)=2\\\ M_{7}(0,2)=1\qquad
M_{7}(1,0)=2\\\ M_{8}(1,1)=1\qquad M_{8}(1,0)=1\qquad M_{8}(0,1)=1\\\
M_{9}(1,0)=2\qquad M_{9}(0,1)=2\end{gathered}$

Thus

$\displaystyle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{h_{2,2}\right\\}$
$\displaystyle=\underbrace{h_{1}}_{M_{1}}+\underbrace{h_{2}}_{M_{2}}+\underbrace{h_{1}^{2}}_{M_{3}}+\underbrace{h_{1}^{2}}_{M_{4}}+\underbrace{h_{1}^{2}}_{M_{5}}+\underbrace{h_{1}h_{2}}_{M_{6}}+\underbrace{h_{1}h_{2}}_{M_{7}}+\underbrace{h_{1}^{3}}_{M_{8}}+\underbrace{h_{2}^{2}}_{M_{9}}$
$\displaystyle=h_{1}+h_{2}+3h_{1,1}+2h_{2,1}+h_{1,1,1}+h_{2,2}.$

######  Example 5.2.

Let us use Section 1(b) to compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{e_{5,3}\right\\}$.
First, we list all the functions $M\colon\\{0,1\\}^{2}\to\mathbb{N}$ such that
$M(0,0)=0$ and $\sum_{j\in\\{0,1\\}^{2}}jM(j)=(5,3)$. There are four such
functions $M_{1},M_{2},M_{3},M_{4}$. Here are all of their nonzero values.

$\begin{gathered}M_{1}(1,0)=5\qquad M_{1}(0,1)=3\\\ M_{2}(1,1)=1\qquad
M_{2}(1,0)=4\qquad M_{2}(0,1)=2\\\ M_{2}(1,1)=2\qquad M_{2}(1,0)=3\qquad
M_{1}(0,1)=1\\\ M_{2}(1,1)=3\qquad M_{2}(1,0)=2\\\ \end{gathered}$

Thus

$\displaystyle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{e_{5,3}\right\\}$
$\displaystyle=\underbrace{e_{5}e_{3}}_{M_{1}}+\underbrace{h_{1}e_{4}e_{2}}_{M_{2}}+\underbrace{h_{2}e_{3}e_{1}}_{M_{3}}+\underbrace{h_{3}e_{2}}_{M_{4}}.$

######  Example 5.3.

Let us use Section 1(c) to compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{p_{15,10,6}\right\\}$.
Take $\lambda=(15,10,6)$ and $\ell=3$. There are five partitions of $[\ell]$
into nonempty sets: $\\{\\{1,2,3\\}\\}$, $\\{\\{1,2\\},\\{3\\}\\}$,
$\\{\\{1,3\\},\\{2\\}\\}$, $\\{\\{2,3\\},\\{1\\}\\}$, and
$\\{\\{1\\},\\{2\\},\\{3\\}\\}$. Thus,

$\displaystyle\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{p_{\lambda}\right\\}=$
$\displaystyle\left(\sum_{d\mid\gcd(\lambda_{1},\lambda_{2},\lambda_{3})}d^{2}p_{d}\right)$
$\displaystyle+\left(\sum_{d\mid\gcd(\lambda_{1},\lambda_{2})}dp_{d}\right)\left(\sum_{d\mid\lambda_{3}}p_{d}\right)$
$\displaystyle+\left(\sum_{d\mid\gcd(\lambda_{1},\lambda_{3})}dp_{d}\right)\left(\sum_{d\mid\lambda_{2}}p_{d}\right)$
$\displaystyle+\left(\sum_{d\mid\gcd(\lambda_{2},\lambda_{3})}dp_{d}\right)\left(\sum_{d\mid\lambda_{1}}p_{d}\right)$
$\displaystyle+\left(\sum_{d\mid\lambda_{1}}p_{d}\right)\left(\sum_{d\mid\lambda_{2}}p_{d}\right)\left(\sum_{d\mid\lambda_{3}}p_{d}\right)$
$\displaystyle=$
$\displaystyle\,\,p_{1}+(p_{1}+5p_{5})(p_{1}+p_{2}+p_{3}+p_{6})$
$\displaystyle+(p_{1}+3p_{3})(p_{1}+p_{2}+p_{5}+p_{10})$
$\displaystyle+(p_{1}+2p_{2})(p_{1}+p_{3}+p_{5}+p_{15})$
$\displaystyle+(p_{1}+p_{3}+p_{5}+p_{15})(p_{1}+p_{2}+p_{5}+p_{10})(p_{1}+p_{2}+p_{3}+p_{6}).$

###### Proof of Section 1.

(a) Let $t_{1},\ldots,t_{\ell}$ be variables and let

$f=H(t_{1})\cdots H(t_{\ell})\in\Lambda\llbracket
t_{1},\ldots,t_{\ell}\rrbracket.$

We will use Proposition 3.12 to compute

$\mathscr{F}\\{f\\}\in\overline{\Lambda}\llbracket
t_{1},\ldots,t_{\ell}\rrbracket.$

For any partition $\mu$, we have

$\displaystyle f(\Xi_{\mu})$
$\displaystyle=\prod_{i=1}^{\ell}H(t_{i})(\Xi_{\mu})$
$\displaystyle=\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}\prod_{k=0}^{\mu_{j}-1}\frac{1}{1-t_{i}\exp(2\pi
ik/\mu_{j})}$
$\displaystyle=\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}\frac{1}{1-t_{i}^{\mu_{j}}}.$

Hence, by Proposition 3.12, we have

$\mathscr{F}\\{f\\}=\sum_{\mu}\left(\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}\frac{1}{1-t_{i}^{\mu_{j}}}\right)\frac{p_{\mu}}{z_{\mu}}.$

We may recognize the right-hand side as a product of exponentials, and then
evaluate the product as follows.

$\displaystyle\mathscr{F}\\{f\\}$
$\displaystyle=\prod_{k}\exp\left(\frac{p_{k}}{k}\prod_{i=1}^{\ell}\frac{1}{1-t_{i}^{k}}\right)$
$\displaystyle=\prod_{k}\exp\left(\frac{p_{k}}{k}\sum_{j\in\mathbb{N}^{\ell}}(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}})^{k}\right)$
$\displaystyle=\prod_{j\in\mathbb{N}^{\ell}}\exp\left(\sum_{k}\frac{p_{k}}{k}(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}})^{k}\right)$
$\displaystyle=\prod_{j\in\mathbb{N}^{\ell}}H(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}}).$

Hence,

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}=\frac{\mathscr{F}\\{f\\}}{H}=\prod_{j\in\mathbb{N}^{\ell}\setminus\\{(0,\ldots,0)\\}}H(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}}).$

Finally, the result follows from taking the coefficient of
$t_{1}^{\lambda_{1}}\cdots t_{\ell}^{\lambda_{\ell}}$ on both sides.

(b) Let $t_{1},\ldots,t_{\ell}$ be variables and let

$f=E(t_{1})\cdots E(t_{\ell})\in\Lambda\llbracket
t_{1},\ldots,t_{\ell}\rrbracket.$

We will use Proposition 3.12 to compute

$\mathscr{F}\\{f\\}\in\overline{\Lambda}\llbracket
t_{1},\ldots,t_{\ell}\rrbracket.$

For any partition $\mu$, we have

$\displaystyle f(\Xi_{\mu})$
$\displaystyle=\prod_{i=1}^{\ell}E(t_{i})(\Xi_{\mu})$
$\displaystyle=\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}\prod_{k=0}^{\mu_{j}-1}(1+t_{i}\exp(2\pi
ik/\mu_{j}))$
$\displaystyle=\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}(1-(-t_{i})^{\mu_{j}}).$

Hence, by Proposition 3.12, we have

$\mathscr{F}\\{f\\}=\sum_{\mu}\left(\prod_{i=1}^{\ell}\prod_{j=1}^{\ell(\mu)}(1-(-t_{i}))^{\mu_{j}}\right)\frac{p_{\mu}}{z_{\mu}}.$

We may recognize the right-hand side as a product of exponentials, and then
evaluate the product as follows.

$\displaystyle\mathscr{F}\\{f\\}$
$\displaystyle=\prod_{k}\exp\left(\frac{p_{k}}{k}\prod_{i=1}^{\ell}(1-(-t_{i})^{k})\right)$
$\displaystyle=\prod_{k}\exp\left(\frac{p_{k}}{k}\sum_{j\in\\{0,1\\}^{\ell}}(-1)^{j_{1}+\cdots+j_{\ell}}((-t_{1})^{j_{1}}\cdots(-t_{\ell})^{j_{\ell}})^{k}\right)$
$\displaystyle=\prod_{j\in\\{0,1\\}^{\ell}}\exp\left((-1)^{j_{1}+\cdots+j_{\ell}}\sum_{k}\frac{p_{k}}{k}((-1)^{j_{1}+\cdots+j_{\ell}}t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}})^{k}\right)$
$\displaystyle=\prod_{j\in\\{0,1\\}^{\ell}}\begin{cases}H(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}})&\mbox{if $j_{1}+\cdots+j_{\ell}$ is even}\\\
E(t_{1}^{j_{1}}\cdots t_{\ell}^{j_{\ell}})&\mbox{if $j_{1}+\cdots+j_{\ell}$ is
odd}\end{cases}$

Hence,

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}\left\\{f\right\\}=\frac{\mathscr{F}\\{f\\}}{H}=\prod_{j\in\\{0,1\\}^{\ell}\setminus\\{(0,\ldots,0)\\}}\begin{cases}H(t_{1}^{j_{1}}\cdots
t_{\ell}^{j_{\ell}})&\mbox{if $j_{1}+\cdots+j_{\ell}$ is even}\\\
E(t_{1}^{j_{1}}\cdots t_{\ell}^{j_{\ell}})&\mbox{if $j_{1}+\cdots+j_{\ell}$ is
odd}\end{cases}.$

Finally, the result follows from taking the coefficient of
$t_{1}^{\lambda_{1}}\cdots t_{\ell}^{\lambda_{\ell}}$ on both sides.

(c) By Proposition 3.12, we have

(8)
$\mathscr{F}\\{p_{\lambda}\\}=\sum_{\mu}p_{\lambda}(\Xi_{\mu})\frac{p_{\mu}}{z_{\mu}}.$

For all $r$, we have

$p_{r}(\Xi_{\mu})=\sum_{d\mid r}dm_{d}(\mu),$

so (8) becomes

(9)
$\mathscr{F}\\{p_{\lambda}\\}=\sum_{\mu}\prod_{i=1}^{\ell}\left(\sum_{d\mid\lambda_{i}}dm_{d}(\mu)\right)\frac{p_{\mu}}{z_{\mu}}.$

Let us say that a function $\mathbf{d}\colon[\ell]\to\mathbb{N}$ is
_permissible_ if $\mathbf{d}(i)\mid\lambda_{i}$ for all $i$. The product on
the right-hand side of (9) can be expanded into a sum over all permissible
functions:

$\displaystyle\mathscr{F}\\{p_{\lambda}\\}$
$\displaystyle=\sum_{\mu}\sum_{\text{$\mathbf{d}$
permissible}}\left(\prod_{i=1}^{\ell}\mathbf{d}(i)m_{\mathbf{d}(i)}(\mu)\right)\frac{p_{\mu}}{z_{\mu}}$
(10) $\displaystyle=\sum_{\mu}\sum_{\text{$\mathbf{d}$
permissible}}\left(\prod_{d\in\mathbb{N}}(dm_{d}(\mu))^{|\mathbf{d}^{-1}(d)|}\right)\frac{p_{\mu}}{z_{\mu}}.$

For any $k$, let $(x)_{k}=x(x-1)\cdots(x-k+1)$ denote the falling factorial.
It is well-known [6, Chapter 6.1] that for any $n$, the monomial $x^{n}$ can
be written as a linear combination

$x^{n}=\sum_{k}\genfrac{\\{}{\\}}{0.0pt}{}{n}{k}(x)_{k}$

of falling factorials, where the coefficient
$\genfrac{\\{}{\\}}{0.0pt}{}{n}{k}$ (a _Stirling number of the second kind_)
is the number of partitions of $[n]$ into $k$ nonempty sets. Let us use this
to rewrite the factor $(m_{d}(\mu))^{{|\mathbf{d}^{-1}(d)|}}$ appearing in
(10). We obtain

(11) $\mathscr{F}\\{p_{\lambda}\\}=\sum_{\mu}\sum_{\text{$\mathbf{d}$
permissible}}\left(\prod_{d\in\mathbb{N}}d^{|\mathbf{d}^{-1}(d)|}\sum_{k}\genfrac{\\{}{\\}}{0.0pt}{}{|\mathbf{d}^{-1}(d)|}{k}(m_{d}(\mu))_{k}\right)\frac{p_{\mu}}{z_{\mu}}.$

Given a permissible function $\mathbf{d}\colon[\ell]\to\mathbb{N}$ and a
partition $\pi$ of $[\ell]$ into nonempty sets, let us say that $\pi$ is
_level_ with respect to $\mathbf{d}$ if $\mathbf{d}(i)=\mathbf{d}(j)$ whenever
$i$ and $j$ are in the same part of $\pi$. If $\pi$ is level with respect to
$\mathbf{d}$, we may define the function
$\widetilde{\mathbf{d}}\colon\pi\to\mathbb{N}$ by taking
$\widetilde{\mathbf{d}}(U)$ to be the common value of $\mathbf{d}(i)$ for
$i\in U$.

Suppose that $\mathbf{d}$ is a fixed permissible function and
$\\{k_{d}\\}_{d}$ is any sequence. Then it is easy to see that the product

$\prod_{d\in\mathbb{N}}\genfrac{\\{}{\\}}{0.0pt}{}{|\mathbf{d}^{-1}(d)|}{k_{d}}$

is equal to the number of set partitions $\pi$ that are level with respect to
$U$ and which satisfy $|\widetilde{\mathbf{d}}^{-1}(d)|=k_{d}$ for all $d$.
Using this fact, we may expand the product in (11) into a sum over all
functions that are level with respect to $\mathbf{d}$:

$\mathscr{F}\\{p_{\lambda}\\}=\sum_{\mu}\sum_{\text{$\mathbf{d}$
permissible}}\sum_{\text{$\pi$
level}}\left(\prod_{d\in\mathbb{N}}d^{|\mathbf{d}^{-1}(d)|}(m_{d}(\mu))_{|\widetilde{\mathbf{d}}^{-1}(d)|}\right)\frac{p_{\mu}}{z_{\mu}}$

We will simplify this expression by switching the order of summation so that
the sum over $\pi$ is all the way on the outside. To do so, given a partition
$\pi$ of $[\ell]$ into nonempty sets, we will now describe the set of all
permissible functions $\mathbf{d}\colon[\ell]\to\mathbb{N}$ such that $\pi$ is
level with respect to $\mathbf{d}$. These are exactly the functions given by
$\mathbf{d}(i)=\widetilde{\mathbf{d}}(U)$ for all $U\in\pi$ and $i\in U$,
where $\widetilde{\mathbf{d}}\colon\pi\to\mathbb{N}$ is any function
satisfying

$\widetilde{\mathbf{d}}(U)\mid\gcd\\{\lambda_{i}\colon i\in U\\}$

for all $U\in\pi$. Let us call such functions _$\pi$ -permissible_. Then

$\mathscr{F}\\{p_{\lambda}\\}=\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\sum_{\mu}\left(\prod_{d\in\mathbb{N}}d^{|\mathbf{d}^{-1}(d)|}(m_{d}(\mu))_{|\widetilde{\mathbf{d}}^{-1}(d)|}\right)\frac{p_{\mu}}{z_{\mu}}.$

Now, let us evaluate the inner sum over $\mu$. Expanding $p_{\mu}$ and
$z_{\mu}$ gives

$\displaystyle\mathscr{F}\\{p_{\lambda}\\}$
$\displaystyle=\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\sum_{\mu}\left(\prod_{d\in\mathbb{N}}d^{|\mathbf{d}^{-1}(d)|}(m_{d}(\mu))_{|\widetilde{\mathbf{d}}^{-1}(d)|}\frac{p_{d}^{m_{d}(\mu)}}{d^{m_{d}(\mu)}(m_{d}(\mu))!}\right)$
$\displaystyle=\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{d\in\mathbb{N}}\left(d^{|\mathbf{d}^{-1}(d)|}\sum_{m=0}^{\infty}(m)_{|\widetilde{\mathbf{d}}^{-1}(d)|}\frac{p_{d}^{m}}{d^{m}m!}\right)$
$\displaystyle=\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{d\in\mathbb{N}}\left(d^{|\mathbf{d}^{-1}(d)|}\sum_{m=|\widetilde{\mathbf{d}}^{-1}(d)|}^{\infty}\frac{p_{d}^{m}}{d^{m}(m-|\widetilde{\mathbf{d}}^{-1}(d)|)!}\right)$
$\displaystyle=\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{d\in\mathbb{N}}\left(d^{|\mathbf{d}^{-1}(d)|}\left(\frac{p_{d}}{d}\right)^{|\widetilde{\mathbf{d}}^{-1}(d)|}\exp\left(\frac{p_{d}}{d}\right)\right)$
$\displaystyle=\left(\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{d\in\mathbb{N}}\left(d^{|\mathbf{d}^{-1}(d)|}\left(\frac{p_{d}}{d}\right)^{|\widetilde{\mathbf{d}}^{-1}(d)|}\right)\right)\cdot
H$ $\displaystyle=\left(\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{d\in\mathbb{N}}\left(d^{|\mathbf{d}^{-1}(d)|-|\widetilde{\mathbf{d}}^{-1}(d)|}\cdot
p_{d}^{|\widetilde{\mathbf{d}}^{-1}(d)|}\right)\right)\cdot H.$

We will rewrite the remaining product as a product over $U\in\pi$ instead of
over $d\in\mathbb{N}$. Each $U\in\pi$ with $\widetilde{\mathbf{d}}(U)=d$
contributes $1$ to $|\widetilde{\mathbf{d}}^{-1}(d)|$ and contributes $|U|$ to
$|\mathbf{d}^{-1}(d)|$. So we obtain

$\displaystyle\mathscr{F}\\{p_{\lambda}\\}=\left(\sum_{\pi}\sum_{\text{$\widetilde{\mathbf{d}}$
$\pi$-permissible}}\prod_{U\in\pi}(\widetilde{\mathbf{d}}(U))^{|U|-1}p_{\widetilde{\mathbf{d}}(U)}\right)\cdot
H.$

Given that $\widetilde{\mathbf{d}}$ is $\pi$-permissible, the possible values
of $\widetilde{\mathbf{d}}(U)$ are exactly those $d\in\mathbb{N}$ that divide
$\gcd\\{\lambda_{i}\colon i\in U\\}$. The choice of
$\widetilde{\mathbf{d}}(U)$ can be made independently for each $U\in\pi$.
Hence, we may factor the innermost sum, which finally yields

$\mathscr{F}\\{p_{\lambda}\\}=\left(\sum_{\pi}\prod_{U\in\pi}\left(\sum_{d\mid\gcd\\{\lambda_{i}\colon
i\in U\\}}d^{|U|-1}p_{d}\right)\right)\cdot H.$

The result follows from dividing both sides of this equation by $H$. ∎

######  Remark 5.4.

Section 1(a) has an alternate, “purely combinatorial” proof. It involves
exhibiting a combinatorial species
$E_{\lambda}\colon\mathrm{Bij}\to\mathrm{Fun}$ with
$\operatorname{ch}(\mathbb{C}E_{\lambda})=h_{\lambda}$, and then computing
$\iota^{*}\iota_{!}E_{\lambda}$ in terms of known combinatorial species. The
details for this proof are forthcoming in a separate paper.

######  Remark 5.5.

One consequence of Section 1(c) which is not obvious _a priori_ is that the
matrix entries of
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}$
in the power sum basis are all nonnegative integers. The same is not true of
$\mathscr{F}$; for example,
$\mathscr{F}\\{1\\}=H=1+p_{1}+\frac{1}{2}(p_{2}+p_{1}^{2})+\cdots$ certainly
has some non-integer coefficients.

To illustrate the utility of Section 1, we now prove Section 1 about the
vanishing of restriction coefficients. Recall from the introduction that
$D(\mu)$ is the size of the Durfee square of $\mu$; that is, the largest
integer $d$ such that $\mu_{d}\geq d$.

*

Before proceeding to the proof, we need a few lemmas.

######  Lemma 5.6.

For any $k\geq 0$, we have

$\operatorname{span}\\{e_{\lambda}\colon\ell(\lambda)\leq
k\\}=\operatorname{span}\\{s_{\lambda}\colon\lambda_{1}\leq k\\}.$

(Here $\operatorname{span}(S)$ refers to the additive subgroup of $\Lambda$
generated by $S$.)

###### Proof.

By the Pieri rule, every $e_{\lambda}$ with $\ell(\lambda)\leq k$ is a linear
combination of Schur functions $s_{\mu}$, where $\mu$ is the union of at most
$k$ vertical strips. Hence

$\operatorname{span}\\{e_{\lambda}\colon\ell(\lambda)\leq
k\\}\subset\operatorname{span}\\{s_{\lambda}\colon\lambda_{1}\leq k\\}.$

By the dual Jacobi-Trudi identity, every $s_{\lambda}$ with $\lambda_{1}\leq
k$ can be written as a determinant of a $k\times k$ matrix whose entries are
elementary symmetric functions $e_{r}$. Hence

$\operatorname{span}\\{s_{\lambda}\colon\lambda_{1}\leq
k\\}\subset\operatorname{span}\\{e_{\lambda}\colon\ell(\lambda)\leq k\\}.$

The result follows. ∎

######  Definition 5.7.

Given a symmetric function $f$, we say that the Schur function $s_{\lambda}$
_appears_ in $f$ if $\langle f,s_{\lambda}\rangle\neq 0$.

######  Lemma 5.8.

Let $\lambda$ be a partition and let $d\geq 0$. Then $D(\lambda)\leq d$ if and
only if there exist partitions $\mu$, $\nu$ with $\ell(\mu),\ell(\nu)\leq d$
such that $s_{\lambda}$ appears in $h_{\mu}e_{\nu}$.

###### Proof.

For the “only if” direction, assume that $D(\lambda)\leq d$. Take

$\displaystyle\mu$ $\displaystyle=(\lambda_{1},\ldots,\lambda_{D(\lambda)})$
$\displaystyle\nu$
$\displaystyle=(\lambda^{T}_{1}-D(\lambda),\ldots,\lambda^{T}_{D(\lambda)}-D(\lambda)).$

Clearly $\ell(\mu),\ell(\nu)\leq d$. Also, the Young diagram of $\lambda$ can
be decomposed into a union of horizontal strips of lengths
$\mu_{1},\ldots,\mu_{\ell(\mu)}$ and vertical strips of lengths
$\nu_{1},\ldots,\nu_{\ell(\nu)}$, so $s_{\lambda}$ appears in $h_{\mu}e_{\nu}$
by the Pieri rule.

For the “if” direction, it easily follows from induction on $\ell(\nu)$ and
the Pieri rule that if $s_{\lambda}$ appears in $h_{\mu}e_{\nu}$, then
$\lambda_{\ell(\mu)+1}<\ell(\nu)+1$ (where we take $\lambda_{i}=0$ for
$i>\ell(\lambda)$). Assuming that $\ell(\mu),\ell(\nu)\leq d$, we get
$\lambda_{d+1}<d+1$, so $D(\lambda)\leq d$, as desired. ∎

###### Proof of Section 1.

First, observe that by multiplying both sides by $H$, Section 1(b) can be
written in the following form, using the Frobenius transform instead of the
surjective Frobenius transform. For any partition $\lambda$ with
$\ell(\lambda)\leq k$, we have

(12)
$\mathscr{F}\\{e_{\lambda}\\}=\sum_{M}\prod_{j\in\\{0,1\\}^{k}}\begin{cases}h_{M(j)}&\mbox{if
$j_{1}+\cdots+j_{\ell}$ is even}\\\ e_{M(j)}&\mbox{if $j_{1}+\cdots+j_{\ell}$
is odd}\end{cases}$

where the sum is over all functions $M\colon\\{0,1\\}^{k}\to\mathbb{N}$ such
that $\sum_{j\in\\{0,1\\}^{\ell}}j_{i}M(j)=\lambda_{i}$ for $i=1,\ldots,\ell$
(where we take $\lambda_{i}=0$ for $i>\ell(\lambda)$). We do not require that
$M(0,\ldots,0)=0$, so the sum in (12) is infinite.

Consider the following statements:

  * (A)

There exists a partition $\lambda$ such that $\lambda_{1}\leq k$ and
$r_{\lambda}^{\mu}>0$.

  * (C1)

There exists a partition $\lambda$ such that $\lambda_{1}\leq k$ and
$\langle\mathscr{F}\\{s_{\lambda}\\},s_{\mu}\rangle\neq 0$.

  * (C2)

There exists a partition $\lambda$ such that $\ell(\lambda)\leq k$ and
$\langle\mathscr{F}\\{e_{\lambda}\\},s_{\mu}\rangle\neq 0$.

  * (C3)

There exists a function $M\colon\\{0,1\\}^{k}\to\mathbb{N}$ such that
$s_{\mu}$ appears in

$\prod_{j\in\\{0,1\\}^{k}}\begin{cases}h_{M(j)}&\mbox{if
$j_{1}+\cdots+j_{\ell}$ is even}\\\ e_{M(j)}&\mbox{if $j_{1}+\cdots+j_{\ell}$
is odd}\end{cases}.$

  * (C4)

There exist partitions $\nu,\nu^{\prime}$ such that
$\ell(\nu),\ell(\nu^{\prime})\leq 2^{k-1}$ and $s_{\mu}$ appears in
$h_{\nu}e_{\nu^{\prime}}$.

  * (B)

$D(\mu)\leq 2^{k-1}$.

By the definition of the Frobenius transform, (A) is equivalent to (C1). By
the linearity of $\mathscr{F}$ and Lemma 5.6, (C1) is equivalent to (C2).
Since each term of (12) is Schur positive, $s_{\mu}$ appears in the sum if and
only if it appears in one or more of its terms, so (C2) is equivalent to (C3).
By taking the parts of $\nu$ to be the nonzero values of $M(j)$ for
$j_{1}+\cdots+j_{\ell}$ even and taking the parts of $\nu^{\prime}$ to be the
nonzero values of $M(j)$ for $j_{1}+\cdots+j_{\ell}$ odd, we see that (C3) is
equivalent to (C4). By Lemma 5.8, (C4) is equivalent to (B). Putting it all
together, (A) is equivalent to (B), as desired. ∎

##  6\. Computations of the inverse surjective Frobenius transform

We will now compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}$
and
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{h_{\lambda}\right\\}$.
In order to state our formulas, first we must recall some definitions from
combinatorics on words. For a more in-depth introduction, see [11, Chapter 5].

######  Definition 6.1.

Let $A$ be a set. A _word_ over the alphabet $A$ is a sequence $w=w_{1}\cdots
w_{n}$ with $w_{1},\ldots,w_{n}\in A$. Given a letter $a\in A$, we write
$m_{a}(w)$ to mean the number of times the letter $a$ appears in $w$.

######  Definition 6.2 ([12]).

Let $A$ be a totally ordered set. We say that a nonempty word $w=w_{1}\cdots
w_{n}$ over the alphabet $A$ is a _Lyndon word_ if it is lexicographically
less than its suffix $w_{i}\cdots w_{n}$ for $i=2,\ldots,n$. Let
$\operatorname{Lyndon}(A)$ be the set of all Lyndon words over the alphabet
$A$.

######  Theorem 6.3 (Witt’s Formula [20]).

Let $\ell>0$ and let $t_{1},\ldots,t_{\ell}$ be variables. For any word
$w=i_{1}\cdots i_{n}$ over $[\ell]$, denote by $t^{w}$ the product

$t_{i_{1}}\cdots t_{i_{n}}=\prod_{i=1}^{\ell}t_{i}^{m_{i}(w)}.$

Then the evaluation $(L_{1}+L_{2}+L_{3}+\cdots)(t_{1},\ldots,t_{\ell})$ is
equal to

$\sum_{w\in\operatorname{Lyndon}([\ell])}t^{w}.$

######  Theorem 6.4 (Chen-Fox-Lyndon Theorem [4]).

Let $A$ be a totally ordered set. Any word $w$ over the alphabet $A$ has a
unique _Lyndon factorization_ ; that is, an expression as a
(lexicographically) non-increasing concatenation of Lyndon words.

######  Definition 6.5.

Let $w$ be a word over a totally ordered alphabet. Define $\pi(w)$ to be the
partition obtained by listing the number of times each Lyndon word appears in
the Lyndon factorization of $w$, and then sorting the resulting positive
numbers in decreasing order.

######  Example 6.6.

If $A=\\{1,2\\}$ and $w=21212121111$, then the Lyndon factorization of $w$ is
$w=(2)(12)(12)(12)(1)(1)(1)(1)$. The Lyndon words appearing in this
factorization are $2$, $12$, and $1$, which appear once, three times, and four
times, respectively, so $\pi(w)=(4,3,1)$.

Now, we are ready to compute
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}$
and
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{h_{\lambda}\right\\}$.
The cleanest way to state our results is to use the series $H(t)$ from
Definition 2.3.

######  Theorem 6.7.

Let $\ell>0$ and let $t_{1},\ldots,t_{\ell}$ be variables.

  1. (a)

For any word $w=i_{1}\cdots i_{n}$ over $[\ell]$, denote by $t^{w}$ the
product

$t_{i_{1}}\cdots t_{i_{n}}=\prod_{i=1}^{\ell}t_{i}^{m_{i}(w)}.$

Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{\frac{1}{\prod_{i=1}^{\ell}H(t_{i})}\right\\}=\frac{1}{\prod_{w\in\operatorname{Lyndon}([\ell])}H(t^{w})}.$

  2. (b)

For any word $w=(i_{1},j_{1})\cdots(i_{n},j_{n})$ over $[\ell]^{2}$ (ordered
lexicographically), denote by $t^{w}$ the product

$(t_{i_{1}}t_{j_{1}})\cdots(t_{i_{n}}t_{j_{n}})=\prod_{i,j=1}^{\ell}(t_{i}t_{j})^{m_{(i,j)}(w)}.$

Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{\prod_{i=1}^{\ell}H(t_{i})\right\\}=\frac{\prod_{w\in\operatorname{Lyndon}([\ell])}H(t^{w})}{\prod_{w\in\operatorname{Lyndon}([\ell]^{2})}H(t^{w})}.$

Before we proceed to the proof of Theorem 6.7, we will prove a few corollaries
that illustrate how to use it.

######  Corollary 6.8.

For any $r$, we have

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{h_{r}\right\\}=\sum_{k=0}^{\lfloor
r/2\rfloor}(-1)^{k}h_{r-2k}e_{k}.$

###### Proof.

Take $\ell=1$ in Theorem 6.7(b). In this case, $1$ is the only Lyndon word
over $[1]$ and $(1,1)$ is the only Lyndon word over $[1]^{2}$. So, with
$t=t_{1}$,

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{H(t)\right\\}=\frac{H(t)}{H(t^{2})}=H(t)E(-t^{2}).$

The result follows from taking the coefficient of $t^{r}$. ∎

######  Corollary 6.9.

Let $\lambda$ be a partition and let $\ell=\ell(\lambda)$. Then

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}=\sum_{\mu}(-1)^{|\lambda|-|\mu|}L_{\lambda}^{\mu}e_{\mu},$

where $L_{\lambda}^{\mu}$ is the number of words $w$ over $[\ell]$ such that

  1. (a)

$m_{i}(w)=\lambda_{i}$ for $i\in[\ell]$, and

  2. (b)

$\pi(w)=\mu$.

###### Proof.

First, the right-hand side of Section 1(a) can be written

(13)
$\prod_{w\in\operatorname{Lyndon}(\ell)}\left(\sum_{r=0}^{\infty}(-1)^{r}e_{r}(t^{w})^{r}\right)=\sum_{\mathbf{r}}\prod_{w\in\operatorname{Lyndon}(\ell)}(-1)^{\mathbf{r}(w)}e_{\mathbf{r}(w)}(t^{w})^{\mathbf{r}(w)}$

where the sum is over all finitely supported functions
$\mathbf{r}\colon\operatorname{Lyndon}([\ell])\to\mathbb{N}$. By the Chen-Fox-
Lyndon theorem (Theorem 6.4), there is a bijection

$\\{\text{words over
$[\ell]$}\\}\underset{\phi}{\longleftrightarrow}\\{\text{finitely supported
functions $\operatorname{Lyndon}([\ell])\to\mathbb{N}$}\\},$

where $(\phi(w))(w^{\prime})$ is the number of times that $w^{\prime}$ appears
in the Lyndon factorization of $w$. Using this bijection, we can rewrite (13)
as a sum over words $w$ over $[\ell]$:

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{\frac{1}{\prod_{i=1}^{\ell}H(t_{i})}\right\\}=\sum_{w}(-1)^{|\pi(w)|}e_{\pi(w)}t^{w}.$

The result follows from taking the coefficient of $t_{1}^{\lambda_{1}}\cdots
t_{\ell}^{\lambda_{\ell}}$. ∎

######  Corollary 6.10.

For any $r$, we have

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{1}^{\ell}\right\\}=e_{1}(e_{1}-1)\cdots(e_{1}-\ell+1).$

###### Proof.

Take $\lambda=(1^{\ell})$ in Corollary 6.9. The words over $[\ell]$ that
satisfy condition (a) are exactly the permutations $w\in\mathfrak{S}_{\ell}$.
The Lyndon factorization of any $w\in\mathfrak{S}_{\ell}$ contains only
distinct Lyndon words, one beginning with each left-to-right minimum of $w$
(that is, each letter $w_{i}$ such that $w_{i}<w_{j}$ for all $j<i$). So
$\pi(w)=(1^{k})$, where $k$ is the number of left-to-right minima of $w$. It
follows that

$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{1}^{\ell}\right\\}=\sum_{k}(-1)^{\ell-k}\genfrac{[}{]}{0.0pt}{}{\ell}{k}e_{1}^{k},$

where $\genfrac{[}{]}{0.0pt}{}{\ell}{k}$ (a _Stirling number of the first
kind_) is the number of permutations $w\in\mathfrak{S}_{\ell}$ with $k$ left-
to-right minima. This is well-known [6, Chapter 6.1] to be equal to
$e_{1}(e_{1}-1)\cdots(e_{1}-\ell+1)$. ∎

######  Question 6.11.

Are there any partitions $\lambda$ other than $(1^{\ell})$ for which
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{e_{\lambda}\right\\}$
has a simple factorization like the one in Corollary 6.10?

We are almost ready to prove Section 1. First, we will restate some lemmas
from Loehr and Remmel’s 2011 “exposé” on plethysm [10].

######  Lemma 6.12 ([10, Example 1]).

There is a unique binary operation
$\bullet[\bullet]\colon\Lambda\times\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket\to\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$ satisfying the following properties.

  1. (i)

For any fixed $g$, the map $\bullet[g]\colon\Lambda\to\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$ is a ring homomorphism.

  2. (ii)

For any fixed $r$, the map $p_{r}[\bullet]\colon\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket\to\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$ is a ring homomorphism which preserves
convergent infinite sums.

  3. (iii)

For any $r$ and $i$, we have $p_{r}[t_{i}]=t_{i}^{r}$.

We refer to the operation from Lemma 6.12 as _plethysm_ , because it is
closely related to the plethysm of symmetric functions
$\bullet[\bullet]\colon\Lambda\times\Lambda\to\Lambda$ mentioned in Section 2.
For example, the two operations are related by a kind of associative property:

######  Lemma 6.13 (Associativity of Plethysm, [10, Theorem 5]).

Let $f,g\in\Lambda$ and $h\in\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$. Then

$f[g[h]]=(f[g])[h]\in\mathbb{Z}\llbracket t_{1},\cdots,t_{\ell}\rrbracket.$

If $g$ has positive integer coefficients, then the plethysm $f[g]$ can be
described as an evaluation, as follows:

######  Lemma 6.14 (Monomial Substitution Rule, [10, Theorem 7]).

Let $f\in\Lambda$ and let $M_{1},M_{2},M_{3},\ldots\in\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$ be a (finite or infinite) sequence of monic
monomials. Then

$f\left[\sum_{n}M_{n}\right]=f(M_{1},M_{2},M_{3},\ldots),$

where the right-hand side denotes the evaluation of $f$ at
$M_{1},M_{2},M_{3},\ldots$.

We also prove one lemma of our own.

######  Lemma 6.15.

Let $M_{1},M_{2},M_{3},\ldots\in\mathbb{Z}[t_{1},\cdots,t_{\ell}]$ be a
(finite or infinite) sequence of monic monomials and let
$a_{1},a_{2},a_{3},\ldots\in\mathbb{Z}$ be a sequence of the same length.
Suppose that the sum $\sum_{n}a_{n}M_{n}$ converges in $\mathbb{Z}\llbracket
t_{1},\cdots,t_{\ell}\rrbracket$. Then for any $f\in\Lambda$, we have

(14) $f\left[\sum_{n}a_{n}M_{n}\right]=\left\langle
f,\prod_{n}H(M_{n})^{a_{n}}\right\rangle.$

###### Proof.

Fix $M_{1},M_{2},M_{3},\ldots$ and let $a_{1},a_{2},a_{3},\ldots$ vary. Given
any fixed monomial $M\in\mathbb{Z}[t_{1},\ldots,t_{\ell}]$, it is easy to see
that the coefficient of $M$ on either side of (14) is a polynomial in
$a_{1},a_{2},a_{3},\ldots$. Hence, we may assume that $a_{n}\geq 0$ for all
$n$. Then by replacing each $M_{n}$ with $a_{n}$ copies of $M_{n}$, we may
make the even stronger assumption that $a_{n}=1$ for all $n$. In other words,
we wish to prove

(15) $f\left[\sum_{n}M_{n}\right]=\left\langle
f,\prod_{n}H(M_{n})\right\rangle.$

By linearity, it suffices to show Equation 15 in the case that $f=m_{\lambda}$
is a monomial symmetric function. Then, since $\langle
m_{\lambda},h_{\mu}\rangle=\delta_{\lambda\mu}$, the right-hand side of the
equation becomes $m_{\lambda}(M_{1},M_{2},M_{3},\ldots)$. The result follows
from the monomial substitution rule (Lemma 6.14). ∎

###### Proof of Theorem 6.7.

In what follows, let

$L=L_{1}+L_{2}+L_{3}+\cdots\in\overline{\Lambda}$

and

$\tilde{L}=\omega(L_{1})-\omega(L_{2})+\omega(L_{3})-\cdots\in\overline{\Lambda}.$

(a) Let $\overline{\omega}\colon\Lambda\to\Lambda$ be the involution given by

$\overline{\omega}(f)=f[-p_{1}]=(-1)^{\deg f}\omega(f)$

for all homogeneous $f\in\Lambda$. Clearly, $\overline{\omega}$ is a ring
automorphism and

$\overline{\omega}(H(t))=E(-t)=\frac{1}{H(t)}.$

We wish to show that

(16)
$(\overline{\omega}\circ\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\circ\overline{\omega})\left\\{\prod_{i=1}^{\ell}H(t_{i})\right\\}=\prod_{w\in\operatorname{Lyndon}([\ell])}H(t^{w}).$

To do so, let $f\in\Lambda$ be arbitrary. It is sufficient to show that each
side of (16) has the same Hall inner product with $f$. By Theorem 4.1(c),
$\overline{\omega}\circ\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\circ\overline{\omega}$
is adjoint to plethysm by $L$, so

$\displaystyle\left\langle
f,(\overline{\omega}\circ\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\circ\overline{\omega})\left\\{\prod_{i=1}^{\ell}H(t_{i})\right\\}\right\rangle$
$\displaystyle=\left\langle f[L],\prod_{i=1}^{\ell}H(t_{i})\right\rangle$
$\displaystyle=(f[L])[t_{1}+\cdots+t_{\ell}]$
$\displaystyle=f[L[t_{1}+\cdots+t_{\ell}]]$
$\displaystyle=f[L(t_{1},\ldots,t_{\ell})]$
$\displaystyle=f\left[\sum_{w\in\operatorname{Lyndon}([\ell])}t^{w}\right]$
$\displaystyle=\left\langle
f,\prod_{w\in\operatorname{Lyndon}([\ell])}H(t^{w})\right\rangle,$

where we used Lemma 6.15, Lemma 6.13, Lemma 6.14, and Theorem 6.3. This
completes the proof of (16) and of part (a) of the theorem.

(b) Again, let $f\in\Lambda$ be arbitrary. It is sufficient to show that each
side of the equation has the same Hall inner product with $f$. By Theorem
4.1(c),
$\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}$
is adjoint to plethysm by $\tilde{L}$, so

$\displaystyle\left\langle
f,\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{\prod_{i=1}^{\ell}H(t_{i})\right\\}\right\rangle$
$\displaystyle=\left\langle
f[\tilde{L}],\prod_{i=1}^{\ell}H(t_{i})\right\rangle$
$\displaystyle=(f[\tilde{L}])[t_{1}+\cdots+t_{\ell}]$ (17)
$\displaystyle=f[\tilde{L}[t_{1}+\cdots+t_{\ell}]]$

Now, let us describe the monomials that appear in the plethysm
$\tilde{L}[t_{1}+\cdots+t_{\ell}]$. By the definition of Lyndon symmetric
functions, we have

$\displaystyle\tilde{L}$ $\displaystyle=\sum_{n}(-1)^{n-1}\omega(L_{n})$
$\displaystyle=\sum_{n}\frac{(-1)^{n-1}}{n}\sum_{d\mid
n}\mu(d)\omega(p_{d}^{n/d})$
$\displaystyle=\sum_{n}\frac{(-1)^{n-1}}{n}\sum_{d\mid
n}\mu(d)(-1)^{(d-1)n/d}p_{d}^{n/d}$
$\displaystyle=\sum_{n}\frac{1}{n}\sum_{d\mid
n}\mu(d)(-1)^{n/d-1}p_{d}^{n/d}.$

Now, the term $(-1)^{n/d-1}$ is equal to $-1$ if $d$ divides $\frac{n}{2}$ and
$1$ otherwise. Hence,

$\tilde{L}=\sum_{n}\frac{1}{n}\left(\sum_{d\mid
n}\mu(d)p_{d}^{n/d}-2\sum_{d\mid\frac{n}{2}}\mu(d)p_{d}^{n/d}\right)$

where the sum over $d\mid\frac{n}{2}$ is understood to be empty if $n$ is odd.
Splitting this into two sums and then performing the change of variables
$\frac{n}{2}\to n$ in the second, we obtain

$\displaystyle\tilde{L}$ $\displaystyle=\sum_{n}\frac{1}{n}\sum_{d\mid
n}\mu(d)p_{d}^{n/d}-\sum_{n}\frac{2}{n}\sum_{d\mid\frac{n}{2}}\mu(d)p_{d}^{n/d}$
$\displaystyle=\sum_{n}\frac{1}{n}\sum_{d\mid
n}\mu(d)p_{d}^{n/d}-\sum_{n}\frac{1}{n}\sum_{d\mid n}\mu(d)p_{d}^{2n/d}$
$\displaystyle=L-L[p_{1}^{2}].$

By Lemma 6.13 and Theorem 6.3,

$\displaystyle\tilde{L}[t_{1}+\cdots+t_{\ell}]$
$\displaystyle=L[t_{1}+\cdots+t_{\ell}]-(L[p_{1}^{2}])[t_{1}+\cdots+t_{\ell}]$
$\displaystyle=L[t_{1}+\cdots+t_{\ell}]-L[p_{1}^{2}[t_{1}+\cdots+t_{\ell}]]$
$\displaystyle=L[t_{1}+\cdots+t_{\ell}]-L[(t_{1}+\cdots+t_{\ell})^{2}]$
$\displaystyle=\sum_{w\in\operatorname{Lyndon}([\ell])}t^{w}-\sum_{w\in\operatorname{Lyndon}([\ell]^{2})}t^{w}.$

Substituting into (17) and using Lemma 6.15 one last time, we finally obtain

$\displaystyle\left\langle
f,\mathscr{F}_{\mathchoice{\operatorname{Sur}}{\operatorname{Sur}}{\mathrm{Sur}}{\mathrm{Sur}}}^{-1}\left\\{\prod_{i=1}^{\ell}H(t_{i})\right\\}\right\rangle$
$\displaystyle=f\left[\sum_{w\in\operatorname{Lyndon}([\ell])}t^{w}-\sum_{w\in\operatorname{Lyndon}([\ell]^{2})}t^{w}\right]$
$\displaystyle=\left\langle
f,\frac{\prod_{w\in\operatorname{Lyndon}([\ell])}H(t^{w})}{\prod_{w\in\operatorname{Lyndon}([\ell]^{2})}H(t^{w})}\right\rangle,$

completing the proof. ∎

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