that sends each formal power series $\sum\limits_{n\geq
0}a_{n}t^{n}\in\Lambda\left[\left[t\right]\right]$ (with $a_{n}\in\Lambda$) to
$\sum\limits_{n\geq 0}\alpha_{k}\left(a_{n}\right)t^{n}$. Consider this
$\mathbf{k}\left[\left[t\right]\right]$-algebra homomorphism
$\alpha_{k}\left[\left[t\right]\right]$.

Applying the map $\alpha_{k}\left[\left[t\right]\right]$ to both sides of the
equality (51), we obtain

$\displaystyle\left(\alpha_{k}\left[\left[t\right]\right]\right)\left(1\right)$
$\displaystyle=\left(\alpha_{k}\left[\left[t\right]\right]\right)\left(\left(\sum\limits_{n\geq
0}\left(-1\right)^{n}e_{n}t^{n}\right)\left(\sum\limits_{n\geq
0}h_{n}t^{n}\right)\right)$
$\displaystyle=\underbrace{\left(\alpha_{k}\left[\left[t\right]\right]\right)\left(\sum\limits_{n\geq
0}\left(-1\right)^{n}e_{n}t^{n}\right)}_{\begin{subarray}{c}=\sum\limits_{n\geq
0}\alpha_{k}\left(\left(-1\right)^{n}e_{n}\right)t^{n}\\\ \text{(by the
definition of
}\alpha_{k}\left[\left[t\right]\right]\text{)}\end{subarray}}\cdot\underbrace{\left(\alpha_{k}\left[\left[t\right]\right]\right)\left(\sum\limits_{n\geq
0}h_{n}t^{n}\right)}_{\begin{subarray}{c}=\sum\limits_{n\geq
0}\alpha_{k}\left(h_{n}\right)t^{n}\\\ \text{(by the definition of
}\alpha_{k}\left[\left[t\right]\right]\text{)}\end{subarray}}$ $\displaystyle\
\ \ \ \ \ \ \ \ \ \left(\text{since
}\alpha_{k}\left[\left[t\right]\right]\text{ is a
}\mathbf{k}\left[\left[t\right]\right]\text{-algebra homomorphism}\right)$
$\displaystyle=\left(\sum\limits_{n\geq
0}\underbrace{\alpha_{k}\left(\left(-1\right)^{n}e_{n}\right)}_{\begin{subarray}{c}=\left(-1\right)^{n}\alpha_{k}\left(e_{n}\right)\\\
\text{(since }\alpha_{k}\text{ is
}\mathbf{k}\text{-linear)}\end{subarray}}t^{n}\right)\cdot\left(\sum\limits_{n\geq
0}\underbrace{\alpha_{k}\left(h_{n}\right)}_{\begin{subarray}{c}=\left[n<k\right]\\\
\text{(by (\ref{pf.thm.G.main.alk}))}\end{subarray}}t^{n}\right)$
$\displaystyle=\left(\sum\limits_{n\geq
0}\left(-1\right)^{n}\alpha_{k}\left(e_{n}\right)t^{n}\right)\cdot\underbrace{\left(\sum\limits_{n\geq
0}\left[n<k\right]t^{n}\right)}_{=t^{0}+t^{1}+\cdots+t^{k-1}=\dfrac{1-t^{k}}{1-t}}$
$\displaystyle=\left(\sum\limits_{n\geq
0}\left(-1\right)^{n}\alpha_{k}\left(e_{n}\right)t^{n}\right)\cdot\dfrac{1-t^{k}}{1-t}.$

Comparing this with

$\left(\alpha_{k}\left[\left[t\right]\right]\right)\left(1\right)=1\ \ \ \ \ \
\ \ \ \ \left(\text{since }\alpha_{k}\left[\left[t\right]\right]\text{ is a
}\mathbf{k}\left[\left[t\right]\right]\text{-algebra homomorphism}\right),$

we obtain

$\left(\sum\limits_{n\geq
0}\left(-1\right)^{n}\alpha_{k}\left(e_{n}\right)t^{n}\right)\cdot\dfrac{1-t^{k}}{1-t}=1.$

Hence,

$\displaystyle\sum\limits_{n\geq
0}\left(-1\right)^{n}\alpha_{k}\left(e_{n}\right)t^{n}$
$\displaystyle=\dfrac{1-t}{1-t^{k}}=\left(1-t\right)\cdot\underbrace{\dfrac{1}{1-t^{k}}}_{=1+t^{k}+t^{2k}+t^{3k}+\cdots}$
$\displaystyle=\left(1-t\right)\cdot\left(1+t^{k}+t^{2k}+t^{3k}+\cdots\right)$
$\displaystyle=1-t+t^{k}-t^{k+1}+t^{2k}-t^{2k+1}+t^{3k}-t^{3k+1}\pm\cdots$
$\displaystyle=\sum\limits_{n\geq
0}\left(-1\right)^{n\%k}\left[n\%k\in\left\\{0,1\right\\}\right]t^{n}$

(here, we have used that $k>1$, since for $k=1$ there would be cancellations
in the sum $1-t+t^{k}-t^{k+1}+t^{2k}-t^{2k+1}+t^{3k}-t^{3k+1}\pm\cdots$).
Comparing coefficients before $t^{m}$ on both sides of this equality, we
obtain

$\left(-1\right)^{m}\alpha_{k}\left(e_{m}\right)=\left(-1\right)^{m\%k}\left[m\%k\in\left\\{0,1\right\\}\right]$
(55)

for each $m\in\mathbb{N}$.

Multiplying both sides of this equality by $\left(-1\right)^{m}$, we obtain

$\alpha_{k}\left(e_{m}\right)=\left(-1\right)^{m+m\%k}\left[m\%k\in\left\\{0,1\right\\}\right].$
(56)

We must prove that

$\alpha_{k}\left(e_{r}\right)=\left(-1\right)^{r+r\%k}\left[r\%k\in\left\\{0,1\right\\}\right].$

If $r\in\mathbb{N}$, then this follows by applying (56) to $m=r$. Hence, for
the rest of this proof, we WLOG assume that $r\notin\mathbb{N}$. Thus, $r$ is
negative, so that $r\in\left\\{-k+2,-k+3,\ldots,-1\right\\}$ (since $r>-k+1$).
Hence, $r\%k\in\left\\{2,3,\ldots,k-1\right\\}$, so that
$r\%k\notin\left\\{0,1\right\\}$. Consequently,
$\left[r\%k\in\left\\{0,1\right\\}\right]=0$. Also, $e_{r}=0$ (since $r$ is
negative) and thus $\alpha_{k}\left(e_{r}\right)=\alpha_{k}\left(0\right)=0$.
Comparing this with
$\left(-1\right)^{r+r\%k}\underbrace{\left[r\%k\in\left\\{0,1\right\\}\right]}_{=0}=0$,
we obtain
$\alpha_{k}\left(e_{r}\right)=\left(-1\right)^{r+r\%k}\left[r\%k\in\left\\{0,1\right\\}\right]$.
This concludes the proof of Lemma 3.28.

######  Proof 3.30 (Proof of Theorem 2.14.).

(a) Assume that $\mu_{k}\neq 0$. But $\mu=\lambda^{t}$, whence

$\mu_{k}=\left(\lambda^{t}\right)_{k}=\left|\left\\{j\in\left\\{1,2,3,\ldots\right\\}\
\mid\ \lambda_{j}\geq k\right\\}\right|\ \ \ \ \ \ \ \ \ \ \left(\text{by the
definition of }\lambda^{t}\right).$

Hence,

$\left|\left\\{j\in\left\\{1,2,3,\ldots\right\\}\ \mid\ \lambda_{j}\geq
k\right\\}\right|=\mu_{k}\neq 0.$

In other words, the set $\left\\{j\in\left\\{1,2,3,\ldots\right\\}\ \mid\
\lambda_{j}\geq k\right\\}$ is nonempty. Hence, there exists some
$j\in\left\\{1,2,3,\ldots\right\\}$ satisfying $\lambda_{j}\geq k$. Consider
this $j$. We have $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$ (since
$\lambda\in\operatorname*{Par}$) and thus $\lambda_{1}\geq\lambda_{j}$ (since
$1\leq j$). Hence, $\lambda_{1}\geq\lambda_{j}\geq k$. Thus, Proposition 2.11
yields $\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)=0$.
This proves Theorem 2.14 (a).

Now, let us prepare for the proof of parts (b) and (c).

Consider the $\mathbf{k}$-algebra homomorphism
$\alpha_{k}:\Lambda\rightarrow\mathbf{k}$ from Definition 3.10.

For each $i\in\left\\{1,2,\ldots,k-1\right\\}$, we have
$\left(\beta_{i}-1\right)\%k\in\left\\{0,1,\ldots,k-1\right\\}$ (by the
definition of a remainder) and thus
$\gamma_{i}\in\left\\{1,2,\ldots,k\right\\}$ (by (8)). In other words,
$\gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}$ are $k-1$ elements of the set
$\left\\{1,2,\ldots,k\right\\}$.

Assume that $\mu_{k}=0$. Thus,
$\mu=\left(\mu_{1},\mu_{2},\ldots,\mu_{k-1}\right)$ (since
$\mu\in\operatorname*{Par}$).

It is known that taking the transpose of the transpose of a partition returns
the original partition. Thus, $\left(\lambda^{t}\right)^{t}=\lambda$. In view
of $\mu=\lambda^{t}$, this rewrites as $\mu^{t}=\lambda$. Hence,
$\lambda=\mu^{t}$. Therefore,

$s_{\lambda}=s_{\mu^{t}}=\det\left(\left(e_{\mu_{i}-i+j}\right)_{1\leq i\leq
k-1,\ 1\leq j\leq k-1}\right)$

(by (6), applied to $\mu$ and $k-1$ instead of $\lambda$ and $\ell$), because
$\mu=\left(\mu_{1},\mu_{2},\ldots,\mu_{k-1}\right)$. Applying the map
$\alpha_{k}$ to both sides of this equality, we find

$\displaystyle\alpha_{k}\left(s_{\lambda}\right)$
$\displaystyle=\alpha_{k}\left(\det\left(\left(\underbrace{e_{\mu_{i}-i+j}}_{\begin{subarray}{c}=e_{\beta_{i}+j}\\\
\text{(since (\ref{eq.thm.petk.explicit.beti=}) yields
}\mu_{i}-i=\beta_{i}\text{)}\end{subarray}}\right)_{1\leq i\leq k-1,\ 1\leq
j\leq k-1}\right)\right)$
$\displaystyle=\alpha_{k}\left(\det\left(\left(e_{\beta_{i}+j}\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)\right)$
$\displaystyle=\det\left(\left(\alpha_{k}\left(e_{\beta_{i}+j}\right)\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)$

(since $\alpha_{k}$ is a $\mathbf{k}$-algebra homomorphism, and thus commutes
with taking determinants of matrices). On the other hand,

$\alpha_{k}\left(\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\right)=\alpha_{k}\left(s_{\lambda/\varnothing}\right)=\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)$

(by (23), applied to $\varnothing$ instead of $\mu$). Comparing these two
equalities, we obtain

$\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)=\det\left(\left(\alpha_{k}\left(e_{\beta_{i}+j}\right)\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right).$ (57)

But each $i\in\left\\{1,2,\ldots,k-1\right\\}$ and
$j\in\left\\{1,2,\ldots,k-1\right\\}$ satisfy $k>1$ 202020Indeed, if
$i\in\left\\{1,2,\ldots,k-1\right\\}$, then $1\leq i\leq k-1$ and thus
$k-1\geq 1>0$, so that $k>1$. and

$\underbrace{\beta_{i}}_{\begin{subarray}{c}=\mu_{i}-i\\\ \text{(by
(\ref{eq.thm.petk.explicit.beti=}))}\end{subarray}}+j=\underbrace{\mu_{i}}_{\geq
0}-\underbrace{i}_{\leq k-1}+\underbrace{j}_{>0}>0-\left(k-1\right)+0=-k+1$

and thus

$\alpha_{k}\left(e_{\beta_{i}+j}\right)=\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]$
(58)

(by (54), applied to $r=\beta_{i}+j$).

Furthermore, each $i\in\left\\{1,2,\ldots,k-1\right\\}$ and
$j\in\left\\{1,2,\ldots,k-1\right\\}$ satisfy

$\displaystyle\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]$
$\displaystyle=\left(-1\right)^{\beta_{i}}\left(-1\right)^{j}\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right].$
(59)

[Proof of (59): Let $i\in\left\\{1,2,\ldots,k-1\right\\}$ and
$j\in\left\\{1,2,\ldots,k-1\right\\}$. The definition of $\gamma_{i}$ yields

$\gamma_{i}=1+\underbrace{\left(\beta_{i}-1\right)\%k}_{\begin{subarray}{c}\equiv\beta_{i}-1\operatorname{mod}k\\\
\text{(since }u\%k\equiv u\operatorname{mod}k\text{ for any
}u\in\mathbb{Z}\text{)}\end{subarray}}\equiv
1+\left(\beta_{i}-1\right)=\beta_{i}\operatorname{mod}k.$

Hence,
$\underbrace{\gamma_{i}}_{\equiv\beta_{i}\operatorname{mod}k}+j\equiv\beta_{i}+j\operatorname{mod}k$.
But if $u\in\mathbb{Z}$ and $v\in\mathbb{Z}$ satisfy $u\equiv
v\operatorname{mod}k$, then $u\%k=v\%k$. Applying this to $u=\gamma_{i}+j$ and
$v=\beta_{i}+j$, we obtain
$\left(\gamma_{i}+j\right)\%k=\left(\beta_{i}+j\right)\%k$. Hence,

$\displaystyle\left(-1\right)^{\beta_{i}}\left(-1\right)^{j}\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]$
$\displaystyle=\underbrace{\left(-1\right)^{\beta_{i}}\left(-1\right)^{j}\left(-1\right)^{\left(\beta_{i}+j\right)\%k}}_{=\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]$
$\displaystyle=\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right].$

This proves (59).]

Now, (57) becomes

$\displaystyle\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)$
$\displaystyle=\det\left(\left(\underbrace{\alpha_{k}\left(e_{\beta_{i}+j}\right)}_{\begin{subarray}{c}=\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]\\\
\text{(by (\ref{pf.thm.petk.explicit.akem1a}))}\end{subarray}}\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)$
$\displaystyle=\det\left(\left(\underbrace{\left(-1\right)^{\left(\beta_{i}+j\right)+\left(\beta_{i}+j\right)\%k}\left[\left(\beta_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]}_{\begin{subarray}{c}=\left(-1\right)^{\beta_{i}}\left(-1\right)^{j}\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]\\\
\text{(by (\ref{pf.thm.petk.explicit.akem2}))}\end{subarray}}\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)$
$\displaystyle=\det\left(\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{j}\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)$
$\displaystyle=\left(\prod\limits_{i=1}^{k-1}\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{i}\right)\right)\cdot\det\left(\left(\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}\right)$

(by Lemma 3.22 (b), applied to $m=k-1$ and $R=\mathbf{k}$ and
$a_{i,j}=\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]$
and $u_{i}=\left(-1\right)^{\beta_{i}}$ and $v_{j}=\left(-1\right)^{j}$).

Define a $\left(k-1\right)\times\left(k-1\right)$-matrix $G$ as in Lemma 3.24.
Then, this becomes

$\displaystyle\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)$
$\displaystyle=\left(\prod\limits_{i=1}^{k-1}\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{i}\right)\right)\cdot\det\left(\underbrace{\left(\left(-1\right)^{\left(\gamma_{i}+j\right)\%k}\left[\left(\gamma_{i}+j\right)\%k\in\left\\{0,1\right\\}\right]\right)_{1\leq
i\leq k-1,\ 1\leq j\leq k-1}}_{\begin{subarray}{c}=G\\\ \text{(by the
definition of }G\text{)}\end{subarray}}\right)$
$\displaystyle=\left(\prod\limits_{i=1}^{k-1}\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{i}\right)\right)\cdot\det
G.$ (60)

Now, we can readily prove parts (b) and (c) of Theorem 2.14:

(b) Assume that the $k-1$ numbers $\gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}$
are not distinct. Then, Lemma 3.24 (a) yields $\det G=0$. Hence, (60) yields

$\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)=\left(\prod\limits_{i=1}^{k-1}\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{i}\right)\right)\cdot\underbrace{\det
G}_{=0}=0.$

This proves Theorem 2.14 (b).

(c) The equality (60) becomes

$\displaystyle\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)$
$\displaystyle=\underbrace{\left(\prod\limits_{i=1}^{k-1}\left(\left(-1\right)^{\beta_{i}}\left(-1\right)^{i}\right)\right)}_{=\left(\prod\limits_{i=1}^{k-1}\left(-1\right)^{\beta_{i}}\right)\left(\prod\limits_{i=1}^{k-1}\left(-1\right)^{i}\right)}\cdot\underbrace{\det
G}_{\begin{subarray}{c}=\left(-1\right)^{g+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right)-\left(1+2+\cdots+\left(k-1\right)\right)}\\\
\text{(by Lemma \ref{lem.petk.explicit.2} {(c)})}\end{subarray}}$
$\displaystyle=\underbrace{\left(\prod\limits_{i=1}^{k-1}\left(-1\right)^{\beta_{i}}\right)}_{=\left(-1\right)^{\beta_{1}+\beta_{2}+\cdots+\beta_{k-1}}}\underbrace{\left(\prod\limits_{i=1}^{k-1}\left(-1\right)^{i}\right)}_{=\left(-1\right)^{1+2+\cdots+\left(k-1\right)}}\cdot\left(-1\right)^{g+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right)-\left(1+2+\cdots+\left(k-1\right)\right)}$
$\displaystyle=\left(-1\right)^{\beta_{1}+\beta_{2}+\cdots+\beta_{k-1}}\left(-1\right)^{1+2+\cdots+\left(k-1\right)}\cdot\left(-1\right)^{g+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right)-\left(1+2+\cdots+\left(k-1\right)\right)}$
$\displaystyle=\left(-1\right)^{\left(\beta_{1}+\beta_{2}+\cdots+\beta_{k-1}\right)+\left(1+2+\cdots+\left(k-1\right)\right)+g+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right)-\left(1+2+\cdots+\left(k-1\right)\right)}$
$\displaystyle=\left(-1\right)^{\left(\beta_{1}+\beta_{2}+\cdots+\beta_{k-1}\right)+g+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right)}.$

This proves Theorem 2.14 (c).

The proof of Proposition 2.15 relies on the following known fact:

######  Proposition 3.31.

Let $\lambda\in\operatorname*{Par}$. Let $\mu=\lambda^{t}$. Then:

(a) If $i$ and $j$ are two positive integers satisfying $\lambda_{i}\geq j$,
then $\mu_{j}\geq i$.

(b) If $i$ and $j$ are two positive integers satisfying $\lambda_{i}<j$, then
$\mu_{j}<i$.

(c) Any two positive integers $i$ and $j$ satisfy
$\lambda_{i}+\mu_{j}-i-j\neq-1$.

For each positive integer $i$, set $\alpha_{i}=\lambda_{i}-i$. For each
positive integer $j$, set $\beta_{j}=\mu_{j}-j$ and $\eta_{j}=-1-\beta_{j}$.
Then:

(d) The two sets $\left\\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right\\}$
and $\left\\{\eta_{1},\eta_{2},\eta_{3},\ldots\right\\}$ are disjoint, and
their union is $\mathbb{Z}$.

(e) Let $p$ be an integer such that $p\geq\lambda_{1}$. Then, the two sets
$\left\\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right\\}$ and
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{p}\right\\}$ are disjoint, and their
union is

$\left\\{\ldots,p-3,p-2,p-1\right\\}=\left\\{k\in\mathbb{Z}\ \mid\
k<p\right\\}.$

(f) Let $p$ and $q$ be two integers such that $p\geq\lambda_{1}$ and
$q\geq\mu_{1}$. Then, the two sets
$\left\\{\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right\\}$ and
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{p}\right\\}$ are disjoint, and their
union is

$\left\\{-q,-q+1,\ldots,p-1\right\\}=\left\\{k\in\mathbb{Z}\ \mid\ -q\leq
k<p\right\\}.$

Note that Proposition 3.31 (f) is a restatement of [Macdon95, Chapter I,
(1.7)].

######  Proof 3.32 (Proof of Proposition 3.31 (sketched).).

Left to the reader (see [Grinbe20b] for a detailed proof). The easiest way to
proceed is by proving (a) and (b) first, then deriving (c) as their
consequence, then deriving (f) from it, then concluding (d) and (e).

######  Proof 3.33 (Proof of Proposition 2.15 (sketched).).

Let $\mu=\lambda^{t}$. Then, the number of parts of $\mu$ is $\lambda_{1}$.
Hence, from $\lambda_{1}<k$, we conclude that $\mu$ has fewer than $k$ parts.
Thus, $\mu_{k}=0$.

For each positive integer $i$, set $\alpha_{i}=\lambda_{i}-i$. Hence,

$\displaystyle\left\\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right\\}$
$\displaystyle=\left\\{\underbrace{\alpha_{i}}_{=\lambda_{i}-i}\ \mid\
i\in\left\\{1,2,3,\ldots\right\\}\right\\}=\left\\{\lambda_{i}-i\ \mid\
i\in\left\\{1,2,3,\ldots\right\\}\right\\}$ $\displaystyle=B\ \ \ \ \ \ \ \ \
\ \left(\text{by the definition of }B\right).$

For each positive integer $j$, set $\beta_{j}=\mu_{j}-j$ and
$\eta_{j}=-1-\beta_{j}$. Note that
$\left(\beta_{1},\beta_{2},\ldots,\beta_{k-1}\right)\in\mathbb{Z}^{k-1}$ is
thus the same $\left(k-1\right)$-tuple that was called
$\left(\beta_{1},\beta_{2},\ldots,\beta_{k-1}\right)$ in Theorem 2.14. It is
easy to see that $\beta_{1}>\beta_{2}>\cdots>\beta_{k-1}$ and
$\lambda_{1}-1>\lambda_{2}-2>\lambda_{3}-3>\cdots$.

From $\lambda_{1}<k$, we obtain $k-1\geq\lambda_{1}$. Hence, Proposition 3.31
(e) (applied to $p=k-1$) yields that the two sets
$\left\\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right\\}$ and
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}$ are disjoint, and their
union is

$\left\\{\ldots,\left(k-1\right)-3,\left(k-1\right)-2,\left(k-1\right)-1\right\\}=\left\\{\text{all
integers smaller than }k-1\right\\}=W.$

Since $\left\\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right\\}=B$, we can
restate this as follows: The two sets $B$ and
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}$ are disjoint, and their
union is $W$. Hence,
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}=W\setminus B$.

It is also easy to see that $\beta_{1}>\beta_{2}>\cdots>\beta_{k-1}$, so that
$\eta_{1}<\eta_{2}<\cdots<\eta_{k-1}$. Hence,
$\eta_{1},\eta_{2},\ldots,\eta_{k-1}$ are the elements of the set
$\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}$ listed in increasing
order (with no repetition).

Let us define a $\left(k-1\right)$-tuple
$\left(\gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\right)\in\left\\{1,2,\ldots,k\right\\}^{k-1}$
as in Theorem 2.14. Now, we have the following chain of logical equivalences:

$\displaystyle\
\left(\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)\neq
0\right)$ $\displaystyle\Longleftrightarrow\ \left(\text{the }k-1\text{
numbers }\gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\text{ are distinct}\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by parts
{(b)} and {(c)} of Theorem \ref{thm.petk.explicit}}\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{the }k-1\text{ numbers
}\left(\beta_{1}-1\right)\%k,\left(\beta_{2}-1\right)\%k,\ldots,\left(\beta_{k-1}-1\right)\%k\text{
are distinct}\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\left(\text{since }\gamma_{i}=1+\left(\beta_{i}-1\right)\%k\text{ for each
}i\right)$ $\displaystyle\Longleftrightarrow\ \left(\text{no two of the
}k-1\text{ numbers }\beta_{1}-1,\beta_{2}-1,\ldots,\beta_{k-1}-1\text{ are
congruent modulo }k\right)$ $\displaystyle\Longleftrightarrow\ \left(\text{no
two of the }k-1\text{ numbers }\beta_{1},\beta_{2},\ldots,\beta_{k-1}\text{
are congruent modulo }k\right)$ $\displaystyle\Longleftrightarrow\
\left(\text{no two of the }k-1\text{ numbers
}-1-\beta_{1},-1-\beta_{2},\ldots,-1-\beta_{k-1}\text{ }\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left.\text{are congruent modulo }k\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{no two of the }k-1\text{
numbers }\eta_{1},\eta_{2},\ldots,\eta_{k-1}\text{ are congruent modulo
}k\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\left(\text{since }\eta_{j}=-1-\beta_{j}\text{ for each }j\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{no two of the }k-1\text{
elements of }\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}\text{ are
congruent modulo }k\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\left(\begin{array}[c]{c}\text{since
}\eta_{1},\eta_{2},\ldots,\eta_{k-1}\text{ are the elements of the set
}\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}\\\ \text{listed in
increasing order (with no repetition)}\end{array}\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{no two of the }k-1\text{
elements of }W\setminus B\text{ are congruent modulo }k\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{since
}\left\\{\eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\\}=W\setminus B\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{each congruence class
}\overline{i}\text{ has at most }1\text{ element in common with }W\setminus
B\right)$ $\displaystyle\Longleftrightarrow\ \left(\text{each
}i\in\left\\{0,1,\ldots,k-1\right\\}\text{ satisfies
}\left|\overline{i}\cap\left(W\setminus B\right)\right|\leq 1\right)$
$\displaystyle\Longleftrightarrow\ \left(\text{each
}i\in\left\\{0,1,\ldots,k-1\right\\}\text{ satisfies
}\left|\left(\overline{i}\cap W\right)\setminus B\right|\leq 1\right)$

(since $\overline{i}\cap\left(W\setminus B\right)=\left(\overline{i}\cap
W\right)\setminus B$ for each $i$). This proves Proposition 2.15.

###  3.12 Proof of Theorem 2.18

######  Proof 3.34 (Proof of Theorem 2.18.).

In this proof, the word “monomial” may refer to a monomial in any set of
variables (not necessarily in $x_{1},x_{2},x_{3},\ldots$).

In the following, an $i$_-monomial_ (where $i\in\mathbb{N}$) shall mean a
monomial of degree $i$.

We shall say that a monomial is $k$_-bounded_ if all exponents in this
monomial are $<k$. In other words, a monomial is $k$_-bounded_ if it can be
written in the form $z_{1}^{a_{1}}z_{2}^{a_{2}}\cdots z_{s}^{a_{s}}$, where
$z_{1},z_{2},\ldots,z_{s}$ are distinct variables and
$a_{1},a_{2},\ldots,a_{s}$ are nonnegative integers $<k$. Thus, the
$k$-bounded monomials in the variables $x_{1},x_{2},x_{3},\ldots$ are
precisely the monomials of the form $\mathbf{x}^{\alpha}$ for
$\alpha\in\operatorname*{WC}$ satisfying $\left(\alpha_{i}<k\text{ for all
}i\right)$. Hence, the $k$-bounded $m$-monomials in the variables
$x_{1},x_{2},x_{3},\ldots$ are precisely the monomials of the form
$\mathbf{x}^{\alpha}$ for $\alpha\in\operatorname*{WC}$ satisfying
$\left|\alpha\right|=m$ and $\left(\alpha_{i}<k\text{ for all }i\right)$.

Now, the definition of $G\left(k,m\right)$ yields

$\displaystyle G\left(k,m\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=m;\\\ \alpha_{i}<k\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}$ $\displaystyle=\left(\text{the sum of
all }k\text{-bounded }m\text{-monomials in the variables
}x_{1},x_{2},x_{3},\ldots\right)$ (61)

(since the $k$-bounded $m$-monomials in the variables
$x_{1},x_{2},x_{3},\ldots$ are precisely the monomials of the form
$\mathbf{x}^{\alpha}$ for $\alpha\in\operatorname*{WC}$ satisfying
$\left|\alpha\right|=m$ and $\left(\alpha_{i}<k\text{ for all }i\right)$).

Let us now substitute the variables
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$ for the variables
$x_{1},x_{2},x_{3},\ldots$ on both sides of the equality (61). (This means
that we choose some bijection
$\phi:\left\\{x_{1},x_{2},x_{3},\ldots\right\\}\rightarrow\left\\{x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right\\}$,
and substitute $\phi\left(x_{i}\right)$ for each $x_{i}$ on both sides of
(61).) The left hand side of (61) turns into
$\left(G\left(k,m\right)\right)\left(\mathbf{x},\mathbf{y}\right)$ upon this
substitution212121because this is how
$\left(G\left(k,m\right)\right)\left(\mathbf{x},\mathbf{y}\right)$ was
defined, whereas the right hand side turns into

$\left(\text{the sum of all }k\text{-bounded }m\text{-monomials in the
variables }x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right)$
222222Indeed, the substitution can be regarded as simply renaming the
variables $x_{1},x_{2},x_{3},\ldots$ as
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$ (in some order). Thus, it
turns the $k$-bounded $m$-monomials in the variables
$x_{1},x_{2},x_{3},\ldots$ into the $k$-bounded $m$-monomials in the variables
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$.

. Thus, our substitution transforms the equality (61) into

$\displaystyle\left(G\left(k,m\right)\right)\left(\mathbf{x},\mathbf{y}\right)$
$\displaystyle=\left(\text{the sum of all }k\text{-bounded }m\text{-monomials
in the variables }x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right).$
(62)

But any monomial $\mathfrak{m}$ in the variables
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$ can be uniquely written as
a product $\mathfrak{np}$, where $\mathfrak{n}$ is a monomial in the variables
$x_{1},x_{2},x_{3},\ldots$ and where $\mathfrak{p}$ is a monomial in the
variables $y_{1},y_{2},y_{3},\ldots$. Moreover, if $\mathfrak{m}$ is written
in this form, then:

  * •

the degree of $\mathfrak{m}$ equals the sum of the degrees of $\mathfrak{n}$
and $\mathfrak{p}$;

  * •

thus, $\mathfrak{m}$ is an $m$-monomial if and only if there exists some
$i\in\left\\{0,1,\ldots,m\right\\}$ such that $\mathfrak{n}$ is an
$i$-monomial and $\mathfrak{p}$ is an $\left(m-i\right)$-monomial;

  * •

furthermore, $\mathfrak{m}$ is $k$-bounded if and only if both $\mathfrak{n}$
and $\mathfrak{p}$ are $k$-bounded.

Thus, any $k$-bounded $m$-monomial $\mathfrak{m}$ in the variables
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$ can be uniquely written as
a product $\mathfrak{np}$, where $i\in\left\\{0,1,\ldots,m\right\\}$, where
$\mathfrak{n}$ is a $k$-bounded $i$-monomial in the variables
$x_{1},x_{2},x_{3},\ldots$ and where $\mathfrak{p}$ is a $k$-bounded
$\left(m-i\right)$-monomial in the variables $y_{1},y_{2},y_{3},\ldots$.
Conversely, every such product $\mathfrak{np}$ is a $k$-bounded $m$-monomial
$\mathfrak{m}$ in the variables
$x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots$. Thus, we obtain a
bijection

$\displaystyle\bigsqcup_{i\in\left\\{0,1,\ldots,m\right\\}}\left(\vphantom{\frac{1}{2}}\left\\{k\text{-bounded
}i\text{-monomials in the variables }x_{1},x_{2},x_{3},\ldots\right\\}\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \
\left.\vphantom{\frac{1}{2}}\times\left\\{k\text{-bounded
}\left(m-i\right)\text{-monomials in the variables
}y_{1},y_{2},y_{3},\ldots\right\\}\right)$
$\displaystyle\rightarrow\left\\{k\text{-bounded }m\text{-monomials in the
variables }x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right\\}$

that sends each pair $\left(\mathfrak{n},\mathfrak{p}\right)$ to
$\mathfrak{np}$. Hence,

$\displaystyle\left(\text{the sum of all }k\text{-bounded }m\text{-monomials
in the variables }x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right)$
$\displaystyle=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\sum\limits_{\begin{subarray}{c}\mathfrak{n}\text{
is a }k\text{-bounded}\\\ i\text{-monomial in the}\\\ \text{variables
}x_{1},x_{2},x_{3},\ldots\end{subarray}}\sum\limits_{\begin{subarray}{c}\mathfrak{p}\text{
is a }k\text{-bounded}\\\ \left(m-i\right)\text{-monomial in the}\\\
\text{variables }y_{1},y_{2},y_{3},\ldots\end{subarray}}\mathfrak{np}$
$\displaystyle=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\left(\sum\limits_{\begin{subarray}{c}\mathfrak{n}\text{
is a }k\text{-bounded}\\\ i\text{-monomial in the}\\\ \text{variables
}x_{1},x_{2},x_{3},\ldots\end{subarray}}\mathfrak{n}\right)\left(\sum\limits_{\begin{subarray}{c}\mathfrak{p}\text{
is a }k\text{-bounded}\\\ \left(m-i\right)\text{-monomial in the}\\\
\text{variables }y_{1},y_{2},y_{3},\ldots\end{subarray}}\mathfrak{p}\right)$
$\displaystyle=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\left(\text{the
sum of all }k\text{-bounded }i\text{-monomials in the variables
}x_{1},x_{2},x_{3},\ldots\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\cdot\left(\text{the sum of all }k\text{-bounded
}\left(m-i\right)\text{-monomials in the variables
}y_{1},y_{2},y_{3},\ldots\right).$ (63)

Now, let $i\in\left\\{0,1,\ldots,m\right\\}$. The same reasoning that gave us
(61) can be applied to $i$ instead of $m$. Thus we obtain

$G\left(k,i\right)=\left(\text{the sum of all }k\text{-bounded
}i\text{-monomials in the variables }x_{1},x_{2},x_{3},\ldots\right).$ (64)

Also, $i\in\left\\{0,1,\ldots,m\right\\}$, so that
$m-i\in\left\\{0,1,\ldots,m\right\\}\subseteq\mathbb{N}$. Hence, the same
reasoning that gave us (61) can be applied to $m-i$ instead of $m$. Thus we
obtain

$\displaystyle G\left(k,m-i\right)$ $\displaystyle=\left(\text{the sum of all
}k\text{-bounded }\left(m-i\right)\text{-monomials in the variables
}x_{1},x_{2},x_{3},\ldots\right).$

Renaming the variables $x_{1},x_{2},x_{3},\ldots$ as
$y_{1},y_{2},y_{3},\ldots$ in this equality, we obtain

$\displaystyle\left(G\left(k,m-i\right)\right)\left(\mathbf{y}\right)$
$\displaystyle=\left(\text{the sum of all }k\text{-bounded
}\left(m-i\right)\text{-monomials in the variables
}y_{1},y_{2},y_{3},\ldots\right).$ (65)

Forget that we fixed $i$. We thus have proved (64) and (65) for each
$i\in\left\\{0,1,\ldots,m\right\\}$.

Now, (62) becomes

$\displaystyle\left(G\left(k,m\right)\right)\left(\mathbf{x},\mathbf{y}\right)$
$\displaystyle=\left(\text{the sum of all }k\text{-bounded }m\text{-monomials
in the variables }x_{1},x_{2},x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right)$
$\displaystyle=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\underbrace{\left(\text{the
sum of all }k\text{-bounded }i\text{-monomials in the variables
}x_{1},x_{2},x_{3},\ldots\right)}_{\begin{subarray}{c}=G\left(k,i\right)\\\
\text{(by (\ref{pf.thm.DeltaGkm.Gki=}))}\end{subarray}}$ $\displaystyle\ \ \ \
\ \ \ \ \ \ \cdot\underbrace{\left(\text{the sum of all }k\text{-bounded
}\left(m-i\right)\text{-monomials in the variables
}y_{1},y_{2},y_{3},\ldots\right)}_{\begin{subarray}{c}=\left(G\left(k,m-i\right)\right)\left(\mathbf{y}\right)\\\
\text{(by (\ref{pf.thm.DeltaGkm.Gkm-iy=}))}\end{subarray}}$ $\displaystyle\ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.thm.DeltaGkm.prod1})}\right)$
$\displaystyle=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\underbrace{G\left(k,i\right)}_{=\left(G\left(k,i\right)\right)\left(\mathbf{x}\right)}\cdot\left(G\left(k,m-i\right)\right)\left(\mathbf{y}\right)=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}\left(G\left(k,i\right)\right)\left(\mathbf{x}\right)\cdot\left(G\left(k,m-i\right)\right)\left(\mathbf{y}\right).$

Hence, (10) holds for $f=G\left(k,m\right)$,
$I=\left\\{0,1,\ldots,m\right\\}$, $\left(f_{1,i}\right)_{i\in
I}=\left(G\left(k,i\right)\right)_{i\in\left\\{0,1,\ldots,m\right\\}}$ and
$\left(f_{2,i}\right)_{i\in
I}=\left(G\left(k,m-i\right)\right)_{i\in\left\\{0,1,\ldots,m\right\\}}$.
Therefore, (9) (applied to these $f$, $I$, $\left(f_{1,i}\right)_{i\in I}$ and
$\left(f_{2,i}\right)_{i\in I}$) yields

$\Delta\left(G\left(k,m\right)\right)=\sum\limits_{i\in\left\\{0,1,\ldots,m\right\\}}G\left(k,i\right)\otimes
G\left(k,m-i\right)=\sum\limits_{i=0}^{m}G\left(k,i\right)\otimes
G\left(k,m-i\right).$

This proves Theorem 2.18.

###  3.13 Proof of Theorem 2.20

######  Proof 3.35 (Proof of Theorem 2.20.).

Consider the ring
$\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]$
of formal power series in one indeterminate $t$ over
$\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]$. We equip this
ring with the topology that is obtained by identifying it with
$\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots,t\right]\right]$ (or,
equivalently, which is obtained by considering
$\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]$ itself as
equipped with the standard topology on a ring of formal power series, and then
adjoining the extra indeterminate $t$).

Now, consider the map

$\displaystyle\mathbf{F}_{k}:\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]$
$\displaystyle\rightarrow\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right],$
$\displaystyle a$ $\displaystyle\mapsto
a\left(x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right).$

This map $\mathbf{F}_{k}$ is a continuous $\mathbf{k}$-algebra homomorphism
(since it is an evaluation homomorphism)232323It is well-defined, since $k$ is
positive.. Hence, it induces a continuous242424Continuity is defined with
respect to the topology that we defined on
$\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]$.
$\mathbf{k}\left[\left[t\right]\right]$-algebra homomorphism

$\mathbf{F}_{k}\left[\left[t\right]\right]:\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]\rightarrow\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]$

that sends each formal power series $\sum\limits_{n\geq
0}a_{n}t^{n}\in\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]$
(with $a_{n}\in\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]$)
to $\sum\limits_{n\geq 0}\mathbf{F}_{k}\left(a_{n}\right)t^{n}$. Consider this
$\mathbf{k}\left[\left[t\right]\right]$-algebra homomorphism
$\mathbf{F}_{k}\left[\left[t\right]\right]$. In particular, it satisfies

$\left(\mathbf{F}_{k}\left[\left[t\right]\right]\right)\left(t^{i}\right)=t^{i}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\mathbb{N}.$

The definition of $\mathbf{F}_{k}$ yields

$\mathbf{F}_{k}\left(x_{i}\right)=x_{i}^{k}\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{1,2,3,\ldots\right\\}.$ (66)

Also, for each $a\in\Lambda$, we have

$\displaystyle\mathbf{F}_{k}\left(a\right)$
$\displaystyle=a\left(x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)\ \ \ \ \ \ \
\ \ \ \left(\text{by the definition of }\mathbf{F}_{k}\right)$
$\displaystyle=\mathbf{f}_{k}\left(a\right)$ (67)

(since the definition of $\mathbf{f}_{k}$ yields
$\mathbf{f}_{k}\left(a\right)=a\left(x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)$).
Thus, in particular, each $n\in\mathbb{N}$ satisfies

$\mathbf{F}_{k}\left(e_{n}\right)=\mathbf{f}_{k}\left(e_{n}\right)$ (68)

(by (67), applied to $a=e_{n}$).

Applying the map $\mathbf{F}_{k}\left[\left[t\right]\right]$ to both sides of
the equality (53), we obtain

$\left(\mathbf{F}_{k}\left[\left[t\right]\right]\right)\left(\prod\limits_{i=1}^{\infty}\left(1+x_{i}t\right)\right)=\left(\mathbf{F}_{k}\left[\left[t\right]\right]\right)\left(\sum\limits_{n\geq
0}e_{n}t^{n}\right)=\sum\limits_{n\geq
0}\mathbf{F}_{k}\left(e_{n}\right)t^{n}$

(by the definition of $\mathbf{F}_{k}\left[\left[t\right]\right]$). Hence,

$\displaystyle\sum\limits_{n\geq 0}\mathbf{F}_{k}\left(e_{n}\right)t^{n}$
$\displaystyle=\left(\mathbf{F}_{k}\left[\left[t\right]\right]\right)\left(\prod\limits_{i=1}^{\infty}\left(1+x_{i}t\right)\right)=\prod\limits_{i=1}^{\infty}\underbrace{\left(\mathbf{F}_{k}\left[\left[t\right]\right]\right)\left(1+x_{i}t\right)}_{\begin{subarray}{c}=1+\mathbf{F}_{k}\left(x_{i}\right)t\\\
\text{(by the definition of
}\mathbf{F}_{k}\left[\left[t\right]\right]\text{)}\end{subarray}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since
}\mathbf{F}_{k}\left[\left[t\right]\right]\text{ is a continuous
}\mathbf{k}\left[\left[t\right]\right]\text{-algebra homomorphism,}\\\
\text{and thus respects infinite products}\end{array}\right)$
$\displaystyle=\prod\limits_{i=1}^{\infty}\left(1+\underbrace{\mathbf{F}_{k}\left(x_{i}\right)}_{\begin{subarray}{c}=x_{i}^{k}\\\
\text{(by
(\ref{pf.thm.G.frob.Fkxi}))}\end{subarray}}t\right)=\prod\limits_{i=1}^{\infty}\left(1+x_{i}^{k}t\right).$

Substituting $-t^{k}$ for $t$ in this equality, we find

$\displaystyle\sum\limits_{n\geq
0}\mathbf{F}_{k}\left(e_{n}\right)\left(-t^{k}\right)^{n}$
$\displaystyle=\prod\limits_{i=1}^{\infty}\underbrace{\left(1+x_{i}^{k}\left(-t^{k}\right)\right)}_{\begin{subarray}{c}=1-\left(x_{i}t\right)^{k}\\\
=\left(1-x_{i}t\right)\left(\left(x_{i}t\right)^{0}+\left(x_{i}t\right)^{1}+\cdots+\left(x_{i}t\right)^{k-1}\right)\\\
\text{(since
}1-u^{k}=\left(1-u\right)\left(u^{0}+u^{1}+\cdots+u^{k-1}\right)\\\ \text{for
any element }u\text{ of any ring)}\end{subarray}}$
$\displaystyle=\prod\limits_{i=1}^{\infty}\left(\left(1-x_{i}t\right)\left(\left(x_{i}t\right)^{0}+\left(x_{i}t\right)^{1}+\cdots+\left(x_{i}t\right)^{k-1}\right)\right)$
$\displaystyle=\left(\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)\right)\left(\prod\limits_{i=1}^{\infty}\left(\left(x_{i}t\right)^{0}+\left(x_{i}t\right)^{1}+\cdots+\left(x_{i}t\right)^{k-1}\right)\right).$

We can divide both sides of this equality by
$\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)$ (since the formal power
series $\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)$ has constant term
$1$ and thus is invertible), and thus obtain

$\displaystyle\dfrac{\sum\limits_{n\geq
0}\mathbf{F}_{k}\left(e_{n}\right)\left(-t^{k}\right)^{n}}{\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)}$
$\displaystyle=\prod\limits_{i=1}^{\infty}\underbrace{\left(\left(x_{i}t\right)^{0}+\left(x_{i}t\right)^{1}+\cdots+\left(x_{i}t\right)^{k-1}\right)}_{=\sum\limits_{u=0}^{k-1}\left(x_{i}t\right)^{u}}=\prod\limits_{i=1}^{\infty}\sum\limits_{u=0}^{k-1}\left(x_{i}t\right)^{u}$
$\displaystyle=\underbrace{\sum\limits_{\begin{subarray}{c}\alpha=\left(\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)\in\left\\{0,1,\ldots,k-1\right\\}^{\infty};\\\
\alpha_{i}=0\text{ for all but finitely many
}i\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\alpha\in\left\\{0,1,\ldots,k-1\right\\}^{\infty}\\\
\text{is a weak composition}\end{subarray}}\\\
=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all }i\end{subarray}}\end{subarray}}\ \
\underbrace{\left(x_{1}t\right)^{\alpha_{1}}\left(x_{2}t\right)^{\alpha_{2}}\left(x_{3}t\right)^{\alpha_{3}}\cdots}_{\begin{subarray}{c}=\left(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots\right)\left(t^{\alpha_{1}}t^{\alpha_{2}}t^{\alpha_{3}}\cdots\right)\end{subarray}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we have expanded the
product}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all }i\end{subarray}}\ \
\underbrace{\left(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots\right)}_{\begin{subarray}{c}=\mathbf{x}^{\alpha}\\\
\text{(by the definition of }\mathbf{x}^{\alpha}\text{)}\end{subarray}}\ \
\underbrace{\left(t^{\alpha_{1}}t^{\alpha_{2}}t^{\alpha_{3}}\cdots\right)}_{\begin{subarray}{c}=t^{\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots}=t^{\left|\alpha\right|}\\\
\text{(since
}\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots=\left|\alpha\right|\text{)}\end{subarray}}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}t^{\left|\alpha\right|}.$

Hence,

$\displaystyle\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}t^{\left|\alpha\right|}$
$\displaystyle=\dfrac{\sum\limits_{n\geq
0}\mathbf{F}_{k}\left(e_{n}\right)\left(-t^{k}\right)^{n}}{\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)}=\left(\sum\limits_{n\geq
0}\underbrace{\mathbf{F}_{k}\left(e_{n}\right)}_{\begin{subarray}{c}=\mathbf{f}_{k}\left(e_{n}\right)\\\
\text{(by (\ref{pf.thm.G.frob.Fken}))}\end{subarray}}\ \
\underbrace{\left(-t^{k}\right)^{n}}_{=\left(-1\right)^{n}t^{kn}}\right)\cdot\underbrace{\prod\limits_{i=1}^{\infty}\left(1-x_{i}t\right)^{-1}}_{\begin{subarray}{c}=\sum\limits_{n\geq
0}h_{n}t^{n}\\\ \text{(by (\ref{pf.lem.h-e-reciprocity-t.h}))}\end{subarray}}$
$\displaystyle=\left(\sum\limits_{n\geq
0}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}t^{kn}\right)\cdot\underbrace{\left(\sum\limits_{n\geq
0}h_{n}t^{n}\right)}_{=\sum\limits_{j\geq
0}h_{j}t^{j}}=\left(\sum\limits_{n\geq
0}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}t^{kn}\right)\cdot\left(\sum\limits_{j\geq
0}h_{j}t^{j}\right)$ $\displaystyle=\underbrace{\sum\limits_{n\geq
0}\sum\limits_{j\geq
0}}_{=\sum\limits_{\left(n,j\right)\in\mathbb{N}^{2}}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}\underbrace{t^{kn}h_{j}t^{j}}_{=h_{j}t^{kn+j}}=\sum\limits_{\left(n,j\right)\in\mathbb{N}^{2}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}h_{j}t^{kn+j}.$

This is an equality between two power series in
$\left(\mathbf{k}\left[\left[x_{1},x_{2},x_{3},\ldots\right]\right]\right)\left[\left[t\right]\right]$.
If we compare the coefficients of $t^{m}$ on both sides of it (where
$x_{1},x_{2},x_{3},\ldots$ are considered scalars, not monomials), we obtain

$\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all }i;\\\
\left|\alpha\right|=m\end{subarray}}\mathbf{x}^{\alpha}=\underbrace{\sum\limits_{\begin{subarray}{c}\left(n,j\right)\in\mathbb{N}^{2};\\\
kn+j=m\end{subarray}}}_{=\sum\limits_{n\in\mathbb{N}}\ \
\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}h_{j}=\sum\limits_{n\in\mathbb{N}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}\cdot\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}.$

Now, the definition of $G\left(k,m\right)$ yields

$\displaystyle G\left(k,m\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=m;\\\ \alpha_{i}<k\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\alpha_{i}<k\text{ for all }i;\\\
\left|\alpha\right|=m\end{subarray}}\mathbf{x}^{\alpha}$
$\displaystyle=\sum\limits_{n\in\mathbb{N}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}\cdot\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}.$ (69)

But the right hand side of this equality can be simplified. Namely, for each
$n\in\mathbb{N}$, we have

$\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}=h_{m-kn}.$ (70)

[Proof of (70): Let $n\in\mathbb{N}$. We must prove the equality (70). If
$m-kn<0$, then $h_{m-kn}=0$ and
$\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}=\left(\text{empty sum}\right)=0$. Thus, if
$m-kn<0$, then (70) boils down to $0=0$, which is obviously true. Therefore,
for the rest of the proof of (70), we WLOG assume that $m-kn\geq 0$. Hence,
the sum $\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}$ has exactly one addend, namely the addend for
$j=m-kn$. Therefore, $\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}=h_{m-kn}$. This proves (70).]

Now, (69) becomes

$\displaystyle G\left(k,m\right)$
$\displaystyle=\sum\limits_{n\in\mathbb{N}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}\cdot\underbrace{\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
kn+j=m\end{subarray}}h_{j}}_{\begin{subarray}{c}=h_{m-kn}\\\ \text{(by
(\ref{pf.thm.G.frob.16}))}\end{subarray}}=\sum\limits_{n\in\mathbb{N}}\mathbf{f}_{k}\left(e_{n}\right)\left(-1\right)^{n}\cdot
h_{m-kn}$
$\displaystyle=\sum\limits_{n\in\mathbb{N}}\left(-1\right)^{n}h_{m-kn}\cdot\mathbf{f}_{k}\left(e_{n}\right)=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)$

(here, we have renamed the summation index $n$ as $i$). This proves Theorem
2.20.

###  3.14 Proofs of the results from Section 2.8

We shall now prove the results from Section 2.8. We begin with Lemma 2.22.
This will rely on the Verschiebung endomorphisms $\mathbf{v}_{n}$ introduced
in Definition 2.24, and on Proposition 2.25 and the equality (11).

######  Proof 3.36 (Proof of Lemma 2.22.).

Applying (11) to $n=k$, we obtain

$\mathbf{v}_{k}\left(p_{m}\right)=\begin{cases}kp_{m/k},&\text{if }k\mid m;\\\
0,&\text{if }k\nmid m\end{cases}\ \ \ \ .$ (71)

Applying Proposition 2.25 to $n=k$, $a=p_{m}$ and $b=e_{j}$, we obtain

$\left(p_{m},\mathbf{f}_{k}\left(e_{j}\right)\right)=\left(\mathbf{v}_{k}\left(p_{m}\right),e_{j}\right).$
(72)

Now, we are in one of the following three cases:

Case 1: We have $m=kj$.

Case 2: We have $k\nmid m$.

Case 3: We have neither $m=kj$ nor $k\nmid m$.

Let us first consider Case 1. In this case, we have $m=kj$. Thus, $k\mid m$
(since $j\in\mathbb{N}\subseteq\mathbb{Z}$) and $m/k=j$. Hence, $j=m/k$, so
that the integer $j$ is positive (since $m$ and $k$ are positive). But (71)
becomes

$\displaystyle\mathbf{v}_{k}\left(p_{m}\right)$
$\displaystyle=\begin{cases}kp_{m/k},&\text{if }k\mid m;\\\ 0,&\text{if
}k\nmid m\end{cases}\ \ \ \ =kp_{m/k}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}k\mid m\right)$ $\displaystyle=kp_{j}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}m/k=j\right).$

Thus, (72) becomes

$\displaystyle\left(p_{m},\mathbf{f}_{k}\left(e_{j}\right)\right)$
$\displaystyle=\left(\underbrace{\mathbf{v}_{k}\left(p_{m}\right)}_{=kp_{j}},e_{j}\right)=\left(kp_{j},e_{j}\right)$
$\displaystyle=\left(e_{j},kp_{j}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{since
the Hall inner product is symmetric}\right)$
$\displaystyle=k\underbrace{\left(e_{j},p_{j}\right)}_{\begin{subarray}{c}=\left(-1\right)^{j-1}\\\
\text{(by Proposition \ref{prop.ejpj}, applied to
}n=j\text{)}\end{subarray}}=k\left(-1\right)^{j-1}=\left(-1\right)^{j-1}k.$

Comparing this with

$\left(-1\right)^{j-1}\underbrace{\left[m=kj\right]}_{\begin{subarray}{c}=1\\\
\text{(since }m=kj\text{)}\end{subarray}}k=\left(-1\right)^{j-1}k,$

we obtain
$\left(p_{m},\mathbf{f}_{k}\left(e_{j}\right)\right)=\left(-1\right)^{j-1}\left[m=kj\right]k$.
Thus, Lemma 2.22 is proven in Case 1.

A similar (but simpler) argument can be used to prove Lemma 2.22 in Case 2.
The main “idea” here is that $k\nmid m$ implies $m\neq kj$. The details are
left to the reader.

Let us finally consider Case 3. In this case, we have neither $m=kj$ nor
$k\nmid m$. In other words, we have $m\neq kj$ and $k\mid m$. From $k\mid m$,
we conclude that $m/k$ is a positive integer252525Indeed, it is positive since
$m$ and $k$ are positive.. From $m\neq kj$, we obtain $m/k\neq j$. Thus, the
symmetric functions $p_{m/k}$ and $e_{j}$ are homogeneous of different
degrees262626since $p_{m/k}$ is homogeneous of degree $m/k$, whereas $e_{j}$
is homogeneous of degree $j$, and therefore satisfy
$\left(p_{m/k},e_{j}\right)=0$ (by (2), applied to $f=p_{m/k}$ and $g=e_{j}$).

Now, (71) becomes

$\mathbf{v}_{k}\left(p_{m}\right)=\begin{cases}kp_{m/k},&\text{if }k\mid m;\\\
0,&\text{if }k\nmid m\end{cases}\ \ \ \ =kp_{m/k}\ \ \ \ \ \ \ \ \ \
\left(\text{since }k\mid m\right).$

Thus, (72) becomes

$\left(p_{m},\mathbf{f}_{k}\left(e_{j}\right)\right)=\left(\underbrace{\mathbf{v}_{k}\left(p_{m}\right)}_{=kp_{m/k}},e_{j}\right)=\left(kp_{m/k},e_{j}\right)=k\underbrace{\left(p_{m/k},e_{j}\right)}_{=0}=0.$

Comparing this with

$\left(-1\right)^{j-1}\underbrace{\left[m=kj\right]}_{\begin{subarray}{c}=0\\\
\text{(since }m\neq kj\text{)}\end{subarray}}k=0,$

we obtain
$\left(p_{m},\mathbf{f}_{k}\left(e_{j}\right)\right)=\left(-1\right)^{j-1}\left[m=kj\right]k$.
Thus, Lemma 2.22 is proven in Case 3.

We have thus proven Lemma 2.22 in all three Cases 1, 2 and 3. Thus, Lemma 2.22
always holds.

Next, let us prove a simple property of Hall inner products:

######  Lemma 3.37.

Let $m$, $\alpha$ and $\beta$ be positive integers. Let $a$ be a homogeneous
symmetric function of degree $\alpha$. Let $b$ be a homogeneous symmetric
function of degree $\beta$. Then, $\left(p_{m},ab\right)=0$.

######  Proof 3.38 (Proof of Lemma 3.37.).

For each $n\in\mathbb{N}$, let $\Lambda_{n}$ denote the $n$-th homogeneous
component of the graded $\mathbf{k}$-algebra $\Lambda$. Thus,
$a\in\Lambda_{\alpha}$ and $b\in\Lambda_{\beta}$ (since $a$ and $b$ are
homogeneous symmetric functions of degrees $\alpha$ and $\beta$).

But it is known that the family
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}}$ is a graded basis
of the graded $\mathbf{k}$-module $\Lambda$; this means that for each
$n\in\mathbb{N}$, its subfamily
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}\nolimits_{n}}$ is a
basis of the $\mathbf{k}$-module $\Lambda_{n}$. 272727This follows, e.g., from
[GriRei20, Corollary 2.5.17(a)] (which shows that
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}}$ is a basis of
$\Lambda$) and from the fact that each $h_{\lambda}$ is homogeneous of degree
$\left|\lambda\right|$. Applying this to $n=\alpha$, we conclude that the
subfamily
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}\nolimits_{\alpha}}$
is a basis of $\Lambda_{\alpha}$. Hence, $a$ is a $\mathbf{k}$-linear
combination of this family
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}\nolimits_{\alpha}}$
(since $a\in\Lambda_{\alpha}$).

We must prove the equality $\left(p_{m},ab\right)=0$. Both sides of this
equality depend $\mathbf{k}$-linearly on $a$. Thus, in proving it, we can WLOG
assume that $a$ belongs to the family
$\left(h_{\lambda}\right)_{\lambda\in\operatorname*{Par}\nolimits_{\alpha}}$
(because we know that $a$ is a $\mathbf{k}$-linear combination of this
family). In other words, we can WLOG assume that $a=h_{\lambda}$ for some
$\lambda\in\operatorname*{Par}\nolimits_{\alpha}$. Assume this. For similar
reasons, we can WLOG assume that $b=h_{\mu}$ for some
$\mu\in\operatorname*{Par}\nolimits_{\beta}$. Assume this, too. Consider these
two partitions $\lambda$ and $\mu$.

We have $\lambda\in\operatorname*{Par}\nolimits_{\alpha}$ and thus
$\left|\lambda\right|=\alpha>0$, so that $\lambda\neq\varnothing$. Hence, the
partition $\lambda$ has at least one part. Likewise, the partition $\mu$ has
at least one part.

Now, let $\lambda\sqcup\mu$ be the partition obtained by listing all parts of
$\lambda$ and of $\mu$ and sorting the resulting list in weakly decreasing
order.282828For example: If $\lambda=\left(5,3,2\right)$ and
$\mu=\left(6,4,3,1,1\right)$, then
$\lambda\sqcup\mu=\left(6,5,4,3,3,2,1,1\right)$. Using Definition 3.1, we can
easily see that $h_{\lambda\sqcup\mu}=h_{\lambda}h_{\mu}$. Comparing this with
$\underbrace{a}_{=h_{\lambda}}\underbrace{b}_{=h_{\mu}}=h_{\lambda}h_{\mu}$,
we obtain $ab=h_{\lambda\sqcup\mu}$.

But the partition $\lambda\sqcup\mu$ has as many parts as $\lambda$ and $\mu$
have combined. Thus, the partition $\lambda\sqcup\mu$ has at least $2$ parts
(since $\lambda$ has at least one part, and $\mu$ has at least one part).
Therefore, $\lambda\sqcup\mu\neq\left(m\right)$ (since the partition
$\lambda\sqcup\mu$ has at least $2$ parts, while the partition
$\left(m\right)$ has only $1$ part). Now, recall that
$p_{m}=m_{\left(m\right)}$ (where, of course, the two “$m$”s in
“$m_{\left(m\right)}$” mean completely unrelated things). Thus,

$\displaystyle\left(\underbrace{p_{m}}_{=m_{\left(m\right)}},\underbrace{ab}_{=h_{\lambda\sqcup\mu}}\right)$
$\displaystyle=\left(m_{\left(m\right)},h_{\lambda\sqcup\mu}\right)=\left(h_{\lambda\sqcup\mu},m_{\left(m\right)}\right)\
\ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since the Hall inner
product}\\\ \text{is symmetric}\end{array}\right)$
$\displaystyle=\delta_{\lambda\sqcup\mu,\left(m\right)}\ \ \ \ \ \ \ \ \ \
\left(\text{by (\ref{eq.hlam-mlam-dual})}\right)$ $\displaystyle=0\ \ \ \ \ \
\ \ \ \ \left(\text{since }\lambda\sqcup\mu\neq\left(m\right)\right).$

This proves Lemma 3.37.

See [Grinbe20b] for a different proof of Lemma 3.37, using the graded dual
$\Lambda^{o}$ of the Hopf algebra $\Lambda$ and the primitivity of the element
$p_{m}\in\Lambda$.

We can now prove Proposition 2.23:

######  Proof 3.39 (Proof of Proposition 2.23.).

Theorem 2.20 yields
$G\left(k,m\right)=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)$.
Hence,

$\displaystyle\left(p_{m},G\left(k,m\right)\right)$
$\displaystyle=\left(p_{m},\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}\left(p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)$
(73)

(since the Hall inner product is $\mathbf{k}$-bilinear).

Now, we claim that every $i\in\mathbb{N}\setminus\left\\{0,m/k\right\\}$
satisfies

$\left(p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)=0.$ (74)

[Proof of (74): Let $i\in\mathbb{N}\setminus\left\\{0,m/k\right\\}$. Thus,
$i\in\mathbb{N}$ and $i\notin\left\\{0,m/k\right\\}$. From
$i\notin\left\\{0,m/k\right\\}$, we obtain $i\neq 0$ and $i\neq m/k$. From
$i\neq m/k$, we obtain $ki\neq m$, so that $m-ki\neq 0$.

We must prove the equality (74). If $m-ki<0$, then $h_{m-ki}=0$, and therefore

$\left(p_{m},\underbrace{h_{m-ki}}_{=0}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)=\left(p_{m},0\right)=0.$

Hence, the equality (74) is proven if $m-ki<0$. Thus, for the rest of this
proof, we WLOG assume that $m-ki\geq 0$. Combining this with $m-ki\neq 0$, we
obtain $m-ki>0$. Thus, $m-ki$ is a positive integer. Also, $i$ is a positive
integer (since $i\in\mathbb{N}$ and $i\neq 0$), and thus $ki$ is a positive
integer (since $k$ is a positive integer).

The map $\mathbf{f}_{k}:\Lambda\rightarrow\Lambda$ operates by replacing each
$x_{i}$ by $x_{i}^{k}$ in a symmetric function (by the definition of
$\mathbf{f}_{k}$). Thus, if $g\in\Lambda$ is any homogeneous symmetric
function of some degree $\gamma$, then $\mathbf{f}_{k}\left(g\right)$ is a
homogeneous symmetric function of degree $k\gamma$. Applying this to $g=e_{i}$
and $\gamma=i$, we conclude that $\mathbf{f}_{k}\left(e_{i}\right)$ is a
homogeneous symmetric function of degree $ki$ (since $e_{i}$ is a homogeneous
symmetric function of degree $i$). Also, $h_{m-ki}$ is a homogeneous symmetric
function of degree $m-ki$.

Hence, Lemma 3.37 (applied to $\alpha=m-ki$, $a=h_{m-ki}$, $\beta=ki$ and
$b=\mathbf{f}_{k}\left(e_{i}\right)$) yields
$\left(p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)=0$. This
proves (74).]

Note that $e_{0}=1$ and thus
$\mathbf{f}_{k}\left(e_{0}\right)=\mathbf{f}_{k}\left(1\right)=1$ (by the
definition of $\mathbf{f}_{k}$).

Note that $m/k>0$ (since $m$ and $k$ are positive). Hence, $m/k\neq 0$. Now,
we are in one of the following two cases:

Case 1: We have $k\mid m$.

Case 2: We have $k\nmid m$.

Let us consider Case 1 first. In this case, we have $k\mid m$. Hence, $m/k$ is
a positive integer (since $m$ and $k$ are positive integers). Thus, $0$ and
$m/k$ are two distinct elements of $\mathbb{N}$ (indeed, they are distinct
because $m/k\neq 0$). Lemma 2.22 (applied to $j=m/k$) yields

$\left(p_{m},\mathbf{f}_{k}\left(e_{m/k}\right)\right)=\left(-1\right)^{m/k-1}\underbrace{\left[m=k\left(m/k\right)\right]}_{\begin{subarray}{c}=1\\\
\text{(since
}m=k\left(m/k\right)\text{)}\end{subarray}}k=\left(-1\right)^{m/k-1}k.$

Now, (73) becomes

$\displaystyle\left(p_{m},G\left(k,m\right)\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}\left(p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)$
$\displaystyle=\underbrace{\left(-1\right)^{0}}_{=1}\left(p_{m},\underbrace{h_{m-k\cdot
0}}_{=h_{m}}\cdot\underbrace{\mathbf{f}_{k}\left(e_{0}\right)}_{=1}\right)+\left(-1\right)^{m/k}\left(p_{m},\underbrace{h_{m-k\cdot
m/k}}_{\begin{subarray}{c}=h_{0}\\\ \text{(since }m-k\cdot
m/k=0\text{)}\end{subarray}}\cdot\mathbf{f}_{k}\left(e_{m/k}\right)\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i\in\mathbb{N}\setminus\left\\{0,m/k\right\\}}\left(-1\right)^{i}\underbrace{\left(p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\right)}_{\begin{subarray}{c}=0\\\
\text{(by (\ref{pf.prop.hall-pmGkm.zeroes}))}\end{subarray}}$ $\displaystyle\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we
have split off the addends for }i=0\text{ and for }i=m/k\\\ \text{from the sum
(since }0\text{ and }m/k\text{ are two distinct elements of
}\mathbb{N}\text{)}\end{array}\right)$
$\displaystyle=\underbrace{\left(p_{m},h_{m}\right)}_{\begin{subarray}{c}=\left(h_{m},p_{m}\right)\\\
\text{(since the Hall inner}\\\ \text{product is
symmetric)}\end{subarray}}+\left(-1\right)^{m/k}\left(p_{m},\underbrace{h_{0}}_{=1}\cdot\mathbf{f}_{k}\left(e_{m/k}\right)\right)+\underbrace{\sum\limits_{i\in\mathbb{N}\setminus\left\\{0,m/k\right\\}}\left(-1\right)^{i}0}_{=0}$
$\displaystyle=\underbrace{\left(h_{m},p_{m}\right)}_{\begin{subarray}{c}=1\\\
\text{(by Proposition \ref{prop.hjpj},}\\\ \text{applied to
}n=m\text{)}\end{subarray}}+\left(-1\right)^{m/k}\underbrace{\left(p_{m},\mathbf{f}_{k}\left(e_{m/k}\right)\right)}_{=\left(-1\right)^{m/k-1}k}$
$\displaystyle=1+\underbrace{\left(-1\right)^{m/k}\left(-1\right)^{m/k-1}}_{=-1}k=1-k.$

Comparing this with

$1-\underbrace{\left[k\mid m\right]}_{\begin{subarray}{c}=1\\\ \text{(since
}k\mid m\text{)}\end{subarray}}k=1-k,$

we obtain $\left(p_{m},G\left(k,m\right)\right)=1-\left[k\mid m\right]k$.
Hence, Proposition 2.23 is proven in Case 1.

Case 2 is similar to Case 1, but simpler because we no longer need to split
off the addend for $i=m/k$ from the sum (since $m/k\notin\mathbb{N}$ in this
case). We leave it to the reader.

We have now proven Proposition 2.23 both in Case 1 and in Case 2. Hence,
Proposition 2.23 always holds.

Theorem 2.21 will follow from Proposition 2.23 using the following general
criterion for generating sets of $\Lambda$:

######  Proposition 3.40.

For each positive integer $m$, let $v_{m}\in\Lambda$ be a homogeneous
symmetric function of degree $m$.

Assume that $\left(p_{m},v_{m}\right)$ is an invertible element of
$\mathbf{k}$ for each positive integer $m$.

Then, the family $\left(v_{m}\right)_{m\geq
1}=\left(v_{1},v_{2},v_{3},\ldots\right)$ is an algebraically independent
generating set of the commutative $\mathbf{k}$-algebra $\Lambda$.

######  Proof 3.41 (Proof of Proposition 3.40.).

Proposition 3.40 is [GriRei20, Exercise 2.5.24].

######  Proof 3.42 (Proof of Theorem 2.21.).

Let $m$ be a positive integer. Proposition 2.23 yields that

$\displaystyle\left(p_{m},G\left(k,m\right)\right)$
$\displaystyle=1-\underbrace{\left[k\mid m\right]}_{=\begin{cases}1,&\text{if
}k\mid m;\\\ 0,&\text{if }k\nmid m\end{cases}}k=1-\begin{cases}1,&\text{if
}k\mid m;\\\ 0,&\text{if }k\nmid m\end{cases}\ \ \ \cdot k$
$\displaystyle=\begin{cases}1-1\cdot k,&\text{if }k\mid m;\\\ 1-0\cdot
k,&\text{if }k\nmid m\end{cases}\ \ \ \ =\begin{cases}1-k,&\text{if }k\mid
m;\\\ 1,&\text{if }k\nmid m\end{cases}\ \ \ \ .$

Hence, $\left(p_{m},G\left(k,m\right)\right)$ is an invertible element of
$\mathbf{k}$ (because both $1-k$ and $1$ are invertible elements of
$\mathbf{k}$).

Forget that we fixed $m$. We thus have showed that
$\left(p_{m},G\left(k,m\right)\right)$ is an invertible element of
$\mathbf{k}$ for each positive integer $m$. Also, clearly, for each positive
integer $m$, the element $G\left(k,m\right)\in\Lambda$ is a homogeneous
symmetric function of degree $m$. Thus, Proposition 3.40 (applied to
$v_{m}=G\left(k,m\right)$) shows that the family
$\left(G\left(k,m\right)\right)_{m\geq
1}=\left(G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots\right)$
is an algebraically independent generating set of the commutative
$\mathbf{k}$-algebra $\Lambda$. This proves Theorem 2.21.

###  3.15 Proof of Theorem 2.27

######  Proof 3.43 (Proof of Theorem 2.27.).

The $\mathbf{k}$-Hopf algebra $\Lambda$ is both commutative and cocommutative
(by [GriRei20, Exercise 2.3.7(a)]). Thus, its antipode $S$ is a
$\mathbf{k}$-Hopf algebra homomorphism292929Proof. The antipode of a Hopf
algebra is an algebra anti-endomorphism (by [GriRei20, Proposition 1.4.8]).
Thus, $S$ is an algebra anti-endomorphism (since $S$ is the antipode of the
Hopf algebra $\Lambda$). But since $\Lambda$ is commutative, an algebra anti-
endomorphism of $\Lambda$ is the same thing as an algebra endomorphism of
$\Lambda$. Hence, $S$ is an algebra endomorphism of $\Lambda$ (since $S$ is an
algebra anti-endomorphism of $\Lambda$). The antipode of a Hopf algebra is a
coalgebra anti-endomorphism (by [GriRei20, Exercise 1.4.25]). Thus, $S$ is a
coalgebra anti-endomorphism (since $S$ is the antipode of the Hopf algebra
$\Lambda$). But since $\Lambda$ is cocommutative, a coalgebra anti-
endomorphism of $\Lambda$ is the same thing as a coalgebra endomorphism of
$\Lambda$. Hence, $S$ is a coalgebra endomorphism of $\Lambda$ (since $S$ is a
coalgebra anti-endomorphism of $\Lambda$). We now know that $S$ is an algebra
endomorphism of $\Lambda$ and a coalgebra endomorphism of $\Lambda$ at the
same time. In other words, $S$ is a bialgebra endomorphism of $\Lambda$.
Hence, $S$ is a $\mathbf{k}$-Hopf algebra endomorphism of $\Lambda$. In other
words, $S$ is a $\mathbf{k}$-Hopf algebra homomorphism..

(a) The map $\mathbf{f}_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (by
[GriRei20, Exercise 2.9.9(d)], applied to $n=k$). The map $\mathbf{v}_{k}$ is
a $\mathbf{k}$-Hopf algebra homomorphism (by [GriRei20, Exercise 2.9.10(e)],
applied to $n=k$). Thus, we have shown that all three maps $\mathbf{f}_{k}$,
$S$ and $\mathbf{v}_{k}$ are $\mathbf{k}$-Hopf algebra homomorphisms. Hence,
their composition $\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}$ is a
$\mathbf{k}$-Hopf algebra homomorphism as well. In other words, $U_{k}$ is a
$\mathbf{k}$-Hopf algebra homomorphism (since $U_{k}=\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}$). This proves Theorem 2.27 (a).

(b) Recall (from [GriRei20, Exercise 1.5.9(a)]) the following fact:

> Claim 1: If $H$ is a $\mathbf{k}$-bialgebra and $A$ is a commutative
> $\mathbf{k}$-algebra, then the convolution $f\star g$ of any two
> $\mathbf{k}$-algebra homomorphisms $f,g:H\rightarrow A$ is again a
> $\mathbf{k}$-algebra homomorphism.

The following fact is dual to Claim 1:

> Claim 2: If $H$ is a $\mathbf{k}$-bialgebra and $C$ is a cocommutative
> $\mathbf{k}$-coalgebra, then the convolution $f\star g$ of any two
> $\mathbf{k}$-coalgebra homomorphisms $f,g:C\rightarrow H$ is again a
> $\mathbf{k}$-coalgebra homomorphism.

(See [GriRei20, solution to Exercise 1.5.9(h)] for why exactly Claim 2 is dual
to Claim 1, and how it can be proved.)

Theorem 2.27 (a) yields that the map $U_{k}$ is a $\mathbf{k}$-Hopf algebra
homomorphism. Hence, $U_{k}$ is both a $\mathbf{k}$-algebra homomorphism and a
$\mathbf{k}$-coalgebra homomorphism.

Now, recall that $\Lambda$ is commutative, and that
$\operatorname*{id}\nolimits_{\Lambda}$ and $U_{k}$ are two
$\mathbf{k}$-algebra homomorphisms from $\Lambda$ to $\Lambda$. Hence, Claim 1
(applied to $H=\Lambda$, $A=\Lambda$,
$f=\operatorname*{id}\nolimits_{\Lambda}$ and $g=U_{k}$) shows that the
convolution $\operatorname*{id}\nolimits_{\Lambda}\star U_{k}$ is a
$\mathbf{k}$-algebra homomorphism. In other words, $V_{k}$ is a
$\mathbf{k}$-algebra homomorphism (since
$V_{k}=\operatorname*{id}\nolimits_{\Lambda}\star U_{k}$).

Next, recall that $\Lambda$ is cocommutative, and that
$\operatorname*{id}\nolimits_{\Lambda}$ and $U_{k}$ are two
$\mathbf{k}$-coalgebra homomorphisms from $\Lambda$ to $\Lambda$. Hence, Claim
2 (applied to $H=\Lambda$, $C=\Lambda$,
$f=\operatorname*{id}\nolimits_{\Lambda}$ and $g=U_{k}$) shows that the
convolution $\operatorname*{id}\nolimits_{\Lambda}\star U_{k}$ is a
$\mathbf{k}$-coalgebra homomorphism. In other words, $V_{k}$ is a
$\mathbf{k}$-coalgebra homomorphism (since
$V_{k}=\operatorname*{id}\nolimits_{\Lambda}\star U_{k}$).

So we know that the map $V_{k}$ is both a $\mathbf{k}$-algebra homomorphism
and a $\mathbf{k}$-coalgebra homomorphism. Thus, $V_{k}$ is a
$\mathbf{k}$-bialgebra homomorphism, thus a $\mathbf{k}$-Hopf algebra
homomorphism. This proves Theorem 2.27 (b).

(c) The map $\mathbf{v}_{k}$ is a $\mathbf{k}$-algebra homomorphism; thus,
$\mathbf{v}_{k}\left(1\right)=1$. Now, we have

$\mathbf{v}_{k}\left(h_{m}\right)=\begin{cases}h_{m/k},&\text{if }k\mid m;\\\
0,&\text{if }k\nmid m\end{cases}$ (75)

for each $m\in\mathbb{N}$. (Indeed, if $m>0$, then this follows from the
definition of $\mathbf{v}_{k}$. But if $m=0$, then this follows from
$\mathbf{v}_{k}\left(1\right)=1$, since $h_{0}=1$.)

We have

$S\left(h_{n}\right)=\left(-1\right)^{n}e_{n}\ \ \ \ \ \ \ \ \ \ \text{for
each }n\in\mathbb{N}.$ (76)

(This follows from [GriRei20, Proposition 2.4.1(iii)].)

Each $i\in\mathbb{N}$ satisfies

$\displaystyle\mathbf{v}_{k}\left(h_{ki}\right)$
$\displaystyle=\begin{cases}h_{ki/k},&\text{if }k\mid ki;\\\ 0,&\text{if
}k\nmid ki\end{cases}\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.thm.Uk.main.c.vh}), applied to }m=ki\right)$ $\displaystyle=h_{ki/k}\
\ \ \ \ \ \ \ \ \ \left(\text{since }k\mid ki\right)$ $\displaystyle=h_{i}\ \
\ \ \ \ \ \ \ \ \left(\text{since }ki/k=i\right)$ (77)

and

$\displaystyle U_{k}\left(h_{ki}\right)$
$\displaystyle=\left(\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}\right)\left(h_{ki}\right)\ \ \ \ \ \ \ \ \ \
\left(\text{since }U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}\right)$
$\displaystyle=\mathbf{f}_{k}\left(S\left(\underbrace{\mathbf{v}_{k}\left(h_{ki}\right)}_{\begin{subarray}{c}=h_{i}\\\
\text{(by
(\ref{pf.thm.Uk.main.c.vkhj1}))}\end{subarray}}\right)\right)=\mathbf{f}_{k}\left(\underbrace{S\left(h_{i}\right)}_{\begin{subarray}{c}=\left(-1\right)^{i}e_{i}\\\
\text{(by
(\ref{pf.thm.Uk.main.c.Sh}))}\end{subarray}}\right)=\mathbf{f}_{k}\left(\left(-1\right)^{i}e_{i}\right)$
$\displaystyle=\left(-1\right)^{i}\mathbf{f}_{k}\left(e_{i}\right)$ (78)

(since the map $\mathbf{f}_{k}$ is $\mathbf{k}$-linear).

On the other hand, if $j\in\mathbb{N}$ satisfies $k\nmid j$, then

$\displaystyle\mathbf{v}_{k}\left(h_{j}\right)$
$\displaystyle=\begin{cases}h_{j/k},&\text{if }k\mid j;\\\ 0,&\text{if }k\nmid
j\end{cases}\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.Uk.main.c.vh}),
applied to }m=j\right)$ $\displaystyle=0\ \ \ \ \ \ \ \ \ \ \left(\text{since
}k\nmid j\right)$ (79)

and

$\displaystyle U_{k}\left(h_{j}\right)$
$\displaystyle=\left(\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}\right)\left(h_{j}\right)\ \ \ \ \ \ \ \ \ \
\left(\text{since }U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}\right)$
$\displaystyle=\left(\mathbf{f}_{k}\circ
S\right)\left(\underbrace{\mathbf{v}_{k}\left(h_{j}\right)}_{\begin{subarray}{c}=0\\\
\text{(by
(\ref{pf.thm.Uk.main.c.vkhj2}))}\end{subarray}}\right)=\left(\mathbf{f}_{k}\circ
S\right)\left(0\right)$ $\displaystyle=0$ (80)

(since the map $\mathbf{f}_{k}\circ S$ is $\mathbf{k}$-linear).

Let $\Delta_{\Lambda}$ be the comultiplication
$\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ of the $\mathbf{k}$-coalgebra
$\Lambda$. Let $m_{\Lambda}:\Lambda\otimes\Lambda\rightarrow\Lambda$ be the
$\mathbf{k}$-linear map sending each pure tensor $a\otimes
b\in\Lambda\otimes\Lambda$ to $ab\in\Lambda$. Definition 2.26 then yields
$\operatorname*{id}\nolimits_{\Lambda}\star
U_{k}=m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta_{\Lambda}$. Thus,

$\displaystyle V_{k}$
$\displaystyle=\operatorname*{id}\nolimits_{\Lambda}\star
U_{k}=m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\underbrace{\Delta_{\Lambda}}_{=\Delta}$
$\displaystyle=m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta.$ (81)

Let $m\in\mathbb{N}$ (not to be mistaken for the map $m_{\Lambda}$). Then,
[GriRei20, Proposition 2.3.6(iii)] (applied to $n=m$) yields

$\displaystyle\Delta\left(h_{m}\right)$
$\displaystyle=\sum\limits_{i+j=m}h_{i}\otimes h_{j}$ $\displaystyle\ \ \ \ \
\ \ \ \ \ \left(\text{where the sum ranges over all pairs
}\left(i,j\right)\in\mathbb{N}\times\mathbb{N}\text{ with }i+j=m\right)$
$\displaystyle=\sum\limits_{j=0}^{m}h_{m-j}\otimes h_{j}$

(here, we have substituted $\left(m-j,j\right)$ for $\left(i,j\right)$ in the
sum). Applying the map $\operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}$ to
both sides of this equality, we obtain

$\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(\Delta\left(h_{m}\right)\right)=\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(\sum\limits_{j=0}^{m}h_{m-j}\otimes
h_{j}\right)=\sum\limits_{j=0}^{m}h_{m-j}\otimes U_{k}\left(h_{j}\right).$

Applying the map $m_{\Lambda}$ to both sides of this equality, we find

$\displaystyle
m_{\Lambda}\left(\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(\Delta\left(h_{m}\right)\right)\right)$
$\displaystyle=m_{\Lambda}\left(\sum\limits_{j=0}^{m}h_{m-j}\otimes
U_{k}\left(h_{j}\right)\right)=\sum\limits_{j=0}^{m}h_{m-j}U_{k}\left(h_{j}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by the definition of }m_{\Lambda}\right)$
$\displaystyle=\underbrace{\sum\limits_{j=0}^{\infty}}_{=\sum\limits_{j\in\mathbb{N}}}h_{m-j}U_{k}\left(h_{j}\right)-\sum\limits_{j=m+1}^{\infty}\underbrace{h_{m-j}}_{\begin{subarray}{c}=0\\\
\text{(since }m-j<0\\\ \text{(because }j\geq
m+1>m\text{))}\end{subarray}}U_{k}\left(h_{j}\right)$
$\displaystyle=\sum\limits_{j\in\mathbb{N}}h_{m-j}U_{k}\left(h_{j}\right)-\underbrace{\sum\limits_{j=m+1}^{\infty}0U_{k}\left(h_{j}\right)}_{=0}$
$\displaystyle=\sum\limits_{j\in\mathbb{N}}h_{m-j}U_{k}\left(h_{j}\right)=\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
k\mid
j\end{subarray}}h_{m-j}U_{k}\left(h_{j}\right)+\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
k\nmid
j\end{subarray}}h_{m-j}\underbrace{U_{k}\left(h_{j}\right)}_{\begin{subarray}{c}=0\\\
\text{(by (\ref{pf.thm.Uk.main.c.Ukhj2}))}\end{subarray}}$ $\displaystyle\ \ \
\ \ \ \ \ \ \ \left(\text{since each }j\in\mathbb{N}\text{ satisfies either
}k\mid j\text{ or }k\nmid j\text{ (but not both)}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\ k\mid
j\end{subarray}}h_{m-j}U_{k}\left(h_{j}\right)+\underbrace{\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
k\nmid
j\end{subarray}}h_{m-j}0}_{=0}=\sum\limits_{\begin{subarray}{c}j\in\mathbb{N};\\\
k\mid
j\end{subarray}}h_{m-j}U_{k}\left(h_{j}\right)=\sum\limits_{i\in\mathbb{N}}h_{m-ki}\underbrace{U_{k}\left(h_{ki}\right)}_{\begin{subarray}{c}=\left(-1\right)^{i}\mathbf{f}_{k}\left(e_{i}\right)\\\
\text{(by (\ref{pf.thm.Uk.main.c.Ukhj1}))}\end{subarray}}$ $\displaystyle\ \ \
\ \ \ \ \ \ \ \left(\text{here, we have substituted }ki\text{ for }j\text{ in
the sum}\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right).$

Comparing this with

$G\left(k,m\right)=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m-ki}\cdot\mathbf{f}_{k}\left(e_{i}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by Theorem \ref{thm.G.frob}}\right),$

we obtain

$G\left(k,m\right)=m_{\Lambda}\left(\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(\Delta\left(h_{m}\right)\right)\right)=\underbrace{\left(m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta\right)}_{\begin{subarray}{c}=V_{k}\\\ \text{(by
(\ref{pf.thm.Uk.main.c.conv-
def}))}\end{subarray}}\left(h_{m}\right)=V_{k}\left(h_{m}\right).$

This proves Theorem 2.27 (c).

(d) Let us recall a few facts from [GriRei20].

From [GriRei20, Exercise 2.9.10(a)], we know that every positive integers $n$
and $m$ satisfy

$\mathbf{v}_{n}\left(p_{m}\right)=\begin{cases}np_{m/n},&\text{if }n\mid m;\\\
0,&\text{if }n\nmid m\end{cases}\ \ .$ (82)

On the other hand, it is easy to see (directly using the definition of
$\mathbf{f}_{n}$) that every positive integers $n$ and $m$ satisfy

$\mathbf{f}_{n}\left(p_{m}\right)=p_{nm}.$ (83)

Finally, [GriRei20, Proposition 2.4.1(i)] yields that every positive integer
$n$ satisfies

$S\left(p_{n}\right)=-p_{n}.$ (84)

Now, let $n$ be a positive integer. We first claim the following:

> Claim 1: We have $U_{k}\left(p_{n}\right)=-\left[k\mid n\right]kp_{n}$.

[Proof of Claim 1: We are in one of the following two cases:

Case 1: We have $k\mid n$.

Case 2: We have $k\nmid n$.

Let us first consider Case 1. In this case, we have $k\mid n$. Hence, $n/k$ is
a positive integer. Now, (82) (applied to $k$ and $n$ instead of $n$ and $m$)
yields

$\mathbf{v}_{k}\left(p_{n}\right)=\begin{cases}kp_{n/k},&\text{if }k\mid n;\\\
0,&\text{if }k\nmid n\end{cases}\ \ \ \ =kp_{n/k}\ \ \ \ \ \ \ \ \ \
\left(\text{since }k\mid n\right).$

Applying the map $S$ to both sides of this equality, we find

$\displaystyle S\left(\mathbf{v}_{k}\left(p_{n}\right)\right)$
$\displaystyle=S\left(kp_{n/k}\right)=k\underbrace{S\left(p_{n/k}\right)}_{\begin{subarray}{c}=-p_{n/k}\\\
\text{(by (\ref{pf.thm.Uk.main.Sp}),}\\\ \text{applied to }n/k\text{ instead
of }n\text{)}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{since the map
}S\text{ is }\mathbf{k}\text{-linear}\right)$
$\displaystyle=k\left(-p_{n/k}\right)=-kp_{n/k}.$

Applying the map $\mathbf{f}_{k}$ to both sides of this equality, we find

$\displaystyle\mathbf{f}_{k}\left(S\left(\mathbf{v}_{k}\left(p_{n}\right)\right)\right)$
$\displaystyle=\mathbf{f}_{k}\left(-kp_{n/k}\right)=-k\underbrace{\mathbf{f}_{k}\left(p_{n/k}\right)}_{\begin{subarray}{c}=p_{k\left(n/k\right)}\\\
\text{(by (\ref{pf.thm.Uk.main.short.fp}),}\\\ \text{applied to }k\text{ and
}n/k\\\ \text{instead of }n\text{ and }m\text{)}\end{subarray}}\ \ \ \ \ \ \ \
\ \ \left(\text{since the map }\mathbf{f}_{k}\text{ is
}\mathbf{k}\text{-linear}\right)$
$\displaystyle=-kp_{k\left(n/k\right)}=-kp_{n}.$

Now, the definition of $U_{k}$ yields $U_{k}=\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}$. Hence,

$U_{k}\left(p_{n}\right)=\left(\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}\right)\left(p_{n}\right)=\mathbf{f}_{k}\left(S\left(\mathbf{v}_{k}\left(p_{n}\right)\right)\right)=-kp_{n}.$

Comparing this with

$-\underbrace{\left[k\mid n\right]}_{\begin{subarray}{c}=1\\\ \text{(since
}k\mid n\text{)}\end{subarray}}kp_{n}=-kp_{n},$

we obtain $U_{k}\left(p_{n}\right)=-\left[k\mid n\right]kp_{n}$. Hence, Claim
1 is proved in Case 1.

Let us now consider Case 2. In this case, we have $k\nmid n$. But (82)
(applied to $k$ and $n$ instead of $n$ and $m$) yields

$\mathbf{v}_{k}\left(p_{n}\right)=\begin{cases}kp_{n/k},&\text{if }k\mid n;\\\
0,&\text{if }k\nmid n\end{cases}\ \ \ \ =0\ \ \ \ \ \ \ \ \ \
\left(\text{since }k\nmid n\right).$

But the definition of $U_{k}$ yields $U_{k}=\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}$. Hence,

$U_{k}\left(p_{n}\right)=\left(\mathbf{f}_{k}\circ
S\circ\mathbf{v}_{k}\right)\left(p_{n}\right)=\left(\mathbf{f}_{k}\circ
S\right)\left(\underbrace{\mathbf{v}_{k}\left(p_{n}\right)}_{=0}\right)=\left(\mathbf{f}_{k}\circ
S\right)\left(0\right)=0$

(since the map $\mathbf{f}_{k}\circ S$ is $\mathbf{k}$-linear). Comparing this
with

$-\underbrace{\left[k\mid n\right]}_{\begin{subarray}{c}=0\\\ \text{(since
}k\nmid n\text{)}\end{subarray}}kp_{n}=0,$

we obtain $U_{k}\left(p_{n}\right)=-\left[k\mid n\right]kp_{n}$. Hence, Claim
1 is proved in Case 2.

We have now proved Claim 1 in both Cases 1 and 2. Thus, Claim 1 always holds.]

Theorem 2.27 (a) shows that the map $U_{k}$ is a $\mathbf{k}$-Hopf algebra
homomorphism. Hence, $U_{k}$ is a $\mathbf{k}$-algebra homomorphism. Thus,
$U_{k}\left(1\right)=1$.

Now, let $\Delta_{\Lambda}$ be the comultiplication
$\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ of the $\mathbf{k}$-coalgebra
$\Lambda$. Let $m_{\Lambda}:\Lambda\otimes\Lambda\rightarrow\Lambda$ be the
$\mathbf{k}$-linear map sending each pure tensor $a\otimes
b\in\Lambda\otimes\Lambda$ to $ab\in\Lambda$. Then, the definition of $V_{k}$
yields $V_{k}=\operatorname*{id}\nolimits_{\Lambda}\star
U_{k}=m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta_{\Lambda}$ (by the definition of convolution).

But [GriRei20, Proposition 2.3.6(i)] yields
$\Delta_{\Lambda}\left(p_{n}\right)=1\otimes p_{n}+p_{n}\otimes 1$. Now,

$\displaystyle\underbrace{V_{k}}_{=m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta_{\Lambda}}\left(p_{n}\right)$
$\displaystyle=\left(m_{\Lambda}\circ\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\circ\Delta_{\Lambda}\right)\left(p_{n}\right)$
$\displaystyle=m_{\Lambda}\left(\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(\underbrace{\Delta_{\Lambda}\left(p_{n}\right)}_{=1\otimes
p_{n}+p_{n}\otimes 1}\right)\right)$
$\displaystyle=m_{\Lambda}\left(\underbrace{\left(\operatorname*{id}\nolimits_{\Lambda}\otimes
U_{k}\right)\left(1\otimes p_{n}+p_{n}\otimes
1\right)}_{=\operatorname*{id}\nolimits_{\Lambda}\left(1\right)\otimes
U_{k}\left(p_{n}\right)+\operatorname*{id}\nolimits_{\Lambda}\left(p_{n}\right)\otimes
U_{k}\left(1\right)}\right)$
$\displaystyle=m_{\Lambda}\left(\operatorname*{id}\nolimits_{\Lambda}\left(1\right)\otimes
U_{k}\left(p_{n}\right)+\operatorname*{id}\nolimits_{\Lambda}\left(p_{n}\right)\otimes
U_{k}\left(1\right)\right)$
$\displaystyle=\underbrace{\operatorname*{id}\nolimits_{\Lambda}\left(1\right)}_{=1}\cdot\underbrace{U_{k}\left(p_{n}\right)}_{\begin{subarray}{c}=-\left[k\mid
n\right]kp_{n}\\\ \text{(by Claim
1)}\end{subarray}}+\underbrace{\operatorname*{id}\nolimits_{\Lambda}\left(p_{n}\right)}_{=p_{n}}\cdot\underbrace{U_{k}\left(1\right)}_{=1}\
\ \ \ \ \ \ \ \ \ \left(\text{by the definition of }m_{\Lambda}\right)$
$\displaystyle=-\left[k\mid n\right]kp_{n}+p_{n}=\left(1-\left[k\mid
n\right]k\right)p_{n}.$

This proves Theorem 2.27 (d).

###  3.16 Proof of Corollary 2.28

######  Proof 3.44 (Proof of Corollary 2.28.).

Recall that the family $\left(h_{n}\right)_{n\geq
1}=\left(h_{1},h_{2},h_{3},\ldots\right)$ generates $\Lambda$ as a
$\mathbf{k}$-algebra. Hence, each $g\in\Lambda$ can be written as a polynomial
in $h_{1},h_{2},h_{3},\ldots$. Applying this to $g=p_{n}$, we conclude that
$p_{n}$ can be written as a polynomial in $h_{1},h_{2},h_{3},\ldots$. In other
words, there exists a polynomial
$f\in\mathbf{k}\left[x_{1},x_{2},x_{3},\ldots\right]$ such that

$p_{n}=f\left(h_{1},h_{2},h_{3},\ldots\right).$ (85)

Consider this $f$. We shall show that this $f$ satisfies (12). This will
clearly prove Corollary 2.28.

Consider the map $V_{k}$ defined in Theorem 2.27. Theorem 2.27 (c) yields that
$V_{k}\left(h_{m}\right)=G\left(k,m\right)$ for each positive integer $m$. In
other words,

$\left(V_{k}\left(h_{1}\right),V_{k}\left(h_{2}\right),V_{k}\left(h_{3}\right),\ldots\right)=\left(G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots\right).$
(86)

The map $V_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (by Theorem 2.27
(b)), and thus is a $\mathbf{k}$-algebra homomorphism. Hence, it commutes with
polynomials over $\mathbf{k}$. Thus,

$\displaystyle V_{k}\left(f\left(h_{1},h_{2},h_{3},\ldots\right)\right)$
$\displaystyle=f\left(V_{k}\left(h_{1}\right),V_{k}\left(h_{2}\right),V_{k}\left(h_{3}\right),\ldots\right)$
$\displaystyle=f\left(G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.cor.p-via-G.1})}\right).$

Now, applying the map $V_{k}$ to both sides of the equality (85), we obtain

$V_{k}\left(p_{n}\right)=V_{k}\left(f\left(h_{1},h_{2},h_{3},\ldots\right)\right)=f\left(G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots\right).$

Comparing this with

$V_{k}\left(p_{n}\right)=\left(1-\left[k\mid n\right]k\right)p_{n}\ \ \ \ \ \
\ \ \ \ \left(\text{by Theorem \ref{thm.Uk.main} {(d)}}\right),$

we obtain

$\left(1-\left[k\mid
n\right]k\right)p_{n}=f\left(G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots\right).$

Thus, we have shown that our $f$ satisfies (12). As we said, this proves
Corollary 2.28.

##  4 Proof of the Liu–Polo conjecture

Let us recall a well-known partial order on the set of partitions of a given
$n\in\mathbb{N}$:

######  Definition 4.1.

Let $n\in\mathbb{N}$. We define a binary relation $\triangleright$ on the set
$\operatorname*{Par}\nolimits_{n}$ as follows: Two partitions
$\lambda,\mu\in\operatorname*{Par}\nolimits_{n}$ shall satisfy
$\lambda\triangleright\mu$ if and only if we have

$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}\
\ \ \ \ \ \ \ \ \ \text{for each }k\in\left\\{1,2,\ldots,n\right\\}.$

This relation $\triangleright$ is the greater-or-equal relation of a partial
order on $\operatorname*{Par}\nolimits_{n}$, which is known as the _dominance
order_ (or the _majorization order_).

This definition is precisely [GriRei20, Definition 2.2.7]. Note that if we
replace “for each $k\in\left\\{1,2,\ldots,n\right\\}$” by “for each
$k\in\left\\{1,2,3,\ldots\right\\}$” in this definition, then the relation
$\triangleright$ does not change.

Our goal in this section is to prove the conjecture made in [LiuPol19, Remark
1.4.5]. We state this conjecture as follows:303030Note that
$\left(n-1,n-1,1\right)$ is a partition whenever $n>1$ is an integer.

######  Theorem 4.2.

Let $n$ be an integer such that $n>1$. Then,

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1,n-1-i,1^{i+1}\right)}.$

######  Example 4.3.

For this example, let $n=3$. Then, $n-1=2$ and $2n-1=5$. Hence, the left hand
side of the equality in Theorem 4.2 is

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{5};\\\
\left(2,2,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=m_{\left(2,2,1\right)}+m_{\left(2,1,1,1\right)}+m_{\left(1,1,1,1,1\right)}.$

Meanwhile, the right hand side of the equality in Theorem 4.2 is

$\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1,n-1-i,1^{i+1}\right)}=\sum\limits_{i=0}^{1}\left(-1\right)^{i}s_{\left(2,2-i,1^{i+1}\right)}=s_{\left(2,2,1\right)}-s_{\left(2,1,1,1\right)}.$

Thus, Theorem 4.2 claims that
$m_{\left(2,2,1\right)}+m_{\left(2,1,1,1\right)}+m_{\left(1,1,1,1,1\right)}=s_{\left(2,2,1\right)}-s_{\left(2,1,1,1\right)}$
in this case.

We will pave our way to the proof of Theorem 4.2 by several lemmas. We begin
with a particularly simple one:

######  Lemma 4.4.

Let $n$ be an integer such that $n>1$. Let
$\lambda\in\operatorname*{Par}\nolimits_{2n-1}$. Then,
$\left(n-1,n-1,1\right)\triangleright\lambda$ if and only if all positive
integers $i$ satisfy $\lambda_{i}<n$.

######  Proof 4.5.

This simple proof (an exercise in following Definition 4.1) is left to the
reader.

The next lemma identifies the left hand side of Theorem 4.2 as the Petrie
symmetric function $G\left(n,2n-1\right)$:

######  Corollary 4.6.

Let $n$ be an integer such that $n>1$. Then,

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=G\left(n,2n-1\right).$

######  Proof 4.7.

Proposition 2.3 (c) (applied to $k=n$ and $m=2n-1$) yields

$G\left(n,2n-1\right)=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=2n-1;\\\ \alpha_{i}<n\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par};\\\
\left|\lambda\right|=2n-1;\\\ \lambda_{i}<n\text{ for all
}i\end{subarray}}m_{\lambda}.$ (87)

But Lemma 4.4 yields the following equality of summation signs:

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\lambda_{i}<n\text{ for all
}i\end{subarray}}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par};\\\
\left|\lambda\right|=2n-1;\\\ \lambda_{i}<n\text{ for all }i\end{subarray}}.$

Hence,

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par};\\\
\left|\lambda\right|=2n-1;\\\ \lambda_{i}<n\text{ for all
}i\end{subarray}}m_{\lambda}.$

Comparing this with (87), we obtain

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=G\left(n,2n-1\right).$

This proves Corollary 4.6.

It was Corollary 4.6 that led the author to introduce and study the Petrie
symmetric functions $G\left(k,m\right)$ in general, even if little of their
general properties has proven relevant to Theorem 4.2.

The next proposition gives a simple formula for certain kinds of Petrie
symmetric functions:

######  Proposition 4.8.

Let $n$ be a positive integer. Let $k\in\left\\{0,1,\ldots,n-1\right\\}$.
Then,

$G\left(n,n+k\right)=h_{n+k}-h_{k}p_{n}.$

Proposition 4.8 can be viewed as a particular case of Theorem 2.20 (applied to
$n$ and $n+k$ instead of $k$ and $m$), after realizing that in the sum on the
right hand side of Theorem 2.20, only the first two addends will (potentially)
be nonzero in this case. However, let us give an independent proof of the
proposition.

######  Proof 4.9 (Proof of Proposition 4.8.).

From $k\in\left\\{0,1,\ldots,n-1\right\\}$, we obtain $k<n$ and thus
$n+k<n+n$. Thus we conclude:

> Observation 1: A monomial of degree $n+k$ cannot have more than one variable
> appear in it with exponent $\geq n$ (since this would require it to have
> degree $\geq n+n>n+k$).

Let $\mathfrak{M}_{k}$ be the set of all monomials of degree $k$. The
definition of $h_{k}$ shows that $h_{k}$ is the sum of all monomials of degree
$k$. In other words,

$h_{k}=\sum\limits_{\mathfrak{m}\in\mathfrak{M}_{k}}\mathfrak{m}.$ (88)

Let $\mathfrak{M}_{n+k}$ be the set of all monomials of degree $n+k$. The
definition of $h_{n+k}$ shows that $h_{n+k}$ is the sum of all monomials of
degree $n+k$. In other words,

$h_{n+k}=\sum\limits_{\mathfrak{n}\in\mathfrak{M}_{n+k}}\mathfrak{n}.$ (89)

Let $\mathfrak{N}$ be the set of all monomials of degree $n+k$ in which all
exponents are $<n$. These monomials are exactly the $\mathbf{x}^{\alpha}$ for
$\alpha\in\operatorname*{WC}$ satisfying $\left|\alpha\right|=n+k$ and
$\left(\alpha_{i}<n\text{ for all }i\right)$. Hence,

$\sum\limits_{\mathfrak{n}\in\mathfrak{N}}\mathfrak{n}=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=n+k;\\\ \alpha_{i}<n\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}.$ (90)

But Proposition 2.3 (c) (applied to $n$ and $n+k$ instead of $k$ and $m$)
yields

$G\left(n,n+k\right)=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=n+k;\\\ \alpha_{i}<n\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}=\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par};\\\
\left|\lambda\right|=n+k;\\\ \lambda_{i}<n\text{ for all
}i\end{subarray}}m_{\lambda}.$

Hence,

$G\left(n,n+k\right)=\sum\limits_{\begin{subarray}{c}\alpha\in\operatorname*{WC};\\\
\left|\alpha\right|=n+k;\\\ \alpha_{i}<n\text{ for all
}i\end{subarray}}\mathbf{x}^{\alpha}=\sum\limits_{\mathfrak{n}\in\mathfrak{N}}\mathfrak{n}$
(91)

(by (90)).

Clearly, the set $\mathfrak{N}$ is a subset of $\mathfrak{M}_{n+k}$, and
furthermore its complement $\mathfrak{M}_{n+k}\setminus\mathfrak{N}$ is the
set of all monomials of degree $n+k$ in which at least one exponent is $\geq
n$. Hence, the map

$\displaystyle\mathfrak{M}_{k}\times\left\\{1,2,3,\ldots\right\\}$
$\displaystyle\rightarrow\mathfrak{M}_{n+k}\setminus\mathfrak{N},$
$\displaystyle\left(\mathfrak{m},i\right)$
$\displaystyle\mapsto\mathfrak{m}\cdot x_{i}^{n}$

is well-defined (because if $\mathfrak{m}$ is a monomial of degree $k$, and if
$i\in\left\\{1,2,3,\ldots\right\\}$, then $\mathfrak{m}\cdot x_{i}^{n}$ is a
monomial of degree $k+n=n+k$, and the variable $x_{i}$ appears in it with
exponent $\geq n$). This map is furthermore surjective (for simple reasons)
and injective (in fact, if
$\mathfrak{n}\in\mathfrak{M}_{n+k}\setminus\mathfrak{N}$, then $\mathfrak{n}$
is a monomial of degree $n+k$, and thus Observation 1 yields that there is at
most one variable $x_{i}$ that appears in $\mathfrak{n}$ with exponent $\geq
n$; but this means that the only preimage of $\mathfrak{n}$ under our map is
$\left(\dfrac{\mathfrak{n}}{x_{i}^{n}},i\right)$). Hence, this map is a
bijection. We can thus use it to substitute $\mathfrak{m}\cdot x_{i}^{n}$ for
$\mathfrak{n}$ in the sum
$\sum\limits_{\mathfrak{n}\in\mathfrak{M}_{n+k}\setminus\mathfrak{N}}\mathfrak{n}$.
We thus obtain

$\displaystyle\sum\limits_{\mathfrak{n}\in\mathfrak{M}_{n+k}\setminus\mathfrak{N}}\mathfrak{n}$
$\displaystyle=\sum\limits_{\left(\mathfrak{m},i\right)\in\mathfrak{M}_{k}\times\left\\{1,2,3,\ldots\right\\}}\mathfrak{m}\cdot
x_{i}^{n}=\underbrace{\left(\sum\limits_{\mathfrak{m}\in\mathfrak{M}_{k}}\mathfrak{m}\right)}_{\begin{subarray}{c}=h_{k}\\\
\text{(by
(\ref{pf.prop.h2n-1.hk=}))}\end{subarray}}\cdot\underbrace{\sum\limits_{i\in\left\\{1,2,3,\ldots\right\\}}x_{i}^{n}}_{=p_{n}}$
$\displaystyle=h_{k}p_{n}.$ (92)

But (89) becomes

$\displaystyle h_{n+k}$
$\displaystyle=\sum\limits_{\mathfrak{n}\in\mathfrak{M}_{n+k}}\mathfrak{n}=\underbrace{\sum\limits_{\mathfrak{n}\in\mathfrak{N}}\mathfrak{n}}_{\begin{subarray}{c}=G\left(n,n+k\right)\\\
\text{(by
(\ref{pf.prop.h2n-1.su1pet}))}\end{subarray}}+\underbrace{\sum\limits_{\mathfrak{n}\in\mathfrak{M}_{n+k}\setminus\mathfrak{N}}\mathfrak{n}}_{\begin{subarray}{c}=h_{k}p_{n}\\\
\text{(by (\ref{pf.prop.h2n-1.su2}))}\end{subarray}}\ \ \ \ \ \ \ \ \ \
\left(\text{since }\mathfrak{N}\subseteq\mathfrak{M}_{n+k}\right)$
$\displaystyle=G\left(n,n+k\right)+h_{k}p_{n}.$

In other words,

$G\left(n,n+k\right)=h_{n+k}-h_{k}p_{n}.$

This proves Proposition 4.8.

######  Corollary 4.10.

Let $n$ be an integer such that $n>1$. Then,

$\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}=h_{2n-1}-h_{n-1}p_{n}.$

######  Proof 4.11.

Corollary 4.6 yields

$\displaystyle\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}$
$\displaystyle=G\left(n,2n-1\right)=G\left(n,n+\left(n-1\right)\right)\ \ \ \
\ \ \ \ \ \ \left(\text{since }2n-1=n+\left(n-1\right)\right)$
$\displaystyle=h_{n+\left(n-1\right)}-h_{n-1}p_{n}\ \ \ \ \ \ \ \ \ \
\left(\text{by Proposition \ref{prop.h2n-1}, applied to }k=n-1\right)$
$\displaystyle=h_{2n-1}-h_{n-1}p_{n}.$

Our next claim is an easy consequence of Proposition 1.1:

######  Corollary 4.12.

Let $n$ be a positive integer. Then,

$h_{n}-p_{n}=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1-i,1^{i+1}\right)}.$

######  Proof 4.13.

Proposition 1.1 yields

$\displaystyle p_{n}$
$\displaystyle=\sum\limits_{i=0}^{n-1}\left(-1\right)^{i}s_{\left(n-i,1^{i}\right)}=\underbrace{\left(-1\right)^{0}}_{=1}\underbrace{s_{\left(n-0,1^{0}\right)}}_{=s_{\left(n-0\right)}=s_{\left(n\right)}=h_{n}}+\sum\limits_{i=1}^{n-1}\left(-1\right)^{i}s_{\left(n-i,1^{i}\right)}$
$\displaystyle=h_{n}+\sum\limits_{i=1}^{n-1}\left(-1\right)^{i}s_{\left(n-i,1^{i}\right)},$

so that

$\displaystyle h_{n}-p_{n}$
$\displaystyle=-\sum\limits_{i=1}^{n-1}\left(-1\right)^{i}s_{\left(n-i,1^{i}\right)}=\sum\limits_{i=1}^{n-1}\underbrace{\left(-\left(-1\right)^{i}\right)}_{=\left(-1\right)^{i-1}}s_{\left(n-i,1^{i}\right)}=\sum\limits_{i=1}^{n-1}\left(-1\right)^{i-1}s_{\left(n-i,1^{i}\right)}$
$\displaystyle=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1-i,1^{i+1}\right)}$

(here, we have substituted $i+1$ for $i$ in the sum).

We shall use the _skewing operators_ $f^{\perp}:\Lambda\rightarrow\Lambda$ for
all $f\in\Lambda$ as defined in [GriRei20, §2.8] or in [Macdon95, Chapter I,
Section 5, Example 3]. The easiest way to define them (following [Macdon95,
Chapter I, Section 5, Example 3]) is as follows: For each $f\in\Lambda$, we
let $f^{\perp}:\Lambda\rightarrow\Lambda$ be the $\mathbf{k}$-linear map
adjoint to the map $L_{f}:\Lambda\rightarrow\Lambda,\ g\mapsto fg$ (that is,
to the map that multiplies every element of $\Lambda$ by $f$) with respect to
the Hall inner product. That is, $f^{\perp}$ is the $\mathbf{k}$-linear map
from $\Lambda$ to $\Lambda$ that satisfies

$\left(g,f^{\perp}\left(a\right)\right)=\left(fg,a\right)\ \ \ \ \ \ \ \ \ \
\text{for all }a\in\Lambda\text{ and }g\in\Lambda.$

It is not hard to show that such an operator $f^{\perp}$ exists313131This is
not completely automatic: Not every $\mathbf{k}$-linear map from $\Lambda$ to
$\Lambda$ has an adjoint with respect to the Hall inner product! (For example,
the $\mathbf{k}$-linear map $\Lambda\rightarrow\Lambda$ that sends each Schur
function $s_{\lambda}$ to $1$ has none.) The reason why the map
$L_{f}:\Lambda\rightarrow\Lambda,\ g\mapsto fg$ has an adjoint is that when
$f$ is homogeneous of degree $k$, this map $L_{f}$ sends each graded component
$\Lambda_{m}$ of $\Lambda$ to $\Lambda_{m+k}$, and both of these graded
components $\Lambda_{m}$ and $\Lambda_{m+k}$ are $\mathbf{k}$-modules with
finite bases. (The case when $f$ is not homogeneous can be reduced to the case
when $f$ is homogeneous, since each $f\in\Lambda$ is a sum of finitely many
homogeneous elements.). The definition of $f^{\perp}$ in [GriRei20, §2.8] is
different but equivalent (because of [GriRei20, Proposition 2.8.2(i)]). One of
the most important properties of skewing operators is the following fact
([GriRei20, (2.8.2)]):

######  Lemma 4.14.

Let $\lambda$ and $\mu$ be any two partitions. Then,

$s_{\mu}^{\perp}\left(s_{\lambda}\right)=s_{\lambda/\mu}.$ (93)

(Here, $s_{\lambda/\mu}$ is a skew Schur function, defined in Subsection 3.3.)

Using skewing operators, we can define another helpful family of operators on
$\Lambda$:

######  Definition 4.15.

For any $m\in\mathbb{Z}$, we define a map
$\mathbf{B}_{m}:\Lambda\rightarrow\Lambda$ by setting

$\mathbf{B}_{m}\left(f\right)=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}e_{i}^{\perp}f\
\ \ \ \ \ \ \ \ \ \text{for all }f\in\Lambda\text{.}$

It is known ([GriRei20, Exercise 2.9.1(a)]) that this map $\mathbf{B}_{m}$ is
well-defined and $\mathbf{k}$-linear.

(Actually, the well-definedness of $\mathbf{B}_{m}$ is easy to check: If
$f\in\Lambda$ has degree $d$, then all integers $i>d$ satisfy
$e_{i}^{\perp}f=0$ for degree reasons, and thus the sum
$\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}e_{i}^{\perp}f$ has
only finitely many nonzero addends. The $\mathbf{k}$-linearity of
$\mathbf{B}_{m}$ is even clearer.)

The operators $\mathbf{B}_{m}$ for $m\in\mathbb{Z}$ have first appeared in
Zelevinsky’s [Zelevi81, §4.20] (in the different-looking but secretly
equivalent setting of a PSH-algebra), where they are credited to J. N.
Bernstein. They have since been dubbed the _Bernstein creation operators_ and
proved useful in various contexts (e.g., the definition of the “dual
immaculate functions” in [BBSSZ13] takes them for inspiration). One of their
most fundamental properties is the following fact (which originates in
[Zelevi81, 4.20, ($\ast\ast$)] and appears implicitly in [Macdon95, Chapter I,
Section 5, Example 29]):

######  Proposition 4.16.

Let $\lambda$ be any partition. Let $m\in\mathbb{Z}$ satisfy
$m\geq\lambda_{1}$. Then,

$\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}e_{i}^{\perp}s_{\lambda}=s_{\left(m,\lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)}.$
(94)

See [GriRei20, Exercise 2.9.1(b)] for a proof of Proposition 4.16. Thus, if
$\lambda$ is any partition, and if $m\in\mathbb{Z}$ satisfies
$m\geq\lambda_{1}$, then

$\displaystyle\mathbf{B}_{m}\left(s_{\lambda}\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}e_{i}^{\perp}s_{\lambda}\
\ \ \ \ \ \ \ \ \ \left(\text{by the definition of }\mathbf{B}_{m}\right)$
$\displaystyle=s_{\left(m,\lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{eq.schur-row-adder-1})}\right).$ (95)

######  Lemma 4.17.

Let $n$ be a positive integer. Let $m\in\mathbb{N}$. Then,
$\mathbf{B}_{m}\left(h_{n}\right)=h_{m}h_{n}-h_{m+1}h_{n-1}$.

######  Proof 4.18 (Proof of Lemma 4.17.).

We have $e_{0}=1$ and thus $e_{0}^{\perp}=1^{\perp}=\operatorname*{id}$.
Hence, $e_{0}^{\perp}\left(h_{n}\right)=h_{n}$.

We shall use the notion of skew Schur functions $s_{\lambda/\mu}$ (as in
Subsection 3.3). Recall that $s_{\lambda/\mu}=0$ when
$\mu\not\subseteq\lambda$.

From $e_{1}=s_{\left(1\right)}$ and $h_{n}=s_{\left(n\right)}$, we obtain

$e_{1}^{\perp}\left(h_{n}\right)=s_{\left(1\right)}^{\perp}\left(s_{\left(n\right)}\right)=s_{\left(n\right)/\left(1\right)}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.lem.Bmhn.skew})}\right).$

But it is easy to see that
$s_{\left(n\right)/\left(1\right)}=s_{\left(n-1\right)}$. (Indeed, this
follows from the combinatorial definition of skew Schur functions, since the
skew Ferrers diagram of $\left(n\right)/\left(1\right)$ can be obtained from
the Ferrers diagram of $\left(n-1\right)$ by parallel shift323232See
[GriRei20, §2.3] for the notions we are using here.. Alternatively, this
follows easily from Theorem 3.7, because $s_{\left(n-1\right)}=h_{n-1}$.)

Thus, we obtain

$e_{1}^{\perp}\left(h_{n}\right)=s_{\left(n\right)/\left(1\right)}=s_{\left(n-1\right)}=h_{n-1}.$

For each integer $i>1$, we have

$\displaystyle e_{i}^{\perp}\left(h_{n}\right)$
$\displaystyle=s_{\left(1^{i}\right)}^{\perp}\left(s_{\left(n\right)}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{since }e_{i}=s_{\left(1^{i}\right)}\text{ and
}h_{n}=s_{\left(n\right)}\right)$
$\displaystyle=s_{\left(n\right)/\left(1^{i}\right)}\ \ \ \ \ \ \ \ \ \
\left(\text{by (\ref{pf.lem.Bmhn.skew})}\right)$ $\displaystyle=0\ \ \ \ \ \ \
\ \ \ \left(\text{since }\left(1^{i}\right)\not\subseteq\left(n\right)\text{
(because }i>1\text{)}\right).$ (96)

Now, the definition of $\mathbf{B}_{m}$ yields

$\displaystyle\mathbf{B}_{m}\left(h_{n}\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}e_{i}^{\perp}\left(h_{n}\right)$
$\displaystyle=\underbrace{\left(-1\right)^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}}\underbrace{e_{0}^{\perp}\left(h_{n}\right)}_{=h_{n}}+\underbrace{\left(-1\right)^{1}}_{=-1}h_{m+1}\underbrace{e_{1}^{\perp}\left(h_{n}\right)}_{=h_{n-1}}+\sum\limits_{i\geq
2}\left(-1\right)^{i}h_{m+i}\underbrace{e_{i}^{\perp}\left(h_{n}\right)}_{\begin{subarray}{c}=0\\\
\text{(by (\ref{pf.lem.Bmhn.3}))}\end{subarray}}$
$\displaystyle=h_{m}h_{n}-h_{m+1}h_{n-1}.$

######  Corollary 4.19.

Let $n$ be a positive integer. Then, $\mathbf{B}_{n-1}\left(h_{n}\right)=0$.

######  Proof 4.20.

Apply Lemma 4.17 to $m=n-1$ and simplify.

######  Lemma 4.21.

Let $m\in\mathbb{N}$. Let $n$ be a positive integer. Then,
$\mathbf{B}_{m}\left(p_{n}\right)=h_{m}p_{n}-h_{m+n}$.

######  Proof 4.22.

This is [GriRei20, Exercise 2.9.1(f)]. But here is a more direct proof: We
will use the comultiplication $\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$
of the Hopf algebra $\Lambda$ (see [GriRei20, §2.3]). Here and in the
following, the “$\otimes$” sign denotes $\otimes_{\mathbf{k}}$. The power-sum
symmetric function $p_{n}$ is primitive333333Recall that an element $x$ of a
Hopf algebra $H$ is said to be primitive if the comultiplication $\Delta_{H}$
of $H$ satisfies $\Delta_{H}\left(x\right)=1\otimes x+x\otimes 1$. (see
[GriRei20, Proposition 2.3.6(i)]); thus,

$\Delta\left(p_{n}\right)=1\otimes p_{n}+p_{n}\otimes 1.$

Hence, for each $i\in\mathbb{N}$, the definition of $e_{i}^{\perp}$ given in
[GriRei20, Definition 2.8.1] (not the equivalent definition we gave above)
yields

$e_{i}^{\perp}\left(p_{n}\right)=\left(e_{i},1\right)p_{n}+\left(e_{i},p_{n}\right)1.$
(97)

Now, the definition of $\mathbf{B}_{m}$ yields

$\displaystyle\mathbf{B}_{m}\left(p_{n}\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}\underbrace{e_{i}^{\perp}\left(p_{n}\right)}_{\begin{subarray}{c}=\left(e_{i},1\right)p_{n}+\left(e_{i},p_{n}\right)1\\\
\text{(by (\ref{pf.lem.Bmpn.eipn}))}\end{subarray}}$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}\left(\left(e_{i},1\right)p_{n}+\left(e_{i},p_{n}\right)1\right)$
$\displaystyle=\underbrace{\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}\cdot\left(e_{i},1\right)p_{n}}_{\begin{subarray}{c}=\left(-1\right)^{0}h_{m+0}\cdot\left(e_{0},1\right)p_{n}\\\
\text{(because the Hall inner product }\left(e_{i},1\right)\\\ \text{equals
}0\text{ whenever }i\neq 0\text{ (by (\ref{eq.Hall.graded})),}\\\ \text{and
thus the only nonzero addend of this}\\\ \text{sum is the addend for
}i=0\text{)}\end{subarray}}+\underbrace{\sum\limits_{i\in\mathbb{N}}\left(-1\right)^{i}h_{m+i}\cdot\left(e_{i},p_{n}\right)1}_{\begin{subarray}{c}=\left(-1\right)^{n}h_{m+n}\cdot\left(e_{n},p_{n}\right)1\\\
\text{(because the Hall inner product }\left(e_{i},p_{n}\right)\\\
\text{equals }0\text{ whenever }i\neq n\text{ (by (\ref{eq.Hall.graded})),}\\\
\text{and thus the only nonzero addend of this}\\\ \text{sum is the addend for
}i=n\text{)}\end{subarray}}$
$\displaystyle=\underbrace{\left(-1\right)^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}}\cdot\underbrace{\left(e_{0},1\right)}_{=\left(1,1\right)=1}p_{n}+\left(-1\right)^{n}h_{m+n}\cdot\underbrace{\left(e_{n},p_{n}\right)}_{\begin{subarray}{c}=\left(-1\right)^{n-1}\\\
\text{(by Proposition \ref{prop.ejpj})}\end{subarray}}1$
$\displaystyle=h_{m}p_{n}+\underbrace{\left(-1\right)^{n}h_{m+n}\cdot\left(-1\right)^{n-1}1}_{=-h_{m+n}}=h_{m}p_{n}-h_{m+n}.$

######  Lemma 4.23.

Let $n$ be a positive integer. Then,

$\mathbf{B}_{n-1}\left(h_{n}-p_{n}\right)=h_{2n-1}-h_{n-1}p_{n}.$

######  Proof 4.24.

The map $\mathbf{B}_{n-1}$ is $\mathbf{k}$-linear. Thus,

$\displaystyle\mathbf{B}_{n-1}\left(h_{n}-p_{n}\right)$
$\displaystyle=\underbrace{\mathbf{B}_{n-1}\left(h_{n}\right)}_{\begin{subarray}{c}=0\\\
\text{(by Corollary
\ref{cor.Bn-1Hn})}\end{subarray}}-\underbrace{\mathbf{B}_{n-1}\left(p_{n}\right)}_{\begin{subarray}{c}=h_{n-1}p_{n}-h_{\left(n-1\right)+n}\\\
\text{(by Lemma \ref{lem.Bmpn},}\\\ \text{applied to
}m=n-1\text{)}\end{subarray}}$
$\displaystyle=-\left(h_{n-1}p_{n}-h_{\left(n-1\right)+n}\right)=\underbrace{h_{\left(n-1\right)+n}}_{=h_{2n-1}}-h_{n-1}p_{n}=h_{2n-1}-h_{n-1}p_{n}.$

######  Lemma 4.25.

Let $n$ be a positive integer. Then,

$\mathbf{B}_{n-1}\left(h_{n}-p_{n}\right)=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1,n-1-i,1^{i+1}\right)}.$

######  Proof 4.26 (Proof of Lemma 4.25.).

We have

$\displaystyle\mathbf{B}_{n-1}\left(h_{n}-p_{n}\right)$
$\displaystyle=\mathbf{B}_{n-1}\left(\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1-i,1^{i+1}\right)}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by Corollary \ref{cor.p-as-sum-2}}\right)$
$\displaystyle=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}\underbrace{\mathbf{B}_{n-1}\left(s_{\left(n-1-i,1^{i+1}\right)}\right)}_{\begin{subarray}{c}=s_{\left(n-1,n-1-i,1^{i+1}\right)}\\\
\text{(by (\ref{eq.schur-row-adder-2}), applied to }m=n-1\\\ \text{and
}\lambda=\left(n-1-i,1^{i+1}\right)\\\ \text{(since }n-i\geq
n-1-i\text{))}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\mathbf{B}_{n-1}\text{ is }\mathbf{k}\text{-linear}\right)$
$\displaystyle=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1,n-1-i,1^{i+1}\right)}.$

Now the proof of Theorem 4.2 is a trivial concatenation of equalities:

######  Proof 4.27 (Proof of Theorem 4.2.).

Corollary 4.10 yields

$\displaystyle\sum\limits_{\begin{subarray}{c}\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\
\left(n-1,n-1,1\right)\triangleright\lambda\end{subarray}}m_{\lambda}$
$\displaystyle=h_{2n-1}-h_{n-1}p_{n}=\mathbf{B}_{n-1}\left(h_{n}-p_{n}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by Lemma \ref{lem.Bm1}}\right)$
$\displaystyle=\sum\limits_{i=0}^{n-2}\left(-1\right)^{i}s_{\left(n-1,n-1-i,1^{i+1}\right)}\
\ \ \ \ \ \ \ \ \ \left(\text{by Lemma \ref{lem.Bm2}}\right).$

##  5 Final remarks

###  5.1 SageMath code

The SageMath computer algebra system [SageMath] does not (yet) natively know
the Petrie symmetric functions $G\left(k,m\right)$; but they can be easily
constructed in it. For example, the code that follows computes
$G\left(k,n\right)$ expanded in the Schur basis:

⬇

m = Sym.m() # monomial symmetric functions

s = Sym.s() # Schur functions

def G(k, n): # a Petrie function

return s(m.sum(m[lam] for lam in Partitions(n, max_part=k-1)))

###  5.2 Understanding the Petrie numbers

Combining Corollary 2.9 with Theorem 2.14 yields an explicit expression of all
coefficients in the expansion of a Petrie symmetric function
$G\left(k,m\right)$ in the Schur basis. It would stand to reason if the
identity in Theorem 4.2 (whose left hand side is $G\left(n,2n-1\right)$) could
be obtained from this expression. Surprisingly, we have been unable to do so,
which suggests that the description of
$\operatorname*{pet}\nolimits_{k}\left(\lambda,\varnothing\right)$ in Theorem
2.14 might not be optimal.

As to $\operatorname*{pet}\nolimits_{k}\left(\lambda,\mu\right)$, we do not
have an explicit description at all, unless we count the recursive one that
can be extracted from the proof in [GorWil74].

###  5.3 MNable symmetric functions

Combining Theorem 2.16 with Proposition 2.7, we conclude that for any $k>0$
and $m\in\mathbb{N}$, the symmetric function $G\left(k,m\right)\in\Lambda$ has
the following property: For any $\mu\in\operatorname*{Par}$, its product
$G\left(k,m\right)\cdot s_{\mu}$ with $s_{\mu}$ can be written in the form
$\sum\limits_{\lambda\in\operatorname*{Par}}u_{\lambda}s_{\lambda}$ with
$u_{\lambda}\in\left\\{-1,0,1\right\\}$ for all
$\lambda\in\operatorname*{Par}$. It has this property in common with the
symmetric functions $h_{m}$ and $e_{m}$ (according to the Pieri rules) and
$p_{m}$ (according to the Murnaghan–Nakayama rule) as well as several others.
The study of symmetric functions having this property – which we call MNable
symmetric functions (in honor of Murnaghan and Nakayama) – has been initiated
in [Grinbe20a, §8], but there is much to be done.

###  5.4 A conjecture of Per Alexandersson

In February 2020, Per Alexandersson suggested the following conjecture:

######  Conjecture 5.1.

Let $k$ be a positive integer, and $n\in\mathbb{N}$. Then,
$G\left(k,m\right)\cdot p_{2}\in\Lambda$ can be written in the form
$\sum\limits_{\lambda\in\operatorname*{Par}}u_{\lambda}s_{\lambda}$ with
$u_{\lambda}\in\left\\{-1,0,1\right\\}$ for all
$\lambda\in\operatorname*{Par}$.

For example,

$G\left(3,4\right)\cdot
p_{2}=s_{\left(1,1,1,1,1,1\right)}+s_{\left(2,2,2\right)}-s_{\left(3,1,1,1\right)}-s_{\left(3,3\right)}+s_{\left(4,2\right)}.$

Conjecture 5.1 has been verified for all $k$ and $n$ satisfying $k+n\leq 30$.

Note that Conjecture 5.1 becomes false if $p_{2}$ is replaced by $p_{3}$. For
example,

$G\left(3,4\right)\cdot
p_{3}=-s_{\left(1,1,1,1,1,1,1\right)}+s_{\left(2,2,1,1,1\right)}-2s_{\left(2,2,2,1\right)}+s_{\left(3,2,1,1\right)}-s_{\left(4,1,1,1\right)}-s_{\left(4,3\right)}+s_{\left(5,2\right)}.$

###  5.5 A conjecture of François Bergeron

An even more mysterious conjecture was suggested by François Bergeron in April
2020:

######  Conjecture 5.2.

Let $k$ and $n$ be positive integers, and $m\in\mathbb{N}$. Let $\nabla$ be
the _nabla operator_ as defined (e.g.) in [Berger19, §3.2.1]. Then, there
exists a sign $\sigma_{n,k,m}\in\left\\{1,-1\right\\}$ such that
$\sigma_{n,k,m}\nabla^{n}\left(G\left(k,m\right)\right)$ is an
$\mathbb{N}\left[q,t\right]$-linear combination of Schur functions.

Using SageMath, this conjecture has been checked for $n=1$ and all
$k,m\in\left\\{0,1,\ldots,9\right\\}$; the signs $\sigma_{1,k,m}$ are given by
the following table:

| $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$
---|---|---|---|---|---|---|---|---|---
$2$ | $+$ | $+$ | $+$ | $+$ | $+$ | $+$ | $+$ | $+$ | $+$
$3$ | $+$ | $-$ | $-$ | $-$ | $+$ | $+$ | $+$ | $-$ | $-$
$4$ | $+$ | $-$ | $+$ | $+$ | $+$ | $-$ | $+$ | $+$ | $+$
$5$ | $+$ | $-$ | $+$ | $-$ | $-$ | $-$ | $+$ | $-$ | $+$
$6$ | $+$ | $-$ | $+$ | $-$ | $+$ | $+$ | $+$ | $-$ | $+$
$7$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $-$ | $-$ | $+$
$8$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $+$ | $+$
$9$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $-$

(where the entry in the row indexed $k$ and the column indexed $m$ is the sign
$\sigma_{1,k,m}$, represented by a “$+$” sign if it is $1$ and by a “$-$” sign
if it is $-1$). I am not aware of a pattern in these signs, apart from the
fact that $\sigma_{1,2,m}=1$ for all $m\in\mathbb{N}$ (a consequence of
Haiman’s famous interpretation of $\nabla\left(e_{m}\right)$ as a character),
and that $\sigma_{1,k,m}$ appears to be $\left(-1\right)^{m-1}$ whenever
$1\leq m<k$ (which would follow from the conjecture that
$\left(-1\right)^{m-1}\nabla\left(h_{m}\right)$ is an
$\mathbb{N}\left[q,t\right]$-linear combination of Schur functions for any
$m\geq 1$).

###  5.6 “Petriefication” of Schur functions

Theorem 2.27 shows the existence of a Hopf algebra homomorphism
$V_{k}:\Lambda\rightarrow\Lambda$ that sends the complete homogeneous
symmetric functions $h_{1},h_{2},h_{3},\ldots$ to the Petrie symmetric
functions $G\left(k,1\right),G\left(k,2\right),G\left(k,3\right),\ldots$. It
thus is natural to consider the images of all Schur functions $s_{\lambda}$
under this homomorphism $V_{k}$. Experiments with small $\lambda$’s may
suggest that these images $V_{k}\left(s_{\lambda}\right)$ all can be written
in the form
$\sum\limits_{\lambda\in\operatorname*{Par}}u_{\lambda}s_{\lambda}$ with
$u_{\lambda}\in\left\\{-1,0,1\right\\}$. But this is not generally the case;
counterexamples include $V_{3}\left(s_{\left(4,4,4\right)}\right)$,
$V_{4}\left(s_{\left(4,4\right)}\right)$ and
$V_{4}\left(s_{\left(5,1,1,1,1\right)}\right)$. (Of course, it is true when
$\lambda$ is a single row, because of
$V_{k}\left(s_{\left(m\right)}\right)=V_{k}\left(h_{m}\right)=G\left(k,m\right)$;
and it is also true when $\lambda$ is a single column, because the Hopf
algebra homomorphism $V_{k}$ commutes with the antipode $S$ that sends
$h_{m}\mapsto\left(-1\right)^{m}e_{m}$ and
$s_{\lambda}\mapsto\left(-1\right)^{\left|\lambda\right|}s_{\lambda^{t}}$.)

## References

  * [AhmMer20] Moussa Ahmia, Mircea Merca, A generalization of complete and elementary symmetric functions, arXiv:2005.01447v1.
  * [BaAhBe18] Abdelghafour Bazeniar, Moussa Ahmia, Hacène Belbachir, Connection between bisnomial coefficients and their analogs and symmetric functions, Turk J Math 42 (2018), pp. 807–818.
https://doi.org/10.3906/mat-1705-27

  * [BBSSZ13] Chris Berg, Nantel Bergeron, Franco Saliola, Luis Serrano, Mike Zabrocki, A Lift of the Schur and Hall–Littlewood Bases to Non-commutative Symmetric Functions, Canadian Journal of Mathematics 66 (2013), Issue 3, pp. 525–565.
https://doi.org/10.4153/CJM-2013-013-0

  * [Berger19] François Bergeron, Symmetric Functions and Rectangular Catalan Combinatorics, 7 September 2019.
http://bergeron.math.uqam.ca/wp-content/uploads/2019/09/Symmetric-
Functions.pdf

  * [Camero94] Peter J. Cameron, Combinatorics: Topics, Techniques, Algorithms, 1st edition, Cambridge University Press 1994.
  * [Crane18] Rixon Crane, Plane Partitions in Number Theory and Algebra, bachelor thesis at the University of Queensland, 5 November 2018.
https://rixonc.github.io/Rixon_Crane_Honours_Thesis.pdf

  * [Egge19] Eric S. Egge, An Introduction to Symmetric Functions and Their Combinatorics, Student Mathematical Library 91, AMS 2019.
  * [FuMei20] Houshan Fu, Zhousheng Mei, Truncated Homogeneous Symmetric Functions, arXiv:2002.02784v1.
  * [GorWil74] Manfred Gordon, E. Martin Wilkinson, Determinants of Petrie matrices, Pacific Journal of Mathematics 51, no. 2, 1974.
  * [Grinbe20a] Darij Grinberg, The Petrie symmetric functions (extended abstract), extended abstract accepted in the FPSAC conference 2020.
https://www.cip.ifi.lmu.de/~grinberg/algebra/fps20pet.pdf

  * [Grinbe20b] Darij Grinberg, Petrie symmetric functions [detailed version], detailed version of the present paper. Also available (possibly in an older edition) as an ancillary file at arXiv:2004.11194v2.
  * [GriRei20] Darij Grinberg, Victor Reiner, Hopf algebras in Combinatorics, version of 19 April 2020, arXiv:1409.8356v6.
(These notes are also available at the URL
http://www.cip.ifi.lmu.de/~grinberg/algebra/HopfComb-sols.pdf . However, the
version at this URL will be updated in the future, and eventually its
numbering will no longer match our references.)

  * [LiuPol19] Linyuan Liu, Patrick Polo, On the cohomology of line bundles over certain flag schemes II, arXiv:1908.08432v2.
  * [Macdon95] Ian G. Macdonald, Symmetric Functions and Hall Polynomials, Oxford Mathematical Monographs, 2nd edition, Oxford Science Publications 1995.
  * [MenRem15] Anthony Mendes, Jeffrey Remmel, Counting with Symmetric Functions, Developments in Mathematics 43, Springer 2015.
  * [Olsson93] Jørn Olsson, Combinatorics and representations of finite groups, Vorlesungen aus dem FB Mathematik der Univ. Essen 20, 1993.
http://web.math.ku.dk/~olsson/manus/comb_rep_all.pdf

  * [SageMath] The Sage Developers, SageMath, the Sage Mathematics Software System (Version 9.0), 2020.
  * [Sagan20] Bruce Sagan, Combinatorics: The Art of Counting, preliminary version, 22 April 2020.
https://users.math.msu.edu/users/bsagan/Books/Aoc/aoc.pdf

  * [Sam17] Steven V. Sam, Notes for Math 470 (Symmetric Functions), 27 April 2017.
https://www.math.wisc.edu/~svs/740/notes.pdf

  * [Stanle01] Richard Stanley, Enumerative Combinatorics, volume 2, First edition, Cambridge University Press 2001\.
See http://math.mit.edu/~rstan/ec/ for errata.

  * [Wildon15] Mark Wildon, A combinatorial proof of a plethystic Murnaghan–Nakayama rule, SIAM J. Discrete Math. 30 (2016), pp. 1526–1533.
A preprint appears at
http://www.ma.rhul.ac.uk/~uvah099/Maths/PlethysticMNExpRev1.pdf .

  * [Zelevi81] Andrey V. Zelevinsky, Representations of Finite Classical Groups: A Hopf Algebra Approach, Lecture Notes in Mathematics 869, Springer 1981.