# Partitions With Designated Summands Not Divisible by $2^{l}$, $2$, and
$3^{l}$ Modulo $2$, $4$, and $3$

Daniel Herden, Mark R. Sepanski, Jonathan Stanfill, Cordell Hammon, Joel
Henningsen, Henry Ickes, Indalecio Ruiz  Department of Mathematics, Baylor
University, Sid Richardson Building, 1410 S.4th Street, Waco, TX 76706, USA
<EMAIL_ADDRESS>
###### Abstract.

Numerous congruences for partitions with designated summands have been proven
since first being introduced and studied by Andrews, Lewis, and Lovejoy. This
paper explicitly characterizes the number of partitions with designated
summands whose parts are not divisible by $2^{\ell}$, $2$, and $3^{\ell}$
working modulo $2,\ 4,$ and $3$, respectively, greatly extending previous
results on the subject. We provide a few applications of our characterizations
throughout in the form of congruences and a computationally fast recurrence.
Moreover, we illustrate a previously undocumented connection between the
number of partitions with designated summands and the number of partitions
with odd multiplicities.

###### Key words and phrases:

Partitions, designated summands, generating functions

###### 2010 Mathematics Subject Classification:

Primary: 11P83; Secondary: 05A17.

###### Contents

  1. 1 Introduction
  2. 2 General Notation and Basic Results
  3. 3 The Case of $\operatorname{pd}_{2^{\ell}}(q)\,\operatorname{mod}2$
  4. 4 An Alternate Characterization of $\operatorname{pd}_{4}(q)\,\operatorname{mod}2$
  5. 5 A Recurrence Relation for $\operatorname{PD}_{k}(n)\,\operatorname{mod}2$
  6. 6 The Case of $\operatorname{pd}_{2}(q)\,\operatorname{mod}4$
  7. 7 A Dissection Proof for Theorem 4.1
  8. 8 The Case of $\operatorname{pd}_{3^{\ell}}(q)\,\operatorname{mod}3$
  9. 9 Open Problems

##  1\. Introduction

Partitions with designated summands have been studied by many authors since
first being introduced by Andrews, Lewis, and Lovejoy in [2]. These partitions
are constructed by taking the ordinary partitions and marking exactly one part
of each part size, often denoted by affixing a prime to the tagged part. For
instance, there are 15 partitions of 5 with designated summands:

$\displaystyle 5^{\prime},\ 4^{\prime}+1^{\prime},\ 3^{\prime}+2^{\prime},\
3^{\prime}+1^{\prime}+1,\ 3^{\prime}+1+1^{\prime},\ 2^{\prime}+2+1^{\prime},$
$\displaystyle 2+2^{\prime}+1^{\prime},\ 2^{\prime}+1^{\prime}+1+1,\
2^{\prime}+1+1^{\prime}+1,\ 2^{\prime}+1+1+1^{\prime},1^{\prime}+1+1+1+1,$
$\displaystyle 1+1^{\prime}+1+1+1,\ 1+1+1^{\prime}+1+1,\ 1+1+1+1^{\prime}+1,\
1+1+1+1+1^{\prime}.$

Hence $\operatorname{PD}(5)=15$, where $\operatorname{PD}(n)$ denotes the
total number of partitions of $n$ with designated summands.

By [2, Theorem 1], the generating function for the number of partitions with
designated summands whose parts belong to a set of positive integers, $S$, is
given by

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{S}(n)q^{n}=\prod_{n\in
S}\dfrac{1-q^{6n}}{(1-q^{n})(1-q^{2n})(1-q^{3n})}.$

Furthermore, by taking $S_{k}$ to be the set of positive integers not
divisible by $k$, one obtains the generating function for the number of
partitions with designated summands whose parts are not divisible by $k$,
denoted by $\operatorname{PD}_{k}(n)$. It is given in [2, Corollary 3] by

(1.1) $\displaystyle\sum_{n\geq
0}\operatorname{PD}_{k}(n)q^{n}=\dfrac{(q^{6};q^{6})_{\infty}(q^{k};q^{k})_{\infty}(q^{2k};q^{2k})_{\infty}(q^{3k};q^{3k})_{\infty}}{(q;q)_{\infty}(q^{2};q^{2})_{\infty}(q^{3};q^{3})_{\infty}(q^{6k};q^{6k})_{\infty}},$

where $(a;q)_{\infty}$ represents the _$q$ -series_ or _$q$ -Pochhammer
symbol_

$\displaystyle(a;q)_{\infty}=\prod_{n\geq 0}(1-aq^{n}).$

The generating function for $\operatorname{PD}_{k}(n)$, which will be the
central object of study for this paper, has been used to prove numerous
congruences for partitions with designated summands in [2, 5, 7, 8, 10, 15].
In some cases, the generating function for $\operatorname{PD}_{k}(n)$ is more
explicitly known. For instance, one has the following complete
characterization of $\operatorname{PD}_{2}(n)$ modulo $2$.

######  Theorem 1.1 ([2, Corollary 10]).

For all $n\geq 0$,

$\displaystyle\operatorname{PD}_{2}(n)\equiv\begin{cases}1\,\operatorname{mod}2,&\text{if
}n=0\text{ or }n=k^{2}\text{ for }k\geq 1,\,3\nmid k\\\
0\,\operatorname{mod}2,&\text{otherwise}.\end{cases}$

Similarly, a complete characterization of $\operatorname{PD}_{3}(n)$ modulo
$3$ exists.

######  Theorem 1.2 ([7, Theorem 2]).

For all $n\geq 0$,

$\displaystyle\operatorname{PD}_{3}(n)\equiv\begin{cases}1\quad\
\,\operatorname{mod}3,&\text{if }n=0,\\\
r(n)\,\operatorname{mod}3,&\text{otherwise},\end{cases}$

where $r(n)$ is the number of solutions to $n=k(k+1)+3m(m+1)+1$ in nonnegative
integers $k,m$.

In this paper, we study further characterizations of
$\operatorname{PD}_{k}(n)$ by means of its generating function. Our main
results include characterizing $\operatorname{PD}_{2^{\ell}}(n)$ mod $2$ in
Section 3, $\operatorname{PD}_{2}(n)$ mod $4$ in Section 6, and
$\operatorname{PD}_{3^{\ell}}(n)$ mod $3$ in Section 8.

In Section 3, Theorem 3.2, we show that

$\operatorname{PD}_{2^{\ell}}(n)\equiv a_{n\ell}\,\operatorname{mod}2$

where

$a_{n\ell}=\Big{|}\\{\text{solutions to
}n=\sum_{m=0}^{\ell-1}2^{m}k_{m}^{2}\,|\,k_{m}\geq 0,\,3\nmid k_{m}\text{ or
}k_{m}=0\\}\Big{|}.$

For example, along with a few auxiliary results, this allows us to validate
numerous congruences in Theorem 3.3 and Corollary 3.6. For $n\geq 0$ and
$r\in\\{5,7,10,13,14,15,20,21,23,26,28,29,30,31\\}$,

$\displaystyle\operatorname{PD}_{4}(32n+r)\equiv 0\,\operatorname{mod}2,$

and

$\displaystyle\operatorname{PD}_{8}(32n+24)\equiv 0\,\operatorname{mod}2.$

Additionally, for $\ell\geq 3$ and $r\in\\{4,6\\}$,

$\operatorname{PD}_{2^{\ell}}(8n+r)\equiv 0\,\operatorname{mod}2,$

for $\ell\geq 4$ and $r\in\\{4,6,10,12,14\\}$

$\operatorname{PD}_{2^{\ell}}(16n+r)\equiv 0\,\operatorname{mod}2,$

and for $\ell\geq 5$ and $r\in\\{4,6,10,12,14,16,20,22,24,26,28,30\\}$

$\operatorname{PD}_{2^{\ell}}(32n+r)\equiv 0\,\operatorname{mod}2.$

In general for $\ell\geq j\geq 3$ and $0\leq s<2^{j-1}$ with $s$ not of the
form $0$ or $4^{a}(8b+1)$ for some $a,b\geq 0$,

$\operatorname{PD}_{2^{\ell}}(2^{j}n+2s)\equiv 0\,\operatorname{mod}2.$

Finally, in Theorem 3.7 we provide the very interesting congruence

$\operatorname{PD}(n)\equiv b_{n}\,\operatorname{mod}2,$

which links the number $\operatorname{PD}(n)$ of partitions of $n$ with
designated summands [18, A077285] to the number $b_{n}$ of partitions of $n$
with odd multiplicities [18, A055922], which may be of use in the study of the
parity of partitions with odd multiplicities as in [12], [13], and [17].

In Sections 4 and 7, we prove the following remarkably explicit alternate
characterization of $\operatorname{PD}_{4}(n)$ mod $2$ in Theorem 4.1: For all
$n\geq 0$,

$\displaystyle\operatorname{PD}_{4}(n)\equiv\begin{cases}1\,\operatorname{mod}2,&\text{if
}n=mk^{2}\text{ for }m,k\geq 0,\,m\mid 6,\\\
0\,\operatorname{mod}2,&\text{otherwise}.\end{cases}$

In Section 5, motivated by [2, Corollaries 6 and 9], [11], and Section 3, for
general $n\geq 0$ and $k\geq 2$, we also provide a computationally fast
recurrence for $\operatorname{PD}_{k}(n)$ mod $2$ in Theorem 5.1 of the form

$\displaystyle\operatorname{PD}_{k}(n)+\sum_{\ell\geq
1,\,3\nmid\ell}\operatorname{PD}_{k}(n-\ell^{2})\equiv\begin{cases}1\,\operatorname{mod}2,&\text{if
}n=0\text{ or }n=km^{2}\text{ for some }3\nmid m,\\\
0\,\operatorname{mod}2,&\text{otherwise}.\end{cases}$

In Section 6, we consider $\operatorname{PD}_{2}(n)\operatorname{mod}4$.
Theorem 6.4 gives the following explicit characterization of
$\operatorname{PD}_{2}(2n+1)$ mod $4$,

$\operatorname{PD}_{2}(2n+1)\equiv d_{n}\,\operatorname{mod}4$

where

$\displaystyle d_{n}=$ $\displaystyle\bigg{|}\\{\text{solutions to
}n=3j(3j-1)+3k(3k-1)=2\big{[}\binom{3j}{2}+\binom{3k}{2}\big{]}\,\Big{|}\,\
k,j\in\mathbb{Z}\\}\bigg{|}.$

Theorem 6.5 provides an explicit characterization of
$\operatorname{PD}_{2}(2n)$ mod $4$ as

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{2}(2n)q^{n}$
$\displaystyle\equiv 1+2\sum_{k\geq 1,\,3\nmid k}q^{k^{2}}+\sum_{k,\ell\geq
1,\,3\nmid k,\ell}q^{k^{2}+\ell^{2}}\,\operatorname{mod}4.$

Thoerem 6.7 gives the following combined expression,

$\displaystyle\operatorname{pd}_{2}(q)\equiv 1+\Big{(}\sum_{k\geq 1,\,3\nmid
k}q^{k^{2}}\Big{)}\Big{(}\sum_{k\in\mathbb{Z}}q^{k^{2}}\Big{)}\,\operatorname{mod}4.$

Finally, in Theorem 6.6, we show that for $n\geq 1,$

$\displaystyle\operatorname{PD}_{2}(3n)\equiv 0\,\operatorname{mod}4$

and for all $n\geq 1$ with $6\nmid n$,

$\displaystyle\operatorname{PD}_{2}(2n+1)$ $\displaystyle\equiv
0\,\operatorname{mod}4.$

In Section 8, Theorem 8.4, we show that

$\operatorname{PD}_{3}(n)\equiv e_{1\ell}\,\operatorname{mod}3$

where

$\displaystyle e_{n1}$ $\displaystyle=\big{|}\\{\text{solutions to
}n=k_{0}^{2}+3k_{1}^{2}\,|\,$
$\displaystyle\quad\qquad\;k_{m}\in\mathbb{Z}\text{ or }\mathbb{N}\text{ when
$k_{m}$ is even or odd, respectively}\\}\big{|}.$

This gives an alternate characterization of Theorem 1.2. See Remark 8.3 for
alternate forms of $e_{nl}$.

In general, Theorem 8.7, for all $n\geq 0$ and $\ell\geq 2$, we show that

$\displaystyle\operatorname{PD}_{3^{\ell}}(3n)\equiv
e^{*}_{n\ell}\,\operatorname{mod}3$

where

$\displaystyle e^{*}_{n\ell}$ $\displaystyle=\big{|}\\{\text{solutions to
}n=k_{0}^{2}+k_{0}^{\prime 2}+3^{\ell-1}(k_{1}^{2}+k_{1}^{\prime 2})\,|\,$
$\displaystyle\quad\qquad\;k_{m},k_{m}^{\prime}\in\mathbb{Z}\text{ or
}\mathbb{N}\text{ when $k_{m},k_{m}^{\prime}$ is even or odd,
respectively}\\}\big{|}.$

As an easy application, Theorem 8.5, we provide very short proofs of the
following congruences for which proofs using dissections can be found in [7,
Theorem 3] and [8, Theorem 3] for $\operatorname{PD}_{3}(9n+6)$ and
$\operatorname{PD}_{3k}(3n+2)$, respectively,

$\displaystyle\operatorname{PD}_{3}(9n+6)\equiv 0\,\operatorname{mod}3$

and

$\displaystyle\operatorname{PD}_{3^{\ell}}(3n+2)\equiv
0\,\operatorname{mod}3.$

As a further application, Theorem 8.6, we show that for $n\geq 1$,

$\displaystyle\operatorname{PD}_{3}(2n)\equiv 0\,\operatorname{mod}3.$

Moreover, for all $n\geq 0$ and $\ell\geq 3$,

$\displaystyle\operatorname{PD}_{3^{\ell}}(27n+9)\equiv
0\,\operatorname{mod}3$

and for all $n\geq 0$ and $\ell\neq 2$,

$\displaystyle\operatorname{PD}_{3^{\ell}}(27n+18)\equiv
0\,\operatorname{mod}3.$

Finally, Corollary 8.9, for all $n\geq 0$, $k\geq 1$ and $\ell\geq 2k+1$, we
show that

$\displaystyle\operatorname{PD}_{3^{\ell}}\big{(}3^{2k}(3n+1)\big{)}\equiv\operatorname{PD}_{3^{\ell}}\big{(}3^{2k}(3n+2)\big{)}\equiv
0\,\operatorname{mod}3.$

In Section 9, we conclude with some remarks and conjectured congruences.

##  2\. General Notation and Basic Results

######  Definition 2.1.

Let

$f_{k}=f_{k}(q)=(q^{k};q^{k})_{\infty}=\prod_{n\geq
1}(1-q^{kn})=f_{1}(q^{k}),$ $\operatorname{pd}(q)=\sum_{n\geq
0}\operatorname{PD}(n)q^{n},$ $\operatorname{pd}_{k}(q)=\sum_{n\geq
0}\operatorname{PD}_{k}(n)q^{n},$

where, once again, $\operatorname{PD}(n)$ denotes the number of partitions of
$n$ with designated summands (ordinary partitions with exactly one part
designated among parts with equal size) and $\operatorname{PD}_{k}(n)$ denotes
the number of partitions of $n$ with designated summands whose parts are not
divisible by $k$.

Using this notation, the generating functions can be written as

$\operatorname{pd}(q)=\frac{f_{6}(q)}{f_{1}(q)f_{2}(q)f_{3}(q)}$

and

$\operatorname{pd}_{k}(q)=\frac{f_{6}(q)}{f_{1}(q)f_{2}(q)f_{3}(q)}\frac{f_{k}(q)f_{2k}(q)f_{3k}(q)}{f_{6k}(q)}=\frac{f_{6}(q)}{f_{1}(q)f_{2}(q)f_{3}(q)}\frac{f_{1}(q^{k})f_{2}(q^{k})f_{3}(q^{k})}{f_{6}(q^{k})}.$

######  Definition 2.2.

Let

$g(q)=\frac{1}{\operatorname{pd}(q)}=\frac{f_{1}(q)f_{2}(q)f_{3}(q)}{f_{6}(q)}.$

We see that

(2.1) $\operatorname{pd}_{k}(q)=\frac{g(q^{k})}{g(q)}.$

Modulo a prime, $p$, the Frobenius automorphism implies that

$f_{pk}(q)=f_{k}(q^{p})\equiv f_{k}(q)^{p}\,\operatorname{mod}p$

so that

$g(q^{p})\equiv g(q)^{p}\,\operatorname{mod}p.$

Therefore, Equation (2.1) gives us

(2.2) $\operatorname{pd}_{p}(q)\equiv g(q)^{p-1}\,\operatorname{mod}p.$

Finally, observe that

(2.3)
$\operatorname{pd}_{p^{\ell}}(q)=\frac{g(q^{p^{\ell}})}{g(q)}=\prod_{m=0}^{\ell-1}\frac{g(q^{p^{m+1}})}{g(q^{p^{m}})}=\prod_{m=0}^{\ell-1}\operatorname{pd}_{p}(q^{p^{m}}).$

##  3\. The Case of $\operatorname{pd}_{2^{\ell}}(q)\,\operatorname{mod}2$

The following can be written more succinctly, but is written to synchronize
better with the general case which follows directly after.

######  Lemma 3.1.

Working modulo $2$,

$\operatorname{pd}_{2}(q)\equiv
g(q)=\frac{1}{\operatorname{pd}(q)}\,\operatorname{mod}2$

and

$\operatorname{pd}_{2}(q)\equiv\sum_{n\geq 0}a_{n1}q^{n}\,\operatorname{mod}2$

where

$a_{n1}=\big{|}\\{\text{solutions to }n=k_{0}^{2}\,|\,k_{0}\geq 0,\,3\nmid
k_{0}\text{ or }k_{0}=0\\}\big{|}.$

###### Proof.

Using $p=2$ in Equation (2.2), we immediately get

$\operatorname{pd}_{2}(q)\equiv
g(q)=\frac{1}{\operatorname{pd}(q)}\,\operatorname{mod}2.$

The second statement is a rephrasing of Theorem 1.1,

(3.1) $\qquad\operatorname{pd}_{2}(q)\equiv 1+\sum_{k\geq 1,\,3\nmid
k}q^{k^{2}}\,\operatorname{mod}2.\qed$

The general case is given by the following.

######  Theorem 3.2.

For all $n\geq 0$ and $\ell\geq 1$,

$\operatorname{PD}_{2^{\ell}}(n)\equiv a_{n\ell}\,\operatorname{mod}2$

where

$a_{n\ell}=\Big{|}\\{\text{solutions to
}n=\sum_{m=0}^{\ell-1}2^{m}k_{m}^{2}\,|\,k_{m}\geq 0,\,3\nmid k_{m}\text{ or
}k_{m}=0\\}\Big{|}.$

###### Proof.

Using Equations (2.3) and (3.1), we see that

$\operatorname{pd}_{2^{\ell}}(q)=\prod_{m=0}^{\ell-1}\operatorname{pd}_{2}(q^{2^{m}})\equiv\prod_{m=0}^{\ell-1}\Big{(}1+\sum_{k\geq
1,\,3\nmid k}q^{2^{m}k^{2}}\Big{)}\,\operatorname{mod}2.$

The result follows. ∎

As a first easy application of Theorem 3.2, we prove the following newly
observed congruences of the form $\operatorname{PD}_{2^{\ell}}(32n+r)$.

######  Theorem 3.3.

For $n\geq 0$ and $r\in\\{5,7,10,13,14,15,20,21,23,26,28,29,30,31\\}$,

$\displaystyle\operatorname{PD}_{4}(32n+r)\equiv 0\,\operatorname{mod}2,$

and for $n\geq 0$,

$\displaystyle\operatorname{PD}_{8}(32n+24)\equiv 0\,\operatorname{mod}2.$

###### Proof.

The first congruence is immediate as the only quadratic residues modulo $32$
are $0,1,4,9,16,17,$ and $25$.

For the second congruence, we need to pair off solutions $(k_{0},k_{1},k_{2})$
of the equation

(3.2) $\displaystyle 32n+24=k_{0}^{2}+2k_{1}^{2}+4k_{2}^{2}.$

Note that we have $k_{0}^{2}\equiv 0$ mod $2$, thus $2\mid k_{0}$ and we can
write $k_{0}=2a$ for some integer $a$. This gives

$32n+24=4a^{2}+2k_{1}^{2}+4k_{2}^{2}.$

Note that with $(2a,k_{1},k_{2})$ also $(2k_{2},k_{1},a)$ is a solution of
(3.2), and for $k_{2}\neq a$ we will pair these two solutions off. This leaves
us with considering solutions of the form $(2a,k_{1},a)$ and the equation

$32n+24=4a^{2}+2k_{1}^{2}+4a^{2}=8a^{2}+2k_{1}^{2}.$

Note that we have $2k_{1}^{2}\equiv 0$ mod $4$, thus $2\mid k_{1}$ and we can
write $k_{1}=2b$ for some integer $b$. This gives the equation

$4n+3=a^{2}+b^{2}$

which is not satisfiable as the only quadratic residues mod $4$ are $0$ and
$1$. Thus, solutions of the form $(2a,k_{1},a)$ do not exist and the pairing
is complete. ∎

Our next result expands on and iterates the idea of pairing off solutions.

######  Theorem 3.4.

For $n\geq 0$, $\ell\geq 3$, and $0\leq s<2^{\ell-1}$,

$\displaystyle\operatorname{PD}_{2^{\ell}}(2^{\ell}n+2s)\equiv
1\,\operatorname{mod}2$

implies that $s$ is a quadratic residue modulo $2^{\ell-1}$. Equivalently, if
$0\leq s<2^{\ell-1}$ and $s$ is quadratic nonresidue modulo $2^{\ell-1}$, then

$PD_{2^{\ell}}(2^{\ell}n+2s)\equiv 0\,\operatorname{mod}2.$

Moreover, for all $n\geq 0$, $\ell\geq 2$, we have

$\displaystyle\operatorname{PD}_{2^{\ell}}(2n)\equiv
a^{*}_{n\ell}\,\operatorname{mod}2$

where

$a^{*}_{n\ell}=\big{|}\\{\text{solutions to
}n=k_{0}^{2}+2^{\ell-1}k_{1}^{2}\,|\,k_{m}\geq 0,\,3\nmid k_{m}\text{ or
}k_{m}=0\\}\big{|}.$

###### Proof.

We will be pairing off the solutions
$(k_{0},k_{1},k_{2},k_{3},\ldots,k_{\ell-1})$ of the equation

(3.3) $\displaystyle
2^{\ell}n+2s=k_{0}^{2}+2k_{1}^{2}+4k_{2}^{2}+8k_{3}^{2}+16k_{4}^{2}+\ldots+2^{\ell-1}k_{\ell-1}^{2}.$

Note that we have $k_{0}^{2}\equiv 0$ mod $2$, thus $2\mid k_{0}$ and we can
write $k_{0}=2a$ for some integer $a$. This gives

$2^{\ell}n+2s=4a^{2}+2k_{1}^{2}+4k_{2}^{2}+8k_{3}^{2}+16k_{4}^{2}+\ldots+2^{\ell-1}k_{\ell-1}^{2}.$

Note that with $(2a,k_{1},k_{2},k_{3},\ldots,k_{\ell-1})$ also
$(2k_{2},k_{1},a,k_{3},\ldots,k_{\ell-1})$ is a solution of (3.3), and for
$k_{2}\neq a$ we will pair these two solutions off. This leaves us with
considering solutions of the form $(2a,k_{1},a,k_{3},\ldots,k_{\ell-1})$ and
the equation

$2^{\ell}n+2s=4a^{2}+2k_{1}^{2}+4a^{2}+8k_{3}^{2}+16k_{4}^{2}+\ldots+2^{\ell-1}k_{\ell-1}^{2}.$

Note that with $(2a,k_{1},a,k_{3},k_{4}\ldots,k_{\ell-1})$ also
$(2k_{3},k_{1},k_{3},a,k_{4}\ldots,k_{\ell-1})$ is a solution of (3.3), and
for $k_{3}\neq a$ we will pair these two solutions off. This leaves us with
considering solutions of the form $(2a,k_{1},a,a,k_{4}\ldots,k_{\ell-1})$ and
the equation

$2^{\ell}n+2s=4a^{2}+2k_{1}^{2}+4a^{2}+8a^{2}+16k_{4}^{2}+\ldots+2^{\ell-1}k_{\ell-1}^{2}.$

This process can be iterated until we are left to consider solutions of the
form $(2a,k_{1},a,a,\ldots,a)$ and the equation

$2^{\ell}n+2s=4a^{2}+2k_{1}^{2}+4a^{2}+8a^{2}+16a^{2}+\ldots+2^{\ell-1}a^{2}=2k_{1}^{2}+2^{\ell}a^{2}.$

In particular,

$2^{\ell-1}n+s=k_{1}^{2}+2^{\ell-1}a^{2}.$

Thus, $s\equiv k_{1}^{2}$ mod $2^{\ell-1}$ and the first part of the theorem
follows. The second part of the theorem is now easy. ∎

We continue with a general auxiliary result.

######  Lemma 3.5.

If there exist $j$ and $r$ such that the congruence

$\displaystyle\operatorname{PD}_{2^{j}}(2^{j}n+r)\equiv
0\,\operatorname{mod}2$

holds for all $n\geq 0$, then

$\displaystyle\operatorname{PD}_{2^{\ell}}(2^{j}n+r)\equiv
0\,\operatorname{mod}2$

for all $n\geq 0$ and $\ell\geq j$.

###### Proof.

Let $j,\ell,n$ and $r$ be fixed. With

$L=\Big{\\{}\text{solutions to
}2^{j}n+r=s+\sum_{m=j}^{\ell-1}2^{m}k_{m}^{2}\,|\,s,k_{m}\geq 0,\,3\nmid
k_{m}\text{ or }k_{m}=0\Big{\\}}$

we have

$a_{2^{j}n+r,\ell}=\sum\big{\\{}a_{sj}\,|\,(s,k_{j},k_{j+1},\ldots,k_{\ell-1})\in
L\big{\\}},$

where $s\equiv r$ mod $2^{j}$ for all $(s,k_{j},k_{j+1},\ldots,k_{\ell-1})\in
L$. Hence,

$a_{sj}\equiv\operatorname{PD}_{2^{j}}(s)\equiv 0\,\operatorname{mod}2$

for all $(s,k_{j},k_{j+1},\ldots,k_{\ell-1})\in L$, thus

$\operatorname{PD}_{2^{\ell}}(2^{j}n+r)\equiv a_{2^{j}n+r,\ell}\equiv
0\,\operatorname{mod}2.\qed$

As a demonstration of how Theorem 3.4 and Lemma 3.5 combine to efficiently
prove congruences, we provide the following corollary. Some of these results
were proven with dissections in [8, Equations (3), (5), and (8)].

######  Corollary 3.6.

Let $n\geq 0$. For $\ell\geq 3$ and $r\in\\{4,6\\}$,

$\operatorname{PD}_{2^{\ell}}(8n+r)\equiv 0\,\operatorname{mod}2.$

For $\ell\geq 4$ and $r\in\\{4,6,10,12,14\\}$

$\operatorname{PD}_{2^{\ell}}(16n+r)\equiv 0\,\operatorname{mod}2.$

For $\ell\geq 5$ and $r\in\\{4,6,10,12,14,16,20,22,24,26,28,30\\}$

$\operatorname{PD}_{2^{\ell}}(32n+r)\equiv 0\,\operatorname{mod}2.$

In general for $\ell\geq j\geq 3$ and $0\leq s<2^{j-1}$ with $s$ not of the
form $0$ or $4^{a}(8b+1)$ for some $a,b\geq 0$,

$\operatorname{PD}_{2^{\ell}}(2^{j}n+2s)\equiv 0\,\operatorname{mod}2.$

###### Proof.

We apply Theorem 3.4 first with $\ell=3$. The quadratic nonresidues modulo $4$
are $2$ and $3$. Thus for $n\geq 0$ and $r\in\\{4,6\\}$,

$\operatorname{PD}_{8}(8n+r)\equiv 0\,\operatorname{mod}2.$

For $\ell=4$, the quadratic nonresidues modulo $8$ are $2,3,5,6,$ and $7$. So
for $r\in\\{4,6,10,12,14\\}$

$\operatorname{PD}_{16}(16n+r)\equiv 0\,\operatorname{mod}2.$

For $\ell=5$, the quadratic nonresidues modulo $16$ are
$2,3,5,6,7,8,10,11,12,13,14,$ and $15$. So for
$r\in\\{4,6,10,12,14,16,20,22,24,26,28,30\\}$

$\operatorname{PD}_{32}(32n+r)\equiv 0\,\operatorname{mod}2.$

In general, for $j\geq 3$, the quadratic residues mod $2^{j-1}$ are $0$ and
all numbers of the form $4^{a}(8b+1)$ for $a,b\geq 0$. So if $0\leq s<2^{j-1}$
is not of this form, then

$\operatorname{PD}_{2^{j}}(2^{j}n+2s)\equiv 0\,\operatorname{mod}2.$

The rest follows from Lemma 3.5. ∎

The function
$\operatorname{pd}_{2}(q)\equiv\frac{1}{\operatorname{pd}(q)}\,\operatorname{mod}2$
is well known. We end this section on an interesting new interpretation of its
reciprocal.

######  Theorem 3.7.

Working modulo $2$,

$\operatorname{pd}(q)\equiv\frac{1}{\operatorname{pd}_{2}(q)}\equiv
1+\sum_{n\geq 1}b_{n}q^{n}\,\operatorname{mod}2$

where

$b_{n}=\big{|}\\{\text{partitions of }n\,|\,\text{the multiplicity of each
part is odd}\\}\big{|}.$

###### Proof.

Observe that

$\displaystyle\frac{1-q^{6}}{(1-q)(1-q^{2})(1-q^{3})}$ $\displaystyle\equiv$
$\displaystyle\frac{(1-q^{3})^{2}}{(1-q)(1-q^{2})(1-q^{3})}$ $\displaystyle=$
$\displaystyle\frac{1+q+q^{2}}{1-q^{2}}\equiv 1+\sum_{k\geq
0}q^{2k+1}\,\operatorname{mod}2.$

Using the above equation in conjunction with the definition of
$\operatorname{pd}(q)$ in terms of the $f_{k}$, we get

$\operatorname{pd}(q)\equiv\prod_{n\geq 1}\Big{(}1+\sum_{k\geq
0}q^{(2k+1)n}\Big{)}\,\operatorname{mod}2$

and the result follows. ∎

Theorem 3.7 provides the hitherto undocumented congruence

$\operatorname{PD}(n)\equiv b_{n}\,\operatorname{mod}2,$

which links the number $\operatorname{PD}(n)$ of partitions of $n$ with
designated summands [18, A077285] to the number $b_{n}$ of partitions of $n$
with odd multiplicities [18, A055922]. This may be of use in understanding the
parity of partitions with odd multiplicities as studied in [12], [13], and
[17].

##  4\. An Alternate Characterization of
$\operatorname{pd}_{4}(q)\,\operatorname{mod}2$

In the case of $\ell=2$, Theorem 3.2 says

$\operatorname{PD}_{4}(n)\equiv a_{n2}\,\operatorname{mod}2$

where

$a_{n2}=\big{|}\\{\text{solutions to }n=k_{0}^{2}+2k_{1}^{2}\,|\,k_{m}\geq
0,\,3\nmid k_{m}\text{ or }k_{m}=0\\}\big{|}.$

In this section we give a remarkably explicit formula for
$a_{n2}\,\operatorname{mod}2$ together with a combinatorial proof. A
dissection proof for the same result is provided in Section 7.

######  Theorem 4.1.

For all $n\geq 0$,

$\displaystyle a_{n2}\equiv\begin{cases}1\,\operatorname{mod}2,&\text{if
}n=mk^{2}\text{ for }m,k\in\mathbb{Z}_{\geq 0}\text{ with }m\mid 6,\\\
0\,\operatorname{mod}2,&\text{otherwise}.\end{cases}$

###### Proof.

Let

$L(n)=\big{\\{}(k_{0},k_{1})\,\mid\,n=k_{0}^{2}+2k_{1}^{2},\,k_{m}\geq
0,\,3\nmid k_{m}\text{ or }k_{m}=0\big{\\}}$

so that $a_{n2}=|L(n)|$. Recall that $0,1$ are the only quadratic residues
$\,\operatorname{mod}3$.

Case 1: $n\equiv 1\,\operatorname{mod}3$

For $(a,b)\in L(n)$, $n=a^{2}+2b^{2}\equiv 1\,\operatorname{mod}3$ is only
possible for $b=0$. Thus,

$\displaystyle|L(n)|=\begin{cases}1,&\text{if }n=k^{2}\text{ for some
}k\in\mathbb{Z}_{\geq 0},\\\ 0,&\text{otherwise},\end{cases}$

confirming the theorem for $n\equiv 1\,\operatorname{mod}3$.

Case 2: $n\equiv 2\,\operatorname{mod}3$

For $(a,b)\in L(n)$, $n=a^{2}+2b^{2}\equiv 2\,\operatorname{mod}3$ is only
possible for $a=0$. Thus,

$\displaystyle|L(n)|=\begin{cases}1,&\text{if }n=2k^{2}\text{ for some
}k\in\mathbb{Z}_{\geq 0},\\\ 0,&\text{otherwise},\end{cases}$

confirming the theorem for $n\equiv 2\,\operatorname{mod}3$.

Case 3: $n\equiv 0\,\operatorname{mod}3$

The case $n\equiv 0\,\operatorname{mod}3$ is more involved. For $n=0$, we have
$L(0)=\\{(0,0)\\}$ and $a_{02}=|L(0)|=1$, confirming the theorem. Thus we may
assume $n>0$. Let

$\displaystyle
S(n)=\\{(k_{0},k_{1})\,\mid\,n=k_{0}^{2}+2k_{1}^{2},\,k_{m}\in\mathbb{Z}\\}.$

We define an equivalence relation on $S(n)$ by

$\displaystyle(a,b)\sim(c,d)\text{ iff }|a|=|c|\text{ and }|b|=|d|,\text{ for
}(a,b),(c,d)\in S(n).$

Next, we investigate the equivalence classes of $S(n)/\sim$.

For $(a,b)\in S(n),\ n=a^{2}+2b^{2}\equiv 0\,\operatorname{mod}3$ is only
possible for $a,b\equiv 0\,\operatorname{mod}3$ or $a,b\not\equiv
0\,\operatorname{mod}3$. In particular, in this case,

$\displaystyle
L(n)=\\{(k_{0},k_{1})\,\mid\,n=k_{0}^{2}+2k_{1}^{2},\,k_{m}>0,\,3\nmid
k_{m}\\},$

and $L(n)$ contains exactly one representative for each equivalence class
$[(a,b)]\in S(n)/\sim$ with $a,b\not\equiv 0\,\operatorname{mod}3$.

Noting

$\displaystyle
a^{2}+2b^{2}=n\quad\iff\quad\left(\dfrac{a+4b}{3}\right)^{2}+2\left(\dfrac{2a-b}{3}\right)^{2}=n,$

we define $\Psi:\mathbb{R}^{2}\to\mathbb{R}^{2}$ by

$\displaystyle\Psi(a,b)=\left(\dfrac{a+4b}{3},\dfrac{2a-b}{3}\right).$

Observe that $\Psi$ is an involution, i.e., $\Psi^{2}=\text{id}$.

We define a graph whose vertices are the equivalence classes of $S(n)/\sim$ by
connecting $[(a,b)]$ with $[(c,d)]$ if $\Psi(a^{\prime},b^{\prime})\in[(c,d)]$
for some $(a^{\prime},b^{\prime})\in[(a,b)]$. Since $\Psi$ is an involution,
$[(a,b)]$ is connected with $[(c,d)]$ if and only if $[(c,d)]$ is connected
with $[(a,b)]$, and the resulting graph is undirected. We will further
investigate this graph. We begin by discussing the most common occurrences,
treating exceptions last.

The equivalence classes $[(a,b)]\in S(n)/\sim$ with $a,b\not\equiv
0\,\operatorname{mod}3$ are in one-to-one correspondence with $L(n)$. With
exceptions to follow below, each such equivalence class is usually connected
with exactly one other equivalence class. In particular, for $a\not\equiv
b\,\operatorname{mod}3$, $[(a,b)]$ is connected to $[\Psi(a,b)]$ and for
$a\equiv b\,\operatorname{mod}3$, $[(a,b)]$ is connected to $[\Psi(a,-b)]$.
Similarly, with exceptions to follow, the equivalence classes $[(a,b)]\in
S(n)/\sim$ with $a,b\equiv 0\,\operatorname{mod}3$ are usually connected with
exactly two other equivalence classes, $[\Psi(a,b)]$ and $[\Psi(a,-b)]$.

Thus, the resulting graph will usually separate the equivalence classes
corresponding to $L(n)$ into connected pairs, hence $|L(n)|\equiv
0\,\operatorname{mod}2$.

Exceptions to the above happen if either of the following two cases occurs:

  * $(1)$

One equivalence class is connected to itself, that is, $[\Psi(a,b)]=[(a,b)]$.
This case occurs if and only if

$\displaystyle\left|\frac{2a-b}{3}\right|=|b|\iff a=-b\text{ or
}a=2b\iff(k,k)\text{ or }(2k,k)\in S(n),$

that is, if and only $n=3k^{2}$ or $6k^{2}$.

  * $(2)$

One equivalence class connects to the same equivalence class twice, that is,
$[\Psi(a,b)]=[\Psi(a,-b)]$. This case occurs if and only if

$\displaystyle\left|\frac{2a-b}{3}\right|=\left|\frac{2a+b}{3}\right|\iff
a=0\text{ or }b=0\iff(0,k)\text{ or }(k,0)\in S(n),$

that is, if and only $n=2k^{2}$ or $k^{2}$ with $k\equiv
0\,\operatorname{mod}3$.

This completes the proof of the Theorem. ∎

######  Remark 4.2.

The solution set of the Diophantine equation $a^{2}+2b^{2}=n$ can be further
investigated using unique factorization in the principal ideal domain
$\mathbb{Z}\big{[}\sqrt{-2}\big{]}$.

##  5\. A Recurrence Relation for
$\operatorname{PD}_{k}(n)\,\operatorname{mod}2$

While Theorem 3.2 provides an explicit description of
$\operatorname{PD}_{2^{\ell}}(n)$ mod $2$, for general
$\operatorname{PD}_{k}(n)$ mod $2$ we have the following computationally fast
recurrence.

######  Theorem 5.1.

For $n\geq 0$ and $k\geq 2$,

$\displaystyle\operatorname{PD}_{k}(n)+\sum_{\ell\geq
1,\,3\nmid\ell}\operatorname{PD}_{k}(n-\ell^{2})\equiv\begin{cases}1\,\operatorname{mod}2,&\text{if
}n=0\text{ or }n=km^{2},\,\,m\geq 1,3\nmid m,\\\
0\,\operatorname{mod}2,&\text{otherwise}.\end{cases}$

###### Proof.

Combining Equation (2.1) with Lemma 3.1 gives

$\displaystyle\operatorname{pd}_{k}(q)=\frac{g(q^{k})}{g(q)}\equiv\frac{\operatorname{pd}_{2}(q^{k})}{\operatorname{pd}_{2}(q)}\,\operatorname{mod}2.$

Substituting Equation (3.1), we have

(5.1) $\left(\sum_{n\geq
0}\operatorname{PD}_{k}(n)q^{n}\right)\left(1+\sum_{\ell\geq
1,\,3\nmid\ell}q^{\ell^{2}}\right)\equiv 1+\sum_{m\geq 1,\,3\nmid
m}q^{km^{2}}\,\operatorname{mod}2,$

and the result follows. ∎

##  6\. The Case of $\operatorname{pd}_{2}(q)\,\operatorname{mod}4$

Next we turn to $\operatorname{pd}_{2}(q)\,\operatorname{mod}4$ beginning with
a helpful result.

######  Lemma 6.1.

Working modulo $4$,

(6.1) $f_{1}^{3}\equiv\sum_{n\geq 0}q^{\frac{n(n+1)}{2}}\,\operatorname{mod}4$

and

(6.2) $qf_{8}^{3}\equiv\sum_{n\geq 1,\,2\nmid
n}q^{n^{2}}\,\operatorname{mod}4.$

###### Proof.

For any positive integers $k$ and $m$, we have the identity, [7, Lemma 5],

(6.3) $\displaystyle f_{2k}^{2m}\equiv f_{k}^{4m}\,\operatorname{mod}4.$

Therefore, the theta identity [1, Equation (2.2.13)] provides

$\displaystyle\sum_{n\geq
0}q^{\frac{n(n+1)}{2}}=\dfrac{(q^{2};q^{2})_{\infty}}{(q;q^{2})_{\infty}}=\frac{f_{2}^{2}}{f_{1}}\equiv
f_{1}^{3}\,\operatorname{mod}4,$

and the second identity is an immediate consequence of

$qf_{8}^{3}=q[f_{1}(q^{8})]^{3}\equiv q\sum_{n\geq 0}q^{4n(n+1)}=\sum_{n\geq
0}q^{(2n+1)^{2}}\,\operatorname{mod}4.\qed$

This result allows one to prove the following characterization:

######  Theorem 6.2.

For all $n\geq 0$,

$\operatorname{PD}_{2}(n)\equiv c_{n}\,\operatorname{mod}4$

where

$\displaystyle c_{n}$ $\displaystyle=\Big{|}\\{\text{solutions to
}n=\frac{3k_{0}(k_{0}+1)}{2}+\sum_{m\geq 1}k_{m}(12a_{m}+b_{m})\,|\,$
$\displaystyle\quad\qquad\;k_{m},a_{m}\geq 0,\,b_{m}\in\\{0,1,2,3\\},\;1\leq
k_{1}<k_{2}<k_{3}<\ldots\\}\Big{|}.$

###### Proof.

With Equation (6.3), we have the generating function

$\displaystyle\operatorname{pd}_{2}(q)=\dfrac{f_{4}f_{6}^{2}}{f_{1}f_{3}f_{12}}\equiv\dfrac{f_{3}^{3}f_{4}}{f_{1}f_{12}}\,\operatorname{mod}4.$

The result follows now easily from

$f_{3}^{3}=[f_{1}(q^{3})]^{3}\equiv\sum_{k\geq
0}q^{\frac{3k(k+1)}{2}}\,\operatorname{mod}4$

and

$\frac{1-x^{4}}{(1-x)(1-x^{12})}=\frac{1}{1+x^{4}+x^{8}}\sum_{n\geq
0}x^{n}=\sum_{\begin{subarray}{c}n=12a+b,\\\ a\geq 0,\\\
b\in\\{0,1,2,3\\}\end{subarray}}x^{n}.\qed$

We further prove an explicit characterization of $\operatorname{PD}_{2}(2n+1)$
modulo $4$ related to generalized pentagonal numbers by means of the following
dissection of $\operatorname{pd}_{2}(q)$ into even and odd powers.

######  Lemma 6.3.

Working modulo $4$,

(6.4)
$\displaystyle\operatorname{pd}_{2}(q)\equiv\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{2}+qf_{12}^{2}\,\operatorname{mod}4,$

splitting $\operatorname{pd}_{2}(q)$ into even and odd powers of $q$,
respectively.

###### Proof.

With Equation (6.3), we have

$\displaystyle\operatorname{pd}_{2}(q)=\dfrac{f_{4}f_{6}^{2}}{f_{1}f_{3}f_{12}}\equiv\dfrac{f_{3}^{3}f_{4}}{f_{1}f_{12}}\,\operatorname{mod}4.$

Using the identity (cf., [20, Equation (3.75)])

$\displaystyle\dfrac{f_{3}^{3}}{f_{1}}=\dfrac{f_{4}^{3}f_{6}^{2}}{f_{2}^{2}f_{12}}+q\dfrac{f_{12}^{3}}{f_{4}},$

and repeatedly using (6.3) yields

$\displaystyle\operatorname{pd}_{2}(q)\equiv\dfrac{f_{3}^{3}f_{4}}{f_{1}f_{12}}\equiv\dfrac{f_{4}^{4}f_{6}^{2}}{f_{2}^{2}f^{2}_{12}}+qf_{12}^{2}\equiv\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{2}+qf_{12}^{2}\,\operatorname{mod}4.$

Noting that $f_{2}^{3}/f_{6}$ and $f_{12}$ are power series in $q^{2}$, the
first term on the right-hand side gives all even power contributions while the
second term gives all odd power contributions due to the multiplication by the
extra factor of $q$. ∎

An application of Lemma 6.3 gives the following explicit characterization of
$\operatorname{PD}_{2}(2n+1)$ mod $4$.

######  Theorem 6.4.

For all $n\geq 0$,

$\displaystyle\operatorname{PD}_{2}(2n+1)\equiv d_{n}\,\operatorname{mod}4,$

where

$\displaystyle d_{n}=$ $\displaystyle\bigg{|}\\{\text{solutions to
}n=3j(3j-1)+3k(3k-1)=2\big{[}\binom{3j}{2}+\binom{3k}{2}\big{]}\,\Big{|}\,\
k,j\in\mathbb{Z}\\}\bigg{|}.$

###### Proof.

One has from (6.4),

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{2}(2n+1)q^{n}\equiv
f_{6}^{2}=[f_{1}(q^{6})]^{2}=\sum_{j,k\in\mathbb{Z}}(-1)^{j+k}q^{3[j(3j-1)+k(3k-1)]}\,\operatorname{mod}4,$

where we have used Euler’s pentagonal number theorem [1, Corollary 1.7]. The
result follows from noting that on the right-hand side the only contributions
with coefficient $-1$ occur with different parities of $j$ and $k$, hence will
come in pairs of solutions $(j,k)$, $(k,j)$ each time and thus contribute
$2(-1)^{j+k}\equiv 2\operatorname{mod}4$ to the count. ∎

Similarly, Lemma 6.3 also provides an explicit characterization of
$\operatorname{PD}_{2}(2n)$ mod $4$.

######  Theorem 6.5.

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{2}(2n)q^{n}$
$\displaystyle\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{2}\equiv\Big{(}1+\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}\Big{)}^{2}$ $\displaystyle\equiv 1+2\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}+\sum_{k,\ell\geq 1,\,3\nmid
k,\ell}q^{k^{2}+\ell^{2}}\,\operatorname{mod}4.$

###### Proof.

Note that by [17, Equations (2) and (4)], we have

$\displaystyle\dfrac{f_{1}^{3}}{f_{3}}\equiv 1+q\dfrac{f_{9}^{3}}{f_{3}}\equiv
1+q\sum_{j\in\mathbb{Z}}q^{3j(3j-2)}=1+\sum_{j\in\mathbb{Z}}q^{(3j-1)^{2}}=1+\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}\,\operatorname{mod}2,$

which implies

(6.5)
$\displaystyle\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{2}\equiv\Big{(}1+\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}\Big{)}^{2}\,\operatorname{mod}4.$

Combining (6.4) with Equation (6.5) gives

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{2}(2n)q^{n}$
$\displaystyle\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{2}\equiv\Big{(}1+\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}\Big{)}^{2}\,\operatorname{mod}4.\qed$

As an easy application of these results, we prove a few observed congruences.
Alternative proofs using dissections can be found in [3, Corollary 1.4,
Theorem 1.5].

######  Theorem 6.6.

For $n\geq 1,$ we have

$\displaystyle\operatorname{PD}_{2}(3n)\equiv 0\,\operatorname{mod}4.$

For all $n\geq 1$ with $6\nmid n$, we have

$\displaystyle\operatorname{PD}_{2}(2n+1)$ $\displaystyle\equiv
0\,\operatorname{mod}4.$

###### Proof.

The first congruence is easily seen by noting from [2, Theorem 22] and
Equation (6.3) that

$\displaystyle\sum_{n\geq 0}\operatorname{PD}_{2}(3n)q^{n}$
$\displaystyle=\dfrac{f_{2}^{2}f_{6}^{4}}{f_{1}^{4}f_{12}^{2}}\equiv\dfrac{f_{1}^{4}f_{6}^{4}}{f_{1}^{4}f_{6}^{4}}\equiv
1\operatorname{mod}4.$

The second congruence is an immediate consequence of Theorem 6.4 as

$n=3j(3j-1)+3k(3k-1)=2\big{[}\binom{3j}{2}+\binom{3k}{2}\big{]}$

implies that $n\equiv 0$ mod $6$. ∎

We end this section with yet another nice characterization of
$\operatorname{PD}_{2}(n)$ mod $4$.

######  Theorem 6.7.

$\displaystyle\operatorname{pd}_{2}(q)\equiv 1+\Big{(}\sum_{k\geq 1,\,3\nmid
k}q^{k^{2}}\Big{)}\Big{(}1+2\sum_{k\geq
1}q^{k^{2}}\Big{)}=1+\Big{(}\sum_{k\geq 1,\,3\nmid
k}q^{k^{2}}\Big{)}\Big{(}\sum_{k\in\mathbb{Z}}q^{k^{2}}\Big{)}\,\operatorname{mod}4.$

###### Proof.

Starting from [2, Equation (3.13)], we have

$\displaystyle\operatorname{pd}_{2}(q)$
$\displaystyle=\frac{\sum_{j\in\mathbb{Z}}q^{(3j)^{2}}-\sum_{j\in\mathbb{Z}}q^{(3j+1)^{2}}}{1+2\sum_{k\geq
1}(-1)^{k}q^{k^{2}}}=\frac{1+2\sum_{k\geq 1,\,3\mid k}q^{k^{2}}-\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}}{1+2\sum_{k\geq 1}(-1)^{k}q^{k^{2}}}$
$\displaystyle\equiv\frac{1+2\sum_{k\geq 1,\,3\mid k}q^{k^{2}}+3\sum_{k\geq
1,\,3\nmid k}q^{k^{2}}}{1+2\sum_{k\geq 1}q^{k^{2}}}=\frac{1+2\sum_{k\geq
1}q^{k^{2}}+\sum_{k\geq 1,\,3\nmid k}q^{k^{2}}}{1+2\sum_{k\geq 1}q^{k^{2}}}$
$\displaystyle=1+\frac{\sum_{k\geq 1,\,3\nmid k}q^{k^{2}}}{1+2\sum_{k\geq
1}q^{k^{2}}}\equiv 1+\Big{(}\sum_{k\geq 1,\,3\nmid
k}q^{k^{2}}\Big{)}\Big{(}1+2\sum_{k\geq
1}q^{k^{2}}\Big{)}\,\operatorname{mod}4.\qed$

##  7\. A Dissection Proof for Theorem 4.1

In this section, we will prove Theorem 4.1 with the help of dissections. This
proof will reuse some of the identities from the last section. In addition, we
will also need the following auxiliary result.

######  Lemma 7.1.

Working modulo $2$,

(7.1) $\displaystyle q^{2}f^{6}_{2}f^{6}_{6}\equiv
q^{2}f^{3}_{16}+q^{6}f^{3}_{48}\,\operatorname{mod}2.$

###### Proof.

Using the identity (cf., [20, Equation (3.12)])

$\displaystyle\dfrac{1}{f_{1}f_{3}}=\dfrac{f_{8}^{2}f_{12}^{5}}{f_{2}^{2}f_{4}f_{6}^{4}f_{24}^{2}}+q\dfrac{f_{4}^{5}f_{24}^{2}}{f_{2}^{4}f_{6}^{2}f_{8}^{2}f_{12}},$

we have

$\displaystyle
f^{3}_{1}f^{3}_{3}=\dfrac{f^{4}_{1}f^{4}_{3}f_{8}^{2}f_{12}^{5}}{f_{2}^{2}f_{4}f_{6}^{4}f_{24}^{2}}+q\dfrac{f^{4}_{1}f^{4}_{3}f_{4}^{5}f_{24}^{2}}{f_{2}^{4}f_{6}^{2}f_{8}^{2}f_{12}}\equiv
f_{4}^{3}+qf_{12}^{3}\,\operatorname{mod}2,$

where we make repeated use of the identity $f_{2k}\equiv
f_{k}^{2}\,\operatorname{mod}2$. Raising the last equation to the fourth power
gives

$\displaystyle f^{6}_{2}f^{6}_{6}\equiv
f_{16}^{3}+q^{4}f_{48}^{3}\,\operatorname{mod}2,$

and the result follows. ∎

Alternate proof of Theorem 4.1. For a proof of Theorem 4.1, we start with the
generating function

$\displaystyle\operatorname{pd}_{4}(q)=\dfrac{f_{4}f_{6}f_{8}f_{12}}{f_{1}f_{2}f_{3}f_{24}}\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}\,\operatorname{mod}2.$

Applying Equation (2.3) and Lemma 6.3 one can provide the even and odd power
dissection of $\operatorname{pd}_{4}(q)$,

$\displaystyle\operatorname{pd}_{4}(q)$
$\displaystyle=\operatorname{pd}_{2}(q)\operatorname{pd}_{2}(q^{2})$
$\displaystyle\equiv\left(\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{2}+qf_{12}^{2}\right)\\!\\!\left(\left[\dfrac{f_{4}^{3}}{f_{12}}\right]^{2}+q^{2}f_{24}^{2}\right)\equiv\left(\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{2}+qf_{6}^{4}\right)\\!\\!\left(\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{4}+q^{2}f_{6}^{8}\right)$
$\displaystyle\equiv\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{6}+q^{2}f_{2}^{6}f_{6}^{6}+qf_{2}^{12}+q^{3}f_{6}^{12}$
(7.2)
$\displaystyle\equiv\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{6}+q^{2}f_{16}^{3}+q^{6}f_{48}^{3}+qf_{8}^{3}+q^{3}f_{24}^{3}\,\operatorname{mod}2,$

where the last line is a result of Equation (7.1) and once again applying the
identity $f_{2k}\equiv f_{k}^{2}\,\operatorname{mod}2$. In Equation (7.2),
noting that

$\left[\dfrac{f_{2}^{3}}{f_{6}}\right]^{6}\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{12}\equiv[\operatorname{pd}_{4}(q)]^{4}\equiv\operatorname{pd}_{4}(q^{4})\,\operatorname{mod}2$

and applying Equation (6.2) to $q^{r}f_{8r}^{3}=q^{r}[f_{8}(q^{r})]^{3}$,
$r\in\\{1,2,3,6\\},$ yields

$\displaystyle\operatorname{pd}_{4}(q)\equiv\operatorname{pd}_{4}(q^{4})+\sum_{n\geq
1,\,2\nmid
n}\big{[}q^{n^{2}}+q^{2n^{2}}+q^{3n^{2}}+q^{6n^{2}}\big{]}\,\operatorname{mod}2.$

Raising this equation to the fourth power yields

$\displaystyle\operatorname{pd}_{4}(q^{4})\equiv\operatorname{pd}_{4}(q^{16})+\sum_{n\geq
1,\,2\nmid
n}\big{[}q^{(2n)^{2}}+q^{2(2n)^{2}}+q^{3(2n)^{2}}+q^{6(2n)^{2}}\big{]}\,\operatorname{mod}2,$

and combining the last two equations results in

$\displaystyle\operatorname{pd}_{4}(q)\equiv\operatorname{pd}_{4}(q^{4^{2}})+\sum_{\begin{subarray}{c}n\geq
1,\,2\nmid n,\\\
i\,\in\\{0,1\\}\end{subarray}}\big{[}q^{(2^{i}n)^{2}}+q^{2(2^{i}n)^{2}}+q^{3(2^{i}n)^{2}}+q^{6(2^{i}n)^{2}}\big{]}\,\operatorname{mod}2.$

Therefore, iterating one obtains, for $\ell\geq 1$,

$\displaystyle\operatorname{pd}_{4}(q)\equiv\operatorname{pd}_{4}(q^{4^{\ell}})+\sum_{\begin{subarray}{c}n\geq
1,\,2\nmid n,\\\ 0\leq
i<\ell\end{subarray}}\big{[}q^{(2^{i}n)^{2}}+q^{2(2^{i}n)^{2}}+q^{3(2^{i}n)^{2}}+q^{6(2^{i}n)^{2}}\big{]}\,\operatorname{mod}2,$

where the first summand on the right-hand side contains only powers of
$q^{4^{\ell}}$. For $\ell\to\infty$, this results in the identity

$\displaystyle\operatorname{pd}_{4}(q)$ $\displaystyle\equiv 1+\sum_{n\geq
1,\,2\nmid n,\,i\geq
0}\big{[}q^{(2^{i}n)^{2}}+q^{2(2^{i}n)^{2}}+q^{3(2^{i}n)^{2}}+q^{6(2^{i}n)^{2}}\big{]}$
$\displaystyle\equiv 1+\sum_{n\geq
1}\big{[}q^{n^{2}}+q^{2n^{2}}+q^{3n^{2}}+q^{6n^{2}}\big{]}\,\operatorname{mod}2,$

completing the proof of Theorem 4.1. ∎
We will end this section with a few Rogers-Ramanujan-type identities for
$\operatorname{pd}_{4}(q)$.

Denote by $\varphi(q),\ \psi(q),$ and $f(a,b)$ Ramanujan’s first, second, and
general theta functions, respectively, defined by, cf. [4, Entry 22(i), Entry
22(ii), Equation (18.1) and Entry 19],

$\displaystyle\varphi(q)$
$\displaystyle=f(q,q)=\sum_{n\in\mathbb{Z}}q^{n^{2}}=\frac{(q^{2};q^{2})_{\infty}(-q;q^{2})_{\infty}}{(q;q^{2})_{\infty}(-q^{2};q^{2})_{\infty}}=\frac{f_{2}^{2}(-q;q)_{\infty}}{f_{1}(-q^{2};q^{2})^{2}_{\infty}}=\dfrac{f_{2}^{5}}{f_{1}^{2}f_{4}^{2}},$
$\displaystyle\psi(q)$ $\displaystyle=f(q,q^{3})=\sum_{n\geq
0}q^{n(n+1)/2}=\frac{(q^{2};q^{2})_{\infty}}{(q;q^{2})_{\infty}}=\dfrac{f_{2}^{2}}{f_{1}},$
$\displaystyle f(a,b)$
$\displaystyle=\sum_{n\in\mathbb{Z}}a^{n(n+1)/2}b^{n(n-1)/2}=(-a;ab)_{\infty}(-b;ab)_{\infty}(ab;ab)_{\infty}.$

Then, one can show the following identities by means of Theorem 4.1.

######  Theorem 7.2.

Working modulo $2$, the following identities hold:

$\displaystyle\quad\
\operatorname{pd}_{4}(q)\\!\equiv\\!\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}\\!$
$\displaystyle\equiv\dfrac{f(q^{6},q^{10})}{\psi(q)}+\dfrac{f(q^{18},q^{30})}{\psi(q^{3})}-1$
$\displaystyle\equiv
f_{1}^{13}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}+f_{3}^{13}(q^{18};q^{48})_{\infty}(q^{30};q^{48})_{\infty}-1$
$\displaystyle\equiv
f_{1}^{13}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}+q^{3}f_{3}^{13}(q^{6};q^{48})_{\infty}(q^{42};q^{48})_{\infty}$
$\displaystyle\equiv
f_{3}^{13}(q^{18};q^{48})_{\infty}(q^{30};q^{48})_{\infty}+qf_{1}^{13}(q^{2};q^{16})_{\infty}(q^{14};q^{16})_{\infty}\,\operatorname{mod}2.$

###### Proof.

Theorem 4.1 claims

(7.3)
$\displaystyle\operatorname{pd}_{4}(q)=\frac{f_{4}f_{6}f_{8}f_{12}}{f_{1}f_{2}f_{3}f_{24}}\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}\\!\equiv\dfrac{1}{2}\big{[}\varphi(q)+\varphi(q^{2})+\varphi(q^{3})+\varphi(q^{6})-2\big{]}\,\\!\operatorname{mod}2,$

where we make repeated use of the identity $f_{2k}\equiv
f_{k}^{2}\,\operatorname{mod}2$. The identities [4, Example (iv), p. 51] and
[4, Corollary (ii), p. 49] imply, working modulo $4$ and $2$, respectively,

(7.4) $\displaystyle\varphi(q)+\varphi(q^{2})$
$\displaystyle\equiv\varphi(-q)+\varphi(q^{2})=2\dfrac{f^{2}(q^{3},q^{5})}{\psi(q)}\,\operatorname{mod}4,$
(7.5) $\displaystyle f^{2}(q^{3},q^{5})$ $\displaystyle\equiv
f(q^{6},q^{10})\equiv\psi(q)+qf(q^{2},q^{14})\,\operatorname{mod}2,$

while for $\psi(q)$ we have the identity

(7.6) $\displaystyle\psi(q)=\dfrac{f_{2}^{2}}{f_{1}}\equiv
f_{1}^{3}\,\operatorname{mod}2.$

Substituting Equations (7.4) and (7.6) into the right-hand side of Equation
(7.3) and applying the definition of $f(a,b)$, working modulo $2$ yields

$\displaystyle\operatorname{pd}_{4}(q)\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}$
$\displaystyle\equiv\dfrac{1}{2}\big{[}\varphi(q)+\varphi(q^{2})\big{]}+\dfrac{1}{2}\big{[}\varphi(q^{3})+\varphi(q^{6})\big{]}-1$
$\displaystyle\equiv\dfrac{f^{2}(q^{3},q^{5})}{\psi(q)}+\dfrac{f^{2}(q^{9},q^{15})}{\psi(q^{3})}-1\equiv\dfrac{f(q^{6},q^{10})}{f_{1}^{3}}+\dfrac{f(q^{18},q^{30})}{f_{3}^{3}}-1$
$\displaystyle\equiv
f_{1}^{13}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}+f_{3}^{13}(q^{18};q^{48})_{\infty}(q^{30};q^{48})_{\infty}-1\,\operatorname{mod}2.$

Now applying Equations (7.5) and (7.6) and the definition of $f(a,b)$, working
modulo $2$ one obtains

$\displaystyle\operatorname{pd}_{4}(q)\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}$
$\displaystyle\equiv\dfrac{f^{2}(q^{3},q^{5})}{\psi(q)}+\dfrac{f^{2}(q^{9},q^{15})}{\psi(q^{3})}-1\equiv\dfrac{f(q^{6},q^{10})}{\psi(q)}+q^{3}\dfrac{f(q^{6},q^{42})}{\psi(q^{3})}$
$\displaystyle\equiv
f_{1}^{13}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}+q^{3}f_{3}^{13}(q^{6};q^{48})_{\infty}(q^{42};q^{48})_{\infty}\,\operatorname{mod}2$

and

$\displaystyle\operatorname{pd}_{4}(q)\equiv\left[\dfrac{f_{1}^{3}}{f_{3}}\right]^{3}$
$\displaystyle\equiv\dfrac{f^{2}(q^{3},q^{5})}{\psi(q)}+\dfrac{f^{2}(q^{9},q^{15})}{\psi(q^{3})}-1\equiv\dfrac{f(q^{18},q^{30})}{\psi(q^{3})}+q\dfrac{f(q^{2},q^{14})}{\psi(q)}$
$\displaystyle\equiv
f_{3}^{13}(q^{18};q^{48})_{\infty}(q^{30};q^{48})_{\infty}+qf_{1}^{13}(q^{2};q^{16})_{\infty}(q^{14};q^{16})_{\infty}\,\operatorname{mod}2.\qed$

##  8\. The Case of $\operatorname{pd}_{3^{\ell}}(q)\,\operatorname{mod}3$

######  Definition 8.1.

Let

$h(q)=\frac{f_{1}(q)^{2}}{f_{2}(q)},$

noting by Gauss’s square power identity [1, Equation (2.2.12)],

(8.1) $h(q)=\dfrac{(q;q)_{\infty}}{(-q;q)_{\infty}}=1+2\sum_{m\geq
1}(-1)^{m}q^{m^{2}}.$

Using the previous definition, we prove the following.

######  Theorem 8.2.

Working modulo $3$,

$\operatorname{pd}_{3}(q)\equiv\sum_{n\geq 0}e_{n1}q^{n}\,\operatorname{mod}3$

where

$\displaystyle e_{n1}$ $\displaystyle=\big{|}\\{\text{solutions to
}n=k_{0}^{2}+3k_{1}^{2}\,|\,$
$\displaystyle\quad\qquad\;k_{m}\in\mathbb{Z}\text{ or }\mathbb{N}\text{ when
$k_{m}$ is even or odd, respectively}\\}\big{|}.$

###### Proof.

Using $p=3$ in Equation (2.2), we get

$\operatorname{pd}_{3}(q)\equiv g(q)^{2}\,\operatorname{mod}3.$

As

$g(q)=\frac{f_{1}(q)f_{2}(q)f_{3}(q)}{f_{6}(q)}\equiv
h(q)^{2}\,\operatorname{mod}3,$

it follows, in a similar vein to Equation (2.2), that

(8.2) $\displaystyle\operatorname{pd}_{3}(q)\equiv h(q)^{4}=h(q)h(q)^{3}\equiv
h(q)h(q^{3})\,\operatorname{mod}3.$

By Equation (8.1), we may write

$h(q)\equiv\sum_{m\in\mathbb{Z},\,2\mid m}q^{m^{2}}+\sum_{m\geq 1,\,2\nmid
m}q^{m^{2}}\,\operatorname{mod}3.$

The result follows. ∎

######  Remark 8.3.

Note that $e_{n1}$ gives an interesting alternate characterization of
$\operatorname{PD}_{3}(n)\,\operatorname{mod}3$ to the one given in [7,
Theorem 2] where it was shown that,
for $n\geq 1$,

$\operatorname{PD}_{3}(n)\equiv\big{|}\\{\text{solutions to
}n=k(k+1)+3m(m+1)+1\,|\,k,m\geq 0\\}\big{|}\,\operatorname{mod}3.$

Note also that we have

$n=k(k+1)+3m(m+1)+1\iff 4n=(2k+1)^{2}+3(2m+1)^{2}.$

Thus, for $n\geq 1$, we can also write

$\displaystyle\operatorname{PD}_{3}(n)$
$\displaystyle\equiv\big{|}\\{\text{solutions to
}4n=k_{0}^{2}+3k_{1}^{2}\,|\,k_{m}\geq 0,k_{m}\text{ odd}\\}\big{|}$
$\displaystyle\equiv\big{|}\\{\text{solutions to
}4n=k_{0}^{2}+3k_{1}^{2}\,|\,k_{m}\in\mathbb{Z},k_{m}\text{
odd}\\}\big{|}\,\operatorname{mod}3.$

Theorem 8.2 extends quite naturally as seen by the next result.

######  Theorem 8.4.

For all $n\geq 0$,

$\operatorname{PD}_{3^{\ell}}(n)\equiv e_{n\ell}\,\operatorname{mod}3$

where

$\displaystyle e_{n\ell}$ $\displaystyle=\Big{|}\\{\text{solutions to
}n=k_{0}^{2}+\sum_{m=1}^{\ell-1}3^{m}(k_{m}^{2}+k_{m}^{\prime
2})+3^{\ell}k_{\ell}^{2}\,|\,$
$\displaystyle\quad\qquad\;k_{m},k_{m}^{\prime}\in\mathbb{Z}\text{ or
}\mathbb{N}\text{ when $k_{m},k_{m}^{\prime}$ is even or odd,
respectively}\\}\Big{|}.$

###### Proof.

Using Equations (2.3) and (8.2),

$\operatorname{pd}_{3^{\ell}}(q)=\prod_{m=0}^{\ell-1}\operatorname{pd}_{3}(q^{3^{m}})\equiv
h(q)\,\bigg{[}\prod_{m=1}^{\ell-1}h(q^{3^{m}})^{2}\bigg{]}\,h(q^{3^{\ell}})\,\operatorname{mod}3.$

The result follows. ∎

As an easy application of Theorem 8.4, we provide very short proofs of the
following congruences for which proofs using dissections can be found in [7,
Theorem 3] and [8, Theorem 3] for $\operatorname{PD}_{3}(9n+6)$ and
$\operatorname{PD}_{3k}(3n+2)$, respectively.

######  Theorem 8.5.

For $n\geq 0$,

$\displaystyle\operatorname{PD}_{3}(9n+6)\equiv 0\,\operatorname{mod}3.$

For all $n\geq 0$ and $\ell\geq 1$,

$\displaystyle\operatorname{PD}_{3^{\ell}}(3n+2)\equiv
0\,\operatorname{mod}3.$

###### Proof.

For the first congruence, if $e_{n1}\neq 0$ in Theorem 8.4, then we can write,
for suitable integers $k_{0},k_{1},$

$\displaystyle n=k_{0}^{2}+3k_{1}^{2}.$

A straightforward calculation shows that $n\not\equiv 6\,\operatorname{mod}9$
as $0,1,4,7$ are the only quadratic residues modulo 9. Hence, $e_{n1}=0$ for
all $n\equiv 6\,\operatorname{mod}9$.

The second congruence follows similarly. If $e_{n\ell}\neq 0$ in Theorem 8.4,
then, since $0,1$ are the only quadratic residues modulo 3, we can write

$\displaystyle n=k_{0}^{2}+\sum_{m=1}^{\ell-1}3^{m}(k_{m}^{2}+k_{m}^{\prime
2})+3^{\ell}k_{\ell}^{2}\equiv k_{0}^{2}\not\equiv 2\,\operatorname{mod}3.$

Thus, $e_{n\ell}=0$ for all $n\equiv 2\,\operatorname{mod}3$. ∎

As a further application of Theorem 8.4, we prove the following newly observed
congruences. We have not found the first congruence in the literature, but
expect it to be known.

######  Theorem 8.6.

For $n\geq 1$,

$\displaystyle\operatorname{PD}_{3}(2n)\equiv 0\,\operatorname{mod}3.$

For all $n\geq 0$ and $\ell\geq 3$,

$\displaystyle\operatorname{PD}_{3^{\ell}}(27n+9)\equiv
0\,\operatorname{mod}3.$

For all $n\geq 0$ and $\ell\neq 2$,

$\displaystyle\operatorname{PD}_{3^{\ell}}(27n+18)\equiv
0\,\operatorname{mod}3.$

###### Proof.

The first congruence is an immediate consequence of Theorem 1.2 as
$n=k(k+1)+3m(m+1)+1$ is odd for nonnegative integers $k,m$.

The case $\ell=1$ of the third congruence follows similarly to that of the
first congruence in the previous theorem. In particular, a straightforward
calculation shows that $n=k_{0}^{2}+3k_{1}^{2}\not\equiv
18\,\operatorname{mod}27$ as $0,1,4,7,9,10,13,16,19,22,$ and $25$ are the only
quadratic residues modulo 27. Hence, $e_{n1}=0$ for all $n\equiv
18\,\operatorname{mod}27$.

For $\ell\geq 3$ in the third congruence, we argue as follows. We want to
count the number of solutions to the Diophantine equation

(8.3) $\displaystyle
27n+18=k_{0}^{2}+\sum_{m=1}^{\ell-1}3^{m}(k_{m}^{2}+k_{m}^{\prime
2})+3^{\ell}k_{\ell}^{2}$

modulo 3 with sign constraints as stated in Theorem 8.2. Note that we have
$k_{0}^{2}\equiv 0\,\operatorname{mod}3$, thus $3\mid k_{0}$ and we can write
$k_{0}=3a$ for some integer $a$. This gives

(8.4) $\displaystyle 27n+18=3(k_{1}^{2}+k_{1}^{\prime
2})+9(a^{2}+k_{2}^{2}+k_{2}^{\prime 2})+\ldots+3^{\ell}k_{\ell}^{2}.$

In particular, $3(k_{1}^{2}+k_{1}^{\prime 2})\equiv 0\,\operatorname{mod}9$.
Thus, $3\mid(k_{1}^{2}+k_{1}^{\prime 2})$, but as 0 and 1 are the only
quadratic residues modulo 3, we conclude $3\mid k_{1},k_{1}^{\prime}$. Hence
$3(k_{1}^{2}+k_{1}^{\prime 2})\equiv 0\,\operatorname{mod}27$, and Equation
(8.4) implies

$\displaystyle 9(a^{2}+k_{2}^{2}+k_{2}^{\prime 2})\equiv
18\,\operatorname{mod}27,$

so that we conclude

(8.5) $\displaystyle a^{2}+k_{2}^{2}+k_{2}^{\prime 2}\equiv
2\,\operatorname{mod}3.$

It is immediate that, together with
$(3a,k_{1},k_{1}^{\prime},k_{2},k_{2}^{\prime},k_{3},k_{3}^{\prime},\ldots,k_{\ell})$,
also
$(3k_{2},k_{1},k_{1}^{\prime},k_{2}^{\prime},a,k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
and
$(3k^{\prime}_{2},k_{1},k_{1}^{\prime},a,k_{2},k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
solve Equation (8.3). As a result, the mapping

$\displaystyle\Psi((k_{0},k_{1},k_{1}^{\prime},k_{2},k_{2}^{\prime},k_{3},k_{3}^{\prime},\ldots,k_{\ell}))=(3k_{2},k_{1},k_{1}^{\prime},k_{2}^{\prime},k_{0}/3,k_{3},k_{3}^{\prime},\ldots,k_{\ell}),$

defines a bijection on the solution set of (8.3) with $\Psi^{3}=\text{id}$.
The proof will be finished by showing that each orbit of the solution set
under iterates of $\Psi$ has order $3$. To see this, by way of contradiction,
suppose

$\displaystyle(3a,k_{1},k_{1}^{\prime},k_{2},k_{2}^{\prime},k_{3},k_{3}^{\prime},\ldots,k_{\ell})=(3k_{2},k_{1},k_{1}^{\prime},k_{2}^{\prime},a,k_{3},k_{3}^{\prime},\ldots,k_{\ell}).$

Comparing components yields $a=k_{2}=k_{2}^{\prime}$, hence
$a^{2}+k_{2}^{2}+k_{2}^{\prime 2}\equiv 0\,\operatorname{mod}3$, contradicting
(8.5) and completing the proof of the third congruence.

The second congruence follows analogously by systematically replacing 18 by 9
in the proof for $\ell\geq 3$. ∎

It is noteworthy that we can recreate some of our structural results on
$\operatorname{pd}_{2^{\ell}}(q)$ modulo $2$ from Section 3 in the context of
$\operatorname{pd}_{3^{\ell}}(q)$ modulo $3$. In particular, the grouping
argument of the last theorem once again can be iterated.

######  Theorem 8.7.

For all $n\geq 0$, $\ell\geq 2$, we have

$\displaystyle\operatorname{PD}_{3^{\ell}}(3n)\equiv
e^{*}_{n\ell}\,\operatorname{mod}3$

where

$\displaystyle e^{*}_{n\ell}$ $\displaystyle=\big{|}\\{\text{solutions to
}n=k_{0}^{2}+k_{0}^{\prime 2}+3^{\ell-1}(k_{1}^{2}+k_{1}^{\prime 2})\,|\,$
$\displaystyle\quad\qquad\;k_{m},k_{m}^{\prime}\in\mathbb{Z}\text{ or
}\mathbb{N}\text{ when $k_{m},k_{m}^{\prime}$ is even or odd,
respectively}\\}\big{|}.$

###### Proof.

We will be grouping the solutions
$(k_{0},k_{1},k_{1}^{\prime},k_{2},k_{2}^{\prime},k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
of the equation

(8.6) $\displaystyle 3n=k_{0}^{2}+3(k_{1}^{2}+k_{1}^{\prime
2})+9(k_{2}^{2}+k_{2}^{\prime 2})+27(k_{3}^{2}+k_{3}^{\prime
2})+\ldots+3^{\ell}k_{\ell}^{2}$

into triplets. Note that we have $k_{0}^{2}\equiv 0$ mod $3$, thus $3\mid
k_{0}$ and we can write $k_{0}=3a$ for some integer $a$. This gives

$3n=9a^{2}+3(k_{1}^{2}+k_{1}^{\prime 2})+9(k_{2}^{2}+k_{2}^{\prime
2})+27(k_{3}^{2}+k_{3}^{\prime 2})+\ldots+3^{\ell}k_{\ell}^{2}.$

Note that with
$(3a,k_{1},k_{1}^{\prime},k_{2},k_{2}^{\prime},k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
also
$(3k_{2},k_{1},k_{1}^{\prime},k_{2}^{\prime},a,k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
and
$(3k_{2}^{\prime},k_{1},k_{1}^{\prime},a,k_{2},k_{3},k_{3}^{\prime},\ldots,k_{\ell})$
are solutions of (8.6), which provide us with a triplet of solutions unless
$k_{2}=k_{2}^{\prime}=a$. This leaves us with considering solutions of the
form $(3a,k_{1},k_{1}^{\prime},a,a,k_{3},k_{3}^{\prime},\ldots,k_{\ell})$ and
the equation

$3n=9a^{2}+3(k_{1}^{2}+k_{1}^{\prime
2})+9(a^{2}+a^{2})+27(k_{3}^{2}+k_{3}^{\prime
2})+\ldots+3^{\ell}k_{\ell}^{2}.$

Note that with
$(3a,k_{1},k_{1}^{\prime},a,a,k_{3},k_{3}^{\prime},\ldots,k_{\ell})$ also
$(3k_{3},k_{1},k_{1}^{\prime},k_{3},k_{3},k_{3}^{\prime},a,\ldots,k_{\ell})$
and
$(3k_{3}^{\prime},k_{1},k_{1}^{\prime},k_{3}^{\prime},k_{3}^{\prime},a,k_{3},\ldots,k_{\ell})$
are solutions of (8.6), which provide us with a triplet of solutions unless
$k_{3}=k_{3}^{\prime}=a$. This leaves us with considering solutions of the
form $(3a,k_{1},k_{1}^{\prime},a,a,a,a,k_{4},k_{4}^{\prime},\ldots,k_{\ell})$,
and this process can be iterated until we are left to consider solutions of
the form $(3a,k_{1},k_{1}^{\prime},a,a,\ldots,a,a,k_{\ell})$ and the equation

$\displaystyle 3n$ $\displaystyle=9a^{2}+3(k_{1}^{2}+k_{1}^{\prime
2})+9(a^{2}+a^{2})+27(a^{2}+a^{2})+\ldots+3^{\ell-1}(a^{2}+a^{2})+3^{\ell}k_{\ell}^{2}$
$\displaystyle=3(k_{1}^{2}+k_{1}^{\prime 2})+3^{\ell}(a^{2}+k_{\ell}^{2})$

In particular,

$n=k_{1}^{2}+k_{1}^{\prime 2}+3^{\ell-1}(a^{2}+k_{\ell}^{2}),$

and the result follows. ∎

The next result can be proven the same way as Lemma 3.5

######  Lemma 8.8.

If there exist $j$ and $r$ such that the congruence

$\displaystyle\operatorname{PD}_{3^{j}}(3^{j}n+r)\equiv
0\,\operatorname{mod}3$

holds for all $n\geq 0$, then

$\displaystyle\operatorname{PD}_{3^{\ell}}(3^{j}n+r)\equiv
0\,\operatorname{mod}3$

for all $n\geq 0$ and $\ell\geq j$.

As an application of these results, we generalize Theorem 8.6.

######  Corollary 8.9.

For all $n\geq 0$, $k\geq 1$ and $\ell\geq 2k+1$,

$\displaystyle\operatorname{PD}_{3^{\ell}}\big{(}3^{2k}(3n+1)\big{)}\equiv\operatorname{PD}_{3^{\ell}}\big{(}3^{2k}(3n+2)\big{)}\equiv
0\,\operatorname{mod}3.$

###### Proof.

We will focus on the congruence
$\operatorname{PD}_{3^{\ell}}\big{(}3^{2k}(3n+1)\big{)}\equiv
0\,\operatorname{mod}3$. With Lemma 8.8, it will be sufficient to prove
$\operatorname{PD}_{3^{2k+1}}\big{(}3^{2k}(3n+1)\big{)}\equiv
0\,\operatorname{mod}3$, and according to Theorem 8.7 we can focus on the
equation

$3^{2k-1}(3n+1)=k_{0}^{2}+k_{0}^{\prime 2}+3^{2k}(k_{1}^{2}+k_{1}^{\prime
2}).$

Note that $k_{0}^{2}+k_{0}^{\prime 2}\equiv 0\,\operatorname{mod}3$. We
conclude $3\mid k_{0},k_{0}^{\prime}$ and can write
$k_{0}=3a_{1},k_{0}^{\prime}=3a_{1}^{\prime}$ for integers
$a_{1},a_{1}^{\prime}$. This leads to the new equation

$3^{2k-3}(3n+1)=a_{1}^{2}+a_{1}^{\prime 2}+3^{2k-2}(k_{1}^{2}+k_{1}^{\prime
2}).$

We can iterate this argument to show that
$k_{0}=3^{k}a_{k},k_{0}^{\prime}=3^{k}a_{k}^{\prime}$ with

$3^{2k-1}(3n+1)=3^{2k}(a_{k}^{2}+a_{k}^{\prime 2}+k_{1}^{2}+k_{1}^{\prime
2}).$

This is an apparent contradiction as the right-hand side of this equation is
divisible by $3^{2k}$ while the left-hand side is not. This shows
$e^{*}_{3^{2k}(3n+1),2k+1}=0$. ∎

##  9\. Open Problems

We conclude with some additional conjectured congruences.

######  Conjecture 9.1.

For $n\geq 0,$ we have

$\displaystyle\operatorname{PD}_{2}(16n+12)\equiv 0\,\operatorname{mod}4,$
$\displaystyle\operatorname{PD}_{2}(24n+20)\equiv 0\,\operatorname{mod}4,$
$\displaystyle\operatorname{PD}_{2}(25n+5)\equiv 0\,\operatorname{mod}4,$
$\displaystyle\operatorname{PD}_{2}(32n+24)\equiv 0\,\operatorname{mod}4,$
$\displaystyle\operatorname{PD}_{2}(48n+26)\equiv 0\,\operatorname{mod}4,$

and for $r\in\\{5,11,15,17\\}$,

$\displaystyle\operatorname{PD}_{9}(54n+3r)\equiv 0\,\operatorname{mod}3.$

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# Pushforward measures on homogeneous spaces of non-unimodular groups and
properties of modular functions

Takashi Satomi

(Dedicated to Professor Toshiyuki Kobayashi with gratitude to his continuous
encouragement
)

###### Abstract

This paper shows that a formula for the pushforward measures on the fiber
bundle $G\to G/H$ of homogeneous space of locally compact groups $G$ and $H$
can be written down by using the modular functions. As a result, we obtain the
equality

$\displaystyle\frac{\Delta_{G}(h)}{\Delta_{G/N}(hN)}=\frac{\Delta_{H}(h)}{\Delta_{H/N}(hN)}$
(0.1)

on the modular functions for any closed subgroups $N$ and $H$ of a locally
compact group $G$ with $N\lhd G$ and $N\subset H\subset G$ and any $h\in H$.

Keywords: locally compact group, homogeneous space, modular function,
pushforward measure.

MSC2020: Primary 28C10; Secondary 22D15, 43A85, 46G12, 54C35.

##  1 Introduction

We use the notation $H<G$ and $H\lhd G$ if $H$ is a closed subgroup and a
closed normal subgroup of a locally compact (Hausdorff) group $G$,
respectively. This paper presents one can write down the pushforward measures
on the fiber bundle $p^{G\to G/H}\colon G\to G/H$ of the homogeneous space of
locally compact groups $G$ and $H$ with $H<G$ by using the modular functions
(Theorem 1.4). As a result, we obtain an equality on the modular functions
(Corollary 1.5). That is, we have
$\Delta_{G}(h)/\Delta_{G/N}(hN)=\Delta_{H}(h)/\Delta_{H/N}(hN)$ for any
$N<H<G$ with $N\lhd G$ and any $h\in H$, where $\Delta_{G}$ is the modular
function of $G$. When $H=N$ (i.e., $H\lhd G$), this equality corresponds to a
known result $\Delta_{G}|_{H}=\Delta_{H}$ (Example 1.6).

In this paper, we denote by $C(X)$ the vector space of continuous functions
$X\to\mathbb{C}$, and by $C_{c}(X)\subset C(X)$ the the subspace of continuous
functions with compact support. A complex Radon measure on a locally compact
Hausdorff space $X$ is defined as a linear form of $C_{c}(X)$. When $p\colon
X_{1}\to X_{2}$ is a continuous map between locally compact Hausdorff spaces
$X_{1}$ and $X_{2}$, the pushforward (whose precise definition will be given
in Definition 2.1) of the measure $\mu\in C_{c}(X_{1})^{*}$ is written as
$p_{*}(\mu)\in C_{c}(X_{2})^{*}$. If $p$ is peoper (i.e., the inverse image of
any compact set is also compact), the pushforward $p_{*}(\mu)$ can be defined
for any Radon measure $\mu\in C_{c}(X_{1})^{*}$. On the other hand, it needs
the properness condition on the support of the measure $\mu\in
C_{c}(X_{1})^{*}$ for the existance of the pushforward $p_{*}(\mu)$ if $p$ is
not proper.

######  Definition 1.1.

We denote by $p^{G\to G/H}\colon G\to G/H$ the canonical quotient map for
locally compact groups $G$ and $H$ with $H<G$. The set $M_{H}(G)\subset
C_{c}(G)^{*}$ of complex Radon measures is defined as

$\displaystyle M_{H}(G)\coloneqq\\{\phi(g)dg\mid\phi\in C(G),\;\text{ $p^{G\to
G/H}|_{\mathrm{supp}(\phi)}$ is proper }\\}.$ (1.1)

Here $\mathrm{supp}(\phi)$ is the support of $\phi$, and $dg$ is a left Haar
measure of $G$ (but $M_{H}(G)$ does not depend on the choice of $dg$).

######  Example 1.2.

We have $M_{\\{e\\}}(G)=\\{\phi(g)dg\mid\phi\in C(G)\\}$, where $e\in G$ is
the identity element of $G$.

The following theorem is one of the main theorems of this paper:

######  Theorem 1.3.

Let $\overline{G}\coloneqq G/N$ and $\overline{H}\coloneqq H/N$ for locally
compact groups $G$, $H$, and $N$ with $N<H<G$. We denote by
$p^{G\to\overline{G}}\colon G\to\overline{G}$ and $p^{\overline{G}\to
G/H}\colon\overline{G}\to G/H$ the canonical maps.

  1. (1)

One has $M_{H}(G)\subset M_{N}(G)$.

  2. (2)

The pushforwards $(p^{\overline{G}\to G/H}\circ
p^{G\to\overline{G}})_{*}(\mu)$ and $p_{*}^{\overline{G}\to G/H}\circ
p_{*}^{G\to\overline{G}}(\mu)$ can be defined for any $\mu\in M_{H}(G)$.
Furthermore, one has

$\displaystyle(p^{\overline{G}\to G/H}\circ
p^{G\to\overline{G}})_{*}(\mu)=p_{*}^{\overline{G}\to G/H}\circ
p_{*}^{G\to\overline{G}}(\mu).$ (1.2)

  3. (3)

We assume $N\lhd G$ and denote by $p^{G/H\to\overline{G}/\overline{H}}\colon
G/H\to\overline{G}/\overline{H}$ and
$p^{\overline{G}\to\overline{G}/\overline{H}}\colon\overline{G}\to\overline{G}/\overline{H}$
the canonical maps. Then $p_{*}^{G/H\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to G/H}(\mu)$ and
$p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\mu)$ can be defined. Furthermore, one has

$\displaystyle p_{*}^{G/H\to\overline{G}/\overline{H}}\circ p_{*}^{G\to
G/H}(\mu)=p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\mu).$ (1.3)

  4. (4)

For $N\lhd G$, one has $p_{*}^{G\to\overline{G}}(\mu)\in
M_{\overline{H}}(\overline{G})$ for any $\mu\in M_{H}(G)$.

In addition, we can write down the pushforward measure of the fiber bundle
$p^{G\to G/H}$ by using the modular functions as follows:

######  Theorem 1.4.

Let $G$, $H$, $p^{G\to G/H}$ be as in Theorem 1.3. We write $dg$ as a left
Haar measure and $\Delta_{G}$ as the modular function of $G$. Similarly, we
write $dh$ as a left Haar measure and $\Delta_{H}$ as the modular function of
$H$. Suppose $\phi\in C(G)$ satisfies $p^{G\to G/H}|_{\mathrm{supp}(\phi)}$ is
proper. Then

$\displaystyle\int_{G/H}\int_{H}\nu(gh)dhdp_{*}^{G\to
G/H}(\phi(g)dg)=\int_{G}\nu(g)\int_{H}\phi(gh)\frac{\Delta_{G}(h)}{\Delta_{H}(h)}dhdg$
(1.4)

holds for any $\nu\in C_{c}(G)$.

As a result, we obtain a property of the modular functions as follows:

######  Corollary 1.5.

Let $G$, $H$, $N$, $\overline{G}$, and $\overline{H}$ be as in Theorem 1.3
with $N\lhd G$. Then

$\displaystyle\frac{\Delta_{G}(h)}{\Delta_{\overline{G}}(\overline{h})}=\frac{\Delta_{H}(h)}{\Delta_{\overline{H}}(\overline{h})}$
(1.5)

holds for any $h\in H$, where $\overline{h}\coloneqq p^{G\to\overline{G}}(h)$.

Corollary 1.5 has been essentially known in the case of $H=N$. The proof of
Corollary 1.5 given later can be regarded as a generalization of the following
example:

######  Example 1.6 (Hewitt–Ross [HR79, Remark 15.23]).

When $H=N$ (in particular, $H\lhd G$), we have
$\Delta_{\overline{G}}(\overline{h})=\Delta_{\overline{H}}(\overline{h})=1$.
Thus, Corollary 1.5 implies $\Delta_{G}(h)=\Delta_{H}(h)$ in this case.

Here is the organization of this paper. In Section 2, we see the definition
and properties of the pushforward measure, and we prove Theorem 1.3. In
Section 3, we prove Theorem 1.4 and Corollary 1.5 by using Theorem 1.3.

##  2 Pushforward measure

In this section, we prove Theorem 1.3. In Subsection 2.1, we review the
definition of the pushforward measure (Definition 2.1) on locally compact
Hausdorff spaces in a general setting and show several properties of this
measure to prove Theorem 1.3. In Subsection 2.2, we apply these properties to
the case of homogeneous spaces and complete the proof of Theorem 1.3.

###  2.1 Definition and properties of the pushforward measure

In this subsection, we review the definition of the pushforward measure for
locally compact Hausdorff spaces. In addition, some properties of such a
measure is also discussed. For any continuous map $p\colon X_{1}\to X_{2}$
between locally compact Hausdorff spaces $X_{1}$ and $X_{2}$, the set
$\mathcal{M}(p,X_{1})$ is defined as

$\displaystyle\mathcal{M}(p,X_{1})\coloneqq\\{\mu\in
C_{c}(X_{1})^{*}\mid\text{ $p|_{\mathrm{supp}(\mu)}$ is proper }\\}.$ (2.1)

For any $\mu\in\mathcal{M}(p,X_{1})$ and $\alpha_{2}\in C_{c}(X_{2})$, there
exists $\alpha_{1}\in C_{c}(X_{1})$ such that

$\displaystyle\alpha_{1}(x)=\alpha_{2}\circ p(x)$ (2.2)

for any $x\in\mathrm{supp}(\mu)$ by the Tietze extension theorem since
$\mathrm{supp}(\mu)\cap\mathrm{supp}(\alpha_{2}\circ p)$ is compact. In this
case, the integration

$\displaystyle\int_{X_{1}}\alpha_{1}\;d\mu$ (2.3)

does not depend on the choice of $\alpha_{1}$ by the definition of
$\mathrm{supp}(\mu)$. Thus, the pushforward measure is defined as follows:

######  Definition 2.1.

Let $p\colon X_{1}\to X_{2}$ be a continuous map between locally compact
Hausdorff spaces $X_{1}$ and $X_{2}$. Then the map
$p_{*}\colon\mathcal{M}(p,X_{1})\to C_{c}(X_{2})^{*}$ is defined as

$\displaystyle\int_{X_{2}}\alpha_{2}\;dp_{*}(\mu)\coloneqq\int_{X_{1}}\alpha_{1}\;d\mu$
$\displaystyle(\mu\in\mathcal{M}(p,X_{1}),\;\alpha_{2}\in C_{c}(X_{2})),$
(2.4)

where $\alpha_{1}\in C_{c}(X_{1})$ satisfies (2.2) for any
$x\in\mathrm{supp}(\mu)$. This measure $p_{*}(\mu)$ is called the pushforward
of $\mu$.

The support of the pushforward measure is contained in the image of the
original measure as follows:

######  Lemma 2.2.

Let $X_{1}$, $X_{2}$, and $p$ be as in Definition 2.1. Then

$\displaystyle\mathrm{supp}(p_{*}(\mu))\subset p(\mathrm{supp}(\mu))$ (2.5)

holds for any $\mu\in\mathcal{M}(p,X_{1})$.

###### Proof.

Let $x_{2}\in\mathrm{supp}(p_{*}(\mu))$. Since $p$ is a proper map to the
locally compact Hausdorff space $X_{2}$, the image $p(\mathrm{supp}(\mu))$ of
the closed set $\mathrm{supp}(\mu)$ is also closed. Thus, it suffices to show
$p(\mathrm{supp}(\mu))\cap U\neq\emptyset$ for any open neighborhood $U$ of
$x_{2}$. Since there exists $\alpha_{2}\in C_{c}(X_{2})$ such that
$\mathrm{supp}(\alpha_{2})\subset U$ and

$\displaystyle\int_{X_{2}}\alpha_{2}\;dp_{*}(\mu)\neq 0$ (2.6)

by $x_{2}\in\mathrm{supp}(p_{*}(\mu))$, there exists $\alpha_{1}\in
C_{c}(X_{1})$ such that

$\displaystyle\int_{X_{1}}\alpha_{1}\;d\mu\neq 0$ (2.7)

and (2.2) hold for any $x\in\mathrm{supp}(\mu)$. Thus, there exists
$x_{1}\in\mathrm{supp}(\mu)$ such that $\alpha_{1}(x_{1})\neq 0$ and hence
$\alpha_{2}\circ p(x_{1})\neq 0$. Then

$\displaystyle p(x_{1})\in
p(\mathrm{supp}(\mu))\cap\mathrm{supp}(\alpha_{2})\subset
p(\mathrm{supp}(\mu))\cap U$ (2.8)

holds and hence we obtain (2.5). ∎

Now we show the following lemma to prove Theorem 1.3:

######  Lemma 2.3.

Let $p^{1,2}\colon X_{1}\to X_{2}$ and $p^{2,3}\colon X_{2}\to X_{3}$ be
continuous maps between locally compact Hausdorff spaces.

  1. (1)

One has $\mathcal{M}(p^{2,3}\circ
p^{1,2},X_{1})\subset\mathcal{M}(p^{1,2},X_{1})$.

  2. (2)

One has $p_{*}^{1,2}(\mathcal{M}(p^{2,3}\circ
p^{1,2},X_{1}))\subset\mathcal{M}(p^{2,3},X_{2})$.

  3. (3)

One has $(p^{2,3}\circ p^{1,2})_{*}=p_{*}^{2,3}\circ p_{*}^{1,2}$.

###### Proof.

  1. (1)

Let $K\subset X_{1}$ be a compact set and $\mu\in\mathcal{M}(p^{2,3}\circ
p^{1,2},X_{1})$. Then it suffices to show that
$\mathrm{supp}(\mu)\cap(p^{1,2})^{-1}(K)$ is compact. We have

$\displaystyle\mathrm{supp}(\mu)\cap(p^{1,2})^{-1}(K)\subset\mathrm{supp}(\mu)\cap(p^{2,3}\circ
p^{1,2})^{-1}(p^{2,3}(K)).$ (2.9)

Since $p^{2,3}(K)$ is compact, $\mathrm{supp}(\mu)\cap(p^{2,3}\circ
p^{1,2})^{-1}(p^{2,3}(K))$ is also compact by $\mu\in\mathcal{M}(p^{2,3}\circ
p^{1,2},X_{1})$. In addition, the sets $\mathrm{supp}(\mu)$ and
$(p^{1,2})^{-1}(K)$ are closed and hence
$\mathrm{supp}(\mu)\cap(p^{1,2})^{-1}(K)$ is compact.

  2. (2)

Let $\mu\in\mathcal{M}(p^{2,3}\circ p^{1,2},X_{1})$ and $K\subset X_{3}$ be a
compact set. It suffices to show that
$(p^{2,3})^{-1}(K)\cap\mathrm{supp}(p_{*}^{1,2}(\mu))$ is compact. Then

$\displaystyle(p^{2,3})^{-1}(K)\cap\mathrm{supp}(p_{*}^{1,2}(\mu))$
$\displaystyle\subset(p^{2,3})^{-1}(K)\cap p^{1,2}(\mathrm{supp}(\mu))$ (2.10)
$\displaystyle=p^{1,2}((p^{2,3}\circ p^{1,2})^{-1}(K)\cap\mathrm{supp}(\mu))$
(2.11)

holds by Lemma 2.2. Since $(p^{2,3}\circ
p^{1,2})^{-1}(K)\cap\mathrm{supp}(\mu)$ is compact by
$\mu\in\mathcal{M}(p^{2,3}\circ p^{1,2},X_{1})$, the set
$(p^{2,3})^{-1}(K)\cap\mathrm{supp}(p_{*}^{1,2}(\mu))$ is also compact.

  3. (3)

Let $\mu\in\mathcal{M}(p^{2,3}\circ p^{1,2},X_{1})$ and $\alpha_{3}\in
C_{c}(X_{3})$. Then it suffices to show

$\displaystyle\int_{X_{3}}\alpha_{3}\;d(p^{2,3}\circ
p^{1,2})_{*}(\mu)=\int_{X_{3}}\alpha_{3}\;dp_{*}^{2,3}\circ p_{*}^{1,2}(\mu).$
(2.12)

We note that both sides of this equality can be defined by (1) and (2). Then
there exists $\alpha_{1}\in C_{c}(X_{2})$ such that
$\alpha_{1}(x_{1})=\alpha_{3}\circ p^{2,3}\circ p^{1,2}(x_{1})$ for any
$x_{1}\in\mathrm{supp}(\mu)$. For any
$x_{1},x_{1}^{\prime}\in\mathrm{supp}(\mu)$ with
$p^{1,2}(x_{1})=p^{1,2}(x_{1}^{\prime})$, we have

$\displaystyle\alpha_{1}(x_{1})=\alpha_{3}\circ p^{2,3}\circ
p^{1,2}(x_{1})=\alpha_{3}\circ p^{2,3}\circ
p^{1,2}(x_{1}^{\prime})=\alpha_{1}(x_{1}^{\prime}).$ (2.13)

Thus, there exists $\alpha_{2}\in C(p^{1,2}(\mathrm{supp}(\mu)))$ such that
$\alpha_{2}\circ p^{1,2}(x_{1})=\alpha_{1}(x_{1})$ for any
$x_{1}\in\mathrm{supp}(\mu)$. We can extend $\alpha_{2}$ to
$X_{2}\to\mathbb{C}$ such that $\alpha_{2}\in C_{c}(X_{2})$ by the compactness
of $\mathrm{supp}(\alpha_{1})$ and the Tietze extension theorem. For any
$x_{2}\in\mathrm{supp}(p_{*}^{1,2}(\mu))$, there exists
$x_{1}\in\mathrm{supp}(\mu)$ such that $x_{2}=p^{1,2}(x_{1})$ by Lemma 2.2.
Then

$\displaystyle\alpha_{3}\circ p^{2,3}(x_{2})=\alpha_{3}\circ p^{2,3}\circ
p^{1,2}(x_{1})=\alpha_{1}(x_{1})=\alpha_{2}\circ
p^{1,2}(x_{1})=\alpha_{2}(x_{2})$ (2.14)

holds and hence

$\displaystyle\int_{X_{3}}\alpha_{3}\;dp_{*}^{2,3}\circ
p_{*}^{1,2}(\mu)=\int_{X_{2}}\alpha_{2}\;dp_{*}^{1,2}(\mu)=\int_{X_{1}}\alpha_{1}\;d\mu=\int_{X_{3}}\alpha_{3}\;d(p^{2,3}\circ
p^{1,2})_{*}(\mu)$ (2.15)

is obtained. ∎

###  2.2 Application to homogeneous spaces

By applying the properties of pushforward measures in Subsection 2.1 to
homogeneous spaces, we can prove Theorem 1.3. For this purpose, we use the
following lemma:

######  Lemma 2.4.

Let $dg$ be a left Haar measure of a locally compact group $G$.

  1. (1)

One has $\mathrm{supp}(\phi(g)dg)=\mathrm{supp}(\phi)$ for any $\phi\in C(G)$.

  2. (2)

Let $H<G$ and $p^{G\to G/H}$ be the canonical map. Then

$\displaystyle M_{H}(G)=\mathcal{M}(p^{G\to G/H},G)\cap M_{\\{e\\}}(G).$
(2.16)

In particular, the pushforward $p_{*}^{G\to G/H}(\mu)$ can be defined for any
$\mu\in M_{H}(G)$.

  3. (3)

Let $N$, $\overline{G}$, and $p^{G\to G/H}$ be as in Theorem 1.3 (which does
not necessarily satisfy $N\lhd G$). Then $\mathcal{M}(p^{G\to
G/H},G)\subset\mathcal{M}(p^{G\to\overline{G}},G)$.

###### Proof.

  1. (1)

The conclusion is obtained by $\mathrm{supp}(dg)=G$.

  2. (2)

The conclusion is obtained by (1) and Example 1.2.

  3. (3)

Since $p^{G\to G/H}=p^{\overline{G}\to G/H}\circ p^{G\to\overline{G}}$, we
obtain $\mathcal{M}(p^{G\to G/H},G)\subset\mathcal{M}(p^{G\to\overline{G}},G)$
by Lemma 2.3 (1). ∎

A left Haar measure of quotient groups can be given as the following fact:

######  Fact 2.5 ([Fol16, Theorem 2.51]).

For $G$, $N$, and $\overline{G}$ as in Corollary 1.5, let $dg$ and $dn$ be
left Haar measures of $G$ and $N$, respectively. Then there exists a left Haar
measure $d\overline{g}$ of $\overline{G}$ such that

$\displaystyle\int_{\overline{G}}\int_{N}\beta(gn)dnd\overline{g}=\int_{G}\beta(g)dg$
(2.17)

for any $\beta\in C_{c}(G)$, where $\overline{g}\coloneqq
p^{G\to\overline{G}}(g)$.

By using a left Haar measure $d\overline{g}$ as in Fact 2.5, the pushforward
$p_{*}^{G\to\overline{G}}(\mu)$ (which is well-defined by Lemma 2.4 (2)) can
be written explicitly for any $\mu\in M_{N}(G)$ as follows:

######  Lemma 2.6.

Let $G$, $N$, $\overline{G}$, $dg$, $dn$, and $d\overline{g}$ be as in Fact
2.5. Then

$\displaystyle
p_{*}^{G\to\overline{G}}(\phi(g)dg)=\int_{N}\phi(gn)dnd\overline{g}$ (2.18)

for any $\phi\in C(G)$ for which $p^{G\to\overline{G}}|_{\mathrm{supp}(\phi)}$
is proper. In particular, one has

$\displaystyle p_{*}^{G\to\overline{G}}(M_{N}(G))\subset
M_{\overline{N}}(\overline{G}).$ (2.19)

###### Proof.

Since

$\displaystyle\int_{\overline{G}}\alpha(\overline{g})dp_{*}^{G\to\overline{G}}(\phi(g)dg)$
$\displaystyle=\int_{G}\alpha(\overline{g})\phi(g)dg$ (2.20)
$\displaystyle=\int_{\overline{G}}\int_{N}\alpha(\overline{gn})\phi(gn)dnd\overline{g}$
(2.21)
$\displaystyle=\int_{\overline{G}}\int_{N}\alpha(\overline{g})\phi(gn)dnd\overline{g}$
(2.22)
$\displaystyle=\int_{\overline{G}}\alpha(\overline{g})\int_{N}\phi(gn)dnd\overline{g}$
(2.23)

holds for any $\alpha\in C_{c}(\overline{G})$ by Fact 2.5, we obtain (2.18).
In particular, we have

$\displaystyle\int_{N}\phi(gn)dnd\overline{g}\in
M_{\overline{N}}(\overline{G})$ (2.24)

and hence (2.19) is obtained. ∎

Now we prove Theorem 1.3 by using Lemma 2.4 and Lemma 2.6.

###### Proof of Theorem 1.3.

  1. (1)

By Lemma 2.4 (2) and (3), we obtain

$\displaystyle M_{H}(G)=\mathcal{M}(p^{G\to G/H},G)\cap
M_{\\{e\\}}(G)\subset\mathcal{M}(p^{G\to\overline{G}},G)\cap
M_{\\{e\\}}(G)=M_{N}(G).$ (2.25)

  2. (2)

We have

$\displaystyle\mu\in M_{H}(G)\subset\mathcal{M}(p^{G\to
G/H},G)=\mathcal{M}(p^{\overline{G}\to G/H}\circ p^{G\to\overline{G}},G)$
(2.26)

by Lemma 2.4 (2). Thus, the pushforwards $(p^{\overline{G}\to G/H}\circ
p^{G\to\overline{G}})_{*}(\mu)$ and $p_{*}^{\overline{G}\to G/H}\circ
p_{*}^{G\to\overline{G}}(\mu)$ can be defined and we obtain (1.2) by Lemma
2.3.

  3. (3)

Since

$\displaystyle\mu$ $\displaystyle\in M_{H}(G)$ (2.27)
$\displaystyle\subset\mathcal{M}(p^{G\to G/H},G)$ (2.28)
$\displaystyle=\mathcal{M}((p^{G/H\to\overline{G}/\overline{H}})^{-1}\circ
p^{\overline{G}\to\overline{G}/\overline{H}}\circ p^{G\to\overline{G}},G)$
(2.29)
$\displaystyle\subset\mathcal{M}(p^{\overline{G}\to\overline{G}/\overline{H}}\circ
p^{G\to\overline{G}},G)$ (2.30)

by Lemma 2.3 (1) and Lemma 2.4 (2), the pushforward
$p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\mu)$ can be defined. Thus, it follows from Lemma 2.3
and $p^{\overline{G}\to\overline{G}/\overline{H}}\circ
p^{G\to\overline{G}}=p^{G/H\to\overline{G}/\overline{H}}\circ p^{G\to G/H}$
that $p_{*}^{G/H\to\overline{G}/\overline{H}}\circ p_{*}^{G\to G/H}(\mu)$ can
also be defined, and (1.3) holds.

  4. (4)

It follows from (1) and Lemma 2.6 that

$\displaystyle p_{*}^{G\to\overline{G}}(\mu)\in
p_{*}^{G\to\overline{G}}(M_{H}(G))\subset
p_{*}^{G\to\overline{G}}(M_{N}(G))\subset M_{\overline{N}}(\overline{G}).$
(2.31)

Thus, we obtain

$\displaystyle
p_{*}^{G\to\overline{G}}(\mu)\in\mathcal{M}(p^{\overline{G}\to\overline{G}/\overline{H}},G)\cap
M_{\overline{N}}(\overline{G})=M_{\overline{H}}(\overline{G})$ (2.32)

by (3) and Lemma 2.4 (2). ∎

##  3 Proof of Theorem 1.4 and Corollary 1.5

In this section, we prove Theorem 1.4 and Corollary 1.5.

###### Proof of Theorem 1.4.

We have

$\displaystyle\int_{G/H}\int_{H}\nu(gh)dhdp_{*}^{G\to G/H}(\phi(g)dg)$
$\displaystyle=\int_{G}\int_{H}\nu(gh)dh\phi(g)dg$ (3.1)
$\displaystyle=\int_{H}\int_{G}\nu(gh)\phi(g)dgdh$ (3.2)
$\displaystyle=\int_{H}\int_{G}\frac{\nu(g)\phi(gh^{-1})}{\Delta_{G}(h)}dgdh$
(3.3)
$\displaystyle=\int_{G}\nu(g)\int_{H}\frac{\phi(gh^{-1})}{\Delta_{G}(h)}dhdg$
(3.4)

by Fubini’s theorem. Since

$\displaystyle\int_{H}\frac{\phi(gh^{-1})}{\Delta_{G}(h)}dh=\int_{H}\phi(gh)\frac{\Delta_{G}(h)}{\Delta_{H}(h)}dh$
(3.5)

holds, we obtain (1.4). ∎

We utilize Theorem 1.3 and Theorem 1.4 together with the following fact to
prove Corollary 1.5:

######  Fact 3.1 ([Fol16, Proposition 2.50]).

Let $G$, $H$, $p^{G\to G/H}$, $dh$ be as in Theorem 1.4, and $\alpha\in
C_{c}(G/H)$. Then there exists $\nu\in C_{c}(G)$ such that

$\displaystyle\alpha(gH)=\int_{H}\nu(gh)dh$ (3.6)

holds for any $g\in G$.

Then we complete the proof of Corollary 1.5 by using Theorem 1.3, Theorem 1.4,
and Fact 3.1.

###### Proof of Corollary 1.5.

Suppose $\phi\in C(G)$ satisfies $p^{G\to G/H}|_{\mathrm{supp}(\phi)}$ is
proper. Now we show

$\displaystyle p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(gh^{\prime})dg)=\frac{\Delta_{\overline{H}}(\overline{h^{\prime}})p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(g)dg)}{\Delta_{\overline{G}}(\overline{h^{\prime}})\Delta_{H}(h^{\prime})}.$
(3.7)

Then

$\displaystyle
p_{*}^{G\to\overline{G}}(\phi(gh^{\prime})dg)=\int_{N}\phi(gnh^{\prime})dnd\overline{g}$
(3.8)

holds by Lemma 2.6. Now we let $\alpha\in C_{c}(\overline{G}/\overline{H})$.
By Fact 3.1, there exists $\nu\in C_{c}(\overline{G})$ such that

$\displaystyle\alpha(\overline{gH})=\int_{\overline{H}}\nu(\overline{gh})d\overline{h}$
(3.9)

holds for any $g\in G$, where $\overline{gH}\coloneqq
p^{G/H\to\overline{G}/\overline{H}}(gH)$. Then

$\displaystyle\int_{\overline{G}/\overline{H}}\alpha(\overline{gH})dp_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(gh^{\prime})dg)$
$\displaystyle=\int_{\overline{G}/\overline{H}}\int_{\overline{H}}\nu(\overline{gh})d\overline{h}\;dp_{*}^{\overline{G}\to\overline{G}/\overline{H}}\left(\int_{N}\phi(gnh^{\prime})dnd\overline{g}\right)$
(3.10)
$\displaystyle=\int_{\overline{G}}\nu(\overline{g})\int_{\overline{H}}\int_{N}\phi(ghnh^{\prime})dn\frac{\Delta_{\overline{G}}(\overline{h})}{\Delta_{\overline{H}}(\overline{h})}d\overline{h}d\overline{g}$
(3.11)

holds by (3.8) and Theorem 1.4. Since

$\displaystyle\int_{\overline{H}}\int_{N}\phi(ghnh^{\prime})dn\frac{\Delta_{\overline{G}}(\overline{h})}{\Delta_{\overline{H}}(\overline{h})}d\overline{h}$
$\displaystyle=\int_{\overline{H}}\int_{N}\phi(ghnh^{\prime})\frac{\Delta_{\overline{G}}(\overline{hn})}{\Delta_{\overline{H}}(\overline{hn})}dnd\overline{h}$
(3.12)
$\displaystyle=\int_{H}\phi(ghh^{\prime})\frac{\Delta_{\overline{G}}(\overline{h})}{\Delta_{\overline{H}}(\overline{h})}dh$
(3.13)
$\displaystyle=\int_{H}\phi(gh)\frac{\Delta_{\overline{G}}(\overline{hh^{\prime-1}})}{\Delta_{\overline{H}}(\overline{hh^{\prime-1}})}dh\frac{1}{\Delta_{H}(h^{\prime})}$
(3.14)
$\displaystyle=\int_{H}\phi(gh)\frac{\Delta_{\overline{G}}(\overline{h})}{\Delta_{\overline{H}}(\overline{h})}dh\frac{\Delta_{\overline{H}}(h^{\prime})}{\Delta_{\overline{G}}(h^{\prime})\Delta_{H}(h^{\prime})}$
(3.15)

holds by Fact 2.5, we have

$\displaystyle\int_{\overline{G}/\overline{H}}\alpha(\overline{gH})dp_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(gh^{\prime})dg)$ (3.16) $\displaystyle=$
$\displaystyle\int_{\overline{G}/\overline{H}}\alpha(\overline{gH})dp_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(g)dg)\frac{\Delta_{\overline{H}}(h^{\prime})}{\Delta_{\overline{G}}(h^{\prime})\Delta_{H}(h^{\prime})}$
(3.17)

and hence (3.7).

In particular, we have

$\displaystyle p_{*}^{G\to
G/H}(\phi(gh^{\prime})dg)=\frac{\Delta_{H}(h^{\prime})p_{*}^{G\to
G/H}(\phi(g)dg)}{\Delta_{G}(h^{\prime})\Delta_{H}(h^{\prime})}=\frac{p_{*}^{G\to
G/H}(\phi(g)dg)}{\Delta_{G}(h^{\prime})}$ (3.18)

by applying $N=\\{e\\}$ to (3.7). Thus, it follows from (3.7) and Theorem 1.3
(3) that

$\displaystyle\frac{\Delta_{\overline{H}}(\overline{h^{\prime}})p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(g)dg)}{\Delta_{\overline{G}}(\overline{h^{\prime}})\Delta_{H}(h^{\prime})}$
$\displaystyle=p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(gh^{\prime})dg)$ (3.19)
$\displaystyle=p_{*}^{G/H\to\overline{G}/\overline{H}}\circ p_{*}^{G\to
G/H}(\phi(gh^{\prime})dg)$ (3.20)
$\displaystyle=p_{*}^{G/H\to\overline{G}/\overline{H}}\left(\frac{p_{*}^{G\to
G/H}(\phi(g)dg)}{\Delta_{G}(h^{\prime})}\right)$ (3.21)
$\displaystyle=\frac{p_{*}^{G/H\to\overline{G}/\overline{H}}\circ p_{*}^{G\to
G/H}(\phi(g)dg)}{\Delta_{G}(h^{\prime})}$ (3.22)
$\displaystyle=\frac{p_{*}^{\overline{G}\to\overline{G}/\overline{H}}\circ
p_{*}^{G\to\overline{G}}(\phi(g)dg)}{\Delta_{G}(h^{\prime})}.$ (3.23)

Therefore, we obtain (1.5). ∎

## Acknowledgement

This work was supported by the RIKEN Special Postdoctoral Researcher (SPDR)
program. The author would like to express his sincere gratitude to Prof.
Toshiyuki Kobayashi for repeated discussions and helpful advice for this
paper. The author is also grateful to Toshihisa Kubo and Yuichiro Tanaka for
their careful comments.

## References

  * [Fol16] G. B. Folland  “A course in abstract harmonic analysis”, Textbooks in Mathematics  CRC Press, Boca Raton, FL, 2016, pp. xiii+305
  * [HR79] E. Hewitt and K. A. Ross  “Abstract harmonic analysis. Vol. I” 115, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]  Springer-Verlag, Berlin-New York, 1979, pp. ix+519

Takashi Satomi: RIKEN Interdisciplinary Theoretical and Mathematical Sciences
(iTHEMS), Wako, Saitama 351-0198, Japan.

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