text:
\begin{document}
\maketitle

\vskip 6mm \noindent {\bf Abstract.}  Andrews recently introduced
$k$-marked Durfee symbols which are connected to moments of
Dyson's rank. By these connections, Andrews  deduced  their
generating functions and some combinatorial properties and left
their purely combinatorial proofs as open problems. The primary
goal of this article is to provide combinatorial proofs in answer
to Andrews' request. We obtain a relation between $k$-marked
Durfee symbols and Durfee symbols by constructing bijections, and
all identities on $k$-marked Durfee symbols given by Andrews could
follow from this relation. In a similar manner, we also prove the
identities due to Andrews on $k$-marked odd Durfee symbols
combinatorially, which resemble ordinary $k$-marked Durfee symbols
with a modified subscript and with odd numbers as entries.

\noindent {\bf Keywords}: rank, the moment of rank, the
symmetrized moment of rank, Durfee symbols, $k$-marked Durfee
symbols, odd Durfee symbols, $k$-marked odd Durfee symbols

\noindent {\bf AMS  Classifications}: 05A17, 05A19, 11P83

\section{Introduction}

We will adopt the terminology on partitions in Andrews
\cite{Andrews-1976}.  {\it A partition} $\lambda$ of a positive
integer $n$ is a finite nonincreasing sequence of positive
integers $(\lambda_1,\,\lambda_2,\ldots,\,\lambda_r)$ such that
$\sum_{i=1}^r\lambda_i=n.$ Then $\lambda_i$ are called the parts
of $\lambda$, $\lambda_1$ is its largest part. The number of parts
of  $\lambda$ is called the length of $\lambda$, denoted by
$l(\lambda).$ The weight of $\lambda$ is the sum of parts of
$\lambda$, denoted by $|\lambda|.$

{\it The rank of a partition} $\lambda$ introduced  by Dyson
\cite{Dyson-1944} is defined as the largest part minus the number
of parts, which is usually denoted by
$r(\lambda)=\lambda_1-l(\lambda).$ Let $N(m;n)$ denote the number
of partitions of $n$ with rank $m$, we have

\begin{thm}[Dyson]\label{gf-r}
The generating function for $N(m;n)$ is given by
\begin{equation}
\sum_{n=0}^{+\infty}
N(m;n)q^n=\frac{1}{(q;q)_{\infty}}\sum_{n=1}^{+\infty}(-1)^{n-1}q^{n(3n-1)/2+|m|n}(1-q^n),\quad
|q|<1.
\end{equation}
\end{thm}
where $(a;q)_n=\prod_{j=0}^{n-1}(1-aq^j)$ and
$(a;q)_{\infty}=\lim_{n\rightarrow\infty}(a;q)_n$. The identity
\eqref{gf-r} was first discovered by Dyson \cite{Dyson-1944} in
1944 and first proved by Atkin and Swinnerton-Dyer
\cite{Atkin-Swinnerton-Dyer-1954}. Later Dyson \cite{Dyson-1969}
gave a simple combinatorial argument of it. We refer to
\cite[p.63]{Berkovich-Garvan-2002} for more details.
\begin{defi} For a nonnegative integer $n$, a Durfee symbol of $n$  is a
 two-row array with a subscript
\begin{equation}
(\alpha,\beta)_D=\left(\begin{array}{cccc}
\alpha_1&\alpha_2&\ldots&\alpha_s\\
\beta_1&\beta_2&\ldots&\beta_t
\end{array} \right)_{D}
\end{equation}
where $D\geq \alpha_1\geq \alpha_2\geq \cdots \geq \alpha_s>0$,
$D\geq \beta_1 \geq \beta_2 \geq \cdots \geq \beta_t>0$ and
$n=\sum_{i=1}^s \alpha_i+\sum_{i=1}^t \beta_i+D^2$.
\end{defi}
For example, there are $5$ Durfee symbols of $4$.
\[\left(\begin{array}{ccc}
1&1&1\\
&&
\end{array}
\right)_1\ \left(\begin{array}{cc}
1&1\\
1&
\end{array}
\right)_1\ \left(\begin{array}{cc}
1&\\
1&1
\end{array}
\right)_1\ \left(\begin{array}{ccc}
\\
1&1&1
\end{array}
\right)_1\ \left(\begin{array}{c}
\ \\
\
\end{array}
\right)_2\] The difference of the lengths of $\alpha$ and $\beta$
is called {\it rank} of Durfee symbols $(\alpha,\,\beta)_D$. We
use $\mathcal{D}_1(m;n)$ to denote the number of Durfee symbols of
$n$ with rank $m$. Andrews \cite[Section 3]{Andrews-07-a} showed
that by constructing a bijection
\begin{thm}[Andrews]\label{dy-op}
The number of ordinary partitions of $n$ with rank equal to $m$ is
equal to the number of Durfee symbols of $n$ with rank equal to
$m$, that is
\begin{equation}
N(m;n)=\mathcal{D}_1(m;n).
\end{equation}
\end{thm}
Andrews \cite{Andrews-07-a} introduced the $k$th symmetrized rank
moment $\eta_k(n)$, defined by
\begin{equation}\label{sym-ran-mom}
\eta_k(n)=\sum_{m=-\infty}^{{+\infty}}
{m+\lfloor\frac{k-1}{2}\rfloor \choose k}N(m,n),
\end{equation}
which are linear combinations of the $k$th rank moments $N_k(n)$
\begin{equation}
N_k(n)=\sum_{m=-\infty}^{{+\infty}} m^kN(m,n),
\end{equation}
 considered by Atkin and Garvan \cite{Atkin-Garvan-03}.  To give a
combinatorial explanation of \eqref{sym-ran-mom}, Andrews
\cite{Andrews-07-a} introduced $k$-marked Durfee symbols, which
can be thought of as the generalized Durfee symbols.

\begin{defi}\label{de-MDS}
A $k$-marked Durfee symbol of $n$ is composed of $k$ pairs of
partitions and a subscript,  which is defined as
 \[\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D,\] where $\alpha^i$ {\rm(}resp. $\beta^i${\rm)}
represents a  partition and
$\sum_{i=1}^k(|\alpha^i|+|\beta^i|)+D^2=n$. Furthermore, the
partitions $\alpha^i$ and $\beta^i$ must satisfy the following
three conditions where  $\alpha^i_1$ {\rm(}resp. $\beta^i_1${\rm)}
is the largest part of the partition $\alpha^i$ {\rm(}resp.
$\beta^i${\rm)} and $\alpha^i_{l(\alpha^i)}$ {\rm(}resp.
$\beta^i_{l(\beta^i)}${\rm)} is the smallest part of the partition
$\alpha^i$ {\rm(}resp. $\beta^i${\rm)}.
\begin{itemize}
\item[{\rm (1)}]For $1\leq i<k$, $\alpha^i$ must be non-empty
partition, while $\alpha^k$ and  $\beta^i$ could be empty{\rm ;}

\item[{\rm (2)}] $\beta^{i-1}_1\leq \alpha^{i-1}_1 \leq
\beta^{i}_{l(\beta^{i})}$ for $2\leq i\leq k${\rm ;}

\item[{\rm (3)}]$\beta^k_1,\alpha^k_1 \leq D.$
\end{itemize}
\end{defi}
Clearly,  $1$-marked Durfee symbol is just Durfee symbol.

 Let $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ be a $k$-marked Durfee symbol. The pair of partitions
$\left(\begin{array}{l}
\alpha^i\\
\beta^i
\end{array}
\right)$ is called the $i$th vector of $\eta$. We define
$\rho_i(\eta)$, the $i$th rank of $\eta$ by
\[\rho_i(\eta)=\left\{
\begin{array}{ll}
l(\alpha^i)-l(\beta^i)-1 \ \ &\text{ for }\ 1\leq i<k,\\[5pt]
l(\alpha^k)-l(\beta^k) \ \ &\text{ for }\ i=k.
\end{array}\right.
\]
For example, $\left(\begin{array}{ccccccc}
4_3&4_3&3_2&3_2&2_2&2_1\\
&5_3&&3_2&2_2&2_1
\end{array}\right)_5$ is a $3$-marked Durfee symbol of $55$ where
$\alpha^3=(4, 4),\,\alpha^2=(3,3,2),\alpha^1=(2),$ and
$\beta^3=(5),\beta^2=(3,2),\beta^1=(2).$ The first rank is $-1$,
the second rank is $0$, and the third rank is $1.$

Let $\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$ denote the number of
$k$-marked Durfee symbols of $n$ with $i$th rank equal to $m_i$
and $\mathcal{D}_k(n)$ denote the number of $k$-marked Durfee
symbols of $n$, it's clear to see that
\begin{equation}
\mathcal{D}_k(n)=\sum_{m_1,\ldots,m_{k}=-\infty}^{+\infty}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n).
\end{equation}

In his recent work \cite{Andrews-07-a}, Andrews used the
connections between $k$-marked Durfee symbols and the symmetrized
rank moments \eqref{sym-ran-mom} to find identities relating the
generating function for $\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$, as
well as to deduce some combinatorial properties and Ramanujan-type
congruences for $k$-marked Durfee symbols. At the end of the
paper, Andrews proposed  a variety of serious questions which fall
into 3 basic groups: combinatorial, asymptotic and congruential.
The recent works \cite{Bringmann-1, Bringmann-2} by Kathrin
Bringmann, Frank Garvan, and Karl Mahlburg  focused on the study
relating asymptotical and congruential properties of $k$-marked
Durfee symbols.  They used the automorphic properties to prove the
existence of infinitely many congruences for $k$-marked Durfee
symbols. The primary goal of this article is to answer Andrews'
request on combinatorics (Problems 1-4 and 6-9 on page 39 of
\cite{Andrews-07-a}). We will give combinatorial proofs of the
identities relating the generating function for
$\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$ and combinatorial properties
for $k$-marked Durfee symbols. To be specific, we first derive the
following partition identity \eqref{main-e} by constructing
bijections, which gives a relation between $k$-marked Durfee
symbols and Durfee symbols. We then show that all identities on
$k$-marked Durfee symbols given  by Andrews ( \cite[Problems
1-4]{Andrews-07-a}) could follow from this identity. We then use
the similar method to study the identities of Andrews on
$k$-marked odd Durfee symbols ( \cite[Problems
6-9]{Andrews-07-a}), which resemble ordinary Durfee symbols with a
modified subscript and with odd numbers as entries.

\begin{thm}\label{main}
For $k\geq 2$, we have
\begin{equation}\label{main-e}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)=\sum_{j=0}^{+\infty}{j+k-2
\choose k-2}N\left(\sum_{i=1}^k|m_i|+2j+k-1;n\right).
\end{equation}
\end{thm}
The paper is organized as follows. In Section 2, we consider the
relation between $k$-marked Durfee symbols and Durfee symbols and
prove Theorem \ref{main}. To this end, we introduce a special
class of $k$-marked Durfee symbols, which we call  $k$-marked
strict shifted Durfee symbols since each of their vectors except
for the $k$th vector is a two-line strict shifted plane partition.
We deduce the desired relation by building the connections between
$k$-marked strict shifted Durfee symbols, Durfee symbols, and
$k$-marked Durfee symbols respectively. In Section 3, we give
combinatorial proofs of the identities due to Andrews on
$k$-marked Durfee symbols with the help of Theorem \ref{main}. In
particular, the symmetry of $k$-marked Durfee symbols
(\cite[Corollary 12]{Andrews-07-a}) could be thought of as a
direct consequence  of Theorem \ref{main}. Section 4 is devoted to
the study of  $k$-marked odd Durfee symbols.

\section{$k$-marked strict shifted Durfee symbols} In this section, we will
establish  the relations between $k$-marked strict shifted Durfee
symbols,  Durfee symbols, and $k$-marked Durfee symbols
respectively, and then deduce Theorem \ref{main}. We begin by
defining $k$-marked strict shifted Durfee symbols.

To define these objects, we need recall the concept of strict
shifted plane partitions introduced by Andrews
\cite{Andrews-1979}. Mills, Robbins and Rumsey
\cite{Mills-Robbins-Rumsey} showed that strict shifted plane
partitions whose row lengths are equal to row leaders are
bijective to cyclically symmetric plane partitions.

A strict shifted plane partition of $n$ is an array $\pi=(a_{ij})$
of positive integers defined only for $j\geq i$, which has
non-increasing rows and strictly decreasing columns, such that the
sum of its elements is $n$. Such an array can be written
\[\begin{array}{ccccclll}
a_{11}&a_{12}&a_{13}&a_{14}&\cdots&&&a_{1\mu_1}\\
&a_{22}&a_{23}&a_{24}&\cdots&&a_{2\mu_2}\\
&&&&\cdots\cdots&& \\
&&&a_{rr}&\cdots&a_{r\mu_r}
\end{array}
\]
where $a_{ij}\geq a_{i,j+1}$ and $a_{ij}> a_{i+1,j}$, $\mu_1\geq
\mu_2\geq\cdots\geq \mu_r$ and $\sum_{i,j}a_{ij}=n$.

We can regard a two-lined strict shifted partition of $n$ as a
pair of partitions $(\alpha,\beta)$ of $n$, where
$\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_r),\,\beta=(\beta_1,\beta_2,\cdots,\beta_s),r>
s,$ and $\alpha_{i+1}>\beta_{i}$
 for $i=1,2,\ldots,s$. For example $\left(\begin{array}{cccccc}
3&3&3&2&2&1\\
&2&1&1&1
\end{array}
\right)$ is a two-lined strict shifted partition.

 A $k$-marked Durfee symbol $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ is said to be {\it strict shifted} if all of its
vectors except for the $k$th vector $(\alpha^k,\beta^k)$ are
two-lined strict shifted partitions. For example, $3$-marked
Durfee symbol in \eqref{example} is strict shifted. Let
$\mathcal{D}^{ss}_k(m_1,m_2,\ldots,m_k;n)$ denote the number of
$k$-marked strict shifted Durfee symbols of $n$ with $i$th rank
equal to $m_i$.

 We now build a connection between $k$-marked strict shifted Durfee
symbols of $n$ and Durfee symbols of $n$.
\begin{thm}\label{r-kms-ds}
Given $k$ nonnegative integers $m_1,\,m_2,\ldots,m_{k}$, there is
a bijection $\Phi$ between the set of $k$-marked strict shifted
Durfee symbols of $n$ with $i$th rank equal to $m_i$ and the set
of Durfee symbols of $n$ with rank equal to $\sum_{i=1}^km_i+k-1$.
\end{thm}
\pf {\bf The map $\Phi$:} Let $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ counted by $\mathcal{D}^{ss}_k(m_1,\ldots,m_k;n)$, we
then obtain a Durfee symbol $(\gamma,\delta)_{D}$ when remove all
subscripts of $\eta$. Obviously, the resulting Durfee symbol
$(\gamma,\delta)_{D}$ is enumerated by $
\mathcal{D}_1(\sum_{i=1}^km_i+k-1;n).$

{\bf The reverse map $\Phi^{-1}$:} Let $(\gamma,\delta)_{D}$ be
counted by $\mathcal{D}_1(\sum_{i=1}^km_i+k-1;n)$, we will
construct a $k$-marked Durfee symbol
$\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_{D'}$ whose $i$th rank equal to $m_i$. Let
\[\left(\begin{array}{l}
\gamma\\
\delta
\end{array}\right)_D=\left(\begin{array}{cccc}
\gamma_1&\gamma_2&\ldots&\gamma_l\\[2pt]
\delta_1&\delta_2&\ldots&\delta_s
\end{array}
\right)_D,\] where $D\geq \gamma_1\geq \gamma_2\geq \cdots\geq
\gamma_l$ and $D\geq \delta_1\geq \delta_2\geq \cdots \geq
\delta_s$, we assume that $\delta_{j}=0$ for $j\geq s+1$. Note
that $l-s=\sum_{i=1}^km_i+k-1$.

We now split $(\gamma, \delta)$ to generate  the $k$th vector
$(\alpha^k,\beta^k)$ of $\eta$.
 Let $j$ be largest nonnegative integer such that
$\delta_{j}\geq \gamma_{m_k+j+1}$,  that is for any $i\geq j+1$,
we have $\delta_{i}<\gamma_{m_k+i+1}$. Let
\[\left(\begin{array}{l}
\alpha^k\\
\beta^k
\end{array}\right)=\left(\begin{array}{cccc}
\gamma_1&\gamma_2&\ldots&\gamma_{m_k+j}\\[2pt]
\delta_1&\delta_2&\ldots&\delta_{j}
\end{array}
\right),\] and
\[\left(\begin{array}{l}
\gamma'\\
\delta'
\end{array}\right)=\left(\begin{array}{cccc}
\gamma'_1&\gamma'_2&\ldots&\gamma'_{l'}\\[2pt]
\delta'_1&\delta'_2&\ldots&\delta'_{s'}
\end{array}
\right),\] where $\gamma'_i=\gamma_{m_k+j+i}$,
$\delta'_i=\delta_{j+i}$ for $i\geq 1$. Obviously,
$l(\alpha^k)-l(\beta^k)=m_k$. Furthermore,  $(\gamma',\delta')$ is
a strict shifted partition from the fact that for any $i\geq 1$,
$\delta_{i+j}<\gamma_{m_k+i+j+1}$ and
$l'-s'=\sum_{i=1}^{k-1}m_i+k-1.$

We continue to split $(\gamma', \delta')$ to construct the
$(k-1)$th vector $(\alpha^{k-1},\beta^{k-1})$  of $\eta$. Let $j$
be largest nonnegative integer such that $\delta'_{j}\geq
\gamma'_{m_{k-1}+j+2}$, we then let
\[\left(\begin{array}{l}
\alpha^{k-1}\\
\beta^{k-1}
\end{array}\right)=\left(\begin{array}{cccc}
\gamma'_1&\gamma'_2&\ldots&\gamma'_{m_{k-1}+j+1}\\[2pt]
\delta'_1&\delta'_2&\ldots&\delta'_{j}
\end{array}
\right),\]

and
\[\left(\begin{array}{l}
\gamma''\\
\delta''
\end{array}\right)=\left(\begin{array}{cccc}
\gamma''_1&\gamma''_2&\ldots&\gamma''_{l''}\\[2pt]
\delta''_1&\delta''_2&\ldots&\delta''_{s''}
\end{array}
\right),\] where $\gamma''_i=\gamma'_{m_{k-1}+j+i+1}$,
$\delta''_i=\delta'_{j+i}$ for $i\geq 1$. Clearly,
$l(\alpha^{k-1})-l(\beta^{k-1})=m_{k-1}+1$ and
$(\alpha^{k-1},\beta^{k-1})$ is a strict shifted partition for
$(\gamma',\delta')$ is strict shifted. Observe that
$\delta'_{j+i}< \gamma'_{m_{k-1}+i+j+2}$ for $i\geq1$, so
$(\gamma'',\delta'')$ is also strict shifted and
$l''-s''=\sum_{i=1}^{k-2}m_i+k-2.$

Repeat the above process to generate
$(\alpha^{k-2},\beta^{k-2}),\ldots,(\alpha^{1},\beta^{1})$
respectively and let $D'=D$, it's straightforward to see that the
$k$-marked Durfee symbol \break $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ is counted by
$\mathcal{D}^{ss}_k(m_1,m_2,\ldots,m_k;n)$. \qed

We now illustrate the reverse map $\Phi^{-1}$ by going through an
example in details. Take $m_1=1,m_2=1,m_3=0$, and let
\[\left(\begin{array}{l}
\gamma\\
\delta
\end{array}\right)_D=\left(\begin{array}{ccccccccccc}
6&6&3&3&3&3&2&2&1&1&1\\
5&5&4&2&1&1&1
\end{array}
\right)_6,\]
 we first split $(\gamma,
\delta)$ to get $(\alpha^3, \beta^3)$, note that the divisional
part $\delta_j$ is the smallest part satisfying $\delta_j\geq
\gamma_{j+1}.$
 \[
 \left(\begin{array}{l}
\alpha^3\\
\beta^3
\end{array}\right)=\left(\begin{array}{cccc}
6&6&3\\
5&5&4
\end{array}
\right),\left(\begin{array}{l}
\gamma'\\
\delta'
\end{array}\right)=\left(\begin{array}{cccccccc}
3&3&3&2&2&1&1&1\\
&2&1&1&1&&&
\end{array}
\right),\] we then split $(\gamma', \delta')$ to generate
$(\alpha^2, \beta^2)$, the divisional part $\delta'_j$ is the
smallest part satisfying $\delta'_j\geq \gamma'_{j+3}$. The
remaining part of $(\gamma', \delta')$ is just $(\alpha^1,
\beta^1)$.

\[
 \left(\begin{array}{l}
\alpha^2\\
\beta^2
\end{array}\right)=\left(\begin{array}{cccccc}
3&3&3&2&2&1\\
2&1&1&1
\end{array}
\right), \left(\begin{array}{l}
\alpha^1\\
\beta^1
\end{array}\right)=\left(\begin{array}{cc}
1&1\\
&
\end{array}
\right).
\]
Thus we get
\begin{equation}\label{example}
\eta=\left(\begin{array}{ccccccccccccc}
6_3&6_3&3_3&3_2&3_2&3_2&2_2&2_2&1_2&1_1&1_1\\
5_3&5_3&4_3&&2_2&1_2&1_2&1_2&&
\end{array}
\right)_6.\end{equation}

By Theorems \ref{r-kms-ds} and \ref{dy-op}, we then deduce the
following partition identity:
\begin{core} For $m_i\geq 0$ and $k\geq 1$,  we have
\begin{equation}
\mathcal{D}^{ss}_k(m_1,m_2,\ldots,m_k;n)=N\left(\sum_{i=1}^km_i+k-1;n\right).
\end{equation}
\end{core}

To establish the relation between $k$-marked strict shifted Durfee
symbols and $k$-marked Durfee symbols, we need define a statistic
on $k$-marked Durfee symbols. In the same way, we first define
this statistic on a pair of partitions.

 For $(\gamma,\delta)$, the part $\delta_i$ is said to be balanced if
$\gamma_{i+1}\leq \delta_i$ and the number of parts greater than
$\delta_{i}$ in $\gamma$ ($\gamma_1$ is not counted) {\it is equal
to} the number of unbalanced parts before $\delta_{i}$ in
$\delta$. For example, let
 $\left(\begin{array}{c}\gamma\\ \delta
 \end{array}
 \right)=\left(\begin{array}{ccccc}
4&3&3&1&1\\
&\underline{3}&2&2&
\end{array}\right)$, the first part  $3$ of $\delta$ is balanced while the third part $2$
is not balanced, although it satisfies the first condition
$\gamma_{i+1}\leq \delta_i$, there are two parts greater than 2
($4$ is not counted) in $\gamma$, while there is one unbalanced
part (the second part 2) before the third part 2 in $\delta$.

It should be pointed out that  for any unbalanced part $\delta_i$,
the number of parts greater than $\delta_{i}$ in $\gamma$
($\gamma_1$ is not counted) {\it is greater than} the number of
unbalanced parts before $\delta_{i}$ in $\delta$. We will state
this in the following proposition:
\begin{pro}\label{pro-bp}
Let $(\gamma,\delta)$ be a pair of partitions, for any part
$\delta_j$ of $\delta$, let $d_j$ denote the difference between
the number of parts greater than $\delta_j$ in $\gamma$
{\rm(}except for $\gamma_1${\rm)} and the number of unbalanced
parts before $\delta_j$ in $\delta$, we then have $d_j\geq 0$.
\end{pro}
For the above example, we see that $d_1=0, d_2=2, d_3=1.$

 \pf If $\gamma_{j+1}> \delta_{j}$, it's clear to see that $d_j\geq 1$; If
 $\gamma_{j+1}\leq\delta_{j}$, we consider the following two
 cases:
\begin{kst}
\item[Case 1] When $\delta_{j-1}<\gamma_{j}$, and at this time,
there are $j-1$ parts greater than $\delta_j$ in $\gamma$
($\gamma_1$ is not counted) and the number of unbalanced parts
before $\delta_j$ in $\delta$ is less than or equal to $j-1$, so
$d_j\geq 0$.

\item[Case 2] When $\delta_{j-1}\geq \gamma_{j}$, let $t$ be
largest nonnegative  integer less than $j$ such that
$\delta_{t}<\gamma_{t+1}$. Here we assume that $\delta_0=0$ and
then $t$ must exist.  From Case 1, we know that $d_{t+1}\geq 0$.
We use the induction to prove that $d_i\geq 0$ for $t+2\leq i\leq
j$.

We first prove that $d_{t+2}\geq 0$. Note that
$\gamma_{t+1}>\delta_{t+1}\geq \gamma_{t+2}$.  If
$\delta_{t+2}<\gamma_{t+2}$, then $d_{t+2}\geq d_{t+1}\geq 0$; If
$\delta_{t+2}\geq \gamma_{t+2}$, then $d_{t+2}\geq d_{t+1}-1$. In
particular, when $d_{t+1}=0$, then $\delta_{t+1}$ is balanced,
thus $d_{t+2}=d_{t+1}=0$. So $d_{t+2}\geq 0.$

Assume that $d_i\geq 0$ and $\gamma_p> \delta_i\geq \gamma_{p+1}$.
We will prove that $d_{i+1}\geq 0$. If $\delta_{i+1}\geq
\gamma_{p+1}$, the similar argument on the case for $\delta_{t+2}$
could show  that  $d_{i+1}\geq 0.$ If $\delta_{i+1}<\gamma_{p+1}$,
and then $d_{i+1}\geq d_i\geq 0$. \qed
\end{kst}

We use $b(\gamma,\delta)$ to denote the number of balanced parts
in $(\gamma,\delta).$ Clearly, $0\leq b(\gamma,\delta)\leq
l(\delta)$. Let  $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ be  a $k$-marked Durfee symbol, we define ${\rm
nb}_i(\eta)$, called the $i$th balanced number by
\[{\rm
nb}_i(\eta)=\left\{
\begin{array}{ll}
b(\alpha^i,\beta^i) \ \ &\text{ for }\ 1\leq i<k,\\[5pt]
0 \ \ &\text{ for }\ i=k.
\end{array}\right.
\]
For $\left(\begin{array}{cccccccc}
4_3&4_3&3_2&2_2&2_2&2_1&2_1&1_1\\
5_3&&&\underline{3_2}&\underline{2_2}&&\underline{2_1}&1_1
\end{array}\right)_5$, we have ${\rm
nb}_1=1,\,{\rm nb}_2=2,\,{\rm nb}_3=0.$

We next state a theorem concerning  strict shifted partitions.
\begin{thm}\label{lem-ssp}
Given two nonnegative integers $r,m$, there is a bijection $\psi$
between the set of pairs of partitions $(\alpha,\beta)$ of $n$
with $\beta_1\leq \alpha_1$ where there are  $r$ balanced parts
and the difference of the lengths of $\alpha$ and $\beta$ equals
to $m$ and the set of strict shifted partitions
$(\bar{\alpha},\bar{\beta})$ of $n$ where the difference of the
lengths of $\bar{\alpha}$ and $\bar{\beta}$ equals to $m+2r$.
\end{thm}
\pf {\bf The map $\psi$:} Let $(\alpha,\beta)$ be a pair of
partitions with $r$ balanced parts and $l(\alpha)-l(\beta)=m$. The
strict shifted partitions $(\bar{\alpha},\bar{\beta})$ is
constructed as follows. $\bar{\alpha}$ is composed of all parts of
$\alpha$ and all balanced parts of $\beta$. $\bar{\beta}$ consists
of all unbalanced parts of $\beta$. Take an example, let
$\left(\begin{array}{c}\alpha\\ \beta
 \end{array}
 \right)=\left(\begin{array}{cccccc}
6&5&5&3&3&2\\
&\underline{5}&4&4&\underline{3}
\end{array}\right)$, where the underlined parts in $\beta$ are
balanced. According to the above construction, we then get $\left(\begin{array}{c}\bar{\alpha}\\
\bar{\beta}
 \end{array}
 \right)=\left(\begin{array}{cccccccc}
6&\underline{5}&5&5&\underline{3}&3&3&2\\
&4&4
\end{array}\right)$. It's clear to see that
$|\alpha|+|\beta|=|\bar{\alpha}|+|\bar{\beta}|$,
$l(\bar{\alpha})-l(\bar{\beta})=l(\alpha)+r-(l(\beta)-r)=m+2r.$ By
Proposition \ref{pro-bp}, one can also easily know that
$(\bar{\alpha},\bar{\beta})$ is strict shifted.

{\bf The reverse map $\psi^{-1}$:} Let
$(\bar{\alpha},\bar{\beta})$ be a strict shifted partition where
the difference  of the lengths of $\bar{\alpha}$ and $\bar{\beta}$
is $m+2r$, that is $l(\bar{\alpha})-l(\bar{\beta})=m+2r$. We now
construct a pair of partitions $(\alpha,\beta)$ with $r$ balanced
parts and $l(\alpha)-l(\beta)=m$.

First of all, attach subscript $g_i$ for each part
$\bar{\alpha}_i$ of $\bar{\alpha}$, where $g_i$ denotes the
difference between the number of parts before $\bar{\alpha}_i$ in
$\bar{\alpha}$ ($\bar{\alpha}_1$ is not counted) and the number of
parts greater than or equal to $\bar{\alpha}_i$ in $\beta$. We let
$g_1=0$.

For example,
if $\left(\begin{array}{c}\bar{\alpha}\\
\bar{\beta}
 \end{array}
 \right)=\left(\begin{array}{ccccccccc}
6&5&5&5&3&3&3&2&1\\
&4&4&3
\end{array}\right)$, attach the subscripts for all parts of $\bar{\alpha}$ to
get
$\left(\begin{array}{ccccccccc}
6_0&5_0&5_1&5_2&3_0&3_1&3_2&2_3&1_4\\
&4&4&3
\end{array}\right).$

One could easily know that $g_2=0$ and $g_i\geq 0$ for any $3\leq
i\leq l(\bar{\alpha})$ from the fact that
$(\bar{\alpha},\bar{\beta})$ is strict shifted. Let
$\bar{\alpha}_{t_i}$ be the smallest part in all of parts of
$\alpha$ with subscript equal to $i$, we have the following
conclusion:

\begin{lem}\label{temp}
For $0\leq i \leq m+2r-2$, $\bar{\alpha}_{t_i}$ exists, and
$\bar{\alpha}_{t_0}\geq \bar{\alpha}_{t_1}\geq \cdots \geq
\bar{\alpha}_{t_{m+2r-2}}$.
\end{lem}
In the above example, $m+2r=6$ and
$\bar{\alpha}_{t_0}=3,\bar{\alpha}_{t_1}=3,\bar{\alpha}_{t_2}=3,\bar{\alpha}_{t_3}=2,\bar{\alpha}_{t_4}=1$.

  \pf We use the induction to show that the
sequence of subscripts $\{g_1,g_2,\ldots,g_{l(\bar{\alpha})}\}$
consists of all nonnegative integers less than $m+2r-1$.
Obviously, $0$ is in this sequence. Assume that $i$ is in this
sequence, that is there is a part $\bar{\alpha}_p$ such that
$g_p=i$, we now prove that $i+1$ is also in this sequence. By the
induction hypothesis, we know the subscript of the part
$\bar{\alpha}_p$ is $i$, that is
$\bar{\beta}_{p-i-1}<\bar{\alpha}_p\leq \bar{\beta}_{p-i-2}$. Let
$l(\bar{\beta})=s$ and note that $i\leq m+2r-3$, we have
$s+i+3\leq s+m+2r=l(\bar{\alpha})$. If $\bar{\alpha}_{s+i+3}\leq
\bar{\beta}_{s}$, then the subscript of $\bar{\alpha}_{s+i+3}$ is
$i+1$; Otherwise, there must exist $p+1\leq j\leq s+i+2$ such that
$\bar{\beta}_{j-i-2}<\bar{\alpha}_{j}\leq \bar{\beta}_{j-i-3}$
this is because that $\bar{\alpha}_{p+1}\leq \bar{\beta}_{p-i-2}$
and $\bar{\alpha}_{s+i+2}>\bar{\beta}_s$. Hence, the subscript of
$\bar{\alpha}_{j}$ is $i+1$. Therefore, $\bar{\alpha}_{t_i}$
exists for $0\leq i \leq m+2r-2$ and $\bar{\alpha}_{t_0}\geq
\bar{\alpha}_{t_1}\geq \cdots \geq \bar{\alpha}_{t_{m+2r-2}}$ when
note that given a part $\bar{\alpha}_p$ with subscript $i$, we
could always find a part after $\bar{\alpha}_p$ whose subscript is
$i+1$. \qed

Let $\gamma$ be a partition having $r$ parts whose parts are
 $\bar{\alpha}_{t_0},
\bar{\alpha}_{t_2},\ldots,\bar{\alpha}_{t_{r-1}}$ respectively.
Take $r=2$ in the above example, $\gamma=(3,3)$. We now construct
the partitions $(\alpha,\beta)$. The partition $\alpha$ consists
of all parts in $\bar{\alpha}$, while not in $\gamma$. $\beta$ is
composed of all parts both in $\bar{\beta}$ and $\gamma$.
In the above example, we therefore get $\left(\begin{array}{c}\alpha\\
\beta
 \end{array}
 \right)=\left(\begin{array}{ccccccccc}
6&5&5&5&3&2&1\\
&4&4&3&\underline{3}&\underline{3}
\end{array}\right)$.

It's obvious to see that
$|\bar{\alpha}|+|\bar{\beta}|=|\alpha|+|\beta|$,
$l(\alpha)-l(\beta)=l(\bar{\alpha})-l(\gamma)-[l(\bar{\beta})+l(\gamma)]=m$.
We now show that $(\alpha,\beta)$ has exactly $r$ balanced parts.
From Lemma \ref{temp} and the definition of $\gamma$, we know that
for each  part $\beta_t$ from $\gamma$ in $\beta$, the number of
parts greater than $\beta_t$ in $\alpha$ equals the number of
parts from $\bar{\beta}$ greater than or equal to $\beta_t$ in
$\beta$. Thus we just need to prove that the parts from
$\bar{\beta}$ in $\beta$ are unbalanced. We use induction on the
part from $\bar{\beta}$ in $\beta$. We first verify the largest
part $\bar{\beta}_1$ of $\bar{\beta}$ is unbalanced. Supposed that
there are $t$ parts $\gamma_1,\gamma_2,\ldots,\gamma_t$ from
$\gamma$ greater than $\bar{\beta}_1$, then
$\gamma_i=\bar{\alpha}_{i+1},i=1,2,\ldots,t$ and
$\alpha_2=\bar{\alpha}_{t+2}$. We claim that
$\alpha_2=\bar{\alpha}_{t+2}>\bar{\beta}_1$. Recall that the part
$\gamma_i$ is the smallest part whose subscript is $i$. If
$\alpha_2=\bar{\alpha}_{t+2}\leq \bar{\beta}_1$, the subscript of
$\bar{\alpha}_{t+2}$ is less than $t+1$, this contradicts to the
definition of $\gamma_i$. Clearly, these $t$ parts from $\gamma$
are balanced, and $\bar{\beta}_1$ is not balanced. We now consider
the part $\bar{\beta}_j$ from $\bar{\beta}$ in $\beta$. Assume
that all parts from $\bar{\beta}$ before $\bar{\beta}_j$ in
$\beta$ are not balanced and there are $t$ parts from $\gamma$
before $\bar{\beta}_j$. We next justify
$\alpha_{j+1}>\bar{\beta}_j$ and then by the hypothesis, we know
that there are $j-1$ unbalanced parts before $\bar{\beta}_j$,
while there are at least $j$ parts larger than $\bar{\beta}_j$ in
$\alpha$, so $\bar{\beta}_j$ is unbalanced. Since there are $t$
parts from $\gamma$ before $\bar{\beta}_j$, then
$\alpha_{j+1}=\bar{\alpha}_{t+j+1}$ and if
$\bar{\alpha}_{t+j+1}=\alpha_{j+1}\leq \bar{\beta}_j$, then the
subscript of $\bar{\alpha}_{t+j+1}$ is less than $t$, which
contradicts to the definition of the parts of $\gamma$, so
$\alpha_{j+1}>\bar{\beta}_j$ and we therefore complete the proof.
\qed

The next theorem gives  a relation between $k$-marked strict
shifted Durfee symbols and $k$-marked Durfee symbols.

\begin{thm}\label{r-km-kms}
Given $2k$ nonnegative integers $m_1,m_2,\ldots,m_k,$ and
$t_1,t_2,\ldots,t_k$ where $t_k=0$. There is a bijection $\Psi$
between the set of $k$-marked Durfee symbols of $n$ with $i$th
rank equal to $m_i$ and $i$th balanced number equal to $t_i$ and
the set of $k$-marked strict shifted Durfee symbols of $n$ with
$i$th rank equal to $m_i+2t_i$.
\end{thm}
\pf Let $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ be a $k$-marked Durfee symbol with $i$th rank equal to
$m_i$ and $i$th balanced number
 equal to $t_i$. We now apply the
bijection $\psi$ in Theorem \ref{lem-ssp} on each vector
$(\alpha^i,\beta^i)$ of $\eta$ except for $k$th vector
$(\alpha^k,\beta^k)$, to generate
$(\bar{\alpha}^i,\bar{\beta}^i)$. From Theorem \ref{lem-ssp}, we
know that $(\bar{\alpha}^i,\bar{\beta}^i)$ is strict shifted and
$l(\bar{\alpha}^i)-l(\bar{\beta}^i)=l(\alpha^i)-l(\beta^i)+2t_i$.
Let $\bar{\eta}=\left(\begin{array}{cccc}
\alpha^k,&\bar{\alpha}^{k-1},&\ldots,&\bar{\alpha}^1\\[2pt]
\beta^k,&\bar{\beta}^{k-1},&\ldots,&\bar{\beta}^1
\end{array}
\right)_D$ which has the same subscript and the same $k$th vector
with $\eta$. It's obvious to see that $\bar{\eta}$ is a $k$-marked
strict shifted Durfee symbol with $i$th rank equal to $m_i+2t_i$.
\qed

By Theorem \ref{r-km-kms}, one can derive the following identity
readily.
\begin{core}\label{c-km-kms}
For $m_i\geq 0$ and $k\geq 2$, we have
\begin{equation}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)=\sum_{t_1,\ldots,\,t_{k-1}=0}^{{+\infty}}\mathcal{D}^{ss}_k(m_1+2t_1,\ldots,m_{k-1}+2t_{k-1},m_k;n).
\end{equation}
\end{core}
Combine  Corollaries \ref{r-kms-ds} and \ref{c-km-kms} to get:
\begin{thm}\label{c-km-ds-1}
For $m_i\geq 0$ and $k\geq 2$, we have
\begin{equation}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)=\sum_{t_1,\ldots,\,t_{k-1}=0}^{+\infty}
N\left(\sum_{i=1}^km_i+2\sum_{i=1}^{k-1}t_i+k-1;n\right).
\end{equation}
\end{thm}

The following compact form  of Theorem \ref{c-km-ds-1} can be
easily obtained upon utilizing the fact that the number of
solutions to $t_1+t_2+\cdots+t_{k-1}=j$ in nonnegative integers is
$j+k-2 \choose k-2$.
\begin{thm}\label{c-km-ds-2}
For $m_i\geq 0$ and $k\geq 2$, we have
\begin{equation}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)=\sum_{j=0}^{+\infty}{j+k-2
\choose k-2}N\left(\sum_{i=1}^km_i+2j+k-1;n\right).
\end{equation}
\end{thm}

We next generalize  Theorem \ref{c-km-ds-2} to give Theorem
\ref{main} which holds for any integer $m_i$.  To do this, we
prove the following conclusion by constructing a simple bijection
$\Theta$.
\begin{thm} \label{r-sym}
For $k\geq 1$ and $1\leq p\leq k$, we have
\begin{equation}
\mathcal{D}_k(m_1,\ldots,m_p,\ldots
m_k;n)=\mathcal{D}_k(m_1,\ldots,-m_p,\ldots m_k;n).
\end{equation}
\end{thm}
\pf  Let $\eta=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D$ be a $k$-marked Durfee symbol with $i$th rank equal to
$m_i$. We will construct another $k$-marked Durfee symbol
$\bar{\eta}$ with $i$th rank equal to $\bar{m}_i$ such that
$\bar{m}_p=-m_p$ and $\bar{m}_i=m_i$ for $i\neq p$.

Define \[\bar{\eta}=\left(\begin{array}{ccccccc}
\alpha^k,&\ldots,&\alpha^{p+1}&\bar{\alpha}^p,&\alpha^{p-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\ldots,&\beta^{p+1},&\bar{\beta}^p,&\beta^{p-1},&\ldots,&\beta^1
\end{array}
\right)_D,\] where $\bar{\alpha}^k=\beta^k$ and
$\bar{\beta}^k=\alpha^k$ for $p=k$; When $p\neq k$,
$\bar{\alpha}^p$ consists of all parts of $\beta^p$ and the
largest part $\alpha^p_1$ of $\alpha^p$. $\bar{\beta}^p$ consists
of all parts of $\alpha^p$ except for the largest part
$\alpha^p_1$. It's clear to see that
$\bar{m}_p=l(\bar{\alpha}^p)-l(\bar{\beta}^p)-1=l(\beta^p)+1-[l(\alpha^p)-1]-1=-[l(\alpha^p)-l(\beta^p)-1)]=-m_p$
for $p\neq k$ and
$\bar{m}_k=l(\bar{\alpha}^k)-l(\bar{\beta}^k)=-[l(\alpha^k)-l(\beta^k)]=-m_k,$
so $\bar{\eta}$ is desired. \qed

By Theorem \ref{r-sym}, we could generalize Theorem
\ref{c-km-ds-1} to give the following theorem which is useful to
prove a relationship between $k$-marked Durfee symbols and the
symmetrized rank moment given by Andrews (see Theorem \ref{int}).

\begin{thm}\label{c-km-ds-1'}
For $k\geq 2$, we have
\begin{equation}
\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)=\sum_{t_1,\ldots,\,t_{k-1}=0}^{+\infty}
N\left(\sum_{i=1}^k|m_i|+2\sum_{i=1}^{k-1}t_i+k-1;n\right).
\end{equation}
\end{thm}
Theorem \ref{main} is a compact form of Theorem \ref{c-km-ds-1'}
which immediately follows from Theorems \ref{c-km-ds-2} and
\ref{r-sym}.

\section{Andrews' identities on $k$-marked Durfee symbols }

In this section, we aim to  show  the identities  on $k$-marked
Durfee symbols given by Andrews with the help of Theorem
\ref{main}. Recall that $\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$
denotes the number of $k$-marked Durfee symbols of $n$ with $i$th
rank equal to $m_i$. Andrews considered the following generating
function for $\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$:
\begin{align}\label{def-g-km}
R_k(x_1,\ldots,x_k;q)=\sum_{m_1,\ldots,m_k=-\infty}^{+\infty}\sum_{n=0}^{+\infty}
\mathcal{D}_k(m_1,\ldots,m_k;n)x_1^{m_1}\cdots x_k^{m_k}q^n,\,
|q|<1.
\end{align}
By  applying the $k$-fold generalization of Watson's
transformation between a very-well-poised $_8\phi_7$-series and a
balanced $_4\phi_3$-series \cite[p.199, Theorem 4]{Andrews-1975},
Andrews gave the generating function $R_k(x_1,x_2,\ldots,x_k;q)$
in the following theorem.

\begin{thm}[Corollary 11, Andrews \cite{Andrews-07-a}]\label{gf-km}
\begin{align}\label{gf-km-e}
&R_k(x_1,x_2,\ldots,x_k;q)
=\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n-1}q^{3n(n-1)/2+kn}
\frac{(1+q^n)(1-q^n)^2}{\prod_{j=1}^k(1-x_jq^n)(1-\frac{q^n}{x_j})}.
\end{align}
\end{thm}
\pf We will reformulate this identity  as the partition identity
\eqref{main-e} in Theorem \ref{main}. The key step is to give a
partition interpretation of the right side hand of
\eqref{gf-km-e}. We will show that it is the generating function
for the summation on the right side of \eqref{main-e}.

First, the right hand side of \eqref{gf-km-e} can be written as
the difference of the following two terms:
\begin{align*}
&\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n}q^{n(3n-1)/2+kn}
\frac{(1-x_1)(1-x_1^{-1})(1+q^n)}{(1-x_1q^n)(1-x_1^{-1}q^n)}\times\frac{1}{\prod_{j=2}^k(1-x_jq^n)(1-\frac{q^n}{x_j})}\nonumber \\
&\hskip
-0.5cm-\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n}q^{n(3n-1)/2+(k-1)n}
\frac{1+q^n}{\prod_{j=2}^k(1-x_jq^n)(1-\frac{q^n}{x_j})}.\nonumber
\end{align*}
We next expand each term of the above two terms, note that
\begin{align*}
\frac{(1-x_1)(1-x_1^{-1})(1+q^n)}{(1-x_1q^n)(1-x_1^{-1}q^n)}&=\frac{(1-x_1)}{(1-x_1q^n)}
+\frac{(1-x_1^{-1})}{(1-x_1^{-1}q^n)}
=2+\sum_{\stackrel{m_1=-\infty}{m_1\neq
0}}^{{+\infty}}x_1^{m_1}[q^{n|m_1|}-q^{n(|m_1|-1)}],\\
\frac{1}{(1-x_jq^n)(1-x_j^{-1}q^n)}&=\sum_{a=0}^{+\infty}
x_j^aq^{na}\sum_{b=0}^{+\infty}
x_j^{-b}q^{nb}=\sum_{m_j=-\infty}^{{+\infty}}x_j^{m_j}\sum_{t_j=0}^{{+\infty}}q^{n(|m_j|+2t_j)}.
\end{align*}
Given $k$ integers $m_1\ldots,m_k$, it's clear to see that the
coefficients of $x_1^{m_1}x_2^{m_2}\cdots x_k^{m_k}$ on the series
expansion of  the right  hand side of \eqref{gf-km-e} are
\begin{align*}
&\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n}q^{n(3n-1)/2+kn}[q^{n|m_1|}-q^{n(|m_1|-1)}]
\sum_{t_2,\ldots,t_k=0}^{{+\infty}}q^{n(\sum_{i=2}^k|m_i|+2\sum_{i=2}^kt_i)}\\
&=\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n-1}q^{n(3n-1)/2+kn}(1-q^n)q^{n(|m_1|-1)}
\sum_{j=0}^{{+\infty}}{j+k-2\choose
k-2}q^{n(\sum_{i=2}^k|m_i|+2j)}\\
&=\sum_{j=0}^{{+\infty}}{j+k-2\choose
k-2}\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n-1}q^{n(3n-1)/2}(1-q^{n})q^{n(\sum_{i=1}^k|m_i|+2j+k-1)}\\
&=\sum_{j=0}^{{+\infty}}{j+k-2\choose k-2}\sum_{n=0}^{+\infty}
N(\sum_{i=1}^k|m_i|+2j+k-1;n)q^n\\
&=\sum_{n=0}^{+\infty} q^n[\sum_{j=0}^{{+\infty}}{j+k-2\choose
k-2}N(\sum_{i=1}^k|m_i|+2j+k-1;n)],
\end{align*}
where the penultimate identity  follows from Theorem \ref{gf-r}
and we then obtain the following combinatorial interpretation:
\begin{align}\label{com-int}
&\frac{1}{(q;q)_\infty}\sum_{n=1}^{+\infty}(-1)^{n-1}q^{3n(n-1)/2+kn}
\frac{(1+q^n)(1-q^n)^2}{\prod_{j=1}^k(1-x_jq^n)(1-\frac{q^n}{x_j})}\nonumber \\[2pt]
&\hskip
-0.5cm=\sum_{m_1,\ldots,m_k=-\infty}^{+\infty}\sum_{n=0}^{+\infty}
\left[\sum_{j=0}^{+\infty}{j+k-2 \choose
k-2}N\left(\sum_{i=1}^k|m_i|+2j+k-1;n\right)\right]x_1^{m_1}\cdots
x_k^{m_k}q^n.
\end{align}
Combining  \eqref{def-g-km} and \eqref{com-int},  we reach our
conclusion that the identity \eqref{gf-km-e} can be restated as
the partition identity \eqref{main-e}. Thus, we have obtained a
combinatorial proof of \eqref{gf-km-e} based on Theorem
\ref{main}. \qed

Recently, Bringmann, Lovejoy, and Osbur defined a two-parameter
generalization of $k$-marked Durfee symbols in
\cite{Bringmann-lovejoy-osbur}. They deduced  the generating
function \cite[Theorem 2.2]{Bringmann-lovejoy-osbur} for the
two-parameter generalization of $k$-marked Durfee symbols using
the similar argument of Andrews, which reduces to the identity
\eqref{gf-km-e} when $d=e=0.$

From the generating function $R_k(x_1,x_2,\ldots,x_k;q)$ in
Theorem \ref{gf-km}, Andrews immediately found the following
symmetry of $k$-marked Durfee symbols.
\begin{thm}[Corollary 12, Andrews \cite{Andrews-07-a}]\label{sym-mds}
$\mathcal{D}_k(m_1,m_2,\ldots,m_k;n)$ is symmetric in
$m_1,\,m_2,\ldots,m_k$.
\end{thm}
\pf This symmetry can also immediately  follow from Theorem
\ref{main}.
 \qed

In fact, the composite of the bijections on Section 2 provides a
bijection for this symmetry. We take an example to explain this
process. Let
\begin{equation*}
\eta=\left(\begin{array}{ccccccccccccc}
6_3&3_2&3_2&2_2&2_2&1_2&1_1\\
5_3&3_2&3_2&1_2&&&1_1&1_1
\end{array}
\right)_6 \text{counted by }
\mathcal{D}_3(-2,1,0;68),\end{equation*} we aim to construct a
$3$-marked Durfee symbol $\bar{\eta}$ counted by
$\mathcal{D}_3(1,-2,0;68)$. We will first combine all subscripts
of $k$-marked Durfee symbol $\eta$ to get a Durfee symbol, and
then split this Durfee symbol over again to get our desired
$k$-marked Durfee symbol $\bar{\eta}$.

First, applying the bijection $\Theta$ in Theorem \ref{r-sym} into
$\eta$ to get $\eta'$ enumerated by $\mathcal{D}_3(2,1,0;68)$,
\begin{equation*}
\eta'=\left(\begin{array}{ccccccccccccc}
6_3&3_2&3_2&2_2&2_2&1_2&1_1&1_1&1_1\\
5_3&&\underline{3_2}&\underline{3_2}&1_2&
\end{array}
\right)_6, \end{equation*} we now  utilize the bijection $\Psi$ in
Theorem \ref{r-km-kms} on $\eta'$ to get a $k$-marked strict
shifted Durfee symbol. Observe that there are two balanced parts
in the second vector of $\eta'$, and there is no balanced part in
other vectors of $\eta'$. So we will get a $k$-marked strict
shifted Durfee symbol $\eta''$ which counted by
$\mathcal{D}^{ss}_3(2,1+2\times2,0;68)$,
\begin{equation*}
\eta''=\left(\begin{array}{ccccccccccccc}
6_3&3_2&3_2&3_2&3_2&2_2&2_2&1_2&1_1&1_1&1_1\\
5_3&&   1_2
\end{array}
\right)_6, \end{equation*} applying the bijection $\Phi$ in
Theorem \ref{r-kms-ds} to get $\eta'''$ which counted by
$\mathcal{D}_1(9;68)$
\begin{equation*}
\eta'''=\left(\begin{array}{ccccccccccccc}
6&3&3&3&3&2&2&1&1&1&1\\
5&1
\end{array}
\right)_6. \end{equation*} Thus we complete the first step. We
next split the Durfee symbol $\eta'''$ over again.

First, apply the reverse map $\Phi^{-1}$ in Theorem \ref{r-kms-ds}
on $\eta'''$ to get $\bar{\eta}''$ which counted by
$\mathcal{D}^{ss}_3(1,2+2\times2,0;68)$,
\begin{equation*}
\bar{\eta}''=\left(\begin{array}{ccccccccccccc}
6_3&3_2&\underline{3_2}&\underline{3_2}&3_2&2_2&2_2&1_2&1_2&1_1&1_1\\
5_3&1_2
\end{array}
\right)_6, \end{equation*} using the reverse map $\Psi^{-1}$ in
Theorem \ref{r-km-kms} on $\bar{\eta}''$, we get $k$-marked Durfee
symbol $\bar{\eta}'$ counted by $\mathcal{D}_3(1,2,0;68)$
\begin{equation*}
\bar{\eta}'=\left(\begin{array}{ccccccccccccc}
6_3&3_2&3_2&2_2&2_2&1_2&1_2&1_1&1_1\\
5_3&&3_2&3_2&1_2
\end{array}
\right)_6, \end{equation*} Finally, we obtain the desired
$k$-marked Durfee symbol $\bar{\eta}$ counted by
$\mathcal{D}_3(1,-2,0;68)$ when applying the bijection $\Theta$ in
Theorem \ref{r-sym} on $\bar{\eta}'$.
\begin{equation*}
\bar{\eta}=\left(\begin{array}{ccccccccccccc}
6_3&3_2&3_2&3_2&1_2&&1_1&1_1\\
5_3&3_2&2_2&2_2&1_2&1_2
\end{array}
\right)_6.\end{equation*}

By the generating function $R_k(x_1,x_2,\ldots,x_k;q)$ and the
generating function for $\eta_{2k}(n)$, Andrews  showed that the
number of $(k+1)$-marked Durfee symbols of $n$ equals the
symmetrized $(2k)$-th moment function at $n$ in
\cite{Andrews-07-a}, that is

\begin{thm}[Corollary 13, Andrews \cite{Andrews-07-a}]\label{int}
For $k\geq 1$,
\begin{equation}\label{int-e}
\mathcal{D}_{k+1}(n)=\eta_{2k}(n).
\end{equation}

\end{thm}

\pf Recall that
\begin{align*}
\eta_{2k}(n)&=\sum_{m=-\infty}^{{+\infty}}{m+k-1 \choose
2k}N(m;n)=\sum_{m=1}^{{+\infty}}\left[{m+k-1 \choose 2k}+{m+k
\choose 2k}\right]N(m;n),
\end{align*}
where the second equality follows from the
 rank symmetry $N(-m;n)=N(m;n)$ and the fact ${-m+k-1 \choose 2k}={m+k \choose 2k}$.
\begin{align*}
\mathcal{D}_{k+1}(n)&=\sum_{m_1,\ldots,m_{k+1}=-\infty}^{+\infty}
\mathcal{D}_{k+1}(m_1,m_2,\ldots,m_{k+1};n)\\
&=\sum_{m_1,\ldots,m_{k+1}=-\infty}^{+\infty}\sum_{t_1,\ldots,\,t_{k}=0}^{+\infty}
N\left(\sum_{i=1}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k;n\right),\end{align*}
where the second equality follows from  Theorem \ref{c-km-ds-1'}.
So it suffices to show that the number of solutions to
$|m_1|+\cdots+|m_{k+1}|+2t_1++\cdots+2t_k=m-k$ where $m_i$ is
integer, and $t_i$  is nonnegative integer equals to ${m+k-1
\choose 2k}+{m+k \choose 2k}$.

Let $c(n)$ denote the number of solutions to
$|m_1|+|m_2|+\cdots+|m_{k+1}|+2t_1+2t_2+\cdots+2t_k=n$ where $m_i$
is integer and $t_i$ is nonnegative integer. It's easy to know
that the generating function for $c(n)$ is
\begin{align*}
\sum_{n=0}^{{+\infty}}c(n)q^n&=(1+2q+2q^2+2q^3+\cdots)^{k+1}(1+q^2+q^4+q^6+\cdots)^k \\
&=\frac{(1+q)^{k+1}}{(1-q)^{k+1}}\times\frac{1}{(1-q^2)^k} \\
&=(1+q)\times\frac{1}{(1-q)^{2k+1}}\\
&=(1+q)\sum_{n=0}^{{+\infty}}{2k+n \choose n}q^n\\
&=\sum_{n=0}^{{+\infty}}{2k+n \choose
2k}q^n+\sum_{n=1}^{{+\infty}}{2k+n-1 \choose 2k}q^n.
\end{align*}
Comparing coefficients of $q^n$ in the above expression, we obtain
\[c(n)={2k+n \choose
2k}+{2k+n-1 \choose 2k}.\] Thus we reach our conclusion. \qed

By partial fraction expansion, Andrews \cite{Andrews-07-a} also
gave the following relationship between the generating function
for $k$-marked Durfee symbols and the generating function for
Durfee symbols, which plays an important role in the study of
Ramanujan-type congruences for $k$-marked Durfee symbols.

\begin{thm}[Theorem 7, Andrews \cite{Andrews-07-a}]\label{thm7}
\begin{align}\label{thm-7}
R_k(x_1,x_2,\ldots,x_k;q)=\sum_{i=1}^k\frac{R_1(x_i;q)}{\prod_{\stackrel{j=1}{
j\neq i}}^k(x_i-x_j)(1-\frac{1}{x_ix_j})}.
\end{align}
\end{thm}
\pf Similarly, we will restate this identity as the partition
identity \eqref{main-e} in Theorem \ref{main}.  We first consider
the series expansion of  the right hand side of \eqref{thm-7}.  To
do this, we need to work in a larger ring: the field of iterated
Laurent series $K\ll
x_k,x_{k-1},\ldots,x_1\gg=K((x_k))((x_{k-1}))\cdots((x_1))$ where
$K=\mathbb{C}(q)$, in which all series are regarded first as
Laurent series in $x_1$, then as Laurent series in $x_2$, and so
on. For more detailed account of the properties of this field,
with other applications, see \cite{Xin-04} and \cite{Xin-05}.

Every element of $K\ll x_k,x_{k-1},\ldots,x_1\gg$ has a unique
iterated Laurent series expansion. In particular, the series
expansion of $1/(1-\frac{1}{x_ix_j})$ is:
\begin{equation}
\frac{1}{1-\frac{1}{x_ix_j}}=-\frac{x_ix_j}{1-x_ix_j}=-\sum_{l=1}^{+\infty}x_i^{l}x_j^{l}.
\end{equation}
The series expansions of $1/(x_i-x_j)$ will be especially
important. If $j<i$, then
\begin{equation*}
\frac{1}{x_i-x_j}=\frac{x^{-1}_i}{1-\frac{x_j}{x_i}}=\sum_{l=0}^{+\infty}x_i^{-l-1}x_j^{l}.
\end{equation*}
However, if $j>i$ then this expansion is not valid and instead we
have the expansion:
\begin{equation*}
\frac{1}{x_i-x_j}=\frac{-x^{-1}_j}{1-\frac{x_i}{x_j}}=-\sum_{l=0}^{+\infty}x_i^{l}x_j^{-l-1}.
\end{equation*}
Thus for $j<i$, the series expansion of
$\frac{1}{(x_i-x_j)(1-x^{-1}_ix^{-1}_j)}$ is
\begin{equation}\label{serexp-1}
\frac{1}{(x_i-x_j)(1-x^{-1}_ix^{-1}_j)}=-\sum_{l=1}^{+\infty}x_i^{l}x_j^{l}\sum_{m=0}^{+\infty}x_i^{-m-1}x_j^{m}
=-\sum_{m_j=1}^{+\infty}x_j^{m_j}\sum_{t_j=0}^{m_j-1}x_i^{2t_j-m_j+1},
\end{equation}
and for $j>i$, we have the following expansion:
\begin{equation}\label{serexp-2}
\frac{1}{(x_i-x_j)(1-x^{-1}_ix^{-1}_j)}=\sum_{l=1}^{+\infty}x_i^{l}x_j^{l}\sum_{m=0}^{+\infty}x_i^{m}x_j^{-m-1}
=\sum_{m_j=-\infty}^{+\infty}x_j^{m_j}\sum_{t_j=0}^{+\infty}x_i^{|m_j|+2t_j+1}.
\end{equation}
We now consider the series expansion of the $i$th term of the
right hand side of \eqref{thm-7}.
\begin{equation}\label{ithterm-i}
\frac{R_1(x_i;q)}{\prod_{\stackrel{j=1}{ j\neq
i}}^k(x_i-x_j)(1-\frac{1}{x_ix_j})}=\frac{\sum_{n=0}^{+\infty}\sum_{m_i=-\infty}^{+\infty}\mathcal{D}_1(m_i;n)x_i^{m_i}q^n}{\prod_{\stackrel{j=1}{
j\neq i}}^k(x_i-x_j)(1-\frac{1}{x_ix_j})}.
\end{equation}
 Observe that the numerator in the above term is a series expansion in $x_i$ and
 $q$ and by the series expansions \eqref{serexp-1} and \eqref{serexp-2},  we obtain a series expansion of \eqref{ithterm-i},
 in which the exponents of $x_1,x_2,\ldots,x_{i-1}$
must be positive  and the coefficients of $x_1^{m_1}\cdots
x_k^{m_k}q^n$ for $m_1,\ldots,m_{i-1}\geq1$ are
\begin{align} \label{coef-iterm}
(-1)^{i-1}\sum_{\mathbf{t}_i}\mathcal{D}_1
\left(\sum_{j=1}^{i-1}m_j+m_i-\sum_{j=i+1}^k|m_j|-2|\mathbf{t}_i|-(k-1);n\right),
\end{align}
where the sum ranges over all sequences
$\mathbf{t}_i=(t_1,\ldots,\hat{t}_i,\ldots,t_k)$ (omitting $t_i$)
where $0\leq t_j<|m_j|$ for $j<i$ and $t_j$ could be arbitrary
nonnegative integer for $j>i$. Define
$|\mathbf{t}_i|=\sum_{\stackrel{j=1}{j\neq i}}^kt_j.$

Thus, we obtain a series expansion of the right hand of
\eqref{thm-7}:
\begin{align}\label{rhand}
&\sum_{i=1}^k\frac{R_1(x_i;q)}{\prod_{\stackrel{j=1}{ j\neq
i}}^k(x_i-x_j)(1-\frac{1}{x_ix_j})}=\sum_{i=1}^{k}
\sum_{m_1,\ldots,m_{i-1}=1}^{+\infty}
\sum_{m_i,\ldots,m_{k}=-\infty}^{+\infty}\sum_{n=0}^{+\infty}x_1^{m_1}\cdots
x_{k}^{m_k}q^n \\
 &\hskip 1cm \times \left[(-1)^{i-1}\sum_{\mathbf{t}_i}\mathcal{D}_1
\left(\sum_{j=1}^{i-1}m_j+m_i-\sum_{j=i+1}^k|m_j|-2|\mathbf{t}_i|-(k-1);n\right)\right]\nonumber.
\end{align}

Let $\mathbf{m}_i=(m_1,\ldots,m_k)$ where  $m_j\geq 1$  for $j<i$,
$m_i\leq 0$, and others could be arbitrary integers. Define
$\mathbf{x}^{\mathbf{m}_i}=x_1^{m_1}\cdots x_k^{m_k}$. Obviously,
the term $\mathbf{x}^{\mathbf{m}_i}q^n$ would be appeared in the
series expansions of the first $i$ terms of \eqref{rhand}. We next
use the induction to prove the coefficients of
$\mathbf{x}^{\mathbf{m}_i}q^n$ in the series expansion of
\eqref{rhand} equal to
\begin{equation}\label{rghthandthm7}
\sum_{j=0}^{+\infty}{j+k-2 \choose
k-2}N\left(\sum_{i=1}^k|m_i|+2j+k-1;n\right).
\end{equation}

 Let $\daleth_i(m_1,\ldots,m_{i-1})$ denote the set
of all sequences $\mathbf{t}_i$ of nonnegative integers (omitting
the $i$th vector $t_i$) such that  $t_j$  less than $|m_j|$ for
$j<i$. The following two lemmas are useful in our argument.

\begin{lem}\label{seq-lem-2}
\begin{itemize}
\item[(1)]The  number of sequences $\mathbf{t}'_{p+1} \in
\daleth_{p+1}(m_1,\ldots,m_{p}) $  equals the number of sequences
$\mathbf{t}_p \in \daleth_p(m_1,\ldots,m_{p-1})$ where the
$(p+1)$th vector $t_{p+1}<|m_p|$ such that
$|\mathbf{t}_p|=|\mathbf{t}'_{p+1}|$.

\item[(2)]The number of sequences $\mathbf{t}_p \in
\daleth_p(m_1,\ldots,m_{p-1})$ where the $(p+1)$th vector
$t_{p+1}\geq |m_p|$ is equal to the  number of sequences
$\mathbf{t}'_{p}\in \daleth_p(m_1,\ldots,m_{p-1})$ such that
$|\mathbf{t}_p|=|\mathbf{t}'_{p}|+|m_p|$.
\end{itemize}
\end{lem}

\pf Given a sequence
$\mathbf{t}_p=(t_1,\ldots,\hat{t}_p,\ldots,t_k)$ (omitting $t_p$)
where $0\leq t_j<|m_j|$ for $j<i$ and others could be arbitrary
nonnegative integers.
\begin{itemize}
\item[(1)]If $t_{p+1}<|m_p|$, we define
$\mathbf{t}'_{p+1}=(t'_1,\ldots,\hat{t'}_{p+1},\ldots,t'_k)$ where
$t_j'=t_j$ for $j\neq p,p+1$ and $t_p'=t_{p+1}$. Obviously,
$\mathbf{t}'_{p+1} \in \daleth_{p+1}(m_1,\ldots,m_{p})$ and
$|\mathbf{t}'_{p}|=|\mathbf{t}'_{p+1}|$.

\item[(2)]If $t_{p+1}\geq|m_p|$, we define
$\mathbf{t}'_{p}=(t'_1,\ldots,\hat{t'}_{p},\ldots,t'_k)$ where
$t_j'=t_j$ for $j\neq p+1$ and $t'_{p+1}=t_{p+1}-|m_p|$, it's
clear to see that $\mathbf{t}'_{p} \in
\daleth_p(m_1,\ldots,m_{p-1})$ and
$|\mathbf{t}_p|=|\mathbf{t}'_{p}|+|m_p|.$
\end{itemize}
Furthermore, one can easily see that the above two  processes are
reservable. \qed

We now consider the coefficients of $\mathbf{x}^{\mathbf{m}_1}q^n$
in the series expansion of \eqref{rhand}. It's known that  only
the series expansion of the first term of \eqref{rhand} contains
the term $\mathbf{x}^{\mathbf{m}_1}q^n$, and the coefficients of
$\mathbf{x}^{\mathbf{m}_1}q^n$ are
\begin{align*}
&\sum_{\mathbf{t}_1}\mathcal{D}_1\left(m_1-\sum_{i=2}^k|m_i|-2|\mathbf{t}_1|-(k-1);n\right)\\
&=\sum_{j=0}^{+\infty}{j+k-2 \choose
k-2}N\left(\sum_{i=1}^k|m_i|+2j+k-1;n\right),
\end{align*}
where the equality follows from  the fact that $m_1\leq 0$,
$\mathcal{D}_1(-m;n)=N(m;n)$ and the number of solutions to
$t_2+t_3+\cdots+t_{k}=j$ in nonnegative integers is $j+k-2 \choose
k-2$.

Assume that the coefficients of $\mathbf{x}^{\mathbf{m}_p}q^n$ in
the series expansion of \eqref{rhand} equal to
\eqref{rghthandthm7}, we now show that the coefficients of
$\mathbf{x}^{\mathbf{m}_{p+1}}q^n$ are also equal to
\eqref{rghthandthm7}. Observe that $\mathbf{x}^{\mathbf{m}_p}q^n$
appears in the series expansions of the first $p$ terms of
\eqref{rhand} and the term $\mathbf{x}^{\mathbf{m}_{p+1}}q^n$
appears in the series expansions of the first $(p+1)$ terms.
Furthermore, the coefficients of $\mathbf{x}^{\mathbf{m}_p}q^n$
and $\mathbf{x}^{\mathbf{m}_{p+1}}q^n$ are the same in the series
expansions of the first $(p-1)$ terms. Therefore, if we verify the
sum of the coefficients of $\mathbf{x}^{\mathbf{m}_{p+1}}q^n$ in
the series expansions of the $p$th term and $(p+1)$th term of
\eqref{rhand} equal to the coefficients of
$\mathbf{x}^{\mathbf{m}_p}q^n$ in the series expansions of the
$p$th term, we could reach our conclusion by the induction
hypothesis.

By \eqref{coef-iterm}, it's known that the coefficients of
$\mathbf{x}^{\mathbf{m}_p}q^n$  (where $m_p\leq 0$) in the series
expansion of the $p$th term are
\begin{equation}\label{coef-pterm}
(-1)^{p-1}\sum_{\mathbf{t}_p}\mathcal{D}_1\left(\sum_{j=1}^{p-1}m_j-\sum_{j=p}^k|m_j|-2|\mathbf{t}_p|-(k-1);n\right),
\end{equation}
and the coefficients of $\mathbf{x}^{\mathbf{m}_{p+1}}q^n$ (where
$m_p\geq 1$ ) in the series expansions of the $p$th term and
$(p+1)$th term are
\begin{align*}
&(-1)^{p-1}\sum_{\mathbf{t}_p}\mathcal{D}_1\left(\sum_{j=1}^{p}m_j-\sum_{j=p+1}^k|m_j|-2|\mathbf{t}_p|-(k-1);n\right)\nonumber \\
&+(-1)^{p}\sum_{\mathbf{t}_{p+1}}\mathcal{D}_1\left(\sum_{j=1}^{p}m_j-\sum_{j=p+1}^k|m_j|-2|\mathbf{t}_{p+1}|-(k-1);n\right),
\end{align*}
which equal to \eqref{coef-pterm} by Lemma \ref{seq-lem-2}. Thus
we get our conclusion, and by the definition of
$R_k(x_1,x_2,\ldots,x_k;q)$ , we could recast \eqref{thm-7} as the
partition identity \eqref{main-e}.
 \qed

\section{$k$-marked odd Durfee symbols}
This section is devoted to solve the problems raised by Andrews
(\cite[Problems 6-9]{Andrews-07-a}) on $k$-marked odd Durfee
symbols. We begin this section by defining odd Durfee symbols
which resemble ordinary Durfee symbols with a modified subscript
and with odd numbers as entries.
\begin{defi}
An odd Durfee symbol of $n$  is a
 two-row array with subscript
\begin{equation}
(\alpha,\beta)_D=\left(\begin{array}{cccc}
\alpha_1&\alpha_2&\ldots&\alpha_s\\
\beta_1&\beta_2&\ldots&\beta_t
\end{array} \right)_{D}
\end{equation}
where $\alpha_i$ and $\beta_i$ are all odd numbers, $2D+1\geq
\alpha_1\geq \alpha_2\geq \cdots \geq \alpha_s>0$, $2D+1\geq
\beta_1 \geq \beta_2 \geq \cdots \geq \beta_t>0$, and
$n=\sum_{i=1}^s \alpha_i+\sum_{i=1}^t \beta_i+2D^2+2D+1$.
\end{defi}

The odd rank of an odd Durfee symbol is defined as the number of
parts of $\alpha$ minus the number of parts of $\beta$, let
$\mathcal{D}_1^0(m;n)$ denote the number of odd Durfee symbols of
$n$ with odd rank $m$, we then have

\begin{thm}\label{gf-r-odd}
The generating function for $\mathcal{D}_1^0(m;n)$ is given by
\begin{equation}
\sum_{n=0}^{+\infty}
\mathcal{D}_1^0(m;n)q^n=\frac{1}{(q^2;q^2)_{\infty}}\sum_{n=0}^{+\infty}(-1)^{n}q^{3n^2+3n+1+|m|(2n+1)}.
\end{equation}
\end{thm}
This result can easily follow by comparing the coefficients of
$z^m$ in  \eqref{gf-ds-odd} given by Andrews
\cite[(8.4)-(8.5)]{Andrews-07-a}:
\begin{align}
\sum_{n=1}^{+\infty}\sum_{m=-\infty}^{+\infty}\mathcal{D}_1^0(m;n)z^mq^n&=\sum_{n\geq
0}\frac{q^{2n(n+1)+1}}{(zq;q^2)_{n+1}(z^{-1}q;q^2)_{n+1}}\nonumber\\
&=\frac{1}{(q^2;q^2)_{\infty}}\sum_{n=-\infty}^{+\infty}\frac{(-1)^nq^{3n^2+3n+1}}{1-zq^{2n+1}},\label{gf-ds-odd}
\end{align}
where the first equality follows by direct combinatorial argument.
The second equality is given by \cite[p.66]{Watson-1936}.

Andrews also \cite{Andrews-07-a} defined the $k$th symmetrized odd
rank moment by
\begin{equation}\label{sym-oran-mom}
\eta^0_k(n)=\sum_{m=-\infty}^{{+\infty}}
{m+\lfloor\frac{k-1}{2}\rfloor \choose k}\mathcal{D}_1^0(m;n),
\end{equation}
and introduced $k$-marked odd Durfee symbols, whose definition is
almost identical to that of $k$-marked Durfee symbols (Definition
\ref{de-MDS}).

\begin{defi} A
$k$-marked odd Durfee symbol of $n$ is composed of $k$ pairs of
partitions into odd parts with the subscript,  which is defined as
 \[\eta^0=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D,\] where $\alpha^i$ {\rm(}resp. $\beta^i${\rm)} are all
partitions with odd parts and
$\sum_{i=1}^k(|\alpha^i|+|\beta^i|)+2D^2+2D+1=n$. Furthermore, the
partitions $\alpha^i$ and $\beta^i$ must satisfy almost the same
conditions with $k$-marked Durfee symbols  expect for the third
term in Definition \ref{de-MDS} where for the $k$th vector
$(\alpha^k, \beta^k)$ of $k$-marked odd Durfee symbol,
$\beta^k_1,\alpha^k_1 \leq 2D+1$.
\end{defi}
Following $k$-marked Durfee symbol, Andrews defined the $i$th odd
rank for  $k$-marked odd Durfee symbol.  For  a $k$-marked odd
Durfee symbol $\eta^0$,  we define $\rho_i(\eta^0)$, the $i$th odd
rank of $\eta^0$ by
\[\rho_i(\eta^0)=\left\{
\begin{array}{ll}
l(\alpha^i)-l(\beta^i)-1 \ \ &\text{ for }\ 1\leq i<k,\\[5pt]
l(\alpha^k)-l(\beta^k) \ \ &\text{ for }\ i=k.
\end{array}\right.
\]

Let $\mathcal{D}^0_k(m_1,m_2,\ldots,m_k;n)$ denote the number of
$k$-marked odd Durfee symbols of $n$ with $i$th odd rank equal to
$m_i$ and $\mathcal{D}^0_k(n)$ denote the number of $k$-marked odd
Durfee symbols of $n$. Define $R^0_k(x_1,x_2,\ldots,x_k;q)$ by
\begin{equation*}
R^0_k(x_1,x_2,\ldots,x_k;q)=\sum_{m_1,\ldots,m_k=-\infty}^{+\infty}\sum_{n=0}^{+\infty}
\mathcal{D}^0_k(m_1,m_2,\ldots,m_k;n)x_1^{m_1}x_2^{m_2}\cdots
x_k^{m_k}q^n.
\end{equation*}

Andrews deduced the following four identities on $k$-marked odd
Durfee symbols which are much similar with $k$-marked Durfee
symbols.

\begin{thm}[Corollary 27, Andrews \cite{Andrews-07-a}]\label{gf-km-odd}
\begin{align}\label{gf-km-odd-e}
R^0_k(x_1,x_2,\ldots,x_k;q)=\frac{1}{(q^2;q^2)_\infty}\sum_{n=0}^{+\infty}(-1)^{n}q^{3n^2+(2k+1)n+k}
\frac{1-q^{4n+2}}{\prod_{j=1}^k(1-x_jq^n)(1-\frac{q^n}{x_j})}.
\end{align}
\end{thm}

\begin{thm}[Corollary 28, Andrews \cite{Andrews-07-a}]\label{sym-odd}
$\mathcal{D}^0_k(m_1,m_2,\ldots,m_k;n)$ is symmetric in
$m_1,\,m_2,\ldots,m_k$.
\end{thm}
\begin{thm}[Corollary 29, Andrews \cite{Andrews-07-a}]\label{int-odd}
For $k\geq 1$,
\begin{equation}
\mathcal{D}^0_{k+1}(n)=\eta^0_{2k}(n).
\end{equation}

\end{thm}

\begin{thm}[Theorem 25, Andrews \cite{Andrews-07-a}]\label{thm-7-odd}
\begin{align}
R^0_k(x_1,x_2,\ldots,x_k;q)=\sum_{i=1}^k\frac{R^0_1(x_i;q)}{\prod_{\stackrel{j=1}{
j\neq i}}^k(x_i-x_j)(1-\frac{1}{x_ix_j})}.
\end{align}
\end{thm}

We  now give a brief expository of how to prove these four
conclusions combinatorially. First of all, it's straightforward to
see that the bijection $\Phi$ in Theorem \ref{r-kms-ds}, $\Psi$ in
Theorem \ref{r-km-kms} and the bijection $\Theta$ in Theorem
\ref{r-sym} on Section 2 are valid for $k$-marked odd Durfee
symbols, one then easily deduces the same result for $k$-marked
odd Durfee symbols as Theorem \ref{main}.
\begin{thm}\label{main-odd}
For $k\geq 2$, we have
\begin{equation}\label{main-odd-e}
\mathcal{D}^0_k(m_1,m_2,\ldots,m_k;n)=\sum_{j=0}^{+\infty}{j+k-2
\choose
k-2}\mathcal{D}_1^0\left(\sum_{i=1}^k|m_i|+2j+k-1;n\right).
\end{equation}
\end{thm}
Thus, Theorems \ref{sym-odd}, \ref{int-odd}, and \ref{thm-7-odd}
can be deduced from Theorem \ref{main-odd}  by the precisely same
progressions as Theorems \ref{sym-mds}, \ref{int}, and \ref{thm7}
on Section 3. To prove Theorem \ref{gf-km-odd}, it suffices to
prove that the right side hand of \eqref{gf-km-odd-e} is the
generating function for the summation on the right side of
\eqref{main-odd-e}, which can be easily derived by Theorem
\ref{gf-r-odd}, following the same progression as Theorem
\ref{gf-km}.

 \noindent{\bf Acknowledgments.} I would like to thank  Guoce Xin and Yue Zhou for helpful
discussions, and I am grateful to George E. Andrews for valuable
comments. This work was supported by the 973 Project, the PCSIRT
Project of the Ministry of Education, the Ministry of Science and
Technology, and the National Science Foundation of China.


TITLE: Expected value of product of random variables
QUESTION [0 upvotes]: If I have two dependent random variables (RVs) $X$ and $Y$ where $X$ is a Gaussian RV with mean $0$ and known variance. Is there a way to lower bound the quantity $ \mathbb{E}[X Y]$?

REPLY [2 votes]: Since $\mathbb{E}[X]=0,$ we have $\mathbb{E}[XY]=\mathbb{Cov}(X,Y).$ Now, it follows from Cauchy-Schwarz inequality that $$ \left| \mathbb{Cov} (X, Y) \right | \le \sqrt{\mathbb{Var}[X]\mathbb{Var}[Y]},$$ which gives the lower bound $$ \mathbb{E}[XY]=\mathbb{Cov} (X, Y) \ge - \sqrt{\mathbb{Var}[X]\mathbb{Var}[Y]}.$$ Note that here equality holds when $Y= aX$ for some $a<0.$


TITLE: Dependence of spin on classical vs non-classical physics?
QUESTION [25 upvotes]: Textbook derivations often state that spin can be derived by adding relativity to quantum mechanics. The general argument comes in several steps :

Schrödinger first tried to describe quantum particles by quantizing $E^2 = p^2 + m^2$, which leads to the Klein-Gordon equation (KGE). The density associated to solutions of KGE can be negative, which is problematic if density is interpreted as a probability density. Ultimately, Schrödinger dropped the relativistic constraint and derived his equation using semi-classical approximations to the Hamilton-Jacobi equation.
Pauli heuristically modified the Schrödinger equation to account for spin, leading to the so-called Pauli equation
Dirac understood that treating time and space on the same footing is the only constraint one should impose on a field equation, even if this leads to using operators as coefficients. By linearizing KGE, he obtained the Dirac equation. Interpreting the equation solutions as quantum wave functions allows to explain spin, but also leads to negative energy states.
Imposing anti-commutation relations on the solution to the Dirac equation, and hence promoting it into the field, solves the negative energies problem.

The conclusion of these steps is generally that spin is derived by going to a relativistic setting.
However, we know since the work of Lévy-Leblond that this is not true. Indeed, if we interpret irreducible representations of the Galilei group as spin states, the wave equation obtained by imposing Galilei invariance is the Schrödinger equation. But this is not the only invariant equation one can derive ! Factoring the Schrödinger equation to have a linear equation with first order time and space derivatives leads to a non-relativistic equation describing spin 1/2 particles, which is exactly the Pauli equation, and the non-relativistic limit of the Dirac equation ! The Schrödinger equation can thus be seen as an exact evolution equation for spin 0 non-relativistic particles or as an approximation of the evolution of all non-relativistic quantum particles when neglecting spin.
Therefore, the conclusion is that spin (especially spin 1/2) comes from imposing a linear relation on time and space derivatives.
My questions will start here :

Am I wrong when thinking that spin has nothing to do with either relativity nor quantum mechanics ? One can define linear equations for classical fields, leading to spinor solutions, which can be non-relativistic.
We know that the heat equation is related to the Schrödinger one by a Wick rotation. What is the equivalent of a Wick-rotated Pauli or Dirac equation ? Of course, it is expected to be a classical field equation constraining the evolution of macroscopic variables of a thermodynamic system, but which one ?
Is there a notion of spin attached to the classical field modes solving the previous equation ? How can we test a spin 1/2 property in such case ?
Is there some deep relationship between linearizing an equation and the Wick rotation ? Both somehow correspond to taking the "square-root of a geometry", using respectively space derivative order reduction and complex time.

Thanks in advance for your help !
EDIT : Thanks for the answers, which helped me focus a bit more my questions into a single one : do you know of any paper about the statistical system obtained by Wick-rotating the Dirac or Pauli equation ? My goal is to check if spin 1/2 classical modes exist for such a system and how they could be detected.

REPLY [5 votes]: Regarding your first question,

Am I wrong when thinking that spin has nothing to do with eitherrelativity nor quantum mechanics?

let us take spin to mean 'intrinsic angular momentum'. Then we can ask, is there a consistent classical Galilean model of spin? Is there a consistent classical Lorentz-covariant theory of spin? The answer is yes, to both questions.
For the Galilean case, it is quite simple. You can do Hamiltonian mechanics on any manifold that has what is called a symplectic structure. Fortunately, the two-sphere $S^2$ admits precisely such a structure. Thus the dynamics can be defined by, e.g., the Hamiltonian function $$H = \frac{(\mathbf p - q\mathbf  A)^2}{2m} + q\phi + \mu \mathbf B(\mathbf x) \cdot \mathbf S $$
and the Poisson brackets $$\{x_i, p_j\} = \delta_{ij} \quad \{s_i, s_j \} = \epsilon_{ijk} s_k.$$
(See how the Poisson bracket for the spin is the same as the commutation relation for the Pauli matrices? It's of course not a coincidence - it's because $\mathfrak{su}(2) \sim \mathfrak{so}(3)$.) The equations of motion are $$\dot f = \{ f, H\}$$
as usual. You can work out that you get the usual Lorentz force plus a gradient force, and the spin precesses around $\mathbf B$.
Now what about the Lorentz case? Actually, such a theory precedes Pauli's introduction of his famous matrices. It was published in covariant form Frenkel in 1926 [1] and later that year Thomas derived his eponymous precession [2]. Thomas's equation of motion for the spin was later rediscovered by Bargmann, Michel, and Telegdi [3]. Thomas considered only homogeneous fields. R.H. Good [4] and I. Y. Tamm (unfortunately I don't have the original reference at the moment, but the paper is in Russian anyway) extended this to inhomogeneous fields. The Frenkel model and the Tamm-Good equations are reviewed in Refs. [5, 6].
Now the question is of course can you distinguish between classical and quantum models? Well, the Stern-Gerlach experiment [7] does that. The real story is a bit more complicated according to historians of science [8-10]. (I should add a disclaimer that I have not read these papers in detail as I've not had the time. But I think you will find them interesting.) On the other hand, Bell and GHZ experiments are rather convincing that we do live in a quantum world.
We can also look for Stern-Gerlach effects in the relativistic regime by using the Frenkel model/Tamm-Good equations and comparing to a calculation based on the Dirac equation. (To avoid having to use full QED, one should use the Foldy-Wouthuysen transformation [11-14].) This was done recently by Weng, Bauke, and Keitel [15], showing that strongly relativistic electrons in extremely intense laser fields have different motions according to the Frenkel and Foldy-Wouthuysen models. According to Weng, Bauke, and Keitel, the difference is plausibly within experimental limits at least with next-generation lasers.
References
[1] J. Frenkel, Nature 117, 653 (1926).
[2] L. H. Thomas, Nature 117, 514 (1926).
[3] V. Bargmann, L. Michel, and V. L. Telegdi, Phys. Rev. Lett. 2, 435 (1959).
[4] R. H. Good, Phys. Rev. 125, 2112 (1962).
[5] V. G. Bagrov and V. A. Bordovitsyn, Sov. Phys. J. 23, 128 (1980).
[6] I. M. Ternov and V. A. Bordovitsyn, Sov. Phys. Uspekhi 23, 679 (1980).
[7] B. Friedrich and D. Herschbach, Phys. Today 56, 53 (2003).
[8] F. Weinert, Stud. Hist. Philos. Mod. Phys. 26, 75 (1995).
[9] D. Giulini, Stud. Hist. Philos. Mod. Phys. 39, 557 (2008).
[10] M. Morrison, Stud. Hist. Philos. Mod. Phys. 38, 529 (2007).
[11] L. L. Foldy and S. A. Wouthuysen, Phys. Rev. 78, 29 (1950).
[12] J. P. Costella and B. H. J. McKellar, Am. J. Phys. 63, 1119 (1995).
[13] A. J. Silenko, Phys. Rev. A 77, 12116 (2008).
[14] D.-W. Chiou and T.-W. Chen, 1 (2015).
[15] M. Wen, H. Bauke, and C. H. Keitel, Sci. Rep. 6, 31624 (2016).


TITLE: Chiral perturbation theory: what is the Quark Condensate? why expand in $U$ rather than Goldstone fields?
QUESTION [12 upvotes]: I'm studying Chiral Perturbation Theory ($\chi PT$) from Scherer's Introduction to Chiral Perturbation Theory.
What I am currently having some trouble understanding are two things:

The quark condensate. What is this and why is it a sufficient condition for spontaneous breaking of chiral symmetry? What I do not really understand is where the operator $S_a=\bar{q} \lambda_a q$ comes from ($\lambda_a$ are the Gell-Mann matrices) and why the expectation value of this (which I gather is zero) gives us this thing called the quark condensate.
The formulation of the effective Lagrangian. There is some stuff in Scherer about the coset $G/H$ where in this case G is the full chiral group and $H$ is the vector subgroup which is left after spontaneous symmetry breaking, but I do not really follow how this discussion explains why the Lagrangian is given in terms of the SU(3) matrix $U=\exp{\frac{i}{F_0} \Phi} = \exp{\frac{i}{F_0} \phi_a\lambda_a}$ for (individual) Goldstone fields $\phi_a$? Why can't we write down the effective Lagrangian in terms of the actual degrees of freedom in the theory, i.e. the Goldstone fields? I've read something about them not transforming non-linearly (and the $U$ transforming linearly) but could not really follow so if someone could elaborate on this I would be very glad.

A big thanks in advance for all help given!
And another thing - if anyone has another tip for an introductory reference to $\chi PT$, I would be very grateful. Scherer works decently but it's always good to read about things from a different viewpoint.

REPLY [5 votes]: The operator $S=\bar{q}\lambda_a q$ is the so-called scalar quark density, and together with its pseudoscalar counterpart, it enters the expressions for the divergence of the vector and axial-vector currents (see section 2.3.6).
Spontaneous symmetry breaking occurs if $n$ generators of a symmetry transformation do not annihilate the ground state, resulting in the existence of $n$ massless Goldstone bosons. As derived in section 4.1.2, the action of the generator on the ground state, $Q_a^A\mid0\rangle$, is related to the scalar quark condensate $\langle\bar{q}q\rangle$. This relation indicates that a non-vanishing condensate is a sufficient condition for spontaneous symmetry breaking.
The main point here is that the Lagrangian formulated in terms of U is invariant (and therefore transforms) under global $SU(3)_L\times SU(3)_R\times U(1)_V$ (corresponding to $G$) while the fields $\phi$ only transform as an octet under the subgroup $SU(3)_V$ (corresponding to $G/H$). In terms of U, it can also be easily shown (section 4.2.2) that the ground state is invariant under vector but not under axial transformations.


TITLE: Completion of $C_c(X)$ with respect to $\|\cdot\|_\infty$
QUESTION [4 upvotes]: Notation:
All functions here are from $X$ to $\mathbb R$.
$C_c(X)$ = compactly supported continuous functions.
$C_b(X)$ = bounded continuous functions.
$B(X)$ = bounded functions.
$C_0(X)$ = continuous functions that tend to zero (so $X$ has to be locally compact and Hausdorff)

Today I proved that $C_c(\mathbb R)$ is not complete with respect to $\|\cdot\|_\infty$. One can do this by taking $g_n$ to be the function that is zero on $(-\infty,-n]$, linear on $[-n, -n+1]$, $1$ on $[-n+1, n-1]$ and symmetric with respect to the $y$-axis. For $f(x) = e^{-x^2}$ one can show that $\|fg_n - f\|_\infty \to 0$ but $f \notin C_c(\mathbb R)$.
Then I read about completions and wanted to work out the completion of $C_c$(X). (I did this all for $X = \mathbb R$). In any case, my thoughts were as follows: $C_c(X)$ is contained in $B(X)$. But it is a proper subset because the uniform limit of continuous functions is continuous but there are discontinuous bounded functions. The next candidate then seems to be $C_b(X)$ but $f(x) = 1$ is in there and not the uniform limit of $f_n$ in $C_c$. (cannot be because if $f_n$ is zero outside a compact set then $\|f_n - 1\|_\infty = 1$ for all $n$ so this doesn't converge in norm).
The next candidate then is $C_0(X)$ and I'm quite sure that that's the completion of $C_c$ with respect to $\|\cdot\|_\infty$ in $B(X)$. But now I need to show this by showing that $C_0(X)$ is isomorphic to the space of Cauchy sequences in $C_c(X)$ quotient Cauchy sequences that tend to the zero function and I don't really know how to think about this. Can someone please show me how to prove this? Thank you. I want to see this quotient construction and an isomorphism but if there are other ways to show that $C_0$ is the completion of $C_c$ then go ahead and post it, I will upvote it.

REPLY [1 votes]: In a locally compact space, the $C_0$ functions are complete in the $\|\cdot\|_\infty$ norm; they constitute a Banach Space in this norm.    The space $C_c(X)$ is dense in $C_0$ if $X$ is locally compact and Hausdorff.  This is sufficient to tell you that $C_0(X)$ is the completion of $C_c(X)$ in a locally compact Hausdorff space.


TITLE: If $G$ a finite $p$-group s.t $G/[G;G]$ cyclic then $G$ abelian.
QUESTION [5 upvotes]: Let $G$ be a finite $p$-group and $[G,G]$ its commutator sub-group, I
need to show that if the group quotient $G/[G,G]$ is cyclic then $G$
is an abelian group.
My attempt is to let $g\in G$ s.t $g$ modulo $[G,G]$ be a generator of
the group quotient, and thus for all elements $y$ of $G$ we have $y=g^m$
modulo $[G,G]$, that is $y=g^mu$ for some $u\in [G,G]$. So for
arbitrary $x,y\in G$, I must to prove that $xy=yx$. In writing
$x=g^nv$ and  $y=g^mu$,  I get $xy=g^nvg^mu$  and $yx=g^mug^nv$
Here I find myself stuck and I don't know what can I do, also I may be stuck because I do not see where I can use  the
assumption that $G$ is $p$-group, it may be that this path is solely
based on the definitions and the universal property of the
commutator group fails, could you help me to achieve this? Thanks in advance for all  participation.

REPLY [3 votes]: This comes because of important (characterizing) property of $p$-groups (nilpotent groups): in a $p$-group, every maximal subgroup is of prime index and normal.
Suppose $G$ is non-abelian. We show that $G/[G,G]$ is not cyclic. Let $M$ be a maximal subgroup of $G$. It can not be unique (otherwise, choosing $x\in G\setminus M$, the subgroup $\langle x\rangle$ will be certainly in some maximal subgroup, but by uniqueness of $M$, we will have $\langle x\rangle=G$, but $G$ is non-cyclic).
Let $M'$ be another maximal subgroup. Now, both $M,M'$ are normal with $G/M$ and $G/M'$ isomorphic to cyclic group of order $p$, hence $[G,G]$ is contained in $M$ as well as $M'$.
Thus, in $G/[G,G]$, there are at least two maximal subgroups - $M/[G,G]$ and $M'/[G,G]$. This means $G/[G,G]$ can not be cyclic.


TITLE: Is the set of continuous function with Lebesgue zero set a Borel set in continuous space?
QUESTION [3 upvotes]: Let $D$ be a domain in $\mathbb{R^d}$ and denote the continuous function space on $D$ as $X := C(\overline{D})$ where we can define the $\sigma$-algebra $\mathscr{B}(X)$ of $X$, that is sets in $X$ containing all open sets of $X$ and closed under countable unions and taking complements. We define the zero set of $u$ as $Z_u = \{x\in \overline{D}: u(x) = 0\}$ and denote $m(A)$ as the Lebesgue measure of $A\subset \overline{D}$. If $F: = \{u\in C(\overline{D}): m(Z_u) = 0\}$, can we prove that $F\in \mathscr{B}(X)$?
It is easy to see that $F$ is not a closed set in $X$. But can we write $F$ as countable union of closed set or other form of Borel set? Otherwise, is there any counterexample to make that false?

REPLY [1 votes]: Your set $F$ is a $G_\delta$ set (countable intersection of open sets).
I assume that your domain $D$ is bounded, so that $K:=\overline D$ is compact;  and that the topology of $X=\mathcal C(K)$ is accordingly given by the sup norm.
Observe first that a set $A\subset K$ has Lebesgue measure $0$ if and only if, for any $\varepsilon >0$, one can find an open set $O$ containing $A$ such that $m(O)<\varepsilon$. Denoting (for each $\varepsilon >0$)by $\mathcal O_\varepsilon$ the family of all open sets $O\subset \mathbb R^d$, it follows that $$F=\bigcap_{k\in\mathbb N} \bigcup_{O\in\mathcal O_{1/k}}\mathbf U_{O}\, , $$
where
$$\mathbf U_{O}=\{ u\in X;\; Z_u\subset O\}\, . $$
So, the only thing to check is that for any fixed ope set $O$, the set $\mathbf U_O$ is open in $X=\mathcal C(K)$.
Let us fix an open set $O$, and let $u\in\mathbf U_O$. Then $u(x)\neq 0$ for all $x\in K\setminus O$. Since $u$ is continuous and $K\setminus O$ is compact, it follows that one can find $\alpha >0$ such that $u(x)\geq\alpha$ for all $x\in K\setminus O$. Then, any function $v\in X$ such that $\Vert v-u\Vert_\infty<\alpha/2$ will satisfy $v(x)>\alpha/2$ on $K\setminus O$, so that in particular $Z_v\subset O$. In other words, the open ball $B(u,\alpha/2)$ in $X$ is contained in $\mathbf U_O$. This shows that $\mathbf U_O$ is open in $X$, as required.


TITLE: Checking Injectivity and Surjectivity on set of functions
QUESTION [1 upvotes]: Let $C_0$ be the set of all continuous functions $f: [0,1] \rightarrow R$ and $C_1$ be the set of all differentiable functions $g:[0,1] \rightarrow R$ such that the derivative $g'$ is continuous. If   $T: C_1 \rightarrow C_0$ is defined by $T(g)=g'$, the
(a) T is one-to-one and onto
(b) T is one-to-one but not onto
(c) T is onto but not one-to-one
(d) T is neither one-to-one nor onto
Source: ISI QROR 2021 18th July
In my view, T cannot be one-to-one as two functions who differ by a constant will be mapped to the same function.

REPLY [2 votes]: $T$ is onto as according to Fundamental theorem of calculus for any $f \in \mathcal C^0$,
$$F(x) = \int_0^x f(t) \ dt$$ is an inverse image of $f$ under $T$.
However $T$ is not one-to-one as any two functions that differ by a constant have the same image under $T$.
Finally, the correct answer is (c).


TITLE: locally finite + quasi compact vs finite
QUESTION [3 upvotes]: An elementary question of algebraic geometry which was posted on math.stackexchange but received no answer there.
Let $f: X \rightarrow Y$ be a morphism of schemes.
Assume that
(i) $f: X \rightarrow Y$ is locally finite, in this sense: $Y$ can be covered by affine open
sets $U_i=spec\  A_i$, and $f^{-1}(U_i)$ can be in turn covered by affine open sets $V_{i,j} = spec \ B_{i,j}$, in
such a way that each $B_{i,j}$ is finitely generated as an $A_i$-module.
(ii) $f$ is quasi-compact.
Then does it imply that $f$ is a finite morphism?

REPLY [4 votes]: $\newcommand\Spec{\mathrm{Spec}}$
$ $
Since finiteness is local in the base, we may
assume that $Y = \Spec(A)$ is affine.
The assumption then becomes that $f: X \rightarrow \Spec(A)$ is
quasi-compact, and that $X$ may be covered by finitely many
(by quasi-compactness)
open affines $\Spec(B)$ which are finite over $A$.
Suppose that $f$ is an open immersion. We show that $X$ is closed
and hence finite in $Y$.
Since conditions (i) and (ii) are local on the base, we may
assume that $Y = \Spec(A)$ is connected.
Then, for each $\Spec(B)$ in our cover of $X$,
$\Spec(B)$ is open in $X$ and hence $Y$, and yet finite and thus closed in $Y$.
Since $Y$ is connected, it follows that either $\Spec(B) = X = Y$,
or $X$ is the empty set.
Suppose that $f$ is separated. Then, by Zariski's main theorem, there
exists a factorization
$$X \rightarrow \Spec(A') = Y' \rightarrow \Spec(A) = Y$$
where the first map is an open immersion and the second is finite.
It follows that for each $B$ there is a map $A \rightarrow A'
\rightarrow B$ which makes $B$ finite over $A$. It follows that $B$
is also finite over $A'$. Hence, by the argument above, $X$ is finite
over $Y'$,
and thus $X \rightarrow Y$ is also finite.
Suppose, however, that $f$ is not separated. Then take $X$ to be the
affine line with the origin doubled and $Y$ to be the affine line.
Then the map $X \rightarrow Y$ is not finite because it is not even affine.


TITLE: How do I perform a change in order of integration here?
QUESTION [2 upvotes]: I have a function $$\int^1_{y=0}\int^1_{x=y}e^{x^2}dx\ dy$$
Which I want to perform a change in order of integration. I have plotted the graph:

And it seems it's the area bounded by the y-axis and x-axis. The answer I know is $(e-1)/2$ but it doesn't make sense since a quick check can tell the area of the triangle under is $1/2.$
The limits to be changed to is:
$$\int^1_{x=0}\int^x_{y=0}e^{x^2}dy\ dx$$
Giving
$$ \left[ \frac{1}{2}e^{x^2} \right]^1_0 = \frac{1}{2}(e-1)$$

REPLY [1 votes]: $$
\int^1_0 \left( \int^1_y e^{x^2} \, dx\right) dy = \iint\limits_{\{\,(x,y)\,:\, 0 \,\le\,y\,\le\,x\,\le\,1\,\}} e^{x^2} \, d(x,y) = \int_0^1 \left( \int_0^x e^{x^2} \, dy \right) dx
$$
Look carefully at these inequalities:
$$
0\le y\le x \le 1.
$$
Here's another way to say what those inequalities say:
$y$ goes from $0$ to $1.$ For any fixed value of $y$ between $0$ and $1,\,\,$ $x$ goes from $y$ to $1.$
$x$ goes from $0$ to $1.$ For any fixed value of $x$ between $0$ and $1,\,\,$ $y$ goes from $0$ to $x.$


TITLE: On what condition $trace(A) \ge trace(AB)$?
QUESTION [1 upvotes]: Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) \ge trace(AB)$?

REPLY [2 votes]: Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \\
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $\lambda_1,.., \lambda_n$ are the eigenvalues of $A$ and $\beta_1,.., \beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $\sigma$ such that
$$\tr(AB)=\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $\lambda_1,.., \lambda_n$  and $\beta_1,.., \beta_n$, (and a permutation $\sigma$ which can actually be ignored by reordering $\beta$) under which condition do we have
$$\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}\leq \lambda_1+..+\lambda_n ?$$
Note that for any such $\lambda, \beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $\leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.


TITLE: The uniqueness of a $K$-fixed vector in a spinor representation
QUESTION [5 upvotes]: Consider $G=SO(2n)$ and $K=U(n)$. $(G,K)$ is a symmetric pair. I'm interested in (zonal) spherical functions on $G/K$ which are matrix elements with respect to $K$-fixed vectors in irreducible representations of $G$.

A classical result by E. Cartan says that if $(G, K)$ is a Riemannian symmetric pair then the algebra of integrable functions on $G$ which are bi-invariant under $K$ is commutative, that is, $(G, K)$ is a Gelfand pair.

(source)
Then this book cited by Wikipedia says that for a Gelfand pair, the space of $K$-fixed vectors in every irreducible unitary representation of $G$ is at most one-dimensional. But an irreducible spinor representation of $SO(2n)$ has more than one orthogonal $U(n)$-fixed vectors. (I only know how to phrase it in physicists' language: The subgroup $U(n)$ preserves the number of fermions, so it fixes the state with no fermion and the fully filled state.) Is this because a spinor representation really is a representation of $Spin(2n)$ and not $SO(2n)$? Do I miss something here?
EDIT: Assume everything is over $\mathbb{C}$.

REPLY [5 votes]: There is no contradiction. To use your physical language, the fully filled state is not $K$-fixed as a vector. Rather, the group ${\rm U}(n)$ acts on it by the determinantal character. The subtle point here is that in quantum mechanics, one considers projective representations, so the action by a character cannot be distinguished from the trivial one. The full decomposition of the (half-)spinor representation into $K$-types uses exterior algebra and is well known — see for example Goodman and Wallach's book. In particular, the spinor representation is multiplicity-free: in fact, every $K$-type that occurs has multiplicity 1, and not just the trivial one. This, of course, is a much stronger property.
Incidentally, for odd $n$, the vacuum (1, the state with fermion number $0$) and the fully filled state (the volume form, fermion number $n$) have different parity. So they appear in different halves of the spinor representation.


TITLE: Are all monotonically normal manifolds of dimension at least two metrizable?
QUESTION [8 upvotes]: Alan Dow and Frank Tall recently proved the consistency of the statement Every hereditarily normal manifold of dimension at least two is metrizable.
See: Dow, Alan; Tall, Franklin D., Hereditarily normal manifolds of dimension greater than one may all be metrizable, Trans. Am. Math. Soc. 372, No. 10, 6805-6851 (2019). ZBL1427.54006.
This suggests a natural question:

Is it true in ZFC that every monotonically normal manifold of dimension at least two is metrizable?

The Long Line is an example of a one-dimensional non-metrizable monotonically normal manifold.
NOTE: By "manifold" I mean a Hausdorff space which is locally euclidean.

REPLY [6 votes]: It would appear that that it is true. The result is due to Z. Balogh and E. Rudin and appears in their paper Monotone Normality, Top. App. 47,  (1992), 115-127. The statement to quote is the following.

Corollary 2.3.(e). A manifold of dimension $\geq2$ is metrizable if and only if it is monotonically normal.

This follows easily from the first of the two main results in the paper.

Theorem I. A monotonically normal space is paracompact if and only if it does not have a closed subspace homeomorphic to a stationary subset of a regular uncountable cardinal.

It's worth mentioning that with a little thought it's not hard to repackage the dimension hypothesis so as to include all manifolds.

A manifold is metrisable if and only if it is separable and monotonically normal.


TITLE: Need help solving a question that includes ratios
QUESTION [0 upvotes]: Three groups, Group A, Group B, Group C, of Sam's friends decided to buy a watch as a memorable gift for his birthday. They contributed 1200 AED in the ratio of 3:4:5. How much did each group contribute?

REPLY [0 votes]: The total amount of 1200 AED was devided in 12 parts. 3 parts are paid by group $A$, 4 parts by group $B$ and 5 parts by group $C$ (because the ratio is 3:4:5).
Each part is
$$
\frac{1200}{12} = 100 \text{ AED}.
$$
(The 12 parts of 100 AED make 1200 AED together.)
So group $A$ paid 3 parts of 100 AED, that is $3 \times 100$ AED $= 300$ AED.
You can calculate the amount paid by $B$ and $C$ the same way (try it!).


TITLE: Chain rule and vector-matrix calculus
QUESTION [2 upvotes]: I'm trying to figure out the chain rule in relation to a vector-matrix calculation. I calculate derivatives of several vector-functions:
$q_1=x^Tx$, $q_2=x \cdot x$, $q_3=xx^T$, $q_4=xx^Tx$, $q_5=(xx^T)(xx^T)$
We use a vector $x$ for differentiation, and the above functions $q_{1...5}$ are various combinations of the vector $x$ and the resulting objects:
$q_1,q_2 \rightarrow$ scalars
$q_3 \rightarrow$ matrix
$q_4 \rightarrow$ vector
$q_5 \rightarrow$ matrix
The derivative of a vector with respect to a vector will be the identity matrix, i.e. $\frac{dx}{dx}=\boldsymbol{1}$:
Now let's see the results obtained through the chain rule:

$\frac{dq_1}{dx}=\frac{dx}{dx}^Tx+x^T\frac{dx}{dx}=\boldsymbol{1}^Tx+x^T\boldsymbol{1}$

$\frac{dq_2}{dx}=\boldsymbol{1}x+x\boldsymbol{1}$

$\frac{dq_3}{dx}=\boldsymbol{1}x^T+x\boldsymbol{1}^T$

$\frac{dq_4}{dx}=\boldsymbol{1}x^Tx+x\boldsymbol{1}^Tx+xx^T\boldsymbol{1}$

$\frac{dq_5}{dx}=\boldsymbol{1}x^T(xx^T)+x\boldsymbol{1}^T(xx^T)+(xx^T)\boldsymbol{1}x^T+(xx^T)x\boldsymbol{1}^T$

Now let's briefly analyze the results:

sum of a row-vector and a column-vector. To get the result, we need to transpose either a row-vector or a column-vector

a similar situation, only this time in one of the terms we need to swap $x$ and $\boldsymbol{1}$ manually

none of the terms is computable, but logically, as a result of differentiation, a tensor should be obtained, therefore, ordinary products must be replaced by Kronecker products

first and third terms are matrices, which corresponds to the logic of the result, but the second has a non-computable structure, and it is not known how to convert it to a computable one

logically, a tensor should be obtained, but the logic of permutations in the terms is also difficult to disclose

My question is: there must be rules for transforming "chain" expressions obtained by differentiating complex vector-matrix expressions by the chain rule to obtain computable results. Are they known? I would be happy and grateful for help in understanding the solution to this problem.
Some example:

EDIT NUMBER 3:

REPLY [1 votes]: $
\newcommand\DD[2]{\frac{\mathrm d#1}{\mathrm d#2}}
\newcommand\tDD[2]{\mathrm d#1/\mathrm d#2}
\newcommand\diff{\mathrm D}
\newcommand\R{\mathbb R}
$
Let's change perspectives. Your rule $\tDD xx = \mathbf 1$ tells me that what you want is the total derivative; this rule is equivalent to saying that the total derivative $\diff f_x$ at any point $x \in \R^n$ of the function $f(x) = x$ is the identity, i.e. $\diff f_x(v) = v$ for all $v \in \R^n$. Your transposes are essentially stand-ins for inner products. Let $\cdot$ be the standard inner product on $\mathbb R^n$. Then we may write each of your $q$'s as
$$
  q_1(x) = q_2(x) = x\cdot x,\quad
  q_3(x; w) = x(x\cdot w),\quad
  q_4(x) = (x\cdot x)x,\quad
  q_5(x; w) = x(x\cdot x)(x\cdot w).
$$
I've interpreted the outer products $xx^T$ as functions $w \mapsto x(x\cdot w)$, and in $q_5$ I've used the associativity of matrix multiplication to get
$$
  (xx^T)(xx^T) = x(x^Tx)x^T.
$$
When taking a total derivative $\diff f_x$, we may leave the point of evaluation $x$ implicit and write e.g. $\diff[f(x)]$ or even just $\diff f$ if $f$ is implicitly a function of $x$. If we want to differentiate a variable other than $x$, e.g. $y$, we will write e.g. $\diff_y[x + 2y](v) = 2v$.
The total derivative has three fundamental properties:

The derivative of the whole is the sum of the derivative of the parts. For example,
$$
  \diff[f(x,x)] = \dot\diff[f(\dot x,x)] + \dot\diff[f(x,\dot x)].
$$
The overdots specify precisely what is being differentiated, and anything without a dot is held constant. A more verbose notation would be
$$
  \diff_x[f(x,x)] = \diff_y[f(y,x)]_x + \diff_y[f(x,y)]_x,
$$
or even more verbose
$$
    \diff_x[f(x,x)] = \bigl[\diff_y[f(y,x)]\bigr]_{y=x} = \bigl[\diff_y[f(x,y)]\bigr]_{y=x}.
$$
The chain rule says the derivative of a composition is the composition of derivatives:
$$
  \diff[f\circ g]_x = (\diff f_{g(x)})\circ(\diff g_x).
$$
We don't need to use the chain rule directly for any of the $q$'s, but property 1 above is actually a consequence of the chain rule.
The derivative of a linear function is itself. If $f(x)$ is linear, then
$$
  \diff f_x(v) = f(v).
$$
To make it clear, if say $f(x, y)$ is a function linear in $x$ then the above means that
$$
  \diff[f(x,x)](v) = \dot\diff[f(\dot x,x)](v) + \dot\diff[f(x,\dot x)] = f(v,x) + \dot\diff[f(x,\dot x)],
$$
and if $f(x, y)$ is additionally linear in $y$ then we can continue in the same fashion to get
$$
  \diff[f(x,x)](v) = f(v,x) + f(x,v).
$$

Lets apply this to each $q$:
$$
  \diff[q_1](v) = \diff[x\cdot x](v) = \dot\diff[\dot x\cdot x](v) + \dot\diff[x\cdot\dot x](v) = 2\dot\diff[\dot x\cdot x](v) = 2v\cdot x,
$$$$
  \diff[q_3](v) = \diff[x(x\cdot w)](v) = \dot\diff[\dot x(x\cdot w)](v) + \dot\diff[x(\dot x\cdot w)](v) = v(x\cdot w) + x(v\cdot w),
$$$$
  \diff[q_4](v) = 2(v\cdot x)x + (x\cdot x)v,
$$$$
  \diff[q_5](v) = v(x\cdot x)(x\cdot w) + 2x(v\cdot x)(x\cdot w) + x(x\cdot x)(v\cdot w),
$$
in summary
$$
  \diff[q_1](v) = 2v\cdot x,\quad \diff[q_3(x; w)](v) = v(x\cdot w) + x(v\cdot w),\quad \diff[q_4](v) = 2(v\cdot x)x + (x\cdot x)vm
$$$$
  \diff[q_5(x; w)](v) = v(c\cdot x)(x\cdot w) + 2x(v\cdot x)(x\cdot w) + x(x\cdot x)(v\cdot w).
$$
Note how $\diff[q_3]$ and $\diff[q_5]$ end up with two extra vector parameters $v, w$; this is indicating that these derivatives are higher-order tensors (where by "tensor" we mean a multilinear map). The tensor types of each of the above are

Tensor type

$\diff[q_1]$
(0, 1)

$\diff[q_3]$
(1, 2)

$\diff[q_4]$
(1, 1)

$\diff[q_5]$
(1, 2)

In this case, $(p, q)$ says that $q$ vectors are inputs and $p$ vectors are outputs. We call $p + q$ the degree of the tensor. We can translate these back into index/tensor notation as follows:
$$
  (\diff[q_1])_i = 2x_i \sim 2x^T,
$$$$
  (\diff[q_3])_{ij}^k = \delta^k_ix_j + \delta_{ij}x^k \sim \mathbf1\otimes x^T + x\otimes\mathbf g,
$$$$
  (\diff[q_4])_i^j = 2x_ix^j + x_kx^k\delta_i^j \sim 2x\otimes x^T + |x|^2\mathbf1,
$$$$
  (\diff[q_5])_{ij}^k = \delta_i^kx_lx^lx_j + 2x^kx_ix_j + x^kx_lx^l\delta_{ij} \sim |x|^2\mathbf1\otimes x^T + 2x\otimes x^T\otimes x^T + x\otimes\mathbf g.
$$
In this context, $x^T$ is best thought of as the $(0,1)$ tensor dual to $x$. $\mathbf1$ is the (1,1)-identity tensor, which can be thought of as the identity matrix. Closely related is the metric tensor $\mathbf g(v, w) = v\cdot w$. Only $\diff[q_1]$ and $\diff[q_2]$ can be written in matrix notation, since they are the only degree $\leq2$ tensors; for $\diff[q_2]$ we could write
$$
  \diff[q_2] \sim 2xx^T + |x|^2\mathbf1.
$$
We can see from the above precisely where your equations fail

The total derivative always takes a $(p,q)$-tensor and produces a $(p,q+1)$-tensor.
More over, this means that in using a matrix derivative positioning matters,
and it only makes sense to matrix-differentiate scalar and vector expressions.
Allow $\tDD{}x$ to act in both directions;
we may treat it like it's a row vector.
Then there are both left and right derivatives:
$$
  \DD{}xx = 1,\quad x\DD{}x = \mathbf 1.
$$
In the first equation, $1$ is scalar; in the second equation, $\mathbf 1$ is a matrix.
The correct derivation of your equation (1) would look like
$$
  (x^Tx)\DD{}x
    = (\dot x^Tx)\DD{}{\dot x} + (x^T\dot x)\DD{}{\dot x}
    = x^T\left(\dot x\DD{}{\dot x}\right) + x^T\left(\dot x\DD{}{\dot x}\right)
    = 2x^T\mathbf1
    = 2x^T.
$$
Note that $\DD{}x(x^Tx)$ doesn't make sense,
being row vector $\times$ row vector $\times$ vector.
If we interpret $x^Tx$ as a scalar $x\cdot x$,
then we will simply reproduce the derivation above.
There's needs to be a distinction between between e.g. the $(1,1)$-tensor $\mathbf1$
and the $(0,2)$-tensor $\mathbf g$.
These have the same components
$$
  (\mathbf1)_i^j = \delta_i^j,\quad (\mathbf g)_{ij} = \delta_{ij}
$$
but act very differently:
$\mathbf1$ is a function $\R^n \to \R^n$ where $\mathbf1(v) = v$,
and $\mathbf g$ is a function $\R^n\times\R^n \to \R$ where
$\mathbf g(v, w) = v\cdot w$.


TITLE: prime numbers in an interval, binomial coefficients
QUESTION [0 upvotes]: I have two combinatoric questions:

Let $p(n)$ denote the number of unordered sets of positive integers whose sum is $n$. I proved that:
$$p(n) \geq \max_{1\leq k\leq n}\left\{\frac{1}{k!}{n-1 \choose k-1} \right\}.$$
I need to use it to show that there is an absolute constant $c > 0$ for which:
$$p(n) \geq e^{c\sqrt{n}}$$
I don't know about any lower bound that involves $e$ and I'm really stuck.
Let $\pi(m, n)$ denote the set of prime numbers in the interval $[m,n]$.
I showed that the product of all the prime numbers between $1$ to $n$ less than or equal to $4^n$.
I need to show that $|\pi(1, n)|$ = $O(n/\log n)$.
I thought of maybe showing it by induction, dividing the interval of $(1,n)$ to $(1,\sqrt{n})$ and$(\sqrt n,n)$ but it did'nt work.

I know this is a lot but I can really use some help.
Thank you so much!!

REPLY [1 votes]: Well, for starters, for what value of $k$ is $f(n,k)=\frac1{k!}\binom{n-1}{k-1}$ maximized? Consider the ratio
$$
\frac{p(n,k+1)}{p(n,k)}=\frac{n-k}{k(k+1)}
$$
This ratio starts above one when $k$ is small, and ends below one when $k$ is close to $n$. The time when the ratio goes from above one to below one is where $p(n,k)$ is maximized. We can  see this occurs at about $k\approx -1+\sqrt[]{n+1}$, suggesting that plugging in $\sqrt{n}$ for $k$ into that inequality should give the best bound on $p(n)$. In other words, you should use
$$
p(n)\ge \frac1{(\sqrt{n})!}\binom{n-1}{\sqrt n-1}
$$
and show the right hand side to be grows faster than $e^{c\sqrt{n}}$. To do this, use Stirling's approximation.

The product of $\pi(n)$ distinct numbers positive integers is at least $\pi(n)$!. Since you know the product of the $\pi(n)$ primes in $[1,n]$ is at most $4^n$, this tells you
$$
\pi(n)!\le 4^n
$$
If $\pi(n)$ was not $O(n/\log n)$, then it follow that $\pi(n)=C(n)n/\log n$, where $C(n)$ is some function which is not bounded, meaning $C(n)$ can be made arbitrarily large by plugging in an appropriate $n$. So show that
$$
(C(n)n/\log n)!\le 4^n
$$
leads to a contradiction for such a function $C(n)$. Again, Stirling's approximation is indispensable.


TITLE: Fourier transform of exp(cos)
QUESTION [1 upvotes]: How do I calculate the Fourier transform ($t \rightarrow \omega$) of the following:
$\exp(A\cos(\omega_0 t))$
$A$ is a real constant, and $\omega_0$ is a real and positive constant. I know that this gives a Bessel function, but how can I show this?

REPLY [0 votes]: The modified Bessel Function of second kind $K_z(a)$ can be expressed as a Fourier transform see here:
$$K_{z}(a)=\frac{1}{2}\int_{-\infty}^{\infty}\exp(-a\cosh t)\cosh(zt){\rm d}t=K_{-z}(a)$$