text:
\section{Computing PAC-Bayes Control Policies}
\label{sec:computing pac-bayes controllers}

We now describe how to tackle the optimization problem in Algorithm \ref{a:pac bayes control} for minimizing the upper bound on the true expected cost. We will first discuss the setting where the control policy space $\Pi$ is finite (Section \ref{sec:finite policy space}). For this setting, the optimization problem can be solved to global optimality via Relative Entropy Programming. We then tackle the more general setting where $\Pi$ is continuously parameterized in Section \ref{sec:continuous policy space}.

\subsection{Finite Control Policy Space}
\label{sec:finite policy space}

Let the space of policies be $\Pi = \{\pi_1, \dots, \pi_L\}$. Our goal is then to optimize a \emph{discrete} probability distribution $P$ (with corresponding probability vector $p$) over the space $\Pi$. Thus, $p[j]$ denotes the probability assigned to policy $\pi_j$. Define a matrix $\hat{C}$ of costs, where each element
\begin{equation}
\hat{C}[i,j] = C(r_{\pi_j}; E_i)
\end{equation}
corresponds to the cost incurred on environment $E_i \in S$ by policy $\pi_j \in \Pi$ (recall that Assumption \ref{assumption} implies that we can compute each $\hat{C}[i,j]$). The training cost from inequality \eqref{eq:pac inequality control} can then be written as:
\begin{equation}
\frac{1}{N} \sum_{E \in S} \underset{\pi \sim P}{\EE} [C(r_\pi; E)] = \frac{1}{N} \sum_{i=1}^N \sum_{j=1}^L \hat{C}[i,j] p[j] := \bar{C}p,
\end{equation}
where the matrix $\bar{C}$ is defined as:
\begin{equation}
\bar{C} := \frac{1}{N} {\bf 1}^T \hat{C}.
\end{equation}
Here, ${\bf 1}$ is the all-ones vector of size $N \times 1$. We note that finding a vector $p$ that minimizes the training cost corresponds to solving a \emph{linear program}.

Minimizing the PAC-Bayes upper bound $C_{\textrm{PAC}}(P)$ corresponds to solving the following optimization problem:

\begin{flalign}
		\label{opt:finite}
		& & \underset{p \in \RR^L}{\textrm{min}} \hspace*{1cm} & \bar{C}p + \sqrt{\frac{\KL(p \| p_0) + \log(\frac{2 \sqrt{N}}{\delta})}{2N}} && \\
		& & \text{s.t.} \hspace*{1cm} & 0 \leq p \leq 1, \ \sum_j p[j] = 1. && \nonumber
\end{flalign}
This optimization problem can be \emph{equivalently} reformulated via an \emph{epigraph constraint} \citep{Boyd04} as:
\begin{flalign}
		& & \underset{p \in \RR^L, \tau}{\textrm{min}} \hspace*{1cm} & \tau && \nonumber \\
		& & \text{s.t.} \hspace*{1cm} & \tau \geq \bar{C}p + \sqrt{\frac{\KL(p \| p_0) + \log(\frac{2 \sqrt{N}}{\delta})}{2N}} && \nonumber \\
		& & &  0 \leq p \leq 1, \ \sum_j p[j] = 1. \nonumber
\end{flalign}
We further rewrite the problem as:
\begin{flalign}
		& & \underset{p \in \RR^L, \tau, \lambda}{\textrm{min}} \hspace*{1cm} & \tau && \label{opt:pac bayes final}  \\
		& & \text{s.t.} \hspace*{1cm} & \lambda^2 \geq \frac{\KL(p \| p_0) + \log(\frac{2 \sqrt{N}}{\delta})}{2N} && \nonumber \\
		& & & \lambda = \tau - \bar{C}p, \ \lambda \geq 0 && \nonumber \\
		& & &  0 \leq p \leq 1, \ \sum_j p[j] = 1. \nonumber
\end{flalign}

Our key observation here is that for a \emph{fixed} $\lambda = \lambda_0$, the above problem is a Relative Entropy Program (REP) since it consists of minimizing a linear cost function subject to linear equality and inequality constraints and an additional inequality constraint of the form $\KL(p \| p_0) \leq \textrm{constant}$.

We note that $\lambda \in [0,1]$ since $\lambda = \tau - \bar{C}p$, where $\tau \in [0,1]$ (because $\tau$ upper bounds the true expected cost) and $\bar{C}p \in [0,1]$ (recall that we assumed that costs are bounded between $0$ and $1$). In order to solve problem \eqref{opt:pac bayes final} to global optimality, we can thus simply search over the one-dimensional parameter $\lambda \in [0,1]$  (e.g., by simply discretizing the interval $[0,1]$, performing a bisection search, etc.) and find the setting of $\lambda$ that leads to the lowest optimal value for the corresponding REP.

\subsection{Continuously-Parameterized Control Policy Space}
\label{sec:continuous policy space}

We now consider policies $\pi_w$ parameterized by the vector $w \in \RR^d$ (e.g., neural networks parameterized by weights). We will consider stochastic policies defined by probability distributions over the parameters $w$. Here, we choose Gaussian distributions $w \sim \mathcal{N}(\mu, \Sigma)$ with diagonal covariance $\Sigma = \text{diag}(s)$ (with $s \in \RR_+^d$) and use the shorthand $\N_{\mu, s} := \N(\mu, \text{diag}(s))$. Using Gaussians makes computations easier since we can express the KL divergence between Gaussians in closed form (see Section \ref{sec:kl divergence}). We can then apply Algorithm \ref{a:pac bayes control} and choose $\mu, s$ to minimize the PAC-Bayes upper bound $C_{\textrm{PAC}}(\N_{\mu, s})$. In order to turn this into a practical algorithm, there are two primary issues we need to address.

First, in order to minimize the bound $C_{\textrm{PAC}}(\N_{\mu, s})$, one would like to apply gradient-based methods (e.g., stochastic gradient descent). However, the cost function may not be a differentiable function of the parameters $w$.

For example, in the case of designing obstacle avoidance policies, a natural (but non-differentiable) cost function is the one that assigns a cost of $1$ if the robot collides (and 0 otherwise).
To tackle this issue, we employ a differentiable surrogate for the cost function during optimization (note that the final bound is still evaluated for the original cost function). This surrogate will necessarily depend on the application at hand; we present examples in the contexts of obstacle avoidance and grasping in Section \ref{sec:examples}.

The second challenge is the fact that computing the training cost $C_S(\N_{\mu,s})$ requires computing the following expectation over policies:
\begin{equation}
\label{eq:contcase expectation over controllers}
\underset{w \sim \N_{\mu, s}}{\EE} [C(r_w; E)].
\end{equation}
For most realistic settings, this expectation cannot be computed in closed form. We address this issue in a manner similar to \citep{Dziugaite17}. In particular, in order to optimize $\mu$ and $s$ using gradient descent, we take gradient steps with respect to the following unbiased estimator of $C_S(\N_{\mu,s})$:
\begin{equation}
\label{eq:unbiased estimate}
\frac{1}{N} \sum_{E \in S} C(r_{\mu + \sqrt{s} \odot \xi} ; E), \quad \xi \sim  \N_{0, I_d}.
\end{equation}
In other words, in each gradient step we use an i.i.d. sample of $\xi$ and compute the gradient of \eqref{eq:unbiased estimate} with respect to $\mu$ and $s$.

At the end of the optimization procedure, we fix the optimal $\mu^\star$ and $s^\star$ and estimate the training cost $C_S(P) = C_S(\N_{\mu^\star,s^\star})$ by producing a large number of samples $w_1, \dots, w_L$ drawn from $\N_{\mu^\star, s^\star}$:
\begin{equation}
\hat{C}_S(\N_{\mu^\star,s^\star}) := \frac{1}{NL} \sum_{E \in S} \sum_{i=1}^L C(r_{w_i} ; E).
\end{equation}
We can then use a sample convergence bound (see \citep{Langford02}) to bound the error between $\hat{C}_S(\N_{\mu^\star, s^\star})$ and $C_S(\N_{\mu^\star, s^\star})$. In particular, the following bound is an application of the relative entropy version of the Chernoff bound for random variables (i.e., costs) bounded in $[0,1]$ and holds with probability $1 - \delta'$:
\begin{equation}
\label{eq:sample bound}
C_S(\N_{\mu^\star, s^\star}) \leq \bar{C}_S(\N_{\mu^\star, s^\star}; L, \delta') := \KL^{-1}(\hat{C}_S(\N_{\mu^\star, s^\star}) \| \frac{1}{L} \log(\frac{2}{\delta'})).
\end{equation}
Combining inequalities \eqref{eq:pac bayes kl} and \eqref{eq:sample bound} using the union bound, we see that the following bound holds with probability at least $1 - \delta - \delta'$:
\begin{equation}
\label{eq:contcase KL inv}
C_\D(\N_{\mu^\star,s^\star}) \leq C_{\textrm{bound}}^\star:= \KL^{-1} \Bigg(\bar{C}_S(\N_{\mu^\star,s^\star}; L, \delta') \| \frac{\KL(\N_{\mu^\star,s^\star} \| P_0) + \log(\frac{2 \sqrt{N}}{\delta})}{N}\Bigg).
\end{equation}
This is the final version of our bound on the expected performance of policies (drawn from $\N_{\mu^\star,s^\star}$).

Algorithm \ref{a:pac bayes SGD} summarizes our approach from this section. Note that in order to ensure positivity of $s \in \RR_+^d$, we perform the optimization with respect to $\eta := \log(s)$.

\begin{algorithm}[h]
	\algsetup{linenosize=\normalsize}
  \small
  \caption{{\small PAC-Bayes Policy Learning via Gradient Descent}}
  \label{a:pac bayes SGD}
  \begin{algorithmic}[1]
    \STATE {\bf Inputs:}
    \STATE $S = \{E_1, \dots, E_N\}$: Training environments
    \STATE $\delta, \delta' \in (0,1)$: Probability thresholds
    \STATE $P_0$: Prior over policies
    \STATE $\mu, s \in \RR^d$: Initializations for $\mu$ and $s$
    \STATE $\gamma$: step size for gradient descent

    \STATE {\bf Outputs:}  \\
    \STATE $\mu^\star, s^\star$: Optimal $\mu, s$
    \STATE $C_{\textrm{bound}}^\star:= \KL^{-1} \Big(\bar{C}_S(\N_{\mu^\star,s^\star}; L, \delta') \| \frac{\KL(\N_{\mu^\star,s^\star} \| P_0) + \log(\frac{2 \sqrt{N}}{\delta})}{N}\Big)$

    \STATE {\bf Procedure:}

    \STATE $B(\mu, s, w) := \frac{1}{N} \sum_{E \in S} C(r_w ; E) + \sqrt{\frac{\KL(\N_{\mu^\star,s^\star} \| P_0) + \log(\frac{2 \sqrt{N}}{\delta})}{2N}}$
    \WHILE{$\neg$converged}
    \STATE Sample $\xi \sim  \N_{0, I_d}$ and set $w \leftarrow \mu + \sqrt{s} \odot \xi$
    \STATE $\mu \leftarrow \mu - \gamma \nabla_\mu B(\mu, \exp(\eta), w)$
    \STATE $\eta \leftarrow \eta - \gamma \nabla_\eta B(\mu, \exp(\eta), w)$
    \STATE $s \leftarrow \exp(\eta)$
    \ENDWHILE
  \end{algorithmic}
\end{algorithm}

config: arxiv


TITLE: Explanation for the uniformity of the distance between a Gaussian variable to its nearest integer?
QUESTION [0 upvotes]: earlier I asked the question Expected distance for a gaussian variable to its nearest integer. and got a good answer. The expected distance is highly close to $1/4$, which is very similar to the uniform case.
Then I went on examine the distribution of $|X-R(X)|$, where $R(X)$ is the nearest integer of the variable.
As user Did and leonboy pointed out, for $\sigma \rightarrow 0 $, the distance are concentrated. On the other side, for $\sigma \rightarrow \infty$, $|X-R(X)|\rightarrow UNIFORM(0,0.5) $
The computer simulation I ran implies the variance doesn't need to go to infinite  to see the uniformity. See the figures below.
My question is how fast does  $|X-R(X)|$ converge to uniform distribution with the increases of $\sigma$.

Thanks

REPLY [2 votes]: For $0 < y < 1/2$,  $|x - R(x)| = y$ if $x = n \pm y$ for integer $n$, with
$\dfrac{d}{dx} |x - R(x)| = \pm 1$ there, so the PDF of $Y = |X - R(X)|$ is
$f_Y(y) = \sum_{n = -\infty}^\infty (f_X(n + y) + f_X(n-y))$ where $f_X$ is the PDF of $X$.
By the Poisson summation formula,
$$ \sum_{n=-\infty}^\infty f_X(n+y) = \sum_{k=-\infty}^\infty e^{2 \pi i k y} \widehat{f_X}(k)$$
where in this case
$$\widehat{f_X}(k) = e^{-2 \pi^2 k^2 \sigma^2 - 2 \pi i k \mu}$$
Thus for $0 < y < 1/2$,
$$ f_Y(y) = 2 + 4 \sum_{k=1}^\infty \cos(2\pi k y) \cos(2 \pi k \mu) e^{-2 \pi^2 k^2 \sigma^2} $$
The  term $2$ is the uniform distribution, and the others decay rapidly as $\sigma$ increases.


TITLE: Show that $T_n=\min \{t>T_{n-1} :|B(t)-B(T_{n-1})|=1\}$ is a stopping time for BM and that the $T_n-T_{n-1}$ are IID.
QUESTION [2 upvotes]: We define $T_0=0$ and $T_n=\min \{t>T_{n-1} :|B(t)-B(T_{n-1})|=1\}$, where $B $ deonotes Brownian motion. I have the following two questions:

(1) Show that for every $n ,\  \ T_n $ is a stopping time and  (2) the waiting times $T_n - T_{n-1 }  $  are independent and identically distributed.

For (1) we of course proceed by induction. We have to show that given any $0<s$, $\{T_n \le s \} \in \mathcal F_s$ where $F_s $ is a filtration for the brownian motion. Then I'm a bit lost: I can see the following. By sample continuity we could reduce the minimum over $\mathbb R $ to a minimum over a countable set. Also from sample continuity we have that for any $t $,  $\{t< T_{n-1 } \}  \in \mathcal F_{n-1 } $. And $\omega \mapsto |B(t)(\omega)-B(T_{n-1 })(\omega) |$ is   $\mathcal F_{T_{n-1 } }  $-measurable.
For (2) I know that from the strong Markov property for Brownian motion $T_n $ must be distributed as $T_{n-1 } $. Does the independence of the increments also follow from the Strong markov property? And how do we find the distribution?
Thanks in advance!

REPLY [2 votes]: For (1), we can use this proposition.

Proposition. Let $X=(X_t)_{t\geq0}$ be an adapted process with values in a metric space $(E, d)$. Assume that the sample paths of $X$ are continuous, and $F$ be a closed subset of $E$. Then
  $$T_F=\inf\{t\geq0: X_t \in F\}$$
  is a stopping time.

$T_1<\infty$ is a stopping time by this proposition. Then we use induction. Note that $T_n-T_{n-1}=\inf\{t\geq0: |B_{t+T_{n-1}}-B_{T_{n-1}}|=1\}$. Let $X_t=B_{t+T_{n-1}}-B_{T_{n-1}}$, then for any $\omega\in\Omega$, $X_t(\omega)$ is a continuous function in $t$. An application of this proposition implies that $T_n-T_{n-1}<\infty$ is a stopping time, so $T_n$ is a stopping time by induction. (If $S$ and $T$ are stopping tmes then $S+T$ is a stopping time.
For (2), we use the strong Markov property, which aseerts that $(X_t)_{t\geq 0}$ defined above is a Brownian motion independent of $\mathscr F_{T_{n-1}}$( Here we need to use that $T_{n-1}<\infty$ a.s.) Since $T_{n-1},T_{n-2}\in \mathscr F_{T_{n-1}}$, we have $T_n-T_{n-1}\perp T_{n-1}-T_{n-2}$. Since $X_t$ is a Brownian motion, we have $T_n-T_{n-1}\overset{d}{=}T_1$. This completes the proof.


TITLE: Density of states for graphene
QUESTION [2 upvotes]: I have seen a lot of plots for the density of states for graphene:

but have been unable to find the calculation explicetely. I know the dispersion relation for graphene is
$E_{\pm} (\textbf{k}) =\pm t \sqrt{1+4 \cos^2 (k_y a/2)+4\cos (k_y a/2) \cos (\sqrt{3} k_x a/2)}$
and I would like to calculate the DoS myself, but how do I begin?

REPLY [2 votes]: Not sure if its too late for this answer, but just in case.
The density of states is defined as $D(E)=\int_{1st BZ}\delta(E-E(\mathbf{k}) d\mathbf{k}$.
For analytic results the following paper was the first one to calculate the DOS in a honeycomb structure with nearest-neighbor hopping (whose dispersion relation is the one that you mention): http://journals.aps.org/pr/abstract/10.1103/PhysRev.89.662 .
For numerical results just apply the definition of $D(E)$. Explicitly you should create two arrays containing values of $k_x$ and $k_y$ (say 100 for each going from their minimum values to their maximum values). Calculate its associated energies. Cut the 1st BZ into same-sized $\mathbf{k}$ regions. Finally calculate the number of states in each area (sum 1 for every $E(\mathbf{k})$ lying inside a given area bounded by the values of $E(\mathbf{k})$ evaluated at the boundaries of each of the $\mathbf{k}$ regions in the 1st BZ). Obviously, the dimensions of your $\mathbf{k}$ area should be greater than the difference between consecutive $k_x$ and $k_y$. Finally normalize your result.


TITLE: Evaluate the following limit: $\lim\limits_{ n\to\infty}\frac{(2n)!\sqrt n}{2^{2n}\cdot (n!)^{2}}$
QUESTION [1 upvotes]: Evaluate
$\lim_{n \to +\infty} \frac{(2n)!\sqrt n}{2^{2n}\cdot (n!)^{2}}$.
Please help with steps, Dont know how to break it down to cancel out terms.

REPLY [4 votes]: Hint: using the Stirling approximation:$$
n!=\left(\frac n e\right)^n\sqrt{2\pi n}$$
one easily finds that the limit is$$
\frac1{\sqrt\pi}.$$


TITLE: Countable set can be listed in a sequence
QUESTION [1 upvotes]: Let $S$ be a countable set, i.e. exists bijection between $\mathbb{N}$ and $S$. Why elements of $S$ can be listed in a sequence?
EDIT: I guess that bijection is not crucial. It's sufficient that mapping (namely $f$) is surjective, i.e. for any $T\in S$ exists $n\in \mathbb{N}$ such that $T=f(n)$. Hence $S$ can be arrabged in sequence $\{f(1), f(2), \dots\}$. Am I right?

REPLY [2 votes]: Because the way we formalize the concept of "infinite sequence" in set theory is as a function with domain $\mathbb N$.
Therefore "$S$ can be listed in a sequence" means, by definition, that there is a function $f:\mathbb N\to S$ that hits each element of $S$ -- and hits it exactly once, because we suppose every element of $S$ to appear only once in the sequence.
But such an $f$ is exactly a bijection between $\mathbb N$ and $S$.


TITLE: Why adding fan to dry cabinet not improve dry performance?
QUESTION [1 upvotes]: My dry cabinet is basically rubber sealed cabinet with peltier module to condense moisture and drain it out. The drying  process run quit slow from ambient humidity around 50% to 70% it take about 3 hour to decrease to 20% which I turn the knob to maximum power (the power Controlled by voltage regulator). So I decide to add fan in front of peltier module as shown. I thing more air flow will help more moiser to condense and faster the drying process but the result is wrost. the relative humidity stick around 35% I left it a night and still 35%
Any ideas what happen and how to improve the performance?
Note: I cut the fan wire befor take this picture
Note: The black plate behide is big heatsink that a quit warm
My theory:

Air flow from fane make the water that one condense to vaporize
Airflow make peltier temperature rise and moisture not condense anymore from 35%
Maybeif I can add more power to peltier It can catch moiser better.

REPLY [2 votes]: Do I understand correctly that the fan is inside a sealed box, and the Peltier device is mounted in the wall of the box?  If so, then...

Airflow make peltier temperature rise and moisture not condense anymore from 35%

...That sounds about right. It sounds like the purpose of the Peltier device is to create a localized cold spot where the water can condense.  When you switch on the fan, you're suddenly trying to chill the entire box. The Peltier device probably isn't able to move enough heat to do that.


TITLE: Birkhoff's representation theorem vs matroid-geometric lattice correspondence
QUESTION [13 upvotes]: This question is motivated by the superficial observation that Birkhoff's representation theorem and the cryptomorphism between matroids and geometric lattices are sort of similar. The former says that for a finite distributive lattice $L$ with set of join-irreducible elements $P$ the subsets $\{p\in P|p\le a\}\subset P, a\in L$ form the set of order ideals of a partial order on $P$ (which uniquely defines the partial order). The latter says that if $L$ is, instead, a geometric lattice, then the same family of subsets is the set of flats for a matroid structure on $P$ (for a geometric lattice join-irreducible elements are precisely its atoms). The former provides a bijection between finite distributive lattices and finite posets, the latter provides a bijection between geometric lattices and finite simple matroids. Both bijections have functorial interpretations.
Broadly speaking, my question is: are there other interesting correspondences of this form and do these two phenomena have some interesting common generalization? More specifically, here is an example of what such a result may look like. We may consider an arbitrary (finite?) lattice $L$ with set of join-irreducible elements $P$ and define a family of subsets of $P$ in the same way. Does this family of subsets define some interesting algebraic structure on $P$? Perhaps, for some specific classes of lattices, maybe classes that include both geometric and distributive $L$? (Apparently, in full generality we can obtain any family which is closed under intersection, has least common supersets and for any $p,q\in P$ includes a subset containing exactly one of $p$ and $q$. But I'm not sure where to go from here.)
Update. Two further nice examples of such correspondences were given by Richard Stanley and Sam Hopkins.

For finite join-distributive $L$ this family of subsets is the family of feasible sets of an antimatroid on $P$. This generalizes Birkhoff's representation theorem. See details here.
In their 2019 paper Reading, Speyer and Thomas introduce the notion of a finite two-acyclic factorization system: a finite set equipped with a specific kind of binary relation. For a finite semidistributive lattice our family of subsets is the set of first components of maximal orthogonal pairs of such a binary relation on $P$ and the relation is recovered from this data.

I've accepted the first answer but I still very much hope to see other examples!

REPLY [1 votes]: The notion of a matroid can be formulated in terms of geometric lattices, closure systems, and closure operators. Furthermore, the finite pre-ordered sets can be considered to be precisely the finite topological spaces. The duality between geometry lattices and the closure system/closure operator formulation of matroids along with the duality between finite posets and finite distributive lattices are both special cases of a duality between closure systems, closure operators, and maps into posets.
A closure system on a set $X$ is a collection $C\subseteq P(X)$ which is closed under arbitrary intersection (including the empty intersection which means that $X\in C$). A closure operator on a set $X$ is a function $C:P(X)\rightarrow P(X)$ such that
$R\subseteq C(R)=C(C(R))$ for $R\subseteq X$ and where $R\subseteq S\Rightarrow C(R)\subseteq C(S).$
If $C$ is a closure system, then define $C^*:P(X)\rightarrow P(X)$ by letting $C^*(R)$ be the smallest set in $C$ with $R\subseteq C^*(R)$. Then $C^*$ is a closure operator. If $C$ is a closure operator, then define $C^*=\{R\subseteq X\mid R=C(R)\}=\{C(R)\mid R\subseteq X\}$. Then $C$ is a closure system. If $C$ is a closure operator, then $C=C^{**}$, and if $C$ is a closure system, then $C=C^{**}$.
If $C$ is a closure system on a set $C$, then define a mapping $I(C):X\rightarrow C$ by letting $\iota(C)(x)=C^*(\{x\})$ for $x\in X$.
Recall that a subset $A$ of a complete lattice $L$ is said to be join dense if $\{\bigvee R\mid R\subseteq A\}=L$.
If $L$ is a complete lattice, and $\iota:X\rightarrow L$ is a function. Then define $M(\iota)=\{\iota^{-1}[\downarrow l]\mid l\in L\}$. Then $M(\iota)$ is a closure system on the set $X$.  Then $I(C)[X]$ is join-dense in $C$.
Theorem:

Suppose that $C$ is a closure system on a set $X$. Then $M(I(C))=C$.

Suppose that $L$ is a complete lattice, and $\iota:X\rightarrow L$ is a function, then define a mapping $\phi:L\rightarrow M(\iota)$ by letting $\phi(l)=\iota^{-1}[\downarrow l]$. Define a mapping $\psi:M(\iota)\rightarrow L$ by letting $\psi(R)=\bigvee\iota[R]$ whenever $R\in C$. Then

i. $\psi(R)\leq\ell$ if and only if $R\subseteq\phi(\ell)$.
ii. if $\iota[X]$ is join-dense in $L$, then $\psi,\phi$ are inverses.
Proof:

If $\ell\in C$, then $x\in I(C)^{-1}[\downarrow\ell]$ iff $I(C)(x)\in\downarrow \ell$ iff $C^*(\{x\})\subseteq\ell$ iff $x\in\ell$. Therefore, $I(C)^{-1}(\downarrow \ell)=\ell$. We conclude that $ M(I(C))=\{I(C)^{-1}[\downarrow\ell]\mid \ell\in C\}=C.$

i. $x\in\phi(\bigwedge_{i\in I}\ell_i)$ iff $\ell(x)\in\downarrow\bigwedge_{i\in I}\ell_i$ iff $\ell(x)\leq\bigwedge_{i\in I}\ell_i$ iff $\ell(x)\leq\ell_i$ for $i\in I$ iff $x\in\ell^{-1}[\downarrow \ell_i]=\phi(\ell_i)$ for $i\in I$ iff $x\in\bigwedge_{i\in I}\phi(\ell_i)$. Therefore, $\phi(\bigwedge_{i\in I}\ell_i)=\bigwedge_{i\in I}\phi(\ell_i)$, so $\phi$ preserves arbitrary meets.
ii. We have $\psi(\phi(\ell))=\psi(\ell^{-1}[\downarrow\ell])=\bigvee\iota[\iota^{-1}[\downarrow\ell]]=\bigvee\{\ell(x)\mid\ell(x)\leq\ell\}=\ell$. Therefore $\phi$ is injective. $\phi$ is also surjective by definition.
Q.E.D.
Posets and distributive lattices
We say that a closure system $C$ is a topological closure system if $C$ is closed under finite unions (including the empty union, so $\emptyset\in C$).
We observe that if $L$ is a complete lattice, and $\iota:X\rightarrow L$ is a function with $\iota[X]$ join dense in $L$, then it is not too hard to show that $M(\iota)$ is a topological closure system if and only if $\iota(x)$ is join-prime for each $x\in X$. However, the finite topological closure systems can be put into a one-to-one correspondence with the finite pre-ordered sets; to get a pre-ordered set from a closure system, just take the specialization ordering on that closure system, and the downwards closed sets in a pre-ordered set form a closure system.
Thus, we obtain a duality between the finite pre-ordered sets $(X,\leq)$ and the functions $\iota:X\rightarrow L$ where $X$ is a finite set, $\iota[X]$ is join-dense in $L$, and each element in $\iota[X]$ is join-prime. This duality restricts to a duality between the finite posets $(X,\leq)$ and the collection of all functions $\iota:X\rightarrow L$ where $L$ is a distributive lattice, and $\iota$ is a bijection between $X$ and the set of all join-irreducible elements in $L$.


TITLE: Is the result true when the valuation is trivial?
QUESTION [2 upvotes]: Here I proved the following result:
Proposition: Let $(K,|\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\cdot|)$. Two norms $p_1,p_2$ on $X$ are equivalent (i.e., they induce the same topology) iff there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$
Is the proposition true when the valuation is trivial? (The valuation $|\cdot|$ is trivial when $|x|=1$, for each $x\in K^*$)

REPLY [2 votes]: Here's a counterexample.  Let $K$ be any field with the trivial valuation and let $X$ be an infinite-dimensional vector space with basis $B$.  We can define one norm $p_1$ on $X$ by $p_1(x)=1$ for all nonzero $x\in X$.  We can define another norm $p_2$ by saying $p_2(x)$ is the number of nonzero coefficients there are when we write $x$ as a linear combination of elements of $B$.  Then both norms induce the discrete topology, but $p_2/p_1$ is unbounded.


TITLE: General Linear Group over the quaternions is a topological group
QUESTION [2 upvotes]: How to show that General Linear Group over the quaternions is a a topological group?

REPLY [3 votes]: The classical group $GL_n(\mathbb{H})$ is in fact a Lie group, hence also a topological group. We have $$\mathrm{GL}(n, \mathbb{H}) = \{g \in \mathrm{GL}(2n, \mathbb{C})|Jg = \overline{g}J, \mathrm{det}( g) \ne 0\} \equiv \mathrm{U}^*(2n),$$
which is a Lie group, see here.


TITLE: How to find the minimum of $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ without using the derivative?
QUESTION [2 upvotes]: Find the minimum of function $$f(x)=\frac{4x^2}{\sqrt{x^2-16}}$$ without using the derivative.

In math class we haven't learnt how to solve this kind of problems (optimization) yet. I already know that is solvable using derivatives, but there should be another way. Thanks in advance!

REPLY [1 votes]: The solution can be arrived at by using the quadratic formula.
To begin, we want to analyze how $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ on the domain $x \gt 4$ can take on any value $h \gt 0$:
$\tag 1 f(x) = h \text{ iff } 16 x^4 - h^2 x^2 + 16 h^2 = 0$
For this quadratic equation in $x^2$ to have any solutions the discriminant must be greater than or equal to zero:
$\tag 2 h^4 - (4)(16)(16 h^2) \ge 0 \text{ iff }  h^2 \ge (4)(16)(16)  \text{ iff } h \ge (2)(16)  \text{ iff } h \ge 32$.
So the range of the function is the interval $[32, +\infty)$ and so the minimum value hit by $f$ is $32$.
Using $\text{(1)}$ and $\text{(2)}$ above and the quadratic formula when the discriminant is zero, we see that
$\tag 3 x^2 = \text{[negative } b \text{ divided by } 2 a \text{]} = (32^2)/(2)(16) \text{ iff } x = 32/(4 \sqrt 2) = 4 \sqrt 2$ so
$\tag 4 f(4 \sqrt 2) = 32$

See Ethan Horsfall's answer.


text:
\begin{document}

\title{Beck-type identities for Euler pairs of order $r$}
\author{Cristina Ballantine}\address{Department of Mathematics and Computer Science\\ College of the Holy Cross \\ Worcester, MA 01610, USA \\}
\email{cballant@holycross.edu}
\author{Amanda Welch} \address{Department of Mathematics and Computer Science\\ College of the Holy Cross \\ Worcester, MA 01610, USA \\} \email{awelch@holycross.edu}

\maketitle

\begin{abstract}
Partition identities are often statements asserting that the set $\P_X$ of partitions of $n$ subject to condition $X$ is equinumerous to the set $\P_Y$ of partitions of $n$ subject to condition $Y$. A Beck-type identity is a companion  identity to $|\P_X|=|\P_Y|$ asserting that the difference $b(n)$ between the number of parts in all partitions in $\P_X$ and the number of parts in all partitions in $\P_Y$ equals a $c|\P_{X'}|$ and also $c|\P_{Y'}|$, where $c$ is some constant related to the original identity, and $X'$, respectively $Y'$, is a condition on partitions that is a very slight relaxation of  condition $X$, respectively  $Y$. A second Beck-type identity involves the difference $b'(n)$ between the total number of different parts in all partitions in $\P_X$ and the total number of different parts in all partitions in $\P_Y$. We extend these results to Beck-type identities accompanying all identities given by Euler pairs of order $r$ (for any $r\geq 2$). As a consequence, we obtain many families of new Beck-type identities. We give analytic and bijective proofs of our results.

\end{abstract}

{\bf Keywords:} partitions, Euler pairs, Beck-type identities, words

{\bf MSC 2010:}  05A17, 11P81, 11P83

\section{Introduction}

The origin of this article is rooted in  two conjectures by  Beck which appeared in The On-Line Encyclopedia of Integer Sequences \cite{B1} on the pages for sequences A090867 and A265251. The conjectures, as formulated by Beck, were proved by Andrews in \cite{A17} using generating functions. Certain generalizations and  combinatorial proofs appeared in \cite{FT17} and \cite{Y18}. Combinatorial proofs of the original conjectures were also given in \cite{BB19}. Several additional similar identities were proved in the last two years. See for example \cite{AB19, LW19, LW19b,LW19c}. In order to define Beck-type identities, we first introduce the necessary terminology and notation.

In this article $\N$ denotes the set of positive integers. Given a non-negative integer $n$, a \textit{partition} $\lambda$ of $n$ is a non-increasing sequence of positive integers $\lambda=(\lambda_1, \lambda_2, \ldots, \lambda_k)$ that add up to $n$, i.e., $\ds\sum_{i=1}^k\lambda_i=n$. Thus, if $\ll=(\lambda_1, \lambda_2, \ldots, \lambda_k)$ is a partition, we have $
\ll_1\geq \ll_2\geq \ldots \geq \ll_k$. The numbers $\lambda_i$ are called the \textit{parts} of $\lambda$ and $n$ is called the \textit{size} of $\lambda$. The number of parts of the partition is called the \textit{length} of $\lambda$ and is denoted by $\ell(\lambda)$.

If $\ll, \m$ are two arbitrary partitions, we denote by $\ll\cup \m$ the partition obtained by taking all  parts  of $\ll$ and all parts of $\m$ and rearranging them to form a partition. For example, if $\ll=(5, 5, 3, 2, 2, 1)$ and $\m=(7, 5, 3, 3)$, then $\ll\cup \m=(7,5,5,5,3,3,3,2,2,1)$.

When convenient,  we use the exponential notation for parts in a partition. The exponent of a part is the multiplicity of the part in the partition. For example, $(7, 5^2, 4, 3^3, 1^2)$ denotes the partition $(7, 5,5, 4, 3, 3, 3, 1, 1)$. It will be clear from the context when exponents refer to multiplicities and when they are exponents in the usual sense.

\medskip

Let $S_1$ and $S_2$ be subsets of the positive integers. We define
$\OO(n)$ to be  the set of partitions of $n$ with all parts from the set $S_2$ and  $\D(n)$ to be the set of partitions of $n$ with  parts in $S_1$ repeated at most $r-1$ times.  Subbarao \cite{S71} proved the following theorem.

\begin{theorem} \label{sub-thm} $|\OO(n)|=|\D(n)|$ for all non-negative integers $n$ if and only if $rS_1\subseteq S_1$ and $S_2=S_1\setminus rS_1$. \end{theorem} Andrews \cite{A69}  first discovered this result for $r=2$ and called a pair $(S_1,S_2)$ such that $|\mathcal{O}_2(n)|=|\mathcal{D}_2(n)|$  an \textit{Euler pair} since the pair $S_1=\N$ and $S_2=2\N-1$ gives  Euler's identity. By analogy, Subbarao called a pair $(S_1,S_2)$ such that $|\OO(n)|=|\D(n)|$  an \textit{Euler pair of order $r$}.

\begin{example}[Subbarao \cite{S71}]\label{eg:epair}
Let
\begin{align*}
S_1 &= \{ m \in \N : \ m \equiv 1 \ (mod \ 2) \};\\
S_2 &= \{ m \in \N : \ m \equiv \pm 1 \ (mod \ 6) \}.
\end{align*}

\noindent Then $(S_1, S_2)$ is an Euler pair of order 3.
\end{example}

Note that Glaisher's bijection  used to prove $|\OO(n)|=|\D(n)|$ when $S_1=\N$ and $S_2=2\N-1$ can be generalized to any Euler pair of order $r$. If $(S_1,S_2)$ is an Euler pair of order $r$,  let $\varphi_r$ be the map from $\OO(n)$ to $\D(n)$ which repeatedly merges $r$ equal parts into a single part until there are no parts repeated more than $r-1$ times. The map $\varphi_r$ is a bijection and we refer to it as Glaisher's bijection.

Given $(S_1,S_2)$, an Euler pair of order $r$, we refer to the elements in $S_2=S_1\setminus rS_1$ as \textit{primitive} elements and to the elements of $rS_1=S_1\setminus S_2$ as \textit{non-primitive} elements. We usually denote primitive parts by bold lower case letters, for example $\bf a$. Non-primitive parts are denoted by (non-bold) lower case letters. If $a$ is a non-primitive part of a partition and we want to emphasize the largest power $k$ of $r$ such that $a/r^k\in S_1$, we write $a=r^k\bf a$ with ${\bf a}\in S_2$ and $k \geq 1$.

 Let $\A(n)$ be the set of partitions of $n$ with parts in $S_1$ such that the set of parts in $rS_1$ has exactly one element. Thus, a partition in $\A(n)$ has exactly one  non-primitive part (possibly repeated).   Let $\C(n)$ be the set of partitions of $n$ with  parts in $S_1$ in which exactly one part is repeated at least $r$ times.

Let $a_r(n)=|\A(n)|$ and $c_r(n)=|\C(n)|$. Let $b_r(n)$ be the difference between the number of parts in all partitions in $\OO(n)$ and the number of parts in all partitions in $\D(n)$.  Thus, $$b_r(n)=\sum_{\lambda\in \OO(n)} \ell(\lambda)-\sum_{\lambda\in \D(n)} \ell(\lambda).$$

Let $\T(n)$ be the subset of $\C(n)$ consisting of partitions of $n$ in which  one part is repeated more than $r$ times but less than $2r$ times. Let $c'_r(n)=|\T(n)|$. Let $b'_r(n)$ be the difference between the total number of different parts  in all partitions in $\D(n)$ and the total number of different parts in all partitions in $\OO(n)$ (i.e., in each partition, parts are counted without multiplicity). If we denote by $\bar\ell(\ll)$ the number of different parts in $\ll$, then $$b'_r(n)=\sum_{\lambda\in \D(n)} \bar\ell(\lambda)-\sum_{\lambda\in \OO(n)} \bar\ell(\lambda).$$

In \cite{B1}, Beck conjectured that, if $S_1=\N$ and $S_2=2\N-1$, then $$a_2(n)=b_2(n)=c_2(n)$$ and $$c'_2(n)=b'_2(n).$$ Andrews proved these identities in \cite{A17}  using generating functions.  Combinatorial proofs were given in \cite{BB19}.  For the case $r\geq 2$,  $S_1=\N$, and  $S_2=\{k\in \N: k \not \equiv 0 \pmod r\}$, Fu and Tang \cite{FT17} gave generating function proofs for \begin{equation}\label{beck1} a_r(n)=\frac{1}{r-1}b_r(n)=c_r(n)\end{equation} and \begin{equation} \label{beck2} c'_r(n)=b'_r(n).\end{equation} They also proved combinatorially that  $a_r(n)=c_r(n)$.  In \cite{Y18},  Yang gave combinatorial proofs of \eqref{beck1} and \eqref{beck2} in the case $r\geq 2$,  $S_1=\N$, and  $S_2=\{k\in \N:  k \not \equiv 0 \pmod r\}$.

Our main theorems establish the analogous result for all Euler pairs. We will prove the theorems both analytically and  combinatorially. We refer to the results in Theorem \ref{T1} as first Beck-type identities and to the result in Theorem \ref{T2} as second Beck-type identity.

\begin{theorem} \label{T1} If $n,r $ are integers  such that $n \geq 0$ and $r\geq 2$, and $(S_1, S_2)$ is an Euler pair of order $r$, then
\begin{enumerate}

\item[(i)] $a_r(n)=\ds\frac{1}{r-1}b_r(n)$ \\ \ \\

\item[(ii)] $c_r(n)=\ds\frac{1}{r-1}b_r(n)$. \end{enumerate}\end{theorem}

\begin{theorem} \label{T2} If $n,r $ are integers  such that $n \geq 0$ and $r\geq 2$, and $(S_1, S_2)$ is an Euler pair of order $r$, then
$c'_r(n)=b'_r(n)$.
\end{theorem}

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}.
We have
$$\mathcal{O}_{3}(7) = \{ (7), (5, 1^2), (1^7) \}; \ \mathcal{D}_{3}(7) = \{(7), (5, 1^2), (3^2, 1)\};$$
and
$$\mathcal{O}_{1, 3}(7) = \{ (3^2, 1), (3, 1^4) \}; \ \mathcal{D}_{1, 3}(7) = \{(1^7), (3, 1^4)\}.$$

Glaisher's bijection gives us

\begin{center}
\begin{tabular}{ccc}
$(7)$ & $\stackrel{\varphi_3}{\longrightarrow}$ & $(7)$\\ \ \\ $(5, 1, 1)$ &$\longrightarrow$ & $(5, 1, 1)$ \\ \ \\ $(\underbrace{1, 1, 1}, \underbrace{1, 1, 1}, 1)$ &$\longrightarrow$ & (3, 3, 1)
\end{tabular}
\end{center}

We note that $$a_3(7) = |{O}_{1, 3}(7)| = 2, c_3(7) = |{D}_{1, 3}(7)| = 2, \mbox{ and } b_3(7) = 11 - 7 = 4.$$ Thus, $$\frac{1}{3-1}b_3(7) = a_3(7) = c_3(7).$$

If we restrict to counting different parts in partitions, we see that there are a total of $4$ diferent parts in the partitions of $\mathcal{O}_{3}(7)$ and a total of $5$ different parts in the partitions of $\mathcal{D}_{3}(7)$. Since $\mathcal{T}_3(7) = \{ (3, 1^4) \}$, we have $$b'_3(7) = 5 - 4 = 1 = |\mathcal{T}_3(7)|.$$
\end{example}

The analytic proofs of Theorems \ref{T1} and \ref{T2} are similar to the proofs in \cite{A17} and \cite{FT17}, while the combinatorial proofs follow the ideas of \cite{BB19}. However, the generalizations of the proofs in the aforementioned articles to Euler pairs of order $r\geq 2$ are important as establishing the theorems in such generality leads to a multitude of new Beck-type identities. We  reproduce  several Euler pairs listed in \cite{S71}. For each identity $|\OO(n)|=|\D(n)|$ holding for the pair below, there are companion Beck-type identities as in Theorems \ref{T1} and \ref{T2}.

The following pairs $(S_1, S_2)$ are Euler pairs (of order $2$).

\begin{enumerate}

\item[(i)] $S_1=\{m\in N : m \not \equiv 0 \pmod 3\}$;

\noindent $S_2=\{m\in N : m  \equiv 1,5 \pmod 6\}$.

In this case, the identity $|\mathcal{O}_2(n)|=|\mathcal{D}_2(n)|$ is known as Schur's identity.\medskip

\item[(ii)] $S_1=\{m\in N : m  \equiv 2,4,5 \pmod 6\}$;

\noindent $S_2=\{m\in N : m  \equiv 2,5, 11 \pmod{12}\}$.

In this case, the identity $|\mathcal{O}_2(n)|=|\mathcal{D}_2(n)|$ is known as G\"ollnitz's identity.\medskip

\item[(iii)] $S_1=\{m\in N : m=x^2+2y^2 \mbox{ for some } x,y\in \Z\}$;

\noindent $S_2=\{m\in N : m \equiv 1 \pmod 2 \mbox{ and } m=x^2+2y^2 \mbox{ for some } x,y\in \Z\}$.

 \end{enumerate}\medskip

 The following is an Euler pair of order $3$.

 \begin{enumerate}

 \item[(iv)] $S_1=\{m\in N : m=x^2+xy+y^2 \mbox{ for some } x,y\in \Z\}$;

\noindent $S_2=\{m\in N : \gcd(m,3)=1 \mbox{ and } m=x^2+xy+y^2 \mbox{ for some } x,y\in \Z\}$. \end{enumerate}\medskip

The following pairs $(S_1, S_2)$ are Euler pairs of order $r$.

\begin{enumerate}\medskip

\item[(v)] $S_1=\{m\in N : m  \equiv \pm r \pmod{r(r+1)}\}$;

\noindent $S_2=\{m\in N : m  \equiv \pm r \pmod{r(r+1)} \mbox{ and } m  \not \equiv \pm  r^2 \pmod{r^2(r+1)} \}$.\medskip

\item[(vi)] $S_1=\{m\in N : m  \equiv \pm r, -1 \pmod{r(r+1)}\}$;

\noindent $S_2=\{m\in N : m \equiv \pm r, -1 \pmod{r(r+1)} \mbox{ and } m  \not \equiv \pm r^2, -r \pmod{r^2(r+1)}\}$.

If $r=2$, this Euler pair becomes G\"ollnitz's pair in (ii) above. \medskip

\item[(vii)] Let $r+1$ be a  prime.

\noindent $S_1=\{m\in N : m \not \equiv 0 \pmod{r+1}\};$

\noindent $S_2=\{m\in N : m \not \equiv tr, t(r+1) \pmod{r^2+r} \mbox{ for } 1\leq t\leq r \}.$

If $r=2$, this Euler pair becomes Schur's pair in (i) above. \medskip

\item[(viii)] Let $p$ be a prime and $r$ a quadratic residue $\pmod p$.

\noindent $S_1=\{m \in \N: \mbox{$m$ quadratic residue} \pmod p\};$

\noindent $S_2=\{m \in \N: m \not \equiv 0 \pmod r \mbox{ and $m$ quadratic residue} \pmod p\}.$

 \end{enumerate}
 Note that each case (v)-(viii) gives infinitely many Euler pairs and therefore leads to infinitely many new Beck-type identities.  We also note that in (vii) we corrected a slight error in (3.4) of  \cite{S71}.

 \begin{example} Consider the Euler pair in  (vii) above with $r=4$.
We have

$S_1 = \{ m \in N  :  m \not\equiv 0 \ \pmod 5 \};$

\medskip

$S_2=\{m\in N : m \not \equiv 4t, 5t \pmod{20} \mbox{ for } 1\leq t\leq 4 \}.$ \\

 Then $(S_1, S_2)$ is an Euler pair of order $4$ and we have\\

$\mathcal{O}_{4}(7) = \{ (7), (6, 1), (3^2, 1), (3, 2^2), (3, 1^4), (3, 2, 1^2), (2^3, 1), (2^2, 1^3), (2, 1^5), (1^7) \};$

\medskip

$\mathcal{D}_{4}(7) = \{ (7), (6,1), (3^2, 1), (3, 2^2), (4, 3), (3, 2, 1^2), (2^3, 1), (2^2, 1^3), (4, 2, 1), (4, 1^3)  \}.$

\medskip

We have $\mathcal{O}_{1, 4}(7) = \{ (4, 1^3), (4, 2, 1), (4, 3) \};$
\ \ \  $\mathcal{D}_{1, 4}(7) = \{ (1^7), (2, 1^5), (3, 1^4) \}.$\medskip

We note that $a_4(7) = |{O}_{1, 4}(7)| = 3$, $c_4(7) = |{D}_{1, 4}(7)| = 3$, and $b_4(7) = 40 - 31 = 9$, so $\frac{1}{3}b_4(7) = a_4(7) = c_4(7)$.
\medskip

If we restrict to counting different parts, we see that there are 19 different parts in the partitions of $\mathcal{O}_{4}(7)$ and 21 different parts in the partitions of $\mathcal{D}_{4}(7)$. So $b'_4(7) = 21 - 19 = 2 = |\mathcal{T}_4(7)|$ since $\mathcal{T}_4(7) = \{ (1^7), (2, 1^5) \}$.

\end{example}

\section{Proofs of Theorem \ref{T1} }

\subsection{Analytic Proof}

In this article, whenever we work with $q$-series, we assume that $|q| < 1$. When working with two-variable generating functions, we assume both variables are complex numbers less than $1$ in absolute value. Then all series converge absolutely. The generating functions for $|\D(n)|$ and $|\OO(n)|$ are given by

\begin{align*}
\sum_{n = 0}^{\infty}|\D(n)|q^n & = \prod_{a \in S_1} (1 + q^a + q^{2a} + \dotsm +q^{(r-1)a})\\
& = \prod_{a \in S_1} \frac{1 - q^{ra}}{1 - q^a};
\end{align*}
and
\begin{align*}
\sum_{n = 0}^{\infty} |\OO(n)|q^n & = \prod_{{\bf b} \in S_2} \frac{1}{1 - q^{\bf b}}.
\end{align*}

To keep track of the number of parts used, we introduce a second  variable $z$, where $|z| < 1$. Let  $$\D(n; m)=\{\ll\in \D(n) \mid \ll \mbox{ has exactly $m$ parts}\}$$ and $$\OO(n; m)=\{\ll\in \OO(n) \mid \ll \mbox{ has exactly $m$ parts}\}.$$
Then, the generating functions for $|\D(n; m)|$ and $ |\OO(n; m)|$ are
\begin{align*}
f_{\D}(z,q):=\sum_{n = 0}^{\infty}\sum_{m = 0}^{\infty}|\D(n; m)|z^mq^n & = \prod_{a \in S_1} (1 + zq^a + z^{2}q^{2a} + \dotsm +z^{(r-1)}q^{(r-1)a})\\
& = \prod_{a \in S_1} \frac{1 - z^rq^{ra}}{1 - zq^a};
\end{align*}
and
\begin{align*}
f_{\OO}(z,q):=\sum_{n = 0}^{\infty}\sum_{m = 0}^{\infty} |\OO(n; m)| z^mq^n & = \prod_{{\bf b} \in S_2} \frac{1}{1 - zq^{\bf b}}.
\end{align*}

To obtain the generating function for  the total number of parts in all partition in $\D(n)$  (respectively $\OO(n)$),
we take the derivative with respect to $z$ of $f_{\D}(z,q)$ (respectively  $f_{\OO}(z,q)$),  and set $z = 1$. We obtain

\begin{align*}
&  \left.\frac{\partial}{\partial z}\right|_{z=1}f_{\D}(z,q)    \\   & \hspace*{1cm}  =  \prod_{a \in S_1} \frac{1 - q^{ra}}{1 - q^a}\sum_{a \in S_1} \frac{-rq^{ra}(1-q^a)+q^a(1-q^{ra})}{(1-q^a)(1-q^{ra})} \\ & \hspace*{1cm}  =  \prod_{a \in S_1} \frac{1 - q^{ra}}{1 - q^a}\sum_{a \in S_1} \left(\frac{q^a}{1-q^a}-\frac{q^{ra}}{1-q^{ra}}-(r-1)\frac{q^{ra}}{1-q^{ra}} \right)\\ & \hspace*{1cm}  =  \prod_{a \in S_1} \frac{1 - q^{ra}}{1 - q^a}\left(\sum_{a \in S_1}  \sum_{\substack{ k=1\\ r\nmid k}}^{\infty}q^{ka}-\sum_{a \in S_1}(r-1)\frac{q^{ra}}{1-q^{ra}} \right)
;
\end{align*}
and

\begin{align*}
\left.\frac{\partial}{\partial z}\right|_{z=1}f_{\OO}(z,q)=\prod_{{\bf b} \in S_2} \frac{1}{1 - q^{\bf b}} \sum_{{\bf b} \in S_2} \frac{q^{\bf b}}{1 - q^{\bf b}}.
\end{align*}

Since $|\D(n)| = |\OO(n)|$, we have

\begin{align*}
\sum_{n=0}^{\infty} b_r(n) q^n & = \prod_{{\bf b} \in S_2} \frac{1}{1 - q^{\bf b}} \bigg( \sum_{{\bf b} \in S_2} \frac{q^{\bf b}}{1 - q^{\bf b}} -   \sum_{\substack{ a\in S_1\\ k\in \N\\ r\nmid k}}q^{ka}+\sum_{a \in S_1}(r-1)\frac{q^{ra}}{1-q^{ra}}  \bigg).
\end{align*}

Next we see that

 \begin{align} \label{sets}
 \sum_{{\bf b} \in S_2} \frac{q^{\bf b}}{1 - q^{\bf b}}= \sum_{\substack{ a\in S_1\\ k\in \N\\ r\nmid k}}q^{ka}.
 \end{align}

 We have

 $$\sum_{\substack{ a\in S_1\\ k\in \N\\ r\nmid k}}q^{ka}=\sum_{\substack{ a\in S_1\\ k\in \N}}q^{ka}-\sum_{\substack{ a\in S_1\\ k\in \N}}q^{rka}= \sum_{a\in S_1}\frac{q^a}{1-q^a}- \sum_{a\in S_1}\frac{q^{ra}}{1-q^{ra}}=\sum_{{\bf b}\in S_2}\frac{q^{\bf b}}{1-q^{\bf b}}.$$ The last equality holds because $S_2=S_1\setminus rS_1$. Therefore, \eqref{sets} holds.

Then, the  generating function for $b_r(n)$ becomes

\begin{align*}
\sum_{n=0}^{\infty} b_r(n) q^n &  = \prod_{{\bf b} \in S_2} \frac{1}{1 - q^{\bf b}}  \bigg( (r-1) \sum_{a \in S_1} \frac{q^{ra}}{1 - q^{ra}} \bigg)\\
& = \prod_{a \in S_1} (1 + q^a + q^{2a} + \dotsm +q^{(r-1)a}) \bigg( (r-1) \sum_{a \in S_1} \frac{q^{ra}}{1 - q^{ra}} \bigg).
\end{align*}

Therefore

\begin{align*}
\sum_{n=0}^{\infty} b_r(n) q^n & = \sum_{n = 0}^{\infty} (r-1) |\A(n)|q^n = \sum_{n = 0}^{\infty} (r-1) |\C(n)|q^n.
\end{align*}

\noindent Equating coefficients results in $a_r(n)=\ds\frac{1}{r-1}b_r(n)$ and $c_r(n)=\ds\frac{1}{r-1}b_r(n)$.

\subsection{Combinatorial Proof}
\subsubsection{$b_r(n)$ as the cardinality of a set of marked partitions}\label{b}

We start with another example of Glaisher's bijection.

\begin{example}\label{eg:gl}
We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}, but this time use $n = 11$.\\

$\mathcal{O}_{3}(11) = \{ (11), (7, 1^4), (5^2, 1),  (5, 1^6), (1^{11})   \};$

$\mathcal{D}_{3}(11) = \{(11), (9, 1^2),  (7, 3, 1), (5^2, 1), (5, 3^2)  \}.$

\medskip

Thus, $b_3(11)= 27-13=14$. \medskip

Glaisher's  bijection gives us

\begin{center}
\begin{tabular}{ccc}
$(11)$ & $\stackrel{\varphi_3}{\longrightarrow}$ & $(11)$\\ \ \\ $(7, \underbrace{1, 1, 1}, 1)$ &$\longrightarrow$ & $(7, 3, 1)$ \\ \ \\ $(5,5, 1)$ & ${\longrightarrow}$ & $(5,5, 1)$\\ \ \\ $(5, \underbrace{1, 1, 1,} \underbrace{1, 1, 1})$ & ${\longrightarrow}$ & $(5, 3,3)$\\ \ \\ $(\underbrace{\underbrace{1, 1, 1}, \underbrace{1, 1, 1}, \underbrace{1, 1, 1,}} 1, 1)$ & ${\longrightarrow}$ & $(9, 1,1)$
\end{tabular}
\end{center}
\end{example}
From Glaisher's bijection, it is clear that each partition  $\lambda\in \OO(n)$ has at least as many parts as its image $\varphi_r(\lambda)\in \D(n)$.

When calculating $b_r(n)$,  we sum up the differences in the number of parts in each pair  ($\lambda,\varphi_r(\lambda))$. Write each part $\mu_j$ of  $\mu=\varphi_r(\lambda)$ as  $\mu_j=r^{k_j} \bf a$. Then, $\mu_j$ was obtained by merging $r^{k_j}$ parts equal to $\bf a$ in $\lambda$ and thus contributes an excess of $r^{k_j}-1$ parts to $b_r(n)$. Therefore, the difference between the number of parts of $\lambda$ and the number of parts of $\varphi_r(\lambda)$ is $\ds \sum_{j=1}^{\ell(\varphi_r(\lambda))} (r^{k_j}-1)$.

\medskip

\begin{example} In the setting of  Example  \ref{eg:gl}, we see that $(7,3,1)$ contributes $2$ to $b_3(11)$, $(5,3,3)$ contributes $2+2$ to $b_3(11)$, and $(9,1,1)$ contributes $8$ to $b_3(11)$. Thus, $b_3(11)=2+4+8=14$. \end{example}

\begin{definition} \label{marked} Given $(S_1,S_2)$, an Euler pair of order $r$, we define the set $\MD_{1,r}(n)$ of \textit{marked partitions} of $n$ as the set  of partitions in $\D(n)$ such that exactly one part of the form $r^k\bf a$ with $k\geq 1$  has as index an integer $t$ satisfying $1\leq t\leq r^k-1$. If $\mu\in\D(n)$ has parts $\mu_i=\mu_j=r^k\bf a$, $k\geq 1$, with $i\neq j$, the marked partition in which $\m_i$ has index $t$ is considered different from the marked partition in which $\m_j$ has index $t$.
\end{definition}
Note that marked partitions, by definition, have a non-primitive part.
Then, from the discussion above we have the following interpretation for $b_r(n)$.

\begin{proposition} Let $n,r$ be integers such that $n\geq 1$ and $r\geq 2$. Then, $$b_r(n)=|\MD_{1,r}(n)|.$$ \end{proposition}

 \begin{definition} An \textit{$r$-word} $w$ is a sequence of letters from the alphabet $\{0,1, \ldots r-1\}$. The \textit{length} of an $r$-word $w$, denoted $\ell(w)$, is the number of letters in $w$. We refer to \textit{position} $i$ in $w$ as the $i$th entry from the right, where the most right entry is counted as position $0$.
\end{definition}

Note that leading zeros are allowed and are recorded. For example, if $r=5$, the $5$-words $032$ and $32$ are different even though in base $5$ they both represent $17$. We have $\ell(032)=3$ and $\ell(32)=2$. The empty bit string has length $0$ and is denoted by $\emptyset$.

\begin{definition} Given $(S_1,S_2)$, an Euler pair of order $r$, we define the set  $\MDS(n)$ of \textit{$r$-decorated partitions} as the set of   partitions  in $\D(n)$ with at least one non-primitive part such that exactly one non-primitive part  $r^k\bf a$  is decorated with an index $w$, where $w$ is an $r$-word satisfying   $0\leq\ell(w)\leq k-1$. As in Definition \ref{marked},  if $\mu\in\D(n)$ has  non-primitive parts $\mu_i=\mu_j=r^k\bf a$ with $i\neq j$, the decorated partition in which $\m_i$ has index $w$ is considered different from the decorated partition in which $\m_j$ has index $w$.\end{definition}

Thus, for each  part $\mu_i=r^{k_i}\bf a$ of $\mu\in \MDS(n)$ there are $\ds \frac{r^{k_i}-1}{r-1}$ possible indices and for each partition $\mu\in \MDS(n)$ there are precisely $\ds\frac{1}{r-1} \sum_{j=1}^{\ell(\mu)} (r^{k_j}-1)$ possible decorated partitions with the same parts as $\mu$. \medskip

The discussion above proves the following interpretation for  $\ds\frac{1}{r-1}b_r(n)$.

\begin{proposition} Let $n,r$ be integers such that $n\geq 1$ and $r\geq 2$. Then, $$\ds\frac{1}{r-1}b_r(n)=|\MDS(n)|.$$

\end{proposition}

While it is obvious that $|\MD_{1,r}(n)|=(r-1)|\MDS(n)|$, to see this combinatorially, consider the map $\psi_r:\MD_{1,r}(n) \to \MDS(n)$ defined as follows. If $\ll \in \MD_{1,r}(n)$, then $\psi_r(\ll)$ is the partition in $\MDS(n)$ in which the $r$-decorated part is the same as the marked part in $\ll$. The index of the part of  $\psi_r(\ll)$ is obtained from the index of the part of $\ll$ by writing it in base $r$ and removing the leading digit.
Clearly, this is a $r-1$ to $1$ mapping.

\subsubsection{A combinatorial proof for $a_r(n)=\ds\frac{1}{r-1}b_r(n)$}\label{an=bn}

To prove combinatorially   that $a_r(n)=\ds\frac{1}{r-1}b_r(n)$ we establish a one-to-one correspondence between $\A(n)$ and $\MDS(n)$.\medskip

\noindent \textit{From $\MDS(n)$ to $\A(n)$:}\label{part i}

Start with an $r$-decorated partition $\mu \in \MDS(n)$. Suppose the non-primitive part $\mu_i=r^k \bf a$ is decorated with an $r$-word $w$ of length $\ell(w)$. Then, $0\leq\ell(w)\leq k-1$. Let $d_w$ be the decimal value of $w$. We set $d_\emptyset=0$. We  transform $\m$ into a partition $\ll \in \A(n)$ as follows. \medskip

Define $\bar \m$ to be the partition whose parts are all non-primitive parts of $\m$ of the form $\m_j=r^t\bf a$ with $j\leq i$, i.e., all parts $r^t\bf a$ with $t>k$ and, if $\m_i$ is the $p$th part of size $r^k\bf a$ in $\m$, then $\bar \m$ also has $p$ parts equal to $r^k\bf a$.

Define $\tilde \m$ to be the partition whose parts are all parts of $\m$ that are not in $\bar \m$.

\begin{enumerate}
\item  In $\bar \m$, split one part of size $r^k\bf a$ into $d_w+1$ parts of size $r^{k-\ell(w)}\bf a$ and  $r^k-(d_w+1)r^{k-\ell(w)}$ primitive parts of size $\bf a$.
 Every other part in $\bar \m$ splits completely into parts of size $r^{k-\ell(w)}\bf a$. Denote the resulting partition by $\bar \ll$.

\item  Let $\tilde \ll=\varphi_r^{-1}(\tilde \m)$.  Thus, $\tilde \ll$ is obtained by splitting all parts in $\tilde \m$  into primitive parts.
\end{enumerate}

Let $\ll=\bar\ll \cup \tilde \ll$. Then $\ll\in \A(n)$ and its set of non-primitive parts is $\{r^{k-\ell(w)}\bf a\}$.

\begin{remark} Since $d_w+1\leq r^{\ell(w)}$,  in step 1, the resulting number of primitive parts equal to $\bf a$ is non-negative. Moreover, in $\bar \ll$ there is at least one non-primitive part.
\end{remark}

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Consider the decorated partition

\begin{align*}
\mu & =(1215, 135_{02}, 135, 51, 35, 15, 15, 3)\\
& =(3^5 \cdot 5, (3^3 \cdot 5)_{02}, 3^3 \cdot 5, 3\cdot17, 35, 3\cdot 5, 3 \cdot 5, 3 \cdot 1)\in \mathcal{DD}_3(1604).
\end{align*}
We have $k=3, \ell(w)=2, d_w=2$, and \medskip

$\bar{\mu} = (3^5 \cdot 5, 3^3 \cdot 5);$

\medskip

$\tilde{\mu} = (3^3 \cdot 5, 3\cdot17, 35, 3\cdot 5, 3 \cdot 5, 3 \cdot 1).$\\

To create $\bar{\lambda}$ from $\bar{\mu}$:

\begin{enumerate}
\item Part $135=3^3\cdot 5$ splits into three parts of size $15$ and eighteen parts of size $5$. \medskip

\item Part $1215=3^5\cdot 5$ splits into eighty one parts of size $15$. \medskip
\end{enumerate}

This results in $\bar{\lambda} = (15^{84}, 5^{18}).$\\

To create $\tilde{\lambda}$ from $\tilde{\mu}$:\\

All parts in $\tilde{\mu}$ are split into primitive parts. Thus, part $3^3 \cdot 5$ splits into twenty seven parts of size 5, part $3\cdot 17$ splits into three parts of size $17$,  both parts of $3\cdot5$ split into three parts of size $5$ each, and part $3\cdot 1$ splits into three parts of size $1$. Part $35$ is already primitive so remains unchanged.

This results in $\tilde{\lambda} = (35, 17^3, 5^{33}, 1^3)$. Then, setting $\lambda = \bar{\lambda} \cup \tilde{\lambda}$ results in $\lambda=(35, 17^3, 15^{84}, 5^{51}, 1^3)\in\mathcal{O}_{1,3}(1604)$. The non-primitive part is $15=3\cdot 5$. \medskip

\end{example}

\noindent \textit{From $\A(n)$ to $\MDS(n)$:}

Start with a partition $\lambda\in \A(n)$. In $\ll$ there is one and only one non-primitive part  $r^k \bf a$. Let $f$ be the multiplicity of the non-primitive part of $\lambda$. We  transform $\ll$ into an $r$-decorated partition in $\MDS(n)$ as follows.

Apply Glaisher's bijection to $\ll$ to obtain $\m=\varphi_r(\ll)\in \D(n)$.  Since $\lambda$ has a non-primitive  part, $\mu$ will  have at least one non-primitive part.

Next, we determine the $r$-decoration of $\mu$. Consider the non-primitive parts $\m_{j_i}$ of $\m$ of the form $r^{t_i} \bf a$  (same $\bf a$ as in the non-primitive part of $\lambda$) and  $t_i\geq k$. Assume $j_1<j_2<\cdots$. For notational convenience, set $\m_{j_0}=0$. Let $h$ be the positive integer such that
\begin{equation}\label{ineq-h}\sum_{i=0}^{h-1}\m_{j_i}<f\cdot r^k {\bf a} \leq \sum_{i=0}^{h}\m_{j_i}.\end{equation} Then, we will decorate part $\m_{j_h}=r^{t_h}\bf a$.
To determine the decoration, let
$$N=\frac{\ds\sum_{i=0}^{h-1}\m_{j_i}}{r^k\bf a}.$$ Then, \eqref{ineq-h} becomes $$r^k{\bf a}N<f\cdot r^k{\bf a}\leq r^k{\bf a}N+r^{t_h}{\bf a},$$ which implies $0<f-N\leq r^{t_h-k}$.

Let $d=f-N-1$ and $\ell=t_h-k$. We have $0\leq \ell \leq t_h-1$. Consider the  representation of $d$ in base $r$ and insert leading zeros to form an $r$-word $w$ of length $\ell$. Decorate $\m_{j_h}$ with $w$. The resulting decorated partition is in $\MDS(n)$.

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Consider the partition $\lambda=(35, 17^3, 15^{84}, 5^{51}, 1^3)\in\mathcal{O}_{1,3}(1604)$. The non-primitive part is $15$. We have $k=1, f=84$.

Glaisher's bijection produces the partition $\mu=(1215, 135^2, 51, 35, 15^2, 3) = (3^5\cdot5, 3^3 \cdot 5, 3^3 \cdot 5, 3 \cdot 17, 35, 3\cdot 5, 3\cdot 5, 3\cdot 1)\in \mathcal D\mathcal D_{3}(1604)$. The parts of the form $3^{r_i}\cdot 5$ with $r_i\geq 1$ are $1215, 135, 135, 15,15$. Since $1215<84(3^1 \cdot 5) \leq 1215 + 135,$ the decorated part will be the first part equal to $135=3^3\cdot 5$. We have $N=1215/15=81$.

To determine the decoration, let $d=84-81-1=2$ and $\ell=3-1=2$. The base $3$ representation of $d$ is $2$. To form a $3$-word of length $2$, we introduce one leading $0$. Thus, the decoration is $w=02$ and the resulting decorated partition is $(1215, 135_{02}, 135, 51, 35, 15, 15, 3)=(3^5 \cdot 5, (3^3 \cdot 5)_{02}, 3^3 \cdot 5, 3\cdot17, 35, 3\cdot 5, 3 \cdot 5, 3 \cdot 1)\in \mathcal{DD}_{1,3}(1604)$.
\end{example}

\subsubsection{A combinatorial proof for $c_r(n)=\ds\frac{1}{r-1}b_r(n)$}

We note that one can  compose the bijection of section \ref{part i} with the bijection of \cite{FT17} to obtain a combinatorial proof of part (ii) of Theorem \ref{T1}. However, we give an alternative  proof
 that $c_r(n)=\ds\frac{1}{r-1}b_r(n)$ by establishing a one-to-one correspondence between $\C(n)$ and $\MDS(n)$.  This proof does not involve the bijection of \cite{FT17} and it mirrors the proof of section \ref{an=bn}.\medskip

\noindent \textit{From $\MDS(n)$ to $\C(n)$:}

Start with an $r$-decorated partition $\mu \in \MDS(n)$. Suppose the non-primitive part $\mu_i=r^k \bf a$ is decorated with an $r$-word $w$ of length $\ell(w)$ and decimal value $d_w$. Then, $0\leq\ell(w)\leq k-1$.

 We transform $\m$ into a partition  $\lambda\in \C(n)$ as follows. \medskip

 Let $\bar{\bar \m}$ be the partition whose parts are all non-primitive parts of $\m$ of the form $\m_j=r^t\bf a$ with $j\geq i$, and $k-\ell(w)-1< t\leq k$,
 i.e., all parts $r^t\bf a$ with $k-\ell(w)-1< t<k$ and, if there are $p-1$ parts of size $r^k\bf a$ in $\m$ after the decorated part, then $\bar{\bar \m}$  has $p$ parts equal to $r^k\bf a$.

 Let $\tilde{\tilde \m}$ be the partition whose parts are all parts of $\m$ that are not in $\bar{\bar \m}$.

 In $\bar{\bar \m}$, perform the following steps.
  \begin{enumerate}
\item  Split one part equal to $r^k\bf a$ into $r(d_w+1)$ parts of size $r^{k-\ell(w)-1}\bf a$ and $m$ primitive parts of size $\bf a$, where $m=r^k-r(d_w+1)r^{k-\ell(w)-1}$.
Apply Glaisher's bijection $\varphi_r$ to the partition consisting of $m$ parts equal to $\bf a$.

\item Split all remaining parts of $\bar{\bar \m}$  completely  into parts of size $r^{k-\ell(w)-1}\bf a$.
\end{enumerate}
 Denote  by $\bar{\bar \ll}$ the partition with parts resulting from steps 1 and 2 above.

Let $\ll=\bar{\bar \ll} \cup \tilde{\tilde \m}$.
Since $r(d_w+1)\geq r$, it follows that $\lambda\in \C(n)$. The  part repeated at least $r$ times is $r^{k-\ell(w)-1}\bf a$.

\medskip

\begin{remark} (i) Since $d_w+1\leq r^{\ell(w)}$, the splitting in step 1 can be performed.

(ii) Note that $r \mid m=r^{k-\ell(w)}(r^{\ell(w)}-(d_w+1))$. Thus, if $w\neq \emptyset$, after applying Glaisher's bijection $\varphi_r$ to the partition consisting of $m$ parts equal to $\bf a$, we obtain parts $r^j\bf a$ with $k-\ell(w)\leq j<k$. Since in $\tilde{\tilde \mu}$, all parts of the form $r^i\bf a$ have $i \geq k$ or $i\leq k-\ell(w)-1$, Glaisher's bijection in step (1) does not create parts equal to any parts in $\tilde{\tilde \mu}$.

(iii) If $w=\emptyset$, then, in step (1), we have $m=0$ and  $r^k\bf a$ splits into $r$ parts equal to $r^{k-1}\bf a$.
\end{remark}

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Consider the partition $\mu = (32805, (10935)_{0120}, 10935, 1215, 45, 45, 25, 9, 3) = (3^8 \cdot 5, (3^7\cdot 5)_{0120}, 3^7 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1) \in \mathcal{DD}_3(56017)$. Then the decorated part is $\mu_2 = 3^7 \cdot 5$ and the decoration is $w = 0120$. We have $k=7$, $\ell(w) = 4$, $d_w = 15$. So\\

$\bar{\bar{\mu}} = (3^7 \cdot 5, 3^7 \cdot 5, 3^5 \cdot 5);$

\medskip

$\tilde{\tilde{\mu}} = (3^8 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1).$

\begin{enumerate}

\item $3^7 \cdot 5$ splits into

\begin{itemize}
\item $r(d_w + 1) = 48$ parts of $3^2 \cdot 5$ and
\item $m = r^k - r(d_w + 1)r^{k- \ell(w) - 1} = 3^7 - 48(3^2) = 1755$ parts of $5$.
\end{itemize}

The 1755 parts of 5 merge into two parts of 3645, one part of 1215, and two parts of 135.

\item $3^7 \cdot 5$ splits into two hundred and forty three parts of $3^2 \cdot 5$ and $3^5 \cdot 5$ splits into twenty seven parts of $3^2 \cdot 5$.

\end{enumerate}

This results in\\

$\bar{\bar{\lambda}} = (3645^2, 1215, 135^2, 45^{318});$

\medskip

$\lambda = \bar{\bar{\lambda}} \cup \tilde{\tilde{\mu}} = (32805, 3645^2, 1215,135^2, 45^{320}, 25, 9, 3) \in \mathcal D_{1,3}(56017)$. The part repeated at least three times is $45=3^2\cdot 5$.

\end{example}\medskip

\noindent \textit{From $\C(n)$ to $\MDS(n)$:}

Start with a partition $\lambda\in \C(n)$. Then, among the parts of $\lambda$, there is one and only one  part  that is repeated at least $r$ times. Suppose this  part is  $r^k \bf a$, $k\geq 0$,  and denote by  $f\geq r$ its  multiplicity in $\lambda$. As in  Glaisher's bijection, we merge repeatedly parts of $\ll$ that are repeated at least $r$ times to obtain  $\m\in \D(n)$. Since $\lambda$ has a part repeated at least $r$ times, $\mu$ will  have at least one  non-primitive  part.

Next, we determine the decoration of $\mu$.  In this case, we want to work with the parts of $\mu$ from the right to the left (i.e., from smallest to largest part). Let $\tilde{\mu}_q=\mu_{\ell(\mu)-q+1}$.
 Consider the parts $\tilde\m_{j_i}$ of the form $r^{t_i} \bf a$ with   $t_i\geq k$. If $t_1\leq t_2\leq\cdots$, we have $j_1<j_2<\cdots$.

As before, we set $\tilde{\m}_{j_0}=0$. Let $h$ be the positive integer such that
\begin{equation}\label{ineq1-h}\sum_{i=0}^{h-1}\tilde\m_{j_i}<f\cdot r^k {\bf a} \leq \sum_{i=0}^{h}\tilde\m_{j_i}.\end{equation} Then, we will decorate part $\tilde\m_{j_h}=r^{t_h}\bf a$.
To determine the decoration, let   \begin{equation} \label{Nh} N=\frac{\ds\sum_{i=0}^{h-1}\tilde\m_{j_i}}{r^k\bf a}.\end{equation} Then, \eqref{ineq1-h} becomes $$r^k{\bf a}N<f\cdot r^k{\bf a}\leq r^k{\bf a}N+r^{t_h}{\bf a},$$ which implies $0<f-N\leq r^{t_h-k}$.

Let $\ds d=\frac{f-N}{r}-1$ and $\ell=t_h-k-1$. Note that, by construction, $t_h>k$, and therefore   $0\leq \ell \leq t_h-1$. Consider the  representation of $d$ in base $r$ and insert leading zeros to form an $r$-word $w$ of length $\ell$. Decorate $\tilde\m_{j_h}$ with $w$. The resulting decorated partition (with parts written in non-increasing order) is in $\MDS(n)$. \medskip

\begin{remark} To see that $f-N$ above is always divisible by $r$, note that if $f=qr+t$ with $q,t \in \Z$ and $0\leq t<r$, then there are $t$ terms equal to $r^k\bf a$ in the numerator of $N$. All other terms, if any,  are divisible by $r^{k+1}\bf a$. Therefore, the remainder of $N$ upon division by $r$ is $t$.
\end{remark}

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Consider the partition $\lambda = (32805, 3645^2, 1215, 135^2, 45^{320}, 25, 9, 3) \in \mathcal{D}_{1,3}(56017)$. The part repeated at least three times is $45=3^2\cdot 5$. We have $k=2$ and $f = 320$.

Applying Glaisher's bijection to $\lambda$ results in\\

$\mu = \varphi_{3}(\lambda) = (3^8 \cdot 5, 3^7\cdot 5, 3^7 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2, 3 \cdot 1) \in \mathcal{D}_{3}(56017)$.\\

The parts of the form $3^{t_i}\cdot 5$ with $t_i \geq 2$ are $3^2\cdot 5, 3^2\cdot 5, 3^5\cdot 5, 3^7\cdot 5, 3^7\cdot 5, 3^8\cdot 5$. Since $3^2\cdot 5 + 3^2\cdot 5 + 3^5\cdot 5 + 3^7\cdot 5 < 320 \cdot 3^2 \cdot 5 \leq 3^2\cdot 5 + 3^2\cdot 5 + 3^5\cdot 5 + 3^7\cdot 5 + 3^7\cdot 5 $, the decorated part will be the second part (counting from the right) equal to $3^7 \cdot 5 = 10935$. We have  $N = \ds\frac{3^2\cdot 5 + 3^2\cdot 5 + 3^5\cdot 5 + 3^7\cdot 5}{3^2 \cdot 5} = 272$. Thus $d = \ds\frac{320-272}{3} - 1 = 15$ and $\ell = 7 - 2 - 1 = 4$. The base $3$ representation of $d$ is $120$. To form a $3$-word of length $4,$ we introduce one leading $0$. Thus, the decoration is $w=0120$ and the resulting decorated partition is

\begin{align*}
\mu = & (32805, (10935)_{0120}, 10935, 1215, 45, 45, 25, 9, 3)\\
= & (3^8 \cdot 5, (3^7\cdot 5)_{0120}, 3^7 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1) \in \mathcal{DD}_3(56017).
\end{align*}

\end{example}

\section{Proofs of Theorem \ref{T2} }

\subsection{Analytic Proof}

We create a bivariate generating function to keep track of the number of different parts in partitions in $\OO(n)$, respectively $\D(n)$.

We denote by $\mathcal O'_r(n;m)$ the set of partitions of $n$ with parts from $S_2$ using $m$ different parts. We denote by $\mathcal D'_r(n;m)$ the set of partitions of $n$ with parts from $S_1$ using $m$ different parts and allowing parts to repeat no more than $r-1$ times. Then,

\begin{align*}
f_{\mathcal O'_r}(z,q):=\displaystyle \sum_{n = 0}^{\infty} \sum_{m=0}^\infty| \mathcal O'_r(n;m)|z^mq^n & = \prod_{{\bf b} \in S_2} (1 + zq^{\bf b} + zq^{2\bf b} + \dotsm)\\
& = \prod_{{\bf b}  \in S_2} \bigg( 1 + \frac{zq^{\bf b} }{1- q^{\bf b} } \bigg),
\end{align*}
and \begin{align*}
f_{\mathcal D'_r}(z,q):=\displaystyle \sum_{n = 0}^{\infty}\sum_{m=0}^\infty |\mathcal D'_r(n;m)|z^mq^n & = \prod_{a \in S_1} (1 + zq^a + \dotsm +zq^{(r-1)a})\\ & = \prod_{a \in S_1} \bigg( 1 + \frac{zq^a-zq^{ra}}{1- q^a } \bigg).
\end{align*}

To obtain the generating function for  the total number of different parts in all partitions in $\OO(n)$  (respectively $\D(n)$),
we take the derivative with respect to $z$ of $f_{\mathcal O'_r}(z,q)$ (respectively  $f_{\mathcal D'_r}(z,q)$),  and set $z = 1$. We obtain

\begin{align*}
 \left.\frac{\partial}{\partial z}\right|_{z=1}f_{\mathcal O'_r}(z,q) &=\sum_{{\bf b}  \in S_2} \frac{q^{\bf b} }{1 - q^{\bf b} } \prod_{{\bf c} \in S_2, {\bf c} \neq {\bf b} } \bigg(1 + \frac{q^{\bf c}}{1 - q^{\bf c}} \bigg)\\ &=\sum_{{\bf b}  \in S_2} \frac{q^{\bf b} }{1 - q^{\bf b} } \prod_{{\bf c} \in S_2, {\bf c} \neq {\bf b} } \bigg( \frac{1}{1 - q^{\bf c}} \bigg)\\
& = \prod_{{\bf b}  \in S_2} \frac{1}{1 - q^{\bf b} } \sum_{{\bf b}  \in S_2} q^{\bf b},
\end{align*}

and

\begin{align*}
 \left.\frac{\partial}{\partial z}\right|_{z=1}f_{\mathcal D'_r}(z,q) &= \sum_{a \in S_1} \frac{q^a-q^{ra}}{1-q^a} \prod_{d \in S_1, d\neq a} \bigg(1 + \frac{q^d-q^{rd}}{1-q^d}\bigg)\\ & = \sum_{a \in S_1} \frac{q^a-q^{ra}}{1-q^a} \prod_{d \in S_1, d\neq a}\frac{1-q^{rd}}{1-q^d}\\
& =\prod_{a \in S_1}\frac{1-q^{ra}}{1-q^a} \sum_{a \in S_1} \frac{q^a - q^{ra}}{1 - q^{ra}}.
\end{align*}

Since $|\D(n)| = |\OO(n)|$, we have

\begin{align*}
 \sum_{n = 0}^{\infty} b'_r(n) q^n
= \prod_{a \in S_1}\frac{1-q^{ra}}{1-q^a}\bigg( \sum_{a \in S_1} \frac{q^a}{1 - q^{ra}} -\sum_{a \in S_1} \frac{q^{ra}}{1 - q^{ra}} - \sum_{{\bf b}  \in S_2} q^{\bf b}  \bigg).
\end{align*}

Moreover,
\begin{align*}
 \sum_{a \in S_1} \frac{q^a}{1 - q^{ra}} -\sum_{a \in S_1} \frac{q^{ra}}{1 - q^{ra}}  & = \bigg( \sum_{a \in S_1} q^a +  \sum_{a \in S_1} \frac{q^{(r+1)a}}{1 - q^{ra}} \bigg) - \bigg( \sum_{a \in rS_1} q^a + \sum_{a \in S_1} \frac{q^{2ra}}{1 - q^{ra}} \bigg)  \\
 & = \bigg( \sum_{a \in S_1} q^a - \sum_{a \in rS_1} q^a \bigg) + \bigg(\sum_{a \in S_1} \frac{q^{(r+1)a}}{1 - q^{ra}} - \sum_{a \in S_1} \frac{q^{2ra}}{1 - q^{ra}}\bigg)\\
 & = \sum_{{\bf b}  \in S_2} q^{\bf b} + \sum_{a \in S_1} \frac{q^{(r + 1)a} - q^{2ra}}{1 - q^{ra}},
 \end{align*}
 the last equality occurring because $S_1 = S_2 \sqcup rS_1$.

 Therefore,
\begin{align*}
 \sum_{n = 0}^{\infty} b'_r(n) q^n & =  \prod_{a \in S_1}\frac{1-q^{ra}}{1-q^a}\sum_{a \in S_1} \frac{q^{(r + 1)a} - q^{2ra}}{1 - q^{ra}}\\ & = \sum_{a \in S_1} \frac{q^{(r + 1)a} + q^{(r + 2)a} + \dotsm +q^{(2r-1)a}}{1 + q^a + \dotsm + q^{(r-1)a}} \prod_{a \in S_1} (1 + q^a + \dotsm + q^{(r-1)a})\\ & = \sum_{a \in S_1} (q^{(r + 1)a} + q^{(r + 2)a} + \dotsm +q^{(2r-1)a}) \prod_{d \in S_1, c\neq a} (1 + q^d + \dotsm + q^{(r-1)d})\\
& = \sum_{n = 0}^{\infty} c'_r(n) q^n.
 \end{align*}

\medskip

\subsection{Combinatorial Proof}

\subsubsection{$b'_r(n)$ as the cardinality of a set of overpartitions} \label{b1}

As in section \ref{b}, we use Glaisher's bijection and calculate $b'_r(n)$ by summing up the difference between the number of different parts of $\varphi_r(\lambda)$ and the number of different parts of $\lambda$ for each partition $\lambda\in \OO(n)$. For a given ${\bf a} \in S_2$, each part in $\varphi_r(\ll)$ of the form $r^k\bf a$, $k\geq 0$, is obtained from $\ll$ by merging $r^k$ parts equal to $\bf a$. Therefore, the contribution to $b'_r(n)$ of each $\m\in \D(n)$ equals $$\sum_{\substack{{\bf a}\in S_2 \\ {\bf a} \mbox{ \small{part of} } \varphi_r^{-1}(\m)}}(m_{\m}({\bf a})-1),$$ where
$$m_{\m}({\bf a})=|\{t\geq 0 \mid r^t{\bf a} \mbox{ is a part of } \m \}|. $$

Next, we define a set of overpartitions. An overpartition is a partition in which the last appearance of a part may be overlined. For example,  $(5, \bar 5, 3,3, \bar 2, 1, 1, \bar 1)$ is an overpartition of $21$. We denote by  $\OD(n)$  the set of overpartitions of $n$ with  parts in $S_1$ repeated at most $r-1$ times in which \textit{exactly one} part is overlined and such that  part $r^s\bf a$ with $s \geq 0$  may be overlined only if there is a part $r^t\bf a$ with $t<s$. In particular, no primitive part can be overlined. Note that when we count parts in an overpartition, the overlined part contributes to the multiplicity.  The discussion above proves the following interpretation of $b'_r(n)$.

\begin{proposition} Let $n\geq 1$. Then, $b'_r(n)=|\OD(n)|.$ \end{proposition}

\subsubsection{A combinatorial proof for $c'_r(n)=b'_r(n)$} We establish a one-to-one correspondence between $\OD(n)$ and $\T(n)$.\medskip

\noindent \textit{From  $\OD(n)$ to $\T(n)$:}

Start with an overpartition $\mu \in \OD(n)$. Suppose the overlined part is $\mu_i=r^s\bf a$. Then there is a part $\mu_p=r^t\bf a$ of $\mu$ with $t<s$. Let $k$ be the largest non-negative integer such that $r^k\bf a$ is a part of $\mu$ and $k<s$. To obtain $\lambda \in \T(n)$ from $\mu$, split $\mu_i$ into $r$ parts equal to $r^k\bf a$ and $r-1$ parts equal to $r^j\bf a$  for each $j=k+1, k+2, \ldots, s-1$.

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Let\\

$\mu = (3^8 \cdot 5, 3^7 \cdot 5, \overline{3^7 \cdot 5}, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1) \in \ODT(56017)$.\\

Then $k=5$ and $3^7 \cdot 5$ splits into three parts equal to $3^5\cdot 5$ and two parts equal to $3^6 \cdot 5$. Thus, we obtain the partition
\begin{align*}
\lambda & = (3^8 \cdot 5, 3^7 \cdot 5, 3^6 \cdot 5, 3^6 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot  1)\\
& \in \mathcal{T}_3(56017).
\end{align*}
The part repeated more than three times but less than six times is $3^5\cdot 5$.

\end{example}

\noindent \textit{From $\T(n)$ to $\OD(n)$:}

Start with a partition $\lambda \in \T(n)$. Suppose $r^k\bf a$ is the part repeated more than $r$ times but less than $2r$ times.   Let $\m=\varphi_r(\ll)\in \D(n)$. Overline the smallest part of $\mu$ of form $r^t \bf a$ with $t>k$.  The resulting overpartition is  in $\OD(n)$.

\begin{example} We continue  with the Euler pair of order $3$ from Example \ref{eg:epair}. Let
\begin{align*}
\lambda & = (3^8 \cdot 5,  3^7 \cdot 5,  3^6 \cdot 5, 3^6 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot  1)\\
& \in \mathcal{T}_3(56017).
\end{align*}The part repeated more than three times but less than six times is $3^5\cdot 5$.
We have $k = 5$. Merging by Glaisher's bijection, we obtain\\

$\mu = (3^8 \cdot 5, 3^7 \cdot 5, 3^7 \cdot 5, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1) \in \mathcal{D}_3(56017).$\\

The smallest part of $\mu$ of the form $r^t\bf{a}$ with $t > k = 5$ is $3^7\cdot 5$. Thus we obtain the overpartition\\

$\mu = (3^8 \cdot 5, 3^7 \cdot 5, \overline{3^7 \cdot 5}, 3^5 \cdot 5, 3^2 \cdot 5, 3^2 \cdot 5, 25, 3^2 \cdot 1, 3 \cdot 1) \in \ODT(56017).$

\end{example}

\begin{remark} \label{rem}  We could have obtained the  transformation above from the combinatorial proof of part (ii) of Theorem \ref{T1}. In the transformation from $\C(n)$ to $\MDS(n)$, if part $r^k\bf a$ is the part repeated more than $r$ times but less than $2r$ times, we have $f=r+s$ for some $1\leq s\leq r-1$, $h=s+1$, and $N=s$. Thus $d=0$ and the decorated part is the last occurrence of smallest part in the transformed partition $\mu$ that is of the form $r^t\bf a$ with $t>k$.  Thus, in $\mu$, the decorated part $r^{t} \bf a$ is decorated with an $r$-word consisting of all zeros and of length $t-k-1$, one less than the difference in exponents of $r$ of the decorated part and the next smallest part with the same $\bf a$ factor. Since in this case the decoration of a partition in $\MDS(n)$ is completely determined by the part being decorated, we can simply just overline the part.  \end{remark}

\section{Concluding remarks}

In this article we proved first and second Beck-type identities for all Euler pairs $(S_1,S_2)$ of order $r\geq 2$. Euler pairs of order $r$ satisfy $rS_1\subseteq S_1$ and $S_2=S_1\setminus rS_1$.

Thus, we established Beck-type identities accompanying all partition identities of the type given in Theorem \ref{sub-thm}.

At the end of \cite{S71}, Subbarao mentions that the characterization of Euler pairs of order $r$ given by Theorem \ref{sub-thm} can be extended to vector partitions.  The corrected statement for partitions of multipartite numbers is given in  \cite[Theorem 12.2]{A98} and indeed it has Beck-type companion identities as we explain below.

A multipartite (or $s$-partite) number $\frak n=(n_1, n_2, \ldots, n_s)$ is an $s$ tuple of non-negative integers, not all $0$. We view multipartite numbers as vectors and refer to $n_1, n_2, \ldots, n_s$ as the entries of $\frak n$.

A multipartition (or vector partition) $\xi=(\xi^{(1)}, \xi^{(2)},\ldots, \xi^{(t)})$ of $\frak n$ is a sequence of multipartite numbers in non-increasing lexicographic order satisfying   $$\frak n=\xi^{(1)}+ \xi^{(2)}+\cdots +\xi^{(t)}.$$ We refer to $\xi^{(i)}$, $1\leq i\leq t$ as the multiparts (or vector parts) of the multipartition $\xi$ and to the number of multiparts $t$ of $\xi$ as the length of $\xi$ which we denote by $\ell(\xi)$.

Let $S_1$ and $S_2$ be sets of positive integers. Given a multipartition $\xi$ of $\frak n$ with all entries of all multiparts in $S_1$, we say that a multipart $\xi^{(i)}$ of $\xi$ is \textit{primitive} if at least one entry of $\xi^{(i)}$ is in $S_2$. Otherwise, the multipart is called \textit{non-prmitive}. We denote by  $\mathcal{VD}_r(\frak n)$ the set of multipartitions $\xi=(\xi^{(1)}, \xi^{(2)},\ldots, \xi^{(t)})$ of $\frak n=(n_1, n_2, \ldots, n_s)$ with all entries of all multiparts in $S_1$ and such that all multiparts are repeated at most $r-1$ times. We denote by  $\mathcal{VO}_r(\frak n)$ the set of multipartitions $\eta=(\eta^{(1)}, \eta^{(2)}, \ldots, \eta^{(u)})$ of $\frak n=(n_1, n_2, \ldots, n_s)$ such that each multipart $\eta^{(i)}$ of $\eta$ is primitive.

Then, Andrews \cite{A98} gives the following theorem mentioning that its proof can be constructed similar to the proof using ideals of order $1$ for the analogous result for regular partitions.

\begin{theorem} \label{SA} Let $S_1$ and $S_2$ be sets of positive integers. Then $$|\mathcal{VD}_r(\frak n)|=|\mathcal{VO}_r(\frak n)|$$ if and only if $(S_1,S_2)$ is an Euler pair of order $r$, i.e., $rS_1\subseteq S_1$ and $S_2=S_1-rS_1$.

\end{theorem}

We note that Glaisher's bijection can be extended to prove Theorem \ref{SA} combinatorially. The Glaisher type transformation, $v\varphi_r$, from $\mathcal{VO}_r(\frak n)$ to $\mathcal{VD}_r(\frak n)$ repeatedly merges $r$ equal multiparts (as addition of vectors) until there are no multiparts repeated more than $r-1$ times. The transformation from $\mathcal{VD}_r(\frak n)$  to $\mathcal{VO}_r(\frak n)$ takes each non-primitive multipart  (all its entries are from $rS_1$) and splits it into $r$ equal multiparts, repeating the process until the obtained multiparts are primitive. The remark below is the key to adapting the combinatorial proofs of Theorems \ref{T1} and \ref{T2} to proofs of Beck-type identities for multipartitions.

\begin{remark} Let $\eta \in \mathcal{VO}_r(\frak n)$ and $\xi=v\varphi_r(\eta)\in \mathcal{VD}_r(\frak n)$. Then multipart $\xi^{(i)}$ of $\xi$ was obtained by merging $r^k$ multiparts of $\eta$ if and only if, when writing all entries of $\xi^{(i)}$ in the form $r^j{\bf a}$, the smallest exponent of $r$ in all entries of $\xi^{(i)}$ is $k$.
\end{remark}

To formulate Beck-type identities for multipartitions, let  $vb_r(\frak n)$ be the difference between the number of multiparts in all multipartitions in $\mathcal{VO}_r(\frak n)$ and the number of multiparts in all multipartitions in $\mathcal{VD}_r(\frak n)$. Similarly,  let  $vb'_r(\frak n)$ be  the difference in the total number of different multiparts in all multipartitions in $\mathcal{VD}_r(\frak n)$  and the total number of different multiparts in all multipartitions in $\mathcal{VO}_r(\frak n)$. Then, we have the following Beck-type identities for multipartions.

\begin{theorem} \label{last} Suppose $(S_1,S_2)$ is an Euler pair of order $r\geq 2$  and let $\frak n$ be a multipartite number. Then

\begin{enumerate}

\item[(i)] $\ds \frac{1}{r-1}vb_r(\frak n)$ equals the number of multipartitions $\xi=(\xi^{(1)}, \xi^{(2)},\ldots, \xi^{(t)})$ of $\frak n$ with all entries of all multiparts in $S_1$ and such that exactly one multipart is repeated at least $r$ times. Moreover, $\ds \frac{1}{r-1}vb_r(\frak n)$ equals the number of multipartitions $\eta=(\eta^{(1)}, \eta^{(2)}, \ldots, \eta^{(u)})$ of $\frak n$ such that each multipart $\eta^{(i)}$ of $\eta$ has all entries in $S_1$ and only one multipart, possibly repeated, is non-primitive. \medskip

\item[(ii)]  $vb'_r(\frak n)$ equals   the number of multipartitions $\xi=(\xi^{(1)}, \xi^{(2)},\ldots, \xi^{(t)})$ of $\frak n$ with all entries of all multiparts in $S_1$ and such that exactly one multipart is  repeated more than $r$ times but less than $2r$ times. \end{enumerate}\end{theorem}

The combinatorial proofs of these statements follow the combinatorial proofs of Theorems \ref{T1} and \ref{T2} with  all references to expression of the form ``part $r^k{\bf a}$ in the partition $\mu$" changed to ``multipart of multipartition $\mu$ in which $r^k{\bf a}$ is the entry with the smallest exponent of $r$ (among the entries of the multipart)."

To our knowledge, there is no  analytic proof of Theorem \ref{last}.

\section*{Acknowledgements}
We are grateful to the anonymous referees for suggestions that improved the exposition of the article. In particular, one referee suggested the short proof of \eqref{sets}, and another referee alerted us to the correct statement of Subbarao's theorem for vector partitions.


TITLE: Energy-momentum tensor from variation of coordinates in quantum field theory
QUESTION [2 upvotes]: I am a beginner in QFT, and I have been reading about the energy-momentum tensor from making "infinitesimal transformations" in the Poincare group. (not variation wrt metric tensor). But I think I got caught in a few technical details that I cannot solve on my own. I would appreciate it if you could use some simple Lagrangians as an example.

What is invariant under infinitesimal transformations, especially translation? Fradkin's QFT an integrated approach requires that the action remains unchanged, so are a bunch of other books on gauge theory, but Greiner's relativistic quantum mechanics as well as Peskin all requires invariance of the Lagrangian itself. The problem is that I can show that the energy-momentum tensor is locally conserved if I require the invariance of Lagrangian, but I cannot show that if I require the invariance of action. I think the invariance of action is more general but I failed to see how I can get a locally conserved energy-momentum tensor from that.
Here is my derivation: For Lagrangian $\mathcal{L}=\left(D_{\mu} \phi(x)\right)^{*}\left(D^{\mu} \phi(x)\right)-m_{0}^{2}\phi(x)\phi^*(x)-\frac{\lambda}{2}\left(\phi(x)\phi^*(x)\right)^2-\frac{1}{4} F^{\mu \nu} F_{\mu \nu}$,
Requiring the system being transnational invariant, $\delta S =0$, we have for arbitrary uniform displacement
\begin{align*}
\int d^4 x   \partial_\mu \left[ \left( g ^\mu_\nu  \mathcal L  - \frac{\delta{\mathcal L }}{\delta \partial_\mu \phi}  \partial_\nu \phi -  \frac{\delta{\mathcal L }}{\delta \partial_\mu \phi^*}\partial_\nu \phi^*   + \frac{\delta{\mathcal L }}{\delta \partial_\nu A_\sigma}  \partial_\nu A_\sigma  \right)  \delta x^\nu\right] = 0
\end{align*}
But that does not guarantee the conservation of the Neother current, because the $ \delta x^\nu$ is a constant (uniform diaplacement), and at best we can say that the integral is zero, but not the integrand. I can still arrive at a conserved charge by doing Stokes theorem on that integral, but I lost the locally conserved tensor.
This problem is not present when we consider rotations, since in that case, $\delta x^\nu$ would be truly arbitrary.

How do we do the variation exactly? In Fradkin's book as well as some gauge theory books, they all talk about the variation of integration measure $d^4 x$ under a coordinate transformation.
Since the derivative of a determinant is gven by
\begin{align*}
\newcommand{\D}[2]{\frac{d #1}{d #2}}
\newcommand{\tr}{\mathrm tr}
  \D{\det J(x)}{x} = \det J \tr \left(J^{-1} \D{J(x)}{x} \right)
 \end{align*}
And if I do the variation of this measure, seeing the integral measure as a functional of coordinate unction $x_\mu$, which is a function of some fixed coordinate $\chi_\lambda$, is given by
\begin{align*}
  d^4 x [x_\mu] = \det J[x_\mu] d^4 x[\chi] = \det J[x_\mu (\chi)] d^4 \chi
 \end{align*}

Then the functional derivative of the measure with respect to the coordinate function is
\begin{align*}
\newcommand{\P}[2]{\frac{\partial #1}{\partial #2}}
  \frac{\delta d^4 x [x_\mu]}{\delta x_\sigma}  =& \frac{\delta \det J[x_\mu (\chi)] d^4 \chi}{\delta x_\sigma (\chi)} \\
  =& \det J[x_\mu] \tr\left(J^{-1}[x_\mu] \frac{\delta J[x_\mu]}{\delta x_\sigma (\chi)} \right)  d^4 \chi\\
  =& \det J[x_\mu] \delta_{\nu \lambda}\P{\chi^\nu}{x_\mu} \left(  \frac{J^{\mu\lambda}[x_\mu]}{\delta x_\sigma (\chi)} \right)   d^4 \chi\\
  =&\delta_{\nu \lambda}\P{\chi^\nu}{x_\mu} \left(  \frac{\delta}{\delta x_\sigma (\chi)} \P{x^\mu}{\chi_\lambda}\right)  \det J[x_\mu]  d^4 \chi\\
  =&\P{\chi^\nu}{x_\mu} \left(  \frac{\delta}{\delta x_\sigma (\chi)} \P{x^\mu}{\chi_\nu}\right)  \det J[x_\mu]  d^4 \chi\\
  =&\P{\chi^\nu}{x_\mu} \P{ g^{\mu \sigma}}{\chi_\nu} \det J[x_\mu]  d^4 \chi\\
  =&\P{g^{\mu \sigma}}{x_\mu} \det J[x_\mu]  d^4 \chi\\
  =& \left(\P{}{x_\sigma} \right)\det J[x_\mu]  d^4 \chi
 \end{align*}
But I am having a hard time understanding this result. If I write
\begin{align*}
  \delta d^4 x [x_\mu] = \left(\P{}{x_\sigma}  \right) \delta x_\sigma \det J[x_\mu]  d^4 \chi = (\partial^\sigma \delta x_\sigma) d^4 x_\mu
 \end{align*}
I got confused. Normally we do $\delta \partial = \partial \delta$, and that would give me nonsense. I am not sure how to properly interpret this result. I think that $\delta x$ here might not be a functional change so I am not allowed to do that? But I am not sure.

There are other ways to do this variation as well, what is the right one to use? Some book says that
\begin{align*}
\delta S =& \int d^4 x' \mathcal L'(\phi ' (x'))-\int d^4 x \mathcal L(\phi (x))\\
=&\int d^4 x \mathcal L'(\phi ' (x))-\int d^4 x \mathcal L(\phi (x))\\
=&\int d^4 x (\mathcal L'(\phi ' (x))- \mathcal L(\phi (x)))\\
\end{align*}
because the $x'$ in the first integral is a dummy index. Therefore there's no variation on the integration measure at all. I am not sure if that works since the boundary might be shaped differently. But the result is the same if we allow that.

There is also another way of doing it by allowing a $\partial \mathcal L/\partial x$ term. Some book says that this term arises from the fact that the Lagrangian density is a scaler itself and should transform, others avoided it, and some other book says that is comes from the freedom of choice of an total derivative of the Lagrangian (and hints super symmetry). I am not sure if they are all equivalent. But allowing a $\partial \mathcal L/\partial x$ term doesn't make sense to me.
I can provide more details if needed.

REPLY [2 votes]: There is a lot of notation above that I'm having trouble parsing but an infinitesimal translation is $x^\prime_\mu = x_\mu + \delta a_\mu$. To compute $\delta S$ under this transformation, you change $x^\mu$ to $x^\mu + \delta a^\mu$ everywhere, expand in $\delta a^\mu$ and keep the linear term. Let's start with the massless free scalar.
\begin{align}
S &= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \\
&\to \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x + \delta a) \partial_\mu \phi(x + \delta a) \\
&= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) + \int d^4 x \, \partial_\nu \partial_\mu \phi(x) \partial_\mu \phi(x) \delta a_\nu + O(\delta a^2) \\
&= \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) + \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu + O(\delta a^2)
\end{align}
Therefore $\delta S$ is the integral of a total derivative, in other words zero. This is reassuring (since we expect translation to be a symmetry) but it answers one of your questions. The variation of the Lagrangian isn't zero, only the variation of the action.
Now how do we get the Noether current for this symmetry? There is a helpful trick due to either Glashow or Gel-Mann which involves promoting the constant $\delta a_\mu$ to a local $\delta a_\mu(x)$. This is not a symmetry anymore (unless we are working with a diffeomorphism invariant theory like GR) but it is still helpful for studying the original symmetry. Proceeding from the last line above, we have
\begin{align}
\delta S &= \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu(x) \\
&= \int d^4 x \, \partial_\nu \left [ \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \delta a_\nu(x) \right ] - \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \partial_\nu \delta a_\nu(x) \\
&= - \int d^4 x \, \frac{1}{2} \partial_\mu \phi(x) \partial_\mu \phi(x) \partial_\nu \delta a_\nu(x) \\
&= \int d^4 x \, \partial_\mu \partial_\nu \phi(x) \partial_\mu \phi(x) \delta a_\nu(x) \\
&= \int d^4 x \, \partial_\mu  \left [ \frac{1}{2} \partial_\nu \phi(x) \partial_\mu \phi(x) \right ] \delta a_\nu(x)
\end{align}
We have now written this more general $\delta S$ as the integral of $\partial_\mu T_{\mu\nu} \delta a_\nu$. Even though $\delta S$ need not be zero because we have promoted our symmetry to a non-symmetry, $\delta S$ on shell does need to be zero since the equations of motion are defined by leading to stationary actions under any variation. Then, since $\delta a_\nu$ is arbitrary, $\partial_\mu T_{\mu\nu} = 0$ on shell. In other words, the $T_{\mu\nu}$ is indeed the stress tensor and we can read it off above.
I hope this explains your confusion in point number 1. It doesn't have to do with translations vs rotations. Translations are $x^\prime_\mu = x_\mu + \delta a_\mu$ specified by four constant parameters and have Noether current $T_{\mu\nu}$. Rotations are $x^\prime_\mu = x_\mu + \delta \omega_{[\mu\nu]} x_\nu$ specified by six constant parameters and have Noether current $M_{\mu\nu\rho} = x_\mu T_{\nu\rho} - x_\nu T_{\mu\rho}$. It's just that, for either one of these symmetries, the constant parameter becomes arbitrary when we want to use the above trick.
These calculations were done for the massless free scalar but a good exercise is to repeat them for the scalar QED theory you wrote or indeed abstract theories where $\mathcal{L}$ is a polynomial in a bunch of fields and their derivatives. All of these will have conserved stress tensors derivable with the same steps. Conversely, Lagrangians that involve further integrals of the fields, such as the famous
\begin{align}
\mathcal{L}[\phi(x)] = \int d^4 y\, \frac{\phi(x) \phi(y)}{|x - y|^\gamma}
\end{align}
will lead to the phenomenon you worried about: global but not local conservation. In other words, translation is still a symmetry but there is no longer an associated conserved current. This is the hallmark of a nonlocal field theory.
Once you are comfortable with this Noether procedure for getting the stress tensor, there are optional steps which involve adding improvement terms which vanish on shell but make the symmetries more manifest. These are usually accounted for automatically by the Hilbert procedure you mention which comes from varying with respect to the metric.


TITLE: Help me derive a formula for finding a formula for sum of x^n
QUESTION [0 upvotes]: I wanna find the sum of divisors of a number defined as:$$2^n$$
Let the series be S, $$ so, S = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 +.....2^n $$
Noww, if we factor out the 2 from 2nd term to last term, S can be written like :$$ S= 2^0 + 2(2^0 + 2^1 +2^2 + 2^{(n-1)})  $$
so. $$S = 2^0 + 2S^{(n-1)}$$
I don't know where to go from here :/

REPLY [1 votes]: Hint: If you let $$S = 2^0+2^1+\cdots+2^{n-1}+2^n$$ then $$2S = 2^1+2^2+\cdots+2^n+2^{n+1}.$$
Subtract the first equation from the second to get $$2S-S = (2^1+2^2+\cdots+2^n+2^{n+1}) - (2^0+2^1+\cdots+2^{n-1}+2^n)$$
You should notice that a lot of terms cancel out. Can you figure out the formula from here?