TITLE: Does Lebesgue's dominated convergence theorem depend on measure
QUESTION [1 upvotes]: I know  Lebesgue's dominated convergence theorem holds for Lebesgue measure. But, in general, does it hold for an general measure? For example, I know it holds for the counting measure, so we obtain the LDCT for infinite series.

REPLY [1 votes]: Yes, the theorem is valid for any measure.  If you read the proof, nothing is used except countable additivity.


text:
\section{Maximum Likelihood Estimation}
\label{sec:mle}

In this section, we develop an approach to experiment design and estimation based on a maximum likelihood approach.  Given a policy, we develop the maximum likelihood estimator (MLE) for the treatment effect $\alpha$.  In particular, we take a nonparametric approach in this paper, as we make no parametric assumptions on the Markov chains being studied.  Thus our approach involves maximum likelihood estimation of the transition matrices, followed by inversion to obtain an MLE for the steady state distribution.

Let $\Gamma_n(\ell,x)$ to be the number of times action $i$ at state $x$ is sampled by time $n$:
\begin{align}
    \Gamma_n(\ell,x):=\sum_{j=0}^{n-1} I(X_j=x, A_j=\ell),\ x\in S,\ \ell=1,2.
\end{align}

Now define:
\begin{align}
\hat{P}_n(\ell,x,y)&=\frac{\sum_{j=0}^{n-1} I(X_j=x, A_j=\ell, X_{j+1}=y)}{\max\{\Gamma_n(\ell,x),1\}};\\
\hat{P}_n(\ell)&=(\hat{P}_n(\ell,x,y):x,y\in S).
\end{align}

The estimators $\hat{P}(1)$ and $\hat{P}(2)$ are standard maximum likelihood estimators (MLE) for the corresponding transition matrices $P(1)$ and $P(2)$.

Define the stopping time $J=\min\{n\ge 0: \hat{P}_n(\ell)$ is irreducible for $\ell=1,2\}$. Note that $\hat{P}_n(\ell)$ will remain irreducible for $n\ge J$, since any path with positive probability under $\hat{P}_J$ will have positive probability under $\hat{P}_n$ for all $n\ge J$.  Thus for each $n\ge J$, $\hat{P}_n(\ell)$ has a unique stationary distribution $\hat{\pi}_n(\ell)$ satisfying
\begin{gather}
    \hat{\pi}_n(\ell)=\hat{\pi}_n(\ell)\hat{P}_n(\ell);\\
    \hat{\pi}_n(\ell,x)\ge 0, x\in S;\\
    \sum_{x \in S}\hat{\pi}_n(\ell,x)=1.
\end{gather}
Note that by equivariance of the MLE, since stationary distributions are functionals of the transition matrices, each $\hat{\pi}_n(\ell)$ is also the MLE for $\pi(\ell)$.

Define
\begin{align}
\hat{r}_n(\ell,x)&=\frac{\sum_{j=0}^{n-1} I(X_j=x, A_j=\ell)R_{j+1}}{\max\{\Gamma_n(\ell,x), 1\}}.
\end{align}
The preceding is the MLE of $r(\ell,x)$ along the realized sample path.

Finally, for $n\ge J$, again by equivariance of the MLE, we conclude that the resulting nonparametric MLE $\hat{\alpha}_n$ for $\alpha$ is given by the following:
\begin{align}
    \hat{\alpha}_n=\hat{\pi}_n(2)\hat{r}_n(2)-\hat{\pi}_n(1)\hat{r}_n(1).
\end{align}
For $n< J$, $\ell = 1,2$, and $x \in S$, we arbitrarily define $\pi_n(\ell,x)=1 / |S|$, $\hat{r}_n(\ell,x)=0$, and $\hat{\alpha}_n = 0$.

config: arxiv


TITLE: Rational functions are decomposed in polynomial products
QUESTION [1 upvotes]: I'm trying to understand why this is true:

Since $K(x)$ is a field, $K(x)$ is an UFD, then $K(x)$ can be written uniquely as products of irreducible elements of $K(x)$.
I didn't understand why we can write this product as polynomials in $K[x]$ instead of just elements in $k(x)$.
Thanks in advance.

REPLY [1 votes]: Let $z\in K(x)$, then there exist $p,q\in K[x]$ with $z=\frac pq$. Since $K[x]$ is a UFD (because $K$ is a field), there is a unique factorization for $p$ and $q$ resp. Let $p = a\cdot\prod_{i\in I}p_i^{k_i}$ and $q=b\cdot\prod_{j\in J}q_j^{n_j}$ be the unique factorizations.
So we can write $$z =  \frac{a\cdot \prod_{i\in I}p_i^{k_i}}{b\cdot \prod_{j\in J} q_j^{n_j}}$$
Let $I'\subseteq I$ and $J'\subseteq J$ be the subset of indices that remain after simplifying the fraction. Then we can write
$$ z = c\cdot \frac{\prod_{i\in I'}p_i^{k_i}}{\prod_{j\in J'}q_j^{n_j}} = c\cdot \prod_{i\in I'} p_i^{k_i} \cdot \prod_{j\in J'} q_j^{-n_j}$$
with $c\in K\setminus\{0\}$. If you combine the last two products, you get exactly the type of equation that you want.


text:
import measure_theory.measure.measure_space
import measure_theory.constructions.borel_space
import measure_theory.integral.bochner

open topological_space (second_countable_topology)
open measure_theory set classical filter

open_locale classical topological_space filter interval

variables {α β E F : Type*} [measurable_space α] {μ : measure α} [normed_add_comm_group E]
          [complete_space E] [normed_space ℝ E]

namespace measure_theory
lemma ae_restrict_eq_iff {s : set α} {f g : α → β} (h : measurable_set {x | f x = g x}) :
  f =ᵐ[μ.restrict s] g ↔ ∀ᵐ x ∂μ, x ∈ s → f x = g x :=
ae_restrict_iff h

/-
MOVE next to ae_restrict_of_ae_restrict_of_subset
-/
lemma ae_mem_imp_of_ae_restrict_of_subset {α : Type*} {m0 : measurable_space α}
  {μ : measure α} {s t : set α} {p : α → Prop} (hst : s ⊆ t) (hp : ∀ᵐ (x : α) ∂μ.restrict t, p x) :
  (∀ᵐ x ∂μ, x ∈ s → p x) :=
ae_imp_of_ae_restrict (ae_restrict_of_ae_restrict_of_subset hst hp)
end measure_theory

lemma ae_restrict_of_forall_mem {α : Type*} [measurable_space α] {μ : measure α} {s : set α}
  {p : α → Prop} (hs : measurable_set s) (h : ∀ x ∈ s, p x) : ∀ᵐ (x : α) ∂μ.restrict s, p x :=
by { rw ae_restrict_iff' hs, exact ae_of_all _ h }

open measure_theory

lemma ae_interval_oc {ν : measure ℝ} {P : ℝ → Prop} {a b : ℝ} :
  (∀ᵐ t ∂(ν.restrict $ Ι a b), P t) ↔
  (∀ᵐ t ∂(ν.restrict $ Ioc a b), P t) ∧ ∀ᵐ t ∂(ν.restrict $ Ioc b a), P t :=
begin
  cases le_or_lt a b with h h,
  { simp [interval_oc_of_le h, Ioc_eq_empty_of_le h] },
  { simp [interval_oc_of_lt h, Ioc_eq_empty_of_le h.le] }
end

lemma is_compact.integrable_const {α : Type*} [measurable_space α] [topological_space α]
  {E : Type*} [normed_add_comm_group E] [measurable_space E]
  {s : set α} (hs : is_compact s) (c : E) (μ : measure α) [is_locally_finite_measure μ] :
  integrable (λ (x : α), c) (μ.restrict s) :=
by simp_rw [integrable_const_iff, measure.restrict_apply_univ, hs.measure_lt_top, or_true]

subset_name: curated


TITLE: Prove that for all $x\in\Bbb R$, precisely one of the statements $x>0$, $x=0$, or $x<0$ holds
QUESTION [0 upvotes]: I came across the following exercise while studying Terrence Tao's book Analysis I:

Exercise 5.4.1 Let $x\in\Bbb R$. Show that precisely one of the statements $x>0$, $x=0$, or $x<0$ holds

My Attempt: First let us prove that at least one of these statements must hold. Suppose $x=\operatorname{LIM}_{n\to\infty}a_n$ is a nonzero real number (where $\operatorname{LIM}_{n\to\infty}$ denotes the formal limit of the sequence that follows).  Then the sequence $(a_n)_{n\in\Bbb N}$ cannot be equivalent to $(0)_{n\in\Bbb N}$, i.e. $$ \neg(\forall\varepsilon >0 \ \exists N\in\Bbb N \ \forall n\ge N:|a_n|\le\varepsilon)\equiv \exists\varepsilon >0 \ \forall N\in\Bbb N \ \exists n\ge N:|a_n|>\varepsilon.$$
The author hints to conclude that $(a_n)_{n\in\Bbb N}$ is either eventually positively or negatively bounded away from zero; namely
$$\exists c>0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\ge c \quad \lor \quad \exists c<0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\le c.$$
Clearly, there are infinitely many $n\in\Bbb N$ such that $|a_n|>\varepsilon$, but there is no certainty that this holds for all $a_n$. Nor is there any certainty that the claim holds for all $n\ge N$ for some $N$. So how can I conclude? I have no problems showing that only one of these can hold at once.
Thank you for your time, and happy 4th for those who celebrate :)

REPLY [2 votes]: Well, remember that the sequences in question are Cauchy, so since there are infinitely-many $a_n\in(-\infty,-\varepsilon)\cup(\varepsilon,\infty),$ we can actually show that either (i) there are all but finitely-many $a_n\in(0,\infty)$ or (ii) there are all but finitely-many $a_n\in(-\infty,0).$
Added:
One difference in approach that I would take is to say that $$\neg\left(\forall\varepsilon>0,\exists N\in\Bbb N:\forall n\ge N,\left|a_n\right|\le\varepsilon\right)\equiv\exists\varepsilon_0>0:\forall N\in\Bbb N,\exists n\ge N:|a_n|>\varepsilon_0.$$
Aside from some fiddly notational stuff (like comma usage and colon placement), the big change is using $\varepsilon_0$ to denote a particular value of $\varepsilon.$ Since we're still hoping to prove something that holds for all $\varepsilon,$ it behooves us to make clear which one that is.
Now, use the definition of Cauchy sequence to show that there exists some $N_0\in\Bbb N$ such that for all $m,n\ge N_0,$ we have $|a_m-a_n|<\varepsilon_0.$ We already know that there is some $n\ge N)$ such that $\left|a_n\right|>\varepsilon_0.$ From this, we can show that for all $m\ge N_0,$ we must have $a_m\in\left(a_n-\varepsilon_0,a_n+\epsilon_0\right),$ and since $|a_n|>\epsilon_0,$ then we're done.


text:
\section{Arens-Fort Space is Completely Hausdorff}
Tags: Completely Hausdorff Spaces, Arens-Fort Space

\begin{theorem}
Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
Then $T$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
\end{theorem}

\begin{proof}
We have:
:[[Fort Space is Completely Normal]]
:[[Arens-Fort Space is Expansion of Fort Space]].
From [[Sequence of Implications of Separation Axioms]] we have that a [[Definition:Fort Space|Fort space]] is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ space]].
Then in [[Separation Properties Preserved by Expansion]], it follows that as a [[Definition:Fort Space|Fort space]] is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ space]], then so is the [[Definition:Arens-Fort Space|Arens-Fort space]].
{{qed}}
\end{proof}

config: wiki


TITLE: Proving the sequence $1/n^4 + 2^3/n^4+\cdots+n^3/n^4$ converges
QUESTION [0 upvotes]: As the title states, I'm trying to prove that the sequence $a_n$ converges to $1/4$ where: $$a_n=\dfrac{1}{n^4}+\dfrac{2^3}{n^4}+\cdots+\dfrac{n^3}{n^4}$$
Is there any way to do this without relying on previous sum identities or integrals? I'm having trouble doing this with just the squeeze theorem.

REPLY [0 votes]: Apply Stolz-Cesaro to
$$\tag 1 \frac{1^3 + 2^3 + \cdots + n^3}{n^4}.$$
We are led to
$$\frac{(n+1)^3}{(n+1)^4-n^4} = \frac{n^3 + 3n^2 + 3n + 1}{4n^3 + 6n^2 + 4n + 1} \to \frac{1}{4}.$$
By S-C, the limit of $(1)$ is $1/4.$


TITLE: Limits of Incommensurable quantities
QUESTION [0 upvotes]: By taking the unit of measure sufficiently small, however, the ratio of two incommensurable quantities can be expressed as nearly to the true value as may be desired. Thus an approximate value of the ratio can be found which shall differ from the true value by an amount less than any specified amount, however small.

I don't understand this. Can somebody explain it to me simply and probably give some examples as well? This is an excerpt from the book Geometry for the practical man by J.E. Thompson to those of you curious which book it came from.

REPLY [1 votes]: This is an ancient conundrum going back to the era of classical Greece.
The Greeks first believed that for any two geometric magnitudes there would exist a common unit which would divide evenly into both magnitudes.
So given, for example, the diagonal of a square and its side they thought there would be a tiny unit which would divide a whole number of times (let is say $m$ times) into the diagonal and a whole number of times (let us say $n$ times) into the side. Then if $d$ is the magnitude of the diagonal and $s$ the magnitude of the side, then
$$ d=\frac{m}{n}\times s $$
However it was learned that in the case of the diagonal and side of a square, there was no such common unit. The actual formula is
$$ d=\sqrt{2}\times s $$
and $\sqrt{2}$ is an irrational number, meaning it is not equal to the ratio of any two whole numbers $m$ and $n$.
However, they realized that if one made the units sufficiently small one could approximate, in this case, $\sqrt{2}$ or any such irrational magnitude, to approximate the magnitude with a rational number to any degree of precision required.
For example, $\dfrac{239}{169}\approx1.414201183$ whereas $\sqrt{2}\approx1.4142135562$. So the rational number $\dfrac{239}{169}$ is approximately equal to $\sqrt{2}$ to four decimal place precision. So if we divide the side of a square into $169$ equal units, that unit will go approximately, but not exactly, $239$ times into the diagonal of the square giving an approximation of the ratio of the diagonal to the side of the square.


TITLE: Why does diamond-graphite phase digram not contain a critical point?
QUESTION [1 upvotes]: I was curious as to why in all the diagrams I have seen that the diamond-graphite phase diagram does not display a critical point, do all substances eventually reach a critical point?

REPLY [6 votes]: I am unsure as to why you think there should be a critical point. Having a critical point means the phase boundary between the phases abrubtly ends and implies it is thermodynamically possible to go from one phase to the other smoothly without passing through a phase transition.
The most common and well-known critical point is the one from liquid to gas. There are no symmetry differences between the liquid and gas phases, unlike the solid phase where translation and rotation symmetriesare broken. Therefore you can go from the liquid to the gas phase smoothly by tuning pressure and temperature strategically.
Graphite and diamond on the other hand have different symmetries, so there is no way to smoothly from one to the other without a phase transition (diamond is cubic, graphite is hexagonal). The same is generically true for all solid-to-solid transitions which differ in crystal symmetry.


TITLE: Please help me understand this. $\frac{dx}{dt} = S x (a-x)$. What does it mean for some constant $S$? How to find $x$ for fastest/slowest growth?
QUESTION [3 upvotes]: I am having some trouble understanding this problem. There is this function that calculates reaction rate of a substance for some constant positive $S$.
$a$ = original amount of the first substance
$x$ = some amount of substance
First question, what does $\frac{dx}{dt} = S x (a-x)$ mean? Does this mean that the rate of change of $x$ in the equation $(S x (a-x))$ is affected by the change in time? So if there was no '$x$' in the equation, than change in time would not affect the equation right? *This is the first time I am encountering '$dt$' in my derivative assignments. When finding derivatives for simple equations, its mostly been of d/dx notation.
When I graphed $S x (a-x)$, I substituted random numbers for $S$ and $a$. Does this tell me anything about the function for rate of increase and decrease? Should I have solved for $a$? I notice the function increases and then decreases as $x$ moves away from $0$. I also know that when the derivative crosses the $x$ axis, the original function (which I don't know) will start to decrease.
What do I need to do to determine the fastest/slowest growth rate using the derivative?
Thanks

REPLY [1 votes]: The explanation given by @user7530 is so excellent that I have little to add.  That said, in case you are curious as to the actual solution of the differential equation you posted, it is actually not at all hard to solve with a little Calc II knowledge.  I will not derive unless you ask, (and I am sure that @user7530 has done it, given he has graphed it) and the result is
$$x(t) = \frac{a x(0)}{x(0) + [a-x(0)] e^{-a S t}}$$


text:
\begin{document}
\title{Exact variance of von Neumann entanglement entropy \\ over the Bures-Hall measure}
\author{Lu Wei}
\email{luwe@umich.edu}
\affiliation{Department of Electrical and Computer Engineering \\ University of Michigan - Dearborn, MI 48128, USA}
\date{\today}

\begin{abstract}
The Bures-Hall distance metric between quantum states is a unique measure that satisfies various useful properties for quantum information processing. In this work, we study the statistical behavior of quantum entanglement over the Bures-Hall ensemble as measured by von Neumann entropy. The average von Neumann entropy over such an ensemble has been recently obtained, whereas the main result of this work is an explicit expression of the corresponding variance that specifies the fluctuation around its average. The starting point of the calculations is the connection between correlation functions of the Bures-Hall ensemble and these of the Cauchy-Laguerre ensemble. The derived variance formula, together with the known mean formula, leads to a simple but accurate Gaussian approximation to the distribution of von Neumann entropy of finite-size systems. This Gaussian approximation is also conjectured to be the limiting distribution for large dimensional systems.
\end{abstract}

\maketitle

\section{Introduction and the main result}
Quantum information theory aims at understanding the theoretical underpinnings of quantum technologies such as quantum computing and quantum communications. Crucial to successful exploitation of the quantum revolutionary advances is the understanding of the non-classical phenomenon of quantum entanglement. Entanglement is the most fundamental characteristic trait of quantum mechanics, which is also the resource and medium that enable quantum technologies.

In this work, we aim to understand the statistical behavior of entanglement of quantum bipartite systems over Bures-Hall measure~\cite{Hall98,Zyczkowski01}. The Bures-Hall measure enjoys the property that, without any prior knowledge on a density matrix, the optimal way to estimate the density matrix is to generate a state at random with respect to this measure~\cite{Zyczkowski01,Sommers04,Osipov10}. In particular, we study the degree of entanglement as measured by the von Neumann entropy over such a measure. The mean value of von Neumann entropy over the Bures-Hall measure has been recently obtained~\cite{Sarkar19,Wei20a}. As an important step towards understanding its statistical distribution, we derive the corresponding variance in this paper. The variance describes the fluctuation of the entropy around its mean value, which also provides crucial information such as whether the average entropy is typical.

The density matrix formulism~\footnote{For a comprehensive treatment of the density matrix formulism, we refer readers to Refs.~\cite{BZ17,Majumdar} and references therein.}, introduced by von Neumann, that has led to the Bures-Hall ensemble is described as follows. Consider a composite (bipartite) system that consists of two subsystems $A$ and $B$ of Hilbert space dimensions $m$ and $n$, respectively. The Hilbert space $\mathcal{H}_{A+B}$ of the composite system is given by the tensor product of the subsystems, $\mathcal{H}_{A+B}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. A random pure state of the composite system $\mathcal{H}_{A+B}$ is defined as a linear combination of the random coefficients $z_{i,j}$ and the complete basis $\left\{\Ket{i^{A}}\right\}$ and $\left\{\Ket{j^{B}}\right\}$ of $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$~\cite{Majumdar},
\begin{equation}\label{eq:S0}
\Ket{\psi}=\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\Ket{i^{A}}\otimes\Ket{j^{B}},
\end{equation}
where each $z_{i,j}$ follows the standard complex Gaussian distribution. We now consider a superposition of the state~(\ref{eq:S0}),
\begin{equation}\label{eq:SB}
\Ket{\varphi}=\Ket{\psi}+\left(\mathbf{U}\otimes\mathbf{I}_{m}\right)\Ket{\psi},
\end{equation}
where $\mathbf{U}$ is an $m\times m$ unitary random matrix with the measure proportional to $\det\left(\mathbf{I}_{m}+\mathbf{U}\right)^{2\alpha+1}$~\cite{Sarkar19}. The corresponding density matrix of the pure state~(\ref{eq:SB}) is
\begin{equation}\label{eq:rho}
\rho=\Ket{\varphi}\Bra{\varphi},
\end{equation}
which has the natural probability constraint
\begin{equation}\label{eq:del}
\tr(\rho)=1.
\end{equation}
We assume without loss of generality that $m\leq n$. The reduced density matrix $\rho_{A}$ of the smaller subsystem $A$ is computed by partial tracing (purification) of the full density matrix~(\ref{eq:rho}) over the other subsystem $B$ (environment) as
\begin{equation}\label{eq:rhoB}
\rho_{A}=\tr_{B}\rho.
\end{equation}
The resulting density of eigenvalues of $\rho_{A}$ ($\lambda_{i}\in[0,1]$, $i=1,\dots,m$) is the (generalized) Bures-Hall measure~\cite{Hall98,Zyczkowski01,Sarkar19}
\begin{equation}\label{eq:BH}
f\left(\bm{\lambda}\right)=\frac{1}{C}~\delta\left(1-\sum_{i=1}^{m}\lambda_{i}\right)\prod_{1\leq i<j\leq m}\frac{\left(\lambda_{i}-\lambda_{j}\right)^{2}}{\lambda_{i}+\lambda_{j}}\prod_{i=1}^{m}\lambda_{i}^{\alpha},
\end{equation}
where the parameter $\alpha$ takes half-integer values
\begin{equation}\label{eq:aBH}
\alpha=n-m-\frac{1}{2},
\end{equation}
and the constant $C$ is
\begin{equation}\label{eq:cBH}
C=\frac{2^{-m(m+2\alpha)}\pi^{m/2}}{\Gamma\left(m(m+2\alpha+1)/2\right)}\prod_{i=1}^{m}\frac{\Gamma(i+1)\Gamma(i+2\alpha+1)}{\Gamma(i+\alpha+1/2)}.
\end{equation}
In Eq.~(\ref{eq:BH}), the presence of the Dirac delta function $\delta(\cdot)$ reflects the constraint~(\ref{eq:del}). Note that another approach to obtain the measure~(\ref{eq:BH}) is by introducing a distance metric (Bures-Hall metric) over reduced density matrices. The Bures-Hall metric is the only monotone metric that is simultaneously Fisher adjusted and Fubini-Study adjusted~\cite{Zyczkowski01,Sommers04}. It is also a function of fidelity~\cite{BZ17}, which is a key performance indicator in quantum information processing.

The above bipartite model is useful in describing the entanglement between the two subsystems of different bipartite systems, in which one subsystem represents a physical object (such as spins) and the other subsystem is the environment (such as a heat bath). The degree of entanglement of subsystems can be measured by entanglement entropies, which are functions of eigenvalues (entanglement spectrum) of the reduced density matrix. We consider the standard measure of von Neumann entropy of the subsystem~\footnote{Note that since the composite system is in a random pure state, the von Neumann entropy of the full system is zero~\cite{Page93}.}
\begin{equation}\label{eq:vN}
S=-\tr\left(\rho_{A}\ln\rho_{A}\right)=-\sum_{i=1}^{m}\lambda_{i}\ln\lambda_{i},
\end{equation}
supported in $S\in\left[0,\ln{m}\right]$, which achieves the separable state ($S=0$) when $\lambda_{1}=1$, $\lambda_{2}=\dots=\lambda_{m}=0$ and the maximally-entangled state ($S=\ln{m}$) when $\lambda_{1}=\dots=\lambda_{m}=1/m$. Statistical information of entropies is encoded through the moments. In particular, the first moment (average value) implies the typical behavior of entanglement and the second moment (variance) specifies the fluctuation around the typical value. The average von Neumann entropy, valid for any subsystem dimensions $m\leq n$, has been recently obtained as~\cite{Sarkar19,Wei20a}
\begin{equation}\label{eq:vNm}
\mathbb{E}_{f}\!\left[S\right]=\psi_{0}\left(mn-\frac{m^2}{2}+1\right)-\psi_{0}\left(n+\frac{1}{2}\right),
\end{equation}
where the expectation $\mathbb{E}_{f}\!\left[\cdot\right]$ is taken over the Bures-Hall ensemble~(\ref{eq:BH}). Here,
$\psi_{0}(x)=\dd\ln\Gamma(x)/\dd x$ is the digamma function~\cite{Prudnikov86} and for a positive integer $l$,
\begin{subequations}
\begin{eqnarray}
\psi_{0}(l)&=&-\gamma+\sum_{k=1}^{l-1}\frac{1}{k},\\
\psi_{0}\left(l+\frac{1}{2}\right)&=&-\gamma-2\ln2+2\sum_{k=0}^{l-1}\frac{1}{2k+1},
\end{eqnarray}
\end{subequations}
where $\gamma\approx0.5772$ is the Euler's constant. The main result of this work is the following formula, proved in Sec.~\ref{sec:vNv}, of the exact variance of von Neumann entropy over the Bures-Hall ensemble
\begin{equation}\label{eq:vNv}
\mathbb{V}\!_{f}\!\left[S\right]=-\psi_{1}\left(mn-\frac{m^2}{2}+1\right)+\frac{2n(2n+m)-m^{2}+1}{2n(2mn-m^2+2)}\psi_{1}\left(n+\frac{1}{2}\right),
\end{equation}
where $\psi_{1}(x)=\dd^{2}\ln\Gamma(x)/\dd x^{2}$ is the trigamma function~\cite{Prudnikov86} and for a positive integer $l$,
\begin{equation}\label{eq:tg}
\psi_{1}(l)=\frac{\pi^{2}}{6}-\sum_{k=1}^{l-1}\frac{1}{k^{2}}.
\end{equation}
Under the Bures-Hall measure, other entropies such as the quantum purity have also been studied in the literature. In particular, the first a few exact moments of quantum purity~\cite{Sommers04,Osipov10,Sarkar19,Wei20a} as well as its asymptotic distribution~\cite{Nadal11} are known~\footnote{The results~\cite{Sommers04,Osipov10} are valid for the special case $m=n$.}. Besides the Bures-Hall measure, the exact moments of von Neumann entropy~\cite{Page93,Foong94,Ruiz95,VPO16,Wei17,Wei20} and quantum purity~\cite{Lubkin78,Giraud07} have been well-investigated over the arguably less complicated~\footnote{The Hilbert-Schmidt measure corresponds to the density without the interaction terms $\prod_{1\leq i<j\leq m}(\lambda_{i}+\lambda_{j})$ in~(\ref{eq:BH}).} Hilbert-Schmidt measure~\cite{BZ17}. Finally, we note that tools from asymptotic geometric analysis have also been utilized to characterize entanglement of large dimensional quantum systems~\cite{ASY14,AS17}.

With the expressions of the mean~(\ref{eq:vNm}) and variance~(\ref{eq:vNv}), simple approximations can be constructed to understand the distribution of von Neumann entropy. For convenience, we standardize the von Neumann entropy as
\begin{equation}\label{eq:X}
X=\frac{S-\mathbb{E}_{f}\!\left[S\right]}{\sqrt{\mathbb{V}\!_{f}\!\left[S\right]}}
\end{equation}
so that the random variable $X$, supported in $X\in(-\infty,\infty)$, has zero mean and unit variance. Thus, a natural approximation to the distribution of $X$ would be a standard Gaussian distribution
\begin{equation}\label{eq:Gau}
\varphi_{X}(x)=\frac{1}{\sqrt{2\pi}}\e^{-\frac{1}{2}x^{2}},
\end{equation}
i.e., the distribution of $S$ is approximated by a Gaussian distribution with mean $\mathbb{E}_{f}\!\left[S\right]$ and variance $\mathbb{V}\!_{f}\!\left[S\right]$.
\begin{figure}[!h]
\centering
\includegraphics[width=0.7\linewidth]{p1.pdf}
\caption{Probability densities of standardized von Neumann entropy~(\ref{eq:X}) of subsystem dimensions $m=4$ and $n=6$: A comparison of the
simulated true distribution (dashed line in black) and the Gaussian approximation~(\ref{eq:Gau}) (solid line in blue).}
\label{fig:p1}
\end{figure}
In Fig.~\ref{fig:p1}, we compare the simulated true distribution of standardized von Neumann entropy~(\ref{eq:X}) to the Gaussian approximation~(\ref{eq:Gau}), where the dimensions of the subsystems are $m=4$ and $n=6$. As opposed to the Gaussian distribution, we see from Fig.~\ref{fig:p1} that the true distribution of von Neumann entropy is non-symmetric, which appears to be left-skewed (a negative skewness). With the knowledge of higher order moments, the Gaussian approximation~(\ref{eq:Gau}) can be systematically improved to provide more accurate approximations to finite-size systems. On the other hand, motivated by the case of Hilbert-Schmidt measure~\cite{Wei20}, here we also conjecture that the first two moments Eqs.~(\ref{eq:vNm}) and~(\ref{eq:vNv}) are sufficient to fully describe the distribution of von Neumann entropy as the dimensions of subsystems become large. Formally, in the limit
\begin{equation}\label{eq:lim}
m\to\infty,~~~~n\to\infty,~~~~\frac{m}{n}=c\in(0,1],
\end{equation}
\begin{figure}[!h]
\centering
\includegraphics[width=0.7\linewidth]{p2.pdf}
\caption{Probability densities of standardized von Neumann entropy~(\ref{eq:X}) of subsystem dimensions $m=16$ and $n=24$: A numerical support to the conjectured Gaussian limit. The dashed line in black and the solid line in blue represents the simulated true distribution and the standard Gaussian distribution~(\ref{eq:Gau}), respectively.}
\label{fig:p2}
\end{figure}
we conjecture that the standardized von Neumann entropy~(\ref{eq:X}) converges in distribution to a Gaussian random variable with zero mean and unit variance. Note that the high-dimensional asymptotic regime~(\ref{eq:lim}) is different from the classical asymptotic regime~\cite{Page93}, where the dimension $m$ is fixed as $n$ goes to infinity. One way to prove the above conjecture is to show that all the higher order (beyond the first two) moments of the random variable~(\ref{eq:X}) vanish in the limit~(\ref{eq:lim}). A numerical evidence to support the conjecture is provided in Fig.~\ref{fig:p2}, where we simultaneously increase the subsystem dimensions to $m=16$ and $n=24$ with their ratio $c=m/n=2/3$ kept the same as in Fig.~\ref{fig:p1}. Comparing Fig.~\ref{fig:p1} with Fig.~\ref{fig:p2}, it is seen that the distribution of von Neumann entropy approaches rather rapidly to the conjectured limiting Gaussian distribution.

The rest of the paper is organized as follows. In Sec.~\ref{sec:vNv}, we derive the main result~(\ref{eq:vNv}) on the exact variance of von Neumann entropy over the Bures-Hall measure. Specifically, in Sec.~\ref{sec:rela} we relate the computation of the variance to that over a more convenient ensemble with no delta function constraint. Calculating the corresponding variance boils down in Sec.~\ref{sec:uvar} to computing four integrals over the correlation functions of the unconstraint ensemble. In Sec.~\ref{sec:IABCD}, the four integrals are evaluated into terms involving polygamma functions by utilizing recent results on the unconstraint ensemble as well as some summation formulas of polygamma functions. We outline potential future works in Sec.~\ref{sec:con} after summarizing the main findings of the paper. The polygamma summation formulas utilized in Sec.~\ref{sec:IABCD} are listed and discussed in Appendix~\ref{app:1}.

\section{Variance Calculation}\label{sec:vNv}
\subsection{Variance relation}\label{sec:rela}
Finding moment relations is a rather standard calculation, see, e.g., Refs.~\cite{Sarkar19,Wei20a,Osipov10,Page93,Ruiz95,Wei17,Wei20}, that relates moment computation to that over an ensemble without the constraint $\delta\left(1-\sum_{i=1}^{m}\lambda_{i}\right)$. As will be seen, the unconstrained ensemble of the Bures-Hall measure~(\ref{eq:BH}) is~\cite{Sarkar19}
\begin{equation}\label{eq:BHu}
h\left(\bm{x}\right)=\frac{1}{C'}\prod_{1\leq i<j\leq m}\frac{\left(x_{i}-x_{j}\right)^{2}}{x_{i}+x_{j}}\prod_{i=1}^{m}x_{i}^{\alpha}\e^{-x_{i}},
\end{equation}
where $x_{i}\in[0,\infty)$, $i=1,\dots,m$, and the constant $C'$ depends on the constant~(\ref{eq:cBH}) as
\begin{equation}\label{eq:cBHu}
C'=C~\Gamma\left(d\right)
\end{equation}
with $d$ denoting
\begin{equation}\label{eq:d}
d=\frac{1}{2}m\left(m+2\alpha+1\right).
\end{equation}
Despite being only interested in the physically relevant $\alpha$ values in Eq.~(\ref{eq:aBH}), the results hereafter, in particular the expression~(\ref{eq:Tva}), are valid for any $\alpha>-1$ that the density~(\ref{eq:BHu}) is defined.

We first derive the density $g(\theta)$ of trace
\begin{equation}\label{eq:tr}
\theta=\sum_{i=1}^{m}x_{i},~~~~\theta\in[0,\infty),
\end{equation}
of the unconstraint ensemble~(\ref{eq:BHu}) as
\begin{eqnarray}
g(\theta)&=&\int_{\bm{x}}h(\bm{x})\delta\left(\theta-\sum_{i=1}^{m}x_{i}\right)\prod_{i=1}^{m}\dd x_{i}\\
&=&\frac{C}{C'}\e^{-\theta}\theta^{d-1}\int_{\bm{\lambda}}f(\bm{\lambda})\prod_{i=1}^{m}\dd\lambda_{i}\\
&=&\frac{1}{\Gamma\left(d\right)}\e^{-\theta}\theta^{d-1},\label{eq:g}
\end{eqnarray}
where we have employed the change of variables
\begin{equation}\label{eq:cv}
x_{i}=\theta\lambda_{i},~~~~i=1,\ldots,m.
\end{equation}
The above calculation implies that the density $h(\bm{x})$ is factored as~\footnote{A similar factorization in the Hilbert-Schmidt measure also exists~\cite{Page93}.}
\begin{equation}\label{eq:g2ft}
h(\bm{x})\prod_{i=1}^{m}\dd x_{i}=f(\bm{\lambda})g(\theta)\dd\theta\prod_{i=1}^{m}\dd\lambda_{i},
\end{equation}
which leads to the fact that $\theta$ is independent of each $\lambda_{i}$ (hence independent of $S$).

To exploit this independence in calculating the variance, we first write by the change of variables~(\ref{eq:cv}) that
\begin{equation}\label{eq:ST}
S^{2}=\theta^{-2}T^{2}+2S\ln\theta-\ln^{2}\theta,
\end{equation}
where
\begin{equation}\label{eq:TvN}
T=\sum_{i=1}^{m}x_{i}\ln x_{i}
\end{equation}
defines the induced von Neumann entropy over the unconstrained ensemble~(\ref{eq:BHu}). The second moment relation can now be found, by multiplying an appropriate constant (cf.~Eq.~(\ref{eq:g}))
\begin{equation}
1=\int_{0}^{\infty}\frac{1}{\Gamma\left(d+2\right)}\e^{-\theta}\theta^{d+1}\dd\theta
\end{equation}
as
\begin{eqnarray}
\mathbb{E}_{f}\!\left[S^{2}\right]&=&\int_{0}^{\infty}\!\!\int_{\bm{\lambda}}\frac{\e^{-\theta}\theta^{d+1}}{\Gamma\left(d+2\right)}S^{2}f(\bm{\lambda})\dd\theta\prod_{i=1}^{m}\dd\lambda_{i}\\
&=&\frac{\Gamma(d)}{\Gamma(d+2)}\mathbb{E}_{h}\!\left[T^{2}\right]+2\mathbb{E}_{f}\!\left[S\right]\mathbb{E}_{g}\!\left[\ln\theta\right]-\mathbb{E}_{g}\!\left[\ln^{2}\theta\right],\nonumber\\
&=&\frac{1}{d(d+1)}\mathbb{E}_{h}\!\left[T^{2}\right]+2\psi_{0}(d+2)\mathbb{E}_{f}\!\left[S\right]-\psi_{0}^{2}(d+2)-\psi_{1}(d+2),\label{eq:S2T}
\end{eqnarray}
where we have used the results Eqs.~(\ref{eq:g2ft}),~(\ref{eq:ST}) and the identities (valid for $\Re(a)>0$)
\begin{subequations}
\begin{eqnarray}
\int_{0}^{\infty}\!\!\e^{-\theta}\theta^{a-1}\ln{\theta}\dd\theta&=&\Gamma(a)\psi_{0}(a),\\
\int_{0}^{\infty}\!\!\e^{-\theta}\theta^{a-1}\ln^{2}{\theta}\dd\theta&=&\Gamma(a)\left(\psi_{0}^{2}(a)+\psi_{1}(a)\right).
\end{eqnarray}
\end{subequations}
By the known mean formulas~(\ref{eq:vNm}) and~\cite{Wei20a}
\begin{equation}\label{eq:Tm}
\mathbb{E}_{h}\!\left[T\right]=\frac{m(m+2\alpha+1)}{2}\psi_{0}(m+\alpha+1),
\end{equation}
the derived moment relation~(\ref{eq:S2T}) translates showing the claimed variance formula~(\ref{eq:vNv}) to proving an induced variance formula
\begin{flalign}\label{eq:Tv}
\mathbb{V}\!_{h}\!\left[T\right]=m(2n-m)\Bigg(\psi_{0}\left(n+\frac{1}{2}\right)+\frac{1}{2}\psi_{0}^{2}\left(n+\frac{1}{2}\right)+\frac{4n^{2}+2mn-m^{2}+1}{8n}\psi_{1}\left(n+\frac{1}{2}\right)\!\Bigg),&&
\end{flalign}
where we have also used the fact $\mathbb{V}\!\left[X\right]=\mathbb{E}\!\left[X^{2}\right]-\mathbb{E}^{2}\!\left[X\right]$ and the identities
\begin{subequations}\label{eq:pgre}
\begin{eqnarray}
\psi_{0}(l+n)&=&\psi_{0}(l)+\sum_{k=0}^{n-1}\frac{1}{l+k},\label{eq:dgre}\\
\psi_{1}(l+n)&=&\psi_{1}(l)-\sum_{k=0}^{n-1}\frac{1}{(l+k)^2}.\label{eq:tgre}
\end{eqnarray}
\end{subequations}

\subsection{Variance of unconstraint ensemble}\label{sec:uvar}
Calculating $\mathbb{E}_{h}\!\left[T^{2}\right]$ requires one and two arbitrary eigenvalue densities, denoted respectively by $h_{1}(x)$ and $h_{2}(x,y)$, of the unconstraint Bures-Hall ensemble~(\ref{eq:BHu}) as
\begin{equation}\label{eq:i2m}
\mathbb{E}_{h}\!\left[T^{2}\right]=m\int_{0}^{\infty}\!\!x^{2}\ln^{2}x~h_{1}(x)\dd x+m(m-1)\int_{0}^{\infty}\!\!\int_{0}^{\infty}\!\!xy\ln{x}\ln{y}~h_{2}\left(x,y\right)\dd x\dd y.
\end{equation}
In general, the density of $k$ arbitrary eigenvalues ($k$-point correlation function) of the ensemble~(\ref{eq:BHu}) is described by a Pfaffian point process of a $2k\times 2k$ anti-symmetric matrix~\cite{FK16}. The corresponding correlation kernels are written in terms of these of the Cauchy-Laguerre biorthogonal ensemble~\cite{Bertola14}, which is a determinantal point process. In particular, the needed eigenvalue densities in Eq.~(\ref{eq:i2m}) are written as~\cite{FK16,Bertola14}
\begin{eqnarray}
h_{1}(x)&=&\frac{1}{2m}\left(K_{01}(x,x)+K_{10}(x,x)\right),\label{eq:h1} \\
h_{2}(x,y)&=&\frac{1}{4m(m-1)}(\left(K_{01}(x,x)+K_{10}(x,x)\right)\left(K_{01}(y,y)+K_{10}(y,y)\right)-2K_{01}(x,y)K_{01}(y,x) \nonumber\\
&&-2K_{10}(x,y)K_{10}(y,x)-2K_{00}(x,y)K_{11}(x,y)-2K_{00}(y,x)K_{11}(y,x)),\label{eq:h2}
\end{eqnarray}
where the correlation kernels are
\begin{subequations}\label{eq:ker}
\begin{eqnarray}
K_{00}(x,y)&=&\sum_{k=0}^{m-1}p_{k}(x)q_{k}(y), \label{eq:K00} \\
K_{01}(x,y)&=&-x^{2\alpha+1}y^{-\alpha-1}e^{-y}\sum_{k=0}^{m-1}p_{k}(x)Q_{k}(-y), \label{eq:K01} \\
K_{10}(x,y)&=&-x^{-\alpha}y^{2\alpha+1}e^{-x}\sum_{k=0}^{m-1}P_{k}(-x)q_{k}(y), \label{eq:K10} \\
K_{11}(x,y)&=&x^{\alpha}y^{\alpha+1}e^{-x-y}\sum_{k=0}^{m-1}P_{k}(-y)Q_{k}(-x)-w(x,y) \label{eq:K11}
\end{eqnarray}
\end{subequations}
with the weight function $w(x,y)$ of the biorthogonal polynomials $p_{k}(x)$, $q_{l}(y)$,
\begin{equation}\label{eq:oc}
\int_{0}^{\infty}\!\!\int_{0}^{\infty}p_{k}(x)q_{l}(y)w(x,y)\dd x\dd y=\delta_{kl}
\end{equation}
given by
\begin{equation}\label{eq:w}
w(x,y)=\frac{x^{\alpha}y^{\alpha+1}e^{-x-y}}{x+y}.
\end{equation}
The functions in Eq.~(\ref{eq:ker}) are further related by~\cite{FK16,Bertola14}
\begin{subequations}\label{eq:PQ}
\begin{eqnarray}
P_{k}(x)&=&\int_{0}^{\infty}\frac{v^{\alpha}\e^{-v}}{x-v}p_{k}(v)\dd v, \\
Q_{k}(y)&=&\int_{0}^{\infty}\frac{w^{\alpha+1}\e^{-w}}{y-w}q_{k}(w)\dd w,
\end{eqnarray}
\end{subequations}
which, together with the orthogonality condition~(\ref{eq:oc}), can be employed to verify that the functions~(\ref{eq:h1}) and~(\ref{eq:h2}) are indeed probability density functions. Moreover, these function are expressed explicitly via Meijer G-functions as~\cite{FK16,Bertola14}
\begin{subequations}\label{eq:kerM}
\begin{eqnarray*}
p_{j}(x)&=&\sqrt{2}(-1)^{j}G_{2,3}^{1,1}\left(\!\begin{array}{c}-2\alpha-1-j;j+1\\0;-\alpha,-2\alpha-1\end{array}\bigg|x\Big.\right),\\
q_{j}(x)&=&\sqrt{2}(-1)^{j}(j+\alpha+1)G_{2,3}^{1,1}\left(\!\begin{array}{c}-2\alpha-1-j;j+1\\0;-\alpha-1,-2\alpha-1\end{array}\bigg|x\Big.\right),\\
P_{j}(x)&=&\sqrt{2}(-1)^{j+1}\e^{-x}G_{2,3}^{2,1}\left(\!\begin{array}{c}-\alpha-j-1;\alpha+j+1\\0,\alpha;-\alpha-1\end{array}\bigg|-\!x\Big.\right),\\
Q_{j}(x)&=&\sqrt{2}(-1)^{j+1}(j+\alpha+1)\e^{-x}G_{2,3}^{2,1}\left(\!\begin{array}{c}-\alpha-j;\alpha+j+2\\0,\alpha+1;-\alpha\end{array}\bigg|-\!x\Big.\right),
\end{eqnarray*}
\end{subequations}
where the Meijer G-function is defined by the contour integral~\cite{Prudnikov86}
\begin{flalign}\label{eq:MG}
G_{p,q}^{m,n}\left(\begin{array}{c} a_{1},\ldots,a_{n}; a_{n+1},\ldots,a_{p} \\ b_{1},\ldots,b_{m}; b_{m+1},\ldots,b_{q} \end{array}\bigg|x\Big.\right)=\frac{1}{2\pi\imath}\int_{\mathcal{L}}{\frac{\prod_{j=1}^m\Gamma\left(b_j+s\right)\prod_{j=1}^n\Gamma\left(1-a_j-s\right)x^{-s}}{\prod_{j=n+1}^p \Gamma\left(a_{j}+s\right)\prod_{j=m+1}^q\Gamma\left(1-b_j-s\right)}}\dd s &&
\end{flalign}
with the contour $\mathcal{L}$ separating the poles of $\Gamma\left(1-a_j-s\right)$ from the poles of $\Gamma\left(b_j+s\right)$. Beside the summation form~(\ref{eq:ker}), the kernels also admit integral representation~\cite{FK16,Bertola14}
\begin{subequations}\label{eq:kerI}
\begin{eqnarray}
K_{00}(x,y)&=&\int_{0}^{1}t^{2\alpha+1}H_{\alpha}(tx)H_{\alpha+1}(ty)\dd t, \label{eq:K00I} \\
K_{01}(x,y)&=&x^{2\alpha+1}\int_{0}^{1}t^{2\alpha+1}H_{\alpha}(tx)G_{\alpha+1}(ty)\dd t, \label{eq:K01I} \\
K_{10}(x,y)&=&y^{2\alpha+1}\int_{0}^{1}t^{2\alpha+1}G_{\alpha}(tx)H_{\alpha+1}(ty)\dd t, \label{eq:K10I} \\
K_{11}(x,y)&=&(xy)^{2\alpha+1}\int_{0}^{1}t^{2\alpha+1}G_{\alpha+1}(tx)G_{\alpha}(ty)\dd t-\frac{x^{\alpha}y^{\alpha+1}}{x+y}, \label{eq:K11I}
\end{eqnarray}
\end{subequations}
where we denote
\begin{subequations}
\begin{eqnarray}
H_{q}(x)&=&G_{2,3}^{1,1}\left(\!\begin{array}{c}-m-2\alpha-1;m\\0;-q,-2\alpha-1\end{array}\bigg|x\Big.\right),\\
G_{q}(x)&=&G_{2,3}^{2,1}\left(\!\begin{array}{c}-m-2\alpha-1;m\\0,-q;-2\alpha-1\end{array}\bigg|x\Big.\right).
\end{eqnarray}
\end{subequations}

Finally, inserting Eqs.~(\ref{eq:h1}) and~(\ref{eq:h2}) into Eq.~(\ref{eq:i2m}), the induced variance is represented as
\begin{equation}\label{eq:Tvi}
\mathbb{V}\!_{h}\!\left[T\right]=\frac{1}{2}\left(I_{A}-I_{B}-I_{C}+2I_{D}\right),
\end{equation}
where
\begin{eqnarray}
\!\!\!\!\!\!\!\!I_{A}&=&\int_{0}^{\infty}x^{2}\ln^{2}x\left(K_{01}(x,x)+K_{10}(x,x)\right)\dd x, \label{eq:IA} \\
\!\!\!\!\!\!\!\!I_{B}&=&\int_{0}^{\infty}\!\!\int_{0}^{\infty}xy\ln x\ln y~K_{01}(x,y)K_{01}(y,x)\dd x\dd y, \label{eq:IB} \\
\!\!\!\!\!\!\!\!I_{C}&=&\int_{0}^{\infty}\!\!\int_{0}^{\infty}xy\ln x\ln y~K_{10}(x,y)K_{10}(y,x)\dd x\dd y, \label{eq:IC} \\
\!\!\!\!\!\!\!\!I_{D}&=&\int_{0}^{\infty}\!\!\int_{0}^{\infty}xy\ln x\ln y~K_{00}(x,y)K_{11}(x,y)\dd x\dd y, \label{eq:ID}
\end{eqnarray}
and we have used the fact (cf.~Eq.~(\ref{eq:Tm}))
\begin{equation}
\int_{0}^{\infty}x\ln x\left(K_{01}(x,x)+K_{10}(x,x)\right)\dd x=2\mathbb{E}_{h}\!\left[T\right].
\end{equation}
To show Eq.~(\ref{eq:Tv}), the remaining task is to compute the four integrals~(\ref{eq:IA})--(\ref{eq:ID}).

\subsection{Computing the integrals $I_{A}$, $I_{B}$, $I_{C}$, and $I_{D}$}\label{sec:IABCD}
\subsubsection{Computation of $I_{A}$}
The evaluation of $I_{A}$ follows a similar procedure as in Ref.~\cite[Sec.~2.2]{Wei20a}. The key is to compute the integral
\begin{equation}\label{eq:Aq}
A_{q}(t)=\int_{0}^{\infty}x^{\beta}(tx)^{2\alpha+1}H_{2\alpha+1-q}(tx)G_{q}(tx)\dd x
\end{equation}
and its derivatives with respect to $\beta$ for $q=\alpha, \alpha+1$. This integral has been obtained in Ref.~\cite{Wei20a} by using the Mellin transform of Meijer G-function~\cite{Prudnikov86}
\begin{flalign}\label{eq:iMG}
\int_{0}^{\infty}x^{s-1}G_{p,q}^{m,n}\left(\begin{array}{c} a_{1},\ldots,a_{n}; a_{n+1},\ldots,a_{p} \\ b_{1},\ldots,b_{m}; b_{m+1},\ldots,b_{q} \end{array}\bigg|\eta x\Big.\right)\dd x=\frac{\eta^{-s}\prod_{j=1}^m\Gamma\left(b_j+s\right)\prod_{j=1}^{n}\Gamma\left(1-a_j-s\right)}{\prod_{j=n+1}^{p}\Gamma\left(a_{j}+s\right)\prod_{j=m+1}^q\Gamma\left(1-b_j-s\right)}&&
\end{flalign}
and the fact that the Meijer G-function $G_{2,3}^{1,1}$ of a negative parameter $a_{i}$ ($i\leq n$) is a terminating hypergeometric function~\cite{Bertola14,Wei20a}~\footnote{This fact is also the starting point in computing $I_{B}$, $I_{C}$, and $I_{D}$.} as
\begin{equation}\label{eq:Ibs0}
A_{q}(t)=t^{-\beta-1}A_{q},
\end{equation}
where
\begin{flalign}\label{eq:Ibs}
A_{q}=\sum_{k=0}^{m-1}\frac{(-1)^{k+m}\Gamma(k+2\alpha+m+2)\Gamma(k+\beta+1)}{\Gamma(k+2\alpha+2)\Gamma(k+2\alpha+2-q)\Gamma(m-k)k!}\frac{\Gamma(k+\beta+2\alpha+2)\Gamma(k+\beta+2\alpha+2-q)}{\Gamma(k+\beta+2\alpha+m+2)\Gamma(k+\beta-m+1)}.&&
\end{flalign}
Using a different representation of Eqs.~(\ref{eq:K01I}) and~(\ref{eq:K10I}) obtained in Ref.~\cite{Wei20a},
\begin{subequations}
\begin{eqnarray*}
K_{01}(x,y)&=&-x^{2\alpha+1}\int_{1}^{\infty}t^{2\alpha+1}H_{\alpha}(tx)G_{\alpha+1}(ty)\dd t,\\
K_{10}(x,y)&=&-y^{2\alpha+1}\int_{1}^{\infty}t^{2\alpha+1}G_{\alpha}(tx)H_{\alpha+1}(ty)\dd t, \\
\end{eqnarray*}
\end{subequations}
and changing the order of integrations, $I_{A}$ is calculated as
\begin{equation}\label{eq:IAi}
I_{A}=-\frac{1}{4}\left(A_{\alpha}+A_{\alpha+1}\right)+\frac{1}{2}\left(H_{\alpha}^{(1)}+H_{\alpha+1}^{(1)}\right)-\frac{1}{2}\left(H_{\alpha}^{(2)}+H_{\alpha+1}^{(2)}\right),
\end{equation}
where we denote
\begin{equation}\label{eq:H12}
H_{q}^{(1)}=\frac{\dd}{\dd\beta}A_{q}^{(\beta)}\Big|_{\beta=2},~~~~~~H_{q}^{(2)}=\frac{\dd}{\dd\beta^{2}}A_{q}^{(\beta)}\Big|_{\beta=2},
\end{equation}
and the integrals over $t$ have been evaluated first by the fact
\begin{equation*}
\int_{1}^{\infty}\frac{1}{t^{3}}\dd t=\frac{1}{2},~~~~~~\int_{1}^{\infty}\frac{\ln t}{t^{3}}\dd t=\frac{1}{4},~~~~~~\int_{1}^{\infty}\frac{\ln^{2} t}{t^{3}}\dd t=\frac{1}{4}.
\end{equation*}
The first term $A_{\alpha}+A_{\alpha+1}$ in Eq.~(\ref{eq:IAi}) has been obtained in Ref.~\cite[Eq.~(50)]{Wei20a}. By resolving indeterminacy in the limit $\epsilon\to0$,
\begin{subequations}\label{eq:pgna}
\begin{eqnarray}
\Gamma(-l+\epsilon)&=&\frac{(-1)^{l}}{l!\epsilon}\left(1+\psi_{0}(l+1)\epsilon+o\left(\epsilon^2\right)\right),\label{eq:pgna1}\\
\psi_{0}(-l+\epsilon)&=&-\frac{1}{\epsilon}\left(1-\psi_{0}(l+1)\epsilon+o\left(\epsilon^2\right)\right),\label{eq:pgna2}\\
\psi_{1}(-l+\epsilon)&=&\frac{1}{\epsilon^2}\left(1+o\left(\epsilon^2\right)\right),\label{eq:pgna3}
\end{eqnarray}
\end{subequations}
the terms~(\ref{eq:H12}) are evaluated into finite sums involving polygamma functions. Computing these summations by the identities in Appendix~\ref{app:1}, we obtain $I_{A}$ as shown in Eq.~(\ref{eq:IAf}), where the list of coefficients can be found in Table~\ref{t:IA} of Appendix~\ref{app:2}. Note that as a result of employing the semi-closed-form identities~(\ref{eq:A6})--(\ref{eq:A8}), the obtained $I_{A}$ expression~(\ref{eq:IAf}) still contains five summations that may not be further simplified. These unsimplifiable sums eventually cancel with the ones in $I_{B}$ and $I_{C}$ as will be seen. Similar phenomena have also been observed in the higher order moment computations over the Hilbert-Schmidt measure~\cite{Wei17,Wei20}.
\begin{eqnarray}\label{eq:IAf}
I_{A}&=&\frac{1}{36\alpha(m+\alpha)(m+2\alpha)(2m+2\alpha+1)^3}\Bigg(\!\!-2a_{0}\Bigg(\sum_{k=1}^{m}\frac{\psi_{0}(k+\alpha)}{k}+\sum_{k=1}^{m}\frac{\psi_{0}(k+2\alpha)}{k}\nonumber\\
&&-\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k}+\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k+\alpha}+\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k+2\alpha}\Bigg)+a_{1}+a_{2}(\psi_{0}(1)\nonumber\\
&&-\psi_{0}(m+1))+a_{3}\psi_{0}(\alpha+1)+a_{4}\psi_{0}(2\alpha+1)+a_{5}\psi_{0}(m+\alpha+1)+a_{6}\psi_{0}(m+2\alpha+1)\nonumber\\
&&+a_{7}\psi_{0}(2m+2\alpha+1)+a_{0}\Big(\!-2\psi_{0}(1)\psi_{0}(\alpha+1)-2\psi_{0}(1)\psi_{0}(2\alpha+1)+2\psi_{0}(1)\nonumber\\
&&\times\psi_{0}(m+2\alpha+1)+\psi_{0}^{2}(\alpha+1)+\psi_{0}^{2}(2\alpha+1)+2\psi_{0}(\alpha+1)\psi_{0}(m+1)-2\psi_{0}(\alpha+1)\nonumber\\
&&\times\psi_{0}(m+\alpha+1)-2\psi_{0}(\alpha+1)\psi_{0}(m+2\alpha+1)+2\psi_{0}(2\alpha+1)\psi_{0}(m+1)-4\psi_{0}(2\alpha+1)\nonumber\\
&&\times\psi_{0}(m+2\alpha+1)-2\psi_{0}(m+1)\psi_{0}(m+2\alpha+1)-2\psi_{0}(m+\alpha+1)\psi_{0}(m+2\alpha+1)\nonumber\\
&&+4\psi_{0}(m+\alpha+1)\psi_{0}(2m+2\alpha+1)-\psi_{0}^{2}(m+2\alpha+1)+8\psi_{0}(m+2\alpha+1)\psi_{0}(2m\nonumber\\
&&+2\alpha+1)-4\psi_{0}^{2}(2m+2\alpha+1)-\psi_{1}(\alpha+1)-\psi_{1}(2\alpha+1)+\psi_{1}(m+2\alpha+1)\Big)\!\Bigg).
\end{eqnarray}

\subsubsection{Computation of $I_{B}$ and $I_{C}$}
The steps in calculating $I_{B}$ and $I_{C}$ are identical. The starting point is the integral form of the kernels~(\ref{eq:K01I}),~(\ref{eq:K10I}) as well as finite sum representation~\cite{Bertola14,Wei20a} of the Meijer G-functions $G_{2,3}^{1,1}$. Instead of changing the order of summations as in $I_{A}$, here we directly evaluate the integrals over $t$ by the identity~\cite{Prudnikov86}
\begin{eqnarray}
&&\int_{0}^{1}\!x^{a-1}G_{p,q}^{m,n}\left(\begin{array}{c} a_{1},\ldots,a_{n}; a_{n+1},\ldots,a_{p} \\ b_{1},\ldots,b_{m}; b_{m+1},\ldots,b_{q} \end{array}\bigg|\eta x\Big.\right)\dd x \nonumber\\
&=&G_{p+1,q+1}^{m,n+1}\left(\begin{array}{c}1-a,a_{1},\ldots,a_{n};a_{n+1},\ldots,a_{p}\\b_{1},\ldots,b_{m};b_{m+1},\ldots,b_{q},-a\end{array}\bigg|\eta\Big.\right).
\end{eqnarray}
This leads $I_{B}$ to
\begin{equation}
I_{B}=\sum_{j,k=0}^{m-1}f_{j,k}f_{k,j},
\end{equation}
where we denote
\begin{eqnarray*}
&&f_{j,k}=\frac{(-1)^{j}\Gamma(m+2\alpha+j+2)}{\Gamma(j+1)\Gamma(\alpha+j+1)\Gamma(2\alpha+j+2)\Gamma(m-j)}\nonumber\\
&&\times\int_{0}^{\infty}\!x\ln x~G_{3,4}^{2,2}\left(\!\begin{array}{c}j-k,j-m;m+2\alpha+j+1\\2\alpha+j+1,\alpha+j;j,j-k-1\end{array}\bigg|x\Big.\right)\dd x.
\end{eqnarray*}
The above integral can be similarly evaluated as in Eq.~(\ref{eq:Aq}) by first utilizing Eq.~(\ref{eq:iMG}) before taking the derivative with respect to $\beta$. We then set $\beta=1$ and resolve the resulting indeterminacy by Eq.~(\ref{eq:pgna}), $I_{B}$ becomes a double sum involving polygamma functions. The summations are evaluated with the help of the identities in Appendix~\ref{app:1}, which completes the calculation of $I_{B}$. Since $I_{C}$ is computed to the same form as $I_{B}$, for convenience we provide the corresponding result of $I_{B}+I_{C}$ as shown in Eq.~(\ref{eq:IBCf}), where the coefficients can be found in Table~\ref{t:IBC}.

\begin{eqnarray}\label{eq:IBCf}
I_{B}+I_{C}&=&\frac{1}{36\alpha(m+\alpha)(m+\alpha+1)(m+2\alpha)(2m+2\alpha+1)^4}\Bigg(2\left(b_{0}+c_{0}\right)\Bigg(\sum_{k=1}^{m}\frac{\psi_{0}(k+\alpha)}{k}\nonumber\\
&&+\sum_{k=1}^{m}\frac{\psi_{0}(k+2\alpha)}{k}-\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k}+\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k+\alpha}\nonumber\\
&&+\sum_{k=1}^{m}\frac{\psi_{0}(k+m+2\alpha)}{k+2\alpha}\Bigg)+b_{1}+c_{1}+\left(b_{2}+c_{2}\right)\left(\psi_{0}(1)-\psi_{0}(m+1)\right)+(b_{3}+c_{3})\nonumber\\
&&\times\psi_{0}(\alpha+1)+(b_{4}+c_{4})\psi_{0}(2\alpha+1)+(b_{5}+c_{5})\psi_{0}(m+\alpha+1)+(b_{6}+c_{6})\nonumber\\
&&\times\psi_{0}(m+2\alpha+1)+(b_{7}+c_{7})\psi_{0}(2m+2\alpha+1)+(b_{0}+c_{0})\Big(2\psi_{0}(1)\psi_{0}(\alpha+1)\nonumber\\
&&+2\psi_{0}(1)\psi_{0}(2\alpha+1)-2\psi_{0}(1)\psi_{0}(m+2\alpha+1)-\psi_{0}^{2}(\alpha+1)-\psi_{0}^{2}(2\alpha+1)\nonumber\\
&&-2\psi_{0}(\alpha+1)\psi_{0}(m+1)+2\psi_{0}(\alpha+1)\psi_{0}(m+\alpha+1)+2\psi_{0}(\alpha+1)\psi_{0}(m+2\alpha+1)\nonumber\\
&&-2\psi_{0}(2\alpha+1)\psi_{0}(m+1)+4\psi_{0}(2\alpha+1)\psi_{0}(m+2\alpha+1)+2\psi_{0}(m+1)\nonumber\\
&&\times\psi_{0}(m+2\alpha+1)+\psi_{1}(\alpha+1)+\psi_{1}(2\alpha+1)-\psi_{1}(m+\alpha+1)-3\psi_{1}(m+2\alpha+1)\nonumber\\
&&+2\psi_{1}(2m+2\alpha+1)\Big)+(b_{8}+c_{8})\psi_{0}^{2}(m+\alpha+1)+(b_{9}+c_{9})\psi_{0}(m+\alpha+1)\nonumber\\
&&\times\psi_{0}(m+2\alpha+1)+(b_{10}+c_{10})\Big(\psi_{0}(m+\alpha+1)\psi_{0}(2m+2\alpha+1)+2\psi_{0}(m+2\alpha\nonumber\\
&&+1)\psi_{0}(2m+2\alpha+1)-\psi_{0}^{2}(2m+2\alpha+1)\Big)+(b_{11}+c_{11})\psi_{0}^{2}(m+2\alpha+1)\Bigg).
\end{eqnarray}

\subsubsection{Computation of $I_{D}$}
We define the integral
\begin{equation}\label{eq:IDg}
D\left(\beta_{1},\beta_{2}\right)=\int_{0}^{\infty}\!\!\int_{0}^{\infty}x^{\beta_{1}}y^{\beta_{2}}~K_{00}(x,y)K_{11}(x,y)\dd x\dd y
\end{equation}
so that the desired $I_{D}$ integral~(\ref{eq:ID}) can be obtained as
\begin{equation}\label{eq:IDr}
I_{D}=\frac{\partial^{2}}{\partial\beta_{1}\partial\beta_{2}}D\left(\beta_{1},\beta_{2}\right)\Big|_{\beta_{1}=1,\beta_{2}=1}.
\end{equation}
To compute the integral~(\ref{eq:IDg}), one uses the summation form of the kernels~(\ref{eq:ker}) instead of the integral representation~(\ref{eq:kerI}). The corresponding integrals over $x$ and $y$ can then be separately evaluated by the formula~(\ref{eq:iMG}) and explicit expressions of the polynomials $p_{k}(x)$ and $q_{k}(y)$ in Ref.~\cite{Bertola14}. Now taking the partial derivatives~(\ref{eq:IDr}) gives
\begin{equation}
I_{D}=I_{D1}-I_{D2},
\end{equation}
where
\begin{equation}\label{eq:ID1}
I_{D1}=\lim_{\beta\to1}\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^{j}\sum_{s=0}^j(j+\alpha+1)(k+\alpha+1)~g_{\alpha,i}~g_{\alpha+1,s},
\end{equation}
\begin{eqnarray}\label{eq:ID2}
I_{D2}&=&\sum_{j=0}^{m-1}\sum_{i=0}^{j}\sum_{s=0}^{j}\frac{2(i+\alpha+1)(s+\alpha+2)h_{i}h_{s}}{(j+\alpha+1)^{-1}(i+s+2\alpha+4)}\bigg(\!\psi_{0}(i+\alpha+2)\psi_{0}(s+\alpha+3)\nonumber\\
&&+\frac{2}{(i+s+2\alpha+4)^2}-\frac{\psi_{0}(i+\alpha+2)+\psi_{0}(s+\alpha+3)}{i+s+2\alpha+4}\bigg)
\end{eqnarray}
with the shorthand notations
\begin{eqnarray}
h_{r}&=&\frac{(-1)^{r}\Gamma(r+j+2\alpha+2)}{\Gamma(r+1)\Gamma(j-r+1)\Gamma(r+2\alpha+2)},\nonumber\\
g_{p,r}&=&\frac{2h_{r}\Gamma(r+\beta+1)\Gamma(p+r+\beta+1)\Gamma(r+2\alpha+\beta+2)}{\Gamma(r-k+\beta+1)\Gamma(r+k+2\alpha+\beta+3)\Gamma(p+r+1)}(\psi_{0}(p+r+\beta+1)\nonumber\\
&&+\psi_{0}(r+2\alpha+\beta+2)+\psi_{0}(r+\beta+1)-\psi_{0}(r-k+\beta+1)-\psi_{0}(r+k+2\alpha+\beta+3)). \nonumber
\end{eqnarray}
For the $I_{D1}$ sums~(\ref{eq:ID1}), the summation over $j$ is evaluated first by the identity~\cite[Lemma~4.1]{Bertola14}
\begin{flalign}\label{eq:L41}
\sum_{j=i}^{m-1}(j+\alpha+1)\frac{\Gamma(j+i+2\alpha+2)\Gamma(j+s+2\alpha+2)}{\Gamma(j-i+1)\Gamma(j-s+1)}=\frac{\Gamma(i+m+2\alpha+2)\Gamma(s+m+2\alpha+2)}{2(i+s+2\alpha+2)\Gamma(m-i)\Gamma(m-s)}.&&
\end{flalign}
After determining the limits when $\beta\to1$, the summation over $k$ is evaluated next by the identity
\begin{flalign}\label{eq:3t2}
\sum_{k=0}^{m}(k+\alpha+1)\frac{\left(\Gamma(s-k+2)\Gamma(k+s+2\alpha+4)\right)^{-1}}{\Gamma(i-k+2)\Gamma(k+i+2\alpha+4)}=\frac{\left(\Gamma(i+2\alpha+3)\Gamma(s+2\alpha+3)\right)^{-1}}{2\Gamma(i+2)\Gamma(s+2)(i+s+2\alpha+4)}&&
\end{flalign}
as well as three additional identities obtained by taking derivatives of Eq.~(\ref{eq:3t2}) with respect to $i$, $s$, and both $i$ and $s$. Now the $I_{D1}$ quadruple sum~(\ref{eq:ID1}) reduces to double sums in $i$ and $s$. Similarly, for the $I_{D2}$ sums~(\ref{eq:ID2}) we evaluate the summation over $j$ first by using Eq.~(\ref{eq:L41}), which also leads to a double sum form for $I_{D2}$. We observe substantial cancellations among the obtained double sums of $I_{D1}$ and $I_{D2}$. With the remaining sums evaluated by the formulas in Appendix~\ref{app:1}, we arrive at a closed-form expression of $I_{D}$ as shown in~(\ref{eq:IDf}), where the coefficients are listed in Table~\ref{t:ID}.
\begin{eqnarray}\label{eq:IDf}
\!\!\!\!\!\!\!\!\!I_{D}&=&\frac{m}{8(2m+2\alpha+1)^4}\Big(d_0+d_{1}\psi_{0}(m+\alpha+1)+d_{2}\psi_{0}(m+2\alpha+1)+d_{3}\psi_{0}(2m+2\alpha+1)\nonumber\\
&&+d_{4}\big(\psi_{0}(m+2\alpha+1)-\psi_{0}(2m+2\alpha+1)\big)\big(\psi_{0}(m+2\alpha+1)-\psi_{0}(2m+2\alpha+1)\nonumber\\
&&+\psi_{0}(m+\alpha+1)\big)+d_{5}\psi_{0}^{2}(m+\alpha+1)+d_{6}\big(\psi_{1}(m+2\alpha+1)-\psi_{1}(2m+2\alpha+1)\big)\Big).
\end{eqnarray}

Finally, inserting Eqs.~(\ref{eq:IAf}),~(\ref{eq:IBCf}),~(\ref{eq:IDf}) into Eq.~(\ref{eq:Tvi}), one observes cancellations of all but three terms
\begin{eqnarray}\label{eq:Tva}
\mathbb{V}\!_{h}\!\left[T\right]&=&m(m+2\alpha+1)\psi_{0}(m+\alpha+1)+\frac{m(m+2\alpha+1)}{2}\psi_{0}^{2}(m+\alpha+1)\nonumber\\
&&+\frac{m(m+2\alpha+1)}{4(2m+2\alpha+1)}\big(5m^2+5m+10\alpha m+4\alpha^2+4\alpha+2\big)\psi_{1}(m+\alpha+1).
\end{eqnarray}
Upon specializing the above expression with the $\alpha$ value in Eq.~(\ref{eq:aBH}) establishes the induced variance formula~(\ref{eq:Tv}). This completes the proof of the main result~(\ref{eq:vNv}).

\section{Summary and outlook}\label{sec:con}
As an important step towards quantifying the statistical performance of bipartite systems, we derived the exact variance of von Neumann entanglement entropy over the Bures-Hall measure in this work. The result is based on recent progress in understanding the correlation functions of the Bures-Hall random matrix ensemble.

Although the Bures-Hall ensemble attains a more involved functional form, the expressions of its first two moments turn out simpler than the ones over the Hilbert-Schmidt ensemble. Further understanding of this counterintuitive fact requires the higher order moments of von Neumann entropy of both ensembles. The results may as well help prove the conjectured Gaussian limit for large dimensional quantum systems. Future work also includes the study of other performance indicators relevant for quantum information processing, such as the fidelity, over the Bures-Hall measure.

\begin{acknowledgments}
The author thanks Shi-Hao Li and Peter Forrester for correspondence. The author also wishes to thank Santosh Kumar for providing the simulation codes.
\end{acknowledgments}

\appendix

\section{List of summation identities}\label{app:1}
In this Appendix, we list the closed-form~(\ref{eq:A1})--(\ref{eq:A5}) and semi-closed-from~(\ref{eq:A6})--(\ref{eq:A8}) finite sum identities useful in simplifying the summations in Sec.~\ref{sec:IABCD}. The identities~(\ref{eq:A1})--(\ref{eq:A3}) and~(\ref{eq:A4})--(\ref{eq:A8}) can be found in Ref.~\cite{Wei20} and Ref.~\cite{Milgram17}, respectively. Note that it is sufficient to assume $a,b\ge0, a\neq b$ in Eqs.~(\ref{eq:A1})--(\ref{eq:A4}),~(\ref{eq:A6}) and $a>m$ in Eqs.~(\ref{eq:A7}),~(\ref{eq:A8}).

\begin{flalign}\label{eq:A1}
\sum_{k=1}^{m}\psi_{0}(k+a)=(m+a)\psi_{0}(m+a+1)-a\psi_{0}(a+1)-m &&
\end{flalign}
\begin{eqnarray}\label{eq:A2}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{k=1}^{m}k\psi_{0}(k+a)&=&\frac{1}{2}\left(m^2+m-a^2+a\right)\psi_{0}(m+a+1)+\frac{1}{2}(a-1)a\psi_{0}(a+1)+\frac{1}{4}m\nonumber\\
&&\times(2a-m-3)
\end{eqnarray}
\begin{eqnarray}\label{eq:A3}
\!\!\sum_{k=1}^{m}k^{2}\psi_{0}(k+a)&=&\frac{1}{6}\left(2m^3+3m^2+m+2a^3-3a^2+a\right)\psi_{0}(m+a+1)-\frac{1}{6}a\left(2a^2-3a+1\right)\nonumber\\
&&\times\psi_{0}(a+1)-\frac{1}{36}m\left(4m^2+15m-6ma+12a^2-24a+17\right)
\end{eqnarray}
\begin{flalign}\label{eq:A4}
\sum_{k=1}^{m}\frac{\psi_{0}(k+a)}{k+a}=\frac{1}{2}\left(-\psi_{0}^{2}(a+1)+\psi_{0}^{2}(m+a+1)-\psi_{1}(a+1)+\psi_{1}(m+a+1)\right) &&
\end{flalign}
\begin{flalign}\label{eq:A5}
\sum_{k=1}^{m}\frac{\psi_{0}(m+1-k)}{k}=-\psi_{0}(1)\psi_{0}(m+1)+\psi_{0}^{2}(m+1)-\psi_{1}(1)+\psi_{1}(m+1) &&
\end{flalign}
\begin{flalign}\label{eq:A6}
\sum_{k=1}^{m}\frac{\psi_{0}(k+b)}{k+a}=&-\sum_{k=1}^{m}\frac{\psi_{0}(k+a)}{k+b}+\psi_{0}(m+a+1)\psi_{0}(m+b+1)-\psi_{0}(a+1)\psi_{0}(b+1)\nonumber\\
&+\frac{1}{a-b}(\psi_{0}(m+a+1)-\psi_{0}(m+b+1)-\psi_{0}(a+1)+\psi_{0}(b+1)) &&
\end{flalign}
\begin{eqnarray}\label{eq:A7}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{k=1}^{m}\frac{\psi_{0}(k)}{a+1-k}&=&\sum_{k=1}^{m}\frac{\psi_{0}(k)}{k+a-m}+\frac{1}{2}\left(\psi_{0}(a-m+1)-\psi_{0}(a+1)\right)^{2}-\frac{1}{2}(\psi_{1}(a-m+1)\nonumber\\
&&-\psi_{1}(a+1))
\end{eqnarray}
\begin{flalign}\label{eq:A8}
\sum_{k=1}^{m}\frac{\psi_{0}(a+1-k)}{k}=&-\sum_{k=1}^{m}\frac{\psi_{0}(k+a-m)}{k}+\left(\psi_{0}(m+1)-\psi_{0}(1)\right)\left(\psi_{0}(a-m)+\psi(a+1)\right)&&\nonumber\\
&+\frac{1}{2}\left((\psi_{0}(a-m)-\psi_{0}(a+1))^2+\psi_{1}(a+1)-\psi_{1}(a-m)\right) &&
\end{flalign}

\section{Coefficient lists of $I_{A}$, $I_{B}$, $I_{C}$, and $I_{D}$}\label{app:2}
\begingroup
\squeezetable
\begin{longtable}[!h]{rl}
\caption{Coefficients of $I_{A}$ in Eq.~(\ref{eq:IAf})}\\
\hline\hline\rule{0pt}{4ex}
$a_{0}=$ & $-18\alpha m(m+\alpha)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^2\left(5m^2+10\alpha m+5m+4\alpha^2+4\alpha+2\right)$ \\
$a_{1}=$ & $-\alpha m(m+\alpha)(m+2\alpha)\big(1756m^5+8760\alpha m^4+4464m^4+15900\alpha^{2}m^3+16152\alpha m^3+3941m^3+12736\alpha^{3}m^2+19320\alpha^{2}m^2$\\
&$+9288\alpha m^{2}+1370m^2+4032\alpha^4 m+8112\alpha^3 m+5604\alpha^{2}m+1500\alpha m+147m+192\alpha^5+480\alpha^4+320\alpha^3-60\alpha-14\big)$\\
$a_{2}=$ & $-18m(m+\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{2}(3m+4\alpha)\left(5m^2+10\alpha m+5m+4\alpha^2+4\alpha+2\right)$ \\
$a_{3}=$ & $12(m+2\alpha)(2m+2\alpha+1)^2\big(15m^5-30\alpha^{2}m^4+60\alpha m^4+30m^4-120\alpha^{3}m^3+27\alpha^2 m^3+72\alpha m^3+21m^3-154\alpha^{4}m^2$ \\
&$-75\alpha^{3}m^2+19\alpha^{2}m^2+24\alpha m^2+6m^2-68\alpha^{5}m-50\alpha^{4}m-16\alpha^{3}m-\alpha^{2}m-4\alpha^6+4\alpha^5+\alpha^4-\alpha^3\big)$ \\
$a_{4}=$ & $6(m+\alpha)(m+2\alpha)(2m+2\alpha+1)^{2}\big(15m^4-120\alpha^{2}m^3+30\alpha  m^3+30m^3-360\alpha^{3}m^2-228\alpha^{2}m^2+12\alpha m^2+21m^2$ \\
&$-256\alpha^{4}m-336\alpha^{3}m-176\alpha^{2}m-18\alpha m+6m-16\alpha^5-32\alpha^4-68\alpha^3-52\alpha^2-12\alpha\big)$ \\
$a_{5}=$ & $6(m+2\alpha)(2m+2\alpha+1)\big(106\alpha m^6-60m^6+636\alpha^{2}m^5-65\alpha m^5-150m^5+1506\alpha^{3}m^4+503\alpha^{2}m^4-324\alpha m^4-144m^4$ \\
&$+1784\alpha^{4}m^3+1300\alpha^{3}m^3+60\alpha^{2}m^3-235\alpha m^3-66m^3+1080\alpha^{5}m^2+1120\alpha^{4}m^2+468\alpha^{3}m^2+15\alpha^{2}m^2-58\alpha m^2$ \\
&$-12m^2+288\alpha^{6}m+320\alpha^{5}m+216\alpha^{4}m+96\alpha^{3}m+16\alpha^{2}m+16\alpha^7-8\alpha^6-12\alpha^5+2\alpha^4+2\alpha^3\big)$ \\
$a_{6}=$ & $6(m+2\alpha+1)(2m+2\alpha+1)\big(212\alpha m^6-30m^6+1512\alpha^{2}m^5+198\alpha m^5-45m^5+4212\alpha^{3}m^4+1820\alpha^{2}m^4+111\alpha m^4$ \\
&$-27m^4+5760\alpha^{4}m^3+4344\alpha^{3}m^3+1238\alpha^{2}m^3+31\alpha m^3-6m^3+3936\alpha^{5}m^2+4304\alpha^{4}m^2+2296\alpha^{3}m^2+440\alpha^{2}m^2$ \\
&$+6\alpha m^2+1152\alpha^{6}m+1680\alpha^{5}m+1480\alpha^{4}m+584\alpha^{3}m+72 \alpha^{2}m+64\alpha^7+128\alpha^6+272\alpha^5+208\alpha^4+48\alpha^3\big)$ \\
$a_{7}=$&$-12\alpha\big(212m^8+1908\alpha m^7+636m^7+7252\alpha^{2} m^6+4920\alpha m^6+823m^6+15148\alpha^{3}m^5+15660\alpha^{2}m^5+5545\alpha m^5+601m^5$\\
&$+18888\alpha^{4}m^4+26400\alpha^{3}m^4+14738\alpha^{2}m^4+3552\alpha m^4+255m^4+14192\alpha^{5}m^3+25104\alpha^{4}m^3+19608\alpha^{3}m^3+7724\alpha^{2}m^3$\\
&$+1329\alpha m^3+59m^3+6080\alpha^{6}m^2+13056\alpha^{5} m^2+13480\alpha^{4}m^2+7696\alpha^{3}m^2+2250\alpha^{2}m^2+272\alpha m^2+6m^2+1248\alpha^{7}m$\\
&$+3168\alpha^{6}m+4304\alpha^{5}m+3432\alpha^{4}m+1494\alpha^{3}m+316\alpha^{2}m+24\alpha m+64\alpha^8+192\alpha^7+416\alpha^6+512\alpha^5+324\alpha^4+100\alpha^3$\\
&$+12\alpha^2\big)$\\
\\\hline\hline
\label{t:IA}
\end{longtable}
\endgroup

\begingroup
\squeezetable
\begin{longtable}[!h]{rl}
\caption{Coefficients of $I_{B}+I_{C}$ in Eq.~(\ref{eq:IBCf})}\\
\hline\hline\rule{0pt}{4ex}
$b_{0}+c_{0}=$ & $18\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{3}\left(5m^2+10\alpha m+5m+4\alpha^2+4\alpha+2\right)$\\
$b_{1}+c_{1}=$ & $-2\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)\big(1756m^6+10516 \alpha m^5+5504m^5+24660\alpha^{2}m^4+25608\alpha m^4+6479m^4$\\
&$+28636\alpha^{3}m^3+44304\alpha^{2}m^3+22151\alpha m^3+3480m^3+16768\alpha^{4}m^2+34376\alpha^{3}m^2+25380\alpha^{2}m^2+7802\alpha m^2$\\
&$+805m^2+4224\alpha^{5}m+10752\alpha^{4}m+10268\alpha^{3}m+4464\alpha^{2}m+819\alpha m+37m+192\alpha^6+576\alpha^5+560\alpha^4+160\alpha^3$\\
&$-60\alpha^2-44\alpha-7\big)$\\
$b_{2}+c_{2}=$ & $-18m(m+\alpha)(m+\alpha+1)(m+2\alpha+1)(2m+2\alpha+1)^{3}(3m+4\alpha)\left(5m^2+10\alpha m+5m+4\alpha^2+4\alpha+2\right)$\\
$b_{3}+c_{3}=$ & $12(m+\alpha+1)(m+2\alpha)(2m+2\alpha+1)^{3}\big(15m^5-30\alpha^{2}m^4+60\alpha m^4+30 m^4-120\alpha^{3}m^3+27\alpha^{2}m^3+72\alpha m^3+21m^3$\\
&$-154\alpha^{4}m^2-75\alpha^{3}m^2+19\alpha^{2}m^2+24\alpha m^2+6m^2-68\alpha^{5}m-50\alpha^{4} m-16\alpha^{3}m-\alpha^{2}m-4\alpha^6+4\alpha^5+\alpha^4-\alpha^3\big)$\\
$b_{4}+c_{4}=$ & $6(m+\alpha)(m+\alpha+1)(m+2\alpha)(2m+2\alpha+1)^{3}\big(15m^4-120\alpha^2 m^3+30\alpha m^3+30m^3-360\alpha^{3}m^2-228\alpha^{2}m^2$\\
&$+12\alpha m^2+21m^2-256\alpha^{4}m-336\alpha^{3}m-176\alpha^{2}m-18\alpha m+6m-16\alpha^5-32\alpha^4-68\alpha^3-52\alpha^2-12\alpha\big)$\\
$b_{5}+c_{5}=$ & $12(m+\alpha+1)(m+2\alpha)(2m+2\alpha+1)\big(106\alpha m^7-60m^7+742\alpha^2 m^6-141\alpha m^6-180m^6+2142\alpha^{3}m^5+378\alpha^{2}m^5$\\
&$-670\alpha m^5-219m^5+3290\alpha^{4}m^4+1743\alpha^{3}m^4-698\alpha^{2}m^4-679\alpha  m^4-138m^4+2864\alpha^{5}m^3+2448\alpha^{4}m^3$\\
&$+104\alpha^{3}m^3-613\alpha^{2}m^3-291\alpha m^3-45m^3+1368\alpha^{6}m^2+1524\alpha^{5}m^2+500\alpha^{4}m^2-84\alpha^{3}m^2-132\alpha^{2}m^2-47\alpha m^2$\\
&$-6m^2+304\alpha^{7}m+360\alpha^{6}m+172\alpha^{5}m+62\alpha^{4}m+18\alpha^{3}m+2\alpha^{2}m+16\alpha^8-16\alpha^6-4\alpha^5+3\alpha^4+\alpha^3\big)$\\
$b_{6}+c_{6}=$ & $6(2m+2\alpha+1)\big(424\alpha m^9-60m^9+4720\alpha^{2}m^8+1036\alpha m^8-240m^8+22640\alpha^{3}m^7+13764\alpha^{2}m^7+716\alpha m^7$\\
&$-399m^7+61184\alpha^{4}m^6+62444\alpha^3 m^6+17280\alpha^{2}m^6-80\alpha m^6-357m^6+102120\alpha^{5}m^5+148488\alpha^{4}m^5+74852\alpha^{3}m^5$\\
&$+13467\alpha^{2}m^5-265\alpha m^5-183m^5+108240 \alpha^{6}m^4+206232\alpha^{5}m^4+152936\alpha^{4}m^4+53828\alpha^{3}m^4+7650\alpha^{2}m^4$\\
&$-74\alpha m^4-51m^4+71744\alpha^{7}m^3+170208\alpha^6 m^3+169160\alpha^{5}m^3+90024\alpha^{4}m^3+25492\alpha^{3}m^3+3057\alpha^{2}m^3+13\alpha m^3$\\
&$-6m^3+27776\alpha^{8}m^2+79328\alpha^{7}m^2+100928\alpha^{6}m^2+74144\alpha^{5}m^2+32140\alpha^{4}m^2+7480\alpha^{3}m^2+722\alpha^{2}m^2+6\alpha  m^2$\\
&$+5248\alpha^{9}m+17664\alpha^{8}m+28608\alpha^{7}m+28512\alpha^{6}m+17544\alpha^{5}m+6216\alpha^{4}m+1112\alpha^{3}m+72\alpha^{2}m+256\alpha^{10}$\\
&$+1024\alpha^9+2432\alpha^8+3712\alpha^7+3344\alpha^6+1696\alpha^5+448\alpha^4+48\alpha^3\big)$\\
$b_{7}+c_{7}=$ & $-12\alpha(2m+2\alpha+1)\big(212m^9+2120\alpha m^8+758m^8+9160\alpha^{2}m^7+6726\alpha m^7+1162m^7+22400\alpha^{3}m^6+25294\alpha^{2}m^6$\\
&$+9200\alpha m^6+1049m^6+34036\alpha^{4}m^5+52450\alpha^{3}m^5+29970\alpha^{2}m^5+7475\alpha m^5+631m^5+33080\alpha^{5}m^4+65124\alpha^{4}m^4$\\
&$+51812\alpha^{3}m^4+20908\alpha^{2}m^4+4038 \alpha m^4+251m^4+20272\alpha^{6}m^3+48896\alpha^{5}m^3+50800\alpha^{4}m^3+29310\alpha^{3}m^3$\\
&$+9392\alpha^{2}m^3+1417\alpha m^3+59m^3+7328\alpha^{7}m^2+21056\alpha^{6}m^2+27624\alpha^{5}m^2+21476\alpha^{4}m^2+10018\alpha^{3}m^2+2550\alpha^{2}m^2$\\
&$+284\alpha m^2+6m^2+1312\alpha^{8}m+4416\alpha^{7} m+7312\alpha^{6}m+7576\alpha^{5}m+4866\alpha^{4}m+1802\alpha^{3}m+340\alpha^{2}m+24\alpha m$\\
&$+64\alpha^9+256\alpha^8+608\alpha^7+928\alpha^6+836\alpha^5+424\alpha^4+112\alpha^3+12\alpha^2\big)$\\
$b_{8}+c_{8}=$ & $-18\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{2}\left(7m^2+14\alpha m+7m+8\alpha^2+8\alpha+2\right)$\\
$b_{9}+c_{9}=$ & $36\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{2}\big(10m^3+30\alpha m^2+9m^2+28\alpha^{2}m+8\alpha^3+4\alpha^2$\\
&$+16\alpha m+3m\big)$\\
$b_{10}+c_{10}=$ & $-72\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{2}\big(10m^3+30\alpha m^2+12m^2+28\alpha^{2}m+22\alpha m$\\
&$+6m+8\alpha^3+8\alpha^2+4\alpha+1\big)$\\
$b_{11}+c_{11}=$ & $18\alpha m(m+\alpha)(m+\alpha+1)(m+2\alpha)(m+2\alpha+1)(2m+2\alpha+1)^{2}\big(10m^3+30\alpha m^2+3m^2+28\alpha^{2}m+4\alpha m-3m$\\
&$+8\alpha^3-4\alpha^2-8\alpha-2\big)$\\
\\\hline\hline
\label{t:IBC}
\end{longtable}
\endgroup

\begingroup
\squeezetable
\begin{longtable}[!h]{rl}
\caption{Coefficients of $I_{D}$ in Eq.~(\ref{eq:IDf})}\\
\hline\hline\rule{0pt}{4ex}
$d_{0}=$ & $m\big(36m^4+136\alpha m^3+68m^3+196\alpha^{2}m^2+188\alpha m^2+31m^2+128\alpha^{3}m+184\alpha^{2}m+64\alpha m-6m+32\alpha^4+64\alpha^3+36\alpha^2$\\
&$-4\alpha-5\big)$\\
$d_{1}=$ & $2m(2m+2\alpha+1)\left(14m^3+46\alpha m^2+29m^2+48\alpha^{2}m+60\alpha m+20m+16\alpha^3+32\alpha^2+22\alpha+5\right)$\\
$d_{2}=$ & $4(2m+2\alpha+1)\big(32\alpha^4+64\alpha^3+48\alpha^2+16\alpha+30m^4+126\alpha m^3+69m^3+192\alpha^{2}m^2+204\alpha m^2+56m^2+128\alpha^{3}m$\\
&$+200\alpha^{2}m+106\alpha m+19m+2\big)$\\
$d_{3}=$ & $-4\big(60m^5+312\alpha m^4+168m^4+636\alpha^{2}m^3+672\alpha m^3+181m^3+640 \alpha^{3}m^2+1000\alpha^{2}m^2+528\alpha m^2+94m^2+320\alpha^{4}m$\\
&$+656\alpha^{3}m+508\alpha^{2}m+176\alpha m+23m+64\alpha^5+160\alpha^4+160\alpha^3+80\alpha^2+20\alpha+2\big)$\\
$d_{4}=$ & $8(m+2\alpha+1)(2m+2\alpha+1)^{2}\left(3m^2+6\alpha m+3m+4\alpha^2+4\alpha+1\right)$\\
$d_{5}=$ & $-2m(m+2\alpha+1)^{2}(2m+2\alpha+1)^{2}$\\
$d_{6}=$ & $4(m+2\alpha+1)(2m+2\alpha+1)^{3}\left(5m^2+10\alpha m+5m+4\alpha^2+4\alpha+2\right)$\\
\\\hline\hline
\label{t:ID}
\end{longtable}
\endgroup

\bibliography{BHv}
\end{document}


TITLE: Basis of tensor product of vector spaces
QUESTION [1 upvotes]: Suppose $V,W$ are vector spaces with bases $\{e_1,e_2,\dots ,e_m\}$ and $\{f_1,f_2,\dots ,f_n\}$. I know that $V\otimes W=F(V\times W)/H$, where $F(V\times W)$ is the free vector space on $V\times W$ and $H$ is the subspace generated by all elements satisfying certain relations such that bi-linearity holds in the quotient space. I'm writing $v\otimes w$ for the congruency class of the pair $(v,w)\in V\times W$. I can easily show that the pairs $e_i\otimes f_j$ span $V\otimes W$, but I'm having trouble showing linear independence i.e. that if $\sum_{i,j}\lambda_{ij}e_i\otimes f_j=0$ then for all $i,j$, $\lambda_{ij}=0$. Could anybody point me in the right direction? Thanks for any replies!

REPLY [1 votes]: $\def\Hom{\mathrm{Hom}}$
So, since you already know that the product of the bases spans $V \otimes W$, you just need to show that the dimension of $V \otimes W$, which is the dimension of its dual, is the product of the dimension of $V$ and the dimension of $W$.
We first reduce the problem to showing that
$$(V \otimes W)^{*} = \Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)) = \Hom_{k}(V, W^{*}).$$
Note that if the above relationship holds, then the dimension of $V \otimes W$ is the dimension of the space of linear maps from $V$ to $W^{*}$, which is just the product of the dimensions of $V$ and $W^{*}$, and is hence the product of the dimensions of $V$ and $W$.
So, to show the above equalities, note that the leftmost and rightmost equalities are by definition of the dual. So we need to show that
$$\Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)).$$
This is actually true if $k$ is any ring and $V$ and $W$ are any $k$-modules. This is the adjunction between $\Hom$ and tensor product. But let's go through the details for the vector space case.
Suppose $\phi$ is a linear map from $V \otimes W$ to $k$. We want to define $f(\phi)$ which takes a vector in $v$ and spits out a linear map from $W$ to $k$. To do so, we define
$$(f(\phi) (v)) (w) = \phi(v \otimes w).$$
You should check that this map is linear and well-defined. It is not too hard. We next construct an inverse map. Given $\psi$ which to each $v$ assigns a linear map from $W$ to $k$, we define $g(\psi)$ as a linear map from $V \otimes W$ to $k$ by first defining it on the simple tensors as
$$g(\psi)(v \otimes w) = \psi(v)(w)$$
noting that this is bilinear in the $v$ and $w$ and then noting that bilinear maps extend to all of $V \otimes W$.
Again, $g$ can be checked to be linear and you can also easily see that $f$ and $g$ are inverses.


TITLE: Existence of an embedding?
QUESTION [2 upvotes]: Let $\mathcal{L}=\{R\}$ a language of the first order. We consider the set of graphs (which are irreflixive and symmetric). Then we define for a graph $G$ the following property $P$ : for $S_1$ and $S_2$ finite and disjoint $\subseteq G$, there exists $z\in G$ such that for all $x\in S_1$, $y\in S_2$ we have $R(x,z)$ and $\neg R(y,z)$. We denote $T$ the first order theory which satisfies these three properties.
First, I proved that there exists a countable model. Then I must prove that for a countable graph $K$ and a model $\mathcal{H}=(H,R)$ of $T$ (not necessary countable) I can find an embedding between $K$ and $H$.
Here is my attempt : I think that if $K$ could satisfy $P$, I could find an embedding between $K$ (which would become a countable model of $T$) and $\mathcal{H}$ by Lowenheim-Skolem. That means that every countable graph satisfies $P$.
First thing I noticed is that if $K$ is a countable graph with the property of irreflexivity I can find a least $2$ elements in $K$ such that $R(x_1,x_2)$. After, maybe I can use the fact that I can find an injective map $\iota$ between $\{x_1\}$ and a bigger finite set $\{\iota(x_1),...,\iota(x_k)\}$ which respects $\bigwedge_{1\le j\le k} R(\iota(x_j),x_2)$ (I'm not sure for $x_2$) . How can I arrive to the second part of the property $P$. Or maybe it's not the right way to proceed.
Thanks in advance !

REPLY [2 votes]: First, a minor comment: the first line of your last paragraph is incorrect - a countable irreflexive graph might have no edges at all!
OK, now the real answer: work in stages. (This is an intuition similar to that behind the back-and-forth method for building an isomorphism between appropriate countable structures, if you've seen that before.) Specifically, what you want is an embedding $f: K\rightarrow H$. $K$ is countable, so let $K=\{k_0, k_1, k_2, . . .\}$ (I'm assuming WLOG that $K$ is infinite, for notational simplicity). So let's build $f$ in stages!
Let $G_n$ be the subgraph of $K$ consisting of $\{k_0, k_1, . . . , k_n\}$. A stage $n$ embedding will be an embedding $f_n$ of $G_n$ into $H$. In order to build $f$, what do we need? Well,

for each $n$, a stage $n$ embedding $f_n$

would be nice. But we want to be able to "glue" the $f_n$s together, so we also want

$f_n(k_i)=f_m(k_i)$ whenever $i\le m, n$.

Do you see why if we have such a set of $f_n$s, the map $\bigcup f_n$ is an embedding from $K$ into $H$?

OK, now let's build our $f_n$s. On the face of it, this might seem hard; however, the point of $T$ is that any model of $T$ "looks the same everywhere," so no matter how we've defined $f_n$, it will always be possible to extend it to $f_{n+1}$.
Here's a sketch of why: suppose I've defined $f_{17}$ - so I have an embedding from $G_{17}$ to $H$. Now I want to extend this to also include $k_{18}$. Bad news: I have to be careful where I send $k_{18}$! Maybe $k_{18}$ is connected to $k_1, k_3, k_{11},$ and $k_{16}$, and not connected to all the other $k_i$s (for $i<18$).

Why is there some $h\in H$ such that $h$ is connected to $f_{17}(k_1), f_{17}(k_3), f_{17}(k_{11})$, and $f_{17}(k_{16})$, and not connected to $f_{17}(k_i)$ for any other $i<18$?
Do you see how to use this to define $f_{18}$?


TITLE: How can I calculate the torque on a body resulting from a spinning propeller?
QUESTION [1 upvotes]: If a quadcopter has all of its propellers spinning on the same direction, the body of the quad will obviously start spinning in the another direction.
I know how to calculate the torque acting on a object but in this case  the torque is caused by the propeller spinning on a constrained axis that is not the same as the axis of rotation of the body. How can the three dimensional torque vector acting on the body (the black rectangle in the figure with center of mass at CM) caused by the accelerating propeller (blue and spinning counterclockwise) be calculated?

REPLY [1 votes]: If the blade was spinning in a vacuum (and there was no friction and no motor), there would be no torque. The blade could spin by itself forever.
Torque comes from spinning in air and pushing air around. Air is viscous. It resists the motion of the blades. It exerts a torque on the blades. The torque on all 4 blades is the torque on the quadcopter.
Figuring out the torque on a blade requires knowing the shape of the blade. It is possible, but not trivial. You would want software to help.
There is an easier way. The motor has to exert an equal torque on the blades to keep them moving. If you know the power of the motor and the angular velocity of the blades, you can use
$P = \tau\omega$
As for how the torque exerted at the end of a stick affects the stick, you haven't drawn everything that is important. If you twisted a free propeller as shown, it would spin without exerting a torque on the stick. But there is a motor at the center and a belt connecting it to the propeller. Air resistance at the blades is transmitted to the belt, which is transmitted to the motor, which is bolted to the quadcopter.


TITLE: How to think about multiplication with a number $0 <x < 1$?
QUESTION [4 upvotes]: To clarify my confusion:
$88 \cdot 0.732$ is the same as $0.732 \cdot 88$
The latter being: 0.732 eighty-eight times: $0.732 + 0.732 + 0.732 + \cdots + 0.732$
But how to think about $88 \cdot 0.732$? that's $88\quad0.732$ times?
Can you see what's bothering me? I apologize if these are weird questions, but it's really bothering me.

REPLY [4 votes]: One way to think about real numbers is to see them as lengthes. In that setting, the product $a \times b$ represents the area of a rectangle whose sides are $a$ and $b$ (I'm not defining that precisely, but just relying on your intuition here). This explain easily why $a \times b = b \times a$.
Now in the special case where $n$ is an integer (and only in that case), we can check that $n \times x$ also has the meaning
$$n \times x = \underbrace{x + \ldots + x}_{n \textrm{ times}}$$
Indeed, it's easy to see why $(a + b) \times c = a \times c + b \times c$ (split a rectangle in two along the $a+b$ side) and $1 \times a = a$. So using this properties, we get
$$2 \times x = (1+1) \times x = x + x$$
$$3 \times x = (2+1) \times x = x + x + x$$
$$4 \times x = (3+1) \times x = x + x + x + x$$
$$5 \times x = (4+1) \times x = x + x + x + x + x$$
And so on.


TITLE: Singularities of exponential function $e^{-1/z^2}$
QUESTION [1 upvotes]: I want to know about the singularities of the function $$f(z) = e^{-1/z^2}, \ z \neq 0, \qquad \\ 0, \quad z = 0$$. Explain in detail please.
For poles i need to look at the points where $e^{1/z^2} = 0$. But Exponential are never $0$, so no poles. Now if i expand its series expansion, then there will be infinitely many terms containing negative powers of $z$, so there must be non isolated essential singularity at $0$. Am i correct?

REPLY [5 votes]: We have $e^z= \sum_{n=0}^{\infty}\frac{z^n}{n!}$, hence, for $z \ne 0$,
$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!z^{2n}}$.
Consequence: $f$ has at $0$ an essential singularity. You wrote "non isolated essential singularity". But the singularity at $0$ is isolated !
A generalization: if $f$ is an entire function and if $g$ is defined by $g(z)=f(1/z)$, then we have:
$g$ has at $0$ an essential singularity $ \iff f$ is not a polynomial.
Try a proof !
In the above case we have $f(z)=e^{-z^2}$


TITLE: Canonical sheaf of product of two smooth projective curves
QUESTION [3 upvotes]: Let $C_1, C_2$ be two smooth projective curves of genus atleast $2$. Let $X:=C_1 \times C_2$. If $P_1, P_2$ are denoted as the first and second projections, then it is known that the canonical bundle relation is given as follows : $K_X := P_1^*(K_{C_1}) \otimes P_2^*(K_{C_2})$.
From this relation is it elementary to see that $h^1(K_X)=h^1(\mathcal O_X) >0$?
Any hint or explaination is appreciated.

REPLY [2 votes]: If you know the Kunneth formula (ref Stacks or MO), yes! The Kunneth formula says that for quasi-compact schemes $X$ and $Y$ over a field $k$ with affine diagonal (separated implies affine diagonal, for instance) and for $\mathcal{F}$ a quasi-coherent $\mathcal{O}_X$-module, $\mathcal{G}$ a quasi-coherent $\mathcal{O}_Y$-module, we have a canonical isomorphism $$H^n(X\times_{\operatorname{Spec} k} Y, p_1^*\mathcal{F}\otimes_{X\times_{\operatorname{Spec} k} Y} p_2^*\mathcal{G}) \cong \bigoplus_{p+q=n} H^p(X,\mathcal{F})\otimes_k H^q(Y,\mathcal{G}).$$
This gives us that $H^1(X,K_X) \cong H^1(C_1,K_{C_1}) \otimes_k H^0(C_2,K_{C_2}) \oplus H^0(C_1,K_{C_1}) \otimes_k H^1(C_2,K_{C_2})$ in our case. As $H^1(C_i,K_{C_i})\cong k$ by Serre duality and $h^0(C_i,K_{C_i}) = g$ by Riemann-Roch, we have that $h^1(K_X)>0$. Noting that $h^1(K_X)=h^1(\mathcal{O}_X)$ and this is calculated by the same expression by Serre duality, we're done.