TITLE: Equivalent form of prime number theorem
QUESTION [3 upvotes]: In the book The elementary proof of Prime Number Theorem it says that

The prime number theorem $\psi (x)\sim x$ is equivalent to $\int_1^\infty \frac{\psi (t)-t}{t^2} =-\gamma -1$, where $\psi (x)=\sum_{n\le x} \Lambda (n)$, $\gamma $ is the Euler constant.

The book also gives a hint to prove $\sum_{n\le x}\frac{\Lambda (n)}{n}=\int_1^x \frac{\psi(t)}{t^2} dt+\frac{\psi (x)}{x}$ first, but I have no idea on how to prove this hint or prove the original equivalence.
All I know now is the formula in this question $\sum_{n\le x}\frac{\Lambda (n)}{n}=\ln x-\gamma +o(1)$ which is exactly another form of prime number theorem and the relation $\psi (x)=x-1+o(x)-\gamma -\int_1^x o(t)dt$ proved before in the book.
Can the proof obtain from these two identites or I need to find a new approach? Thanks in advance.

REPLY [2 votes]: From $\sum_{n \le x} \frac{\Lambda(n)}{n} = \ln x+C+o(1)$ you know that $$\sum_{n \le x} \frac{\Lambda(n)-1}{n} = -H_n+\ln x+C+o(1) = C-\gamma+o(1) $$
With the Abel summation formula you also have for $x \in \mathbb{N}$ :
$$\sum_{n \le x} \frac{\Lambda(n)-1}{n}
= \int_1^x \frac{\psi(t)-\lfloor t\rfloor}{t^2} dt+  \frac{\psi(x)}{x}-1$$
And with $\int_1^x \frac{\psi(t)-t}{t^2}dt = A+o(1)$ then
$$\int_1^x \frac{\psi(t)-\lfloor t\rfloor}{t^2}dt =  A+o(1)+\int_1^x \frac{t-\lfloor t\rfloor}{t^2}dt = A+o(1)+1-\gamma$$
since $\displaystyle\int_1^x \frac{t-\lfloor t\rfloor}{t^2}dt = \int_1^\infty \frac{t-\lfloor t\rfloor}{t^2}dt+o(1) = 1-\gamma+o(1)$
($\zeta(s) = \frac{s}{s-1}-s\int_1^\infty (t-\lfloor t\rfloor)t^{-s-1}dt=\frac{1}{s-1}+1-s\int_1^\infty (t-\lfloor t\rfloor)t^{-s-1}dt$ and $\lim_{s \to 1}\zeta(s) - \frac{1}{s-1} = \gamma$)

Put altogether
$$C-\gamma+o(1) =\frac{\psi(x)}{x}-1 +A+1-\gamma+o(1) $$
which means $\frac{\psi(x)}{x} = C-A+o(1)$
and with $C = -\gamma,A = -1-\gamma$ you get the prime number theorem
$$\frac{\psi(x)}{x} = 1+o(1)$$


TITLE: One Pion Exchange Potential properties for a two-nucleon system
QUESTION [4 upvotes]: I'm going through my Nuclear Physics book, and has come across a section called "Properties of OPEP for the two-nucleon system".
It start out by considering the n-p system in a singlet spin state $(S=0)$. It says that from the generalized Pauli principle, the n-p system must be in a triplet isotopic spin state. Thus, $\overrightarrow{\sigma}_1 \cdot \overrightarrow{\sigma}_2 = -3$, while $\overrightarrow{\tau}_1 \cdot \overrightarrow{\tau}_2 = 1$ - which I assume is just the Pauli spin and isospin matrices.
Then this is used to determine the OPEP via a big equation, where stuff vanishes and such.
My question is: How is these determined ? Why is it a triplet isospin state, where do the -3 and 1 come from ?
I've looked through the whole chapter, but I can't seem to figure it out :/

REPLY [1 votes]: The only thing I have to add is to answer your question about where the -3 and 1 come from. First, the spin operator is related to the Pauli matrices by
$$s_1 = \frac{1}{2}\sigma_1$$
and so we have the relation
$$\sigma_1 \cdot \sigma_2 = 4 s_1\cdot s_2$$
Next, we can evaluate $s_1\cdot s_2$ using
$$ S^2 = (s_1+s_2)^2 = s_1^2 + s_2^2 + 2 s_1\cdot s_2$$
$$ s_1 \cdot s_2 = \frac{1}{2} (S^2 - s_1^2 - s_2^2)$$
and so we have
$$ \sigma_1\cdot \sigma_2 = 2(S^2 - \frac{3}{2} ) = 2S(S+1) - 3$$
which yields -3 for $S=0$ and 1 for $S=1$.


TITLE: numerical evaluation of an integral
QUESTION [7 upvotes]: I have this integral:
$$\int_{-1}^{1} \frac{e^x}{\sqrt{1-x^2}}\,dx$$
How can I get rid of the infinities at the ends of the interval so that I can evaluate this integral numerically? I tried to make some substitutions but didn't succeed.

REPLY [7 votes]: There are many ways to evaluate this numerically. Let $$I = \int_{-1}^1 \dfrac{e^x}{\sqrt{1-x^2}}dx$$
One way, I can think of is as follows. Set $x = \sin(\theta)$. We then get that
$$I = \int_{-\pi/2}^{\pi/2} e^{\sin(\theta)} d \theta$$ Once you have it in this form, you could recognize this is the modified Bessel function evaluated at $1$. To evaluate this numerically, expand $e^{\sin(\theta)}$ in a series.
We have
$$I = \int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\infty} \dfrac{\sin^k(\theta)}{k!} d \theta = \sum_{k=0}^{\infty} \int_{-\pi/2}^{\pi/2} \dfrac{\sin^k(\theta)}{k!} d \theta = \sum_{k=0}^{\infty} \int_{-\pi/2}^{\pi/2} \dfrac{\sin^{2k}(\theta)}{(2k)!} d \theta\\
 = 2\sum_{k=0}^{\infty} \int_{0}^{\pi/2} \dfrac{\sin^{2k}(\theta)}{(2k)!} d \theta$$
Now recall that $$\int_0^{\pi/2} \sin^{2k}(\theta) d \theta = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}{2} = \dfrac{(2k)!}{4^k k! k!} \dfrac{\pi}2$$
Hence, we have
$$I = \pi \sum_{k=0}^{\infty} \dfrac1{4^k \cdot (k!)^2}$$
The series converges at an exponential rate, hence truncating the series after a few terms should give you a very good accuracy.

REPLY [5 votes]: $$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx $$
$$ x = \sin t; dx=\cos t dt $$
$$\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}} \frac{e^{\sin t}}{\sqrt{1-\sin^2 t}} \cos t dt$$
$$\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}} e^{\sin t} dt$$
No more infinities.


TITLE: How do I describe a partition determined by $R$ by listing pieces?
QUESTION [1 upvotes]: $R$ is the relation $\{(1,1),(2,3),(2,2),(3,2),(3,3)\}$ on the set $\{1, 2, 3\}$. It then is reflexive, symmetric, transitive and thus an equivalence relation. How do I describe the partition formed by $R$ by listing the pieces in the partition?

REPLY [0 votes]: The "pieces of the partition" that arise from an equivalence relation $R$ given on a set are usually called the equivalence classes. To describe an equivalence class of an equivalence relation $R$, it is enough to specify any one member of the equivalence class.

For your question, there are two equivalence classes, namely,
$$
\{ 1 \} \text{ and } \{ 2, 3 \}.
$$
An equivalence class is usually denoted using box brackets, $[\cdot]$. So, one can list the partition of $\{ 1,2,3 \}$ under the given equivalence relation $R$ as
$$
[1] \text{ and } [2]
$$
or
$$
[1] \text{ and } [3].
$$
To elaborate, if $R$ is an equivalence relation on a set $S$, then for each $x \in S$, the notation $[x]$ is defined as follows:
$$
[x] = \{ y \in S : xRy \}.
$$

Note that the equivalence classes will change as your equivalence relation changes. Based on the discussion in the comments, it appears that you were uncertain whether the partition
$$
[1],[2],[3]
$$
would work as an answer to the current question. It does not, because they describe a partition arising from a different equivalence relation on the set $\{ 1,2,3\}$, namely
$$
xRy \iff x = y.
$$


text:
\section{Experiment Validation}

\subsection{Experiment Setting}
To simulate the profitability of demand optimization, we used taxi operation data from 2006 to 2018 in New York City. The LSTM models are implemented using Python 3.7 under the platform CUDA 10.2 GPU, 16GB RAM, Ubuntu 18.04. The entire data set contains more than 22 million taxi trip records, because huge amount of data set is always beneficial to get better performance using deep learning models. To model LSTM, we split the total data set by train and validation data set. We first separate a test set by selecting the data after 2016-06-01. The training and validation data set are further separated from the data before 2016-6-1 so that the time order is preserved. The features for training models includes latitudes and longitudes for both pickup and drop-off locations, as well as pickup time stamp and number of passengers. These features are necessary for a taxi dispatching system when a passenger call for a demand. First, we hope to build predictive models to estimate the taxi duration trip. In the historical training data, the taxi trip duration is known and therefore the target variable for this task. Although the online APIs such as google map can estimate the taxi trip duration by calculate the distance between the pick-up and drop-off coordinates, it is not automation processing for a dispatching system, and moreover, the taxi trip duration depends on many other factors such as the traffic jam affected by time stamps, the road conditions and so on. The GRU-MDN are implemented using Python 3.7 under the platform CUDA 10.2 GPU, 16GB RAM, Ubuntu 18.04, the mathematical models are solved by Gurobi \footnote{https://www.gurobi.com/downloads/gurobi-software/} 9.0 academic version under the platform Intel i7, 32GB RAM, Windows 10.

The charing station information can be found from \footnote{https://data.ny.gov/Energy-Environment/Electric-Vehicle-Charging-Stations-in-New-York/7rrd-248n}

The EVDP can be formulated as follows.

$$
\text{maximize} \sum_{t \in T} \sum_{v \in V} \sum_{q \in Q} x_{v q t} d_{q}-\lambda \sum_{t \in T} \sum_{v \in V} t w_{v t}
$$

\subsection{Performance Measures}

config: arxiv


TITLE: Attempt to generalize suprema and infima
QUESTION [0 upvotes]: Let $(\mathfrak{A};\leq)$ be a poset. Let $A$ be its subset.
I will denote

$\operatorname{Up} A = \{ x\in\mathfrak{A} \mid \forall a\in A: x \geq a \}$;
$\operatorname{Down} A = \{ x\in\mathfrak{A} \mid \forall a\in A: x \leq a \}$.

I remind that $\sup A = \min\operatorname{Up} A$.
Question: Does it make sense to repeat these operations more than once, like: $$\max\operatorname{Down}\operatorname{Up} A$$ or further $$\min\operatorname{Up}\operatorname{Down}\operatorname{Up} A,$$ etc.?

REPLY [0 votes]: Attempt to generalize suprema and infima
The upper set of A, written [small up arrow] A or up A
= { x | some a in A with a <= x }.
The over or above set of A, the set of upper bounds of A,
above A = { x | A <= x } = { x | for all a in A, a <= x }.
down A and below A are the order dual concepts.
Above and below can be used for an algebraic approach
to inf and sup.  sup A = min above A, for example.
It's an awkward method as compared to the usual point method.
Accordingly I've little use for it.  As for compounding above
and below, the following may be of interest.
below A = below up A;  above A = above down S
A subset down A subset below above A
A subset up A subset above below A
below A = below above below A;  above A = above below above A
below below A = {bottom} if bottom exists, otherwise it's empty.
above above A = {top} if top exists, otherwise it's empty.
Let S be the set that is (partially) ordered.
below above above A = above below below A = S
below empty set = above empty set = S
below S = {bottom} if bottom exists, otherwise it's empty.
above S = {top} if top exists, otherwise it's empty.
Proofs, if needed, upon request.
What would you suggest for generalized supremum?


TITLE: Virtual particles in EM interaction and Weak interaction
QUESTION [1 upvotes]: I know that a real electron has a probability (which depends on the intensity of the EM force) of emitting a photon, changing his 4-momentum. The photon should be virtual. Now, my teacher says that this photon can decay in an electron and a positron, but "obviously not all the electrons are real". I'm referring to the E1 process in the image below.

What does this mean? Does it mean that the electron and positron created are virtual? And if this is the case they should annichilate themselves to create again the virtual photon, am I right? This is because if they are virtual, being massive particles, they can't maintain their "virtuality" as long as they want (differently from the photon, which is massless and can exist as a virtual particle without time limits).
But my doubts arise when the analogous process, the beta decay, occurs via weak interaction. So instead of having an electron changing his 4-momentum I have a neutron becoming a proton and emitting a $W^{-}$ virtual boson. Then the virtual boson decays into an electron and an antineutrino (real particles because they can be observed).
My final question, IF what I said 'till now is right, which I'm not sure, is:
Why does the photon decays into two virtual particles and the boson into real ones?

REPLY [2 votes]: Why does the photon decays into two virtual particles and the boson into real ones?

The E1 diagram does not conserve energy  in the center of mass system. The incoming center of mass energy  is the mass of the electron, the outgoing is at least three masses of electrons, and mind you, the electron is an elementary particle and does not decay.
Also photons interact, they do not decay. Within correct feynman diagrams they are called virtual. Similar diagrams exist for off mass shell Z boson exchange.Outgoing lines from an interaction vertex are real, not virtual. Only internal lines are virtual.
The correct pair production diagram from a photon can be seen here, where the center of mass has to take into account the nucleus also:

The virtual photon in your E1 has to interact with the field of a nucleus, by a virtual electron exchange and then a real electron positron pair can appear,
Correspondingly there will be a diagram with a Zboson exchange from your E1, but unless the energies involved are very high, the Z propagator and the weak coupling constant make the contribution to the crossection of this diagram tiny.


TITLE: Given a set of integers find two disjoint subsets $I$ and $J$ so $|I|+|J|=k$ and $\sum\limits_{i \in I}x_i = \sum\limits_{j \in J}x_j = n^2$
QUESTION [0 upvotes]: Given a set of integers $1 \le x_1,x_2\dots,x_n \le n^2$ and a number $k \le n$.
describe an algorithm that will determine if there exists two disjoned subsets $I$ and $J$ such that $|I|+|J|=k$ and $\sum\limits_{i \in I}x_i = \sum\limits_{j \in J}x_j = n^2$
this question is form exam i had . my intuition tells me its related to linear-programming or dynamic-programming but i'm not sure .(after some search online for similar questions i'm almost sure its dynamic-programming)

REPLY [1 votes]: Here is a naive algorithm.

Verify that $\sum{x_i}\leq2*n²$

If yes, then

possible_sets = [(0,0,0)] # ($\sum{x_i}$, $\sum{x_j}$, |I|+|J|)
for x in {$x_k$}

for $s_i$,$s_j$,$nb$ in possible_sets

if $nb<k$

if $s_i+x<n²$

if $nb+1=k$ and $s_i+x=s_j=n²$

return True

possible_sets += (s_i+x, s_j, nb+1)

if $s_j+x<n²$

if $nb+1=k$ and $s_i=s_j+x=n²$

return True

possible_sets += (s_i, s_j+x, nb+1)

return False

This can be optimized using symmetry between I and J


TITLE: Double Negation in Type Theory
QUESTION [2 upvotes]: How do I derive a proof of double negation in type theory for Exercise 1.11 of the book HoTT? That is a proof of type $(((A \rightarrow \textbf{0}) \rightarrow \textbf{0}) \rightarrow \textbf{0}) \rightarrow (A \rightarrow \textbf{0})$, where objects of type $\textbf{0}$ are contradictions.
Is this equivalent to proving $((A \rightarrow \textbf{0}) \rightarrow \textbf{0}) \rightarrow A$?

REPLY [1 votes]: It is not equivalent to $((A \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow A$; in fact, that statement (for arbitrary types $A$) is inconsistent in HoTT.
To prove $(((A \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow (A \rightarrow \mathbf{0})$ is by definition to provide a function from $((A \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow \mathbf{0}$ and $A$ to $\mathbf{0}$—that is, a value of the form $\lambda f . \lambda a . …$.  What should go in place of the ellipses?  Well, we need a value of type $\mathbf{0}$, so it seems prudent to apply $f$ to something of type $(A \rightarrow \mathbf{0}) \rightarrow \mathbf{0}$.  In order to construct such a thing, it suffices to provide a value of the form $\lambda g . …$.  This second set of ellipses needs to be replaced with a value of type $\mathbf{0}$ again, but we now have access to $g : A \rightarrow \mathbf{0}$, so we can apply it to $a$ and we are completely done.  Consequently, a proof (and, I might add, the intended proof) of $(((A \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow \mathbf{0}) \rightarrow (A \rightarrow \mathbf{0})$ is $\lambda f . \lambda a . f(\lambda g . g(a))$.


TITLE: Directional derivative conflicts with normal derivative?
QUESTION [1 upvotes]: I am learning about directional derivative and the definition bothered me so much.
$\frac{\partial f}{\partial \vec{v}} = \vec{\nabla}f . \vec{v}$
which by extension can be written as
$\frac{\partial f}{\partial \vec{v}} = (\vec{\nabla}f . \hat{v})||\vec{v}||$
So that would mean derivative of $f$ along $\vec{2x}$ will be
$\frac{\partial f}{\partial \vec{2x}} = (\vec{\nabla}f . \hat{i})2 = 2\frac{\partial f}{\partial x}$
but $\frac{\partial f}{\partial(2x)}$ should be $\frac{1}{2}\frac{\partial f}{\partial x}$
Why there is a difference?
Note: I am beginner to multivariate calculus, so the question is probably a silly misunderstanding of the concepts. I will be glad if you can clarify it. Thank you.

REPLY [2 votes]: The way you should think about the directional derivative is this: Take any parametrized curve through $p$ with velocity vector $\vec v$ at $p$. Compute the rate of change of $f$ along this curve at the instant you pass through $p$. (Then it's clear that if you're moving twice as fast, the function is changing twice as fast at that instant.)
COMMENT: Personally, I prefer the notation $D_{\vec v}f(p)$ for the directional derivative.


text:
\begin{document}

	\begin{abstract} Let $X$ be a compact K\"ahler manifold. We prove that if $X$ admits a smooth Hermitian metric $\omega$ with \emph{quasi-positive} second Chern-Ricci curvature $\mathrm{Ric}^{(2)}(\omega)$, then $X$ is projective and rationally connected. In particular, $X$ is simply connected.
	\end{abstract}

	\maketitle

	\tableofcontents

	\vspace{-0.2cm}

	\section{Introduction}
    The geometry of complex manifolds are characterized by various positivity notions in complex differential geometry and algebraic geometry. Since the seminal works of Mori   and Siu-Yau on the solutions to
	Hartshorne
	conjecture   and Frankel conjecture
	(\cite{Mori1979}, \cite{SiuYau1980}) on the characterizations of projective spaces,   many remarkable generalizations  have been
	established,   for instances, Mok's uniformization theorem on compact K\"ahler manifold
	with non-negative holomorphic bisectional curvature (\cite{Mok1988}) and  the  works of Campana,
	Demailly, Peternell and Schneider (\cite{CampanaPeternell1991},
	\cite{DemaillyPeternellSchneider1994})  on the
	structure of projective manifolds with nef tangent bundles. For more geometric characterizations, we refer to \cite{Mok1988,CampanaPeternell1991,DemaillyPeternellSchneider1994,DemaillyPeternellSchneider2001,Wu2002,Chau2012,LiWuZheng2013,Liu2014b,FengLiuWan2017,HuangLeeTamTong2018,NiZheng2019b,Liu2019,LiOuYang2019,LiuOuYang2020} and the references therein.\\

	    The  holomorphic sectional curvature also carries much geometric information of   complex manifolds. Indeed, thanks to the breakthrough work \cite{WuYau2016} of Wu-Yau, it is well-known that  a compact K\"ahler manifold $X$ with negative or quasi-negative holomorphic sectional curvature is algebraic and has ample canonical bundle (\cite{TosattiYang2017,WuYau2016a,Diverio2019}), which settles down a long-standing conjecture of S.-T. Yau affirmatively. For more recent works on  non-positive holomorphic sectional curvature, we refer to \cite{Wong1981,Heier2010,Heier2016,Heier2018,Nomura2018,Guenancia2018,YangZheng2019TAMS,WuYau2020} and the references therein. On the other hand, in his “Problem section”,
	    S.-T. Yau proposed the well-known conjecture \cite[Problem~47]{Yau1982} that compact K\"ahler manifolds with positive holomorphic sectional curvature must be projective and rationally connected. Recently, we solved this conjecture affirmatively in \cite{Yang2018} by introducing the concept of RC-positivity for abstract vector bundles, and many properties of such bundles are developed in \cite{Yang2018, Yang2018a,Yang2018b,Yang2021}. For instance, we proved in \cite{Yang2018b}  that a compact K\"ahler manifold with uniformly RC-positive tangent bundle must be projective and rationally connected.  By using the ideas of RC-positivity and some deep analytical techniques in algebraic geoemtry, Shin-ichi Matsumura established in \cite{Matsumura2018a,Matsumura2018b,Matsumura2018} a structure theorem for projective manifolds  with non-negative holomorphic sectional curvature, which is analogous to fundamental works in \cite{Mok1988, Campana2015a,Cao2017a,Cao2019} for manifolds with various non-negative properties (see also an approach in \cite{HeierWong2015}). In the same spirit, Lei Ni and Fangyang Zheng introduced in \cite{Ni2018,NiZheng2018a} various notions of Ricci curvature and scalar curvature to obtain rational connectedness of compact K\"ahler manifolds.\\

	In this paper, we investigate the geometry characterized by the first and second Chern-Ricci curvatures on Hermitian manifolds. Recall that for a Hermitian metric $\omega_g$,  its Chern curvature tensor has components $R_{i\bar j k\bar\ell}$. The first Chern-Ricci curvature is $$\mathrm{Ric}^{(1)}(\omega)=\sq \left(g^{k\bar\ell}R_{i\bar j k\bar \ell}\right)dz^i\wedge d\bar z^j=-\sq \p\bp\log\det(h_{k\bar\ell}) $$
	and the second Chern-Ricci curvature is
	$$\mathrm{Ric}^{(2)}(\omega)=\sq \left(g^{k\bar\ell}R_{ k\bar \ell i\bar j}\right)dz^i\wedge d\bar z^j.$$
	When the Hermitian metric $\omega_g$ is not K\"ahler,  $\mathrm{Ric}^{(1)}(\omega)$ and $\mathrm{Ric}^{(2)}(\omega)$ are not necessarily the same. It is well-known that $\mathrm{Ric}^{(1)}(\omega)$ represents the first Chern class of the complex manifold. However, the geometry of $\mathrm{Ric}^{(2)}(\omega)$ is still mysterious.\\

	Thanks to the celebrated Calabi-Yau theorem (\cite{Yau1978}),  we know that  a
	compact K\"ahler manifold $X$ has a Hermitian metric with positive first Chern-Ricci curvature $\mathrm{Ric}^{(1)}(\omega)$
	if and only if $X$ is Fano. As an analog, we proved in \cite{Yang2018} that if a compact K\"ahler manifold
	admits a smooth Hermitian metric with positive second Chern-Ricci curvature $\mathrm{Ric}^{(2)}(\omega)$, then $X$ is projective and rationally connected. This result is also a generalization of the classical result of Campana \cite{Campana1992} and Koll\'ar-Miyaoka-Mori \cite{KollarMiyaokaMori1992} that Fano manifolds are rationally connected. The main result of this paper is the following theorem.

		\btheorem\label{thm:quasipositivesecondchernriccirationallyconnected} Let $X$ be a compact K\"ahler manifold. If there exist a smooth Hermitian metric $\omega$ on $X$ and a smooth Hermitian metric $h$ on the holomorphic tangent bundle $TX$ such that  $$\mathrm{tr}_\omega R^{(TX,h)}\in \Gamma(X, \mathrm{End}(TX))$$ is quasi-positive. Then $X$ is projective and rationally connected. In particular, $X$ is simply connected.
	\etheorem
\noindent Here quasi-positive means non-negative everywhere and strictly positive at some point.  We follow the ideas in \cite{Campana2015a} and \cite{Graber2003} in the proof of Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected}, and the key new ingredient is an integration argument  for singular Hermitian metrics instead of the pointwise maximum principle for RC-positive vector bundles employed in \cite{Yang2018}, since the latter does not work for manifolds with quasi-positive curvature tensors. It is easy to see that many compact K\"ahler manifolds with stable tangent bundle and positive slope can support smooth Hermitian metrics with positive or quasi-positive second Chern-Ricci curvature as required in Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected} (e.g. \cite{UhlenbeckYau1986,Donaldson1987}). On the other hand, the K\"ahler condition in Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected} is necessary (\cite[Section~6]{LiuYang2017}).
As a special case of Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected}, one has

	\bcorollary \label{quasipositivesecondchernriccirationallyconnected} Let $X$ be a compact K\"ahler manifold. If $X$
	admits a smooth Hermitian metric $\omega$ with quasi-positive second Chern-Ricci curvature $\mathrm{Ric}^{(2)}(\omega)$, then $X$ is projective and rationally connected. In particular, $X$ is simply connected.
	\ecorollary

\noindent	By using  Corollary \ref{quasipositivesecondchernriccirationallyconnected}  and Yau's theorem \cite{Yau1978}, one has the following generalization of the result of  Campana \cite{Campana1992} and Koll\'ar-Miyaoka-Mori \cite{KollarMiyaokaMori1992}.

		\bcorollary \label{weakfano} Let $X$ be a compact K\"ahler manifold. If $X$
	admits a smooth Hermitian metric $\omega$ with quasi-positive first Chern-Ricci curvature $\mathrm{Ric}^{(1)}(\omega)$, then $X$ is projective and rationally connected. In particular, $X$ is simply connected.
	\ecorollary

	  \noindent Similarly,	for quasi-negative first Chern-Ricci curvature $\mathrm{Ric^{(1)}}(\omega)$, one has

	\btheorem \label{quasinegative} Let $(X,\omega)$ be a compact Hermitian manifold with quasi-negative first Chern-Ricci curvature $\mathrm{Ric^{(1)}}(\omega)$.
	If $X$ contains no rational curve, then $X$ is  projective  and $K_X$ is ample.
	\etheorem

		\noindent  This result is a straightforward consequence of deep results in complex analytical and algebraic geometry (\cite{Mori1982,Siu1984,Demailly1985,Kawamata1985,Birkar2010,Cascini2013}), which would be of independent interest from the viewpoint of complex differential geometry. It is known that  compact complex manifolds with quasi-negative (or quasi-positive) first Chern-Ricci curvature
	$\mathrm{Ric^{(1)}}(\omega)$ are Moishezon (\cite{Siu1984,Demailly1985}), which are not necessarily projective (e.g. \cite{Ma2007}).
	There are also many compact complex manifolds containing no rational curves, for instances,
	hyperbolic manifolds and Hermitian manifolds with non-positive holomorphic sectional curvature. Theorem \ref{quasinegative} also provides another proof of \cite[Corollary~1.1]{Lee2018} that  compact Hermitian manifolds with non-positive holomorphic bisectional curvature and quasi-negative first Chern-Ricci curvature are projective manifolds with ample canonical bundles,  which was established by purely analytical method.\\

	\noindent
	The following conjecture on quasi-positive holomorphic sectional curvature is well-known and still widely open (e.g. \cite[Conjecture~1.9]{Yang2018b}) and in the special case when $X$ is projective it was confirmed affirmatively by Matsumura  (\cite{Matsumura2018b}).

	\begin{conjecture}\label{quasipositiveHSC} Let $(X,\omega)$ be a compact K\"ahler manifold. If it has quasi-positive holomorphic sectional curvature, then $X$ is projective and rationally connected.
	\end{conjecture}

	\noindent{\bf Acknowledgements.}  The author would like to thank Jie Liu, Wenhao Ou, Valentino Tosatti and S.-T. Yau for inspiring discussions and useful communications.

	\vskip 2\baselineskip

	\section{Singular Hermitian metrics on vector bundles}

	Let $X$ be a complex manifold and $\omega$ be a smooth Hermitian metric on $X$. Locally, we can write the curvature tensor of $(T_X, \omega)$ as
	$$R_{i\bar j k\bar \ell}=-\frac{\p^2 g_{k\bar\ell}}{\p z^i\p\bar z^j}+g^{p\bar q}\frac{\p g_{k\bar q}}{\p z^i}\frac{\p g_{p\bar \ell}}{\p\bar z^j}.$$ where $\omega=\sq g_{i\bar j}dz^i\wedge d\bar z^j$.
	The \emph{first Chern-Ricci curvature} $\mathrm{Ric}^{(1)}(\omega)=\sq R_{i\bar j}dz^i\wedge d\bar z^j$  has components
	$R^{(1)}_{i\bar j}=g^{k\bar\ell}R_{i\bar j k\bar \ell}=-\frac{\p^2\log \det (g_{k\bar\ell})}{\p z^i\p\bar z^j}.$
	The \emph{second Chern-Ricci curvature} is  $\mathrm{Ric}^{(2)}(\omega)=\sq R^{(2)}_{i\bar j}dz^i\wedge d\bar z^j$ where
	$R^{(2)}_{i\bar j}=g^{k\bar\ell}R_{ k\bar \ell i\bar j}$. The scalar curvature $s$ of the Chern connection is $\mathrm{tr}_\omega \mathrm{Ric^{(1)}}(\omega)$ which is also the same as $\mathrm{tr}_\omega \mathrm{Ric^{(2)}}(\omega)$.

		Let $E\>X$ be a holomorphic vector bundle and $h$ be a smooth Hermitian metric on $E$. The curvature $R^E$ of the Chern connection $\nabla $ on $(E,h)$ has a similar formula $$R^E_{i\bar j \alpha\bar\beta}=-\frac{\p^2 h_{\alpha\bar\beta}}{\p z^i\p\bar z^j}+h^{\gamma\bar\delta}\frac{\p h_{\alpha\bar \delta}}{\p z^i}\frac{\p h_{\gamma\bar\beta}}{\p \bar z^j}.$$
	We define
	$R^{(1)}_{i\bar j}=h^{\alpha\bar\beta} R_{i\bar j \alpha\bar\beta}$ and $R^{(2)}_{\alpha\bar\beta}=g^{i\bar j}R_{i\bar j \alpha\bar\beta}$.
	We also call $\mathrm{tr}_hR^E=\sq R_{i\bar j}^{(1)}dz^i\wedge d\bar z^j$ and $\mathrm{tr}_\omega R^E= \sq R_{\alpha\bar \beta}^{(2)}e^\alpha \ts \bar e^\beta$ the first and second Chern-Ricci curvature of $(E,h)$ with respect to the Hermitian manifold $(X,\omega)$ respectively. When $(E,h)=(TX,\omega)$, they are exactly the same as those curvatures of $(X,\omega)$.   A smooth Hermitian $(1,1)$-form $A=\sq A_{i\bar j}dz^i\wedge d\bar z^j$ on $X$ is call \emph{quasi-positive}, if $(A_{i\bar j})$ is non-negative everywhere and positive at some point of $X$. Similarly, we can define it for a  tensor $A \in \Gamma(X,\mathrm{End}(E))$.\\

	 Singular Hermitian metrics on line bundles are introduced in \cite{Demailly1992a} by Demailly. Let $L$ be a holomorphic vector bundle. A singular metric $h^L$ on $L$ can be written locally as $h^L=e^{-\phi}$ for some $\phi\in L^1_{\mathrm{Loc}}(X,\R)$, and the curvature $R^L=-\sq\p\bp\log h^L$ is defined in the sense of distributions. For singular Hermitian metrics on vector bundles, the definition would be very subtle, and we refer to \cite[Section~2]{Paun2018a} for a detailed discussion (see also \cite{Cataldo1998,Raufi2015,DengWangZhangZhou2020}). Recall that, for a smooth Hermitian metric $h$ on $E$, its curvature also takes the local form $R^E=\bp(h^{-1}\p h)$. For the specified purpose in this paper, we only consider singular metrics such that $h^{-1}\p h$ is locally integrable, and then we can use standard theory on distributions (e.g. \cite{Demailly2012}) to define the notion of weak positivity.

	 \bdefinition \label{definitionforsingularmetric} Let $X$ be a  complex manifold. A vector bundle $E$ is called to have positive second Chern-Ricci curvature in the sense of distributions, if there exist a smooth metric $\omega$ on $X$ and a singular Hermitian metric $h^E$ on $E$ such that
	 \beq \mathrm{tr}_\omega R^E \in \Gamma(X,\mathrm{End}(E))\eeq
	 is strictly positive in the sense of distributions. The definition for non-negativity, quasi-positivity and etc. can be defined similarly.
	 \edefinition

	 \bremark Of course, the notion of second Chern-Ricci curvature can be defined in a broader way by using similar constructions as in \cite{Cataldo1998}.  So far, it is not clear to the author whether the notions of Griffiths positivity or Nakano positivity for singular metrics defined in  \cite{Paun2018a,Raufi2015,DengWangZhangZhou2020} can imply the positivity of the second Chern-Ricci curvature in suitable sense, though it is obvious for smooth Hermitian metrics.
	\eremark

	\vskip 2\baselineskip

	\section{Vanishing theorems for singular Hermitian metrics}

	The main result of this section is the following theorem.

	\btheorem\label{vanishingforquasipositive} Let $E$ be a holomorphic vector bundle over a compact complex manifold $X$.
	Suppose there exist a smooth Hermitian metric $\omega$ on $X$ and a smooth Hermitian metric $h$ on  $E$ such that  $$\mathrm{tr}_\omega R^{(E,h)}\in \Gamma(X, \mathrm{End}(TX))$$
	is quasi-positive. We have the following assertions.

	\bd \item Any invertible \textbf{subsheaf} $L$ of $\sO(\ts^k E^*)$ $(k\geq 1)$ is not pseudo-effective.
	\item $\det E^*$ is not pseudo-effective. \ed
	\etheorem

	\noindent The key difficulty in the proof of Theorem \ref{vanishingforquasipositive} is to deal with a line bundle $L$ which is only a subsheaf of $\sO(\ts^k E^*)$. If it is indeed a subbundle, then its follows from a simple observation (c.f. \cite{Campana2015a}).

	\blemma\label{monotonicityofsecondchernricci} The second Chern-Ricci curvature is decreasing in subbundles and increasing in quotient bundles.
	\elemma

	\noindent It is well-known that the first Chern-Ricci curvature is not necessarily monotone as described in Lemma \ref{monotonicityofsecondchernricci}.
	The proof follows from a standard computation  and we include details here since we need it for the distribution case.

	\bproof  Let $(E,h)$ be a Hermitian vector bundle and $S$ be a holomorphic subbundle of $E$.
	Let $r$ be the rank of $E$ and $s$ the rank of $S$. Without
	loss of generality, we can assume,  at a fixed point $p\in X$, there
	exists a local holomorphic frame $\{e_1,\cdots,e_r\}$ of $E$
	centered at point $p$ such that $\{e_1,\cdots, e_s\}$ is a local
	holomorphic frame of $S$. Moreover, we can assume that
	$h(e_\alpha,e_\beta)(p)=\delta_{\alpha\beta}, \text{for } 1\leq
	\alpha, \beta \leq r.$ Hence, the curvature tensor of
	$S$ at point
	$p$ is $$ R^S_{i\bar j \alpha\bar\beta}=-\frac{\p^2
		h_{\alpha\bar\beta}}{\p z^i\p\bar z^j}+\sum_{\gamma=1}^{s}\frac{\p
		h_{\alpha\bar\gamma}}{\p z^i}\frac{\p h_{\gamma\bar\beta}}{\p\bar
		z^j}$$ where $1\leq \alpha,\beta\leq s$. For any Hermitian metric $\omega=\sq g_{i\bar j}dz^i\wedge d\bar z^j$ on $X$, we have
	\beq (\mathrm{tr}_\omega R^E)|_{S}-\mathrm{tr}_\omega R^S= \mathrm{tr}_\omega(R^E|_{S})-\mathrm{tr}_\omega R^S=\sq
	\sum_{\alpha,\beta=1}^sg^{i\bar j}\left(\sum_{\gamma=s+1}^r\frac{\p
		h_{\alpha\bar\gamma}}{\p z^i}\frac{\p h_{\gamma\bar\beta}}{\p\bar
		z^j}\right) e^\alpha\ts
	\bar e^\beta. \label{secondchernriccimonotone}\eeq
	It is easy to see that the right hand side of (\ref{secondchernriccimonotone}) is non-negative. The proof for quotient bundles is similar.
	\eproof

	\bcorollary\label{functorialforsecondricci} Let $E$ be a holomorphic vector bundle over a complex manifold $X$.

	\bd \item  If $E$ has  positive (resp, non-negative) second Chern-Ricci curvature, then each quotient bundle has    positive (resp, non-negative) second Chern-Ricci curvature.

	\item If $E$ has negative (resp, non-positive) second Chern-Ricci curvature, then each subbundle has   negative (resp, non-positive) second Chern-Ricci curvature.

	\item If $E$ has  positive (resp, non-negative, quasi-positive,  negative, non-positive, quasi-negative) second Chern-Ricci curvature, then so is $\ts^k E$ for each $k\geq 1$.

	\item  If $E$ has  positive (resp, non-negative, quasi-positive,  negative, non-positive, quasi-negative) second Chern-Ricci curvature, then so is $\mathrm{Sym}^{\ts k} E$ $(k\geq 1)$ and $\Lambda^pE$ $ (1\leq p\leq \rk(E))$.

	\ed
	\ecorollary

	\bproof We only need to prove $(3)$. From the expression of the induced curvature formula of
	$(\ts ^kE^{},\ts ^kh^{})$, one has
	$$R^{(\ts ^kE^{},\ts ^kh^{})}=\ts ^k R^{(E,h)}\in \Gamma\left(X,\Lambda^{1,1}T^*X\ts \mathrm{End}(\ts^kE)\right),$$
	and
	$$\mathrm{tr}_\omega R^{(\ts ^kE^{},\ts ^kh^{})}=\mathrm{tr}_\omega\left(\ts ^k R^{(E,h)}\right)=\ts^k\left(\mathrm{tr}_\omega R^{(E,h)}\right)\in \Gamma\left(X,\mathrm{End}(\ts^k E)\right).$$
	Hence, the result follows.
	\eproof

	\bremark By standard distribution theory, one has similar results as in   Lemma \ref{monotonicityofsecondchernricci}
and Corollary \ref{functorialforsecondricci}  for singular Hermitian metrics  in  Definition \ref{definitionforsingularmetric}.
	\eremark

	\blemma\label{nonwherevanishing} Suppose that $E$ has non-negative second Chern-Ricci curvature in the  sense of distributions and $L$ is a pseudo-effective line bundle.
	If $\sigma\in H^0(X,E^*\ts L^*)$ is a non-trivial holomorphic section, then $\sigma$ does not vanish everywhere.
	\elemma

	\bproof Let $\omega$ be a smooth Hermitian metric on $X$ and $h^E$ be a singular metric such that $\mathrm{tr}_\omega R^E$ is non-negative in the sense of distributions.
	Since $$\mathrm{tr}_{e^f\omega} R^{E}=e^{-f}\mathrm{tr}_\omega R^E,$$
	we can assume further that $\omega$ is a Gauduchon metric (\cite{Gauduchon1977}), i.e. $\p\bp\omega^{n-1}=0$
	where $\dim X=n$. Let $h^L$ be a singular metric on $L$ such that its curvature $\theta=-\sq\p\bp\log h^L\geq 0$
	in the sense of distributions. For any $\sigma\in H^0(X,E^*\ts L^*)$, we have
	\beq\mathrm{tr}_\omega\left(\sq \p\bp |\sigma|^2_{h^{E^*}\ts h^{L^*}}\right)
	=|\nabla \sigma|^2_{\omega,h^{E^*},h^{L^*}}+(\mathrm{tr}_\omega R^E\cdot h^{L^*}+\mathrm{tr}_\omega\theta\cdot h^{E^*})(\sigma,\sigma).\label{bochnerformula}\eeq
	Note that we are dealing with positive currents, and the product makes sense.
	An integration by part argument with respect to the Gauduchon metric $\omega$  shows that $\nabla\sigma=0$ a.e., and
	\beq (\mathrm{tr}_\omega\sq\p\bp) |\sigma|^2_{h^{E^*}\ts h^{L^*}}\geq 0 \eeq
	in the sense of distributions.  We know $|\sigma|^2_{h^{E^*}\ts h^{L^*}}$ is a global constant on the compact base $X$.
	If $\sigma$ is non-trivial, this constant is nonzero, i.e. $\sigma$ does not vanish everywhere.
	\eproof

	\bcorollary\label{subsheafsubbundle} Suppose that $E$ has non-negative second Chern-Ricci curvature  in the  sense of distributions, and  $L$ is an invertible
	subsheaf of $\sO(E^*)$. If $L$ is pseudo-effective, then $L$ is a line subbundle of $E^*$.
	\ecorollary

	\bproof Note that the subsheaf morphism $f: L\> \sO(E^*)$ induces a nonzero section $\sigma \in H^0(X,E^*\ts L^*)$. By Lemma \ref{nonwherevanishing},
	$\sigma$ does not vanish everywhere. This gives a trivial line subbundle  $\underline{\C}$ of the vector bundle $E^*\ts L^*$, and so $L$ is a line subbundle of $E^*$.
	\eproof

	\btheorem\label{distributionvanishing} Suppose that $E$ has quasi-positive second Chern-Ricci curvature  in the  sense of distributions, and  $L$ is a pseudo-effective line bundle. Then
	\beq H^0(X,E^*\ts L^*)=0.\eeq
	\etheorem

	\bproof By using a similar argument as in the proof of Lemma \ref{nonwherevanishing}, for any holomorphic section $\sigma\in H^0(X,E^*\ts L^*)$,
	we deduce  from (\ref{bochnerformula}) that the holomorphic section $\sigma$ vanishes on an open subset of $X$. By Aronszajn's principle (\cite{Aronszajn1957}), $\sigma$ is a zero section.
	\eproof

\begin{proof}[Proof of Theorem \ref{vanishingforquasipositive}]
	Let $L$ be an invertible subsheaf of $\sO(\ts^k E^*)$ for some
	$k\geq 1$. Suppose-to the contrary-that $L$ is pseudo-effective. By
	Corollary \ref{functorialforsecondricci}, $\ts^k E$ has
	quasi-positive second Chern-Ricci curvature. By Corollary
	\ref{subsheafsubbundle}, the pseudo-effective invertible subsheaf
	$L$ is actually a line subbundle of $\ts^k E^*$. It is easy to see
	that if $\ts^k E$ has quasi-positive second Chern-Ricci curvature,
	then its dual bundle  $\ts^k E^*$ has quasi-negative second
	Chern-Ricci curvature. By decreasing principle in Lemma
	\ref{monotonicityofsecondchernricci}, the second Chern-Ricci
	curvature of $L$ is quasi-negative. Since $L$ is a line bundle,
	that means, the induced metric $h^L$  has quasi-negative Chern
	scalar curvature $s=\mathrm{tr}_\omega (-\sq\p\bp\log h^L)$. Without
	loss of generality, we can assume $\omega$ is Gauduchon. Therefore \beq
	\int_X s\omega^n=n \int_X c_1^{\mathrm{BC}}(L)\wedge
	\omega^{n-1}<0.\eeq  Hence, $L$ can not be
	pseudo-effective (\cite[Proposition~3.1]{Yang2019CM}, ). This is a contradiction.

	On the other hand, since $\det E^*$ is a subbundle of $\ts^k E^*$ for some large $k$, we deduce $\det E^*$ can not be pseudo-effective. \end{proof}

	\vskip 2\baselineskip

	\section{The proof of main theorems}

In this section, we prove Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected}, Corollary \ref{weakfano} and Theorem \ref{quasinegative}.

	\bproof[{Proof of Theorem \ref{thm:quasipositivesecondchernriccirationallyconnected}}]  We set $(E,h)=(TX,h)$.  By Corollary \ref{functorialforsecondricci} and Theorem \ref{distributionvanishing}, we have
	$$H^0(X,(T^*X)^{\ts k})=0.$$
	It is well-known that for any $p$ satisfying $1\leq p\leq \dim X$, $\Lambda^pT^*X$ is a direct summand of $(T^*X)^{\ts k}$ for some large $k$. Therefore, we have
	$$H_{\bp}^{p,0}(X,\C)\cong H^{0}(X,\Lambda^p T^*X)=0.$$
	In particular,  $H_{\bp}^{2,0}(X,\C)=0$. Since $X$ is K\"ahler, we deduce that $X$ is projective (\cite{Kodaira1954}).
	Now we following the ideas in \cite{Graber2003,Peternell2006,Campana2015a} to get the rational connectedness.
	By using Theorem \ref{vanishingforquasipositive}, we conclude that $K_X=\det T^*X$ is not pseudo-effective.
	Thanks to \cite{BoucksomDemaillyPuaunPeternell2013}, $X$ is actually uniruled.  Let $\pi:X\dashrightarrow Z$ be the
	associated MRC fibration of $X$. After possibly resolving the
	singularities of $\pi$ and $Z$, we may assume that $\pi$ is a proper
	morphism and $Z$ is smooth. By \cite[Corollary~1.4]{Graber2003}, it
	follows that the target of the MRC fibration is either a point or a
	positive dimensional variety which is not uniruled. Suppose $X$ is
	not rationally connected, then $\dim Z\geq 1$ and $Z$ is not
	uniruled. By using \cite{BoucksomDemaillyPuaunPeternell2013} again, $K_Z$ is indeed pseudo-effective.
	Since $K_Z=\det(T^*Z)$ is  a direct summand of the vector bundle $(T^*Z)^{\ts k}$ for some large $k$ and $\sO((T^*Z)^{\ts k})$ is a subsheaf of $ \sO\left((T^*X)^{\ts k}\right)$,  we obtain a pseudo-effective invertible subsheaf $K_Z$ of $\sO((T^*X)^{\ts k})$, which contradicts to part $(1)$ of Theorem \ref{vanishingforquasipositive}.
	\eproof

\bproof[Proof of Corollary \ref{weakfano}] By the celebrated Calabi-Yau theorem, there exists a smooth \emph{K\"ahler metric} $\omega_0$ on $X$ such that
\beq \mathrm{Ric}^{(1)}(\omega_0)=\mathrm{Ric}^{(1)}(\omega).\eeq
Since $\omega_0$ is K\"ahler, one can deduce
$$\mathrm{Ric}^{(2)}(\omega_0)=\mathrm{Ric}^{(1)}(\omega_0).$$
Therefore, Corollary \ref{weakfano} follows from Corollary \ref{quasipositivesecondchernriccirationallyconnected}.
\eproof

	\bproof[Proof of Theorem \ref{quasinegative}] By using the criterion of Siu-Demailly (\cite{Siu1984,Demailly1985}), we know $X$ is a Moishezon manifold and $K_X$ is  big. On the other hand, by the deep result  \cite[Corollary~1.4.6]{Birkar2010}, it is proved in \cite[Theorem~3.1]{Cascini2013} that a Moishezon manifold without rational curve must be projective. Moreover, by the base-point-free theorem and the relative cone theorem \cite{Mori1982,Kawamata1985,KollarMiyaokaMori1992}, one can deduce that $K_X$ is ample since $X$ is a projective manifold of general type  without rational curve.
	\eproof

\vskip 2\baselineskip

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\end{document}


text:
\section{Example: Flocking via Second-Order Consensus}
\label{sec:example}

Flocking is a behaviour exhibited by certain multi-agent systems that are coordinating their motion into a cohesive formation, for example birds or stampeding buffalo.
A consensus-type algorithm can be proposed that allows a system of $n$ agents to agree on their velocity vector while maintaining a separation from their neighbours~\cite{Chapman2015}.

\subsection{Matrix-Valued Double Integrator Consensus}

In the case of vector-valued states, we can write the dynamics as
\begin{align}
 \begin{bmatrix}
    \dot{\mathbf{x}} \\ \E\ddot{\mathbf{x}}
  \end{bmatrix} =
  \begin{bmatrix}
    \mathbf{0} & I \\
    - \lap_w & - \lap_w
  \end{bmatrix}
                 \begin{bmatrix}
                   \mathbf{x}\\\dot{\mathbf{x}}
                 \end{bmatrix}+
  \begin{bmatrix}
    \mathbf{0} & \mathbf{0}\\
    I & - \D_\graph
  \end{bmatrix}
        \begin{bmatrix}
          \omega \\ v
        \end{bmatrix}.
\end{align}
\begin{theorem}
  The double-integrator edge consensus model is given by
  \begin{align}
  \begin{bmatrix}
    \dot{\mathbf{x}}_e\\ \ddot{\mathbf{x}}_e
  \end{bmatrix} &=
  \begin{bmatrix}
    \mathbf{0} & 0 & ~~~~I & \\
      -\lap_e \R\W\R^T & \mathbf{0} & -\lap_e \R\W\R^T & \mathbf{0}\\
      \mathbf{0} & 0 & \mathbf{0} & 0
  \end{bmatrix}
\begin{bmatrix}
  \mathbf{x}_e \\ \dot{\mathbf{x}}_e
\end{bmatrix}\\
&+
  \begin{bmatrix}
    \mathbf{0} & \mathbf{0}\\
    \D_\tau^T\E^{-1} & -\lap_e \R\\
    \Xi^{-1} \mathbb{1}^T & \mathbf{0}
  \end{bmatrix}
        \begin{bmatrix}
          \omega \\ v
        \end{bmatrix},
  \end{align}
and so the double-integrator consensus on the edge states of the chosen spanning tree $\tau$ is given by
\begin{align}\label{eq:6}
  \begin{split}
  \begin{bmatrix}
    \dot{\mathbf{x}}_\tau \\ \ddot{\mathbf{x}}_\tau
  \end{bmatrix}&=
                 \begin{bmatrix}
                   \mathbf{0} & I \\
                   -\lap_e \R\W\R^T & -\lap_e \R\W\R^T
                 \end{bmatrix}
                                   \begin{bmatrix}
                                     \mathbf{x}_\tau \\ \dot{\mathbf{x}}_\tau
                                   \end{bmatrix}\\
&+
  \begin{bmatrix}
    \mathbf{0} & \mathbf{0}\\
    \D_\tau^T & -\lap_e \R
  \end{bmatrix}
               \begin{bmatrix}
                 \omega \\ v
               \end{bmatrix}.
               \end{split}
\end{align}
\end{theorem}
\begin{proof}
Appling the coordinate transform $\mathbf{x}_e = S_v \mathbf{x}$ and following a similar calculation as in Theorem~\ref{thm:similarity} yields the result.
\end{proof}

We can also explicitly compute the form of the $\mathcal{H}_2$ norm for the matrix-weighted double-integrator consensus.
\begin{theorem}
  \label{thr:2}
  The controllability gramian for the time scaled double-integrator consensus is given by
  \begin{align}
    \X^* &= \dfrac{1}{2}
    \begin{bmatrix}
      \X_1& \mathbf{0}\\
      \mathbf{0}& \X_2
    \end{bmatrix}\\
   \X_1 &=\sigma_w^2 \left[\left(\R\W\R^T\right)^{-1}\lap_e^{-1}\left(\R\W\R^T\right)^{-1} \right] +\sigma^2_v\left(\R\W\R^T\right)^{-1}\\
    \X_2 &= \sigma_w^2\left(\R\W\R^T\right)^{-1} + \sigma_v^2 \lap_e.
  \end{align}
  Furthermore, the blocks $\X_1,\X_2$ of $\X^*$ correspond to the position and velocity states, respectively.
  Hence, one can consider the $\Htwo$ performance of the position and velocity states separately or aggregately by examining the $\Htwo$ norms
  \begin{align}
    \Htwo^{(1)} = \frac{1}{2}\mathbf{tr}&\left(\R^T\X_1\R\right),~\Htwo^{(2)} = \frac{1}{2}\mathbf{tr}\left(\R^T\X_2\R\right)\\\Htwo^{(1)} &= \frac{1}{2}\mathbf{tr}\left(\R^T(\X_1+ \X_2)\R\right).
  \end{align}
\end{theorem}
\begin{proof}
  From the dynamics~\eqref{eq:6}, the controllablity gramian is given by the positive semi-definite solution to the Lyapunov equation
  \begin{align}
    A\X^* + \X^*A^T + BB^T = \mathbf{0},\label{eq:7}
  \end{align}
  where $\X^*,A,$ and $BB^T$ are given by,
  \begin{align}
    A &=
        \begin{bmatrix}
          \mathbf{0} & I \\
          -\lap_e \R\W\R^T & -\lap_e \R\W\R^T
        \end{bmatrix},\\
    BB^T &=
           \begin{bmatrix}
             \mathbf{0}&\mathbf{0}\\
             \mathbf{0}&\sigma^2_w \lap_e + \sigma^2_v\lap_e \R\R^T \lap_e
           \end{bmatrix},\\
    \X^* &=
          \begin{bmatrix}
            \X_1 & \X_3 \\ \X_3& \X_2
          \end{bmatrix}.
  \end{align}
  Solving Equation~\eqref{eq:7} yields
  \begin{align}
    \X_3 &= \mathbf{0},\ \X_2 = \sigma_w^2\left(\R\W\R^T\right)^{-1} + \sigma_v^2 \lap_e\\
    \X_2 &= \X_1\R\W\R^T\lap_e,
  \end{align}
  and solving for $\X_1$ in the last display yields
  \begin{align}
    \resizebox{\columnwidth}{!}
    {
    $\X_1 = \sigma_w^2 \left[\left(\R\W\R^T\right)^{-1}\lap_e^{-1}\left(\R\W\R^T\right)^{-1} \right] +\sigma^2_v\left(\R\W\R^T\right)^{-1}.$
    }
  \end{align}
To measure the position, velocity, or both states for consideration in the $\Htwo$ norm, one can choose the observation matrices
  \begin{align}
    \begin{bmatrix}
      \R^T & \mathbf{0}
    \end{bmatrix},~
    \begin{bmatrix}
       \mathbf{0} & \R^T
    \end{bmatrix},~
    \begin{bmatrix}
      \R^T & \mathbf{0}\\
      \mathbf{0} & \R^T
    \end{bmatrix},
  \end{align}
  respectively.
\end{proof}

\subsection{Numerical Example}

The weight update scheme applied to the second-order consensus problem was implemented numerically.
The task assigned to the agents was to use the second-order consensus protocol to achieve the formation shown in Figure~\ref{fig:formation} -- a formation assigned by sampling discrete points on a 2D spiral.

\begin{figure}
  \centering
  \includegraphics[width=0.6\columnwidth]{./figs/formation.eps}

  \caption{Relative formation of the agents, defined by discrete points on a spiral.}
  \label{fig:formation}
\end{figure}

Consensus on the formation is achieved by the second-order protocol with a constant signal $\mathbf{d}_i$ specifying the position in the formation:
\begin{align}
\resizebox{\columnwidth}{!}
{
  $\ddot{\mathbf{x}}_i = -\sum_{j\in \mathcal{N}(i)} \W_{ij}(\mathbf{x}_i- \mathbf{x}_j - \mathbf{d}_i) - \sum_{j \in \mathcal{N}(i)} \W_{ij}(\dot{\mathbf{x}}_i - \dot{\mathbf{x}_j}).$
}\label{eq:8}
\end{align}

\noindent Since this is a constant signal, the $\mathcal{H}_2$ performance of the edge states remains the same as in the previous section.

The initial selection of weights was chosen at random using the generator
\begin{align}
W_{\{i,j\}} = \alpha\left[\dfrac{G_{\{i,j\}} + G_{\{i,j\}}^T}{2} + 2 I_2\right],\label{eq:5}
\end{align}
where $G_{\{i,j\}}$ is a $2\times 2$ matrix with entries distributed according to a zero-mean standard Gaussian, and $\alpha=0.3$ was chosen arbitrarily to yield a suboptimal initial selection of weights.
The upper and lower bounds on the weights were chosen with the same generator in Equation~\eqref{eq:5}, but with $\alpha_l = 0.05$ and $\alpha_u = 10$. The initial time scale parameters were taken to be identically unity. The penalty parameter was chosen as $h=0.01$.

The gradient descent algorithm from Proposition~\ref{prop:1} converges quickly, and intermediate graph weights are visualized in Figure~\ref{fig:weights}.
None of the optimal edge weights saturated the upper and lower bounds in this setup.
The minimizing time scale assignment was calculated using Proposition~\ref{prop:2}.

\begin{figure}
  \centering
  \includegraphics[width=0.9\columnwidth]{./figs/4graphs.eps}

  \vspace{-7mm}
  \caption{Visualization of edge weights over iterations $k=1$ (top left), 2 (top right), 3 (bottom left) \& 4 (bottom right). Each of the 3 independent parameters of the $2\times 2$ matrix-valued weight is visualized in a multigraph.}
  \label{fig:weights}
\end{figure}

These weights and scaling paramters were then used in a simulation of the dynamics in~\eqref{eq:8} over a time span of 30 seconds.
At $10\leq t\leq 20$, the formation is subject to a `gust' of noise on the nodes and edges with covariance $\sigma_w^2 I$ and $\sigma_v^2 \W$, with $\sigma_v=\sigma_w = 5$.
Simulations were performed with no updates (NUD), updates to edge weights (WUD) or time scales (TUD) during the wind gusts, and with both updates (BUD) during gusts.
The edge states for the $x$ and $y$ directions for these four cases are shown in Figure~\ref{fig:edgeStates}, and the variance away from the consensus value $x_e(t_f)$, $\mathrm{Var}[x_e(t)]:= \left[x_e(t) - x_e(t_f)\right]^2$, is shown in Figure~\ref{fig:edgeVar}.
The updated weights and scales outperform their initial, suboptimal values.

\begin{figure}
  \centering
  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesNUDx.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesNUDy.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesWUDx.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesWUDy.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesTUDx.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesTUDy.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesBUDx.eps}\\

  \includegraphics[width=0.9\columnwidth]{./figs/edgeStatesBUDy.eps}

  \caption{Edge states in $x,y$ directions over time subjected to gust at $10\leq t\leq 20$; cases from top to bottom are with no updates (NUD), weight updates (WUD), time scale update (TUD), and both updates (BUD). }
  \label{fig:edgeStates}
\end{figure}

\begin{figure}
  \centering
  \includegraphics[width=\columnwidth]{./figs/edgeVariance-14.eps}

  \caption{Variance from consensus for a representative edge $[x_e(t) - x_e(t_f)]^2$ for the case with the weight update (WUD), time scale update (TUD), both updates (BUD), and with no updates (NUD).}
  \label{fig:edgeVar}
\end{figure}


text:
\subsection{Summary of the Approach}
The analysis in Sections \ref{sec:infinite_2_controllers} and  \ref{sec:model_N_controllers} suggests a general approach   for solving infinite horizon decentralized control/DNCS problems. This can be summarized as follows:

\begin{enumerate}

\item Solve the finite horizon version of the DNCS/decentralized control problem (for instance by using the common information approach \cite{nayyar2013decentralized}). Suppose the optimal strategies  are characterized by matrices $P_t^m, m \in \mathcal M = \{1,2,\dots,M\}$ which satisfy $M$ coupled Riccati recursions
\begin{align}
P_t^m =  \Omega(\sum_{j} \theta^{mj} P_{t+1}^j,Q^m,R^m,A^m,B^m),
\label{eq:P_finite_general}
\end{align}
for some matrices $Q^m,R^m,A^m,B^m$ and positive numbers $\theta^{mj}$ for $m,j \in\mathcal M$. Note that we can scale the $\theta$'s such that $\sum_{j\in\mathcal M}\theta^{mj} = 1$ by appropriately scaling $A^m$ and $R^m$ for all $m\in\mathcal M$.

\item Construct a $M$-mode auxiliary MJLS with transition probabilities $\theta^{mj}$ and system matrices $Q^m,R^m,A^m,B^m$
so that the Riccati recursions associated with optimal control of the MJLS coincide with the Riccati recursions \eqref{eq:P_finite_general}.

\item Analyze stability criteria of the auxiliary MJLS to find conditions under which the Riccati recursions of the DNCS reach a steady state.

\item Verify  that the decentralized strategies characterized by the steady state DNCS Riccati equations are optimal.
\end{enumerate}

\subsection{The Information Switching}
 Even though the auxiliary MJLS we used in our analysis is an artificial system without apparent physical meaning (see Remark \ref{rem:fictitious}), a general DNCS with random packet drops (or random packet delays) does have some aspects of a switched system. In particular, the information at a controller (e.g, the remote controller in our problem) switches between different patterns  based on the state of the underlying communication network.   The information of the remote controller in Section \ref{sec:model_2_controllers} clearly switches between two patterns: no observation when the packet is dropped, and perfect observation when the transmission is successful. The number of such patterns seems  related to (but not always equal to) the number of modes in the MJLS used to analyze the DNCS. For the two-controller DNCS of Section  \ref{sec:model_2_controllers}, the number of patterns between which the remote controller's information switches is the same as the number of modes in its auxiliary MJLS.  For the $N+1$ controller DNCS in Section \ref{sec:model_N_controllers}, the remote controller's information appears to switch between $2^N$ patterns (depending on the state of the $N$ links) but its auxiliary MJLS has only $N+1$ modes. This difference between the number of information patterns and the number of modes is due to the nature of the plant dynamics which ensure that the remote controller's estimate of the $n$th plant is not affected by the state of the $m$th link if $m\neq n$.


TITLE: Avg time for human to reach a destination
QUESTION [0 upvotes]: A man near a junction has to reach a house of his relative which is 100m away northwards from the junction.He does not know the location of house.So he travels randomly every minute at 100 m/min speed in some random direction(i.e, north,south,East or west)with equal probability.What is the average time of this travel?
My workout is that he can reach the destination in only odd minutes.But the real difficulty is regarding finding probability,as after every minute the probability is getting split into many like branches of a tree. Can someone please help me out how to approach it in a easy way. Thanks in advance
Edit:The path is like a cordinate system, for example he can be at (1,1) or (-1,1) or (2,0),so on .
In the first minute the probability that he reach is 1/4.
Next minute he can't,the third minute he can reach with certain probabilty and so on.There is a pattern in it for sure but I could not get it.I also tried using permutations,that is lets say that he has reached the destination on 17th minute.
It means he has to travel eight pairs of north-south or east-west or combination with one north step.But again there are severe restrictions within this calculations.

REPLY [0 votes]: an analytical solution using geometric series is the following
$$\mathbb{E}[T]=\frac{1}{4}\sum_{x=0}^{\infty}(2x+1)\left(  \frac{3}{4}\right)^x=2\cdot\sum_{x=0}^{\infty} x\frac{1}{4}\left(  \frac{3}{4}\right)^x+\frac{1}{4}\sum_{x=0}^{\infty}\left(  \frac{3}{4}\right)^x=$$
$$=2\cdot\frac{3}{4}\underbrace{\sum_{x=1}^{\infty}x\frac{1}{4}\left(  \frac{3}{4}\right)^{x-1}}_{=4}+\frac{1}{4}\cdot\frac{1}{1-\frac{3}{4}}=2\times3+1=7$$
the expression in the "underbrace" is the expectation of a geometric rv with parameter $\frac{1}{4}$, thus it is 4


TITLE: Steps and explanation for solving $z^2-(3-2i)z+(5-5i)=0$?
QUESTION [1 upvotes]: I am reading a book and current I am doing some exercises. The exercise that I am having trouble with is this:

Solve the following equation for $z$:
$z^2-(3-2i)z+(5-5i)=0$

I know that the solution is $2+i$ and $1-3i$ but I do not know the steps to get to this result. I have tried using the polar form of complex number to calculate the roots using the solution formula for quadratic equations. However, what I get is not even near the correct results :/
So if anyone could give a user-friendly walkthrough of how to solve that quadratic polynomial - I would be very very happy!
Oh, and here is my general results, if you manage to read it :P

REPLY [1 votes]: The OP made two errors in the exercise:

As the quadratic formula is $\lambda= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, the $-(3-2i)$ should be replaced by $(3-2i)$ for $-b$.
The calculation of $\Delta=b^2-4ac$ is incorrect.

We have
$$\Delta =b^2-4ac=(3-2i)^2-4(1)(5-5i)=(5-12i)-(20+20i)=-15+8i$$
so that
$$\lambda=\frac{3-2i\pm \sqrt{-15+8i}}{2}$$
where we can find the square roots $\pm (a+bi)$ of $-15+8i$ by solving the equation
$$(a+ib)^2=-15+8i$$
We have that $(1+4i)^2=-15+8i$, hence the two square roots are $\pm(1+4i).$ Therefore
$$\lambda=\frac{3-2i + (1+4i)}{2}=2+i$$
or
$$\lambda=\frac{3-2i - (1+4i)}{2}=1-3i$$


TITLE: Evaluate the Nabla Value for Function
QUESTION [1 upvotes]: You are given $\nabla f(0, 1) = (3, 4)$, let $h(x, y) = f(y - x - 1, y - x)$, what is $\nabla h(4, 3)?$

To my understanding cant we just do
$$\nabla h(4, 3) = \nabla f(0, 1) = (3, 4)$$
Or do we have to go through the process of finding $h_x(0, 1)$ and $h_y(0, 1)$?

REPLY [1 votes]: You just need to use the chain rule appropriately. I have seen many fellas who become confused when things come to chain rule. My experience tells me that the confusion is mostly due to inappropriate notations that are used.
Note that $f:\mathbb{R}^2\to\mathbb{R},\,\nabla f:\mathbb{R}^2\to\mathbb{R}^2$. Also, we have a function which is not defined explicitly in the question, i.e.
\begin{align*}
g\,:&\mathbb{R}^2\to\mathbb{R}^2 \\
&(x,y)\to(y-x-1,y-x)
\end{align*}
next, we have the following
$$h=f \circ g$$
Consequently, by the chain rule we have
$$\nabla h=\nabla[f\circ g]=\Big[[\nabla f]\circ g\Big]\cdot\nabla g$$
when we write the above functional equation at $(x,y)$, we get
\begin{align*}
\nabla h(x,y)&=\Big[[\nabla f]\circ g\Big](x,y)\cdot\nabla g(x,y) \\
&=\nabla f\big(g(x,y)\big)\cdot\nabla g(x,y) \\
&=\nabla f\big(y-x-1,y-x\big)\cdot\nabla g(x,y)
\end{align*}
I think you can proceed now.


text:
\begin{document}
\maketitle

\begin{abstract}
	We propose a semantic representation of the standard quantum logic QL within
		a classical, normal modal logic, and this via
			a \emph{lattice-embedding} of
				orthomodular lattices into
				Boolean algebras with one modal operator.

	Thus our classical logic is a \emph{completion} of the quantum logic QL.

	In other words,
		we refute Birkhoff and von Neumann's classic thesis that
			the logic (the formal character) of Quantum Mechanics would be non-classical as well as
		Putnam's thesis that quantum logic (of his kind) would be the correct logic for propositional inference in general.

	The propositional logic of Quantum Mechanics is modal but classical, and
		the correct logic for propositional inference need not have an extroverted quantum character.

	One normal necessity modality $\Box$
	suffices
		to capture the subjectivity of observation in quantum experiments, and this
			thanks to its failure to distribute over classical disjunction.

	The key to our result is the translation of
		\emph{quantum negation as classical negation of observability.}

	\medskip
	\noindent
	\textbf{Keywords:}
		Boolean Algebras with Operators (BAOs);
		algebraic, normal modal logic;
		lattice representation theory;
		ordered algebraic structures;
		orthomodular lattices;
		quantum logic;
		quantum structures.
\end{abstract}

\section{Introduction}
The idea for this paper originates in an observation that
	I made in the plenum of
		a talk delivered by Newton da Costa at
			the 4th World Congress and School on Universal Logic in Rio de Janeiro in 2013.

The observation is about a presentation of a quantum experiment that
	has been put forward
		as a counter-example to the suitability of classical logic for reasoning about quantum phenomena and
		as a motivation for alternative logics such as quantum logics, over and over again.

The presentation usually involves three statements, say $P$, $Q$, and $R$,
	each one being about an elementary quantum phenomenon produced by the experiment,
		but such that
			\begin{equation}
				\underbrace{\text{$\text{$\underbrace{\text{$P$ is observed to be true}}_{\text{$\Box P$ is true}}$ and
						$\underbrace{\text{($Q$ or $R$) is observed to be true}}_{\text{$\Box(Q\lor R)$ is true}}$}$}}_{
							\text{$\Box P\land\Box(Q\lor R)$ is true}}.
			\end{equation}

Notice that the observation of the truth of a disjunction does not imply
	the observation of the truth of one of its disjuncts.

That is,
	$\Box(Q\lor R)\rightarrow(\Box Q\lor\Box R)$ is \emph{not} a valid principle.

This is an essential uncertainty.\footnote{This is also an important difference to
	Intuitionistic Logic (and others), where the validity of a disjunction does imply
		the validity of at least one disjunct (cf.\ Disjunction Property).}

(On the other hand,
	the converse $(\Box Q\lor\Box R)\rightarrow\Box(Q\lor R)$ \emph{is} a valid principle.)

Hence, and in fact,
	\begin{equation}
		\underbrace{\text{neither ($P$ and $Q$) nor ($P$ and $R$) is observed to be true}.}_{\text{
		$\neg\Box(P\land Q)\land\neg\Box(P\land R)$ is true\quad
		(or: $\neg(\Box(P\land Q)\lor\Box(P\land R))$ is true)}}
	\end{equation}

The quantum-mechanical details need not concern us here.

However,
	what does need concern us here is that
		the presentation of the experiment concludes that
			$$\text{($P$ and ($Q$ or $R$)) is true but
				not (($P$ and $Q$) or ($P$ and $R$)).}$$

That is,
	\begin{equation}
		\text{``($P$ and ($Q$ or $R$))'' and ``(($P$ and $Q$) or ($P$ and $R$))'' are not equivalent.}
	\end{equation}

Apparently,
	the distributivity of
		classical conjunction and disjunction fails!

Whence arises the motivation for special \emph{quantum} conjunction and disjunction.

Now, my observation is that
	the experiment---though classic---is not well presented, that is,
		the formalisation of the experimental observations is unfortunate.

The point is that
	the fact of \emph{observing} the fact $P$ and the fact ($Q$ or $R$)
		should be made explicit in the formalisation too, for example as
			$\Box P\land\Box(Q\lor R)$.

Hence in the experiment,
	$\Box P\land\Box(Q\lor R)$ is true (1) but not $\Box(P\land Q)\lor\Box(P\land R)$ (2).

That is,
	$(\Box P\land\Box(Q\lor R))\rightarrow(\Box(P\land Q)\lor\Box(P\land R))$ is false.

On the other hand,
	the converse $(\Box(P\land Q)\lor\Box(P\land R))\rightarrow(\Box P\land\Box(Q\lor R))$ is true, because:
		$$\begin{array}{@{}r@{\;}c@{\;}l@{}}
			 (\Box(P\land Q)\lor\Box(P\land R))&\rightarrow& \Box((P\land Q)\lor(P\land R)) \text{ is true}\\[2\jot]
			 	&\leftrightarrow& \Box(P\land (Q\lor R)) \text{ is true}\\[2\jot]
			 	&\leftrightarrow& (\Box P\land\Box(Q\lor R)) \text{ is true}
		\end{array}$$
(As noticed above,
	$\Box$ distributes over $\land$ in both directions, but over $\lor$ only in one direction.)

Thus, and in close correspondence with (3),
	\begin{equation}
		\text{$(\Box P\land\Box(Q\lor R))\leftrightarrow(\Box(P\land Q)\lor\Box(P\land R))$ is false.}
	\end{equation}

Hence,
	if we \emph{\textbf{make explicit the fact of observing facts}} (for example by means of a modal operator $\Box$)
	then we do not need to introduce the special-purpose formalism of Quantum Logic with
		special and possibly counter-intuitive quantum operators
			to account for quantum phenomena (due to the apparent failure of classical conjunction to distribute over classical disjunction), but
				can get by with intuitive classical (Boolean) logic at the small price of
					adding a single, classical modal operator $\Box$\,.

This operator is characterised by the following
	two axioms K (normality) and BQ plus the single deduction rule N (normality):
	\begin{enumerate}
		\item $\Box(A\rightarrow B)\rightarrow(\Box A\rightarrow\Box B)$\qquad(Kripke's law, K)
		\item $\Box\Diamond A\leftrightarrow A$, where $\Diamond := \neg\Box\neg$\qquad(BQ)
		\item from $A$ infer $\Box A$\qquad(necessitation rule, N)
	\end{enumerate}

BQ has the encouraging meaning that quantum truth is equivalent to
	\begin{itemize}
		\item observing the possibility of that truth, or, equivalently,
		\item the possibility of observing that truth.
	\end{itemize}

(Consider that
	$\Box\Diamond A\leftrightarrow A$ is true if and only if
	$\neg\Box\Diamond A\leftrightarrow\neg A$ is true if and only if
	$\Diamond\neg\Diamond A\leftrightarrow\neg A$ is true if and only if
	$\Diamond\Box\neg A\leftrightarrow\neg A$ is true if and only if
	$\Diamond\Box A\leftrightarrow A$ is true.)

The laws of Boolean logic plus
the three modal laws K, BQ, and N, which are embodied in our classical modal logic BQ,
suffice
	to model the logical essence of Quantum Mechanics as captured by
		the standard quantum logic QL \cite[Page~32]{HBQuantumLogic}.

To appreciate the smallness of the price of understanding BQ,
	consider the price of understanding QL.

This standard quantum logic is characterised by
	the following 12 axioms plus one deduction rule \cite[Page~32]{HBQuantumLogic}:
\begin{enumerate}
	\item $( A \equiv B )\rightarrow_{0}(( B \equiv C )\rightarrow_{0}( A \equiv C ))$
	\item $( A \equiv B )\rightarrow_{0}(\acompl A \equiv\acompl B )$
	\item $( A \equiv B )\rightarrow_{0}(( A \curlywedge C )\equiv( B \curlywedge C ))$
	\item $( A \curlywedge B )\equiv( B \curlywedge A )$
	\item $( A \curlywedge ( B \curlywedge C ))\equiv(( A \curlywedge B )\curlywedge C )$
	\item $( A \curlywedge ( A \curlyvee B ))\equiv A $
	\item $(\acompl  A \curlywedge A )\equiv((\acompl A \curlywedge A )\curlywedge B )$
	\item $A \equiv\acompl\acompl A $
	\item $\acompl( A \curlyvee B )\equiv(\acompl A \curlywedge\acompl B )$\qquad(De-Morgan law)
	\item $( A \equiv B )\equiv( B \equiv A )$
	\item $( A \equiv B )\rightarrow_{0}( A \rightarrow_{0} B )$
	\item $( A \rightarrow_{0} B )\rightarrow_{3}( A \rightarrow_{3}( A \rightarrow_{3} B ))$
	\item from $A$ and $A \rightarrow_{3} B$ infer $B$
\end{enumerate}
with the three abbreviations:
	$$\begin{array}[t]{@{}r@{\ \ }c@{\ \ }l@{}}
				 A \rightarrow_{0} B &:=&\acompl A \curlyvee B\\[\jot]
				 A \rightarrow_{3} B &:=&(\acompl A \curlywedge B )\curlyvee(\acompl A \curlywedge\acompl B )\curlyvee( A \curlywedge(\acompl  A \curlyvee B ))\\[\jot]
				 A \equiv B &:=&( A \curlywedge B )\curlyvee(\acompl A \curlywedge\acompl B )
		\end{array}$$
and $\curlywedge$, $\curlyvee$, and $\acompl$ symbolising
	quantum conjunction, quantum disjunction, and quantum negation, respectively
		(my choice, to distinguish the quantum operators clearly from their classical counterparts).

Observe that
	this axiom system for QL employs
		two notions of implication, symbolised as $\rightarrow_{0}$ and $\rightarrow_{3}$, as well as
		one notion of equivalence, symbolised as $\equiv$\,, but that
			is defined in terms of neither notion of implication!

Proving the adequacy of this axiom system for
	the standard quantum structures of orthomodular lattices
	must have been a real \emph{tour de force,} which
	must be appreciated as such.

Fortunately,
	we do not need to understand nor use this axiom system in order to
		understand the logical essence of Quantum Mechanics.

All we need to understand is that
	QL and its corresponding, standard quantum-structure semantics of
		orthomodular lattices satisfy the De-Morgan law, and thus
			quantum disjunction is definable in terms of quantum negation and quantum conjunction.

Hence in essence,
	the culprit for the failure of quantum conjunction
		to distribute over quantum disjunction boils down to quantum negation!

We thus must find a translation of quantum negation in terms of
	the modal operator $\Box$ and Boolean operators.

The translation that we have found and shall now present and explicate is to
	$$\boxed{\text{translate quantum negation $\acompl$ as $\neg\Box$\,.}}$$

That is,
	we translate \emph{quantum negation as classical negation of observability.}

Recall from classical normal modal logic that
	$\neg\Box$ (``not necessarily'') is the same as $\Diamond\neg$ (``possibly not'').

Hence,
	the classical negation of observability is
		classically equivalent to the possibility of observing classical negation.

Thus,
	we can also view \emph{quantum negation as
		the possibility of observing classical negation.}

\section{The technical details of the translation}
We shall carry out our translation from QL into BQ semantically, by
	producing a \emph{lattice-embedding} from
		the standard orthomodular-lattice (OML-)model of QL into
		the standard Boolean-algebra-with-one-operator (BAO-)model of BQ \cite{AlgebrasAndCoalgebras}.

So, not only logicians but also mathematicians will be able to appreciate our result.

For simplicity,
	we will reuse some of the symbols
		for the (syntactic) operators of QL from the introduction
		for their corresponding (semantic) operators of the OML-model of QL, and
	only use different symbols for
		the (syntactic) operators of BQ and
		their corresponding (semantic) operators of our BAO.

Both algebraic models involve a set of subsets of a set of states as carrier together with
	algebraic operations on this carrier set.

So,
	let
		$\states$ designate our \emph{state space,} that is,
			the set of all possible worlds, points, or states,
		$\mathrm{H}(\states)$ the set of subsets of $\states$ that
			is algebraically closed under the OML-operations for
			the carrier of the OML-model, and
		$\mathrm{P}(\states)$ the powerset of $\states$ for
			the carrier of our BAO-model.

Then, translate
	the \emph{\textbf{orthomodular}} (and thus \emph{De Morgan}) \emph{\textbf{lattice}} \cite{HBQuantumLogic}
		$$\OML\ :=\
			\langle\,\mathrm{H}(\states),0,\curlywedge, \curlyvee,1,\ocompl{\cdot},\preccurlyeq\,\rangle$$
	on $\states$ to a corresponding (inclusion-ordered, complete) \emph{\textbf{Boolean} (powerset) \textbf{algebra} (lattice)}
		$$\BAO\ :=\
			\langle\,\mathrm{P}(\states), \emptyset, \cap, \cup, \states, \compl{\,\cdot\,},\posop{R},\subseteq\,\rangle$$
	 \emph{\textbf{with one operator}}
		$\posop{R}:2^\states\to2^\states$
		$$\posop{R}(S)\ :=\
				\{\,s\in\states\mid\text{there is $s'\in\states$ such that $s\mathrel{R}s'$ and $s'\in S$}\,\}$$ for an---only for now---arbitrary (see the end of this section) accessibility relation\footnote{
					Accessibility relations are at the heart of Kripke-semantics for modal logics \cite{ModalLogicSemanticPerspective}.} $R\subseteq\states\times\states$ \cite{AlgebrasAndCoalgebras}, that is,
						a binary relation on $\states$ of no particularity, and with
	dual operator
			$\necop{R}:2^\states\to2^\states$ \cite[Definition~3.8.2]{PracticalFoundationsOfMathematics}
			$$\necop{R}(S)\ :=\
				\{\,s\in\states\mid\text{for all $s'\in\states$, if $s\mathrel{R}s'$ then $s'\in S$}\,\}$$
		by means of
		an \emph{injective} mapping $\rho:\OML\rightarrowtail\BAO$ such that
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{}}
	\rho(\ocompl{H}) &=& \acompl\rho(H)\qquad\text{($\acompl$-homomorphism)}\\[\jot]
	\rho(H\curlywedge H') &=& \rho(H)\cap\rho(H')\qquad\text{(meet homomorphism)}
\end{array}$$
	where $\acompl:=\posop{R}\circ\compl{\,\cdot\,}$.
As usual, $\circ$ designates function composition.

Note that $\rho$ is induced (and thus exists) by
	its syntactic counterpart mentioned in the introduction
		(translate $\acompl$ as $\neg\Box$, or, equivalently,  as $\Diamond\neg$), and this via
			the standard Lindenbaum-Tarski-algebra construction \cite{DaveyPriestley}
				on the language of QL and on the language of BQ.

(See also Corollary~\ref{corollary:reduction} for this translation.)

The operator $\necop{R}$ is the semantic analog of the modality $\Box$, and
	is related to $R$ as asserted by the following fact.
\begin{fact}[{\cite[Exercise~3.65]{PracticalFoundationsOfMathematics}}]
$$\text{$s\mathrel{R}s'$ if and only if
	for all $S\subseteq\states$, $s\in\necop{R}(S)$ implies $s'\in S$}$$
\end{fact}
\noindent
As opposed to the operator $\necop{R}$,
	its dual $\posop{R}$ does distribute over set union $\cup$,
		as asserted by the following well-known, but to our development crucial fact.
\begin{fact}[Property of $\posop{R}$]\label{fact:PropertiesOfPosOp}
	$$\posop{R}(S\cup S')=\posop{R}(S)\cup\posop{R}(S')$$
\end{fact}
\noindent
Now, recall the following laws of orthomodular lattices:
\begin{itemize}
	\item De Morgan:
			$$H\curlyvee H' = \ocompl{(\ocompl{H}\curlywedge\ocompl{H'})}$$
	\item orthocomplementarity:
		\begin{itemize}
			\item involution: $\ocompl{\ocompl{H}}=H$
			\item disjointness: $H\curlywedge\ocompl{H}=0$
			\item exhaustiveness: $H\curlyvee\ocompl{H}=1$
			\item antitonicity: $H\preccurlyeq H'\Rightarrow\ocompl{H'}\preccurlyeq\ocompl{H}$
		\end{itemize}
	\item orthomodularity (OM):
			$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\ }r@{\ }l@{}}
				H\preccurlyeq H' &\text{:iff}& H&=&H\curlywedge H'\\[\jot]
								 &\Leftrightarrow& H'&=&H\curlyvee H'\\[\jot]
								 &\Rightarrow& H'&=&H\curlyvee(H'\curlywedge\ocompl{H})\qquad\text{(OM)}
			\end{array}$$
\end{itemize}

\begin{proposition}\label{proposition:Completion}
	The complete lattice $\BAO$ is a \emph{completion} \cite[Definition~7.36]{DaveyPriestley} of
		the lattice (and thus partially ordered set) $\OML$ via
			the \emph{\textbf{order}-embedding} $\rho$, that is,
	$$\text{for all $H,H'\in\mathrm{H}(\states)$,
		$H\preccurlyeq H'$ if and only if $\rho(H)\subseteq\rho(H')$\,.}$$
\end{proposition}
\begin{proof}
	Let $H,H'\in\mathrm{H}(\states)$.

	Suppose that $H\preccurlyeq H'$.

	By definition of $\preccurlyeq$,
		$H=H\curlywedge H'$.

	Further suppose that $s\in\rho(H)$.

	Thus $s\in\rho(H\curlywedge H')$.

	Hence $s\in\rho(H)\cap\rho(H')$ by the meet-homomorphism property of $\rho$.

	Thus $s\in\rho(H')$.

	Conversely, suppose that $\rho(H)\subseteq\rho(H')$.

	Thus $\rho(H)=\rho(H)\cap\rho(H')$.

	Hence $\rho(H)=\rho(H\curlywedge H')$ by the meet-homomorphism property of $\rho$.

	Hence $H=H\curlywedge H'$ by the injectivity of $\rho$.

	Thus $H\preccurlyeq H'$.
\end{proof}
\noindent
Notice that
	we have not used any property of $\acompl$ in the proof of Proposition~\ref{proposition:Completion}.

The interesting properties of $\acompl$ are enumerated in the next proposition, which
	asserts that $\acompl$ can mimic the orthocomplement $\ocompl{}$ of $\OML$ in our $\BAO$ via $\rho$.

The essential ones are the first and the second
	(see the proofs of the others).
\begin{proposition}[Properties of $\acompl{}$]\label{proposition:PropertiesOfSim}\
	\begin{enumerate}
		\item \colorbox[gray]{0.75}{$\acompl\acompl\rho(H)=\rho(H)$\qquad(involutive interaction with itself)}
		\item \colorbox[gray]{0.75}{$\acompl(S\cap S')=\acompl S\cup\acompl S'$\qquad(De-Morgan interaction with meet and join)}

				(thus $\acompl(\rho(H)\cap\rho(H'))=\acompl\rho(H)\cup\acompl\rho(H')$)
		\item $\acompl(\rho(H)\cup\rho(H'))=\acompl\rho(H)\cap\acompl\rho(H')$
		\item $\rho(H)\cap\acompl{\rho(H)}=\rho(0)$
		\item $\rho(H)\cup\acompl{\rho(H)}=\rho(1)$
		\item $(H\preccurlyeq H'\text{ or }\rho(H)\subseteq\rho(H'))$ implies
				\begin{enumerate}
					\item $\acompl\rho(H')\subseteq\acompl\rho(H)$ and
					\item $\rho(H')=\rho(H)\cup(\rho(H')\cap\acompl{\rho(H)})$
				\end{enumerate}
		\item $\acompl\rho(0)=\rho(1)$
		\item $\acompl\rho(1)=\rho(0)$
	\end{enumerate}
\end{proposition}
\begin{proof}
	For (1),
		let $H\in\mathrm{H}(\states)$ and
		recall that $\ocompl{\ocompl{H}}=H$.

	Thus $\rho(\ocompl{\ocompl{H}})=\rho(H)$.

	Hence $\acompl{\acompl{\rho(H)}}=\rho(H)$.

	For (2),
		let $S,S'\in\mathrm{P}(\states)$ and
		consider:
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
		\acompl(S\cap S')	&=& \acompl\left(\,\compl{\compl{S}\cup\compl{S'}}\,\right) & \text{([Boolean] De Morgan)}\\[2\jot]
							&=& (\posop{R}\circ\compl{\,\cdot\,}\circ\compl{\,\cdot\,}\,)\left(\,\compl{S}\cup\compl{S'}\,\right) & \text{(definition)}\\[\jot]
							&=& \posop{R}\left(\,\compl{S}\cup\compl{S'}\,\right) & \text{([Boolean] involution)}\\[\jot]
							&=& \posop{R}\left(\,\compl{S}\,\right)\cup\posop{R}\left(\,\compl{S'}\,\right) & \text{(Fact~\ref{fact:PropertiesOfPosOp})}\\[\jot]
							&=& (\posop{R}\circ\compl{\,\cdot\,}\,)(S)\cup(\posop{R}\circ\compl{\,\cdot\,}\,)(S') & \text{(definition)}\\[\jot]
							&=& \acompl S\cup\acompl S' & \text{(definition)}
	\end{array}$$
	For (3),
		let $H,H'\in\mathrm{H}(\states)$ and
		consider:
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
		\acompl(\rho(H)\cup\rho(H'))	&=& \acompl(\acompl(\acompl\rho(H))\cup\acompl(\acompl\rho(H'))) & \text{(1)}\\[\jot]
							&=& \acompl\acompl(\acompl\rho(H)\cap\acompl\rho(H')) & \text{(2)}\\[\jot]
							&=& \acompl\rho(H)\cap\acompl\rho(H') & \text{(1)}
	\end{array}$$

	For (4) and (5), let $H\in\mathrm{H}(\states)$.

	For (4), recall that $H\curlywedge\ocompl{H}=0$.

	Thus $\rho(H\curlywedge\ocompl{H})=\rho(0)$.

	Hence,
		$\rho(H)\cap\rho(\ocompl{H})=\rho(0)$ and then
		$\rho(H)\cap\acompl{\rho(H)}=\rho(0)$.

	For (5), recall that $H\curlyvee\ocompl{H}=1$.

	Hence:
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
		\rho(1) &=& \rho(H\curlyvee\ocompl{H}) & \\[2\jot]
				&=& \rho\left(\ocompl{\left(\ocompl{H}\curlywedge\ocompl{\ocompl{H}}\right)}\right) & \text{(definition)}\\[4\jot]
				&=& \rho\left(\ocompl{(\ocompl{H}\curlywedge H)}\right) & \text{(involution)}\\[2\jot]
				&=& \acompl{\rho(\ocompl{H}\curlywedge H)} & \text{($\acompl$-homomorphism)}\\[\jot]
				&=& \acompl{(\rho(\ocompl{H})\cap\rho(H))} & \text{(meet-homomorphism)}\\[\jot]
				&=& \acompl{\rho(\ocompl{H})}\cup\acompl{\rho(H)} & \text{(2)}\\[\jot]
				&=& \acompl{\acompl{\rho(H)}}\cup\acompl{\rho(H)} & \text{($\acompl$-homomorphism)}\\[\jot]
				&=& \rho(H)\cup\acompl{\rho(H)} & \text{(1)}
	\end{array}$$
	For (6.a) and (6.b),
		let $H,H'\in\mathrm{H}(\states)$ and
		suppose that $H\preccurlyeq H'\text{ or }\rho(H)\subseteq\rho(H')$.

	For (6.a),
		first suppose that $H\preccurlyeq H'$.

	Thus $\ocompl{H'}\preccurlyeq\ocompl{H}$ by antitonicity.

	Hence $\rho(\ocompl{H'})\subseteq\rho(\ocompl{H})$ by Proposition~\ref{proposition:Completion}.

	Hence $\acompl{\rho(H')}\subseteq\acompl{\rho(H)}$ by $\acompl$-homomorphism.

	Now suppose for (6.a) that $\rho(H)\subseteq\rho(H')$.

	Hence $H\preccurlyeq H'$ by Proposition~\ref{proposition:Completion}, and
		proceed like in the first case.

	For (6.b),
		first suppose that $H\preccurlyeq H'$.

	Thus $H'=H\curlyvee(H'\curlywedge\ocompl{H})$ by orthomodularity.

	Hence:
		$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
			\rho(H')	&=& \rho(H\curlyvee(H'\curlywedge\ocompl{H})) & \\[2\jot]
						&=& \rho\left(\ocompl{\left(\ocompl{H}\curlywedge\ocompl{\left(H'\curlywedge\ocompl{H}\right)}\right)}\right) & \text{(definition)}\\[4\jot]
						&=& \acompl\rho\left(\ocompl{H}\curlywedge\ocompl{\left(H'\curlywedge\ocompl{H}\right)}\right) & \text{($\acompl$-homomorphism)}\\[3\jot]
						&=& \acompl\left(\rho(\ocompl{H})\cap\rho\left(\ocompl{\left(H'\curlywedge\ocompl{H}\right)}\right)\right) & \text{(meet homomorphism)}\\[3\jot]
						&=& \acompl\rho(\ocompl{H})\cup\acompl\rho\left(\ocompl{\left(H'\curlywedge\ocompl{H}\right)}\right) & \text{(2)}\\[2\jot]
						&=& \acompl\acompl\rho(H)\cup\acompl\acompl\rho(H'\curlywedge\ocompl{H}) & \text{($\acompl$-homomorphism)}\\[\jot]
						&=& \rho(H)\cup\rho(H'\curlywedge\ocompl{H}) & \text{(1)}\\[\jot]
						&=& \rho(H)\cup(\rho(H')\cap\rho(\ocompl{H})) & \text{(meet-homomorphism)}\\[\jot]
						&=& \rho(H)\cup(\rho(H')\cap\acompl\rho(H)) & \text{($\acompl$-homomorphism)}
		\end{array}$$
	Now suppose for (6.b) that $\rho(H)\subseteq\rho(H')$.

	Hence $H\preccurlyeq H'$ by Proposition~\ref{proposition:Completion}, and
		proceed like in the first case.

	For (7), consider (4).

	Hence:
		$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
			\acompl\rho(0)	&=& \acompl(\rho(H)\cap\acompl\rho(H)) & \text{}\\[\jot]
							&=& \acompl\rho(H)\cup\acompl\acompl\rho(H) & \text{(2)}\\[\jot]
							&=& \acompl\rho(H)\cup\rho(H) & \text{(1)}\\[\jot]
							&=& \rho(1) & \text{(5)}
		\end{array}$$

	For (8), consider (7).

	Thus $\acompl\acompl\rho(0)=\acompl\rho(1)$.

	Hence  $\rho(0)=\acompl\rho(1)$ by (1).
\end{proof}
\noindent
In spite of Proposition~\ref{proposition:PropertiesOfSim} having only the status of a proposition,
	its proof actually contains more information than
		the proof of the following (main) theorem.
\begin{theorem}[Representation Theorem for Orthomodular Lattices]\label{theorem:Representation}
	The structure
		$\mathfrak{S}:=\langle\,\{\,\rho(H) \mid H\in\mathrm{H}(\states)\,\},\rho(0),\cap, \cup,\rho(1),\acompl,\subseteq\,\rangle$ is a \emph{sublattice of sets} of the powerset lattice $\BAO$ that
			is \emph{isomorphic} to $\OML$ via
			the \emph{\textbf{lattice}-embedding} $\rho$, that is,
				$\rho$ is a bijection between $\OML$ and $\mathfrak{S}$, and preserves the structure:
			$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{}}
			\rho(0) &=& \rho(0) \\[\jot]
			\rho(H\curlywedge H') &=& \rho(H)\cap\rho(H') \\[\jot]
			\rho(H\curlyvee H')	&=& \rho(H)\cup\rho(H') \\[\jot]
			\rho(1) &=& \rho(1) \\[\jot]
			\rho(\ocompl{H}) &=& \acompl{\rho(H)}
		\end{array}$$
\end{theorem}
\begin{proof}
	By definition,
		$\rho$ is an injection from $\OML$ into $\BAO$ and thus also into $\mathfrak{S}$, and
		preserves the orthocomplement $\ocompl{}$ and the orthomodular meet $\curlywedge$.

	By definition of $\mathfrak{S}$,
		$\rho$ is also a surjection from $\OML$ onto $\mathfrak{S}$, and
		preserves also the orthomodular bounds $0$ and $1$.

	For the preservation of the orthomodular join $\curlyvee$ consider that:
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{\qquad}l@{}}
		\rho(H\curlyvee H') &=&
			\rho\left(\ocompl{\left(\ocompl{H}\curlywedge\ocompl{H'}\right)}\right) & \text{(definition)}\\[\jot]
			&=& \acompl\rho(\ocompl{H}\curlywedge\ocompl{H'}) & \text{($\acompl$-homomorphism)}\\[\jot]
			&=& \acompl(\rho(\ocompl{H})\cap\rho(\ocompl{H'})) & \text{(meet homomorphism)}\\[\jot]
			&=& \acompl\rho(\ocompl{H})\cup\acompl\rho(\ocompl{H'}) & \text{(Proposition~\ref{proposition:PropertiesOfSim}.2)}\\[\jot]
			&=& \acompl\acompl\rho(H)\cup\acompl\acompl\rho(H') & \text{($\acompl$-homomorphism)}\\[\jot]					&=& \rho(H)\cup\rho(H') & \text{(Proposition~\ref{proposition:PropertiesOfSim}.1)}
		\end{array}$$
\end{proof}
\noindent
Let us take stock, and record
	which properties of the accessibility relation $R$ were actually required to prove our theorem.

Observe that only
	Proposition~\ref{proposition:PropertiesOfSim}.1 and
	\ref{proposition:PropertiesOfSim}.2 require such properties.

The proof of Proposition~\ref{proposition:PropertiesOfSim}.2 requires
	Fact~\ref{fact:PropertiesOfPosOp} and only that one as such a property.

Less obviously,
	because somehow hidden in plain sight,
		Proposition~\ref{proposition:PropertiesOfSim}.1
			\emph{itself} is actually another such property.

It stipulates that
	for all $S\in\{\,\rho(H) \mid H\in\mathrm{H}(\states)\,\},$
	$$\begin{array}{@{}r@{\ \ }c@{\ \ }l@{}}
				S &=&	(\posop{R}\circ\compl{\,\cdot\,}\circ\posop{R}\circ\compl{\,\cdot\,})(S)\\[\jot]
				  &=&	(\posop{R}\circ\necop{R}\circ\compl{\,\cdot\,}\circ\compl{\,\cdot\,})(S)\\[\jot]
				  &=&	(\posop{R}\circ\necop{R})(S)\,.
		\end{array}$$
It is well known that
	the inclusion $(\posop{R}\circ\necop{R})(S)\subseteq S$ stipulates the \emph{symmetry} of $R$,
		corresponding to the so-called \emph{B-axiom} $A\rightarrow\Box\Diamond A$
			(see for example \cite{ModalProofTheory}).

However to our knowledge,
	the inclusion $S\subseteq(\posop{R}\circ\necop{R})(S)$ has no name yet;
		we shall call it the \emph{\textbf{Q-property}} of $R$,
		its corresponding axiom $\Box\Diamond A\rightarrow A$ the \emph{\textbf{Q-axiom,}} and
			the resulting normal modal logic the logic \emph{\textbf{BQ}} (= K+B+Q).

Observe that
	both the B-axiom as well as the Q-axiom are Sahlqvist-formulas \cite{Sahlqvist}, and
		so correspond to (satisfiable) first-order-logically (FOL-)definable classes of Kripke-frames.

Of course,
	symmetry is FOL-definable:
		$\forall s\forall s'(s\mathrel{R}s' \rightarrow s'\mathrel{R}s)$.

And so is the class of Kripke-frames corresponding to the Q-axiom \cite{SQEMA}:
	$\forall s\exists s'(s\mathrel{R}s'\land\forall s''((s'\mathrel{R}s'')\rightarrow(s''=s)))$, which
		implies the seriality $\forall s\exists s'(s\mathrel{R}s')$ of $R$, which in turn
			corresponds to the D-axiom $\Box A\rightarrow\Diamond A$ (or equivalently, $\neg\Box\bot$).

That is,
	falsehood can never be observed to be true in a quantum experiment (consistency of observability).

Note that the satisfiability of these FOL-translations implies
	the existence of our accessibility relation $R$, which
		is characterised in terms of these translations.

In the light of \cite{QLnonelementary},
	the FOL-definability of our accessibility relation $R$ may seem contradictory.

However,
	consider that \cite{QLnonelementary} applies to
		Kripke-semantics of QL and not to
			Kripke-semantics of some modal logic (such as BQ) into which QL embeds.

The properties of the two accessibility relations need not coincide.

For a well-known counter-example consider
	Intuitionistic Logic (IL), whose
		Kripke-semantics has an accessibility relation that is a \emph{partial} order \cite{KripkeSemanticsIL}, and
	the normal modal logic S4, into which IL embeds via the G\"odel-McKinsey-Tarski translation, and
		whose accessibility relation is only a \emph{pre-}order.

In the following corollary,
	we apply $\rho$ tacitly on QL-formulas $A$ rather than
		on their semantic counterparts $H$ (their Lindenbaum-Tarski algebra quotient).
\begin{corollary}\label{corollary:reduction}
	Syntactically,
		$\rho$ is a \emph{linear-time reduction} from QL to BQ:
			$$\text{$A\in QL$ if and only if $\rho(A)\in BQ$.}$$
\end{corollary}
\begin{proof}
	Assuming that any quantum disjunction in a given QL-formula $A$ has been expanded by
	its definition in terms of quantum negation and quantum conjunction,
we just substitute
	any occurrence of the quantum-negation symbol $\acompl$ in
	$A$ with $\neg\Box$ or with $\Diamond\neg$ (as already said in the introduction), in
		order to obtain the corresponding BQ-formula.

When performed on $A$ represented as a (linear) string of symbols,
	this substitution procedure obviously takes linear time (no back-tracking required).
\end{proof}

\section{Conclusion}
We have demonstrated that
	Quantum Logic (QL) is a fragment of the classical normal modal logic BQ, which
		in turn is a fragment of classical (in every sense of the word) first-order logic
			(via Standard Translation \cite{ModalLogicSemanticPerspective}).

	In other words,
		we have refuted Birkhoff and von Neumann's classic thesis that
			the logic (the formal character) of Quantum Mechanics would be non-classical \cite{BirkhoffVonNeumann} as well as
		Putnam's thesis that quantum logic (of his kind) would be the correct logic for propositional inference in general \cite{PutnamQL}.

	The propositional logic of Quantum Mechanics has turned out to be modal but classical, and
		the correct logic for propositional inference need not have an extroverted quantum character.

The philosophical key to our result has been to
	\emph{internalise observability} into our (logical) system (by means of a normal necessity modality),
		which in some sense is what Quantum Mechanics has always told us to do.

With that,
	the mystery of the failure of classical conjunction to distribute over classical disjunction
		has dissolved and
	an elementary-logical solution for this (weak)\footnote{
		As opposed to a strong paradox (a formal contradiction/inconsistency),
			a weak paradox dissolves upon (proper) formalisation.} paradox has emerged.

(Other paradoxes of Quantum Mechanics may subsequently dissolve too.)

In a formal sense,
	we have reduced Quantum Logic (QL) to Classical Logic,
		within a simple modal logic (BQ).

Translations of QL in a similar spirit but to more complex modal systems can be found in \cite{QLintoBr} and \cite{DynamicQL}.

Then, translations of ortholattices
	(lattices with an orthocomplement but,
		as opposed to our orthomodular lattices,
			without the orthomodular law) can be found in
				\cite{OrthologicSemantically} and \cite{QuantumLogics}.

Finally, first-order extensions (with quantifiers) as well as
	dynamic and fixpoint extensions are possible
		(see for example \cite{FOModalLogic}, \cite{DynamicQL}, and \cite{ModalMuCalculi}, respectively).

\paragraph{Acknowledgements}
I thank Norman Megill and  Mladen Pavi\v{c}i\'{c} for
	pointing out an erroneous Corollary~\ref{corollary:reduction} in
			the first arXiv-version of this paper.

I take the sole responsibility for this error.

Then,
	I also thank
		Robert Goldblatt, Denis Saveliev, and Ronnie Hermens for discussions that
			have lead to improvements of the second arXiv-version of this paper.

\bibliographystyle{plain}


TITLE: What happens when the basis vectors of an integer lattice are not linearly independent?
QUESTION [4 upvotes]: The definition of a lattice requires basis vectors that are linearly independent.
Why?
For example, the following three vectors are linearly independent and form the basis of a lattice:
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 2 & -2 \\
1 & -2 & 1 \end{array}
But what if we add a fourth vector such that they're not linearly independent anymore. For example:
\begin{array}{ccc}
0 & 0 & 1 & 4\\
0 & 2 & -2 & 2\\
1 & -2 & 1 & 3\end{array}
Are the four vectors the basis of a lattice? Why or why not? And, if it is, why does the definition require linear independence? Is there an equivalent basis that is linearly independent?

REPLY [4 votes]: The definition of a lattice is that it is a discrete additive subgroup of $\mathbb{R}^n$. The requirement that it be discrete gives us the answer to your question!
The "discrete" bit means that there is some $\epsilon > 0$ such that for any two distinct lattice points $x, y \in \Lambda$, $||x-y|| > \epsilon$.
In English: there has to be some minimum distance by at least which all lattice points are separated.
What would happen if we tried to define a lattice with the following basis:
$\begin{pmatrix}
   1 & 0 & \sqrt{2}\\
   0 & 1 & \sqrt{2}
\end{pmatrix}$
There's no way we're ever going to get $C_3$ from integer multiples of $C_1$ and $C_2$, so why isn't this a valid lattice basis?
Note that $\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1\end{pmatrix}$ will be in $\Lambda$, as will $\begin{pmatrix} 0.4142135...\\ 0.4142135...\end{pmatrix} = \begin{pmatrix} \sqrt{2} \\ \sqrt{2} \end{pmatrix} - \begin{pmatrix} 1 \\ 1\end{pmatrix}$
as will $ \begin{pmatrix}0.5857864... \\ 0.5857864... \end{pmatrix}= \begin{pmatrix} 1 \\ 1\end{pmatrix} - \begin{pmatrix} 0.4142135...\\ 0.4142135...\end{pmatrix}$ and so on and so forth.
So if I came to you claiming that our set of 3 vectors forms a basis for a 2-D lattice, I would have to offer up some minimum distance between lattice points, $\epsilon$, per the definition of a lattice. However, you would always be able to find two lattice points $x, y \in \Lambda$ such that $|| x - y || < \epsilon$ for any value of $\epsilon$ that I proposed, proving that I was a liar and our set of vectors doesn't form a lattice basis after all.


TITLE: Prove that $\Gamma'(1)=-\gamma$
QUESTION [5 upvotes]: Use the product formula for $1/\Gamma(z)$ to prove that $$\Gamma'(1)=-\gamma$$

I know that for Euler constant $\gamma$,
$$\frac{1}{\Gamma(z)} =ze^{\gamma z}\prod _{k=1}^{\infty} (1+\frac{z}{k})e^{-z/k}$$
But I cannot prove it properly. Please show me explicitly. Thank you:)

REPLY [10 votes]: $$\log\Gamma(z)=-\log z-\gamma z-\sum_{k=1}^\infty\left[\log\left(1+\frac zk\right)-\frac zk\right]\implies$$
$$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{k=1}^\infty\left(\frac1k\frac k{k+z}-\frac1k\right)=-\frac1z-\gamma-\sum_{k=1}^\infty\left(\frac1{k+z}-\frac1k\right)\implies$$
$$\Gamma'(1)=\frac{\Gamma'(1)}{\Gamma(1)}=-1-\gamma-\left(\frac12-1+\frac13-\frac12+\ldots\right)=-1-\gamma-(-1)=-\gamma$$


TITLE: Why is there so little data-driven fluid dynamics research?
QUESTION [1 upvotes]: With the insane growth of data science, I notice that there's hardly any data-driven fluid dynamics research out there.  What could explain this phenomenon?
I have heard that fluid dynamicists generally want to do their research based on first principles, i.e. use physics, and not use data-driven methods that might not care for domain knowledge.
Having said that, I'm still somewhat surprised that not more fluid dynamicists have done data-driven work.

REPLY [1 votes]: Well, for people doing physics research, especially mathematical analysis, data-driven approaches are basically useless. A deep NN can't really guarantee that its output is realistic nor can it guarantee that it will generalize to situations never seen in the training data. If one wants to understand the underlying mechanisms, then ML can't really help usually.
I know there are also people in physics-based animation that don't use it either, since for some applications deep NNs are too slow and/or hard to train. Maybe they are less interested in it as well. Either way, this could likely change soon.
On the other hand, there are plenty of data-driven fluid dynamics work! For example:

Kutz, Deep learning in Fluid Dynamics, 2017
Zhang & Duraisamy, Machine Learning Methods for Data-Driven Turbulence Modeling, 2015
Beck et al, Deep Neural Networks for Data-Driven Turbulence Models
Wiewel et al, Latent-space Physics: Towards Learning the Temporal Evolution of Fluid Flow, 2018


TITLE: Supplementary notes to Mumford's The Red Book of Varieties and Schemes
QUESTION [2 upvotes]: I am a graduate student with good mathematical maturity (I took advanced courses like category theory, commutative algebra...). I want to study algebraic geometry from Mumford's red book. I find it difficult to keep motivated. Is there any class notes supplements Mumford?

REPLY [7 votes]: You could have a look at Vakil's "Foundations of algebraic geometry" ? [Available on his Stanford homepage]
Eisenbud&Harris' "Geometry of schemes" is quite good to get the points too (pun intended)


TITLE: Diffraction pattern from a controlled single-slit experiment
QUESTION [1 upvotes]: I'm returning to the old thought experiemnt of a controlled electron beam pointing at one or two slits. By controlled I mean that the electrons are released with sufficient time between firings that no two subsequent electrons should interact.
I found the this paper (https://iopscience.iop.org/article/10.1088/1367-2630/15/3/033018) where they perform exactly this experiment - but the results surprised me and I wondered if anyone could help me clarify.
When both slits are un-masked we see the good ol' inteference pattern - a big sinc for the single slit modulated by narrow fringes from the interaction. But in the edge cases, where only one slit is unmasked, I'm surprised that we still seem to see narrow fringes (maybe clearest in the penultimate couple of panels).
How are these occuring?

REPLY [2 votes]: By looking at the paper we can see that they used a setup where the 'mask' appears after the double slit (by $240\mu \text{m}$). So, we would still expect interference effects. We can also see that there is a difference between when the right slit is covered vs when the left is covered, and that is that the interference effect is stronger on the opposite side (what you'd expect if the mask was post the double slit).


TITLE: homologies of some subsets of ${R}^{n}$
QUESTION [4 upvotes]: This might be something well-known.
For $1\le k\le n$, let $A(n,k)\subset\mathbb{R}^{n}$ be the set  of points $x=(x_{1},...,x_{n}%
)$ with at least $k$ distinct coordinates. Then what are the homologies of
$A(n,k)$?
For example, $A(n,2)$ is equal to $\mathbb{R}^{n}\backslash\Delta$,
where $\Delta$ is the diagonal, thus it is homotopy equivalent to
$\mathbb{S}^{n-2}$.

REPLY [3 votes]: Let me try again after deleting an answer based on a misunderstanding of the question. (Thanks to @YCor for pointing this out.)
Edited thanks to observations of Dan Petersen.
Your space $A(n,k)$ is obtained from ${\mathbb R}^n$ by deleting a bunch of subspaces.  Thus the theorem of Goresky and McPherson dealing with complements of unions of subspaces applies.  This says that
$$\widetilde{H}^d(A(n,k)) \cong \bigoplus_X \widetilde{H}_{codim(X)-2-d}(\Delta(0,X)).$$
Here $X$ runs through the set of intersections of collections deleted subspaces, which is partially ordered by reverse inclusion, $\Delta(0,X)$ is the order complex of the interval $(0,X)$ ($0$ is the ambient space ${\mathbb R}^n$, which is not included in the sum), and $codim(X)$ is the usual codimension of the subspace $X$ in ${\mathbb R}^n$.
The intersection poset for this union of subspaces is isomorphic with the poset of set partitions of $\{1,\ldots,n\}$ having less than $k$ parts, ordered by reverse inclusion.  As this is a rank selected subposet of the lattice of all set partitions, it is a Cohen-Macaulay poset. This means that the homology of $\Delta(0,X)$ is concentrated in degree $rank(X)-2$.  On the other hand, if $X$ has $j$ parts, then $X$ has rank $k-j$ (in the intersection post) and codimension $n-j$.
It follows now from direct computation that the reduced homology of your space is concentrated in degree $n-k$.  As Dan Petersen pointed out to me, if $k \leq n-2$, then $A(n,k)$ is simply connected, and it follows from the Hurewicz Theorem that $A(n,k)$ is homotopy equivalent to a wedge of spheres of dimension $n-k$.
Calculating the number of such spheres involves determining the M\"obius function on the intersection poset, which might or might not be tough.