TITLE: Singular ideal of an idealization
QUESTION [0 upvotes]: Let $S$ be a commutative ring, and let $A$ be a faithful $S$-module. Through idealization, we can make the abelian group $R=S⊕A$ into a commutative ring using the multiplication $(s,a)(s',a')=(ss',sa'+s'a)$. By Problem 16, section D, Chapter 1 of the book Ring Theory. Nonsingular Rings and Modules by Kenneth Goodearl, the singular ideal $Z(R)=T⊕A$, where $T=\{s\in S\mid Bs=0$ for some essential submodule $B$ of $A$}. My question:

If $z\in Z(S)$, the singular ideal of $S$, and $a\in A$ what condition may be imposed on $z$ and/or $a$ to guarantee that $(z,a)\in Z(R)$?

From $z\in Z(S)$ we get that the annihilator ideal  of $z$ in $S$ is essential in $S$ meaning that its intersection with any nonzero ideal of $S$ is nonzero; and I search for an essential submodule $B$ of $A$ which annihilates the element $z$. But I can't get any good idea for introducing $B$. Thanks for any cooperation!

REPLY [1 votes]: You have asked,
If $z\in Z(S)$, the singular ideal of $S$,
and $a\in A$, what condition may be imposed on
$z$ and/or $a$ to guarantee that $(z,a)\in Z(R)$?
From the exercise, $Z(R)=T\oplus A$. Thus $(z,a)\in Z(R)=T\oplus A$ iff
$z\in T$. This is the correct
restriction on $z$ and there is no restriction on $a$.
Since you are asking about the additional effect of assuming $z\in Z(S)$,
it may be that you want to know how $Z(S)$ and $T$ are related.
Here are some examples that show that they are not very related.

$S = \mathbb Z$, $A = \bigoplus_{n=1}^{\infty} \mathbb Z/p^n\mathbb Z$.
In this example, $A$ is a faithful $S$-module,
$Z(S) = (0)$, and $T = p\mathbb Z$.
Therefore $T$ contains $Z(S)$ and depends on $p$,
while $Z(S)$ does not depend on $p$.
$S$ is a commutative ring whose singular ideal properly
contains the nilradical,
$A=S\oplus \left(\bigoplus_{P\lhd S, P\;\textrm{prime}} S/P\right)$.
In this example, $A$ is a faithful $S$-module
because $S$ is a summand of $A$. Any element that
lies outside some prime ideal $P$ cannot lie
in $T$ because $S/P$ is a summand of $A$. Thus $T$ must
equal the nilradical, which is properly contained inside $Z(S)$.


TITLE: What does it mean for a phase to be unstable due to quantum fluctuations?
QUESTION [11 upvotes]: Generally in the literature on quantum critical phenomena (as opposed to ordinary critical phenomena in statistical mechanics), there is the idea that quantum fluctuations can prevent ordering of a phase. My very basic question is: formally speaking, what does it mean for a phase to be unstable (or melted) due to quantum fluctuations? I want to emphasize that I am looking for a formal definition that isn't tied to a specific model Hamiltonian.
I have seen this idea discussed in the case of the Heisenberg antiferromagnet in 1D, where the classical antiferromagnet state (all spins alternating spin up and down) is unstable towards the creation of domains. One way I hear people describe this is that the "quantum spin fluctuations melte the classical antiferromagnet phase", but that seems odd to me, because I can't really put my finger on how this idea generalizes. Is it the fact that spin is a non-commuting operator that is important here? Otherwise what makes this destruction by "quantum fluctuations" as opposed to a phase destroyed by classical fluctuations? After all, you could say similar things about classical spins not ordering in the 1D Ising model due to spin fluctuations, right?
For example, one may look at the following links where the author(s) all use language suggesting that "quantum fluctuations" prevent ordering of a system in various ways:
Example 1: Antiferromagnetism
Example 2: Quantum Paraelectricity
Example 3: Quantum Pendulum

REPLY [3 votes]: Yes (in addition to the other answers), the point is that if multiple terms in the Hamiltonian don't commute with one another, the ground state of the total Hamiltonian is no longer the same as the ground state of each individual term; and therefore can no longer qualify as a 'phase' in the traditional sense.
To speak of a 'phase', you (generally) want a many-body state that

Is robust to small decoherence, i.e. you can look at a macroscopic 'order parameter' that makes classical sense.
Does not manifestly change in time

The clearest example is perhaps crystallization of atoms. The Hamiltonian consists of a Lennart-Jones-like potential $\sum_{ij}V(\hat{r}_i-\hat{r}_j)$ , and a kinetic part $\sum_i \frac{\hat{p}^2}{2m}$. If this were a classical system, we could just treat $r_i$ and $p_i$ as classical numbers and minimize these terms separately. The first one results in the formation of a periodic lattice, the second one confirms that they all stand still, and there you have the ground state.
However, you cannot do this, because $\hat{r_i}$ and $\hat{p}_i$ are in fact conjugate operators, and according to quantum mechanics, they don't share a ground state. Now of course, the total Hamiltonian will still have a ground state, but it is not so clearly interpretable. The result is that atoms 'keep wiggling' according to QM, even at zero temperature (where everything is in the ground state). For some elements, such as Helium, these 'fluctuations' even prevent the proper formation of a solid phase in a natural setting.
A more general way to phrase this is that the quantum harmonic oscillator, unlike the classical one, must have an energy of at least 1/2. The question is then for each system, is this zero-point energy small enough to maintain macroscopic order?
For another perspective, the introductionary chapter of Quantum Noise may be relevant.
Note: as LorenzMayer has reminded me in the comments, also in a classical system can it be that the ground state doesn't minimize all Hamiltonian terms separately (frustrated systems). However, the point here is rather that for a given system, quantumness makes things worse than for the classical limit and as such can let systems reach a tipping point where order is destroyed.


text:
\begin{document}

\title{Stronger classes of  sum uncertainty and reverse uncertainy relations}

\author{Chiranjib Mukhopadhyay}
\author{Arun Kumar Pati}
\address{Quantum Information and Computation Group, Harish Chandra Research Institute, Homi Bhabha National Institute, Allahabad 211019, India}

\begin{abstract}
\noindent
\pacs{03.65 Mn}

Uncertainty relations are old, yet potentially rewarding to explore. By introducing a quantity called the \emph{uncertainty matrix}, we provide a link between purity and observable incompatibility, and  derive several stronger uncertainty relations in both forward and reverse directions for arbitrary quantum states, i.e., mixed as well as pure, and arbitrary incompatible quantum observables, none of which suffer from the problem of triviality. Besides the tightness, the interpretations of terms in these uncertainty relations may be of independent interest. We provide the possible generalization of stronger uncertainty relations to sum of variances of more than two observables. We also demonstrate  applications of techniques used here to, firstly, obtain a simple reverse quantum speed limit for quantum states undergoing Markovian dynamical evolution, and secondly, to provide a lower bound for fidelity between two quantum states.

\end{abstract}

\maketitle

\emph{Introduction-} Uncertainty relations are at the heart of quantum theory. Almost a century after the discovery of quantum theory, one may thus be surprised by the recent resurgence in research on preparation uncertainty relations \citep{mpur, maziero, urchin1, urchin2, urchin3, branciardur, mdur, debasis, bialynicki, maassen, wehner_review, huang, schwonnek}, as well as on the exact interpretation and formulation of Heisenberg's uncertainty principle \citep{ozawa,experimental, pekka, ozawa_mixed, branciard, namrata, busch_review}. The problem of triviality of the uncertainty relation was recently addressed in \cite{mpur}, but there is still ample scope for  devising stronger uncertainty relations. In addition to the fundamental importance of these developments, they allow us to come up with applications, such as quantum speed limits \citep{tamm, anandan, del_campo,  seventh, deffner_review} or, detection of nonclassical correlations \citep{hofmann, guehne, cavalcanti} among many others. Thus, the discovery of any new uncertainty relation may have important potential implications for quantum technology.

In this letter, we outline several new uncertainty relations. Firstly, we extend the formalism of Ref. \cite{mpur} to obtain non-trivial lower bounds for sum of variance-based uncertainties for all states, not just pure states or \emph{some} mixed states. Subsequently, we formulate yet another sum uncertainty relation, which is optimization-free, and nicely separates out the quantum and the classical contributions to uncertainty. Next, we obtain a very strong uncertainty relation for arbitrary dimensional systems, perhaps one of the strongest variance based state-dependent uncertainty relations in literature.  This is followed by an extension of these ideas to the case of sum uncertainty relations for arbitrary number of observables. In addition to the uncertainty relations, we also formulate a family of reverse uncertainty relations, and mention two applications of these relations. The first is to formulate a lower bound for speed of quantum evolution, as opposed to the upper bound which has been the subject of considerable interest  throughout the history of quantum theory \citep{deffner_review}, for an initially mixed qubit system undergoing Markovian evolution. The second aims at figuring out a lower bound on the fidelity between two arbitrary states.

\emph{Uncertainty matrix -} Let $A$ and $B$ be two observables. Let us introduce the following Hermitian operators $C = A - \bra A \ket, D = B - \bra B \ket$. We term the following matrix, $K$ as the \emph{uncertainty matrix}.
\begin{equation}
K = (C \pm iD)(C \mp iD)
\label{urmat}
\end{equation}
 Since $K$ is expressed in the form $X^{\dagger}X$, where $X = C \pm iD$, it is guaranteed to be Hermitian and positive semi-definite. For qubit systems, it has been shown in the supplementary material, that the purity of the density matrix thus constructible from the uncertainty matrix via normalization, captures the intrinsic incompatibility of the two operators. This provides a clue to link the resource theory of purity \citep{purity} with quantum uncertainty.  Let us note the following identity for the uncertainty matrix, derivable from the parallelogram law \citep{mpur},  which holds for any pure state $|\psi\rangle$.
\begin{equation}
 || (C \mp iD) |\psi \rangle ||^2  = \Delta A^{2} + \Delta B^{2} \mp  i \langle [A,B] \ket |
\label{urequality}
\end{equation}

\emph{Stronger uncertainty relation for arbitrary mixed states-} The problem of possible triviality of the variance based Robertson-type sum uncertainty relation even in the case of non-commuting observables was recently resolved in Ref. \citep{mpur} for pure states, but still left the case of arbitrary mixed states open. Let us now formulate an analogus sum uncertainty relation valid for arbitrary mixed states utilizing the vectorization operation \citep{supp}, which entails constructing a vector $|\rho\rangle$ from a matrix $\rho$ by stacking columns of $\rho$ on top of each other. The supplementary material contains relevant properties of vectorization, as well as an intriguing result linking entanglement, coherence, purity, and imaginarity of qubit systems via the vectorization technique.

As before, let $A - \langle A \rangle = C$, and $B - \langle B \rangle = D$. Now, $\Delta A = \sqrt{\text{Tr} (\rho C^2)} = \sqrt{\text{Tr} (\sqrt{\rho} C^2 \sqrt{\rho})} = \sqrt{\langle \sqrt{\rho} | \mathbb{I} \otimes C^2 | \sqrt{\rho} \rangle}$. Thus, $\Delta A =  || (\mathbb{I} \otimes C ) |\sqrt{\rho} \rangle ||$. Similarly one can show that $\Delta B =   || (\mathbb{I} \otimes iD ) |\sqrt{\rho} \rangle||$. The vectors here can be normalized to unity and, thus be considered as pure states in a $d^2$-dimensional Hilbert space, if the original system was $d$-dimensional. Since square root of a matrix is non-unique, we can simply choose that square root which is Hermitian. A possible prescription for constructing such a matrix is to first diagonalize $\rho$ via an unitary $U$, then take the positive square roots of the population elements, then finally apply the reverse unitary $U^{\dagger}$.  From \eqref{vecinnerpdt}, it follows that the vectorization of this matrix is normalized, and thus, a quantum state vector.  Now, we are in the position to prove the following mixed state generalization to the first inequality in \citep{mpur}.

\begin{theorem}
If $|k^{\perp}\rangle$ is a (normalized) state vector perpendicular to the vectorization $|k\rangle$, the following variance based sum uncertainty relation holds-
\begin{equation}
\Delta A^2 + \Delta B^2 \geq \pm i \langle [A,B] \rangle + |\bra\sqrt{\rho}|  \mathbb{I} \otimes (A \pm iB)  |\sqrt{\rho}^{\perp} \rangle |^2.
\end{equation}
\label{mixed_ur_thm}
\end{theorem}
\emph{Proof-} Upon simplification, like before, $|| (C \pm iD |\sqrt{\rho}) \rangle ||^2 = \Delta A^2 + \Delta B^2 \mp i \langle [A,B] \rangle$. Now, by applying the Cauchy-Schwarz inequality on two vectors $(C\pm iD)|\rho\rangle$ and $|\rho^{\perp}\ket$, the latter being a (normalized) quantum state vector orthogonal to $|\sqrt{\rho}\rangle$, and using the parallelogram law,we have $\Delta A^2 + \Delta B^2 \mp i \langle [A,B] \rangle   \geq  |\langle \sqrt{\rho}| A\pm iB - (\langle A \rangle + \langle iB \rangle)| \sqrt{\rho}^{\perp}\ket|^2 = |\langle\sqrt{\rho}|  \mathbb{I} \otimes (A \pm iB) |\sqrt{\rho^{\perp}} \rangle |^2$. This completes the proof.

 \qed

An important point to note here is that, unlike the vector $|\sqrt{\rho}\ket$, the orthogonal vector $|\sqrt{\rho}^{\perp}\ket$ need not come via vectorization from a Hermitian matrix.  Therefore, optimizing over all such orthogonal vectors turns out to be much tighter than if we restrict to all orthogonal vectors derived from the vectorization of a $d-$dimensional Hermitian matrix.

\noindent \emph{An optimization-free uncertainty relation-} The uncertainty relations formulated above all depend on optimization over an infinite set of states. While this lets one considerably tighten the corresponding inequalities, it is perhaps desirable to search for a sum uncertainty relation which retains the advantage of non-triviality, but, in addition, is also free from any optimization. Thus, we now formulate the following optimization-free variance based sum uncertainty relation.

\begin{theorem}
For arbitrary mixed state $\rho$ and observables $A$ and $B$, if the corresponding uncertainty matrix is denoted by $K$, then the following optimization-free variance based sum uncertainty relation holds -
\begin{equation}
\Delta A^{2} + \Delta B^{2} \geq |\langle [A,B] \rangle | + S(\rho) - \ln \tr \left( e^{-K} \right), \label{peierls}
\end{equation}
where $S(\rho)$ is the von Neumann entropy of the quantum state $\rho$.
\end{theorem}

\noindent \emph{Proof- } For any mixed state $\rho$,  \eqref{urequality} may be written as \[\tr (\rho K) = \Delta A^2 + \Delta B^2 \mp i \langle [A,B]\rangle. \]  Now let us recall the Peierls-Bogoliubov inequality \citep{bhatia} for the positive observable $K$ and density matrix $\rho$ \beq \tr(\rho K) + \tr (\rho \ln \rho) \geq - \ln \tr e^{-K}. \eeq Combining the above results immediately lead to the proof. \qed\\
This inequality is saturated provided $\rho = e^{-K}$. Since the uncertainty matrix $K$ implicitly depends on the choice of density operator $\rho$, if one fixes the observables, this condition leads to a transcendental equation for the density matrix $\rho$. In the supplementary material \citep{supp}, we comment on the solution of this equality condition in mixed qubit systems. The uncertainty relation above can also be extended in terms of generalized entropies utilizing a recent extension of the Peierls-Bogoliubov inequality for deformed exponentials \citep{hansen}. Let us now comment on the terms that appear in the RHS of the uncertainty relation above.

The uncertainty, quantified by variances, for a mixed quantum state can be thought as originating from two different sources of randomness. One is the intrinsic randomness native to the quantum formalism, the other is the randomness introduced by the classical statistical mixture in forming the mixed state. Thus, when we consider the uncertainty relations as a pillar of quantum mechanics, it is desirable to separate out these two contributions explicitly. This was first attempted by Luo, who introduced the Wigner-Yanase skew information as a quantifier of the `quantum' part of the uncertainty \citep{luo}.  The uncertainty relation \eqref{peierls} above offers an alternative view of approaching this problem. The first term in RHS is the commutator and can be thought of as the intrinsic quantum contribution. The second term is the entropy of the state, which arises from randomness entirely due to the classical mixing. The final term depends only on the uncertainty matrix, which is free from explicit state dependence if the classical averages, i.e., expectation values, of operators $A$ and $B$ are already specified. Thus, the uncertainty lower bound in \eqref{peierls} can be thought of as sum of three distinct contributions. The first being that due to inherent non-commutativity of quantum mechanics, the second entirely due to the classical randomness introduced via mixing, and the third representing, in some sense, a  state-independent contribution to the uncertainty.

\begin{figure}
 \includegraphics[width=0.45\textwidth, keepaspectratio]{PB_plus_reverse_qutrit_p_with_max_coh.pdf}
\caption{(Color online) Demonstration of the optimization-free sum uncertainty relation \eqref{peierls} (golden dash-dotted line) and the first reverse uncertainty relation (red dotted line)  \eqref{reverse_eqn} with respect to the sum of variances (green line) and the lower bound in Robertson's uncertainty relation (blue dashed line) for the family of qutrit states $\rho = p |\psi \ket \bra \psi| + \frac{1-p}{3} \mathbb{I}$, where $|\psi\ket = \frac{1}{\sqrt{3}} \left(|0\ket + |1\ket + |2\ket  \right)$. The observables are $J_x$ and $J_y$ respectively. }
\label{pb_reverse_fig}
\end{figure}

\noindent \emph{Even stronger uncertainty relation for arbitrary pure  states-} Having dealt with the triviality problem of Heisenberg uncertainty relation for arbitrary quantum states, let us now present a further tightening of the stronger uncertainty relation originally enunciated in \citep{mpur}, which holds for pure states.
\begin{theorem}
For arbitrary pure state $|\psi\ket$  and  observables $A$ and $B$, if the corresponding uncertainty matrix is denoted by $K$, then the following variance based sum uncertainty relation holds for all pure states $|\phi\ket$

\begin{equation}
\Delta A^{2} + \Delta B^{2} \geq |\langle [A,B] \rangle | + \frac{1}{\langle K \rangle_{\phi}} \frac{|\langle \phi | K | \psi \rangle |^2}{\cos \left( 2 \cot^{-1} (\alpha \cot \frac{\theta}{2}) \right)},
\label{thm1}
\end{equation}
where $\theta$ is the inner product angle between $|\psi\ket$ and $|\phi\ket$, and $\alpha = \sqrt{\lambda_{\max} / \lambda_{\min}}$, where $\lambda_{\max}$ and $\lambda_{\min}$ are, respectively, the maximum and minimum eigenvalues of the observable $K$.
\end{theorem}
\emph{Proof- } Let us recall the Bauer-Householder inequality for arbitrary vectors $\vec{x}, \vec{y}$ making an inner product angle $\xi$ between themselves and an invertible matrix $A$ of the commensurate dimension \beq |(A\vec{y})^{\dagger} (A \vec{x}) | \leq ||A \vec{x} || \ ||A \vec{y}|| \cos \Upsilon, \eeq where $\Upsilon = 2 \cot^{-1}\left[\frac{\lambda_{\text{max}} (A)}{\lambda_{\text{min}} (A)} \cot (\xi /2 )  \right]$. Applying this inequality with the uncertainty matrix $K$ chosen as the matrix $A$ above, along with the uncertainty equality \eqref{urequality} leads to the uncertainty relation stated above. \qed \\

The conditions for equality in the Bauer-Householder inequality \citep{liu} for vectors $ \lbrace |\psi\ket, |\phi\ket \rbrace$ and positive semi definite operator $K$ with eigenvectors $|\max\ket$ and $|\min\ket$ corresponding to its maximum and minimum eigenvalues respectively  are any of the following

\begin{enumerate}
\item $|\psi\ket$ and $|\phi\ket$ are equal upto some phase.

\item The states $|\psi\ket = \frac{1}{\sqrt{2}} \left( \xi \sqrt{1 + |m |} |\lambda_{\max}\ket - \eta \sqrt{1 - |m|}|\lambda_{\min}\ket \right)$, and  $|\phi\ket = \frac{\epsilon}{\sqrt{2}} \left( \xi \sqrt{1 + |m|} |\lambda_{\max}\ket +  \eta \sqrt{1 - |m|} |\lambda_{\min}\ket \right)$ , where $|\xi| = |\eta| = |\epsilon| = 1$, $m = \cos \theta$, and $|\lambda_{\max}\ket$, $|\lambda_{\min}\ket$ are eigenvectors of the uncertainty matrix $K$ corresponding to the maximum and minimum eigenvalues of $K$ respectively.
\end{enumerate}

Thus, this inequality is tight if one chooses $|\psi\ket = |\phi\ket$. However, this is experimentally not very economical since so choosing $|\phi\ket$  requires one to perform the complete state tomography on the system state anyway. Selecting an orthogonal state to the original state is easier. Thus, this inequality can be tight even if $|\psi\ket$ is orthogonal to $|\phi\ket$, provided the second condition above is met. The key improvement over Ref. \citep{mpur} lies in the fact that if we only have access to arbitrary states, not necessarily orthogonal to the system state, then it is possible to obtain a non-trivial tightening of the Robertson UR vide Eq. \eqref{thm1}, which was not possible via Ref. \citep{mpur}. In particular, for qubit systems, the \emph{maximization over all states orthogonal to $|\psi\ket$} prescription doesn't work simply because once a state is specified  in two dimensions, the orthogonal pure state is automatically uniquely determined. However, Eq. \eqref{thm1}, written in terms of states whose Bloch vectors make an arbitrary angle with the Bloch vector of the system qubit, still works. The generalization of Eq. \eqref{thm1} for arbitrary mixed states may be similarly shown via the vectorization procedure in a similar way to the proof of \eqref{mixed_ur_thm}.

\emph{Extension for arbitrary number of observables -} In this work, we have considered the sum uncertainty relations for two observables so far. However, we are often interested in sum of variances for more number of observables, e.g., for entanglement or nonlocality detection \citep{hofmann, guehne, cavalcanti}. Thus, it is imperative that we try to extend our formalism for sum of variances of arbitrary number of observables $A_1, A_2,..., A_n$. A geometrical result, which generalizes the parallelogram law for $n$-vectors \citep{douglas} helps us formulate stronger uncertainty relations in these cases. As the simplest generalization, the case of three-observable sum uncertainty relations has been studied in the supplementary material \citep{supp}. These results allow us to derive the many-observable versions of the uncertainty relations derived in \citep{mpur} as well as the present work\citep{supp}. We hope they turn out to be useful, among many other tasks, in witnessing non-classicality.

\emph{Reverse uncertainty relations - } While uncertainty relations guarantee the existence of intrinsic fluctuations in quantum theory, they usually do not let us estimate an upper bound on such fluctuations. Thus, the problem of devising reverse uncertainty relations is one of considerable theoretical interest \citep{puchala,debasis}. Below we formulate such a reverse uncertainty relation for observables.

\begin{theorem}
For arbitrary mixed state $\rho$ and observables $A$ and $B$, if the corresponding uncertainty matrix is denoted by $K = \lambda \sigma$, where $\lambda = \tr(K)$, the following  variance based reverse sum uncertainty relations hold.
\begin{equation}
\Delta A^{2} + \Delta B^{2} \leq |\langle [A,B] \rangle | + \lambda \mathcal{F}^{2}(\rho,\sigma) \leq |\langle [A,B] \rangle | + \lambda \left( 1 - S(\rho || \sigma) \right)
\label{reverse_eqn}
\end{equation}
where $\mathcal{F}$ is the fidelity between two quantum states and $S(\rho||\sigma)$ the relative entropy.
\end{theorem}
\emph{Proof-} The first inequality follows from applying the Araki-Lieb-Thirring inequality to the second term of the RHS in the uncertainty equality \eqref{urequality}. The subsequent inequality follows from the first one in two steps - first using the Fuchs van de Graaf inequality linking fidelity and trace distance, followed by Audenaert and Eisert's relation \citep{audenaert} between trace distance and relative entropy.  \qed

The problem of finding the reverse uncertainty relations in our framework turns out to be very closely related to the famous problem of determining the numerical radius of an operator in Matrix analysis. Based on the latest mathematical advances \footnote{See, for example Ref. \citep{dragomir_book}}, we derive several such reverse uncertainty relations  in the supplementary material \citep{supp}.

\emph{Reverse quantum speed limit -} Suppose the Markovian evolution of a quantum system represented by the density matrix $\rho (t)$ is given by \begin{equation}
\dot{\rho} (t) = L_t (\rho(t)),
\end{equation} We term $L_t$ as the generator of the dynamics. Suppose further that during the dynamics, the system evolves from an initial state $\rho_0$ to a final state which makes a Bures angle $\mathcal{L}$ with the initial state. The lower bound on the evolution time (equivalently the quantum speed limit) for this situation has already been explored \citep{deffner_lutz} in some detail. However, in the scenario above, we may prove \citep{supp} the following non-trivial inverse bound, i.e. upper bound on the evolution time $\tau$

\begin{equation}
\tau \leq \frac{\sin^2 \mathcal{L}_{\tau}}{\Lambda_{\text{reverse}}}
\label{pbrur_main}
\end{equation}
where $\Lambda_{\text{reverse}} = 1/ \tau \int_{0}^{\tau}  dt \left[ S(\rho_0) - \ln \tr e^{\mp L_t (\rho(t))} \right]$, and $\mathcal{L}_{\tau}$ is the Bures angle between the initial and final state. To prove this, we will follow the setting and notation laid down in Deffner and Lutz's recent work \citep{deffner_lutz}. If the initial state is $\rho_0$, the Bures angle of the state $\rho(t)$ at time $t$ with respect to the initial state is given by $\mathcal{L}= \cos^{-1} \left(\sqrt{\tr(\sqrt{\rho_0} \rho(t) \sqrt{\rho_0}} \right)$. Taking time derivative of both sides, we have the following equation \beq 2 \cos \mathcal{L} \sin \mathcal{L} |\frac{d \mathcal{L}}{dt}| = \vert \tr \left[ \rho_0 L_t (\rho(t)) \right] \vert \eeq

\noindent Now, assuming that the fidelity decreases, i.e. Bures angle $\mathcal{L}$ increases monotonically over time, as well as assuming that the quantity  $\tr \left[ \rho_0 L_t (\rho(t)) \right]$ in the RHS of the above equation is positive, we have the following equality condition

\beq \sin 2\mathcal{L} \frac{d\mathcal{L}}{dt} = \tr \left[ \rho_0 L_t (\rho(t)) \right]  \eeq

\noindent Applying the Peierls-Bogoliubov inequality \citep{bhatia} to the RHS of this equality leads to

\begin{equation}
\sin 2 \mathcal{L} \frac{d \mathcal{L}}{dt} \geq S(\rho_0) - \ln \tr e^{-L_t (\rho(t))}
\label{pbrur}
\end{equation}

\noindent Now integrating \eqref{pbrur}, the following reverse speed limit on evolution time $\tau$ is obtained

\begin{equation}
\tau \leq \frac{\sin^2 \mathcal{L}_{\tau}}{\Lambda_{\text{reverse}}}
\end{equation}
where $\Lambda_{\text{reverse}} = 1/ \tau \int_{0}^{\tau}  dt \left[ S(\rho_0) - \ln \tr e^{-L_t (\rho(t))} \right]$, and $\mathcal{L}_{\tau}$ is the Bures angle between the initial and final state.

If the opposite case is true, i.e.,  the quantity $\tr \left[ \rho_0 L_t (\rho(t)) \right]$ is always negative, then the reverse speed limit is given by
\begin{equation}
\tau \leq \frac{\sin^2 \mathcal{L}_{\tau}}{\Lambda_{\text{reverse}}}
\end{equation}
where $\Lambda_{\text{reverse}} = 1/ \tau \int_{0}^{\tau}  dt \left[ S(\rho_0) - \ln \tr e^{L_t (\rho(t))} \right]$, and $\mathcal{L}_{\tau}$ is the Bures angle between the initial and final state.
\begin{figure}[h]
\includegraphics[width = 0.45 \textwidth, keepaspectratio]{Diag_fidelity.pdf}
\caption{(Color online) Diagram for the scheme to measure a lower bound to fidelity between states $\rho$ and $\sigma$.  }
\label{fid_diagram}
\end{figure}
\\

\emph{Measuring fidelity - }  Exactly measuring the fidelity between two quantum states, say $\rho$ and $\sigma$, necessitates performing complete state tomography on both of them. The computational cost for this grows exponentially with Hilbert space dimension. However, for a given state $\rho$, if we choose the observables $A$ and $B$ judiciously such that the normalized uncertainty matrix $K/ \tr(K)$ is the second state $\sigma$, then the reverse uncertainty relation \eqref{reverse_eqn} guarantees that we find a lower bound to the fidelity between $\rho$ and $\sigma$ by from simply measuring the variances  of $A$ and $B$ experimentally. The number of such measurements required does not scale exponentially with Hilbert space dimension, therefore this represents an economic way to estimate the minimum fidelity between two quantum states.

\emph{Conclusion -} We have provided several stronger uncertainty and reverse uncertainty relations, and mentioned a few of their applications in this work. However, we still think there is ample scope for discovering newer and better uncertainty relations. As an illustration, applying the methods of Ref. \citep{debasis} to our uncertainty relations would yield even tighter bounds. We hope these new uncertainty and reverse uncertainty relations can stimulate new thoughts on the so called uncertainty principle debate, as well as quantum metrology. Investigating the optimization-free uncertainty relation with the aim of relating it to the Wigner Yanase skew information may also give us new insight into the nature of uncertainty relations. These new relations have also been illustrated to result in two applications, viz., coming up with a lower bound for fidelity between two states as well as an inverse quantum speed limit for Markovian evolution. We hope that the stronger uncertainty and reverse uncertainty relations derived and discussed in this letter should empower physicists to reveal further new facts about the quantum world. In particular, we are optimistic that the reverse uncertainty relations may turn out  to be of some practical utility for metrologists, as well as be connected to some form of a reverse data-processing inequality.

\emph{Acknowledgement-} CM acknowledges Department of Atomic Energy, Governent of India, for granting of graduate research fellowship. We thank A.K. Rajagopal and L. Maccone  for discussions and comments.
\bibliographystyle{apsrev4-1}
\bibliography{ur_ref}

\newpage

\section{Supplementary Material for ``Stronger classes of sum uncertainty and reverse uncertainty relations"}
This supplementary material is organized as follows.  The first section  deals with the possible implications of vectorization technique on  establishing another link between the resource theories of entanglement an coherence. The second section touches briefly on the uncertainty matrix and its relation with the purity of the underlying quantum system. Two subsequent sections discuss the conditions for which the stronger uncertainty relations enunciated in the paper are saturated. This is followed by a section dedicated to deriving stronger uncertainty relations for three observables starting from a generalization of the parallelogram law. Reverse uncertainty relations have been treated subsequently. We finally elaborate on the method for finding a  lower bound on fidelity, which has been mentioned in the main text.
\subsection{Vectorization}
A tool of matrix analysis useful in various applications ranging from diffusion MRI techniques \citep{mri} to finite element analysis \citep{finel} to  studying generalized Hooke's law \citep{voigt} is vectorization, which has also been used in quantum information theory in the context of quantum maps \citep{gilchrist,modi} and calculating the quantum Fisher information \citep{qfi_vec}. For any matrix $A$, this entails stacking columns of that matrix on top of one another to construct a one-dimensional array, hence called the \emph{vectorization} of the matrix $A$, which we shall denote by $|A\rangle$, and the corresponding dual vector by $\langle A|$. We mention below the following relevant relations which hold for vectorization.

\begin{enumerate}
\item \emph{Vectorization of product of matrices: } \begin{equation}
|AB\rangle= (\mathbb{I} \otimes A) |B \rangle
\end{equation}
\item \emph{Inner product of matrices: }\begin{equation}
\text{Tr} (A^{\dagger} B) =  \bra A|B \ket
\label{vecinnerpdt}
\end{equation}
\end{enumerate}

\subsubsection{Using vectorization to link quantum resources}
\label{append_sub_1}

The link between quantum correlations and coherence has been established in several past works \citep{uttament,mile,yao}. Let us consider a quantum state $\rho$ and ask the following question - what can the entanglement of the vectorized (and adequately normalized) state $|\rho\ket$ tell about the coherence of the original state $\rho$ ? Let us confine ourselves to the qubit case and suppose $\rho$ is a general qubit mixed state such that $\rho = \frac{1}{2} \left( \mathbb{I} + \sum \vec{r}. \vec{\sigma} \right)$. The corresponding vectorized pure two-qubit state equals = $\alpha [1+r_z,r_x + i r_y, r_x - i r_y, 1 - r_z]^T$, where $\alpha$ is a normalization constant = $1/ \sqrt{2 + r_x^2 - r_y^2 + 2r_z^2} = 1/ \sqrt{2 + 2 P^2 - C_{l_1}^{2} - 2 r_y^{2}}$, where $P = \sqrt{r_x^2 + r_y^2 + r_z^2}$ is a measure of purity of the original state, and $C_{l_1}$ is the $l_1$-norm of coherence. The concurrence of the vectorized state now reads as $E_C = 2 |\alpha|^2 |1 + r_{z}^2 - r_x^2 - r_y^2|$, which, for real qubit states  implies \beq E_c =\frac{|1+P^2 - 2  C_{l_1}^{2}|}{|1 +  P^2 - \frac{C_{l_1}^{2}}{2}|} \eeq

This is an exact equality relating three quintessentially quantum features, viz. purity content, entanglement and coherence, which holds for arbitrary real qubit states. In fact, the general expression in terms of $r_y$ also includes a fourth quantum feature, viz. imaginarity, a resource theory for which has recently been constructed \citep{imaginarity}. This may turn out to be useful when considering the conversion of one quantum resource to another.

\subsection{Uncertainty matrix and purity}
\label{append_sub_2}

Suppose a qubit state $\rho$ is of the form \beq \rho = \frac{1}{2} \left(\mathbb{I} + \vec{r}.\vec{\sigma} \right) \eeq and the operators in question are $A = \vec{\sigma}.\hat{n_1}$ and $B =\vec{\sigma}.\hat{n_2}$ respectively. Assuming that the positive sign holds in the definition of the uncertainty matrix in the main paper - the expression for the density matrix obtained through normalizing the uncertainty matrix can be shown after some tedious algebra as

\beq
\frac{K}{\tr{K}} = \frac{\mathbb{I}}{2} + \frac{\vec{\sigma}. \left[(\hat{n_1}\times \hat{n_2}) - (p_1\hat{n_1} + p_2\hat{n_2})\right]}{2 + p_1^2 +  p_2^2}
\label{purity_ur}
\eeq
where $p_i = \vec{r}.\hat{n_i}$.
The length of Bloch vector $\vec{R}$ for this density matrix, which is a measure of purity, can now be shown to be equal to \beq |\vec{R}| = \frac{\sqrt{\alpha \beta - \gamma^2}}{(\alpha + \beta)/2} \eeq  where $\alpha = 1 + p_1^2$, $\beta = 1 + p_2^2$, and $\gamma = (p_1 p_2 - \hat{n_1}.\hat{n_2})$. If the state $\rho$ is highly mixed, i.e., $|\vec{r}|$ is very small, then upto leading order \beq |\vec{R}|\approx \sqrt{1 - (\hat{n_1}.\hat{n_2})^2} \label{almost_pure}\eeq
\begin{figure}[h]
\includegraphics[width = 0.35 \textwidth, keepaspectratio]{purity_um.pdf}
\caption{(Color online) Plot of purity of the (normalized) uncertainty matrix vs. incompatbility captured through inner product angle between the directions $\hat{n_1}$ and $\hat{n_2}$ corresponding to observables $A$ and $B$ respectively for 60,000 iterations of random quantum states $\rho$ and random unit-vector directions $\hat{n_1}$ and $\hat{n_2}$ corresponding to observables $A$ and $B$.}
\label{purity}
\end{figure}
Clearly, the angle between the two observables $A$ and $B$ captures the incompatibility between them. If the vectors $\hat{n_1}, \hat{n_2}$ are collinear or anti-collinear, then the corresponding operators are compatible. The more the deviation from (anti)-collinearity, the more incompatible they are. From \eqref{almost_pure}, we may see in this case, that the purity of the density matrix constructed from the uncertainty matrix nicely quantifies the incompatibility of these operators. For arbitrary qubit cases, the situation is not as clear-cut. Nonetheless, Fig. \ref{purity} confirms that the general trend that, increasing purity is accompanied by increasing incompatibility between the two observables, continue to hold by and large.

\subsection{Condition for tightness of \eqref{thm1} for orthogonal states in qutrit systems}

For the Cauchy Schwarz inequality to be saturated, the corresponding vectors must be co-linear. This is not the case for the Wielandt inequality for vectors , when the saturation condition may be met even if the vectors are non-collinear, even orthogonal. If we only pick states $|\phi\ket$ as orthogonal states to $|\psi\ket$ (denoted by $|\psi^{\perp}\ket )$ in \eqref{thm1}, the condition for saturation of that uncertainty relation is given in terms of the eigenvectors of the uncertainty matrix K by the following conditions\citep{liu} together -

\begin{enumerate}
\item $|\psi\ket$ is expressible as $|\psi\ket = \frac{1}{\sqrt{2}} \left( \xi |\lambda_{\text{max}}\ket + \eta |\lambda_{\text{min}}\ket \right)$.

\item and, the perpendicular state $|\psi^{\perp}\ket$ is expressible as $|\psi^{\perp}\ket = \frac{\epsilon}{\sqrt{2}} \left( \xi |\lambda_{\text{max}}\ket - \eta |\lambda_{\text{min}}\ket \right)$.
\end{enumerate}

Let us illustrate this for the case of a general qubit pure state and observables $A = \sigma_x$ and $B = \sigma_y$. Let us a qubit pure state $|\psi\ket = \cos \theta/2 |0\ket + e^{i \phi}\sin \theta/2  |1\ket$ in the parameter regime $\theta \in [0,\pi /4]$ \footnote{The case of $\theta \in [\pi/4 , \pi]$ can be analogusly treated.}. In this case, the uncertainty matrix $K = (C - iD) (C + iD) = C^2 + D^2 + i [C, D] $. The corresponding eigenvectors are given by
\begin{equation} |\lambda_{\text{max}}\ket = \frac{1}{\sqrt{1 + \left( \frac{\sin \theta}{1 -\sqrt{1+ \sin ^2 \theta}}\right)^2}} \left[  1   , \frac{e^{i\phi} \sin \theta}{1 - \sqrt{1 + \sin ^2 \theta}}      \right]^{T}
\end{equation}and, \begin{equation} |\lambda_{\text{min}}\ket = \frac{1}{\sqrt{1 + \left( \frac{\sin \theta}{1 + \sqrt{1+ \sin ^2 \theta}}\right)^2}} \left[  1   , \frac{e^{i\phi} \sin \theta}{1 + \sqrt{1 + \sin ^2 \theta}}      \right]^{T}
\end{equation}.

Now if $|\psi\ket$ and $|\psi^{\perp}\ket$ can be represented in the form given above, the inequality is tight.

\subsection{Condition for tightness of \eqref{peierls} for single qubit mixed states and fixed observables}

Suppose we again choose the operators $A$ and $B$ as  $\vec{\sigma}.\hat{n_1}$ and $\vec{\sigma}.\hat{n_1}$ respectively . The uncertainty matrix $K$ can now be written as  \beq K =  \left[ 2 + p_1^2 + p_2^2 \right] \mathbb{I} + 2 \vec{\sigma}.\left[ (\hat{n_1} \times \hat{n_2}) - p_1\hat{n_1} - p_2 \hat{n_2} \right] \eeq

We write $t=2+p_1^2+p_2^2 $ and $\vec{R}=2\left[ (\hat{n_1} \times \hat{n_2}) - p_1\hat{n_1} - p_2 \hat{n_2} \right]$, which yields the following expression for $e^{-K}$

\beq e^{-K} = t' \mathbb{I} + \vec{R'}.\vec{\sigma} \eeq

where $t' = e^{-t} \cosh |\vec{R}|$, and $\vec{R'} = - e^{-t} \frac{\sinh |\vec{R}|}{|\vec{R}|} \vec{R}$. For equality condition to be satisfied, this has to equal $\rho = \frac{1}{2} \left( \mathbb{I} + \vec{r}.\vec{\sigma} \right)$. That is, the following sets of non-linear equations have to be simultaneously satisfied -

\begin{eqnarray}
 e^{-t} \cosh |\vec{R}| = \frac{1}{2} \\
e^{-t} \frac{\sinh |\vec{R}|}{|\vec{R}|} \vec{R} = -\frac{\vec{r}}{2}
\end{eqnarray}

If these set of equations do not have a solution - this inequality is not saturated.

\subsection{Uncertainty relation for three incompatible observables}
Let us confine ourselves to three observables $A, B, C$. Let us denote $P = A - \bra A\ket , Q = B - \bra B\ket, R = C - \bra C\ket$. Now, let us consider the parallelo-hexagon, i.e. the hexagon ABCDEF whose opposite pairs are parallel and equal in length, in Fig. \ref{hexagon}. Let us choose vectors $\vec{AB} = |\psi_1\ket = (P  + ik_1 Q) |\psi\ket, \vec{BC} = |\psi_2\ket = (Q + i k_2 R) |\psi\ket, \vec{CD} = |\psi_3 \ket = (P +  i k_3 R) |\psi\ket$. The numbers $k_1,k_2,k_3 \in \lbrace -1, +1 \rbrace$,  are chosen based on whether the corresponding commutators give rise to positive or negative real numbers, in a similar way to Ref. \citep{mpur}. Now, the following equality holds \cite{douglas} for the parallelo-hexagon.

\begin{figure}
\includegraphics[width = 0.45 \textwidth, keepaspectratio]{schematic_hexagon.pdf}
\caption{(Color online) One generalization of the parallelogram law for parallelo-hexagons.}
\label{hexagon}
\end{figure}

\begin{figure}
\includegraphics[width = 0.45 \textwidth, keepaspectratio]{hexagon_2.pdf}
\caption{(Color online) Another generalization of the parallelogram law for parallelo-hexagons.}
\label{hexagon2}
\end{figure}
\begin{equation}
||\vec{AB}||^2 + ||\vec{BC}||^2 + ||\vec{CD}||^2 = \frac{1}{4} \left( ||\vec{AD}||^2 + ||\vec{BE}||^2 + ||\vec{CF}||^2 \right)
\end{equation}

The LHS quantity of this equality can now be shown to equal $\Delta A^2 + \Delta B^2 + \Delta C^2 - |\langle [A,B] \rangle|- |\langle [B,C] \rangle| - |\langle [A,C] \rangle| =\frac{1}{2} \left[ \Delta A^2 + \Delta B^2 - |\langle[A,B] \rangle| \right] + \frac{1}{2} \left[ \Delta A^2 + \Delta C^2 - |\langle[A,C] \rangle| \right] + \frac{1}{2} \left[ \Delta B^2 + \Delta C^2 - |\langle[B,C] \rangle| \right]$. That is, broken down into sums of two-observable Robertson uncertainty relations. Thus, since the RHS is always positive semi-definite, applying the generalized Wielandt inequality to the norms of each of the vector $\vec{AD}, \vec{BE}, \vec{CF}$, coupled with any qutrit pure state vectors, give rise to stronger uncertainty relations of the type similar to \eqref{thm1}.

Is this the only form of uncertainty relation derivable ? It turns out, there exists another \citep{douglas} geometric identity (please see Fig. \ref{hexagon2}).

\begin{equation}
||\vec{AB}||^2 + ||\vec{BC}||^2 + ||\vec{CD}||^2 = \frac{1}{3} \left( ||\vec{AC}||^2 + ||\vec{CE}||^2 + ||\vec{EA}||^2 \right)
\end{equation}

Proceeding similarly as before, it is possible to obtain another set of uncertainty relations for observables $A$, $B$, $C$ from this equality too.

config: arxiv


TITLE: Integral with respect to $x +$ constant
QUESTION [1 upvotes]: Is this a valid expression:
$$\int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$\int (u - 5)du= \frac{u^2}{2} - 5u + C = \frac{(x+5)^2}{2} - 5(x+5) +C = \frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$\int xd(x+5) = \int xdx$$

REPLY [1 votes]: Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$\begin{align}
\int xe^{-x^2}dx &= \int \frac{xe^{-x^2}dx}{1}\\
&= \int \frac{xe^{-x^2}dx}{1}\cdot\frac{\frac{d(-x^2)}{dx}}{\frac{d(-x^2)}{dx}}\\
&= \int \frac{xe^{-x^2}dx\cdot\frac{d(-x^2)}{dx}}{-2x}\\
&= \int \frac{xe^{-x^2}d(-x^2)}{-2x}\\
&= \color{red}{-\frac{1}{2}\int e^{-x^2}d(-x^2)} &[1]\\
&= \color{green}{-\frac{1}{2}\int e^udu} &[2]\\
&= -\frac{1}{2}e^u + C\\
&= -\frac{1}{2}e^{-x^2} + C\\
\end{align}$$


TITLE: Projection map from TM to M is smooth
QUESTION [0 upvotes]: Given smooth manifold $M$ how do you prove that the projection map $\pi : TM\to M$, $(p,v)\mapsto p$ is smooth?

REPLY [3 votes]: There are nice charts for $M$ and $TM$, given a chart $x$ for $M$:
$$
x:U\rightarrow V \text{ and the induced } Tx:TU\rightarrow U \times \mathbb R^n
$$
One way to show that a map is smooth, is to express it in charts, which typically is written $y\circ f \circ x^{-1}$ or something like that. And it is enough to show smoothness for one pair of charts around $p$ and $f(p)$ for every point $p$. For $\pi$ and the nice charts this is just
$$
x \circ \pi \circ (Tx)^{-1}
$$
which is the projection $U \times \mathbb R^n \rightarrow U$


TITLE: How many $m$-length strings over $\{0,1\}$, with at least $p$ $1$'s there are?
QUESTION [0 upvotes]: How many $m$-length strings over $\{0,1\}$, with at least $p$ $1$'s
  there are?

My answer is $\sum^m_{k=p} \binom{m}{k}$, but is there a close-form answer, without summation?

REPLY [2 votes]: On page 165 of Concrete Mathematics the authors say: "...there is no closed form for the partial sum of a row of Pascal's triangle."
Reference: Concrete Mathematics: A Foundation for Computer Science, by Ronald Graham, Donald Knuth, and Oren Patashnik.


TITLE: If $f$ is an entire function that goes to infinity, $f$ is a polynomial
QUESTION [5 upvotes]: I am asked to show that if $f$ is entire with the property that $\lim_{z\to\infty} f(z)=\infty$, then $f$ must be a polynomial.
However, I feel as if I am missing something.
$e^z$ is entire (holomorphic at all finite $z$), and clearly $\lim_{z\to\infty} f(z)=\infty$ holds. However, $e^z$ is not a polynomial.
This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

REPLY [2 votes]: One important feature about holomorphic functions is the number and position of their zeroes. The function going to $\infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.
Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.


TITLE: An integer $n > 7$ has form $r + 3k,\ r \in \{8,9,10\}\ $ [remainder systems]
QUESTION [2 upvotes]: I want to prove the following statement, related to a coins problem.
All integers greater than 7 can be expressed as 8 + 3k or 9 + 3k or 10 + 3k for some non-negative integer k.
I believe it can be done by strong induction, but I'm not quite ready to tackle that yet and am looking for a more elementary proof. I'm not sure how to approach it. It is obvious from investigating the integer number line, but I want something more formal.
$(\forall n \in \mathbb{N}), n > 7 \implies (n = 8 + 3k  \lor n = 9 + 3k \lor n = 10 +3k) $ for some $k \in Z_{\ge 0}$
Did I get the statement of the proposition right?
What would be a good way to proceed? Contradiction? Cases?

REPLY [1 votes]: The division algorithm says that for any integer $n$ there then there is a unique integer $m$ and an unique $b = 0, 1,2$ so that  $n = 3m + b$.
In other words this says exactly: Any integer can be expressed as $3m + 0$ or $3m + 1$ or $3m + 2$.
Case 1:  $n = 3m$ for some $m$.  And if $n > 7$ then $n \ge 9$ so $m \ge 3$.  Let $k = m-3 \ge 0$.  Then $n = 3m = 3(m-3) + 9 = 3k + 9$.
Case 2:  $n = 3m + 1$ for some $m$.  And if $n > 7$ then $3m > 6$ and $m > 2$.  So $m \ge 3$.  Let $k = m-3 \ge 0$.  Then $n = 3m + 1 = 3(m-3) + 9 + 1 = 3k +10$.
Case 3: $n = 3m +2$ for some $m$.  And if $n > 7$ then $n \ge 8$ and $3m + 2 \ge 8$ so $m \ge 2$. Let $k = m-2 \ge 0$.  Then $n =3m + 2 = 3(m-2) + 6 + 2 = 3k +8$.
And we are done.  If $n$ is an integer greater than $7$ it falls into exactly one of those three cases.


TITLE: Corollary 3.3.15 in Bruns and Herzog, Cohen-Macaulay Rings (self-contained question)
QUESTION [3 upvotes]: Theorem 3.3.14 [Bruns and Herzog, CMR]: Let $(R,m)$ be a CM local ring and $(R,m) \rightarrow (S,n)$ a flat local homomorphism. Then
(a) If $\omega_R$ exists (this is the canonical module of $R$) and $S/mS$ is Gorenstein, then $\omega_S = \omega_R \otimes_R S$.
(b) If $C$ is a finite $R$-module and in addition $C$ is CM with $\omega_S = C \otimes_R S$, then $S/mS$ is Gorenstein and $\omega_R \cong C$.
Corollary 3.3.15 [Bruns and Herzog, CMR]: Let $(R,m) \rightarrow (S,n)$ be a homomorphism of local Noetherian rings. Then $S$ is Gorenstein if and only if $R, S/mS$ are Gorenstein.

Question: I can see the direction $R, S/mS$ are Gorenstein $\Rightarrow S$ is Gorenstein, but i don't see how the reverse direction follows from the Theorem.

In particular, if $S$ is Gorenstein then it is CM and so both $R,S/mS$ are CM. Also, by Theorem 3.3.7 $\omega_S$ exists. But then to apply the theorem we would have to show that $\omega_S = C \otimes S$ for some $R$-finite module $C$, which i don't see why is true.

REPLY [1 votes]: I don't know if this follows from the theorem, but certainly follows from Proposition 1.2.16, Theorem 2.1.7, and Theorem 3.2.10 (from the same book).


text:
\begin{document}

\title{Algorithm for planar 4-Body Problem central configurations with given masses}

\author{E. Pi\~na \\
Department of Physics \\ Universidad Aut\'onoma Metropolitana - Iztapalapa, \\
P. O. Box 55 534. Mexico, D. F., 09340 Mexico \\
e-mail: pge@xanum.uam.mx}

\date{ }
\maketitle

\abstract{
An algorithm to compute the six distances between particles of a planar Four-Body central configuration is presented according to the following schema. An orthocentric tetrahedron is computed as a function of given masses. Each mass is placed at the corresponding vertex of the tetrahedron. The center of mass (and orthocenter) of the tetrahedron is at the origin of coordinates. The tetrahedron is orientated in a particular position function of the masses: with one of the particles placed on axis 3. The tetrahedron is rotated by two angles (to be tuned variables) around the center of mass until a direction orthogonal to the plane of configuration coincides with axis 3. The four coordinates of the vertices of the tetrahedron along this direction are identified with the weighted directed areas of the central configuration. The central configuration corresponding to these weighted directed areas is computed giving rise to four masses and corresponding distances. The given masses are compared with the computed ones. The two angles of the rotation are tuned until the given masses coincide with the computed. The corresponding distances of this last computation determine the central configuration. The case with two equal masses also is considered.
}
\

{\sl Keywords:} Four-Body Problem. Planar central configuration.

\

PACS 45.50.Pk Celestial mechanics 95.10.Ce Celestial mechanics(including n-body problems)

\newpage

\section{Introduction}

The coordinate system used in this paper has been presented in reference  \cite{pi}. The algorithm to compute planar central configurations starting from the weighted areas was published with several applications in \cite{pl}.

To begin I make a review of the new four body coordinates that mimics an important portion of \cite{pi}, expanded at important points and reduced in several comments.

The masses of the four bodies $m_1$, $m_2$, $m_3$ and $m_4$ are positive, generally different, but the values could be repeated.

We transform from the inertial referential, to the frame of principal axes of inertia by means of a three dimensional rotation $\bf G$ parameterized by three independent coordinates.

In addition to this rotation three more coordinates are introduced, as scale factors $R_1$, $R_2$, $R_3$, which are three directed distances closely related to the three principal inertia moments through
\begin{equation}
I_1 = \mu (R_2^2 + R_3^2)\, , \quad I_2 = \mu (R_3^2 + R_1^2) \, , \quad \mbox{and} \quad I_3 = \mu (R_1^2 + R_2^2) \, ,
\end{equation}
where $\mu$ is the mass
\begin{equation}
\mu = \sqrt[3]{\frac{m_1 \, m_2\, m_3\, m_4}{m_1 + m_2 + m_3 + m_4}} \, .
\end{equation}

With the first rotation and the change of scale the resulting four-body configuration has a moment of inertia tensor with the three principal moments of inertia equal. The second rotation $\bf G'$ does not change this property.

The cartesian coordinates of the four particles, with the center of gravity at the origin, written in terms of the new coordinates are
\begin{equation}
\left(\begin{array}{cccc}
x_1 & x_2 & x_3 & x_4 \\
y_1 & y_2 & y_3 & y_4 \\
z_1 & z_2 & z_3 & z_4
\end{array} \right) = {\bf G} \left(\begin{array}{ccc}
R_1 & 0 & 0 \\
0 & R_2 & 0 \\
0 & 0 & R_3
\end{array} \right) {\bf G'}^{\rm T}
\left(\begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4
\end{array} \right)\, ,
\end{equation}
where $\bf G$ and $\bf G'$ are two rotation matrices, each one a function of three independent coordinates such as the Euler angles, and where the column elements of the constant matrix
\begin{equation}
{\bf E} = \left(\begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4
\end{array} \right)\, ,
\end{equation}
are the coordinates of the four vertices of a rigid orthocentric tetrahedron, with the center of mass at the origin of coordinates, namely:
\begin{equation}
\begin{array}{c}
a_1 m_1 + a_2 m_2 + a_3 m_3 + a_4 m_4 = 0 \, , \\
b_1 m_1 + b_2 m_2 + b_3 m_3 + b_4 m_4 = 0 \, , \\
c_1 m_1 + c_2 m_2 + c_3 m_3 + c_4 m_4 = 0 \, .
\end{array}\, . \label{base}
\end{equation}

We introduce the following notation for the matrix
\begin{equation}
{\bf M} = \left( \begin{array}{cccc}
m_1 & 0 & 0 & 0 \\
0 & m_2 & 0 & 0 \\
0 & 0 & m_3 & 0 \\
0 & 0 & 0 & m_4
\end{array} \right) \, .
\end{equation}

An equivalent condition in order to have three equal inertia moments for the rigid tetrahedron is expressed as
\begin{equation}
{\bf E\, M\, E}^{\rm T} = \mu \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right) \label{ort}
\end{equation}

An orthocentric tetrahedron has the property that the perpendicular lines to the faces trough the four vertices intersect at the same point. Orthocentric tetrahedra were considered by Lagrange in 1773 \cite{la}. Other old references on orthocentric tetrahedra are found in a paper by Court \cite{co}, he calls them orthocentric and orthogonal to these tetrahedra. Placing the four masses at the corresponding vertices, the intersection point is actually the center of mass of the four masses, and the moment of inertia tensor of the four particles has the same principal value in any direction. Equation (\ref{ort}) imply that the inertia tensor of the rigid tetrahedron is proportional by a factor $2 \mu$ to the unit matrix.

To show these properties we consider for a while a fourth coordinate for the four vertex of the tetrahedron
\begin{equation}
(d_1, d_2, d_3, d_4) = \sqrt{\frac{\mu}{m}} (1, 1, 1, 1)\, , \label{cuarto}
\end{equation}
where we use the notation $m = m_1 + m_2 + m_3 + m_4$ for the total mass of the system. Then using properties (\ref{base}), (\ref{ort}), and (\ref{cuarto}) we write them in terms of $r = \sqrt{\frac{\mu}{m}}$ in the form
\begin{equation}
\frac{1}{\mu} \left( \begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4 \\
r & r & r & r
\end {array} \right) {\bf M} \left( \begin{array}{cccc}
a_1 & b_1 & c_1 & r \\
a_2 & b_2 & c_2 & r \\
a_3 & b_3 & c_3 & r \\
a_4 & b_4 & c_4 & r
\end {array} \right) = \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end {array} \right)\, .
\end{equation}
Since the inverse matrix from the left is equal to the inverse from the right, this equation transforms into
\begin{equation}
\left( \begin{array}{cccc}
a_1 & b_1 & c_1 & r \\
a_2 & b_2 & c_2 & r \\
a_3 & b_3 & c_3 & r \\
a_4 & b_4 & c_4 & r
\end {array} \right) \left( \begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4 \\
r & r & r & r
\end {array} \right) = \left( \begin{array}{cccc}
\frac{\mu}{m_1} & 0 & 0 & 0 \\
0 & \frac{\mu}{m_2} & 0 & 0 \\
0 & 0 & \frac{\mu}{m_3} & 0 \\
0 & 0 & 0 & \frac{\mu}{m_4}
\end {array} \right)\, .
\end{equation}
This matrix equation is equal to its transposed; therefore it just has ten independent equations. Four of them are
\begin{equation}
a_j^2 + b_j^2 + c_j^2 = \mu \left( \frac{1}{m_j} - \frac{1}{m}\right)\, \quad (j=1, 2, 3, 4)\, , \label{20}
\end{equation}
for the square of the distance from the orthocenter to the vertex $j$.
The other six are
\begin{equation}
a_i a_j + b_i b_j + c_i c_j = - \frac{\mu}{m} \quad \mbox{ ($i \neq j$)}\, , \label{21}
\end{equation}
the internal product of two directed vertices is the same for the six different possible choices of the pair $i, j$.
From these basic equations it is easy to show that the position vector of one vertex is orthogonal to the three vectors between two vertices of the oposite face (to the position vector of the vertex.)
\begin{equation}
a_i (a_j - a_k) + b_i (b_j - b_k) + c_i (c_j - c_k) = 0 \quad \mbox{ ($i,j,k$ different.) }
\end{equation}
In addition, the square of the distance between two vertices is given by
\begin{equation}
(a_i - a_j)^2 + (b_i - b_j)^2 + (c_i - c_j)^2 = \mu \left( \frac{1}{m_i} + \frac{1}{m_j}\right)\, . \label{lados}
\end{equation}
This last is the condition to have a moment of inertia tensor with the same three principal moments of inertia. The six edges of the tetrahedron should be equal (for an arbitrary $\mu$) to the square root of the right-hand side of this equation. The volume of this tetrahedron is equal to 1/6 if $\mu$ is selected as above.

There are other remarkable geometrical properties of an orthocentric tetrahedron. The center of mass of each face is at the orthocenter where the three altitudes of the face intersect. This point is on the same straight line between the opposite vertex and the center of mass. In addition to the orthogonality of the three sets of two opposite edges of the tetrahedron, the two orthogonal edges are also orthogonal to the line joining the center of mass of the two edges.

There are important coordinate systems to fix the origin for measuring the $\bf G'$ rotation; from these I prefer to choose in this paper particle with mass $m_1$ along coordinate axis 3, the other three in a parallel plane to the coordinate plane of 1 and 2, which does not include the  first particle; the particle with mass $m_2$ on an orthogonal coordinate plane that includes the first particle and the center of mass, and the other two particles on a line that is parallel to the coordinate axis 1 and perpendicular to the coordinate plane of the first two particles. Particle 1 has the coordinates
\begin{equation}
(a_1, b_1, c_1) = \left(0, 0, \sqrt{\frac{\mu (m - m_1)}{m_1 m}}\right)\, .
\end{equation}
Particle 2 has the coordinates
\begin{equation}
(a_2, b_2, c_2) = \left(0, \sqrt{\frac{\mu (m_3+m_4)}{m_2 (m - m_1)}}, - \sqrt{\frac{\mu m_1}{(m - m_1) m}}\right)\, .
\end{equation}
Particle 3 has the coordinates
$$
(a_3, b_3, c_3) =
$$
\begin{equation}
\left(\sqrt{\frac{\mu m_4}{m_3 (m_3+m_4)}}, - \sqrt{\frac{\mu m_2}{(m_3+m_4) (m - m_1)}}, - \sqrt{\frac{\mu m_1}{(m - m_1) m}}\right)\, .
\end{equation}
Particle 4 has the coordinates
$$
(a_4, b_4, c_4) =
$$
\begin{equation}
\left(- \sqrt{\frac{\mu m_3}{m_4 (m_3+m_4)}}, - \sqrt{\frac{\mu m_2}{(m_3+m_4) (m - m_1)}}, - \sqrt{\frac{\mu m_1}{(m - m_1) m}}\right)\, .
\end{equation}

This rigid tetrahedron is the generalization of the rigid triangle of the Three-Body problem with the center of mass at the orthocenter discussed previously in \cite{pb}. The same triangle was used with different purposes by C. Sim\'o \cite{si}.

I assume for simplicity that the potential energy is given by the Newton potential (the gravitational constant is equal to 1)
\begin{equation}
V = - \sum_{i<j}^3\frac{m_i m_j}{r_{ij}}\, ,
\end{equation}
although our results may be generalized for any potential with a given power law of the relative distances between particles $r_{ij}$. It follows the relation between the inter-particle distance and the new coordinates. The relative position between particles $i$ and $j$ is
\begin{equation}
\left( \begin{array}{c}
x_j - x_i \\
y_j - y_i \\
z_j - z_i
\end{array} \right) = {\bf G} \left( \begin{array}{ccc}
R_1 & 0 & 0 \\
0 & R_2 & 0 \\
0 & 0 & R_3
\end{array} \right)  {\bf G'}^{\rm T} \left( \begin{array}{c}
a_j - a_i \\
b_j - b_i \\
c_j - c_i
\end{array} \right)\, .
\end{equation}
The square of this vector is not a function of the first rotation $\bf G$, but just of the scale matrix and the second rotation matrix
\begin{equation}
r_{ij}^2 = (a_j - a_i \ b_j - b_i \ c_j - c_i) {\bf A} \left( \begin{array}{c}
a_j - a_i \\
b_j - b_i \\
c_j - c_i
\end{array} \right)\, .
\end{equation}

where $\bf A$ is the symmetric matrix
\begin{equation}
{\bf A}  = \left( \begin{array}{ccc}
A_{11} & A_{12} & A_{13} \\
A_{12} & A_{22} & A_{23} \\
A_{13} & A_{23} & A_{33}
\end{array} \right) = {\bf G'} \left( \begin{array}{ccc}
R_1^2 & 0 & 0 \\
0 & R_2^2 & 0 \\
0 & 0 & R_3^2
\end{array} \right) {\bf G'}^{\rm T} \, .
\end{equation}
The six distances are thus functions of six components of matrix $A$ or equivalently, are functions of the six independent coordinates in the scales $R_i$, and the rotation $\bf G'$.

We also compute the kinetic energy as a function of the new coordinates, which is given by
$$
T = {\mu \over 2} \left[ \sum_{i=1}^3 \dot{R_i}^2 - 4( R_2 R_3 \omega_1 \Omega_1 + R_3 R_1 \omega_2 \Omega_2 + R_1 R_2 \omega_3 \Omega_3) \right. +
$$
$$
\omega^{\rm T} \left(\begin{array}{ccc}
R_2^2 + R_3^2 & 0 & 0 \\
0 & R_3^2 + R_1^2 & 0 \\
0 & 0 & R_1^2 + R_2^2
\end{array} \right) \omega +
$$
\begin{equation}
\left.\Omega^{\rm T} \left(\begin{array}{ccc}
R_2^2 + R_3^2 & 0 & 0 \\
0 & R_3^2 + R_1^2 & 0 \\
0 & 0 & R_1^2 + R_2^2
\end{array} \right) \Omega  \right]\, ,
\end{equation}
where $\omega = (\omega_1, \omega_2, \omega_3)$ is the angular velocity vector of the first rotation $\bf G$, and $\Omega = (\Omega_1, \Omega_2, \Omega_3)$ is the corresponding angular velocity vector of the second rotation $\bf G'$.

\section{Equations of Motion}
The equations of motion follow from the Lagrange equations derived from the Lagrangian $T - V$ as presented in any standard text on Mechanics \cite{ll}, \cite{js}.

However, the three coordinates related to the first rotation produces Lagrange equations that imply, when the potential energy is a function only of the distances, conservation of the angular momentum vector in the inertial system
\begin{equation}
{\bf G} \; \, {\bf L}\, ,
\end{equation}
where $\bf L$ is the angular momentum in the principal moments of inertia frame
$$
{\bf L} = \frac{\partial T}{\partial \omega}
$$
\begin{equation}
= \mu \left(\begin{array}{c}
(R_2^2 + R_3^2) \omega_1 \\
(R_3^2 + R_1^2) \omega_2 \\
(R_1^2 + R_2^2) \omega_3
\end{array} \right) - 2 \mu \left(\begin{array}{c}
R_2 R_3 \Omega_1 \\
R_3 R_1 \Omega_2 \\
R_1 R_2 \Omega_3
\end{array} \right)\, .
\end{equation}

This conservation leads to three first order equations forming, for this four-body problem a generalization of the Euler equations (valid for the rotation of a rigid body), namely
$$
\frac{d}{d t} \left( \begin{array}{c}
\mu (R_2^2 + R_3^2) \omega_1 - 2 \mu R_2 R_3 \Omega_1 \\
\mu (R_3^2 + R_1^2) \omega_2 - 2 \mu R_3 R_1 \Omega_2 \\
\mu (R_1^2 + R_2^2) \omega_3 - 2 \mu R_1 R_2 \Omega_3
\end{array} \right) =
$$
\begin{equation}
\left( \begin{array}{ccc}
\mu (R_3^2 - R_1^2) \omega_2 \omega_3 + 2 \mu R_1 (R_2 \omega_2 \Omega_3 - R_3 \omega_3 \Omega_2) \\
\mu (R_1^2 - R_2^2) \omega_3 \omega_1 + 2 \mu R_2 (R_3 \omega_3 \Omega_1 - R_1 \omega_1 \Omega_3) \\
\mu (R_2^2 - R_3^2) \omega_1 \omega_2 + 2 \mu R_3 (R_1 \omega_1 \Omega_2 - R_2 \omega_2 \Omega_1)
\end{array}\right)\, . \label{euler}
\end{equation}

The so called \textit{elimination of the nodes} in the Three-Body problem \cite{wh}, has a similar representation in this coordinates for the Four-Body problem by means of the equation that equals the angular momentum vector in the principal moments of inertia frame to the rotation of a constant vector, which may be written in terms of two Euler angles
\begin{equation}
\mu \left(\begin{array}{c}
(R_2^2 + R_3^2) \omega_1 \\
(R_3^2 + R_1^2) \omega_2 \\
(R_1^2 + R_2^2) \omega_3
\end{array} \right) - 2 \mu \left(\begin{array}{c}
R_2 R_3 \Omega_1 \\
R_3 R_1 \Omega_2 \\
R_1 R_2 \Omega_3
\end{array} \right) = \ell {\bf G}^{\rm T} \left( \begin{array}{c}
0 \\
0 \\
1
\end{array} \right)\, ,
\end{equation}
where $\ell$ is the magnitude of the conserved angular momentum.

The Lagrangian equations of motion for the three scale coordinates are
\begin{equation}
\mu \frac{d^2}{d t^2} R_1 + 2 \mu [R_2 \omega_3 \Omega_3 + R_3 \omega_2 \Omega_2 ] + \mu R_1 (\omega_2^2 + \omega_3^2 +\Omega_2^2 + \Omega_3^2) = - \frac{\partial V}{\partial R_1} \, ,
\end{equation}
\begin{equation}
\mu \frac{d^2}{d t^2} R_2 + 2 \mu [R_3 \omega_1 \Omega_1 + R_1 \omega_3 \Omega_3 ] + \mu R_2 (\omega_3^2 + \omega_1^2 +\Omega_3^2 + \Omega_1^2) = - \frac{\partial V}{\partial R_2} \, ,
\end{equation}
and
\begin{equation}
\mu \frac{d^2}{d t^2} R_3 + 2 \mu [R_1 \omega_2 \Omega_2 + R_2 \omega_1 \Omega_1 ] + \mu R_3 (\omega_1^2 + \omega_2^2 + \Omega_1^2 + \Omega_2^2) = - \frac{\partial V}{\partial R_3} \, .
\end{equation}

The three equations of motion for the three coordinates associated with the second rotation $\bf G'$ are written as an Euler equation similar to the one found for the first rotation, although the internal angular momentum is not conserved because of the presence of an internal torque
$$
\frac{d}{d t} \left( \begin{array}{c}
\mu (R_2^2 + R_3^2) \Omega_1 - 2 \mu R_2 R_3 \omega_1 \\
\mu (R_3^2 + R_1^2) \Omega_2 - 2 \mu R_3 R_1 \omega_2 \\
\mu (R_1^2 + R_2^2) \Omega_3 - 2 \mu R_1 R_2 \omega_3
\end{array} \right) = \left( \begin{array}{c}
K_1 \\
K_2 \\
K_3
\end{array} \right) +
$$
\begin{equation}
\left( \begin{array}{ccc}
\mu (R_3^2 - R_1^2) \Omega_2 \Omega_3 - 2 \mu R_1 (R_2 \omega_2 \Omega_3 - R_3 \omega_3 \Omega_2) \\
\mu (R_1^2 - R_2^2) \Omega_3 \Omega_1 - 2 \mu R_2 (R_3 \omega_3 \Omega_1 - R_1 \omega_1 \Omega_3) \\
\mu (R_2^2 - R_3^2) \Omega_1 \Omega_2 - 2 \mu R_3 (R_1 \omega_1 \Omega_2 - R_2 \omega_2 \Omega_1)
\end{array}\right)\, ,
\end{equation}
where $K_1, K_2, K_3$ are the components of the internal torque $\bf K$ which is expressed in terms of the derivatives of the potential energy with respect to the three independent coordinates $q_j$ in the rotation $\bf G'$ and the three vectors ${\bf c}_j$ that appear in the expression of the angular velocity in terms of the same coordinates
\begin{equation}
\Omega = \sum_{j=1}^3 {\bf c}_j \dot{q}_j\, ,
\end{equation}
where the vectors ${\bf c}_j$ are generally functions of the coordinates $q_j$.

The internal torque is determined by the equations
\begin{equation}
{\bf K} \cdot {\bf c}_j = \frac{\partial V}{\partial q_j}\, .
\end{equation}

There is one  more constant of motion, namely the total energy
$$
E = T + K = V + {\mu \over 2} \left[ \sum_{i=1}^3 \dot{R_i}^2 - 4 ( R_2 R_3 \omega_1 \Omega_1 + R_3 R_1 \omega_2 \Omega_2 + R_1 R_2 \omega_3 \Omega_3) \right. +
$$
$$
\omega^{\rm T} \left(\begin{array}{ccc}
R_2^2 + R_3^2 & 0 & 0 \\
0 & R_3^2 + R_1^2 & 0 \\
0 & 0 & R_1^2 + R_2^2
\end{array} \right) \omega +
$$
\begin{equation}
\left.\Omega^{\rm T} \left(\begin{array}{ccc}
R_2^2 + R_3^2 & 0 & 0 \\
0 & R_3^2 + R_1^2 & 0 \\
0 & 0 & R_1^2 + R_2^2
\end{array} \right) \Omega  \right]\, .
\end{equation}

 \section{The plane problem}
The case with the four particles in a constant plane is an important and old subject \cite{dz}. Our coordinates are now adapted to that case. The third components of the cartesian coordinates of the four particles are zero. The modification of our coordinates (4) for this case is given by two changes: the first one is a rotation by just one angle in the plane of motion; and the second, the scale associated with the third coordinate is set equal to zero, namely
$$
\left(\begin{array}{cccc}
x_1 & x_2 & x_3 & x_4 \\
y_1 & y_2 & y_3 & y_4 \\
0 & 0 & 0 & 0
\end{array} \right) =
$$
\begin{equation}
\left(\begin{array}{ccc}
\cos \psi & -\sin \psi & 0\\
\sin \psi & \cos \psi & 0 \\
0 & 0 & 1
\end{array} \right)
 \left(\begin{array}{ccc}
R_1 & 0 & 0 \\
0 & R_2 & 0 \\
0 & 0 & 0
\end{array} \right) {\bf G'}^{\rm T}
\left(\begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4
\end{array} \right)\, .
\end{equation}
This equation simplifies to
$$
\left(\begin{array}{cccc}
x_1 & x_2 & x_3 & x_4 \\
y_1 & y_2 & y_3 & y_4
\end{array} \right) =
$$
\begin{equation}
\left(\begin{array}{cc}
\cos \psi & -\sin \psi \\
\sin \psi & \cos \psi
\end{array} \right) \left(\begin{array}{ccc}
R_1 & 0 & 0 \\
0 & R_2 & 0
\end{array} \right) {\bf G'}^{\rm T}
\left(\begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4
\end{array} \right)\, , \label{flat}
\end{equation}
in terms of six degrees of freedom.

We need three independent coordinates (for example three Euler angles) in $\bf G'$ for the two independent vectors in four dimensions expressed in the base of the three constant vectors $\bf a$, $\bf b$, and $\bf c$, orthogonal to the mass vector.

In order to formulate in a mathematical language the conditions for a plane solution we use the equations introduced by Dziobek \cite{dz}, that have been promoted by many years by A. Albouy and coworkers (see \cite{al} and references therein,) which consists in using the four directed areas of the triangles formed by the particles.

The four (double) directed areas are written in terms of the cartesian coordinates as
$$
S_1 = \left| \begin{array}{ccc}
1 & 1 & 1 \\
x_2 & x_3 & x_4 \\
y_2 & y_3 & y_4
\end{array} \right| \, , \quad S_2 = \left| \begin{array}{ccc}
1 & 1 & 1 \\
x_1 & x_4 & x_3 \\
y_1 & y_4 & y_3
\end{array} \right| \, ,
$$
\begin{equation}
S_3 = \left| \begin{array}{ccc}
1 & 1 & 1 \\
x_1 & x_2 & x_4 \\
y_1 & y_2 & y_4
\end{array} \right| \, , \quad S_4 = \left| \begin{array}{ccc}
1 & 1 & 1 \\
x_1 & x_3 & x_2 \\
y_1 & y_3 & y_2
\end{array} \right| \, ,
\end{equation}
which are the four signed 3 $\times$ 3 minors formed from the matrix
\begin{equation}
\left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
x_1 & x_2 & x_3 & x_4 \\
y_1 & y_2 & y_3 & y_4
\end{array} \right)\, .
\end{equation}
From now on I will refer to these areas simply as the directed areas.

The addition to the previous matrix of a row equal to any of its three rows produces a square matrix with determinant zero, that implies that the necessary and sufficient conditions to have a constant plane tetrahedron are
\begin{equation}
\sum_{i=1}^4 S_i = 0\, ,\label{plan}
\end{equation}
and
\begin{equation}
\sum_{i=1}^4 S_i x_i = 0\, , \quad \sum_{i=1}^4 S_i y_i = 0\, .\label{afin}
\end{equation}
The two last equations are summarized by the zero vector condition
\begin{equation}
\sum_{i=1}^4 S_i {\bf r}_i = {\bf 0}\, .\label{agrupo}
\end{equation}

An expression for the three directed areas in terms of the previous coordinates follows
\begin{equation}
\left( \begin{array}{c}
S_1 \\
S_2 \\
S_3 \\
S_4
\end{array} \right) =  C {\bf M E}^{\rm T} {\bf G'} \left( \begin{array}{c}
0 \\
0 \\
1
\end{array} \right) \, , \label{areas}
\end{equation}
where $C$ is a constant with units of area over mass. Equation (\ref{plan}) is satisfied from this expression of the directed areas, upon using that the four-vectors in matrix $\bf E$ are orthogonal to the mass vector. With the substitution of equations (\ref{flat}) and (\ref{areas}) in equations (\ref{afin}) one obtains an identity, independent of coordinates $R_1$, $R_2$, $\psi$, and of the rotation angles of $\bf G'$.

I denote here $\theta$ and $\phi$ the spherical coordinates determining this vector.
\begin{equation}
{\bf G'} \left( \begin{array}{c}
0 \\
0 \\
1
\end{array} \right) = \left( \begin{array}{c}
\sin \theta \cos \phi \\
\sin \theta \sin \phi \\
\cos \theta
\end{array} \right)\, ,
\end{equation}

Given the four masses, the four directed areas of the four particles are functions of this unit vector direction only, up to a multiplicative constant $C$ depending on the choice of physical units. Other form equivalent to the plane condition (\ref{agrupo}) is also published in reference \cite {pl}.

In the plane case the angular momentum has a constant direction orthogonal to the plane and of magnitude
\begin{equation}
P_\psi = \frac{\partial T}{\partial \dot{\psi}} = \mu [\dot{\psi}(R_1^2 + R_2^2) - 2 R_1 R_2 \Omega_3]\, . \label{mom}
\end{equation}
The kinetic energy becomes
$$
T = {\mu \over 2} \left[ \sum_{i=1}^2 \dot{R_i}^2 - 4( R_1 R_2 \dot{\psi} \Omega_3)+ \dot{\psi}^2 (R_1^2 + R_2^2) \right.+
$$
\begin{equation}
\left.\Omega^{\rm T} \left(\begin{array}{ccc}
R_2^2 & 0 & 0 \\
0 & R_1^2 & 0 \\
0 & 0 & R_1^2 + R_2^2
\end{array} \right) \Omega  \right]\, .
\end{equation}

Now I substitute polar coordinates for the $R_1$ and $R_2$ coordinates
\begin{equation}
R_1 = R \cos \alpha \, , \quad R_2 = R \sin \alpha\, .
\end{equation}
Writing the kinetic energy in terms of the angular momentum constant of motion instead of the $\dot{\psi}$ velocity lead us to
\begin{equation}
T = {\mu \over 2} \left[ \dot{R}^2 + R^2 \left(\dot{\alpha}^2 + \Omega_3^2 \cos^2 (2 \alpha) + \Omega_1^2 \sin^2 \alpha + \Omega_2^2 \cos^2 \alpha \right) \right] +\frac{P_\psi^2}{2 \mu R^2}\, .
\end{equation}
The energy conservation is thus expressed as
\begin{equation}
E = {\mu \over 2} \left[ \dot{R}^2 + R^2 \left(\dot{\alpha}^2 + \Omega_3^2 \cos^2 (2 \alpha) + \Omega_1^2 \sin^2 \alpha + \Omega_2^2 \cos^2 \alpha \right) \right] + \frac{P_\psi^2}{2 \mu R^2} + V \, , \label{ene}
\end{equation}
where $V$ represents the potential energy.

\section{Central configurations}

The non collinear planar central configurations are characterized in our coordinates by constant values of the $\bf G'$ matrix and of the coordinate $\alpha$ associated to the constant value of the ratio $R_1/R_2$. For these cases the angular velocity vector $\Omega$ is the null vector, the angular velocity $\dot{\alpha}$ is also zero and the conservation equations of moment and energy, (\ref{mom}) and (\ref{ene}) respectively, become
\begin{equation}
P_\psi = \mu \dot{\psi} R^2 \, .
\end{equation}
and
\begin{equation}
E = {\mu \over 2} \dot{R}^2 +\frac{P_\psi^2}{2 \mu R^2} + V \, .
\end{equation}
Note that the potential function $V$ becomes equal to a constant divided by coordinate $R$.

These equations are identical to similar equations obtained for the Euler and Lagrange central configurations of the Three-Body problem \cite{bp}. They are formally the same as the equations for the conics in the Two-Body problem in terms of the radius $R$ and the true anomaly $\psi$.

 The constant values of the $\bf G'$ matrix and angle $\alpha$ referred to above are not arbitrary but they are determined by three independent quantities as discussed in what follows.

 I begin with the approach by Dziobek \cite{dz}. The Four-Body central configurations are determined as critical points of the potential energy with a fixed total inertia moment. The planar solutions with zero volume but finite area are obtained taking in account that the variational equation is modified adding the restriction of planar motion. This condition is obtained by Dziobek \cite{dz} from the derivative of the Cayley-Menger determinant with respect to $r_{ij}^2$ that he found to be proportional to the product of the directed areas $S_i S_j$.

It follows that the solution is given in terms of Lagrange's multipliers $\lambda$ and $\sigma$
\begin{equation}
r_{jk}^{-3} = \sigma + \lambda A_j A_k\, ,
\end{equation}
where $A_j = S_j/m_j$ are weighted areas, quotient of the directed area divided by the corresponding mass. This equation was presented by Dziobek \cite{dz}. A proof was published by Moeckel \cite{mo}, and using a different approach to the same problem, deduced by Albouy \cite{al}. A new proof of the equation was obtained in a different approach by Pi\~na and Lonngi \cite{pl}.

It follows from (\ref{areas}) that in a planar solution the weighted directed areas are expressed as
\begin{equation}
\left( \begin{array}{c}
A_1 \\
A_2 \\
A_3 \\
A_4
\end{array} \right) =  C {\bf E}^{\rm T} {\bf G'} \left( \begin{array}{c}
0 \\
0 \\
1
\end{array} \right) = C {\bf E}^{\rm T} \left( \begin{array}{c}
\sin \theta \cos \phi \\
\sin \theta \sin \phi \\
\cos \theta
\end{array} \right)\, . \label{sec}
\end{equation}
The weighted directed areas are up to a normalization factor equal to the third rotated coordinate of the rigid tetrahedra. In terms of vectors (15-18) and angles $\theta$ and $\phi$ this equation is expressed as
\begin{equation}
Aj = C (a_j \sin \theta \cos \phi + b_j \sin \theta \sin \phi + c_j \cos \theta) \, .
\end{equation}
The explicit expressions choosing $C = \sqrt{(m-m_1)/\mu}$ are
$$
A_1 = \frac{m - m1}{\sqrt{m_1 m}} \cos \theta \, ,
$$
$$
A_2 = - \sqrt{\frac{m_1}{m}} \cos \theta + \sqrt{\frac{m_3 + m_4}{m_2}} \sin \theta \sin \phi \, ,
$$
$$
A_3 = - \sqrt{\frac{m_1}{m}} \cos \theta - \sqrt{\frac{m_2}{m_3 + m_4}} \sin \theta \sin \phi + \sqrt{\frac{m_4 (m - m_1)}{m_3 (m_3 + m_4)}} \sin \theta \cos \phi
$$
\begin{equation}
A_4 = - \sqrt{\frac{m_1}{m}} \cos \theta - \sqrt{\frac{m_2}{m_3 + m_4}} \sin \theta \sin \phi - \sqrt{\frac{m_3 (m - m_1)}{m_4 (m_3 + m_4)}} \sin \theta \cos \phi\, .\label{sfero}
\end{equation}

Since the lengths and masses are defined up to arbitrary units, we assume \cite{pl}, with no loss of generality, that the parameter $\sigma$ equals unity
\begin{equation}
r_{jk}^{-3} = 1 + \lambda A_j A_k\, \quad (j \neq k).
\end{equation}

This implies, according to \cite{pl} that the unit of distance always obeys the restriction
\begin{equation}
\sigma = \frac{\sum\limits_{j>k} m_j m_k / r_{jk}}{\sum\limits_{j>k} m_j m_k r_{jk}^2} = 1\, .
\end{equation}

In the paper by Pi\~na and Lonngi \cite{pl} the directed weighted areas are known as four given constants. The previous equation then gives the distances as functions of the unknown parameter $\lambda$. Through them, the areas of the four triangles become functions of $\lambda$, that should obey the necessary restrictions (\ref{plan}), (\ref{afin}), to verify that one has a planar solution. This restriction allows in many cases to determine the value of $\lambda$ and hence the values of the six distances and the four masses. This is an implicit way to deduce planar central configurations with four masses.

In this paper we assume the four masses are known from the beginning, the four weighted areas are then determined by expressions (\ref{sfero}) in terms of the two tuning variables $\theta$ and $\phi$. Given particular values of these two angles, become determined the four constants $A_j$, until a multiplicative factor, which in turn produces a computed central configuration with given distances and masses. These computed masses are not in general equal (or proportional) to the starting values used to compute the orthocentric tetrahedron. The two angles are then tuned until a numerical coincidence is produced between the given and the computed masses. The distances between particles, computed for this last central configuration, correspond to the given masses.

\begin{figure}
\centering
\scalebox{0.6}{\includegraphics{proyesf1.ps}}

\caption{Spherographic projection of the hemisphere of the two angles motion of the orthocentric tetrahedron. The great circles represent the positions where three particles are collinear. The four spherical triangles are concave open sets labeled by the particle at the interior of the triangle. The spherical rectangular open sets correspond to convex configurations with the same order that the neighbor triangles. The isolated points are at the angles where a central configuration has been computed. The values of the masses are $m_1=10, m_2=13, m_3=15, m_4=17$. }
\end{figure}

Choosing the values of the masses as $m_1 = 10, m_2 = 13, m_3 = 15, m_4 = 17$, I have been able to compute seven central configurations.

\noindent A concave with $m_1$ at the interior for $\lambda = -2.32656490060845$
$$
\begin{array}{ll}
\theta = 0.139240050165164 & \phi = 4.8453912490189 \\
r_{31} = 0.749352173668766 & r_{42} = 1.28360719430431 \\
r_{41} = 0.730065912777101 & r_{23} = 1.22971324110202 \\
r_{12} = 0.639643905964532 & r_{43} = 1.10420199339901
\end{array}
$$
A concave with $m_2$ at the interior for $\lambda = -1.82536262997443$
$$
\begin{array}{ll}
\theta = 1.03401714345126 & \phi = 4.70428233126543 \\
r_{23} = 0.804939324544205 & r_{41} = 1.36218205283364 \\
r_{42} = 0.798422993763499 & r_{31} = 1.32975926553097 \\
r_{12} = 0.640524087661698 & r_{43} = 1.07615875773082
\end{array}
$$
A concave with $m_3$ at the interior for $\lambda = -1.6355867896085$
$$
\begin{array}{ll}
\theta = 1.00916705438091 & \phi = 2.69053445366272 \\
r_{23} = 0.819159125672848 & r_{41} = 1.41307775730063 \\
r_{43} = 0.807449041118773 & r_{12} = 1.34385908526924 \\
r_{31} = 0.666344917442357 & r_{42} = 1.08738922587974
\end{array}
$$
A concave with $m_4$ at the interior for $\lambda = -1.55004225977105$
$$
\begin{array}{ll}
\theta = 1.33726059387504 & \phi = 0.589297944199426 \\
r_{41} = 0.80384936609868 & r_{23} = 1.44896029979243 \\
r_{43} = 0.768450306675012 & r_{12} = 1.27362211521548 \\
r_{42} = 0.730824120911527 & r_{31} = 1.18407793737811
\end{array}
$$
A convex with $r_{41}$ at the diagonal for $\lambda = -0.59067068058041$
$$
\begin{array}{ll}
\theta = 0.833903746254753 & \phi = 3.74619198490858 \\
r_{41} = 1.26246852646001 & r_{23} = 1.21353489669217 \\
r_{12} = 0.834421340108319 & r_{43} = 0.914579880279118 \\
r_{31} = 0.855040198605932 & r_{42} = 0.900505139131835
\end{array}
$$
A convex with $r_{42}$ at the diagonal for $\lambda = -0.571602583650311$
$$
\begin{array}{ll}
\theta = 0.849784880696665 & \phi = 5.66013081680728 \\
r_{42} = 1.1712284103105 & r_{31} = 1.31084659340392 \\
r_{12} = 0.837603049326044 & r_{43} = 0.916446768964837 \\
r_{41} = 0.874842967906726 & r_{23} = 0.888555880648135
\end{array}
$$
A convex with $r_{43}$ at the diagonal for $\lambda = -0.532931485997706$
$$
\begin{array}{ll}
\theta = 0.861931053448714 & \phi = 1.64118840218233 \\
r_{43} = 1.13115400784062 & r_{12} = 1.36414262641605 \\
r_{31} = 0.863912312413253 & r_{42} = 0.907050181187521 \\
r_{41} = 0.880188793100079 & r_{23} = 0.893570787310162
\end{array}
$$

\section{The case with two equal masses}

When two masses are equal, the other two generally different, the orthocentric tetrahedron become symmetric with respect to a symmetry plane containing the two (possible) different masses and the center of mass. When the tetrahedron is projected orthogonal to a direction contained in the symmetry plane the configuration becomes with the kite symmetry. Central configurations with two equal masses and kite symmetry are allowed if and only if the plane of symmetry is projected in a line. Such particular case is considered rotating the symmetry plane by a single tuning angle until the given masses coincide with the computed ones. It is only relevant half a circle of this angle. Three cases are possible separated in the semicircle by three contiguous different open sectors. Two concave sectors and one convex sector.

One illustrate this case computing central configurations for the masses
$m_1=8, m_2=10, m_3=m_4=9$. Two different symmetric concave central configurations were found with the largest mass in the interior for $\lambda = - 1.938380387574535$
$$
\begin{array}{ll}
\theta = 0.053846821928071 & \phi = \pi/2 \\
r_{12} = 0.771017850970647 & r_{43} = 1.37174365887618 \\
r_{31} = 0.737687261978754 & = r_{41} \\
r_{23} = 1.24798073364597 & = r_{42}
\end{array}
$$
and for $\lambda = -1.9051833373390781$
$$
\begin{array}{ll}
\theta = 0.131677055692469 & \phi = 3 \pi/2 \\
r_{12} = 0.704334104242389 & r_{43} = 1.16065690193433 \\
r_{31} = 0.777736225967895 & = r_{41}  \\
r_{23} = 1.35289892174911 & = r_{42}
\end{array}
$$
and one convex for $\lambda = -0.6629645165685381$$$
\begin{array}{ll}
\theta = 1.00225261834462 & \phi = \pi/2 \\
r_{43} = 1.2388066072406 & r_{12} = 1.24313752444539 \\
r_{31} = 0.891030948126205 & = r_{41}  \\
r_{23} = 0.864175379988576 & = r_{42}.
\end{array}
$$

No concave solution was found for the lowest mass $m_1 = 8$ in the interior.

Non symmetric central configurations do exist for the case of two equal masses. For the given masses $m_1=8, m_2=10, m_3=m_4=9$ one has the non symmetric convex central configuration obtained by searching with two angles as tuning variables with $r_{23}$ at the diagonal, and the two equal masses not on a single diagonal, for $\lambda = -0.65459120484497$
$$
\begin{array}{ll}
\theta = 0.991809150592805 & \phi = 3.71189655731751 \\
r_{23} = 1.28383284893302 & r_{41} = 1.20043491498806 \\
r_{42} = 0.865256456835286 & r_{31} = 0.891936018662304 \\
r_{12} = 0.878638767853787 & r_{43} = 0.879635445676687
\end{array}
$$
and the the non symmetric concave central configuration with one of the equal masses at the interior for $\lambda = -2.0782713344769$
$$
\begin{array}{ll}
\theta = 1.2731702593358 & \phi = 2.4920236726443 \\
r_{23} = 0.681773343374296 & r_{41} = 1.13338465823949 \\
r_{31} = 0.767118374257656 & r_{42} = 1.31028169747428 \\
r_{43} = 0.776744110452106 & r_{12} = 1.35194750927076
\end{array}
$$


TITLE: Expected Value of $Z = min(X, M)$ where M is a constant
QUESTION [0 upvotes]: Let $X \sim Geometric(p)$, and let $M > 0$ be a positive integer. Determine the expected value of $Z = min(X, M)$.

I have done similar exercises where $M$ is a random variable, but not a constant. I don't know how to proceed here, and therefore have no work to show. Hints are very welcome although It would be nice to see more or less of a complete solution for a case like this.
Thank you.

REPLY [1 votes]: Note that
$$
EZ=EXI(X<M)+EMI(X\geq M)\tag{0}
$$
where $I$ is the indicator function since $Z=X$ if $X<M$ and $Z=M$ if $X\geq M$. But
$$
EMI(X\geq M)=MP(X\geq M)\tag{1}
$$
and
$$
EXI(X<M)=\sum_{k=0}^{M-1}kP(X=k).\tag{2}
$$
Since $X$ is geometric we can compute, $(1)$ and $(2)$ and hence $(0)$. I leave this computation to you.


text:
\setcounter{figure}{0}
\renewcommand{\thefigure}{F\arabic{figure}}

\section{COVID-19 CT-Scan Images} \label{sec:appx_covid}

Here, we analyze the contents of the SARS-COV-2 CT-Scan Dataset \citep{angelov2020explainable}. This dataset contains 2,482 images of CT-Scan, 1,230 of which belong to infected patients and 1,252 belong to non-infected patients. Figure~\ref{fig:covid_samples} shows some samples from this dataset.

\begin{figure}[H]
  \centering
  \fbox{
   \includegraphics[width=0.22\linewidth]{images-covid/covid_image1_1.jpg}
   \includegraphics[width=0.22\linewidth]{images-covid/covid_image1_2.jpg}}
   \hfil
  \fbox{
   \includegraphics[width=0.2\linewidth]{images-covid/covid_image2_2.jpg}
   \includegraphics[width=0.2\linewidth]{images-covid/covid_image2_3.jpg}}
   \caption{Sample of images in the SARS-COV-2 CT-Scan dataset. Two images in the left box are from {\em infected} patients and the ones in the right box are from {\em non-infected} patients.}
  \label{fig:covid_samples}
\end{figure}

We form the data into two matrices, $\mathcal{D}_1$ for COVID samples and $\mathcal{D}_2$ for non-COVID samples. For wavelet transformation, we use Daubechies-3 wavelets. We use the RR-QR algorithm to choose a subset of $m=1200$ most influential wavelet coefficients. Figure~\ref{fig:covid_stencil} highlights the pixels corresponding to those wavelet coefficients.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.25\linewidth]{images-covid/covid_stencil.png}
   \caption{Pixels corresponding to the most influential coefficients chosen by the RR-QR algorithm. A radiologist could verify whether the patterns corresponding to infections are expected to be in those regions. RR-QR detects that most of the variations among images are in the white regions, and changes in the black regions are either small or linearly dependent.}
  \label{fig:covid_stencil}
\end{figure}

Using Higher Order GSVD, we decompose the tensor of wavelet coefficients $\mathcal{D}$. The resulting singular values, shown in Figure~\ref{fig:covid_singulars} show a clear separation between the Covid and non-Covid patients.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.95\linewidth]{images-covid/covid_singulars.png}
   \caption{First 200 singular values (in logarithmic scale) show a clear separation between many of the patterns in COVID and non-COVID patients. Nitoce that when singular values are large for the COVID class, singular values are usually small for the non-COVID class, and vice versa.}
  \label{fig:covid_singulars}
\end{figure}

Figure~\ref{fig:covid_patterns} shows the most dominant patterns obtained for the COVID patients and Figure~\ref{fig:noncovid_patterns} shows the most dominant patterns obtained for non-COVID patients. Clearly patterns seem very different for non-COVID and COVID patients.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right1.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right2.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right3.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right4.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right5.png}
   \caption{The most dominant patterns for COVID infected patients. Notice the color variations within the boundaries defined in Figure~\ref{fig:covid_stencil}.}
  \label{fig:covid_patterns}
\end{figure}

\begin{figure}[H]
  \centering
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right9non.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right11non.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right13non.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right14non.png}
   \includegraphics[width=0.25\linewidth]{images-covid/covid_right16non.png}
   \caption{The most dominant patterns for non-COVID patients. Notice the color variations within the boundaries defined in Figure~\ref{fig:covid_stencil}. Also, notice how the color variations are different from the COVID-19 infected patients in Figure~\ref{fig:covid_stencil}.}
  \label{fig:noncovid_patterns}
\end{figure}

We can use these patterns to reconstruct an image. Figure~\ref{fig:covid_reconstruct} shows reconstruction of an image using low-rank approximation defined by Equation~\ref{eq:sum_rank1}.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba50.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba100.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba300.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba400.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba500.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba700.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba1000.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_cl1_im1_ba1200.png}
   \includegraphics[width=0.7\linewidth]{images-covid/covid_cl1_im1_residual.jpg}
   \caption{Reconstruction of an image using the patterns extracted in $V$. We can see the evolution of image as more patterns are added to it. The residual also drops relatively quickly. We have captured most distinctive features of the image by using about 400 patterns. Using 1,000 patterns captures the image almost perfectly.}
  \label{fig:covid_reconstruct}
\end{figure}

The left basis defines how the patterns should be merged together. Images with lower norm in the left basis appear to have fewer distinctive features in them, as some of them are shown in Figure~\ref{fig:covid_simple}. In other word, they are made from fewer patterns in $V$. On the other hand, images with larger norm in the left basis have more distinctive features and many patterns contribute to them, as some of them are shown in Figure~\ref{fig:covid_complex}.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple293.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple286.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple980.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple1181.png}
   \caption{Images with small norm in the left basis appear are made from fewer patterns and appear to have less distinctive features.}
  \label{fig:covid_simple}
\end{figure}

\begin{figure}[H]
  \centering
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple113.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple365.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple409.png}
   \includegraphics[width=0.24\linewidth]{images-covid/covid_simple604.png}

   \caption{Images with large norm in the left basis appear relatively complex, have many distinctive features and many patterns contribute to them.}
  \label{fig:covid_complex}
\end{figure}

We note that evaluating the usefulness of our patterns, from the practical point of view, requires domain expertise. A radiology scientist, for example, could verify whether the patterns in Figure~\ref{fig:covid_patterns} are sensible and useful in practice.

file: 2006.09879/patterns in image classification/a6-covid.tex


TITLE: Formal definition of a complement of a language.
QUESTION [0 upvotes]: I am in the midst of learning decidable and undecidable language and I came across the following theorem:

A language  $A \subseteq \Sigma^*$ is decidable if and only if $A$ as well as $\Sigma^* \setminus A$ are semi-decidable.

I'm not sure what $\Sigma^* \setminus A$, the complement of the language $A$, is supposed to mean. A language is a set of strings, so would its complement then be the set of all strings that are not in $A$? Can anyone provide a more formal explanation and definition?

REPLY [2 votes]: $\Sigma^*\setminus A = \{x\mid x\in\Sigma^*\wedge x\not\in A\}$.


TITLE: Any general algorithm to find $x$ given $y$ for the elliptic curve equation $y^2 = x^3 + ax + b$, in real numbers?
QUESTION [0 upvotes]: Is there a way to solve for $x$ given $y$, in the equation $y^2 = x^3 + ax + b$, in the real number space (not finite field).
Or is the solution too difficult to achieve? I suspect the solution will require factoring a 3rd degree polynomial.
I'm trying to animate a dot going from top to bottom of the screen while following the curve. This will require inputing the $y$ component into some function, and getting the $x$ component in return, so that I can place the dot in the correct spot on the graph.
This question is not to be confused with the question asking about finding $x$ for the elliptic curve equation, in a finite field. I'm trying to solve in the real number space, not in the finite field.

REPLY [1 votes]: Consider the cubic equation $$ x^3 + \alpha x + (\beta-y^2) =0$$ The discriminant
$$\Delta=-4 \alpha ^3-27 \left(y^2-\beta \right)^2$$ is always negative if $\alpha >0$. So, if this is the case, there is only one real root.
Using the hyperbolic method, this root is
$$\color{blue}{x=-2 \sqrt{\frac{\alpha }{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac 32 \sqrt{\frac3 {\alpha }}\,\,\frac{(\beta-y^2)}\alpha\right)\right)}$$
If $\alpha <0$ (still with $\Delta<0$), the solution is
$$\color{blue}{x=-2 \sqrt{-\frac{\alpha }{3}}\,\,\frac{\left| \beta -y^2\right| }{\beta -y^2}\,\cosh \left(\frac{1}{3} \cosh ^{-1}\left(-\frac 32 \sqrt{-\frac3 {\alpha }} \,\,\frac{\left| \beta -y^2\right| }{\alpha }\right)\right)}$$
This is much better than using Cardano formula.


TITLE: Is there an easy way to find orthogonal polynomials w.r.t. $w(x)=-\ln(x)$ on $(0, 1)$?
QUESTION [2 upvotes]: I've been evaluating them recurcively using $$\varphi_{n}=x^n - \sum_{k=0}^{n-1} \frac{\langle x^{k}, \varphi_{k} \rangle_{w}}{\langle \varphi_{k}, \varphi_{k} \rangle_{w}}\cdot\varphi_{k}$$
Where $\langle f,g \rangle_{w}=\int_{0}^{1} -\ln|{x}|f(x)g(x)\: dx$.
The computations get very complicated for $n\ge 3$.

REPLY [3 votes]: Try the substitution $x= e^{-t}$.  Then the inner product $<x^j, x^k>$ of the  monomial pairs $x^j, x^k$  in their transformed  format is $$ \int_{t=0}^{t=\infty} e^{-kt} e^{-jt} (t e^{-t}) dt$$
That integral can be evaluated sneakily by making use of the  identity $ -\frac{d}{da} e^{-at} = t e^{-at}$ and setting $a=1$ afterward.
In detail  $$<x^j, x^k> = (-\frac {d}{da})  \int_0^{\infty}  e^{-(k+j+a) t} \ dt = -\frac {d}{da}( \frac {1}{j+k+a})|_{a=1} $$
$$= \frac{1}{(j+k+1)^2}$$
Then you can use the usual Gram-Schmidt method to convert these monomials into an orthonormal basis.
P.S. Using Mathematica (by another method) I got this small table that you can use to check your work. Good luck


text:
{\bf Problem.} Compute $\sin 0^\circ$.
{\bf Level.} Level 1
{\bf Type.} Geometry
{\bf Solution.} Rotating the point $(1,0)$ by $0^\circ$ counterclockwise about the origin gives us the point $(1,0)$, so $\sin 0^\circ = \boxed{0}$.

set_name: MATH