text:
\begin{document}

\title{Exact Solutions of the DKP Oscillator in 3D Spaces with Extended Uncertainty Principle}
\author{{Mokhtar Falek}$^{1}$ \and {Mustafa Moumni}$^{1,a}$ \and {Mahmoud Merad}$^{2}$ \\

$^{1}$Laboratory of Photonic Physics and Nano-Materials (LPPNNM)\\
Department of Matter Sciences, University of Biskra, ALGERIA \\
$^{2}$Faculty of Exact Sciences, University of Oum-el-Bouaghi, ALGERIA\\
$^{a}$correspondant author m.moumni@univ-biskra.dz}
\maketitle

\begin{abstract}
We present the exact solutions of the three-dimensional Duffin--Kemmer--Petiau oscillator for both spin 0 and spin 1 cases, with the presence of minimal uncertainty in momentum for anti--de Sitter model and we derive the solutions for the case of deSitter space. We use the representation of vector spherical harmonics and the method of Nikiforov--Uvarov to determine exactly the energy eigenvalues and the full expressions of the eigenfunctions in all cases. Our study of the energy spectrum allows us to define a new interpretation of natural and unnatural parity states of the vector particle and we show the crucial role played by the spin--orbit coupling in this differentiation between the parities.

Keywords: DKP oscillator; (Anti--)de Sitter spaces; Three dimensions

P. A. C. S. 03.65.Ge, 03.65.Pm.
\end{abstract}

\section{Introduction}

During recent years, there has been a growing interest in absorbing the divergences appearing in quantum field theory (and other theories aimed at unifying the fundamental interactions) and several scenarios have been proposed to solve this kind of problems. Among these attempts, we find the extension of the quantum field theory to curved space-time, which can be considered as a first approximation of the theory of quantum gravity. In such situation, the usual Heisenberg uncertainty principle can be replaced by the so--called extended uncertainty principle (EUP) which is characterized mainly by the existence of a minimal length scale in the order of the Planck length \cite{Mignemi10}. A fundamental consequence deduced from this extension is that the minimal length uncertainty in quantum gravity can be related to a modification of the standard Heisenberg algebra by adding small corrections to the canonical commutation relations; we quote here Mignemi which has shown that, in a (anti--)de Sitter background, the Heisenberg uncertainty principle is modified by introducing corrections proportional to the cosmological constant \cite{Mignemi10}. Such a scenario is motivated by Doubly Special Relativity (DSR) \cite{Amelino01, Amelino02}, string theory \cite{Capozziello00}, non-commutative geometry \cite{Douglas01}, black hole physics \cite{Scardigli99, Scardigli03} and even from Newton's gravity effects on quantum systems \cite{Kuzmichev20}.

Recently, the introduction of this idea of EUP has drawn great interest and a significant number of papers appeared in the literature to address the effects of the extended commutation relations in quantum mechanics systems. We cite here the studies of thermodynamic properties of the relativistic harmonic oscillators on anti-de Sitter (AdS) space \cite{Hamil18}, the Klein--Gordon oscillator in an uniform magnetic field \cite{Chung19}, the exact solution of (1+1)-dimensional bosonic oscillator subject to the influence of an uniform electric field in AdS space \cite{Moussa18}. In addition, certain problems have also been solved in non-relativistic quantum mechanics despite the fact that, in conventional field theory approach of static de Sitter (dS)\ and AdS space--time models, we cannot derive any nonrelativistic covariant Schr\"{o}dinger--like equation from covariant Klein--Fock--Gordon equation. In this context, we can use the EUP formulation to write the dS and AdS versions of the Schr\"{o}dinger equation. Indeed, we find the treatment of the exact solution of the D--dimensional Schr\"{o}dinger equation for the free particle and the harmonic oscillator in AdS space \cite{Hamil19}, the study, with perturbative methods, of the implications of extended uncertainty principle of dS Space on the spectrums of both harmonic oscillator and hydrogen atom \cite{Ghosh11} and the exact solution of the Schr\"{o}dinger equation for the hydrogen atom in dS and AdS spaces \cite{Falek20}.

In this work, we solve the three--dimensional (3D) Duffin--Kemmer--Petiau (DKP) oscillator for spin 0 and spin 1 particles in AdS models as it was done very recently for the Dirac oscillator \cite{Merad19}. We will show rigorously that the problem admits analytical solutions for both scalar and vector particles; so we will compute the exact expressions of the eigenenergies and write the final forms of the eigenfunctions in all cases. Our interest in the DKP equation \cite{Duffin38, Kemmer39, Petiau36} comes from the fact that it is richer than those of Klein-Gordon and Proca and therefore it has more potential applications, especially for the study of hadrons and nuclei \cite{Clark85, Kalber86, Kozack89}. We also find studies on DKP in Hamiltonian covariant dynamic \cite{Kanatchikov00}, in Galilei covariance \cite{Montigny00} and in different topologies such as non--commutative spaces \cite{Falek08, Hassanabadi12, Achour19} or curved space--times \cite{Lunardi00, Castro15, Hosseinpoor18}. We focus on the harmonic oscillator given the great interest for this potential in quantum systems since it reflects a confinement with a non-zero residual energy. That is why it is the central potential of the nuclear shell model and also of the confining two--body potential for quarks. The relativistic version of the harmonic oscillator generated much interest especially since the work of Moshinsky and Szczepaniak \cite{Moshinsky89} for the Dirac equation and the works of Nedjadi and Barrett \cite{Nedjadi941, Nedjadi942} for the DKP version; we refer the reader here to the works already cited \cite{Hamil18, Chung19, Moussa18, Hamil19} or to \cite{Castro15} and \cite{Hosseinpoor19} where there is a very extensive list of references on both Dirac and DKP oscillators. We mention here that the relativistic oscillator was experimented recently \cite{Franco13, Fujiwara18}.

The outline of this paper is as follows: In the next section \ref{sec:RV}, we give a review on dS and AdS models, while in the third section \ref{sec:NU}, we introduce Nikiforov--Uvarov (NU) method used in our work to solve the system. In section four \ref{sec:S0}, we expose the explicit calculation of the deformed 3D DKP oscillator for spin 0 case in the framework of EUP; we do this in position space representation. By a straightforward calculation, our system will be converted to a Klein--Gordon equation type and we use the representation of vector spherical harmonics to solve it analytically; the corresponding radial wave functions are expressed with the Jacobi polynomials. In the fifth section \ref{sec:S1}, we use the same method and determine the exact solutions of the DKP oscillator for both natural and unnatural parities of spin 1 case. We will derive the solutions for dS models from the ones corresponding to AdS spaces in the penultimate section \ref{sec:dS} and finally, the concluding remarks come in the last section \ref{sec:CC}.

\section{Review of the deformed quantum mechanics relation}
\label{sec:RV}
In three--dimensional case, the deformed Heisenberg algebra leading to EUP of dS and AdS Spaces is defined by the following commutation relations \cite{Mignemi12, Stetsko15}:
\begin{equation}
\left[X_{i},X_{j}\right]=0\text{ , }\left[P_{i},P_{j}\right] =-i\hbar
\lambda \epsilon _{ijk}L_{k}\text{ , }\left[ X_{i},P_{j}\right] =i\hbar
\left( \delta _{ij}+\lambda X_{i}X_{j}\right)  \tag{1}  \label{eqt1}
\end{equation}
where $\lambda $ is a small parameter related to the deformation; it is positive for AdS case and negative for dS one. For example in the context of quantum gravity, this EUP parameter $\lambda$ is determined as the fundamental constant associated to the scale factor of the expanding universe and it is proportional to the cosmological constant $\Gamma=-3\lambda =-3a^{-2}$ where $a$ is the AdS radius \cite{Bolen05}.

$L_{k}$ are the component of the angular momentum expressed as follows:
\begin{equation}
L_{k}=\epsilon _{ijk}X_{i}P_{j}  \tag{2}  \label{eqt2}
\end{equation}
and it satisfies the usual algebra:
\begin{equation}
\left[ L_{i},P_{j}\right] =i\hbar \varepsilon _{ijk}P_{k}\text{ , }\left[
L_{i},X_{j}\right] =i\hbar \varepsilon _{ijk}X_{k}\text{ , }\left[
L_{i},L_{j}\right] =i\hbar \varepsilon _{ijk}L_{k}  \tag{3}  \label{eqt3}
\end{equation}
The deformed algebra of the AdS model from \ref{eqt1} is characterized by the presence of a nonzero minimum uncertainty in momentum and it gives rise to modified Heisenberg uncertainty relations:
\begin{equation}
\Delta X_{i}\Delta P_{i}\geq \frac{\hbar }{2}\left( 1+\lambda \left( \Delta
X_{i}\right) ^{2}\right)  \tag{4}  \label{eqt4}
\end{equation}
where we have chosen the states for which $\left\langle X_{i}\right\rangle =0$.

For simplicity, we assume isotropic uncertainties $X_{i}=X$, therefore, we arrive to a minimal uncertainty in momentum for the AdS model given by:
\begin{equation}
\left( \Delta P_{i}\right) _{\min}^{AdS}=\hbar \sqrt{\lambda }  \tag{5} \label{eqt5}
\end{equation}

\begin{figure}
\centering

\includegraphics[width=0.5\textwidth]{Fig1.eps}
\caption{$\Delta P$ versus $\Delta X$ in both AdS and dS spaces (forbidden regions for AdS case are colored)}
\label{fig1}
\end{figure}

This is shown in fig\ref{fig1} (we use the units $\hbar =c=m=1$ in all figures), where the usual Heisenberg uncertainty relation is plotted along with the modified relation found in \ref{eqt4} and where we have colored the forbidden region for the AdS case $\lambda \neq 0$ to show the limits on $\Delta P$ values.

For dS model, it suffices to invert the sign of $\lambda$ in \ref{eqt1} and \ref{eqt4} to write the corresponding relations. Of course there is no minimal uncertainty in momentum is this case as one can see in fig\ref{fig1}.

In the following sections, we will employ the noncommutative operators $X_{i}$ and $P_{i}$ satisfying the AdS algebra \ref{eqt1} which gives rise to rescaled uncertainty relation \ref{eqt4} in momentum space. In order to study the exact solutions of the deformed DKP oscillator equation in 3D with EUP, we represent these operators as functions of the usual operators $x_{i}$ and $p_{i}$, satisfying the ordinary canonical commutation relations in position space; this is done with the following transformations:
\begin{align}
X_{i}& =\frac{x_{i}}{\sqrt{1-\lambda r^{2}}}  \tag{6a}  \label{eqt6a} \\
P_{i}& =-i\hbar \sqrt{1-\lambda r^{2}}\partial _{x_{i}}  \tag{6b}
\label{eqt6b}
\end{align}
where the variable $r$ vary in the domain $\left] -1/\sqrt{\lambda },1/\sqrt{\lambda }\right[ $.

\section{Nikiforov--Uvarov method}
\label{sec:NU}
The method is based on the hypergeometric differential equation and its aim is to reduce second order differential equations to this type with an appropriate coordinate transformation $s=s\left(x\right)$:
\begin{equation}
\psi ^{\backprime \backprime }\left( s\right) +\frac{\tilde{\tau}\left(
s\right) }{\sigma \left( s\right) }\psi ^{\backprime }\left( s\right) +\frac{
\tilde{\sigma}\left( s\right) }{\sigma ^{2}\left( s\right) }\psi \left(
s\right) =0  \tag{7}  \label{eqt7}
\end{equation}
where $\sigma\left(s\right)$ and $\widetilde{\sigma}\left(s\right)$ are polynomials of degree two at most and $\widetilde{\tau}\left(s\right)$ is a first degree polynomial at most \cite{Nikiforov88, Egrifes99}. If we take the following factorization $\psi\left(s\right)=\phi\left(s\right)y\left(s\right)$ in \ref{eqt7} we get \cite{Nikiforov88}:
\begin{equation}
\sigma \left( s\right) y^{\backprime \backprime }\left( s\right) +\tau
\left( s\right) y^{\backprime }\left( s\right) +\Lambda y\left( s\right) =0
\tag{8}  \label{eqt8}
\end{equation}
where the polynomial factors are given by:
\begin{equation}
\pi \left( s\right) =\sigma \left( s\right) \frac{d}{ds}\left( \ln \phi
\left( s\right) \right) \text{\ and }\tau \left( s\right) =\tilde{\tau}
\left( s\right) +2\pi \left( s\right)  \tag{9}  \label{eqt9}
\end{equation}
and $\Lambda $ is defined with:
\begin{equation}
\Lambda _{n}+n\tau ^{\backprime }+\frac{n\left( n-1\right) \sigma
^{\backprime \backprime }}{2}=0\text{ , }n=0,1,2,...  \tag{10}  \label{eqt10}
\end{equation}
To find the energy eigenvalues, we first have to determine $\pi\left(s\right)$ and $\Lambda$ by defining $k=\Lambda -\pi^{\backprime}\left(s\right)$ and solving the resulting quadratic equation for $\pi\left(s\right)$; so we get $\pi\left( s\right)$ as a polynomial of $s$:
\begin{equation}
\pi \left( s\right) =\left( \frac{\sigma ^{\backprime }-\widetilde{\tau }}{2}
\right) \pm \sqrt{\left( \frac{\sigma ^{\backprime }-\tilde{\tau}}{2}\right)
^{2}-\tilde{\sigma}+\sigma k}  \tag{11}  \label{eqt11}
\end{equation}

The determination of $k$ is the key point in computing $\pi \left( s\right) $ and it is done by setting that the expression in the root must be a square of a polynomial; this gives a quadratic equation for $k$.

To determine the polynomial solutions $y_{n}\left( s\right) $, we use \ref{eqt9} and the Rodrigues relation:
\begin{equation}
y_{n}\left( s\right) =\frac{C_{n}}{\rho \left( s\right) }\frac{d^{n}}{ds^{n}}
\left[ \sigma ^{n}\left( s\right) \rho \left( s\right) \right]  \tag{12}
\label{eqt12}
\end{equation}
where $C_{n}$ is a normalizable constant and the weight function $\rho\left(s\right) $ satisfies the following relation:
\begin{equation}
\frac{d}{ds}\left[ \sigma \left( s\right) \rho \left( s\right) \right] =\tau
\left( s\right) \rho \left( s\right)  \tag{13}  \label{eqt13}
\end{equation}
This last equation refers to classical orthogonal polynomials and we write for $y_{n}\left( s\right)$:
\begin{equation}
\int_{a}^{b}y_{n}\left( s\right) y_{m}\left( s\right) \rho \left( s\right)
ds=0\text{ , }m\neq n  \tag{14}  \label{eqt14}
\end{equation}
This completely determines the solutions with the NU method.

\section{Spin 0 DKP Oscillator}
\label{sec:S0}
We start first with some useful formulas of the DKP equation then we apply them in our system.

The DKP equation describing a free scalar and vector boson \cite{Moussa18, Duffin38, Nedjadi941} is written as:
\begin{equation}
\left[ c\vec{\beta}\cdot\vec{p}+mc^{2}\right] \Psi =i\hbar \beta ^{0}\left(
\frac{\partial \Psi }{\partial t}\right)  \tag{15}  \label{eqt15}
\end{equation}
where $m$ is the mass and $\beta^{\mu}$ are the DKP matrices (with $\mu =0,1,2,3$); one can find their properties listed in several works and we cite for example \cite{Chetouani04, Merad07, Cardoso16, Montigny19}.

We write the DKP oscillator in 3D space by analogy with the Dirac oscillator \cite{Moshinsky89}, so we introduce the non--minimal substitution \cite{Nedjadi941}:
\begin{equation}
\vec{p}\rightarrow \vec{p}-im\omega \eta ^{0}\vec{r}  \tag{16}
\label{eqt16}
\end{equation}
where $\omega$ is the oscillator frequency and $\eta^{0}$ is a matrix defined by $\eta^{0}=2\left(\beta^{0}\right)^{2}-1$, with $\left(\eta^{0}\right)^{2}=1$. With this substitution, we obtain the new equation for the DKP oscillator:
\begin{equation}
\left[ c\vec{\beta}\cdot \left( \vec{p}-im\omega \eta ^{0}\vec{r}\right)
+mc^{2}\right] \Psi =i\hbar \beta ^{0}\left( \frac{\partial \Psi }{\partial t}\right)  \tag{17}  \label{eqt17}
\end{equation}
The non--commuting coordinate and momentum operators due to the presence of EUP are expressed in terms of the commuting operators as following:
\begin{equation}
\vec{r}\rightarrow \frac{\vec{r}}{\sqrt{1-\lambda r^{2}}}\text{ \ and \ }
\vec{p}\rightarrow \sqrt{1-\lambda r^{2}}\vec{p}  \tag{18}  \label{eqt18}
\end{equation}
and using $\Psi\left(\mathbf{r,}t\right)=e^{-iEt/\hbar}\tilde{\Psi}\left( \mathbf{r}\right)$ in \ref{eqt17}, we get the stationary DKP equation:
\begin{equation}
\left[ c\vec{\beta}\cdot \left( \sqrt{1-\lambda r^{2}}\vec{p}-im\omega \eta
^{0}\frac{\vec{r}}{\sqrt{1-\lambda r^{2}}}\right) +mc^{2}\right] \tilde{\Psi}
\left( \mathbf{r}\right) =E\beta ^{0}\tilde{\Psi}\left( \mathbf{r}\right)
\tag{19}  \label{eqt19}
\end{equation}

We note here that, because of the symmetry of the problem, the components of the wave function $\tilde{\Psi}\left( \mathbf{r}\right) $ are also eigenfunctions of both $J^{2}$ and $J_{z}$ with the eigenvalues $J\left(J+1\right)$ and $M$ respectively; here $\vec{J}$ represents the total angular momentum and it is defined as the sum of the orbital angular momentum $\vec{L}$ and the spin $\vec{S}$. This total angular momentum commutes with the external potential because this later in central and with $\beta ^{0}$ and so it is a constant of motion.

For spin $0$ particles, the wave function is a vector with five components. We will use the method of the spherical coordinates in momentum space \cite{Hill54, Edmonds57} to write the wave function because it is more convenient for the symmetry of the problem and because, as we will mention later, this method is applicable for spin 1 case too. In this formulation, the five components wave function of a scalar particle $\Psi_{JM}\left(\mathbf{r}\right)$ is given by \cite{Nedjadi941}:
\begin{equation}
\Psi _{JM}\left( \mathbf{r}\right) =\left(
\begin{array}{c}
F_{nJ}\left( r\right) Y_{JM}\left( \Omega \right) \\
G_{nJ}\left( r\right) Y_{JM}\left( \Omega \right) \\
i\sum\limits_{L}H_{nJL}\left( r\right) Y_{JL1}^{M}\left( \Omega \right)
\end{array}
\right)  \tag{20}  \label{eqt20}
\end{equation}
In this expression $F_{nJ}\left(r\right)$, $G_{nJ}\left(r\right)$ and $H_{nJL}\left(r\right)$ are the radial wave functions, $Y_{JM}\left(\Omega\right)$ are the usual (or scalar) spherical harmonics and $Y_{JL1}^{M}\left(\Omega\right)$ are the normalised vector spherical harmonics.

We insert this new form of $\Psi_{JM}\left(\mathbf{r}\right)$ into \ref{eqt19} and we use the properties of vector spherical harmonics in position space representation (\cite{Hill54, Edmonds57}) to keep only the radial functions in the equations; we obtain the following coupled system:
\begin{align}
H& =0  \tag{21a}  \label{eqt21a} \\
mc^{2}G& =EF  \tag{21b}  \label{eqt21b} \\
mc^{2}H_{-1}& =c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{
dr}+\frac{J+1}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right] F
\tag{21c}  \label{eqt21c} \\
mc^{2}H_{+1}& =-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{
dr}-\frac{J}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right] F
\tag{21d}  \label{eqt21d} \\
mc^{2}F-EG& =c\left[ -\xi _{J}\left( \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}+\frac{J+2}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right) H_{+1}\right.  \notag \\
& \left. +\zeta _{J}\left( \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{dr}-
\frac{J-1}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right) H_{-1}
\right]  \tag{21e}  \label{eqt21e}
\end{align}
where we used the following parameters and notations:
\begin{gather}
\xi _{J}=\sqrt{\left( J+1\right) /\left( 2J+1\right) }\text{ , }\zeta _{J}=
\sqrt{J/\left( 2J+1\right) }  \notag \\
F_{nJ}\left( r\right) =F\text{ , }G_{nJ}\left( r\right) =G\text{ , }
H_{nJJ}\left( r\right) =H\text{ , }H_{nJJ\pm 1}\left( r\right) =H_{\pm 1}
\tag{22}  \label{eqt22}
\end{gather}
If we insert equations \ref{eqt21a}, \ref{eqt21b}, \ref{eqt21c} and \ref{eqt21d} into \ref{eqt21e}, we remain with one equation for $F\left(r\right)$:
\begin{equation}
\left[ \left( \sqrt{1-\lambda r^{2}}\frac{d}{dr}\right) ^{2}+\frac{2\left(
1-\lambda r^{2}\right) }{r}\frac{d}{dr}-\frac{J\left( J+1\right) \left(
1-\lambda r^{2}\right) }{r^{2}}-\frac{\eta r^{2}}{\left( 1-\lambda
r^{2}\right) }+\varepsilon \right] F\left( r\right) =0  \tag{23}
\label{eqt23}
\end{equation}
with:
\begin{equation}
\eta =\frac{m\omega }{\hbar }\left( \frac{m\omega }{\hbar }-\lambda \right)
\text{ and }\varepsilon =\frac{E^{2}-m^{2}c^{4}}{\left( \hbar c\right) ^{2}}+
\frac{3m\omega }{\hbar }  \tag{24}  \label{eqt24}
\end{equation}
We transform \ref{eqt23} using the following transformations $F\left(r\right)=y^{\mu}g\left(y\right)$ and $y=\sqrt{1-\lambda r^{2}}$ to get:
\begin{equation}
\left[ \left( 1-y^{2}\right) \frac{d^{2}}{dy^{2}}+\left( \frac{2\mu }{y}
-\left( 2\mu +3\right) y\right) \frac{d}{dy}-\frac{J\left( J+1\right) y^{2}}{
1-y^{2}}+\frac{\varepsilon}{\lambda} -3\mu \right] g\left( y\right) =0  \tag{25} \label{eqt25}
\end{equation}
where $\mu $ verifies the relation $\eta-\lambda^{2}\mu\left(\mu-1\right)=0$. Solving this later gives us two solutions:
\begin{equation}
\mu _{1}=1-\frac{m\omega }{\lambda \hbar }\text{ and }\mu _{2}=\frac{m\omega
}{\lambda \hbar }  \tag{26}  \label{eqt26}
\end{equation}
The accepted value of $\mu $ in \ref{eqt26} is the second solution because, from the expression of $F\left( r\right) $, the function $g\left( y\right) $ should be nonsingular at $y=\pm 1$; so $\mu =\mu _{2}=m\omega /\lambda \hbar $.

In addition, we note that \ref{eqt25} possesses three singular points $y=0,\pm 1$ and to reduce it to a class of known differential equation with a polynomial solution, we use a new variable $s=2y^{2}-1$:
\begin{equation}
\left[ \left( 1-s^{2}\right) ^{2}\dfrac{d^{2}}{ds^{2}}+\left( 1-s^{2}\right)
\left( \left( \mu -1\right) -\left( \mu +2\right) s\right) \dfrac{d}{ds}
+a_{1}s^{2}+a_{2}s+a_{3}\right] g\left( s\right) =0  \tag{27}  \label{eqt27}
\end{equation}
The parameters $a_{1}$, $a_{2}$ and $a_{3}$ are defined by:
\begin{equation}
a_{1,3}=\frac{-1}{4}\left( J\left( J+1\right) \pm \frac{\varepsilon }{
\lambda }\mp 3\mu \right) \text{ and }a_{2}=\frac{-J\left( J+1\right) }{2}
\tag{28}  \label{eqt28}
\end{equation}
We see that equation \ref{eqt27} for $g\left( s\right) $ is similar to equation \ref{eqt7} for $\psi \left( s\right) $ and this enables us to use the NU method with the following expressions for the NU polynomials:
\begin{equation}
\sigma \left( s\right) =\left( 1-s^{2}\right) \text{ , }\tilde{\tau}\left(
s\right) =\left( \mu -1\right) -\left( \mu +2\right) s\text{ \ and }\tilde{
\sigma}\left( s\right) =a_{1}s^{2}+a_{2}s+a_{3}  \tag{29}  \label{eqt29}
\end{equation}
Substituting them into \ref{eqt11}, we obtain:
\begin{equation}
\pi \left( s\right) =\frac{\left( \mu s-\left( \mu -1\right) \right) }{2}\pm
\sqrt{\left( \frac{\mu ^{2}}{4}-a_{1}-k\right) s^{2}-\left( \frac{\mu \left(
\mu -1\right) }{4}+a_{2}\right) s+\frac{\left( \mu -1\right) ^{2}}{4}-a_{3}+k
}  \tag{30}  \label{eqt30}
\end{equation}
The parameter $k$ is determined as mentioned in the precedent section and we get two values:
\begin{equation}
k_{1}=\frac{1}{4}\left[ \frac{\varepsilon }{\lambda }-3\mu +\left( 2\mu
-1\right) \left( J+1\right) \right] \text{ and }k_{2}=\frac{1}{4}\left[ \frac{
\varepsilon }{\lambda }-3\mu -\left( 2\mu -1\right) J\right]  \tag{31} \label{eqt31}
\end{equation}

For $\pi\left(s\right)$, we obtain the following possible solutions:
\begin{equation}
\pi \left( s\right) =\left\{
\begin{array}{c}
\pi _{1}=\frac{1}{2}\left( \left( 2\mu -J-1\right) s-\left( 2\mu +J-1\right)
\right) \\
\pi _{2}=\frac{1}{2}\left( J+1\right) \left( s+1\right) \\
\pi _{3}=\frac{1}{2}\left( \left( 2\mu +J\right) s-\left( 2\mu -J-2\right)
\right) \\
\pi _{4}=-\frac{1}{2}J\left( s+1\right)
\end{array}
\right.  \tag{32}  \label{eqt32}
\end{equation}
where $\pi_{1}$ and $\pi_{2}$ are related to $k_{1}$ while $\pi_{3}$ and $\pi_{4}$ are linked to $k_{2}$. The correct solution is $\pi_{4}$, so:
\begin{equation}
\tau \left( s\right) =-\left( \mu +J+2\right) s+\left( \mu -J-1\right)
\tag{33}  \label{eqt33}
\end{equation}
From \ref{eqt10}, we obtain:
\begin{equation}
\Lambda =k_{2}-\frac{J}{2}=n\left( n+\mu +J+1\right) \text{ , }n=0,1,2,...
\tag{34}  \label{eqt34}
\end{equation}
Hence, the energy eigenvalues are found as ($N=2n+J$ is the principal quantum number):
\begin{equation}
E_{N,J}^{2}=m^{2}c^{4}+2\hbar m\omega c^{2}N+\lambda \hbar ^{2}c^{2}\left[
\left( N+1\right) ^{2}-J\left( J+1\right) -1\right]  \tag{35}  \label{eqt35}
\end{equation}

We remark that the above expression of the energies contains the usual 3D DKP oscillator term and an additional correction term depending on the deformation; the later is linearly proportional to the deformation parameter $\lambda $. Here it should be noted that the presence of a correction term proportional to $N^{2}$ indicates the appearance of a hard confinement due to the deformation. This is equivalent to the energy of a spinless relativistic quantum particle in a square well potential whose boundaries are placed at $\pm
\pi /2\sqrt{\lambda }$. The second term in the $\lambda$ correction is proportional to $J\left(J+1\right)$, so it appears as some kind of rotational energy and it removes the degeneracy of the usual oscillator spectrum according to this number $J$.

We note here that the spectral corrections due to the EUP are qualitatively different to those associated to the GUP \cite{Falek09, Falek10}. One can also recover the usual spectrum energy using the limit $\lambda \rightarrow 0$ and it coincides with that of the ordinary spinless 3D DKP oscillator \cite{Nedjadi941}.

Another interesting characteristic of this spectrum is the limit of the energy levels spacing:
\begin{equation}
\lim_{N\rightarrow \infty }\left\vert \Delta E_{N,J}\right\vert =\hbar c\sqrt{\lambda }  \tag{36}  \label{eqt36}
\end{equation}
We see from \ref{eqt35} and fig\ref{fig2} that this difference becomes constant for large values of $N$ and the spectrum remains bounded even at the limit $N\rightarrow \infty$. In the ordinary case when the parameter $\lambda$ vanishes, this energy spacing tends to zero for large $N$ and the spectrum becomes almost continuous.

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig2.eps}
\caption{Energy spacing $\Delta E=E(N+1)-E(N)$ with and without deformation}
\label{fig2}
\end{figure}

To obtain an upper bound on the deformation parameter, we expand \ref{eqt35} to the first order in $\lambda $:
\begin{equation}
E_{N,0}=\sqrt{m^{2}c^{4}+2\hbar m\omega c^{2}N}+\frac{\hbar ^{2}c^{2}N\left(
N+2\right) \lambda }{2\sqrt{m^{2}c^{4}+2\hbar m\omega c^{2}N}}  \tag{37} \label{eqt37}
\end{equation}
The deviation of the $N$-th energy level caused by the modified commutation relations \ref{eqt1} is:
\begin{equation}
\frac{\Delta E_{N,0}}{\hbar \omega }=\frac{\lambda \hbar ^{2}N\left(
N+2\right) }{2m\hbar \omega \sqrt{1+2\hbar \omega m^{-1}c^{-2}N}}  \tag{38} \label{eqt38}
\end{equation}
We use experimental results of the cyclotron motion of an electron in a Penning trap \cite{Brown86, Mittleman99} where $\omega_{c}=eB/m_{e}$ is its cyclotron frequency when it is trapped in a magnetic field $B$. Therefore, for a magnetic field of strength $B=6T$ (in IS units), we have $m_{e}\hbar \omega _{c}=e\hbar B=10^{-52}kg^{2}m^{2}s^{-2}$. If we assume that at the level $n=10^{10}$, only a deviation of the $\hbar\omega_{c}$ scale can be detected and by taking $\Delta E_{n}<\hbar\omega_{c}$ (no perturbation is observed for the $n$--th energy level) \cite{Chang02}, we get the following upper bound for the minimal uncertainty in momentum:
\begin{equation}
\Delta P_{\min }=\hbar \sqrt{\lambda }<3.25\times 10^{-36}kgms^{-1}  \tag{39} \label{eqt39}
\end{equation}
Now we focus on the corresponding eigenfunctions. Taking the expression of $\pi _{4}\left( s\right) $ from \ref{eqt32}, the $\phi \left( s\right) $ part is defined from \ref{eqt9} as $\phi \left( s\right) =\left( 1-s\right)^{J/2}$ and according to the form of $\sigma \left( s\right) $ in \ref{eqt29}, the $y\left( s\right) $ part comes from the Rodrigues relation \ref{eqt12}:
\begin{equation}
y_{n}\left( s\right) =\frac{C_{n}}{\rho \left( s\right) }\frac{d^{n}}{ds^{n}}
\left[ \left( 1-s^{2}\right) ^{n}\rho \left( s\right) \right]  \tag{40} \label{eqt40}
\end{equation}
where the weight function $\rho \left( s\right) $ is determined from the expressions of $\tau\left(s\right)$ and $\sigma\left(s\right)$:
\begin{equation}
\rho \left( s\right) =\left( 1+s\right) ^{\mu -1/2}\left( 1-s\right) ^{J+1/2}
\tag{41}  \label{eqt41}
\end{equation}
The relation \ref{eqt40} stands for the Jacobi polynomials, so we get:
\begin{equation}
y_{n}\left( s\right) \equiv P_{n}^{\left( J+1/2,\mu -1/2\right) }\left(
s\right)  \tag{42}  \label{eqt42}
\end{equation}
Hence, $g(s)$ is written from its definition as follows:
\begin{equation}
g(s)=\phi \left( s\right) y_{n}\left( s\right) =C_{n}\left( 1-s\right)
^{J/2}P_{n}^{\left( J+1/2,\mu -1/2\right) }\left( s\right)  \tag{43} \label{eqt43}
\end{equation}
This allows us to write the general form of the component $F$ in terms of the variable $r$ as follows:
\begin{equation}
F_{n}\left( r\right) =C_{n}\left( 1-\lambda r^{2}\right) ^{\frac{\mu }{2}
}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}P_{n}^{\left( J+1/2,\mu
-1/2\right) }\left( 1-2\lambda r^{2}\right)  \tag{44}  \label{eqt44}
\end{equation}
where $C_{n}$ is the normalization constant.

Now to recover $\Psi_{nJ}\left(\mathbf{r}\right)$, we use the following property of Jacobi polynomials \cite{Gradshteyn80}:
\begin{equation}
\frac{dP_{n}^{\left( a,b\right) }\left( y\right) }{dy}=\frac{1}{2}\left(
n+a+b+1\right) P_{n-1}^{\left( a+1,b+1\right) }\left( y\right)  \tag{45} \label{eqt45}
\end{equation}
to obtain its other components from equations 25 and then its final form from \ref{eqt20}:
\begin{align}
H &=0\text{ , }G=C_{n}\frac{E}{mc^{2}}\left( 1-\lambda r^{2}\right) ^{\frac{
\mu }{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}P_{n}^{\left( J+1/2,\mu
-1/2\right) }\left( 1-2\lambda r^{2}\right)  \notag \\
H_{+1} &=-iC_{n}\xi _{J}\frac{\hbar }{mc}\left( 1-\lambda r^{2}\right) ^{
\frac{\mu +1}{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}\left( \frac{
m\omega }{\hbar }+\left( n+J+1\right) \lambda \right) rP_{n-1}^{\left(
J+3/2,\mu +1/2\right) } \tag{46}  \label{eqt46} \\
H_{-1} &=iC_{n}\zeta _{J}\frac{\hbar }{mc}\left( 1-\lambda r^{2}\right) ^{
\frac{\mu +1}{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}\left[ \frac{
\left( 2J+1\right) }{r}P_{n}^{\left( J+\frac{1}{2},\mu -\frac{1}{2}\right) }
\right] -\frac{\zeta _{J}}{\xi _{J}}H_{+1}  \notag
\end{align}

\section{Spin 1 DKP Oscillator}
\label{sec:S1}
In this case, the wave function has ten components and we have to use the spherical spatial form of the ten components wave function $\Psi \left( \mathbf{r}\right) $ in the momentum space, otherwise we will not be able to decouple the system; so we write:
\begin{equation}
\Psi _{JM}\left( \mathbf{r}\right) =\left(
\begin{array}{c}
i\phi _{nJ}\left( r\right) Y_{JM}\left( \Omega \right) \\
\sum\limits_{L}F_{nJL}\left( r\right) Y_{JL1}^{M}\left( \Omega \right) \\
\sum\limits_{L}G_{nJL}\left( r\right) Y_{JL1}^{M}\left( \Omega \right) \\
\sum\limits_{L}H_{nJL}\left( r\right) Y_{JL1}^{M}\left( \Omega \right)
\end{array}
\right)  \tag{47}  \label{eqt47}
\end{equation}
Here, $\Psi _{JM}$, $F_{nJ}\left(r\right)$, $G_{nJ}\left(r\right)$ and $H_{nJL}\left(r\right)$ are the radial wave functions, while $Y_{JM}\left(\Omega\right)$ are the scalar spherical harmonics and $Y_{JL1}^{M}\left(\Omega\right)$ are the normalised vector spherical harmonics.

Putting this form of $\Psi _{JM}\left( \mathbf{r}\right) $ into \ref{eqt19} leads to ten coupled differential radial equations which can be reduced to two classes associated with the two parities $\left( -1\right) ^{J}$ and $\left(-1\right) ^{J+1}$ \cite{Nedjadi941, Kulikov05}.

\subsection{Natural Parity States}

Here the parity is $\left( -1\right) ^{J}$ and the relevant radial differential equations are:
\begin{align}
mc^{2}G_{0} &=EF_{0} \tag{48a} \label{eqt48a} \\
mc^{2}H_{-1} &=-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{
dr}+\frac{J+1}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right]
F_{0} \tag{48b} \label{eqt48b} \\
mc^{2}H_{+1} &=-c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d
}{dr}-\frac{J}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right]
F_{0} \tag{48c} \label{eqt48c} \\
mc^{2}F_{0}-EG_{0} &=-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}-\frac{J-1}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right] H_{-1}  \notag \\
&-c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{dr}+\frac{
J+2}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right] H_{+1}
\tag{48d} \label{eqt48d}
\end{align}
with the following notations $R_{nJJ}\left(r\right)=R_{0}$, $R_{nJJ\pm 1}\left(r\right)=R_{\pm}$ and $R$ designates $F$, $G$ and $H$.

We eliminate $G_{0}$, $H_{+1}$ and $H_{-1}$ in favour of $F_{0}$ to obtain the following differential equation:
\begin{equation}
\left[ \left( \sqrt{1-\lambda r^{2}}\frac{d}{dr}\right) ^{2}+\frac{2\left(
1-\lambda r^{2}\right) }{r}\frac{d}{dr}-\frac{J\left( J+1\right) \left(
1-\lambda r^{2}\right) }{r^{2}}-\frac{\eta r^{2}}{\left( 1-\lambda
r^{2}\right) }+\varepsilon ^{\backprime }\right] F_{0}=0  \tag{49} \label{eqt49}
\end{equation}
where:
\begin{equation}
\eta =\frac{m\omega }{\hbar }\left( \frac{m\omega }{\hbar }-\lambda \right)
\text{ and }\varepsilon ^{\backprime }=\frac{E^{2}-m^{2}c^{4}}{\left( \hbar
c\right) ^{2}}+\frac{m\omega }{\hbar }-\lambda  \tag{50}  \label{eqt50}
\end{equation}

We note that equation \ref{eqt49} is similar to \ref{eqt23} and so we can solve it exactly in the same manner to obtain the $F_{0}$ function and then the other components from the equation system $48$; this enables us to get the complete solutions for the eigenfunctions of natural parity states:
\begin{align}
F_{0} &=C_{n}^{\backprime }\left( 1-\lambda r^{2}\right) ^{\frac{\mu }{2}
}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}P_{n}^{\left( J+1/2,\mu
-1/2\right) }\left( 1-2\lambda r^{2}\right) \tag{51a} \label{eqt51a} \\
G_{0} &=C_{n}^{\backprime }\frac{E}{mc^{2}}\left( 1-\lambda r^{2}\right) ^{
\frac{\mu }{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}P_{n}^{\left(
J+1/2,\mu -1/2\right) }\left( 1-2\lambda r^{2}\right) \tag{51b} \label{eqt51b} \\
H_{+1} &=C_{n}^{\backprime }\zeta _{J}\frac{\hbar }{mc}\left( 1-\lambda
r^{2}\right) ^{\frac{\mu +1}{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}
\left[ \left( \frac{m\omega }{\hbar }+\left( n+J+1\right) \lambda \right)
rP_{n-1}^{\left( J+3/2,\mu +1/2\right) }\right] \tag{51c} \label{eqt51c} \\
H_{-1} &=-C_{n}^{\backprime }\xi _{J}\frac{\hbar }{mc}\left( 1-\lambda
r^{2}\right) ^{\frac{\mu +1}{2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}
\left[ \frac{\left( 2J+1\right) }{r}P_{n}^{\left( J+1/2,\mu -1/2\right) }
\right] +\frac{\xi _{J}}{\zeta _{J}}H_{+1} \tag{51d} \label{eqt51d}
\end{align}
where $C_{n}^{\backprime }$ is a normalization constant and $\mu =m\omega /\lambda \hbar $ from \ref{eqt26}.

The correspondent energy spectrum is given by:
\begin{equation}
E_{N,J}^{2}=m^{2}c^{4}+2\hbar m\omega c^{2}\left( N+1\right) +\lambda \hbar
^{2}c^{2}\left[ \left( N+1\right) ^{2}-J\left( J+1\right) \right]  \tag{52} \label{eqt52}
\end{equation}
We can easily compare this spectrum with the one corresponding to the spin 0 case \ref{eqt35}. We observe the same effects due to deformation, a strong confinement term and a rotational energy term which removes the $J$ degeneracy. Of course, the similarity is valid for the non--relativistic limits too:
\begin{align}
E_{NJ}^{nr}(S=1) &=\hbar \omega \left( N+1\right) +\frac{\lambda \hbar ^{2}}{2m}
\left[ \left( N+1\right) ^{2}-J\left( J+1\right) \right]  \tag{53} \label{eqt53} \\
E_{NJ}^{nr}(S=0) &=\hbar \omega N +\frac{\lambda \hbar ^{2}}{2m}\left[ \left(
N+1\right) ^{2}-J\left( J+1\right) -1\right]  \tag{54}  \label{eqt54}
\end{align}
The corrections have the same behavior in both cases; they increase with $N$ and decrease with $J$. We give an example of this similarity in the behaviors in fig\ref{fig3}, by representing the dependence with the values of $N$ of the relativistic energies for spin 0 particle (from \ref{eqt39}) and also for natural parity of spin 1 particle (from \ref{eqt52}) for the $J=0$ states.

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig3.eps}
\caption{$E_{N,0}$ for both spin 0 case and spin 1 natural states}
\label{fig3}
\end{figure}

\subsection{Unnatural Parity States}

The parity in this case is $\left( -1\right) ^{J+1}$ and the relevant system for radial functions is:
\begin{align}
mc^{2}F_{+1}-EG_{+1}& =-c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}-\frac{J}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right] H_{0}  \tag{55a}  \label{eqt55a} \\
mc^{2}F_{-1}-EG_{-1}& =-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}+\frac{J+1}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right] H_{0}  \tag{55b}  \label{eqt55b} \\
mc^{2}H_{0}& =-c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{
dr}+\frac{J+2}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right]
F_{+1}  \notag \\
& -c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{dr}-\frac{J-1
}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right] F_{-1}
\tag{55c}  \label{eqt55c} \\
mc^{2}G_{+1}-EF_{+1}& =-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}-\frac{J}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right] \phi  \tag{65d}  \label{eqt65d} \\
mc^{2}G_{-1}-EF_{-1}& =c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left(
\frac{d}{dr}+\frac{J+1}{r}\right) -\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}
\right] \phi  \tag{55e}  \label{eqt55e} \\
mc^{2}\phi & =-c\xi _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{dr
}+\frac{J+2}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right]
G_{+1}  \notag \\
& +c\zeta _{J}\left[ \hbar \sqrt{1-\lambda r^{2}}\left( \frac{d}{dr}-\frac{
J-1}{r}\right) +\frac{m\omega r}{\sqrt{1-\lambda r^{2}}}\right] G_{-1}
\tag{55f}  \label{eqt55f}
\end{align}
and we adopt the same notations as those already used for the functions in the system corresponding to natural parity states: $R_{nJJ}\left(r\right)=R_{0}$, $R_{nJJ\pm 1}\left(r\right)=R_{\pm}$ and $R$ designates $F$, $G$ and $H$.

To solve this system, we start by simplifying its writing to this compact shape:
\begin{align}
\left(
\begin{array}{c}
F_{+1} \\
G_{+1}
\end{array}
\right) & =\frac{\hbar }{\epsilon }\left( \sqrt{1-\lambda r^{2}}\left( \frac{
d}{dr}-\frac{J}{r}\right) -\frac{m\omega \hbar ^{-1}r}{\sqrt{1-\lambda r^{2}}
}\right) \left(
\begin{array}{cc}
\xi _{J}E & \zeta _{J}mc^{2} \\
\xi _{J}mc^{2} & \zeta _{J}E
\end{array}
\right) \left(
\begin{array}{c}
\phi \\
H_{0}
\end{array}
\right)  \tag{56a}  \label{eqt56a} \\
\left(
\begin{array}{c}
F_{-1} \\
G_{-1}
\end{array}
\right) & =\frac{\hbar }{\epsilon }\left( \sqrt{1-\lambda r^{2}}\left( \frac{
d}{dr}+\frac{J+1}{r}\right) -\frac{m\omega \hbar ^{-1}r}{\sqrt{1-\lambda
r^{2}}}\right) \left(
\begin{array}{cc}
-\zeta _{J}E & \xi _{J}mc^{2} \\
-\zeta _{J}mc^{2} & \xi _{J}E
\end{array}
\right) \left(
\begin{array}{c}
\phi \\
H_{0}
\end{array}
\right)  \tag{56b}  \label{eqt56b}
\end{align}
and:
\begin{align}
2\sqrt{J\left( J+1\right) }WEH_{0}& =\left[ \left( \sqrt{1-\lambda r^{2}}
\frac{d}{dr}\right) ^{2}+\frac{2\left( 1-\lambda r^{2}\right) }{r}\frac{d}{dr
}-\frac{J\left( J+1\right) \left( 1-\lambda r^{2}\right) }{r^{2}}\right.
\notag \\
& \left. -\frac{\eta r^{2}}{\left( 1-\lambda r^{2}\right) }+\frac{\epsilon c
}{\left( \hbar m\omega c\right) ^{2}}-\frac{3m\omega }{\hbar }\right] \phi
\tag{57a}  \label{eqt57a} \\
2\sqrt{J\left( J+1\right) }WE\phi & =\left[ \left( \sqrt{1-\lambda r^{2}}
\frac{d}{dr}\right) ^{2}+\frac{2\left( 1-\lambda r^{2}\right) }{r}\frac{d}{dr
}-\frac{J\left( J+1\right) \left( 1-\lambda r^{2}\right) }{r^{2}}\right.
\notag \\
& \left. -\frac{\eta r^{2}}{\left( 1-\lambda r^{2}\right) }+\frac{\epsilon c
}{\left( \hbar m\omega c\right) ^{2}}-\frac{m\omega }{\hbar }-\lambda \right]
H_{0}  \tag{57b}  \label{eqt57b}
\end{align}
Here $\epsilon=c^{-1}\left(E^{2}-m^{2}c^{4}\right)$ and $W=\left(1-\lambda\hbar/2m\omega\right)\omega /\hbar c^{2}$.

Now, according to the procedure of diagonalization \cite{Nedjadi941, Nedjadi942}:
\begin{equation}
\left(
\begin{array}{c}
\phi \\
H_{0}
\end{array}
\right) =\frac{1}{\sqrt{2\kappa \left( \kappa +1\right) }}\left(
\begin{array}{cc}
\left( 1+\kappa \right) & k \\
k & -\left( 1+\kappa \right)
\end{array}
\right) \left(
\begin{array}{c}
R_{+} \\
R_{-}
\end{array}
\right)  \tag{58}  \label{eqt58}
\end{equation}
with $k=2\sqrt{J\left( J+1\right) }E/mc^{2}$ and $\kappa =\sqrt{1+k^{2}}$, we can decuple \ref{eqt57a} and \ref{eqt57b} as follows:
\begin{equation}
\left[ \left( \sqrt{1-\lambda r^{2}}\frac{d}{dr}\right) ^{2}+\frac{2\left(
1-\lambda r^{2}\right) }{r}\frac{d}{dr}-\frac{J\left( J+1\right) \left(
1-\lambda r^{2}\right) }{r^{2}}-\frac{\eta r^{2}}{\left( 1-\lambda
r^{2}\right) }+\varepsilon _{+}\right] R_{+}=0  \tag{59}  \label{eqt59}
\end{equation}
\begin{equation}
\left[ \left( \sqrt{1-\lambda r^{2}}\frac{d}{dr}\right) ^{2}+\frac{2\left(
1-\lambda r^{2}\right) }{r}\frac{d}{dr}-\frac{J\left( J+1\right) \left(
1-\lambda r^{2}\right) }{r^{2}}-\frac{\eta r^{2}}{\left( 1-\lambda
r^{2}\right) }+\varepsilon _{-}\right] R_{-}=0  \tag{60}  \label{eqt60}
\end{equation}
where:
\begin{equation}
\varepsilon _{\pm }=\frac{E^{2}-m^{2}c^{4}}{\hbar ^{2}c^{2}}+\frac{\omega }{
\hbar c^{2}}\left( 1-\frac{\lambda \hbar }{2m\omega }\right) \left(
mc^{2}\mp \sqrt{m^{2}c^{4}+4J\left( J+1\right) E^{2}}\right) -\frac{3m\omega
}{\hbar }  \tag{61}  \label{eqt61}
\end{equation}

We remark that equations \ref{eqt59} and \ref{eqt60} are exactly the same as equation \ref{eqt23} corresponding to the spin 0 case and also to equation \ref{eqt49} of natural parity states; so we solve them in the same manner. The relativistic spinor wave functions $R_{\pm }$ are exactly the same as $F(r)$ in \ref{eqt44} and they are given using the Jacobi polynomials as:

\begin{equation}
R_{\pm }\left( r\right) =C_{\pm }\left( 1-\lambda r^{2}\right) ^{\frac{\mu }{
2}}\left( 2\lambda r^{2}\right) ^{\frac{J}{2}}P_{n}^{\left( J+1/2,\mu
-1/2\right) }\left( 1-2\lambda r^{2}\right)  \tag{62}  \label{eqt62}
\end{equation}
with $\mu=m\omega/\lambda\hbar$ and where $C_{\pm}$ are the normalization constants.

The determination of $R_{\pm }$ gives us both $\phi$ and $H_{0}$ from \ref{eqt58} and then, using \ref{eqt56a} and \ref{eqt56b}, the remaining components of the wave function $\Psi\left(\mathbf{r}\right)$ associated with the unnatural parity states $\left( -1\right) ^{J+1}$:
\begin{multline}
\left(
\begin{array}{c}
\phi \\
H_{0} \\
F_{+} \\
G_{+} \\
F_{-} \\
G_{-}
\end{array}
\right) =\frac{\left( 1-\lambda r^{2}\right) ^{\frac{\mu }{2}}\left(
2\lambda r^{2}\right) ^{\frac{J}{2}}}{\epsilon \sqrt{2\kappa \left( \kappa
+1\right) }}\left\{ \left[ C_{+}\left(
\begin{array}{c}
\left( 1+\kappa \right) \epsilon \\
k\epsilon \\
\alpha _{+}\Gamma _{1} \\
\beta _{+}\Gamma _{1} \\
\gamma _{+}\Gamma _{2} \\
\delta _{+}\Gamma _{2}
\end{array}
\right) +C_{-}\left(
\begin{array}{c}
k\epsilon \\
-\left( 1+\kappa \right) \epsilon \\
\delta _{-}\Gamma _{1} \\
\gamma _{-}\Gamma _{1} \\
\beta _{-}\Gamma _{2} \\
\alpha _{-}\Gamma _{2}
\end{array}
\right) \right] P_{n}^{\left( J+\frac{1}{2},\mu -\frac{1}{2}\right) }\right.
\notag \\
\left. -\left[ C_{+}\left(
\begin{array}{c}
0 \\
0 \\
\alpha _{+} \\
\beta _{+} \\
\gamma _{+} \\
\delta _{+}
\end{array}
\right) +C_{-}\left(
\begin{array}{c}
0 \\
0 \\
\delta _{-} \\
\gamma _{-} \\
\beta _{-} \\
\alpha _{-}
\end{array}
\right) \right] \Gamma _{3}P_{n-1}^{\left( J+\frac{3}{2},\mu +\frac{1}{2}
\right) }\right\} \left( 1-2\lambda r^{2}\right)  \tag{63}  \label{eqt63}
\end{multline}
with the following abbreviated notations:
\begin{align*}
\alpha _{\pm }& =\zeta _{J}mc^{2}k\pm \xi _{J}E\left( \kappa +1\right) \text{
, }\beta _{\pm }=-\zeta _{J}Ek\pm \xi _{J}mc^{2}\left( \kappa +1\right) \\
\gamma _{\pm }& =\xi _{J}mc^{2}k\pm \zeta _{J}E\left( \kappa +1\right) \text{
, }\delta _{\pm }=\xi _{J}Ek\pm \zeta _{J}mc^{2}\left( \kappa +1\right) \\
\Gamma _{1}\left( r\right) & =\frac{-2m\omega r}{\sqrt{1-\lambda r^{2}}}
\text{ , }\Gamma _{2}\left( r\right) =\frac{\hbar \left( 2J+1\right) \sqrt{
1-\lambda r^{2}}}{r}+\Gamma _{1}\left( r\right) \text{ , }\Gamma _{3}\left(
r\right) =\lambda r\sqrt{1-\lambda r^{2}}\left( n+\mu +J+1\right)
\end{align*}

We also derive the correspondent equations for the relativistic energies $E_{+}$and $E_{-}$ related to the spinors $R_{+}$ and $R_{-}$ respectively and this for all values of $\lambda $:
\begin{multline}
\frac{E_{\pm }^{2}-m^{2}c^{4}}{\hbar \omega }\mp \left( 1-\frac{\lambda
\hbar }{2m\omega }\right) \sqrt{m^{2}c^{4}+4J\left( J+1\right) E_{\pm }^{2}}=
\notag \\
mc^{2}\left( 2N+5\right) +\frac{\lambda \hbar c^{2}}{\omega }\left[ \left(
N+1\right) ^{2}-J\left( J+1\right) -\frac{1}{2}\right]  \tag{64} \label{eqt64}
\end{multline}

The exact solutions of these eigenvalue equations take the forms:
\begin{multline}
E_{\pm }^{2}=m^{2}c^{4}+2\hbar m\omega c^{2}\left( N+\frac{5}{2}\right)
+2\hbar ^{2}\omega ^{2}\left( 1-\frac{\lambda \hbar }{2m\omega }\right)
^{2}J\left( J+1\right) +  \notag \\
\lambda \hbar ^{2}c^{2}\left( \left( N+1\right) ^{2}-J\left( J+1\right) -
\frac{1}{2}\right) \pm \Delta  \tag{65}  \label{eqt65}
\end{multline}
where $\Delta$ is given by ($b_{0}=\left(2J+1\right)^{2}$ and $b_{1}=4J\left(J+1\right)$):
\begin{multline}
\Delta =\hbar m\omega c^{2}\left( 1-\frac{\lambda \hbar }{2m\omega }\right)
\left( 2J+1\right) \left[ 1+\frac{b_{1}^{2}}{4b_{0}}\left( \frac{\hbar
\omega }{mc^{2}}\right) ^{2}\left( 1-\frac{\lambda \hbar }{2m\omega }\right)
^{2}\right.  \notag \\
\left. +\frac{2b_{1}}{b_{0}}\frac{\hbar \omega }{mc^{2}}\left( N+\frac{5}{2}+
\frac{\lambda \hbar }{2m\omega }\left( \left( N+1\right) ^{2}-J\left(
J+1\right) -\frac{1}{2}\right) \right) \right] ^{1/2}  \tag{66} \label{eqt66}
\end{multline}

The limit $\lambda\rightarrow 0$ gives the ordinary spectrum and it coincides exactly with the one found in \cite{Nedjadi941}.

The non--relativistic spectrum is obtained using the usual approximation mentioned above:
\begin{multline}
E_{\pm }^{nr}=\hbar \omega \left( N+\frac{5}{2}\right) +\frac{\lambda \hbar
^{2}}{2m}\left( \left( N+1\right) ^{2}-J\left( J+1\right) -\frac{1}{2}
\right)  \notag \\
 +\frac{\hbar ^{2}\omega ^{2}}{mc^{2}}\left( 1-\frac{\lambda \hbar }{
2m\omega }\right) ^{2}J\left( J+1\right) \pm \frac{\Delta }{2mc^{2}}
\tag{67}  \label{eqt67}
\end{multline}

Comparing this formula with those of the natural spin 1 case \ref{eqt53} and of the spin 0 case \ref{eqt54}, we see that this spectrum differs by the last two terms in \ref{eqt67} and thus its dependence on the deformation is more pronounced. The three spectra are identical when $\lambda =2m\omega /\hbar $ and this is due, as one can notice from \ref{eqt45}, to the fact that this critical value cancels the spin-orbit term.

By studying the behavior of the energies $E_{\pm }$ as a function of the frequency, we note that their dependencies are almost linear for the states $J=0$ as shown in fig\ref{fig4} for $N=1$. Whereas for the states $J\neq 0$, both energies are linear only for weak frequencies, then their behaviors differ in the high frequencies regime and we see, as in  fig\ref{fig5}, that $E_{+}$ does not cease increasing towards infinity while $E_{-}$ tends towards a constant $\sqrt{\left( N+2\right) \left( N+3\right) /J\left( J+1\right)}$.

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig4.eps}
\caption{$E_{1,0}$ versus $\omega$ for unnatural spin 1 case (the $\pm$ signs near $\lambda$ holds for $E_{\pm}$ energies) }
\label{fig4}
\end{figure}

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig5.eps}
\caption{$E_{2,1}$ versus $\omega$ for unnatural spin 1 case (the $\pm$ signs near $\lambda$ holds for $E_{\pm}$ energies) }
\label{fig5}
\end{figure}

If we focus on the contributions coming only from the deformation in these energies, we notice that, when $J=0$, they decrease with $\omega$ until they cancel out for both $E_{+}$ and $E_{-}$ cases (fig \ref{fig6}). On another hand, when $J\neq 0$, they decrease for the $E_{+}$ energies and tend towards the negative value $-\sqrt{J\left( J+1\right)}\lambda$, while they also decrease for the $E_{-}$ energies until becoming zero for very high frequencies (fig\ref{fig7} as an example for the level $(N,J)=(2,1)$).

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig6.eps}
\caption{$\Delta E_{1,0}=E(\lambda)-E(\lambda=0)$ versus $\omega$ for unnatural spin 1 case
\newline (the $\pm$ signs near $\lambda$ holds for $E_{\pm}$ energies) }
\label{fig6}
\end{figure}

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Fig7.eps}
\caption{$\Delta E_{2,1}=E(\lambda)-E(\lambda=0)$ versus $\omega$ for unnatural spin 1 case
\newline (the $\pm$ signs near $\lambda$ holds for $E_{\pm}$ energies) }
\label{fig7}
\end{figure}

We associate the energies $E_{\pm }$ with the states $L=J\pm 1$ \cite{Nedjadi941, Nedjadi942, Falek10} and we mention that the same matches appear in 2D DKP systems \cite{Falek19} but they are absent for 1D ones \cite{Boumali08, Falek17} since there is neither orbital moment nor spin in this case. We also associate the natural states with the vanishing projection of the $S=1$ case, or $J=L+0$, and this explains the fact that its spectrum \ref{eqt52} is similar to that of the $S=0$ case in \ref{eqt35} (or \ref{eqt53} and \ref{eqt54} as mentioned before) and it also explains the absence of spin--orbit coupling for this parity. By making the analogy with the case of Dirac's equation, this gives a new meaning to natural and non-natural states since the parity is defined by $\left( -1\right) ^{L}$ in relativistic quantum mechanics \cite{Bjorken64} and so natural states correspond to $L=J$ while unnatural ones are related to $L=J+1$ and to $L=(J+2)-1$.

\section{Solutions for deSitter Spaces}
\label{sec:dS}
As we have already mentioned in section \ref{sec:RV}, one only has to change the sign of the deformation parameter $\lambda$ to obtain the expressions in the dS case from those corresponding to the AdS one.

For the modified Heisenberg uncertainty relations, we transform eq.\ref{eqt4} to the following:
\begin{equation}
\Delta X_{i}\Delta P_{i}\geq \frac{\hbar }{2}\left( 1-\lambda \left( \Delta X_{i}\right) ^{2}\right)  \tag{68}  \label{68}
\end{equation}
it means that we have no minimum value for $\Delta P$ in this case as shown in fig.\ref{fig1}.

For spin 0 particle we use eq.\ref{eqt35} to write the energy eigenvalues:
\begin{equation}
E_{N,J}^{2}=m^{2}c^{4}+2\hbar m\omega c^{2}N-\lambda \hbar ^{2}c^{2}\left[
\left( N+1\right) ^{2}-J\left( J+1\right) -1\right]  \tag{69}  \label{69}
\end{equation}

\section{Conclusion}
\label{sec:CC}
In this work, we have exposed an explicit calculation of the relativistic Duffin--Kemmer--Petiau oscillator in three-dimension spaces with the presence of minimal uncertainty in momentum for anti-de Sitter model. We have used the Nikiforov--Uvarov method to solve both spin 1 and spin 0 cases and we have written the wave functions in the representation of vector spherical harmonics. For both scalar and vector bosons, we obtained the complete expressions of the eigenfunctions analytically in terms of the Jacobi polynomials. We also deduced the corresponding energy eigenvalues and we found that the deformation added a hard confinement term ($\propto N^{2}$) and a rotational term ($\propto J(J+1)$ for both spin 0 particle and natural parity states of the spin 1 case. However for unnatural parity states of the spin 1 particle, there was extra additions of a new spin-orbit contribution and a supplementary rotational term to the ones that already existed for this parity in the absence of deformation.

Moreover, in order to see the effect of the deformation on physical systems, we compared them with the experimental results from the cyclotron motion of an electron in a Penning trap and we have determined the upper bound of the minimal momentum uncertainty for AdS space. In addition, our results were tested by deducing the limit formulas of the spectra for $\lambda \rightarrow 0$ and we have obtained the results of the ordinary relativistic quantum harmonic oscillator for both scalar and vector particles.

Our results show that the spectrum of the natural states of spin 1 particle is similar to the one of the scalar particle. On the other hand, that of the non-natural states of the vector particle differs from the previous two by the presence of the spin--orbit coupling. We have associated the natural states with $L=J+0$ of the $S=1$ case and this explains the fact that its spectrum is similar to that of the case $S=0$ and also explains why there is no spin--orbit coupling for them since the projection of the spin is null in these two cases. On another side, the unnatural states are associated with both $L=J+1$ and $L=(J+2)-1$ for the $S=1$ case or when the projection of the spin in either $+1$ or $-1$; this explains the presence of the spin--orbit coupling even without deformation. This new interpretation allows us to redefine the parity with $\left( -1\right) ^{L}$ as it is normally done in relativistic quantum theory.

\section*{Acknowledgment}

This work was done with funding from the DGRSDT of the Ministry of Higher Education and Scientific Research in Algeria as part of the PRFU B00L02UN070120190003.


TITLE: Prove that $f$ is surjection.
QUESTION [8 upvotes]: $f:\mathbb{R}^n\to \mathbb{R}$ be a continuous map. $f$ not be a surjection if for all $c>0$, $f^{-1}(c)$ bounded.
My idea: I guess I can use "if $A$ is a connected set, then $f(A)$ is connected." But I can't prove $f$ is a surjection.

REPLY [6 votes]: Assume that the distance from the origin to any point from $f^{-1}([-1, 1])$ is at most $R$. Let $B_R$ be the ball of radious $R$ with the center at the origin. In particular $f^{-1}([-1, 1]) \subset B_R$. Since $f$ is continuous, this means that $f(B_R)$ is bounded. Let $C > 0$ be such that absolute value of any point in $f(B_R)$ is smaller than $C$. Since $f$ is surjective, there are two points, $a, b\in\mathbb{R}^2$ such that $f(a) = C, f(b) = -C$. These two points ($a$ and $b$) are outside $B_R$. Connect this to points by a continious path which does not intersect $B_R$. There is a point on this path, call it $c$, such that $f(c) = 0$. This is a contradiction, because $c\notin B_R$.
This even means that if $f:\mathbb{R}^2\to\mathbb{R}$ is continuous and surjective, then $f^{-1}(0)$ is unbounded.


TITLE: In Lebesgue world, is there $f=g$?
QUESTION [3 upvotes]: As we develop the theory of Lebesgue integration, it seems every function being equal is no longer the equality in the Riemann world. The notion of "almost everywhere" seems to basically replace the notion of equalit between two real-valued functions. Is this correct?
In other words, when you compare two real-valued functions defined on $E\subset\mathbb{R}^d$, you always compare their equality including all negligible set.
Hence the birth of the notion of equal almost everywhere.
Is this a correct interpretation? So, there is no true equality in the Lebesgue world then?

REPLY [0 votes]: One should typically judge this based on context - typically, sets of measure zero won't matter, and, in such cases it may be understood that, even if $f=g$ literally means that $f(x)=g(x)$ for all $x$, this is interchangeable with $f$ and $g$ being equal almost everywhere. I suspect that some authors don't pay too much attention to the issue because it often doesn't matter.
However, it's worth being aware that, in the context of Lebesgue integration, it's also common to define things based upon almost-everywhere equivalence classes rather than functions. Often, one makes the following definition:

If $f:\mathbb R\rightarrow\mathbb R$ is a measurable function, let $\langle f\rangle$ be the set of all functions $g$ such that $g$ and $f$ are equal almost everywhere.

and then defines integration and whatnot based on these classes - and then authors often abuse notation to forget about the difference between functions and equivalence classes, but it's still there. This comes up particularly strongly when one talks about the $L^p$ function spaces, where you need this definition* to get the properties you want out of this space. Note that $\langle f\rangle = \langle g\rangle$ precisely says that $f$ and $g$ are equal almost everywhere - and if your author is working with these sets instead of functions (openly or secretly), then equality is best understood as "almost everywhere" by the definitions.

*...or completely different ones like $L^1$ being the space of absolutely continuous signed measures of finite total mass - but that's a different story.


text:
\section{Two Stage Radioactive Decay}
Tags: Decay Equation

\begin{theorem}
Let $R_A$ be a [[Definition:Radioactive Isotope|radioactive isotope]] which [[Definition:Radioactive Decay|decays]] into another [[Definition:Radioactive Isotope|radioactive isotope]] $R_B$ with the [[Definition:Rate Constant|rate constant]] $k_1$.
Let $R_B$ [[Definition:Radioactive Decay|decay]] into a third [[Definition:Chemical Element|element]] $R_C$ with the [[Definition:Rate Constant|rate constant]] $k_2$.
Let the initial quantity of $R_A$ be $x_0$.
Let the amounts of $R_A$ and $R_B$ present at [[Definition:Time|time]] $t$ be $x$ and $y$ respectively.
Then:
:$y = \begin{cases}
\dfrac {k_1 x_0} {k_2 - k_1} \left({e^{-k_1 t} - e^{-k_2 t} }\right) & : k_1 \ne k_2 \\
& \\
k_1 x_0 t e^{-k_1 t} & : k_1 = k_2
\end{cases}$
\end{theorem}

config: wiki


TITLE: How similar are the measures if their Fourier transform coincide?
QUESTION [2 upvotes]: Let $\mu$ and $\nu$ be finite Borel nonnegative measures exponentially decreasing at infinity, i.e. there exists $A > 0$ such that
$$
   \int\limits_{0}^{\infty} e^{Ax} \mu(dx) < \infty, \;\;\; \int\limits_{0}^{\infty} e^{Ax} \nu(dx) < \infty
$$
and let supports $\mathrm{supp}(\mu)$ and $\mathrm{supp}(\nu)$ lie in $\mathbb{R}_{+} = [0,\infty)$ and
$$
   \int\limits_{0}^{\infty} e^{-p x}\mu(dx) = \int\limits_{0}^{\infty} e^{-px} \nu(dx), \;\;\; p \geq 0
$$
Functions on the LHS and on the RHS of this equation may be continued analitically to $\mathbb{R}_{+} + i\mathbb{R}$. Hence Fourier transforms of $\mu$ and $\nu$ are well defined and coincide. But what can we say in this case about $\mu$ and $\nu$? How similar they are? Are they equal in general?

REPLY [4 votes]: The conditions imply that the identity
$$
\int_0^\infty e^{-px} \, d\mu(x) = \int_0^\infty e^{-px} \, d\nu(x)
$$
holds for all complex $p$ with $\mathrm{Re}\, p>-A$, so adding up the two identities for purely imaginary $p$ with positive and negative imaginary part we indeed get equality of the Fourier transforms
$$
\int_{-\infty}^\infty e^{-itx}\, d\mu(x) = \int_{-\infty}^\infty e^{-itx} \, d\nu(x)
$$
for all $t \in \mathbb{R}$. Now to show that the measures are actually the same is a somewhat technical standard proof, found in many textbooks. First you use the Weierstrass approximation theorem to show that every continuous periodic function can be uniformly approximated by trigonometric polynomials, so $\mu$- and $\nu$-integrals of continuous periodic functions are the same. Then you can approximate compactly supported continuous continuous functions by periodic functions with periods $x_n \to \infty$, which implies that $\mu$- and $\nu$-integrals agree on compactly supported continuous functions. Lastly you can approximate characteristic functions of intervals by compactly supported continuous functions and you get that $\mu$ and $\nu$ agree on intervals, which shows $\mu = \nu$.


TITLE: Probability the addition rule
QUESTION [0 upvotes]: The percent distribution of the number of grandchildren for a sample of 1904 grandparents is shown in the pie chart. Find each probability.
One- 9%
Two to four-36%
Five to seven- 23%
Eight or nine- 9%
Ten or more- 23%
a. Randomly selecting a grandparent with one grandchild.
b. Randomly selecting a grandparent with less than five grandchildren.
c. Randomly selecting a grandparent with two or more grandchildren.
d. Randomly selecting a grandparent with between two and seven grandchildren, inclusive.

REPLY [0 votes]: P(X=1) = 0.09
P(X < 5) = 0.09 + 0.36
P(X >= 2) = .36 + .23 + .09 + .23
P(2 <= X <= 7) = .36 + .23


TITLE: Why does the limit definition of e fail?
QUESTION [6 upvotes]: I'm trying to calculate the following limit:
$$\lim_{x\to +\infty} \frac{(1+\frac{1}{x})^{x^2}}{e^x}$$
I tried to use the fact that $e^x = \lim_{x\to +\infty}(1+\frac{1}{x})^{x}$, but this gives $\frac {e^x}{e^x} = 1,$ which is not the answer. WolframAlpha says the answer should be $\frac {1}{\sqrt{e}}$. I was able to get the right answer using Taylor series of $\ln(1+ \frac{1}{x})$, but I can't find a good reason why my first approach fails. Can someone please help? Thanks!

REPLY [15 votes]: Your first approach uses the reasoning

$(1+\frac1x)^x$ is approximately $e$, so $(1+\frac1x)^{x^2}=[(1+\frac1x)^x]^x$ behaves like $e^x$.

This is illegal because you're letting $x\to\infty$ in your inner expression while preventing $x\to\infty$ in the outer expression. There's no rule that allows you to do that -- you can't hold $x$ fixed in one part of your expression while letting it tend to infinity in another part.
As a simpler example, it is incorrect to compute $\lim_{n\to\infty}1$ by writing $1=\frac1n\cdot n$, then observing that $\frac1n\to0$ as $n\to\infty$, and deducing that
$$
\lim_{n\to\infty}1 = \lim_{n\to\infty}\left[ \left(\lim_{n\to\infty}\frac 1n\right)\cdot n \right]= \lim_{n\to\infty}(0\cdot n) = 0.
$$
Another simpler, but relevant example: $(1+\frac1x)$ tends to $1$ as $x\to\infty$, but it doesn't follow that
$$\lim_{x\to\infty}\left(1+\frac1x\right)^x = \lim_{x\to\infty}\left[\lim_{x\to\infty}\left(1+\frac1x\right)\right]^x=\lim_{x\to\infty} 1^x = 1.
$$


TITLE: Inequality without BW and Uvw, just AM GM
QUESTION [1 upvotes]: Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$
I guess equality holds: two of them equal $1$ and one equal to $0$.
My approach: The equality equivalent to
$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c)+3(a+b+c)\ge5$
I tried to use Pqr method but it is quite complicated. Any one help me to get nice proof? Thanks.

REPLY [1 votes]: $uvw$ helps!
Let $\sqrt{ab}=z$, $\sqrt{ac}=y$ and $\sqrt{bc}=x$.
Thus, $x^2+y^2+z^2=1$ and we need to prove that:
$$\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\right)(3-x-y-z)+2(xy+xz+yz)\geq4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $w^3$, but
$$\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}=\frac{9v^4-6uw^3}{w^3}=\frac{9v^4}{w^3}-6u$$ decreases respect to  $w^3,$ which by $uvw$ says
(about $uvw$ see here:
https://artofproblemsolving.com/community/c6h278791 )  that it's enough to prove the last inequality for equality case of two variables.
Can you end it now?


text:
\section{The first target}\label{sec:firstTarget}

Since if $G$ has rank $1$ all Newton points are integral and Theorem \ref{thm:translations} applies, from now on we assume that $G$ has rank $n \geq 2$.  Let $x_0 = t^\lambda w_0 \in \eW$ be such that the alcove $\x_0$ is in the shrunken dominant Weyl chamber.
In Section~\ref{sec:minGallery}, we construct a minimal gallery $\gamma:\fa \rightsquigarrow \x_0$, and then  we carry out a PRS-folding sequence on $\gamma$ in Section~\ref{sec:folding}.  We then use these constructions in Section~\ref{sec:conjugationFinal} to prove the following result.

\begin{thm}\label{thm:firstTarget} Let $x_0=t^\lambda w_0\in \eW$, and assume that $\x_0$ is in the shrunken dominant Weyl chamber $\Cfs$.
Suppose that
\[
\proj_i(\lambda - 2\check{\rho}) = \nu,
\]
where $\nu$ is a non-integral Newton point with associated standard spherical parabolic subgroup $P_i$ of rank $1$.
Then for any $b \in \eW$ such that $\nu_b = \nu$ and $\kappa_G(b)=\kappa_G(x_0)$, we have
$X_{x_0}(b) \neq \emptyset.$
\end{thm}

We will also use the constructions from this section later in Section~\ref{sec:remainingTargets}.

\subsection{Construction of minimal gallery}\label{sec:minGallery}

In this section, we construct a minimal gallery $\gamma: \fa \rightsquigarrow \x_0$ using a specific choice of reduced word for the longest element $w_0 \in \sW$.
An example of the construction of $\gamma$ in type $\tilde{A}_2$ is given in Figure~\ref{fig:gamma}. Although our constructions apply equally well to any fixed sheet of extended alcoves, we illustrate all explicit gallery constructions in the 0-sheet.

\begin{figure}[ht]
\centering
\begin{overpic}[width=0.4\textwidth]{gamma}
\put(21.5,94.5){\color{blue}$\lambda$}
\put(19,76){\color{blue}$\lambda - \check{\rho}$}
\put(21,89){\color{blue}$\x_0$}
\put(32,53){\color{blue}$\gamma$}
\put(37,17){\color{black}$\fa$}
\put(37,27){\color{black}$\check{\rho}$}
\put(47,15){\color{red}$\alpha_1^\vee$}
\put(25,15){\color{red}$\alpha_2^\vee$}
\put(54,51){\color{black}$H_{\alpha_2}$}
\put(9,51){\color{black}$H_{\alpha_1}$}
\put(36,4.5){\color{DarkGreen}$\w_0$}
\put(44,8){\color{DarkGreen}$\tilde\tau$}
\put(30,8){\color{LimeGreen}$\tau$}
\put(33,31){\color{LimeGreen}$\tau'$}
\put(47,25){\color{DarkGreen}$H_1$}
\put(53,14){\color{DarkGreen}$H_2$}
\put(45,1){\color{DarkGreen}$H_3$}
\end{overpic}
\caption{An example of the construction of the minimal gallery $\gamma:\fa \rightsquigarrow \x_0$.}
\label{fig:gamma}
\end{figure}

Fix $i \in \{1,\dots,n\}$.  The reduced word for $w_0$ is chosen as follows.  Since conjugation by $w_0$ permutes the (spherical)  simple reflections, $w_0 s_i w_0$ is again a simple reflection, which we denote by $s_{i_1}$.  Thus $w_0 \alpha_i = -\alpha_{i_1}$.  We then denote by $s_{i_2},\dots,s_{i_n}$ the remaining spherical simple reflections $s_j$ for $j \in \{1,\dots,n\} \setminus \{ i_1\}$, written in any order.  Now choose a reduced word $\vec w_0 = s_{k_1} \cdots s_{k_\ell}$ for $w_0$, where $\ell = \ell(w_0)$, which ends with the Coxeter element $s_{i_n}s_{i_{n-1}} \cdots s_{i_1}$.  That is, $\vec w_0$ has last letter $s_{k_\ell} = s_{i_1}$, and $s_{k_{\ell-1}}=s_{i_2}$, \dots, $s_{k_{\ell - n + 1}} = s_{i_n}$ are the remaining spherical simple reflections.  In the example depicted in Figure~\ref{fig:gamma}, we have $i = 1$, and hence $i_1 = 2$, $i_2 = 1$, and $\vec w_0 = s_2 s_1 s_2$.

Recall that $x_0 =t^\lambda w_0$, and suppose throughout the rest of the paper that $\lambda \in \omega + R^\vee$, or equivalently that $\x_0$ is in the $\omega$-sheet of extended alcoves with base alcove $\fa = \fa_\omega$. We construct a minimal gallery $\gamma: \fa \rightsquigarrow \x_0$ as follows. Denote by $\w_0 = w_0\fa$. Let the gallery $\tilde\tau:\fa \rightsquigarrow \w_0$ be of type $(s_{\omega k_1},\dots,s_{\omega k_\ell})$; that is, $\tilde\tau$ has the same type as $\vec w_0$, appropriately relabeled according to the action of $\omega$ on $\fa_0$. Denote by $H_1, H_2, \dots, H_\ell$ the (ordered) list of hyperplanes crossed by $\tilde\tau$, which is independent of $\omega$. One can compute from the chosen word $\vec w_0$ for $w_0$ that $H_1=H_{\alpha_{k_1}}$ and $H_j= s_{k_1}s_{k_2}\cdots s_{k_{j-1}}H_{\alpha_{k_j}}$ for all $j=2, \dots, \ell$.  In Figure~\ref{fig:gamma}, the gallery $\tilde \tau$ is the dark green gallery from $\fa$ to $\w_0$, of $(s_2,s_1,s_2)$, and the hyperplanes $H_1$, $H_2$, and $H_3$ are also labeled in dark green.

Now let $\tau: \w_0\rightsquigarrow \fa$ be the minimal gallery that crosses the same list of hyperplanes as $\tilde\tau,$ in the same order. In other words, the alcoves and panels in $\tau$ are opposite those in $\tilde\tau$ at each step, and $\tau$ can be thought of as the image of $\tilde \tau$ under the opposition map in the star (or link) of the origin. The type of $\tau$ may or may not be the same as the type of $\tilde\tau$, depending on $G$.  For example, if $G$ is of type $\tilde{B}_2$ then $\tau$ will have the same type as $\tilde \tau$.  In Figure~\ref{fig:gamma}, in which $G$ is type $\tilde{A}_2$, the gallery $\tau$ is the light green gallery from $\w_0$ to $\fa$, and $\tau$ has type $(1,2,1)$, which is different from the type of $\tilde \tau$.

Next, let $\tau'$ be the gallery from the $w_0$-position at $\check{\rho}$ to the identity position at $\check{\rho}$ obtained by applying the translation $t^{\check{\rho}}$ to $\tau$.   Note that the translation $t^{\check{\rho}}$ is not necessarily type-preserving; however, for each simple root $\alpha_k$, it takes the hyperplane $H_{\alpha_k,0}$ through the origin to the hyperplane $H_{\alpha_k,1}$ through $\check{\rho}$.  In Figure~\ref{fig:gamma}, the gallery $\tau' = t^{\check{\rho}} \tau$ is also depicted in light green.

Now let $\gamma:\fa \rightsquigarrow \x_0$ be the gallery obtained by concatenating the following five galleries:
\begin{enumerate}
    \item a minimal gallery from $\fa$ to the $w_0$-position at $\check{\rho}$;
    \item the gallery $\tau'$;
    \item a minimal gallery from the identity position at $\check{\rho}$ to the $w_0$-position at $\lambda - \check{\rho}$;
    \item a minimal gallery from the $w_0$-position at $\lambda - \check{\rho}$ to the identity position at $\lambda-\check{\rho}$, whose final crossing is in a panel of the $\alpha_i$-hyperplane passing through $\lambda - \check{\rho}$;
    \item a minimal gallery from the identity position at $\lambda - \check{\rho}$ to the alcove $\x_0$, which is in the $w_0$-position at $\lambda$.
\end{enumerate}
In Figure~\ref{fig:gamma}, such a gallery $\gamma$ is shown in blue.  The resulting gallery $\gamma$ is minimal by similar arguments to those in \cite[Lemma 6.2]{MST1}, and we denote its type by $\vec{x}_0$.  For the constructions in this section we will only need some features of $\gamma$, and the remaining features will be used in Section~\ref{sec:remainingTargets}.

\subsection{Folding the gallery}\label{sec:folding}

In this section we carry out a PRS-folding sequence on the minimal gallery $\gamma:\fa \rightsquigarrow \x_0$ from Section~\ref{sec:minGallery} to obtain a gallery $\gamma_{\check{\rho}}$ of type $\vec{x}_0$, and we determine the first and final alcoves of $\gamma_{\check{\rho}}$.  To illustrate this folding sequence, we continue the example from Figure~\ref{fig:gamma} in Figure~\ref{fig:gamma_rho} below.  We also establish a sufficient condition on certain elements $y \in \eW$ for $\gamma_{\check{\rho}}$ to be positively folded with respect to the $(P_i,y)$-chimney.

\begin{figure}[ht]
\centering
\begin{overpic}[width=0.7\textwidth]{gamma_rho}
\put(29,96){\color{blue}$\lambda$}
\put(24,70){\color{blue}$\lambda - 2\check{\rho}$}
\put(29,91.7){\color{blue}$\x_0$}
\put(37,65){\color{blue}$\gamma$}
\put(68,91){\color{black}$H_{\alpha_2}$}
\put(10,91){\color{black}$H_{\alpha_1}$}
\put(40.5,38.5){\color{black}$\fa$}
\put(37,37){\color{black}$\s_1$}
\put(44,37){\color{black}$\s_2$}
\put(35,33.5){\color{black}$\s_1\s_2$}
\put(43,33.5){\color{black}$\s_2 \s_1$}
\put(40,30){\color{black}$\w_0$}
\put(40.5,46){\color{black}$\check{\rho}$}
\put(49,37){\color{red}$\alpha_1^\vee$}
\put(31,37){\color{red}$\alpha_2^\vee$}
\put(37.5,48){\color{LimeGreen}$\tau'$}
\put(10.5,19){\color{OrangeRed}$\z$}
\put(4.5,14){\color{OrangeRed}$\zeta$}
\put(4.5,57){\color{OrangeRed}$w_0\zeta$}
\put(25,26){\color{OrangeRed}$\gamma_{\check{\rho}}$}
\end{overpic}
\caption{An example of the construction of the folded gallery $\gamma_{\check{\rho}}$.}
\label{fig:gamma_rho}
\end{figure}

To obtain $\gamma_{\check{\rho}}$ from $\gamma$, we fold the first $\ell(w_0) -1$ crossings of the subgallery $\tau' = t^{\check{\rho}} \tau$ of $\gamma$ using a PRS-folding sequence, so that there are $\ell(w_0)  - 1$ folds in total, all occurring in panels of the $w_0$-position at $\check{\rho}$.  (Note that since $G$ has rank $n \geq 2$, we have $\ell(w_0) > 1$.)   This folding sequence consists of exactly the first $\ell(w_0) - 1$ folds of the sequence described in more detail in \cite[Sec.~6.2]{MST1}. In Figure~\ref{fig:gamma_rho}, where $\ell(w_0) - 1 = 2$, recall that the subgallery $\tau'$ is shown in light green and $\gamma$ is blue. The gray gallery is that obtained from $\gamma$ by carrying out the first fold in this folding sequence, and $\gamma_{\check{\rho}}$ is the orange gallery obtained from the gray gallery by carrying out the second and final fold.

We note that $\gamma_{\check{\rho}}$, like $\gamma$, has first alcove $\fa$, since this folding sequence leaves fixed the initial minimal subgallery of $\gamma$ from $\fa$ to the $w_0$-position at $\check{\rho}$. We now determine the final alcove $\z$ of the folded gallery $\gamma_{\check{\rho}}$.

\begin{lemma}\label{lem:sphericalZ}  The spherical direction of $\z$ is $s_{i_1} = w_0 s_i w_0$.
\end{lemma}

\noindent For example, in Figure~\ref{fig:gamma_rho}, we have $i_1 = 2$ and the spherical direction of $\z$ is in fact $s_2$.

\begin{proof} By construction, the final alcove $\x_0$ of $\gamma$ has spherical direction $w_0$.  The subgallery $\tau' = t^{\check{\rho}} \tau$ of $\gamma$ crosses the hyperplanes $t^{\check{\rho}} H_1, \dots, t^{\check{\rho}} H_\ell$, in this order, and all folds of $\gamma_{\check{\rho}}$ occur within $\tau'$.  Thus the spherical direction of $\z$ can be found by first translating $\x_0$ by $-\check{\rho}$, then folding the first $\ell(w_0) - 1$ crossings of $\tau$, then applying the translation $t^{\check{\rho}}$ to the final alcove of the resulting gallery.  Since translations do not alter spherical directions of alcoves, it suffices to consider the effect of the reflections which correspond to   folding the first $\ell(w_0) - 1$ crossings of $\tau$.

The hyperplanes crossed by $\tau$ are $H_1,\dots,H_\ell$, with $H_j= s_{k_1}s_{k_2}\cdots s_{k_{j-1}}H_{\alpha_{k_j}}$.  The reflection corresponding to the first fold of $\tau$ is thus $s_{\alpha_{k_1}}$, and we have $s_{\alpha_{k_1}}H_2 = H_{\alpha_{k_2}}$, so the second fold of $\tau$ occurs in $H_{\alpha_{k_2}}$.  Continuing in this way we obtain that the hyperplanes in which the folds of $\tau$ occur are $H_{\alpha_{k_1}}$, $H_{\alpha_{k_2}}$, \dots, $H_{\alpha_{k_{\ell - 1}}}$, in this order.  Hence the spherical direction of $\z$ is $s_{k_{\ell -1}}\cdots s_{k_1}w_0$.

Using $\vec w_0 = s_{k_1} \cdots s_{k_\ell}$ we then obtain
\[s_{k_{\ell -1}}\cdots s_{k_1}w_0 = s_{k_{\ell -1}}\cdots s_{k_1} \cdot s_{k_1}\cdots s_{k_\ell}=s_{k_\ell}=s_{i_1}=w_0 s_i w_0, \]
as claimed.
\end{proof}

We now determine the translation vertex of the final alcove of the gallery $\gamma_{\check{\rho}}$.

\begin{lemma}\label{lem:translationZ}  The translation vertex of $\z$ is given by
\begin{equation}\label{eq:zeta}
\zeta = \check{\rho} + w_0 s_i (\lambda - \check{\rho}) \;\in \omega + R^\vee.
\end{equation}
\end{lemma}

\noindent For example, in Figure~\ref{fig:gamma_rho}, we have $i = 1$ so $w_0 s_i = s_1 s_2$, and both $\lambda \in R^\vee$ and $\zeta = \check{\rho} + s_1 s_2 (\lambda - \check{\rho}) \in R^\vee$.

\begin{proof}
First, note that although $\check{\rho}$ may not have the same image as $\lambda$ in $Q^\vee/R^\vee$, the final vertex $\zeta$ of $\gamma_{\check{\rho}}$ will, since reflections in hyperplanes through $\check{\rho}$ preserve this image.

Since $\x_0$ has translation vertex $\lambda$, by similar arguments to those in the proof of Lemma~\ref{lem:sphericalZ}, the final vertex of $\gamma_{\check{\rho}}$ can be found by first translating $\lambda$ by $-\check{\rho}$, then carrying out the reflections in the successive hyperplanes $H_{\alpha_{k_1}},\dots,H_{\alpha_{k_{\ell - 1}}}$, and finally translating the result by $\check{\rho}$.  So it suffices to show that $
s_{k_{\ell - 1}}\cdots s_{k_1} = w_0 s_i$.  Using the reverse of our chosen word $\vec{w}_0$, we obtain
\[
s_{k_{\ell - 1}}\cdots s_{k_1} = s_{k_\ell} s_{k_\ell} s_{k_{\ell - 1}}\cdots s_{k_1} = s_{k_\ell} w_0 = s_{i_1} w_0 = w_0 s_i w_0 w_0 = w_0 s_i,
\]
which completes the proof.
\end{proof}

We point out that the gallery $\gamma_{\check{\rho}}$ is not positively folded with respect to the $P_i$-chimney.  For example, compare the orientations for the $P_1$-chimney from Figure \ref{fig:chimneyOrientations} to see that the two folds in $\gamma_{\check{\rho}}$ from Figure \ref{fig:gamma_rho} both occur on the negative side of the respective hyperplane.  The next lemma determines an infinite family of elements $y \in \eW$ such that $\gamma_{\check{\rho}}$ is positively folded with respect to the $(P_i,y)$-chimney.

\begin{lemma}\label{lem:posFolded}
Let $y = t^\mu w_0 s_i \in \eW$, where $\mu \in Q^\vee$.  If $\langle \alpha_{i_1}, \mu \rangle \geq 1$, then the gallery  $\gamma_{\check{\rho}}$ is positively folded with respect to the $(P_i,y)$-chimney.
\end{lemma}

\begin{proof}  Since the spherical direction of $\y$ is $w_0 s_i$, and $w_0 s_i \alpha_i = w_0(-\alpha_i) = \alpha_{i_1}$, any $(P_i,y)$-sector lies between the hyperplanes $H_{\alpha_{i_1},k}$ and $H_{\alpha_{i_1},k+1}$ for some $k \in \Z$.  More precisely, as any $P_i$-sector lies between $H_{\alpha_{i},0}$ and $H_{\alpha_{i},1}$, any $(P_i, y)$-sector lies between $H_{\alpha_{i_1},k}$ and $H_{\alpha_{i_1},k+1}$ where $k = \langle \alpha_{i_1}, \mu \rangle$.

Let $\beta \in \Phi^+ \setminus \{ \alpha_i \}$.  Then by definition the $P_i$-chimney contains all half-apartments bounded by $\beta$-hyperplanes which contain subsectors of the antidominant Weyl chamber.  Write $\beta' = w_0 s_i \beta$.  Then $s_i \beta$ is positive, hence $\beta'$ is negative.  Now since $\alpha_{i_1} = w_0 s_i \alpha_i$, it follows that $-\beta' \in \Phi^+ \setminus \{ \alpha_{i_1} \}$.  Moreover the map $\Phi^+ \setminus \{ \alpha_i \} \to \Phi^+ \setminus \{ \alpha_{i_1} \}$ given by $\beta \mapsto -\beta'$ is a bijection.

Therefore for all $\beta \in \Phi^+ \setminus \{ \alpha_{i_1} \}$ and all $y = t^\mu w_0 s_i$, any $(P_i,y)$-chimney contains all half-apartments bounded by $\beta$-hyperplanes which contain subsectors of the dominant Weyl chamber.  Thus for all $\beta \in \Phi^+ \setminus \{ \alpha_{i_1} \}$ and all $y = t^\mu w_0 s_i$, the periodic orientation on $\beta$-hyperplanes with respect to the opposite standard orientation agrees with the orientation induced by the $(P_i,y)$-chimney.

Now by construction, the gallery $\gamma_{\check{\rho}}$ has all folds positive with respect to the opposite standard orientation, and its only folds in an $\alpha_{i_1}$-hyperplane are in $H_{\alpha_{i_1},1}$.   So $\gamma_{\check{\rho}}$ will be positively folded with respect to any $(P_i,y)$-chimney represented by a sector lying between $H_{\alpha_{i_1},k}$ and $H_{\alpha_{i_1},k+1}$ where $k \geq 1$.  This is exactly the condition $\langle \alpha_{i_1}, \mu \rangle \geq 1$ from the statement.
\end{proof}

\begin{remark}\label{rem:chimney}
Conversely, we point out that if $\langle \alpha_{i_1}, \mu \rangle <1,$ then in fact $\gamma_{\check{\rho}}$ is not positively folded with respect to the $(P_i,y)$-chimney.  In particular, the folds in $H_{\alpha_{i_1},1}$ occur on the negative side of this hyperplane if $\langle \alpha_{i_1}, \mu \rangle <1$.  As we shall see in Section \ref{sec:remainingTargets}, we will need to handle these two cases separately.
\end{remark}

We will need the next result in Section~\ref{sec:conjugationFinal} and in Section~\ref{sec:remainingTargets}, which says that both $\lambda-2\check{\rho}$ and $w_0\zeta$ project to the same point on $H_{\alpha_i}$. Note that since $\sW$ preserves $R^\vee$ and $2\check{\rho} \in R^\vee$ by definition, then $w_0\zeta$ and $\lambda-2\check{\rho}$ have the same image in $Q^\vee/R^\vee$, which equals that of $\lambda$. See Figure~\ref{fig:gamma_rho} for an example, and recall that $i = 1$.

\begin{lemma}\label{lem:top}
For any $\lambda \in Q^\vee$ and $i \in [n]$ as in Theorem \ref{thm:firstTarget}, and $\zeta$ defined by Lemma \ref{lem:translationZ},
\[\proj_i(w_0\zeta) = \proj_i(\lambda - 2\check{\rho}).\]
\end{lemma}

\begin{proof}
By the definition of $\zeta$, properties of $\proj_i$, and the fact that $w_0 \check{\rho} = - \check{\rho}$,
\begin{eqnarray*}
\proj_i(w_0 \zeta) & = & \proj_i(w_0 \check{\rho} + s_i(\lambda - \check{\rho})) \\
& = & \proj_i(w_0 \check{\rho}) + \proj_i(s_i(\lambda - \check{\rho})) \\
& = & \proj_i(-\check{\rho}) + \proj_i(\lambda - \check{\rho}) \\
& = & \proj_i(\lambda -2\check{\rho}),
\end{eqnarray*}
as claimed.
\end{proof}

\subsection{Reaching the first target}\label{sec:conjugationFinal}

We are now prepared to prove Theorem~\ref{thm:firstTarget}.  We thus assume that $\nu = \proj_i(\lambda - 2\check{\rho})$ is a non-integral Newton point with associated parabolic subgroup $P=P_i$ of rank 1.   We will show that the final alcove $\z$ of the folded gallery $\gamma_{\check{\rho}}$ from Section~\ref{sec:folding} is (labeled by) a certain $\eW$-conjugate of the standard representative $b_\nu$.  This allows us to conclude that $\gamma_{\check{\rho}}$ is positively folded with respect to a suitable chimney, such that we can apply Theorem~\ref{thm:ADLVChimneys}. Figure~\ref{fig:firstTarget}, which continues the example from Figures~\ref{fig:gamma} and~\ref{fig:gamma_rho}, illustrates the proof of Theorem \ref{thm:firstTarget}.

\begin{figure}[ht]
\centering
\begin{overpic}[width=0.6\textwidth]{firstTargetFigure}
\put(29,96){\color{blue}$\lambda$}
\put(25,70){\color{blue}$b_\nu$}
\put(29,91.7){\color{blue}$\x_0$}
\put(37,65){\color{blue}$\gamma$}
\put(29.5,66){\color{blue}$\eta$}
\put(23.5,62.6){\color{blue}$\nu$}
\put(69,93){\color{blue}$\cT_\nu$}
\put(40.5,38.5){\color{black}$\fa$}
\put(37,37){\color{black}$\s_1$}
\put(44,37){\color{black}$\s_2$}
\put(35,33.5){\color{black}$\s_1\s_2$}
\put(43,33.5){\color{black}$\s_2 \s_1$}
\put(40,30){\color{black}$\w_0$}
\put(40.5,46){\color{black}$\check{\rho}$}
\put(49,37){\color{red}$\alpha_1^\vee$}
\put(31,37){\color{red}$\alpha_2^\vee$}
\put(10.5,19){\color{OrangeRed}$\z$}
\put(4.5,14){\color{OrangeRed}$\zeta$}
\put(4.5,57){\color{OrangeRed}$w_0\zeta$}
\put(36,77){\color{OrangeRed}$s_i w_0\zeta$}
\put(21.5,33){\color{OrangeRed}$\gamma_{\check{\rho}}$}
\put(25.5,6){\color{OrangeRed}$w_0 s_i \cT_\nu$}
\put(18,11){\color{DarkMagenta}$\mu$}
\put(14.5,7){\color{DarkMagenta}$\y$}
\end{overpic}
\caption{An example illustrating the proof of Theorem~\ref{thm:firstTarget}.}
\label{fig:firstTarget}
\end{figure}

We start by using some results on conjugacy classes from Section~\ref{sec:conjugacy}.  Recall that the subspace transverse to the Newton point $\nu$ with standard representative $b_\nu = t^\eta v$ is defined as $\cT_\nu = \nu + \Ker A_v$.  In Figure~\ref{fig:firstTarget}, in which $i = 1$, the affine subspace $\cT_\nu$ is the blue line, and the orange line is $w_0 s_i \cT_\nu = s_1 s_2 \cT_\nu$.

\begin{lemma}\label{lem:firstTransverse} Suppose that $\nu = \proj_i(\lambda - 2\check{\rho})$ is a non-integral Newton point, with standard representative $b_\nu = t^\eta s_i$, and denote the corresponding Kottwitz point by $\omega$.
Then \[\zeta \in (w_0s_i \cT_\nu) \cap (\omega+R^\vee).\]
\end{lemma}

\begin{proof}
Recall that $A_{s_i} = \proj_i$ by Lemma~\ref{lem:AwProperties}(4), in which case $s_i(\Ker A_{s_i}) = \Ker A_{s_i}$.  Since $\nu \in H_{\alpha_i}$ and hence is fixed by $s_i$, it follows by the definition of $\cT_\nu = \nu + \Ker A_{s_i}$ that $s_i \cT_\nu = \cT_\nu$.  Since $\zeta \in \omega + R^\vee$ by Lemma \ref{lem:translationZ}, it now suffices to show that $w_0 \zeta \in \cT_\nu$.  By Lemma~\ref{lem:top} and assumption, we have $\proj_i(w_0 \zeta) = \proj_i(\lambda - 2\check{\rho}) = \nu$. Combining this observation with  Proposition~\ref{prop:stdReps}(2), we get that $\proj_i(w_0 \zeta) = \nu = \proj_i(\eta)$.  Thus $w_0 \zeta - \eta \in \Ker A_{s_i}$.  From Corollary~\ref{cor:stdRepTransverse} we have $\eta \in \cT_\nu$, and hence $w_0 \zeta \in \cT_\nu$ as required.
\end{proof}

We now identify $z=t^\zeta s_{i_1}$ as a particular $\eW$-conjugate of the standard representative $b_\nu$.

\begin{corollary}\label{cor:firstConjugation}
Let $z = t^\zeta s_{i_1} \in \eW$.  Then:
\begin{enumerate}
    \item there exists an integer $d_i \in \Z$ such that  $s_i w_0 \zeta - \eta = d_i \alpha_i^\vee$; and
    \item if $y = t^\mu w_0 s_i \in \eW$ with $\mu = d_i w_0 s_i \eta \in Q^\vee$ for $d_i \in \Z$ as in (1), then $z=b_\nu^y$.
\end{enumerate}
\end{corollary}

\noindent For example, in Figure~\ref{fig:firstTarget} we have $d_i = 1$, and hence $\mu = w_0 s_i \eta = s_1 s_2 \eta$ with $y = t^\mu s_1 s_2$.  The vertex $\mu \in R^\vee$ and the alcove $\y$ are labeled in purple.  (Note that, coincidentally in this example,  $\lambda - 2\check{\rho} = \eta$.)

\begin{proof}
Put $u = w_0 s_i$.  Then $s_{i_1} = w_0 s_i w_0 = w_0 s_i s_i s_i w_0 = u s_i u^{-1}$. We also have $\zeta \in (u \cT_\nu) \cap (\omega +R^\vee)$ by Lemma~\ref{lem:firstTransverse}.  The remaining claims are then immediate by Proposition~\ref{prop:inverseRank1}.
\end{proof}

Our final corollary in this section combines the previous results to conclude that the gallery $\gamma_{\check{\rho}}$ is positively folded with respect to the chimney suggested by Corollary \ref{cor:firstConjugation}.

\begin{corollary}\label{cor:firstTarget} Let $x_0 =t^\lambda w_0 \in \eW$, and assume that $\x_0$ is in the shrunken dominant Weyl chamber $\Cfs$.
Suppose that
\[
\proj_i(\lambda - 2\check{\rho}) = \nu,
\]
where $\nu$ is a non-integral Newton point with associated standard spherical parabolic subgroup $P_i$ of rank $1$, and denote the corresponding standard representative by $b_\nu = t^\eta s_i$.

Let $\mu = d_i w_0 s_i \eta$ and $y = t^\mu w_0 s_i \in \eW$ be as in the statement of Corollary~\ref{cor:firstConjugation}. Then the gallery $\gamma_{\check{\rho}}:\fa \rightsquigarrow \bb_\nu^y$ has type $\vec{x}_0$ and is positively folded with respect to the $(P_i,y)$-chimney.
\end{corollary}

\noindent For example, a $(P_i,y)$-sector is shaded in purple in Figure~\ref{fig:firstTarget} with the alcove $\y$ based at $\mu$ outlined, and the orange gallery $\gamma_{\check{\rho}}$ is positively folded with respect to the $(P_i,y)$-chimney.

\begin{proof}
By construction, $\gamma_{\check{\rho}}: \fa \rsa \z$ has type $\vec{x}_0$. Corollary \ref{cor:firstConjugation} says that $\z=z\fa_0$ where $z=b^y_\nu$, and so $\bb_\nu^y = b_\nu^y\fa_0$ is indeed the final alcove of $\gamma_{\check{\rho}}$. In order for $\gamma_{\check{\rho}}$ to be positively folded with respect to the $(P_i,y)$-chimney, by Lemma~\ref{lem:posFolded} it suffices to show that $\langle \alpha_{i_1}, \mu \rangle \geq 1$.
We have
\[
\langle \alpha_{i_1}, \mu \rangle = \langle w_0 s_i \alpha_i, d_i w_0 s_i \eta \rangle = d_i \langle \alpha_i, \eta \rangle = d_i,
\]
since $\eta \in H_{\alpha_i,1}$ by Proposition \ref{prop:stdReps}. We thus aim to show that $d_i \geq 1$.

As $s_i w_0 \zeta - \eta = d_i \alpha_i^\vee$ by Corollary \ref{cor:firstConjugation}(1), we obtain that $s_i w_0 \zeta - \eta \in H_{\alpha_i,2d_i}$, and hence $s_i w_0 \zeta \in H_{\alpha_i,2d_i +1}$ again using that $\eta \in H_{\alpha_i,1}$.  Thus
\[
2d_i + 1 = \langle \alpha_i, s_i w_0 \zeta \rangle = \langle -\alpha_i, -\check{\rho} + s_i(\lambda - \check{\rho}) \rangle = \langle -\alpha_i, -\check{\rho} \rangle + \langle \alpha_i, \lambda \rangle + \langle \alpha_i, -\check{\rho} \rangle = \langle \alpha_i, \lambda \rangle.
\]
Now recall by Lemma \ref{lem:NPimageAsi} that $\proj_i(\lambda - 2\check{\rho})$ is a non-integral Newton point if and only if $\langle \alpha_i, \lambda-2\check{\rho} \rangle = 2k+1$ for some $k \in \Z$, which in turn happens if and only if $\lambda$ itself lies in an $\alpha_i$-hyperplane of odd index, since $\langle \alpha_i, 2\check{\rho} \rangle = 2$.  Finally, since $\lambda - 2\check{\rho}$ is dominant, we know that $\langle \alpha_i, \lambda \rangle \geq 2$.  Combining this observation with the fact that $\lambda$ lies in an $\alpha_i$-hyperplane of odd index, in fact we have $\langle \alpha_i, \lambda \rangle \geq 3$.  Therefore $d_i \geq 1$ above,
as required.
\end{proof}

The main result of this section now immediately follows.

\begin{proof}[Proof of Theorem~\ref{thm:firstTarget}] Combine Corollary~\ref{cor:firstTarget} with Theorem~\ref{thm:ADLVChimneys}(1).
\end{proof}

config: arxiv


TITLE: Given integers n,b what (all) the integer solutions for $a_1^2+a_2^2+....+a_n^2 = b.C^2$ ??
QUESTION [0 upvotes]: Given integers b,n, what are integer C,$a_i$ who solves $a_1^2+a_2^2+....+a_n^2 = b.C^2$ ??
Example for n=4 , b=7-> $a_1^2+a_2^2+a_3^2+a_4^2 = 7C^2$
or
for n=3, b=1 -> $a_1^2+a_2^2+a_3^2 = C^2$
Please if any reference, book, author I will appreciate thanks

REPLY [0 votes]: Every positive integer is the sum of four squares, in particular there exist $a_1,a_2,a_3,a_4$ with $a_1^2+\cdots +a_4^2=7C^2$ for every $C$. For two squares and three squares there are well-known theorems as well:
Sum of one, two, and three squares
Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares.


TITLE: What is the difference between field theory and Galois theory
QUESTION [15 upvotes]: I am about to finish the book Galois theory by Harold Edwards. I am planning to study Galois theory at a more advanced level or field theory. I am unable to decide because I don't know the difference between Galois theory and field theory.
Question: What is the difference between field theory and Galois theory ?
My guess is the Galois theory would be a "subset" of field theory, no ? If my guess is true I don't know whether Galois theory is a proper subset of field theory or not. If my guess is true, I'd like to have an idea about motivations about topics in field theory that do not fall under Galois theory.
Thank you

REPLY [12 votes]: Galois theory is the theory of the duality between profinite groups associated to fields and closed subgroups which arise as dual to field extensions of the original field. It's about the algebra of polynomials over a field and how that helps to understand other fields constructed algebraically from the original field, i.e. from roots of polynomials over a field. Some basic examples are
$$\Bbb Q(i)=\Bbb Q[x]/(x^2+1), \quad \Bbb Q[x]/(x^5+x-1)$$
The latter has no simple description in terms of the usual algebraic operations such as $\sqrt[n]\cdot$ or $+,-,\cdot, /$, but is nonetheless an example because it talks about roots of a polynomial with rational coefficients. It was actually the question of solving polynomials using radicals that originally motivated Galois to create the notion which we now call Galois theory.
which is constructed from the rationals by adding in the square root of $-1$, which is a root of a polynomial with coefficients in the base field, $\Bbb Q$.
You are correct in saying it is one aspect of studying fields, at least the classical theory is.
Due to the implications in how Galois groups act, it also has applications in understanding integer rings for global fields including ramification, inertia, and splitting behavior of primes (see Artin reciprocity for the abelian case) and in today's mathematical world the non-commutative Galois theory is connected to the Langlands program, a popular area of study related to things such as automorphic forms and other areas of number theory. For elliptic curves defined over a field, the Galois action is indispensable particularly for the representation on the $\ell$-adic Tate module. Other topics include Galois cohomology from which we derive the invaluable Hilbert Theorem 90 and Kümmer theory, both central to class field theory and the latter helping to describe a wide class of useful algebraic extensions, particularly for radical and abelian ones.
For topics that do not fall under the purvey of Galois theory, you are considering non-algebraic aspects of fields, eg. transcendence, which is the case when you totally lack a polynomial and your extension is of infinite degree, with no way to break it up via finite pieces. The other case is the inseparable one where your polynomials are in some sense "simple," but are too simple to give rise to the structure of the group symmetry because the roots can't be separated from one another.
An example of each is
$$\Bbb F_p(t)[x]/(x^p-t)$$
where the extension is no longer separable because all the roots of the polynomial are the same because of the positive characteristic, in fact, if $z$ is one such, we see $x^p-t=(x-z)^p$.
$$\Bbb{Q}(x_1,\ldots, x_n)/\Bbb Q$$
where the elements adjoined are a formal variables, not roots of a polynomial in $\Bbb Q$, so gives rise to an infinite (transcendental) extension. These extensions are quite useful in that they can help to prove that all finite groups are the Galois group for some field (not to be confused with the inverse Galois problem which deals with extensions only of $\Bbb Q$). This is also part of the crux of the idea that all profinite groups are Galois groups for some field extension.


TITLE: How many distinct ways to climb stairs in 1 or 2 steps at a time?
QUESTION [24 upvotes]: I came across an interesting puzzle:

You are climbing a stair case. It takes $n$ steps to reach to the top.
  Each time you can either climb $1$ or $2$ steps. In how many distinct
  ways can you climb to the top?

Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?

REPLY [2 votes]: We can change this question into: In how many ways is possible to write a number as the ordered sum of 1s and 2s?
We can prove this using direct proof.
Hypothesis: For Fibbonaci number $F_1=1, F_2=1, F_{n}=F_{n-1}+F_{n-2}$, $Q(k)=F_{k+1}$
We say that we have constructed every order of $Q(n)$. Then, we construct $Q(n+1)$ this way:

For all orders, we add $+1$ to change $n$ to $n+1$. (There are now $Q(n)$ numbers.)
Note: all numbers that are constructed here ends in $+1$.
For all orders that end with $+1$, we change it to $+2$. We shall prove that the amount will be $Q(n-1)$ to complete the proof.
Note: all numbers that are constructed here ends in $+2$.

It can be seen that by constructing in this way there contains all orders.
Proof of step 2: If we constructed $Q(n)$ in the way above, then all numbers ended in $+1$ will be the amount $Q(n-1)$ by step 1. above, and we are done.


TITLE: When is $f''(x)$ undefined but $x$ not an inflection point of $f$?
QUESTION [2 upvotes]: This is supposed to be for bullet 5.2 in this question: A rigorous yet intuitive summary of inflection and critical points for beginning calculus?

There are examples for undulation points, where $f: A \to B, A,B \subseteq \mathbb R, f''(x)=0$ but $(x,f(x))$ is not an inflection point of $f$ like $f(t) = \frac{t^4}{12},f''(t)=t^2$ (If $f''(x)=0$ but is not an inflection point, what is it called?)
There are examples where $f''(x)$ is undefined but $(x,f(x))$ is an inflection point like $f'(t)=|t|$ (An inflection point where the second derivative doesn't exist?)

I'm missing something very obvious, but what's an example where $f''(x)$ is undefined, but $(x,f(x))$ is not an inflection point of $f$?

After some thought, I have come up with some examples. Are these correct?
First example: This example is one where $f$ is defined but not continuous at $x$.

$f: \mathbb R \to \mathbb R, f(t)=\text{sign}(t), x=0$. $f'(0)$ is undefined and thus so is $f''(0)$. However $f''(t)=0$ for $t \ne 0$.
In case someone asks further 'Okay, what if we assumed $f$ is continuous? Then $x$ would have to be an inflection point right?', then we proceed:

Second example: This example is one where $f$ is continuous but not differentiable at $x$.

A sharp turn like $g(t)=|t|$ but always increasing at an increasing rate but at different rates $f: \mathbb R \to \mathbb R, f(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$. This gives us $f': \mathbb R \setminus \{0\} \to \mathbb R, f'(t) = \text{sign}(t)+2$. $f'(0)$ is undefined and thus so is $f''(0)$. However $f''(t)>0$ for $t \ne 0$.
In case someone asks further 'Okay, what if we assumed $f$ is differentiable? Then $x$ would have to be an inflection point right?', then we proceed:

Third example: This example is one where $f$ is differentiable at $x$.

$f': \mathbb R \to \mathbb R, f'(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$.

REPLY [1 votes]: I can’t tell what you're asking, so I’ll answer the question in the title. Or, more precisely, I’ll give an example of an $f$, differentiable on the whole real line, with the property you describe in the title. I can’t tell what you’re doing by listing all those cases.
Take $f$ to be $x^2$ when $x$ is negative and $x^4$ when $x$ is non-negative.
Then $f’’(0)$ is undefined, and moreover $f$ does not have an inflection point at the origin.


TITLE: Magnetized nail lifting another nail
QUESTION [1 upvotes]: I have two nails (made of Fe). A and B.
A can not lift B.
If I rub A on the magnets north pole, then it can lift B.
Then if I rub it again but on the south pole, it can not lift B.
why is that? Why cant it lift B anymore?
the title I have chosen for this question is really bad. If someone else got a better idea please edit. sorry.

REPLY [0 votes]: Iron is composed of small magnetic dipoles all with their north south axis pointing randomly.
By rubbing the nail in a consistent specific direction the small dipoles line up with the magnetic field, so when you are rubbing the north pole, the dipoles align
 south-north. Thus you get a small magnet, not permanent because the little dipoles are not frozen in a crystal.
When you take this nail and rub it on the south pole you essentially demagnetize it for a while, if you take the same time you took on the north pole. If you persist in rubbing,  the dipoles will align the other way and you will have a small magnet again.


TITLE: Find the Laurent Series of $\frac{z^2}{(z-1)(z^2+4)}$
QUESTION [1 upvotes]: I have attempted to split this into partial fractions of the form $\frac{A}{(z-1)} + \frac{B}{(z-2i)} + \frac{C}{(z+2i)}$ as well as by $\frac{A}{(z-1)} + \frac{Bz+C}{(z^2+4)}$, but things get very complex (ha!) with i's and z's showing up in the numerator. How should I proceed?
I am told to find the series about 0 and 1, taking into account the different values of the radii.

REPLY [0 votes]: Since you are told to find the series about $0$ and $1$, the first one does not make much trouble using long division.
For the second one, define first $z=x+1$ which makes
$$f=\frac{z^2}{(z-1)(z^2+4)}=\frac{(x+1)^2}{x \left(x^2+2 x+5\right)}=\frac{x^2+2x+1}{x \left(x^2+2 x+5\right)}$$ Perform the long division for $$x \times f=\frac{x^2+2x+1}{x^2+2 x+5}$$ When done, divide the result by $x$ and replace $x$ by $(z-1)$