text:
\section{Introduction}
The theoretical explanation of the ongoing accelerated expansion of the universe\cite{RiessAdamG:1998cb, WMAP:2003elm,WMAP:2006bqn,WMAP:2008lyn,SDSS:2003eyi,Jain:2003tba} is one of the most unsolved cosmological problems. Achieving accelerated expansion by adding a constant to Einstein-Hilbert(E-H) action suffers from fine-tuning problem\cite{Weinberg:1988cp}. There are many ways to explain the accelerated epoch of the current universe.
In this context modified gravity theories have been developed by modifying the E-H action, for example, F(R) theories of gravity\cite{Capozziello:2002rd,Nojiri:2006gh,Sotiriou:2008rp,DeFelice:2010aj,Starobinsky:2007hu,Odintsov:2019evb}.
In recent times another class of most popular modified gravity theories is nonlocal gravity models\cite{Dirian:2014ara, Belgacem:2020pdz}. These models are motivated by Einstein action's ultraviolet (UV) and infrared (IR) corrections. They also provide a theoretical explanation of the accelerated expansion of the current universe. Some salient features of nonlocal gravity models are: (1) they can be employed to study cosmology in both the infrared and the ultraviolet, (2) valid cosmological perturbation theory, and (3) the resulting cosmology has good agreement with most observations.

In this direction, the nonlocal terms with Ricci scalar $R$ and $F(R)$ are extensively studied along with Einstein Hilbert action. These terms involve analytic transcendental functions of the covariant d'Alembert operator $\Box$. Initially, as developed by Wetterich, models with correction terms like $ R\Box^{-1}R $ are shown to be effective IR corrected nonlocal gravity model\cite{Wetterich:1997bz}. Further, Deser and Woodard\cite{Deser:2007jk}  introduce a general form as$ Rf(\frac{1}{\Box}R) $ which can be responsible for the late-time cosmic expansion of the universe. In this line, for a recent review on this topic, refer to\cite{Capozziello:2022lic}.

 The nonlocal theory can have an equivalent scalar-tensor form by introducing auxiliary variables. Higher derivative terms with d'Alembertian operator and curvature can be considered higher derivative field theories. According to Ostrogradsky theorem\cite{Ostrogradsky:1850fid}, non-degenerate higher derivative Lagrangians are cursed with instabilities (popularly known Ostrogradsky instabilities). Generally, these instabilities are easy to identify by linear momentum terms within the Hamiltonian, which make it unbounded above and below depending on the structure\cite{Woodard:2015zca}. However degenerate theories\cite{Horndeski:1974wa,Nicolis:2008in,Deffayet:2011gz,Kobayashi:2011nu,Gleyzes:2014dya,Gleyzes:2014qga,Langlois:2015cwa,Motohashi:2016ftl,Joshi:2019tzr,Joshi:2021azw}(for review, refer to \cite{Langlois:2018dxi,Kobayashi:2019hrl}) in this regard is free from these instabilities by reducing the phase space non trivially \cite{Chen:2012au}

 In order to check the appearance of Ostrogradsky instability in any higher derivative theory, a Hamiltonian analysis would be required. The Hamiltonian analysis of these types of theories can be performed by the Dirac method of constraint system\cite{Wipf:1993xg, Dirac:1958sq, Dirac:1958sc, dirac2013lectures, Matschull:1996up}. The Hamiltonian analysis of the nonlocal model with inverse powers of d'Alembertian operators acting only on  Ricci scalar is performed in\cite{Kluson:2011tb}. A more general analysis of constructing Hamiltonian in this context from a modern perspective is done in \cite{Joshi:2019cyk} including terms with inverse d'Alembertian operators acting on Riemann tensor, Ricci tensor, and Ricci scalar.

 In this work, we extend the analysis of ref. \cite{Joshi:2019cyk} for a non local F(R) gravity models. Earlier works in this line can be found in\cite{DeFelice:2014kma, Nojiri:2019dio}. Authors of ref.\cite{DeFelice:2014kma} consider a class of nonlocal gravity in which the Lagrangian is a general function of $\Box^{-1}R$ and derive the condition for the appearance of ghost fields by considering the kinetic term of the localized Lagrangian.
 In another ref.  \cite{Nojiri:2019dio}, they have found the condition for the nonexistence of ghost degrees of freedom in nonlocal F(R) gravity. Here the ghosts are usually identified through the kinetic matrix of the Lagrangian with a negative determinant, and after appropriate modification in the Lagrangian, ghost-free condition can be obtained. Application of this model to cosmology has also been carried out in ref.\cite{Nojiri:2019dio}.

This work tries to shed some light on the results obtained in \cite{Nojiri:2019dio, DeFelice:2014kma} from the Hamiltonian point of view. Here we construct the Hamiltonian for the localized action and identify the linear momenta term in the Hamiltonian, which is usually responsible for the appearance of Ostrogradsky ghosts in any higher derivative theory. Then we show that by putting an appropriate condition in the structure of constraints, we can get rid of these ghosts analogous to the approach followed in \cite{Nojiri:2019dio, DeFelice:2014kma} by analyzing the kinetic term of the Lagrangian.
For deriving Hamiltonian, first, we separate both spatial and time derivative terms in Lagrangian in terms of ADM variables, similar to the analysis done in \cite{Kluson:2011tb, Joshi:2019cyk}. Then we identify several constraints of our theory, which eventually lead us to count the number of ghost degrees of freedom.

 This paper has many sections. In section II, ADM formalism is reviewed. In section III, we formulate the Hamiltonian for the F(R) nonlocal model and shed light on conditions that can give ghost-free theories. In section IV, we perform Hamiltonian formalism for a general action of $\Box^{-1}R$; we study the ghost structure from the Hamiltonian analysis. Finally, we summarise our results in section 5.

\section{ADM Formalism}
In this section, we review the 3 + 1 formalism's basic features in order to formulate the Hamiltonian analysis of the purposed theory.  Space-time is characterized by $(M,g_{\mu\nu} )$, where M is a 4-dimensional differentiable manifold and $g_{\mu\nu}$ is a Lorentzian metric.  $(M,g_{\mu\nu} )$ can be foliated by the family of space like surface ($\Sigma_{t}$), for more details refer to\cite{Arnowitt:1962hi,Gourgoulhon:2007ue,17b23f5522334b1ca8bb575ecaf5c01e}. Mathematically, four dimensional metric can be written in terms of induced metric $ h_{\mu\nu}$ and normal vector $ n_{\mu}$,
\begin{equation}
g_{\mu\nu}= h_{\mu\nu}- n_{\mu} n_{\nu}. \label{metric}
\end{equation}
The time-like future-directed vector $n_{\mu} $ is  normal to  the three dimensional space like surface, having a property,
$n_{\mu}n^{\mu}=-1$. The time direction vector $t^{a}=\frac{\partial}{\partial t}$, related to $n^{\mu}$ as,
\begin{equation}
t^{\mu}=Nn^{\mu}+N^{\mu},
\end{equation}
where, $N^{\mu}$ is the shift vector, and $N$ is the lapse function.
The line element in the  3+1 decomposition  reads
\begin{equation}
ds^2= -N^{2}dt^{2}+h_{ij}(dx^{i}+N^{i}dt)(dx^{j}+N^{j}dt).
\end{equation}
We used the indices $i,j....$ for showing spatial character of shift vector and spatial metric. The value of N and $ N^{i}$  in terms of metric is given as,
\begin{equation}
N=\frac{1}{\sqrt{-g^{00}}},\qquad N^i=-\frac{g^{0i}}{\sqrt{g^{00}}}.
\end{equation}
Further, the various metric elements can be written in terms of $h_{ij},N^{i}$ and $N$ as,
\begin{eqnarray}
g_{00}=-N^2+h_{ij}N^{i}N^{j},\qquad g_{0i}=N_{i},\qquad g_{ij}=h_{ij}, \nonumber\\ g^{00}=\frac{1}{N^{2}},\qquad g^{0i}=\frac{N^i}{N^2}, \qquad g^{ij}=h^{ij}-\frac{N^i N^j}{N^2}.
\end{eqnarray}
  The normal vector in terms of lapse and shift function are
$$ n_{0}=-N, \qquad n_{i}=0, \qquad n^{0}=\frac{1}{N}, \qquad n^{i}=\frac{N^{i}}{N}.$$
The evolution of the spatial metric, $h_{ij}$ is given by the Lie derivative along $n^{a}$,
\begin{equation}
    \mathcal{L}_{n}h_{ij}=2NK_{ij}
\end{equation}
 where, $K_{ij}$ is the extrinsic curvature which is,
 \begin{eqnarray}
K_{ij}=\frac{1}{2N}\left( \dot{h_{ij}}-\mathcal{D}_{i} N_{j}-\mathcal{D}_{j}N_{i}\right).
\end{eqnarray}
 Here the overdot represents the lie-derivative with respect to the time flow vector $t^{b}$ of the metric, and $\mathcal{D}_{i}$ denotes the spatial derivative.
 Evolution of $K_{ij}$ is related to $L_{n}K_{ij}$ which takes the form,
\begin{eqnarray}
\mathcal{L}_{n}K_{ij}=\frac{1}{N}(\dot{ K_{ij}}-\mathcal{L}_{\Vec{N}}K_{ij}).
\end{eqnarray}
where $\mathcal{L}_{\Vec{N}}$ is lie derivative with respect to shift vector, $\dot{ K_{ij}}$ contains second order time derivative of induced metric $h_{ij}$. The Ricci scalar in (3+1) decomposition takes the form,
\begin{equation}
R= \mathcal{R}+ K^{2}-3 K_{i j} K^{ij}+ 2 h^{ij} L_{n}  K_{ij} -\frac{2}{N}D_{i}D^{i}N. \label{RIS}
\end{equation}
Similarly $\mathcal{L}_{n} \phi$ containing the first order time derivative of any scalar  $\phi$ is defined as,
\begin{equation}
\mathcal{L}_{n}\phi=n^{a}\nabla_{a}\phi=\frac{1}{N}(\dot\phi-\mathcal{L}_{\Vec{N}}\phi). \label{DOS}
\end{equation}
The 3+1 decomposition of our action will help us construct canonical momenta for the Hamiltonian. Next, we derive a Hamiltonian for our model.
\section{Hamiltonian Formalism of Nonlocal F(R) Gravity model}
Let us consider action with nonlocal terms of the form\cite{Nojiri:2019dio},
\begin{equation}
S=\int d^{4}x \sqrt{-g}\left[G(R)+ F(R) \Box^{-k}H(R) \right], \label{action}
\end{equation}
where, $G(R)$, $ F(R) $ and $ H(R) $ are general functions of Ricci scalar, and  $k$ is a positive integer. This model is examined in detail for $k=1$ and $k=2$ from the Lagrangian point of view. It is shown that this theory is ghost free for $k=1.$ However it can be made to be free from ghosts by adding extra fields  for $k=2$. Next, our plan is to analyze the same model in  Hamiltonian formalism. For this we  write the above action in an equivalent  scalar-tensor form by introducing Lagrange multiplier C, which replaces R by Q,
 \begin{equation}
 S_{1}^{eqv} = \int d^{4}x  \sqrt{-g}\left[G(R)+ F(R) \Box^{-k}H(R)+ C(R-Q) \right],\label{s12}
\end{equation}
 introducing different set of auxiliary fields $B_{k}$ and $A_{k}$\cite{Joshi:2019cyk},
 after some straightforward calculation, eq.(\ref{s12}) can be  written as,
\begin{align}
S^{eqv} = \int d^{4}x\sqrt{-g} \bigg[ G(Q) + F(Q)A_{1} + C\left( R -Q\right) -B_{k}H(Q)  \label{ass}\\\nonumber -\sum_{n=1}^{k-1} B_{n} A_{n+1}+\sum_{n=1}^{k} B_{n}\Box A_{n} \bigg].
\end{align}
The last term of action eq.(\ref{ass}) simplifies to
\begin{equation}
\int d^{4}x N\sqrt{h}\sum_{L=1}^{\infty} B_{n }\Box A_{n}=-\int d^{4}x  N\sqrt{h}\sum_{L=1}^{\infty}\nabla _{\rho}B_{n} \nabla^{\rho} A_{n}+\textnormal{surface term}.\label{eq:ss1}
\end{equation}
Further, the  eq.(\ref{ass}) in terms of ADM variables looks,
\begin{eqnarray}
\begin{aligned}
S^{eqv}= \int d^{3}x N\sqrt{h} \bigg [G(Q) +  F(Q)A_{1} + C\left( K_{ij}K^{ij} -K^{2} + \mathcal{R} -Q\right) &\\ -2K \ (n^{a} \nabla_{a}C)  -  \sum_{n= 1}^{k-1}  B_{n} A_{n+1} - B_{k}H(Q) - h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} &\\+ \sum_{n=1}^{k}  (n^{a}\nabla_{a}B_{n})(n^{b}\nabla_{b}A_{n}) +\mathcal{D}_{i}\mathcal{D}^{i}C\bigg].\label{ADMaction}
\end{aligned}
\end{eqnarray}
Now we derive the canonical momenta  with respect to set of variables $(N, N^i, h_{ij}, Q, A_{1},A_{2}...A_{n}, B_{1},B_{2}...B_{n})$, which are
\begin{eqnarray}
\begin{aligned}
\Pi_{N} & \approx 0, \ \ \ \ \
\Pi_{i} \approx 0,\ \ \
\Pi^{ij} = \sqrt{h} C( \mathcal{K}^{ij} -h^{ij}\mathcal{K} ) - \sqrt{h} h^{ij} (n^{a} \nabla_{a}C), \qquad \qquad \qquad \qquad   \\
P_{Q} & \approx  0 , \qquad \qquad
P_{C} = -2\sqrt{h} \mathcal{K}, \qquad\\
P_{A_{1}} &= \sqrt{h} (n^{a}\nabla_{a}B_{1}), \ \ P_{A_{2}} = (n^{a}\nabla_{a}B_{2}) \  ,,,,,,,\ P_{A_{k}}=\sqrt{h} (n^{a}\nabla_{a}B_{k})
\\ P_{B_{1}}& = \sqrt{h}  (n^{a}\nabla_{a}A_{1})\  \ P_{B_{2}} = \sqrt{h}  (n^{a}\nabla_{a}A_{2}),,,,,,,,P_{B_{k}} = \sqrt{h}  (n^{a}\nabla_{a}A_{k}).\label{cmo}
\end{aligned}
\end{eqnarray}
Here we use $\Pi$ notation for denoting momenta for variables appearing from the gravity side $(N, N^i, h_{ij})$ and P for other auxiliary variables $(C, A_{1},A_{2}...A_{n}, B_{1},B_{2}...B_{n}).$ For further calculations, we follow the Dirac method for deriving the Hamiltonian for constraints system, in which variables with vanishing canonical momenta as primary constraints are denoted by the $\approx$ sign in the constraint space $\Gamma$.
Thus, from the eq.(\ref{cmo}), the primary constraints in our theory are,
\begin{equation*}
\Pi_{N}\approx 0 ,\ \ \Pi_{i}\approx 0 ,\ \ P_{Q}\approx 0.
\end{equation*}
After deriving all the canonical momenta and identifying primary constraints, we can write Hamiltonian density for the corresponding Lagrangian eq.(\ref{ADMaction}) as,
\begin{eqnarray}
\mathcal{H}=\Pi^{ij}\dot{h}_{ij}+p^C \dot{C} +\sum_{n = 1}^{k}( P_{B_{n}}\dot{B}_{n} + P_{A_{n}} \dot{A}_{n})-L
\end{eqnarray}
The time derivative of all variables to  their corresponding momenta can be extracted form eq.(\ref{cmo}).
After putting these values, the final  form of Hamiltonian  can be written as,
\begin{equation}
\mathcal{H}= N\mathcal{H}_{N}  + N^{i} \mathcal{H}_{i},
\end{equation}
with
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -\frac{1}{3\sqrt{h}} \Pi P_{C} + \frac{1}{6 \sqrt{h}} C P_{C}^{2} -\sqrt{h} C \mathcal{R}+ \sqrt{h}CQ  \\& - \sqrt{h}  G(Q) - F(Q)A_{1} +\sqrt{h}\sum_{n=1}^{k} A_{n+1}B_{n} +\sqrt{h}B_{k}H(Q) \\&  +\sum_{n = 1}^{k} \frac{P_{A_{n}}P_{B_{n}}}{\sqrt{h}} -\sqrt{h} h^{ij} \sum_{n = 1}^{k} \Big( \mathcal{D}_{i}B_{n} \mathcal{D}_{j}A_{n}\Big),\label{eq201}
\end{aligned}
\end{eqnarray}
and
\begin{equation}
\begin{split}
\mathcal{H}_{i}  = -2 h_{ik}\mathcal{D}_{j}\Pi^{kj} + \sum_{n = 1}^{k} \Big(P_{A_{n}}\mathcal{D}_{i}A_{n} + P_{B_{n}}\mathcal{D}_{i}B_{n} \Big) + P_{C}\mathcal{D}_{i}C.
\end{split}
\end{equation}
The $\Pi$ is defined as, $h_{ij}\Pi^{ij}=\Pi$.  The total Hamiltonian density with primary constraints in terms of Lagrange multipliers $ \lambda^{i},\lambda^{N} $ and $\lambda^{Q}$ takes the form,
\begin{equation}
\mathcal{H}_{tot} =  N\mathcal{H}_{N} + N^{i}\mathcal{H}_{i} +\lambda^{i}\Pi_{i}+\lambda^{N}\Pi_{N} + \lambda^{Q}P_{Q},
\end{equation}
and the total Hamiltonian is expressed as,
\begin{equation}
H_{total} = \int d^{3}x \Big( N \mathcal{H}_{N}  + N^{i} \mathcal{H}_{i} +\lambda^{i}\Pi_{i}+\lambda^{N}\Pi_{N} + \lambda^{Q}P_{Q} \Big),\label{eq;TH}
\end{equation}
 In Dirac method for constraints system, the constraints are classified according to their nature of time evolution. For $\Pi_{N}$ and $\Pi_{i}$, we get
\begin{eqnarray}
\dot{\Pi_{N}}=\{\Pi_{N},\mathcal{H}_{tot}\}=H_N,~~ \qquad \dot{\Pi_{i}}=\{\Pi_{i},\mathcal{H}_{tot}\}=H_i,
\end{eqnarray}
where $H_{N}$ and $H_{i}$ are secondary constraints, are known as Hamiltonian and diffeomorphism constraints respectively. On the constraint space $ \Gamma, $ the time evolution of constraints $H_N$ and $H_i$ vanishes weakly as,
\begin{equation}
\dot{\mathcal{H_N}}=\{\mathcal{H}_N,H_{tot}\}\approx0, \label{hn}
\end{equation}
and
   \begin{equation}
\dot{\mathcal{H}}_i=\{\mathcal{H}_i,H_{tot}\}\approx0. \label{hi}
    \end{equation}
Further, we evaluate time evolution of  primary constraint $P_Q$ with total Hamiltonian,
\begin{eqnarray}
\Xi_{Q} = \partial_{t}P_{Q} =\{ P_{Q}, \mathcal{H}_{tot}\} = N\sqrt{h}\Big\{-C +G'(Q) +F'(Q)A_{1} - H'(Q)B_{k}\Big\}\approx 0. \label{con1}
\end{eqnarray}
where, the dashed sign (') denotes the derivative with respect to Q, for example $G'(Q)=\frac{\partial G(Q)}{\partial Q}$. $\Xi_{Q}\approx 0$, $\Xi_{Q}$ act as secondary constraint corresponding to primary constraint $P_Q\approx 0$.  Now, we check time evolution of $\Xi_{Q}$ and demanding that it vanishes on constraint surface we obtain,
\begin{eqnarray}
\begin{aligned}
\dot{\Xi}_{Q} =& \{ \Xi_{Q},\mathcal{H}_{tot}\}=
N \Bigg[ \frac{1}{3}(\Pi -CP_{C}) + F'(Q)P_{B_{1}}
 -H'(Q)P_{A_{k}}\Bigg] \\& + \sqrt{h}\lambda^{Q}(G''(Q) +F''(Q)A_{1}-H''(Q)B_{k})\approx0. \label{aab}
 \end{aligned}
\end{eqnarray}
where the double dash (' ') show second order derivative with respect to $Q$.
This condition eq.(\ref{aab}) fixes the Lagrange multiplier $\lambda^{Q}$, and hence no tertiary constraint appears in this theory.\\  Now, we have identified all the primary and secondary constraints. Then we categorize
them into first and second-class constraints. For which, we  check the  Poisson bracket of  $P_Q$ and $\Xi_{Q}$,
\begin{equation}
\{P_Q,\Xi_{Q}\}\neq 0,
\end{equation}
and it implies that $P_Q$ and $\Xi_{Q}$ are second class constraints.
All other Poisson brackets are obtained as,
\begin{eqnarray}
\{ \Pi_{N}, \Pi_{i}\}= \{ \Pi_{N}, \Pi_{N}\} = \{ \Pi_{N}, \mathcal{H}_{N}\}=\{ \Pi_{N},\mathcal{H}_{i}\} \approx 0 ,\nonumber \\
 \{ \Pi_{i}, \Pi_{i}\} = \{ \Pi_{i},\mathcal{H}_{N}\} = \{ \Pi_{i},\mathcal{H}_{i}\} \approx 0 ,\nonumber \\
 \{ P_{Q}, P_{Q}\} = \{ P_{Q}, \Pi_{N}\} = \{ P_{Q}, \Pi_{i}\} \approx 0.
 \end{eqnarray}
In summary, we have  eight first class constraints $(\Pi_{N}, \Pi_{i}, \mathcal{H}_{N}, \mathcal{H}_{i})$ and two second class constraints $P_Q $ and $ G_Q.$
Now we can derive results for various choices of general function  $G(R)$, $ F(R) $, and $ H(R) $  from the above analysis.
\subsection{Analysis for Different Choices of \texorpdfstring{$G(R)$, $ F(R) $}{TEXT} and \texorpdfstring{$ H(R) $}{TEXT}}
Here we reiterate that the Ostrogradsky ghost is seen in the Hamiltonian by linear momenta terms. It can be observed from eq.(\ref{eq201}) that there are k terms of linear momenta of auxiliary variables $A_{k}$  and $B_{k}$. We also note that for the general function  $G(R)$, $ F(R) $, and $ H(R) $,  we obtain only two constraint equations. These constraints are not sufficient to remove all the linear momenta terms from eq.(\ref{eq201}). Now we analyze different cases.
\begin{itemize}
    \item For all our analysis we set  $C=G'(Q)$ in eq.(\ref{con1}. Then we obtain the form of constraint $\dot {\Xi}_{Q}$ as,
\begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q} =
N \Bigg[ F'(Q)P_{B_{1}}
 -H'(Q)P_{A_{k}}\Bigg] + \sqrt{h}\lambda^{Q}(F''(Q)A_{1}-H''(Q)B_{k})\approx0, \label{aab10}
\end{eqnarray}
\end{itemize}

\textbf{Case1}: If we choose $ F''(Q)=0$ and $ H''(Q)=0$, then the Lagrange multipliers remain undetermined. And these conditions hold only if $ F(Q)$ and $ H(Q)$ are linear functions of $Q,$ which provides a relation between
$ F(Q)$ and $ H(Q)$ i.e $ H(Q)=\alpha F(Q)+\textnormal{any constant}.$ For this case the eq. \ref{aab10} reads
\begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q}^{} = N \Bigg[  P_{B_{1}} -\alpha P_{A_{k}}\Bigg]\approx0.
\end{eqnarray}
 Due to undetermined nature of Lagrange multiplier, the time evolution  $\Xi_{Q}^{II}$ with Hamiltonian generate new constraint (tertiary constraint),
\begin{eqnarray}
\Xi_{Q}^{2I}=\dot{\Xi}_{Q}^{II} = N \Bigg[  A_{2} -\alpha B_{k-1}\Bigg]\approx0,
\end{eqnarray}
and the time evolution  $\Xi_{Q}^{2I}$ with Hamiltonian becomes,
\begin{eqnarray}
\Xi_{Q}^{2II}=\dot{\Xi}_{Q}^{} = N \Bigg[  P_{B_{2}} -\alpha P_{A_{k-1}}\Bigg]\approx0.
\end{eqnarray}
 Next we repeat the above procedure for n times which yields,
\begin{eqnarray}
\Xi_{Q}^{nI}=\dot{\Xi}_{Q}^{n-1} = N \Bigg[  A_{n} -\alpha B_{1}\Bigg]\approx0.\label{eq:op}
\end{eqnarray}
 Time evolution of  $\Xi_{Q}^{nI}$ eq.(\ref{eq:op}) becomes,
\begin{eqnarray}
\Xi_{Q}^{nII}=\dot{\Xi}_{Q}^{I} = N \Bigg[  P_{B_{n}} -\alpha P_{A_{1}}\Bigg]\approx0.
\end{eqnarray}
Now the chain of constraints ends and we have total $2n$ constraints  ($\Xi_{Q}^{},\Xi_{Q}^{II},\ \Xi_{Q}^{2I},\Xi_{Q}^{2II},\ .....,\ \Xi_{Q}^{nI},\Xi_{Q}^{nII}$).
These constraints relate $A_{1}$ to $ B_{n}$, $A_{2}$ to $B_{n-1}$...$A_{n}$ to $B_{1}$, and similarly relate momenta $ P_{B_{n}}$  to $ P_{A_{1}}$,  $P_{B_{n-1}}$ to $P_{A_{2}}$.....$ P_{B_{1}}$  to $ P_{A_{n}}.$ However for removing the linear momenta terms in Hamiltonian eq.(\ref{eq201}) relations between different momenta, i.e., $ P_{B_{1}}$  to $ P_{A_{1}}$,  $P_{B_{2}}$ to $P_{A_{2}}$.....$ P_{B_{n}}$  to $ P_{A_{n}}$ are required. Because of the absence of these one to one relation in this theory, we get $n$ ghosts.\\
For a special case like $k=1$, eq.(\ref{con1}) becomes,
\begin{eqnarray}
\Xi_{Q}=N\sqrt{h}\big\{\alpha A_{1}-B_{1}\big\}\approx0,\label{se:eq}
\end{eqnarray}
and its time evolution is,
 \begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q}^{} = N \Bigg[  P_{B_{1}} -\alpha P_{A_{1}}\Bigg]\approx0.\label{se1:eq}
\end{eqnarray}
 Eq.\ref{se1:eq}  relates $P_{A_{1}}$ to $P_{B_{1}}$. Consequently, $H_{N}$ in eq.(\ref{eq;TH}) reads,
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2}  -\sqrt{h} C \mathcal{R}+ \sqrt{h}CQ  \\& - \sqrt{h}  G(Q) + \frac{P_{A_{1}}^{2}}{\sqrt{h}} -\sqrt{h} h^{ij} \sum_{n = 1}^{k} \Big( \mathcal{D}_{i}A_{1} \mathcal{D}_{j}A_{1}\Big),
\end{aligned}
\end{eqnarray}
Here all the momenta terms are quadratic, and there is no place for Ostrogradsky ghost to appear.

\begin{itemize}
    \item \textbf{Case2}: Here we choose $ H(Q)=Q$ and $ F(Q)$ is not a linear function of Q. In this case then the  form of $\dot {\Xi}_{Q}$ is,
\begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q} = N \Bigg[ F'(Q) P_{B_{1}} -P_{A_{k}}\Bigg]  + \sqrt{h}\lambda^{Q} F''(Q)A_{1}\approx0.\label{eq27}
\end{eqnarray}
It can be observed from eq.(\ref{eq27}) that by fixing the Lagrange multiplier, we cannot get rid of all the linear momenta terms in the Hamiltonian. A model of this kind is formulated in\cite{Nojiri:2019dio} from the Lagrangian point of view, where authors draw the result that ghosts exit as the determinant of kinetic matrix takes negative values. The same conclusions from our Hamiltonian analysis for this particular model can be drawn here. \\

 \textbf{Case3}: Here we take $ F(Q)$=$ H(Q)$ for which the structure of $\Xi_{Q}^{II}$ becomes
    \begin{eqnarray}
    \Xi_{Q}^{II}=\dot{\Xi}_{Q} = NF'(Q) \Bigg[ P_{B_{1}} -P_{A_{k}}\Bigg] + \sqrt{h}\lambda^{Q}F''(Q)( A_{1}-B_{k})\approx0.\label{eq:aag}
    \end{eqnarray}
    In this case we can fix the Lagrange multiplier which in turn provides only two constraints. However these constraints are not sufficient to remove all the linear momenta term from the Hamiltonian eq.(\ref{eq201}).

    Now, the theory with $k=1$, we can remove the linear momenta term in the Hamiltonian by imposing the condition that $A_{1}$ and $B_{1}$ are  related to each other linearly i.e. $A_{1}=p B_{1},$ (p=constant). Under this condition no ghosts appear for $k=1$ case and the constraint eq.(\ref{con1}) becomes,
    \begin{eqnarray}
    \Xi_{Q} = \partial_{t}P_{Q} =\{ P_{Q}, \mathcal{H}_{tot}\} = N\sqrt{h}\Big\{F'(Q)A_{1}( 1- p)\Big\}\approx 0,
    \end{eqnarray}
    and its time evolution with Hamiltonian reads,
 \begin{eqnarray}
    \Xi_{Q}^{II}=\dot{\Xi}_{Q} = NF'(Q) P_{B_{1}}\Bigg[ 1-p \Bigg]+\sqrt{h}\lambda^{Q}F''(Q)A_{1}(1-p)\approx0.
    \end{eqnarray}

 For a particular value of $p$, i.e., $p=1,$ then $A_{1}= B_{1}$ and their corresponding momenta are equal $ P_{A_{1}} = P_{B_{1}}.$ This implies $ \Xi_{Q}=0.$ This is a exact relation and happens only for special case when $p=1.$ However for other values of $p$ we have to use usual procedure to show the nonexistence of ghosts. Here we conclude that for $k=1$ no Ostrogradsky ghost is present for the condition when auxiliary variables depend linearly on each other .
\end{itemize}
Next we consider $k=2$ and the structure of constraints takes the form,
\begin{eqnarray}
\Xi_{Q} = \partial_{t}P_{Q} =\{ P_{Q}, \mathcal{H}_{tot}\} = N\sqrt{h}\Big\{ F'(Q)(A_{1} -B_{2})\Big\}\approx 0.\label{eq:cca}
\end{eqnarray}
and  its evolution gives,
\begin{eqnarray}
\begin{aligned}
\dot{\Xi}_{Q} =& \{ \Xi_{Q},\mathcal{H}_{tot}\}=
N \Bigg[ F'(Q)(P_{B_{1}}
 -P_{A_{2}})\Bigg] + \sqrt{h}\lambda^{Q}(F''(Q)(A_{1}-B_{2})\approx0. \label{aac}
 \end{aligned}
\end{eqnarray}
Similar to the the analysis performed for $k=1$, the mathematical structure of constraints eq.(\ref{eq:cca}) and  eq.(\ref{aac} relate  $A_{1} $ to $B_{2}$, and $ P_{B_{1}}$ to $P_{A_{2}}$. On top of it a linear relation between $ P_{B_{1}}$ and $P_{A_{1}}$ is required to make the theory ghost free. But there is no way to achieve this for $k=2.$  Generalizing the above analysis for values of $k\geq 2,$ it can be shown that ghosts always appear.

For $k=2$, \cite{Nojiri:2019dio}, suggested a method for developing a ghost-free theory, we need to modify the Lagrangian. For clarity, we again reformulate the Hamiltonian formalism for $k=2$. Now, we start with an action\cite{Nojiri:2019dio},
 \begin{equation}
S=\int d^{4}x \sqrt{-g}\left[R-\frac{1}{2} F(R) \Box^{-2}F(R) \right].\label{eq:aat}
\end{equation}
 Following \cite{Nojiri:2019dio}, after introducing  $B_{1}=\frac{1}{2}\tilde{A}+\tilde{B}$ and  $B_{2}=\tilde{A}-2\tilde{B}$ in eq.(\ref{ass}), the  final scalar tensor form of eq.(\ref{eq:aat}) reads,

\begin{align}
S^{eqv} = \int d^{4}x\sqrt{-g} \bigg[ Q - F(Q)A_{1} + C\left( R -Q\right) -  \tilde{A}^2+A_{1}\Box \tilde{A} \bigg].
\end{align}
Despite these procedures, the Hamiltonian may still include linear momenta terms due to the presence of the term $A_{1}\Box \tilde{A}.$  Next, we adopt a method put forward by authors of ref.\cite{Nojiri:2019dio} where the kinetic term is modified to obtain a ghost-free action. Modified action takes the form,
\begin{eqnarray}
S_{mod} = S^{eqv}+ \int d^{4}x\sqrt{-g} \bigg[\frac{\beta}{2}A_{1}\Box A_{1}+\frac{\alpha}{2}\tilde{A}\Box \tilde{A} \bigg],
\end{eqnarray}
and after putting $S^{eqv}$,
\begin{align}
S_{mod} = \int d^{4}x\sqrt{-g} \bigg[ Q - F(Q)A_{1} + C\left( R -Q\right) -  \tilde{A}^2+A_{1}\Box \tilde{A}+\frac{\beta}{2}A_{1}\Box A_{1}+\frac{\alpha}{2}\tilde{A}\Box \tilde{A}].\label{q;mid}
\end{align}
 3+1 decomposition of $S_{mod}$ becomes,
\begin{eqnarray}
\begin{aligned}
S_{mod} =& \int d^{4}x\sqrt{-g} \bigg[ Q -F(Q)A_{1} + C\left( K_{ij}K^{ij} -K^{2} + \mathcal{R} -Q\right)  -2K \ (n^{a} \nabla_{a}C)-  \tilde{A}^2\\&+(n^{a}\nabla_{a}A_{1})(n^{b}\nabla_{b}\tilde{A})+\frac{\beta}{2}(n^{a}\nabla_{a}A_{1})(n^{b}\nabla_{b}A_{1})+\frac{\alpha}{2}(n^{a}\nabla_{a}\tilde{A})(n^{b}\nabla_{b}\tilde{A})- h^{ij}\mathcal{D}_{i}A_{1}\mathcal{D}_{j}\tilde{A} \\&-\frac{\beta}{2}h^{ij}\mathcal{D}_{i}A_{1}\mathcal{D}_{j}A_{}-\frac{\alpha}{2}h^{ij}\mathcal{D}_{i}\tilde{A}\mathcal{D}_{j}\tilde{A}+\mathcal{D}_{i}\mathcal{D}^{i}C\bigg].
\end{aligned}
\end{eqnarray}
Here, we show the canonical momenta related to $A_{1}$ and $\tilde{A}$ only,  and the canonical momenta relation for $C, N, N^{i}$ can be calculated similarly to the procedure used for previous cases of study. Thus,
\begin{eqnarray}
P_{\tilde{A}}=\frac{\partial L}{\partial \tilde{A}}=\sqrt{h}\left\{\left(n^{a}\nabla_{a}A_{1}\right)+\alpha \left(n^{a}\nabla_{a}\tilde{A}\right)\right\}\nonumber\\P_{A_{1}}=\frac{\partial L}{\partial A_{1}}=\sqrt{h}\left\{(n^{a}\nabla_{a}\tilde{A})+\beta (n^{a}\nabla_{a}A_{1})\right\}
\end{eqnarray}
After solving these equations for the $n^{a}\nabla_{a}\tilde{A}$ and $n^{a}\nabla_{a}A_{1}$ is,
\begin{eqnarray}
n^{a}\nabla_{a}\tilde{A}=\frac{1}{\sqrt{h}(\alpha\beta-1)}(\beta P_{\tilde{A}}- P_{A_{1}})\\
n^{a}\nabla_{a}A_{1}=\frac{1}{\sqrt{h}(\alpha\beta-1)}(\alpha P_{A_{1}}- P_{\tilde{A}})
\end{eqnarray}
The Hamiltonian of action eq.(\ref{q;mid}) is written as,
\begin{eqnarray}
\mathcal{H}=\Pi^{ij}\dot{h}_{ij}+p^C \dot{C} + P_{\tilde{A}} \dot{\tilde{A}}+ P_{A_{1}} \dot{A}_{1}-L.
\end{eqnarray}
After some simplification $\mathcal{H}$ is
\begin{equation}
\mathcal{H}= N\mathcal{H}_{N}  + N^{i} \mathcal{H}_{i},
\end{equation}
with
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -Q + F(Q)A_{1} +  \tilde{A}^2+ h^{ij}\mathcal{D}_{i}A_{1}\mathcal{D}_{j}\tilde{A} \\&+\frac{\beta}{2}h^{ij}\mathcal{D}_{i}A_{1}\mathcal{D}_{j}A_{1}+\frac{\alpha}{2}h^{ij}\mathcal{D}_{i}\tilde{A}\mathcal{D}_{j}\tilde{A}+\frac{1}{2\,\alpha\,\beta-2\,}(\beta P_{\tilde{A}}^{2}-2\,P_{\tilde{A}}\,P_{A_{1}}+\alpha P_{A_{1}}^{2})
\end{aligned}
\end{eqnarray}
and
\begin{equation}
\begin{split}
\mathcal{H}_{i}  = -2 h_{ik}\mathcal{D}_{j}\Pi^{kj} +  \Big(P_{\tilde{A}}\mathcal{D}_{i}\tilde{A} + P_{A_{1}}\mathcal{D}_{i}A_{1} \Big) + P_{C}\mathcal{D}_{i}C.
\end{split}
\end{equation}

As we can see from above Hamiltonian that linear terms of $P_{\tilde{A}}$ and $P_{A_{1}}$ appear. We can remove the linear momenta terms only if $\alpha\beta=1$, but the Hamiltonian is not finite with this choice. Therefore, the above method is not effective in removing linear terms of momenta from the Hamiltonian. In other words, a Hamiltonian formalism for $\alpha\beta=1$ is unphysical. However, we will now show that it is possible to find the viable parameter space where no ghosts appear after diagonalization. To do this, let us start with the kinetic part of the last term of $H_{N}$, which can take the form,
\begin{gather}
 \begin{bmatrix} P_{\tilde{A}} & P_{A_{1}}  \end{bmatrix}
  \begin{bmatrix}
  \alpha  & -1 \\
  -1 & \beta
   \end{bmatrix}
   \begin{bmatrix}
   P_{\tilde{A}} \\ P_{A_{1}}
   \end{bmatrix}
\end{gather}
Let $M= \begin{bmatrix}
  \alpha  & -1 \\
  -1 & \beta
   \end{bmatrix}$ and after diagnolisation, we get the kinetic matrix as
   \begin{equation}
 \left[ \begin {array}{cc} \beta/2+\alpha/2-1/2\,\sqrt {{\alpha}^{2}-2
\,\beta\,\alpha+{\beta}^{2}+4}&0\\ \noalign{\medskip}0&\beta/2+\alpha/
2+1/2\,\sqrt {{\alpha}^{2}-2\,\beta\,\alpha+{\beta}^{2}+4}\end {array}
 \right]
\end{equation}
Both the diagonal term should be positive for non-existence of ghosts we obtain the following conditions
\begin{eqnarray}
 \beta+\alpha> \sqrt {{\alpha}^{2}-2
\,\beta\,\alpha+{\beta}^{2}+4},
\end{eqnarray}
which holds only if
\begin{eqnarray}
\,\alpha\,\beta>1, \, \, \textnormal{and}\, \, \alpha+\beta>0.
\end{eqnarray} This matches with the results obtained in \cite{Nojiri:2019dio}.
Next, we will create a Hamiltonian for some generalized nonlocal gravity models.
\section{Hamiltonian Formalism for \texorpdfstring{$f(R,\Box^{-1}R,\Box^{-2}R...\Box^{-n}R)$}{TEXT} Gravity Model}
In this section, our aim is to perform Hamiltonian analysis  of the   model purposed in\cite{DeFelice:2014kma}. Here Lagrangian has a form as
\begin{eqnarray}
L=\sqrt{-g}\ f(R,\Box^{-1}R,\Box^{-2}R..........\Box^{-n}R),\label{ac4:eq}
\end{eqnarray}
where, f is general function of $\Box^{-n}R$ $(n=1,2,3...)$ and $n$ is finite. Then action eq.(\ref{ac4:eq})  satisfies  the following condition,
 \begin{equation}
     \frac{\partial L}{\partial R \partial (\Box^{-n}R})\neq 0.
 \end{equation}
By defining auxiliary variables $Q, A_{1},..A_{n}, B_{1},...B_{n}, C,$ the above non local action eq.(\ref{ac4:eq}) can be written in localized form,
\begin{eqnarray}
L=\sqrt{-g}\bigg[ f(Q,A_{1},A_{2},.......A_{n})+C(R-Q)+B_{1}(Q-\Box A_{1})\\\nonumber+B_{2}(A_{1}-\Box A_{2}),.........B_{n}(A_{n-1}-\Box A_{n})\bigg],
\end{eqnarray}
and further with straightforward calculation, transformed Lagrangian becomes,
\begin{eqnarray}
L=\sqrt{-g}\bigg[ f(Q,A_{1},A_{2},.......A_{n})+CR+Q( B_{1}-C)+g^{\mu\nu}(\partial_{\mu}B_{1}\partial_{\nu} A_{1}+\\\partial_{\mu}B_{2}\partial_{\nu}A_{2}......... \partial_{\mu}B_{n}\partial_{\nu}A_{n})\nonumber+B_{2}A_{1} +B_{3}A_{2}........B_{n}A_{n-1}\bigg].\label{red:eq}
\end{eqnarray}
Using  3+1 decomposition the eq.(\ref{red:eq}) reads,
\begin{eqnarray}
\begin{aligned}
S^{eqv}= \int d^{3}x N\sqrt{h} \bigg [f(Q,A_{1},A_{2},.......A_{n})  + C\left( K_{ij}K^{ij} -K^{2} + \mathcal{R} -Q\right) &\\ -2K \ (n^{a} \nabla_{a}C)+Q(B_{1}-C)+\mathcal{D}_{i}\mathcal{D}^{i}C  -  \sum_{k= 1}^{n-1}  B_{k} A_{k+1} &\\ + h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} - \sum_{n=1}^{k}  (n^{a}\nabla_{a}B_{n})(n^{b}\nabla_{b}A_{n}) \bigg]
\end{aligned}
\end{eqnarray}
Likewise here the canonical momenta with respect to set of variables  $(N, N^i, h_{ij}, Q, C, A_{L}, B_{L})$ are defined by the following relations,
\begin{eqnarray}
\begin{aligned}
\Pi_{N} & \approx 0, \ \ \ \ \
\Pi_{i} \approx 0,\ \ \
\Pi^{ij} = \sqrt{h} C( \mathcal{K}^{ij} -h^{ij}\mathcal{K} ) - \sqrt{h} h^{ij} (n^{a} \nabla_{a}C), \qquad \qquad \qquad \qquad   \\
P_{Q} & \approx  0 , \qquad \qquad
P_{\Phi} = -2\sqrt{h} \mathcal{K}, \qquad\\
P_{A_{1}} &= \sqrt{h} (n^{a}\nabla_{a}B_{1}), \ \ P_{A_{2}} = (n^{a}\nabla_{a}B_{2}) \  ,,,,,,,\ P_{A_{n}}=\sqrt{h} (n^{a}\nabla_{a}B_{n})
\\ P_{B_{1}}& = \sqrt{h}  (n^{a}\nabla_{a}A_{1})\  \ P_{B_{2}} = \sqrt{h}  (n^{a}\nabla_{a}A_{2}),,,,,,,,P_{B_{n}} = \sqrt{h}  (n^{a}\nabla_{a}A_{n}).\label{Con:eq}
\end{aligned}
\end{eqnarray}
Here, the primary constraints are $\Pi_{N} \approx 0,\Pi_{i} \approx 0, P_{Q}\approx 0$. The final form of Hamiltonian is given as,
\begin{equation}
\mathcal{H}= N\mathcal{H}_{N}  + N^{i} \mathcal{H}_{i,}
\end{equation}
with
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -\frac{1}{3\sqrt{h}} \Pi P_{C} + \frac{1}{6 \sqrt{h}} C P_{C}^{2} -\sqrt{h} C \mathcal{R} \\& - \sqrt{h}  f(Q,A_{1},A_{2},.......A_{n})-Q(B_{1}-C)-\mathcal{D}_{i}\mathcal{D}^{i}C\\&  +  \sum_{k= 1}^{n-1}  B_{k} A_{k+1}  - h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} + \sum_{n=1}^{k}  P_{B_{n}}P_{A_{n}}, \label{HN:eq}
\end{aligned}
\end{eqnarray}
and
\begin{equation}
\begin{split}
\mathcal{H}_{i}  = -2 h_{ik}\mathcal{D}_{j}\Pi^{kj} + \sum_{n = 1}^{k} \Big(P_{A_{n}}\mathcal{D}_{i}A_{n} + P_{B_{n}}\mathcal{D}_{i}B_{n} \Big) + P_{\Phi}\mathcal{D}_{i}\Phi.
\end{split}
\end{equation}
The behaviour of primary constraints $\Pi_{N} \approx 0,
\Pi_{i} \approx 0$ is analogous to analysis discussed in \textsection III. Here we mainly emphasize on  $P_{Q}$ and its evolution with total Hamiltonian gives rise to,
\begin{eqnarray}
\Xi_{Q} = \partial_{t}P_{Q} =\{ P_{Q}, \mathcal{H}_{tot}\} = N\sqrt{h}\Big\{B_{1}-C -f'\Big\}\approx 0. \label{CE:eq}
\end{eqnarray}
where $f'$ denotes derivative of $f$ with respect to to Q. Further the  evolution of $\Xi_{Q}$ yields,
\begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q} = N \Bigg[\frac{1}{3}(\Pi-C P_{C}) +P_{A_{1}} -\frac{\partial f'}{\partial A_{1}}P_{B_{1}}+\frac{\partial f'}{\partial A_{2}}P_{B_{2}}+....\frac{\partial f'}{\partial A_{j}}P_{B_{j}} \\\nonumber+.... \frac{\partial f'}{\partial A_{n}}P_{B_{n}}\Bigg]  + \sqrt{h}\lambda^{Q}( f'')\approx0.\label{CE2:eq}
\end{eqnarray}
Using $\Xi_{Q}^{II}$ Lagrange multiplier $\lambda^{Q}$ can be evaluated which   ensures appearance of no further constraints in this theory. Now, we have found all the constraints. The constraints eq.(\ref{CE:eq}) shows that any field $A_{n}$  can be written in terms of the other fields,
\begin{eqnarray}
A_{n}=A_{n}(Q,A_{j},C-B_{1}).
\end{eqnarray}
After taking time derivatives of $A_{n},$ we get
\begin{eqnarray}
\dot{A_{n}}=\frac{\partial A_{n}}{\partial Q} \dot{Q}+\frac{\partial A_{n}}{\partial A_{j}} \dot{A_{j}}+\frac{\partial A_{n}}{\partial C} \dot{C}-\frac{\partial A_{n}}{\partial B_{1}}\dot{B_{1}}\label{eq:yt}
\end{eqnarray}
Replacing the time derivative terms of eq.(\ref{eq:yt})  by their corresponding momenta and ignoring the spatial derivative terms, then we obtain,
\begin{eqnarray}
P_{B_{n}}=\frac{\partial A_{n}}{\partial Q} \dot{Q}+\frac{\partial A_{n}}{\partial A_{j}} P_{B_{j}}+\frac{\partial A_{n}}{\partial C} \frac{1}{3}(\Pi-C P_{C})-\frac{\partial A_{n}}{\partial B_{1}} P_{A_{1}}\label{SE:eq}
\end{eqnarray}
Whereas the time derivative of Q can be fixed by the equation $\dot{Q}=\frac{\partial H}{\partial P_{Q}}=\lambda_{Q}.$ Then eq.(\ref{SE:eq}),
yields,
\begin{eqnarray}
P_{B_{n}}=\frac{\partial A_{n}}{\partial Q} \lambda_{Q}+\frac{\partial A_{n}}{\partial A_{j}} P_{B_{j}}+\frac{\partial A_{n}}{\partial C} \frac{1}{3}(\Pi-C P_{C})-\frac{\partial A_{n}}{\partial B_{1}} P_{A_{1}}\label{SE:eq1}
\end{eqnarray}
Finally we express $P_{B_{n}}$ in terms of other variables. After putting $P_{B_{n}}$ in eq.(\ref{CE2:eq}), we get
\begin{eqnarray}
\Xi_{Q}^{II} = N \Bigg[\frac{1}{3}(\Pi-C P_{C})\left(1+ \frac{\partial f'}{\partial A_{n}}\frac{\partial A_{n}}{\partial C}\right) +P_{A_{1}}\left(1- \frac{\partial f'}{\partial A_{n}}\frac{\partial A_{n}}{\partial B_{1}}\right) \nonumber\\-\left(\frac{\partial f'}{\partial A_{1}}+ \frac{\partial f'}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{1}}\right)P_{B_{1}}+....+\left(\frac{\partial f'}{\partial A_{j}}+ \frac{\partial f'}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{j}}\right)P_{B_{j}} \\\nonumber+.... \left(\frac{\partial f'}{\partial A_{n-1}}+ \frac{\partial f'}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{n-1}}\right)P_{B_{n-1}}\Bigg]  + \sqrt{h}\lambda^{Q}\left( f''+N\frac{\partial A_{n}}{\partial Q} \frac{\partial f'}{\partial A_{n}}\right)\approx0.
\end{eqnarray}
This constraint $\Xi_{Q}^{II}\approx 0$ has a solution of the form
\begin{eqnarray}
\frac{\partial A_{n}}{\partial C}=& -\frac{1}{ \frac{\partial f'}{\partial A_{n}}}=-\frac{1}{f_{,QA_{n}}}\label{eq:75}
\end{eqnarray}
\begin{eqnarray}
 \frac{\partial A_{n}}{\partial A_{j}}=&-\frac{\frac{\partial f'}{\partial A_{j}}}{\frac{\partial f'(Q)}{\partial A_{n}}}=-\frac{f_{,QA_{j}}}{f_{,QA_{n}}}\\
 \frac{\partial A_{n}}{\partial B_{1}}=&-\frac{1}{ \frac{\partial f'}{\partial A_{n}}}=\frac{1}{f_{,QA_{n}}}\\
N\frac{\partial A_{n}}{\partial Q} =&\ \ \frac{f''}{\frac{\partial f'}{\partial A_{n}}}=\frac{f_{,QQ}}{f_{,QA_{n}}}\label{eq:as}
\end{eqnarray}
Taking the result from eq.(\ref{eq:75}-\ref{eq:as}) then the eq.(\ref{SE:eq1})  becomes,
\begin{eqnarray}
P_{B_{n}}=\frac{f_{,QQ}}{f_{,QA_{n}}} \lambda_{Q}+-\frac{f_{,QA_{j}}}{f_{,QA_{n}}} P_{B_{j}}-\frac{1}{f_{,QA_{n}}} \frac{1}{3}(\Pi-C P_{C})-\frac{1}{f_{,QA_{n}}} P_{A_{1}}
\end{eqnarray}
Replacing $P_{B_{n}}$ by the above expression  in eq.(\ref{HN:eq}), we obtain
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -\frac{1}{3\sqrt{h}} \Pi P_{C} + \frac{1}{6 \sqrt{h}} C P_{C}^{2} -\sqrt{h} C \mathcal{R} \\& - \sqrt{h}  f(Q,A_{1},A_{2},.......A_{n})-Q(B_{1}-C)-\mathcal{D}_{i}\mathcal{D}^{i}C\\&  +  \sum_{k= 1}^{n-1}  B_{k} A_{k+1}  - h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} + \sum_{k=1}^{n-1}  P_{B_{n}}P_{A_{n}}\\& +P_{A_{n}}\left(\frac{f_{,QQ}}{f_{,QA_{n}}} \lambda_{Q}+-\frac{f_{,QA_{j}}}{f_{,QA_{n}}} P_{B_{j}}-\frac{1}{f_{,QA_{n}}} \frac{1}{3}(\Pi-C P_{C})-\frac{1}{f_{,QA_{n}}} P_{A_{1}}\right),
\end{aligned}
\end{eqnarray}

Finally, we arrive at the final form of Hamiltonian after exhausting all the constraints. Various linear momenta terms still exist in this Hamiltonian, which indicates the presence of the Ostrogradsky ghost in this theory.

\subsection{Hamiltonian Formalism for \texorpdfstring{$Rf(R,\Box^{-1}R,\Box^{-2}R...\Box^{-n}R)$}{TEXT}}
In this section, we follow the analysis of previous section for deriving Hamiltonian of the Lagrangian in the action,
\begin{eqnarray}
S=\int d^{3}x \sqrt{-g}\ R f(\Box^{-1}R,\Box^{-2}R..........\Box^{-n}R), \ \ \ \textnormal{with} \ \ \  \ n \neq +\infty, \label{Ac:eq3}
\end{eqnarray}
and its scalar tensor form looks,
\begin{eqnarray}
S=\int d^{3}x\sqrt{-g}\bigg[Q f(A_{1},A_{2},.......A_{n})+C(R-Q)+B_{1}(Q-\Box A_{1})\\\nonumber+B_{2}(A_{1}-\Box A_{2}),.........B_{n}(A_{n-1}-\Box A_{n})\bigg].
\end{eqnarray}
Action eq.(\ref{Ac:eq3}) in a expanded form is,
\begin{eqnarray}
S=\int d^{3}x \sqrt{-g}\bigg[Q f(A_{1},A_{2},.......A_{n})+CR+Q( B_{1}-C)+g^{\mu\nu}(\partial_{\mu}B_{1}\partial_{\nu} A_{1}+\\\partial_{\mu}B_{2}\partial_{\nu}A_{2}......... \partial_{\mu}B_{n}\partial_{\nu}A_{n})\nonumber+B_{2}A_{1} +B_{3}A_{2}........B_{n}A_{n-1}\bigg].
\end{eqnarray}
The above action in 3+1 decomposition can be written as,
\begin{eqnarray}
\begin{aligned}
S^{eqv}= \int d^{3}x N\sqrt{h} \bigg [Q f(A_{1},A_{2},.......A_{n})  + C\left( K_{ij}K^{ij} -K^{2} + \mathcal{R} -Q\right) &\\ -2K \ (n^{a} \nabla_{a}C)+Q(B_{1}-C)+\mathcal{D}_{i}\mathcal{D}^{i}C  -  \sum_{k= 1}^{n-1}  B_{k} A_{k+1} &\\ + h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} - \sum_{n=1}^{k}  (n^{a}\nabla_{a}B_{n})(n^{b}\nabla_{b}A_{n}) \bigg].
\end{aligned}
\end{eqnarray}

 Here canonical momenta with respect to set  $(N, N^i, h_{ij}, Q, C, A_{L}, B_{L})$ are,
\begin{eqnarray}
\begin{aligned}
\Pi_{N} & \approx 0, \ \ \ \ \
\Pi_{i} \approx 0,\ \ \
\Pi^{ij} = \sqrt{h} C( \mathcal{K}^{ij} -h^{ij}\mathcal{K} ) - \sqrt{h} h^{ij} (n^{a} \nabla_{a}C), \qquad \qquad \qquad \qquad   \\
P_{Q} & \approx  0 , \qquad \qquad
P_{\Phi} = -2\sqrt{h} \mathcal{K}, \qquad\\
P_{A_{1}} &= \sqrt{h} (n^{a}\nabla_{a}B_{1}), \ \ P_{A_{2}} = (n^{a}\nabla_{a}B_{2}) \  ,,,,,,,\ P_{A_{k}}=\sqrt{h} (n^{a}\nabla_{a}B_{k})
\\ P_{B_{1}}& = \sqrt{h}  (n^{a}\nabla_{a}A_{1})\  \ P_{B_{2}} = \sqrt{h}  (n^{a}\nabla_{a}A_{2}),,,,,,,,P_{B_{k}} = \sqrt{h}  (n^{a}\nabla_{a}A_{k}).
\end{aligned}
\end{eqnarray}
After some simplification Hamiltonian takes a form,
\begin{equation}
\mathcal{H}= N\mathcal{H}_{N}  + N^{i} \mathcal{H}_{i,} \label{eq:dwa}
\end{equation}
with
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -\frac{1}{3\sqrt{h}} \Pi P_{C} + \frac{1}{6 \sqrt{h}} C P_{C}^{2} -\sqrt{h} C \mathcal{R} \\& - \sqrt{h}   Qf(A_{1},A_{2},.......A_{n})-Q(B_{1}-C)-\mathcal{D}_{i}\mathcal{D}^{i}C\\&  +  \sum_{k= 1}^{n-1}  B_{k} A_{k+1}  - h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} + \sum_{n=1}^{k}  P_{B_{n}}P_{A_{n}},
\end{aligned}
\end{eqnarray}
and
\begin{equation}
\begin{split}
\mathcal{H}_{i}  = -2 h_{ik}\mathcal{D}_{j}\Pi^{kj} + \sum_{n = 1}^{k} \Big(P_{A_{n}}\mathcal{D}_{i}A_{n} + P_{B_{n}}\mathcal{D}_{i}B_{n} \Big) + P_{\Phi}\mathcal{D}_{i}\Phi.
\end{split}
\end{equation}
Here the constraint is
\begin{eqnarray}
\Xi_{Q} = \partial_{t}P_{Q} =\{ P_{Q}, \mathcal{H}_{tot}\} = N\sqrt{h}\Big\{B_{1}-C -f(A_{1},A_{2}......A_{n}\Big\}\approx 0. \label{CE:eq1}
\end{eqnarray}
and its evolution yields,
\begin{eqnarray}
\Xi_{Q}^{II}=\dot{\Xi}_{Q} = N \Bigg[\frac{1}{3}(\Pi-C P_{C}) +P_{A_{1}} -\frac{\partial f}{\partial A_{1}}P_{B_{1}}+\frac{\partial f}{\partial A_{2}}P_{B_{2}}+....\frac{\partial f}{\partial A_{j}}P_{B_{j}} \\\nonumber+.... \frac{\partial f}{\partial A_{n}}P_{B_{n}}\Bigg]\approx0.\label{CE:eq2}
\end{eqnarray}

Equivalently, the eq.(\ref{CE:eq2}) can be written as,
\begin{eqnarray}
\Xi_{Q}^{2I}=N \Bigg[\frac{1}{3}(\Pi-C P_{C}) + -\frac{\partial f}{\partial A_{1}}P_{B_{1}}+\zeta\Bigg]\approx0.\label{CE:eq21}
\end{eqnarray}
where,$$\zeta=\frac{\partial f}{\partial A_{2}}P_{B_{2}}+....\frac{\partial f}{\partial A_{j}}P_{B_{j}} \\\nonumber+.... \frac{\partial f}{\partial A_{n}}P_{B_{n}}$$
$\Xi_{Q}^{2I}\approx0$ acts as a tertiary constraint and its time evolution is,
\begin{eqnarray}
\Xi_{Q}^{2II}=N \bigg[\frac{-\Pi^{ij}}{\sqrt{h}} h_{ik} h_{jl} \Pi^{kl} +\frac{1}{3\sqrt{h} C} \Pi^{2} +\frac{1}{3\sqrt{h}} \Pi P_{C} - \frac{1}{6 \sqrt{h}} C P_{C}^{2} \nonumber \\+\sqrt{h} C \mathcal{R}  - \sqrt{h}C Q -\frac{\partial f}{\partial A_{1}} Q +P_{B_{1}}\{\frac{\partial f}{\partial A_{1}},\mathcal{H}\} +\{\zeta,\mathcal{H}\} \bigg]\approx0.\label{same2}
\end{eqnarray}
Further the time evolution of $\Xi_{Q}^{2II}$ fixes the Lagrange multiplier $\lambda_{Q}.$  Now, all the Lagrange multipliers are fixed, ensuring no further constraint appears in this theory.
Therefore, we have four secondary constraints in our theory which are,
\begin{eqnarray}
 \Xi_{Q}\approx0,\qquad \Xi_{Q}^{II}\approx0,\qquad \Xi_{Q}^{2I}\approx0,\qquad \Xi_{Q}^{2II}\approx0. \label{sc}
\end{eqnarray}
The constraint equation \ref{CE:eq1} field $A_{n}$ is related to other $A_{n}$ fields by the relation
\begin{eqnarray}
A_{n}=A_{n}(A_{j},C-B_{1}).
\end{eqnarray}
By taking the time derivatives of $A_{n},$ we obtain
\begin{eqnarray}
\dot{A_{n}}=\frac{\partial A_{n}}{\partial A_{j}} \dot{A_{j}}+\frac{\partial A_{n}}{\partial C} \dot{C}-\frac{\partial A_{n}}{\partial B_{1}}\dot{B_{1}}\label{kl}
\end{eqnarray}
Eq.(\ref{kl})  can be written in terms of momenta,
\begin{eqnarray}
P_{B_{n}}=\frac{\partial A_{n}}{\partial A_{j}} P_{B_{j}}+\frac{\partial A_{n}}{\partial C} \frac{1}{3}(\Pi-C P_{C})-\frac{\partial A_{n}}{\partial B_{1}} P_{A_{1}}. \label{eq:ss}
\end{eqnarray}
 Replacing $P_{B_{n}}$ in eq.(\ref{CE:eq2}) by using eq.(\ref{eq:ss}), we get
\begin{eqnarray}
\Xi_{Q}^{I} = N \Bigg[\frac{1}{3}(\Pi-C P_{C})\left(1+ \frac{\partial f}{\partial A_{n}}\frac{\partial A_{n}}{\partial C}\right) +P_{A_{1}}\left(1- \frac{\partial f}{\partial A_{n}}\frac{\partial A_{n}}{\partial B_{1}}\right) \nonumber\\-\left(\frac{\partial f}{\partial A_{1}}+ \frac{\partial f}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{1}}\right) \label{eq:swa}P_{B_{1}}+....+\left(\frac{\partial f}{\partial A_{j}}+ \frac{\partial f}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{j}}\right)P_{B_{j}} \\\nonumber+.... \left(\frac{\partial f}{\partial A_{n-1}}+ \frac{\partial f}{\partial A_{n}}\frac{\partial A_{n}}{\partial A_{n-1}}\right)P_{B_{n-1}}\Bigg] \approx0.
\end{eqnarray}
This constraint has a solution of the form,
\begin{eqnarray}
\frac{\partial A_{n}}{\partial C}=& -\frac{1}{ \frac{\partial f(Q)}{\partial A_{n}}}=-\frac{1}{f_{,A_{n}},}\label{eq:80}\\
 \frac{\partial A_{n}}{\partial A_{j}}=&-\frac{\frac{\partial f(Q)}{\partial A_{j}}}{\frac{\partial f'(Q)}{\partial A_{n}}}=-\frac{f_{,A_{j}}}{f_{,A_{n}}},\\
 \frac{\partial A_{n}}{\partial B_{1}}=&-\frac{1}{ \frac{\partial f(Q)}{\partial A_{n}}}=\frac{1}{f_{,A_{n}}}.\label{eq:81}
\end{eqnarray}
Taking the result from eq.(\ref{eq:80}-\ref{eq:81}) then the eq.(\ref{eq:swa}) becomes,
\begin{eqnarray}
P_{B_{n}}=-\frac{f_{,QA_{j}}}{f_{,QA_{n}}} P_{B_{j}}-\frac{1}{f_{,QA_{n}}} \frac{1}{3}(\Pi-C P_{C})+-\frac{1}{f_{,QA_{n}}} P_{A_{1}}. \label{eq:awa}
\end{eqnarray}
After putting eq.(\ref{eq:awa}) into Hamiltonian eq.(\ref{eq:dwa}) takes the form,
\begin{eqnarray}
\begin{aligned}
\mathcal{H}_{N} & = \frac{1}{\sqrt{h} C} \Pi^{ij} h_{ik} h_{jl} \Pi^{kl} -\frac{1}{3\sqrt{h} C} \Pi^{2} -\frac{1}{3\sqrt{h}} \Pi P_{C} + \frac{1}{6 \sqrt{h}} C P_{C}^{2} -\sqrt{h} C \mathcal{R} \\& - \sqrt{h}  f(Q,A_{1},A_{2},.......A_{n})-Q(B_{1}-C)-\mathcal{D}_{i}\mathcal{D}^{i}C\\&  +  \sum_{k= 1}^{n-1}  B_{k} A_{k+1}  - h^{ij} \sum_{n=1}^{k}  \mathcal{D}_{i}B_{n}\mathcal{D}_{j}A_{n} + \sum_{k=1}^{n-1}  P_{B_{n}}P_{A_{n}}\\& +P_{A_{n}}\left(-\frac{f_{,QA_{j}}}{f_{,QA_{n}}} P_{B_{j}}-\frac{1}{f_{,QA_{n}}} \frac{1}{3}(\Pi-C P_{C})-\frac{1}{f_{,QA_{n}}} P_{A_{1}}\right),
\end{aligned}
\end{eqnarray}
The final Hamiltonian form can then be obtained after exhausting all the constraints. This Hamiltonian still contains numerous linear momenta terms, which indicate the presence of the Ostrogradsky ghost.

\section{Conclusion}

There has been investigation of the instability issue in nonlocal theory, arising due to the presence of higher derivatives in the Lagrangian. Studies related to this show that it is possible to overcome the issue of  Ostrogradsky instability in infinite derivative gravity models,\cite{Biswas:2013cha, Biswas:2011ar, Conroy:2014eja, Joshi:2019cyk}.

Further, the nonlocal F(R) gravity model is investigated in  \cite{Nojiri:2019dio} and demonstrates under what conditions the issue of  Ostrogradsky instability can be resolved simply by analyzing the kinetic matrix of the localized action. The final form of a kinetic matrix is obtained by relating the Lagrange multipliers using equations of motion. In order to obtain a consistent theory, it is necessary to examine the same formalism(obtaining no ghost conditions) from a Hamiltonian point of view. It is well known that Hamiltonian of non-degenerate higher derivative Lagrangian contains linear momenta terms, indicating the presence of Ostrogradsky ghost, and possible way to tackle ghost issue in Hamiltonian, first, to find any hidden condition(if any) that can relate both of linear momenta term, and second by adding a constraint. Here, we use the first method.

Our analysis starts with non-local action eq.(\ref{action}) of general functions of Ricci scalar. We derive the Hamiltonian and constraint for eq.(\ref{action}). By analyzing the constraint structure, we showe the theory that the ghost issue cannot be resolved with $F(R)\neq H(R)$. Next, we examine for the case when $F(R)$ and $H(R)$ are linear functions of Ricci scalar and notice that for $k\geq 2$ ghost issue is always present. However, for the case $k=1$, we show that no ghost appears.

 Further we analyze the case when $F(R)=H(R)$ and $G(R)=R$ for two choices $k=1$ and $k=2.$ We can obtain a ghost free theory for $k=1,$  whereas  ghost issue is always present for $k\geq 2.$
 However in \cite{Nojiri:2019dio} it is shown that that by modifying the kinetic term in a particular fashion such that no ghost appears in later choice. When we analyze the  Hamiltonian in this case we get the same ghost-free condition as in \cite{Nojiri:2019dio}.

 For completeness, we also study the Hamiltonian of most generalised non local models containing only Ricci scalars and find the structure of constraints and nature of ghosts in general.

Our model can be generalized for the more complicated form of the $F(R)$ gravity model.

redpajama_set_name: RedPajamaArXiv


The UK is known for a lot of things. It has one of the most hustling and bustling cities in the world. Never far from the sea, the United Kingdom also offers beauty in all forms of nature. Rolling hills, vast farmland, pristine lakes, valleys, greenery, and mountains, you're never far from a peaceful, scenic view. Evident by royal palaces, historically, people in the UK value lush gardens and well-kept gardens. One of the best ways to keep up with your senses is to find retreat locations to spend a week or two. Enjoy a walk in the garden to spot exotic plants and blooming, colorful flowers. Grab a cup of afternoon tea and a dessert or appetizer while taking in the calm, fresh air. The UK has a list of incredible healing spaces for you to choose from. Experience all kinds of wellness from a secluded and stunning countryside retreat venue in Shropshire. Explore the 17th-century-old farmhouse in Berkshire. Detoxify and refresh in a scenic and tranquil retreat house in Scotland. Bask in the sun as you have your wildlife adventure in Devon. Meditate by the lake in the morning, drink your tea in the evening. Browse our retreat centers in the United Kingdom!


text:
'Harold & Kumar' Star John Cho Lands Lead Role on ABC Pilot 'Selfie'
The actor joins the comedy project from "Suburgatory" creator Emily Kapnek
Jethro Nededog | March 13, 2014 @ 10:45 AM	Last Updated: August 4, 2014 @ 10:58 AM
ABC comedy pilot "Selfie" has completed its casting of potential series regulars.
John Cho has joined in the male lead role of Henry, TheWrap has learned.
The character is described as a self-assured, successful marketing expert, Henry is a different breed from today's social media-addicted society. As a challenge, he decides to "re-market" his coworker Eliza (Karen Gillan) — a social media star with a bad reputation.
Also read: 'Doctor Who's' Karen Gillan Lands Lead on ABC Pilot 'Selfie'
The comedy project follows Eliza's social makeover after suffering a very public and humiliating breakup and becoming the subject of a viral video.
Cho is best known as the titular Harold in the "Harold & Kumar" movie franchise. He also starred in "Star Trek Into Darkness" (and will next be seen in the franchise's next film) and "Total Recall." Cho recently played a recurring role on Fox's "Sleepy Hollow." His other television credits include "Go On," "FlashForward" and "Ugly Betty."
In addition to Gillian, Cho joins already announced castmembers David Harewood, Allyn Rachel, Tim Peper and Da'Vine Joy Randolph.
Also read: ABC's 'Suburgatory' Boss Emily Kapnek Signs Overall Deal With Warner Bros. TV
Written and executive-produced by "Suburgatory" creator Emily Kapnek under her overall deal with Warner Bros., the half-hour, single camera pilot will be produced by Warner Bros. TV in association with Piece of Pie Productions.
UTA and firm Morris Yorn Barnes Levine Krintzman Rubenstein Kohner & Gellman represent Cho.
Anne Sweeney's Move to Directing: A Top TV Director on Her Odds of Success
ABC Countersues Prospect Park Over 'All My Children,' 'One Life to Live' Dispute
By Tim Kenneally | March 12, 2014 @ 2:43 PM

redpajama_set_name: RedPajamaCommonCrawl


Aristolochia disticha was originally described based on a specimen collected by R. Spruce in Santarem (Pará, Brazil). Unfortunately, the specimen lacks flowers and fruits, although it has four slender, ramiflorous inflorescences that distinguish the species. Here we fully describe the species, based on the holotype and additional flowering and fruiting specimens that allow us to report a range expansion to the states of Acre, Amazonas, Bahia, Paraíba, and Rio Grande do Norte, Brazil, and also to Venezuela, French Guiana, Surinam, Peru, and Bolivia. Finally, we reduce the recently described A. setulosa to the synonymy of A. disticha.


text:
E-archiving, Electronic Invoicing, Global Tax Determination, LatAm Compliance, Tax Compliance, Value Added Tax
New Global VAT Compliance Report Provides Guidance for Multinationals
Christiaan Van Der Valk
Companies are dealing with a fundamental shift in the way they do business with trading partners. In a rapidly increasing number of countries, there's a third party inserting itself into every transaction. It's the government, and it's wedging its way into every order a company ships or receives.
In an effort to close a massive tax gap and enforce value-added tax (VAT) regulations, tax authorities worldwide are moving to continuous transaction controls (CTCs), including clearance-model electronic invoicing, in which the government actually becomes a real-time third trading partner in business-to-business transactions. What began in Latin America about a decade ago has spread to Europe and is moving into Asia.
Multinationals are rethinking VAT compliance while migrating ERP
As a result, multinational companies are having to rethink their tax compliance strategies in order not only to avoid fines and penalties in a growing number of countries but also simply to keep doing business in those places. Failure to comply with CTCs can lead to more than just financial punishments; in can sour relationships with customers and suppliers, and even lead to the shuddering of the business in some countries.
All of this is happening against a backdrop of large companies moving to new enterprise resource planning (ERP) systems with the goal of taking advantage of new technological capabilities on the path to digital transformation. While tax compliance might not seem like a major issue for ERP migration, attempting to cobble together fragmented point solutions across the world to deal with tax mandates can, in fact, halt or even reverse a company's digital transformation. Multinationals need a centralized, automated tax solution that integrates with ERP.
Tax compliance has to be at the forefront of ERP migration
Tax strategy, then, has to be at the forefront of ERP migrations, and managing mandates for electronic invoicing, electronic archiving and tax determination has to be at the forefront of developing a tax strategy. With more and more countries shifting to CTCs to enforce VAT policies, the complexity involved in keeping up with tax mandates worldwide is not only staggeringly challenging, it's also fraught with risk.
In the 11th edition of "Trends in Continuous Global VAT Compliance," industry experts explain, with both high-level and extremely granular analyses, how multinationals can manage CTCs and tax compliance without derailing digital transformation. In fact, this comprehensive report, unique in the industry, details how organizations can actually use VAT compliance to accelerate ERP migration and shift to adopting new, data-driven processes.
The new report provides guidance on dealing with global VAT mandates
The 2020 report expands on previous efforts by including detailed information about VAT reporting worldwide as well as delving trends in e-invoicing. The report includes:
A complete explanation of how the global trend toward continuous transaction controls affects businesses now;
Strategic recommendations for protecting and accelerating digital transformation while complying with changing tax mandates;
A comprehensive, country-by-country summary of e-invoicing, e-archiving and VAT determination regulations.
The global move to CTCs shows no signs of slowing. Companies that prepare themselves now to meet the current and forthcoming challenges of VAT compliance will mitigate risk and gain an advantage over competitors that choose to remain behind the curve.
Download "Trends in Continuous Global VAT Compliance."
Stay up to date with the latest tax and compliance updates that may impact your business.
Christiaan Van Der Valk is vice president, strategy. Elected a World Economic Forum Global Leader for Tomorrow in 2000, Christiaan is an internationally recognized voice on e-business strategy, law, policy, best practice and commercial issues. Formerly co-founder and president of Trustweaver (acquired by Sovos), Christiaan also holds long-standing leadership roles at the International Chamber of Commerce (ICC) and the European E-invoicing Service Providers Association (EESPA). Over the past 20 years, he has presented at and authored key papers for international meetings at the Organisation for Economic Cooperation and Development (OECD), the Asia Europe Meeting, World Trade Organization and several other UN agencies. Christiaan earned his Master of Laws degree from Erasmus Universiteit Rotterdam.
Latin America VAT & Fiscal Reporting
Sovos Acquires Taxweb, Extends Tax Determination Capabilities in World's Most Challenging Compliance Landscape
Earlier this month Sovos announced its second acquisition of 2020, completing our solution for Brazil with an unparalleled offering that solves tax compliance in the place where it is most challenging to do so. Too many companies doing business in Brazil have been burdened by managing multiple point solutions for continuous transaction controls (CTCs), tax […]
EMEA VAT & Fiscal Reporting
How the EU-UK Trade and Co-operation Agreement Affects Business
The EU-UK Trade and Co-operation Agreement (TCA) was finally agreed on 24 December 2020, a week before the end of the transition period. The TCA has been fully implemented into UK law but remains provisional as it needs to be ratified in the European Parliament. It is therefore being applied on a provisional basis until […]
E-Invoicing Compliance EMEA
Greece's myDATA scheme postponed to April 2021
On 31 December 2020, in a last-minute change, the Greek government once again postponed the go-live date of myDATA to April 2021. This announcement had been anticipated and came as a great relief to Greek taxpayers who were concerned about being ready for the previous go-live date of January 2021. The first quarter of 2021 […]
North America Tax Information Reporting
How to File Unclaimed Property Reports Now That UPExpress Has Been Discontinued
UPExpress, the free tool for companies to eFile unclaimed property reports directly to participating states, was discontinued at the end of 2020 because the current file formats no longer aligned with many of the state's newly implemented technologies. You can still eFile unclaimed property reports to states that accept online reporting. However, if you plan […]
EMEA IPT
In Focus: Hungary – Taxpayers' Categories
Just like in every jurisdiction, settling tax liabilities compliantly is important to avoid penalties and other consequences. But in Hungary there are additional benefits to a taxpayer if tax liabilities are settled as compliantly as possible. As of 1 January 2016, Hungary introduced the so-called taxpayers' qualification system. This means the Hungarian Tax Office (HUTA) […]
North America Sales & Use Tax
What is the Nebraska Economic Nexus?
Prior to the South Dakota v. Wayfair, Inc. decision, Nebraska could not require remote sellers to collect sales tax on sales made to customers in the state. However, there is now a Nebraska economic nexus following the Wayfair decision. The Nebraska Department of Revenue explains that this does not create a new tax, but just […]


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NKU News October 5, 2016.
HIGHLAND HEIGHTS, Ky. – HIGHLAND HEIGHTS, Ky. – Northern Kentucky University's Center for Innovation and Entrepreneurship (CIE) is among the best in the world at launching and supporting student-led startup companies.
The Center was selected as one of the five finalists for the "Outstanding Contributor to Venture Creation" award at the 2016 Global Consortium of Entrepreneurship Centers' (GCEC) annual conference last week in Rochester, New York.
Other schools selected as finalists were Arizona State University, University of Michigan, Northwestern University, and UNC Chapel Hill. The GCEC awards are judged by peer entrepreneurship directors to recognize the very best in entrepreneurship education. This year had a record number of applicants and more than 450 participants from around the world.
The Center for Innovation and Entrepreneurship was formed in 2014 to create, advance, enable, and expand entrepreneurship education opportunities across campus as well as in the community. Its advisory board includes leaders from the Greater Cincinnati/Northern Kentucky startup and technology world; through these connections, NKU students receive unique opportunities to engage with and learn from these and other leaders.
"At NKU, no dream is too big or too bold to be achieved. We work each day to help student entrepreneurs from across campus achieve their dreams, and we are proud that those efforts continue to earn international recognition," said Rodney D'Souza, Fifth Third Bank Endowed Professor of Entrepreneurship and Director of the Center for Innovation and Entrepreneurship.
Last year, the CIE received the Exceptional Activities in Entrepreneurship Across Campus award from the Global Consortium of Entrepreneurship Centers. The GCEC is the nation's premier organization for university-based entrepreneurship centers. With membership from more than 225 entrepreneurship centers across the globe, the organization links universities of all sizes to share best practices and to advance, strengthen, and celebrate university-based entrepreneurship centers and programs.


Its that time of the year, no not summer time (but I do love writing about that I know, I know!) but an even more special time of the year, the countdown of the Seven Days of Samantha!
And in honor of our little one turning TWO, (OHHH MEEE GGGG!), I figured it was time to bring back the SDOS Series.
Our little love, who has turned into quite the little lady, continues to steal our hearts everyday.
Whether it be her screaming "MAAAAA-MAAAAA" with such excitement when I get home in the mornings from working out, or her JOY when Da-Da hits the door in the evenings, Jordan and I just look at each other at the end of the day with "that" look of pure JOY for our little love.
She still, despite her increasing energy and zeal and zest for life, kills us with LOVE!
How did we get so blessed?
Why is she the cutest thing we have ever seen?
Why does she remind us of US so darn much?
How do we someday love another one as much as we love this one?
We know were not alone, that other parents ask these questions as well, and I believe that speaks to the heart of all parents, the parts of us all, beaming with pride, love, and adoration for these littles entrusted to us.
Samantha Grace, you truly are God's best gift to us.
I love, love, love February.
A whole entire month to decorate with pink?
and has been replaced with lots of pink and touches of red.
The red in the month is acceptable…as long as its mostly featuring pink.
I know some of you feature the two colors beautifully "combined" but in this magical Queendom, I like to keep it simple.
And by simple, I mean as much pink as possible.
Around these parts, I have been on the lookout for some new Valentine's Day Decor the past couple of weeks to spruce up this humble abode of ours.
Searched Home Goods, Pier One, Target, and all the usual suspects with a little love to show for the quest.
The below items have all been added to the Goodmama Season of "Love Collection".
Here are some of my favorite fab february finds!
Sides- Gifted From My Crafty Bestie.
b. I let the red Go. They were that Gorg.
c. Please note the sunset in the back. So beautiful!
Also, hanging around our kitchen is this little number. Or shall I say, "these" little numbers. Our Christmas Card Wall! Why, might you ask, are our Christmas Cards up in umm…February?
b. I really love looking at them every day. I think they should be a permanent wall fixture.
So, in honor of the "month of Love" I have decided the way to best live on our little Christmas Card Craze, is to take them down (sigh*) and place them in a basket near our little kitchen table and do what my family used to do (stealing your idea's Gram!) and pray for each of you who sent us a card.
Each night during dinner (when were home), we will pick a card out of the basket and love on you through thoughts, prayers, and conversation.
As Tay Tay Swift would say, that way you will, Long Live!
This past Monday, the Goodmama household celebrated one of my FAVORITE holidays-Valentine's Day! I have always adored the day-I think a lot to do with the fact that my family growing up made it such a special day. Mama always cooked a meal full of "red favorites"- Chicken Parmesan, Stuffed Shells, Salad, Chocolate Desert- the table was always decorated with a gorgeous pink and red themed design. Dad always brought home his girls flowers- I loved this day.
As I got older, I continued to enjoy this day- but once I met J Good-he set the bar for incredible Valentine's Days. Flowers, Chocolates, presents, delicious fancy dinners out-it was a day of spoiling. I always say to Jordan that Valentine's Day is about US-he kindly reminds me its mostly about me. Oh well, a girl gets to be loved on right? (Especially if one of the theme colors is PINK-umm, hello, my fav).
This year, Valentine's Day was especially one of my favorites. The sun was shining-the skies were blue-Samantha Grace & I were in matching colored outfits (her request, of course 🙂 and her little laughter infused our car with smiles and joy as we traveled around town.
With my little valentine in tow-we had breakfast with my little brother (not so little anymore-but will always be my little brother in my eyes), dropped of goodies for family members, decorated the table, prepared dinner, and awaited OUR bigger valentine to come home. I think my favorite part of being a mama right now is watching the PURE love, affection, joy, and happiness that Samantha EXUDES when Jordan walks through the door.
We did have a rather hilarious moment of the afternoon-hubbs sent flowers-flowers arrived not as picture online indicated. Lesson learned-do not order flowers on Valentine's Day from ProFlowers. Or maybe, forgo flowers on valentines day? (Umm, nope).
HE redeemed the day with much better flowers of course.
It was one of my favorite Valentine's Day's ever. (Including the flower mix-up-I laughed all afternoon!) No crazy dressing up or reservations at some fancy restaurant-but camping at home-eating the meal my mom always cooked-sharing stories & dreams for the future & most of all-together hand in hand-loving on the best gift we have ever received from the Lord-our Samantha Grace.
Happy Week of Love to you all.
Including one that SINGS-she truly gets a kick out of this!


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John Lowe the Money Doctor gives invaluable advice on love and marriage – however before setting any big dates or buying any hats, he advises taking time out to discuss your finances.
You may think you are the perfect match, but one of the biggest compatibility tests for couples is bringing the finances together. Will this kill your romance? Are you afraid you might just open a can of worms? Discussing credit cards and spending habits are not generally on the love agenda.
But somewhere between proposing and saying "I do" are some key questions you and your better half should be asking each other.
Not exactly an ice-breaker but if you're going to be marrying this person and sharing their life you should both volunteer this information and hold nothing back. In an age where student loans over €20,000 are not uncommon, this debt discussion is crucial. Entering a marriage with hidden debt can be a disaster and start a marriage off on the wrong footing – mistrust can set in.
2. Do you have savings?
Yes, that's an easy one. You might have been going out a couple of years, gone Dutch on the holidays (paying for yourself) but have you salted any money away for the Rainy Day?
White goods are usually the first purchases for couples – they need their fridges and washing machines. How are you going to pay for this? What about the wedding itself, the 10% deposit for the house, the honeymoon, and of course, what about the children?
3. What about our children?
If you're thinking of becoming parents, and you haven't broken up over the debt and savings questions, this is a good one to ask.
Most couples address the 'when' and 'how many' when it comes to discussing children, but children are expensive – from birth to completion of 3rd level education, about €240,000 per child – so the financial impact should also be considered.
Do both spouses want to pay for private schooling? Who will pay for 3rd level college? Perhaps you will invest the €140 per month Child Benefit from day one for those 17 years when it stops at that point… amounting to c. €28,000 at current deposit interest rates.
Will one spouse stop working – financially – to take care of the children? (The answer to the last may be predicated by the financial commitment of both spouses to the mortgage until maturity).
While we're at it, what's fair pocket money allowance? And are we going to buy a TV for our darling son's bedroom the first time he asks? How much money will our daughter get when the Tooth Fairy visits?
You could go demented raising these issues but you will get a sense of where your partner stands making them even more potentially compatible with your own views.
4. What about our parents?
From an early age, we are so influenced by how our parents spent or saved and how we now spend or save. It's important to examine how we were raised and how our pecuniary habits were formed so that we can make changes now if a change is needed and change can be affected.
5. Who is paying for what?
How will money be handled in your household? Joint current account? Separate current accounts or one joint one? Jointly assessed or separately assessed? Who is in charge of paying the household bills and budgeting?
Is it fair to have one person be responsible for paying the bills for the rest of your married life? On the other hand, what if one person is really organised and the other isn't, does the organised person have to be the one to pay the bills?
This is also the time to ask your partner a few testing questions. For instance, if they have a gambling or shopping problem, even a drinking issue – is this a problem for you now and are you prepared to live with it?
Other questions include what kind of lifestyle do you see us living, and how much will it cost? What is your approach to saving and investing?
Meaning, how important is saving for retirement? Saving for our children's college education? Are you averse to risk or a risk-taker?
Who is in charge of paying the household bills and budgeting?
The point of asking a lot of questions isn't to map out your entire marriage, but to see if you both truly are financially compatible, or you're just actually hoping to give your local friendly solicitor some business in the years ahead!
In fact, if you are marrying, I would contend that the most important person you should consult with as a couple isn't the wedding planner – it's the financial planner.


But in Japan's case, the word change? WHOA. It is really difficult when you think about it because a lot has to be done, a lot has to be accepted by their people and a lot of time is needed for them to get used to it. But the thing is, I think Japan is in dire need of change because of the instability that is happening in their country. That situation is not new, but rather it is been there a long, long time ago. It just so happens that maybe the financial crisis now or the recession showed us more of the weaknesses that has been around Japan that we haven't notice before. I believe that the major factor that causes this instability is Japan's political leaders. That is why Japan was even compared to a dinosaur. Despite the fact that this country is one of the most powerful country in the entire world, Japan's leaders are weak and are not that efficient when it comes to leading a very powerful country. Japan needs more to that than having leaders that cannot solve their country's instabilities.
Moreover, I believe that Japan cannot have a major transformation or reform if they still have the same political leaders in their government.
Same beliefs, same platforms, same ideas, it is just all the same.
Where can we find the change there? How can Japan achieve the change they need in order to rise from their instabilities if the same leaders are still sitting in those political positions?
If Japan wants change, they should also change the people in the government. Aren't people fed up with the same political party or the LDP? If change or reform will not occur, the instabilities in their country may get worse. I also think that the opposition should prove to the people that they deserve a chance to show that they can also make a change in Japan. They should strive hard and show their people their passion for having a reformation in their country. They need to have a say or make a stand in order for them to get those positions they are running for.
That is why change is also a matter of choice.


Made in Korea. Colorverse is created by a group of makers with years of knowhow. The inks are treated for hydration and antibacterial properties, so they don't dry or deteriorate easily in pens. It is also designed to maintain the optimal pH balance of 7-8 (neutral). The ink writes smoothly in various kinds of pens and dries quickly on paper.
The ink is made of globally recognised premium quality colourants. It is highly pure with vivid and clear colours. It also has exceptional solubility and great to use in premium fountain pens. Due to the use of plant-based ingrediants, the inks are free from unpleasant chemically odours, such as alcohol.


It's the Classics season in Europe, a time of cobblestone roads, crosswinds, and fascinating tweaks to the pros bikes to hand the rough farm roads of Belgium and northern France. We're taking a look at some of the little adjustments pros make to their normal bikes to make all those miles bouncing over the cobbles and climbing up the hellingen a bit more bearable.
The cobbled classics are in full swing. The first cobble races of the season start at the beginning of March with the Omloop Het Nieuwsblad and Kuurne-Brussels-Kuurne. Over the next few weeks, races build into a crescendo with smaller semi-Classics, used by pro teams to test their riders and their gears on the roads used in the Monuments that cap off the spring campaign, the Tour of Flanders and Paris-Roubaix. As some of the oldest races on the calendar, the two races have seen plenty of changes, though their identities remain largely unchanged since their inception over 100 years ago. Both send racers of sections of cobblestone, or pave, though each race presents these in different ways.
De Ronde. At just over 100 years old, the Ronde van Vlaanderen features over a dozen small bergs, or climbs, that are still made with cobblestones. Volunteers and local organizations have banded together over the year to protect iconic roads like the Paterberg, the Koppenberg, and the Oude Kwaremont from being resurfaced and to preserve the character of the race. These bergs may not be towering alpine climbs, but their rough surface and short, steep pitches are incredibly taxing on riders' legs, especially over the massive 260 kilometer course. Many of those climbs are taken twice of even three times throughout the race, with fans sticking to one spot to see riders more often, or racing along dusty farm tracks to catch the race passing elsewhere.
The Paris-Roubaix, known as the 'Hell of the North' or the Queen of the Classics, offers a different challenge. It's route from the northern outskirts of Paris to the small industrial town of Roubaix are almost entirely flat; there's hardly a climb in over 250 kilometers. Instead, the 29 cobbles secteurs are littered with bigger, rougher, and more jarring stones than those you'll find at Flanders, and as a result, many riders make more modifications to their bike for this race.
All of that has changed in the past half decade, with more and more brands designing bikes that offer more vertical compliance, more tire clearance, disc brakes, and a geometry designed to offer the rider a bit more support. These bikes often use a taller head tube, slightly longer wheelbase for more stability, and a carbon layup that allows for more vertical flex in the frame to offer more relieve for the rider. Maybe the biggest difference is the ability to run 25, 28, even 30c tires at lower tire pressures for a smoother ride and more grip. With the prevalence of disc brakes, some riders are even able to stick with their normal aerodynamic, and less forgiving, bikes while still being able to fit a wide 27 or 28c tire. When Matt Hayman won the 2016 Paris-Roubaix, he didn't do it on Scott's cobble or gravel-specific frame; he did it on their aero road bike, the Foil.
That said, most teams and riders will hop on their Classics bikes, and most continue to add little tweaks to take that specificity even further. You'll still see double-wrapped handlebars and taped wrists, but new tech means new ideas, too. Many riders will add satellite shifters to their bars, allowing them to shift not only from the hoods, but also from the tops and from the drops with wired or even wireless buttons.
Another Classics favorite is grip tape, the kind you'd see on a skateboard, added to the inside of water bottle cages. Even those get swapped out; instead of light carbon cages, many teams install regular aluminum cages that can be bent down for a tighter fit to prevent losing bottles.
Gearing also gets tweaked at these races. With the steep, sharp climbs of Flanders offering plenty of fatigue, riders will use a 53/39 crankset paired with an 11-30 or 32 cassette to have a little extra room to shift on the 15, 20, even 25% pitches of the hellingen. Alternatively, there's not much need for a 39 at Roubaix; instead, riders will put on a 55 or 54t big chainring and run a 46, 48, or 50 small ring, just to give the legs a bit of a break when necessary. Many will never shift off the big ring, and many run a chain catcher adjusted very close to the chain line to make sure they never drop a chain over the rough cobbles.
Maybe the biggest adjustment any rider can make is to the tires. Many riders opt for 27c tubulars, though there have been sightings of 28s and 30c tires this spring and over the past few seasons. Additionally, they run low tire pressure for more comfort, often down to as low as 55 psi, depending on the weight of the rider. Nearly every brand has tried to find the perfect tire for cobbles, which can offer up kilometer after kilometer of rough, sharp rocks and plenty of debris in the gutters, too.
We love seeing all the nerdy, techy adaptations added to bikes this time of year, and if we see anything cool, we'll share it on our Facebook and Instagram. To see what gravel bikes we've got set-up like the pros, stop in and see us!


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UWL Confronts Diversity
2016-2017 LASO group.
Edith Osorio
UWL's list of values declares, "Diversity, equity, and the inclusion and engagement of all people…within an increasingly integrated and culturally diverse global community." This statement begs the question: How diverse is UWL and how does it affect its students?
Focusing in on racial and ethnic diversity within Midwest Universities is just what a study by U.S. News did. Chief Data Strategist, Robert Morse, reported, "Our formula produces a diversity index that ranges from 0 to 1. The closer a school's number is to 1, the more diverse the student population."
UWL was given a 0.19 on the diversity index. After hearing this Joe Gow, UWL's Chancellor, responded, "We know we must continue our efforts to enhance the diversity of the UWL campus. That is why it is so essential for us to create the position of Vice Chancellor for Diversity and Inclusion, which we look forward to implementing in the Spring Semester of 2018."
With 80% of UWL's students having Wisconsin residence (UWL Admission Fast Facts) and 87.5% of Wisconsin's residents identifying as white (U.S. Census Bureau), UWL's low number on the diversity index is unsurprising. According to the U.S. Census Bureau, La Crosse County, WI itself is 91.7% white.
Nicholas Shankey, Co-Chair of UWL's Asian Student Organization (ASO), commented, "UW-La Crosse has made a pledge to Inclusive Excellence and programs like [Awareness Through Performance], but that seems hypocritical when in 2014-2015 there were 192 reports of hate/bias on campus."
Shankey noted, "If there was abundant diversity, we wouldn't need programs like [Students Educating and Embracing Diversity], and we wouldn't have to require student senators to attend multicultural events."
Shankey expressed that UWL could do more to increase diversity, "I don't think UW-L does enough to promote events done by Multicultural Student Organizations or the Pride Center, because those events are always empty."
"It's hard to walk into a classroom of 40 people and be the only student of color," Shankey said. "It's intimidating."
Shankey added, "When there's no diversity, I feel like there's no support or empathy for what minority students go through. Increasing diversity should be important, because more diversity leads to more open conversations and understanding."
Shankey also spoke of ASO's work within the UWL community. "ASO works to benefit students by giving them an open space to discuss topics such as racism, cultural heritage, social justice, and racial identity. We try to give students a social opportunity to be themselves and have fun."
When asked about UWL's diversity, Edith Osorio, Vice President of UWL's Latin American Student Organization (LASO), also responded, "I think there's not a lot of it." Osorio suggested that UWL should highlight accomplishments by students of color as well as advertise resources and organizations pushing for diversity to welcome future students of color.
Osorio also stated, "I think it's very important for all people to understand all racial and ethnic diversities. It's important for UWL to incorporate that because it's just a life-lesson that people are going to need to know."
Osorio went on to explain, "Once you're out in the workforce, it may not be in the Midwest. You're going to have to work with people of color, so starting now, you need to get rid of these stereotypes."
Osorio mentioned, "I know how I felt, as a freshman, coming into a predominately white school. I was looking for the Latin American students at times because I still wanted to connect with my culture."
Increasing understanding, empathy, and hosting candid discussions are a few of the reasons why organizations like LASO and ASO are on campus.
Osorio said, "Our goal is to include all people, of all backgrounds, and educate them in Latin American culture." Osorio went on to say that another goal of LASO is to increase awareness and respect for Latin American culture.
Osorio emphasized that LASO is not only for Latin American students. "[LASO] is for everyone, people of all backgrounds." Osorio went on to say, "We have a variety of backgrounds represented in our group and I love it. So, it's not just for Latin American students, we like to encourage people of all backgrounds to come and if they have a passion for Latin American countries to come."
Part of LASO's meetings include an open discussion about global issues in Latin American countries. This fall, Mexico experienced a devastating earthquake and Puerto Rico has been recovering from Hurricane Maria; both are in need of all the help they can get.
"We're trying to work really fast right now to start a fundraiser to send to Puerto Rico and to Mexico. We do have a good amount of people of Mexican and Puerto Rican heritage in our organization and we do want to support them as much as we can," Osorio added. "We like to say we're part of a familia."
If you would like to donate hygiene products (shampoo, soap, etc.) and/or new or gently clothing and shoes, you can drop them off in room 3105 in the Student Union on Thursdays at 5 pm.
La Crosse Parking Utility institutes 'Food for Fines' program until the end of the year
Committee Close Up: Student Organization Committee
WGSS student creates website about sexual misconduct resources on and off campus for capstone
Student Association discusses new security systems and the Green Fund Act
Caycee Bean speaks to media about sexual misconduct allegations against Joel Elgin


It is common for employers in the tri-state area to require employees to sign noncompete agreements or other restrictive covenants as a condition of employment. However, these agreements cannot be unlimited in their restrictions. The Law Offices of Tacopina & Seigel assists clients by reviewing noncompete agreements, negotiating the terms of noncompete agreements and entering fierce litigation when a restrictive covenant is unreasonable or unfair. For these types of cases, the firm's skilled attorneys have a reputation for providing New Yorkers with high-quality legal service since 1994.
Your employer fails to give you something of value — A noncompete agreement is a contract where an employer gives you something of value, (for example, the job offer itself) in exchange for you not working for a competitor after you leave. If you sign a noncompete agreement while already working for an employer, the court may feel that you should have been given something of greater value.
The agreement covers an unreasonable span of time — In order for a NYC noncompete agreement to be valid, it must last a reasonable amount of time.
Geographic restrictions are too broad — A noncompete agreement must focus on a specific geographic location, namely, an area where your employer does business.
Pleading requirements — Sometimes, an employee can escape a noncompete agreement by attacking the agreement's validity. In order to convince a court to set aside a noncompete agreement, you must generally prove that the agreement was overly restrictive and unfairly limits your ability to earn a living.
Issues to look for during discovery — The type of discovery sought in a noncompete or restrictive-covenant case depends upon the nature of your agreement with your employer and what you hope to achieve through litigation. In some cases, it may be necessary to prove that the restrictions placed on your future employment are so unreasonable as to make it impossible for you to earn a living in your particular field. In these types of cases, third-party discovery can be essential. If your grounds for setting aside the noncompete agreement involve some malfeasance by your employer, then discovery runs the full gamut including interrogatories, depositions and document production.
Using expert witnesses — It's quite possible to litigate a noncompete or restrictive-covenant case successfully without the use of any expert witnesses. In some cases, however, it may be helpful to rely on the testimony of an expert in your industry who can explain to the fact-finder why the restrictive-covenant or noncompete agreement is unreasonable in light of the nature of your work and available employment opportunities.
The Law Offices of Tacopina & Seigel is a premier employment litigation law firm dedicated to protecting the rights of New York employees. Call today at 212-227-8877 or contact us online to schedule a consultation with a highly qualified attorney at the firm.


Ogier's offices in Jersey, Guernsey and Cayman have assisted Biffa, a UK leading integrated waste management company, with its recent listing on the Main Market of the London Stock Exchange, including a refinancing and pre-IPO restructuring.
The initial public offering resulted in total gross proceeds of around £262 million, making it the UK's biggest IPO since the Brexit vote.
Richard Daggett in Ogier's Jersey office acted as lead partner and was assisted by Alan Stirling and Amy Garrod. The Guernsey team was led by Bryon Rees supported by Michelle Watson Bunn and the Cayman team was led by partner Angus Davison with assistance from Mark Santangeli and Tommy Tuohy.
Linklaters acted as lead advisers to Biffa plc.
In conjunction with the IPO, Ogier's banking and finance team comprising Daniel Richards, Marcus Leese, James Heinicke, Michael Evans, Paul Chanter, Tiny Musesengwa and James Lydeard advised the lending banks on Jersey, Guernsey and Cayman law in relation to a refinancing of Biffa's debt, with Clifford Chance acting as lead advisers.
Richard said: "We are delighted to have worked with Biffa on their successful listing, which is especially pleasing at a time of low activity in equity capital markets following June's referendum.
"It was also an opportunity to showcase our cross-jurisdictional expertise, with teams in Jersey, Guernsey and Cayman providing time critical advice in a co-ordinated manner, which was essential in the lead up to the float and refinancing.


text:
Compete for a spot at Singularity University
Submitted on 13 March 2016 - 11:13pm
FCTI
2016 SingularityU Puerto Rico Global Impact Competition is a challenge to find the most innovative Puerto Ricans with the skills, passion, and ideas or projects that could potentially impact in a positive manner the lives of millions of people around the world through the use of exponential technologies within 5-10 years. Innovators from all academic and professional backgrounds are welcome, especially with Science, Technology, Engineering, and Mathematics backgrounds and who like to solve problems with moonshot thinking. The winner will have a confirmed spot at Singularity University for its flagship 10-week Global Solutions Program (GSP) from June 18 to August 28, 2016 at NASA Research Park in Silicon Valley. The costs for the participation in the GSP will be covered by a scholarship by the Puerto Rico Science, Technology, and Research Trust.
Apply at http://global.singularityu.org/puertorico/gic/. You can also attend one of the Briefing Sessions that the Trust will be offering to explain the simple application process. Attached please find the dates and places of those Briefing Sessions.
Singularity University is a benefit corporation that provides educational programs, innovative partnerships and a startup accelerator to help individuals, businesses, institutions, investors, NGO's and governments understand cutting-edge technologies, and how to utilize these technologies to positively impact billions of people. It was founded by Peter Diamandis, creator of XPrize, and by Ray Kurzweil, current director of Engineering at Google. It's sponsored by NASA and Google, and backed by organizations that are innovation leaders, such as Genentech, IDEO, International Space University, and others.
Its flagship initiative is the Global Solutions Program, which convenes leaders, entrepreneurs, and technologists for 10 weeks to work on team-based technology solutions to widespread global challenges.
Fideicomiso de Ciencia y Tecnología de Puerto Rico
Singularity University
Global Solutions Program
Content Categories:
Creativity: a new skill (?) needed in science and technology
SCIENCE CITY: INNOVATE, WORK, AND LIVE
Puerto Rican student will be part of intensive program of Singularity University in Silicon Valley
Finalists from local competition for global impact are announced
Singularity University in Silicon Valley actively seeks the best exponential innovations in Puerto Rico


I have never known a single trickster who had spent a few years tricking and regretted the time they spent tricking.
Finally, I answer the question with a NUMBER!
Tricking late is much more difficult than building muscle late.
Probably the best program in the universe.


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Q: Building a dynamic querytask 4.11 Learning the js api i tried to make a querytask that will query a layer and update graphics on change.
I think the problem is in my success("succ") function.
I made this codepen:
https://codepen.io/segev-salman/pen/joBwyW
in the 3.28 version of the api I saw a code example building an array and a for loop inserting features into a graphic from the querytask reault, but the 4.11 doc says that querytask returns a graphic so does that mean i can throw the result straight into a new graphic layers ?

A: i had a ton of problems some require inconsistency , some bad syntax but mostly used 3.x function instead of 4.x.
 codepen.io/segev-salman/pen/rgmaGr

redpajama_set_name: RedPajamaStackExchange