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\begin{abstract}
It is shown that the axial and polar perturbations of the
spherically symmetric black hole can be described in a
gauge-invariant way.  The reduced phase space describing
gravitational waves outside of the horizon is described by the
gauge-invariant quantities. Both degrees of freedom fulfill
generalized scalar wave equation. For the axial degree of freedom the
radial part of the equation corresponds to the Regge-Wheeler result
\cite{RW} and for the polar one we get Zerilli result
\cite{Zerilli}, see also  \cite{Chandra}, \cite{Moncrief} for both.
An important ingredient of the analysis is the {\em concept of
quasilocality} which does duty for the separation of the angular variables
in the usual approach. Moreover, there
is no need to represent perturbations by normal modes (with time
dependence $\exp(-ikt)$), we have fields in spacetime and the
Cauchy problem for them is well defined outside of the horizon.
The reduced symplectic structure explains the origin of the axial and
polar invariants. It allows to introduce an energy and angular
momentum for the gravitational waves which is invariant with respect to
the gauge transformations. Both generators
 represent quadratic approximation of the ADM nonlinear formulae
in terms of the perturbations of the Schwarzschild metric.
We also discuss the boundary-initial value problem for the linearized
Einstein equations on a Schwarzschild background outside of the horizon.
\end{abstract}

\section{Introduction}
In section 2 we introduce standard notions for linearized gravity and its
3+1 formulation. In the next section the analysis of constraints and
Killing fields on an initial surface gives charges for the linear field on
the Schwarzschild background.

Section 4 contains the main technical results related to the so-called 2+1
decomposition of the initial data and to the gauge invariant
description of the evolution.
The invariants introduced in this section contain the full
gauge-independent information about initial data. If we ``insert'' the
initial value constraints into the canonical symplectic structure we can
express the symplectic 2-form $\Omega$ in terms of the invariants. This is
precisely shown in Appendix B. This way our invariants play a role of the
reduced
unconstrained initial data which is gauge independent. Moreover, the axial
degree of freedom fulfills four-dimensional counterpart of the radial
equation proposed by Regge and Wheeler \cite{RW}, \cite{Chandra},
\cite{Moncrief} and the
polar invariant is related to the Zerilli equation \cite{Zerilli},
\cite{Chandra}, \cite{Moncrief}.

 Section 5 is devoted to the stationary
solutions, their behaviour on the horizon and precise interpretation of
the ``mono-dipole'' solutions. In
particular, the result of Vishveshwara \cite{Vish} that stationary
axial perturbation can exist only for dipole perturbation can be
easily recovered. However, if we assume more general conditions on a
horizon we may have other solutions.
 The dipole part of the axial
invariant corresponds to the spherical symmetry of the background
metric and represents angular momentum. On the other hand, there is
no dipole invariants in the polar part because the background metric is
not invariant with respect to the boosts and spatial
translations\footnote{Dipole invariants exist in flat Minkowski space and
they represent linear momentum  and  center of mass respectively
 (see \cite{JJspin2}).}. Moreover,
the lack of invariant in the dipole polar part corresponds to the fact
that this part of the metric can be always ``gauged away''. We would
like to stress that mono-dipole perturbations of the Schwarzschild
metric represent different phenomena than the higher-pole
perturbations. It is clear from the approach that $l \geq 2$
represents {\em gravitational wave perturbation} whereas $l=0$ and
$l=1$ correspond to the {\em charges}. For example, $l=0$ in polar
part represents infinitesimal perturbation of the mass (we move in
the space of solutions from one Schwarzschild solution to another),
dipole part in axial part corresponds to the infinitesimal angular
momentum (we move from Schwarzschild to Kerr solution). Formally
also monopole part in axial degree of freedom can be analyzed and represents
Taub-NUT charge but this move is excluded by topology.

 Reduction of the symplectic
form presented in Appendix B allows to introduce invariants from the
symplectic point of view in section 6. In the next section we define
(in a gauge-invariant way)
hamiltonian system\footnote{After separation of the angular variables
several ``hamiltonian'' results proposed by the author can be easily
translated into Moncrief's approach presented in \cite{Moncrief}.}, energy
and angular momentum generators and
boundary-initial value problem for the linearized field on the
Schwarzschild background.
Moreover, we are able to show that the obtained hamiltonian is a quadratic
approximation of the ADM mass defined at spatial infinity for the full
nonlinear Einstein theory and this particular result will be described
elsewhere.

The energy and angular momentum generators are well defined
for regular radiation data
$\underline{\bf x}$, $\underline{\bf X}$, $\underline{\bf y}$,
$\underline{\bf Y}$ which is finite on the horizon and vanishing at spatial
infinity according to the S.A.F.
 Christodolou-Kleinerman condition \cite{Ch-Kl}.

In Appendix A we show how to reconstruct the full four-metric $h_{\mu\nu}$
from the invariants assuming the gauge conditions used in \cite{RW} and
\cite{Vish}. This construction explains the relations between our
invariants and the special form of the metric used in \cite{RW} and
\cite{Vish}.
 In particular, we examine precisely ``mono-dipole'' part of
the metric ($l=0$ and $l=1$) which seems to be not fully analyzed in
literature.

\section{Linearized gravity}
In this section we remind some standard notions related to the Einstein
equations and the initial value problem.

Linearized
Einstein theory (see e.g. \cite{MTW} or \cite{Landau}) can be formulated as
follows.
Einstein equation
\begin{equation}\label{nrE}
 2{\rm G}_{\mu \nu} (g) = 16\pi {\rm T}_{\mu \nu}
 \end{equation}
 after linearization gives
\begin{equation}  \label{lrE}
h_{\mu \alpha}{_{;\nu}}{^{;\alpha}} +
         h_{\nu \alpha}{_{;\mu}}{^{;\alpha}} -
h_{\mu \nu}{^{;\alpha}}{_{\alpha}} -
(\eta ^{\alpha \beta}h_{\alpha \beta}){_{;\mu \nu}} - \eta _{\mu \nu}
[ h_{ \alpha\beta}{^{;\alpha \beta}} -
h_{\alpha}{^{\alpha ; \beta}}{_{\beta}} ] = 16\pi T_{\mu \nu} \end{equation}
\noindent
where pseudoriemannian metric ${\rm g}_{\mu \nu} = \eta _{\mu \nu} +
h_{\mu \nu}$,
$\eta _{\mu \nu}$ is the background metric and
 ``$;$'' denotes four--dimensional covariant derivative with respect
to the metric $\eta_{\mu \nu}$.
Moreover, we assumed that  $\eta _{\mu \nu}$ is a
vacuum solution of Einstein equation (${\rm G}_{\mu\nu}(\eta)=0$).

The (3+1)-decomposition of (\ref{lrE}) gives 6
dynamical equations for the space--space components $h_{kl}$ of the
metric (latin indices run from 1 to 3) and 4 equations which do not
contain time derivatives of $h_{kl}$. It is possible to introduce this
decomposition straightforward. However, a natural way for the formulation of
the (3+1)-splitting for the equation (\ref{lrE}) is to linearize ADM
formulation
of the initial value problem for the full nonlinear Einstein equation
(\ref{nrE}).
We shall introduce canonical variables for the
linearized case. They appear in a natural way if
we start from the ADM formulation of the initial value problem for
Einstein equations \cite{ADM}.

Let $({\rm g}_{kl}, {\rm P}^{kl})$ be the Cauchy data for Einstein
equations on a
three--dimensional space-like surface $\Sigma$. This means
that ${\rm g}_{kl}$ is a Riemannian metric on $\Sigma$ and ${\rm P}^{kl}$
is a symmetric
tensor density, which we identify with the ADM momentum \cite{ADM},
i.e.
\[
{\rm P}^{kl} = \sqrt{\det {\rm g}_{mn}} ({\rm g}^{kl} {\rm Tr} K -  K^{kl})
\]
where $K_{kl}$ is the second fundamental form (external curvature) of the
imbedding of $\Sigma$ into a spacetime $M$. \\
The 12 functions $({\rm g}_{kl}, {\rm P}^{kl})$ must fulfill 4 Gauss--Codazzi
constraints
\begin{equation}
     {\rm P}_i{^l}{_{| l}} = 8\pi\sqrt{\det g_{mn}}\, {\rm T}_{i\mu}n^{\mu}
                \label{ww}
\end{equation}
\begin{equation}
(\det {\rm g}_{mn}){\rm R}- {\rm P}^{kl}{\rm P}_{kl}
 + \frac{1}{2} ({\rm P}^{kl}{g}_{kl})^2  =
 16\pi(\det {\rm g}_{mn}) {\rm T}_{\mu \nu}n^{\mu}n^{\nu}
 \label{ws}
\end{equation}
where ${\rm T}_{\mu\nu}$ is an energy momentum tensor of the matter,
by $\rm R$ we denote the (three--dimensional) scalar curvature of
${\rm g}_{kl}$, $n^\mu$ is a future timelike four--vector normal to the
hypersurface
$\Sigma$ and the calculations have been made with respect to the three--metric
${\rm g}_{kl}$ ("$|$" denotes the covariant derivative, indices are raised and
lowered etc.).

The Einstein equations and the definition of the metric connection imply
the first order (in time) differential equations for ${\rm g}_{kl}$ and
${\rm P}^{kl}$
(see \cite{ADM} or \cite{MTW} p. 525) and contain the lapse function
$N$ and the shift vector $N^k$ as parameters

\begin{equation}
\label{gdot}
\dot{\rm g}_{kl}=\frac{2N}{\sqrt{\rm g}}\left( {\rm P}_{kl} -\frac 12
{\rm g}_{kl} {\rm P} \right) +
N_{k|l} +N_{l|k}
\end{equation}
where ${\rm g}:= \det {\rm g}_{mn}$ and ${\rm P}:= {\rm P}^{kl}{\rm g}_{kl}$
\[
\dot{\rm P}^{kl} = -N\sqrt{\rm g} {\rm R}^{kl}
 + \sqrt{\rm g} \left( N^{|kl} -{\rm g}^{kl}N^{|m}{_{|m}} \right) +\]
\[+\frac 12 N\sqrt{\rm g} {\rm g}^{kl}{\rm R}
  - \frac{2N}{\sqrt{\rm g}}\left( {\rm P}^{km}{\rm P}_{m}{^l}-
\frac{1}{2} {\rm P} {\rm P}^{kl} \right) + \left( {\rm P}^{kl} N^m
\right)_{|m} + \]
\begin{equation}\label{Pdot}
+ \frac{N}{2\sqrt{\rm g}}{\rm g}^{kl} \left( {\rm P}^{kl}{\rm P}_{kl}-
\frac{1}{2}{\rm P}^2 \right)
-N^k{_{|m}} {\rm P}^{ml}-N^l{_{|m}} {\rm P}^{mk}
 + 8\pi N\sqrt{\rm g} {\rm T}_{mn}{\rm g}^{km}{\rm g}^{ln}
\end{equation}

Let us consider an initial value problem for the linearized Einstein
equations on Schwarzschild background $\eta_{\mu\nu}$:
\begin{equation} \label{Sch}
\eta_{\mu\nu}\, {\rm d}x^\mu \, {\rm d}x^\nu\, =
-\left(1-\frac{2m}r \right) {\rm d}t^2 + \left(1-\frac{2m}r\right)^{-1}
{\rm d}r^2  +r^2 {\rm d}\theta^2 + r^2 \sin^2 \theta {\rm d}\varphi^2
\end{equation}
together with the radial coordinates: $x^3=r$,
$x^1=\theta $, $x^2=\varphi$. Moreover, $t=x^0$ denotes the time
coordinate. We consider only the part of the Schwarzschild spacetime outside
of the horizon, $r \geq 2m$.

We use the following convention for indices: greek indices $\mu, \nu,
\ldots$ run from 0 to 3;
$k,l, \ldots$ are spatial coordinates and run from 1 to 3;
$A,B,\ldots$ are spherical
angles $(\theta, \varphi)$ on a two-dimensional sphere $S(r):= \{
r=x^3=\mbox{const} \}$ and run from 1 to 2.
Moreover, let $\eta_{AB}$ denote a two-dimensional metric on $S(r)$.

Let $\displaystyle v:=1-\frac{2m}r$.
There are the following non-vanishing Christoffel symbols for the metric
(\ref{Sch}):
\[  \Gamma^{3}{_{33}}= -\frac{m}{vr^2}  \, ; \;\;
 \Gamma^{3}{_{AB}}= -\frac vr  \eta_{AB} \, ; \;\;
 \Gamma^{A}{_{3B}}= \frac 1r \delta^A{_B} \, ; \;\;
 \Gamma^{3}{_{00}}= \frac{mv}{r^2}  \, ; \;\;
 \Gamma^{0}{_{30}}= \frac{m}{vr^2}  \, ; \;\;  \Gamma^{A}{_{BC}} \]
 \noindent
where $\delta^A{_B}$ is the Kronecker's symbol
 and $\displaystyle \Gamma^{A}{_{BC}}$ are the same as for a standard
unit sphere $S(1)$ (in usual spherical coordinates
$\Gamma^{\theta}{_{\phi\phi}} = -\sin\theta\cos\theta$
 and $\Gamma^{\phi}{_{\phi\theta}}=\cot\theta$).

The curvature of the background metric we denote by
$C^\mu{_{\nu\lambda\kappa}}$ and the following
components of the Riemann tensor are non-vanishing (up to the symmetries
of the indices)
\[ C^0{_{A0B}}=C^3{_{A3B}}=\frac m{r^3}\eta_{AB} \, ; \quad
  C^{AB}{_{CD}}=\frac{2m}{r^3}\left( \delta^A{_C}\delta^B{_D}-
  \delta^A{_D}\delta^B{_C}  \right) \]

  We can introduce the following submanifolds of the Schwarzschild
spacetime $M$:
  \begin{equation} \label{Hs}
   \Sigma_s:=\{ x\in M \; : \; x^0=s, x^3 \geq 2m \} =\bigcup_{r \in
[2m,\infty[} S_s(r)
  \quad \mbox{where} \; \; S_s(r):= \{ x \in \Sigma_s \; : \;
x^3=r \} \end{equation}
and $\Sigma_s$ is a partial Cauchy surface outside of the horizon.

The surface $\Sigma$ carries  the induced
Riemannian metric $\eta_{kl}$:
\begin{equation} \label{eta3}
 \eta_{kl}\rd y^k \rd y^l=
 \frac 1v \rd r^2 +r^2 (\rd \theta^2 +\sin^2\theta \rd \phi^2)
 \end{equation}
Usually it is convenient to change the coordinate $r$ to $r_*$ which is a
solution of an ordinary differential equation (see \cite{Chandra}):
 \begin{equation} \label{r*}  \frac{\rd r_*}{\rd r}=\frac 1v \, ; \; \quad
  r_* :=r+2m\ln \left(\frac r{2m}-1\right) \end{equation}
and moves horizon to $-\infty$.

  The ADM momentum
${\rm P}^{kl}$  for the metric (\ref{eta3}) on each slice
$\Sigma_s$ vanishes (${\rm P}^{kl}=0$).
The shift vector  is also trivial ($N^k=0$) and the lapse $N=\sqrt v$ is
vanishing on the horizon.
Moreover, the Ricci tensor for the three-metric $\eta_{kl}$
 has the following components:
\[ {\rm R}_{kl}=\frac1N N_{|kl} \]
 and the scalar curvature ${\rm R}$ vanishes.

Let us define the linearized variations $(h_{kl},P^{kl} )$ of the full
nonlinear Cauchy data $({\rm g}_{kl}, {\rm P}^{kl} )$ around background data
($\eta_{kl}$, 0)
\begin{equation}  h_{kl}:={\rm g}_{kl}-\eta_{kl}\, ,\quad
 P^{kl}:={\rm P}^{kl}  \end{equation}

We should now rewrite equations (\ref{ww})--(\ref{Pdot}) in a linearized
form in terms of $(h_{kl},P^{kl} )$.
Let us denote $P:=\eta_{kl}P^{kl}$ and  $h:=\eta^{kl}h_{kl}$.
The vector constraint (\ref{ww}) can be linearized as follows
\begin{equation} \label{ww1}
  {\rm P}_i{^l}{_{| l}} \approx P_i{^l}{_{| l}}
\end{equation}
Let us stress that the symbol ``$|$'' has  different meanings on the
left-hand side and on the right-hand side of the above formula.
It denotes the covariant derivative with
respect to the full nonlinear metric ${\rm g}_{kl}$
when applied to the ADM momentum ${\rm P}^{kl}$,
but on the right-hand side it means the covariant derivative with respect to
the background metric $\eta_{kl}$.
The scalar constraint (\ref{ws}) after linearization takes the form
\begin{equation}   \label{ws1}
\sqrt{\rm g}{\rm R}-\frac 1{\sqrt{\rm g}}\left( {\rm P}^{kl}{\rm P}_{kl}
 - \frac{1}{2} ({\rm P}^{kl}{\rm g}_{kl})^2 \right) \approx
 \sqrt\eta \left[ \left( h^{kl}{_{|l}}- h^{|k} \right)_{|k} -
 h^{kl}{\rm R}_{kl} \right]
\end{equation}
where $\sqrt\eta :=\sqrt{\det \eta_{kl}}$.

The linearized constraints for the vacuum ($T_{\mu\nu}=0$) have the form
\begin{equation} \label{wwl}
  P_l{^k}{_{| k}}  =0
\end{equation}
\begin{equation}   \label{wsl}
  \left( h^{kl}{_{|l}}- h^{|k} \right)_{|k} -h^{kl}{\rm R}_{kl}  =0
\end{equation}
The linearization of (\ref{gdot}) leads to the equation
\begin{eqnarray}
\dot{h}_{kl} & = & \frac{2N}{\sqrt\eta} \left( P_{kl} -\frac 12
\eta_{kl} P \right) + h_{0k|l} +h_{0l|k}
\label{hdot}
\end{eqnarray}
where $N:=\frac 1{\sqrt{-\eta^{00}}}=\sqrt v$,
 $N_k=\eta_{0k}=0$ are the lapse and shift
for the background. Let us denote the linearized lapse by $n:=\frac12
{\sqrt v}h^0_0$. The linearization of (\ref{Pdot}) takes the form
\begin{equation}
\frac 1{\sqrt\eta}\dot{P}_{kl} = -n {\rm R}_{kl} -N\delta R{_{kl}}
  + n_{|kl} -N_m\delta\Gamma^m{_{kl}}
   - \eta_{kl}\left( n^{|m}{_{|m}}
-\eta^{ij}\delta\Gamma^m{_{ij}}N_{|m} -h^{mn}N_{|mn}  \right)
\label{pidot} \end{equation}
where
\[ \delta\Gamma^m{_{kl}} :=\frac12\left(
h^m{_{k|l}}+h^m{_{l|k}}-h_{ml}{^{|k}} \right) \]
is the linearized Christoffel symbol and similarly
\[ \delta R{_{kl}} :=\frac12\left(
h^m{_{k|lm}}+h^m{_{l|km}}-h_{kl|m}{^{m}} -h_{|kl} \right) \]
is the linearized Ricci tensor.

It is well known (see for example \cite{JJspin2}) that the linearized Einstein
equations (\ref{lrE}) are invariant with respect to the ``gauge''
transformation:
\begin{equation}
        h_{\mu \nu} \rightarrow h_{\mu \nu} + \xi _{\mu ; \nu} +
\xi _{\nu ; \mu}   \label{gauge4}
\end{equation}
where $\xi _{\mu}$ is a covector field.
 The (3+1)-decomposition of
the gauge acts on the Cauchy data in the following way
\begin{eqnarray}
2v\Lambda ^{-1} P_{kl} & \rightarrow & 2v\Lambda ^{-1} P_{kl}
 +(v \xi^0_{|k})_{|l}+(v \xi^0_{|l})_{|k} - \eta_{kl} (v\xi{^{0|m}})_m
    \label{pixi} \\
 \label{hxi}
 h_{kl}& \rightarrow & h_{kl} + \xi_{l | k}+ \xi_{k | l}
\end{eqnarray}
where  $\Lambda:={\sqrt v}\sqrt{\det \eta_{kl}}$ ($=r^2 \sin\theta$).
It can be easily checked that the scalar constraint (\ref{wsl}) and the vector
constraint (\ref{wwl}) are invariant with respect to the gauge
transformations (\ref{pixi}) and (\ref{hxi}).
The Cauchy data ($h_{kl}$, $P^{kl}$) and ($\overline{h}_{kl}$,
$\overline{P}^{kl}$) on $\Sigma$ are equivalent to each other
 if they can be related by the gauge transformation $\xi_{\mu}$.
 The evolution of canonical variables $P^{kl}$ and $h_{kl}$ given by
equations  (\ref{hdot}), (\ref{pidot})
 is not unique unless the lapse function $n$ and the shift vector
 $h_{0k}$ are specified.

We will show in the sequel that it is possible to define a reduced dynamics
in terms of invariants, which is no longer sensitive on gauge conditions.
The construction is analogous to the analysis given in \cite{GRG}.

\section{``Charges'' on the Schwarzschild background}

The vector constraint (\ref{wwl}) allows to introduce ``charges'' related
to the symmetries of the background metric. There are three generators of
the rotation group, which are simultaneously Killing vectors on the
initial surface $\Sigma$.
Let us denote this Killing field by $Z^k$. It is a solution of the Killing
equation
\begin{equation}\label{Zk} Z_{k|l} + Z_{l|k} = 0 \end{equation}
Let $V\subset \Sigma$ be a compact region in $\Sigma$. For example
$\displaystyle V:=\bigcup_{r\in [r_0,r_1]} S_s(r)$ and $\partial
V=S_s(r_0) \cup S_s(r_1)$.
 From (\ref{wwl}) and (\ref{Zk}) we get
\begin{equation}\label{Lc}
0=\int_V P^{kl}{_{|l}} Z_k= \int_V (P^{kl}Z_k){_{|l}} =
 \int_{\partial V} P^{3k}Z_k
\end{equation}
The equation (\ref{Lc}) expresses the ``Gauss'' law for the angular
momentum charge ``measured'' by the flux integral.
 It is easy to relate this charge to
the dipole part of invariant $\bf y$, which will appear in the sequel
 -- (\ref{y}) in the next section. For example, when $Z=\partial
/\partial\phi$ we have
\[
16\pi s^z:= 16\pi j^{xy} = -2\int_{\partial V} P^3{_\phi } =
 -2\int_{\partial V} P^3{_A} (r^2\varepsilon^{AB}\cos\theta)_{||B} =\]
\begin{equation} \label{sz}
 = 2\int_{\partial V} r^2 P^{3}{_{A||B}}\varepsilon^{AB} \cos\theta
 =\hspace{0.5cm} \int_{\partial V} \Lambda {\bf y} \cos\theta
 \end{equation}

The time translation defines a  mass charge from the scalar constraint
(\ref{wsl}) as follows
 \[
 0=\int_V
 N \sqrt\eta \left[\left( h^{kl}{_{|l}}- h^{|k} \right)_{|k}
-h^{kl}N_{|kl}\right] =
 \]
 \[
  =\int_V
  \left[
  N\sqrt\eta \left( h^{lk}{_{|k}}- h^{|l} \right) +\sqrt\eta
  \left(  N^{|l} h -N_{|k} h^{kl}  \right) \right]_{|l}  = \]
   \begin{equation}   \label{BM}
 = \int_{\partial V}
   \Lambda \left( h^{3k}{_{|k}}-  h^{|3} + \frac1N h N^{|3} - \frac1N
h^{k3}N_{|k}  \right)
\end{equation}
and it can be related to the monopole part of an invariant $\bf x$
introduced in the next section by (\ref{x}).

\[
16\pi p^0=\int_{\partial V}
   \Lambda \left( N h^{3k}{_{|k}}- N h^{|3}  + N_k h^{k3} -N^3 h
\right)= \] \begin{equation} \label{p0}
=\int_{\partial V} \frac{\Lambda}{r} \left(
 2 h^{33} -vr H,{_3} - (1-\frac{3m}r )H  \right) =
  \hspace*{0.5cm} \int_{\partial V} \frac{\Lambda}r {\cal B}^{-1}{\bf x}
\end{equation}

\section{Equations of motion for the invariants}

The radial foliation of the Cauchy surface $\Sigma$ related to the
spherical symmetry allows to perform (2+1)-decomposition of the initial data.
In this section we introduce reduced gauge invariant data on $\Sigma$ for
the gravitational
field, similar to the invariants introduced in \cite{GRG}. For this purpose
we use a spherical foliation of $\Sigma$ (see formulae (\ref{Hs}) and
(\ref{eta3})).

Let ${\mathbf{\Delta}}$ denotes the two--dimensional Laplace--Beltrami operator on a
unit sphere $S(1)$. Moreover, $H:=\eta^{AB}h_{AB}$, $\chi_{AB}:= h_{AB}
-\frac 12 \eta_{AB} H$, $S:=\eta^{AB}P_{AB}$, $S_{AB}:= P_{AB}
-\frac 12 \eta_{AB} S$ according to the notation used in \cite{GRG}.

The spatial gauge (\ref{hxi}) splits in the following way
\begin{eqnarray}
 \label{hxi33}
 h_{33}& \rightarrow & h_{33} + \frac2{\sqrt v}({\sqrt v}\xi_{3})_{,3}
\\ \label{hxi3A}
 h_{3A}& \rightarrow & h_{3A} + \xi_{3,A}+ \xi_{A,3} -\frac2r \xi_A
 \\ \label{hxiAB}
 h_{AB}& \rightarrow & h_{AB} + \xi_{A|| B}+ \xi_{B|| A} +\frac2r
\eta_{AB} \xi^3
\end{eqnarray}
where by ``$||$'' we denote the covariant derivative with respect to the
two--metric $\eta_{AB}$ on $S(r)$.
Similarly, the temporal gauge (\ref{pixi}) can be splitted as follows
\begin{eqnarray}
\Lambda ^{-1} P^{3}{_3} & \rightarrow & \Lambda ^{-1} P^3{_3}
 -\xi^{0||A}{_A} -\frac2r \xi^{0,3}
 \label{pxi33}  \\
 \label{pxi3A}
\Lambda ^{-1} P_{3A} & \rightarrow & \Lambda ^{-1} P_{3A} +
\left( \xi^0_{,3}-\frac1r \xi^{0} +\frac{m}{vr^2}\xi^0\right)_{||A}
  \\
  \label{sxiAB}
\Lambda ^{-1} S_{AB} & \rightarrow & \Lambda ^{-1} S_{AB}
  + \xi^0_{||AB} -\frac12\eta_{AB}\xi^{0||C}{_C}  \\
  \label{sxi}
  \Lambda ^{-1} S & \rightarrow & \Lambda ^{-1} S
 -\frac2N(N\xi^{0,3})_{,3} -\frac2r \xi^{0,3}   -\xi^{0||C}{_C}
 \end{eqnarray}

 It is also quite easy to rewrite the (2+1)-decomposition of (\ref{hdot})
\begin{equation}
\dot{h}_{33} = \Lambda^{-1}(P^3{_3}-S) + \frac 2{\sqrt v}
({\sqrt v} h_{03}),{_3} \label{h33}
\end{equation}
\begin{equation}
\dot{h}_{3A} =2v\Lambda^{-1}P_{3A} + h_{03}{_{||A}} + h_{0A},{_3}
-\frac 2r h_{0A}   \label{h3A}
\end{equation}
\begin{equation}
\dot{h}_{AB}
= 2v\Lambda^{-1}S_{AB}- \eta_{AB}\Lambda^{-1}P^{33} +h_{0A||B} + h_{0B||A} +
 2vr^{-1} \eta_{AB} h_{03}      \label{hAB}
\end{equation}

The dynamical equations (\ref{pidot}) take the following (2+1)-form:

\[
2\Lambda^{-1}\dot{P}_{33}= -\frac 1v h^0{_0}{^{||A}}_A -2r^{-1}h^0{_0},{_3} +
\frac 1v h_3^{3 ||A}{_A} + 2r^{-2}h^3{_3}  + \]
\begin{equation} +  (H,{_3}- 2h_3{^{A}}{_{||A}} -2r^{-1}h_3{^3}),{_3}
+2r^{-1}(H,{_3}-2h_3{^A}{_{||A}}-2r^{-1}h_3{^3})        \label{P33}
\end{equation}
\[
2\Lambda^{-1}\dot{P}_{3C}= \left[\frac 1{\sqrt v} ({\sqrt v} h^0{_0}),{_3}
-r^{-1}h^0{_0}\right]_{ ||C} -\frac m{vr^2} h^3_{3||C} +
\frac{1}{2} (H,{_3}- 2h_3{^A}{_{||A}} -2r^{-1}h^3{_3})_{ ||C} +\]
\begin{equation}
+h_{3C||A}{^{||A}} +\frac 1{r^2} h_{3C} -
 \chi^A{_{C||A}},{_3} \label{P3C}
\end{equation}
\[
2\Lambda^{-1}\dot{S}_{AB} = h^0{_{0||AB}} -\frac12 \eta_{AB}h^0{_0}^{||C}{_C}
+ h^3{_{3||AB}} -\frac12 \eta_{AB}h^3{_3}^{||C}{_C}+\]
\[ - (h^3{_{A||B}}+h^3{_{B||A}}-
\eta_{AB}h^{3C}{_{||C}}),{_3}+(v\chi^C{_B},{_3}\eta_{CA}),{_3} +
\]
\begin{equation}
 + \chi_{AB}{^{||C}}_{||C} -
\chi^C{_{A||BC}} - \chi^C{_{B||AC}} + \eta_{AB}\chi^{CD}{_{||CD}} +
\frac 2{r^2}\chi_{AB}  \label{SAB}
\end{equation}

\[
2\Lambda^{-1} \dot S= -\sqrt{v} (\sqrt{v} h^0_{0,3})_{,3}-\frac 1r h^0_{0,3}
-h_0^{0||A}{_A} + (h^3{_3} +H)^{||A}{_A} +\frac2{r^2}(h^3{_3} +H)
 -\frac{12m}{r^3} h^3{_3}+
\]
\begin{equation}
 +\frac{2m}{r^2}h^3{_{3,3}}
+  v(H,{_3}- 2h_3{^{A}}{_{||A}} -2r^{-1}h_3{^3}),{_3}
+4\frac{v}r (H,{_3}-2h_3{^A}{_{||A}}-2r^{-1}h_3{^3}) -2\chi^{CD}{_{||CD}}
  \label{Sdot}
\end{equation}

 The vector constraint (\ref{wwl}) splits in a similar way
 \begin{equation}
\frac 1{\sqrt v}({\sqrt v}P^3{_3}),{_3} +  P_3{^A}{_{||A}} - r^{-1}S = 0
\label{ww3}
\end{equation}
\begin{equation}
P^3{_A},{_3} + S_A{^B}{_{||B}}+\frac{1}{2} S_{||A} = 0
\label{wwA}
\end{equation}

And finally
the (2+1)-decomposition of the scalar constraint (\ref{wsl}) can be written
in the form
\[
 h ^{|l}{_l} - h^{kl}{_{|kl}} +h^{kl} {\rm R}_{kl}
 = \frac{\sqrt v}{r^3}\left[ {r^2}{\sqrt v}
 (rH,{_3}-2rh_{3A}{^{||A}}-2h_3{^3})\right],{_3} +
\]
\begin{equation}
 +h^3{_{3 ||A}}{^A}+2r^{-2}h^3{_3} -\frac{6m}{r^3}h^3{_3}
 +\frac{1}{2}H^{||C}{_C}+r^{-2}H
  - \chi^{AB}{_{||AB}} = 0   \label{wsk}
\end{equation}

Let us notice that we can split the dynamics into two separate parts
which we call axial and polar respectively.
The axial part of initial data consists of two momentum components
 $P^{3A||B}\varepsilon_{AB}$, $S^C{_{A||CB}}\varepsilon^{AB}$ and two metric
components $h_{3A||B}\varepsilon^{AB}$, $\chi^C{_{A||CB}}\varepsilon^{AB}$. The
only gauge freedom is contained
 in $\xi_{A||B}\varepsilon^{AB}$ which acts on the metric components.
 The gauge invariants $P^{3A||B}\varepsilon_{AB}$ and
 $S^C{_{A||CB}}\varepsilon^{AB}$ are related by the curl part of the vector
constraint  (\ref{wwA}) as follows:
\begin{equation}
(r^2P^{3A||B}\varepsilon_{AB}),{_3}
 + r^2 S^C{_{A||CB}}\varepsilon^{AB} =0
\label{wAB}
\end{equation}
It is easy to verify that
 the following pair of gauge invariants:
 \begin{equation}
 {\bf y}:=2\Lambda^{-1}r^2 P^{3A||B}\varepsilon_{AB} \label{y}
\end{equation}
\begin{equation}
 {\bf Y}:= \Lambda(\mathbf{\Delta} +2)h_{3A||B}\varepsilon^{AB}
 - r^2(\Lambda \chi^C{_{A||CB}}\varepsilon^{AB}),{_3} \label{Y}
\end{equation}
contains the whole information about axial part of initial data up to the
gauge freedom (see also Appendix A).
We will show in the sequel that this is a canonical pair with respect to the
symplectic structure of linearized Einstein equations.

Let us define
the polar invariants as follows
\begin{equation}
 {\bf x} := r^2\chi^{AB}{_{||AB}}-\frac{1}{2}(\mathbf{\Delta}+2) H+
  {\cal B}\left[ 2 h^{33}+2r h^{3C}{_{||C}} -rv H,{_3}  \right]
                \label{x}
\end{equation}
\begin{equation}
 {\bf X} :=
 2r^2 S^{AB}{_{||AB}} +{\cal B}\left[ 2rP^{3A}{_{||A}}
+\mathbf{\Delta} P^3{_3} \right]  \label{X}
\end{equation}
where
\[ {\cal B}:= (\mathbf{\Delta} +2)\left(\mathbf{\Delta} +2-\frac{6m}r\right)^{-1} \]
is a {\em quasilocal} operator -- it is local with respect to the
coordinates $t,r$ but non-local on each sphere $S(r)$.
 The proof that ${\bf x}$ and $\bf X$ contain
the whole information about polar part of initial data
$P^{3A}{_{||A}}$, $S^{AB}{_{||AB}}$, $P^3{_3}$, $S$,
 $h^{3C}{_{||C}}$, $\chi^{AB}{_{||AB}}$, $h^3{_3}$, $H$ up to the gauge
transformation $\xi^0, \xi^3, \xi^A{_{||A}}$ we give in Appendix A.
Moreover, in Appendix B we show that $({\bf x}, {\bf X}, {\bf y}, {\bf Y})$
is the reduced canonical initial data on $\Sigma$.

We can check the reduced field equations for the axial invariants

 \begin{equation}\label{py} \dot{\bf y}=\frac{v}\Lambda {\bf Y}  \end{equation}
\begin{equation}\label{pY} \dot{\bf Y}= \Lambda \left\{
\left[\frac v{r^2}(r^2{\bf y}),{_3}\right],{_3} +\frac1{r^2}(\mathbf{\Delta} +2)
{\bf y}\right\} \end{equation}
More precisely, the curl part of (\ref{P3C}) gives (\ref{py})
 and (\ref{pY}) follows directly from (\ref{h3A}), (\ref{hAB}) and may be
checked by inspection.

 It can be easily verified that the axial invariant $\bf y$ fulfills
the generalized scalar wave equation (as a consequence of the above dynamical
equations)
 \begin{equation} \left(
 \mbox{\msa\symbol{3}} +\frac{8m}{r^3}\right) {\bf y}=0 \label{rfy} \end{equation}
where $\mbox{\msa\symbol{3}}$ denotes the usual wave operator with respect to
the background metric $\eta_{\mu\nu}$.

There exists a simple relation between the equation (\ref{rfy}) and the
so-called {\em Regge-Wheeler equation} \cite{RW}, \cite{Chandra},
\cite{Moncrief}.
Let us rewrite equation (\ref{rfy}) in the following form:
\begin{equation}\label{RWE}
 -\ddot{\bf y} +\frac vr\left[v(r{\bf y})_{,3}\right]_{,3} = V^{(-)}
{\bf y} \end{equation}
where $V^{(-)}$ is a ``spherical operator'' defined as follows:
\[ V^{(-)}:=-\frac v{r^2} \left(\mathbf{\Delta}+\frac{6m}r\right) \]
If we insert the invariant ${\bf y}$ in a special form
 ${\bf y} =\exp(i\sigma t) Y_l(\theta,\phi) Z^{(-)}(r)/r$
into equation (\ref{RWE}), we obtain Regge-Wheeler equation
\[ \left(\frac{d^2}{dr_*^2}+\sigma^2 \right)Z^{(-)}=V^{(-)}Z^{(-)} \]
Here $Y_l$ is a spherical harmonics such that
$\mathbf{\Delta} Y_l =-l(l+1) Y_l$ and coordinate $r_*$ is defined by (\ref{r*}).

The polar invariants fulfill the following equations:
 \begin{equation}\label{px} \dot{\bf x}=\frac{v}\Lambda {\bf X}  \end{equation}
\begin{equation}\label{pX} \dot{\bf X}= \frac\Lambda{r^2} \left\{
\left( v{r^2}{\bf x},{_3}\right),{_3} + \left[\mathbf{\Delta} +v(1-2{\cal B}) +1 \right]
{\cal B}{\bf x}\right\} \end{equation}
To obtain (\ref{px}) we need equations (\ref{h33}), (\ref{h3A}), (\ref{hAB})
and vector constraints (\ref{ww3}), (\ref{wwA}). Similarly, (\ref{pX}) is a
consequence of (\ref{P33}-\ref{SAB}) and scalar
constraint (\ref{wsk}) (see also Appendix A).

There exists also a generalized scalar wave equation for the polar
invariant but it is no longer local, it is only quasilocal:
 \begin{equation} \left[\mbox{\msa\symbol{3}} +\frac{8m}{r^3}(\mathbf{\Delta} -1)
 \left(\mathbf{\Delta} +2-\frac{3m}r\right)\left(
 \mathbf{\Delta} +2-\frac{6m}r\right)^{-2}\right] {\bf x}=0
\label{rfx} \end{equation}

The similar equation to the (\ref{RWE}) can be presented in the analogous form
\begin{equation}\label{ZE}
-\ddot{\bf x} +\frac vr\left[v(r{\bf x})_{,3}\right]_{,3} = V^{(+)}{\bf x}
\end{equation}
but now the operator $V^{(+)}$ is defined ``quasilocally''
\[ V^{(+)}:=-\frac {v}{r^2}  \left[(\mathbf{\Delta}+2)^2\left(\mathbf{\Delta}
-\frac{6m}r\right)+\frac{36m^2}{r^2}\left(\mathbf{\Delta} +2 -\frac{2m}r\right)\right]
\left(\mathbf{\Delta}+2-\frac{6m}r\right)^{-2} \]
If we insert the invariant ${\bf x}$ in an analogous special form
 ${\bf x} =\exp(i\sigma t) Y_l(\theta,\phi) Z^{(+)}(r)/r$
into equation (\ref{ZE}) we obtain {\em Zerilli equation} \cite{Zerilli},
\cite{Chandra}, \cite{Moncrief}
\[ \left(\frac{d^2}{dr_*^2}+\sigma^2 \right)Z^{(+)}=V^{(+)}Z^{(+)} \]
This way we have shown that both equations (Regge-Wheeler and Zerilli)
 posses gauge invariant formulations
and their solutions contain the entire gauge--independent information
about the linearized gravitational field on the Schwarzschild background
(see also \cite{Moncrief}).

Let us notice that $\bf x$ and $\bf y$ are scalars on each sphere $S_t(r)$
with respect to the coordinates $x^A$.
For the scalar $f$ on a sphere we can define a ``monopole'' part mon$(f)$
and a ``dipole'' part dip$(f)$ as a corresponding component with respect
to the spherical harmonics on $S^2$. Similarly, the ``dipole'' part of a
vector $v^A$ corresponds to the dipole harmonics for the scalars $v^A{_{||A}}$
and $\varepsilon^{AB}v_{A||B}$. Let us denote by $\underline f$ the
``mono--dipole--free'' part of $f$. According to this decomposition we have

\[ {\bf x}=\mbox{mon}({\bf x}) +\underline{\bf x}
\]
\[ {\bf y}= \mbox{dip}({\bf y}) +\underline{\bf y}
\]
The dipole part of $\bf x$ and monopole part of $\bf y$ are vanishing\footnote{
This is included in the definitions (\ref{x}) and (\ref{y}). More precisely,
$\bf y$ is a divergence (see also (\ref{dv})) and ${\bf x}=
r^2\chi^{AB}{_{||AB}}+ (\mathbf{\Delta} +2)[...]$. Moreover,
 dip$(\chi^{AB}{_{||AB}})=0$ because
double-divergence of any traceless tensor is ``mono-dipole-free'' (see also
(\ref{ddtt}) in Appendix B).}.
The rest of the mono-dipole part of each scalar can be solved explicitly from
the equations (\ref{py})--(\ref{pX}) and the solution has the form
\[ {\bf x} - \underline{\bf x}= \frac{4{\bf m}}{r-3m}
\]
\[ {\bf y}-\underline{\bf y}=\frac{12{\bf s}}{r^2}
\]
From (\ref{py}), (\ref{px}) and the observation that
$\mbox{mon}({\bf X})=\mbox{dip}({\bf Y})=0$ we obtain
\[ \dot{\bf m}=\dot{\bf s}=0 \]
Moreover, ${\mathbf{\Delta} m}=0$, $(\mathbf{\Delta} +2) {\bf s}=0$, which simply means that
$\bf m$ is a monopole and  $\bf s$ is a dipole, and they are constant with
respect to the coordinates $t,r$. They correspond to the charges
introduced in section 3. More precisely, ${\bf m}=p^0$.
Moreover, the angular momentum charge (\ref{sz}) can be obtained from
 the relation between spatial constant three-vector in
cartesian coordinates and dipole harmonics
\[  {\bf s} =\frac{s^l z_l}{r} \]
 where $z_l$ are cartesian coordinates, $r=\sqrt{\delta^{kl}z_k z_l}$
  and  $s^l$ is a corresponding three-vector representing angular momentum
(see \cite{JJspin2}).

\section{Stationary solutions}
For the axial degree of freedom $\bf y$ we can rewrite equation
 ({\ref{RWE}) using a new coordinate $z:=\frac{2m}r$.
\[ 4m^2\ddot{\bf y}=(1-z)^2 z^4 \frac{\partial^2 {\bf y}}{\partial z^2}
 -(1-z) z^4 \frac{\partial {\bf y}}{\partial z} +
 (1-z) z^2(\mathbf{\Delta} +4z) {\bf y} \]
 This way horizon
corresponds to $z=1$ and spatial infinity to $z=0$. Let us consider
stationary solutions of the above equation which are regular
 at the spatial infinity (corresponding to $z=0$).
 \begin{equation}\label{sty}
  (1-z) z^2 \frac{\partial^2 {\bf y}}{\partial z^2}
 - z^2 \frac{\partial {\bf y}}{\partial z} +
 (\mathbf{\Delta}+4z) {\bf y}=0 \end{equation}
  Moreover, if we separate the
angular variables and include standard asymptotic behaviour at $z=0$:
\[ {\bf y} =z^{l+1} Y_l(x^A) u(z) \]
then the equation for the function $u$ is relatively simple
\[ (1-z) z u'' +[2l+1-(2l+3)z]u' -(l-1)(l+3)u =0 \]
and the solution regular at $z=0$ is given by the hypergeometric function
\[ u=F(l-1,l+3,2l+2;z) \]
In particular for $l=1$ the function $u$ is constant, it represents
 the angular momentum charge solution (\ref{sz}), and
corresponding $\bf y$ is finite on the horizon. On the other hand,
for $l\geq 2$ we obtain logarithmic divergence of the hypergeometric
function $F$ at $z=1$. More precisely,
 \[ F(l-1,l+3,2l+2;z) = z^{-2l-1}[P(z) + \tilde P (z)\ln(1-z)] \]
 where $P$ and $\tilde P$ are polynomials. The solution is not regular at
$z=1$ and it confirms the result of
Vishveshwara \cite{Vish} that the only nontrivial stationary perturbation
is given by the axial perturbation with $l=1$. The interpretation of
this solution is given by (\ref{sz}).
 One can check by direct
computation for the Kerr metric \cite{Kerr}, \cite{B-L}:
\[  \rho ^2 := r^2 + a^2 \cos^2 \theta \quad \triangle := r^2 -2mr +a^2 \]
\[ g_{00} = -1 +{2mr \over \rho ^2}  \quad g_{0\phi} = -{2mra\sin ^2 \theta
\over \rho ^2} \quad g_{rr} = {\rho ^2 \over \triangle} \quad
g_{\theta \theta} = \rho^2 \]
\[ g_{\phi\phi} =
\sin^2\theta\left(r^2+a^2+{2mra^2\sin^2\theta\over\rho^2}\right) \]
 that the infinitesimal
angular momentum gives the same result. More precisely,
 the linear part of the Kerr metric $g_{\mu\nu}$
 with respect to the parameter $a$:
\[ g_{\mu\nu}\rd x^\mu\rd x^\nu =
 \eta_{\mu\nu}\rd x^\mu\rd x^\nu - 4\frac{ma}r\sin^2\theta \rd
t\rd\phi + O(a^2) \]
gives only the one non-vanishing component of the linearized Kerr metric on the
Schwarzschild background:
 \[ h_{0\phi}=-\frac{2ma}r \sin^2\theta \]
From the definition (\ref{y}) and the equation (\ref{h3A}) we can
calculate the invariant ${\bf y}=12ma\cos\theta/r^2$.
It is easy to compare the result with (\ref{sz}) and it gives
angular momentum charge $s^z=ma$.

Can we consider monopole solution in axial invariant?
The definition (\ref{y}) in terms of the initial data does not allow
nontrivial monopole part of the axial invariant.
However, we can consider such situation if we admit singular metric.
Formally this case corresponds to the infinitesimal Taub-NUT solution. The
Taub-Nut metric \cite{Taub}, \cite{NUT}:

\[ g_{\mu\nu}\rd x^\mu\rd x^\nu =
(r^2 +l^2)(\rd\theta^2 +\sin^2\theta\rd\phi^2) +\tilde v^{-1} \rd r^2 -
\tilde v (\rd t +2l\cos\theta\rd\phi)^2 = \] \[
= \eta_{\mu\nu}\rd x^\mu\rd x^\nu
-4vl\cos\theta \rd t \rd\phi + O(l^2) \, ;\quad
 \tilde v := {r^2 -2mr -l^2 \over r^2 + l^2} \]
gives the linearized metric $h_{0\phi}=-2lv\cos\theta$ and the monopole
axial invariant
$\displaystyle {\bf y}=-\frac{4l}r(1-\frac{3m}r)$. We should stress that
this is only
formal calculation because the tensor $h_{0\phi}=-2lv\cos\theta$ is not
well defined along the $z-$axis and is excluded as a global solution.
The monopole charge in $\bf y$ plays a role of the topological
obstruction for the existence of the global metric. It is similar to the
magnetic monopole in electrodynamics (see also \cite{JJspin2}).

For polar degree of freedom $\bf x$ from equation (\ref{ZE}) we
get
$$ 4m^2\ddot{\bf x}=(1-z)^2 z^4 \frac{\partial^2 {\bf x}}{\partial z^2}
 -(1-z) z^4 \frac{\partial {\bf x}}{\partial z} +
 (1-z) z^2\left[\frac13+\frac23(\mathbf{\Delta}-1)(\mathbf{\Delta}+2)(\mathbf{\Delta}+2-3z)^{-2}
\right](\mathbf{\Delta} +2) {\bf x} $$
and the corresponding stationary equation has the following form
\begin{equation}\label{stx}
 (1-z) z^2 \frac{\partial^2 {\bf x}}{\partial z^2}
 - z^2 \frac{\partial {\bf x}}{\partial z} +
 \left[\frac13+\frac23(\mathbf{\Delta}-1)(\mathbf{\Delta}+2)(\mathbf{\Delta}+2-3z)^{-2}
\right](\mathbf{\Delta} +2) {\bf x}=0 \end{equation}
For $l=0$ the solution $\displaystyle {\bf x}=\frac{z}{2-3z}$ is related to
(\ref{p0}) and corresponds to the mass charge. More precisely, if we put
in the metric (\ref{Sch})
$m+\delta m$ instead of parameter $m$ and take the linear part in $\delta
m$ we get
\[ \eta_{\mu\nu}(m+\delta m)\rd x^\mu\rd x^\nu =
 \eta_{\mu\nu}(m)\rd x^\mu\rd x^\nu + \frac{2\delta m}r \rd t^2 +
 \frac{2\delta m}{v^2r} \rd r^2 + O(\delta m^2) \]
 and the invariant $\displaystyle {\bf x}= 2{\cal B} h^{33}=\frac{4\delta m}r
\left(1-\frac{3m}r\right)^{-1}$.
 If we compare with mass charge (\ref{p0}) we obtain
$p^0=\delta m$ (see also \cite{RW}).

For $l\geq 2$ we have the following transformation law:
\[ \mathbf{\Delta}(\mathbf{\Delta}+2){\bf x}=-6z^2(1-z)\frac{\partial {\bf y}}{\partial z} +
\left[ \mathbf{\Delta}(\mathbf{\Delta}+2) +6z(1-z)-18z^2(1-z)(\mathbf{\Delta}+2-3z)^{-1} \right] {\bf y}
\]
which moves the solution of (\ref{sty}) into solutions of (\ref{stx}).
In particular, it is clear that for $l\geq 2$ the stationary solutions of
the equation (\ref{stx}) have the same logarithmic divergence on the
horizon as the solutions of the equation (\ref{sty}).
The explicit stationary solutions (in a specific gauge) were also given by
Zerilli in \cite{Zerilli} and here we present a gauge invariant confirmation
of his result.

\underline{Remark} Although
 $l=1$ is excluded in the definition of $\bf x$,
 we can consider another variable
 \[ (\mathbf{\Delta} +2)^{-1}{\bf x}:= (\mathbf{\Delta} +2)^{-1}r^2\chi^{AB}{_{||AB}} -\frac
12 H +\left(\mathbf{\Delta} +2-\frac{6m}r\right)^{-1}
[2h^{33} +2rh^{3A}{_{||A}}-rvH_{,3}] \]
which is no longer gauge-invariant in its dipole part
(see also (\ref{trg}) in Appendix A). More precisely, it transforms with
respect to the gauge transformation as follows
\[ \mbox{dip}\left(\frac12 H +\frac{r}{6m}vQ\right)=
 \mbox{dip}(-(\mathbf{\Delta} +2)^{-1}{\bf x}) \longrightarrow \mbox{dip}(-(\mathbf{\Delta}
+2)^{-1}{\bf x}) + \mbox{dip}(\xi^A{_{||A}}) \]
where $Q$ is defined in Appendix A by (\ref{defQ}).
Formally, the dipole solution $(\mathbf{\Delta} +2)^{-1}{\bf x}=\ln(1-z) Y_1(x^A)$
  fulfills the same equation (\ref{stx}), and it
corresponds to the Regge-Wheeler infinitesimal translation as a gauge
transformation\footnote{If we assume that the translation corresponds to
the $\xi^3=\cos\theta$ then the component $\xi^\theta$ is uniquely defined
as the polar gauge transformation preserving
the gauge condition $h_3{^A}_{||A}=0$ which has been used by Regge-Wheeler
 \cite{RW} and Vishveshwara \cite{Vish}. }
 (eq. 32 in \cite{RW}):
\[ \xi^3=\cos\theta \, ; \quad \xi^{\theta}=\frac{\sin\theta}{2m}\ln
\left(1-\frac{2m}r\right) \, ; \quad \xi^A{_{||A}}=\frac{\cos\theta}m \ln
\left(1-\frac{2m}r\right)
\]
This way the stationary polar dipole solution corresponds to the
infinitesimal translation gauge and it is also logarithmically divergent
on the horizon.

\section{The symplectic form and its reduction}

In this section we show the relation between the symplectic structure and the
invariants introduced in the section 4.
 Let $(P^{kl},h_{kl})$ be the  Cauchy data on a hypersurface $\Sigma$.
  Let us consider the symplectic structure
  \[ \Omega:= \int_{\Sigma}\delta{P}^{kl}\wedge \delta{h}_{kl} \]
 It is invariant with respect to the spatial gauge transformation which
 is fixed on the boundary ($\left.\delta\xi^k\right|_{\partial {\Sigma}}=0$):
\begin{eqnarray}
 \int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl}
 & \stackrel{\xi^k}{\longrightarrow} &
  \int_{{\Sigma}} \delta{P}^{kl}\wedge\delta{h}_{kl} +
  2 \int_{\partial {\Sigma}}\delta{P}^3{_l}\wedge \delta{\xi}^{l} \nonumber
 \end{eqnarray}
Moreover, it is invariant with respect to the temporal gauge if we assume
that $\xi^0$ and its normal derivative\footnote{You may ``correct'' the
symplectic form by a boundary term in such a way that the result
is gauge invariant for $\xi^{\mu}$ vanishing on the $\partial {\Sigma}$ without an
extra assumption about derivatives (see \cite{Kij-qlh}).}
  are fixed on the boundary:
  \begin{eqnarray}
 \int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl} &
 \stackrel{\xi^0}{\longrightarrow} &
  \int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl} +\nonumber \\
  & &   +\int_{\partial {\Sigma}}\sqrt{\eta}\left[
  \delta (N\xi^0)_{|k}\wedge\delta h^{3k} -\delta(N\xi^0)^{|3}\wedge\delta h
  + N\delta\xi^0\wedge \delta(h^{|3}-h^{3l}{_{|l}}) \right] \nonumber
  \end{eqnarray}
Roughly speaking, the symplectic structure is invariant with
respect to the gauge modulo boundary terms.
 The quadratic form $\int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl}$
can be decomposed into monopole part, dipole part and the remainder in a
natural way.

 From the considerations given in the Appendix B (formulae (\ref{rap}) and
(\ref{rpp})) we can easily see that the ``radiation'' part
\[ \underline{\Omega}=
\int_{{\Sigma}}\delta\underline{P}^{kl}\wedge\delta\underline{h}_{kl} \sim
\int_{{\Sigma}}\delta\underline{\bf X}\wedge
\mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}\delta\underline{\bf x}+
 \delta\underline{\bf Y}\wedge
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta\underline{\bf y} \]
where symbol ``$\sim$'' denotes equality modulo
boundary term. Moreover, the ``mono-dipole'' part has the form
(see (\ref{mpp}) and (\ref{dap}) respectively)
 \begin{equation}\label{mfs}
 \mbox{mon}( \int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl}) =
 \int_{{\Sigma}}
\frac1{2}\delta{P}_{33}\wedge \delta {\cal B}^{-1}\mbox{mon}({\bf x} )+
\frac12 \int_{\partial {\Sigma}} r \delta P^3{_3}\wedge\delta \mbox{mon}(H)
\end{equation}

 \begin{equation}\label{dfs}
 \mbox{dip}(\int_{{\Sigma}}\delta{P}^{kl}\wedge\delta{h}_{kl}) \sim
 -\int_{{\Sigma}} \Lambda \delta\mbox{dip}({\bf y}) \wedge
 \mathbf{\Delta} ^{-1}\delta({h}_{3A||B}\varepsilon^{AB})
\end{equation}

The monopole part of the polar invariant $\mbox{mon}({\bf x} )$ and
the dipole part of the axial one
 $\mbox{dip}({\bf y})$ represent mass and angular momentum respectively, and
 they are supposed to be fixed. These quantities are analogous to the
electric charge in electrodynamics. If we assume that there is no matter
inside volume ${\Sigma}$
then both of them are fixed by the constraints, provided that they are
controlled at $r=2m$ as a boundary condition on the horizon. Let us assume
that $\left. \delta\mbox{mon}({\bf x} )\right|_{S(r=2m)}=0$,
$\left. \delta \mbox{dip}({\bf y})\right|_{S(r=2m)}=0$ then
 the mono-dipole part (\ref{mfs}) and (\ref{dfs})
vanish and the symplectic
structure reduces to the ``mono-dipole-free'' invariants
\begin{equation}\label{ssonS}
 \int_{\Sigma}\delta{P}^{kl}\wedge\delta{h}_{kl} \sim \int_{\Sigma}
\delta\underline{\bf X}\wedge\mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}\delta\underline{\bf x}+
\delta\underline{\bf Y}\wedge\mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta\underline{\bf y}
\end{equation}
This way we obtain $\underline{\bf X}, \underline{\bf x},
 \underline{\bf Y}, \underline{\bf y}$ as the quasi-local canonical
variables describing reduced unconstrained initial data on $\Sigma$.

\underline{Remark} One can ask the question when (\ref{ssonS}) is a strict
equality not only modulo boundary term.
 The symplectic 2-form $\Omega$
reduces to the mono-dipole-free invariants if we assume the following
boundary conditions which fix the gauge freedom on the boundary:
 \[ \left. \delta h_{AB}\right|_{\partial {\Sigma}}=0 \quad
 \left. \delta(2h^3_3+2rh_3{^A}{_{||A}}-rH_{,3})\right|_{\partial{\Sigma}}=0
 \]
 \begin{equation}\label{bd}
  \left. \delta(P^{3A||B}\varepsilon_{AB}) \right|_{\partial {\Sigma}}=0 \quad
   \left. \delta (P^3_3 + 2r{\mathbf{\Delta}}^{-1}P^{3A}{_{||A}} )
    \right|_{\partial {\Sigma}}=0  \end{equation}
 In Appendix A we analyze the possibility when $\underline H$,
 $\underline Q$, $\underline \Pi$ and
$\chi^{AC}{_{||CB}} \varepsilon_A{^B}$ are precisely the gauge conditions
and then $\chi^{AB}{_{||AB}}$ corresponds to the control of
$\underline{\bf x}$.
Roughly speaking, the ``control mode'' given by (\ref{bd}) contains
four boundary conditions related to the gauge freedom plus two boundary
conditions for the unconstrained degrees of freedom which we propose to call
Dirichlet boundary data.

If we introduce quasilocal coordinates:
\[ q^1:=\mathbf{\Delta} ^{-1/2}(\mathbf{\Delta} +2)^{-1/2}\underline{\bf x} \, ; \quad
 p_1:=\mathbf{\Delta} ^{-1/2}(\mathbf{\Delta} +2)^{-1/2}\underline{\bf X} \]
\[ q^2:=\mathbf{\Delta} ^{-1/2}(\mathbf{\Delta} +2)^{-1/2}\underline{\bf y} \, ; \quad
 p_2:=\mathbf{\Delta} ^{-1/2}(\mathbf{\Delta} +2)^{-1/2}\underline{\bf Y} \]
we can rewrite the reduced symplectic structure (\ref{ssonS})
 in the canonical form
\[ \Omega = \int_{\Sigma}
\delta\underline{\bf X}\wedge \mathbf{\Delta} ^{-1}(\mathbf{\Delta}
+2)^{-1}\delta\underline{\bf x}+
 \delta\underline{\bf Y}\wedge \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}
\delta\underline{\bf y} = \sum_{n=1}^2
\int_{\Sigma}  \delta p_n \wedge \delta q^n
 \]
 Unconstrained initial data for the full nonlinear theory (which is
similar to the considerations in this article) has been proposed in
\cite{CQG94}. Moreover, the concept of quasilocality appeared
in \cite{Kyoto91} and has been developed in \cite{Kij-qlh}.  The boundary
data possesses its counterpart in the full nonlinear theory (see
proposition in \cite{Kij-qlh}) and it will be discussed elsewhere.

\section{Energy and angular momentum of the gravitational waves}

The reduction of the symplectic form (\ref{ssonS}) allows to
formulate the hamiltonian relation in terms of the reduced canonical
variables
 \[  \int_{\Sigma} \dot{\bf x}\mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}  \delta {\bf X}
 - \dot{\bf X} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta {\bf x} + \]
 \[  + \int_{\Sigma} \dot{\bf y}\mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}  \delta {\bf Y}
 - \dot{\bf Y} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta {\bf y} =  16\pi\delta
{\cal H} + \]
 \begin{equation}  + \int_{\partial {\Sigma}} \frac{\Lambda}r v (r{\bf x}) ,_{3}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta {\bf x}
 +\frac{\Lambda}{r} v (r{\bf y}) ,_{3}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} \delta {\bf y} \end{equation}
where
\[ 16\pi{\cal H}:= \frac12 \int_{\Sigma} \frac v\Lambda {\bf X}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} {\bf X} + \frac v\Lambda {\bf Y}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} {\bf Y}+
\]
\[  + \frac1{2} \int_{\Sigma} \frac{\Lambda}{r^2}\left[ v (r{\bf x})_{,3}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} (r{\bf x})_{,3} +
  {\bf x} \frac{r^2}v V^{(+)}\mathbf{\Delta} ^{-1} (\mathbf{\Delta} +2)^{-1}{\bf x} \right] +\]
\begin{equation} + \frac1{2} \int_{\Sigma} \frac\Lambda{r^2} \left[ v (r{\bf y})_{,3}
 \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1} (r{\bf y})_{,3} + {\bf y}
\frac{r^2}v V^{(-)}\mathbf{\Delta} ^{-1} (\mathbf{\Delta} +2)^{-1} {\bf y} \right]
 \label{Hxy} \end{equation}
 (see also eq. 4.19 and 5.34 in \cite{Moncrief}). \\
Similarly for angular momentum we propose the following expression
\begin{equation}\label{JZ} 16\pi J(Z)=\int_\Sigma {\bf X} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}
 Z^A\partial_A{\bf x} + {\bf Y} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}
 Z^A\partial_A{\bf y}
\end{equation}
where $Z$ is a Killing field (\ref{Zk}). In particular for $Z=\partial
/\partial\phi$ the $z-$component of the angular momentum takes the form:
\begin{equation}\label{Jxy} 16\pi J_z=\int_\Sigma {\bf X} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}
{\bf x}_{ ,\phi} + {\bf Y} \mathbf{\Delta} ^{-1}(\mathbf{\Delta} +2)^{-1}
{\bf y}_{ ,\phi}
\end{equation}
The conservation laws for the energy and angular momentum
\[ \partial_0 {\cal H} = 0
\]
\[ \partial_0 J(Z) = 0
\]
are fulfilled if we assume appropriate boundary conditions on the horizon
and at the spatial infinity. The natural choice of those conditions is to
assume that $\underline{\bf x}$ and $\underline{\bf y}$ are vanishing on
the boundary $\partial\Sigma$.

After separation of the angular variables in (\ref{Hxy})
 we obtain a hamiltonian which has
been used for the energy method by Wald \cite{Wald} (see also in
\cite{Chandra} eq. 386) and it confirms stability of the Schwarzschild
metric (see also \cite{Moncrief}).
Here we have shown how to combine an energy of different multipoles together.

\subsection{Regular initial data}
If we assume that the invariants ${\bf x}$, ${\bf y}$ are vanishing on
the horizon then we get a nice hamiltonian system outside of the horizon.
We can also assume that $\underline{\bf x}$ and $\underline{\bf y}$ are
fixed and finite on the horizon. More precisely, we propose the following
initial-boundary data:\\
1. mon($\bf x$)=0, because of the singularity at $r=3m$,
moreover this charge indicates that we have chosen wrong parameter $m$ in
the background.\\
2. dip($\bf y$) -- weak internal angular momentum.\\
 3. The radiation data $(\underline{\bf x}$, $\underline{\bf X}$,
$\underline{\bf y}$, $\underline{\bf Y})$
has to be finite on the horizon, moreover $\underline{\bf x}$,
$\underline{\bf y}$  should be controlled as a Dirichlet boundary condition.
We assume also that $h_{\mu\nu}=O(1/r)$ at spatial infinity.
The center of mass contained in dip$[(\mathbf{\Delta}+2)^{-1}{\bf x}]$ and
linear momentum in dip$(\Pi)$ can be always ``gauged out'', see Appendix A.1,
and those gauge transformations correspond to the infinitesimal
translation and boost respectively. Performing those gauge transformations
we pass to the ``center mass rest frame''.  \\
4. The asymptotics of the invariants $\underline{\bf x}$, $\underline{\bf y}$
at spatial infinity should guarantee finite hamiltonian and this can be
achieved for standard asymptotics $h_{\mu\nu}=O(1/r)$.
Unfortunately, the standard asymptotics $\frac 1r$ does not guarantee
finite angular momentum. More precisely, standard asymptotics at spatial
infinity $\underline{\bf x}, \underline{\bf y}=O(1/r)$,
$\underline{\dot{\bf x}}, \underline{\dot{\bf y}}=O(1/r^2)$
gives logarithmically divergent integral in (\ref{Jxy}).
The proposition by Christodolou and Kleinerman \cite{Ch-Kl}, so-called
S.A.F. condition, fits perfectly if we adopt it to the linearized theory.
We propose the following asymptotics at spatial infinity
\[ {\bf x},{\bf y}=O(r^{-3/2})\, , \quad \dot{\bf x},\dot{\bf y}
=O(r^{-5/2}) \]
\underline{Remark}
Christodolou and Kleinerman assume that the full ADM data has better
asymptotics (except conformal factor in the riemannian metric). Let us
notice that if we assume that $P^{kl}=O(r^{-1/2})$ and
$h_{kl}=O(r^{-3/2})$ then all boundary
terms in the symplectic form (analyzed in Appendix B) are vanishing like
$\frac 1r$ at spatial infinity.

\section{Conclusions}
We have shown a natural functionals which represent energy and angular
momentum of the weak gravitational radiation on a Schwarzschild background.
Particularly, we do not have to separate angular variables and the result is
gauge-invariant.
Moreover, the equations of motion for the gauge-invariant degrees of
freedom correspond to the well-known Regge-Wheeler and Zerilli results. We
give also a complete (together with the interpretation) analysis of the
stationary solutions.

After separation of the angular variables we should notice that the
results presented in sections 4, 6 and 7 are very close to the approach
presented by Moncrief in \cite{Moncrief}. However, the invariants used by
Moncrief in polar part are different and non-local.

redpajama_set_name: RedPajamaArXiv


Seminar program has been released!
We released the details of the seminar of Halal Expo Japan 2016. The wide range of topics will be talked by many speakers who will participate from the government, company, and university. This is the best opportunity to learn Halal and Muslim tourism from various angles. Please do not miss it.
← Visitor Pre-registration started! Everyone is welcome!


The story of a nervous collapse and descent into madness. Lost in a nightmare of visions, voices and words, a con artist is forced to search for an elusive figure manhunted by a vicious criminal gang. "You know what life is? It's your last chance. A day never goes by again."


There are three question to answer, to know whether you need KM.
If you are considering whether to invest in Knowledge Management, there are three questions you need to ask. These are listed and explained below.
Is knowledge a key business issue for you?
If knowledge is a key business driver in your organisation, or if knowledge is one of your key products, then knowledge management may well be important for you. Knowledge is defined by Peter Senge as "the ability to take effective action", and knowledge is the basis both of judgment, and of good decision making.
If your organization requires good knowledge based decisions, then knowledge is one of your key assets.
If you are a consulting firm, a contractor, or an educational or professional body that creates and deploys knowledge on behalf of customers and clients, then again knowledge is one of your key assets.
Knowledge will also be a key business issue for you if your staff turnover is large, and you need to transfer knowledge into a new generation of workers.
or if much of your core operational knowledge is held by people approaching retirement age.
or if you are involved in repeat activity, where knowledge from the past can help improve future performance.
orif many dispersed parts of the business are performing the same process, with varying results.
Is there evidence that management of knowledge might be sub-optimal in your organisation?
You need to collect evidence, in the form of anecdotes, examples and social network plots, which show that knowledge management could be improved.
Is it likely that improved knowledge management might add real value?
If you do improve knowledge management, is it likely that this would lead to better performance? Better performance could come through eliminated mistakes, more consistent performance with poor performing teams learning from high performing teams, from faster transfer of knowledge to young staff, and from better retention of knowledge from departing staff. You need to be able to make a good case that there is potential value here; enough value to investigate knowledge management further.
You are not making a full business case; this business case will be made in the next stage. You are however marshalling the anecdotal evidence that performance improvement is possible, and that this is potentially significant enough to take the next step. To make the case for further investment, it is good to have the following evidence.
A list of likely positive business outcomes from better knowledge management.
Once the decision to invest in KM has been made, then you move on to the standard KM implementation steps of assessment, strategy, piloting, roll-out and embedding.
Excellent, spot on, and simple!
Actually, the implied assumption was that an enterprise has asked an individual into the firm because the firm already felt the need to explore KM. Pain must already exist in order for the relevance of Km to be felt.
In the last 35 years, when consulting in IM and KM, I have been invited into firms to discuss, describe, and define KM. Not a single one of those invitations resulted in a consulting initiative. of course, you might say that was my problem. Most organizations still do not comprehend Intellectual Capital, a lot less how KM could increase employee engagement and corporate performance management.
A conversation over what KM is results in a warm fuzzy feeling for us, since we purport to understand the field. A dialogue over enterprise pain, can result in action.
I think thats the case Michael. My assumption was that there was at least someone in the organisation who was wondering whether knowledge management might be useful. This blog already preaches to the choir, as the people who read and subscribe already have an interest in KM, and there may be a choir member in the organisation who might feel the need to "make the case for KM" to their management.
I also agree, that its very difficult to sell KM to a company that doesn't feel the need for it. The first step is to identify a need - a pain point that KM might address.


The truth is, if you know your prescription and have your existing rx glasses convenient, it is very quite simple to purchase glasses online.
your current prescription. I keep mine in medical file in my own filing cabinet. If you can't find your prescription, visit wherever you previously had your sight examined and obtain a copy. Your optician, optometrist, or ophthalmologist is necessary by law to release your eyewear prescription to you. You now need to be aware of a few critical pieces of information from your prescription (see example above–click to enlarge). Among other information, you should see three columns on your prescription–sphere, cylinder, and axis.
In the example prescription above, you will see the Pupillary Distance of the individual is 62. Write down this quantity also, it's the seventh of your seven critical figures.
A slightly less critical number is your temple size. If you're lucky, you'll also visit a temple span and bridge size in your prescription. If not, don't sweating it, you can determine it out yourself. The temple span describes the length of both bars that connect the spectacles to your ears. It really is measured from where the 90-degree curve from your zoom lens structure to the side bars (temples) commences to the end of the temple, like the curve. It is not measured from the flex point of the temples, because the "stub" between the flex point and zoom lens structure is not a standard size.
Now, remove some measuring tape (preferably the kind of tape used for appropriate clothes and sewing, as it's very flexible) and measure an existing pair of spectacles that fit easily. The average temple span for men is 135mm or 145mm. For girls, 130mm to 140mm is standard. Should your tape only has inches, measure to the nearest eighth or sixteenth and multiply that number by 25.4 to obtain the measurement in millimeters. Temple lengths are generally available in 5mm increments, such as 130, 135, 140, 145, etc. For the most part online glasses stores, temple span is shown in just a frame description and is not customizable. Therefore, you will need to find a frame with a proper temple size. This sometimes varies if you are in a site specializing in designer glasses, where you have the choice to source your own temple size.
Seriously, temple size is not a huge deal. I've one pair of eyeglasses with 135mm temples and another with 140mm temples. Both pairs fit perfectly. If for reasons uknown you can't come up with a temple measurement, don't allow that stop you from making an eyewear purchase online. Instead, simply choose 140mm temples if you are a guy and 135mm temples if you are a female. Chances are, this duration will fit easily.


text:
Dredd: Underbelly Comic Book Is The Sequel Fans Might Be Hoping For
By Rudie Obias | 7 years ago
The Dredd reboot from 2012 was a box office disappointment, as it only took in $35.6 million worldwide against a $50 million production budget. While Dredd has seen some success on home video and with online streaming, the reboot hasn't really moved the needle enough for Lionsgate to justify making a sequel. For the meantime, a comic book sequel will have to do.
As reported on Movies.com, Dredd: Underbelly will be the first-ever official U.S. format release from 2000AD publisher Rebellion. The anthology collects Arthur Wyatt and Henry Flint's comic book sequel to Dredd, which was originally serialized earlier in the year in the British publication Judge Dredd Megazine. According to 2000AD editor Matt Smith, the Dredd: Underbelly release is aimed at the American market and will capitalize on the awareness of the Dredd character, played by Karl Urban. Smith says:
"We've noted a real swell of US interest in Dredd off the back of the film. This first foray for 2000AD in producing a US-format book, geared specifically to the [comic book] direct market, and aimed at satisfying those demanding a sequel, is a stumm gas grenade-style lob across the Atlantic to show that you don't have to camp outside your local multiplex to get your next fix of Mega-City One — the Judge will carry on enforcing the Law through the pages of the comic. Arthur and Henry [creators Arthur Wyatt and Henry Flint] have created a terrific story, and a powerful follow-up to the film — Dredd fans won't be disappointed.
When the Judge Dredd comic book launched in 1977, it only had distribution in its native United Kingdom. The publishers have tried to tap into the American market with distribution and licensing deals with Quality Comics, DC Comics, and now IDW Publishing. 2000AD publisher Rebellion has been re-issuing Dredd to American readers since 2010 and has signed an exclusive distribution deal with Simon and Schuster. Aside from Free Comic Book Day releases, Rebellion has stayed away from releasing Dredd in comic book stores, until now.
While a comic book sequel isn't ideal for die-hard Dredd fans, at least the story will continue from the 2012 action film. Fans will get the opportunity to see the next adventures Judge Dredd and Judge Anderson go on after they leave the Peach Trees housing block. A few months ago, fans organized a "Dredd Day of Action" to build awareness for the character and to send Lionsgate a direct message that fans want a sequel to Dredd. September 18 was designated as "Dredd Day," as fans would either buy copies of the movie on Blu-ray/DVD or iTunes, or stream the film on Netflix Instant or Amazon Prime.
There's no telling why Dredd did so poorly in theaters last year. The movie is fun, exciting, and one of the most entertaining action films in 2012. Maybe it was the poor reception of the original Judge Dredd film in 1995 or the plot similarities between Dredd and The Raid: Redemption, which was released earlier in 2012, but a movie like Dredd should've been seen by more people. Now that Karl Urban is on Almost Human, maybe more people will discover Dredd on home video or online streaming.
Dredd: Underbelly will hit physical comic stores and online retailers such as the Amazon Kindle Fire, Barnes & Noble Nook, and 2000AD iPad app in January 2014. If you like to watch a Dredd sequel, then sign the online petition. If Firefly and Veronica Mars both got fan-generated feature films, then why can't Dredd? Keep the faith!!
We Want Karl Urban But Judge Dredd's New TV Series Is Making This Huge Mistake
Dropped From Netflix: There's Only One Marvel Movie Left
Flash Gordon Slapped With An Offensive Content Warning
The Next Terminator Movie Will Switch Genres

redpajama_set_name: RedPajamaCommonCrawl