text:
\section{Introduction}

The metric of the stationary space-time resulting from the gravitational
collapse of (rotating) black hole is diffeomorphic to the Kerr metric
\cite{carter,robinson,chand}.
Because general diffeomorphisms contain also supertranslations
(angular-dependent
time translations at the null infinity) \cite{strom1}, the  metric of the
final state of collapse, in general, contains also supertranslation field
associated with supertranslations.

The Israel theorem states that "among all static, asymptotically flat vacuum
space-times with closed, simply connected equipotential surfaces
$g_{00}=const$, the Schwarzschild solution is the only one which
has a nonsingular infinite-red-shift surface $g_{00}=0$" \cite{isr1,isr2}.
Physically, this means that no static
asymmetric perturbation by sources within the horizon can preserve
regularity of the event horizon. Examples of such sources are quadrupole
perturbations \cite{dor},
 magnetic dipole fields \cite{gin}  inside a black hole,  and
small perturbations of the Schwarzschild black hole \cite{RW}.

A family of vacuum solutions of the Einstein equations diffeomorphic to
Schwarzschild metric and containing a supertranslation field
was constructed in \cite{comp2}.
The metric containing a supertranslation field is
obtained from the Schwarzschild metric
by a diffeomorphism containing a supertranslation field and is
 not spherically symmetric.
The metrics with supertranslation field are physically different
from the Schwarzschild metric, and
transformations of the metric which depend on supertranslation field
are not pure (gauge) transformations, and,  in a general case,
change  superrotation charges
\cite{strom1,comp2,comp1}.

In this note we consider a vacuum axisymmetric solution of the Einstein
equations with a supertranslation field.
The metric is obtained by transformation of the solution of \cite{comp2}
and is chosen so that the horizon  of the metric is located at the
sphere of the radius $2M$, where $M$ is the mass parameter of the
solution (and the original Schwarzschild metric).

The metric satisfies all conditions of the Israel theorem, except for
the condition on the form of the metric at spatial infinity. Nevertheless,
repeating  the steps of the proof of the theorem, we show that
in special coordinates, used in the proof of the theorem,
the metric is spherically symmetric.

We show that a seeming discrepancy is a consequence of the form of
transformation to coordinates in which the theorem is proved.
Transformation from coordinates in which the metric is
initially written to  coordinates used in the proof of the theorem
is not a pure gauge transformation, but a diffeomorphism containing
supertranslation  field.
The combined action
of both "large" transformations - the initial one
from the Schwarzschild solution to an axisymmetric solution
containing supertranslation field and the second to the preferred
coordinates  used in the proof of the theorem -
 produces the metric independent of supertranslation field.

The note is organized as follows. After  short reviews of
the formulation of the Israel theorem and the form of
 the axial symmetric vacuum solution with supertranslation field,
we discuss the coordinate transformation used in the proof of the theorem.
Next, following the steps of the proof of the theorem,
we consider application of the inequalities following from finiteness
of the Kretschmann's scalar \cite{isr1,isr2} to the present case.
Applying the inequalities, we conclude that the (transformed) metric is
spherically symmetric and discuss the source of the seeming discrepancy.
In the
Appendix we  calculate the Kretschmann scalar and
show that it coincides with that for the Schwarzschild solution.

\section{Conditions of the Israel theorem}

The  Israel theorem states that the only vacuum static metric satisfying
the conditions listed below is the Schwarzschild solution.

The conditions are as follows:
Let $\Sigma$ be a hypersurface $t=const$, maximally extended so that
the square of the Killing vector $\xi$ is negative $\xi\cdot\xi<0$.
It is assumed that the 3D hypersurface $\Sigma$ is regular and
non-compact.
In this case the line element can be locally reduced to a form
\begin{equation}
\label{n.1}
ds^2 =g_{\alpha\beta}(x^1 ,x^2 ,x^3 )dx^\alpha dx^\beta -V^2 (x^1 ,x^2, x^3 )dt^2
,\end{equation}
where $|V^2|=\xi_\alpha\xi^\alpha$ (Greek indices run 1-3) .

The metric has the following asymptotic form
\begin{eqnarray}
\nonumber
1.&{}& g_{\alpha\beta} =\delta_{\alpha \beta } +O(r^{-1}),\qquad \partial_\gamma g_{\alpha\beta}=O(r^{-2}),
\qquad  r^2 =g_{\alpha\beta} x^\alpha x^\beta\rightarrow\infty,\\\nonumber
2.&{}& V=1-M/r +\eta ,\qquad \eta =O(r^{-2}), \qquad \partial_\alpha \eta = O(r^{-3}),
\qquad \partial_\alpha\partial_\beta\eta = O(r^{-4}),
\end{eqnarray}
and the surfaces $V(x)=const >0$ are  connected closed regular 2D surfaces.
\newline
\hspace*{1.7cm}3.\,\,\,
 4D invariant $R_{ijkl}R^{ijkl}$ is bounded on $\Sigma$.\newline
\hspace*{1.7cm}4. \,\,
If the greatest lower bound of $V$ on $\Sigma$ is zero, then  geometry of
the equipotential surfaces $V=\varepsilon$ in the limit $\varepsilon\rightarrow 0$ approaches the
geometry corresponding to a closed regular 2-space of finite area.

Because the metric is a vacuum solution, from the Einstein equations it follows
that
\begin{equation}
\label{1}
g^{\alpha\beta}R_{\alpha\beta}=0,\qquad R_{\alpha\beta}+ V^{-1}V_{;\alpha\beta}=0,
\end{equation}
$(;)$ denotes covariant derivative. The consequence of (\ref{1}) is that $V$
is  harmonic function
\begin{equation}
\label{2}
g^{\alpha\beta} V_{;\alpha\beta}=0.
\end{equation}
\section{Vacuum solution with supertranslation field}

We consider a class of vacuum, static, asymptotically flat metrics containing
supertranslation field $C(\theta )$ constructed in \cite{comp2}
\begin{equation}
\label{3}
ds^2 =-\frac{(1-M/2\rho_s )^2}{(1+M/2\rho_s )^2}dt^2 +
(1+M/2\rho_s )^4 \left(d\rho^2 + (((\rho -E)^2 +U)\gamma_{ab} +(\rho -E)C_{ab})dz^a dz^b
 \right).
\end{equation}
The metric (\ref{3}) was obtained from the Schwarzschild metric
by application of a
diffeomorphism containing a supertranslation field.
 Variables $z^a$ are realized as angles $\theta ,\varphi$
on the unit sphere with the metric
 $ds_{(2)}^2 =\gamma_{ab}dz^a dz^b=d\theta^2 +\sin^2\theta d\varphi^2$.
The functions $C_{ab}$ and  $E,U$ depend on $ C(\theta )$ and
its derivatives,
\begin{equation}
\label{3.1}
\rho_s (\rho, C) =\sqrt{(\rho -C )^2 + {C'}^2(\theta ) }.
\end{equation}
The prime is derivative over $\theta$.
The horizon of the metric (\ref{3}) is located at
the surface $\rho_s (\rho ,C) =M/2$.

By the  transformation
\begin{equation}
\label{3.2}
r (\rho,\theta )=\rho_s (\rho, C)\left(1+\frac{M}{2\rho_s (\rho, C)}\right)^2
,\end{equation}
the metric (\ref{3}) is transformed to a form with the horizon located
at the sphere $r=2M$ \cite{iofa}:
\begin{eqnarray}
\nonumber
&{}&ds^2  =-V^2 dt^2 +  \frac{dr^2}{V^2(1-b^2 )} +2drd\theta\frac{br (\sqrt{1-b^2}-b' )}{(1-b^2 )V}+\\
\label{3.4}
&{}& + d\theta^2 r^2\frac{ (\sqrt{1-b^2}-b' )^2}{(1-b^2 )} +d\varphi^2 r^2\sin^2\theta (b\cot\theta
-\sqrt{1-b^2 })^2=\\
\label{4a}
&{}& =-V^2 dt^2 +\frac{dr^2 \bar{g}_{rr}}{V^2} +\frac{2drd\theta \bar{g}_{r\theta}}{V}+
d\theta^2 \bar{g}_{\theta\th}+ d\varphi^2 \sin^2\theta\bar{g}_{\varphi\p}
.\end{eqnarray}
In (\ref{3.4}) we introduced the functions
\begin{equation}
\label{3.6}
V^2=1-\frac{2M}{r},\qquad b=\frac{2C'(\theta )}{K}, \qquad K=r-M +rV.
\end{equation}
In (\ref{4a}) we separated the powers of $V$ and denoted by
$\bar{g}$  the $O(V^0 )$ parts of the metric components.
We  express the  spatial components of the metric and their inverse as
\begin{equation}
\label{3.7}
\begin{array}{c} {}\\gamma_{\alpha\beta}=\\{}\\{}\end{array}
\left|\begin{array}{ccc}
\bar{g}_{rr}/{V^2}  & \bar{g}_{r\theta}/{V}&0\\
 {\bar{g}_{r\theta}}/{V} & {g}_{\theta\th}&0\\
0&0&{g}_{\varphi\p}
\end{array}\right|
\qquad
\begin{array}{c} {}\\gamma^{\alpha\beta}=\\{}\\{}\end{array}
\left|\begin{array}{ccc}
V^2  & -{V\bar{g}_{r\theta}}/{{g}_{\theta\th}}&0\\
 -{V\bar{g}_{r\theta}}/{{g}_{\theta\th}} &{\bar{g}_{rr}}/{ {g}_{\theta\th}}&0\\
0&0&1/{g}_{\varphi\p}
\end{array}\right|
.\end{equation}
The metric components satisfy the identity
\begin{equation}
\label{3.8}
{g}_{rr}{g}_{\theta\th}-{{g}_{r\theta}}^2= \frac{{g}_{\theta\th}}{V^2}
.\end{equation}
The metric (\ref{3.4}) has the time-like Killing vector
$\xi^i =const \delta^i_t$ which becomes null at the horizon.
Solving the geodesic equations for null geodesics, it is possible to show
that the surface $r=2M$ is the surface of infinite redshift \cite{iofa}.

In the limit $r\rightarrow\infty$ the metric (\ref{3.4}) takes a form
\begin{equation}
\label{3.9}
ds^2 =-dt^2 +dr^2 +2C'(\theta )dr d\theta + r^2 (d\theta^2 +\sin^2\theta d\varphi^2 ).
\end{equation}
The metric satisfies all conditions of the theorem except for
 the first condition
on the form of the metric at spatial infinity.

\section {Transformation to coordinate system used in
the proof of the theorem}

To prove the theorem, Israel considered a system of the equipotential surfaces
$V=const,\, t=const$ with intrinsic coordinates $\hat{x}^A,\,\, A=1,2$.
Coordinates $\hat{x}^A$ are chosen
to be constant along the trajectories orthogonal to the surfaces $V=const$,
\begin{equation}
\label{n.2}
g^{\alpha\beta}\partial_\alpha \hat{x}^A\partial_\beta V=0
.\end{equation}
In coordinates $x^A ,V$ the spatial part of the metric is
\begin{equation}
\label{n.2a}
ds^2 =\hat{g}_{AB} d\hat{x}^A d\hat{x}^B +\hat{g}_{VV}dV^2.
\end{equation}
Substituting in (\ref{n.2}) the metric components from (\ref{3.7}), we have
\begin{equation}
\label{n.3}
g^{rr}\partial_r \hat{x}^A \partial_r V +g^{r\theta}\partial_\theta \hat{x}^A\partial_r V =0,
\end{equation}
where we used that $\partial_\varphi V=0$.
The $A=1$ component of Eq. (\ref{n.3}) is
\begin{equation}
\label{n.5}
V^2 \partial_r \hat{x}^1 - V\frac{ \bar{g}_{r\theta}(r,\theta ) }{{g}_{\theta\th}(r,\theta )}
\partial_\theta \hat{x}^1 =0
.\end{equation}
For the coordinate $\hat{x}^2$ we take coordinate $\varphi$. A general solution of the
partial differential equation  (\ref{n.5}) is
$\hat{x}^1 =F(\psi (r,\theta ))$, where $F$
is an arbitrary function and $\psi (r,\theta )= const$ is the integral
of the ordinary differential equation
\begin{equation}
\label{n.6}
dr  =-\frac{ d\theta V {g}_{\theta\th}(r,\theta ) }{\bar{g}_{r\theta}(r,\theta ) }
.\end{equation}
From Eqs.(\ref{n.2})-(\ref{n.3}) it follows that
\begin{equation}
\label{n.7}
\hat{g}^{AV} =\hat{g}_{AV}=0.
\end{equation}
Using the explicit form of $g^{\alpha\beta}$ [Eq. (\ref{3.7})], we have
\begin{eqnarray}
\nonumber
&{}&\hat{g}^{VV}=g^{\alpha\beta}\partial_\alpha V\partial_\beta V = g^{rr}(\partial_r V)^2= (V\partial_r V )^2\\
&{}&
\label{n.8}
\hat{g}^{12}=g^{\alpha\beta}\partial_\alpha \hat{x}^1\partial_\beta \hat{x}^2 =
g^{\theta\varphi}\partial_\theta \hat{x}^1 (\theta ,r)\partial_\varphi \varphi
+g^{r\varphi}\partial_r \hat{x}^1 (\theta ,r)\partial_\varphi \varphi =0,
\end{eqnarray}
where it was used that $g^{\theta\varphi}=g^{r\varphi}=0$. In coordinates $(x^A ,V)$
the metric is
\begin{equation}
\label{n.9}
d\hat{s}^2 = \hat{g}_{VV}dV^2 + \hat{g}_{11}d\hat{x}^1 d\hat{x}^1
+\hat{g}_{22}d\hat{x}^2 d\hat{x}^2.
\end{equation}
Equating  the expressions for the volume elements in  coordinates $(r,\theta,\varphi)$
and $(x^A ,V)$ [Eq. (\ref{n.9})], we have
$$
dr d\theta d\varphi\sqrt{det(g)}=dV d^2\hat{x} \sqrt{det(\hat{g})}
.$$
Using Eqs.(\ref{3.7}), \ref{3.8}), we obtain the volume element in coordinates
$(r,\theta,\varphi)$ as
\begin{equation}
dr d\theta	d\varphi \sqrt{ (g_{rr}g_{\theta\th}-g^2_{r\theta})g_{\varphi\p} }=
dr d\theta d\varphi\sqrt{ {g}_{\varphi\p}{g}_{\theta\th}V^{-2} }
.\end{equation}
In coordinates $(x^A ,V)$ the volume element is
\begin{eqnarray}
dV d^2\hat{x}\sqrt{ \hat{g}_{11}\hat{g}_{22}\hat{g}_{VV} }=dr \frac{\partial V}{\partial r} d^2\hat{x}
\sqrt{\hat{g}_{11}\hat{g}_{22}(V\partial_r V)^{-2} }
,\end{eqnarray}
and thus
\begin{equation}
\label{n.11}
 d\theta d\varphi\sqrt{ g_{\theta\th}g_{\varphi\p} }=d^2 \hat{x} \sqrt{\hat{g}_{11}\hat{g}_{22}V^{-2} }.
\end{equation}

\section{Israel inequalities}

The proof of the theorem is based on the use of relations which follow from
the assumptions of the theorem, notably from finiteness of the Kretschmann's
scalar and harmonic property of $V$ (\ref{2}).

The expressions for the extrinsic curvature of the surface
$V=const$ and related formulas are
\begin{equation}
\label{n.13}
K_{AB} = \frac{1}{2\rho}\frac{\partial \hat{g}_{AB}}{\partial V},\qquad
K=K_{AB}= K_{AB}\hat{g}^{AB},\qquad
\frac{\partial\sqrt{\hat{g}}}{\partial V}=\sqrt{\hat{g}}\rho K, \qquad
\end{equation}
From (\ref{n.13}) and (\ref{2}) follows the equation
\begin{equation}
\label{n.14}
\frac{\partial}{\partial V} \left(\frac{ \sqrt{\hat{g}}}{\rho }\right)=0
,\end{equation}
where $$\hat{g} =\hat{g}_{11}\hat{g}_{22}, \qquad
\rho (r) =(g^{rr}\partial_r V \partial_r V )^{-1/2}=(V \partial_r V )^{-1}. $$

Integrating (\ref{n.14}) over $\Sigma$, we have
\begin{equation}
\label{n.15}
0= \int dV d^2\hat{x}\frac{\partial}{\partial V} \left(\frac{\sqrt{\hat{g} }}{\rho }\right)=
\int_{V\rightarrow 1}d^2 \hat{x}\frac{\sqrt{\hat{g}}}{\rho}-
\int_{V\rightarrow 0}d^2 \hat{x}\frac{\sqrt{\hat{g}}}{\rho}
=4\pi -S_0/\rho_0
.\end{equation}
Here we introduced
$$
S_0 =\int_{V\rightarrow 0}d^2\hat{x}\sqrt{\hat{g}},
\qquad \rho_0 = \lim\limits_{V\to 0} \rho (r)=const
$$
and
$$
\int_{V\rightarrow 1}d^2 \hat{x}\frac{\sqrt{\hat{g}}}{\rho}=
\int_{V\rightarrow 1}d\theta d\varphi\frac{\sqrt{g}}{\rho}=
\int_{r\rightarrow\infty}d\theta d\varphi \frac{\sqrt{ {g}_{\theta\th}{g}_{\varphi\p} } }{r^2}=4\pi
.$$
Here we used that from (\ref{3.4})  it follows that in the limit $r\rightarrow\infty$
$$
{g}_{\theta\th}=r^2 (1+O(r^{-1}),\qquad {g}_{\varphi\p}=r^2 \sin^2\theta (1+O(r^{-1}).
$$
In \cite{isr1} it was shown that in a general case the relation
$\rho_0=\rho (0_+,\hat{x})=const$
follows from condition 3 of the theorem
that the Kretschmann scalar
\begin{equation}
\label{16}
\frac{1}{8}R_{ijkl}R^{ijkl}=\frac{1}{(V\rho )^2}\left[K_{AB}K^{AB} +2\rho^{-2}\rho_{;A}\rho^{;A}+
\rho^{-4}(\partial\rho/\partial V )^2 \right]
\end{equation}
is bounded in $\Sigma $.

Next, in \cite{isr1,isr2} were obtaind the identities
\begin{eqnarray}
\label{17}
\frac{\partial}{\partial V}\left(\sqrt{\frac{\hat{g} }{ \rho } }\frac{K}{V}\right)
=-\frac{ \sqrt{\hat{g} }}{V}\left[ \nabla^2 (\rho ^{1/2})
+\frac{1}{2} \rho^{-3/2} \rho_{;A}\rho^{;A} + \rho^{1/2}(K_{AB}K^{AB}-K^2 /2)\right]\\
\label{18}
\frac{\partial}{\partial V}\left[\frac{\sqrt{ \hat{g} }}{\rho }\left( KV +\frac{4}{\rho}\right)\right]=
-\sqrt{\hat{g} }V\left[\nabla^2(\ln\rho ) +\rho_{;A}\rho^{;A}
+2K_{AB}K^{AB}-K^2 -\hat{R}^{(2)}\right]
.\end{eqnarray}
Integrating the first identity over $\Sigma$ and noting that the r.h.s. of (\ref{17})
not positive,
one obtains the inequality
\begin{equation}
\label{19}
\int d^2\hat{x}dV
\frac{\partial}{\partial V}\left(\sqrt{\frac{\hat{g} }{ \rho }}\frac{K}{V}\right)=
\int\limits_{V\rightarrow 1}d^2 x\sqrt{g}\sqrt{\frac{K}{\rho V}}-
\int\limits_{V\rightarrow 0}d^2 \hat{x}\sqrt{\hat{g}}\sqrt{\frac{K}{\rho V}}\leq 0
\end{equation}
In the limit $r\rightarrow\infty \quad K\simeq 2/r,\,\, \rho\simeq r^2/M$.
The first integral
in the rhs of (\ref{19}) is $8\pi \sqrt{M}$. The second integral, using the relation
$\lim\limits_{V \to 0} K/V=-(1/2)\rho_0 R^{(2)}(0_+, \hat{x})$
which follows from Eq.(\ref{16}),
and
$$\int_\Sigma d^2\hat{x} \sqrt{\hat{g}}\hat{R}^{(2)}=-8\pi ,                                        $$
    is equal to $4\pi \sqrt{\rho_0}$. It follows that
\begin{equation}
\label{20}
2\sqrt{M}-\sqrt{\rho_0}\leq 0.
\end{equation}
The integration of Eq.(\ref{18}) over $\Sigma$ yields
\begin{eqnarray}
\label{21}\nonumber
&{}&\int\limits_{V\rightarrow 1}d^2\hat{x}\sqrt{ {g}}\frac{1}{\rho}\left(K+\frac{4}{\rho}\right)
-\int\limits_{V\rightarrow 0}d^2\hat{x}\sqrt{ \hat{g}}\frac{4}{\rho_0^2}=\\
&{}&
=-\int d^2\hat{x}\sqrt{ \hat{g}}\,\int_0^1 dV V\left[\nabla^2(\ln\rho ) +
\rho_{;A}\rho^{;A} +2K_{AB}K^{AB}-K^2 \right]-4\pi
,\end{eqnarray}
where it was used that
$$
\int d^2\hat{x}\sqrt{ \hat{g}}\int_0^1\,dV V \hat{R}^{(2)}=-4\pi
.$$
The integral in the rhs of Eq.(\ref{21}) is not positive,
the first integral in lhs vanishes leaving the inequality
\begin{equation}
\label{22}
4\pi\leq\frac{4 S_0}{\rho_0^2}
.\end{equation}
Repeating the reasoning of \cite{isr1,isr2}, from the (in)equalities
(\ref{n.15}), (\ref{20}), and (\ref{22}) and harmonic property (\ref{3}),
we conclude that the metric
Eq.(\ref{n.9}) is spherically symmetric, i.e. is
the Schwarzschild metric.

The initial  metric  (\ref{3.4}) contained
a supertranslation field and was axisymmetric.
The metric (\ref{3.4}) was obtained
from the Schwarzschid metric by application of a diffeomorphism containing
a supertranslation field.
Diffeomorphism containing supertranslation field is
not a pure gauge transformation (transformation conserving
all the charges), but changes the superrotation charge \cite{strom1,comp1,comp2}.

Transformation from coordinates $x^\alpha$ in the metric (\ref{3.4})
to coordinates $\hat{x}^A$ again includes
supertranslation field (\ref{n.5}) and is not a pure gauge transformation.
The actions of both diffeomorphisms, the first from the Schwarzshild metric to
Eq.(\ref{3.4}) and the second from Eq.(\ref{3.4}) to Eq.(\ref{n.9})
result in the metric without supertranslation field, and we returned to the
Schwarzschild metric .

\vspace{1cm}
{\large\bf Acknowledgments}

This work was partially supported by the Ministry of Science and
Higher Education of
Russian Federation under the project 01201255504.

\section{Appendix:  calculation of the Kretschmann scalar}
\setcounter{equation}{0}
\renewcommand{\theequation}{A\arabic{equation}}

In this Appendix we calculate the Kretschmann scalar for the metric
(\ref{3.4}). The Kretchmann scalar can be presented in a form \cite{isr1,muller}
\begin{equation}
\label{1a}
\frac{1}{8}R_{ijkl}R^{ijkl}= V^{-2}V_{;\alpha\beta}V^{;\alpha\beta}
.\end{equation}
Here the Latin indices run over $0-3$,the  Greek ones over $1-3$.

Calculating $V_{;\alpha\beta}$, we obtain
\begin{eqnarray}
\nonumber
&{}&V_{;rr}=V_{,rr}-\Gamma^r_{rr}V_{,r}
=-\frac{M(2rV^2  +M)}{r^4 V^3 }-
\frac{1}{2}\left[{g}^{rr}{g}_{rr,r}+
{g}^{r\theta}(2 {g}_{r\theta ,r}-{g}_{rr,\theta})\right]V_{,r}=
\\ \label{3a}
&{}&=-\frac{2M}{Vr^3 }-\frac{M}{2Vr^2}
\left( \bar{g}_{rr ,r}-2\frac{ \bar{g}_{r\theta} \bar{g}_{r\theta,r} }{\bar{g}_{\theta\th} }
+\frac{ \bar{g}_{rr,\theta} \bar{g}_{r\theta} }{V\bar{g}_{\theta\th} }\right),
\\ \label{5a}
&{}&V_{;r\theta}=-\Gamma^r_{r\theta}V_{,r}=-\frac{M}{2Vr^2}\left(\frac{1}{2}g^{rr}g_{rr,\theta}
+\frac{1}{2} g^{r\theta}g_{\theta\th,r}\right),
\\  \label{6a}
&{}&V_{;\theta\th}=-\Gamma^r_{\theta\th}V_{,r}=-\frac{M}{Vr^2}\left(\frac{1}{2}g^{rr}
(2 g_{r\theta} -g_{\theta\th,r})+\frac{1}{2}g^{r\theta}g_{\theta\th,\theta}\right)
 ,\\   \label{7a}
&{}&V_{;\varphi\p}=-\Gamma^r_{\varphi\p}V_{,r}=-\frac{M}{Vr^2}
\left(\frac{1}{2}g^{rr}g_{\varphi\p,r}+\frac{1}{2}g^{r\theta}g_{\varphi\p,\theta}\right)
,\\  \label{8a}
&{}&V_{;r\varphi}=V_{;\theta\varphi}=0.
\end{eqnarray}
Using Eqs.(\ref{2}) and (\ref{3.8}), we rearrange the terms in
$V_{;\alpha\beta}V^{;\alpha\beta}$ and present it in a form
\begin{equation}
\label{9a}
V_{;\alpha\beta}V^{;\alpha\beta}=2(V_{;rr}V_{;\theta\th}-V^2_{;r\theta})(({g}^{r\theta})^2 -
{g}^{rr}{g}^{\theta\th})+2(V_{;\varphi\p}{{g}^{\varphi\p})^2 }
.\end{equation}
Using Eq.(\ref{3.8}), we have
\begin{equation}
\label{10a}
({g}^{r\theta})^2 -
{g}^{rr}{g}^{\theta\th} =
V^2\frac{\bar{g}_{r\theta}^2 -\bar{g}_{rr}{g}_{\theta\th}}{{g}^2_{\theta\th} }=
-\frac{V^2}{{g}_{\theta\th}}.
\end{equation}
Introducing the notations
$$
f= \sqrt{1-b^2 }-b' ,\qquad N= 1-b^2
,$$
and noting the relations
$$
K_{,r}=K/rV , \qquad b_{,r}=-b/rV ,\qquad V_{,r}=M/Vr^2
,$$
where $b$ and $K$ were defined in  (\ref{3.6}),
we calculate Eqs.(\ref{3a}) -(\ref{6a}) as
\begin{eqnarray}
\label{11a}
&{}&V_{;rr}=V_{,r}\left(-\frac{2}{r}+\frac{b^2}{rN}\right) ,\\
\label{12a}
&{}&V_{;\theta\th}=V_{,r}\frac{V^2 r f^2}{N},\\
\label{13a}
&{}&V_{;r\theta}=V_{,r}\frac{Vbf}{N},\\
\label{14a}
&{}&V_{;\varphi\p}=V_{,r}\frac{V^2 g_{\varphi\p}}{r}.
\end{eqnarray}
Substituting these expressions into Eq.(\ref{9a}), we finally obtain
\begin{equation}
\label{15a}
V^{-2}V_{;\alpha\beta}V^{;\alpha\beta}=\frac{6M^2}{r^6}.
\end{equation}
The expression (\ref{15a}) coincides with the Kretschmann scalar for the
Schwarzschild
metric, because the metric with supertranslation field Eq.(\ref{3.8}) and the
Schwarzschild metric are connected by a diffeomorphism.


text:
\section{Introduction}

It is common  knowledge  that the sum of powers of integers
\begin{equation}
S_m(n):=\sum_{q=1}^nq^m
\label{014}
\end{equation}
is a polynomial $\sigma_m(n)$ in $n$ of degree $m+1$. This proposition can be showed  by using Pascal's elementary proof (see, for example \cite{Beardon}).  Polynomials $\sigma_m(n)$ for any integer $m\geq 1$ are defined  in terms of the Bernoulli numbers:
\[
\sigma_m(n)=\frac{1}{m+1}\sum_{q=0}^m(-1)^q{m+1\choose q}B_{q}n^{m-q+1}.
\]
The numbers $B_k$, in turn, are defined by a recursion
\[
\sum_{q=0}^{k}{k+1\choose q}B_q=0,\;\;
B_0=1.
\]
They were discovered by Bernoulli in 1713 but as explained in \cite{Edwards1},  they were known by Faulhaber before this time.

Faulhaber's theorem says that for any odd $m\geq 3$, the sum (\ref{014}) is expressed as a polynomial of $S_1(n)$. It is common knowledge that $S_1(n)=n(n+1)/2$. Let us denote $t:=n(n+1)$. Therefore, the Faulhaber's theorem asserts that for any $k\geq 1$, $S_{2k+1}(n)$ can be expressed as a polynomial in $\mathbb{Q}[t]$. These polynomials are referred as Faulhaber's polynomials. Also, it is worth remembering that it was shown by Jacobi  in \cite{Jacobi} that $S_{2k}(n)$ is expressed as a polynomial in $(2n+1)\mathbb{Q}[t]$.

There exists a variety of different modifications and generalizations of  Faulhaber's theorem in the literature. Faulhaber himself considered $r$-fold sums $S_{m}^{r}(n)$ which are successively defined by
\[
S_{m}^{r}(n)=\sum_{q=1}^{n}S_{m}^{r-1}(q),\;\;
r\geq 1
\]
beginning from $S_{m}^{0}(n):=n^m$. He observed that $S_{m}^{r}(n)$ can be expressed as a polynomial in $n(n+r)$ if $m-r$ is even \cite{Knuth}.

In this paper we consider some class of sums $S_{k, j}(n)$ which include classical sums of powers of integers (\ref{014}) with odd $m=2k+1$.  These sums will be introduced in section \ref{s2} and \ref{s3}.  In the particular case $n=1$, they represent some class of binomial sums studied in \cite{Tuenter}. Our crucial idea comes from this work. We  briefly describe the results of this article in section \ref{s4}. In section \ref{s5}, we formulate our main conjecture concerning a representation of the sums  $S_{k, j}(n)$ in terms of some polynomials whose coefficients rationally depend on $n$. In section \ref{s6}, we discuss the relationship of this representation with a possible generalization of the Faulhaber's theorem. In section \ref{s7}, we show the explicit form of some $n$-dependent coefficients of these polynomials and speculate about the relationship of these coefficients with the number of $q$-points in the simple symmetric $(2k+1)$-Venn diagram.

\section{The sums $s_{k, j}(n)$}
\label{s2}

Let us define the numbers $\{C_{j, r}(n) : j\geq 1, r=0,\ldots, j(n-1)\}$ through the relation
\begin{equation}
\left(\sum_{q=0}^{n-1}a^{q+1}\right)^j=\sum_{q=0}^{j(n-1)}C_{j, q}(n)a^{q+j},
\label{013}
\end{equation}
where $a$ is supposed to be an auxiliary variable, taking into account that $a^la^m=a^{l+m}$. For example, in the case $n=1$, we have only $C_{j, 0}(1)=1$, while  in the case $n=2$, we  get
\[
\left(a+a^2\right)^j=\sum_{q=0}^j{j\choose q}a^{j+q},
\]
and therefore  $C_{j, q}(2)$  is a binomial coefficient.  From
\begin{eqnarray}
\left(\sum_{q=0}^{n-1}a^{q+1}\right)^j&=&\left(\sum_{q=0}^{(j-1)(n-1)}C_{j-1, q}(n)a^{q+j-1}\right)\left(\sum_{q=0}^{n-1}a^{q+1}\right)\nonumber\\
&=&C_{j-1, 0}(n)a^{j}+\left(C_{j-1, 0}(n)+C_{j-1, 1}(n)\right)a^{j+1}+\cdots+\left(C_{j-1, 0}(n)+\cdots\right.\nonumber\\
&&\left.+C_{j-1, n-1}(n)\right)a^{j+n-1}\nonumber\\
&&+\left(C_{j-1, 1}(n)+\cdots+C_{j-1, n)}(n)\right)a^{j+n}+\cdots+C_{j-1, (j-1)(n-1)}(n)a^{jn}\nonumber
\end{eqnarray}
we deduce  that
\begin{equation}
C_{j,r}(n)=\sum_{q=r-n+1}^rC_{j-1, q}(n)
\label{008}
\end{equation}
in assumption that $C_{j-1, q}(n)=0$ for $q<0$ and $q>(j-1)(n-1)$. In the case $n=2$, the relation (\ref{008}) becomes a well-known property for the binomial coefficients:
\[
{j\choose r}={j-1\choose r-1}+{j-1\choose r}.
\]

Clearly, the coefficient $C_{j,r}(n)$ can be represented as a proper sum of multinomial coefficients.
Putting $a=1$ into (\ref{013}), we get the property
\[
\sum_{q=0}^{j(n-1)}C_{j, q}(n)=n^j
\]
for these numbers which evidently generalize the well-known attribute of binomial coefficients. Also it is easy to get the following property:
\[
\sum_{q=0}^{j(n-1)}qC_{j, q}(n)=\frac{j(n-1)}{2}n^j.
\]

With the coefficients $C_{j, q}(n)$ we define the sum $s_{k, j}(n)$ by
\[
s_{k, j}(n):=\sum_{q=0}^{j(n-1)}C_{j, q}(n)x_{j+q},
\]
where $x_r:=r^{2k+1}$. Let us remark that $s_{k, 1}(n)=S_{2k+1}(n)$. Next, we would like to give a suitable expression for the sums $s_{k, 2}(n)$, which will be used in the following section.
It is evident that
\[
C_{2, q}(n)=\left\{
\begin{array}{l}
q+1,\;\; q=0,\ldots, n-2, \\[0.3cm]
2n-q-1,\;\; q=n-1,\ldots, 2n-2
\end{array}
\right.
\]
and therefore
\begin{eqnarray}
s_{k, 2}(n)&:=&\sum_{q=0}^{2n-2}C_{2, q}(n)x_{q+2}\nonumber\\
&=&\sum_{q=0}^{n-2}(q+1)x_{q+2}+\sum_{q=n-1}^{2n-2}(2n-q-1)x_{q+2}.\nonumber
\end{eqnarray}
Shifting $q\rightarrow q-2$, we get
\[
s_{k, 2}(n)=\sum_{q=1}^{n}(q-1)x_{q}+\sum_{q=n+1}^{2n}(2n-q+1)x_{q},
\]
and since $q-1=(2n-q+1)-(2n-2q+2)$, we deduce that
\begin{equation}
s_{k, 2}(n)=-\sum_{q=1}^{n}(2n-2q+2)x_{q}+\sum_{q=1}^{2n}(2n-q+1)x_{q}.
\label{020}
\end{equation}

\section{The sums $S_{k, j}(n)$ and $\tilde{S}_{k, j}(n)$}
\label{s3}

Let us define a sum
\[
\tilde{S}_{k, j}(n):=\sum_{\{\lambda\}\in B_{j, jn}}\left\{\lambda_1^{2k+1}+(\lambda_2-n)^{2k+1}+\cdots+(\lambda_j-jn+n)^{2k+1}\right\},
\]
where $B_{j, jn}:=\{\lambda_k : 1\leq \lambda_1\leq \cdots \leq \lambda_j\leq jn\}$ and
\begin{equation}
S_{k, j}(n)=\sum_{q=0}^{j-1}{j(n+1)\choose q} s_{k, j-q}(n).
\label{003}
\end{equation}
It is obvious that
\[
\tilde{S}_{k, 1}(n)=S_{k, 1}(n)=s_{k, 1}(n)=S_{2k+1}(n).
\]

\begin{conjecture}
\label{con1}
For any $k\geq 0$ and $j\geq 1$, one has
\begin{equation}
\tilde{S}_{k, j}(n)=S_{k, j}(n).
\label{001}
\end{equation}
\end{conjecture}
Let us prove (\ref{001}) in the particular case $j=2$. Using simple arguments we deduce that
\begin{eqnarray}
\tilde{S}_{k, 2}(n)&=&\sum_{\{\lambda\}\in B_{2, 2n}}\left\{\lambda_1^{2k+1}+(\lambda_2-n)^{2k+1}\right\}\nonumber\\
&=&\sum_{q=1}^{2n}(2n-q+1)x_q+\sum_{q=1}^{2n}(2n-q+1)x_{n-q+1}.
\label{002}
\end{eqnarray}
Notice that  we have several $x_r$'s with negative subscript $r$ in (\ref{002}). Clearly, we must put $x_r=-x_{-r}$. Taking into account this rule, we rewrite the second sum in (\ref{002})  as
\[
\sum_{q=1}^{n}(n+q)x_{q}-\sum_{q=1}^{n}(n-q)x_{q}=2\sum_{q=1}^{n}qx_{q}.
\]
Adding to (\ref{002}) the terms
\[
-\sum_{q=1}^{n}(2n-2q+2)x_{q}+\sum_{q=1}^{n}(2n-2q+2)x_{q}
\]
and taking into account (\ref{020}), we finally get
\begin{eqnarray}
\tilde{S}_{k, 2}(n)&=&s_{k, 2}(n)+(2n+2)\sum_{q=1}^{n}x_q\nonumber\\
&=&s_{k, 2}(n)+{2(n+1)\choose 1}s_{k, 1}(n)\nonumber\\
&=&S_{k, 2}(n).\nonumber
\end{eqnarray}
Therefore, we proved conjecture \ref{con1} in the particular case $j=2$.

\section{Special case of the sums $S_{k, j}(n)$}
\label{s4}

In the case $n=1$, the sum (\ref{003}) becomes
\begin{equation}
S_{k, j}(1)=\sum_{q=0}^{j-1}{2j\choose q} (j-q)^{2k+1}.
\label{004}
\end{equation}
This type of the sums was studied in \cite{Bruckman}, \cite{Strazdins}, \cite{Tuenter}. To be precise, the authors of these works considered binomial sums of the form
\[
\mathcal{S}_{m}(j)=\sum_{q=0}^{2j}
{2j\choose q}|j-q|^m.
\]
It is evident that $S_{k, j}(1)=\mathcal{S}_{2k+1}(j)/2$.

Some bibliographical remarks are as follows.  Bruckman  in \cite{Bruckman} asked to prove that $\mathcal{S}_3(j)=j^2\left(
\begin{array}{c}
2j\\
j
\end{array}
\right)$. Strazdins in \cite{Strazdins} solved this problem and conjectured that $\mathcal{S}_{2k+1}(j)=\tilde{P}_{k}(x)|_{x=j}{2j\choose j}$ with some monic polynomial $\tilde{P}_{k}(x)$ for any $k\geq 0$. Tuenter showed in \cite{Tuenter} that it is almost true. More exactly, he proved that there exists a sequence of polynomials $P_k(x)$ such that
\[
\mathcal{S}_{2k+1}(j)=P_{k}(x)|_{x=j}j
{2j\choose j}=P_{k}(x)|_{x=j}\frac{(2j)!}{(j-1)!j!}.
\]
One can see that polynomial $\tilde{P}_{k}(x)$ is monic only for $k=0, 1$. The following proposition \cite{Tuenter} is verified by direct computations.
\begin{proposition}
The sums $S_{k, j}(1)$ satisfy the recurrence relation
\begin{equation}
S_{k, j}(1)=j^2S_{k-1, j}(1)-2j(2j-1)S_{k-1, j-1}(1).
\label{rec1}
\end{equation}
\end{proposition}
\begin{remark}
This was, in fact, proved  for the sums $\mathcal{S}_{2k+1}(j)$ in \cite{Tuenter}.
\end{remark}

Let us introduce the sequence of positive integer numbers:
\[
g_1=1,\;\;g_j=\frac{j}{(j-1)!}\prod_{q=1}^{j-1}(j+q),
\;\;j\geq 2.
\]
As can be checked, this sequence  satisfies the recurrence relation
\begin{equation}
g_{j+1}=2\frac{2j+1}{j}g_j.
\label{016}
\end{equation}
Some of the first numbers $g_j$ are given in the following table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$j$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&9  \\
\hline
$g_{j}$&$1$&6&30&140& 630&2772 &12012 &51480 &218790\\
\hline
\end{tabular}
\end{center}
They are included in the \textrm{A002457} integer sequence in \cite{Sloane}.

Let us write
\[
S_{k, j}(1)=P_{k, j}g_j
\]
with some numbers  $P_{k, j}$ to be calculated. Taking into account (\ref{016}), it can be easily seen that (\ref{rec1}) is valid if the relation
\begin{equation}
P_{k+1, j}=j^2\left(P_{k,j}-P_{k,j-1}\right)+jP_{k, j-1}
\label{007}
\end{equation}
holds. Now we need the polynomials studied in \cite{Tuenter} which are defined by a recurrence relation
\begin{equation}
P_{k+1}(x)=x^2\left(P_k(x)-P_{k}(x-1)\right)+xP_k(x-1)
\label{006}
\end{equation}
with initial condition $P_0(x)=1$. The first eight polynomials yielded by (\ref{006}) are as follows:
\[
P_0(x)=1,\;\;
P_1(x)=x,\;\;
P_2(x)=x(2x-1),
\]
\[
P_3(x)=x(6x^2-8x+3),
\]
\[
P_4(x)=x(24x^3-60x^2+54x-17),
\]
\[
P_5(x)=x(120x^4-480x^3+762x^2-556x+155),
\]
\[
P_6(x)= x(720x^5-4200x^4+10248x^3-12840x^2+8146x-2073).
\]
\begin{eqnarray}
P_7(x)&=& x\left(5040x^6-40320x^5+139440x^4-263040x^3\right. \nonumber\\
&&\left.+282078x^2-161424x+38227\right).\nonumber
\end{eqnarray}

Introducing a recursion operator $R:=x^2\left(1-\Lambda^{-1}\right)+x\Lambda^{-1}$, where $\Lambda$ is a shift operator acting as $\Lambda(f(x))=f(x+1)$, one can write $P_k(x)=R^k(1)$. One could notice that the polynomial $P_k(x)$ has $k!$ as a coefficient at $x^k$. In addition, as was noticed in \cite{Tuenter}, the constant terms of the polynomials $P_k(x)/x$ form a sequence of the Genocchi numbers with opposite sign:
\[
-G_2=1,\;\;
-G_4=-1,\;\;
-G_6=3,\;\;
-G_8=-17,\;\;
-G_{10}=155,\;\;
-G_{12}=-2073,\ldots
\]
Recall that the Genocchi numbers are defined in terms of the generating function
\[
\frac{2x}{e^x+1}=\sum_{q\geq 1}G_{q}\frac{x^q}{q!}
\]
and are related to the Bernoulli numbers as $G_{2k}=2\left(1-2^{2k}\right)B_{2k}$.

Comparing (\ref{006}) with (\ref{007}) we conclude that $P_{k, j}=P_k(x)|_{x=j}$.
\begin{remark}
As was noticed in \cite{Tuenter}, the polynomials $P_{k}(x)$ can be obtained as a special case of Dumont-Foata polynomials of three variables \cite{Dumont}.
\end{remark}
\section{General case of the sums $S_{k, j}(n)$}
\label{s5}

Let us remember that $t:=n(n+1)$. Our main conjecture is as follows.
\begin{conjecture}
There exists some polynomials $P_k(t, x)$ in $x$ of degree $k$ whose coefficients rationally depend on $t$ such that
\[
S_{k,j}(n)=\frac{t^{k+1}}{2^{k+1}}P_k(t, x)|_{x=j}g_j(n),
\]
where
\[
g_1(n)=1,\;\;
g_j(n):=\frac{j}{(j-1)!}\prod_{q=1}^{j-1}(jn+q).
\]
\end{conjecture}
The first eight polynomials $P_k(t, x)$ are
\[
P_0(t, x)=1,\;\;
P_1(t, x)=x,\;\;
P_2(t, x)=x\left(2x-\frac{2(t+1)}{3t}\right),\;\;
\]
\[
P_3(t, x)=x\left(6x^2-\frac{16(t+1)}{3t}x+\frac{4(t+1)^2}{3t^2}\right),\;\;
\]
\[
P_4(t, x)=x\left(24x^3-\frac{40(t+1)}{t}x^2+\frac{24(t+1)^2}{t^2}x-\frac{24(t+1)^3+8t^2}{5t^3}\right),
\]
\begin{eqnarray}
P_5(t, x)&=& x\left(120x^4-\frac{320(t+1)}{t}x^3+\frac{1016(t+1)^2}{3t^2}x^2\right.\nonumber\\
&&\left.-\frac{160(t+1)^3+32t^2}{t^3}x+\frac{80\left((t+1)^4+t^2(t+1)\right)}{3t^4}\right),\nonumber
\end{eqnarray}
\begin{eqnarray}
P_6(t, x)&=& x\left(720x^5-\frac{2800(t+1)}{t}x^4+\frac{13664(t+1)^2}{3t^2}x^3\right.\nonumber\\
&&-\frac{55936(t+1)^3+7632t^2}{15t^3}x^2+\frac{22112(t+1)^4+13664t^2(t+1)}{15t^4}x\nonumber\\
&&\left.-\frac{22112(t+1)^5+44224t^2(t+1)^2}{105t^5}\right),\nonumber
\end{eqnarray}
\begin{eqnarray}
P_7(t, x)&=&x\left(5040x^6-\frac{26880(t+1)}{t}x^5+\frac{185920(t+1)^2}{3t^2}x^4\right.\nonumber\\
&&-\frac{76800(t+1)^3+7680t^2}{t^3}x^3\nonumber\\
&&+\frac{157088(t+1)^4+67968t^2(t+1)}{3t^4}x^2\nonumber\\
&&-\frac{17920(t+1)^5+22528t^2(t+1)^2}{t^5}x\nonumber\\
&&\left. +\frac{6720(t+1)^6+22400t^2(t+1)^3+1344t^4}{3t^6}\right).\nonumber
\end{eqnarray}
Unfortunately, in the general case we do not know a recursion relation for these polynomials except for the case $t=2$. One can check that $P_k(t, x)|_{t=2}=P_k(x)$, where $P_k(x)$ are Tuenter's polynomials introduced above.

It can be noticed that the  coefficients of the  polynomials $P_k(t, x)$ have a special form. Namely, let
\begin{equation}
P_k(t, x)=x\left(p_{k, 0}(t)x^{k-1}-p_{k, 1}(t)x^{k-2}+\cdots+(-1)^{k-1}p_{k, k-1}(t)\right).
\label{005}
\end{equation}
Based on actual calculations, we are led to the following.
\begin{conjecture} \label{4.2}
The coefficients of $t$-dependent polynomial (\ref{005}) are given by
\[
p_{k, j}(t)=\frac{r_{k, j}(t)}{t^j},
\]
where the polynomials $r_{k, j}(t)$ are of the form
\[
r_{k, j}(t)=\sum_{q=0}^{m}\alpha_{k, j, q}t^{2q}(t+1)^{j-3q}
\]
with rational positive  nonzero numbers $\alpha_{k, j, q}$. Here $m\geq 0$, by definition,  is the result of division of the number $j$
by $3$ with some remainder $l$, that is, $j=3m+l$.
\end{conjecture}
It should be noticed that the last conjecture is quite strong. With this conjecture the number $N_k$ of parameters which entirely define the polynomial $P_{k}(t, x)$ for small values of $k$  is given in the following table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$k$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&9  \\
\hline
$N_{k}$&$1$&2&3&5&7&9&12&15 &18\\
\hline
\end{tabular}
\end{center}
These numbers form the \textrm{A001840} integer sequence in \cite{Sloane}. We have, in fact, verified  conjecture \ref{4.2}  up to $k=11$ and calculated all corresponding coefficients $\alpha_{k, j, q}$.

It is evident that for all the polynomials $P_k(t, x)$ we have $p_{k, 0}(t)=k!$. Actual calculations show that
\[
\alpha_{k, 1, 0}=\frac{(k-1)(k+1)}{9}k!,\;\;k\geq 2,
\]
\[
\alpha_{k, 2, 0}=\frac{(k-2)(k+1)(5k^2+k-3)}{810}k!,\;\;k\geq 3,
\]
\[
\alpha_{k, 3, 0}= \frac{(k-3)(k+1)(175k^4-70k^3-724k^2+643k-690)}{765450}k!,\;\;k\geq 4,
\]
\[
\alpha_{k, 3, 1}= \frac{(k-3)(k+1)(2k^2-4k+5)}{1575}k!,\;\;k\geq 4
\]
etc. Looking at these patterns we may guess that
\[
\alpha_{k, j, 0}=(k-j)(k+1)p_j(k)k!,\;\;k\geq j+1
\]
with some polynomial $p_j(k)$ of degree $2j-2$.

\section{Faulhaber's theorem}
\label{s6}

With  the polynomials  $P_k(t, x)$  we are able, for example, to calculate
\[
S_{0,j}(n)=\frac{t}{2}\frac{j}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),\;\;
S_{1,j}(n)=\frac{t^2}{4}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),
\]
\[
S_{2,j}(n)=\frac{t^2}{12}\left\{(3j-1)t-1\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),
\]
\[
S_{3,j}(n)=\frac{t^2}{24}\left\{(9j^2-8j+2)t^2-(8j-4)t+2\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),
\]
\begin{eqnarray}
S_{4,j}(n)&=&\frac{t^2}{20}\left\{(15j^3-25j^2+15j-3)t^3-(25j^2-30j+10)t^2\right.\nonumber\\
           &&\left.+(15j-9)t-3\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),\nonumber
\end{eqnarray}
\begin{eqnarray}
S_{5,j}(n)&=&\frac{t^2}{24}\left\{(45j^4-120j^3+127j^2-60j+10)t^4\right.\nonumber\\
&&-(120j^3-254j^2+192j-50)t^3+(127j^2-180j+70)t^2\nonumber\\
&&\left.-(60j-40)t+10\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q),\nonumber
\end{eqnarray}
\begin{eqnarray}
S_{6,j}(n)&=&\frac{t^2}{840}\left\{(4725j^5-18375j^4+29890j^3-24472j^2+9674j-1382)t^5\right.\nonumber\\
&&-(18375j^4-59780j^3+76755j^2-44674j+9674)t^4\nonumber\\
&&+(29890j^3-73416j^2+64022j-19348)t^3-(24472j^2+38696j-16584)t^2\nonumber\\
&&\left.+(9674j-6910)t-1382\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q).\nonumber
\end{eqnarray}
\begin{eqnarray}
S_{7,j}(n)&=&\frac{t^2}{48}\left\{(945j^6-5040j^5+11620j^4-14400j^3+9818j^2-3360j+420)t^6\right.\nonumber\\
&&-(5040j^5-23240j^4+44640j^3-43520j^2+21024j-3920)t^5\nonumber\\
&&+(11620j^4-43200j^3+63156j^2-42048j+10584)t^4\nonumber\\
&&-(14400j^3-39272j^2+37824j-12600)t^3\nonumber\\
&&+(9818j^2-16800j+7700)t^2\nonumber\\
&&\left.-(3360j-2520)t+420\right\}\frac{j^2}{(j-1)!}\prod_{q=1}^{j-1}(jn+q).\nonumber
\end{eqnarray}
Looking at these patterns we  suggest the following.
\begin{conjecture}
\label{4.3}
$S_{k,j}(n)$ is expressed as a polynomial in $\prod_{q=1}^{j-1}(jn+q)\mathbb{Q}[t]$.
\end{conjecture}

For $j=1$,  conjecture \ref{4.3} becomes the well-known  Faulhaber's theorem \cite{Faulhaber} which was, in fact, proved by Jacobi in  \cite{Jacobi}. The first eight  Faulhaber's polynomials  are as follows:
\[
S_{0, 1}(n)=\frac{t}{2},\;\;
S_{1, 1}(n)=\frac{t^2}{4},\;\;
S_{2, 1}(n)=\frac{t^2}{12}(2t-1),\;\;
S_{3, 1}(n)=\frac{t^2}{24}\left(3t^2-4t+2\right),
\]
\[
S_{4, 1}(n)=\frac{t^2}{20}\left(2t^3-5t^2+6t-3\right),\;\;
S_{5, 1}(n)=\frac{t^2}{24}\left(2t^4-8t^3+17t^2-20t+10\right),
\]
\[
S_{6, 1}(n)= \frac{t^2}{840}\left(60t^5-350t^4+1148t^3-46584t^2+2764t-1382\right),
\]
\[
S_{7, 1}(n)= \frac{t^2}{48}t^2(3t^6-24t^5+112t^4-352t^3+718t^2-840t+420).
\]
One usually writes
\[
S_{k, 1}(n)=\frac{1}{2(k+1)}\sum_{q=0}^{k}A_{q}^{(k+1)}t^{k-q+1},
\]
where $A^{(k)}_0=1$ and $A^{(k)}_{k-1}=0$.  We know quite a lot about the coefficients $A_q^{(k)}$. Jacobi proved that the coefficients $A_{q}^{(k)}$ satisfy the recurrence relation
\[
(2k+2)(2k+1)A_q^{(k)}=2(k-q+1)(2k-2q+1)A_{q}^{(k+1)}+(k-q+1)(k-q+2)A_{q-1}^{(k+1)}
\]
and tabulated some of them. It was shown by Knuth in \cite{Knuth} that these coefficients satisfy the quite simple implicit recurrence relation
\begin{equation}
\sum_{q=0}^r{k-q\choose 2r+1-2q}A_{q}^{(k)}=0,\;\;
r>0
\label{009}
\end{equation}
which yields an infinite triangle system of equations from which one easily obtains
\[
A_{1}^{(k)}=-\frac{(k-2)k}{6},\;\;
A_{2}^{(k)}=\frac{(k-3)(k-1)k(7k-8)}{360},
\]
\[
A_{3}^{(k)}=-\frac{(k-4)(k-2)(k-1)k(31k^2-89k+48)}{15120},
\]
\[
A_{4}^{(k)}= \frac{(k-5)(k-3)(k-2)(k-1)k(127k^3-691k^2+1038k-384)}{604800}
\]
etc. Gessel and Viennot  showed in \cite{Gessel} that the solution of system (\ref{009}) can be represented as a $k\times k$ determinant
\[
A_{q}^{(k)}=\frac{1}{(1-k)\cdots(q-k)}\left|
\begin{array}{ccccc}
{k-q+1 \choose 3}    &{k-q+1 \choose 1}      &0     & \cdots   &    0  \\[0.1cm]
{k-q+2 \choose 5}    &{k-q+2 \choose 3}      &{k-q+2 \choose 1}     & \cdots   &    0  \\[0.1cm]
\vdots               &\vdots                 &\vdots          &          &    \vdots  \\[0.1cm]
{k-1 \choose 2k-1}    &{k-1 \choose 2k-3}      &{k-1 \choose 2k-5}     & \cdots   &  {k-1 \choose 1} \\[0.1cm]
{k \choose 2k+1}    &{k-1 \choose 2k-1}      &{k-1 \choose 2k-3}     & \cdots   &  {k \choose 3}
\end{array}
\right|
\]
(see also \cite{Edwards}). A bivariate generating function for the coefficients $A_{q}^{(k)}$ was obtained in \cite{Gessel1}.

\section{The conjectural relationship of the coefficients $p_{k, k-1}(t)$ to simple symmetric Venn diagrams}
\label{s7}

Let us rewrite the coefficients $p_{k, k-1}(t)$  being expressed via $n$, that is,
\[
p_{k, k-1}(n)=p_{k, k-1}(t)|_{t=n(n+1)}.
\]
For example,
\[
p_{1, 0}(n)=1,\;\;
p_{2, 1}(n)= \frac{2}{3}\frac{n^2+n+1}{n(n+1)},\;\;
p_{3, 2}(n)= \frac{4}{3}\frac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2},
\]
\[
p_{4, 3}(n) = \frac{24}{5}\frac{n^6+3n^5+\frac{19}{3}n^4+\frac{23}{3}n^3+\frac{19}{3}n^2+3n+1}{n^3(n+1)^3},
\]
\[
p_{5, 4}(n)= \frac{80}{3}\frac{n^8+4n^7+11n^6+19n^5+23n^4+19n^3+11n^2+4n+1}{n^4(n+1)^4},
\]
\[
p_{6, 5}(n)= \frac{22112}{105}\frac{n^{10}+5n^{9}+17n^{8}+38n^7+61n^6+71n^5+61n^4+38n^3+17n^2+5n+1}{n^5(n+1)^5}
\]
etc. Looking at these patterns we see that
\begin{equation}
p_{k, k-1}(n)=c_k\frac{v_k(n)}{n^{k-1}(n+1)^{k-1}},
\label{012}
\end{equation}
where $v_k(n)$ is a monic polynomial of degree $2k-2$. All these polynomials are invariant with respect to the transformation
\begin{equation}
v_k(n)\mapsto n^{2k-2}v_k\left(\frac{1}{n}\right).
\label{015}
\end{equation}
Also it is worth to remark that the polynomial $v_4(n)$ unlike the others, has several fractional coefficients.
\begin{conjecture}
The polynomials $v_k(n)$ are given by
\begin{equation}
v_k(n)=\sum_{q=1}^{2k-1}\frac{{2k\choose q}+(-1)^{q+1}}{2k+1}n^{2k-q-1},
\label{010}
\end{equation}
while the coefficients $c_k$ are expressed via the Bernoulli numbers as
\begin{equation}
c_{k}=(-1)^{k+1}(2k+1)2^kB_{2k}.
\label{011}
\end{equation}
\end{conjecture}
Let us notice that if (\ref{010}) is valid then the invariance of the corresponding polynomial  with respect to (\ref{011}) is obvious in virtue of the invariance of binomial coefficients.

Let $p=2k+1$. It is known that if $p$ is prime then
\[
T(p,q)=\frac{{p-1\choose q}+(-1)^{q+1}}{p},\;\;
p\geq 5
\]
is the number of $q$-points on the left side of a crosscut of the simple symmetric $p$-Venn diagram  \cite{Mamakani}. This integer sequence is known as the \textrm{A219539} sequence in \cite{Sloane}.
It is evident that the  row sum
\[
t_p:=\sum_{q=1}^{p-2}T(p,q)=\frac{2^{p-1}-1}{p}
\]
can be calculated as $v_k(n)|_{n=1}$. The Fermat quotients $(2^{p-1}-1)/p$ for prime $p$'s form the integer sequence \textrm{A007663}  in \cite{Sloane}. Taking into account (\ref{012}) and (\ref{011}), we get
\[
p_{k, k-1}(n)|_{n=1}=2(2^{2k}-1)B_{2k}=-G_{2k}.
\]

\section{Discussion}

In the paper we have considered some class of sums $S_{k, j}(n)$ and conjectured a representation of these sums  in terms of a sequence of the polynomials $\{P_k(t, x) : k\geq 0\}$. This assumption  resulted from computational experiments and is supported  by a large amount of actual calculations. For $n=1$, we get the well-known results from \cite{Tuenter}. This also confirms our assumptions. The conjectural relationship of several coefficients of the polynomials $P_k(t, x)$ being expressed via $n$ to simple symmetric Venn diagrams is quite unexpected and requires explanation.

\section*{Acknowledgments}
I wish to thank the referee for carefully reading the manuscript and for remarks which enabled the presentation of the paper to be improved.
This  work  was  supported  in  part  by the Council for Grants of the
 President  of  Russian  Foundation  for  state  support of the leading
 scientific schools, project NSh-8081.2016.9.

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