text:
\section{Introduction}

In this letter we discuss the Weyl gauge symmetry and its spontaneous breaking
together with its implications for model building beyond the Standard Model (SM)
and for inflation.

One phenomenological motivation  relates to the observation  that the SM with a Higgs mass
parameter set to zero has a classical scale symmetry  \cite{Bardeen}.
If this symmetry is preserved at the quantum level
 by  (a scale-invariant) UV regularisation as in \cite{Englert,S1,Armillis,Monin,D1,Tamarit}, and
 is broken spontaneously only,  it can naturally protect {\it at the quantum level} a
hierarchy of  fields vev's  of the  theory \cite{S1,R1,D1,K1,D2}.
The hierarchy  we refer to is that between  the Higgs field vev (electroweak scale) and  that
of ``new physics'' represented  by the vev of  the flat direction  (dilaton)
associated   with global scale symmetry breaking. Such  hierarchy of vev's
can be generated  by a classical hierarchy of the dimensionless
couplings of the theory \cite{Allison,CW-BL}.

A proper study of the hierarchy  problem, based on the above idea,
 demands  including gravity and  generating
spontaneously the Planck scale ($M_p$). This can be  done   in Brans-Dicke-Jordan
theories of gravity \cite{BDJ}
via a non-minimal  coupling between a scalar field(s) and  the scalar
curvature ($R$), when this field(s) develops a non-zero vev.   However, demanding the
theory be conformal invariant and spontaneous-only  breaking of the conformal symmetry,
leads to  a negative kinetic term for the corresponding scalar field,
a ``nuisance'' that is  often  quietly glided over.
{This problem is automatically avoided  in models with
Weyl gauge symmetry \cite{Weyl,Dirac} and motivated our study of this symmetry
in Sections~\ref{WI} and \ref{3}.}

{The Weyl gauge symmetry }
is  the natural extension  for conformal invariant models e.g.
\cite{Weyl,Dirac,Scholz,Scholz2,BDJ,Turok,Smolin,Cheng,Ni,
Moffat1,TW,Davidson,Moffat2,Heisenberg,Oh,Wheeler,Quiros,FRH0,FRH2,FRH1};
the conformal transformation of the metric is extended by the associated
gauge transformation of a Weyl gauge field ($\omega_\mu$) which is of geometric origin.
Section~\ref{3} discusses how
 $\omega_\mu$  undergoes a  Stueckelberg mechanism and becomes massive by ``eating'' the
would-be-Goldstone field (dilaton $\rho$); here, the dilaton is the radial direction in the field
space of scalar fields ($\phi_j$)  of different non-minimal couplings $\xi_j$ to $R$.
The  Weyl gauge symmetry is then spontaneously broken and
 there are no negative kinetic terms in the theory. The  vacuum expectation value
$\langle\rho\rangle$ of the flat direction  (dilaton)  controls the mass of $\omega_\mu$
and $M_p$. After $\omega_\mu$ decouples, the potential depends only on the remaining
angular variables scalar fields which can account for the Higgs
field, inflaton, etc\footnote{
The mechanism of the Weyl gauge symmetry breaking used here differs  from that
in \cite{Oh} where  a complex scalar is considered rather the (real) dilaton,
and  a  Coleman-Weinberg mechanism (unitary gauge)
is used (instead of a Stueckelberg mechanism),
which  breaks explicitly  the Weyl symmetry by  UV regularization.
An explicit (classical) breaking was also considered in \cite{Davidson}
for one scalar field case.
}.
Our analysis  extends previous studies \cite{Smolin,Cheng,Ni,Moffat1,Oh}
to  multiple scalar  fields ($\phi_j$) and different non-minimal  couplings ($\xi_j$).

We also  show (Section~\ref{3.4})  how prior to  this symmetry breaking
the dilaton can enforce a  UV regularization of the quantum corrections that keeps manifest
the Weyl symmetry.  In Weyl-invariant  models the dilaton
replaces the subtraction scale, thus  maintaining this symmetry at the quantum
level \cite{Englert,S1,Armillis,Monin,D1,Tamarit}, after which is
``eaten'' by  $\omega_\mu$ and disappears from the spectrum.
One is left with the potential for angular variables fields (e.g. Higgs field, etc).
 This is relevant for the  hierarchy problem in Weyl-symmetric theories.

While our analysis  (Section~\ref{2.2}) is formulated in Riemannian geometry (RG)
extended by the Weyl gauge  symmetry, the natural framework for this study is
Weyl conformal geometry (WG) \cite{Weyl,Dirac,Scholz}. In the RG case, imposing the Weyl
symmetry {to avoid generating ghosts}, leads  to a SM-like  Lagrangian with the
corresponding current $K_\mu=\partial_\mu K$ where $K_\mu$ interacts with the field
$\omega_\mu$  and $K=\rho^2$.  We show
that this  Lagrangian  is identical, up to a total derivative
term, to the simplest Lagrangian one can build in the Weyl geometry for the same set of
 matter fields, using  the curvature  scalar and curvature tensors of WG
(Section~\ref{2.3}).
This equivalence  is an  interesting  result that
 follows from the relation between $R$ computed in
Riemannian geometry  and its
counterpart $\tilde R$ computed in  Weyl geometry.

We also verify that in the Lagrangian ${\cal L}$ of the SM endowed  with Weyl gauge symmetry,
unlike the Higgs sector, gauge bosons and  fermions do not couple
to $\omega_\mu$ \cite{Moffat1}
(except a possible kinetic mixing of $\omega_\mu$ to  $U(1)_Y$).
${\cal L}$ can be used for further  phenomenological studies of the Weyl gauge symmetry.

For the case of two scalar fields present
with non-minimal couplings, after the Weyl field $\omega_\mu$ decouples,
the  potential depends only on the angular field $\theta$ and  becomes constant for
large  $\tan\theta$. We show that  successful inflation is then possible,
in which the  field  $\theta$ is  playing the role of the inflaton.
This is another  result of this work, discussed in Section~\ref{4}.
Our conclusions are presented in Section~\ref{5}.

\section{Implications of  Weyl gauge symmetry} \label{WI}

We review how  models invariant under conformal transformations  become ghost-free
 while generating spontaneously a positive Newton constant, when a Weyl
 gauge transformation is added. The Lagrangian so obtained is then shown to be
equivalent to that derived
 in Weyl geometry,  up to a total derivative; a SM-like model with this symmetry is
  also constructed.

\subsection{Weyl symmetry or how to obtain  a Lagrangian without ghosts}\label{2.1}

Consider a (local) conformal  transformation of the metric\footnote{
Conventions: metric $(+,-,-,-)$,
$R^\lambda_{\mu\nu\sigma}=
\partial_\nu \Gamma^\lambda_{\mu\sigma}
-\partial_\sigma\Gamma^\lambda_{\mu\nu}
+ \Gamma^\lambda_{\nu\rho}\,\Gamma^\rho_{\mu\sigma}
-\Gamma^\lambda_{\sigma\rho}\,\Gamma^\rho_{\mu\nu}$,
 $R_{\mu\sigma}=R^\lambda_{\mu\lambda\sigma}$, $R=g^{\mu\nu} R_{\mu\nu}$.}  and of a scalar field
$\phi$ and  a  fermion $\psi$, as follows
\begin{eqnarray}
\label{ct}
g_{\mu\nu} &\rightarrow&  g_{\mu\nu}^\prime =  e^{2\,\alpha(x)}\, g_{\mu\nu},
\nonumber\\[3pt]
\phi &\rightarrow& \phi^\prime   =  e^{-\alpha(x)\,\Delta_s}\,\phi,
\quad\quad \psi \,\,\,\rightarrow\,\,\, \psi' = e^{-\alpha(x)\,\Delta_f}\,\psi.
\end{eqnarray}
Then $g^{\mu\nu \prime}= e^{-2\alpha(x)} g^{\mu\nu}$ and
$\sqrt{g^\prime}\!=\! e^{4\alpha(x)}\! \sg$ with
$g\!=\!\vert\det g_{\mu\nu}\vert$.
Here  $\Delta_s\!=\!1$  and $\Delta_f=3/2$.

We would like to generate the Planck scale spontaneously, from the vev of a scalar field $\phi$.
To this purpose one uses that the Lagrangian
\begin{eqnarray}\label{d1}
L_1^\pm=\pm \sg\, \frac{\xi}{2}\,
 \Big\{\,\frac{1}{6}\, \phi^2\,R + g^{\mu\nu} \partial_\mu\phi\partial_\nu\phi\,\Big\}
\end{eqnarray}

\medskip\noindent
 is invariant under transformation (\ref{ct})\footnote{To see this, one uses
that under eq.(\ref{ct}) $R$ transforms as
$R\rightarrow R^\prime= e^{-2\alpha(x)} \big( R-6 \, e^{-\alpha(x)} \Box e^{\alpha(x)}\big)$}.
 $\xi$ is the non-minimal coupling and we assume  $\xi\!>\!0$.

Then one is facing the following issue. To   generate  the Einstein term
\begin{eqnarray}
L_E= -\frac 12\, \sg\, M_p^2 \,R
\end{eqnarray}
after spontaneous breaking of conformal symmetry from a vev of $\phi$ from
the first term in (\ref{d1}), one must  take the minus sign in front of (\ref{d1});  that
means  a negative kinetic term for $\phi$ (ghost) is present in the theory, which
may not be acceptable.
 Alternatively a positive kinetic term leads to $M_p^2\!<\!0$.
One usually sets $M_p=\langle\phi\rangle$ (``gauge fixing'' the Planck scale)
and the ghost  presence is then ignored. Yet, one cannot have the benefit of  conformal symmetry
 but ignore this ``side effect'', therefore  we would like
to understand its meaning.

To avoid this problem, we associate to transformation (\ref{ct}) that of a
(Weyl) vector field $\omega_\mu$ \cite{Weyl}
which, in the light of (\ref{ct}), is of geometric origin
\begin{eqnarray}\label{ct2}
\omega_\mu\rightarrow \omega_\mu^\prime &=&\omega_\mu-\frac{2}{q}\,\partial_\mu\alpha(x),
\end{eqnarray}
then consider adding the kinetic term below, with a suitable normalization coefficient
\medskip
\begin{eqnarray}
L_2=\frac{1}{2} \,(1+\xi)\,\sg\, g^{\mu\nu} \,\tilde D_\mu \phi\,
\tilde D_\nu \phi, \qquad\qquad \tilde D_\mu\equiv \partial_\mu-\frac{q}{2} \,\omega_\mu.
\end{eqnarray}

\medskip\noindent
$L_2$ is  invariant under (\ref{ct}), (\ref{ct2}) since
$\tilde D_\mu\phi\rightarrow e^{-\alpha}\tilde D_\mu\phi$, due to the presence of $\omega_\mu$.
Since $L_1^\pm$ is also invariant under (\ref{ct}), (\ref{ct2}), the sum
$L_1^\pm\!+\!L_2$  is also invariant. Hereafter  we take  $L_1^-$.
One has
$L_1^-\!+\!L_2=(1/2)\, g^{\mu\nu}\, \partial_\mu\phi\,\partial_\nu\phi
-(1/12)\,\xi\, \phi^2 R+\cdots$,
with a canonically normalized kinetic term for $\phi$.
 Thus,  the Planck (mass)$^2$ generated by $\langle\phi^2\rangle$ and  the
 kinetic term of $\phi$ can be simultaneously positive\footnote{This
 is automatic in Weyl geometry, see  Section~\ref{2.3} and
\cite{Smolin,Oh,Moffat1}.}.
 This is made possible by the additional presence of the
 Weyl  field $\omega_\mu$; this is a sufficient condition for the
consistency of the theory (absence of ghosts).

\subsection{SM Lagrangian with Weyl gauge symmetry }\label{2.2}

We use the above observation about $L_1^-+L_2$  to construct a Lagrangian
without ghosts and invariant under eqs.(\ref{ct}), (\ref{ct2}).
For generality,  consider a version of $L_1^-+L_2$ with  more scalar fields $\phi_j$ of non-minimal
couplings $\xi_j$, then a Weyl-invariant Lagrangian is
\medskip
\begin{eqnarray}
L= \sg\,\Big\{ - \frac{\xi_j}{2} \, \Big[\frac{1}{6}\, \, \phi^2_j\,R+ g^{\mu\nu} \,
\partial_\mu\phi_j\, \partial_\nu\phi_j\Big]
\!+\!(1+\xi_j)\, \frac{1}{2} g^{\mu\nu} \tilde D_\mu\phi_j\,\tilde D_\nu\phi_j
-V(\phi_j)\Big\}.
\end{eqnarray}

\medskip\noindent
A summation is understood over repeated index $j=1,2,3\cdots$.
We also added a potential $V(\phi_j)$  for the scalars $\phi_j$; given the conformal symmetry,
$V$ is a homogeneous function, so
\begin{eqnarray}\label{hf}
V(\phi_j)=\phi_k^4\, V(\phi_j/\phi_k), \qquad k={\rm fixed.}
\end{eqnarray}
$L$ can be re-written as
\medskip
\begin{eqnarray}\label{LL}
L=
\sg\,\Big\{\!-\frac{\xi_j}{12} \,\phi_j^2\,R +\!\frac{g^{\mu\nu}}{2}
(\partial_\mu\phi_j)\,(\partial_\nu\phi_j)
- \frac{q}{4} \,g^{\mu\nu}\, \omega_\mu\,K_\nu
+ \!\frac{q^2}{8}\,K\, \omega_\mu\,\omega^\mu\!
-V(\phi_j)\! \Big\},
\end{eqnarray}
where
\begin{eqnarray}
K_{\nu}=\partial_\nu K, \qquad K=(1+\xi_j)\,\phi_j^2.
\end{eqnarray}

\medskip\noindent
 $L$ above is  invariant under (\ref{ct}), (\ref{ct2}),
for {\it all values} of  $\xi_j$, thanks to the $\omega_\mu$-dependent
terms. $L$ has positive kinetic term for $\phi_j$ and $M_p^2>0$ when generated by the vev of
$\langle\phi\rangle$ (assuming $\xi_j>0$).
In the absence of the $\omega_\mu$-dependent part,
 $L$ is not  conformal (unless $\xi_j=-1$),
 but only global conformal.
Unlike in gauge theories, $\omega_\mu$ is a vector under a {\it real}
transformation of the  fields $\phi_j$ (missing the $i$ factor). The associated current $K_\mu$
 is non-zero for $\phi_j$ reals.

Further, we include a  kinetic term for  $\omega_\mu$ with  the ``usual'' (pseudo)Riemannian
definition
\begin{eqnarray}
\label{gauge}
L_g=-\,\frac{\sg}{4} \, g^{\mu\rho}\, g^{\nu\sigma}\,F_{\mu\nu}\,F_{\rho\sigma},
\qquad
F_{\mu\nu}= D_\mu \omega_\nu -D_\nu \omega_\mu,
\quad
 D_\mu \omega_\nu=\partial_\mu \omega_\nu -\Gamma_{\mu\nu}^\rho \omega_\rho,
\end{eqnarray}
$L_g$ is  invariant under (\ref{ct}), (\ref{ct2}), since
 the metric part is invariant and $F_{\mu\nu}$ ($=\!\partial_\mu\omega_\nu\!-\!\partial_\nu\omega_\mu$)
is invariant, too.
The Riemann connection\footnote{The Riemann affine connection used here is
$\Gamma_{\mu\nu}^\rho=(1/2)\,g^{\rho\beta} \big[ \partial_\nu g_{\beta\mu} +\partial_\mu g_{\beta\nu}
-\partial_\beta g_{\mu\nu}\big]$.} $\Gamma_{\mu\nu}^\rho$, symmetric in $\mu,\nu$,
 is  not invariant under (\ref{ct}).

Finally, one can consider the Weyl-invariant Lagrangian $L_f$ for
the massless fermions  of the theory that transform under (\ref{ct}).
$L_f$ has  the usual form in (pseudo)Riemann space
\begin{equation}\label{fermions}
L_f= \sqrt{g}\,  {\bar\psi}\, i\gamma^a\, e^\mu\,_a  { D}_\mu  \psi,
\qquad
{D}_\mu\psi =\Big(\partial_\mu+\frac{1}{2} \omega^{ab}_\mu \sigma_{ab}
\Big)\,\psi
\end{equation}
where $\omega^{ab}_\mu=e^{\lambda b}(-\partial_\mu e_\lambda\,^a+e_\nu\,^a \Gamma^\nu_{\mu\lambda})$
is the spin connection and $\sigma_{ab}=\frac{1}{4}[\gamma_a,\gamma_b]$.
Note that $g_{\mu\nu}=e_\mu\,^a e_\nu\,^b \eta_{ab}$ and $e^\mu\,_a e_\nu\,^a=\delta^\mu_\nu$.
Under  a Weyl transformation of the metric, eq.(\ref{ct}),
the vielbein $e_\mu^a$ transforms as  $e_\mu\,^{a\, \prime}=e^{\alpha(x)} e_\mu\,^a$,
while for the spin connection   we have
$\omega^{ab\,\prime}_\mu=\omega^{ab}_\mu + (e_\mu\,^a e^{\nu b}-e^{\nu a} e_\mu\,^b)\partial_\mu\alpha$.
Then it can be shown that $L_f$ is  invariant  under a Weyl gauge transformation,
eqs.(\ref{ct}), (\ref{ct2}),
and  there is no  coupling of fermions to  the gauge field $\omega_\mu$!

Regarding the SM gauge fields kinetic terms ($L_G$), these are invariant under Weyl
gauge symmetry. Indeed, the gauge fields presence under the covariant derivative
that contains $\partial_\mu$ shows that these are invariant, since coordinates
do not transform under (\ref{ct}).
Therefore, there is no coupling between
SM gauge fields and $\omega_\mu$\footnote{An exception
 is a possible kinetic mixing of the field strength of
$\omega_\mu$ to that of $U(1)_Y$ \cite{Hm}.}.
 For example, for the  $U(1)_Y$ gauge field $A_\mu$,
the covariant derivative can be written as
 $ D_\mu A_\nu=\partial_\mu A_\nu - \Gamma_{\mu\nu}^\rho A_\rho$.
The gauge kinetic terms do not contain the Christoffel symbols
because $F_{\mu\nu}= D_\mu A_\nu-D_\nu A_\mu=\partial_\mu A_\nu-\partial_\nu A_\mu$.

The sum, ${\cal L}=L+L_g+L_f+L_G$, is the total SM-like Lagrangian with
Weyl gauge symmetry\footnote{
One could also add a  Weyl tensor-squared term to the action  which is
invariant under  (\ref{ct}) or a quadratic term in the Weyl scalar curvature $\tilde R^2$,
see \cite{new1,new2} for further  details.}
 which is invariant under (\ref{ct}), (\ref{ct2}).
Here $L$ is immediately adapted to accommodate the Higgs doublet of the SM
with one of the $\phi_j$ fields to account for the Higgs neutral scalar. In conclusion,
we have a SM-like Lagrangian that is invariant under (\ref{ct}), (\ref{ct2}).

\subsection{From  Riemann to Weyl conformal geometry}\label{2.3}

The presence of the Weyl gauge field in our model
in  the Riemannian geometry and invariant under (\ref{ct}), (\ref{ct2})
is natural in  Weyl's conformal geometry  \cite{Weyl,Dirac} (also \cite{Scholz}).
Following \cite{Dirac} we write the SM-like Lagrangian with this symmetry
directly in Weyl geometry and we verify that it agrees  with that
 of the previous section, built in the Riemannian geometry
(with $\omega_\mu$ introduced to avoid ghosts).

Weyl geometry is a scalar-vector-tensor theory of gravity and thus
provides a generalization (to classes of equivalence) of Brans-Dicke-Jordan scalar-tensor theory
\cite{BDJ}
and of other  conformal invariant models  \cite{Turok}. It was  used for model building
\cite{Smolin,Cheng}
with renewed recent interest in  \cite{Moffat1,TW,Davidson,Moffat2,Heisenberg,
Oh,Wheeler,Quiros} and applications to inflation, see e.g.
 \cite{FRH0,FRH2,FRH1,GG1,GG2,GG3,GG4,GG5,higgsdilaton}.
 If the Weyl
field is set to zero, one obtains  (Weyl integrable) models similar to Brans-Dicke-Jordan
 theory \cite{Quiros}.

In Weyl geometry the curvature scalars and tensors and the connection
are  different from the Riemannian case where they are induced by the metric alone.
In Weyl geometry
\medskip
\begin{eqnarray}\label{tGamma}
\tilde\Gamma_{\mu\nu}^\rho=
\Gamma_{\mu\nu}^\rho+
\frac{q}{2}\,\Big[ \delta_\mu^\rho \,\omega_\nu +\delta_\nu^\rho\,\omega_\mu- g_{\mu\nu}\,\omega^\rho\Big],
\end{eqnarray}

\medskip\noindent
where $\Gamma_{\mu\nu}^\rho$ are the connection coefficients in the Riemannian geometry.
Under (\ref{ct}), (\ref{ct2}) the coefficients $\tilde\Gamma_{\mu\nu}^\rho$ are invariant,
as one can easily check. The system is torsion-free.
The Riemann tensor in Weyl geometry is then generated by the ``new'' $\tilde \Gamma_{\mu\nu}^\rho$
\medskip
\begin{eqnarray}
\tilde R^\lambda_{\mu\nu\sigma}=
\partial_\nu \tilde\Gamma^\lambda_{\mu\sigma}
-\partial_\sigma\tilde\Gamma^\lambda_{\mu\nu}
+ \tilde\Gamma^\lambda_{\nu\rho}\,\tilde\Gamma^\rho_{\mu\sigma}
-\tilde\Gamma^\lambda_{\sigma\rho}\,\tilde\Gamma^\rho_{\mu\nu},
\end{eqnarray}

\medskip\noindent
and then $\tilde R_{\mu\sigma}=\tilde R^\lambda_{\mu\lambda\sigma}$, $\tilde R=g^{\mu\nu} \tilde R_{\mu\nu}$.
We can then compute $\tilde R$ and  find
\medskip
\begin{eqnarray}\label{tildeR}
\tilde R& =&
R- 3\,q\,
\Big[\partial_\mu \omega^\mu +\frac12 \omega^\rho\, g^{\lambda\beta}\,\partial_\rho \,g_{\lambda\beta}\Big]
-\frac32 q^2 \omega^\mu \,\omega_\mu
\nonumber\\[-4pt]
&=&
R-3 \,q\,   D_\mu \omega^\mu -\frac32 q^2 \, \omega^\mu \omega_\mu.
\end{eqnarray}
Then  under transformations (\ref{ct}) and (\ref{ct2}),
\begin{eqnarray}
\tilde R\rightarrow \tilde R^\prime=e^{- 2\alpha(x)} \tilde R.
\end{eqnarray}
As a result
\begin{eqnarray}
 L_{1w}=-\sg\, \frac{1}{12}\,\xi_j\,\phi^2_j \,\tilde R,\qquad \textrm{(sum over $j=1,2.$)},
\end{eqnarray}
is invariant under combined transformations (\ref{ct}), (\ref{ct2}).
This is unlike  in the Riemannian case of the previous section
 where  the non-minimal coupling term in the action was not invariant.

Further, we can define a kinetic term for $\phi$ in Weyl geometry,  invariant under (\ref{ct}), (\ref{ct2})
\begin{eqnarray}
L_{2w}=\frac{1}{2}\, \sg\, g^{\mu\nu}\, \tilde D_\mu \phi_j \,\tilde D_\nu \phi_j-\sg \,V(\phi_j).
\end{eqnarray}

\medskip\noindent
We also have a gauge kinetic term ($L_{3w}$) for $\omega_\mu$, now defined by
new coefficients $\tilde \Gamma$ of (\ref{tGamma})
\begin{equation}
L_{3w}=-\frac{\sg}{4} g^{\mu\rho} \,g^{\nu\sigma}\, F_{\mu\nu} F_{\rho\sigma},\qquad
F_{\mu\nu}=\tilde D_\mu \omega_\nu-\tilde D_\nu \omega_\mu, \qquad
\tilde D_\mu \omega_\nu=\partial_\mu\omega_\nu-\tilde\Gamma_{\mu\nu}^\rho \omega_\rho.
\end{equation}
However,  $\tilde \Gamma_{\mu\nu}^\rho$ are symmetric in $\mu\leftrightarrow\nu$
and also invariant under Weyl transformation eqs.(\ref{ct}), (\ref{ct2}).
Thus, $F_{\mu\nu}$ and $L_{3w}$ are  equal to their counterparts in
the previous section, eq.(\ref{gauge}), so $L_{3w}=L_g$.
 The same can be said about the SM gauge  fields kinetic terms.

Further, the fermionic Lagrangian is defined with the Weyl connection, as follows,
\begin{equation}\label{fermions2}
L_{4w}= \sqrt{g}\,  {\bar\psi}\, i\gamma^a\, e^\mu\,_a  {\tilde D}_\mu  \psi,
\qquad
{\tilde D}_\mu\psi =\Big(\partial_\mu+\frac{1}{2} {\tilde\omega}^{ab}_\mu \sigma_{ab}
-\frac{3}{4}q\,\omega_\mu\Big)\,\psi
\end{equation}
where ${\tilde\omega}^{ab}_\mu\!=e^{\lambda b}(
-\partial_\mu e_\lambda\,^a\!+\!e_\nu\,^a\, {\tilde\Gamma}^\nu_{\mu\lambda})$.
However, one shows  \cite{Moffat1} that $L_{4w}\!=L_f$ with $L_f$ of (\ref{fermions}).

Adding together $L_{1w}$, $L_{2w}$, $L_{3w}$, and $L_{4w}$,
each of these  invariant under (\ref{ct}), (\ref{ct2}),
we obtain a total  Lagrangian for the case of Weyl geometry.
It is interesting to see that  this Lagrangian is equal to
$L+L_g+L_f$ of (\ref{LL}), (\ref{gauge}) and (\ref{fermions}),
up to a total derivative term. This follows from the relation
\medskip\noindent
\begin{eqnarray}\label{equiv}
L_{1w}+L_{2w}=L + \frac{q}{4}\, \xi_j\,\partial_\mu\Big[ \sg\, \phi_j^2\,\omega^\mu\Big].
\end{eqnarray}

\medskip\noindent
To show eq.(\ref{equiv}), one uses  the relation between $\tilde R$ and $R$ of eq.(\ref{tildeR})
that relates Weyl and Riemann scalar curvatures and that
 $\partial_\lambda g=g\,g^{\mu\sigma}\,\partial_\lambda g_{\sigma\mu}$.

Eq.(\ref{equiv})  shows that our model  agrees (for two fields case) with that  in
 \cite{Moffat1}  built within Weyl geometry from the onset and following \cite{Dirac}.
We  thus obtained the same Lagrangian in  Riemann and Weyl geometry,
albeit with different initial motivations.
Our motivation  for a consistent, ghost-free conformal action, with this symmetry broken spontaneously,
lead us to introduce a gauge transformation and Weyl gauge field associated to (\ref{ct}).

\section{Spontaneous breaking of Weyl gauge symmetry}\label{3}

 In this section we show how the Weyl conformal symmetry of our model is spontaneously broken
for one or more scalar fields of non-minimal couplings $\xi_j$  to $R$.
  Then, we show that the (radial mode)  would-be Goldstone boson (dilaton $\rho$) of the Weyl symmetry
  decouples from the angular variables  fields   due to a Stueckelberg mechanism for the Weyl
 gauge field which becomes massive.
Before decoupling, the dilaton can provide a scale-invariant ultraviolet (UV)
regularisation for models in which quantum scale invariance  is important.

\subsection{One scalar field and Stueckelberg mechanism for $\omega_\mu$}

Let us first show  how  spontaneous breaking of Weyl symmetry happens for  one
scalar field $\phi$.
Then $L$ of eq.(\ref{LL}) simplifies (no sum over $j$) and we  replace $\phi_j\rightarrow \phi$, then
\begin{eqnarray}
K=(1+\xi)\,\phi^2,\qquad
V=\frac{\lambda}{4!}\phi^4, \label{onefield}
\end{eqnarray}
where $V$ is the only  one allowed by the Weyl symmetry.
To decouple the scalar field fluctuations  from $R$,  we go to the Einstein frame
by rescaling the metric to
\begin{eqnarray}\label{metric1}
\hat g_{\mu\nu}=\Omega\, g_{\mu\nu}, \qquad
\Omega=\frac{\xi}{6}\,\frac{\phi^2}{\langle\phi\rangle^2}.
\end{eqnarray}

\medskip\noindent
Hereafter a hat on a variable denotes the Einstein frame value of that variable.
From eq.(\ref{LL}) for one field and  eq.(\ref{metric1}) we  obtain the tensor-scalar part of
Einstein-frame Lagrangian as
\medskip
\begin{eqnarray}
\LE= \sqrt{\hat g}\, \Big\{-\frac{1}{2}\,\langle\phi\rangle^2\,\RE
+\frac{3}{4}\langle\phi\rangle^2 (\partial_\mu \ln\Omega)^2
+\frac{1}{\Omega}\,\Big[\,\frac{1}{2}(\partial_\mu\phi)^2
+ \frac{q^2}{8}K \omega_\mu\omega^\mu
-\frac{q}{4} \omega^\mu K_\mu \Big]
-\frac{V}{\Omega^2}\Big\}
\end{eqnarray}

\medskip\noindent
giving
\begin{eqnarray}
\LE &= & \sqrt{\hat g}\, \Big[ -\frac{1}{2}\langle\phi\rangle^2 \RE
+3 \langle\phi\rangle^2 \Big(1+\frac{1}{\xi}\Big)\Big(\frac{\partial_\mu\phi}{\phi}\Big)^2
+\frac{3}{4}\,q^2\langle\phi\rangle^2\Big(1+\frac{1}{\xi}\Big)\omega_\mu\,\omega^\mu
 \nonumber \\
&&\quad
-3\, q\,\langle\phi\rangle^2\Big(1+\frac{1}{\xi}\Big) \omega^\mu\,{\partial_\mu\ln \phi}
-\frac{3\lambda}{2\,\xi^2} \langle\phi\rangle^4  \Big].
\end{eqnarray}

\medskip\noindent
where all contractions are with the new metric $\hat g_{\mu\nu}$. Finally, we introduce
\medskip
\begin{equation}
\omega^\prime_\mu =\omega_\mu -\frac{2}{q} \partial_\mu\ln\phi.
\label{weyl1}
\end{equation}
giving
\begin{eqnarray}
\LE= \sqrt{\hat g}\, \Big[-\frac{1}{2}\langle\phi\rangle^2\,\RE
+\frac{3}{4}\, q^2\,\langle\phi\rangle^2\,
\Big(1+\frac{1}{\xi}\Big)\,\omega^\prime _\mu \omega^{\prime\mu}
-\frac{3\lambda}{2\,\xi^2}\langle\phi\rangle^4  \Big]
\end{eqnarray}

\medskip\noindent
As a result, the scalar (dilaton) field $\phi$
is ``eaten'' by the Weyl gauge boson $\omega_\mu$. The  mass of $\omega_\mu$ is
$m_\omega^2=(3 q^2/2)(1+1/\xi)\langle\phi\rangle^2$. Therefore, conformal symmetry is broken
spontaneously as in the Stueckelberg formulation for a massive $U(1)$ {\it without} a
corresponding Higgs mode.
The number of degrees of freedom remains the same (three): in Jordan frame we had
a real scalar  and a massless vector, while in Einstein frame, after breaking
there is no scalar field but a massive vector boson.
Also note that the gauge kinetic term $L_g$ of $\omega_\mu$, see eq.(\ref{gauge}),
 is invariant under (\ref{metric1}), (\ref{weyl1}).
The scalar potential becomes a cosmological
constant, $V_0=3\lambda\langle\phi\rangle^4/(2\xi^2)$, in Einstein frame.

Transformation (\ref{weyl1}) may be seen as
a Weyl gauge transformation (\ref{ct2}) with $\alpha=\ln \sqrt{\Omega}$
corresponding to  (\ref{metric1}).
Then the scalar field $\phi$ transforms  according to eq.(\ref{ct}) into
\medskip
\begin{eqnarray}
 \phi^\prime = e^{-\ln\sqrt\Omega}\phi
=\sqrt{6/\xi}\,\, \langle\phi\rangle,
\end{eqnarray}

\medskip\noindent
so $\phi^\prime$ is not dynamical anymore.
Therefore spontaneous breaking of conformal symmetry fixing
 the Planck scale (to $M_p=\langle\phi\rangle$) and  Stueckelberg mechanism
are related to  a Weyl transformation to a special ``unitary'' gauge
(``gauge fixing'').

 While we  used
$\langle\phi\rangle\!\not=\!0$ in the definition of $\Omega$ and subsequent equations,
 this is actually not needed and an arbitrary  mass scale ${\cal M}$
can be used instead of $\langle\phi\rangle$, corresponding to a
 different ``gauge fixing'' (and different Planck scale!).
Indeed, Stueckelberg mechanism is
a re-arrangement of the degrees of freedom (that does  not
 require $\langle\phi\rangle\not=0$).
Using an arbitrary ${\cal M}$
 is consistent with the fact that for a single scalar field
in a Weyl-invariant theory $\langle\phi\rangle$ cannot
be determined from  the condition  $4\,V(\phi)-\phi V^\prime(\phi)=0$
 which is automatically respected, hence $\langle\phi\rangle$ remains a parameter (unknown).
This condition is also related to the conservation of the current $D^\mu K_\mu=0$, which
 for  a FRW metric leads to a constant  solution $\langle\phi\rangle$ \cite{FRH0,FRH2} that is
 not fixed by the theory.

\subsection{Two scalar fields and Stueckelberg mechanism for $\omega_\mu$}
\label{section3.2}

Let us consider now the more interesting case of two scalar fields in eq.(\ref{LL}), (j=1,2).
Then
\medskip
\begin{equation}\label{two}
K\!=(1+\xi_1)\,\phi_1^2+(1+\xi_2)\,\phi_2^2,
\end{equation}

\medskip\noindent
Since  $V$ is a homogeneous function of fields, one can have
\medskip
\begin{equation}\label{Vh1}
V(\phi_1,\phi_2)=\frac{\lambda_1}{4!}\phi_1^4
+\frac{\lambda_{12}}{4}\phi_1^2\,\phi_2^2
+\frac{\lambda_2}{4!}\,\phi_2^4.
\end{equation}
In particular, if \, $3 \lambda_{12}=-\sqrt{\lambda_1\lambda_2}$, then
\begin{equation}\label{Vh2}
V(\phi_1,\phi_2)=\frac{\lambda_1}{4!}\,\Big(
\phi_1^2 -\frac{\sqrt \lambda_2}{\sqrt\lambda_1}\,\phi_2^2\Big)^2.
\end{equation}

\medskip\noindent
$V$ can also contain terms like $\phi_1^6/\phi_2^2$, etc \cite{D1}.
The results below are for a  general homogeneous function $V(\phi_{1,2})$
i.e. it has a flat direction:
 $V(\phi_{1,2})=(\phi_1^2-k_0\, \phi_2^2)^2\,f(\phi_1/\phi_2)$, ($k_0$=constant).

To decouple  $R$ from the fluctuations of $\phi_{1,2}$, we consider a transformation
to the Einstein frame.
Let us   perform a metric rescaling of $L$ eq.(\ref{LL}),  to
\medskip
\begin{eqnarray}\label{Omega2fields}
\sgh_{\mu\nu}=\Omega\, g_{\mu\nu},\qquad
\Omega=\frac{1}{6\,v^2}\,(\xi_1\phi^2_1+\xi_2\phi^2_2),
\qquad v^2\equiv\langle\xi_1\phi_1^2+\xi_2\phi_2^2\rangle.
\end{eqnarray}

\medskip\noindent
Here $v$ ensures that $\Omega$ is dimensionless\footnote{{
As for the one-field case we could use instead of $v$ an arbitrary mass scale.}}.
From eq.~(\ref{LL}) for two fields and  with (\ref{Omega2fields}),
 we obtain the corresponding  Einstein-frame Lagrangian as
\medskip
\begin{eqnarray}
\LE\!&=&\!\! \sqrt{\hat g} \,\Big[ -\frac{1}{2}\,v^2\,\RE
+\frac{3}{4}\,v^2\,(\partial_\mu \ln\Omega)^2
\nonumber\\
&+&\frac{1}{\Omega}\,
\Big(\,\frac{1}{2}(\partial_\mu\phi_1)^2
+\frac{1}{2}(\partial_\mu\phi_2)^2
+ \frac{q^2}{8}K\, \omega_\mu\omega^\mu
-\frac{q}{4} \omega^\mu K_\mu \Big)
-\VE
\Big]
\end{eqnarray}

\medskip\noindent
where all contractions are with the new metric $\sgh_{\mu\nu}$;  $\Omega$,
 $K$ and $\VE$ are  functions of $\phi_{1,2}$ with
\medskip
\begin{eqnarray}\label{VE}
\VE(\phi_1,\phi_2)=\frac{1}{\Omega^2} \,V(\phi_1,\phi_2).
\end{eqnarray}
Then
\begin{eqnarray}
\LE
&=& \sqrt{\hat g}\, \Big[ -\frac{1}{2}\,v^2\,\RE
+\frac{1}{2}G_{ij}\, \partial_\mu\phi_i \,\partial^\mu \phi_j
+\frac{q^2}{8} \frac{K}{\Omega} \omega_\mu\omega^\mu
-\frac{q}{4}  \omega^\mu \frac{K_\mu}{\Omega}-\VE  \Big]
\label{twofield2}
\end{eqnarray}
where
\begin{eqnarray}\label{Gmatrix}
G_{ij} =
\frac{1}{6\,v^2\,\Omega^2}
\left(\begin{array}{cc}
\xi_1(1+\xi_1)\phi^2_1
+\xi_2\phi^2_2 &  \xi_1 \xi_2\, \phi_1\phi_2 \\
 \xi_1 \xi_2\, \phi_1\phi_2 &  \xi_2(1+\xi_2)\phi^2_2+\xi_1\phi^2_1
\end{array}\right), \quad  i,j=1,2.
\end{eqnarray}

\medskip\noindent
The kinetic terms in $\LE$ become diagonal (no mixing)
 in a new  fields basis  of ($\rho$, $\theta$) where
\medskip
\begin{eqnarray}\label{phi1}
\phi_1&=& \frac{1}{\sqrt{1+\xi_1}}\,\rho\,\sin\theta, \nonumber\\[-2pt]
\phi_2&=& \frac{1}{\sqrt{1+\xi_2}}\, \rho \,\cos\theta.
\end{eqnarray}

\medskip\noindent
It is more  illustrative however to first bring  the Weyl terms in $\LE$ to a
quadratic form using
\medskip
\begin{eqnarray}\label{om}
\omega'_\mu=\omega_\mu -\frac{1}{q}\partial_\mu\ln  K,
\end{eqnarray}

\medskip\noindent
where notice that   $K=\rho^2$.
Adding $\hat L_g$ of eq.(\ref{gauge}) with eq.(\ref{Omega2fields})  for the Weyl field $\omega_\mu$
 then
\medskip
\begin{eqnarray}
\LE+\hat L_g\!\!\!& = &\!\!\!  \sqrt{\hat g} \,\Big[-\frac{1}{2}\,v^2\,\RE
+\frac{1}{2}\,G_{ij} \partial_\mu\phi_i \,\partial^\mu \phi_j
-\frac{1}{8K \Omega} (\partial_\mu K)^2
-\frac{1}{4} F'_{\mu\nu}F^{\prime \mu\nu}
+ \frac{K}{8\Omega}\, q^2 \omega'_\mu \omega^{\prime \mu}
- \VE\Big]
\nonumber \\
&=&\!\! \sqrt{\hat g} \, \Big[-\frac{1}{2}\,v^2\,\RE
+\frac{1}{2}\, T_{ij}\, \partial_\mu\phi_i \partial^\mu \phi_j
-\frac{1}{4} F'_{\mu\nu}F^{\prime \mu\nu}
+ \frac{K}{8\Omega}\, q^2 \omega'_\mu \omega^{\prime \mu}
-\VE \Big]  \label{twofield3}
\end{eqnarray}

\medskip\noindent
where  $F'_{\mu\nu}=\tilde D_\mu \omega'_\nu-\tilde D_\nu\omega'_\mu$
is invariant under (\ref{om}). Above we denoted   $T_{ij}=G_{ij}+H_{ij}$, ($i,j=1,2$),  with:
\medskip
\begin{equation}\label{Hmatrix}
H_{ij} =
-\frac{1}{\Omega}
\, \frac{1}{K}
\left(\!\begin{array}{cc}
 (1+\xi_1)^2\phi^2_1 & (1+\xi_1)(1+\xi_2)\phi_1 \phi_2
 \\
(1+\xi_1)(1+\xi_2)\phi_1 \phi_2  & (1+\xi_2)^2 \phi^2_2
\end{array}\! \right).
\end{equation}

\medskip\medskip\noindent
In the new basis (\ref{phi1}) the scalar kinetic terms in
eq.~(\ref{twofield3}) are reduced to a single term and
\medskip
\begin{eqnarray}\label{Lfinal}
\LE+\hat L_g=\sqrt{\hat g} \,\Big[ \frac{-1}{2} \,v^2\,\RE
+\frac12\,
F(\theta)\,
\, v^2\,(\partial_\mu\tan\theta)^2
-\frac{1}{4} \,F'_{\mu\nu} F^{' \mu\nu}+ \frac{1}{2} m^2(\theta) \,\sgh^{\mu\nu}  w_\mu' w_\nu'
-\VE\Big]
\end{eqnarray}
with
\begin{eqnarray}
F(\theta)=\frac{6\,b}{\xi_2} \frac{\tan^2\theta+\xi_2/\xi_1}{(1+\tan^2\theta)(\tan^2\theta+b)^2},
\qquad
 b= \frac{\xi_2(1+\xi_1)}{\xi_1(1+\xi_2)}.
\end{eqnarray}

\medskip
Therefore,  we are  left with the ``angular'' kinetic term for $\theta$ only.
The kinetic term of the  radial (Goldstone) coordinate $\rho$\, (where $\rho^2= K$)
has disappeared, via Stueckelberg mechanism, as it was ``eaten'' by  the Weyl gauge
 boson $\omega'$ in  eq.(\ref{om}). This is  similar to the case with one scalar field in
eq.~(\ref{weyl1}). Thus, in the Einstein frame we  have a massive vector boson and one
(real) scalar field left ($\theta$), while in Jordan frame we had two (real) scalar fields
and a massless $\omega_\mu$, so the number of degrees of freedom is again conserved.

Further, the function $m^2(\theta)$ in (\ref{Lfinal}) is given by
\medskip
\begin{eqnarray}\label{thetaf}
m^2(\theta)=\frac{q^2\, K}{4\Omega}=
 \frac{3\,q^2}{2}\, \frac{v^2 \,(1+\xi_1) (1+\xi_2)(1+\tan^2\theta)}{
\xi_1 (1+\xi_2)\,\tan\theta^2  +\xi_2\,(1+\xi_1)},
\end{eqnarray}
with
\begin{eqnarray}\label{vvv}
v^2=
\langle\rho\rangle^2 \Big[
\frac{\xi_1}{1+\xi_1}
\sin^2\langle\theta\rangle
+ \frac{\xi_2}{1+\xi_2}
\cos^2\langle\theta\rangle\Big].
\end{eqnarray}

\medskip\noindent
Notice that if $\xi_1=\xi_2$ or if  $\tan\theta$ is large,
 the function $m^2(\theta)$ is actually independent
of $\theta$ and then the  Weyl gauge field ($\omega_\mu^\prime$) and the field $\theta$  decouple.

On the ground state $\theta=\langle\theta\rangle$ and the mass of $\omega_\mu$ is
\begin{eqnarray}\label{mass}
m^2(\langle\theta\rangle) =\frac{3}{2}\,q^2 \langle\rho\rangle^2.
\end{eqnarray}
The  mass of  $\omega_\mu$ is thus determined by $\langle\rho\rangle$ alone; unlike $\theta$
whose vev is determined from $\VE$ (see below),
$\langle\rho\rangle$  cannot be predicted  by the theory and is a free
parameter (flat direction)\footnote{$\langle\rho\rangle$
may be fixed by quantum corrections; however, in quantum scale invariant theories
only ratios of field vev's (scales) can be determined (in terms of dimensionless
couplings), so it remains a free parameter.}.

The Planck  scale $M_p^2=v^2$, eq.(\ref{vvv}),  depends in general on $\langle\theta\rangle$.
This is not a problem since unlike $\rho$, the field variable $\theta$ does not change under a
Weyl transformation, eq.(\ref{ct}). However, if the theory has an $\cO(2)$ symmetry, i.e.
identical non-minimal couplings  $\xi_1=\xi_2$,
 then $M_p$ is determined by the vev of the dilaton alone
$M_p^2=v^2=\xi_1\langle\rho\rangle^2/(1+\xi_1)$; in this case, the would-be Goldstone
(dilaton) field  $\rho$ ``eaten'' by $\omega_\mu$  and ``fixing'' its mass also  fixes
the Planck scale. The same is true in the limit of
large $\tan\theta\rightarrow \infty$,  when $M_p^2=v^2=\xi_1\langle\rho\rangle^2/(1+\xi_2)$.

Regarding the potential $\VE$  in eq.(\ref{Lfinal}), it is given by eq.(\ref{VE})
expressed in terms of the new field variables $\rho,\theta$. With eq.(\ref{hf})
and  $V(\phi_1, \phi_2)$ the initial potential in Jordan frame, then
\medskip
\begin{eqnarray}
\VE= 36\, v^4\,\frac{b^2}{\xi_2^2}\, \frac{V(c\,\tan\theta,1)}{(\tan^2\theta+b)^2},
\qquad \textrm{where}\qquad
c=\sqrt{\frac{1+\xi_2}{1+\xi_1}},
\end{eqnarray}

\medskip\noindent
which depends on $\theta$ only.
Finally, another mechanism to decouple the dilaton was studied in \cite{FRH0}
using  a  global version of  the  Weyl  symmetry studied here
(and assuming  this survives black hole physics \cite{Banks}).

\subsection{More fields and Stueckelberg mechanism}

The  Stueckelberg mechanism for $\omega_\mu$ can be extended for  more
scalar fields with non-minimal couplings,  using general coordinates. For
three fields
$\phi_1=(1/\sqrt{1+\xi_1})\rho\sin\theta\cos\zeta$,
$\phi_2=(1/\sqrt{1+\xi_2})\rho\sin\theta\sin\zeta$,
$\phi_3=(1/\sqrt{1+\xi_3})\rho\cos\theta$.
As before, the kinetic term of radial field $\rho$ is the Goldstone  eaten by the
vector boson $\omega_\mu$ of mass $q^2 K/(4\Omega)\vert_{\theta=\langle\theta\rangle}$.
 One is left with  kinetic terms
for the angular-coordinates fields $\theta$, $\zeta$; similarly, the scalar potential
will depend only on these fields. This generalization is useful in cases where
one of the scalar fields left is a Higgs field, while the other is a second
Higgs-like scalar, inflaton, etc. The scalar potential is then
\medskip
\begin{eqnarray}
\VE(\theta,\phi)=\frac{1}{\Omega^2} \,V(\phi_1,\phi_2,\phi_3)=
\frac{36\,v^4\,V(z_1,z_2,z_3)}{(\xi_1 z_1^2+\xi_2 z_2^2+\xi_3 z_3^2)^2}
\end{eqnarray}

\medskip\noindent
where $V(\phi_1,\phi_2,\phi_3)$ is the initial potential in the Jordan frame
and  $z_j=\phi_j/\rho$ are functions of $\theta$, $\zeta$ only.
If  $\xi_1=\xi_2=\xi_3$, then
the Planck scale is also determined by the same $\rho$ field.
The extension to more scalar fields  is straightforward. This study can also be extended to include
additional (Weyl gauge invariant) terms  quadratic in the scalar curvature \cite{new1,new2}.

\subsection{Other implications: UV scale-invariant regularization}\label{3.4}

The above results  have  implications for
models with (global)  scale  invariance at the quantum level. Such
models are important since they can have a  quantum stable hierarchy between two scalar
fields vev's (higgs and dilaton), which is  relevant for the SM hierarchy problem, as we detail below.

Consider first a classical scale invariant model. The SM with a vanishing higgs mass parameter is an
example. This symmetry can  be preserved at the quantum level, by ensuring
that the UV regularization  respects it.
This is done by replacing the subtraction scale $\mu$ by the dilaton
field  $\rho$ \cite{Englert}.  After spontaneous
breaking of this symmetry, $\mu\sim \,\langle\rho\rangle$.  In this way one
obtains scale invariant results at the quantum  level \cite{S1,Armillis,D1,D2,Monin}. After the quantum
calculation one can expand the result (e.g. the scalar potential) about the vev of the
dilaton to recover standard results (e.g. Coleman-Weinberg potential) plus additional
higher dimensional operators suppressed by the dilaton vev \cite{D1}. Such models have
only spontaneous breaking  of the scale symmetry,  thus there is no dilatation   anomaly
\cite{Englert,Armillis,Tamarit,D2}.

The relation to the hierarchy problem is that the dilaton vev is fixing $M_p$
and so it must be much higher than the Higgs vev. Such hierarchy can be the result of
 one initial {\it classical} tuning of the (dimensionless) couplings.
This tuning remains stable at the quantum level, due to quantum scale
invariance and a shift symmetry of the dilaton (Goldstone mode)~\cite{K1}.
However, the dilaton remains in the spectrum  as a flat direction,
even  at  quantum level. {One can then ask  what happens
to this flat direction for a  more general, local Weyl symmetry.}

The result of this paper answers this question.  As we saw, the  dilaton is ``eaten''
by the Weyl field $\omega_\mu$ which becomes massive, decouples from the spectrum and
{leaves a potential   function of the angular variables fields
only (which can be the Higgs field, inflaton, etc).}

{
For example, in a two-field case and assuming a potential $V$ of eq.(\ref{Vh2}),
after decoupling the potential $\hat V$
depends only on $h\equiv v\cot\theta$ which can be  the neutral Higgs field;
taking for simplicity $\xi_2=0$ and large $\xi_1$, after dilaton decoupling,
the scalar potential (for a canonical kinetic term for $h$) becomes in the Einstein frame
\medskip
\begin{eqnarray}
\hat V=\frac{\lambda_2}{4!}
\Big( \,h^2- \frac{\sqrt\lambda_1}{\sqrt\lambda_2}\frac{v^2}{(\xi_1/6)}\Big)^2,
\end{eqnarray}}

\medskip\noindent
{which is indeed that  of the SM, with $m_h^2=\sqrt{\lambda_1\lambda_2}\,v^2/\xi_1
\!\ll\! v^2\sim \langle\rho\rangle^2 $
for small (ultra-weak) couplings and large $\xi_1$.
What happens with this hierarchy at the quantum level? }

{Before the Stueckelberg mechanism, the dilaton can enforce a scale (or Weyl) invariant
UV regularisation \cite{Englert}
of the quantum corrections to potential (\ref{Vh2}), as described above.
In this way one can construct a quantum scale (Weyl) invariant theory,
dilatation (conformal) anomaly-free, respectively.
 In this case the mentioned classical hierarchy
 between $m_h^2$ and $v^2\sim \langle\rho\rangle^2$  remains stable at the  quantum level
(for more details  see discussion in \cite{D1,D2}).}

\section{Inflation from Weyl gauge symmetry}\label{4}

In this section we study  inflation in  models with spontaneously  broken Weyl gauge
symmetry.  We consider the case of two scalar fields of Section~\ref{section3.2}
 and regard the potential for the angular-variable field $\theta$, obtained after
the Stueckelberg mechanism,  as being responsible for inflation. The potential becomes
constant at large $\tan\theta$.
In this limit, from eqs.(\ref{Lfinal}), (\ref{thetaf}), $\theta$ and $\omega_\mu^\prime$ decouple
 and the action for the inflaton ($\theta$)  can be written as
\begin{eqnarray}
L_{\rm infl}
=  \sqrt{\hat g} \Big[-\frac{1}{2}\,v^2\, \RE  +  \frac{3b\,v^2}{\xi_2}\, \frac{\tau^2
+\xi_2/\xi_1}{(1+\tau^2) (\tau^2+b)^2}\, (\partial_\mu\tau)^2-\VE(\tau)\Big],
\qquad
\tau\equiv \tan\theta
\label{inflaton}
\end{eqnarray}

\medskip\noindent
For $V$ of (\ref{Vh1}), the Einstein-frame potential  expressed in terms of $\tau$ is:
\medskip
\begin{eqnarray}
\VE &=&\frac{36\,v^4\,}{(\xi_1\phi^2_1+\xi_2\phi^2_2)^2}\, \Big(\frac{\lambda_1}{4!}
\,\phi_1^4+\frac{\lambda_{12}}{4}\,\phi^2_1\,\phi^2_2
+\frac{\lambda_2}{4!}\,\phi_2^4 \Big)
\nonumber \\
&=&\frac{36\,v^4\, (1+\xi_1)^2}{\xi^2_1(\tau^2+b)^2} \,
\Big(c_1 \tau^4+c_{12}\tau^2+c_2 \Big), \label{inflatonpot}
\end{eqnarray}
with
\begin{eqnarray}
c_1= \frac{\lambda_1}{4!(1+\xi_1)^2}, \qquad
c_{12} = \frac{\lambda_{12}}{4(1+\xi_1)(1+\xi_2)}, \qquad
c_2 = \frac{\lambda_2}{4!(1+\xi_2)^2}.
\end{eqnarray}

\medskip\noindent
If $c_{12}=-2\sqrt{c_1 c_2}$ ($c_{1,2}>0$), then $\hat V=0$ at the minimum.
The potential is similar to that in  Higgs portal inflation \cite{lebedev,higgsdilaton},
but in our case  the angular field $\theta$ is the dynamical field responsible for a slow-roll inflation
 (instead of being frozen). For the case of  global Weyl invariant models inflation was
already studied in \cite{FRH2,GG1}.

In the following we consider the case   $\xi_1=\xi_2$,
since then $\theta$ and $\omega_\mu^\prime$ are
decoupled from each other for all values of $\theta$ (c.f.(\ref{Lfinal})). Then  $b=1$,  and  the kinetic
term acquires a canonical form  $(1/2)\,(\partial_\mu\chi)^2$,
 with  the  actual   inflaton  field $\chi$  defined by
\medskip
\begin{eqnarray}
\chi
=v\,\theta \,\sqrt{\frac{6}{\xi_1}}.  \label{canon}
\end{eqnarray}

\medskip\noindent
Then the inflaton potential in eq.~(\ref{inflatonpot}) becomes
\medskip
\begin{eqnarray}
{\hat V}=
V_0 \cos^4\theta\Big(\tan^4\theta+\frac{c_{12}}{c_1}\,\tan^2\theta+\frac{c_2}{c_1} \Big),
\quad\text{with}\quad
V_0=\frac{36\, v^4}{\xi^2_1}\,c_1\,(1+\xi_1)^2 \,
\end{eqnarray}

\medskip\noindent
 with $\theta\equiv\sqrt{\xi_1/6}\,\chi/v $.  The potential  is illustrated
in Figure~\ref{pot} for some choices of the quartic couplings, with  $\hat V=0$ at the minimum.
Inflation takes place at $\tan\theta\gg 1$  (or $\theta\sim\frac{\pi}{2}$)
for which $V=V_0\, [1+ (c_{12}/c_1-2)\cot^2\theta +(3 -2 \,c_{12}/c_1 +c_2/c_1)\cot^4\theta+\cdots]$.
 For $\xi_1\neq \xi_2$ with $\tan\theta\gg 1$
 we note that there is an approximate relation between $\theta$
 and the canonical inflaton field $\chi$  similar to that in eq.~(\ref{canon}).
  \begin{figure}[t!]
   \begin{center}
     \includegraphics[height=0.40\textwidth]{a-plot1-png.eps} \quad
     \includegraphics[height=0.40\textwidth]{b-plot1-png.eps}
       \end{center}
  \caption{\small Inflaton potential (divided by constant $V_0$) as a function of $\tau=\tan\theta$,
  for $\xi_1=\xi_2$ and  ratios  $c_{12}/c_1$, $c_2/c_1$ of values shown above.
  $V=0$ at the minimum;
note $\theta=(\chi/v)\sqrt{\xi_1/6}$ with $\chi$ the actual inflaton. The
spectral index and tensor-to-scalar ratio for these  cases are shown in Figure~\ref{plot}.}
  \label{pot}
\end{figure}

The slow-roll parameters (for $\xi_1=\xi_2$) are given by
\medskip
\begin{eqnarray}
\epsilon &=& \frac{1}{3}\,\xi_1 \,\frac{\Big(c_2-c_1+(c_1-c_{12}+c_2)\cos(2\theta)\Big)^2\sec^4\theta\tan^2\theta}{(c_2+c_{12}\tan^2\theta+c_1\tan^4\theta)^2}, \label{eps} \\
\eta&=& -\frac{1}{3}\, \xi_1\, \frac{\Big((c_2-c_1)\cos(2\theta)+(c_1-c_{12}+c_2)\cos(4\theta)\Big)\sec^4\theta}{c_2+c_{12}\tan^2\theta+c_1\tan^4\theta}.  \label{eta}
\end{eqnarray}

\medskip\noindent
We find that $\epsilon$ and $\eta$  can be small simultaneously for $\xi_1\ll 1$.
 \begin{figure}[t!]
  \begin{center}
    \includegraphics[height=0.40\textwidth]{a-plot2-png.eps} \quad
     \includegraphics[height=0.40\textwidth]{a-plot3-png.eps} \\
    \includegraphics[height=0.40\textwidth]{b-plot2-png.eps} \quad
     \includegraphics[height=0.40\textwidth]{b-plot3-png.eps}
  \end{center}
  \caption{\small{
Left: The spectral index $n_s$ as a function of the number of e-foldings $N$ for two different
cases.
Right:  The spectral index $n_s$ versus tensor-to-scalar ratio $r$ for the corresponding cases.
The green region in
all the plots corresponds to Planck2018 values for $n_s$ within $1\sigma$ (see text).
The cyan region with $r>0.07$
is excluded by Planck at 95\% CL. We chose $\xi_1=\xi_2$ for all the plots and
$c_2/c_1=1$ and $c_{12}/c_1=(-2)$ (upper panels) and $c_2/c_1=0.25$ and $c_{12}/c_1=(-1)$
(lower panels). Here $c_{12}$ is fixed by the condition ${\hat V}=0$ at the minimum.
 }}
  \label{plot}
\end{figure}

If we choose $c_2=c_1$, $c_{12}=-2 c_1$ then the expressions of the
slow-roll  parameters simplify further:
\begin{eqnarray}
\epsilon &=& \frac{4}{3} \, \xi_1 \tan^2(2\theta),  \label{eps1} \\
\eta&=& \frac{4}{3} \, \xi_1\Big(-1+\tan^2(2\theta) \Big),
\end{eqnarray}

\medskip\noindent
which leads to the scalar spectral index and the tensor-to-scalar ratio as
\medskip
\begin{eqnarray}
n_s&=&1 +2\,\eta_*
-6\epsilon_*= 1-\frac{8}{3}\xi_1 \Big(1+2\tan^2(2\theta_*) \Big),
\end{eqnarray}
and
\begin{eqnarray}
r&=& 16\,\epsilon_* =\frac{64}{3} \,\xi_1 \tan^2(2\theta_*).
\end{eqnarray}

\medskip\noindent
Further, the  number of e-foldings  during inflation is also given by
\begin{eqnarray}
N&=&  \, v^{-1}\int^{\chi_*}_{\chi_{\rm end}} \frac{{\rm sign}({\hat V}') d\chi}{\sqrt{2\epsilon(\chi)}} \nonumber \\
&=& \frac{3}{4\xi_1}\, \log\bigg|\frac{\sin(2\theta_{\rm end})}{\sin(2\theta_*)}\bigg|
\end{eqnarray}

\medskip\noindent
where $\theta_*$ is evaluated at the horizon exit and $\theta_{\rm end}$ is the inflaton
value at the end of inflation. Inflation ends at $\epsilon=1$, i.e. $|\tan(2\theta_{\rm end})|=(\frac{3}{4\xi_1})^{1/2}$ from eq.~(\ref{eps1}).

The normalization of the CMB anisotropies,
$V_0/(24\pi^2 v^4\,\epsilon_*)=2.1\times 10^{-9}$  \cite{planck2018}, constrains $c_1$
(or equivalently the quartic coupling $\lambda_1$) and the non-minimal
 couplings $\xi_{1,2}$ to satisfy
\begin{eqnarray}
\frac{c_1(1+\xi_1)^2}{\xi_1^3}= 1.8\times 10^{-8} \tan^2(2\theta_*).
\end{eqnarray}
This constraint is respected  by choosing small values  of $c_1$ (or $\lambda_1$),
for given $\xi_1$.

We have that $n_s=0.9670\pm 0.0039$ ($68\%$ CL) and $r < 0.07$ ($95\%$ C.L.)
 from Planck 2018 (TT, TE, EE + low E + lensing + BK14 + BAO) \cite{planck2018}.
In Figure~\ref{plot}, we illustrated the relation between the spectral index versus
 the number of e-foldings in the left plots and showed the spectral index versus the
tensor-to-scalar ratio in the right plots. Here we have fixed $\xi_1=\xi_2=0.007$,
 $c_2=c_1$ and $c_{12}=-2c_1$ in the upper panel, and $\xi_1=\xi_2=0.01$, $c_2=0.25c_1$,
 $c_{12}=-c_1$ in the lower panel. As a result, we find  that our model of inflation
is consistent
with the observed spectral index and the bound on $r$, for small non-minimal couplings.

\section{Conclusions}\label{5}

In this work we discussed the Weyl conformal symmetry and its spontaneous breaking
and some implications for model  building beyond the SM and inflation.

In models with conformal symmetry (of the Brans-Dicke-Jordan type)
with scalar fields with non-minimal couplings to the Ricci scalar,  one can generate
spontaneously the Planck scale from the vev of a scalar field (or a combination of them).
However, a positive (negative) Newton constant is accompanied by a negative (positive) kinetic
term  for this field, respectively.
This situation  is naturally avoided in models
with an  additional Weyl gauge symmetry and a gauge field $\omega_\mu$ which is of geometric origin,
 with a gauge transformation  dictated by  the conformal  transformation of the metric.

We showed that the Weyl field $\omega_\mu$ couples only to the scalar sector but not to the fermionic
sector of a SM-like Lagrangian in curved space-time, which  is interesting for model building.
Further, the field $\omega_\mu$ undergoes a Stueckelberg mechanism and
becomes massive after ``eating'' the radial mode $\rho\sim \sqrt K$  (in field space)
and would-be-Goldstone mode (dilaton).
The Weyl gauge symmetry is then spontaneously broken (and there are no negative kinetic
terms  in the theory). Further, the  vev $\langle\rho\rangle$
determines the mass of  $\omega_\mu$ and the Planck scale $M_p$ (up to possible
additional angular-variables field dependence). The mass of $\omega_\mu$ can be larger or
smaller than $M_p$ depending on  the scalar fields charge and non-minimal couplings.
 After decoupling  of $\omega_\mu$ the potential  depends on the angular variables fields only
which can play the role of the neutral Higgs field, inflaton, etc.
For two scalar fields of equal non-minimal couplings,
the field $\omega_\mu$   decouples  from the action even if it is light.

For the case with two scalar fields, the scalar potential generally depends only on the angular variable
 field $\theta$, and it is nearly constant at large $\tan\theta$, when $\omega_\mu$ also decouples.
 Therefore, the potential can be relevant  for a single-field inflation.
Investigating the details of the inflaton potential, we found that successful inflation
is possible, with values of $n_s$ and $r$ consistent with Planck2018 constraints,
  for perturbative values of the couplings.

While this  study was  formulated in (pseudo)Riemannian  geometry extended with a
real Weyl field (undergoing a gauge transformation dictated by conformal transformation
of the metric),
the natural framework is that of Weyl conformal geometry where this symmetry is manifest.
In the Riemannian case imposing this symmetry avoids the ghost kinetic term of conformal theory
and leads to a Lagrangian with a current $\partial_\mu K$ that interacts with the Weyl field.
This  Lagrangian was shown to be  identical, up to a total derivative term, to that obtained in
Weyl geometry (WG) where the Weyl symmetric Lagrangian is naturally
built-in,  with  curvature scalar, tensors  and affine connection of Weyl geometry.
This equivalence is showed for a SM-like Lagrangian endowed with Weyl gauge symmetry,
using the relation between $R$ computed in Riemann geometry
with Levi-Civita connection and its  counterpart in Weyl geometry.
This Lagrangian can be used for examining the phenomenological constraints on the
SM extended with Weyl gauge symmetry.

\vspace{1cm}
\noindent
{\bf Acknowledgements:\,\,\,\,}
The authors are grateful for hospitality and  support to CERN Theory Division
where this work was done.
The work of H.M.L. is supported in part by Basic Science Research Program through the National
Research Foundation of Korea (NRF) funded by the Ministry of Education,
Science and Technology (NRF-2016R1A2B4008759 and NRF-2018R1A4A1025334).
D.M.G. is supported by National Programme PN 18090101.

{\small

redpajama_set_name: RedPajamaArXiv


Education's purpose is to equip students with the skill sets they require to deal with future challenges and opportunities that we can't yet even imagine. The forward-learning education is so much important for the budding generation to help them find their own paths to their own dreams. It is to create lifelong learners who are prepared for both the contemporary world and the future. Also to empower them who will make a positive impact in the world.
The world has changed at a rapid pace fuelled by the advent of the internet and related technology. This radical transformation is forcing people to learn new theories, business models, and applications that simply did not exist even a decade ago. Recognizing such real-world challenges, students are being prepared is dynamic with exponential changes especially in technology. Think about how much technology has changed and progressed over the years. To adequately prepare students for an untold future, schools today must focus on programs that place the greatest value on creativity and the ability to engage in independent, critical thinking. It will be the independent creative thinking students who will continue to push us forward with their skills, defining the future.
Among various other top international schools in India, Oakridge International School is one that earned accreditation from Fortune's Future 50 Schools Shaping Success for progressive schooling and sharing best learning practices. "Future 50 Schools Shaping Success" is a one-of-its-kind recognition, by Univariety as Education Partner and Fortune India as Media Partner that recognizes schools that prepare students for the 21st century and demonstrate a strong vision towards crafting student success. This certification recognizes and honour schools that are going beyond the regular. The evaluation process demanded schools to submit their claims through digital forms covering activities and efforts done in ensuring student success.
On the basis of evaluation parameters around the efforts put in by the school and the results garnered by the students that include student exposure, university connect & admissions, exceptional activities, school counseling activities, teacher capabilities & training, alumni relationship building, academic results, scholarships, test prep activities, extracurricular activities and past student success, the final list of top schools in India to be certified as 'Future 50 Schools Shaping Success' was confirmed by the juries.
Oakridge International School firmly believes in the impact of discovering the inner strengths of each child and then working with them to help them discover success in that area. Success for the school is equipping the students with 21st-century skills such that their global outlook is intertwined with local connections. The school gives almost equal emphasis to activities as it does to the academics and focus on skills required not only to do well in tertiary education but also in life after leaving academia. Well-known universities around the world visit the school and offer the students good placements. Apart from providing top-class educational services with world-class infrastructure that features state-of-art Indoor and Outdoor Amenities, exceptional student case studies implemented by the school to go that extra mile. Oxford Summer Program, Student Global Leadership Program, Columbia Business School Program, TED Talks, MUN (Model United Nations) are a few to name.
At Oakridge, students get opportunities to take part in a corporate internship where they experience the look and feel of corporate work culture. Robotics is another platform that belongs to the present as well as more of future, where Oakridgers are setting benchmarks in inventing incredible models of future robots. In a nutshell, the school has left no stone unturned to ensure that its students are future-ready to succeed in every field no matter what obstacles they encounter.


Rose Cottage is the perfect place for a relaxing drink, or meal away from the hustle and bustle of day to day life..
We pride ourselves with serving fresh food sourced from local suppliers, outstanding service in beautiful surroundings, with a classic coffee menu and a fantastic range of draught lagers, bottled ciders, ales and a fine selection of wines.
We offer an extensive gin menu, as we hold over 70 Gins Ranging from traditional Dry Gins to a wide range of flavoured Gins.
We are open Monday to Saturday 12pm – 9pm for food & Sunday 12pm – 8pm.
We cater for all occasions large or small & tailor them to your needs.


text:
Girl's inoperable brain tumor disappears, parents claim 'God healed Roxli'
Roxli Doss' tumor was characterized as extremely rare and inoperable, but then the MRI scans showed it disappeared completely and doctors can't explain how.
When 11-year-old Roxli Doss from Hays County, Texas, was diagnosed with diffuse intrinsic pontine glioma (DIPG), a rare and inoperable tumor, her family was left heartbroken.
She experienced headaches, nausea, and double vision for some time and her parents brought her to the hospital for a check-up. But, little did they know the news would be devastating.
They were informed how Roxli's tumor was very aggressive and difficult to treat which is why they claim God had his share in making the cancer disappear completely after she received her treatments.
Speaking of the time they learned of Roxli's condition, her father, Scott Doss, said: "It was the worst news we could have had."
"At Dell Children's, Texas Children's, at Dana-Farber, at John Hopkins, and MD Anderson, all agreed it was DIPG."
All of the doctors who saw Roxli's scans agreed how this tumor was very rare.
"It is a devastating disease. You have decreased ability to swallow, sometimes vision loss, decreased ability to talk, eventually difficulty with breathing," said Dr. Virginia Harrod of Dell Children's Hospital. "We don't have a cure. This is considered a terminal diagnosis."
Soon after the diagnosis, Roxli underwent 11 weeks of radiation and during the whole time, her mommy and daddy prayed to God and asked from him to save their girl.
Finally, their prayers were answered!
"The doctors guide us, but really the true guidance that we're looking for is from God because that's who we need our help from right now, and we're really just praying for a miracle," said Scott.
The latest MRI scans showed that Roxli was cancer-free. The doctors who believed the tumor could not be operated on and said Roxli's chances of surviving were bleak were now left in complete shock.
"When I first saw Roxli's MRI scan, it was actually unbelievable. The tumor is undetectable on the MRI scan, which is really unusual," Dr. Harrod said.
But Roxli's parents were not really surprised from the positive outcome because their trust in God is endless and they were convinced their faith would work its magic.
"A Miracle!! We all walked out in shock even though we had been begging, praying, asking and believing. Thank you Jesus just wasn't enough to say," they wrote on CaringBridge.org:
"God is using precious Roxli as a vessel for His mighty power! REVIVAL is springing up in hearts everywhere!" Gena Doss wrote on Facebook.
The dedicated medical professionals, along with Doss' strong faith in God, made the unbelievable happen.
Although the scans show Roxli is cancer-free, she's currently undergoing precautionary immune treatments to ensure every single cancer cell is eradicated from her brain.
Dr. Harrod said: "We are cautiously optimistic and will continue to monitor her on a regular basis along with the joy of celebrating in her recovery."
This story is so inspiring because it gives a ray of hope to other people and children fighting cancer and reminds them to never stop believing in miracles, because if you trust with all your hearts, they do happen.
11-yr-old girl
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Jon Bon Jovi is helping to feed the needy with his chain of community kitchens

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text:
You are here: Home / NewsCenter / Feds Unveil Plan to Grow Wind Power While Sparing Rare Whale
Feds Unveil Plan to Grow Wind Power While Sparing Rare Whale
Aerial survey of North Atlantic Right Whales courtesy of New England Aquarium, taken under NEFSC permit #25739
PORTLAND, Maine (AP) — The federal government has outlined a strategy to try to protect an endangered species of whale while also developing offshore wind power off the East Coast.
President Joe Biden's administration has made a priority of encouraging offshore wind along the Atlantic coast as the U.S. pursues greater energy independence.
Those waters are also home to the declining North Atlantic right whale, which numbers about 340 in the world.
The National Oceanic and Atmospheric Administration and Bureau of Ocean Energy Management released a draft plan this month to conserve the whales while also building wind projects.
Local experts recently reported they are still hopeful the population could rebound despite new challenges the whales face.
BY PATRICK WHITTLE, THE ASSOCIATED PRESS
Filed Under: Local News, NewsCenter, Top Story Tagged With: Associated Press, noaa, North Atlantic Right Whale
About Brian Engles
Brian Engles is a longtime local of the Cape. He studied Film & TV at Boston University and in addition to his role at Cape Cod Broadcasting Media, he also works as a music instructor and records original songs.


text:
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GSK in Hot Water
Kevin Gopal
Kevin Gopal is Pharmaceutical Executive's international correspondent, covering pharma and regulatory issues around the word. He is also a political columnist for North West Business Insider, one of the UK's leading regional business magazines. He started his career as a journalist at SiYu, the UK's Chinese community magazine, before joining the PE staff.
Pharmaceutical Executive, Pharmaceutical Executive-03-01-2002,
Geneva-The International Federation of Pharmaceutical Manufacturers Associations (IFPMA) has upheld a complaint against GlaxoSmithKline that one of its executives made inappropriate claims about Paxil (paroxetine). Charles Medawar of Social Audit-an activist charity that examines inappropriate use of medicines-filed the allegation in August 2000.
Medawar claims that a GSK executive's comments in the program, "A Painful Withdrawal," screened in the United States, contravenes sections of the IFPMA code of pharmaceutical marketing practices. The executive said, "What we have seen in terms of the anecdotal reports [of withdrawal from Paxil] is that it happens very rarely." A voiceover added that withdrawal or discontinuation syndrome "occurs in only two out of every 1,000 patients who are discontinued appropriately. Even then, the symptoms are mild and short-lived."
Medawar says scientific evidence does not support those claims, which were repeated in promotional material. So far, the company has made no response to his requests for information about the statements.
Because GSK is headquartered in the United Kingdom, the complaint was referred to the Prescription Medicines Code of Practice Authority (PMCPA), which initially found in favor of the company. But on appeal, IFPMA overturned the decision and ruled that GSK had contravened sections 1.3 and 1.7 of the code, covering scientific evidence and communication to the public, respectively.
The appeal board says, although GSK did not positively promote Paxil in the program, the employee's defense of the product was itself a form of promotion and therefore section 1.3 of the code applied. Notes to section 1.3 state: "Companies should deal objectively with requests for information made in good faith and should provide data appropriate to the source of the enquiry."
It further noted the difference between the executive's claims that withdrawal symptoms happened "very rarely" and US product information describing them as "rare." Thus the board concluded that there had been a further breach of section 1.3, as well as a breach of section 1.7, because the claims were made directly to the public. A public reprimand is the only consequence.
Last December, the company changed the medicine's US labeling when it was approved for post-traumatic stress disorder. The new labeling states that there have been reports of adverse events, "which may have no causal relationship to the drug," including dizziness, agitation, and anxiety. The company now recommends tapering off the dose of the medicine rather than abrupt withdrawal.
Meanwhile, a powerful UK parliamentary committee is considering summoning the company to explain comments attributed to its R&D head, Tachi Yamada, who reportedly said GSK might sell parts of its research operation to boost productivity. At the time of the merger between Glaxo Wellcome and SmithKlineBeecham, Sir Richard Sykes, then GW executive chairman, claimed that scale was a critical factor in the merger.
Dr. Ian Gibson, Labour chairman of the Commons Science and Technology Select Committee, says the company misled Parliament members about the merger's rationale and announced that he would summon senior executives before the committee. He says Yamada's comments open up questions about what the merger promised and what it really delivered and adds, "I always smelt a rat about what they were really up to."


text:
Q: Obtain standard error of mean as computed in seaborn.catplot I am using seaborn.catplot with kind='point' to plot my data. I would like to calculate the standard error of the mean (SEM) for each hue var and each category using the same method as seaborn in order to make sure that my computed values exactly match the plotted error bars. The default solution for calculating the SEM and the 95%-confidence intervals (CIs) contains a bootstrapping algorithm, where the mean is bootstrapped 1000 times in order to calculate the SEM/CIs. In an earlier post, I saw a method that might offer functions for that (using seaborn source code functions like seaborn.utils.ci() and seaborn.algorithms.bootstrap()) but I am not sure how to implement it. Since the bootstrapping uses random sampling it would also be necessary to make sure that the same array of 1000 means is produced both for plotting and for obtaining the SEM.
Here is a code example:
import numpy as np
import pandas as pd
import seaborn as sns

# simulate data
rng = np.random.RandomState(42)
measure_names = np.tile(np.repeat(['Train BAC','Test BAC'],10),2)
model_numbers = np.repeat([0,1],20)
measure_values = np.concatenate((rng.uniform(low=0.6,high=1,size=20),
                                rng.uniform(low=0.5,high=0.8,size=20)
                                ))
folds=np.tile([1,2,3,4,5,6,7,8,9,10],4)

plot_df = pd.DataFrame({'model_number':model_numbers,
                        'measure_name':measure_names,
                        'measure_value':measure_values,
                        'outer_fold':folds})

# plot data as pointplot
g = sns.catplot(x='model_number',
                y='measure_value',
                hue='measure_name',
                kind='point',
                seed=rng,
                data=plot_df)

which produces:

I would like to obtain the SEM for all train and test scores for both models. That is:
# obtain SEM for each score in each model using the same method as in sns.catplot
model_0_train_bac = plot_df.loc[((plot_df['model_number'] == 0) & (plot_df['measure_name'] == 'Train BAC')),'measure_value']
model_0_test_bac = plot_df.loc[((plot_df['model_number'] == 0) & (plot_df['measure_name'] == 'Test BAC')),'measure_value']
model_1_train_bac = plot_df.loc[((plot_df['model_number'] == 1) & (plot_df['measure_name'] == 'Train BAC')),'measure_value']
model_1_test_bac = plot_df.loc[((plot_df['model_number'] == 1) & (plot_df['measure_name'] == 'Test BAC')),'measure_value']

A: I'm not sure I get the requirement that you have the exact same samples taken. By definition, bootstrapping works by taking a random sample and therefore there will be a bit of variability from one run to the next (unless I'm mistaken).
You could calculate the CI the same way seaborn does like so:
# simulate data
rng = np.random.RandomState(42)
measure_names = np.tile(np.repeat(['Train BAC','Test BAC'],10),2)
model_numbers = np.repeat([0,1],20)
measure_values = np.concatenate((rng.uniform(low=0.6,high=1,size=20),
                                rng.uniform(low=0.5,high=0.8,size=20)
                                ))
folds=np.tile([1,2,3,4,5,6,7,8,9,10],4)

plot_df = pd.DataFrame({'model_number':model_numbers,
                        'measure_name':measure_names,
                        'measure_value':measure_values,
                        'outer_fold':folds})

x_col = 'model_number'
y_col = 'measure_value'
hue_col = 'measure_name'
ci = 95
est = np.mean
n_boot = 1000

for gr,temp_df in plot_df.groupby([hue_col,x_col]):
    print(gr,est(temp_df[y_col]), sns.utils.ci(sns.algorithms.bootstrap(temp_df[y_col], func=est,
                                          n_boot=n_boot,
                                          units=None,
                                          seed=rng)))

which ouputs:
('Test BAC', 0) 0.7581071363371585 [0.69217109 0.8316217 ]
('Test BAC', 1) 0.6527812067134964 [0.59523784 0.71539669]
('Train BAC', 0) 0.8080546943810699 [0.73214414 0.88102816]
('Train BAC', 1) 0.6201161718490218 [0.57978654 0.66241543]

Notice that If you run the loop a second time, you'd get CIs that are similar, but not exactly the same.
If you really want to get the exact values that were used in the plot by seaborn (note that, again, those values will differ slightly if you plot the same data a second time), then you could extract the values directly from the Line2D artists used to draw the error-bars:
g = sns.catplot(x=x_col,
                y=y_col,
                hue=hue_col,
                kind='point',
                ci=ci,
                estimator=est,
                n_boot=n_boot,
                seed=rng,
                data=plot_df)
for l in g.ax.lines:
    print(l.get_data())

output:
(array([0., 1.]), array([0.80805469, 0.62011617]))
(array([0., 0.]), array([0.73203808, 0.88129836])) # <<<<
(array([1., 1.]), array([0.57828366, 0.66300033])) # <<<<
(array([0., 1.]), array([0.75810714, 0.65278121]))
(array([0., 0.]), array([0.69124145, 0.83297914])) # <<<<
(array([1., 1.]), array([0.59113739, 0.71572469])) # <<<<

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Life can seem meaningless and a burden at times. One may feel there is no reason to live. The life ahead scares us. Many think to end their life at such difficult times as if death is the only solution to get rid of all the problems.
Our collective effort can stop the tendency of committing suicide.
International Association for Suicide Prevention observes World Suicide Prevention Day on 10 September every year.
Human beings are capable to live in adverse situations, no matter how hard it is. We even cope up with the loss of children and live. Therefore, suicide cannot be a normal reaction in any circumstances.
Study says those who commited suicide were going through psychological troubles.
About 50 per cent among them have flawed personality, 6 per cent have depressions, 7 per cent were drug addicts and the remaining others have complicated mental disorders including schizophrenia. Less skilled and socially isolated individuals are more prone to suicide than others.
About 800,000 people commit suicide every year around the globe, according to World Health Organisation. Most of them are above 70 years while the people who commit suicide in several countries including Bangladesh are mostly young.
A recent study says 39.9 people commit suicide among every 100,000 people every year.
The suicide rate is higher among males worldwide, but the rate is higher among women in Bangladesh, expecially among young girls.
The study also shows that lack of education, poverty and unhappy marital life are related to suicide while another study shows complications in relationships, financial crisis, disease and family disputes are also reasons for commiting suicide.
There is no specific reason that prompts people to commit suicide, rather it can be said that, the instincts of an individual, mental condition, socio-cultural-economic atmosphere and many other vulnerable situation are responsible for suicide.
When human beings feel isolated they are more prone to commiting suicides, according to renowned socilogist Emile Durkheim.
News of suicide published in media often instigate mentally vulnerable individuals. The incidents of suicide increase shortly after the suicide of any celebrity or significant person or release of cinemas or TV plays based on suicide.
There are some misconception about suicide among people. They say 'those who threatens to commit suicide, actually seek attentions of others', 'a survivor of a suicide sttempt will never repeat it', 'talking about suicide only increase suicidal tendency among people' and 'those who commit suicide never show any signs of commiting it'.
However, most of those who commit suicide express their feeling to others before doing it. About 40 to 60 per cent of victims attempt suicide more than once and one among every six people attempt suicide for the second time within a year of the first attempt.
Discussing about suicide does not instigate the tendency rather it helps the victim. The victim finds some warmth and emotional support while discussing about it. In most of the cases, the victim shows some signs before committing suicide.
· Mental health service should be ensured for all. Also, it is possible to reduce the number of suicide by raising awareness, identifying the problems as soon as possible and providing psychological treatment.
· The possible equipment that can be used to commit suicide should be kept away. Medicines should not be provided without prescription and insecticides should be kept beyond reach of vulnerable victims.
· Everyone has to be aware of signs like threats to commit suicide, suicide attempts, depression and creating distance from people or isolation.
· Newspapers and media have to be more careful while publishing such news.
·Initiatives must be taken to build crisis centres and telephone hotline throughout the country.
A suicide is a mere death; it is also a loss to the friends and family of the victim. It is possible to stop suicide if we try by all means.


My Frickin' DVDs are FAULTY!
fnordboy wrote: Pioneer multi regions are just way too overpriced IMO.
Multregion players are pretty much the same price as single region over here. Good thing too!
arteitle wrote: I find it fascinating that despite the long-time difference of the US and Japan using NTSC while much of Europe used PAL, the abundance of NTSC material has made it so better European TVs include support for NTSC natively. Very interesting. What's next, will Europeans start putting 120 volt outlets in their homes too?
Main reason is that its a very simple thing to do. NTSC has less lines than PAL and changing the colour carrier is simple processing.
120v outlets would be good for you USers visiting us, but it would end up being less useful, although slightly safer. :-) (Thus the reason no outlets in UK bathrooms).
I'd suggest something as simple as checking to see if your SCART cable is plugged into the RGB compliant socket on yout TV.
Most TVs with more than one SCART/EURO socket only offer one capable of playing RGB, and thus NTSC signals.
Calhoun07 wrote: It's due to copy protection on the discs. Not all DVDs are created equal, so it's possible to play a hundred DVDs and not have the same problem, but when you run your DVD player thru your VCR, you will run the risk of getting DVDs with copy protection that detect the VCR, and it sends a singal back to the DVD player and tells it that you're trying to copy something on to VHS tape, and will either scramble the singal or turn the image black and white. I knew from the instant I read your first post, that was what the problem was. It's quite common, actually. I work part time in a video store and I guarantee you every person who brings back DVDs claiming they were "defective" due to picture image problems like that (I even had customers bring them back saying the DVD played black and white), every one of them had the DVD player hooked up thru the VCR and easilly fixed their delimma by hooking it up thru the TV. But like I said, not all DVDs are encoded with the same kind of copy protection, DVD companies may use various types of copy protection to thwart bootleggers, but from time to time you will find some that don't like your VCR. Just a word to the wise if anybody has any future problems with any other DVDs.
I'm pretty sure that can't be the case here. The My So-Called Life DVDs don't seem to have any copyright protection.
The reason I noticed this is that I have an old TV that only has cable input and no component inputs (the three jacks: yellow (video), red, and white (L and R audio signal)). This means that I need an RF modulator to take the output from my DVD player and convert it to a cable output for my TV. Well, a thunderstorm had fried my RF modulator (thankfully leaving DVD player and TV intact), so I was unable to hook my player directly to the TV. As someone said in another post, Macrovision (and color striping, etc.) means putting the DVD output to the VCR (which has component input and cable output) then to the TV gives you a distorted picture.
Before I got around to getting a new RF modulator, I hooked up my DVD player through the VCR to the TV to see which discs would work. All the My So-Called Life discs worked (and for anyone interested Mad About You also worked). A friend explained that some niche products don't have copyright/copying protection because it costs the company extra to add it, so on products where limited sales are expected, they don't bother.
Long story short, you can make VCR copies of all the discs (for you personal viewing pleasure of course) without distortion.
Just a note: Some VCR's are 'smart' enough to let the dvd signal through to the TV until you hit the 'record' button. Mine is like that, and plays dvds fine, but as soon as you hit record, the picture fades in and out, and gets fuzzy.
I can't believe it but my DVD-player (Samsung DVD 611) refuses to play the first DVD. Not even the old one from before the set. I have both but neither will play. The other volumes will. I donno what can be wrong with the player. It has worked fine before. We have another player in our PC and that one plays anything.


It's goodbye sadness, hello Los Campesinos! after a barnstorming performance at the Phoenix.
In terms of the live music circuit, Exeter is often perceived as being too 'out of the way' by most tour bookers.
Thankfully this was not the case for the masterminds between Cardiff-based indie outfit Los Campesinos!, who kicked off their UK tour on the back of the November release of their fourth album, Hello Sadness.
Any preconceived doubts about how new members Jason, Rob and Kim would fare were shattered as LC! blew through set and album opener "By Your Hand".
The new set-up proved themselves to be one of those live acts who reproduce their recorded work with a kick of adrenaline shot straight through the cochlea and into the heart, hammering through third album title track "Romance is Boring" and new single "Songs About Your Girlfriend".
Frontman Gareth Campesinos! was the archetypal showman, committing himself to every melody and punching raw feeling into every lyric. In between songs, he was wry, asking the obligatory gig question: "Any couples in?", before following up with "…odd decision to bring them to a Los Campesinos! concert…" When a heckler revealed she'd met her partner at the last one, Gareth quipped "Seems like an appropriate book-end…"
Much of the appeal of LC! is their ability to switch between euphoria and desolation. Nowhere was this more evident than in their performance of a lachrymose "The Sea is a Good Place to Think About the Future", the audience shouting every word back at the band, mirroring the angst of the frontman.
After the song's climax, Gareth commented on how different songs resonate in different locations, with particular reference to the lyric "But you could never kiss a Tory boy without wanting to cut off your tongue again", one which was bellowed by the crowd with utter ferocity: "If there's anything anyone should be persecuted for, it's voting Tory – the next time you're in a club putting your hands down someone's pants, make sure you check…because those cunts don't deserve it."
The band attempted to sneak off after a furious rendition of "Baby I've Got the Death Rattle", but the crowd were having none of it, beckoning them back for an encore of "In Medias Res" and their debut album finale "Sweet Dreams, Sweet Cheeks".
Limbs flailed, lager flew, and hearts burned: and then it was over. LC! were down to earth enough to meet the fans after at the merchandise stand, and whilst my friend was utilising Gareth's inhaler following an asthma attack (thanks again for that), I was lucky enough to grab a brief word.
MMc: Gareth, I'll be honest: I'm a huge fan. No-one else is currently writing lyrics that are as resonant as yours.
GC!: …I really don't respond well to compliments.
MMc: OK…maybe one day you'll be as good a writer as I am…is that better?
MMc: A lot of your songs use football as an allegory. I've got to ask, which team do you support?
GC!: I'm a fan of Welton Rovers (a Somerset side currently mid-table in the Toolstation Western Football League Division 1) as that's my local side. But I'm also of that generation where I can follow Manchester United.
MMc: I'm young enough to hate Chelsea.
GC!: Now I feel old, ha!
All in all then, a rip-roaring return to the South West for Los Campesinos!. It's a crying shame that this was the first night of a UK tour and not a residency at the Phoenix.


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HomeCD ReviewsAztek - 'Dream Dealer'
Aztek - 'Dream Dealer' http://rocktopia.co.uk/media/reviews/photos/thumbnail/200x200s/b0/b1/13/aztek-dream-dealer-63-1522354443.jpg Hot
Added by Central Electronic Brain March 29, 2018
Dream Dealer
Mighty Music / Target
An "okay" album that won't set the world on fire, but there is room for them to grow. Maybe next time.
I quote, "Aztek are a Progressive Rock band from Aalborg, Denmark. They play music influenced by Opeth, Jeff Buckley and Led Zeppelin, but their music is actually rooted in the Classic Rock sound". Another highly accurate bio write-up?! Progressive? No. Classic Rock? Maybe. Seventies retro gloomy Doom Rock? Deffo!!
Now I have no problem with gloomy, Doomy music and when done well it can be very good indeed. My problem with this album is that, although it does have some strong moments, it also has moments where your attention wanders and you find yourself going off and doing something else, forgetting that you were supposed to be listening to an album. Not good. Also, comparing a new act to giants like Zeppelin and Opeth doesn't help.
The moments to look out for, if you're looking to give this a go, are the enigmatically entitled 'Nick Cave' where Aztek singer Benjamin Vestergaard tries to do his best Cave impression. The track itself is a dark and meandering piece with weird sounds and strange arrangements that does bring to mind Cave's own 'Where The Wild Roses Grow' (probably why they did it). 'Stay With The Word' wouldn't sound out of place on a Quentin Tarantino gangster movie soundtrack, 'Preaching' has hauntingly dramatic riffs and odd vocals and 'Cold Wars And Hot Bombs' has a jarring guitar sound (in a good way) that helps the song to weave and stroll wistfully along from start to finish.
I'm usually a sucker for this retro stuff and for the life of me I can't figure out why the music of Aztek, at times, leaves me cold. They can play, that much is obvious. They can also write great songs. Maybe then, it's the consistency that I find lacking. Whatever it is, I'm sorry to say Aztek have made an "okay" album that won't set the world on fire, but there is room for them to grow. Maybe next time.
Ian Johnson


Microsoft has a nice business collecting license revenue from Android OEMs. That's not Apple's motivation. Apple "started" the patent war – and yes, may have escalated it, in some cases unreasonably – because Android was such a blatant rip-off of the iPhone.
Maybe it's the use of "blame" in a sentence structure that seems to exonerate Google – I don't know. Other than that incredibly shallow gripe, I heartily recommend it.


Richard Laughlain well known as "Lochlyn" Munro is a actor from Canada.He gained fame after appearing in movies as White Chicks, scary.movie and others. Currently he is playing in supernatural drama Charmed.
He took lessons of drama and improvisational comedy. He left ice hokey because of serious sports injury and pursued his career in acting.
He has a net worth of $400 thousand and it is growing significantly. Even at the age of 49 he has a well toned.physique. He played music in various clubs in his teenage.
He keep his personal life private. As is much more dedicated in his acting career, he kept his relationship status low. He is not gay. He might be searching a perfect match for himself.
Edward John "Eddie" Izzard (born February 7, 1962) is an English stand-up comedian, actor, and writer. He is best known for his role in The Riches, Ocean's Twelve, Shadow of the Vampire, The Cat's Meow, Across the Universe, and Valkyrie.
Karim Zeroual is an actor, known for The Sparticle Mystery (2011), Da Vinci's Demons (2013) and EastEnders (1985) but mostly notable for his role as Sadiq in the CBBC fantasy television series The Sparticle Mystery.
Scott Elrod is an actor. He is famous for his appearance in the movie Lone Survivor where he starred the role of Peter Musselman.
David Hart is an actor who is also gifted with a loving voice. He is known for his role in the TV series In the Heat of the Night.


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People who drink heavily may increase their risk of dying in house fires that should otherwise have been escapable, a new study suggests.
The findings, reported in the September issue of the Journal of Studies on Alcohol and Drugs, highlight one of the less-recognized dangers of downing too much alcohol - especially in combination with smoking.
Looking at coroners' records for 95 fire victims, Australian researchers found that 58% had positive results on blood alcohol tests, often with very high alcohol levels. And those intoxicated victims were less likely than sober ones to have had obstacles preventing their escape from the fire - like barred windows or an exit route that was blocked by the fire.
The implication, the researchers say, is that at least some intoxicated victims might have survived had they been roused in time.
Most of the victims in the study were alone at the time of the fire, and many were asleep (including close to half of intoxicated victims). So it's possible that well-placed smoke detectors - or having other, sober people in the house - would have protected some, according to lead researcher Dorothy Bruck, Ph.D., of Victoria University in Melbourne.
But probably the biggest issue is the combination of heavy drinking and smoking. In this study, victims who had been drinking were about 4.5 times more likely to have died in fires that involved "smoking materials," like discarded cigarettes.
"A key message is that smoking and drinking together constitute a high-risk activity, even in your own home," Bruck says.
Aside from not combining drinking and smoking, people may be able to lessen their fire risks in other ways, according to Bruck.
One way would be to always have someone else in your home who stays sober, even if you're having a few drinks. Also, it's a good idea to have smoke detectors in bedrooms or living areas, rather than just hallways, Bruck says. In general, smoke alarms are less effective at rousing people who've been drinking heavily.
"Our research has also shown that certain changes to improve the sound that smoke alarms make, and better alarm locations, may arouse more sleeping intoxicated people in the case of a fire," Bruck says. But, she adds, experts don't know how loud smoke alarms, even with the improved signal, need to be to wake a person with very high blood alcohol levels.
Another potential way to prevent fire deaths is to use so-called fire-safe cigarettes, Bruck notes. The cigarettes self-extinguish if a smoker sets one down and forgets about it or falls asleep while smoking.
All of the fatalities in the current study were among adults who died in house fires in the Australian state of Victoria between 1998 and 2006. The high percentage of intoxicated victims (58%) is consistent with what's been seen in studies from the United States, Canada and Europe.
In recent U.S. studies, anywhere from 40% to 55% of people who died in fires were intoxicated, based on blood alcohol tests.
Bruck, D., Ball, M., & Thomas, I. R. (September 2011). Fire fatality and alcohol intake- Analysis of key risk factors. Journal of Studies on Alcohol and Drugs, 72(5), 731-736.


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Q: Is isolated points always boundary points? Is Discrete Topology consisting of isolated points? Is isolated points always boundary points?
What happens in the discrete topology space? If the space is R with discrete topology s.t f(x) = 0 if x is odd natural number, f(x) = 1, x is even natural number, f(x) = 2, if x = 1.5.
Is this a discrete topology? Is x= 1.5 an isolated point? Is x = 1.5 a boundary point?
Thanks very much for help! I am very confused here..

A: Yes, isolated points are always boundary points. The reason is as follow:
Isolated points are defined as $A^i=A-A'$, where $A'$ is the limit points of $A$. So isolated points are those points that are in $A$ but not in $A'$.
Boundary points are defined as $\partial A=\overline{A}\cap \overline{A^c}=\overline{A}-A^o$, where $A^o$ is the interior of $A$. So boundary points are those points that are in $\overline{A}$ but not in $A^o$.
Now if $x\in A^i$, then $x\notin A^o$. If not so, then there is always open set $G\subset A$ such that $x\in G$. This means $x\in A'$, a contradiction.
In your example, $1, 1.5, 2$ are isolated points and thus boundary points.


Scrapbook - Announcing... | WTF!?
Congrats Jane on your beautiful baby girl. Also congrats to your husband.
3 am feedings are the best!
All the hard work pays off in the end and now a lifetime of memories begins. Congratulation's to all three of you for the effort involved and the effort to come. Parenthood is the toughest job you'll ever love.
Felicidades guapa. Here's to you and the new soul you brought into this earth!
Omg I love that face!
Gracias guapo! Ser Madre es algo que no puedo explicar!
Yeah cute face… the cat never made it though.
Congratulation's Mrs. Deere. For the next 18 years (at least) you don't get to have a life of your own! Woo hoo!
Seriously though, she's adorable. You have spawned well.
Omfg I didn't even notice the cat somehow!!!!!! Wtf, Jane?!
Congratulations for creating an angel.
As for a name, how about ... Zohan.
Kinda? Hey, wait a minute. I was spot on.
Zoë + Angelyn = You Don't Mess With The Zoë Angelyn!
Thanks for letting me name your baby and congratulations on your little one!
Bahahaha! Thank you! I'll have to find the right moment to say that to my husband.
She is so pretty Jane .....awwww .


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3 soldiers killed in Afghan attack were from Ohio
US & World // U.S. & World
Associated Press April 5, 2012
COLUMBUS, Ohio - Three U.S. soldiers killed in a suicide attack this week in Afghanistan were from Ohio, as were several others seriously wounded in the bombing, the Ohio National Guard said Thursday.
The soldiers killed in the attack Wednesday in Maimanah, the capital of Faryab province, were from the Guard's 37th Infantry Brigade Combat Team, according to the Ohio Adjutant General's Department.
The brigade is based in Columbus but includes soldiers from across Ohio, the Guard said.
Among the victims was Sgt. 1st Class Shawn Hannon, 44, who also worked as chief legal counsel for the Ohio Department of Veterans Services.
"Shawn felt it was a privilege to serve his country and did so honorably for almost 20 years. He was proud to be a soldier and all who loved him knew it," his family said in a statement released to media. Survivors include wife and their 9-month-old son.
Hannon joined the state veterans agency last year after working for a Columbus law firm.
"He was one of the most well respected guys I ever met," said Steve Palmer, a lawyer who worked with Hannon.
Police sergeant remains hospitalized after clinic shooting
Deadly California house explosion damaged neighboring homes


In the morning we pick you up at your hotel and start our journey to the village of Upis, the starting point of the trek, passing several traditional villages and enjoying some of the most beautiful sceneries that Peru has to offer. We arrive in Upis in the afternoon and check in our hotel where you can relax and enjoy a relaxing bath in the thermal waters near the hotel.
Today we start our hiking to the Arapa pass at 4,850m / 15,900ft. Then we continue the hike down the valley in the afternoon, passing the green lake of Puqa Q'ocha, until we reach our first camp at the beautiful turquoise Lake Jatun Puqa Q'ocha where we spend the night. At night you might hear the sound of ice breaking away from a nearby glacier and falling into the water of the lake – a really special atmosphere.
In the morning we'll ascend the second pass of Apuchata (4,900m / 16,070ft) with the turquoise Laguna Ausangate below which is fed from the melting snow of Ausangate. From the lake we'll climb to the Palomani pass, the highest point of the trek, at 5,200m / 17,060ft. We follow the trail into the next valley and arrive our campsite at Pampacancha.
We start hiking with the snow-capped peaks of the Tres Picos and Puca Punta in sight. Along the trail between the broad green valleys of Pampa Jutunpata and the Rio Q'ampa valley you can observe the delightful viscacha, a small rodent that looks like a cross between a squirrel and a rabbit. You may also be lucky enough to see groups of Andean camelids like vicuñas. The trail continues North West to the final pass of this hike, the Q'ampa pass at 5,000m / 16,400ft. We'll descend for a couple of hours and camp besides the green turquoise lake Q'oma Q'ocha..
Today we start our last trek for about 2½ hours to the small Andean village of Pacchanta. There you can rest and relax in the hot thermal springs for a couple of hours, maybe have a well-deserved beer and enjoy a nice meal in this interesting Andean village. In the afternoon we'll return to Cusco by bus.