text:
\section{Introduction}
\label{sec:intro}
It is of paramount importance to establish a symmetric subtraction strategy
for the divergences of
the \underline{massive} nonlinear sigma model. In fact this model provides a  viable
infrared regulator, a useful phenomenological theory
and finally a template to approach more complex theories, as
the nonabelian gauge theories in the 't Hooft gauge.
Moreover a  simulation on a lattice is free from the artifacts related
to the presence of the zero modes.
\par
The subtraction strategy for the ultraviolet divergences,
devised for the massless nonlinear sigma model \cite{Ferrari:2005ii},
can be successfully extended to the model with non zero mass, although  chirality is broken.
\par
A powerful  Local Functional
Equation (LFE) for the generating functionals (of the Green-
and one-particle-irreducible-functions (1-PI)) ensues from
the invariance properties of the
path integral measure. By using the LFE, one
can derive the complete hierarchy relations among the ancestor amplitudes
(i.e. with no pion fields) and the descendant amplitudes (i.e.
with at least one pion field) and get the full control of the
divergences by a finite number of divergent amplitudes
at each order in the loop expansion. The counterterms are chosen according to the
symmetry properties of the effective action (not of the action!) fixed by the LFE.
Pure pole subtraction in dimensional
regularization turns out to be the right thing to do, in order
to satisfy the LFE.
\par
The  physical parameters of the model are those of
the classical action augmented by the scale of the radiative
corrections $\Lambda$ (introduced later on).
\par
To compare with the Chiral Perturbation Theory (ChPT)
\cite{Weinberg:1978kz}-\cite{Bijnens:2009zi}, we stress
that the subtracted amplitudes depend on
a fixed number of parameters. Moreover another important
feature characterizes the present approach: $\Lambda$ is a physical parameter,
while in ChPT
the requirement of independence from
$\Lambda$ leads to a Renormalization Group Equation.
\section{The Model}
\label{sec:model}
The generating functionals $Z$ and $W$ are introduced via the path integral
\begin{eqnarray}&&
Z[\vec K,K_0,\vec J_\mu,N_{AB}] = \exp iW
\simeq \int {\cal D}[\phi]\frac{1}{2\phi_0}
\exp\biggl[i S
\nonumber\\&&
+ \Lambda^{D-4} \int d^D x \Bigl(\phi_0 K_0 +\phi_A N_{AB}\phi_B\Bigr) + \int d^D x\phi_a K_a
\biggr].
\label{model.1}
\end{eqnarray}
The classical action is
\begin{eqnarray}&&
 S=\Lambda^{D-4} \int d^D x \frac{v^2}{2}\Bigl[\frac{1}{2}~Tr~\Bigl\{ (F_{\mu}-J_{\mu})^2 \Bigr\}
 - m^2 \vec\phi^2 \Bigr]
\nonumber\\&&
= \Lambda^{D-4} \int d^D x \frac{v^2}{2}\Bigl[\frac{1}{4}\Bigl(F_{a\mu}-J_{a\mu}\Bigr)^2
 - m^2 \vec\phi^2 \Bigr]
\nonumber\\&&
= \Lambda^{D-4} \int d^D x \frac{v^2}{2}\Bigl(\partial_\mu\phi_A\partial^\mu\phi_A -
\frac{1}{2} F_{a\mu}J_{a}^\mu+ \frac{1}{4} J^2 - m^2 \vec\phi^2\Bigr).
\label{model.2}
\end{eqnarray}
where
\begin{eqnarray}&&
\phi_0= \sqrt{1-\vec\phi\,^2}
\nonumber\\&&
\Omega = \phi_A\tau_A=\phi_0 +i \phi_a\tau_a , \quad \Omega \in SU(2)
\nonumber\\&&
F_\mu = \frac{\tau_a}{2}F_{a\mu}\equiv i \Omega\partial_\mu\Omega^\dagger
\nonumber\\&&
F_{a\mu} = 2(\phi_0\partial_\mu\phi_a
-\partial_\mu\phi_0\phi_a + \epsilon_{abc}\partial_\mu\phi_b \phi_c)
\nonumber\\&&
\tau_A\equiv \{1, i \vec\tau\}.
\label{model.3}
\end{eqnarray}
Capital letter indexes run over $\{0,1,2,3\}$ while lower cases over $\{1,2,3\}$.
The fact that
the fields $\phi_0, \vec\phi$ have null canonical dimension is only a matter of choice.
$N_{AB}$ are  real independent sources introduced in order to account for the
extra composite operators generated by the local chiral transformations. By construction
we have
\begin{eqnarray}
Z[\vec K,K_0,\vec J_\mu,N_{AB}] =
Z[\vec K,K_0,\vec J_\mu,N_{BA}],
\label{model.1.1}
\end{eqnarray}
i.e. $Z$ does not depend on the antisymmetric part of the
matrix $N_{BA}$.
\section{The Local Functional Equation}
\label{sec:lfe}
The measure in the path integral (\ref{model.1}) is invariant under the
local left-transformation
\begin{eqnarray}
\Omega'=U(\vec \omega)\Omega
\label{lfe.1}
\end{eqnarray}
i.e.
\begin{eqnarray}&&
\delta \phi_0 = -  \frac{\omega_a(x)}{2}\phi_a
\nonumber\\&&
\delta \phi_a = \frac{ \omega_a(x)}{2}\phi_0 + \frac{
\omega_c(x)}{2}
\epsilon_{abc}\phi_b
\label{lfe.2}
\end{eqnarray}
for infinitesimal parameters $\vec\omega$.
\par\noindent
The path integral is invariant under the field-coordinate
transformation (\ref{lfe.2}), thus we get an identity for  the
connected amplitude functional
\begin{eqnarray}&&
\int d^Dx\Biggl\langle
\Lambda^{D-4} \Biggl\{v^2\Bigl[
\frac{1}{4}(F_{a\mu}- J_{a\mu})(\epsilon_{abc}F_b^\mu\omega_c+\partial^\mu \omega_a )
-m^2 \frac{\omega_a}{2}\phi_a \phi_0 \Bigr]
\nonumber\\&&
-\frac{\omega_a}{2}\phi_a K_0 - \frac{\omega_a}{2}\phi_a  N_{0B}\phi_B
+ \frac{\omega_a}{2} \phi_0 N_{aB}\phi_B
+\frac{\omega_c}{2}\epsilon_{abc}\phi_b  N_{aB}\phi_B
\nonumber\\&&
 -  \phi_A  N_{A0}\frac{\omega_b}{2}\phi_b
+ \phi_A  N_{Aa}\frac{\omega_a}{2}\phi_0
+ \phi_A  N_{Ab}\frac{\omega_c}{2}\epsilon_{bb'c}\phi_{b'}
\Biggr\}
\nonumber\\&&
+\frac{\omega_a}{2} \phi_0 K_a  +\frac{\omega_c}{2}\epsilon_{abc}\phi_b K_a
\Biggr\rangle_C  =0.
\label{lfe.3}
\end{eqnarray}
The brackets $\langle  \cdot \rangle$ denote the mean value over the paths according to
eq. (\ref{model.1}).
Then by using the symmetry (\ref{model.1.1})
\begin{eqnarray}&&
\Biggl\{
(-\epsilon_{abc}J_c^\mu+\delta_{ab}\partial^\mu  )\frac{\delta}{\delta J_b^\mu}
-m^2 \frac{v^2}{2}\frac{\delta}{\delta N_{a0}}
\nonumber\\&&
- \frac{1}{2} N_{0B}\frac{\delta}{\delta N_{aB}}
+  \frac{1}{2} N_{aB}\frac{\delta}{\delta N_{0B}}
- \frac{1}{2}\epsilon_{abc} N_{cB}\frac{\delta}{\delta N_{bB}}
\nonumber\\&&
- \frac{1}{2} N_{A0}\frac{\delta}{\delta N_{Aa}}
+  \frac{1}{2} N_{Aa}\frac{\delta}{\delta N_{A0}}
- \frac{1}{2}\epsilon_{abc} N_{Ac}\frac{\delta}{\delta N_{Ab}}
\nonumber\\&&
-\frac{\Lambda^{D-4} }{2} K_0 \frac{\delta}{\delta K_a}
+\frac{1}{2\Lambda^{D-4} } K_a \frac{\delta}{\delta K_0} +\frac{1}{2}\epsilon_{abc} K_b\frac{\delta}{\delta K_c}
\Biggr\} \, W  =0.
\label{lfe.4}
\end{eqnarray}
We introduce the generators $ L_{cAB}$ of the transformations (\ref{lfe.2})
\begin{eqnarray}&&
\delta\phi_A = \frac{1}{2}\omega_c L_{cAB} \phi_B,\quad
L_{c0b}= -\delta_{cb}, L_{ca0}= \delta_{ca}, L_{cab}=\epsilon_{abc}
\nonumber\\&&
\omega_c L_c=
\left(
\begin{array}{llll}
0&-\omega_1 & - \omega_2 & - \omega_3 \\
\omega_1&0&\omega_3&-\omega_2\\
\omega_2&-\omega_3&0&\omega_1\\
\omega_3&\omega_2&-\omega_1&0\\
\end{array}
\right)
\nonumber\\&&
[L_a,L_b] = - 2 \epsilon_{abc}L_c
\label{lfe.4.1}
\end{eqnarray}
and we rewrite eq. (\ref{lfe.4}) in the compact form
\begin{eqnarray}&&
\Biggl\{
(-\epsilon_{abc}J_c^\mu+\delta_{ab}\partial^\mu  )\frac{\delta}{\delta J_b^\mu}
-m^2 \frac{v^2}{2}\frac{\delta}{\delta N_{a0}}
\nonumber\\&&
- \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\frac{\delta}{\delta N_{AB}}
\nonumber\\&&
-\frac{\Lambda^{D-4} }{2} K_0 \frac{\delta}{\delta K_a}
+\frac{1}{2\Lambda^{D-4} } K_a \frac{\delta}{\delta K_0} +\frac{1}{2}\epsilon_{abc} K_b\frac{\delta}{\delta K_c}
\Biggr\} \, W  =0.
\label{lfe.4.2}
\end{eqnarray}
For the effective action (1-PI generating functional) one gets
\begin{eqnarray}&&
-\partial^\mu \Gamma_{J_a^\mu} - \epsilon_{abc}J_b^\mu \Gamma_{J_c^\mu}
+ \frac{v^2}{2} m^2 \Gamma_{N_{0a}}
+ \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\Gamma_{ N_{AB}}
\nonumber\\&&
+ \frac{\Lambda^{D-4} }{2} K_0 \phi_a
+\frac{1}{2\Lambda^{D-4} }  \Gamma_{ K_0} \Gamma_{\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \Gamma_{\phi_c}
 \,   =0,
\label{lfe.5}
\end{eqnarray}
where we use the notation
\begin{eqnarray}
\Gamma_X=
 \frac{\delta }{\delta X} \Gamma .
\label{lfe.5.1}
\end{eqnarray}

\section{The Subtraction Strategy at $D=4$}
\label{sec:lin}
The LFE (\ref{lfe.5}) is the tool we use in order
to make finite the massive nonlinear sigma model.
We work in the framework of the loop expansion of $\Gamma$ in eq. (\ref{lfe.5}).
\par
This strategy has been employed with success for the massless nonlinear sigma model
\cite{Ferrari:2005va}-\cite{Bettinelli:2007kc}, for the massive Yang-Mills in the Landau gauge
\cite{Bettinelli:2007tq},\cite{Bettinelli:2007cy}, for a Higgsless Electroweak model also
in the Landau gauge  \cite{Bettinelli:2008ey}-\cite{Bettinelli:2009wu}.
Therefore in models where dimensional regularization allows
to drop the tadpoles involving the scalars fields $\vec\phi$. In the massive nonlinear sigma model
the tadpoles play an important r\^ole and therefore it is necessary to extend the
formalism to this case.
\par
At the tree level  $\Gamma^{(0)}$ is a solution of eq. (\ref{lfe.5}) by construction.
Dimensional regularization yields radiative corrections which do not generate any
anomaly. The proof of this property is sketched in Ref. \cite{Ferrari:2005ii}
and displayed in Ref. \cite{Bettinelli:2007zn}.
The Feynman rules are provided
by the classical action $\Gamma^{(0)}$, while the counterterms $\hat\Gamma^{(k)}$
are introduced via the effective action which must obey the LFE (\ref{lfe.5}).
Therefore what matters are the symmetry properties of $\Gamma$ and not those
of the action.
\par
If the procedure
of subtraction of infinities by means of the counterterms $\hat\Gamma^{(k)}$
has been carried out successfully up to order $n-1$, then
\begin{eqnarray}&& \!\!\!\!\!\!\!\!\!\!\!\!
-\partial^\mu \Gamma^{(k)}_{J_a^\mu} - \epsilon_{abc}J_b^\mu \Gamma^{(k)}_{J_c^\mu}
+ \frac{v^2}{2} m^2 \Gamma^{(k)}_{N_{0a}}
+ \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\Gamma^{(k)}_{ N_{AB}}
\nonumber\\&&
\!\!\!\!\!\!\!\!\!\!\!\!
+\frac{1}{2\Lambda^{D-4} }  \Gamma^{(0)}_{\phi_a} \Gamma^{(k)}_{ K_0}
+\frac{1}{2} \phi_0 \Gamma^{(k)}_{\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \Gamma^{(k)}_{\phi_c}
 \,    +\sum_{j=1}^{k-1}
\frac{1}{2\Lambda^{D-4} }  \Gamma^{(k-j)}_{ K_0} \Gamma^{(j)}_{\phi_a} = 0,\,\, \forall k<n.
\label{lin.1}
\end{eqnarray}
At order $n$ we expect a violation of eq. (\ref{lfe.5})
\begin{eqnarray}&&
-\partial^\mu \Gamma^{(n)}_{J_a^\mu} - \epsilon_{abc}J_b^\mu \Gamma^{(n)}_{J_c^\mu}
+ \frac{v^2}{2} m^2 \Gamma^{(n)}_{N_{0a}}
+ \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\Gamma^{(k)}_{ N_{AB}}
\nonumber\\&&
+\frac{1}{2\Lambda^{D-4} }  \Gamma^{(0)}_{\phi_a} \Gamma^{(n)}_{ K_0}
+\frac{1}{2} \phi_0 \Gamma^{(n)}_{\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \Gamma^{(n)}_{\phi_c}
 \,    +\sum_{j=1}^{n-1}
\frac{1}{2\Lambda^{D-4} }  \Gamma^{(n-j)}_{ K_0} \Gamma^{(j)}_{\phi_a}
\nonumber\\&&
 =
\sum_{j=1}^{n-1}
\frac{1}{2\Lambda^{D-4}} \widehat{ \Gamma}^{(n-j)}_{ K_0} \widehat{ \Gamma}^{(j)}_{\phi_a}.
\label{lin.1.1}
\end{eqnarray}
By assumption $\Gamma^{(k<n)}$ is finite,
thus the removal of the poles in the Laurent expansion of
\begin{eqnarray}
\frac{1}{\Lambda^{D-4} } \Gamma^{(n)}
\label{lin.1.1.1}
\end{eqnarray}
is a strategy that maintains the validity of the LFE,
since the RHS of eq. (\ref{lin.1.1}) is a pure pole part, when the
normalization (\ref{lin.1.1.1}) is used.
The finite part in the limit
$D=4$ is the subtracted  amplitude. It should be stressed that no further
finite subtraction is allowed (even those keeping eq. (\ref{lin.1.1}) unchanged). In fact
by adding extra counterterms one modifies the pure pole structure of the breaking terms.
For instance on-shell renormalization is not a doable procedure.
Finally the spotted  counterterms $\widehat{ \Gamma}^{(n)}$
obey the equation
\begin{eqnarray}&&
-\partial^\mu \widehat{ \Gamma}^{(n)}_{J_a^\mu} - \epsilon_{abc}J_b^\mu \widehat{ \Gamma}^{(n)}_{J_c^\mu}
+ \frac{v^2}{2} m^2 \widehat{ \Gamma}^{(n)}_{N_{0a}}
+ \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\widehat{\Gamma}^{(k)}_{ N_{AB}}
\nonumber\\&&
+\frac{1}{2\Lambda^{D-4} }  \widehat{ \Gamma}^{(0)}_{\phi_a} \widehat{ \Gamma}^{(n)}_{ K_0}
+\frac{1}{2} \phi_0 \widehat{ \Gamma}^{(n)}_{\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \widehat{ \Gamma}^{(n)}_{\phi_c}
 +
 \sum_{j=1}^{n-1}
\frac{1}{2\Lambda^{D-4} } \widehat{ \Gamma}^{(n-j)}_{ K_0} \widehat{ \Gamma}^{(j)}_{\phi_a}=0.
\label{lin.1.2}
\end{eqnarray}
The last equation (\ref{lin.1.2}) provides a {\sl a posteriori} explanation of why the
breaking term takes the form exhibited in (\ref{lin.1.1}).
A direct proof is provided in Ref. \cite{Bettinelli:2007zn}. The subtraction strategy has been
tested for a solvable model in Ref. \cite{Ferrari:2009uj}.
\section{The Algebraic Aspects of the LFE}
\label{sec:alg}
It is convenient to define a new functional
\begin{eqnarray}
\TG[\phi_a, J_a^\mu,K_0,N_{00},N_{0a}, N_{ab}]
\equiv \Gamma[\phi_a, J_a^\mu,K_0,N_{00}{-}\frac{1}{2}v^2m^2,N_{0a}, N_{ab}]
\label{alg.2}
\end{eqnarray}
in presence of the \underline{nontrivial background}
\begin{eqnarray}&&
\phi_a = J_a^\mu = K_0 = N_{0a}= N_{ab}=0
\nonumber\\&&
N_{00} =\frac{1}{2} m^2 v^2
\label{alg.2.1}
\end{eqnarray}
and of the boundary conditions
\begin{eqnarray}&&
\frac{\delta \TG^{(0)}}{\delta K_0}\biggl |_{\phi_a = J_a^\mu = K_0 = N_{0a}= N_{ab}=[N_{00} {-}\frac{1}{2}v^2 m^2]=0} = \Lambda^{D-4}
\nonumber\\&&
\frac{\delta \TG^{(0)}}{\delta N_{00}}\biggl |_{\phi_a = J_a^\mu = K_0 = N_{0a}= N_{ab}=[N_{00} {-}\frac{1}{2}v^2 m^2]=0} = \Lambda^{D-4}.
\label{alg.2.1.bou}
\end{eqnarray}
By using $ \TG$ we get rid of the mass term in eq. (\ref{lfe.5}) and all the subsequent equations.
For the counterterms $\widehat\TG^{(n)}$ eq. (\ref{lin.1.2}) gives
\begin{eqnarray}
{\cal S}_a\widehat\TG^{(n)}
 \,   = - \sum_{j=1}^{n-1}
\frac{1}{2\Lambda^{D-4} } \widehat \TG^{(n-j)}_{ K_0} \widehat\TG^{(j)}_{\phi_a},
\label{alg.3}
\end{eqnarray}
where
\begin{eqnarray}&&
{\cal S}_a \equiv
-\partial^\mu \frac{\delta}{\delta J_a^\mu} - \epsilon_{abc}J_b^\mu
\frac{\delta}{\delta J_c^\mu}
+ \frac{1}{2} \Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\frac{\delta}{\delta N_{AB}}
\nonumber\\&&
+\frac{1}{2\Lambda^{D-4}}  \TG^{(0)}_{\phi_a} \frac{\delta}{K_0}
+\frac{1}{2} \phi_0 \frac{\delta}{\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \frac{\delta}{\phi_c}.
\label{alg.4}
\end{eqnarray}
It is convenient to introduce the operators
\begin{eqnarray}&&
{\cal S}^J_a \equiv
-\partial^\mu \frac{\delta}{\delta J_a^\mu} - \epsilon_{abc}J_b^\mu
\frac{\delta}{\delta J_c^\mu}
\nonumber\\&&
{\cal S}^N_a \equiv
\frac{1}{2} \Bigl( L_{aAC}N_{CB}+ L_{aBD}N_{AD}\Bigr)\frac{\delta}{\delta N_{AB}}
\nonumber\\&&
{\cal S}^\phi_a \equiv
\frac{1}{2} \phi_0 \frac{\delta}{\delta\phi_a}
-
\frac{1}{2}\epsilon_{abc}\phi_b \frac{\delta}{\delta\phi_c}
\nonumber\\&&
{\cal S}^\TG_a \equiv
\frac{1}{2\Lambda^{D-4}} \TG^{(0)}_{\phi_a}
\frac{\delta}{\delta K_0}.
\label{alg.4.1}
\end{eqnarray}
It is straightforward to verify the following relations
\begin{eqnarray}&&
[{\cal S}^J_a(x) , {\cal S}^J_{a'}(y)]
=   \epsilon_{a a'c}\delta(x-y){\cal S}^J_c(x)
\nonumber\\&&
[{\cal S}^N_a(x) , {\cal S}^N_{a'}(y)]
=   \epsilon_{a a'c}\delta(x-y){\cal S}^N_c(x)
\nonumber\\&&
[{\cal S}^\phi_a(x) , {\cal S}^\phi_{a'}(y)]
=   \epsilon_{a a'c}\delta(x-y){\cal S}^\phi_c(x)
\label{alg.10.1.1}
\end{eqnarray}
and finally
\begin{eqnarray}
[{\cal S}_a(x) , {\cal S}_{a'}(y)]
=   \epsilon_{a a'c}\delta(x-y){\cal S}_c(x).
\label{alg.10.1}
\end{eqnarray}
\subsection{The Consistency Condition}
From eqs. (\ref{alg.3}) and  (\ref{alg.10.1}) we get the consistency condition
that has to be obeyed by the counterterms
\begin{eqnarray}&&
\Delta_a(x) \equiv  - \sum_{j=1}^{n-1}\,
 \widehat \TG^{(n-j)}_{ K_0(x)}\,  \widehat\TG^{(j)}_{\phi_a(x)}
\nonumber\\&&
{\cal S}_a(x) \Delta_{a'}(y)
 -
{\cal S}_{a'}(y) \Delta_a(x)
=\delta(x-y) \epsilon_{aa'c}~\Delta_c(x).
\label{alg.8.1}
\end{eqnarray}
\subsection{The Local  Solutions: the bleaching Method}
The counterterms $\widehat \TG^{(n)}$ are given by linear combinations of local monomials
constructed in terms of fields $\vec\phi$ and sources $K_0, J_a^\mu, N_{AB}$.
The construction of the counterterms proceeds via the evaluation
of the pole parts of the amplitudes as  in eq. (\ref{lin.1.1.1})
and consequently the fixing of the coefficients of the general solution
of eq. (\ref{alg.3}). For one-loop corrections one needs just the solution
for the associated homogeneous equation.
\par
This task is made easy if we replace the above mentioned variables by
suitably chosen composite local invariants. The general procedure
in Ref. \cite{Bettinelli:2007kc} is here applied straightforwardly.
\par
The local invariant solutions can be constructed with the following fields
and sources
\begin{eqnarray}&&
F_a^\mu- J_a^\mu
\nonumber\\&&
D^\mu[F] (F-J)^\nu|_{ab} =(\partial_\mu \delta_{ab}-\epsilon_{abc}F_c^\mu) (F_b^\nu- J_b^\nu)
=\partial_\mu \delta_{ab}-\epsilon_{abc}J_c^\mu (F_b^\nu- J_b^\nu)
\nonumber\\&&
\phi_A \equiv\{\phi_0,\phi_a\}
\nonumber\\&&
K_A \equiv \{ K_0,-\frac{1}{\Lambda^{D-4} }\TG^{(0)}_{\phi_a}\},
\qquad -\frac{1}{\Lambda^{D-4} }\TG^{(0)}_{\phi_a}\Big|_{\vec\phi=0}
= - \frac{v^2}{2}\partial_\mu J_a^\mu - N_{0a} - N_{a0}
\nonumber\\&&
N_{AB}.
\label{alg.5}
\end{eqnarray}
(Notice $D^\mu[F] (F-J)^\nu|_{ab}-D^\mu[J] (F-J)^\nu|_{ab}
=-\epsilon_{abc}(F_c^\mu- J_c^\mu)(F_b^\nu- J_b^\nu)$).
\par\noindent
$K_A$ and $\phi_A$ transform in the same way. In fact
\begin{eqnarray}&&
{\cal S}_a (x)\TG^{(0)}_{\phi_b(y)}
 =
\frac{1}{2}\delta(x-y) \Bigl\{-\delta_{ab}\Lambda^{D-4}  K_0 (y)
+\epsilon_{bca} \TG^{(0)}_c(y)
\Bigr \}
\nonumber\\&&
{\cal S}_a (x)K_0(y)=\delta(x-y)\frac{1}{2\Lambda^{D-4} }
\TG^{(0)}_{\phi_a(y)}.
\label{alg.5.1}
\end{eqnarray}
The bleached  variables can be constructed
according to the transformation
properties of the fields  in eq. (\ref{alg.5}). For instance
{
\begin{eqnarray}&&
{\mathfrak J}_{\mu} \equiv \Omega^\dagger (J_{\mu}-F_{\mu})\Omega
=\Omega^\dagger J_{\mu}\Omega + i \Omega^\dagger\partial_\mu\Omega\, ,
\nonumber\\&&
\partial_\mu{\mathfrak J}_{\nu} =\Omega^\dagger \left(\partial_\mu+
\Omega\partial_\mu\Omega^\dagger
\right)(J_{\nu}-F_{\nu})\Omega=\Omega^\dagger{\cal D}_\mu[F](J_{\nu}-F_{\nu})\Omega\, ,
\nonumber\\&&
{\mathfrak K}_0 \equiv K_A \phi_A =
K_0\phi_0 - \frac{1}{\Lambda^{D-4} }\TG^{(0)}_{\phi_a} \phi_a\, ,
\nonumber\\&&
{\mathfrak N}_{\alpha \beta, \rho \sigma}= \Omega^\dagger_{\alpha\alpha'}(\tau_A)_{\alpha' \rho}
(\tau_B^\dagger)_{\beta\beta' }\Omega_{\beta'\sigma} N_{AB}.
\label{alg.6}
\end{eqnarray}
}
The bleaching yields ${\cal S}_a$-invariant local variables. Moreover the mapping is
invertible.
\par
The actual construction of the counterterms can profit of further properties of
the LFE that limit the number of independent divergent amplitudes.  The next Section
deals with this feature.

\section{Hierarchy and Weak Power Counting}
\label{sec:h}
Eq. (\ref{lfe.5}) is nonlinear in $\Gamma$. This allows us to grade
the 1PI functions in a hierarchical way according to the
number of external $\phi$ - legs. In fact we have
\begin{eqnarray}&&
\Gamma_{\phi_a} = \frac{\Lambda^{D-4} } { \Gamma_{ K_0} }
\Biggl\{
\epsilon_{abc}\phi_b \Gamma_{\phi_c}
+2\partial^\mu \Gamma_{J_a^\mu} + 2\epsilon_{abc}J_b^\mu \Gamma_{J_c^\mu}
\nonumber\\&&
-v^2 m^2 \Gamma_{N_{0a}}
- \Lambda^{D-4} K_0 \phi_a
-\Bigl( L_{aAA'}N_{A'B}+L_{aBB'} N_{AB'}\Bigr)\Gamma_{ N_{AB}}
\Biggr\}
\label{h.1}
\end{eqnarray}
The derivative of eq.(\ref{h.1}) with respect to any
ancestor variable ($J_a^\mu, K_0, N_{AB}  $) yields all
the descendant amplitudes involving one $\phi-$ field.
In a recursive way one obtains all the descendant
amplitudes from the ancestor ones (hierarchy).
\par
This result is
very important because, at fixed order in the loop
expansion, the number of independent divergent ancestor amplitudes
is finite. In fact,
by simple dimensional analysis, one can show that the superficial degree
of divergence of a 1-PI graph $G$ for ancestor and descendant variables
is bounded by
\begin{eqnarray}
\delta(G)\leq n_L(D-2)+2 -  N_J-2(N_K+N_N),
\label{h.2}
\end{eqnarray}
where $n_L$ is the number of loops and $N_J,N_K,N_N$
are the numbers of insertions of the ancestor variables
$J_a^\mu, K_0, N_{AB}  $. Thus for fix $n_L$ the number
of independent divergent ancestor amplitudes is finite.
The bound in eq. (\ref{h.2}) does not depend on the number of external
$\phi$ - legs; therefore, if the ancestor amplitudes are
divergent, an infinite number of descendant will also be divergent.
The divergent parts of the descendant amplitudes will not be independent,
due to the hierarchy property.
\par\noindent
The bound (\ref{h.2}) comes from the following two relations
\begin{eqnarray}&&
\delta(G)= n_L D -2 I  +  N_J + \sum_k k V_k
\nonumber\\&&
n_L = I - N_J - (N_K+N_N) -\sum_k  V_k + 1
\label{h.2.7}
\end{eqnarray}
where $V_k$ is the number of vertexes with $k$ derivatives
and $I$ the number of internal lines.
The inequality in eq. (\ref{h.2}) comes from the fact that for the
unsubtracted theory $k\le 2$.
\par
The WPC criterion consists in building the classical
action $\Gamma^{(0)}$ such that the bound (\ref{h.2}) is obeyed.
It is not necessary to introduce in $\Gamma^{(0)}$ all possible terms
that are allowed by the WPC criterion. For instance it is not necessary
to introduce a $\phi^4$ interaction in the model considered in
the present paper. This approach is at variance with the algebraic
renormalization procedure where the Power Counting theorem requires
that all allowed couplings should enter with independent parameters.
\par
The WPC theorem says that the bound (\ref{h.2})
is stable under the subtraction procedure described
in Section \ref{sec:lin}. The proof goes as follows.
The counterterm  $\widehat\Gamma^{(k)}$,
of order $k$ in the loop expansion,
is a finite sum of local monomials built with  $J_a^\mu, K_0, N_{AB}  $
sources, space-time derivatives and $N_\phi$  $\phi$ - fields.
Each monomial ${\cal M}^{(k)}(J_a^\mu, K_0, N_{AB},N_\phi,n_d)$
entering in   $\widehat\Gamma^{(k)}$ is constructed from graphs that
obey eq. (\ref{h.2}) and therefore it satisfies the condition
\begin{eqnarray}
 k (D-2)+2  - n_J- 2(n_K+n_N)- n_d \ge 0 ,
\label{h.3}
\end{eqnarray}
where $n_d$ is the number of derivatives and $n_J,n_K,n_N$ are
the numbers of times the sources enter in the monomial.
When one of this monomial is inserted in a graph, then the
relations in eqs. (\ref{h.2.7}) are modified by an extra vertex
with $n_d$ derivatives. Moreover the numbers of external
sources become $N_J-n_J, N_K-n_K, N_N-n_N$. The superficial
degree of divergence becomes
\begin{eqnarray}&&
\delta(G)\leq (n_L-k)(D-2)  + n_d -  (N_J-n_J)-2[(N_K-n_K)+(N_N-n_N)]
\nonumber\\&&
\leq n_L(D-2)+2 -  N_J-2(N_K+N_N)
\label{h.4}
\end{eqnarray}
by using eq. (\ref{h.3}) in the last step.
Thus WPC is stable under subtraction of divergences.
\section{Hierarchical Relations and Feynman Rules}
\label{sec:h2}
In this Section we discuss some of the hierarchical relations
for the two-point functions. This is an example. More
relations and Feynman rules are given in Appendix \ref{app:A}.
\par
By successive differentiation of eq. (\ref{lfe.5}) we get
\begin{eqnarray}&&
\partial^\mu \Gamma_{J_a^\mu J_b^\nu}
- \frac{v^2}{2}m^2 \Gamma_{N_{0a}J_b^\nu}
-\frac{1}{2\Lambda^{D-4} }  \Gamma_{ K_0} \Gamma_{\phi_aJ_b^\nu}
 \,   =0.
 \nonumber\\&&
\partial^\mu \Gamma_{J_a^\mu N_{0b}}
- \frac{v^2}{2}m^2 \Gamma_{N_{0a} N_{0b}}
{-\frac{1}{2} \Gamma_{ N_{ab}}}
+ \frac{1}{2} \delta_{ab}\Gamma_{ N_{00}}
+ \frac{1}{2}\epsilon_{abc} \Gamma_{ N_{0c}}
 \nonumber\\&&
-\frac{1}{2\Lambda^{D-4} }  \Gamma_{ K_0} \Gamma_{\phi_a  N_{0b}}
 \,   =0,
\label{h2.1}
\end{eqnarray}
\begin{eqnarray}
\partial^\mu \Gamma_{J_a^\mu \phi_b}
- \frac{v^2}{2}m^2 \Gamma_{N_{0a} \phi_b}
-\frac{1}{2\Lambda^{D-4} }  \Gamma_{ K_0} \Gamma_{\phi_a  \phi_b}
 \,   =0
\label{h2.2}
\end{eqnarray}
and finally ($\partial^\mu \Gamma_{ K_0}=0$)
\begin{eqnarray}&&
\!\!\!\!\!\!\!\!\!\!\!\!
\Gamma_{\phi_a  \phi_b} = \frac{2\Lambda^{D-4} }{\Gamma_{ K_0} }
\Biggl(
\partial^\mu \Gamma_{J_a^\mu \phi_b}
- \frac{v^2}{2}m^2 \Gamma_{N_{0a} \phi_b}
\Biggr)
=\biggl(\frac{2\Lambda^{D-4} }{\Gamma_{ K_0} }\biggr)^2
\Biggl(
{
\partial^\mu \Bigl[-
\partial^\sigma \Gamma_{J_a^\mu J_b^\sigma}
-\frac{v^2}{2}m^2 \Gamma_{J_a^\mu N_{0b}}\Bigr]

\nonumber\\&&
- \frac{v^2}{2}m^2 \Bigl[
{
- \partial^\mu \Gamma_{ N_{0a}J_b^\mu}
- \frac{v^2}{2}m^2 \Gamma_{N_{0a} N_{0b} }
{-\frac{1}{2} \Gamma_{ N_{ab}}}+ \frac{1}{2} \delta_{ab}\Gamma_{ N_{00}}

\Bigr]
\Biggr).
\label{h2.3}
\end{eqnarray}
In the massless case all the terms present in eq. (\ref{h2.3}) are zero.
For nonzero mass the expected contributions should come from
the tadpole integral
\begin{eqnarray}&&
B_0 \equiv  \frac{1}{ v^2} \int \frac{d^D q}{(2\pi)^D}
\frac{1}{(q^2-m^2)};
\nonumber\\&&
\frac{1}{\Lambda^{D-4}  v^2} \int \frac{d^D q}{(2\pi)^D}
\frac{1}{(q^2-m^2)}
=
- \frac{i}{v^2} \frac{1}{(4\pi)^2}\Gamma(1-\frac{D}{2})m^2
\Bigl(
\frac{m}{\Lambda\sqrt{4\pi}}
\Bigr)^{D-4}
\nonumber\\&&
=  i \frac{m^2}{v^2} \frac{1}{(4\pi)^2} \Bigl[-\frac{2}{D-4}
+1 - \gamma + \ln(\frac{m^2}{\Lambda^2(4\pi)})
\Bigr] + {\cal O}(D-4) .
\label{h2.4.1}
\end{eqnarray}
In fact one has
\begin{eqnarray}&&
\Gamma^{(1)}_{K_0} = -\frac{3}{2} i B_0
\label{h2.4.2.1}
\\&&
\Gamma^{(1)}_{N_{00}} = 2 \Gamma^{(1)}_{K_0}= -3 i B_0
\label{h2.4.2.2}
\\&&
p^\mu
\Gamma^{(1)}_{J^\mu_a  J^\nu_{b}} =
-i \frac{\delta_{ab}}{4}p^\mu\int \frac{d^D q}{(2\pi)^D}
\frac{(2q+p)_\mu(2q+p)_\nu}{[(p+q)^2-m^2][q^2-m^2]}
=
-i \frac{v^2 \delta_{ab}}{2} p_\nu B_0
\label{h2.4.2.3}
\\&&
\Gamma^{(1)}_{ N_{ab}} = i  \delta_{ab}  B_0
\label{h2.4.2.4}
\\&&
\Gamma^{(1)}_{N_{0a} N_{0b} }= \Gamma^{(1)}_{J_a^\mu N_{0b}}
=\Gamma^{(1)}_{ N_{0b}}=0
\label{h2.4.2.6}
\\&&
\Gamma^{(1)}_{ J_{a\mu}(p)\phi_{b}  } =  \frac{v^2}{4} B_0p^\mu \delta_{ab}
\label{h2.4.2.7}
\\&&
\Gamma^{(1)}_{\phi_{a} N_{0b} } =  \delta_{ab}\frac{5}{3} \Gamma^{(1)}_{K_0}
= -  \frac{5}{2}i B_0\delta_{ab}
\label{h2.4.2.8}
\\&&
\Gamma^{(1)}_{\phi_a \phi_b} = i \delta_{ab}v^2(p^2+m^2 )  B_0.
\label{h2.4.2.5}
\end{eqnarray}

\subsection{Two-point Feynman Rules}
In the zero loop approximation the boundary conditions (\ref{alg.2.1}) and
eqs. (\ref{h2.1}) and  (\ref{h2.2}) provide the Feynman rules
\begin{eqnarray}&&
\Gamma^{(0)}_{J_a^\mu \phi_b} =
ip_\mu \delta_{ab}\frac{\Lambda^{D-4} v^2}{2}
\nonumber\\&&
\Gamma^{(0)}_{N_{0a} \phi_b} =
\delta_{ab}\Lambda^{D-4}
\nonumber\\&&
\Gamma^{(0)}_{\phi_a \phi_b} =
 \delta_{ab}\Lambda^{D-4}v^2(p^2-m^2).
\label{h2.4}
\end{eqnarray}
\section{On the One-loop Counterterms}
\label{sec:one}
The results of the previous sections allow to extract finite quantities
from dimensional regularized amplitudes. Moreover local counterterms
can be constructed so that one has a recursive process of subtraction.
Perturbative unitarity is guaranteed by construction, under the
form of cutting equation (see for instance Ref. \cite{'tHooft:1973pz}).
Moreover the LFE (\ref{alg.3}) and the consistency condition (\ref{alg.8.1})
allows a order-by-order check of the counterterms.
\par
It is of some interest to look for the local solutions of the LFE (\ref{alg.3})
and of its associated homogeneous equation. In fact, when the solutions
are known, one can obtain all the counterterms for the descendant
amplitudes counterterms. Further on we provide an example for the one-loop
approximation. On one side this subject is very instructive, but on the other
side, as it will be clear later on,
the study of the local invariant solutions becomes very complex
at higher order in the perturbative expansion.
\par
We list some of the  monomials that can be associated to one-loop divergent
amplitudes.
According to the eq. (\ref{h.2}) they are expected to have
dimension  4 or less, if we consider the subtraction procedure at $D=4$.
Any monomial that does not contain the $\phi$-field (also implicitly as in
$F_{a\mu}$ or in $K_A$) is a sterile term, since it cannot be the ancestor
of any descendant amplitude. These sterile terms cannot be neglected:
some of them take care of the counterterms associated to tadpoles.
Here are few examples of dimension four constructed by using the bleached
operators (i.e. invariant under the local transformations generated
by ${\cal S}_a (x)$). Moreover global $SU(2)_R$ invariance is imposed.
The antisymmetric part of $N_{AB}$ does
not appear in the counterterms.
\par
Let us list first the possible counterterms  present in the massless case
\cite{Ferrari:2005va}
{
\begin{eqnarray}
&& {\cal I}_1 = \int d^Dx \, \Big [ D_\mu[F] ( J- F )_\nu \Big ]_a \Big [ D^\mu[F] ( J- F )^\nu \Big ]_a  \, ,
\nonumber \\
&& {\cal I}_2 = \int d^Dx \, \Big [ D_\mu[F] ( J- F )^\mu \Big ]_a \Big [ D_\nu[F] ( J- F )^\nu \Big ]_a  \, ,
\nonumber \\
&& {\cal I}_3 = \int d^Dx \, \epsilon_{abc} \Big [ D_\mu[F] ( J- F )_\nu \Big ]_a \Big ( J^\mu_b - F^\mu_b \Big ) \Big ( J^\nu_c -F^\nu_c \Big ) \, ,  \nonumber \\
&& {\cal I}_4 =
\int d^Dx \, \Big (K_A\phi_A\Big )^2
\nonumber \\&&
~~~~~=\int d^Dx \, \Big ( K_0\phi_0
-\frac{\phi_a}{\Lambda^{D-4} }\TG^{(0)}_{\phi_a}\Big )^2 \, ,
\nonumber \\
&& {\cal I}_5 = \int d^Dx \, \Big ( K_A\phi_A\Big ) \Big (J^\mu_c -F^\mu_c\Big )^2
\nonumber \\
&&
~~~~~=\int d^Dx \, \Big (  K_0\phi_0  -\frac{\phi_a}{\Lambda^{D-4} }\TG^{(0)}_{\phi_a} \Big )
 \Big (J^\mu_c -F^\mu_c\Big )^2 \, ,
\nonumber \\
&& {\cal I}_6 = \int d^Dx \, \Big ( J^\mu_a -F^\mu_a\Big  )^2
 \Big ( J^\nu_b -F^\nu_b \Big )^2 \, , \nonumber \\
&& {\cal I}_7 = \int d^Dx \, \Big ( J^\mu_a -F^\mu_a\Big  )
   \Big ( J^\nu_a -F^\nu_a\Big  )
   \Big ( J_{b\mu} -F_{b\mu} \Big  )
   \Big ( J_{b\nu} -F_{b\nu} \Big  )
\, .
\label{alg.7}
\end{eqnarray}
}
Notice the identity
\begin{eqnarray}&&
2({\cal I}_1 -{\cal I}_2 )-4{\cal I}_3+({\cal I}_6-{\cal I}_7 )
= \int d^D x {\cal G}_{a\mu\nu}[{\mathfrak J}]{\cal G}_a^{\mu\nu}[{\mathfrak J}]
= \int d^D x {\cal G}_{a\mu\nu}[J]{\cal G}_a^{\mu\nu}[J],
\label{alg.8}
\end{eqnarray}
where
\begin{eqnarray}
{\cal G}_{a\mu\nu}[J] \equiv \partial_\mu J_{a\nu}-\partial_\nu J_{a\mu}
+\epsilon_{abc}J_{b\mu}J_{c\nu}.
\label{alg.8.1.2}
\end{eqnarray}
The right hand term in eq. (\ref{alg.8}) is sterile: no descendant terms are generated.
The calculation \cite{Ferrari:2005va} for the massless nonlinear sigma model gives
\begin{eqnarray}&&
\widehat \G^{(1)}\Bigr|_{\rm MASSLESS}
= \frac{1}{D-4} \frac{\Lambda^{D-4}}{(4\pi)^2}\Big [
- \frac{1}{12}   \Big (
{\cal I}_1 - {\cal I}_2 -  {\cal I}_3 \Big )
+  \frac{1}{48}
\Big ( {\cal I}_6 + 2 {\cal I}_7 \Big )
\nonumber \\& &
+
 \frac{3}{2} \frac{1}{v^4 } {\cal I}_4
+  \frac{1}{2} \frac{1}{v^2} {\cal I}_5 \Big ].
\label{alg.7.1}
\end{eqnarray}
These counterterms are expected to be present in the same combination in the massive case,
since no tadpoles contribute to the invariants ${\cal I}_1-{\cal I}_7$.
\par
For later use we display the local invariant
\begin{eqnarray}&&
K_A\phi_A \Bigr|_{N_{00} \to N_{00}-\frac{m^2v^2}{2}}
=\frac{ K_0}{\phi_0} - \phi_a \biggl(-v^2\Box\phi_a
\nonumber \\& &
- v^2 \int d^D x \Bigl[\frac{J_b^\mu}{4}\frac{\delta F_{b\mu}}{\delta \phi_a}\Bigr]
+N_{aB}\phi_B + \phi_BN_{Ba}-2\phi_a (N_{00}{+}\frac{m^2v^2}{2})
\biggr)
\label{alg.7.2}
\end{eqnarray}
and its ancestor content
\begin{eqnarray}
K_A\phi_A\Bigr|_{\vec\phi=0} = K_0 .
\label{alg.7.3}
\end{eqnarray}
\par
The massive case requires the introduction of the sources $N_{AB}$ thus we get
a numerous set of new invariants. At the one-loop level we have the
 local invariants
candidates for $\widehat\TG^{(1)}$
\begin{eqnarray}
&&
{\cal I}_{8} = \int d^Dx \,\,\,N_{CC}
   \nonumber \\&&
{\cal I}_{9} = \int d^Dx \,\,\,K_A \phi_A
   \nonumber \\&&
{\cal I}_{10} = \int d^Dx \,\,\,\Big (J^\mu_c -F^\mu_c\Big )^2
   \nonumber \\&&
{\cal I}_{11} = \int d^Dx \,\,\, \phi_A N_{AB} \phi_B
   \nonumber \\&&
{\cal I}_{12} = \int d^Dx \, \,\, \phi_A \Big (N_{AB}+N_{BA}\Big  )\Big (N_{BC}+N_{CB}\Big  ) \phi_C
   \nonumber \\&&
{\cal I}_{13} = \int d^Dx \,  \,\, \phi_A N_{AB}\phi_B \phi_C N_{CD}\phi_D
   \nonumber \\&&
{\cal I}_{14} = \int d^Dx \, \,\, K_C \phi_C  \phi_A N_{AB}\phi_B
   \nonumber \\&&
{\cal I}_{15} = \int d^Dx \,\,\, K_A  \Big (N_{AB}+N_{BA}\Big  )  \phi_B ,
   \nonumber \\&& \,\,\,\,
{K_A  \Big (N_{AB}+N_{BA}\Big  )\Bigr|_{\vec\phi=0}=2K_0N_{00}
-(\frac{v^2}{2}\partial_\mu J_a^\mu + N_{0a} + N_{a0})(N_{0a} + N_{a0})
}
   \nonumber \\&&
{\cal I}_{16} = \int d^Dx \,\,\,  \phi_A N_{AB} \phi_B\Big (J^\mu_c -F^\mu_c\Big )^2
   \nonumber \\&&
{\cal I}_{17} = \int d^Dx \, \,\, K_C \phi_C   N_{AA}
   \nonumber \\&&
{\cal I}_{18} = \int d^Dx \, \,\,   \Big ( N_{AB} N_{AB}+ N_{AB} N_{BA}\Big )
   \nonumber \\&&
{\cal I}_{19} = \int d^Dx \, \,\, N_{AA} \Big (J^\mu_c -F^\mu_c\Big )^2
   \nonumber \\&&
{\cal I}_{20} = \int d^Dx \, \,\, (N_{AB}+N_{BA}) \, Tr\, \Bigl\{\Omega ^\dagger (F^\mu-J^\mu)\Omega
\tau^\dagger_A (F_\mu-J_\mu)\tau_B \Bigr\}
   \nonumber \\&&
{{\cal I}_{21} = \int d^Dx \, \,\, K_A K_A,\qquad
K_A K_A\Bigr|_{\vec\phi=0}=K_0^2
+(\frac{v^2}{2}\partial_\mu J_a^\mu + N_{0a} + N_{a0})^2
}
\, .
\label{alg.9}
\end{eqnarray}
Let us find the ancestor variables content of the invariant ${\cal I}_{20}$ by using
\begin{eqnarray}
[\tau_A^\dagger,\frac{\tau_b}{2} ]= \delta_{Ax}\epsilon_{xbc}\tau_c,\qquad (N_{AB}+N_{BA})\tau_A^\dagger\tau_B
= 2 N_{AB}.
\label{alg.9.2}
\end{eqnarray}
We have
\begin{eqnarray}&&
{\cal I}_{20}\Bigr|_{\vec\phi=0} =(N_{AB}+N_{BA}) \, Tr\, \Bigl\{ J^\mu
\tau^\dagger_A J_\mu\tau_B \Bigr\}
\nonumber \\&&
= (N_{AB}+N_{BA}) \biggl(\frac{1}{2}
J_a^\mu J_{a\mu} \delta_{AB}
+
\delta_{Ax}\epsilon_{xyz}J_{y\mu} \, Tr\, \Bigl\{J^\mu \tau_z
\tau_B \Bigr\}
\biggr)
\nonumber \\&&
= (N_{AB}+N_{BA}) \biggl(\frac{1}{2}
J_a^\mu J_{a\mu} \delta_{AB}
-
\delta_{Ax}\epsilon_{xyz}J_{y\mu} \epsilon_{y' z x'}\delta_{x'B}\, J_{y'}^\mu
\biggr)
\nonumber \\&&
= (N_{AB}+N_{BA}) \biggl(\frac{1}{2}
J_a^\mu J_{a\mu} \delta_{AB}
+
\delta_{Ax}J_{y\mu}(\delta_{xy'}\delta_{yx'}-\delta_{xx'}\delta_{yy'}) \delta_{x'B}\, J_{y'}^\mu
\biggr)
\nonumber \\&&
=
N_{AA}J_a^\mu J_{a\mu} + 2 N_{ab} J_{a\mu} J_{b}^\mu - 2 N_{bb}  J_a^\mu J_{a\mu}
\nonumber \\&&
=
(N_{00}-N_{bb} )J_a^\mu J_{a\mu} + 2 N_{ab} J_{a\mu} J_{b}^\mu .
\label{alg.9.3}
\end{eqnarray}
\par
In Appendix \ref{app:eva} we evaluate the coefficients of the
invariants listed in eq. (\ref{alg.9}).
Finally the tadpoles, originating from the mass term,
necessitate the following counterterms at one-loop, that add to those
for the massless nonlinear sigma model in eq. (\ref{alg.7.1})
\begin{eqnarray}
\widehat \G^{(1)}\Bigr|_{\rm TADPOLES}
&= &\frac{1}{D-4} \frac{\Lambda^{D-4}}{(4\pi)^2}\Big [
-2\frac{m^2}{v^2}   {\cal I}_{8}
+2 \frac{m^2}{v^2}{\cal I}_{11}-\frac{1}{v^4}   {\cal I}_{12}
+8 \frac{1}{v^4} {\cal I}_{13}
\nonumber \\& &
+8 \frac{1}{v^4}  {\cal I}_{14}
+\frac{2}{v^2} {\cal I}_{16}-2 \frac{1}{v^4}   {\cal I}_{17}
+\frac{1}{v^4}   {\cal I}_{18}-\frac{1}{2v^2} {\cal I}_{19}
-\frac{1}{2v^2} {\cal I}_{20}\Big ].
\label{alg.10.2.18}
\end{eqnarray}
From the expression in eqs. (\ref{alg.7.1}) and (\ref{alg.10.2.18}) one can get
all the one-loop counterterms by taking the relevant functional
derivatives. As an example one gets
\begin{eqnarray}&&
\widehat \G^{(1)}_{\phi_a\phi_b}
= \delta_{ab}\frac{1}{D-4} \frac{\Lambda^{D-4}}{(4\pi)^2}\Big [
-2  m^4
+ 2  m^4
-8 m^4
-8  m^2(p^2 {-m^2})
\nonumber \\& &
+8 m^2  p^2
+2  m^2(p^2 {-m^2})
-2m^2  p^2
-2m^2  p^2\Big ]
\nonumber \\& &
= \delta_{ab}\frac{1}{D-4} \frac{\Lambda^{D-4}}{(4\pi)^2}\Big [
- 2  m^2 (p^2 {+m^2})\Big ].
\label{alg.10.2.19}
\end{eqnarray}
which agrees with the direct calculation in eq. (\ref{h2.4.2.5}).

\section{Conclusions}
\label{sec:conc}
The subtraction strategy, recently developed for
the nonlinear sigma model and for the nonabelian gauge theories,
is implemented here for the {\sl massive} nonlinear sigma model
(without fermions in this work).
In the present paper the technique has been applied to the
simple case of $SU(2)\otimes SU(2)$. The extension to other groups
of transformations
is expected to be straightforward.
\par
The main tool is the LFE for the
effective action, derived from the invariance properties of the
path integral measure. The presence of a mass term requires
the introduction of more  sources coupled to additional composite
operators. However this fact brings only to a more complex algebra,
without diminishing the power of the LFE. The hierarchy still
works so that all 1PI-amplitudes with external  field-parameters
$\vec\phi$ can be derived from those with only composite operators.
The hierarchy allows to organize, at every order of the loop
expansion, the infinite set of divergent amplitudes, so that their
divergent parts can be expressed in term of a finite number
of divergent amplitudes.
The subtraction algorithm exploits
this powerful property and it is based on the dimensional regularization
and on the subtraction of the {\sl sole} pole parts (no finite adjustments
are allowed). The WPC is shown to be stable under the subtraction
procedure. The linearized  LFE suggests the use of powerful
{\sl local} $SU(2)$ symmetry properties (gauge-type) in order to study the
form of the counterterms. The bleaching technique is very
useful since it maps all external source monomials into
invariant quantities, hence very handy objects for the final aim:
the construction of the counterterms.
\par
The resulting perturbative expansion yields amplitudes
that depend on the mass $m$, on the Spontaneous Symmetry Breaking
parameter $v$ and on the scale of the radiative corrections $\Lambda$.

\section*{Acknowledgments}
The author is pleased to thank the Center for Theoretical Physics at MIT,
Massachusetts, where he had the possibility to work on this research.

redpajama_set_name: RedPajamaArXiv


The Rev. Franklin Graham says just mentioning the name Jesus Christ in the public square is increasingly frowned upon and warns: "The spirit of anti-Christ is everywhere."
The Samaritan's Purse founder and son of beloved evangelist Dr. Billy Graham voiced his dismay at the rapid secularization of society during a Tuesday chapel service at John Brown University, a private Christian university based in Siloam Springs, Ark.
The president and CEO of the Billy Graham Evangelistic Association criticized the recent memorial service for the Tucson shooting victims.
Unlike the memorials held after the Oklahoma City bombing and 9/11, he said, the Tucson program did not include an official prayer or mention of God. Graham noted gratefully, however, that President Barack Obama did quote scripture from the biblical book of Job.
Earlier this week, Graham penned an op-ed in the Washington Times saying it was "a shame" that the University of Arizona had failed to put God anywhere on its program.
Graham told the university students what was missing from the memorial: "There was no call for God to put His loving arms around those who were hurting," said Graham. "Why did they leave him out? They scoff at the name of Jesus Christ."
Graham predicted the persecution of Christians will get worse in coming years, and he encouraged students to use the new media and the Internet to spread the gospel.
"I'm going to build an army of young people," Graham said, according to ChristianPost.com. "We're going to find out how do you infiltrate every language of the world with the Gospel of Jesus Christ and do it on the Internet."


text:
Medjugorje phenomenon
About Parish
Glasnik mira
Bishops in Medjugorje duringthe year 2005
During 2005, 15 bishops from 12 countries came to Medjugorje:
- Mons. Hieronymus Herculanus Bumbun, O.F.M. Cap., Archbishop of Pontianak (Indonesia)
- Mons. Salvatore Boccaccio, Bishop of Frosinone (Italy)
- Mons. Geevarghese Mar Divannasios Ottathengil, syro-malankar Bishop of Bathery (Kerala, India)
- Mons. José Luís Azcona Hermoso, O.A.R., Augustinian, Bishop of Marajó (Para, Brazil)
- Mons. Bruno Tommasi, retired Archbishop of Lucca (Italy)
- Mons. Hermann Raich, S.V.D., Bishop of Wabag (Papua New Guinea)
- Mons. Issam John Darwish, BSO, DD., Eparch of the Melkite Catholic Church in Australia and New Zealand
- Mons. Domenico Pecile, retired Bishop of Latina (Italy)
- Mons José Domingo Ulloa Mendieta, O.S.A., Auxiliary Bishop of Panama (Panama)
- Mons. Stanislas Lukumwena Lumbala, O.F.M., Bishop of Kole (Democratic Republic of Congo)
- Mons. Pearce Lacey, retired Auxiliary Bishop of Toronto (Canada)
- Mons. Thomas Msusa, Bishop of Zomba (Malawi)
- Mons. Allan Chamgwera, retired Bishop of Zomba (Malawi)
- Mons. Remi Joseph Gustave Saint-Marie, Bishop of Dedza (Malawi)
- Mons. Joseph Faber MacDonald, Bishop of Saint John (New Brunswick, Canada)
Izrada i održavanje: NOVENA HRV | ENG | DEU | FRA | ITA | POL | ESP | MAG | العربية | © Sva prava pridržana: Udruga Međugorje - MIR, Split;1995. - 2021.

redpajama_set_name: RedPajamaCommonCrawl


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text:
namespace Ravelin.Models
{
	public class Venue
	{
		/// <summary>
		/// Name of the venue.
		/// </summary>
		public string Name { get; set; }

		/// <summary>
		/// Location of the venue.
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		public Location Location { get; set; }
	}
}

redpajama_set_name: RedPajamaGithub


text:
HomeMainItaly launches antibody tests for virus immunity
Italy launches antibody tests for virus immunity
Italy began conducting antibody tests in the northern region of Lombardy on Thursday, seeking information about coronavirus immunity to help guide authorities as they reopen the long locked-down country.
Lombardy, the region hardest-hit by the coronavirus crisis in Europe's worst-affected country, is betting that the science about "herd immunity" derived from the blood tests will help the prosperous industrial region return to work faster and safer.
Nearly 13,000 people have already died of the virus in densely populated Lombardy, whose capital is Milan — more than half of Italy's total dead.
Although Germany has already started nationwide antibody tests and countries such as Finland and Britain have announced plans to roll them out, many questions remain about how reliable data derived from the tests will be.
Health authorities have said 20,000 tests would be performed every day in Lombardy. First to be tested are those in the worst-hit provinces: health workers, those under quarantine showing coronavirus symptoms and those they have been in contact with, as well as others with mild symptoms.
Authorities hope to roll out the tests to the wider region after April 29.
The head of Italy's National Health Council, Franco Locatelli, said last month that antibody tests would help authorities determine the spread of the coronavirus.
Data would also provide "very relevant information on herd immunity" which would useful in developing strategies to help restart the country, he said, such as who could be allowed to go back to work.
The kits, made by Italian biotech firm DiaSorin, look for the presence of antibodies in the blood. Such antibodies indicate that the person has been exposed to the virus, pointing to some level of immunity.
They differ from the more common swab tests, which test molecules from nasal secretions to determine whether a person currently has the virus.
Lombardy's swab testing has revealed that 24 percent of those tested have the virus.
Immunity to the virus is little understood and hopes about its efficacy possibly exaggerated.
Lacking data, virologists and epidemiologists must extrapolate information from past coronaviruses, such as the SARS outbreak in 2002-2003, to make predictions.
Experts believe at least 60 to 70 percent of a population must be immune to the virus in order to gradually wipe it out.
But recent studies, such as one conducted in March and April by France's Institut Pasteur, have found that so-called "herd immunity" was harder to attain than believed.
At a high school in the Oise department of France, site of one of the country's first outbreaks, researchers found only 26 percent of students, teachers and their families carried antibodies.
Moreover, it is not known for how long immunity to coronavirus lasts, meaning there is a risk those deemed "immune" may be re-infected and pass along the virus to others.
"There's no guarantee that these antibodies protect from a new infection. We can only hope so for the moment. We'll know in the future," Guido Marinoni, president of Bergamo's surgical and dental association, told AFP on Thursday.
Marinoni, who organised local testing in the badly affected province of Bergamo and plans to study results, also cautioned that a lack of antibodies might mean the disease was still its early stages and antibodies had not kicked in yet.
Even more risky, a person who has developed antibodies can still carry traces of the virus, and be contagious. Therefore, experts such as Italy's Locatelli say antibody tests should be accompanied by swab testing.
Immunologist Jean-Francois Delfraissy, who heads France's scientific council formed to fight coronavirus, said many doubts remain.
"We're currently asking the question whether someone who has had COVID-19… is as protected as we think," he said.
Scientists must wait until more reliable data is available, said Saad Omer, director of the Yale Institute for Global Health.
"It's too premature," Omer told AFP. "We should be able to get clearer data very quickly — in a couple of months — when there will be reliable antibody tests with sensitivity and specificity."
Moody's projects Pakistan's first annual recession in 2020 over pandemic
Time to heed health experts' advice on Covid-19 measures: Bilawal


text:
Ionic conductivity studies of poly(vinyl alcohol) alkaline solid polymer electrolyte and its use in nickel–zinc cells
Mohamad , A. A. and Mohamed, N. S. and Yahya, M. Z. A. and Othman, Raihan and Ramesh, S. and Alias, Y. and Arof, A. K. (2003) Ionic conductivity studies of poly(vinyl alcohol) alkaline solid polymer electrolyte and its use in nickel–zinc cells. Solid State Ionics, 156 (1-2). pp. 171-177. ISSN 0167-2738
PDF (Ionic conductivity studies of poly(vinyl)) - Published Version
Official URL: http://dx.doi.org/10.1016/S0167-2738(02)00617-3
X-ray diffraction (XRD) pattern reveals that potassium hydroxide (KOH) disrupts the crystalline nature of poly(vinyl alcohol) (PVA)-based polymer electrolytes and converts them into an amorphous phase. The PVA–KOH alkaline solid polymer electrolyte (ASPE) system with PVA/KOH wt.% ratio of 60:40 exhibits the highest room temperature ionic conductivity of 8.5�10 exp(-4) S/cm. This electrolyte was used in the fabrication of a nickel–zinc (Ni–Zn) cell. The cell was charged at a constant current of 10 mA for 1 h providing it with 1.6 V. The cell was cycled 100 times. At the end of the last cycle, the cell still contained a capacity of 5.5 mA h.
poly(vinyl alcohol), potassium hydroxide, alkaline solid polymer electrolyte, nickel–zinc cell
T Technology > TJ Mechanical engineering and machinery > TJ807 Renewable energy sources
T Technology > TK Electrical engineering. Electronics Nuclear engineering > TK2896 Production of electricity by direct energy conversion
T Technology > TP Chemical technology > TP250 Industrial electrochemistry
Kulliyyah of Engineering > Department of Science
Raihan Othman


text:
#ifndef _LINUX_MIGRATE_H
#define _LINUX_MIGRATE_H

#include <linux/mm.h>

typedef struct page *new_page_t(struct page *, unsigned long private, int **);

/* Check if a vma is migratable */
static inline int vma_migratable(struct vm_area_struct *vma)
{
	if (vma->vm_flags & (VM_IO|VM_HUGETLB|VM_PFNMAP|VM_RESERVED))
		return 0;
	return 1;
}

#ifdef CONFIG_MIGRATION
extern int isolate_lru_page(struct page *p, struct list_head *pagelist);
extern int putback_lru_pages(struct list_head *l);
extern int migrate_page(struct address_space *,
			struct page *, struct page *);
extern int migrate_pages(struct list_head *l, new_page_t x, unsigned long);

extern int fail_migrate_page(struct address_space *,
			struct page *, struct page *);

extern int migrate_prep(void);
extern int migrate_vmas(struct mm_struct *mm,
		const nodemask_t *from, const nodemask_t *to,
		unsigned long flags);
#else

static inline int isolate_lru_page(struct page *p, struct list_head *list)
					{ return -ENOSYS; }
static inline int putback_lru_pages(struct list_head *l) { return 0; }
static inline int migrate_pages(struct list_head *l, new_page_t x,
		unsigned long private) { return -ENOSYS; }

static inline int migrate_pages_to(struct list_head *pagelist,
			struct vm_area_struct *vma, int dest) { return 0; }

static inline int migrate_prep(void) { return -ENOSYS; }

static inline int migrate_vmas(struct mm_struct *mm,
		const nodemask_t *from, const nodemask_t *to,
		unsigned long flags)
{
	return -ENOSYS;
}

/* Possible settings for the migrate_page() method in address_operations */
#define migrate_page NULL
#define fail_migrate_page NULL

#endif /* CONFIG_MIGRATION */
#endif /* _LINUX_MIGRATE_H */


This failure comes on the heels of another abysmal vote in which the package put together by Senators Ted Cruz (R-TX) and Rob Portman (R-OH) to allow insurance companies the freedom to sell non-Obamacare compliant plans coupled with an increase in state funding was also voted down.
This refusal to make good on the Republican establishment's promises should really surprise no one, especially after two of its architects, former Speaker of the House John Boehner and former House Majority Whip Eric Cantor admitted that the promise to repeal Obamacare was a lie.
But now the Piper whose music created that fantasy must be paid.
A week ago we observed that this vote would be the last chance Capitol Hill's establishment Republicans had to fulfill their promise to repeal Obamacare and get right with the GOP's conservative majority.
Now voters have watched this circus for six months and come to realize that there is no excuse for not passing the agenda President Trump and most Republicans in Congress campaigned on.
And with that realization has come recognition of another undeniable truth: The only sure path to repeal Obamacare, reform the tax code, rebuild our economy and build the border wall lies through the Republican Primary Elections.
So far, the battle for the repealing Obamacare and the rest of the Trump agenda has been conducted on the basis of a polite family disagreement. President Trump and his supporters have negotiated with the nasty establishment Republican children in the family and assumed that politeness (and perhaps some pork barrel-type bribery) would eventually bring them around, but it hasn't worked.
And the reason the family disagreement strategy hasn't worked is because the Susan Collins, Lisa Murkowski, John Kasich-type "Republicans" aren't, and don't want to be, in the same family with us conservatives – they are on the other side.
They don't want to liberate their fellow citizens from the chains of Obamacare and government control of health care and health insurance – they want to keep the power to tell other people what to do firmly in their own hands.
The good news in all of this is that government by show vote is over.
*Senators Heller, Capito, McCain, Portman, Alexander and Murkowski all were NO votes on the repeal plan. All six Senators voted YES back in 2015.
Submitted by email_registrat... on Thu, 07/27/2017 - 15:27.
Senator's Heller, Capito, Portman, McCain, Alaxander, and Merkowski, all need to go. The showed their colors in voting for this bill in 2015 but would not vote for it this time. Byebye to each of them the next time around. Not only that no money for Republicans until the get on our side, that's the voters side.
Submitted by McFerguson on Thu, 07/27/2017 - 14:32.
This was an incredibly revealing article. And I get it now. The lying bastards played us for a bunch of fools and these six liars, Alexander, Capito, Heller, McCain, Murkowski and Portman need to be primaried if/when they run again. And not to be cruel, but McCain probably won't be among the group seeking re-election because of other issues. Bye, bye GOP it was nice knowing ya.
Submitted by Daniel Spickard on Thu, 07/27/2017 - 14:12.
We had a health-care before Obama-care, and we'll still have one after Obama-care! But if Obama-care implodes in on itself, it's going to take the whole health care industry with it! Then the only thing that is going to be left is the single payer system in which the federal government dictates what health-care you get! Let me remind you that both Russia and China have tried this and almost went bankrupt in the process! Single payer systems only work for smaller countries, and countries with small populations! This is just another way for the federal government to control your life! Just imagine the next time you go to see your doctor and they give you what you think is another shot, but all the time it's a tracking chip! This is what if your read the Bible is the beginning of Armageddon! This would be the equivalent of the government telling you whether you live or die, or the 666, or a bar-code to identify you!
Submitted by email_registrat... on Thu, 07/27/2017 - 12:37.
The insider/establishment GOP has been killing the Conservative Republicans since 1988 when an "Up-Easter Republican" George H.W. Bush became President. The entire Congress is nothing short of a "good Ole Boy's club" which is fixed on making themselves rich and powerful and they have found all of the support they need from the Far-left Democrat Party. As Mr. Franklin once said....."We are an Republic, if you can hold it", and guess what, our Congress had done yeoman work to kill it!


The Dash Road Cam HD will be a witness to your driving experience. Easy to mount and starts recording when you start driving, this camera offers several modes of recording. Included is an 4GB SD card to easily transfer videos to your computer. Wide angle camera catches all the action. Sharp 2.7" display allows for instant playback. The set up is simple and has a rechargeable battery that charges via an available 12v socket in your car. Free tech support and includes a 1 year warranty.


text:
A TRAITOR'S LOYALTY
A Novel in the Tradition of the New York Times Best-Seller, Fatherland.
It's 1971 and Hitler has just died . . .
Author: Ian C. Racey
Genres: Thriller, Fiction
Publisher: Ignition Books
Vantage Point Books
Trade Paperback (May 2012)
Ignition Books
eBook & Trade Paperback (July 2013)
In the thrilling tradition of such national best-sellers as SS-GB, Fatherland, and Los Alamos, comes a first novel wherein history's path has taken a different turn. . . .
Twenty-five years have passed since the German victory in World War II. Hitler has just died, unleashing a conspiracy that threatens the future of the world . . .
Simon Quinn walked away from a brilliant career with MI6, the British Secret Intelligence Service, but now MI6 has blackmailed him into returning to Berlin. His mission: locate Richard Garner, a British spy who has disappeared and is suspected of defecting. He enlists the help of Ellie Voss, a Third Reich dissident who opposes Nazi rule and wishes for a different future for Germany.
Pursued by both the Gestapo and MI-6, Simon Quinn must choose, not between his country and treason, but between two brutal Nazi leaders battling for the succession: Reinhard Heydrich, the key architect of the Final Solution, and Heinrich Himmler, chief of the SS and Gestapo. For this British spy, it is a choice that will test even . . .
Ian C. Racey
Born in England, Ian C. Racey is an author of thrillers that he hopes recapture the excitement he felt as a boy, when his grandfather would tell him stories of bombing missions over occupied Europe. He studied many things at the University of Florida before eventually settling on Classics, an education that gave him a solid grounding to work as a bookseller for a national chain.
He and his wife live in Northern Virginia, where they're raising two children on condition at least one of them take care of them in their old age.
We are a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for us to earn fees by linking to Amazon.com and affiliated sites. We also participate in other affiliate advertising programs associated with other online booksellers. This means if you click on the link and purchase the item, The Zack Company, Inc. will receive an affiliate commission.


text:
Obama To Congress: 'Do The Right Thing,' Make Deal President Obama told Republicans that it's time to "rip off the Band-Aid" on a spending deal. He'll meet again Monday with congressional leaders in hopes of ending the impasse over the deficit, taxes and raising the debt ceiling.
Obama To Congress: 'Do The Right Thing,' Make Deal
Obama To Congress: 'Do The Right Thing,' Make Deal 30:19
Heard on Talk of the Nation
President Obama told Republicans that it's time to "rip off the Band-Aid" on a spending deal. He'll meet again Monday with congressional leaders in hopes of ending the impasse over the deficit, taxes and raising the debt ceiling.
Ron Elving, senior Washington editor, NPR
Matt Continetti, opinion editor, The Weekly Standard
Michael Ettlinger, vice president for economic policy, Center for American Progress
NEAL CONAN, host: This is TALK OF THE NATION. I'm Neal Conan in Washington. The president, the vice president and Congressional leaders are meeting in the White House this hour as the clock ticks down to default. This morning, the president declared he's willing to take significant political heat from his party to reach a deficit reduction deal and expects Republicans to do the same.
His idea of a balanced agreement includes cuts to Medicare and Social Security that Democrats would find hard to swallow and tax increases that Republicans utterly reject. The shared political pain would be worth it, the president argues, to reach a grand bargain to cut four trillion from the deficit over the next 10 years.
Over the weekend, John Boehner, the speaker of the House, said Republicans will not support such an agreement and called for a smaller deal, two trillion or so, but that agreement appears to be no closer than the big one.
The whole debate alienates the conservative wing in the GOP and the liberal flank of the Democratic Party. If you vote in your party's primary, what do you want your member of Congress to do? Give us a call, 800-989-8255. Email us, talk@npr.org. You can also join the conversation on our website. Go to npr.org. Click on TALK OF THE NATION.
Later in the program, media critic Howard Kurtz on the tactics of tabloid journalism, but first the debt ceiling. We're joined here in Studio 3A by NPR senior Washington editor Ron Elving. Nice to have you on the program as always, Ron.
RON ELVING: Good to be with you, Neal.
CONAN: And did the president say anything new today?
ELVING: The president did not say anything terribly new today. The newness about today would be the circumstances of the president's remarks. He is saying what he's been saying for some while, and John Boehner, who came out a little bit later and spoke for about - and took questions for about eight minutes, said much what he had said in the past.
The circumstance that has changed is that what appeared to be an effort toward that big deal, $4 trillion over 10 years, has apparently collapsed over the weekend, and John Boehner was explaining that in his view nothing had really changed. He had always said he was against any kind of tax increases, there may have been some misunderstanding, but that there were not votes in the House for the kind of deal that at least the White House's description implied was in negotiation between Obama and Boehner.
So we are hearing essentially things that we have heard many times. I'm sure the average listener straining to hear anything a little bit new is a little disappointed. But the truth is that what's changed is that the effort that was made during last week has apparently come to naught.
And there's some kabuki here. So I'm trying to add the word apparently. It seems to have come to naught, and we will see where we go from here. There had been hope we would get a deal this week, and they would be able to write something into law over the days between July 15 and August 2, but you know, it would be easier to make the case that we're further away from a deal than we were a week or two ago than to say we were closer.
CONAN: If you look at the congressional mathematics, there are some amongst the Republican majority in the House of Representatives who will not vote to increase the debt ceiling under almost any condition.
ELVING: Eighty to 100 is one estimate that I've heard. I don't know what the actual number is, and we - probably not a knowable number.
CONAN: So they will need some Democrats if that bill is going to go through, whatever it is. In the Senate, there are 53 Democrats. Anything important in the Senate needs 60 votes. So you're going to need at least seven Republicans in the United States Senate. We have divided government, as the president points out today. Nobody is going to get everything they want.
ELVING: No one would seem to be likely to get everything they want. But whichever of these two sides more successfully games the negotiations will get more of what it wants, is my guess. We've seen this before. We have seen many negotiations in the past between Republicans and Democrats, between the House and the Senate.
And by the way, those are four different groups of people in a very real sense. The House Republicans are not the Senate Republicans, et cetera. These negotiations wind up, when they come to the brink, with one side or the other yielding in some important way, and even if you perfectly nuance it and get everyone to give up equal amounts, the perception that then takes over, over time, can be quite surprising and often quite devastating.
Take the budget deal of 1990, which was thought to be a balanced deal by many at the time. President George H.W. Bush, the first President Bush, thought it was going to help him get re-elected to have put together such a great deal in the summer of 1990, even though there were a lot of House Republicans, especially, against it, and they voted it down once.
He thought it was really going to help him politically. A lot of Democrats thought it was going to help him politically, and in the end it brought on the challenge from Pat Buchanan, it brought on a big rebellion by a lot of anti-tax Republicans, probably had something to do with Ross Perot's strength in 1992, and labeled him the guy who violated his no-new-taxes pledge.
So the perception of that deal in 1990 turned out to be quite different from what the perceived reality at the moment of the deal had been.
CONAN: The one thing the president said everybody does agree on is the debt ceiling - at least the leadership - the debt ceiling does need to be raised. There seems to be these two diametrically opposed camps, or you can say both sides have painted themselves into opposite corners, and as we saw in a quote about the government shutdown in Minnesota, that every once in a while a game of chicken - well, they don't blink at the end and reach an agreement, it turns out to be a car crash.
ELVING: Yes, and we are seeing something of a car crash in Minnesota, but I think we can all observe, in the case in Minnesota, that not only have they had the car crash, but now that the car crash has happened, they continue to drive at each other.
They are both continuing to blame the other side, and attempts that have been made in some degree by bipartisan elements of the Minnesota political scene to broker some kind of compromise have not been successful and have not even really been taken very seriously.
So we have to look at the political situation in the larger context of the summer of 2011 and say we may not have the kind of politics we had in 1990, when we got a big deal. We may not have the kind of politics that we had in 1996 when President Clinton worked together with a Republican House, a Republican Senate, on a number of large bills, including the welfare overhaul.
We may not have the kind of politics we had for some of the budget deals and compromises we've seen over the last 10 years. So we're in a new world. We're in the post-election of 2010 world. And the government shutdown negotiations that kept the government open in March and April are not regarded kindly by many of the activists on the right.
Many Tea Party people feel they were snookered in that deal. They didn't get the budget cuts they wanted as fast as they wanted them, and they're bringing those concerns to this negotiation and holding John Boehner's feet to the fire.
CONAN: Some people on the left weren't happy with that either. They wanted those Bush tax cuts rescinded for the wealthiest among us too. In any case, let's bring a couple of other voices into the conversation. And let me introduce Matt Continetti. He's with us here in Studio 3A, opinion editor for the Weekly Standard. Always nice to have you on the program with us, Matt.
MATT CONTINETTI: Thanks for having me, Neal.
CONAN: And also with us here in Studio 3A is Michael Ettlinger. He's the vice president for economic policy at the Center for American Progress. And thanks for joining us.
MICHAEL ETTLINGER: Appreciate it.
CONAN: And just to start with Matt Continetti, was it a good idea for Speaker Boehner to seemingly embrace the idea of the grand bargain and then pull the plug on it over the weekend?
CONTINETTI: Well, like Ron referred earlier, there's a lot of shadow boxing and kabuki theater going on. And I think Boehner is open to a large deal, just as the president is, knowing that there are certain red lines that he won't cross, just like there are certain lines the president won't cross.
So you know, all these deals remember, we had the Gang of Six that failed, and we had the Biden talks, and Eric Cantor walked out of that, and then we have this new round of talks, and then all of a sudden the White House says no, we want...
CONAN: The Big Eight, yes.
CONTINETTI: Right, well - and remember, even these eight talks started out with a smaller deal, the $2.4 trillion deal. Then the White House says, well, we want the $4 trillion deal. Now it's back to the 2.4 trillion, and we might not even get any deal at all.
So until it actually, you know, comes down to the wire, I don't think any of us have really any idea when it's going to happen.
CONAN: Well, in any negotiation, nothing is final until everything is final. But what is - are taxes completely off the table?
CONTINETTI: Well, you look it from a Republican legislative perspective, right - if you're a House Republican or even a Senate Republican, you had the election of 2010, which was - Republicans ran on a small-government platform of, you know, keep taxes where they are and reduce the size of government. And they won big. In fact, they won a historic victory in terms of the history of the Republican Party.
So then you have in December of 2010, when a Democratic Congress and a Democratic president agreed to maintain the current tax rates through 2012. And so if you're a House Republican, why on earth would you say now, only six months, you know, seven months after that agreement: Well, you're right. We'll give them up. Especially considering - you know, Ron also mentioned the 1990 budget deal.
And you saw what happened there, where George Bush's broken promise ended up with a lot of - with a primary challenge. And all these Tea Party candidates, of course, many of them primaried Republicans who had voted for the last big bipartisan bill, which was the TARP in 2008. So that's a red line, taxes.
CONAN: And let's bring Michael Ettlinger into the conversation here. Are cuts to Social Security, Medicare, off the table for most Democrats?
ETTLINGER: Well, no. I mean, I think that, you know, there's a real distinction here, sort of, you know, this has been portrayed by a lot of people as, well, the Republicans are going to have to give on taxes, and the Democrats are going to have to give on entitlements, and they're both, you know, being absolutely rigid on that.
But in truth, you know, the president has signaled that he's willing to look at entitlements, including Social Security, which is probably the harder thing to touch. And yes, there are many Democrats who draw a hard line on Social Security. But you know, I think Medicare has already been touched in the Affordable Care Act, and in fact one of the things Matt didn't mention that many, particularly House Republicans, ran on was criticizing the president for cutting Medicare.
So you know, I actually think we sort of - there's one side who's being absolutely rigid and doctrinaire, to the point where, you know, closing loopholes on corporate jets are even off the table and things like that, whereas the other side is actually, you know, willing to engage and has some flexibility.
There's a line they won't cross, but...
CONAN: Former Speaker Pelosi, still the Democratic leader in the House, said no cuts to Medicare benefits.
ETTLINGER: Right, and she and, and - but there are ways to cut spending on Medicare without it being cuts in benefits. There's a lot of room within that.
CONAN: Well...
ETTLINGER: And she doesn't speak for every Democrat as well.
CONAN: Well, she's the leader, but we think that that means something. But do you see the prospect for compromise here?
ETTLINGER: I think - I mean, I think there's various paths this could take, as particularly Ron was saying. There's many different ways this could play out. I, you know, I think there is room for compromise. I think we will raise the debt limit. So by definition, if I think that, I think there's going to be some sort of compromise at it.
CONAN: Which will include taxes?
ETTLINGER: I think it will probably include some taxes. It depends on the scope of the deal. You know, I think there is a shorter-term deal that's possible without taxes. I think that would be a mistake, but I think that that's...
CONAN: Shorter-term, another - requiring another vote before the election?
ETTLINGER: Not necessarily. But also, more important than shorter term, a smaller deal. I mean, you can have a smaller deal that loads up this with less in terms of deficit reduction and just extends the debt limit, and that, that - you know, is much easier.
CONAN: The sound you hear is a can being kicked down the road. Stay with us. We're talking about the push for a deal in Washington to cut the deficit and raise the debt ceiling. If you vote in your party's primary, what do you want your member of Congress to do? Give us a call, 800-989-8255. Email us, talk@npr.org. Stay with us. I'm Neal Conan. It's the TALK OF THE NATION from NPR News.
(SOUNDBITE OF MUSIC)
CONAN: This is TALK OF THE NATION from NPR News. I'm Neal Conan. Before he headed to the White House for talks with President Obama this afternoon, House Speaker John Boehner told reporters that yes, he agrees with the president that the debt limit needs to be raised but also said the White House is not serious enough about overhauling Social Security and Medicare.
He also said the House Republicans will not back any deal that raises taxes. Speaker Boehner is pushing for a smaller, $2 trillion deal or thereabouts. President Obama wants something more along the lines of $4 trillion in savings over the next 10 years and promises to meet with leaders from both parties every day until they can come to agreement.
All of this as the Treasury warns the country is three weeks from default. If you vote in your party's primaries, what do you want your member of Congress to do? 800-989-8255. Email talk@npr.org. Our guests are Ron Elving, who is NPR senior Washington editor; Matt Continetti, opinion editor of the Weekly Standard; and Michael Ettlinger, the vice president for economic policy at the Center for American Progress. And let's start with Jane(ph), and Jane's with us from Morrison in Colorado.
JANE: Hi.
CONAN: Hi, Jane, go ahead please.
JANE: Yeah, I'm - I participate in our local caucus, and I vote in the primaries, as well. And it just seems to me that...
CONAN: In which party, Jane?
JANE: Well, I'm registered as Republican, and so I participate in that caucus and primary. I don't vote a strict party line, though. And I just really feel like the parties are fighting against each other and not necessarily for the American people.
I am fine with them raising taxes as long as we start paying for the things that we really need, like education and other things like that, that we're falling so far behind in this country in education and other areas. And it's time we step up and pay for those things.
CONAN: Thank you very much, Jane, appreciate the phone call. Matt Continetti, a Republican or Republican primary voter who says we need to raise taxes as long as they are directed towards the right things, and she emphasizes education. We're seeing education cuts being imposed all across the country.
CONTINETTI: And I would advise Jane to vote in the Democratic primary next cycle because the views that I heard from her are more in line with that. The Republican Party is going in exactly the opposite direction and has been since the financial crisis and the recession in 2008.
And they don't want to pay any more money. They think that the tax rates that we have now, we're getting about 14 percent GDP right now, which is...
CONAN: The lowest percentage, I think, since 1950 or something like that.
CONTINETTI: A lot of that is the economic recession, as well. The same rates in the middle of the decade were making about 18 percent. So they think it's just a question of spending, and they see a surge in spending, and they want it to stop. And that's where we have - that's where we're coming.
And this is why I'm very skeptical of a deal being reached is I just think the two parties are highly ideological, have become more ideological over the past few election cycles as a result of the cascading losses, first on the Republican side in two cycles, then on the Democratic side in the last cycle.
And they both have extremely different priorities about where the country should go in the future. And that makes me, like I say, skeptical that they can agree on an increase in the debt ceiling this time around.
CONAN: Let's go next to Steve(ph), Steve on the line from San Francisco.
STEVE: Hi, Neal, thanks for taking my call. My take, if you don't mind, I'll point to it real quickly, is that the compromise under Bush 1 may have been a loss of sorts looking at the '92 presidential election. Clinton did get elected.
But, you know, for a lot of reasons, I think a Republican agenda, you know, that included NAFTA, welfare reform, increasing privatization, you know, that all worked to the Republicans' advantage, as did the, you know, the real growth of a coalition that now dominates the Republican Party.
And I think that happened in part because the ideological purity resonates with the base, and...
CONAN: And which party do you vote in, Steve?
STEVE: I'm a Democrat, and I vote Democratic. I see that under Obama, pharma, banking, big ag, the military-industrial complex all have gotten their - they've all gotten a share of the pie. But the unions, the unemployed, the young, kids who need education, now that, you know, they're the ones who stand to benefit the most from either continued government investment or in particular growth in government investment, they're being told hey, there is no more.
And, you know, the Bush tax cuts have been extended. Jackie Speier, my rep, I'm going to ask her to stand with the progressive caucus, not give in. And if there's a default, you know, it's not, again, the Democratic base that stands to lose. You know, the strength of the...
CONAN: Doesn't everybody - I'm sorry, Steve, just listening to the economists, doesn't everybody stand to lose?
STEVE: I don't think so, I really don't. Maybe short-term, yeah, there'll be some losses, but unless we have a party that stops giving in to this ideological - I mean even David Brooks calls almost a religious belief that you can have a government without some kind of revenue-generating capacity.
I mean, I live in the state of California, and we're still trying to figure out how to get out from under a Prop 13 that, you know, is seen generally as some kind of an extension of grandmothers who would otherwise be taxed out of their homes but actually provides incredible tax incentives, you know, to commercial establishments who are paying tax rates based on property valuations that are, you know, almost 30 years old.
CONAN: Steve, I just wanted...
STEVE: You know, (unintelligible) some backbone.
CONAN: Okay, thank you very much, Steve, appreciate the phone call. And Michael Ettlinger, I think he speaks for some substantial part of the Democratic Party. Just last week, when President Obama started talking about the grand bargain, got emails from moveon.org saying there would be fewer contributions and fewer workers for any Democrat who might support such an idea.
ETTLINGER: Well, I think there's a lot of frustration out there because, you know, Matt was saying that both sides are ideological. But on the other hand, you know, you look at the Democratic side, honestly, you know, if saying that you're willing to touch Social Security, but you want to preserve it as a solid benefit based on its - the way it's run in the past, you want to keep Medicare even though you're willing to look at it for some savings - if that's an ideological position, they're ideological.
But I don't think of that as a strongly ideological position, and I think one of the things that sort of weakened the hand from a Democratic standpoint is you do have, you know, these 80, however many ideological Republicans who are saying: You know what? We'll let the debt limit - we'll keep the debt limit the same. We'll cause default. And we don't care because we're going to hold out for our things. But you don't see Democrats doing that...
CONAN: On the other hand, doesn't that give Democrats in the House power? In other words, John Boehner is going to need Democratic votes. He's going to have to get some Democrats aboard, thereby give them something they can vote for.
ETTLINGER: No, absolutely, that could - but I think what's happened is that the no-tax line in particular has infected beyond those 80 to 100 people. I mean, you've got 80 to 100 who won't vote for any debt limit, but even you have other people...
CONAN: Unless there's a balanced budget amendment with it, but that's not going to happen.
ETTLINGER: Exactly.
CONAN: Ron Elving, let me ask you: Matt Continetti was talking before about the new dynamic in the Republican Party, to use the verb primary, to be primaried, and these are challenges almost entirely from Tea Party-type candidates, this is not a Democratic Party phenomenon, not yet.
ELVING: Not yet, and that's partly I think what the caller was referring to a moment ago, when he was admiring, in a sense, the move of the Republican Party over the last 20 years, since the George H.W. Bush 1990 deal. And many people have said the same thing about the '64 defeat, the 1964 defeat of Barry Goldwater, that out of that defeat, which was devastating at the time and elected enormous majorities of Democrats in the Congress and LBJ, but things that happened over the next four years established the basis for not only Richard Nixon's presidency but later Ronald Reagan's presidency and the conservative movement in general, and that couldn't have happened without 1964 - maybe, maybe not. But a lot of people have made that observation. You could, I suppose, say something of the same thing about 1990.
It seems to me the caller is, in a sense, seeking a Democratic moment of truth. Maybe that would have to come from a disaster. I think that's the implication, that at least an electoral disaster would be necessary in which the Democrats lost everything, and presumably there were other consequences, as well, perhaps economic consequences from a default.
I think to some degree there is a minority within both parties that is not entirely unwilling to roll the dice and see, in the long run, which party would benefit if the two parties fail to come to the agreement that most conventional observers in politics and economics have always assumed would be done.
Look, I mean, since 1917, when this debt ceiling idea was established, and it's entirely voluntarily on the part of Congress to establish this debt ceiling, it has nothing to do with the market, it has to do only with the political will to raise more money through borrowing.
And since 1917, we've raised this thing almost 100 times. And most of the time it's not even been particularly noticed, and sometimes it's been politically controversial, other times it's been kind of swept under the rug. Sometimes it's been combined with other votes. But it's not been this big crisis, certainly not very often. And so now for this to come along at this particular moment I think illustrates that we're in a different moment in our politics and that people are willing to have this kind of Armageddon-like confrontation and risk, and maybe they're bluffing, maybe they're not - Matt seems to think maybe they're not bluffing - suggest that they're willing to drive over the cliff to see where we land.
CONAN: And, Matt Continetti, there are Republican presidential candidates, I think, perhaps with one exception, say don't raise the debt ceiling.
CONTINETTI: That's right, because they're catering to the grassroots of the party. And may I also point out, they're also entirely conforming to public opinion, which, as of the questioners at the president's news conference said today, you know, you look at these polls, and even when you introduce them to the question, you know, the risks involved that the economists say will happen if we don't raise the debt ceiling come August, the American people are still against raising the debt ceiling and...
CONAN: The American people are also overwhelmingly against cutting Medicare and Social Security.
CONTINETTI: What I'm saying is, politicians go where the signals are. And right now, Republican politicians running, you know, running in a presidential primary, have absolutely no incentive to call for raising the debt ceiling.
CONAN: Here's an email from Ben in Monmouth, Oregon. I'm a 30-year-old strong Democrat who votes in every primary. I volunteer and donate money to Democrats. Since my high school economics class, myself and every friend of mine has been warned not to count on Social Security, so we don't. I have no problem with Social Security being changed for my generation. I expect Congressman Schrader to support any change like this, but will not - it will not change my support for him in the primary as we are in a swing district.
So, there a very tactical voter. Why do you think so far, Michael Ettlinger, there have no significant numbers of challenges from the left to Democratic congressmen?
ETTLINGER: Well, I think there have been challenges to the left. I mean, people have certainly complained. These callers and emailers have done that. You know, I don't know. I mean, I think that - I think part of it is that a lot of the organization on the left has been, in the past, through unions. And this is an old story, but there was a point in time where, because in response to trade agreements, that the unions organized a big response against Democratic congressmen who'd supported traded agreements and that was seen as a big error because lots of Democrats lost, and they lost a lot of friends in Congress. So I think, at least among the organized, sort of, left of center community, I think there's some concern about that.
CONAN: We're talking with Michael Ettlinger, vice president for economic policy at the Center for American Progress; Matt Continetti, who's opinion editor at The Weekly Standard; and Ron Elving, NPR senior Washington editor. You're listening to TALK OF THE NATION from NPR News.
And let's go to Bruce, and Bruce on the line from Ann Arbor.
BRUCE: Hi.
CONAN: Hi.
BRUCE: Boy, you know, I just want to say I am so incredibly opposed to the president giving any additional concessions to the Republicans. The fact that he's even remotely considering additional cuts to any social program just puts me over the edge. I mean, I'm a life-long Democrat, and this is the kind of thing that would literally push me to consider a third party candidate. To me, what the Republicans are doing is not about creating a smaller government, it's about eliminating a budget of revenue so that they can finally get rid of the social programs that they've been against for decades.
CONAN: The Republicans control the House of Representatives. If there's going to be a new debt limit, a new debt ceiling, you're going to have to get their votes.
BRUCE: Not according to the 14th Amendment.
CONAN: Ah, 14th Amendment. Ron Elving, would you help us out on the 14th Amendment?
ELVING: The 14th Amendment, which is a result of the Civil War and the end of Reconstruction, has a section in it that says that no debt of the United States should be ignored. No debt of the United States should be repudiated. One of the political motivations at the time was that there was a fear that once the Southern states were represented again by people who may have been sympathetic with the Southern cause in the Civil War, that they might try to repudiate some of the debts for pensions. So this fourth section of the 14th Amendment specifically mentions including pensions, and they were concerned about the pensions for the Civil War veterans on the Northern side.
This has not been applied, in conventional thinking, or in recent years, by any of the people that I know who have had anything to do...
CONAN: Recent years, we're talking a century.
ELVING: We're talking since the 1917 debt ceiling in the context of a discussion of the debt ceiling. If the debt ceiling is going to have any meaning, if any of these things are going to have any meaning, it would appear that we're talking about apples and oranges. But there is this language in the 14th Amendment that would seem to indicate that no debt can be ignored.
What does that, then, practically mean for a president on August 2nd? We've already reached 14.3 trillion and some in borrowing. We need more borrowing to pay the bills of the government. Do we either not pay the obligations of the government to, among others, the VA, pensioners, if you will, of our day, of the Civil War veterans, their inheritors in our era? Or do we not pay for the continuing wars in Iraq and Afghanistan? Do we not pay the obligations that come due?
Even if you could, in some sense or another, bring this 14th Amendment language to bear in some sort of an intellectual argument, how do you actually pay any bills with it? And how do you force the appearance of money that doesn't otherwise exist on the basis of that language? It would have to be a matter taken to the courts, decided by the courts. That always takes time. It's always a very disputed process. This is an immediate crisis that needs a political solution.
CONAN: Bruce, thanks very much for the call. And looking for immediate solutions, in just a few seconds, Matt Continetti, you looking forward to that car crash?
CONTINETTI: No, not at all. That's why I'm going around saying pass the Zoloft. It's - when you look at the political situation in America today, the economic situation, the geopolitical situation, there is very little cause for cheer.
CONAN: And, Michael Ettlinger, do you think we're going to raise the debt ceiling?
ETTLINGER: I think we're going to raise the debt ceiling.
CONAN: And how are we going to do that?
ETTLINGER: They'll come up with a narrower set of things that they have to agree on. They'll be less ambitious on deficit reduction.
CONAN: Matt?
CONTINETTI: And if Michael is right, and he very well maybe, that means Obama will cave. And I think then you have a whole host of dynamics that Bruce alluded to, including a possible a primary challenge.
CONAN: Thank you very much for joining us today. I'm afraid this is a subject to which we may return before the next three weeks, or maybe four weeks, are up. Michael Ettlinger, of the Center for American Progress and Matt Continetti of The Weekly Standard, and as always, our thanks as well, to Ron Elving, NPR senior Washington editor, all of them here with us in Studio 3A.
Coming up, scandal brought down Britain's tabloid News of the World. Howard Kurtz argues that America's news media often succumbs to tabloid tactics as well. He'll join us on the Opinion Page, next. Stay with us. I'm Neal Conan. It's the TALK OF THE NATION from NPR News.
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A Federal High Court in Abuja has ordered the Economic and Financial Crimes Commission (EFCC), the Department of State Services (DSS) and other security agencies to arrest Diezani Alison-Madueke, the former minister of petroleum, within 72 hours.
According to The Cable, the order was issued by Justice Valentine Ashi based on a motion filed by the counsel to the EFCC, Msuur Denga.
Denga had earlier prayed the court to issue a warrant for the former minister's arrest to enable the anti-graft agency arraign her in court.
Mrs Alison-Madueke and Mr Omokore are also alleged to have accepted and given gifts in properties located at Penthouse 22, Block B, Admiralty Estate, Ikoyi, and Penthouse 21, Building 5, Block C, Banana Island, Lagos.


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