text:
\section{Introduction}

The origin of magnetic fields in the Universe at various scales is one of the mysteries in modern cosmology. There thus have been a large number of attempts to find a theoretical framework in which magnetic fields in the Universe are generated in the course of the cosmic history (see \cite{Widrow:2002ud,Kandus:2010nw,Durrer:2013pga,Subramanian:2015lua} for reviews). The author recently proposed a new scenario of inflationary magnetogenesis in which magnetic fields in the Universe at all scales may originate from a classical, homogeneous magnetic field sustained during inflation~\cite{Mukohyama:2016npi}. It was found that a $U(1)$ gauge theory nonminimally coupled to scalar-tensor gravity admits a cosmological attractor solution that represents a de Sitter universe with a homogeneous magnetic field, fully taking into account the backreaction of the magnetic field on the geometry.

In the standard Einstein-Maxwell theory, a homogeneous magnetic field would introduce anisotropies in the geometry through the stress-energy tensor, making it impossible to admit a de Sitter solution. One thus needs to modify the standard Einstein-Maxwell theory one way or another to make a homogeneous magnetic field and an isotropic expansion of the universe consistent with each other. In the solution found in \cite{Mukohyama:2016npi}, a homogeneous magnetic field has a preferred spatial direction but its stress-energy tensor is isotropic and is of the form of an effective cosmological constant. In this sense the magnetic field in the solution is ``stealth''.

The basic idea behind the model of \cite{Mukohyama:2016npi} is as follows. For simplicity let us first consider an action for a $U(1)$ gauge field $A_{\mu}$ of the form
\begin{equation}
 I_w = \int d^4x \sqrt{-g}\, f(w)\,, \quad w = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\,,
\end{equation}
where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is the field strength and $g_{\mu\nu}$ is the spacetime metric. By taking the variation of the action with respect to $g_{\mu\nu}$, one obtains the corresponding stress-energy tensor as
\begin{equation}
 T_w^{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta I_w}{\delta g_{\mu\nu}} = f'(w)F^{\mu}_{\ \rho}F^{\nu\rho} + f(w)g^{\mu\nu}\,.
\end{equation}
If $f'(w)=0$ then the stress-energy tensor is proportional to $g^{\mu\nu}$ and thus admits an exact de Sitter solution. For example, this would be the case if $f(w) = f_0 + f_2 (w-w_0)^2$ with $f_0$, $f_2$ and $w_0$ constant and if $w=w_0$. However, in Friedmann-Lema\^{i}tre-Robertson-Walker (FLRW) backgrounds (including the de Sitter spacetime), $w$ for a homogeneous magnetic field decays as $1/a^4$, where $a$ is the scale factor, and thus the only constant value of $w$ that is consistent with the expansion of the universe is zero. For this reason, this simple action does not work. Let us next consider an action of the form
\begin{equation}
 I_W = \int d^4x \sqrt{-g}\, f(W)\,, \quad W = -\frac{1}{4}e^{2\phi}F_{\mu\nu}F^{\mu\nu}\,,
  \label{eqn:simpleaction}
\end{equation}
where $\phi$ is a scalar field and it is understood that some kinetic terms for $\phi$ are also added to the action. In this case the corresponding stress-energy tensor is again proportional to $g^{\mu\nu}$ if $f'(W)=0$. The main difference from the previous case is that $W$ for a homogeneous magnetic field is proportional to $e^{2\phi}/a^4$ in FLRW backgrounds and thus is constant if $e^{\phi}\propto a^2$. This is the basic idea behind the model of \cite{Mukohyama:2016npi}.

The model action of \cite{Mukohyama:2016npi} however includes not only a nonlinear function of $W$ but also other terms such as the Horndeski's nonminimal vector coupling and the Horndeski scalar terms. This is because, while the simple action (\ref{eqn:simpleaction}) supplemented by kinetic terms for $\phi$ in principle admits a de Sitter solution with a homogeneous magnetic field, the stability of such a solution requires inclusion of additional terms in the action.

The purpose of the present paper is to show a detailed stability analysis for the solution representing a de Sitter spacetime with a stealth homogeneous magnetic field. In particular we study the stability of the solution against linear perturbations in the subhorizon and superhorizon limits. We show that the stability of the solution requires inclusion of the Horndeski's nonminimal vector coupling as well as the Horndeski scalar terms. We then present some explicit examples that satisfy all stability conditions.

The rest of the present paper is organized as follows. In section~\ref{sec:model} we briefly review the model and the solutions studied in \cite{Mukohyama:2016npi}, focusing on those without the electric field. In section~\ref{sec:linearperturbation} we formulate the linear perturbation analysis and study the stability of the de Sitter solution with the stealth magnetic field against subhorizon perturbations. In section~\ref{sec:longwavelengthperturbations} we then study the stability of the solution against superhorizon perturbations. In section~\ref{sec:examples}, after showing a couple of classes of models that are unstable, we present explicit models that satisfy all stability conditions. Section~\ref{sec:summary} is then devoted to a summary of the paper and some discussions. Appendices~\ref{app:submatrices}-\ref{app:calA123} show explicit expressions of some matrices and coefficients.

\section{Review of the model}
\label{sec:model}

In this section we briefly review the model and the de Sitter solution with a homogeneous magnetic field studied in \cite{Mukohyama:2016npi}.

\subsection{Action}

The model consists of a metric $g_{\mu\nu}$, a $U(1)$ gauge field $A_{\mu}$ and a scalar field $\phi$, described by the action
\begin{equation}
 I = \int d^4x \sqrt{-g}
  \left[ L + L_3 + L_4 + L_5 + L_{\rm H}\right]\,, \label{eqn:action}
\end{equation}
where $L=L(X,W,Y,Z)$ is an arbitrary function of
\begin{align}
 &
  X \equiv  -\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\,,\quad
  W \equiv  -\frac{1}{4}\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}\,,\nonumber\\
 &
  Y \equiv  \mathcal{F}_{\mu\nu}\tilde{\mathcal{F}}^{\mu\nu}\,,\quad
 Z \equiv  \mathcal{F}^{\rho\mu}\mathcal{F}_{\rho}^{\ \nu}\partial_{\mu}\phi\partial_{\nu}\phi\,;
\end{align}
$\mathcal{F}_{\mu\nu}$ and $\tilde{\mathcal{F}}^{\mu\nu}$ are defined by
\begin{align}
& \mathcal{F}_{\mu\nu} \equiv e^{\phi}F_{\mu\nu}\,,
  \quad
  \tilde{\mathcal{F}}^{\mu\nu} \equiv e^{\phi}\tilde{F}^{\mu\nu}\,, \nonumber\\
&
 F_{\mu\nu} \equiv \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}\,, \quad
  \tilde{F}^{\mu\nu} \equiv \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\,,
\end{align}
and $\epsilon^{0123}=-1/\sqrt{-g}$;
\begin{align}
 L_3 =& -G_3(X)\Box\phi\,, \nonumber\\
 L_4 =& G_4(X) R + G_{4X}(X)\left[(\Box\phi)^2-(\nabla^{\mu}\nabla_{\nu}\phi)(\nabla^{\nu}\nabla_{\mu}\phi)\right]\,,\nonumber\\
 L_5 =& G_5(X)G^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi
  - \frac{1}{6}G_{5X}(X)
  \left[ (\Box\phi)^3 \right.\nonumber\\
&  - 3(\Box\phi)(\nabla^{\mu}\nabla_{\nu}\phi)(\nabla^{\nu}\nabla_{\mu}\phi) \nonumber\\
& \left.+ 2(\nabla^{\mu}\nabla_{\nu}\phi)(\nabla^{\nu}\nabla_{\rho}\phi)(\nabla^{\rho}\nabla_{\mu}\phi) \right]\,,
\end{align}
are shift-symmetric Horndeski scalar terms~\cite{Horndeski:1974wa,Deffayet:2011gz}; and
\begin{equation}
 L_{\rm H} = \xi\tilde{\mathcal{F}}^{\mu\nu}\tilde{\mathcal{F}}^{\rho\sigma}R_{\mu\nu\rho\sigma}\,,
\end{equation}
is a simple modification ($\tilde{F}^{\mu\nu} \to \tilde{\mathcal{F}}^{\mu\nu}$) of the Horndeski's nonminimal coupling of the $U(1)$ gauge field to the Riemann tensor $R^{\mu}_{\ \nu\rho\sigma}$ of the metric $g_{\mu\nu}$~\cite{Horndeski:1976gi}. Here, the scalar field $\phi$ is normalized so that its mass dimension is zero, $G_{3,4,5}(X)$ are arbitrary functions of $X$, the subscript $X$ denotes derivative with respect to $X$, and $\xi$ is an arbitrary constant.

The action is diffeomorphism invariant and enjoys the $U(1)$ gauge symmetry
\begin{equation}
 A_{\mu} \to A_{\mu} + \partial_{\mu} \lambda\,,
\end{equation}
where $\lambda$ is an arbitrary function. Furthermore, the action respects the global symmetry
\begin{equation}
 \phi \to \phi + \phi_0\,, \quad A_{\mu} \to e^{-\phi_0} A_{\mu}\,,
\end{equation}
where $\phi_0$ is an arbitrary constant. We assume that the function $L(X,W,Y,Z)$ is even with respect to $Y$
\begin{equation}
 L(X,W,Y,Z) = L(X,W,-Y,Z)\,. \label{eqn:Parity}
\end{equation}
This assumption ensures that the equations of motion admit a solution without the electric field~\cite{Mukohyama:2016npi}.

\subsection{Bianchi I solution with magnetic field}
\label{subsec:ansatz}

We first consider a Bianchi I spacetime
\begin{align}
g_{\mu\nu}dx^{\mu}dx^{\nu} =& -N(t)^2dt^2 + a(t)^2
  \left[ e^{4\sigma(t)}dx^2\right. \nonumber\\
& \left. + e^{-2\sigma(t)}(dy^2+dz^2)\right]\,,
\end{align}
with
\begin{equation}
 \frac{\dot{a}}{Na} = const. \equiv H_0\,, \quad  \frac{\dot{\sigma}}{N} = const. \equiv \Sigma_0\,,
\end{equation}
where an over-dot represents derivative with respect to $t$. We further assume that the scalar field is homogeneous, $\phi = \phi(t)$, and has a constant ``velocity'' as
\begin{equation}
 \frac{\dot{\phi}}{N} = const. \,,
\end{equation}
and that the $U(1)$ gauge field represents a homogeneous magnetic field as
\begin{equation}
 A_t = A_x = 0\,,\quad
 A_y = \frac{1}{2}Bz\,,\quad  A_z = -\frac{1}{2}By\,,  \label{eqn:ansatz-Amu}
\end{equation}
where $B$ is a constant.

Assuming that not only $H_0$, $\Sigma_0$, $B$ and $\dot{\phi}/N$ but also $\chi\equiv e^{\phi}/a^2$ are constant (see the sentences just after (\ref{eqn:simpleaction}) for the motivation for this extra assumption) and rescaling the spatial coordinates so that $\chi=1$, the equations of motion are reduced to
\begin{equation}
 0 = bL_Y\,, \label{eqn:LY=0}
\end{equation}
and
\begin{align}
 0 =& 16(1-s)^6H_0^6G_{5X}-6G_4(1-s)^2 H_0^2 - L\nonumber\\
 &  + 4\left[4\xi b^2+6(1-s)^2G_{4X}\right](1-s)^2H_0^4\,, \nonumber\\
 0 =& 8G_{5XX}(1+2s)(1-s)^6H_0^6\nonumber\\
 & + 6(1-s)^4\left[ 4(1+s)G_{4XX}+(1+4s)G_{5X}\right]H_0^4\nonumber\\
 & + 2\left\{-(1-4s)\xi b^2 + 3(1-s)^2[G_{3X} \right.\nonumber\\
 & \left.+(1+3s)G_{4X}]\right\}H_0^2-3sG_4 + (1-s)L_X \,,\nonumber\\
 0 =& 4\left[18s (1-s)^2G_{4X} -\xi(5-4s+8s^2)b^2 \right]H_0^2 \nonumber\\
 & + 72G_{5X}(1-s)^4sH_0^4 -b^2L_W - 18sG_4 \,, \label{eqn-Bianchi}
\end{align}
where $b\equiv B/H_0$ and $s\equiv\Sigma_0/H_0$. Hereafter, subscripts $W$, $Y$ and $Z$ represent derivatives with respect to them. Since $Y=0$ for the ansatz (\ref{eqn:ansatz-Amu}) and we have assumed (\ref{eqn:Parity}), the first equation (\ref{eqn:LY=0}) is automatically satisfied. Therefore we have three algebraic equations (\ref{eqn-Bianchi}) to be solved with respect to the three parameters ($H_0$, $b$, $s$) of the ansatz. Generically they admit a solution (or a set of solutions).

\subsection{de Sitter solution with magnetic field}
\label{subsec:dSnoE}

By fine-tuning one of parameters in the action, one can then take the limit $s\to 0$ so that the spacetime becomes a de Sitter,
\begin{equation}
g_{\mu\nu}dx^{\mu}dx^{\nu} = -N(t)^2dt^2 + a(t)^2(dx^2+dy^2+dz^2)\,.
\end{equation}
Then, under the assumption (\ref{eqn:Parity}), the independent equations of motion are
\begin{align}
0 =&   16G_{5X}H_0^6 + 8(2\xi b^2+3G_{4X})H_0^4 -6G_4H_0^2 - L\,,\nonumber\\
 0 =&   8G_{5XX}H_0^6 + 6(4G_{4XX}+G_{5X})H_0^4 \nonumber\\
 & + 2\left[ -\xi b^2 + 3(G_{3X}+G_{4X})\right]H_0^2+ L_X\,,\nonumber\\
0 =&   20\xi H_0^2 + L_W\,. \label{eqn:dSnoE}
\end{align}
These are three algebraic equations. One of the three equations (or one combination of them) simply represents the fine-tuning of one of parameters in the action and the remaining two equations can generically be solved with respect to the two parameters ($H_0$, $b$) of the ansatz.

If we relax the fine-tuning without abandoning the discrete symmetry (\ref{eqn:Parity}) then the solution goes back to the Bianchi I solution with the three parameters ($H_0$, $b$, $s$) described in the previous subsection.

\subsection{Attractor condition}
\label{subsec:attractor}

Ref.~\cite{Mukohyama:2016npi} found the necessary and sufficient condition under which the de Sitter solution with a homogeneous magnetic field introduced in subsection~\ref{subsec:dSnoE} is stable against homogeneous linear perturbations, under the assumption (\ref{eqn:Parity}). The stability condition, or the attractor condition, is $\mathcal{A}/\mathcal{N}>0$, where
\begin{align}
 \mathcal{N} =& 2\zeta_3g_h(\zeta_3-8\zeta_1)b^2+\zeta_1(\zeta_1\zeta_2+3\zeta_3^2)\,,\nonumber\\
 \mathcal{A} =& 56b^6g_h^3-4(9\zeta_1+\zeta_2+15\zeta_3)g_h^2b^4-2g_h\zeta_4(\zeta_1-\zeta_3)b^3\nonumber\\
 &
  +\left[6(-\zeta_1^2+\zeta_1\zeta_2+2\zeta_1\zeta_3+2\zeta_3^2)g_h\right.\nonumber\\
 & \left.+\zeta_5(\zeta_1-\zeta_3)^2\right]b^2+\frac{3}{2}\zeta_1\zeta_4(\zeta_1-\zeta_3)b \,,
 \label{eqn:def-calN-calA}
\end{align}
$g_h$ and $\zeta_i$ ($i=1,\cdots 5$) are constants defined by
\begin{align}
 g_h =& \xi H_0^2/M_{\rm Pl}^2\,, \quad
  \zeta_1 = 2b^2g_h+g_4-4g_{4x}-4g_{5x}\,,\nonumber\\
 \zeta_2 =& 2b^2g_h+6g_{3x}+24g_{3xx}+72g_{4xx}+96g_{4xxx}\nonumber\\
& +6g_{5x}+48g_{5xx}+32g_{5xxx}+4l_{xx}  \,,\nonumber\\
 \zeta_3 =& 4b^2g_h+2g_{3x}+4g_{4x}+16g_{4xx}+6g_{5x}+8g_{5xx}\,,\nonumber\\
 \zeta_4 =& -4(g_h+l_{xw})b  \,,\quad
  \zeta_5 = -12g_h-b^2l_{ww}\,, \label{eqn:def-gh-zeta}
\end{align}
and
\begin{align}
 &
  L_{XX} = l_{xx}\frac{M_{\rm Pl}^2}{H_0^2}\,, \
  L_{XW} = l_{xw}\frac{M_{\rm Pl}^2}{H_0^2}\,, \
  L_{WW} = l_{ww}\frac{M_{\rm Pl}^2}{H_0^2}\,, \nonumber\\
 &
  G_{3X} = g_{3x}\frac{M_{\rm Pl}^2}{H_0^2}\,, \
  G_{3XX} = g_{3xx}\frac{M_{\rm Pl}^2}{H_0^4}\,, \nonumber\\
 &
  G_4 = g_4M_{\rm Pl}^2\,, \
  G_{4X} = g_{4x}\frac{M_{\rm Pl}^2}{H_0^2}\,, \
  G_{4XX} = g_{4xx}\frac{M_{\rm Pl}^2}{H_0^4}\,, \nonumber\\
 &
  G_{4XXX} = g_{4xxx}\frac{M_{\rm Pl}^2}{H_0^6}\,,\
  G_{5X} = g_{5x}\frac{M_{\rm Pl}^2}{H_0^4}\,, \nonumber\\
 &
  G_{5XX} = g_{5xx}\frac{M_{\rm Pl}^2}{H_0^6}\,, \
  G_{5XXX} = g_{5xxx}\frac{M_{\rm Pl}^2}{H_0^8}\,. \label{eqn:dimensionlessparameters}
\end{align}
It is understood that the left hand sides of (\ref{eqn:dimensionlessparameters}) are evaluated at the de Sitter solution with a homogeneous magnetic field and thus are constant.

We shall later see that the absence of ghosts implies that $\mathcal{N}>0$. Therefore the attractor condition is
\begin{equation}
 \mathcal{A}>0\,. \label{eqn:attractor}
\end{equation}

\section{Linear perturbation}
\label{sec:linearperturbation}

 In this section we consider inhomogeneous linear perturbations around the de Sitter attractor solution with a homogeneous magnetic field ($H_0>0$, $s=0$, $b\ne 0$) and seek the no-ghost conditions and the sound speeds in the subhorizon limit.

 \subsection{Decomposition of linear perturbation}

 Around the general background introduced in subsection \ref{subsec:ansatz} we introduce the components $\Phi$, $V_i$ and $h_{ij}$ ($i,j=1,2,3$) of the metric perturbation $\delta g_{\mu\nu}$, the components $\delta A_a$ ($a=0,\cdots,3$) of the perturbation of the vector field $\delta A_{\mu}$ and the perturbation $\pi$ of the scalar field $\phi$ as
\begin{align}
&  \delta g_{\mu\nu}dx^{\mu}dx^{\nu} =
   -2\Phi\, e^0e^0 + V_i\, e^0e^i + V_i\, e^ie^0 + h_{ij}\, e^ie^j\,, \nonumber\\
  &  \delta A_{\mu}dx^{\mu} = \delta A_a e^a\,,\quad
   \delta\phi = \pi\,,
 \end{align}
 where
\begin{align}
& e^0 = N(t)dt\,, \quad e^1 = a(t) e^{2\sigma(t)}dx\,, \nonumber\\
&  e^2 = a(t) e^{-\sigma(t)}dy\,, \quad   e^3 = a(t) e^{-\sigma(t)}dz\,,
\end{align}
and the Einstein's summation rule is employed. We then decompose each component of the perturbation into spatial Fourier modes. For this purpose, we first introduce scalar harmonics $Y_n$ ($n = cc, cs, sc, ss$) as
\begin{align}
  Y_{cc} =& \cos(k_x x)\cos(k_y y + k_z z)\,,\nonumber\\
  Y_{cs} =& \cos(k_x x)\sin(k_y y + k_z z)\,,\nonumber\\
  Y_{sc} =& \sin(k_x x)\cos(k_y y + k_z z)\,,\nonumber\\
  Y_{ss} =& \sin(k_x x)\sin(k_y y + k_z z)\,,
\end{align}
and then define odd and even vector harmonics $Y^{\rm odd/even}_{n, p}$ ($n = cc, cs, sc, ss$; $p, q = 2, 3$) as
\begin{equation}
  Y^{\rm odd}_{n, p} = \epsilon_{pq}\frac{\partial}{\partial x^q}Y_n\,, \quad
  Y^{\rm even}_{n, p} =\frac{\partial}{\partial x^p}Y_n\,,
\end{equation}
where $x^2=y$, $x^3=z$, and $\epsilon_{pq}$ is the two-dimensional Levi-Civita symbol ($\epsilon_{23}=-\epsilon_{32}=1$ and $\epsilon_{22}=\epsilon_{33}=0$), and odd and even tensor harmonics $Y^{\rm odd/even}_{n, pq}$ ($n = cc, cs, sc, ss$; $p, q = 2, 3$) as
\begin{equation}
  Y^{\rm odd/even}_{n, pq} =\frac{1}{2}\left(\frac{\partial}{\partial x^p}Y^{\rm odd/even}_{n,q}+\frac{\partial}{\partial x^q} Y^{\rm odd/even}_{n,p}\right)\,.
\end{equation}
The components of the perturbations are then decomposed as
\begin{align}
& \pi = \int\frac{d^3k}{(2\pi)^3} \pi_n\, Y_n\,,\quad
 \Phi = \int\frac{d^3k}{(2\pi)^3} \Phi_n\, Y_n\,,\nonumber\\
&
  V_1 = \int\frac{d^3k}{(2\pi)^3} \chi_n\, Y_n\,,\quad
  h_{11} = \int\frac{d^3k}{(2\pi)^3}\psi_n\, Y_n\,,\nonumber\\
&
   \delta A_0 = \int\frac{d^3k}{(2\pi)^3} A_{0,n}\, Y_n\,,\quad
   \delta A_1 = \int\frac{d^3k}{(2\pi)^3} A_{1,n}\, Y_n\,,\nonumber\\
&  V_p = \int\frac{d^3k}{(2\pi)^3} (V^{\rm odd}_nY^{\rm odd}_{n,p}+V^{\rm even}_nY^{\rm even}_{n,p})\,,\nonumber\\
 &
   h_{1p} = h_{p1} = \int\frac{d^3k}{(2\pi)^3} (h^{\rm odd}_{1,n}Y^{\rm odd}_{n,p}+\beta_nY^{\rm even}_{n,p})\,,\nonumber\\
& \delta A_p = \int\frac{d^3k}{(2\pi)^3} (A^{\rm odd}_nY^{\rm odd}_{n,p}+A^{\rm even}_nY^{\rm even}_{n,p})\,, \nonumber\\
&
 h_{pq} = \int\frac{d^3k}{(2\pi)^3} (h^{\rm odd}_nY^{\rm odd}_{n,pq}+E_nY^{\rm even}_{n,pq}+\tau_nY_n\delta_{pq})\,,
\end{align}
where $p, q = 2, 3$ and the Einstein's summation rule for $n = cc, cs, sc, ss$ is understood.

We fix gauge freedom associated with the spacetime coordinate transformation and the $U(1)$ gauge transformation as
\begin{equation}
 \beta_n=h^{\rm odd}_n = E_n=\tau_n= 0\,,\quad
  A^{\rm even}_n=0\,.
\end{equation}

\subsection{General quadratic action}

Substituting the decomposition of perturbations to the action, imposing the gauge condition and expanding the action up to the quadratic order in perturbations, we obtain the quadratic action for the linear perturbations around a general background introduced in subsection \ref{subsec:ansatz} as
\begin{align}
 I^{(2)} = & \frac{1}{2} \int\frac{d^3k}{(2\pi)^3} \int Na^3dt
  \left[ \frac{1}{N^2}\dot{\bf Y}_n^{\intercal} {\bf K}\dot{\bf Y}_n
   + \frac{1}{N}\left(\dot{{\bf Y}}_n^{\intercal}{\bf M} {\bf Y}_n \right.\right.\nonumber\\
 & \left.
 + {\bf Y}_n^{\intercal}{\bf M}^{\intercal}\dot{{\bf Y}}_n\right)
 - {\bf Y}_n^{\intercal}{\bf V}{\bf Y}_n + \left({\bf Z}_n^{\intercal}{\bf A} {\bf Y}_n + {\bf Y}_n^{\intercal}{\bf A}^{\intercal} {\bf Z}_n\right) \nonumber\\
 & \left. + \frac{1}{N}\left({\bf Z}_n^{\intercal}{\bf B}\dot{{\bf Y}}_n+\dot{{\bf Y}}_n^{\intercal}{\bf B}^{\intercal}{\bf Z}_n\right) + {\bf Z}_n^{\intercal}{\bf C}{\bf Z}_n  \right]\,,
  \label{eqn:generalI2}
\end{align}
where the superscript ${\intercal}$ represents the transpose operation,
\begin{equation}
 {\bf Y}_n  =
  \left(
   \begin{array}{c}
    k_{\perp}a^2A^{\rm odd}_n \\
    k_{\perp}h^{\rm odd}_{1,n}\\
    a^2A_{1,n}\\
    \pi_n\\
    \psi_n
   \end{array}
		 \right),
\end{equation}
represent dynamical degrees of freedom,
\begin{align}
&
 {\bf Z}_{cc}  =
  \left(
   \begin{array}{c}
    k_{\perp}V^{\rm odd}_{sc}\\
    k_{\perp}V^{\rm even}_{cc}\\
    a^2A_{0,sc}\\
    \chi_{sc}\\
    \frac{a}{k_{\perp}}H_0\Phi_{cc}\\
   \end{array}
		 \right),\,
 {\bf Z}_{cs}  =
  \left(
   \begin{array}{c}
    k_{\perp}V^{\rm odd}_{ss}\\
    k_{\perp}V^{\rm even}_{cs}\\
    a^2A_{0,ss}\\
    \chi_{ss}\\
    \frac{a}{k_{\perp}}H_0\Phi_{cs}\\
   \end{array}
		 \right),\nonumber\\
&
 {\bf Z}_{sc}  =
  \left(
   \begin{array}{c}
    -k_{\perp}V^{\rm odd}_{cc}\\
    k_{\perp}V^{\rm even}_{sc}\\
    -a^2A_{0,cc}\\
    -\chi_{cc}\\
    \frac{a}{k_{\perp}}H_0\Phi_{sc}\\
   \end{array}
		 \right),\,
 {\bf Z}_{ss}  =
  \left(
   \begin{array}{c}
    -k_{\perp}V^{\rm odd}_{cs}\\
    k_{\perp}V^{\rm even}_{ss}\\
    -a^2A_{0,cs}\\
    -\chi_{cs}\\
    \frac{a}{k_{\perp}}H_0\Phi_{ss}\\
   \end{array}
		 \right),
\end{align}
represent non-dynamical degrees of freedom, $k_{\perp} = \sqrt{(k_y)^2 + (k_z)^2}$ and the Einstein's summation rule for $n = cc, cs, sc, ss$ is again understood. Important points are that the matrices ${\bf K}$, ${\bf M}$, ${\bf V}$, ${\bf A}$, ${\bf B}$ and ${\bf C}$ are common for all $n = cc, cs, sc, ss$ and that there are no coupling between ($Y_n$, $Z_n$) and ($Y_{n'}$, $Z_{n'}$) with $n \ne n'$. In general, however, even and odd sectors within ($Y_n$, $Z_n$) couple with each other.

Integrating out the non-dynamical variables ${\bf Z}_n$, one obtains the quadratic action for the dynamical variables ${\bf Y}_n$ as
\begin{align}
 \tilde{I}^{(2)} =& \frac{1}{2} \int\frac{d^3k}{(2\pi)^3} \int Na^3dt
  \left[ \frac{1}{N^2}\dot{\bf Y}_n^{\intercal} \bar{\bf K}\dot{\bf Y}_n
   + \frac{1}{N}\left(\dot{{\bf Y}}_n^{\intercal}\bar{\bf M} {\bf Y}_n \right.\right.\nonumber\\
& \left.\left.
 + {\bf Y}_n^{\intercal}\bar{\bf M}^{\intercal}\dot{{\bf Y}}_n\right)
 - {\bf Y}_n^{\intercal}\bar{\bf V}{\bf Y}_n \right]\,,
  \label{eqn:reducedaction}
\end{align}
where
\begin{align}
& \bar{\bf K} = {\bf K} - {\bf B}^{\intercal}{\bf C}^{-1}{\bf B}\,,\quad
 \bar{\bf M} = \frac{1}{2}\left(\bar{\bf m}-\bar{\bf m}^{\intercal}\right)\,,\nonumber\\
 & \bar{\bf V} = {\bf V} + {\bf A}^{\intercal}{\bf C}^{-1}{\bf A} + \frac{1}{2Na^3}\partial_t\left[a^3\left(\bar{\bf m}+\bar{\bf m}^{\intercal}\right)\right]\,,
\end{align}
and
\begin{equation}
 \bar{\bf m} = {\bf M} - {\bf B}^{\intercal}{\bf C}^{-1}{\bf A}\,,\label{eqn:reducedmatrices}
\end{equation}

It is then straightforward to diagonalize the kinetic matrix $\bar{\bf K}$ as
 \begin{equation}
  \bar{\bf K} \to \tilde{\bf K} = {\bf U}^{\intercal}\bar{\bf K}{\bf U}\,,\quad
   {\bf U} = {\bf U}_1{\bf U}_2{\bf U}_3{\bf U}_4{\bf U}_5\,,
   \label{eqn:tildeK}
 \end{equation}
 where
\begin{align}
 {\bf U}_1 =&  \left(\begin{array}{ccccc}
1 & -\frac{\bar{K}_{12}}{\bar{K}_{11}} & -\frac{\bar{K}_{13}}{\bar{K}_{11}} & -\frac{\bar{K}_{14}}{\bar{K}_{11}} & -\frac{\bar{K}_{15}}{\bar{K}_{11}} \\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
\end{array}
			  \right)\,, \nonumber\\
 {\bf U}_2 =&  \left(\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\
0 & 1 & -\frac{\bar{K}_{23}}{\bar{K}_{22}} & -\frac{\bar{K}_{24}}{\bar{K}_{22}} & -\frac{\bar{K}_{25}}{\bar{K}_{22}} \\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
\end{array}
			  \right)\,, \nonumber\\
 {\bf U}_3 =&  \left(\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & -\frac{\bar{K}_{34}}{\bar{K}_{33}} & -\frac{\bar{K}_{35}}{\bar{K}_{33}} \\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
\end{array}
			  \right)\,, \nonumber\\
 {\bf U}_4 =&  \left(\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & -\frac{\bar{K}_{45}}{\bar{K}_{44}} \\
0 & 0 & 0 & 0 & 1\\
\end{array}
			  \right)\,,
\end{align}
and ${\bf U}_5={\rm diag}(1,k/k_{\perp},k/k_{\perp},1,k^2/k^2_{\perp})$. Here $k=\sqrt{(k_x)^2+(k_{\perp})^2}$. Correspondingly, the friction matrix $\bar{\bf M}$ and the mass matrix $\bar{\bf V}$ transform as
\begin{align}
\bar{\bf M} \to& \tilde{\bf M}
= \frac{1}{2}\left(\tilde{\bf m}-\tilde{\bf m}^{\intercal}\right)\,,\nonumber\\
\bar{\bf V} \to& \tilde{\bf V} = {\bf U}^{\intercal}\bar{\bf V}{\bf U} - \frac{1}{N}\left(\dot{{\bf U}}^{\intercal}\bar{\bf M}{\bf U}+{\bf U}^{\intercal}\bar{\bf M}^{\intercal}\dot{{\bf U}}\right)\nonumber\\
 & - \frac{1}{N^2}\dot{{\bf U}}^{\intercal}\bar{\bf K}\dot{{\bf U}} + \frac{1}{2Na^3}\partial_t\left[a^3\left(\tilde{\bf m}+\tilde{\bf m}^{\intercal}\right)\right]\,,
\end{align}
where
\begin{equation}
 \tilde{\bf m} = {\bf U}^{\intercal}\bar{\bf M}{\bf U} + \frac{1}{N}{\bf U}^{\intercal}\bar{\bf K}\dot{{\bf U}}\,.\label{eqn:def-mtilde}
\end{equation}
The quadratic action is now written as
\begin{align}
 \tilde{I}^{(2)} = & \frac{1}{2} \int\frac{d^3k}{(2\pi)^3} \int Na^3dt
  \left[ \frac{1}{N^2}\dot{\tilde{\bf Y}}_n^{\intercal} \tilde{\bf K}\dot{\tilde{\bf Y}}_n \right.\nonumber\\
& \left.
   + \frac{1}{N}\left(\dot{{\tilde{\bf Y}}}_n^{\intercal}\tilde{\bf M} \tilde{\bf Y}_n + \tilde{\bf Y}_n^{\intercal}\tilde{\bf M}^{\intercal}\dot{{\tilde{\bf Y}}}_n\right)
 - \tilde{\bf Y}_n^{\intercal}\tilde{\bf V}\tilde{\bf Y}_n \right]\,,
  \label{eqn:reducedaction-diagonal}
\end{align}
where the new variables $\tilde{\bf Y}_n$ are related to the original variables ${\bf Y}_n$ via ${\bf Y}_n={\bf U}\tilde{\bf Y}_n$.

\subsection{Quadratic action around de Sitter background with magnetic field}

In the de Sitter background with a homogeneous magnetic field, the matrices in (\ref{eqn:generalI2}) are greatly simplified and all their components are written in terms of $p_{\perp}\equiv k_{\perp}/a$, $p_x\equiv k_x/a$, $b$, $g_h$, $g_4$ and $\zeta_{\alpha}$ ($\alpha=1,\cdots, 7$), where $g_h$ and $\zeta_{\alpha}$ are defined in (\ref{eqn:def-gh-zeta}) and
\begin{equation}
\zeta_6 = 4(-3g_h+2l_z)\,, \
  \zeta_7 = 4(-3g_h+2l_z+4b^2l_{yy})\,.
\end{equation}
Here, $l_z$ and $l_{yy}$ are defined as
\begin{equation}
 L_{Z} = l_{z}\frac{M_{\rm Pl}^2}{H_0^2}\,, \quad
     L_{YY} = l_{yy}\frac{M_{\rm Pl}^2}{H_0^2}\,,  \label{eqn:def-lz-lyy}
\end{equation}
where it is understood that the left hand sides of (\ref{eqn:def-lz-lyy}) are evaluated at the de Sitter solution with a homogeneous magnetic field. It is convenient to decompose the matrices into sub-matrices as
\begin{align}
& {\bf K} = \left(\begin{array}{cc}
      {\bf K}_{\rm odd}&  {\bf 0}\\
	    {\bf 0} &  {\bf K}_{\rm even}
	     \end{array}\right),\,
 {\bf M} = \left(\begin{array}{cc}
      {\bf 0} &  {\bf M}_{\rm mix}\\
	    -{\bf M}_{\rm mix}^{\intercal}& {\bf M}_{\rm even}
	     \end{array}\right),\nonumber\\
&
 {\bf V} = \left(\begin{array}{cc}
      {\bf V}_{\rm odd}&  {\bf V}_{\rm mix}\\
	    {\bf V}_{\rm mix}^{\intercal}&  {\bf V}_{\rm even}
	     \end{array}\right),\,
{\bf A} = \left(\begin{array}{cc}
      {\bf A}_{\rm odd}&  {\bf A}_{\rm mixI}\\
	    {\bf A}_{\rm mix II}&  {\bf A}_{\rm even}
	     \end{array}\right),\nonumber\\
&
 {\bf B} = \left(\begin{array}{cc}
      {\bf B}_{\rm odd}&  {\bf 0}\\
	    {\bf B}_{\rm mix}&  {\bf B}_{\rm even}
	     \end{array}\right),\,
 {\bf C} = \left(\begin{array}{cc}
      {\bf C}_{\rm odd}&  {\bf C}_{\rm mix}\\
	    {\bf C}_{\rm mix}^{\intercal}&  {\bf C}_{\rm even}
	     \end{array}\right), \label{eqn:submatrices}
\end{align}
where explicit expressions for the sub-matrices are given in Appendix~\ref{app:submatrices}.

If $b=0$ (we have already set $s=0$) then, since $L$ is assumed to be even with respect to $Y$ in (\ref{eqn:Parity}), the theory and the background respect the parity invariance and, as a result, the mixing matrices ${\bf M}_{\rm mix}$, ${\bf V}_{\rm mix}$, ${\bf A}_{\rm mix I}$, ${\bf A}_{\rm mix II}$, ${\bf B}_{\rm mix}$ and ${\bf C}_{\rm mix}$ vanish. In general they do not vanish and thus the even and odd perturbations do mix.

The quadratic action for the dynamical variables ${\bf Y}_n$ is then given by the formulas (\ref{eqn:reducedaction})-(\ref{eqn:reducedmatrices}). After diagonalization of the kinetic matrix, one obtains (\ref{eqn:reducedaction-diagonal}) with (\ref{eqn:tildeK})-(\ref{eqn:def-mtilde}).

\subsection{Subhorizon limit}

For the theoretical consistency, one needs to analyze the stability of the background in the UV, i.e. in the subhorizon limit $p^2\gg H_0^2$, where $p=\sqrt{p_x^2+p_{\perp}^2}=k/a$. This ensures the absence of instability whose timescale is parametrically shorter than the cosmological timescale $\sim 1/H_0$. The stability of the solution against perturbations with $p^2\ll H_0^2$ shall be studied in the next section.

In the subhorizon limit $p^2\gg H_0^2$, introducing a small bookkeeping parameter $\epsilon$ so that $H_0^2/p^2=\mathcal{O}(\epsilon^2)$, components of the matrices in the quadratic action (\ref{eqn:reducedaction-diagonal}) are simplified as
 \begin{align}
 \tilde{\bf K} =&
  \left(\begin{array}{ccccc}
   \tilde{K}_1 &  0 &  0 & 0 & 0 \\
	 0 &  \tilde{K}_2 &  0 & 0 & 0 \\
	 0 &  0 &  \tilde{K}_3 & 0 & 0 \\
	 0 &  0 &  0 & \tilde{K}_4 & 0 \\
	 0 &  0 &  0 & 0 & \tilde{K}_5
	  \end{array}
 \right) + \mathcal{O}(\epsilon^2)\,, \nonumber\\
 \tilde{\bf M} =&
  \left(\begin{array}{ccccc}
   0 &  0 &  0 & \tilde{M}_1 & \tilde{M}_2\\
	 0 &  0 &  \tilde{M}_3 & 0 & 0\\
	 0 &  -\tilde{M}_3 &  0 & 0 & 0\\
	 -\tilde{M}_1 &  0 &  0 & 0 & 0\\
 	 -\tilde{M}_2 &  0 &  0 & 0 & 0
	  \end{array}
 \right) + p\mathcal{O}(\epsilon^2)\,, \nonumber\\
 \tilde{\bf V} =&
  \left(\begin{array}{ccccc}
   \tilde{V}_1 &  0 &  0 & 0 & 0 \\
	 0 &  \tilde{V}_2 &  0 & 0 & 0 \\
	 0 &  0 &  \tilde{V}_3 & 0 & 0 \\
	 0 &  0 &  0 & \tilde{V}_4 & \tilde{V}_6 \\
	 0 &  0 &  0 & \tilde{V}_6 & \tilde{V}_5
	  \end{array}
 \right)+ p^2\mathcal{O}(\epsilon^2)\,,
\end{align}
where
 \begin{align}
\tilde{K}_1 =& \zeta_6\,,\nonumber\\
\tilde{K}_2 =& \left(\frac{p_{\perp}^2}{\zeta_1}+ \frac{p_x^2}{\zeta_1-2b^2g_h}\right)^{-1}p^2\,,\nonumber\\
\tilde{K}_3 =& \left(\frac{p_{\perp}^2}{\zeta_7}+\frac{p_x^2}{\zeta_6}\right)^{-1}p^2\,,\nonumber\\
\tilde{K}_4 =& \frac{c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4}{[(\zeta_1-\zeta_3)p^2+3b^2g_hp_{\perp}^2]^2}\,,\nonumber\\
\tilde{K}_5 =& \frac{(\zeta_1-2b^2g_h)\mathcal{N}p^4}{4(c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4)}\,,\nonumber\\
\tilde{M}_1 =& \frac{p_{\perp}(c_3p_{\perp}^2+c_4p_{x}^2)}{(\zeta_1-\zeta_3)p^2+3b^2g_hp_{\perp}^2}\,,\nonumber\\
\tilde{M}_2 =& \frac{bg_hp_{\perp}p^2(\zeta_1-2b^2g_h)(c_5p_{\perp}^2+c_6p_{x}^2)}{\zeta_1(c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4)}\,,\nonumber\\
\tilde{M}_3 =& -\frac{2bg_hp_{\perp}}{\zeta_1}\tilde{K}_2\,,\nonumber\\
\tilde{V}_1 =& \zeta_5p_{\perp}^2+4g_h\left(\frac{4b^2g_h}{\zeta_1}-3\right)p_{x}^2\,,\nonumber\\
\tilde{V}_2 =& g_4p^2 \,,\nonumber\\
\tilde{V}_3 =& \frac{4g_hp^2[3(2b^2g_h-\zeta_1)p_{\perp}^2+(4b^2g_h-3\zeta_1)p_{x}^2]}{(\zeta_1-2b^2g_h)p_{\perp}^2+\zeta_1p_{x}^2}\,,\nonumber\\
\tilde{V}_4 =& \frac{c_7p_{\perp}^6+c_8p_{\perp}^4p_{x}^2+c_9p_{\perp}^2p_{x}^4+c_{10}p_{x}^6}{[(\zeta_1-\zeta_3)p^2+3b^2g_hp_{\perp}^2]^2}\,,\nonumber\\
\tilde{V}_5 =& \frac{p^4(c_{11}p_{\perp}^6+c_{12}p_{\perp}^4p_{x}^2+c_{13}p_{\perp}^2p_{x}^4+c_{14}p_{x}^6)}{[c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4]^2}\,,\nonumber\\
\tilde{V}_6 =& \frac{b^2g_hp_{\perp}^2p^2(c_{15}p_{\perp}^4+2c_{16}p_{\perp}^2p_{x}^2+c_{17}p_{x}^4)}{[(\zeta_1-\zeta_3)p^2+3b^2g_hp_{\perp}^2][c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4]}\,. \label{eqn:tildeK-tildeM-tildeV}
\end{align}
Here, $\mathcal{N}$ is defined in (\ref{eqn:def-calN-calA}) and $c_1, \cdots, c_{17}$ are shown in Appendix~\ref{app:coefficients}.

\subsection{No-ghost conditions}

All components $\tilde{K}_{1,2,3,4,5}$ of the diagonalized kinetic matrix in the subhorizon limit are positive for ${}^{\forall}p_{\perp}$ and ${}^{\forall}p_x$ such that $p^2\ne 0$, if and only if $\mathcal{N}_a>0$ ($a=1,\cdots,7$), where
\begin{align}
 & \mathcal{N}_1 = \zeta_1\,,\quad \mathcal{N}_2 = \zeta_1-2b^2g_h\,,
  \nonumber\\
 &  \mathcal{N}_3 = \zeta_6\,,\quad \mathcal{N}_4=\zeta_7\,,\quad \mathcal{N}_5=\mathcal{N}\,, \label{eqn:def-calN1to5}
\end{align}
and
\begin{equation}
 \mathcal{N}_6 = c_1\,,\quad \mathcal{N}_7 = c_2 + \sqrt{c_1\mathcal{N}}\,.
\end{equation}
Here, the positivity of the last three $\mathcal{N}_{5,6,7}$ is necessary and sufficient for $c_1p_{\perp}^4+2c_2p_{\perp}^2p_{x}^2+\mathcal{N}p_{x}^4$ to be positive for ${}^{\forall}p_{\perp}$ and ${}^{\forall}p_x$ such that $p^2\ne 0$.

One can actually show that
\begin{align}
 \mathcal{N}_6 =& \frac{\mathcal{N}_2}{4\mathcal{N}_1^2}(\mathcal{N}_1-\mathcal{N}_2)^2(4\mathcal{N}_1-\zeta_3)^2+\frac{\mathcal{N}_5}{4\mathcal{N}_1^2}(\mathcal{N}_1+\mathcal{N}_2)^2\,,\nonumber\\
 \mathcal{N}_7 =& \frac{\mathcal{N}_5}{2\mathcal{N}_1}(\mathcal{N}_1+\mathcal{N}_2) + \sqrt{\mathcal{N}_5\mathcal{N}_6}\,.
\end{align}
Therefore the positivity of $\mathcal{N}_6$ and $\mathcal{N}_7$ follows from other conditions and the no-ghost condition is simply
\begin{equation}
 \mathcal{N}_a >0\,, \quad (a=1,\cdots,5)\,. \label{eqn:noghost}
\end{equation}

\subsection{Sound speeds}

The squared sound speeds $c_s^2$ of the five modes are determined by
\begin{align}
 0 =& \lim_{\epsilon\to 0}\det \left[ c_s^2 \tilde{\bf K} -2i\sqrt{c_s^2}\tilde{\bf M}/p - \tilde{\bf V}/p^2 \right]\nonumber\\
 =& \tilde{K}_1\tilde{K}_2\tilde{K}_3\tilde{K}_4\tilde{K}_5
  \times \left[(c_s^2)^2-2\alpha_1c_s^2+\alpha_2\right]\nonumber\\
 & \times \left[(c_s^2)^3-3\alpha_3(c_s^2)^2+3\alpha_4c_s^2-\alpha_5\right]\,,
 \label{eqn:dispersionrelation}
\end{align}
where
\begin{align}
 \alpha_1 =&
  \frac{b^2\mathcal{N}_1\mathcal{N}_3\mathcal{S}_1+2\mathcal{N}_2(\mathcal{N}_1+2\mathcal{N}_2)\mathcal{S}_2}{2b^2\mathcal{N}_1\mathcal{N}_2\mathcal{N}_3}\frac{p_x^2}{p^2} \nonumber\\
 &
  + \frac{b^2\mathcal{N}_4\mathcal{S}_1+2(\mathcal{N}_1+2\mathcal{N}_2)\mathcal{S}_2}
  {2b^2\mathcal{N}_1\mathcal{N}_4}\frac{p_{\perp}^2}{p^2}\,, \nonumber\\
 \alpha_2 =&
  \frac{2[(\mathcal{N}_1+2\mathcal{N}_2)p_x^2+3\mathcal{N}_2p_{\perp}^2](\mathcal{N}_3p_{\perp}^2+\mathcal{N}_4p_x^2)}{b^2\mathcal{N}_1\mathcal{N}_2\mathcal{N}_3\mathcal{N}_4 p^4}\mathcal{S}_1\mathcal{S}_2\,,\nonumber\\
 \alpha_3 =&
  \frac{\mathcal{N}_1\mathcal{S}_3p_x^2+\mathcal{N}_2\mathcal{S}_4p_{\perp}^2}
  {3b^2\mathcal{N}_1^2\mathcal{N}_2\mathcal{N}_3\mathcal{N}_5 p^2}\,,\nonumber\\
 \alpha_4 =&
  \frac{\mathcal{S}_5p_x^4+2(\mathcal{S}_7-\sqrt{\mathcal{S}_5\mathcal{S}_6})p_x^2p_{\perp}^2+\mathcal{S}_6p_{\perp}^4}
  {3b^2\mathcal{N}_1\mathcal{N}_2\mathcal{N}_3\mathcal{N}_5 p^4}\,,\nonumber\\
 \alpha_5 =&
  \frac{\mathcal{S}_9p_x^4+2(\mathcal{S}_{11}-\sqrt{\mathcal{S}_9\mathcal{S}_{10}})p_x^2p_{\perp}^2+\mathcal{S}_{10}p_{\perp}^4}
  {b^2\mathcal{N}_1\mathcal{N}_2\mathcal{N}_3\mathcal{N}_5 p^6}\nonumber\\
&
  \times
  [2(\mathcal{N}_1+\mathcal{N}_2)\mathcal{S}_2p_x^2+b^2\mathcal{N}_1\mathcal{S}_8p_{\perp}^2]\,,\label{eqn:def-alphas}
\end{align}
and $\mathcal{S}_{1,\cdots,11}$ are given in Appendix~\ref{app:calS}.

For the absence of gradient instabilities, it is necessary and sufficient to impose
\begin{align}
&\alpha_1 >0\,, \
  \alpha_2 >0\,, \
  \alpha_3 >0\,, \
  \alpha_4 >0\,, \
  \alpha_5 >0\,, \nonumber\\
 & \mbox{ for }
 {}^{\forall}p_{\perp} \mbox{ and } {}^{\forall}p_x \mbox{ such that } p^2\ne 0\,,
  \label{eqn:stability1pre}
\end{align}
and
\begin{align}
 & \alpha_1^2 - \alpha_2 > 0\,,\quad
  \alpha_3^2 - \alpha_4 > 0\,,\nonumber\\
 & -\alpha_5^2 + 2(3\alpha_4-2\alpha_3^2)\alpha_3\alpha_5 + (3\alpha_3^2-4\alpha_4)\alpha_4^2 > 0\,,\nonumber\\
 & \mbox{ for }
 {}^{\forall}p_{\perp} \mbox{ and } {}^{\forall}p_x \mbox{ such that } p^2\ne 0\,.
 \label{eqn:stability2}
\end{align}
Here, we supposed that the no-ghost condition (\ref{eqn:noghost}) is satisfied. The condition (\ref{eqn:stability1pre}) is equivalent to
\begin{equation}
 \mathcal{S}_a >0\,, \quad (a=1,\cdots,11)\,. \label{eqn:stability1}
\end{equation}

\section{Long wavelength perturbations}
\label{sec:longwavelengthperturbations}

In the previous section we studied the stability of the de Sitter solution with a homogeneous magnetic field in the UV limit, i.e. the limit where the wavelengths of perturbations are sufficiently shorter than the size of the de Sitter horizon. In this section we study the stability of the solution in the opposite limit, namely the limit where the wavelengths of perturbations are sufficiently longer than the size of the horizon.

As argued in \cite{Gumrukcuoglu:2016jbh}, instabilities for modes with $p^2=\mathcal{O}(H_0^2)$ can be as harmless as the standard Jeans instability. It is nonetheless necessary to impose the stability for those modes with $p^2\ll H_0^2$ if one wants the background under investigation to realize dynamically as an attractor of the system. In general relativity, the standard Jeans instability prevents the matter-dominated FLRW background from being an attractor of the system, but this is not a problem since it is not the matter-dominated epoch but the inflationary epoch that sets the initial condition of our local patch of the universe. On the other hand, for the system under investigation in the present paper, we would like the homogeneous magnetic field to realize dynamically and to be sustained during a de Sitter phase representing the inflationary epoch. For this reason, we require the solution to be an attractor of the system. Namely, we require the solution to be stable against perturbations with $p^2\ll H_0^2$.

For this purpose we send $p_{\perp}/H_0$ and $p_x/H_0$ to zero after obtaining equations of motion for properly normalized dynamical variables. Since the vector harmonics $Y^{\rm odd/even}_{n, p}$ ($n = cc, cs, sc, ss$; $p, q = 2, 3$) vanish in the limit and $Y^{\rm odd/even}_{n, p}/k_{\perp}$ remain finite, we introduce $\tilde{h}^{\rm odd}_{1,n} \equiv k_{\perp}h^{\rm odd}_{1,n}$, $\tilde{A}^{\rm odd}_n \equiv k_{\perp}A^{\rm odd}_n$ and $\tilde{V}^{\rm odd/even}_n \equiv k_{\perp}V^{\rm odd/even}_n$. The stability of the background solution against anisotropic perturbations with long wavelengths can then be analyzed by studying the equations of motion for the dynamical variables ($\tilde{h}^{\rm odd}_{1,n}$, $A_{1,n}$, $\pi_n$, $\psi_n$, $\tilde{A}^{\rm odd}_n$) in the limit where $p_{\perp}/H_0$ and $p_x/H_0$ are sent to zero, after eliminating non-dynamical variables. For simplicity we set $N(t)=1$ in the rest of this section.

The equation of motion for $\tilde{h}^{\rm odd}_{1,n}$ in the long wavelength limit does not involve other variables and is
\begin{equation}
 \ddot{\tilde{h}}^{\rm odd}_{1,n} + 3H_0\dot{\tilde{h}}^{\rm odd}_{1,n} = 0\,,
\end{equation}
where an over-dot represents derivative with respect to the time variable $t$. This gives the solution
\begin{equation}
 \tilde{h}^{\rm odd}_{1,n} = C_1 + C_2 e^{-3H_0 t}\,,
\end{equation}
where $C_1$ and $C_2$ are constants. The first solution, which is constant in space and time, is actually a pure gauge. Therefore, $\tilde{h}^{\rm odd}_{1,n}$ is stable.

The equation of motion for $A_{1,n}$ in the long wavelength limit also does not involve other variables and is
\begin{equation}
 \ddot{A}_{1,n} + 7H_0\dot{A}_{1,n} + 6H_0^2A_{1,n} = 0\,,
\end{equation}
which gives the solution
\begin{equation}
 A_{1,n} = C_3e^{-H_0t} + C_4e^{-6H_0t}\,,
\end{equation}
where $C_3$ and $C_4$ are constants. Therefore, $A_{1,n}$ is stable.

The equations of motion for $\pi_n$ and $\psi_n$ in the long wavelength limit are coupled but can be solved easily by setting
\begin{equation}
 \pi_n = \pi_{n}^{0} e^{\Gamma H_0 t}\,, \quad
   \psi_n = \psi_{n}^{0} e^{\Gamma H_0 t}\,, \label{eqn:pi-psi-homogeneous-anisotropic}
\end{equation}
where $\pi_{n}^{0}$ and $\psi_{n}^{0}$ are constants. The general solution is of course a linear combination of solutions of this form for all allowed values of $\Gamma$. By substituting (\ref{eqn:pi-psi-homogeneous-anisotropic}) to the equations of motion for $\pi_n$ and $\psi_n$, one obtains
\begin{equation}
 \Gamma (\Gamma+3)\left(\Gamma^2 + 3\Gamma + \frac{\mathcal{A}}{\mathcal{N}}\right) = 0\,,
\end{equation}
where $\mathcal{A}$ and $\mathcal{N}$ are the same as those defined in (\ref{eqn:def-calN-calA}) and thus in particular $\mathcal{N}=\mathcal{N}_5$ (see (\ref{eqn:def-calN1to5})). For $\Gamma=0$, $\pi_{n}^{0}=0$ and $\psi_n$ is a pure gauge. Therefore, $\pi_n$ and $\psi_n$ are also stable, provided that the attractor condition (\ref{eqn:attractor}) and the no-ghost condition (\ref{eqn:noghost}) are satisfied.

The equation of motion for $\tilde{A}^{\rm odd}_n$ in the long wavelength limit does not involve other variables and is
\begin{align}
& \ddot{\tilde{A}}^{\rm odd}_n + 5H_0\dot{\tilde{A}}^{\rm odd}_n
 + \mathcal{B}H_0^2 \tilde{A}^{\rm odd}_n = 0\,, \nonumber\\
& \mathcal{B} = \frac{\mathcal{A}_1 k_x^4 + 2(\mathcal{A}_3-\sqrt{\mathcal{A}_1\mathcal{A}_2})k_x^2k_{\perp}^2 + \mathcal{A}_2k_{\perp}^4}{[(\zeta_1-\zeta_3)k_x^2+(3g_hb^2+\zeta_1-\zeta_3)k_{\perp}^2]^2}\,, \label{eqn:eom-tildeAodd-longwavelength}
\end{align}
where $\mathcal{A}_{1,2,3}$ are given in Appendix~\ref{app:calA123}. Therefore, $\tilde{A}^{\rm odd}_n$ is stable if and only if
\begin{equation}
 \mathcal{A}_a > 0\,, \quad (a=1,2,3)\,. \label{eqn:stability-anisotropy}
\end{equation}

\section{Examples}
\label{sec:examples}

In this section we first show a couple of classes of models that always violate either (\ref{eqn:noghost}) or (\ref{eqn:stability1}). After that, we show some concrete examples of stable models with specific choices of parameters, which satisfy not only (\ref{eqn:noghost}) and (\ref{eqn:stability1}) but also (\ref{eqn:stability2}), (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}).

\subsection{Unstable classes of models}

Let us first consider the case without the Horndeski's non-minimal vector coupling ($\xi = 0$). Setting $g_h = 0$, we obtain $\mathcal{S}_2 = 0$. This implies that $\alpha_2 = 0$ and thus one of the five modes has vanishing sound speed.

Let us next consider the case with the Horndeski's non-minimal vector coupling ($\xi \ne 0$) but without Horndeski scalar terms. Setting $g_{3x} = g_{3xx} = g_{4x} = g_{4xx} = g_{5x} = g_{5xx} = 0$, we obtain $\mathcal{S}_9/\mathcal{S}_2 = -(2\mathcal{N}_2-\mathcal{N}_1)^2\mathcal{N}_2$, which is non-positive if we impose the no-ghost condition. Therefore, either $\mathcal{S}_9$ or $\mathcal{S}_2$ is non-positive if the no-ghost condition is imposed.

These two unstable examples imply that the stability requires both the Horndeski's nonminimal vector coupling and Horndeski scalar terms.

\subsection{Stable models}

 Our strategy here is to find a set of parameters that satisfies (\ref{eqn:noghost}) and (\ref{eqn:stability1}), and then to check if (\ref{eqn:stability2}), (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are also satisfied.

We have already assumed (\ref{eqn:Parity}). For simplicity, in all the examples considered in this subsection we further set
\begin{align}
0 =& g_{3xx} = g_{4xx} = g_{4xxx} = g_{5x} \nonumber\\
 =&  g_{5xx} = g_{5xxx} = l_{yy} = l_{xw} = l_{z}\,. \label{eqn:simplicity}
\end{align}
In each of the following examples we specify ($g_h$, $g_4$, $g_{4x}$, $g_{3x}$, $b$, $l_{xx}$, $l_{ww}$). We then solve the background equations of motion (\ref{eqn:dSnoE}) with respect to ($l$, $l_x$, $l_w$), where
\begin{equation}
 L = l\, M_{\rm Pl}^2H_0^2\,, \
  L_X = l_x\, M_{\rm Pl}^2\,, \
  L_W = l_w\, M_{\rm Pl}^2\,, \label{eqn:def-l-lx-lw}
\end{equation}
and it is understood that the left hand sides of (\ref{eqn:def-l-lx-lw}) are evaluated at the de Sitter solution with a homogeneous magnetic field. After that, we confirm that (\ref{eqn:noghost}) and (\ref{eqn:stability1}) as well as (\ref{eqn:stability2}), (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are satisfied.

For the set of parameters ($g_h$, $g_4$, $g_{4x}$, $g_{3x}$, $b$, $l_{xx}$, $l_{ww}$; $l$, $l_x$, $l_w$) and the assumptions (\ref{eqn:Parity}) and (\ref{eqn:simplicity}), one can easily reconstruct the Lagrangian assuming a simple ansatz and noting that $X=2H_0^2$ and $W=-b^2H_0^2/2$ on the de Sitter solution parameterized by ($H_0$, $b$). For example, if we assume that $L$ is quadratic in $X$ and $W$ and independent of $Y$ and $Z$ then
\begin{align}
 \frac{L}{M_{\rm Pl}^2} =&
  \left(l-2 l_x+2 l_{xx}+\frac{1}{2} b^2 l_w+\frac{1}{8} b^4 l_{ww}\right) H_0^2 \nonumber\\
&
  +(l_x-2 l_{xx})X + \left(l_w+\frac{1}{2} b^2 l_{ww}\right)W \nonumber\\
&
  +\frac{1}{2H_0^2}( l_{xx} X^2 + l_{ww} W^2)\,.
\end{align}
The simplest choice of $G_{3,4,5}$ is
\begin{equation}
 G_3 = g_{3x}\frac{M_{\rm Pl}^2}{H_0^2}X\,,\
 G_4 = g_4M_{\rm Pl}^2 + g_{4x}\frac{M_{\rm Pl}^2}{H_0^2}X\,,\
 G_5 = 0\,.
\end{equation}
The value of $\xi$ is
\begin{equation}
 \xi = g_h \frac{H_0^2}{M_{\rm Pl}^2}\,.
\end{equation}

\subsubsection{Example 1}

Let us consider the following choice of parameters.
\begin{align}
&
  g_h = -1\,, \
  g_4 = \frac{1}{2}\,, \
  g_{4x} = 0\,,\
  g_{3x} = \frac{1}{10}\,, \nonumber\\
&
  b = \frac{1}{10}\,, \
  l_{xx} = -\frac{1}{10}\,, \
  l_{ww} = -1\,,
\end{align}
for which the background equations of motion (\ref{eqn:dSnoE}) give
\begin{equation}
 l = -\frac{79}{25}\,,\ l_x = -\frac{31}{50}\,,\ l_w = 20\,. \label{eqn:l-lx-lw-ex1}
\end{equation}

In this case we have
\begin{align}
& \mathcal{N}_1 = \frac{12}{25}\,, \
  \mathcal{N}_2 = \frac{1}{2}\,, \
  \mathcal{N}_3 = 12\,, \nonumber\\
& \mathcal{N}_4 = 12\,, \
  \mathcal{N}_5 = \frac{1408}{15625}\,,
\end{align}
and
\begin{align}
 &
  \mathcal{S}_1 = \frac{1}{2}\,,\
  \mathcal{S}_2 = \frac{1}{50}\,, \
  \mathcal{S}_3 = \frac{59456}{9765625}\,, \nonumber\\
 &
  \mathcal{S}_4 = \frac{35469972}{6103515625}\,, \
  \mathcal{S}_5 = \frac{1696}{390625}\,, \
  \mathcal{S}_6 = \frac{7940769}{1953125000}\,, \nonumber\\
 &
  \mathcal{S}_7 = \frac{2060321}{488281250}+\sqrt{\mathcal{S}_5\mathcal{S}_6}\,,\
  \mathcal{S}_8 = \frac{1201}{100}\,, \
  \mathcal{S}_9 = \frac{224}{15625}\,, \nonumber\\
 &
  \mathcal{S}_{10} = \frac{29}{2500}\,, \
 \mathcal{S}_{11} = \frac{823}{62500}+\sqrt{\mathcal{S}_9\mathcal{S}_{10}}\,.
  \label{eqn:calS-example1}
\end{align}
Thus (\ref{eqn:noghost}) and (\ref{eqn:stability1}) are satisfied.

The conditions (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are also satisfied as
\begin{align}
 &
  \mathcal{A} = \frac{5979}{390625}\,,\
  \mathcal{A}_1 = \frac{256}{625}\,,\
  \mathcal{A}_2 = \frac{87451}{250000}\,,\nonumber\\
&
  \mathcal{A}_3 = \frac{47493}{125000}+\sqrt{\mathcal{A}_1\mathcal{A}_2}\,.
\end{align}

After substituting (\ref{eqn:calS-example1}) into (\ref{eqn:def-alphas}) to obtain explicit expressions of $\alpha_{1,\cdots,5}$, it is straightforward to compute the three expressions on the left hand sides of the inequalities in (\ref{eqn:stability2}) and to show that they all have the form $A/B$, where $A$ and $B$ are polynomials of $p_{\perp}^2$ and $p_x^2$ with positive coefficients. Therefore, (\ref{eqn:stability2}) is also satisfied.

\subsubsection{Example 2}

Let us consider the following choice of parameters.
\begin{align}
&
  g_h = -1\,, \
  g_4 = \frac{1}{2}\,, \
  g_{4x} = 0\,,\
  g_{3x} = \frac{1}{10}\,, \nonumber\\
&
  b = \frac{1}{10}\,, \
  l_{xx} = -\frac{1}{10}\,, \
  l_{ww} = 0\,,
\end{align}
for which the background equations of motion (\ref{eqn:dSnoE}) give the same values of $l$, $l_x$ and $l_w$ as shown in (\ref{eqn:l-lx-lw-ex1}).

In this case we have
\begin{align}
& \mathcal{N}_1 = \frac{12}{25}\,, \
  \mathcal{N}_2 = \frac{1}{2}\,, \
  \mathcal{N}_3 = 12\,, \nonumber\\
& \mathcal{N}_4 = 12\,, \
  \mathcal{N}_5 = \frac{1408}{15625}\,,
\end{align}
and
\begin{align} \label{eqn:calS-example3}
&
  \mathcal{S}_1 = \frac{1}{2}\,,\
  \mathcal{S}_2 = \frac{1}{50}\,, \
  \mathcal{S}_3 = \frac{59456}{9765625}\,, \nonumber\\
&
  \mathcal{S}_4 = \frac{1418292}{244140625}\,, \
  \mathcal{S}_5 = \frac{1696}{390625}\,, \
  \mathcal{S}_6 = \frac{79353}{19531250}\,, \nonumber\\
&
  \mathcal{S}_7 = \frac{16477}{3906250}+\sqrt{\mathcal{S}_5\mathcal{S}_6}\,,\
  \mathcal{S}_8 = 12\,, \
  \mathcal{S}_9 = \frac{224}{15625}\,, \nonumber\\
&
  \mathcal{S}_{10} = \frac{29}{2500}\,, \
  \mathcal{S}_{11} = \frac{823}{62500}+\sqrt{\mathcal{S}_9\mathcal{S}_{10}}\,.
\end{align}
Thus (\ref{eqn:noghost}) and (\ref{eqn:stability1}) are satisfied.

The conditions (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are also satisfied as
\begin{align}
&
  \mathcal{A} = \frac{239}{15625}\,,\
  \mathcal{A}_1 = \frac{256}{625}\,,\
  \mathcal{A}_2 = \frac{1749}{5000}\,,\nonumber\\
&
  \mathcal{A}_3 = \frac{5699}{15000}+\sqrt{\mathcal{A}_1\mathcal{A}_2}\,.
\end{align}

After substituting (\ref{eqn:calS-example2}) into (\ref{eqn:def-alphas}) to obtain explicit expressions of $\alpha_{1,\cdots,5}$, it is straightforward to compute the three expressions on the left hand sides of the inequalities in (\ref{eqn:stability2}) and to show that they all have the form $A/B$, where $A$ and $B$ are polynomials of $p_{\perp}^2$ and $p_x^2$ with positive coefficients. Therefore, (\ref{eqn:stability2}) is also satisfied.

\subsubsection{Example 3}

Let us consider the following choice of parameters.
\begin{align}
&
  g_h = -\frac{1}{20}\,, \
  g_4 = \frac{1}{2}\,, \
  g_{4x} = 0\,,\
  g_{3x} = \frac{1}{10}\,, \nonumber\\
&
  b = \frac{1}{10}\,, \
  l_{xx} = -\frac{1}{10}\,, \
  l_{ww} = 0\,,
\end{align}
for which the background equations of motion (\ref{eqn:dSnoE}) give
\begin{equation}
 l = -\frac{376}{125}\,,\ l_x = -\frac{601}{1000}\,,\ l_w = 1\,.
\end{equation}

In this case we have
\begin{align}
& \mathcal{N}_1 = \frac{499}{1000}\,, \
  \mathcal{N}_2 = \frac{1}{2}\,, \
  \mathcal{N}_3 = \frac{3}{5}\,, \nonumber\\
& \mathcal{N}_4 = \frac{3}{5}\,, \
  \mathcal{N}_5 = \frac{108990799}{1000000000}\,,
\end{align}
and
\begin{align} \label{eqn:calS-example2}
&
  \mathcal{S}_1 = \frac{1}{2}\,,\
  \mathcal{S}_2 = \frac{1}{1000}\,, \
  \mathcal{S}_3 = \frac{371370384707}{1000000000000000}\,, \nonumber\\
&
  \mathcal{S}_4 = \frac{92615485451247}{250000000000000000}\,, \
  \mathcal{S}_5 = \frac{50621001581}{200000000000000}\,, \nonumber\\
&
  \mathcal{S}_6 = \frac{252272242903}{1000000000000000}\,, \nonumber\\
&
  \mathcal{S}_7 = \frac{50538124821}{200000000000000}+\sqrt{\mathcal{S}_5\mathcal{S}_6}\,,\
  \mathcal{S}_8 = \frac{3}{5}\,, \nonumber\\
&
  \mathcal{S}_9 = \frac{29949199}{2000000000}\,, \
    \mathcal{S}_{10} = \frac{59301}{4000000}\,, \nonumber\\
&
  \mathcal{S}_{11} = \frac{59601699}{4000000000}+\sqrt{\mathcal{S}_9\mathcal{S}_{10}}\,.
\end{align}
Thus (\ref{eqn:noghost}) and (\ref{eqn:stability1}) are satisfied.

The conditions (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are also satisfied as
\begin{align}
&
  \mathcal{A} = \frac{304101}{500000000}\,,\
  \mathcal{A}_1 = \frac{90601}{250000}\,,\
  \mathcal{A}_2 = \frac{718809}{2000000}\,,\nonumber\\
&
  \mathcal{A}_3 = \frac{90045653}{249500000}+\sqrt{\mathcal{A}_1\mathcal{A}_2}\,.
\end{align}

After substituting (\ref{eqn:calS-example3}) into (\ref{eqn:def-alphas}) to obtain explicit expressions of $\alpha_{1,\cdots,5}$, it is straightforward to compute the three expressions on the left hand sides of the inequalities in (\ref{eqn:stability2}) and to show that they all have the form $A/B$, where $A$ and $B$ are polynomials of $p_{\perp}^2$ and $p_x^2$ with positive coefficients. Therefore, (\ref{eqn:stability2}) is also satisfied.

\subsubsection{Example 4}

Let us consider the following choice of parameters.
\begin{align}
&
  g_h = -\frac{1}{20}\,, \
  g_4 = \frac{1}{2}\,, \
  g_{4x} = -\frac{1}{10}\,,\
  g_{3x} = 0\,, \nonumber\\
&
  b = \frac{1}{10}\,, \
  l_{xx} = -\frac{1}{10}\,, \
  l_{ww} = 0\,,
\end{align}
for which the background equations of motion (\ref{eqn:dSnoE}) give
\begin{equation}
 l = -\frac{676}{125}\,,\ l_x = \frac{599}{1000}\,,\ l_w = 1\,.
\end{equation}

In this case we have
\begin{align}
& \mathcal{N}_1 = \frac{899}{1000}\,, \
  \mathcal{N}_2 = \frac{9}{10}\,, \
  \mathcal{N}_3 = \frac{3}{5}\,, \nonumber\\
& \mathcal{N}_4 = \frac{3}{5}\,, \
  \mathcal{N}_5 = \frac{108704599}{1000000000}\,,
\end{align}
and
\begin{align} \label{eqn:calS-example4}
&
  \mathcal{S}_1 = \frac{1}{2}\,,\
  \mathcal{S}_2 = \frac{1}{1000}\,, \
  \mathcal{S}_3 = \frac{9143271446151}{5000000000000000}\,, \nonumber\\
&
  \mathcal{S}_4 = \frac{456009833453647}{250000000000000000}\,, \
  \mathcal{S}_5 = \frac{9305798288929}{5000000000000000}\,, \nonumber\\
&
  \mathcal{S}_6 = \frac{9284212649127}{5000000000000000}\,, \nonumber\\
&
  \mathcal{S}_7 = \frac{9295017384133}{5000000000000000}+\sqrt{\mathcal{S}_5\mathcal{S}_6}\,,\
  \mathcal{S}_8 = \frac{3}{5}\,,  \nonumber\\
&
  \mathcal{S}_9 = \frac{207508199}{2000000000}\,, \
    \mathcal{S}_{10} = \frac{10351541}{100000000}\,, \nonumber\\
&
  \mathcal{S}_{11} = \frac{414541899}{4000000000}+\sqrt{\mathcal{S}_9\mathcal{S}_{10}}\,.
\end{align}
Thus (\ref{eqn:noghost}) and (\ref{eqn:stability1}) are satisfied.

The conditions (\ref{eqn:attractor}) and (\ref{eqn:stability-anisotropy}) are also satisfied as
\begin{align}
&
  \mathcal{A} = \frac{9185101}{500000000}\,,\
  \mathcal{A}_1 = \frac{1692601}{250000}\,,\
  \mathcal{A}_2 = \frac{13530009}{2000000}\,,\nonumber\\
&
  \mathcal{A}_3 = \frac{3042086553}{449500000}+\sqrt{\mathcal{A}_1\mathcal{A}_2}\,.
\end{align}

After substituting (\ref{eqn:calS-example4}) into (\ref{eqn:def-alphas}) to obtain explicit expressions of $\alpha_{1,\cdots,5}$, it is straightforward to compute the three expressions on the left hand sides of the inequalities in (\ref{eqn:stability2}) and to show that they all have the form $A/B$, where $A$ and $B$ are polynomials of $p_{\perp}^2$ and $p_x^2$ with positive coefficients. Therefore, (\ref{eqn:stability2}) is also satisfied.

\section{Summary and discussion}
\label{sec:summary}

We have presented a detailed stability analysis for the de Sitter solution with a homogeneous magnetic field that was recently found in \cite{Mukohyama:2016npi} in the context of a $U(1)$ gauge theory nonminimally coupled to scalar-tensor gravity. The magnetic field is ``stealth'' in the sense that the corresponding stress-energy tensor is of the form of an effective cosmological constant and thus is isotropic despite the fact that the magnetic field has a preferred spatial direction. We have studied the stability of the solution against linear perturbations in the subhorizon and superhorizon limits and have shown some explicit examples that satisfy all stability conditions. Stable models include both Horndeski's nonminimal vector coupling and Horndeski scalar terms.

The stable de Sitter solution with a homogeneous magnetic field opens up a new possibility for inflationary magnetogenesis, in which magnetic fields in the Universe at all scales may originate from a classical, homogeneous magnetic field sustained during inflation. Towards such a new scenario of inflationary magnetogenesis, an important step forward is to show a graceful exit from the de Sitter solution with constant values of ($X$, $W$, $Y$, $Z$, $\chi$). In particular, the scalar field $\phi$ should be stabilized around a local minimum of its potential with a sufficiently large mass at late time after inflation in order to recover the standard Einstein-Maxwell theory. (To be more precise, what is recovered is the Einstein-Maxwell theory with the Horndeski's nonminimal vector coupling.) Also, as already pointed out in \cite{Mukohyama:2016npi}, a field other than $\phi$ needs to be introduced as the main source of curvature perturbation, i.e. either an inflaton or a curvaton, in order to avoid too large statistical anisotropies and non-Gaussianities. One eventually needs to establish the stability of the whole system all the way from the inflationary epoch to the standard expansion history at late time through the graceful exit. The present paper has established the stability of the inflationary epoch and thus can be considered as the first step towards the new inflationary magnetogenesis scenario based on a classical, homogeneous magnetic field sustained during inflation.

The recent multi-messenger detection of binary neutron stars put a strong constraint on the deviation of the propagation speed of gravitational waves from that of light at the present~\cite{TheLIGOScientific:2017qsa,Monitor:2017mdv}. If the Horndeski scalar-tensor theory is used as a model of the late-time acceleration of the universe then the functions $G_4$ and $G_5$ are strongly constrained. On the other hand, the propagation speed of gravitational waves in the early universe such as the inflationary epoch is not constrained at all by such observations. Therefore, the model considered in \cite{Mukohyama:2016npi} and in the present paper is consistent with the multi-messenger detection of binary neutron stars as far as the scalar field is stabilized around a local minimum of its potential with a sufficiently large mass at late time after inflation, as required anyway for the recovery of the Einstein-Maxwell theory (with the Horndeski's nonminimal vector coupling).

After inflation and the stabilization of $\phi$ around a local minimum of its potential, the homogeneous magnetic field is no longer stealth and thus the background spacetime is then described by a Bianchi I geometry (instead of FLRW one). It is therefore important to investigate phenomenology of the Bianchi I universe with a homogeneous magnetic field and the standard content of the universe (radiation, matter and the cosmological constant) in the context of the Einstein-Maxwell theory with the Horndeski's nonminimal vector coupling~\cite{nextproject}.

\begin{acknowledgments}
 The author thanks Axel Brandenburg, Ruth Durrer, Yiwen Huang, Tina Kahniashvili and Sayan Mandal for helpful discussions. A part of the work was completed during the author's visits to Simon Fraser University, University of Alberta and University of Victoria. The author is therefore grateful to Andrei Frolov, Valeri Frolov, Werner Israel, Maxim Pospelov and Andrei Zelnikov for their warm hospitality. The work was supported in part by Japan Society for the Promotion of Science (JSPS)  Grants-in-Aid for Scientific Research (KAKENHI) No. 17H02890, No. 17H06359, and by World Premier International Research Center Initiative (WPI), MEXT, Japan.
\end{acknowledgments}

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The Dalai Lama: Leader, Peacemaker, and Author
By Kristin Masters. Mar 27, 2012. 3:14 AM.
Topics: Rare Books, Book Collecting, History
In March 1959, the people of Tibet were in revolt, rallying for independence from China. But the Chinese government squashed the revolt. The Dalai Lama fled his country and sought refuge in India, where he has resided ever since. Jump forward about fifty years, to March 2011: the Dalai Lama officially announced his retirement as the head of the Tibetan government.
New Arrival Spotlight: Shanty Bay Press Books
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Ever hear of "The Turnip Princess"? This unlikely fairy tale is one of over 500 recently discovered in Germany. Compiled by Franz Xaver von Schönwerth in the mid-1800's, the stories have been locked away in an archive for over 150 years.
Though "The Turnip Princess" may not be the next "Cinderella," the stories certainly represent hidden gems of cultural history and illustrate the rich tradition of the fairy tale. The canon of children's literature and folklore are always evolving; our canon will soon undoubtedly include several outstanding authors and illustrators, most notably Charles van Sandwyk.
Charles van Sandwyk''s Exceptional Illustrated Literature
Happy Birthday, Gabo!
By Kristin Masters. Mar 5, 2012. 12:28 AM.
Topics: Literature, Nobel Prize Winners, Modern First Editions
Today is the 85th birthday of legendary author Gabriel García Márquez. The Nobel laureate began his writing career as a journalist but established himself as one of the most important authors of the twentieth century. Here's a look at a few facts you probably didn't know about Gabriel García Márquez.

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text:
Cardinal Sarah: Pope's Lenten Message is About More than Material Poverty
Says Message Makes an Important Distinction Between Poverty and Destitution
February 04, 2014 00:00ZENIT StaffSpirituality
Pope Francis' first Lenten Message focuses on poverty and Christ's poverty in particular, according to the head of the Vatican dicastery that is responsible for disseminating it.
Cardinal Robert Sarah, president of the Pontifical Council "Cor Unum", told reporters at a Vatican press conference Feb. 4 that the concept of poverty is "very dear" to Pope Francis, who since the beginning of his pontificate has attempted to emphasise this dimension of Christian life.
But the Guinean cardinal stressed that the Christian vision of poverty is not the same as that which is commonly held. "Too often we consider poverty from a sociological perspective, and it is understood as a lack of material goods," he said. "Furthermore, the concept of a "poor Church for the poor" is often evoked as a sort of challenge to the Church, unfortunately also setting a Church of the poor, "a good Church … against a Church of preaching and truth, a Church dedicated to prayer and to the defence of doctrine and morals".
He continued: "The first point of reference for a Christian to understand poverty is indeed Christ, who made himself poor so that he could enrich us through his poverty." But Cardinal Sarah said Christ's choice of poverty suggests "a positive dimension" to poverty which resonates throughout the Gospel, and which proclaims that the poor are blessed.
"It is clear that in this dimension of poverty there is an aspect of despoliation and sacrifice," he said, but this is possible because Jesus' wealth lies in his being the Son'.
"We cannot set our bourgeois consciences at rest, the Pope means, by denouncing material lack on the part of others or denouncing poverty as a system," the Vatican official explained. "The Lenten Message we are presenting here today makes an important distinction between poverty and destitution.
"It is not poverty, which is an evangelical attitude, but rather destitution that we wish to combat," he stressed.
The cardinal explained that the Holy Father lists three forms of destitution in his Message: material, moral and spiritual. The first affects those living in conditions opposed to human dignity.
"Faced with this form of destitution, the Church offers her service, 'her diakonia, in meeting these needs and binding these wounds which disfigure the face of humanity'," the cardinal explained. "Moral destitution consists in slavery to vice and sin. This form of destitution is also the cause of economic ruin, and is always linked to spiritual destitution, which occurs when we drift away from God and refuse His love".
He said he believed this broader view of poverty, of destitution, offers a more "complete vision" of man and his needs," without falling in the trap of anthropological reductionism which claims to resolve all the problems of the human person simply by resolving the problems of physical and material well-being".
The president of Cor Unum recalled that in the Apostolic Exhortation "Evangelii Gaudium", Pope Francis writes that the Church's preferential option for the poor must mainly translate into a "privileged and preferential religious care". He affirmed that this concept is fundamental "so as not to transform the Church into that non-governmental organisation" – a point Pope Francis made in his first homily as Pontiff.
"It would be a great pity if our gaze upon those in need failed to acknowledge the spiritual poverty that often lurks in the heart of man and pains him deeply, even though he may be in a condition of material comfort," Cardinal Sarah said. "But if we wish to fully grasp Pope Francis' Message, we must not consider it only in terms of its anthropological value. Man is by nature the son of God. This is his wealth!"
He added that the "great flaw" of modern culture is that it has imagined mankind capable of being happy without God, thus denying that which is most profound in the human person. "It is a crime to deprive the poor of the presence of God, just as it is a crime to consider man and allow man to live as if God did not exist, to negate his being as a creation and therefore his fundamental belonging and affiliation with God," he said. "Therefore, work in development cannot be simply that of creating new needs, but rather taking a serious look at what the person truly is".
Before the presentation, Cardinal Sarah announced that in March he will visit Haiti again, in order to open a school financed on behalf of the Pope as a sign of his closeness to the Haitian population, afflicted in 2010 by an earthquake which claimed more than 220,000 victims and affected a total of more than 3 million people.
On ZENIT's webpage:
Pope's Lenten Message
On the NET:
February 04, 2014 00:00Spirituality
European Parliament Passes Lunacek Report
Pope Francis: Have Complete Trust in God in Times of Trial


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City of Greenwood Calling for Safety Improvements After Multiple Accidents.
Officials of the city of Greenwood, Indiana, are calling for safety improvements along State Road 135 after three serious accidents including a fatal crash within the last ten days. The Mayor of Greenwood, Mark Myers is among those who believe safety measures need to be improved along State Road 135. The mayor believes that there […]
Wrong-way Crash on Northeast Side Leaves Two Dead
On the early morning of January 30, 2018, two automobiles crashed on the northeast side of Indianapolis. The Indianapolis Metro Police Department believe that the crash happened on the Shadeland Avenue ramp to I-465 southbound on the northeast side of Indianapolis. This led to the closure of the ramp, and because access to Shadeland was […]
Tow Truck Companies State Fewer Interstate Accidents, More Local Slide Offs
In Indiana, the mixture of ice, sleet, and snow on the roads are making it difficult for some drivers during this chaotic winter. Multiple Tow Truck companies have reported that there appeared to be fewer drivers on the roads during the morning commute, particularly on Indy-area highways. Tow truck companies say that dip in volume […]
Multiple Vehicular Accidents Slow Traffic on 1-74 near the Indiana/Illinois State Line
The new year started with Indiana State police reporting to the scenes of several multiple vehicle accidents along I-74, west of the Covington area on the morning of January 3, 2018. The initial problem, according to investigating officers, began when three semi-trucks crashed while traveling east bound along I-74 at the mile 4 marker, east […]
Two People Killed in Multiple Vehicle Crash
On Tuesday, December 5, 2017, two motorists were killed in a multi-vehicle crash on Interstate 65 in Tippecanoe County. The crash occurred at about 3:20 P.M. in the southbound lane near State Road26 in Tippecanoe County according to investigating officers. Officers are still working on how exactly the accident happened, but the preliminary investigation revealed […]
Child Injured in Off-road Vehicle Accident
In the afternoon of Sunday, December 3, 2017, an 11-year old child was injured while driving an off-road vehicle after colliding with a motorcyclist. The child was riding an on a Monster Moto mini-bike along Whetstine Road in Morgan County, as he followed behind his Grandfather who was on a lawnmower. The child attempted to […]
Pedestrian is Killed after Being Struck by Vehicle
Around 8:00 P.M. on Wednesday, November 15, 2017, a car traveling east near U.S. 52 at South Creekside Drive in New Palestine struck a pedestrian who was crossing the road. The pedestrian was taken to an Indianapolis-area hospital, where she was later pounced dead. No explanation as to how the collision occurred has been ascertained […]
Crash involving Car and Semi-Truck leaves Driver Seriously Injured
On Monday, November 27, 2017, a driver was seriously injured after a crash that involved a semi-truck in Madison County. According to officials with the Madison County Sheriff's Department, the crash occurred around 5:55 a.m. at State Road 13 and State Road 37, six miles south of the city of Elwood. Several witnesses to the […]
Firefighters Sent to Hospital after Fire Engine struck by Dump Truck.
At about 11:00 AM on October 18th, 2017 a dump truck collided with an Indianapolis Fire Department engine near the intersection of 43rd street and Capitol Avenue. The fire engine was on an emergency medical service run near the 600 block of West 43rd Street when it was struck by a dump truck and pushed […]
Excavating Crew Damages Underground Gas Line, Leads to House Explosion
Investigators with the Indiana Utility Regulatory Commission Pipeline Safety Division believe that a gas leak that lead to the explosion of a House in the City of Camby was caused by an excavating crew that damaged an underground gas line. The investigators believe that the service line going into the house was hit by a […]
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12.20.16 December 20, 2016
Brendon Chung's latest game makes a mockery of RPGs
by Chris Priestman
@CPriestman
I missed the latest game by Brendon Chung (creator of Thirty Flights of Loving, Quadrilateral Cowboy) when he released it last month, but it's certainly worth highlighting. Called Acre 6, it's a deconstruction of the classic RPG, full of jest, made for the Procedural Generation Jam.
It starts you out with a map marked only with icons for areas of forest and landmarks like temples, ruins, and villages. The only interaction available is to move the mouse to guide a circle, representing you as the player character, who journeys across this map leaving a dashed line behind them.
open-world games rarely leave players to explore a map alone
After a short distance, a text bubble should pop up at a point in your journey, informing you of some event that happened there. You might find an item, defeat an enemy, purchase or be rewarded something, perhaps complete a quest or level up. The names of these events and items are procedurally generated and represent the typically absurd use of adverbs and nouns of the RPG (and fantasy literature too). I first found a "Famous Brick," later hypnotized a "Nutty-Tank Fox," and then purchased a "Faucet of Wicked Lo." Here, Chung makes an effective joke out of the scrambled language that procedural generation is so conducive to.
He goes further. As each of the events burst into view, you gain experience, represented by a bar that fills up at the bottom, while the top of the screen hosts a bar that depicts the progress of each act. In the top left corner you're informed of the date and how many years your journey has taken. Chung writes in the readme file for Acre 6, that it's inspired by one of his favorite games, Progress Quest. It's a similar game that removes all the depicted action and larger systems of the RPG and leaves only progress bars in their wake.
Acre 6 essentially does the same as Progress Quest, but it also adds to that the map, which starts out as a site of exploration, but is overwhelmed by the pop-up events of the player's journey. It brings to mind the way that open-world games rarely leave players to explore a map alone, instead dividing them into various landmarks that can be navigated to with waypoints. It's become something of a running joke that every open-world Ubisoft game will decimate its map with an atlas of icons showing the locations of side quests, dialogue opportunities, and viewpoints. This is what Acre 6 satirizes with its own map of messy and overblown annotations.
You can download Acre 6 over on itch.io.
Chris Priestman
Chris is interested in videogames, architecture, and data errors. You'll find him rummaging around in lesser known corners of the internet.
acre 6Blendo GamesBrendon Chungdeconstructionfreewareprocedural generationrpg
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text:
Thursday January 16
Led by Emmy nominated, DGA and BAFTA® award-winning director Michael Apted throughout the decades, this groundbreaking documentary anthology has now reached 63 Up, gaining further illuminating insight into its premise of asking whether or not our adult lives are predetermined by our earliest influences and the social class in which we are raised. An issue as relevant to our society now as when the series first appeared. The documentary film reveals more life-changing decisions, more shocking announcements and joy and tears in equal measure.
The original 7 Up was broadcast as a one-off World in Action Special inspired by the founding editor Tim Hewat's passionate interest in the Jesuit saying, "Give me the child until he is seven, and I will give you the man," and his anger at what he saw as the rigidity of social class in England. 7 Up featured the children talking about their hopes and dreams for the future. As members of the generation who would be running the country by the year 2000, what did they think they would become? The result was ground-breaking and the follow-up films every seven years have documented the group as they became adults and entered middle-age, dealing with everything life has thrown at them in between. Now, as the group reaches retirement age, the film is back to discover what they are doing as this new phase of life approaches. NR / 180 min.


text:
Q: Initializing AutoMapper in an Azure Function I am trying to create an Azure function in which I am using some code with AutoMapper. I am quite new to C#, Azure and AutoMapper and I'm having some trouble finding the correct way of initializing my AutoMapper configuration.
MapInitializer.cs:
public static class MapInitializer
    {
        public static void Activate()
        {
            Mapper.Initialize(cfg =>
            {
                // initialize mappings here
            });
        }
    }

Then in my function, I am trying to do the following:
Function.cs:
public static class ProcessQueueForIntercom
    {

        [FunctionName("ProcessQueue")]
        public static void Run([QueueTrigger("messages")]string myQueueItem, TraceWriter log)
        {
            MapInitializer.Activate();

            // rest of the code
        }
    }

Now the problem is, the first time I processed a message with this function, everything went smoothly and the code ran as I expected. However, from the second time on, I get an error saying my configuration is already initialized. But I don't really have an idea on how to do this properly with an Azure Function, since normally you would initialize this in the App Startup, but I don't think there is such a thing for Azure Functions (CMIW), and I'm not finding much information about how to exactly do this. I was thinking about just surrounding the Activate() call with a try catch and just log a warning that the configuration is already loaded, but that doesn't seem very clean...

A: You only need to call Activate once. You can do it from a static constructor:
public static class ProcessQueueForIntercom
{
    static ProcessQueueForIntercom()
    {
        MapInitializer.Activate();
    }

    [FunctionName("ProcessQueue")]
    public static void Run([QueueTrigger("messages")]string myQueueItem, TraceWriter log)
    {
        // rest of the code
    }
}

Or just make a static constructor on MapInitializer itself.
See also this answer.

A: As the static Mapper.Initialize API is now officially dead, I think the best solution is to inject a mapper in the function class constructor. So, assuming you want to map User model to UserResponse DTO, you need to first configure an automapper in your Startup.cs like this
using AutoMapper;
using Microsoft.Azure.Functions.Extensions.DependencyInjection;
using Microsoft.Extensions.DependencyInjection;
using YourNamespace.Models;

[assembly: FunctionsStartup(typeof(YourNamespace.Startup))]

namespace YourNamespace
{
    class Startup : FunctionsStartup
    {
        public override void Configure(IFunctionsHostBuilder builder)
        {
            var mappperConfig = new MapperConfiguration(cfg =>
                {
                cfg.CreateMap<User, UserResponse>();
                // you can have the maps in a dedicated Profile subclass if you like
            });
            builder.Services.AddSingleton(mappperConfig.CreateMapper());
        }
    }
}

Then, you inject the automapper to your functions class like this:
public class UsersApi
{
    private readonly IMapper _mapper;

    public UsersApi(IMapper mapper)
    {
        _mapper = mapper;
    }

    [FunctionName(nameof(UsersApi.GetUser))]
    public async Task<IActionResult> GetUser(
        [HttpTrigger(AuthorizationLevel.Anonymous, "get", Route = "users/{id}")] HttpRequest req,
        ILogger log,
        string id)
    {
        User user = await UserRepository.GetUser(id);
        UserResponse res = _mapper.Map<UserResponse>(user);
        return new OkObjectResult(res);
    }
}

redpajama_set_name: RedPajamaStackExchange


Richard Whittle at the New York Post offers a piece on a surreal day in America's war against ISIS.
One hundred and fifty years ago, in 1866, the town of Waterloo, NY, celebrated the first Memorial Day, remembering the fallen of the Civil War. Fifty years ago, President Lyndon B. Johnson honored Waterloo and reaffirmed the Memorial Day celebration, then centered around the heroes of World War II.
This Memorial Day, we're fighting a different kind of war — one that would be unrecognizable to our forefathers.
Though it involves thousands of Americans, and a good chunk of our economy, the war against the Islamic State is fought mostly in front of computer screens, in places like Langley, Va., and upstate Syracuse.
It is, in many ways, a surreal conflict, one where some combatants can get a latte after the battle.
For that reason, it can be easy for Americans to forget we're even fighting a war. Though our airmen and women fly hundreds of missions a day, it's not uncommon for the conflict to go unmentioned on the news for weeks.
But the stakes are just as high, as ISIS brutalizes the people of Iraq and Syria and preaches global jihad. Also, while the United States mostly conducts airstrikes at a remove, there are still 4,000 US troops in Iraq. Three have lost their lives in this war that we don't call a war. It's called Operation Inherent Resolve.


text:
The Immortal (Review)
nother from an outstanding collection of space warfare stories, Space Fighters, accumulated by Joe Haldeman way back in 1988. This time, it's Gordon Dickson's The Immortal.
So a grissled space pilot gets a wakeup call. He needs to head to HQ (on Earth). Here, he's told that he's going to fly a top secret mission with a "specialist" civillian (who is also trained as a gunner). Turns out we're in an endless war with a bunch of baddies, out towards Polaris. They can't get around us, we can't get around us, so it's endless grinding war between us. We've got ships that can jump (limited to calculations). And he's on his way out to the fronteer with another four ships. And why? Because a ship lost decades ago, a French-Canadian scout supposedly lost to some special bomb, is returning from deep in enemy space, flying its endless way home. Is it haunted? Is the guy alive? We won't know if the enemy finds him first, so off our hero goes on his suicide mission.
I thought the story really rolled well. Of course the hero dosen't like his guy-in-back (it made for interesting tension, them bickering at each jump point). And when they do find the ship, La Chasse Gallerie, chugging along, of course here come the enemy fighters. And then it's grab-him-in-our-tractor-beam and jump! But of course, deep in enemy territory, the hosiles can rely on planet-based computers to recalculate the jumps (as opposed to the heroes, who have their dinky ship's comps). And so, with every jump, the enemy is there a little sooner, the heroes lose ships, and the La Chasse Gallerie continues along, singing softly to itself, pitted from the ages and slashed from enemy fire, a flying wreck but still homeward bound, still moving Earthward.
Yes, spookie and interesting and tense, a great read. I checked online and couldn't find any copies so this one you'll have to hunt down yourself. But I liked it – very good storytelling from a time when scifi was really hardcore.
>>>HISTORICAL FICTION AND TIME-TRAVEL BOOKS HERE, MY OWN HUMBLE OFFERINGS. HAVE A LOOK. WHAT'S IT GOING TO COST YOU?<<<


text:
Trail planned to round White Bear Lake gets big…
Trail planned to round White Bear Lake gets big chunk of funding
By Bob Shaw | bshaw@pioneerpress.com | Pioneer Press
PUBLISHED: June 1, 2017 at 5:26 pm | UPDATED: June 1, 2017 at 5:43 pm
A plan for a trail around White Bear Lake has received a boost from the state.
The state's bonding bill includes $225,000 for the Lake Links project, which seeks to build a 10-mile walking and biking trail.
Out of the total, $130,000 will go to the city of White Bear Lake to help pay for an extension of city trails from Lion's Park to South Shore Boulevard. That work should be completed in 2018.
The rest of the money will pay for engineering in several areas: South Shore Boulevard in White Bear Lake and White Bear Township, Birchwood Road and Hall Avenue between Mahtomedi and Birchwood, and Mahtomedi and Dellwood avenues on the northeast side of the lake.
'Give Kids a Smile' program offers free dental care Feb. 7 & 8
Ice-castle installation opens Friday in New Brighton. It's expected to be popular.
St. Anthony City Council appoints member to fill vacant seat
State orders 3M to form new plan to contain PFAS at Oakdale site after water contamination found
"It feels good to reactivate a project that has sat dormant for so long, yet everyone seems to want to succeed," said Mike Brooks, Ramsey County co-chair of the Links Trail Association in an email.
The funding is a success for the association, which was formed four months ago.
"This is amazingly rapid progress on a project which has languished for 16 years," said Lake Links Washington County co-chair Steve Wolgamot in a news release.
"It simply demonstrates the enormous public support for walking, for biking, for the safety of our children and for the mobility rights of disabled Minnesotans."
Bob is a 40-year veteran (yes, he is grizzled) who edited one Pulitzer Prize winner and wrote two that were nominated. He has also worked in Des Moines, Colorado Springs and Palo Alto. He writes about the suburbs, the environment, housing, religion -- anything but politics. Secret pleasures: Kayaking on the Mississippi on the way to work, doughnuts brought in by someone else. Best office prank: Piling more papers onto Fred Melo's already trash-covered desk.
Follow Bob Shaw	bob.shaw.980967 Follow Bob Shaw	@BShawPP


text:
Alla Pugacheva showed pranks son
Russian pop star doted in their three year old twins Lisa and Harry. The singer has shared with fans a new video with the participation of the son, which caused a storm of emotion among its subscribers.
Thousands of followers of the page of Alla Pugacheva in the "instagrame" watching what is happening in the life of a star, her husband Maxim Galkin and their young children Harry and Lisa. Diva decided to show how children dabble and play at home.
The star has posted on his page a video in which her three year old son tries to climb into a wicker basket with Christmas decorations. The boy was old enough, but he manages to fit in such a small area. Climbing with the feet in the basket, the boy along with her lover. Standing next to a nanny who watches the antics of the baby and helps him to his feet. Harry laughs so that his cheeks are dimpled as the Pope.
Alla Pugacheva has shared a touching family photo
Fans Pugacheva and Galkin noted that the boy is very grown up. "Gorgeous baby, beautiful", "What is the wisdom of perception – not to interfere, but to observe and gently suggest", "What a beautiful boy, a clever boy, a real man doesn't cry" – such nice comments left by subscribers of the pages of prima donnas.
Video posted by ALLA PUGACHEVA PHOTOS & VIDEOS (@allaprimadonna) May 23 2016 1:27 PDT
The children Pugacheva and Galkina are growing very talented. Star dad repeatedly told that son is a very independent boy. "Stood and patched the roof of a toy house, so she does not leak. All their actions, he always accompanies the review, however, not all words (except me and Alla) understandable to others. And here's a picture: Lisa on bike rides to the Garik, long stares at him and says: "Garik works, and Lisa's beautiful!" Rarely hear such a paradoxical sentence from the mouth of a child", – said Maxim in an interview.
According to the circle of the couple, Lisa is very similar to the mother. When the familiar stars, Dmitry Dibrov with his wife, visited castle Pugacheva and Galkin in the village of Mud, they noted that there is indescribable atmosphere of warmth and kindness. "Lisa sociable and musical, Harry brooding. Wonderful family. Real", – Dibrova so the wife described the children of Alla and Maxim. Children Dubrovich met with Alla Pugacheva
Lisa and Harry was born in September 2013. Parents loved in a charming babies. In interviews, Pugacheva has repeatedly said that as soon as Lisa and Harry grow up, they will definitely going to study at her school.
The soloist SEREBRO Pauline Tabor gave the car
Resident "Comedy Club" for the first time, told how he survived the death of his daughter


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Benzodiazepines are medications that are usually prescribed to patients who suffer from anxiety. However, they can be used to treat other conditions, such as insomnia and seizures. These medications work by depressing the activity of the nerves inside of the central nervous system. The effects of benzodiazepine can last anywhere from six to 24 hours. Xanax, Valium and Ativan are examples of some of the most commonly prescribed benzodiazepines.
Benzodiazepines are among the oldest anti-anxiety drugs. In fact, these medications have been around since the 1960s.
Physicians recommend a dosage based on the patient's health and the condition that is being treated. A benzodiazepine can be taken with or without food. However, if this drug causes an upset stomach, then people are encouraged to take it with milk or food. If this medication is prescribed to treat insomnia, then doctors typically recommend taking 30 to 60 minutes before going to bed. A physician may gradually decrease the dosage if one has been taking it for an extended period of time. Physicians may also adjust the dosage if a patient starts to develop side effects to this medication.
What Are Some of the Possible Side Effects of Benzodiazepines?
Some of the most common side effects of benzodiazepines include things like confusion, sedation, vomiting, nausea, dry mouth, decreased libido and fatigue. It is also important to note that benzodiazepines may interact with certain medications, such as antiacids, antidepressants and narcotics. That is why people are advised to inform their doctor of any medications that they are taking. Women who are pregnant or nursing will also need to inform their doctors.
Furthermore, a small percentage of people may experience an allergic reaction to this medication. Breathing problems, itching, swelling and hives are some of the possible signs of an allergic reaction. People who suspect they are having an allergic reaction should seek medical help as soon as possible.
It is important to note that it is very easy to become dependent on benzodiazepines. The longer a person takes a benzodiazepine, the more likely he or she is to become dependent on it. In fact, studies have shown that a person's risk of becoming addicted to a benzodiazepine increases even if he or she only takes the medication for four weeks.
People who take this medication for an extended period of time are likely to develop a tolerance to it. This means the benzodiazepine may become less effective over time. One will have to increase the dosage in order to experience the same effects. That is why people who take benzodiazepines for an extended period of time are more likely to develop an addiction to it.
People who abuse benzodiazepines are also at an increased risk of becoming addicted to it. Any time a person takes this medication in a way that is not intended by a physician he or she is abusing it. Some of the signs of benzodiazepine abuse include blurred vision, slurred speech, lack of coordination and breathing problems.
In many cases, people who abuse benzodiazepines experience worsening anxiety. A benzodiazepine addiction can be very dangerous. People who abuse benzodiazepines can increase their chances of developing dementia by up to 50 percent. It is estimated that nearly 350,000 emergency room visits in 2010 were linked to benzodiazepine addiction and abuse.
Furthermore, a benzodiazepine overdose can result in death. It is estimated that benzodiazepine overdose caused nearly 6,500 deaths in 2010. Even though benzodiazepine addiction is dangerous, people who withdraw from this medication may experience serious side effects. Some of those side effects include insomnia, depression and anxiety. People may also experience musculoskeletal and neurological problems.
Even though benzodiazepines are among the most commonly prescribed anti-anxiety medications, the number of prescriptions written for these drugs have decreased within the past few years. Many doctors are reluctant to prescribe this medication because of the risk of abuse and addiction.
Benzodiazepine addiction has increased drastically within the past few years. The number of people admitted to a drug treatment center for a benzodiazepine addiction increased by 79 percent from 1992 to 2002.
Why Inpatient Rehab is the Best Choice for Benzodiazepine Addiction?
Benzodiazepine addiction is nearly impossible to break without medical help. If you or one of your loved ones is battling benzodiazepine addiction you should consider enrolling in inpatient rehab. One of the benefits of being in inpatient rehab is that the addicted person will get the constant support and supervision that he or she needs.
The skilled and compassionate professionals at a inpatient rehab will help a person develop the skills and confidence he or she needs to live a drug-free life.


Situated just a 5 metres away from the iconic Kuta Beach, Kutabex Boutique Hotel & Spa offers modern beachfront accommodation with free WiFi access in all areas of the property.
It takes about a 5-minute walk from the property to Beachwalk Shopping Centre, and also a 5-minute walk in another direction to Hard Rock Cafe. Getting to Bali Denpasar International Airport will take around 20 minutes by car. Airport shuttle service is available at an additional fee.
Each room here will provide you with air conditioning, flat-screen cable TV, and an electric kettle with tea/coffee making facilities. There is also an en suite bathroom featuring hot and cold shower facility and free toiletries. Towels are included in the room, whilst hairdryer and ironing facility are available upon request.
At Kutabex Boutique Hotel & Spa you will find a 24-hour front desk with concierge service, luggage storage, and currency exchange service. Other facilities at this property include car hire, laundry and dry cleaning service meeting/banquet facility are available with extra charges. Guests can also indulge in relaxing massage treatments at the on-site spa. Activities like windsurfing, diving, snorkeling, fishing, and cycling can be arranged at the property.
For dining option the on-site Kutabex Restaurant serves Indonesian and western dishes. Room service is also available.