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Home » Music » Britney Spears' tragic family life – graveside suicide to bitter custody battle
Britney Spears' tragic family life – graveside suicide to bitter custody battle
Britney Spears has suffered more than her fair share of heartbreak and devastation in her life, but it didn't just start with the 40-year-old pop sensation.
From her grandmother's tragic death to Britney's own struggles to have rights to her children and her own life, it's been a challenge for Britney.
Britney has only recently managed to get her life back under her own control following her public meltdown in 2007, and after years of fighting, celebrations erupted in November as she welcomed freedom.
While Britney is surrounded by the more positive scenes now with her engagement and seemingly happy life, it's not been smooth sailing for the Spears family – even before Britney was born.
Grandmother's tragic suicide
The Spears name faced tragedy far before Britney's own tough experiences, with her paternal grandmother suffering from depression and taking her own life at the age of 31.
US Magazine previously reported that Emma Jean Spears had killed herself at the grave of her three-day-old son eight years after his untimely death.
It was reported at the time that Emma had tried to kill herself three times before her fatal incident, leaving behind four children including Britney's father, Jamie.
Jamie Lynn Spears abortion 'pushing'
Jamie Lynn Spears famously got pregnant aged 16 which ultimately caused her hit TV show, Zoey 101, to end its reign on Nickelodeon.
However, years later, in her forthcoming memoir, Jamie is now claiming that her parents repeatedly tried to convince to her consider abortion or adoption.
In extracts from the memoir obtained by TMZ, she recalls her parents telling her that having a baby at that point in her life was a "terrible idea".
She alleged that her parents said: "It will kill your career. You are just too young. You don't know what you're doing. There are pills you can take. We can help you take care of this problem … I know a doctor."
She added: "Everyone around me just wanted to make this 'issue' disappear…everyone was certain that termination would be the best course of action."
Ultimately, Jamie Lynn ignored the alleged pushing and went on to have her daughter, Maddie, in 2008.
Britney's custody battle
Britney and Kevin Federline met in 2004 and after three months they announced their engagement, before getting married in September of that same year.
The following year, Britney had her first son Sean and welcomed Jayden a year after that.
By November 2006, Britney filed for divorce and it was settled in March 2007.
She lost custody of her two children in 2007 following her public breakdown which saw her shave her own head before attacking a photographer's car with an umbrella.
In 2008, Britney was arrested after she locked herself in a bathroom with Jayden which only ended when she was taken home in an ambulance following a four-hour standoff.
After that, the pair agreed to 50-50 custody, but after an altercation between Sean and his grandfather Jamie, Kevin was granted 70% custody of both children.
Father's illness
Britney, who previously relied heavily on her father, was hit with concern when Jamie Spears fell ill in late 2018, with the star pressing pause on her Domination Las Vegas residency.
Jamie was said to have suffered a ruptured colon and was put in hospital for nearly a month.
At the time, a source told Us Weekly that Britney was left devastated over the ordeal and chose to enter a treatment facility.
They said: "She's been depressed and hasn't been able to find a way out of it. After Jamie's recent surgery, she found herself so full of anxiety and panic, she didn't know who to turn to or what to do. She's been spiralling and feeling lost."
Conservatorship battle
One year after Britney's public breakdown in 2007, her father Jamie was placed in control of the star's assets, finances and rights.
Now, 13 years after the reigns were handed over to Jamie, a judge has granted the termination of Britney's conservatorship, allowing the star freedom to make her own decisions.
Britney told a court in June: "I want to be able to get married and have a baby. I was told right now in the conservatorship I am not able to get married and have a baby."
The conservators wouldn't let her have her birth control removed, she claimed.
She stated that she was forced to go on tour, made to take medications she didn't want and go into rehab.
Jamie has since said that he loves his daughter "unconditionally" and has "tried to do what is in her best interests".
He agreed to step down as her conservator and it was officially approved this year.
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\section{Introduction}

Since the remarkable paper of S. K. Donaldson [D1] appeared in 1983, Yang-Mills gauge theory has been one of central themes in geometry and mathematical physics. There are plenty of papers on this subject, and many interesting theories have been developed.

But most of them study only \textit{finite energy} ASD connections and their \textit{finite dimensional} moduli spaces.
 We know little about \textit{infinite energy} ASD connections and their moduli spaces.
In fact I don't know even whether such connections exist in general situations.
This paper is one step in this direction.
We will show that we can construct a huge amount of infinite energy ASD connections by using a ``gluing construction''.

Gluing ASD connections is a well-known technique in Yang-Mills theory [D1], [D2], [DFK], [DK], [FU], [T1], [T2], but we usually study a gluing of two (or a finite number of) instantons.
In this paper we study a gluing of infinitely many instantons, and show that
this new gluing technique produces the desired infinite energy ASD connections.
Moreover we show that the construction of ``gluing an infinite number of instantons"
has the following interesting new feature;
To define the gluing of ASD connections, we need to introduce a ``gluing parameter space",
which is isomorphic to $SU(2)$ in the case of gluing two $SU(2)$ instantons.
In the case of gluing infinitely many $SU(2)$ instantons, the gluing parameter space becomes
the \textit{infinite product} of the copies of $SU(2)$.
This gigantic parameter space is the interesting new feature of our approach.

Although the above ``infinite dimensional gluing parameter space" is the source of the richness
of the construction, it also causes several serious difficulties in analysis.
In particular, we cannot use the usual Fredholm deformation theory
because of infinite degrees of freedom of the parameter space.
We overcome this difficulty by adapting Donaldson's ``alternating method'' in [D2] to this situation. This decomposes the problem into an infinite number of small pieces where we can use the usual Fredholm deformation theory.

I think that the technique of ``gluing an infinite number of instantons"
probably can be applied to other equations
\footnote{After I posted a preprint version of this paper to the arXiv in 2005,
I was informed that there is a paper of M. Macr\`{i}, M. Nolasco and T. Ricciardi [MNR] which
studies an ``infinite energy" situation for selfdual vortices in $\mathbb{R}^2$.
From [MNR] I learned that there is a paper of S. Angenent [A]
which develops a gluing technique for infinitely many solutions of an elliptic PDE
of the form $\Delta u + f(x, u) = 0$ in $\mathbb{R}^n$.
[MNR] adapts this gluing technique to an elliptic equation with singular sources.
I think that our alternating method is different from the analytic technique in [A].} , e.g., the Seiberg-Witten equation, pseudo-holomorphic curves, etc.
In fact the classical Mittag-Leffler theorem in complex analysis can be considered as a variant of
this gluing technique.
Singularities of a meromorphic function are instanton-like objects, and
the Mittag-Leffler theorem constructs a meromorphic function which has
an infinite number of singularities given in the complex plane.
In other words the result of this paper is a Yang-Mills analogue of the Mittag-Leffler theorem.

The organization of the paper is as follows.
In Section 2 we present notation and state  main theorems. We establish basic estimates in Section 3.
The details of the gluing construction are given in Section 4. We discuss a moduli problem in Section 5.

\vspace{2mm}

\textbf{Acknowledgement.}\hspace{0.7mm}
 I wish to thank Professor Kenji Fukaya for his suggestion and help during the course of this work.

\section{Main Results}
\subsection{Brief review of Yang-Mills theory}\label{subsection: review of Yang-Mills}
To begin with, we review some basic notions of Yang-Mills theory. For the details, see [DK] or [FU].
Let $G$ be a Lie group $SU(2)$ or $SO(3)$.  Let $C(G)$ be the center of $G$,
 i.e., $C(SU(2)) = \{\pm 1\}$ and $C(SO(3)) = 1$.
Suppose that $X$ is a closed oriented Riemannian 4-manifold and that
$E \rightarrow X$ is a principal $G$ bundle over $X$.
Let $\mathrm{ad}E$ be the Lie algebra bundle associated with $E$.
A connection $A$ on $E$ is said to be irreducible if the isotropy group of $A$ in the gauge group
is equal to $C(G)$. ($C(G)$ acts trivially on all connections.)

A connection $A$ is said to be anti-self-dual, or ASD, if the  self-dual part of its curvature is zero:
\[ F^+_A = 0.\]
If $A$ is an ASD connection, then we have the Atiyah-Hitchin-Singer complex for $A$:
\[ 0 \rightarrow \Omega^0(\mathrm{ad}E) \xrightarrow{d_A} \Omega^1(\mathrm{ad}E) \xrightarrow{d^+_A} \Omega^+(\mathrm{ad}E) \rightarrow 0,
\]
where $\Omega^+$ is the space of self-dual 2-forms and $d^+_A$ denotes the self-dual part of $d_A$.
An ASD connection $A$ is said to be acyclic if the above complex is exact (acyclic).
The Atiyah-Hitchin-Singer complex is exact if and only if
\[d^*_A + d^+_A : \Omega^1(\mathrm{ad}E) \rightarrow
\Omega^0(\mathrm{ad}E) \oplus \Omega^+(\mathrm{ad}E) \]
is isomorphism. Here $d^*_A$ is the formal adjoint of $d_A : \Omega^0(\mathrm{ad}E) \rightarrow \Omega^1(\mathrm{ad}E)$.
$A$ is said to be regular if $A$ is irreducible and $d^*_A + d^+_A$ is surjective.
\begin{Exa}
Let $X$ be the 4-sphere $S^4$ with the usual round metric.
All non-flat $G$ ASD connections on $S^4$ are regular.
This follows from the Weitzenb\"{o}ck formula. For example, see [FU, Chapter 6].
\end{Exa}
\begin{Exa}
$K3$ surface has a non-flat, irreducible, acyclic $SO(3)$ ASD connection. See [DK, Section 9.1.3].
\end{Exa}

\subsection{Gluing construction}\label{subsection: gluing construction}
Next we introduce the following convenient definition.
 \begin{Def}\label{def: gluing data}
A 5-tuple $(X,x^L,x^R,E,A)$ is called a gluing data if it satisfies the following (i)$\sim$(v):
     \begin{enumerate}
       \item $X$ is a closed oriented Riemannian 4-manifold.
       \item $x^L$ and $x^R$ are two distinct points in $X$.
       \item $E$ is a principal $G$ bundle over $X$.
       \item $A$ is an irreducible acyclic ASD connection on $E$.
(We can replace this condition with the weaker condition that $A$ is a regular ASD connection.
See the remark at the end of Section 5.)
       \item The metric on $X$ is flat in some neighborhoods of $x^L$ and $x^R$. (See the remark below.)
    \end{enumerate}

\end{Def}
\begin{Rem}
The condition (v) in the Definition \ref{def: gluing data} is artificial and just for simplicity.
We can construct our gluing argument without this condition; see [D2, (4.48) Proposition].
Actually, this condition is allowable in the following sense.
Let $g$ be a Riemannian metric on $X$ which does \textit{not} satisfy the condition (v).
Let $A$ be an irreducible acyclic (resp. regular) $g$-ASD connection on $E$.
Then small perturbation gives a new metric $g'$ satisfying the condition (v) and
an irreducible acyclic (resp. regular) $g'$-ASD connection $A'$.
This fact follows from the usual implicit function theorem.
\end{Rem}
\begin{Def}\label{def: finite type}
   Let $(X_i,x^L_i,x^R_i,E_i,A_i)$, $(i \in \mathbb{Z})$, be a sequence of gluing data indexed by integers.
 This sequence is said to be \textit{of finite type} if it satisfies the following:
\[ \{(X_i,x^L_i,x^R_i,E_i,A_i) |i \in \mathbb{Z}\} \text{ is a  finite set}. \]
(Of course, the sequence itself is an \textit{infinite} sequence.)
This condition assures the uniformity of analysis and geometry on each data.
\end{Def}
We will define a connected sum of gluing data. This construction is a generalization of the ``conformal connected sum'' construction in [D2, pp. 306-307] and [DK, Section 7.2.1].

 Let $(X,x^L,x^R,E,A)$ be a gluing data. By the condition (v) in Definition \ref{def: gluing data}, there are an oriented Euclidean coordinate $\xi$ centered on $x^R$ and
an oriented Euclidean coordinate $\eta$ centered on $x^L$.
We set $k := 0.9$, and
we define the annular region $\Omega^R$ and the ``shells'' $R^-$ and $R^+$ in the neighborhood of $x^R$  by
\begin{equation*}
 \begin{split}
  \Omega^R &:= \{ \xi \,|\, kN^{-1}\sqrt{\lambda}< |\xi| < k^{-1}N\sqrt{\lambda}\} ,\\
  R^- &:= \{\xi \,|\, kN^{-1}\sqrt{\lambda}< |\xi| < N^{-1}\sqrt{\lambda}\} ,\\
  R^+ &:= \{\xi \,|\, N\sqrt{\lambda}< |\xi| < k^{-1}N\sqrt{\lambda}\} .
 \end{split}
\end{equation*}
Here $N$ and $\lambda$ are positive parameters.
We define $\Omega^L$, $L^-$ and $L^+$ in the same way by using the Euclidean coordinate $\eta$ centered on $x^L$.
So $L^-$ is the inner ``shell" of $\Omega^L$ and $L^+$ is the outer ``shell";
 see Figure \ref{fig: gluing data}.
(Figure \ref{fig: gluing data} is a variant of [D2, Diagram (4.13)].)
We define the open set $U$ by
\[ U := X \setminus (\Bar{B}(x^R, kN^{-1}\sqrt{\lambda}\,)\cup \Bar{B}(x^L,kN^{-1}\sqrt{\lambda}\,)). \]
Here $\Bar{B}(x^R, kN^{-1}\sqrt{\lambda}\,)$ and $\Bar{B}(x^L,kN^{-1}\sqrt{\lambda}\,)$
are the balls $|\xi|\leq kN^{-1}\sqrt{\lambda}$ and $|\eta|\leq kN^{-1}\sqrt{\lambda}$.

In the later gluing argument, ``error terms" will be supported in the shells $R^{\pm}$ and $L^{\pm}$.
The volumes of these shells are determined by the parameters $\lambda$ and $N$.
We will choose $\lambda$ small and $N$ large.
The value of $N$ will be fixed in the end of Section 3.
We choose $\lambda$ so small that
\begin{equation}
\lambda N^{100} \ll 1. \label{lambda is small}
\end{equation}
Here $``100"$ does not have a particular meaning. The point is that $\lambda$ is very small.
(In the above, we set $k = 0.9$. This value $``0.9"$ does not have a particular meaning, either.
The point is that $k$ is smaller than 1.)

   \begin{figure}[h]
      \psfrag{1}{$X$}
      \psfrag{2}{$x^L$}
      \psfrag{3}{$x^R$}
      \psfrag{4}{$U$}
      \psfrag{5}{$\Omega^L$}
      \psfrag{6}{$\Omega^R$}
      \psfrag{7}{$L^-$}
      \psfrag{8}{$L^+$}
      \psfrag{9}{$R^+$}
      \psfrag{10}{$R^-$}
      \psfrag{11}{$U_i$}
      \psfrag{12}{$U_{i+1}$}
      \psfrag{13}{$R^+_i = L^-_{i+1}$}
      \psfrag{14}{$R^-_i = L^+_{i+1}$}
      \psfrag{15}{$\Omega^R_i = \Omega^L_{i+1}$}
      \psfrag{16}{$x^R$}
      \psfrag{17}{$R^-$}
      \psfrag{18}{$R^+$}
      \psfrag{19}{$\Omega^R$}
      \psfrag{20}{$ k N^{-1} \sqrt{\lambda}$}
      \psfrag{21}{$ N^{-1} \sqrt{\lambda}$}
      \psfrag{22}{$N \sqrt{\lambda}$}
      \psfrag{23}{$ k^{-1} N \sqrt{\lambda}$}
      \psfrag{24}{$\sharp_{i\in\mathbb{Z}}X_i$}

     \begin{flushleft}
       \includegraphics[scale = 1]{gluing-data.eps}
     \end{flushleft}
     \caption{A gluing data and a connected sum.}  \label{fig: gluing data}
     \end{figure}

 Let $(X_i,x^L_i,x^R_i,E_i,A_i)_{i\in \mathbb{Z}}$ be a sequence of gluing data of finite type.
We have $\Omega^R_i$, $R^-_i$, $R^+_i$, and $\Omega^L_i$, $L^-_i$, $L^+_i$, and $U_i$ for each $i\in\mathbb{Z}$.
 We define the connected sum
\[\sharp_{i\in\mathbb{Z}}X_i :=  \cdots \sharp X_{-1} \sharp X_0 \sharp X_1 \sharp
\cdots.
\]
by identifying $\Omega^R_i$ and $\Omega^L_{i+1}$.
The identification map between $\Omega^R_i$ and $\Omega^L_{i+1}$ is defined by
\begin{equation}
 \eta = \lambda \bar{\xi}/|\xi|^2.  \label{conformal map}
\end{equation}
Here $\xi$ is an oriented Euclidean coordinate centered on $x^R_i$,
and $\eta$ is an oriented Euclidean coordinate centered on $x^L_{i+1}$.
$\xi$ $\mapsto$ $\bar{\xi}$ is any reflection.
$R^-_i$ and $R^+_i$ are identified with $L^+_{i+1}$ and $L^-_{i+1}$, respectively.
The connected sum space $\sharp_{i\in\mathbb{Z}}X_i$ is covered by the open sets $U_i$:
\[ \sharp_{i\in\mathbb{Z}}X_i = \bigcup_{i\in \mathbb{Z}}U_i. \]
Since the identification map (\ref{conformal map}) is conformal,
$\sharp_{i\in\mathbb{Z}}X_i$ has a natural conformal structure which is compatible with the metrics of the given manifolds $X_i$.
(Note that ASD condition is conformally invariant.)

Next we will define a gluing of the principal $G$ bundles $E_i$.
\begin{Def}
We define the \textit{gluing parameter space} GlP
and the \textit{effective gluing parameter space} EGlP by
\begin{equation*}
 \begin{split}
  \mathrm{GlP} &:= \prod_{i\in\mathbb{Z}}\mathrm{Hom}_{G}((E_i)_{x^R_i},(E_{i+1})_{x^L_{i+1}})
  \; \cong G^{\mathbb{Z}},\\
  \mathrm{EGlP} &:= \prod_{i\in\mathbb{Z}}(\mathrm{Hom}_{G}((E_i)_{x^R_i},(E_{i+1})_{x^L_{i+1}})/C(G))
  \; \cong (G/C(G))^{\mathbb{Z}}.
 \end{split}
\end{equation*}
Here
$\mathrm{Hom}_{G}((E_i)_{x^R_i},(E_{i+1})_{x^L_{i+1}})$ is the set of $G$-isomorphisms between the fibres $(E_i)_{x^R_i}$ and $(E_{i+1})_{x^L_{i+1}}$, and it is isomorphic to $G$.
\end{Def}
If $x\in X_i$ is a point in a small neighborhood of $x^R_i$,
then the fibre $(E_i)_{x}$ can be identified with $(E_i)_{x^R_i}$
by using the parallel transport along the radial line from $x^R_i$ to $x$.
This ``exponential gauge" centered on $x_i^R$ defines a local trivialization of $E_i$ over $\Omega^R_i$.
In the same way
we define a trivialization of $E_i$ over $\Omega^L_i$ by the exponential gauge centered on $x_i^L$.
(For the detail of exponential gauge, see [FU, Chapter 9] or [DK, Section 2.3.1].)
Let $\rho = (\rho_i)_{i\in \mathbb{Z}}\in \mathrm{GlP}$ be a gluing parameter.
$\rho_i$ is an isomorphism between the fibres $(E_i)_{x^R_i}$ and $(E_{i+1})_{x^L_{i+1}}$.
Since we have the identification $\Omega^R_i \cong \Omega^L_{i+1}$ in (\ref{conformal map}) and
the bundle trivializations $E_i|_{\Omega^R_i} \cong \Omega^R_i \times (E_i)_{x^R_i}$ and
$E_{i+1}|_{\Omega^L_{i+1}} \cong \Omega^L_{i+1} \times (E_{i+1})_{x^L_{i+1}}$,
$\rho_i$ defines an identification map between $E_i|_{\Omega^R_i}$ and $E_{i+1}|_{\Omega^L_{i+1}}$
covering the base space identification (\ref{conformal map}).
(See the diagram below.)
We define a principal $G$ bundle $\bm{E}(\rho)$ over $\sharp_{i\in\mathbb{Z}}X_i$ by using this identification map.
\begin{equation}\label{bundle gluing}
\begin{CD}
E_i|_{\Omega^R_i} @>{\text{exp. gauge}}>>  \Omega^R_i \times (E_i)_{x^R_i}\\
@VVV @VV{\rho_i}V \\
E_{i+1}|_{\Omega^L_{i+1}} @>{\text{exp. gauge}}>>  \Omega^L_{i+1} \times (E_{i+1})_{x^L_{i+1}}
\end{CD}
\end{equation}

The gluing construction in Section 4 will give an ASD connection $\bm{A}(\rho)$ on $\bm{E}(\rho)$ which is ``close'' to $A_i$ over each $U_i$:
\begin{Thm}\label{thm: construction}
For any positive number $\varepsilon >0$ and $p \ge 1$,
we can choose the parameters $N$ and $\lambda$ so that
there exists an ASD connection $\bm{A}(\rho)$ on $\bm{E}(\rho)$ for all $\rho \in \mathrm{GlP}$
such that
\[ |\!|\bm{A}(\rho) - A_i|\!|_{L^p(U_i)} < \varepsilon \quad \text{for all}\ i \in \mathbb{Z}.  \]
(Since there is a natural identification between $\bm{E}(\rho)$ and $E_i$ over $U_i$,
$\bm{A}(\rho) - A_i$ is well defined over $U_i$.)
Moreover, if there are infinitely many non-flat ASD connections $A_i$ in the given sequence of gluing data,
then the ASD connection $\bm{A}(\rho)$ has infinite $L^2$-energy:
\[\norm{F(\bm{A}(\rho))}_{L^2} = + \infty. \]
\end{Thm}
In the above, the norm $\norm{\cdot}_{L^p(U_i)}$ is defined by using the Riemannian metric on $X_i$.
Since the $L^2$-norm of a 2-form is conformally invariant, the $L^2$-norm of the curvature $\norm{F(\bm{A}(\rho))}_{L^2}$
is well-defined. ($\sharp_{i\in\mathbb{Z}}X_i$ has the natural conformal structure.)
We will prove a more precise result in Section 4.

In Section 5 we will discuss a moduli problem. Our main result is the following.
(Here the parameters $N$ and $\lambda$ are chosen appropriately.)
\begin{Thm}\label{thm: injectivity}
Let $\rho$ and $\rho'$ be two gluing parameters.
Then $\bm{A}(\rho)$ is gauge equivalent to $\bm{A}(\rho')$ if and only if $[\rho] = [\rho']$ in $\mathrm{EGlP}$.
Here the gauge equivalence is defined by usual smooth bundle maps between the principal $G$ bundles $\bm{E}(\rho)$ and $\bm{E}(\rho')$.
\end{Thm}
The effective gluing parameter space $\mathrm{EGlP} = (G/C(G))^{\mathbb{Z}}$ has infinite degrees of freedom. Hence we can consider that our ASD connections $\bm{A}(\rho)$ have infinite dimensional parameter space.

\subsection{Outline of the alternating method}\label{subsection: outline}
We will construct the ASD connection $\bm{A}(\rho)$ by applying
``Donaldson's alternating method" in [D2]. Here we explain the basic idea of it.
The alternating method is an iterative process for solving the ASD equation.
First we construct the initial approximate solution $\bm{A}^{(0)}(\rho)$ on $\bm{E}(\rho)$
by using cut-off functions.
Of course, this $\bm{A}^{(0)}(\rho)$ is not ASD in general, i.e., $F^{+}(\bm{A}^{(0)}(\rho)) \neq 0$.
The alternating method iteratively improves this $\bm{A}^{(0)}(\rho)$ as follows;
\begin{description}
\item[Step 0]
To cancel the ``error term" $F^{+}(\bm{A}^{(0)}(\rho))$,
we slightly perturb $\bm{A}^{(0)}(\rho)$ over each $X_{2i}$
and construct $\bm{A}^{(1)}(\rho)$ on $\bm{E}(\rho)$
by using these perturbations and cut-off functions.
The new error term $F^{+}(\bm{A}^{(1)}(\rho))$ becomes smaller than
the old error term $F^{+}(\bm{A}^{(0)}(\rho))$.
\item[Step 1]
To cancel the error term $F^{+}(\bm{A}^{(1)}(\rho))$,
we slightly perturb $\bm{A}^{(1)}(\rho)$ over each $X_{2i+1}$
and construct $\bm{A}^{(2)}(\rho)$ on $\bm{E}(\rho)$
by using these perturbations and cut-off functions.
The new error term $F^{+}(\bm{A}^{(2)}(\rho))$ becomes smaller than
the old error term $F^{+}(\bm{A}^{(1)}(\rho))$.
\end{description}
Step 2 is the same as Step 0, and Step 3 is the same as Step 1. In general,
\begin{description}
\item[Step $(2n)$]
We slightly perturb $\bm{A}^{(2n)}(\rho)$ over each ``even" manifold $X_{2i}$
and construct $\bm{A}^{(2n +1)}(\rho)$.
\item[Step $(2n+1)$]
We slightly perturb $\bm{A}^{(2n+1)}(\rho)$ over each ``odd" manifold $X_{2i+1}$
and construct $\bm{A}^{(2n+2)}(\rho)$.
\end{description}
The error terms $F^{+}(\bm{A}^{(n)}(\rho))$ go to $0$ as $n$ goes to infinity,
and the sequence of the approximate solutions $\bm{A}^{(n)}(\rho)$ converges to
the desired ASD connection $\bm{A}(\rho)$.
This is the outline of the alternating method in Section 4.

\section{Main Estimates}
In this section we establish basic estimates for the proofs of Theorem \ref{thm: construction} and
\ref{thm: injectivity}.
These are essentially the reproduction of the estimates in [D2, (4.22) Lemma and (4.24) Remarks].
We follow the arguments of [D2, IV (iii)].

Fix $p >6$ and let $(X, x^L, x^R, E, A)$ be a gluing data.
We have open sets $L^{\pm}$, $R^{\pm}$ and $U$ on $X$.
Let us consider the following map over $X$:
\begin{equation}\label{map: perturb}
 \begin{split}
 L^p_1(\Omega^1 (\mathrm{ad}E)) &\times L^{2p}(\Omega^1 (\mathrm{ad}E))
 \rightarrow L^p(\Omega^0 (\mathrm{ad}E) \oplus  \Omega^+ (\mathrm{ad}E)),  \\
 &( b , a ) \mapsto D_A b + [a \wedge b]^+ + (b \wedge b)^+.
 \end{split}
\end{equation}
Here $D_A$ is the differential operator $d^*_A + d^+_A : \Omega^1(\mathrm{ad}E)
 \rightarrow \Omega^0 (\mathrm{ad}E) \oplus \Omega^+(\mathrm{ad}E)$.
Since we suppose that $A$ is acyclic in Definition \ref{def: gluing data},
$D_A$ becomes an invertible map from $L^p_1$ to $L^p$.
Hence if $a \in \Omega^1(\mathrm{ad}E)$ is small in $L^{2p}$
and $\sigma \in \Omega^+(\mathrm{ad}E)$ is small in $L^{p}$,
then we can solve the following equation for $b\in \Omega^1(\mathrm{ad}E)$:
\begin{equation} \label{eq: perturb}
D_A b + [a \wedge b]^+ +(b \wedge b)^+ = - \sigma,
\end{equation}\label{apriori estimate for b}
and we get a solution $b$ which satisfies
\begin{equation}
|\!|b|\!|_{L^p_1(X)} \leq  \mathrm{const} \cdot  |\!|\sigma|\!|_{L^p(X)}. \label{implicit}
\end{equation}
(This is because $b = 0$ is a solution to (\ref{eq: perturb}) if $\sigma = 0$.)

Suppose that $a$ is a smooth form defined on $U$ and that the self-dual part of the curvature
$\sigma := F^+(A + a)$ is supported in $L^+ \cup R^+$. ($\sigma$ is also defined on $U$.)
We extend $a$ and $\sigma$ to $X$ by zero, and set
\[ \delta := |\!|\sigma|\!|_{L^\infty (X)} .\]
The volume of $(L^+\cup R^+)$ which contains the support of $\sigma$ is bounded by
$\mathrm{const}\cdot  N^4 \lambda^2$.
Hence if $a$ is small in $L^{2p}$ and $\delta$ is bounded by some constant,
then we have a small solution $b$ to (\ref{eq: perturb}) for sufficiently small $\lambda$.
In the later gluing argument $\sigma$ will be the ``error term" and
$b$ will be the ``perturbation" described in Section \ref{subsection: outline}.

Let us introduce a cut-off function $\psi$ which satisfies $\mathrm{supp}(d\psi) \subset L^-\cup R^-$ and
 $\psi =0$ on $X\setminus U$ and $\psi = 1$ on $U\setminus(L^-\cup R^-)$.
We can choose $\psi$ such that
\begin{equation}
|d\psi| \leq  \frac{20 N}{\sqrt{\lambda}}. \label{cutoff}
\end{equation}
(Note that $20N/\sqrt{\lambda}$ is a large number because we choose $\lambda$ very small.)
We define the ``new error term" $\tau$ by $\tau := F^+(A + a + \psi b)$ on $U$.
\begin{eqnarray*}
\tau &=& F^+(A + a) + d^+_A(\psi b) + \psi [a\wedge b]^+ + \psi^2(b\wedge b)^+\\
     &=& \sigma + (d\psi \wedge b)^+ + \psi (d^+_A b + [a\wedge b]^+ + (b\wedge b)^+) + \psi(\psi - 1)(b\wedge b)^+\\
     &=& (1-\psi)\sigma + (d\psi \wedge b)^+ + \psi(\psi - 1)(b\wedge b)^+\\
     &=& (d\psi \wedge b)^+ + \psi(\psi - 1)(b\wedge b)^+, \quad
\text(\mathrm{supp}(\sigma)\cap \mathrm{supp}(1-\psi) = \emptyset).
\end{eqnarray*}
Hence $\tau$ is supported in $L^- \cup R^-$, and we have
\begin{equation}\label{estimate of tau by b}
|\!|\tau|\!|_{L^{\infty}} \leq \frac{20N}{\sqrt{\lambda}}\ |\!| b|_{L^-\cup R^-}|\!|_{L^\infty} + |\!| b|_{L^-\cup R^-}|\!|_{L^\infty}^2.
\end{equation}
Note that the ``old" error term $\sigma$ is supported in $L^+ \cup R^+$
and that the ``new" error term $\tau$ is supported in $L^- \cup R^-$.
\begin{Prop}\label{prop: estimate for construction}
For any $\kappa > 0$, there is $\eta > 0$ such that if $\delta \leq \kappa$, $|\!|a|\!|_{L^{2p}} \leq \eta$ and $\lambda \ll 1$,
then we have the following:
\begin{eqnarray*}
&\mathrm{(i)}&
 |\!| b |\!|_{L^{2p}}\leq C_1(N)\cdot \lambda^{(p+2)/{2p}} \cdot \delta.\\
&\mathrm{(ii)}&
 |\!| \tau |\!|_{L^\infty} \leq C_2 \cdot K_N \cdot N^4 \cdot \delta.
\hspace{6mm} \text{Here we put }\hspace{1mm} K_N := N/(N-N^{-1})^3.\\
&\mathrm{(iii)}&
 |\!| \psi \cdot  b |\!|_{L^p_1} \leq C_3(N) \cdot \lambda^{2/p} \cdot \delta.\\
&\mathrm{(iv)}&
 |\!|F(A + a + \psi \cdot b) - F(A + a)|\!|_{L^2(U)} \leq C_4(N) \cdot \lambda \cdot \delta.
\end{eqnarray*}
Here $C_1(N)$, $C_3(N)$, $C_4(N)$ are constants depending on N,
and $C_2$ is a constant independent of $N$.
(The fact that $C_2$ is independent of $N$ is essential to the later argument.)
\end{Prop}
\begin{proof}
The proof follows [D2, (4.22) Lemma]. In our case, some care must be paid to interactions between $L^{\pm}$ and $R^{\mp}$.

\noindent
(i) This follows from (\ref{implicit}) and the Sobolev embedding $L^{4p/(p+2)}_1 \hookrightarrow L^{2p}.$ (Note that this Sobolev embedding \textit{does not} hold at p = $\infty.$)
\begin{eqnarray*}
\norm{b}_{L^{2p}} &\leq& \mathrm{const} \cdot |\!| b |\!|_{L^{4p/(p+2)}_1}\\
                   &\leq& \mathrm{const} \cdot |\!| \sigma |\!|_{L^{4p/(p+2)}}\\
                   &\leq& C_1(N) \cdot \lambda^{(p+2)/2p}\cdot \delta.
\end{eqnarray*}
\noindent
(ii) The estimate of the $L^{\infty}$ norm of $\tau$ can be derived from the $L^{\infty}$ norm of $b$
over $L^- \cup R^-$. (See (\ref{estimate of tau by b}).)
First we will estimate a ``leading term" of $b$.
Let $\tilde{b}$ be the solution to
\[ D_A \tilde{b} = - \sigma. \]
Since $D_A = d^*_A \oplus d^+_A$ is invertible,
we have the Green kernel $G(x,y)$ for $D_A$:
\[\tilde{b}(x) = -\int_{X} G(x,y)\cdot \sigma (y)\, dy.\]
$G(x,y)$ has a singularity along the diagonal such that
\[ |G(x,y)| \leq \mathrm{const} \cdot \frac{1}{d(x,y)^3}. \]
Here $d(x,y)$ is the distance between $x$ and $y$.
This is because $D_A$ is a \textit{first order} elliptic differential operator.
(For the details, see [D2, (4.22) Lemma].)
In the complement of a neighborhood of the diagonal, $G(x,y)$ is bounded.

 Let us consider the case $x \in R^-$. Since $\sigma$ is supported in $L^+ \cup R^+$,
\[\tilde{b}(x) = - \int_{L^+} G(x,y)\cdot \sigma(y) \, dy - \int_{R^+} G(x,y)\cdot \sigma(y)\, dy.\]
 Because the distance between $R^-$ and $L^+$ is  sufficiently large, the first term can be easily estimated:
\[
\abs{\int_{L^+} G(x,y)\cdot \sigma(y) \, dy}\leq \mathrm{const}\cdot \mathrm{vol}(L^+)\cdot \delta
                           \leq \mathrm{const}\cdot N^4\cdot \lambda^2\cdot \delta.
\]
This implies that the interaction between $R^-$ and $L^+$ is comparatively small.
The second term is more delicate.
Since $d(R^+, R^-) = N\sqrt{\lambda}-N^{-1} \sqrt{\lambda}$, we have
\begin{eqnarray*}
\abs{\int_{R^+} G(x,y)\cdot \sigma(y) \, dy} &\leq& \mathrm{const}\cdot \mathrm{vol}(R^+)\cdot \frac{\delta}{(N\sqrt{\lambda} - N^{-1} \sqrt{\lambda})^3}\\
&\leq& \mathrm{const}\cdot K_N \cdot N^3 \cdot \sqrt{\lambda} \cdot \delta ,
\quad (K_N = \frac{N}{(N-N^{-1})^3}).
\end{eqnarray*}
Using (\ref{lambda is small}) in Section 2.2, we get
\[ |\tilde{b}(x)|\leq \mathrm{const}\cdot K_N \cdot N^3 \cdot \sqrt{\lambda}\cdot \delta.\]
We have the same estimate in the case $x \in L^-$.

Next we will estimate the ``lower term".
Set $b=\tilde{b}+\beta$. Then $\beta$ satisfies
\[D_A\beta + [a\wedge b]^+ +(b\wedge b)^+ = 0.\]
From the Sobolev embeddings,
\begin{eqnarray*}
|\!| \beta |\!|_{L^{\infty}} &\leq& \mathrm{const}\cdot \norm{D_A\beta}_{L^6},
 \quad (L^6_1 \hookrightarrow L^{\infty}),\\
&\leq& \mathrm{const} \cdot(|\!|a|\!|_{L^{12}}\cdot|\!|b|\!|_{L^{12}} + |\!|b|\!|^2_{L^{12}})\\
&\leq& \mathrm{const} \cdot(|\!|a|\!|_{L^{2p}}\cdot |\!|b|\!|_{L^3_1} + |\!|b|\!|^2_{L^3_1}),
 \quad (L^3_1\hookrightarrow L^{12}),\\
&\leq& \mathrm{const}\cdot(\eta \cdot |\!|b|\!|_{L^3_1} + |\!|b|\!|^2_{L^3_1}).
\end{eqnarray*}
From (\ref{implicit}) we have
\[|\!|b|\!|_{L^3_1}\leq \mathrm{const}\cdot |\!|\sigma|\!|_{L^3} \leq \mathrm{const}\cdot N^{4/3}\cdot \lambda^{2/3}\cdot \delta.\]
Hence
\[|\!| \beta |\!|_{L^{\infty}} \leq C(N)\cdot \lambda^{2/3}\cdot \delta ,
 \quad(C(N) \text{ is a constant depending on N}).
\]
Since $C(N)\cdot \lambda^{2/3} \ll \sqrt{\lambda}$, $\tilde{b}$
is the leading term of $b$ over $R^- \cup L^-$:
\begin{equation}\label{L^infty of b}
\norm{ b|_{R^-\cup L^-}}_{L^{\infty}} \leq \mathrm{const}\cdot K_N\cdot N^3\cdot \sqrt{\lambda}\cdot \delta.
\end{equation}
Substituting this into (\ref{estimate of tau by b}), we conclude that
\[|\!|\tau|\!|_{L^{\infty}}\leq \frac{20N}{\sqrt{\lambda}}\cdot |\!|b|_{R^-\cup L^-}|\!|_{L^{\infty}} + |\!|b|_{R^-\cup L^-}|\!|^2_{L^{\infty}} \leq C_2\cdot K_N\cdot N^4\cdot \delta.\]

\vspace{1mm}

\noindent
(iii) Since $\nabla_A(\psi \cdot b) =  d\psi \otimes  b + \psi \cdot \nabla_Ab$
and $|d\psi| \leq 20N/\sqrt{\lambda}$,
\begin{eqnarray*}
|\!|\psi \cdot b|\!|_{L^p_1}&\leq& \frac{20N}{\sqrt{\lambda}}\cdot
\mathrm{vol}(L^-\cup R^-)^{1/p} \cdot |\!|b|_{L^-\cup R^-}|\!|_{L^{\infty}} + |\!|b|\!|_{L^p_1} \\
&\leq& \mathrm{const}\cdot N^{1+4/p}\cdot \lambda^{-1/2 + 2/p}\cdot |\!|b|_{L^-\cup R^-}|\!|_{L^{\infty}} + |\!|b|\!|_{L^p_1}.
\end{eqnarray*}
From (\ref{implicit}) and (\ref{L^infty of b})
\[|\!|\psi \cdot b|\!|_{L^p_1} \leq C_3(N)\cdot \lambda^{2/p}\cdot \delta.\]

\noindent
(iv)\[F(A + a + \psi b) = F(A + a) + d_A(\psi b) + \psi[a \wedge b] + \psi^2 (b\wedge b).\]
Hence
\[\norm{F(A + a + \psi b) - F(A + a)}_{L^2(U)} \leq
\norm{\psi b}_{L^2_1} + \sqrt{2} \norm{a}_{L^4}\cdot \norm{b}_{L^4} + \sqrt{2} \norm{b}_{L^4}^2. \]
The arguments in (i) and (iii) show
\[ \norm{\psi b}_{L^2_1} \leq C(N) \cdot \lambda \cdot \delta, \quad \norm{b}_{L^4} \leq C'(N) \cdot \lambda \cdot \delta.\]
It follows that
\[\norm{F(A + a + \psi b) - F(A + a)}_{L^2(U)} \leq C_4(N) \cdot \lambda \cdot \delta.\]
\end{proof}
Proposition \ref{prop: estimate for construction} is sufficient to prove Theorem \ref{thm: construction}, but Theorem \ref{thm: injectivity} needs more.
Next we will establish  a ``parametrized version" of Proposition \ref{prop: estimate for construction},
which is the key estimate for the proof of Theorem \ref{thm: injectivity}.
Here we follow the arguments in [D2, (4.24) Remarks].

Suppose that $a$ and $\sigma$ smoothly depend on a parameter $t \in [0,1]$: $a=a(t)$, $ \sigma = \sigma(t)$. (To be precise, it is enough to assume that $a : [0,1] \rightarrow L^{2p}$ and $ \sigma : [0,1] \rightarrow L^{\infty}$ are smooth maps.)
Set $\delta = \delta(t) = |\!|\sigma(t) |\!|_{L^{\infty}(X)}$.
If $\delta \leq \kappa$ and $|\!|a|\!|_{L^{2p}(X)} \leq \eta$ for all $t$,
($\kappa$ and $\eta$ are the constants in Proposition \ref{prop: estimate for construction}),
then we have the following proposition.
\begin{Prop}
If we choose $\eta$ and $\lambda$ sufficiently small,
then we have the following estimates:
\begin{eqnarray*}
&\mathrm{(i)}&
 \norm{\frac{\partial b}{\partial t}}_{L^{2p}(X)} \leq C_5(N)\cdot \lambda^{(p+2)/{2p}} \left( \norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}(X)} + \delta \cdot \norm{\frac{\partial a}{\partial t}}_{L^{2p}(X)} \right).\\
&\mathrm{(ii)}&
\norm{\frac{\partial \tau}{\partial t}}_{L^{\infty}(X)} \leq C_6 \cdot K_N \cdot N^4 \cdot \norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}(X)} + C_7(N)\cdot \lambda^{1/p} \cdot \delta \cdot \norm{\frac{\partial a}{\partial t}}_{L^{2p}(X)}.
\end{eqnarray*}
Here $C_5(N)$ and $C_7(N)$ are constants depending on $N$ and
$C_6$ is a constant independent of $N$.
\end{Prop}
\begin{proof}
(i) It is clear that $b : [0,1] \rightarrow L^p_1$ is smooth.
Differentiating the equation $D_Ab + [a\wedge b]^+ +(b\wedge b)^+ = -\sigma$ with respect to $t$,
we get
\[D_A\left(\frac{\partial b}{\partial t}\right) + \left[\frac{\partial a}{\partial t} \wedge b\right]^+ +\left[a\wedge \frac{\partial b}{\partial t}\right]^+ +\left(\frac{\partial b}{\partial t}\wedge b\right)^+ +\left(b\wedge \frac{\partial b}{\partial t}\right)^+ = - \frac{\partial \sigma }{\partial t} .\]
Put $D'_A(\cdot) = D_A(\cdot) + [a\wedge \cdot]^+ +(\cdot \wedge b)^+ +(b\wedge \cdot)^+.$
 Then
\[D'_A\left(\frac{\partial b}{\partial t}\right) = -\left[\frac{\partial a}{\partial t} \wedge b\right]^+ -\frac{\partial \sigma}{\partial t}.\]
 Since $D_A$ is invertible and $\norm{a}_{L^{2p}}< \eta$ and $\norm{b}_{L^{2p}}< C_1(N)\cdot \lambda^{(2+p)/2p} \delta$ are sufficiently small,
\begin{eqnarray*}
\norm{\frac{\partial b}{\partial t}}_{L^{2p}} &\leq& \mathrm{const} \cdot \norm{D'_A\left(\frac{\partial b}{\partial t}\right)}_{L^{4p/(p+2)}}\\
&\leq& \mathrm{const} \left( \norm{\frac{\partial \sigma }{\partial t}}_{L^{4p/(p+2)}} + \> \norm{\left[ \frac{\partial a}{\partial t} \wedge b\right]^+}_{L^{4p/(p+2)}}\right) \\
&\leq& C(N) \left(\lambda^{(2+p)/2p}\norm{\frac{\partial \sigma }{\partial t}}_{L^{\infty}} + \>
 \norm{\frac{\partial a}{\partial t}}_{L^{2p}} |\!|b|\!|_{L^{2p}}\right)\\
&\leq& C_5(N)\cdot \lambda^{(2+p)/2p}\left(\norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}} +  \>  \delta \cdot \norm{\frac{\partial a}{\partial t}}_{L^{2p}}\right).
\end{eqnarray*}
Here we have used the fact that $\partial \sigma /\partial t$ is supported in
$L^+\cup R^+$ whose volume is $O(\lambda^2)$.

\vspace{1mm}
\noindent
(ii) Differentiating $\tau = (d\psi \wedge b)^+ + \psi(\psi -1)(b\wedge b)^+$ with respect to $t$,
we have
\[ \frac{\partial \tau}{\partial t} = \left(d\psi \wedge \frac{\partial b}{\partial t}\right)^+ +\psi(\psi -1)\{\left(\frac{\partial b}{\partial t} \wedge b\right)^+ + \left(b \wedge \frac{\partial b}{\partial t}\right)^+\}.\]
From (\ref{cutoff}) and (\ref{L^infty of b})
\begin{equation}\label{estimate of tau/t}
 \begin{split}
 \norm{\frac{\partial \tau}{\partial t}}_{L^{\infty}} &\leq \> \frac{20N}{\sqrt{\lambda}}
 \norm{\left.\frac{\partial b}{\partial t}\right|_{L^- \cup R^-}}_{L^{\infty}}
 + 2\sqrt{2}\cdot \norm{b|_{L^- \cup R^-}}_{L^{\infty}}
 \norm{\left.\frac{\partial b}{\partial t}\right|_{L^- \cup R^-}}_{L^{\infty}} \\
 & \leq \> \frac{20N}{\sqrt{\lambda}}
 \norm{\left.\frac{\partial b}{\partial t}\right|_{L^- \cup R^-}}_{L^{  \infty}}
 + C(N)\cdot \sqrt{\lambda}\cdot \delta \cdot
 \norm{\left.\frac{\partial b}{\partial t}\right|_{L^- \cup R^-}}_{L^{\infty}}.
 \end{split}
\end{equation}
Set $b = \tilde{b} + \beta$ with $D_A \tilde{b} = - \sigma$
as in the proof of Proposition 3.1 (ii).
Then
\begin{eqnarray*}
\dfrac{\partial \tilde{b}}{\partial t} (x) &=& -\int_{X} G(x,y)\cdot \dfrac{\partial \sigma}{\partial t} (y)\, dy, \quad (G(x,y) \text{ is the Green kernel}),\\
&=& - \int_{L^+} G(x,y)\cdot \dfrac{\partial \sigma}{\partial t} (y)\, dy
-\int_{R^+}  G(x,y)\cdot \dfrac{\partial \sigma}{\partial t} (y) \, dy.
\end{eqnarray*}
This can be estimated in the same way as before and we get
\[\norm{\left.\frac{\partial \tilde{b}}{\partial t}\right|_{L^- \cup R^-}}_{L^{\infty}}\leq \mathrm{const}\cdot K_N \cdot N^3 \cdot \sqrt{\lambda}\cdot \norm{\frac{\partial \sigma}{ \partial t}}_{L^{\infty}}.\]
We have $\partial b/\partial t = \partial \tilde{b}/\partial t + \partial \beta/\partial t$
and $\partial \beta/\partial t$ satisfies
\[D_A\left(\frac{\partial \beta}{\partial t}\right) + \left[\frac{\partial a}{\partial t} \wedge b\right]^+ + \left[a\wedge \frac{\partial b}{\partial t}\right]^+ +\left(\frac{\partial b}{\partial t} \wedge b\right)^+ +\left(b\wedge \frac{\partial b}{\partial t}\right)^+ = 0.\]
Using the above (i) and the Sobolev embedding, we get
\begin{eqnarray*}
\norm{\frac{\partial \beta}{\partial t}}_{L^{\infty}} &\leq& \mathrm{const}\cdot \norm{D_A\left(\frac{\partial \beta}{\partial t}\right)}_{L^6}, \quad (L^6_1\hookrightarrow L^{\infty}),\\
&\leq& C(N)\cdot \lambda^{(2+p)/2p}\left(\norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}}+ \delta \cdot \norm{\frac{\partial a}{\partial t}}_{L^{2p}}\right).
\end{eqnarray*}
Hence
\[\norm{\left.\frac{\partial b}{\partial t}\right|_{L^- \cup R^-}}_{L^{\infty}} \leq C\cdot K_N  N^3 \sqrt{\lambda}\norm{\frac{\partial \sigma }{\partial t}}_{L^{\infty}} + C(N)\cdot \lambda^{(2+p)/2p} \cdot \delta \norm{\frac{\partial a}{\partial t}}_{L^{2p}}.\]
Substituting this into (\ref{estimate of tau/t}), we conclude that
\[\norm{\frac{\partial \tau}{\partial t}}_{L^{\infty}} \leq C_6 \cdot K_N N^4 \norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}} + C_7(N)\cdot \lambda^{1/p}\cdot \delta  \norm{\frac{\partial a}{\partial t}}_{L^{2p}}.\]
\end{proof}
Here we fix the value of N. Any value of N with
\[C_2\cdot K_N \leq \frac{1}{2}, \quad C_6\cdot K_N \leq \frac{1}{2},\]
will do. (Note that $K_N=N/(N-N^{-1})^3$ converges to zero as $N \rightarrow \infty$.)
So let N be a large positive number which satisfies the above.
(This choice depends on the gluing data $(X, x^L, x^R, E, A)$.)
Then Proposition 3.1 (ii) and Proposition 3.2 (ii) become:
\begin{eqnarray*}
&\text{(ii)$'$}&
 \norm{\tau}_{L^{\infty}} \leq \frac{1}{2}\cdot N^4 \cdot \delta ,\\
&\text{(ii)$'$}&
 \norm{\frac{\partial \tau}{\partial t}}_{L^{\infty}} \leq \frac{1}{2}\cdot N^4 \cdot \norm{\frac{\partial \sigma}{\partial t}}_{L^{\infty}} + C_7(N)\cdot \lambda^{1/p} \cdot \delta \cdot
 \norm{\frac{\partial a}{\partial t}}_{L^{2p}}.
\end{eqnarray*}
\begin{Rem}
In fact, the parameter $N$ can be chosen independent of the gluing data $(X, x^L, x^R, E, A)$.
See [D2, p. 308, (4.14)] and the proof of [D2, (4.22) Lemma].
But we don't need this fact in this paper.
\end{Rem}

\section{Donaldson's alternating method}
In this section we prove Theorem \ref{thm: construction}. Let $(X_i, x^L_i, x^R_i, E_i, A_i)_{i \in \mathbb{Z}}$ be a sequence of gluing data of finite type as in Section 2.
The finite type condition means that
we can apply the estimates in Section 3 to all $(X_i, x^L_i, x^R_i, E_i, A_i)$
with the \textit{same} constants.
(This is the reason for introducing this notion.)

To begin with,
we make one remark on the conformal transformation (\ref{conformal map})
which identifies $\Omega^R_i$ and $\Omega^L_{i+1}$:
\begin{Rem}\label{rem: conformal map}
The coordinate transformation (\ref{conformal map}) is only conformal and not an isometry.
Hence the metric of $X_i$ is different from the metric of $X_{i+1}$
over the identified region $\Omega^R_i \cong \Omega^L_{i+1}$.
The derivative of (\ref{conformal map}) over
$\Omega^R_i = \{kN^{-1}\sqrt{\lambda} < |\xi| < k^{-1}N\sqrt{\lambda}\}$
has the following bound independent of $\lambda$:
\[\norm{\frac{\partial\eta}{\partial \xi}}= \frac{\lambda}{\abs{\xi}^2}\leq \frac{N^2}{k^2},\quad
 \norm{\frac{\partial \xi}{\partial \eta}} = \frac{\lambda}{\abs{\eta}^2} \leq
\frac{N^2}{k^2}.
\]
Here $\norm{\cdot}$ is the operator norm.
In particular, if $\xi \in R^-_i = L^+_{i+1}$, then
\[ \norm{\frac{\partial \xi}{\partial \eta}} \leq \frac{1}{N^2}.\]
It follows that
\[ \norm{\omega}_{L^{\infty}(L^+_{i+1})}\leq \frac{1}{N^4}\norm{\omega}_{L^{\infty}(R^-_i)}\]
 for a 2-form $\omega$ under the identification $R^-_{i} = L^+_{i+1}$.
\end{Rem}
Let $\rho=(\rho_i)_{i\in \mathbb{Z}} \in \mathrm{GlP}$ be a gluing parameter,
($\rho_i \in \mathrm{Hom}_{G}((E_i)_{x^R_i},(E_{i+1})_{x^L_{i+1}})$),
and $\bm{E}(\rho)$ be the principal $G$ bundle over $\sharp_{i\in\mathbb{Z}}X_i$ defined in
Section \ref{subsection: gluing construction}.
As we explained in Section \ref{subsection: outline},
we will inductively construct a sequence of connections
$\bm{A}^{(0)}(\rho)$, $\bm{A}^{(1)}(\rho)$, $\bm{A}^{(2)}(\rho)$, $ \cdots$, on $\bm{E}(\rho)$ which converges to an ASD connection $\bm{A}(\rho)$.
This is a generalization of the alternating method in [D2, IV (iv)].
At the $n$-th stage of the iteration we will have a connection $\bm{A}^{(n)}(\rho)$ which is expressed by
\[\bm{A}^{(n)}(\rho) = (A_i + a^{(n)}_i(\rho))_{i \in \mathbb{Z}}.\]
Here $a^{(n)}_i(\rho)$ is an element of $\Omega^1(\mathrm{ad}E_i)$ over $U_i$ and we have
$A_i + a^{(n)}_i(\rho) = A_{i+1}+a^{(n)}_{i+1}(\rho)$ over $U_i\cap U_{i+1}$
under the identification $\rho_i$.
Then they compatibly define the connection $\bm{A}^{(n)}(\rho)$ on $\bm{E}(\rho)$.

Let us consider the following ``inductive hypotheses":
\begin{Hyp}\label{inductive hypotheses}
(i) Let $\sigma^{(n)} := F^{+}(\bm{A}^{(n)}(\rho))$ be the self-dual curvature of $\bm{A}^{n}(\rho)$,
and set
\[\delta_n := \begin{cases}
\mathrm{sup}_{i \in \mathbb{Z}} |\!|\sigma^{(n)}_{2i}|\!|_{L^{\infty}(X_{2i})} & \text{if }n \text{ is even.}\\
\mathrm{sup}_{i \in \mathbb{Z}}|\!|\sigma^{(n)}_{2i+1}|\!|_{L^{\infty}(X_{2i+1})} & \text{if }n \text{ is odd.}
\end{cases}
\]
Here $\sigma^{(n)}_{i}$ is the restriction of $\sigma^{(n)}$ on $U_i$.
(We extend $\sigma^{(n)}_i$ to $X_i$ by zero.)
We suppose that $\sigma^{(n)}$ is supported as follows:
\[\mathrm{supp}(\sigma^{(n)}) \subset \begin{cases}
\coprod_{i\in \mathbb{Z}}(L^+_{2i}\cup R^{+}_{2i}) & \text{if }n\text{ is even.}\\
\coprod_{i\in \mathbb{Z}}(L^+_{2i+1}\cup R^{+}_{2i+1}) & \text{if }n\text{ is odd.}
\end{cases}
\]
And we suppose that $\delta_n$ has the following bound:
\[ \delta_n \leq \kappa , \quad (\kappa \text{ is the constant in Proposition 3.1 and will be given later}).\]

\noindent
(ii) We suppose that $a^{(n)}_i$ has the following bound:
\[ |\!|a^{(n)}_i|\!|_{L^{2p}(U_i)} \leq \eta , \quad
(\eta \text{ is also the constant in Proposition 3.1}) .\]
\end{Hyp}
If these conditions hold at the $n$-th stage of the iteration and $\lambda \ll 1$,
then we can pass to the $(n+1)$-th stage as follows.
If $n$ is even, then we use the construction in Section 3 on each $X_{2i}$ and get $b^{(n)}_{2i}$.
We define $a^{(n)}_{i}(\rho)$ over $U_i$ by
\begin{equation}\label{iteration1}
 \begin{split}
   &a^{(n+1)}_{2i}(\rho) :=  a^{(n)}_{2i}(\rho) + \psi_{2i}\cdot b^{(n)}_{2i}, \\
   &a^{(n+1)}_{2i+1}(\rho) :=  a^{(n)}_{2i+1}(\rho) + \psi_{2i}\cdot \rho_{2i}b^{(n)}_{2i}
\rho_{2i}^{-1} + \psi_{2i+2}\cdot \rho_{2i+1}^{-1}b^{(n)}_{2i+2}\rho_{2i+1}.
 \end{split}
\end{equation}
Here $\psi_i$ is a cut-off over $X_i$ as in Section 3. Note that $\psi_{2i}\cdot \rho_{2i}b^{(n)}_{2i}\rho_{2i}^{-1}$ and $\psi_{2i+2}\cdot \rho_{2i+1}^{-1}b^{(n)}_{2i+2}\rho_{2i+1}$ are well-defined over $U_{2i+1}$.
From the arguments in Section 3 and Proposition \ref{prop: estimate for construction} (ii)$'$ at the end of Section 3, we have
\[
\mathrm{supp}(\sigma^{(n+1)}) \subset \coprod_{i \in \mathbb{Z}}(R^-_{2i}\cup L^-_{2i})
= \coprod_{i \in \mathbb{Z}}(L^+_{2i+1}\cup R^+_{2i+1}), \]
\[ |\!|\sigma^{(n+1)}_{2i}|\!|_{L^{\infty}(X_{2i})} \leq \frac{1}{2}\cdot N^4 \delta_n. \]
Taking account of Remark \ref{rem: conformal map}, we have
\[|\!|\sigma^{(n+1)}_{2i+1}|\!|_{L^{\infty}(X_{2i+1})} \leq \frac{1}{2}\delta_n.\]
Thus
\begin{equation}
\delta_{n+1} \leq \frac{1}{2}\delta_n. \label{er.term}
\end{equation}
If $n$ is odd, then we use Proposition \ref{prop: estimate for construction} over each $X_{2i+1}$
and construct a similar procedure to pass to the $(n+1)$-th stage.
The estimate (\ref{er.term}) holds as long as we can continue this iteration.
Hence
\begin{equation}
\delta_n \leq \frac{1}{2^n}\delta_0. \label{geo.decay}
\end{equation}

We start the iteration by defining $\bm{A}^{(0)}(\rho) = (A_i + a^{(0)}_i(\rho))_{i \in \mathbb{Z}}$ by
\begin{equation}\label{ini.value1}
 \begin{split}
 &a^{(0)}_{2i+1}(\rho) :=
 (\psi_{2i+1}-1)(A_{2i+1} - \rho_{2i}A_{2i}\rho_{2i}^{-1} -\rho_{2i+1}^{-1}A_{2i+2}\rho_{2i+1}), \\
 &a^{(0)}_{2i}(\rho) :=
 \psi_{2i+1}(-A_{2i} + \rho_{2i}^{-1}A_{2i+1}\rho_{2i}) +
 \psi_{2i-1}(-A_{2i} + \rho_{2i-1}A_{2i-1}\rho_{2i-1}^{-1}).
 \end{split}
\end{equation}
Here each $A_i$ is the connection matrix in the exponential gauge centered at $x^L_i$ and $x^R_i$.
These expressions are well-defined on $U_{2i+1}$ or $U_{2i}$ and compatible over the overlap regions.
The self-dual curvature of $\bm{A}^{(0)}(\rho)$ is supported in $\coprod (L^-_{2i+1}\cup R^-_{2i+1}) = \coprod (L^+_{2i}\cup R^+_{2i})$.
Since each $\abs{A_i}$ is $O(\sqrt{\lambda})$ and
$\abs{d\psi_i}$ is $O(1/\sqrt{\lambda})$ (cf. (\ref{cutoff})),
we have a constant $\kappa$ independent of $\lambda$ such that
\begin{equation}
 |\!|\sigma^{(0)}_{i}|\!|_{L^{\infty}(X_{i})} \leq \kappa. \label{ini.error}
\end{equation}
We take this $\kappa$ as the constant in Proposition \ref{prop: estimate for construction}.
The condition (i) in Hypotheses \ref{inductive hypotheses} is satisfied at $n=0$.
Since $a^{(0)}_i$ is supported in the overlap whose volume is $O(\lambda^2)$, we have
\begin{equation}
|\!|a^{(0)}_i|\!|_{L^{2p}(U_i)} \leq \mathrm{const}\cdot \lambda^{(2+p)/2p}. \label{ini.estimate1}
\end{equation}
Therefore the condition (ii) is also satisfied at $n=0$ if we choose $\lambda \ll 1$.
Then we can start the iteration.

The geometric decay (\ref{geo.decay}) shows that the condition (i) in Hypotheses 4.2 is always satisfied as long as the condition (ii) is valid.
At the $n$-th stage of the iteration, if $n$ is even, Proposition 3.1 (i) shows \[|\!|a^{(n+1)}_{2i} - a^{(n)}_{2i}|\!|_{L^{2p}(U_{2i})} \leq \mathrm{const}\cdot \lambda^{(2+p)/2p}\delta_n \leq \mathrm{const}\cdot \lambda^{(2+p)/2p}\frac{\delta_0}{2^n}.\]
From Remark \ref{rem: conformal map},
\[ |\!|a^{(n+1)}_{2i+1} - a^{(n)}_{2i+1}|\!|_{L^{2p}(U_{2i+1})} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p}\frac{\delta_0}{2^n}.\]
We have similar estimates when $n$ is odd, and
\begin{equation}\label{total error}
 \mathrm{const}\cdot \lambda^{(2+p)/2p} + \mathrm{const}\cdot \lambda^{(2+p)/2p} \sum_{n\geq 0}
 \frac{\delta_0}{2^n} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p}\ll 1.
\end{equation}
Hence the condition (ii) in Hypotheses 4.2 is satisfied for any $n$ if we choose $\lambda$ small enough. Thus we can continue the iteration indefinitely and
the sequence $\{ a^{(n)}_i(\rho) \}_{n=0}^{\infty}$ has a limit $ a_i(\rho)$ in $L^{2p}(U_i)$.

Proposition 3.1 (iii) and an argument similar to the above show that $\{ a^{(n)}_i(\rho) \}_{n=0}^{\infty}$ converges to $a_i(\rho)$ also in $L^p_1(U_i)$.
From the geometric decay (\ref{geo.decay}), the self-dual curvature of $A_i + a_i(\rho)$ becomes $0$.
Thus we get an ASD connection
\[\bm{A}(\rho) = (A_i + a_i(\rho))_{i\in \mathbb{Z}}.\]
(\ref{total error}) shows that this is close to $A_i$ over each $U_i$:
\begin{equation}
|\!|a_i|\!|_{L^{2p}(U_i)} \leq \mathrm{const}\cdot \lambda^{(2+p)/2p}. \label{finalterm}
\end{equation}
 So we get a conclusion:
\begin{Thm}
For all sufficiently small $\lambda$, $\bm{A}(\rho)$ is an ASD connection on $\bm{E}(\rho)$ such that
\[|\!|\bm{A}(\rho) - A_i|\!|_{L^{2p}(U_i)} \leq C_8 \cdot \lambda^{(2+p)/2p}
\quad \text{for all } i \in \mathbb{Z}.\]
Here $C_8$ is a constant independent of $i$, $\rho$, $\lambda$.
\end{Thm}
In general this ASD connection $\bm{A}(\rho)$ has infinite energy:
\begin{Thm}
If there are infinitely many non-flat ASD connections $A_i$ in the given sequence of gluing data,
then the $L^2$-energy of $\bm{A}(\rho)$ is infinite: $\norm{F(\bm{A}(\rho))}_{L^2}= + \infty$.
\end{Thm}
\begin{proof}
We will show that the loss of energy over each $U_i$ during the iteration is sufficiently small.
At the $n$-th stage of the iteration, if $n$ is even, we can apply Proposition 3.1 (iv) to (\ref{iteration1}) and get
\[ |\!|F(A_{2i} + a^{(n+1)}_{2i}) - F(A_{2i} + a^{(n)}_{2i})|\!|_{L^2(U_{2i})} \leq \mathrm{const}\cdot \lambda \cdot \delta_n \leq \mathrm{const} \cdot \frac{\lambda}{2^n}.\]
Since $F(A_{2i+1} + a^{(n+1)}_{2i+1}) - F(A_{2i+1} + a^{(n)}_{2i+1})$ is supported in the overlaps
and the $L^2$-norm of a 2-form is conformally invariant,
\[|\!|F(A_{2i+1} + a^{(n+1)}_{2i+1}) - F(A_{2i+1} + a^{(n)}_{2i+1})|\!|_{L^2(U_{2i+1})} \leq \mathrm{const}\cdot \frac{\lambda}{2^n}.\]
We have the same estimates in the case that $n$ is odd. Hence
\[|\!| F(A_{i} + a_{i}) - F(A_{i} + a^{(0)}_{i})|\!|_{L^2(U_{i})} \leq \mathrm{const} \cdot \lambda \sum_{n\geq 0}\frac{1}{2^n} \leq \mathrm{const} \cdot \lambda. \]
Next we will show that $|\!| F(A_{i} + a^{(0)}_{i}) - F(A_{i})|\!|_{L^2(U_{i})}$ is also small.
An estimate similar to (\ref{ini.error}) shows
\[|F(A_i + a_i^{(0)}) - F(A_i)| \leq \mathrm{const}, \quad (\text{independent of }\lambda).\]
Since this is supported in the overlaps, we get
\[|\!|F(A_i + a_i^{(0)}) - F(A_i)|\!|_{L^2(U_i)} \leq \mathrm{const} \cdot \lambda.\]
Hence
\[|\!| F(A_{i} + a_{i}) - F(A_{i})|\!|_{L^2(U_{i})} \leq \mathrm{const} \cdot \lambda.\]

From the finite type condition, we have a positive constant $M$ independent of $\lambda$ such that $|\!|F(A_i)|\!|_{L^2(U_i)} \geq M$ if $A_i$ is not flat and $\lambda$ is sufficiently small.
Therefore if we take $\lambda \ll 1$, then
\[|\!|F(A_i + a_i)|\!|_{L^2(U_i)} \geq \frac{1}{2}M \]
for non-flat $A_i$.
 Since there are infinitely many non-flat $A_i$, we conclude that
\[|\!|F(\bm{A}(\rho))|\!|_{L^2} = + \infty.\]
\end{proof}

\section{Moduli Problem}
We will prove Theorem \ref{thm: injectivity} in this section. The proof is a generalization of [D2, (4.31) Lemma]. We always suppose that $\lambda$ is sufficiently small.

We need the following elementary result.
\begin{Lem}\label{lem:recurrence}
Let $K$ be a positive number and $\varepsilon$ be a positive number with $\varepsilon \leq 1/100$.
Let $\{ \alpha_n\}, \{ s_n\}$, $(n=0,1,2,\cdots)$, be sequences of non-negative numbers which satisfy
the following recurrence inequalities:
\begin{equation*}
 \begin{split}
 \alpha_{n+1} - \alpha_n &\leq \varepsilon(s_n + \frac{1}{2^n}\alpha_n +\frac{1}{2^n} K),  \\
 s_{n+1} &\leq \frac{1}{2}s_n + \frac{1}{2^n} \alpha_n + \frac{1}{2^{n+1}} K.
 \end{split}
\end{equation*}
Suppose that
\[\alpha_0 \leq \varepsilon \cdot K , \quad s_0 \leq K.\]
Then we get
\[ \alpha_n \leq 10 \varepsilon \cdot K \quad \text{for all } n.\]
\end{Lem}
\begin{proof}
Set $t_n := 2^n s_n$. Then we have $t_0 \leq K$, and the above recurrence inequalities become
\begin{equation*}
 \begin{split}
 \alpha_{n+1} - \alpha_n &\leq \frac{\varepsilon}{2^n}(t_n + \alpha_n + K), \\
 t_{n+1} &\leq t_n + 2 \alpha_n + K.
 \end{split}
\end{equation*}
We will inductively prove $\alpha_n \leq 10\varepsilon \cdot K$ for all $n$.
Suppose that this is true if $n \leq n_0$ for some $n_0$.
Then
\[t_n \leq K + 2 (10 \varepsilon K)n + K n  =
  K \{ 1 + (1 + 20 \varepsilon)n \},  \quad(n \leq n_0 + 1).  \]
Hence
\begin{eqnarray*}
\alpha_{n_0 +1} - \alpha_0 &=& \sum_{n=0}^{n_0} (\alpha_{n+1} - \alpha_{n})\\
     &\leq& \sum_{n\geq 0} \frac{\varepsilon}{2^n}\left[K \{ 1 + (1 + 20 \varepsilon)n \} + 10 \varepsilon K + K \right] \\
     &=& \varepsilon K (6 + 60\varepsilon) \leq 7 \varepsilon K.
\end{eqnarray*}
Thus
\[ \alpha_{n_0 + 1} \leq \varepsilon K + 7 \varepsilon K \leq 10 \varepsilon K. \]
\end{proof}

\begin{Prop}
For two \textit{gluing parameters} $\rho = (\rho_i)_{i\in\mathbb{Z}}$ and $\rho' = (\rho'_i)_{i\in \mathbb{Z}}$, let $K:= \mathrm{sup}_{i\in \mathbb{Z}}|\rho_i - \rho'_i|$. Then we have
\[ |\!|a_i(\rho) - a_i(\rho')|\!|_{L^{2p}(U_i)} \leq C_9 \cdot \lambda^{(2+p)/2p} K
\quad \text{for all } i \in \mathbb{Z}. \]
Here $C_9$ is a constant independent of $i$, $\lambda$, $\rho$, $\rho'$.
\end{Prop}
\begin{proof}
Let $\bar{\rho}(t) = (\bar{\rho_i}(t))_{i\in\mathbb{Z}}$, $(0\leq t \leq 1)$, be a path in $\mathrm{GlP}$ such that each $\bar{\rho_i}(t)$ is the geodesic from $\rho_i$ to $\rho'_i$ in $G$.
Since the distance $d(\rho_i, \rho'_i)$ is equivalent to $|\rho_i - \rho'_i|$:
\[ C^{-1} |\rho_i - \rho'_i| \leq d(\rho_i, \rho'_i) \leq C |\rho_i - \rho'_i| \quad \text{for some constant } C, \]
we have $|d \bar{\rho_i}/dt| = d(\rho_i, \rho'_i) \leq C\cdot K$.

Let $\bm{A}^{(n)}(\bar{\rho}(t)) = ( A_i + a_i^{(n)}(\bar{\rho}(t)))_{i \in \mathbb{Z}}$
be the connection on $\bm{E}(\bar{\rho}(t))$ constructed by the iterative process in Section 4; this is parametrized by $t$ and we can apply Proposition 3.2.
We define $\alpha_n = \alpha_n(t)$ and $s_n = s_n(t)$ by
\[ \alpha_n := \mathrm{sup}_{i \in \mathbb{Z}} \norm{\frac{\partial}{\partial t} a^{(n)}_i}_{L^{2p}(X_i)},\quad
   s_n := \begin{cases}
          \mathrm{sup}_{i \in \mathbb{Z}} \norm{\frac{\partial}{\partial t} \sigma_{2i}^{(n)}}_{L^{\infty}(X_{2i})}          & \text{ if } n \text{ is even.} \\
          \mathrm{sup}_{i \in \mathbb{Z}} \norm{\frac{\partial}{\partial t} \sigma_{2i + 1}^{(n)}}_{L^{\infty}(X_{2i+1})}  & \text{ if } n \text{ is odd.}
          \end{cases}
\]
From the definition of $a_i^{(0)}$ in (\ref{ini.value1}),
\[ \abs{\frac{\partial}{\partial t} a^{(0)}_i} \leq \mathrm{const}\cdot \sqrt{\lambda} \cdot K. \]
Here we used the fact that the connection matrices in the exponential gauges are $O(\sqrt{\lambda})$.
Since $\partial a^{(0)}_i / \partial t$ is supported in the overlaps, we get
\[ \norm{\frac{\partial}{\partial t} a^{(0)}_i}_{L^{2p}(X_i)}  \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K.\]
In a similar way, we have
\[ \norm{\frac{\partial}{\partial t} \sigma^{(0)}_i}_{L^{\infty}(X_i)} \leq \mathrm{const} \cdot K.\]
Hence
\begin{equation}
 \alpha_0 \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K, \quad s_0 \leq \mathrm{const} \cdot K.
\label{ini.value3}
\end{equation}

Next we will establish recurrence inequalities on $\alpha_n$ and $s_n$.
If $n$ is even, then Proposition 3.1 and 3.2 give bounds for $|\!| b^{(n)}_{2i} |\!|_{L^{2p}(X_{2i})}$ and $|\!| \partial b^{(n)}_{2i}/ \partial t |\!|_{L^{2p}(X_{2i})}$.
Then, from the inductive definitions (\ref{iteration1}) and the geometric decay (\ref{geo.decay}),
\begin{equation*}
 \begin{split}
 &\norm{\frac{\partial}{\partial t} a^{(n+1)}_{2i}}_{L^{2p}(X_{2i})} \leq \alpha_n + \mathrm{const} \cdot
 \lambda^  {(2+p)/2p}(s_n + \frac{1}{2^n} \alpha_n), \\
 &\norm{\frac{\partial}{\partial t} a^{(n+1)}_{2i+1}}_{L^{2p}(X_{2i+1})} \leq \alpha_n + \mathrm{const} \cdot
 \frac{\lambda^{(2+p)/2p}}{2^n} K + \mathrm{const} \cdot \lambda^{(2+p)/2p}
 (s_n + \frac{1}{2^n} \alpha_n).
 \end{split}
\end{equation*}
It follows that
\begin{equation}
\alpha_{n+1} - \alpha_{n} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p}(s_n + \frac{1}{2^n} \alpha_n + \frac{K}{2^n}).   \label{iteration3}
\end{equation}
Similarly, from Proposition 3.2 (ii)$'$ at the end of Section 3,
\[\norm{\frac{\partial}{\partial t} \sigma^{(n+1)}_{2i}}_{L^{\infty}(X_{2i})} \leq \frac{N^4}{2} s_n + \mathrm{const} \cdot \frac{\lambda^{1/p}}{2^n} \alpha_n. \]
Since $\sigma^{(n+1)}_{2i+1} = \bar{\rho}_{2i} \sigma^{(n+1)}_{2i} \bar{\rho}_{2i}^{-1}$ on
$U_{2i} \cap U_{2i+1}$ and
$\sigma^{(n+1)}_{2i+1} = \bar{\rho}_{2i+1}^{-1} \sigma^{(n+1)}_{2i+2} \bar{\rho}_{2i+1}$
on $U_{2i+1}\cap U_{2i+2}$, we get
\begin{eqnarray*}
\norm{\frac{\partial}{\partial t} \sigma^{(n+1)}_{2i+1}}_{L^{\infty}(X_{2i+1})} &\leq& \mathrm{const} \cdot \delta_{n+1} \cdot K + \frac{1}{2} s_n + \mathrm{const} \cdot \frac{\lambda^{1/p}}{2^n} \alpha_n \\
&\leq& \mathrm{const} \cdot \frac{K}{2^{n+1}}  + \frac{1}{2} s_n + \mathrm{const} \cdot \frac{\lambda^{1/p}}{2^n} \alpha_n.
\end{eqnarray*}
Here we have used Remark \ref{rem: conformal map}.
This shows
\begin{equation}
s_{n+1} \leq \frac{1}{2} s_n + \mathrm{const} \cdot \frac{\lambda^{1/p}}{2^n} \alpha_n + \mathrm{const} \cdot \frac{K}{2^{n+1}}.  \label{iteration4}
\end{equation}
The inequalities (\ref{iteration3}), (\ref{iteration4}) hold also in the case that $n$ is odd.
So we can apply Lemma \ref{lem:recurrence}
with initial values (\ref{ini.value3}) and $\varepsilon = \mathrm{const} \cdot \lambda^{(2+p)/2p}$.
Then we get
\[ \alpha_n \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K. \]
Therefore
\[ |\!| a^{(n)}_i(\rho) - a^{(n)}_i(\rho')|\!|_{L^{2p}(U_i)} = \norm{\int_{0}^{1} \frac{\partial}{\partial t} a^{(n)}_i dt}_{L^{2p}(U_i)} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K,\]
for all $i$ and $n$.
So we get a conclusion:
\[ |\!|a_i(\rho) - a_i(\rho')|\!|_{L^{2p}(U_i)} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K. \]
\end{proof}

The following is Theorem \ref{thm: injectivity}.
\begin{Thm}
For two \textit{gluing parameters} $\rho$ and $\rho'$, $\bm{A}(\rho)$ is gauge equivalent to $\bm{A}(\rho')$ if and only if $[\rho] = [\rho']$ in $\mathrm{EGlP}$.
\end{Thm}
\begin{proof}
If $[\rho] = [\rho']$ in $\mathrm{EGlP}$, we have $\varepsilon_i \in C(G)$
with $\rho'_i = \varepsilon_i \rho_i$ for all $i\in \mathbb{Z}$.
Since the center $C(G)$ acts trivially on connections,
it is clear from the construction in Section 4 that
we have $a_i(\rho) = a_i(\rho')$.
And we can choose $\gamma_i \in C(G)$ such that $\gamma_{i+1} = \varepsilon_i \gamma_i$.
Then these $\gamma_i$ give a gauge transformation
$h = (\gamma_i)_{i\in \mathbb{Z}} :\bm{E}(\rho) \rightarrow \bm{E}(\rho')$ with
$h. \bm{A}(\rho) = \bm{A}(\rho')$.

Conversely, suppose that there is a gauge transformation $ g: \bm{E}(\rho) \rightarrow \bm{E}(\rho')$
with $ g. \bm{A}(\rho) = \bm{A}(\rho')$.
This means that there is a gauge transformation $g_i: E_i \rightarrow E_i$ over each $U_i$ such that
\begin{equation*}
 \begin{split}
 &d_{A_i}{g_i} = g_i a_i(\rho) - a_i(\rho' )g_i, \\
 &g_{i+1}\rho_i = \rho'_i g_i.
 \end{split}
\end{equation*}
Because $A_i$ is irreducible, we have a constant C independent of $\lambda$ such that
\[\min_{\gamma \in C(G)}|\!|\gamma g_i - 1|\!|_{C^0(U_i)}
\leq C \cdot |\!|d_{A_i} g_i|\!|_{L^{2p}(U_i)}. \]
The proof of this inequality is standard. For example, see  [D2, (4.31) Lemma].
We can suppose that $|\!|g_i - 1|\!|_{C^0(U_i)} \leq C \cdot |\!|d_{A_i} g_i|\!|_{L^{2p}(U_i)}$
without loss of generality.
From (\ref{finalterm}) and Proposition 5.2,
\begin{eqnarray*}
|\!|g_i - 1|\!|_{C^0} &\leq& C\cdot |\!|g_i a_i(\rho) - a_i(\rho')g_i|\!|_{L^{2p}} \\
&\leq& C\cdot |\!|g_i - 1|\!|_{C^0}(|\!|a_i(\rho)|\!|_{L^{2p}} + |\!|a_i(\rho')|\!|_{L^{2p}}) + C \cdot |\!|a_i(\rho) - a_i(\rho')|\!|_{L^{2p}} \\
&\leq& \mathrm{const} \cdot\lambda^{(2+p)/2p} |\!|g_i - 1|\!|_{C^0} + \mathrm{const}\cdot
\lambda^{(2+p)/2p} K.
\end{eqnarray*}
Since $\lambda$ is sufficiently small, we get
\[|\!|g_i - 1|\!|_{C^0} \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K.\]
On the other hand, the ``compatibility condition" $g_{i+1}\rho_i = \rho'_i g_i$ means
\[\rho'_i - \rho_i = (g_{i+1}-1)\rho_i g^{-1}_i + \rho_i g^{-1}_i (1-g_i). \]
Hence
\[ |\rho'_i - \rho_i| \leq \mathrm{const} \cdot (|\!|g_{i+1} -1|\!|_{C^0(U_{i+1})} + |\!|g_i - 1 |\!|_{C^0(U_{i})})  \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K. \]
Since this holds for all $i$, it follows
\[K \leq \mathrm{const} \cdot \lambda^{(2+p)/2p} K.\]
If we choose $\lambda$ small enough, then this shows $K=0$ and $\rho = \rho'$.
\end{proof}

\begin{Rem}
So far we have studied only acyclic connections.
But this is just for simplicity.
The condition (iv) in Definition \ref{def: gluing data} can be replaced with
\[ \text{(iv)' } A \text{ is a regular ASD connection on } E. \]
If we glue an infinite number of regular ASD connections,
then the resulting ASD connections are parametrized by
\[ \mathrm{EGlP} \times \prod_{i\in \mathbb{Z}}B_i
\subset \mathrm{EGlP} \times \prod_{i\in \mathbb{Z}}H^1_{A_i}.\]
Here $B_i$ is a small ball centered at the origin in $H^1_{A_i} := \mathrm{ker} D_{A_i}$. This generalization is a simple application of the methods in this paper and [D2, pp. 324-325].
The proof is routine, so I omit the details.
\end{Rem}

redpajama_set_name: RedPajamaArXiv


The South Glens Falls Central School District, in collaboration with the Community Coalition for Family Wellness, will be holding a Parent University program "Social Media Safety" on Wednesday, Oct. 24, 2018, in the High School LGI. The program will be held from 6:30-7:30 p.m.
Attendees will learn about social media safety, safety rules and how to take ownership of online accounts.
The presenter will be David J. DeCelle, community educator at the National Center for Missing & Exploited Children.


Does Beauty and the Beast need to be remade? Heck, you can extend that question to all of Disney's recent and upcoming live-action remakes. Do any of them need to be remade? The short answer, of course, no. Those films generally hold up okay, and there's a reason why they've lasted decades, but that's not to say they can't be remade. We live in a different time, and as such, there are always aspects of the narrative that can be improved on or shifted to meet more modern sensibilities.
However, with the upcoming Beauty and the Beast film, I've wrestled with this question on several occasions. Given how recently the other movie came out (20+ years, which is comparatively low for Disney's animated features), and how modern their take was on Belle, it didn't seem essential. Even more discouraging from a creative standpoint seemed to be the visuals. Rather than take the movie in a new direction, the filmmakers seemed to have one goal: to transliterate what was in animation to live-action.
As stated above, the original film actually did a pretty great job of giving Belle that sort of modern mindset, but with the way our movies have advanced nowadays — mostly on the visual front — I can see the appeal that this new version would have over the original. Despite my misgivings, the fact is that to a lot of kids, these old 2D films are probably akin to black and white movies when I was growing up. Not exactly the most appealing thing to watch.
But in addition to transitioning it to live-action, they're giving the film a longer runtime, which will allow it to delve deeper into Belle's character and her relationships with her father, family, and the rest of the town.
As much as I loved the characters in the films, there is little denying that those things are brisk, 90-minute fares. There's little time to dilly-dally or get too involved in the characters. Adding an extra layer to it and giving the story time to breathe may actually do quite a bit to justify this remake, even if it is similar on a surface level.
What do you think of Condon's comments? Is Beauty and the Beast a movie worth remaking? Let us know your thoughts down below!


She has four children, he has two dogs and they are moving in together.
Staffordshire Bull Terrier Suzie is on the left and Ben below They are beautiful, friendly dogs but too excitable and not a little unruly!
Both dogs jump all over the sofas and all over people. Ben jumps up at the children and can knock the younger ones over.
The 3-year-old child has to be watched because she wants to hug the dogs and they don't like it.
What worries their mother most is Suzie's way of snatching food. She is rightly concerned that if a child is walking around with a biscuit for example, that it could unintentionally be bitten.
When I held a tiny treat in my fist, Suzie was absolutely frantic to get it. It's like she's starving hungry. I have seldom seen anything quite like it.
I'm sure there is more to this although a vet has previously said she's okay.
Although Suzie eats just the same amount as solid Buster, she is extremely thin. They have much more to pick up after her in the garden than from Ben, so food must be passing through her. She is given very little exercise so she shouldn't really be so skinny. If a change in diet doesn't work quickly, then another visit to the vet is called for.
Dog walks have become more and more infrequent for various reasons. These two dogs will be a lot more settled with regular exercise and healthy stimulation to compensate for the hours they spend alone during the day.
I showed the couple how to start a walk without the dog charging out of the door and pulling frantically down the passage. Suzie caught on really fast. This is not easy when a walk has become an exciting rarity. The couple will make themselves a do-able schedule of 'little and often' for the dogs involving walks and basic training.
The children also have their own things to learn. The youngest needs to be taught not to hug dogs. Safety first, a gate should be between kitchen and sitting room so that dogs and children can be separated either when any food is about or when an adult hasn't their eye on them.
With work, I'm sure that these lovely dogs will then calm down and learn some self-control. It's all up to their humans. It's great that they love the children.
NB. The precise protocols to best use for your own dog may be different to the approach I have worked out for Sasha and Ben, which is why I don't go into all exact details here of our plan. Finding instructions on the internet or TV that are not tailored to your own dog can do more harm than good, most particularly where children are involved also. One size does not fit all. If you live in my own area I would be very pleased to help with strategies specific to your own dogs (see my Get Help page).


text:
Cork Man Pleads Guilty To Gay Dating Site Blackmail
Neil Duggan pled guilty to the use of threatening behaviour and blackmail in order to extract thousands of euro from victim.
News 3 May, 2018 . Written by Peter Dunne.
Mr. Duggan of Cobh, Co. Cork, was originally charged following an investigation into the alleged blackmail of men on gay dating site, Fabguy. A complainant stated the blackmail had arisen with accusations he had linked up with a young man on the site only to discover the young man was a minor. This accusation then progressed into Duggan demanding thousands of euro from his victim through the use of threatening behaviour towards him.
Duggan admitted the charges of demanding money by menace, meaning no trial was needed to secure hisconviction. This involved demanding €2500 from the 60-year-old victim on April 11, 2017, and a further €5000 two days later at Topaz Filling Station in Douglas. He was also charged with threatening to kill the same man and with making threats to damage his property and van in the car park of the Silver Key Pub in Ballinlough. He had previously plead guilty to the latter charges, the fact of which was withheld from publishing until the outcome of the trial.
Duggan's barrister, Brendan Kelly, speaking at the Cork Circuit Criminal Court, stated his client had resorted to such desperate measures due to a gambling addiction. He said, "I would ask you to extend legal aid for a report from a counsellor Mr Duggan is attending. He has a severe gambling addiction. That is the genesis of his offending behaviour."
Judge Seán Ó Donnabháin agreed to extend the legal aid but did not agree to Mr. Kelly's request for Mr. Duggan's sentencing being adjourned up until October to allow him time to raise €7,500 in compensation for his victim. Sentencing in the case has been scheduled for June 14 with Duggan remanded on bail. As part of the conditions of bail, Duggan is to have no contact with the injured party and abstain from intoxicants as well as signing on with his local Garda station.
#gay dating site
#internet safety
3 May, 2018 . Written by Peter Dunne.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Tag Archives: thinkprogress
EcoOptimistic News for the End of an Environmentally Crappy Year
We can all use some positive news these days, especially on the environmental front in which science is considered political opinion, denial is an alternative fact and the word "protection" in the Environmental Protection Agency's name is a cynical leftover from its original mission. And while I don't want to gloss over the issues – there isn't enough paint in the world to do that – I offer here The Distillery, a weekly (or thereabouts) selection of posts to help offset the PTSD of our current nightmare.
Yeah, we've all heard about the UN IPCC report that gives us 12 years to get our shit together. And then we got that surprisingly frank White House "National Climate Assessment" that, despite the Trump regime's best efforts to bury it, made headlines. Plus, of course, there was devastating evidence of climate change already rearing its head in the form of a sometimes record-breaking series of hurricanes and typhoons.
But fortunately for the holidays – and perhaps for our sanity and our therapy sessions – we can snag some happier news. So I started compiling EcoOptimistic articles a few months ago, though some of it is from earlier in the year, when I realized that, now more than ever, we need to counterbalance the daily litany of the-end-is-nigh headlines.
It's not that I don't believe those headlines. Rather, it's that I won't give in to the fatalism of them. Many of us, sometimes – OK, often – including me, feel the despair coupled with the frustration and anger at those who avert their eyes, who won't listen to fact or reason, who pursue blind self-interest, or who rationalize it in desperate ways. (No, it won't cost jobs.) But letting them rule the news is infuriating and letting them determine our future is unacceptable.
FIRST, A SUMMARY…
From Earther
"The Rare Environmental Victories of 2018"
EcoOptimism's take: The headline, I think, is self-explanatory.
source: https://ocasio2018.com/
Perhaps one of the most (eco)optimistic events of the year was the surprise election of Alexandria Ocasio-Cortez to the House of Representatives. Her "Green New Deal," which combines environmental issues with job issues, is creating a politically appealing scenario that's garnering more and more support with both local officials and voters.
And combined with this poll, perhaps the political "climate" may have turned a corner.
From Yale Environment 360:
"Americans Who Accept Climate Change Outnumber Those Who Don't 5 to 1"
EcoOptimism's take: Chew on that, Fox News.
From the Washington Post:
"The Energy 202: Why 2020 candidates will be talking a lot more about climate change"
"Environment, Jaffa Cakes and Kylie Jenner star in statistics of the year"
EcoOptimism's take: Um, OK. Not entirely sure what to make of this, but nevermind.
While it's a bit cliché to refer to the younger generation taking the reins, there has been some notable news on that front, too. The suit by a group of teenagers against the federal government is continues to move forward despite the administration's efforts to get it thrown out of court. A Swedish 15-year-old made headlines at the recent UN climate change conference, lecturing the officials, "You are not mature enough to tell it like it is."
I've mentioned before that Teen Vogue has been consistently taking up environmental topics, most recently taking up the issue of ocean plastics here and here. (See more on that topic below.)
RENEWABLE ENERGY IS COMPETITIVE WITH – AND SOMETIMES CHEAPER THAN – FOSSIL FUELS…
One of the first items in that Earther post above notes the falling prices of renewable energy. "Beautiful, clean coal" is not cutting it financially despite the administration's best efforts. And there are some significant milestones accompanying it. It's been happening all year, not just in the last few months.
From Business Insider:
"One simple chart shows why an energy revolution is coming — and who is likely to come out on top"
From Earther:
"Solar Just Hit a Record Low Price In the U.S."
A Same-Day Twofer from Forbes:
"Plunging Prices Mean Building New Renewable Energy Is Cheaper Than Running Existing Coal"
"Coal Power Plants Lose Their Cost Advantage Over Clean Energy"
EcoOptimism's take: So much for "The sun don't always shine and wind don't always blow."
By David Dixon, CC BY-SA 2.0
We're seeing some of the results
"Britain Achieves the 'Unthinkable' as Renewables Leapfrog Fossil Fuel Capacity"
"10 States Now Get At Least 20 Percent of Their Electricity from Solar and Wind"
From Think Progress:
AND RENEWABLE ENERGY GOALS ARE BEING MET AHEAD OF SCHEDULE…
From Gizmodo:
"California Is Way Ahead of Schedule for Cutting Greenhouse Gas Emissions"
From Treehugger:
"Sweden to reach its 2030 renewables target 12 years early!"
EcoOptimism's take: Yes, we may need our rose-colored glasses here, but it's evidence of – don't get too choked up here – "yes, we can."
SINGLE-USE PLASTICS ARE IN THE CROSSHAIRS…
In some previous years, I've nominated a word of the year. (2012, 2013, 2014) This year, Collins Dictionary did it for me, choosing "single-use."
Spurred by a graphic and very disturbing video of a turtle having a plastic straw removed from its nose, the nascent movement to regulate or ban SUPs got a jump start. EcoOptimism has been charting the international movement.
Along with bans have come alternatives. We're not talking about bioplastics, which while interesting have their own issues, but about reducing or replacing demand.
"Corona Becomes First Big Beer Brand to Trial Plastic-Free Rings"
And from The Guardian:
"Carlsberg to replace plastic ring can holders with recyclable glue"
Images: Beverage Daily
EcoOptimism's take: We know Brett Kavanaugh "likes beer," though probably not for this reason.
The subtitle of this blog is "Finding the Future We Want." A great example of that is "turning lemons into lemonade."
"Where does your recycled plastic go? Perhaps into future highways."
EcoOptimism's take: There isn't a much bigger lemon than all that plastic waste and, while we may have mixed feelings about roads (unless they're for non-fossil-fueled vehicles and don't encourage more sprawl), here's some lemonade.
"A Former UK Coal Plant is Being Redeveloped Into an Eco-Village"
EcoOptimism's take: Re-use of decommissioned power plants may be emblematic of the possibilities, making this a good story to end on.
This entry was posted in climate change, Economics, Energy, Policy, Politics, Recycling, Resilience, Technology, The Distillery and tagged Alexandria Ocasio-Cortez, climate change, Earther, ecowatch, fossil fuels, Guardian, plastics, renewable energy, thinkprogress, Treehugger, Yale Environment 360 on December 26, 2018 by David Bergman.
Plastics: A Combined Distillery and EcoOptimism Post, Part 2
Beyond the Ban:
Plastics Alternatives and Mitigation
In just the short span since our recent post on the scourge of plastics, there have been more announcements of new or proposed bans on plastic straws and other types of plastics. (This week, Starbucks announced it will stop using plastic straws.) Rather than keep enumerating these individually, I've created a page, "Keeping Tabs on Plastics Bans," with a list organized by type of entity – country, locality, company – and type of plastics – bags, straws, packaging, microbeads. The list provides an easily graspable view of the extent of the movement, and will be updated as additional bans are set.
As promised in that previous post, we're going to focus here a bit on some alternative proposals and materials. Dealing with plastics – as with most environmental issues – can be broken down into two approaches: what to do once the problem is happening versus how to prevent the problem in the first place. These are commonly referred to respectively as adaptation versus mitigation. Years ago, I also heard this described as "front of tailpipe" and "back of tailpipe," the metaphor being pollution from cars, which can be dealt with either by filtering it in the exhaust pipe (that would be the "end of tailpipe") by, for example, a catalytic converter, or by modifying the engine so that the pollution is prevented or at least diminished before it occurs. As you can imagine, heading off the problem is preferable to fixing it afterward.
In the case of plastics, we have a combination of damage already done along with a continuing stream of new plastics adding to the damage. Where plastic refuse has accumulated, such as in ocean gyres, the only remedy is to somehow, laboriously, retrieve it. Another adaptive after-the-fact approach is recycling. That at least keeps it out of the waste stream. (In theory, anyway. Less than 10% of plastic in the US is actually recycled. And, as others have noted, recycling shifts the responsibility – environmental and economic – from the actual producers of the plastic onto us, the consumers)
Image: Ocean Cleanup Project via EcoWatch
EcoOptimism's take: Whether this type of ocean plastic reclamation would actually have a significant impact, given the scale of the problem and size of the oceans, is a topic of debate. But in any case, upstream prevention would be a much better solution, at least in terms of addressing a continuing problem.
The better solution would have been to not produce the plastic in the first place. We're well beyond that option, obviously. But that doesn't mean we shouldn't attempt to avoid further exacerbating the problem through mitigation, meaning let's not make more plastic if we can avoid it.
Plastic is so ubiquitous at this point that it's hard to imagine a world without it. But there are indeed alternatives, both mitigative and adaptative.
Some of those solutions exist right in front of us, or at least in other places we can adapt from. A well-known one in some environmental circles is the Indian tiffin box. The tiffin involves a system in which lunches for workers and school children are packed at home in the morning and then distributed through a remarkable system by dabbawalas. The relevant idea here is that the home prepared meals arrive in tiffin boxes made of stacked metal dishes rather than disposable take out containers and, after the meal, are picked up and returned to each family's kitchen for reuse.
Two dabbawalas in Mumbai delivering meals packed in tiffin carriers. Image credit: Wikipedia
This system has found a modern interpretation in Brussels where the city has introduced the Tiffin Project. People sign up for the project, purchase tiffin containers and bring them when they purchase take out foods. They even get a 5% discount on their orders.
From Treehugger
"Brussels has an ingenious solution to wasteful takeout containers"
Image: Tiffin Belgium
EcoOptimism's take: as optimistic as we like to be, it's hard to imagine such a system succeeding here. As the Treehugger post notes, the system works best with small, local restaurants and "helps people discover new places to eat." Takeout food in much of this country tends to come from large chains with familiar menus.
Tiffin boxes, if not the delivery system, have found their way west. You can even buy them on Amazon and elsewhere, with a Western interpretation:
A better-known example of a plastic substitute is, of course, the reusable cloth bag. They've become so ubiquitous that you'll find them for sale (branded, of course) in all kinds of stores – not just the eco-oriented ones like Whole Foods, but conventional stores, too. And they're an almost inevitable part of events. We have a (reusable) bag full of reusable bags that we've accumulated from fundraisers, trade shows and promotions. Too many, in fact.
Image credit: David Bergman
But there are times when a cloth bag just won't cut it. For those, there are forms of plastic that are not made from oil and that can decompose after use.
The first plastics invented were made from plants. (The word cellophane refers to the fact that it was made of cellulose: plant fiber.) A famous photo shows Henry Ford swinging an axe against the bioplastic trunk of a Model T to show the material's strength. (The axe was actually covered with fabric but made for an impressive display nonetheless.) As the story goes, he wanted to make his cars with bioplastic, but the steel industry had other ideas about that.
Image from HemmingsDaily
Bioplastics are getting renewed interest for applications such as plastic utensils made from potato starch. The main caveats here are how well they decompose or recycle and that the plant starches not be taken from foods. The solution to the latter is to use crop byproducts such as wheat chaff as opposed to the grain.
Image credit: www.ecoproducts.com
EcoOptimism's take: In addition to the points above, bioplastics, for now, are not quite as versatile as synthetic plastics, but applications are broadening and have wide potential.
But there's now a renewed interest in bioplastics. A case in point:
From Engadget:
"Lego will soon make bricks out of sugarcane bioplastics"
EcoOptimism's take: While it's exciting to think that all those future Lego creations might not end up buried forever in landfills, this announcement is a bit misleading because, for now, it's only Lego's landscape elements, comprising about 1% – 2% of their production, not the iconic bricks.
But it's not likely that all synthetic plastics can be substituted with bioplastics. And that leads us to a back-of-tailpipe types of mitigation. Among them are technologies that break down plastics.
"Scientists accidentally create mutant enzyme that eats plastic bottles"
Credit: Still image from video in The Guardian
From Grist:
"Mealworms munch on Styrofoam without dying, shock scientists"
Image credit: Geek.com
EcoOptimism's take: In our Parsons School of Design Sustainable Systems course, we have the students try this out. Though they tend to be grossed out by the mealworms (see photo above!), they get to see that it actually works.
Still, these last two are after-the-fact approaches and, not to belabor the point, we'd be much better off not incurring the problem of more plastics on the first place.
This entry was posted in Recycling, Technology, The Distillery, Waste and tagged Bioplastic, consumption, Engadget, externalities, Grist, Plastic straws, plastics, thinkprogress, Treehugger on July 13, 2018 by David Bergman.
The Distillery: April 22, 2018
Here on this anniversary of Earth Day, it seems appropriate to update a topic I first wrote about in 2012 in a post I titled "Planets Are People, My Friends." It was a reference at the time to a statement by then-presidential candidate Mitt Romney who told attendees at a rally that "corporations are people, my friend." While the statement was actually in response to a comment about taxes, it also could be seen as being about the infamous Citizens United Supreme Court case that basically said corporations have the same free speech rights as people, and that spending on political campaigns is a form of free speech. That court decision has had a disastrous affect on our elections ever since.
While that Supreme Court case was about establishing the rights of corporations, my post drew a parallel with the equally odd-sounding idea of nature having rights. It talked about movements to give rights to South American forests, a New Zealand River and apes in Spain. Since then the movement has spread further.
"Group files suit to recognize the Colorado River as a person"
EcoOptimism's take: New Zealand has a river with rights and now the US may get one, too.
"The Colombian Amazon Is Now a 'Person', and You Can Thank Actual People"
EcoOptimism's take: In addition to being about recognizing nature's rights, this also ties into some EcoOptimism posts including a recent Distillery post on the topic of intergenerational rights, meaning the right of young generations to grow up with a healthy environment. The Colombian Supreme Court case that decided this was brought by Colombian youth.
"Florida kids are taking their climate-denying governor to court"
And also in Teen Vogue:
"Florida Governor Rick Scott Is Getting Sued by Teens for His Environmental Polices"
EcoOptimism's take: More evidence of the growing trend of youth suing their unresponsive government. In this case, the suit is directed toward adamant climate change denier Governor Rick Scott. Scott has also been the subject of another teen-led suit. That one is over gun control in the aftermath of the Parkland High School shooting and has grown into an example of what galvanized youth can do.
This entry was posted in climate change, Messaging, Nature, Policy, Politics, The Distillery and tagged climate change, Earther, media, Teen Vogue, thinkprogress, Treehugger on April 22, 2018 by David Bergman.
The Distillery: February 27, 2018
A recent EcoOptimism Distillery post was on the theme of "good news disguised as bad news." Here, perhaps, is the ultimate example of that. In spite of and maybe because of all the astoundingly bad news about Trump's environmental "witch hunt" (to redirect his term)….
Poll reveals Americans are hitting their breaking point on the environment
(The previous high point – 2006 – was shortly after Al Gore's An Inconvenient Truth came out.)
Image source: ThinkProgress
EcoOptimism's take: Sure, it'd be better if public support for environmental actions was due to something positive, but this support as a reaction to the absolute illogic of Trump is encouraging. And even better, must piss off the denier-in-chief.
Here's hoping it makes his twitter finger sore.
And in a similar vein, though media coverage of climate change still lags ridiculously behind other topics…
Also from ThinkProgress:
Trump's climate denial backfires, drives more media coverage of the issue
Image source: MediaMatters
EcoOptimism's take: The post's subtitle kinda says it all. "How the president is getting more people to think and talk about climate change." The post then explains: "Trump is driving TV coverage of climate change, and as a result, he is raising the profile of the issue. Last year's spike in coverage of climate change corresponded with an uptick in public concern. Worry about climate change is now at an all-time high across several polls." (As shown in the article and graph above.) But it also goes on to say "News outlets gave an uncontested platform to climate deniers." And Media Matters, the data source for the post, said "The networks undercovered or ignored the ways that climate change had real-life impacts…."
I guess that makes it a qualified "good news disguised as bad news."
This entry was posted in climate change, Messaging, Policy, Politics, The Distillery and tagged climate change, Joseph Romm, media, thinkprogress, Trump on February 27, 2018 by David Bergman.
The Distillery: February 2, 2018
The theme here: good news disguised as bad news
2017's costly climate change-fueled disasters are the 'new normal,' warns major reinsurer
and from The New York Times:
2017 Set a Record for Losses from Natural Disasters. It Could Get Worse
And the "Bomb Cyclone" wasn't even in 2017, so we're off to an inauspicious start. But there's optimism here ….
Photo: U.S. Air Force photo/Tech. Sgt. Ben Bloker
EcoOptimism's take: On the face of it, this doesn't sound like an EcoOptimistic post, but it in fact emphasizes the economic incentive to mitigating and adapting to climate change. Insurance companies – especially reinsurers, the ones who insure the insurance companies – have been concerned with this for a while. The Times article references Munich Re, but Swiss Re has also studied the potential costs of covering insurance losses due to climate change and has been ringing alarm bells.
Arguably, since politicians (American, that is) aren't onboard, it may be the business world that spurs, finally, US action. Ironically, politicians cling to the belief that environmental action is bad for business.
And if the business case for climate action doesn't work, maybe the military case can…
Climate Impacts Nearly Half of U.S. Military Bases
Photo: Michael Lavender / U.S. Navy / Flickr
EcoOptimism's take: The US military is a surprisingly staunch advocate for adaptation to climate change. Though Trump ridiculed Obama for saying so, climate change is a national security threat that could both create or exacerbate geopolitical and affect military readiness. This post, though, emphasizes the potential direct cost of climate change. As a military policy, it might even have Trump's ear and bring him to his senses. But maybe that's my EcoOptimism speaking.
China is going to stop accepting plastic for recycling, so…
America Needs a Plastics Intervention. Now's the Time.
photo source: Scrap Monster
EcoOptimism's take: So China's going to stop taking the world's plastic waste for recycling. That problem, though, creates the impetus for better recycling, plastics that are more recyclable, and/or plastics bans here and elsewhere. And, of course, to stop shipping our problems elsewhere.
A new twist to "Read My Lips" and taxes…
From Evolution News:
Would a Beef Tax Reduce Greenhouse Gas Emissions?
Photo source: Evolution News
EcoOptimism's take: : Taxes are bad, right? Not if they discourage consumption of things that aren't good for us or the environment – and meat is both. Think about, for instance, taxes on cigarettes or alcohol. Or sugary drinks, as is catching on in some places. (Though NYC's proposed tax didn't survive a lawsuit. On the other hand, the city's proposal for congestion pricing which, arguably, is a tax, is getting some traction after a false start.)
Now if we can only institute carbon taxes. Or if not that, maybe at least an increase in the decades old gasoline tax, which has been 18.4 cents since 1993. That means it's actually decreased significantly due to inflation.
But taxes are bad, right?
This entry was posted in climate change, Economics, Energy, Natural Resources, Policy, Recycling, Resilience, Resources, The Distillery and tagged carbon pricing, climate change, ecowatch, greenhouse gas, natural disasters, plastics, thinkprogress on February 5, 2018 by David Bergman.


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Lilly Maier
Lilly Maier, an intern for PolitiFact's national staff for Fall 2013, was a student at American University. Originally from Europe, she switches between writing in English and German.
Other publications include the Huffington Post and the Austrian daily newspaper Kurier. Previously, Maier worked for German and Austrian magazines, including News, Presse and Die Stadtspionin ("Cityspottergirl").
In 2007, she founded a bimonthly European youth magazine, Schnipsel ("papercuts"), that was awarded with funding by the European Union.
Featured Statements
Latest Fact-checks by Lilly Maier
stated on December 5, 2013 in a tweet:
"FACT: Repealing Obamacare would be easier than repealing Prohibition."
By Lilly Maier • December 9, 2013
stated on December 5, 2013 in a meme on the Internet:
Says Vladimir Putin said, "President Obama and his fellow Democrats are either idiots or deliberately trying to destroy their own economy."
Matthew Dowd
stated on November 24, 2013 in comments on ABC's "This Week":
"Harry Reid was against (the nuclear option) before he was for it. Mitch McConnell was for it before he was against it. President Obama was against it before he was for it."
By Julie Kliegman • November 24, 2013
stated on November 21, 2013 in speeches:
On changing the rules for filibusters on presidential nominees
By Lilly Maier • November 22, 2013
stated on October 1, 2013 in viral posts on the Internet:
Says a man opted out of Obamacare after the marketplace launched on Oct. 1, 2013, and was informed he owed a $4,000 fine.
By Lilly Maier • October 4, 2013
Ted Yoho
stated on September 4, 2013 in an interview on Fox Business Network:
Chemical weapons have been used "probably 20 times" since the Persian Gulf War.
By Becky Bowers • September 10, 2013


So it started with Self-Service BI Tools: Round Up and Deep Dive Webcast Recap which I posted on the way to Nashville. Many of the products mentioned in the blog were topics at the BI+Analytics conference.
Veselina Peykova writes about workarounds in Accessing sap.com with different users (at the same time). She has received 17 likes already so this is a popular blog on a popular (but frustrating) topic.
Jitendra Kansal, SAP Mentor a new SAP employee, Learn about and get hands dirty on latest & greatest of SAP Mobile offerings @SAPTechEd 2017 -don't you wish you could clone yourself and attend all sessions at all three SAP TechEd's?
I invite everyone to the Fall Forward and Join the #ASUG Virginia Chapter Fall Meeting in October in historic Richmond, Virginia. With a mix of partner and customer stories, and great hosts (Altria), the event is not to be missed. See you there? Did I say Altria serves the best ASUG Chapter food I've ever had? They do.
Think about whether you want to dedicate your free time to the community, whether you have that passion for what you do, whether you enjoy learning constantly, and whether you have availability, because being a Mentor is not just partying, going to events and earning a cool shirt.
Jocelyn Dart writes Fiori elements – Status Icons and Semantic Colors and offers sessions for SAP TechEd. Jocelyn continues focusing on accessibility, and important topic. She also writes Fiori elements – Percentage Progress Indicators. Jocelyn's blogs are in-depth, sharing her knowledge, and not to be missed.
I recapped the BI+Analytics Conference in BI+Analytics Conference – It's a Wrap #BIAConf – what stays with me are the people connections and Jessica Buchanan's keynote. Someone counted 10 SAP Mentors in attendance too, with Clint Vosloo traveling 53 hours to get there. I failed to get pictures of dust bins, something Jelena or Jim Spath might do.
SAP Mentor Program Lead Jason Cao writes More Voices from the Community at SAP TechEd Barcelona 2017 – I'm looking forward to Moya Watson's upcoming blog about SAP Community Sessions at Las Vegas.
What other moments did I miss?