text:
\begin{document}
\setlength{\baselineskip}{17pt}

\title{Small coupling limit and multiple solutions to the Dirichlet
Problem for Yang Mills connections in $4$ dimensions - Part III}
\author{Takeshi Isobe \thanks {
Tokyo Institute of Technology; email:
isobe.t.ab@m.titech.ac.jp} $\,$ and Antonella Marini \thanks {
University of L'Aquila / Yeshiva University; email:
marini@dm.univaq.it}}

\date{}
\maketitle

\begin{abstract}
{\footnotesize \baselineskip 4mm} In this paper, the third of this
series, we prove that the spaces $\mathcal{A}^{*,p}_k(A_0;\q)$ and
$\mathcal{B}^{*,p}_{0,k}(A_0;\q)$ which contain $L^p_k$-approximate
solutions to the Dirichlet problem for the $\epsilon$-Yang Mills
equations on a four dimensional disk $B^4$, carry a natural manifold
structure (more precisely a natural structure of Banach bundle), for
$p(k+1)> 4$. All results apply also if $B^4$ is replaced by a
general compact manifold with boundary, and $SU(2)$ is replaced by
any compact Lie group. We also construct bases for the tangent space
to the space of approximate solutions, thus showing that this space
is 8-dimensional for $\epsilon$ sufficiently small, and prove some
technical results used in Parts I and II for the proof of the
existence of multiple solution and, in particular, non-minimal ones,
for this non-compact variational problem.
\end{abstract}

\section{Introduction}
We consider connections $A$ over principal bundles $P$ over the four
dimensional disk, with fiber isomorphic to $SU(2)_\epsilon$, for
$\epsilon >0$, where $SU(2)_\epsilon=SU(2)$ as a set, but it is
endowed with the \emph{$\epsilon$-deformed} multiplication, i.e.,
its Lie algebra is
$(\mathfrak{su}(2)_{\epsilon},[\cdot,\cdot]_{\epsilon})
:=(\mathfrak{su}(2),\epsilon[\cdot,\cdot])$, where $[\cdot,\cdot]$
is the ordinary Lie bracket on $\mathfrak{su}(2)$. We recall that
for a given smooth boundary value $A_0$, the Dirichlet problem for
the $\epsilon$-Yang Mills equations is obtained via a variational
principle from the $SU(2)_{\epsilon}$-Yang Mills functional
\begin{equation}
\label{YMe} \YMe(A)=\int_{B^4}|{F_A}^\epsilon|^2\,dx\,,
\end{equation}
where
${F_A}^\epsilon=dA+\frac{1}{2}[A,A]_{\epsilon}:=dA+\frac{\epsilon}{2}[A,A]$,
and consists of the system $$
(\mathcal{D}_{\epsilon})\quad\qquad\left\{\begin{array}{ll}
{d_A^{\ast}}^\epsilon {F_A}^\epsilon=0\quad&\mbox{in } B^4\\
\iota^{\ast}A\sim A_0\quad&\mbox{at } \partial B^4\,,
\end{array}\right.
$$
where, $\iota:\partial B^4\to\overline{B}^4$ is the inclusion, the
symbol $\sim$  stands for gauge equivalence via a gauge
transformation that extends smoothly to the interior, and
${d_A^{\ast}}^\epsilon:=\ast
d\ast+\ast[A,\ast\cdot]_{\epsilon}:=\ast
d\ast+\epsilon\ast[A,\ast\cdot]$,  where $\ast$ is the Hodge star
operator with respect to the flat metric on $\R^4$.

An absolute minimum, say $\underline{A}_{\epsilon}$, for the
Yang-Mills functional is known to exist by \cite{Marini}. Moreover,
in \cite{IM}, it is shown that the space of connections with
boundary value $A_0$, denoted by $\mathcal{A}(A_0)$, has countable
connected components, i.e.,
$\mathcal{A}(A_0)=\bigsqcup_{j=-\infty}^{\infty}\mathcal{A}_j(A_0)$,
where $\mathcal{A}_j(A_0)$ is the space of connections with relative
2nd Chern number with respect to $\underline{A}_{\epsilon}$ equal to
$j$, and that there always exists a minimizer in
$\mathcal{A}_{+1}(A_0)$ (or $\mathcal{A}_{-1}(A_0)$) if $A_0$ is not
flat. In \cite{IM1, IM2} we have found solutions to $
(\mathcal{D}_{\epsilon})$ in $\mathcal{A}_{+1}(A_0),$ by first
constructing approximate solutions, for small values of the
parameter $\epsilon >0. $ However, some of our proofs utilize
technical results proved in the present paper.

Throughout this paper, we assume $\lambda_0,$ $D_1,$ $D_2,$
$\epsilon$, $\q$ be fixed, with $0<2\lambda_0<d_0$, $0<D_1<D_2$,
$\q:=(p,[g],\lambda)\in\PP(d_0,\lambda_0;D_1,D_2;\epsilon):=\{\q:=(p,[g],\lambda)\in
\PP(d_0,\lambda_0):D_1\epsilon<\lambda^2<D_2\epsilon\},$ where
$\PP(d_0,\lambda_0):= B^4_{1-d_0}\times SO(3)\times(0,\lambda_0)$ is
the parameter space used in \cite{IM1} in the gluing procedure to
construct approximate solutions to the $\epsilon-$Dirichlet problem,
that look like the connected sum of
$\underline{A}_{\epsilon}\#\frac{1}{\epsilon}\text{(1-bubble)}.$

Although all our arguments apply to the general case of $M$, any
smooth compact manifold with boundary, and $G$, any compact Lie
group, we focus on $M=B^4$ and $G= SU(2)$.

We recall (cf. \cite{IM1}) that, for $\q:=(p,g,\lambda)$ fixed,
\begin{equation}
\label{A(q)} A(\q)=\left\{\begin{array}{ll}
(1-\beta_{\lambda,p})\underline{A}_{\epsilon}+\frac{1}{\epsilon}\beta_{\lambda/4,p}\,g\,I^2_{\lambda,p}\,g^{-1}+
\frac{1}{\epsilon}(1-\beta_{\lambda/4,p})gPI^2_{\lambda,p}g^{-1}&\quad\mbox{in } B^4\setminus\{p\}\\
\frac{1}{\epsilon}gI^1_{\lambda,p}g^{-1}&\quad\mbox{in }
B^4_{\lambda/4}(p)\,,
\end{array}\right.
\end{equation}
where
$I^1_{\lambda,p}(x)=\text{Im}\frac{(x-p)d\overline{x}}{\lambda^2+|x-p|^2}$
in $B^4 (p):=\{x\in\R^4:|x-p|<1\}$, together with
$I^2_{\lambda,p}(x)=\text{Im}\frac{\lambda^2(\overline{x}-\overline{p})dx}{|x-p|^2(\lambda^2+|x-p|^2)}$
in $\R^4\setminus\{p\}$ compose the 1-instanton solution to the Yang
Mills equations on $\R^4$,  $\beta(x)=\beta(|x|)\in
C^{\infty}_0(\R^4)$
 is such that $\beta=1$ for $|x|\le 1$, $\beta(x)=0$ for
$|x|\ge 2$ and $0\le\beta(x)\le 1$, and
$\beta_{\lambda,p}(x):=\beta(\lambda^{-1}(x-p)).$

\noindent These connections form the space
\begin{equation}
\label{N}
\mathcal{N}(d_0,\lambda_0):=\{A(\q):\q\in\PP(d_0,\lambda_0)\}\,,
\end{equation}
of approximate solutions to $(\mathcal{D}_{\epsilon})$ with relative
second Chern class equal to +1.

\medskip \noindent Let $g_{12,p}(x)=\frac{x-p}{|x-p|}$ be the
transition maps for the $1$-instanton solution defined above. The
connections $A(\q)$ live on the bundles
\begin{equation}
\label{Pq} P(\q):= \biggl(B^4_{\lambda/4}(p), B^4\setminus\{p\},
g\,g_{12,p}\,g^{-1}\biggr)\,.
\end{equation}
(Notice that $P(p, g, \lambda)= P(p, -g, \lambda)$ and $A(p, g,
\lambda)= A(p, -g, \lambda)$, that's why we take $g\in SU(2)/\{\pm1\}
\cong SO(3)$).

We define $\mathcal{A}^*(A_0;\q)$ as the space of all connections
over $P(\q)$ satisfying $\iota^{\ast}A\sim A_0$ on $\partial B^4$,
via a gauge transformation $g\in \mathcal{G}^*(\q)$ (the group of
all smooth gauge transformations at $\partial B^4$ such that
$g\bigl((1,0,0,0)\bigr)=\mathbf{1}$), which extends smoothly to the
interior, and
$\mathcal{B}(A_0;\q)^*:=\mathcal{A}^*(A_0;\q)/\mathcal{G}^*(\q)$. We
define also the Sobolev counterparts of these spaces:
$\mathcal{A}^{\ast,p}_k(A_0;\q)$, as the space of all connections
over $P(\q)$ satisfying $\iota^{\ast}A\sim A_0$ on $\partial B^4$,
via a gauge transformation $g\in \mathcal{G}^{\ast,p}_{k+1-1/p}(\q)$
(the group of all gauge transformations in $L^p_{k+1-1/p}(\partial
B^4)$ such that $g\bigl((1,0,0,0)\bigr)=\mathbf{1}$), which admits
an $L^p_{k+1}$ extension to the interior, and
$\mathcal{B}^{\ast,p}_k(A_0;\q):=\mathcal{A}^{\ast,p}_k(A_0;\q)/\mathcal{G}^{\ast,p}_{k+1}(\q)$.

\noindent We assume $p(k+1)>4$ throughout the present paper.

\section{A manifold structure for $\mathcal{A}^{*,p}_k(A_0;\q)$ and
$\mathcal{B}^{\ast,p}_k(A_0;\q)$} The space of all connections on a
principal bundle $P$ is an affine space and carries a natural
differentiable structure, while $\mathcal{A}^*(A_0; \q))$ is not an
affine space and it is non-trivial to show that it carries a natural
differentiable structure. The purpose of this section is to show
that $\mathcal{A}^*(A_0; \q)$, its quotient $\mathcal{B}^{\ast}(A_0;
\q)$, and, more in particular, their Sobolev counterparts, possess a
natural differentiable structure.

\noindent We start with  $\mathcal{A}^{\ast,p}_k(A_0; \q)$. In what
follows, we omit the dependence on $\q$ in the notation.
\newtheorem{proposition}{Proposition}[section]
\begin{proposition} \label{P5.1} The space of connections
$\mathcal{A}^{\ast,p}_k(A_0)$  carries a natural structure of Banach
bundle over $\mathcal{G}^{\ast,p}_{k+1-1/p}$. More precisely, for
$A\in\mathcal{A}^{\ast,p}_k(A_0)$ with $\iota^{\ast}A=g^{\ast}A_0$
on $\partial B^4$, for some $g\in\mathcal{G}^{\ast,p}_{k+1-1/p}$,
define
$\pi:\mathcal{A}^{\ast,p}_k(A_0)\to\mathcal{G}^{\ast,p}_{k+1-1/p}$
by $\pi(A)=g$. The map $\pi$ is a well-defined projection, and
$\pi:\mathcal{A}^{\ast,p}_k(A_0)\to\mathcal{G}^{\ast,p}_{k+1-1/p}$
is a vector bundle with fiber isomorphic to
$L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$.
\end{proposition}
\textit{Proof:} By the Sobolev embedding $L^p_{k+1-1/p}(\partial
B^4)\subset C^0(\partial B^4)$ for $p(k+1)>4$,
$\mathcal{G}^{\ast,p}_{k+1-1/p}$ is a Banach Lie group. Since
$\mathcal{G}^{\ast,p}_{k+1-1/p}$ acts freely on
$\mathcal{A}^{\ast,p}_k(A_0)$ (the restriction of $g$ to the
boundary is uniquely determined), the map
$\pi:\mathcal{A}^{\ast,p}_k(A_0)\to\mathcal{G}^{\ast,p}_{k+1-1/p}$
is well-defined.

\noindent The group $\mathcal{G}^{\ast,p}_{k+1-1/p}$ possesses a
natural manifold structure. In fact, the space
$$L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4})):=\{u\in
L^p_{k+1-1/p}(\Ad(P|_{\partial B^4})):u((1,0,0,0))=0\}$$  is a
closed subspace of the Banach space $L^p_{k+1-1/p}(\Aut(P|_{\partial
B^4}))$, which is isomorphic to the Lie algebra of
$\mathcal{G}^{\ast,p}_{k+1-1/p}$. For $u\in
L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))$, and fixed
$s\in\mathcal{G}^{\ast,p}_{k+1-1/p}$, define $(\Exp_su)(x)=s(x)\exp
u(x)$, where ``$\exp$" denotes the exponential map for the Lie group
$SU(2)$. A local chart of $\mathcal{G}^{\ast,p}_{k+1-1/p}$ at $s$ is
then constructed as follows. Set
\begin{equation}
\label{5.1}
U:=\{u\in L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial
B^4})):\|u\|_{k+1-1/p,p}<\sigma\},
\end{equation}
where $0<\sigma<\frac{r}{3}$, where $r$ is the injectivity radius of
$SU(2)$. Let $\tilde{U}_s:=\Exp_s(U)$. With this definition,
$\tilde{U}_s$ is a neighborhood of $s$ in
$\mathcal{G}^{\ast,p}_{k+1-1/p}$ and $\Exp_s:U\to\tilde{U}_s$ yields
a local chart near $s$. See~\cite{FU} for more details.

We construct the `extension operator'
$T:\mathcal{G}^{\ast,p}_{k+1-1/p}\to\mathcal{G}^{\ast,p}_{k+1}$, in
such a way that for any $s\in\mathcal{G}^{\ast,p}_{k+1}$, there
exists a neighborhood $U$  (as in \eqref{5.1}) of $\iota^{\ast}s$ in
$\mathcal{G}^{\ast,p}_{k+1-1/p}$ such that its restriction
$T|U=:T_s:U\to\mathcal{G}^{\ast,p}_{k+1}$ is smooth and satisfies
$T_s(\iota^{\ast}s)=s$.

We take a smooth connection which extends $A_0$ to $P$. For
simplicity, we also denote this connection by $A_0$. Set
$\Delta_{A_0}=\nabla_{A_0}^{\ast}\nabla_{A_0}:L^p_{k+1}(\Ad(P))\to
L^p_{k-1}(\Ad(P))$. Let $U\subset\mathcal{G}^{\ast,p}_{k+1-1/p}$ be
a small neighborhood of $\iota^{\ast}s$. We first define the linear
operator $T_0:U\to L^{\ast,p}_{k+1}(\Ad(P))$ as follows (here we
identify locally $\mathcal{G}^{\ast,p}_{k+1-1/p}$ with its Lie
algebra $L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))$ via the
exponential map): for given $\xi\in U$, $u=T_0(\xi)$ is the unique
solution of $\Delta_{A_0}u=0$ in $B^4$ with boundary value $u=\xi$
on $\partial B^4$. Elliptic
estimates yield smoothness of $T_0$. Moreover,
$\|T_0(\xi)\|_{k+1,p}\le C\|\xi\|_{k+1-1/p,p}\le C\sigma$, for some
$C>0$ independent of $\xi$. Thus, for small $\sigma$,
$T_s(\xi):=\Exp_s(T_0(\xi))$ is well-defined, smooth with respect to
$s$ and $\xi$, and $T_s(0)=s$.

We next define local charts for $\mathcal{A}^{\ast,p}_k(A_0)$. Let
$U$ be as in \eqref{5.1}. We observe that any
$A\in\mathcal{A}^{\ast,p}_k(A_0)$ can be written as
$A=g^{\ast}(A_0+\alpha)$, for some $g\in\mathcal{G}^{\ast,p}_{k+1}$
and some $\alpha\in L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$. Hence,
for $s\in\mathcal{G}^{\ast,p}_{k+1-1/p}$, we can define the map
\begin{align}
&\Phi_s:U\times
L^p_{0^T,k}(T^{\ast}\otimes\Ad(P))\to\mathcal{A}^{\ast,p}_k(A_0),\notag\\
&\qquad\Phi_s(u,\alpha)=T_s(u)^{\ast}(A_0+\alpha).\notag
\end{align}

\noindent In order to complete this proof we need the following two
lemmas.

\newtheorem{lemma}{Lemma}[section]
\begin{lemma} \label{L5.1} $\Phi_s$ is one to one.
\end{lemma}
\textit{Proof:} One has that $\Phi_s(u,\alpha)=\Phi_s(v,\beta)$ for
$(u,\alpha),(v,\beta)\in U\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$ if and only if
$$T_s(u)^{\ast}A_0+T_s(u)^{-1}\alpha T_s(u)=T_s(v)^{\ast}A_0+T_s(v)^{-1}\beta T_s(v)\,.$$
The restriction of this formula to $\partial B^4$ yields
$\iota^{\ast}T_s(u)^{\ast}A_0=\iota^{\ast}T_s(v)^{\ast}A_0$, and,
since the $\mathcal{G}^{\ast,p}_{k+1-1/p}$-action is free,
$\iota^{\ast}T_s(u)=\iota^{\ast}T_s(v)$ on $\partial B^4$. Thus,
$\Exp_su=\Exp_sv$, by definition of $T_s$. From this, $u=v$ and
$\alpha=\beta$. \hfill$\Box$

\medskip

\begin{lemma} \label{L5.2}
$\{\Phi_s,U\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))\}_{s\in\mathcal{G}^{\ast,p}_{k+1-1/p}}$
is a differentiable vector bundle structure for
$\mathcal{A}^{\ast,p}_k(A_0)$.
\end{lemma}
\textit{Proof:} Let $s,t\in\mathcal{G}^{\ast,p}_{k+1-1/p}$ with
$\|s-t\|_{k+1-1/p,p}<\epsilon$ for some small $\epsilon>0$. Here and
in what follows, we replace $\iota^{\ast} s$ by $s$ in our notation.

We need to show that $\Phi_s^{-1}\circ\Phi_t:U\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))\to U\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$ is smooth for small positive
$\epsilon$. Setting $\Phi_s^{-1}\circ\Phi_t(u,\alpha)=(v,\beta)$,
for $(u,\alpha),(v,\beta)\in U\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$, one has
$v=\Exp_s^{-1}\circ\Exp_tu$ (cf. Lemma \ref{L5.1}). Then, $\beta$ is given
by
$$\beta=T_s(\Exp_s^{-1}\circ\Exp_tu)\bigl(T_t(u)^{\ast}A_0+T_t(u)^{-1}\alpha
T_t(u)-T_s(\Exp_s^{-1}\circ\Exp_tu)^{\ast}A_0\bigr)\,T_s\,(\Exp_s^{-1}\circ\Exp_tu)^{-1}.$$
Thus, $v$ and $\beta$ depend smoothly on $u$ and $\alpha$. This
completes the proof. \hfill$\Box$

\medskip

\textit{Completion of the proof of Proposition \ref{P5.1}:} We observe that for $s\in\mathcal{G}^{\ast,p}_{k+1-1/p}$
\begin{align*}
\pi^{-1}(s)&=\{A\in\mathcal{A}^{\ast,p}_k(A_0):\iota^{\ast}A=s^{\ast}A_0\}\\
&=\{A={\tilde{s}}^{\ast}(A_0+\alpha):\alpha\in
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))\}\cong
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P)),
\end{align*}
where $\tilde{s}\in\mathcal{G}^{\ast,p}_{k+1}$ satisfies
$\iota^{\ast}\tilde{s}=s$ and
\begin{align*}
&\pi^{-1}(\tilde{U}_s)=\{A\in\mathcal{A}^{\ast,p}_k(A_0):\iota^{\ast}A=g^{\ast}A_0,~g\in\tilde{U}_s\}\\
&=\{A=T_s(u)^{\ast}(A_0+\alpha):u\in U,~\alpha\in
L^2_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))\} \cong\tilde{U}_s\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P)).
\end{align*}
Combining these results with Lemma \ref{L5.2}, one sees that
$\pi:\mathcal{A}^{\ast,p}_k(A_0)\to\mathcal{G}^{\ast,p}_{k+1-1/p}$
is a vector bundle with fiber isomorphic to
$L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$. This completes the proof.
\hfill$\Box$

\medskip

The following proposition holds for the quotient space
$\mathcal{B}^{\ast,p}_k(A_0)=\mathcal{A}^{\ast,p}_k(A_0)/\mathcal{G}^{\ast,p}_{k+1}$:
\begin{proposition}
\label{P5.2} The space $\mathcal{B}^{\ast,p}_k(A_0)$ defined in $\S1$ has a differentiable manifold structure.
Moreover, the canonical projection
$\mathcal{A}^{\ast,p}_k(A_0)\to\mathcal{B}^{\ast,p}_k(A_0)$ yields a
principal $\mathcal{G}^{\ast,p}_{k+1}$-bundle.
\end{proposition}
\textit{Proof:} The argument is standard (cf.~\cite{DK},~\cite{FU}).
One first seeks a candidate for the slice in
$\mathcal{A}^{\ast,p}_k(A_0)$ at a given
$A_1\in\mathcal{A}^{\ast,p}_k(A_0)$.

For $\xi\in L^{\ast,p}_{k+1}\Ad(P))$ and $t\in\R$ with $|t|$ small,
define $g_t:=\Exp_{\mathbf{1}}(t\xi)\in\mathcal{G}^{\ast,p}_{k+1}$.
There exists $g_0\in\mathcal{G}^{\ast,p}_{k+1}$ and $\alpha_0\in
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$ such that
$A_1=g_0^{\ast}(A_0+\alpha_0)$. From
\begin{align*}
g_t^{\ast}A_1&=(\Exp_{g_0}(t\xi))^{\ast}(A_0+\alpha_0)\\
&=
T_{g_0}(t\xi)^{\ast}[A_0+\alpha_0+T_{g_0}(t\xi)(\Exp(t\xi)^{\ast}(A_0+\alpha_0)
-T_{g_0}(t\xi)^{\ast}(A_0+\alpha_0))T_{g_0}(t\xi)^{-1}],
\end{align*}
one has
$$
\Phi_{g_0}^{-1}(g_t^{\ast}A_1)=(t\xi|_{\partial B^4},\alpha_0+
T_{g_0}(t\xi)(\Exp_{g_0}(t\xi)^{\ast}(A_0+\alpha_0)-T_{g_0}(t\xi)^{\ast}(A_0+\alpha_0))T_{g_0}(t\xi)^{-1}).
$$
Thus a tangent vector to the orbit $\{g_t^{\ast}A_1\}_t$ at $A_1$ is
given by
$$\frac{d}{dt}\Big|_{t=0}\Phi_{g_0}^{-1}(g_t^{\ast}A_1)=(\xi|_{\partial B^4},g_0(d_{A_1}\xi-d_{A_1}T_0(\xi))g_0^{-1}).$$
Therefore, a vector $(u,\alpha)\in
L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$ in the slice at $A_1$ must
satisfy
\begin{equation}
\label{5.2}
(\xi|_{\partial B^4},u)_{L^2(\partial
B^4)}+(g_0(d_{A_1}(\xi-T_0(\xi))g_0^{-1},\alpha)_{L^2(B^4)}=0\quad\forall
\xi\in L^{\ast,p}_{k+1}(\Ad(P)).
\end{equation}
By taking $\xi|_{\partial B^4}=0$, one has $T_0(\xi)=0$, and \eqref{5.2}
yields
\begin{equation}
\label{5.3}
(g_0(d_{A_1}\xi)g_0^{-1},\alpha)_{L^2(B^4)}=0,
\end{equation}
thus
\begin{equation}
\label{5.4}
d_{A_0+\alpha_0}^{\ast}\alpha=0\quad\text{in $M$}\;.
\end{equation}
On the other hand, \eqref{5.2}, \eqref{5.4} yield $(\xi|_{\partial
B^4},u)_{L^2(\partial B^4)}=0$ for all $\xi\in
L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))$, thus $u=0$.
Therefore, a suitable candidate for the slice at $A_1$ is
$$
N=\{(0,\alpha)\in L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P)):d_{A_0+\alpha_0}^{\ast}\alpha=0\}.
$$
We must now prove rigorously that $\mathcal{A}^{\ast,p}_k(A_0)$ is
indeed locally diffeomorphic to $N\times\mathcal{G}^{\ast,p}_{k+1}$,
i.e., that  there exists a neighborhood of
$A_1\in\mathcal{A}^{\ast,p}_k(A_0)$, say $\mathcal{O}(A_1)$,  and
$g:\mathcal{O}(A_1)\to\mathcal{G}^{\ast,p}_{k+1}$ such that
$g(A)^{\ast}A\in N$, for all $A\in\mathcal{O}(A_1)$. This way, the
map $\mathcal{A}^{\ast,p}_k(A_0)\ni A\mapsto (g(A)^{\ast},g(A))\in
N\times\mathcal{G}^{\ast,p}_{k+1}$ would provide a local
diffeomorphism. To prove this, we work in local coordinates near
$A_1$, that is, we express the connections
$A\in\mathcal{A}^{\ast,p}_k(A_0)$ near $A_1$ as
$A=T_{g_0}(u)^{\ast}(A_0+\alpha_0+\alpha)$, for some $(u,\alpha)\in
L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$, with $\|u\|_{k+1-1/p,p}$
and $\|\alpha\|_{k,p}$ small, and prove the following lemma.
\begin{lemma}
\label{L5.3}
For $(u,\alpha)\in L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial
B^4}))\times L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$, with
$\|u\|_{k+1-1/p,p}$ and $\|\alpha\|_{k,p}$ suitably small, there
exists $g(u,\alpha)\in\mathcal{G}^{\ast,p}_{k+1}$ such that
$g(u,\alpha)^{\ast}A\in N$. Moreover, $g(u,\alpha)$ depends smoothly
on $u$ and $\alpha$, and $g(0,0)=\mathbf{1}$.
\end{lemma}
\textit{Proof:} Since we look for $g$ near the identity, we may
assume that $g$ is of the form $g=\Exp_{\mathbf{1}}\xi$ for $\xi\in
L^{\ast,p}_{k+1}(\Ad(P))$, with $\|\xi\|_{k+1,p}$ small. There
exists $\epsilon>0$ such that the map

\noindent $f:\{u\in L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial
B^4})):\|u\|_{k+1-1/p,p}<\epsilon\} \times\{\xi\in
L^{\ast,p}_{k+1}(\Ad(P|_{\partial B^4})):\|\xi\|_{k+1,p}<\epsilon\}
\to L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))$,
 defined via
$\Exp_{\mathbf{1}}u\cdot\Exp_{\mathbf{1}}\xi=\Exp_{\mathbf{1}}(f(u,\xi))$,
is smooth.

One has
\begin{align*}
g^{\ast}A&=T_{g_0}(f(u,\xi))^{\ast}(A_0+\alpha_0+\alpha)+(T_{g_0}(u)g)^{\ast}(A_0+\alpha_0+\alpha)-T_{g_0}(f(u,\xi))^{\ast}(A_0+\alpha_0+\alpha)\notag\\
&=T_{g_0}(f(u,\xi))^{\ast}(A_0+\alpha_0+\alpha+\beta),
\end{align*}
where
$\beta=\beta(u,\xi)=T_{g_0}(f(u,\xi))[(T_{g_0}(u)g)^{\ast}(A_0+\alpha_0+\alpha)-
T_{g_0}(f(u,\xi))^{\ast}(A_0+\alpha_0+\alpha)]T_{g_0}(f(u,\xi))^{-1}$\;.
Hence,
$$\Phi_{g_0}^{-1}(g^{\ast}A)=(f(u,\xi),\alpha_0+\alpha+\beta(u,\xi)).$$
Thus, we must find $\xi\in L^{\ast,p}_{k+1}(\Ad(P))$ which satisfies
the following equations for small $\|u\|_{k+1-1/p,p}$ and
$\|\alpha\|_{k,p}$:
\begin{align}
&f(u,\xi)=0\quad\text{on $\partial B^4$,}\label{5.5}\\
&d_{A_0+\alpha_0}^{\ast}(\alpha+\beta(u,\xi))=0\quad\text{in $B^4$.}\label{5.6}
\end{align}
The implicit function theorem yields the existence of a solution for
the system \eqref{5.5}-\eqref{5.6} as follows: define
$$F:
L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial B^4}))\times
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))\times
L^{\ast,p}_{k+1}(\Ad(P))\to L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial
B^4}))\times L^p_{k-1}(\Ad(P))$$
by
$$F(u,\alpha,\xi):=(f(u,\xi)|_{\partial B^4},d_{A_0+\alpha_0}^{\ast}(\alpha+\beta(u,\xi))),$$
where $g=\Exp_{\mathbf{1}}\xi$. For $\eta\in
L^{\ast,p}_{k+1}(\Ad(P))$, a simple calculation yields
\begin{align*}
\Big\langle\frac{\partial
F}{\partial\xi}(0,0,0),\eta\Big\rangle=\frac{d}{dt}\Big|_{t=0}F(0,0,t\eta)
=(\eta|_{\partial
B^4},d_{A_0+\alpha_0}^{\ast}d_{A_0+\alpha_0}(g_0(\eta-T_0(\eta))g_0^{-1})).
\end{align*}
We claim that $\frac{\partial F}{\partial\xi}(0,0,0):
L^{\ast,p}_{k+1}(\Ad(P))\to L^{\ast,p}_{k+1-1/p}(\Ad(P|_{\partial
B^4}))\times L^p_{k-1}(\Ad(P))$ is an isomorphism. To see this, let
$\eta\in\ker\frac{\partial F}{\partial\xi}(0,0,0)$. Then
$\eta|_{\partial B^4}=0$ and, also,
$d_{A_0+\alpha_0}^{\ast}d_{A_0+\alpha_0}(g_0(\eta-T_0(\eta))g_0^{-1})=0$.
These entail $T_0(\eta)=0$ and $\eta=0$. Thus, $\frac{\partial
F}{\partial\xi}(0,0,0)$ is one-to-one. Furthermore, $\frac{\partial
F}{\partial\xi}(0,0,0)$ is onto, since:
\begin{enumerate}[(i)]
\item For all $\varphi\in L^p_{0,k+1}(\Ad(P))$ and $\eta\in L^{\ast,p}_{k+1}(\Ad(P))$, we have
$T_0((\eta+\varphi)|_{\partial B^4})=T_0(\eta|_{\partial B^4})$.
(Thus, $(\varphi+\eta)-T_0((\varphi+\eta)|_{\partial
B^4})=\varphi+\eta-T_0(\eta|_{\partial B^4})$).
\item $\Ad(g_0^{-1}):L^p_{0,k+1}(\Ad(P))\to L^p_{0,k+1}(\Ad(P))$ is an isomorphism.
\item $\Delta_{A_0+\alpha_0}:=d_{A_0+\alpha_0}^{\ast}d_{A_0+\alpha_0}:L^p_{0,k+1}(\Ad(P))\to L^p_{k-1}(\Ad(P))$ is an isomorphism.
\end{enumerate}
Thus  $\frac{\partial F}{\partial\xi}(0,0,0)$ is an isomorphism and,
 by the implicit function theorem, there exists a neighborhood
$\mathcal{U}$ of $(0,\alpha_0)\in U\times
L^p_{0^T,k}(T^{\ast}\otimes\Ad(P))$ such that for all
$(u,\alpha_0+\alpha)\in\mathcal{U}$, there exists
$g(u,\alpha)\in\mathcal{G}^{\ast,p}_{k+1}$ satisfying
$g(u,\alpha)^{\ast}A\in N$, for
$A=T_{g_0}(u)^{\ast}(A_0+\alpha_0+\alpha)$. This completes the
proof. \hfill$\Box$

\medskip

The next lemma can be proved similarly to the corollary at p. 50 of
\cite{FU}, so we omit its proof.

\begin{lemma} \label{L5.4} Assume $p(k+1)>4$. Then $\mathcal{B}^{\ast,p}_k(A_0)$ is Hausdorff.
\end{lemma}

In order to complete the proof of Proposition \ref{P5.2}, we also need the
following:

\begin{lemma}\label{L5.5}
For any given $(0,\alpha_0)\in N$, there exists $\epsilon>0$ such
that the $\epsilon$-ball of $N$ centered at $(0,\alpha_0)$ injects
into $\mathcal{B}^{\ast,p}_k(A_0)$.
\end{lemma}
\textit{Proof:} Let $A_1=g_0^{\ast}(A_0+\alpha_0)$ (where
$g_0\in\mathcal{G}^{\ast,p}_{k+1}$, and $\alpha_0\in
L^p_{0^T,k}(T^{\ast}B^4\otimes\Ad(P))$). Let
$(0,\alpha_0+\alpha),(0,\alpha_0+\beta)\in N$, with
$\|\alpha\|_{k,p}<\epsilon$ and $\|\beta\|_{k,p}<\epsilon$, where
$\epsilon>0$ is a small number. Set
$A'=g_0^{\ast}(A_0+\alpha_0+\alpha)$,
$A''=g_0^{\ast}(A_0+\alpha_0+\beta)$. Assume that there exists
$g\in\mathcal{G}^{\ast,p}_{k+1}$ such that $g^{\ast}A'=A''$. Under
these assumptions, we need to show that $g=\mathbf{1}$ and
$\alpha=\beta$. To prove this, we rewrite the condition
$g^{\ast}A'=A''$ as
\begin{equation}
\label{5.7}
d_{A_1}g=g(g_0^{-1}\alpha g_0)-(g_0^{-1}\beta g_0)g.
\end{equation}
Let us now write $g:=\tilde{g}+c\in L^p_{k+1}(\text{End}(P))$, where
$c\in\ker d_{A_1}$, $\tilde{g}\in(\ker d_{A_1})^{\perp}$ (here,
$(\ker d_{A_1})^{\perp}$ is the $L^2$-orthogonal complement of $\ker
d_{A_1}$ in $L^p_{k+1}(\text{End}(P)).$ By \eqref{5.7}, there exists
a constant $C>0$ such that
\begin{equation}
\label{5.8}
\|\tilde{g}\|_{p,k+1}\le C\epsilon.
\end{equation}
On the other hand, since $g((0,0,0,1))=\mathbf{1}$,
$c\simeq\mathbf{1}$ and, by \eqref{5.8},
$\|g-\mathbf{1}\|_{k+1,p}<C\epsilon$. By Lemma \ref{L5.3}, one has
$g=\mathbf{1}$ and $\alpha=\beta$. This completes the proof.
\hfill$\Box$

\section{Interaction estimates and construction of orthonormal bases for
$T_{A(\q)}\mathcal{N}(d_0,\lambda_0)$ and
$T_{\tilde{A}(\q)}\tilde{N}(d_0,\lambda_0)$}

In \cite{IM1}, for technical reasons, to the purpose of transforming
the non compact $\epsilon-$Dirichlet problem into a finite
dimensional problem, we introduced the extensions $\tilde{P}(\q)$ to
$\R^4$ of the bundles $P(\q)$,
 \begin{equation}
 \label{Ptilde(q)}
 \tilde{P}(\q):=\biggl(\{B_{\lambda/4}(p), \;\R^4\setminus\{p\}\},
 \;\{gg_{12,p}g^{-1}\}\biggr),\quad \mbox{ for }\,\q\in\PP(d_0,\lambda_0),
 \end{equation}
and the spaces of connections
\begin{equation}
\label{tildeN} \mathcal{\tilde
 N}(d_0,\lambda_0):=\{\tilde{A}(\q):\q\in\PP(d_0,\lambda_0)\}\,,
 \end{equation}
where
\begin{equation}
\label{tildeA(q)} \tilde{A}(\q)=\left\{\begin{array}{ll}
 \frac{1}{\epsilon}gI^2_{\lambda,p}g^{-1}&\quad\mbox{in } \R^4\setminus\{p\}\\
 \frac{1}{\epsilon}gI^1_{\lambda,p}g^{-1}&\quad\mbox{in }
 B^4_{\lambda/4}(p)\,.
 \end{array}\right.
 \end{equation}

 In this section, we prove some technical
lemmas on the tangent space to approximate solutions
$T_{A(\q)}\mathcal{N}(d_0,\lambda_0)$ (cf. $\S1$) and on the tangent
space $T_{\tilde{A}(\q)}\tilde{N}(d_0,\lambda_0).$ These lemmas have
been used in sections $\S3.5$-$\S3.8$ of \cite{IM1}, where, by a
standard technique for non-compact variational problems, the Yang
Mills equation is first solved in
$T_{A(\q)}\mathcal{N}(d_0,\lambda_0)^\perp$ (the orthogonal
complement to the tangent space to approximate solutions),  i.e.,
essentially, orthogonally to the kernel of the Hessian of the
$\epsilon$-Yang Mills functional (cf., in particular, Lemma 3.9
in~\cite{IM1}). This result allowed us to turn the problem into a
finite dimensional one (cf. Proposition 3.2 in \cite{IM1}).

\medskip\noindent
In the following, we choose $\bigl(\xi_1, \xi_2, \xi_3\bigr)$, where
$\xi_1=\begin{pmatrix}
0&0&0\\ 0&0&-1\\ 0&1&0\end{pmatrix}$, $\xi_2=\begin{pmatrix} 0&0&1\\
0&0&0\\ -1&0&0\end{pmatrix}$, $\xi_3=\begin{pmatrix} 0&-1&0\\
1&0&0\\ 0&0&0\end{pmatrix}$, as basis for the Lie algebra
 $\mathfrak{so}(3)$. Right translation by $g$ yields a basis for
$T_{[g]}SO(3)$, which we denote by $\bigl(\xi_1[g], \xi_2[g],
\xi_3[g]\bigr)$.

\begin{lemma} \label{L5.7} The following estimates hold for small $\epsilon>0$:
\begin{align}
&\Big\|\frac{\partial A}{\partial
p_i}(\q)\Big\|_{A(\q);1,2;B^4}\simeq\epsilon^{-3/2}\quad(1\le i\le
4),\\
&\Big\|\frac{\partial
A}{\partial\xi_i[g]}(\q)\Big\|_{A(\q);1,2;B^4}\simeq\epsilon^{-1}\quad(1\le
i\le 3),\\
& \Big\|\frac{\partial
A}{\partial\lambda}(\q)\Big\|_{A(\q);1,2;B^4}\simeq\epsilon^{-3/2},\\
&\Big|\Big(\frac{\partial A}{\partial p_i}(\q),\frac{\partial
A}{\partial
p_j}(\q)\Big)_{A(\q);1,2;B^4}\Big|\lesssim\epsilon^{-3/2}\quad(1\le
i\ne j\le 4),\\
&\Big|\Big(\frac{\partial A}{\partial\xi_i[g]}(\q),\frac{\partial
A}{\partial\xi_j[g]}(\q)\Big)_{A(\q);1,2;B^4}\Big|\lesssim\epsilon^{-1}\quad(1\le
i\ne j\le 3),\\
&\Big|\Big(\frac{\partial A}{\partial p_i}(\q),\frac{\partial
A}{\partial\xi_j[g]}(\q)\Big)_{A(\q);1,2;B^4}\Big|\lesssim\epsilon^{-1}\quad(1\le
i\le 4,~1\le j\le 3),\\
&\Big|\Big(\frac{\partial A}{\partial p_i}(\q),\frac{\partial
A}{\partial\lambda}(\q)\Big)_{A(\q);1,2;B^4}\Big|\lesssim\epsilon^{-2}\quad(1\le
i\le 4),\\
&\Big|\Big(\frac{\partial A}{\partial\xi_i[g]}(\q),\frac{\partial
A}{\partial\lambda}(\q)\Big)_{A(\q);1,2;B^4}\Big|\lesssim\epsilon^{-3/2}\quad(1\le
i\le 3).
\end{align}
\end{lemma}
\textit{Proof:} For  $A(\q)$ in the same gauge as described in $\S1$, one has  (modulo
the action of infinitesimal gauge transformations of $P(\q))$:
\begin{align}
\frac{\partial A}{\partial p_i}(\q)&=-\frac{\partial\beta_{\lambda,p}}{\partial
p_i}\underline{A}_{\epsilon}+
\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial
p_i}gI^2_{\lambda,p}g^{-1}+
\frac{1}{\epsilon}\beta_{\lambda/4,p}g\frac{\partial I^2_{\lambda,p}}{\partial p_i}g^{-1}+\qquad\qquad\qquad\notag\\
&\quad-\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial p_i}gPI^2_{\lambda,p}g^{-1}+
\frac{1}{\epsilon}(1-\beta_{\lambda/4,p})g\frac{\partial PI^2_{\lambda,p}}{\partial p_i}g^{-1}\quad\text{in $B^4\setminus\{p\}$},\label{5.19}\\
\frac{\partial A}{\partial p_i}(\q)&=\frac{1}{\epsilon}g\frac{\partial I^1_{\lambda,p}}{\partial
p_i}g^{-1}\quad\text{in
$B_{\lambda/4}(p)$},\\
\frac{\partial A}{\partial\xi_i[g]}(\q)&=\frac{1}{\epsilon}\beta_{\lambda/4,p}[\xi_i,gI^2_{\lambda,p}g^{-1}]+
\frac{1}{\epsilon}(1-\beta_{\lambda/4,p})[\xi_i,gPI^2_{\lambda,p}g^{-1}]\quad\text{in $B^4\setminus\{p\}$},\\
\frac{\partial A}{\partial\xi_i[g]}(\q)&=\frac{1}{\epsilon}[\xi_i,gI^1_{\lambda,p}g^{-1}]\quad\text{in
$B_{\lambda/4}(p)$},\\
\frac{\partial A}{\partial\lambda}(\q)&=-\frac{\partial\beta_{\lambda,p}}{\partial\lambda}\underline{A}_{\epsilon}+
\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial\lambda}gI^2_{\lambda,p}g^{-1}+
\frac{1}{\epsilon}\beta_{\lambda/4,p}g\frac{\partial I^2_{\lambda,p}}{\partial\lambda}g^{-1}+\qquad\qquad\qquad\notag\\
&\quad-\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial\lambda}gPI^2_{\lambda,p}g^{-1}+
\frac{1}{\epsilon}(1-\beta_{\lambda/4,p})g\frac{\partial PI^2_{\lambda,p}}{\partial\lambda}g^{-1}\quad\text{in $B^4\setminus\{p\}$},\\
\frac{\partial A}{\partial\lambda}(\q)&=\frac{1}{\epsilon}g\frac{\partial
I^1_{\lambda,p}}{\partial\lambda}g^{-1}\quad\text{in
$B_{\lambda/4}(p)$}.
\end{align}
The result follows from these formulas by direct calculation.
\hfill$\Box$

\medskip

\noindent From Lemma \ref{L5.7}, it follows that
$T_{A(\q)}\mathcal{N}(d_0,\lambda_0)$ is an $8$-dimensional linear space, if
$\epsilon>0$ is suitably small.

Let us recall that, for a given connection $A$ on the bundle
$P(\q)$, the inner product $(\alpha, \beta)_{A;1,2;B^4}$ for
one-forms $\alpha$, $\beta$ in $L^2_{1; A}(T^{\ast}
B^4\otimes\Ad(P))$ is defined as:
\begin{equation}
\label{innerprod} (\alpha,
\beta)_{A;1,2;B^4}:=\int_{B^4}\bigl({\nabla_A}^{\epsilon}\alpha,
{\nabla_A}^{\epsilon}\beta\bigr)\,dx + \int_{B^4}\bigl(\alpha,
\beta\bigr)\,dx\
\end{equation}
with ${\nabla_A}^{\epsilon}=\nabla+\epsilon[A,\,\cdot\,]$.

\noindent As a corollary of Lemma \ref{L5.7},
one obtains the following orthonormal basis for $T_{A(\q)}\mathcal{N}(d_0,\lambda_0)$,
the tangent space to approximate solutions.
\begin{lemma}
\label{L5.8}
There exists a basis
$\langle\A_1(\q),\A_2(\q),\ldots,\A_8(\q)\rangle$ for
$T_{A(\q)}\mathcal{N}(d_0,\lambda_0)$, orthonormal with respect to the inner
product $(\cdot,\cdot)_{A(\q);1,2;B^4}$, of the following form:
\begin{align}
\A_1(\q)&=a_{11}(\q)\frac{\partial A}{\partial p_1}(\q),\quad \hbox{
with }\;
a_{11}(\q)\simeq\epsilon^{3/2},\label{5.25}\\
\A_2(\q)&=a_{22}(\q)\frac{\partial A}{\partial
p_2}+a_{21}(\q)\A_1(\q),\quad \hbox{ with }\;
a_{22}(\q)\simeq\epsilon^{3/2},
|a_{21}(\q)|\lesssim\epsilon^{3/2},\\
\A_3(\q)&=a_{33}(\q)\frac{\partial A}{\partial
p_3}(\q)+a_{32}(\q)\A_2(\q)+a_{31}(\q)\A_1(\q),\notag\\
&\quad \hbox{ with }\; a_{33}(\q)\simeq\epsilon^{3/2},~|a_{3i}(\q)|\lesssim\epsilon^{3/2}~(i=1,2),\\
\A_4(\q)&=a_{44}(\q)\frac{\partial A}{\partial
p_4}(\q)+a_{43}(\q)\A_3(\q)+a_{42}(\q)\A_2(\q)+a_{41}(\q)\A_1(\q),\notag\\
&\quad \hbox{ with }\; a_{44}(\q)\simeq\epsilon^{3/2},~
|a_{4i}(\q)|\lesssim\epsilon^{3/2}~(1\le i\le 3),\\
 \A_5(\q)&=a_{55}(\q)\frac{\partial
 A}{\partial\xi_1[g]}(\q)+a_{54}(\q)\A_4(\q)+a_{53}(\q)\A_3(\q)+a_{52}(\q)\A_2(\q)+
 a_{51}(\q)\A_1(\q)\,,\notag\\
 &\quad \hbox{ with }\; a_{55}(\q)\simeq\epsilon,~|a_{5i}(\q)|\lesssim\epsilon^{3/2}~(1\le
 i\le 4),\\
 \A_6(\q)&=a_{66}(\q)\frac{\partial
 A}{\partial\xi_2[g]}(\q)+a_{65}(\q)\A_5(\q)+a_{64}(\q)\A_4(\q)+a_{63}(\q)\A_3(\q)\notag\\
 &\quad+a_{62}(\q)\A_2(\q)+
 a_{61}(\q)\A_1(\q),\notag\\
&\quad \hbox{ with
}a_{66}(\q)\simeq\epsilon,~|a_{65}(\q)|\lesssim\epsilon,~|a_{6i}(\q)|\lesssim\epsilon^{3/2}\quad~(1\le
 i\le 4),\\
 \A_7(\q)&= a_{77}(\q)\frac{\partial
 A}{\partial\xi_3[g]}(\q)+a_{76}(\q)\A_6+a_{75}(\q)\A_(\q)+a_{74}(\q)\A_4(\q)\notag\\
&\quad+ a_{73}(\q)\A_3(\q)+a_{72}(\q)\A_2(\q)+a_{71}(\q)\A_1(\q),\notag\\
&\quad \hbox{ with }\;
a_{77}(\q)\simeq\epsilon,~|a_{76}(\q)|,\,|a_{75}(\q)|\lesssim\epsilon,
 ~|a_{7i}(\q)|\lesssim\epsilon^{3/2}~(1\le i\le 4),\\
 \A_8(\q)&=a_{88}(\q)\frac{\partial
 A}{\partial\lambda}(\q)+a_{87}(\q)\A_7(\q)+a_{86}(\q)\A_6(\q)+a_{85}(\q)\A_5(\q)+a_{84}(\q)\A_4(\q)\notag\\
&\quad+a_{83}(\q)\A_3(\q)+a_{82}(\q)\A_2(\q)+a_{81}(\q)\A_1(\q),\notag\\
&\quad \hbox{ with }\;
a_{88}(\q)\simeq\epsilon^{3/2},~|a_{8i}(\q)|\lesssim\epsilon\quad
 ~(1\le i\le 7).
\end{align}
 \end{lemma}
\textit{Proof:} We apply the Gram-Schmidt's orthogonalization
procedure to construct $\A_i(\q)$ ($1\le i\le 8$) from
$\frac{\partial A}{\partial p_i}(\q)$ ($1\le i\le 4$),
$\frac{\partial A}{\partial\xi_i[g]}(\q)$ ($1\le i\le 3$), and
$\frac{\partial A}{\partial\lambda}(\q)$. The asserted result
follows from Lemma \ref{L5.7} by direct calculation. \hfill$\Box$

\medskip

\noindent We construct the following vector fields $\q_i=\q_i(\q)$ ($1\le i\le
8$) inductively on $\PP(d_0,\lambda_0)$, in such a way that the derivative of $A(\q)$ in the direction $\q_i(\q)$
yields $\A_i(\q)$, i.e., $\A_i(\q)=A_{\q_i}(\q)$ for $1\le i\le
8$. More precisely,
\begin{align}
\q_1(\q)&=a_{11}(\q)\frac{\partial}{\partial p_1},\\
\q_2(\q)&=a_{22}(\q)\frac{\partial}{\partial p_2}+a_{21}\q_1(\q),\\
\q_3(\q)&=a_{33}(\q)\frac{\partial}{\partial
p_3}+a_{32}(\q)\q_2(\q)+a_{31}(\q)\q_1(\q),\\
\q_4(\q)&=a_{44}(\q)\frac{\partial}{\partial
p_4}+a_{43}(\q)\q_3(\q)+a_{42}(\q)\q_2(\q)+a_{41}(\q)\q_1(\q),\\
\q_5(\q)&=a_{55}(\q)\xi_1[g]+a_{54}(\q)\q_4(\q)+a_{53}(\q)\q_3(\q)+a_{52}(\q)\q_2(\q)+a_{51}(\q)\q_1(\q),\\
\q_6(\q)&=a_{66}(\q)\xi_2[g]+a_{65}(\q)\q_5(\q)+a_{64}(\q)\q_4(\q)+a_{63}(\q)\q_3(\q)\notag\\
&\quad+a_{62}(\q)\q_2(\q)+a_{61}(\q)\q_1(\q)\;,\\
\q_7(\q)&=
a_{77}(\q)\xi_3[g]+a_{76}(\q)\q_6(\q)+a_{75}(\q)\q_5(\q)+a_{74}(\q)\q_4+a_{73}(\q)\q_3(\q)\notag\\
&\quad+a_{72}(\q)\q_2(\q)+a_{71}(\q)\q_1(\q)\;,\\
\q_8(\q)&=a_{88}(\q)\frac{\partial}{\partial\lambda}+a_{87}(\q)\q_7(\q)+
a_{86}(\q)\q_6(\q)+a_{85}(\q)\q_5(\q)+a_{84}(\q)\q_4(\q)\notag\\
&\quad+a_{83}(\q)\q_3+a_{82}(\q)\q_2(\q)+a_{81}(\q)\q_1(\q).
\end{align}

\noindent Let now $\tilde{\A}_i(\q)=\tilde{A}_{\q_i}(\q)$ (for $1\le
i\le 8$), where $\tilde{A}(\q)$ are the instanton solutions
introduced at the beginning of this section. In the following lemma,
we construct a special basis for
$T_{\tilde{A}(\q)}\tilde{N}(d_0,\lambda_0),$ orthonormal with
respect to the weighted inner product on
$L^2_{1;\tilde{A}(\q)}(T^{\ast}\R^4\otimes\Ad(\tilde{P}))$ defined
by
\begin{equation}
\label{wip}
(\alpha,\beta)_{1,2;\tilde{A}(\q)}:=\int_{\R^4}\bigl({\nabla_{\tilde{A}(\q)}}^{\epsilon}\alpha,{\nabla_{\tilde{A}(\q)}}^{\epsilon}\beta\bigr)\,dx
+ \int_{\R^4} w(x)\bigl(\alpha,\beta\bigr)\,dx,
\end{equation}
where $w(x)=1$ for $|x|\le 1$, and $w(x)=1/(1+|x|^2)^2$ for $|x|>
1$.

\begin{lemma}
\label{L5.9} There exists a basis $\langle\hat{\A}_1(\q),\hat{\A}_2(\q),\ldots,\hat{\A}_8(\q)\rangle$ for
$T_{\tilde{A}(\q)}\tilde{N}(d_0,\lambda_0)$, orthonormal with
respect to the inner product
$(\cdot,\cdot)_{\tilde{A}(\q);1,2;\R^4}$, of the following form:
\begin{align}
\hat{\A}_1(\q)&=b_{11}(\q)\tilde{\A}_1(\q),\quad \hbox{ with }\;
|b_{11}(\q)-1|\lesssim\epsilon,\\
\hat{\A}_2(\q)&=b_{22}(\q)\tilde{\A}_2(\q)+
b_{21}(\q)\hat{\A}_1(\q),\quad \hbox{ with }\;|b_{22}(\q)-1|\lesssim\epsilon,~|b_{21}(\q)|\lesssim\epsilon,\\
\hat{\A}_3(\q)&=b_{33}(\q)\tilde{\A}_3(\q)+b_{32}(\q)\hat{\A}_2(\q)+b_{31}(\q)\hat{\A}_1(\q),\notag\\
&\quad \hbox{ with }\;|b_{33}(\q)-1|\lesssim\epsilon,~|b_{3i}(\q)|\lesssim\epsilon~(i=1,2),\\
\hat{\A}_4(\q)&=b_{44}(\q)\tilde{\A}_4(\q)+b_{43}(\q)\hat{\A}_3(\q)+b_{42}(\q)\hat{\A}_2(\q)+
b_{41}(\q)\hat{\A}_1(\q),\notag\\
&\quad \hbox{ with
}|b_{44}(\q)-1|\lesssim\epsilon,~|b_{4i}(\q)|\lesssim\epsilon~(1\le
i\le 3),\\
\hat{\A}_5(\q)&=b_{55}(\q)\tilde{\A}_5(\q)+b_{54}(\q)\hat{\A}_4(\q)+b_{53}(\q)\hat{\A}_3(\q)+
b_{52}(\q)\hat{\A}_2(\q)+b_{51}(\q)\hat{\A}_1(\q),\notag\\
&\quad \hbox{ with
}|b_{55}(\q)-1|\lesssim\epsilon,~|b_{54}(\q)|\lesssim\epsilon~(1\le
i\le 4),\\
\hat{\A}_6(\q)&=b_{66}(\q)\tilde{\A}_6(\q)+b_{65}(\q)\hat{\A}_5(\q)+b_{64}(\q)\hat{\A}_4(\q)+
b_{63}(\q)\hat{\A}_3(\q)\notag\\
&\quad+b_{62}(\q)\hat{A}_2(\q)+ b_{61}(\q)\hat{\A}(\q),~ \hbox {
with
}\;|b_{66}(\q)-1|\lesssim\epsilon,~|b_{6i}(\q)|\lesssim\epsilon~(1\le
i\le 5),\\
\hat{\A}_7(\q)&=b_{77}(\q)\tilde{\A}_7(\q)+b_{76}(\q)\hat{\A}_6(\q)+b_{75}(\q)\hat{\A}_5(\q)+b_{74}(\q)\hat{\A}_4(\q)+b_{73}(\q)\hat{\A}_3(\q)
\notag\\
&\quad+b_{72}(\q)\hat{\A}_2(\q) +b_{71}(\q)\hat{\A}(\q),~\hbox {
with
}\;|b_{77}(\q)-1|\lesssim\epsilon,~|b_{7i}(\q)|\lesssim\epsilon~(1\le
i\le 6),\\
\hat{\A}_8(\q)&=b_{88}(\q)\tilde{\A}_8(\q)+b_{87}(\q)\hat{\A}_7(\q)+b_{86}(\q)\hat{\A}_6(\q)+b_{85}(\q)\hat{\A}_5(\q)+
b_{84}(\q)\hat{\A}_4(\q)\notag\\
&\quad+b_{83}(\q)\hat{\A}_3(\q)
+b_{82}(\q)\hat{\A}_2(\q)+b_{81}(\q)\hat{\A}_1(\q),~\hbox { with
}\;|b_{88}(\q)-1|\lesssim\epsilon,~|b_{8i}(\q)|\lesssim\epsilon~(1\le
i\le 7).
\end{align}
\end{lemma}
\textit{Proof:} We apply the Gram-Schmidt's orthogonalization
procedure to construct $\hat{\A}_i(\q)$ ($1\le i\le 8$) from
$\tilde{\A}_i(\q)$ ($1\le i\le 8$). In the gauge used for $\tilde
A(\q)$ (cf. beginning of $\S3$), the derivatives of $\tilde{A}(\q)$
are written as (modulo the action of the infinitesimal gauge
transformation of $\tilde{P}(\q)$):
\begin{align}
\frac{\partial\tilde{A}}{\partial p_i}(\q)=
\frac{1}{\epsilon}g\frac{\partial I^2_{\lambda,p}}{\partial
p_i}g^{-1}\quad\text{in $\R^4\setminus\{p\}$},
\quad\frac{\partial\tilde{A}}{\partial
p_i}(\q)=\frac{1}{\epsilon}g\frac{\partial I^1_{\lambda,p}}{\partial
p_i}g^{-1} \quad\text{in $B_{\lambda/4}(p)$},\label{5.49}\\
\frac{\partial\tilde{A}}{\partial\xi_i[g]}(\q)=\frac{1}{\epsilon}[\xi_i,gI^2_{\lambda,p}g^{-1}]
\quad\text{in
$\R^4\setminus\{p\}$},\quad\frac{\partial\tilde{A}}{\partial\xi_i[g]}(\q)=\frac{1}{\epsilon}[\xi_i,gI^1_{\lambda,p}g^{-1}]
\quad\text{in $B_{\lambda/4}(p)$},\label{5.50}\\
\frac{\partial\tilde{A}}{\partial\lambda}(\q)=\frac{1}{\epsilon}\frac{\partial\tilde{A}}{\partial\lambda}(\q)
\quad\text{in
$\R^4\setminus\{p\}$},\quad\frac{\partial\tilde{A}}{\partial\lambda}(\q)=
\frac{1}{\epsilon}\frac{\partial\tilde{A}}{\partial\lambda}(\q)\quad\text{in
$B_{\lambda/4}(p)$}\label{5.51}
\end{align}
From \eqref{5.49}-\eqref{5.51}, by direct computation, one has
\begin{equation}
\label{5.52}
\|\tilde{\A}_i(\q)\|_{\tilde{A}(\q);1,2;\R^4\setminus B^4}\lesssim\epsilon^{3/2}~(1\le i\le 4),\quad
\|\tilde{\A}_i(\q)\|_{\tilde{A}(\q);1,2;\R^4\setminus B^4}\lesssim\epsilon~(5\le i\le 8).
\end{equation}
On the other hand,
\begin{equation}
\label{5.53}
(\tilde{\A}_i(\q),\tilde{\A}_j(\q))_{\tilde{A}(\q);1,2}=\delta_{ij}+O(\epsilon),
\end{equation}
for $1\le i,j\le 8$. In fact,
\begin{align}
\label{5.54}
(\tilde{\A}_i(\q),\tilde{\A}_j(\q))_{\tilde{A}(\q);1,2}&=(\tilde{\A}_i(\q),\tilde{\A}_j(\q))_{\tilde{A}(\q);1,2;B^4}+
(\tilde{\A}_i(\q),\tilde{\A}_j(\q))_{\tilde{A}(\q);1,2;\R^4\setminus B^4}\notag\\
&=(\tilde{\A}_i(\q),\tilde{\A}_j(\q))_{\tilde{A}(\q);1,2;B^4}+O(\epsilon^2)~(\text{by (3.47)})\notag\\
&=(\A_i(\q)+\tilde{\A}_i(\q)-\A_i(\q),\A_j(\q)+\tilde{\A}_j(\q)-\A_j(\q))_{\tilde{A}(\q);1,2;B^4}+O(\epsilon^2)\notag\\
&=(\A_i(\q),\A_j(\q))_{\tilde{A}(\q);1,2;B^4}+O(\epsilon)~(\text{by
Lemma \ref{L3.6} below}).
\end{align}
Here, by setting $b(\q):=\tilde{A}(\q)-A(\q)$, one obtains
\begin{align}
\label{5.55}
(\A_i(\q),\A_j(\q))_{\tilde{A}(\q);1,2;B^4}&=(\A_i(\q),\A_j(\q))_{A(\q);1,2;B^4}+
\int_{B^4}({\nabla_{\tilde{A}(\q)}}^{\epsilon}\A_i(\q),{\nabla_{\tilde{A}(\q)}}^{\epsilon}\A_j(\q))\,dx\notag\\
&\qquad-\int_{B^4}({\nabla_{A(\q)}}^{\epsilon}\A_i(\q),{\nabla_{A(\q)}}^{\epsilon}\A_j(\q))\,dx\notag\\
&=\delta_{ij}+\epsilon\int_{B^4}({\nabla_{A(\q)}}^{\epsilon}\A_i(\q),[b(\q),\A_j(\q)])\,dx+
\epsilon\int_{B^4}([b(\q),\A_i(\q)],{\nabla_{A(\q)}}^{\epsilon}\A_j(\q))\,dx\notag\\
&\qquad+\epsilon^2\int_{B^4}([b(\q),\A_i(\q)],[b(\q),\A_j(\q)])\,dx
=\delta_{ij}+O(\epsilon),
\end{align}
since $\epsilon\|b(\q)\|_{\infty}\lesssim\epsilon$.

Combining \eqref{5.54}, \eqref{5.55}, one obtains \eqref{5.53}, hence the
$\hat{\A}_i(\q)$'s ($1\le i\le 8$) satisfy the assertion of the
lemma. \hfill$\Box$

\medskip
\begin{lemma}
\label{L3.6}
For $\q\in\PP(d_0,\lambda_0;D_1,D_2;\epsilon)$,
\begin{align*}\|\A_i(\q)-\tilde{\A}_i(\q)\|_{A(\q);1,2;B^4}\lesssim\epsilon^{3/2}\quad(1\le
i\le 4),\\
\|\A_i(\q)-\tilde{\A}_i(\q)\|_{A(\q);1,2;B^4}\lesssim\epsilon\quad(5\le
i\le 8).\end{align*}
\end{lemma}

\textit{Proof:} The connections $A(\q)$ and $\tilde{A}(\q)$,
represented in the same gauge as in the previous lemmas,
 differ only on the domain $B^4\setminus
B_{\lambda/4}(p)$. In particular,
$A(\q)-\tilde{A}(\q)=(\beta_{\lambda,p})\underline{A}_{\epsilon}+
\frac{1}{\epsilon}(\beta_{\lambda/4,p}-1)gh_{\lambda,p}g^{-1}$. One
has (modulo infinitesimal gauge transformations):
\begin{align}
\frac{\partial A}{\partial
p_i}(\q)-\frac{\partial\tilde{A}}{\partial p_i}(\q)&=-
\frac{\partial\beta_{\lambda,p}}{\partial
p_i}\underline{A}_{\epsilon}+
\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial
p_i}gh_{\lambda,p}g^{-1}+
\frac{1}{\epsilon}(\beta_{\lambda/4,p}-1)g\frac{\partial
h_{\lambda,p}}{\partial p_i}g^{-1},\\
\frac{\partial
A}{\partial\xi_i[g]}(\q)-\frac{\partial\tilde{A}}{\partial\xi_i[g]}(\q)&=
\frac{1}{\epsilon}(\beta_{\lambda/4,p}-1)[\xi_i,gh_{\lambda,p}g^{-1}],\\
\frac{\partial
A}{\partial\lambda}(\q)-\frac{\partial\tilde{A}}{\partial\lambda}(\q)&=
-\frac{\partial\beta_{\lambda,p}}{\partial\lambda}\underline{A}_{\epsilon}+\frac{1}{\epsilon}\frac{\partial\beta_{\lambda/4,p}}{\partial\lambda}gh_{\lambda,p}g^{-1}+\frac{1}{\epsilon}(\beta_{\lambda/4,p}-1)g\frac{\partial
h_{\lambda,p}}{\partial\lambda}g^{-1}.
\end{align}
The asserted result follows from Lemma \ref{L5.8} by direct computation.
\hfill$\Box$

\medskip
We recall that the Hessian of $\YMe$, denoted by $\nabla^2\YMe(A)$,
is defined by
\begin{equation}
\label{3.58}
\langle\nabla^2\YMe(A)a,b\rangle:=2\int_{B^4}({d_A}^{\epsilon}a,{d_A}^{\epsilon}b)
+2\int_{B^4}(F_A,\epsilon[a,b])\;\hbox{ for all } a, b\in
L^2_{1,0}(T^{\ast}B^4\otimes\Ad(P))\,,
\end{equation}
 where
$\langle\cdot,\cdot\rangle$ denotes the pairing between
$L^2_{1,0}(T^{\ast}B^4\otimes\Ad(P))$ and its dual and ${d_A}^{\epsilon}=d+\epsilon[A,\,\cdot\,]$.
\begin{lemma}
\label{L3.7}
For $\q\in\PP(d_0,\lambda_0;D_1,D_2;\epsilon)$,
\begin{align*}\|\big(\nabla^2\YMe(A(\q))-\nabla^2\YMe(\tilde{A}(\q))\bigr)\A_i(\q)\|_{A(\q);1,2,\ast}&
\lesssim\epsilon^{3/2}\quad(1\le
i\le 8),\\
\|(d_{A(\q)}d_{A(\q)}^{\ast}-d_{\tilde{A}(\q)}d_{\tilde{A}(\q)}^{\ast})\A_i(\q)\|_{A(\q);1,2,\ast}&
\lesssim\epsilon^{3/2}\quad(1\le
i\le 8).
\end{align*}
\end{lemma}

\textit{Proof:} We perform the calculation for $i=1$ (the
remaining cases are analogous).  For $b(\q):=\tilde{A}(\q)-A(\q)$,
$\alpha\in L^2_1(T^{\ast}B^4\otimes\Ad(P(\q)))$, $\beta\in
L^2_{0,1}(T^{\ast}B^4\otimes\Ad(P(\q)))$, one has
\begin{align}
\label{5.59}
&\frac{1}{2}(\nabla^2\YMe(\tilde{A}(\q))-\nabla^2\YMe(A(\q))(\alpha,\beta)\notag\\
&=\epsilon\int_{B^4}({d_{A(\q)}}^{\epsilon}\alpha,[b(\q),\beta])\,dx+\epsilon\int_{B^4}([b(\q),\alpha],{d_{A(\q)}}^{\epsilon}\beta)\,dx\notag\\
&\quad+\epsilon\int_{B^4}({d_{A(\q)}}^{\epsilon}b(\q),[\alpha,\beta])\,dx+\epsilon^2\int_{B^4}([b(\q),\alpha],[b(\q),\beta])\,dx\notag\\
&\quad+\frac{\epsilon^2}{2}\int_{B^4}([b(\q),b(\q)],[\alpha,\beta])\,dx.
\end{align}
We set $\alpha=\A_1(\q)$ in \eqref{5.59}, and estimate each term. In the
following, we write $r:=|x-p|$. Since
\begin{equation}
\label{5.60}
|\A_1(\q)|=a_{11}(\q)\frac{\partial A}{\partial p_i}(\q)
\lesssim\epsilon|\nabla\beta(\lambda^{-1}(\cdot-p))|+\frac{\lambda^3}{r^2(\lambda^2+r^2)}+|\nabla\beta(\lambda^{-1}(\cdot-p))|\lambda^2
+\lambda^3
\end{equation}
and
\begin{equation}
\label{5.61}
|\epsilon
{d_{A(\q)}^{\ast}}^{\epsilon}[b(\q),\beta]|\lesssim(\epsilon\lambda^{-1}|\nabla(\lambda^{-1}(\cdot-p))|+\lambda^2+\epsilon|I^2_{\lambda,p}|)|\beta|+
\epsilon|\nabla\beta|,
\end{equation}
one obtains
\begin{align}
\label{5.62}
&\biggl|\epsilon\int_{B^4}({d_{A(\q)}}^{\epsilon}\alpha,[b(\q),\beta]) \biggr|=\biggl|\epsilon\int_{B^4}(\alpha,{d_{A(\q)}^{\ast}}^{\epsilon}[b(\q),\beta]) \biggr|\notag\\
&\lesssim\int_{B_{2\lambda}(p)\setminus
B_{\lambda}(p)}\Big(\epsilon^2\lambda^{-1}+\epsilon\lambda+\lambda^4+
\frac{\epsilon^2\lambda^2}{r(\lambda^2+r^2)}+\frac{\epsilon\lambda^2}{r^2(\lambda^2+r^2)}+
\frac{\epsilon\lambda^4}{r(\lambda^2+r^2)}\Big)|\beta|\,dx\notag\\
&\quad+\int_{B_{2\lambda}(p)\setminus B_{\lambda}(p)}(\epsilon^2+\epsilon\lambda^2)|\nabla\beta|\,dx\notag\\
&\quad+\int_{B^4\setminus B_{\lambda/4}(p)}\Big(\lambda^5+\frac{\lambda^5}{r^2(\lambda^2+r^2)}+\frac{\epsilon\lambda^5}{r^3(\lambda^2+r^2)^2}+
\frac{\epsilon\lambda^5}{r(\lambda^2+r^2)}\Big)|\beta|\,dx\notag\\
&\quad+\int_{B^4\setminus B_{\lambda/4}(p)}\Big(\epsilon\lambda^3+\frac{\epsilon\lambda^3}{r^2(\lambda^2+r^2)}\Big)|\nabla\beta|\,dx\notag\\
&\lesssim (\epsilon^2\lambda^2+\epsilon\lambda^4+\lambda^7+\epsilon^2\lambda^2+\epsilon\lambda+\epsilon\lambda^4)
\|\beta\|_{L^4(B_{2\lambda}(p)\setminus B_{\lambda}(p))}+(\epsilon^2\lambda^2+\epsilon\lambda^4)\|\nabla\beta\|_{L^2(B_{2\lambda}(p)\setminus B_{\lambda}(p))}\notag\\
&\quad+(\lambda^5+\lambda^4+\epsilon\lambda+\epsilon\lambda^5|\log\lambda|)\|\beta\|_{L^4(B^4\setminus B_{\lambda/4}(p))}+(\epsilon\lambda^3+\epsilon\lambda^2)\|\nabla\beta\|_{L^2(B^4\setminus B_{\lambda/4}(p))}\notag\\
&\lesssim\epsilon^{3/2}\|\beta\|_{A(\q);1,2}.\qquad\qquad\qquad
\end{align}
From \eqref{5.62}, the first part of Lemma \ref{L3.7} (for $i=1$) follows.

To prove the second part, for any given $\beta\in
L^2_{0,1}(T^{\ast}B^4\otimes\Ad(P(\q)))$ we estimate
\begin{align}
\label{5.63}
&\langle({d_{A(\q)}}^{\epsilon}{d_{A(\q)}^{\ast}}^{\epsilon}-{d_{\tilde{A}(\q)}}^{\epsilon}{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon})\A_i(\q),\beta\rangle
=\int_{B^4}({d_{A(\q)}^{\ast}}^{\epsilon}\A_i(\q),{d_{A(\q)}^{\ast}}^{\epsilon}\beta)\,dx-
\int_{B^4}({d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_i(\q),{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\beta)\,dx\notag\\
&=\int_{B^4}(({d_{A(\q)}^{\ast}}^{\epsilon}-{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon})\A_i(\q),{d_{A(\q)}^{\ast}}^{\epsilon}\beta)\,dx+
\int_{B^4}({d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_i(\q),({d_{A(\q)}^{\ast}}^{\epsilon}-{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon})\beta)\,dx\notag\\
&=\int_{B^4\setminus B_{\lambda/4}(p)}(\epsilon\ast[b(\q),\ast\A_i(\q)],{d_{A(\q)}^{\ast}}^{\epsilon}\beta)\,dx+
\int_{B^4\setminus B_{\lambda/4}(p)}({d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_i(\q),\epsilon\ast[b(\q),\ast\beta])\,dx\notag\\
&\lesssim\epsilon\biggl(\int_{B^4\setminus
B_{\lambda/4}(p)}|\A_i(\q)|^2\,dx\biggr)^{1/2}\|\beta\|_{A(\q);1,2}+\epsilon\biggl(\int_{B^4\setminus
B_{\lambda/4}(p)}|{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_i(\q)|^{4/3}\,dx\biggr)^{3/4}\|\beta\|_{A(\q);1,2}.
\end{align}
By \eqref{5.60}, the first integral in \eqref{5.63} is estimated as
\begin{align}
\label{5.64}
\int_{B^4\setminus B_{\lambda/4}(p)}|\A_1(\q)|^2\,dx&\lesssim\epsilon^2\lambda^4+\lambda^2+\lambda^8+\lambda^6\lesssim\epsilon\,.
\end{align}
As for the second integral, since
$|{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_1|\le|d^{\ast}\A_1|+\epsilon|[\tilde{A}(\q),\ast\A_1(\q)]|$,
from \eqref{5.19}, \eqref{5.25}, \eqref{5.49}, \eqref{5.60}, it follows that
\begin{align}
\label{5.65}
|{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_1(\q)|&\lesssim\epsilon\lambda^{-1}|\nabla^2\beta(\lambda(\cdot-p))|+
\epsilon|\nabla\beta(\lambda(\cdot-p))|+
\frac{\lambda^3}{r^3(\lambda^3+r^2)}\notag\\
&\quad+\lambda|\nabla^2\beta(\lambda/4(\cdot-p))|+\lambda^2|\nabla\beta(\lambda(\cdot-p))|+\lambda^3\notag\\
&\quad+\frac{\lambda^2}{r(\lambda^2+r^2)}\Big(\epsilon|\nabla\beta(\lambda(\cdot-p))|+
\frac{\lambda^3}{r^2(\lambda^2+r^2)}+
\lambda^2|\nabla\beta(\lambda/4(\cdot-p))|+\lambda^3\Big).
\end{align}
Thus,
\begin{align}
\label{5.66}
&\int_{B^4\setminus B_{\lambda/4}(p)}|{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon}\A_1(\q)|^{4/3}\,dx\lesssim\epsilon^{4/3}\lambda^{8/3}+
\epsilon^{4/3}\lambda^4+\int_{\lambda/4}^2\frac{\lambda^4}{r(\lambda^2+r^2)^{4/3}}\,dr\notag\\
&\quad+\lambda^{16/3}+\lambda^{20/3}+\lambda^4+
\epsilon^{4/3}\lambda^{8/3}\int_{\lambda}^{2\lambda}\frac{r^{5/3}}{(\lambda^2+r^2)^{4/3}}\,dr
+
\lambda^{20/3}\int_{\lambda/4}^2\frac{1}{r(\lambda^2+r^2)^{8/3}}\,dr\notag\\
&\quad+\lambda^{16/3}\int_{\lambda}^{2\lambda}\frac{r^{5/3}}{(\lambda^2+r^2)^{4/3}}\,dr+
\lambda^{20/3}\int_{\lambda/4}^2\frac{r^{5/3}}{(\lambda^2+r^2)^{4/3}}\,dr\notag\\
&\lesssim\,\epsilon^{4/3}\lambda^{8/3}+\epsilon^{4/3}\lambda^4+\lambda^{4/3}+\lambda^{16/3}+\lambda^{20/3}+
\lambda^4
+\epsilon^{4/3}\lambda^{8/3}+\lambda^{4/3}+\lambda^{16/3}+\lambda^{20/3}|\log\lambda|
\lesssim \,\epsilon^{2/3}.
\end{align}
Combining \eqref{5.63}, \eqref{5.64}, \eqref{5.66}, one obtains
\begin{equation}
\label{5.67}
|\langle({d_{A(\q)}}^{\epsilon}{d_{A(\q)}^{\ast}}^{\epsilon}-{d_{\tilde{A}(\q)}}^{\epsilon}{d_{\tilde{A}(\q)}^{\ast}}^{\epsilon})\A_i(\q),\beta\rangle|
\lesssim\epsilon^{3/2}\|\beta\|_{A(\q);1,2},
\end{equation}
i.e., the second assertion of Lemma \ref{L3.7} for $i=1$. \hfill$\Box$

\medskip
\begin{lemma} \label{L3.10} Let $\A_i(\q)$ be the elements of the orthonormal basis constructed in Lemma \ref{L5.8}.  The following estimates hold:
\begin{equation}
\label{140}
\|{\A_i}_{\q_j}(\q)^{\perp}\|_{A(\q);1,2;B^4}\lesssim\epsilon
\end{equation}
for $1\le i,j\le 8$, where ${\A_i}_{\q_j}(\q)$ denotes the
directional derivative of $\A_i(\q)$ in the direction $\q_j$.
\end{lemma}
\textit{Proof:} We prove the lemma for $i=1$, $j=1$.
The remaining cases are similar. For $A(\q)$ as represented  in $\S1$, we first observe that
$$({\A_1}_{\q_1}(\q))^{\perp}=
\Big(a_{11}(\q)\frac{\partial}{\partial p_1}\Big(a_{11}(\q)\frac{\partial A}{\partial p_1}(\q)\Big)\Big)^{\perp}
=a_{11}(\q)^2\Big(\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big)^{\perp}\,$$
so, we need to estimate
$\Big\|a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big\|_{A(\q);1,2}$.

On $B^4\setminus B_{\lambda/4}(p)$,
\begin{align}
\label{5.68}
\epsilon\frac{\partial^2 A}{\partial p_1^2}(\q)&=
\,-\epsilon\lambda^{-2}\frac{\partial^2\beta}{\partial x_1^2}(\lambda^{-1}(\cdot-p))\underline{A}_{\epsilon}+
g\frac{\partial^2 I^2_{\lambda,p}}{\partial p_1^2}g^{-1}+
16\lambda^{-2}\frac{\partial^2\beta}{\partial x_1^2}(4\lambda^{-1}(\cdot-p))gh_{\lambda,p}g^{-1}\notag\\
&\quad-4\lambda^{-1}\frac{\partial\beta}{\partial
x_1}(4\lambda^{-1}(\cdot-p))g\frac{\partial h_{\lambda,p}}{\partial
p_1}g^{-1}+(\beta_{\lambda/4,p}-1)g\frac{\partial^2
h_{\lambda,p}}{\partial p_1^2}g^{-1}.
\end{align}
Hence,
\begin{align}
\label{5.69}
\Big|\epsilon\frac{\partial^2 A}{\partial p_1^2}(\q)&\Big|\lesssim\epsilon\lambda^{-2}|\nabla^2\beta(\lambda^{-1}(\cdot-p))|+
\frac{\lambda^2}{r^3(\lambda^2+r^2)}+
|\nabla^2\beta(\lambda^{-1}(\cdot-p))|\notag\\
&\quad+\lambda|\nabla\beta(\lambda^{-1}(\cdot-p))|+\lambda^2,
\end{align}
\begin{align}
\label{5.70}
\Big|\epsilon\nabla\Big(\frac{\partial^2 A}{\partial p_1^2}(\q)\Big)&\Big|\lesssim\epsilon\lambda^{-3}|\nabla^3\beta(\lambda^{-1}(\cdot-p))|+
\epsilon\lambda^{-2}|\nabla^2\beta(\lambda^{-1}(\cdot-p))|+\frac{\lambda^2}{r^4(\lambda^2+r^2)}\notag\\
&\quad+\lambda^{-1}|\nabla^3\beta(\lambda^{-1}(\cdot-p))|+|\nabla^2\beta(\lambda^{-1}(\cdot-p))|+
\lambda|\nabla\beta(\lambda^{-1}(\cdot-p))|+
\lambda^2,
\end{align}
and
\begin{align}
\label{5.71}
\Big|\epsilon\Big[A(\q),\epsilon\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big]&\Big|\lesssim
\epsilon^2\lambda^{-2}|\nabla^2\beta(\lambda^{-1}(\cdot-p))|+\frac{\epsilon\lambda^2}{r^3(\lambda^2+r^2)}+
\epsilon|\nabla^2\beta(4\lambda^{-1}(\cdot-p))|\notag\\
&\quad+\epsilon\lambda|\nabla\beta(\lambda^{-1}(\cdot-p))|+\epsilon\lambda^2+
\frac{\epsilon}{r(\lambda^2+r^2)}|\nabla^2\beta(\lambda^{-1}(\cdot-p))|\notag\\
&\quad+\frac{\lambda^4}{r^4(\lambda^2+r^2)^2}+
\frac{\lambda^2}{r(\lambda^2+r^2)}|\nabla^2\beta(4\lambda^{-1}(\cdot-p))|\notag\\
&\quad+\frac{\lambda^3}{r(\lambda^2+r^2)}|\nabla\beta(4\lambda^{-1}(\cdot-p))|+\frac{\lambda^4}{r(\lambda^2+r^2)}.
\end{align}
From \eqref{5.69}--\eqref{5.71}, using $|a_{11}(\q)|\lesssim\epsilon^{3/2}$, one
 obtains
\begin{align}
\label{5.72}
a_{11}(\q)^4\int_{B^4\setminus B_{\lambda/4}(p)}\Big|\frac{\partial^2 A}{\partial p_1^2}(\q)\Big|^2\,dx&\lesssim\epsilon^6+\epsilon^4\int_{B^4\setminus B_{\lambda/4}(p)}\frac{\lambda^4}{r^6(\lambda^2+r^2)^2}\,dx+\epsilon^4\lambda^4\notag\\
&\quad+\epsilon^4\lambda^6+\epsilon^4\lambda^4\lesssim\epsilon^3,
\end{align}
\begin{align}
\label{5.73}
a_{11}(\q)^4\int_{B^4\setminus B_{\lambda/4}(p)}\Big|\nabla\Big(\frac{\partial^2 A}{\partial p_1^2}\Big)\Big|^2\,dx&\lesssim\epsilon^6\lambda^{-2}+\epsilon^6+\epsilon^4\int_{B^4\setminus B_{\lambda/4}(p)}\frac{\lambda^4}{r^8(\lambda^2+r^2)^2}\,dx+\epsilon^6\lambda^2\notag\\
&\quad+\epsilon^4\lambda^4\lesssim\epsilon^2,
\end{align}
and
\begin{align}
\label{5.74}
&\int_{B^4\setminus B_{\lambda/4}(p)}\Big|\epsilon\Big[A(\q),a_{11}(\q)^2\frac{\partial^2 A}{\partial p_1^2}(\q)\Big]\Big|^2\,dx\notag\\
&\lesssim\epsilon^8+\epsilon^6\int_{B^4\setminus B_{\lambda/4}(p)}\frac{\lambda^4}{r^6(\lambda^2+r^2)^2}\,dx+\epsilon^6\lambda^4+
\epsilon^6\lambda^6+\epsilon^6\lambda^4+\epsilon^6\int_{B_{2\lambda}(p)\setminus B_{\lambda}(p)}\frac{1}{r^2(\lambda^2+r^2)^2}\,dx\notag\\
&\quad+\epsilon^4\int_{B^4\setminus B_{\lambda/4}(p)}\frac{\lambda^8}{r^8(\lambda^2+r^2)^4}\,dx+\epsilon^4\int_{B_{\lambda/2}(p)\setminus B_{\lambda/4}(p)}\frac{\lambda^4}{r^2(\lambda^2+r^2)^2}\,dx\notag\\
&\quad+\epsilon^4\int_{B_{\lambda/2}(p)\setminus
B_{\lambda/4}(p)}\frac{\lambda^6}{r^2(\lambda^2+r^2)^2}\,dx+\epsilon^4\int_{B^4\setminus
B_{\lambda/4}(p)}\frac{\lambda^8}{r^2(\lambda^2+r^2)^2}\,dx\lesssim\epsilon^2.
\end{align}
Combining \eqref{5.72}--\eqref{5.74} yields
\begin{equation}
\label{5.75}
\Big\|a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big\|_{A(\q);1,2;B^4\setminus
B_{\lambda/4}(p)}\lesssim\epsilon\;.
\end{equation}
Since $\epsilon A(\q)=gI^1_{\lambda,p}g^{-1}$ and
$\epsilon\frac{\partial^2 A}{\partial p_1^2}=g\frac{\partial^2
I^1_{\lambda,p}}{\partial p_1^2}g^{-1}$ on $B_{\lambda/4}(p)$, one
has
\begin{align}
\Big|a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big|\lesssim\frac{\epsilon^2}{(\lambda^2+r^2)^{3/2}}\;,\notag\\
\Big|\epsilon\Big[A(\q),a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big]\Big|\lesssim\frac{\epsilon^2}{(\lambda^2+r^2)^2}\;,\notag\\
\Big|\nabla\Big(a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big)\Big|\lesssim\frac{\epsilon^2}{(\lambda^2+r^2)^2}\;.\notag\\
\notag
\end{align}
Hence,
\begin{align}
\int_{B_{\lambda/4}(p)}\Big|a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big|^2\,dx\lesssim\epsilon^4\int_{B_{\lambda/4}(p)}
\frac{1}{(\lambda^2+r^2)^3}\,dx\lesssim\epsilon^3,\label{5.76}\\
\int_{B_{\lambda/4}(p)}\Big|\nabla\Big(a_{11}(\q)^2\frac{\partial^2
A}{\partial
p_1^2}(\q)\Big)\Big|^2\,dx\lesssim\epsilon^4\int_{B_{\lambda/4}(p)}\frac{1}{(\lambda^2+r^2)^4}\,dx
\lesssim\epsilon^2,\label{5.77}\\
\int_{B_{\lambda/4}(p)}\Big|\epsilon\Big[A(\q),a_{11}(\q)^2\frac{\partial^2
A}{\partial
p_1^2}(\q)\Big]\Big|^2\,dx\lesssim\epsilon^2
\int_{B_{\lambda/4}(p)}\frac{1}{(\lambda^2+r^2)^4}\,dx\lesssim\epsilon^2.\label{5.78}
\end{align} From \eqref{5.76}-\eqref{5.78}, one obtains
\begin{equation}
\label{5.79}
\Big\|a_{11}(\q)^2\frac{\partial^2 A}{\partial
p_1^2}(\q)\Big\|_{A(\q);1,2;B_{\lambda/4}(p)}\lesssim\epsilon,
\end{equation}
and, finally, from \eqref{5.75} and \eqref{5.79}, the assertion of Lemma \ref{L3.10}
follows, for the case $i=1$, $j=1$. The remaining cases can be
proved similarly.\hfill$\Box$


TITLE: Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$
QUESTION [2 upvotes]: Prove that  $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ in ring $ {\displaystyle \mathbb {R}}$$[x]$ where $m, n, p \in  {\displaystyle \mathbb {N}}$
I have tried to use factor theroem to write down  $x^{3m}+x^{3n+1}+x^{3p+2}$ as a product of $x^2+x+1$ and something. But I cannot find the second factor.
Thanks in Advance.

REPLY [3 votes]: Note:
$$\frac{x^{3m}+x^{3n+1}+x^{3p+2}}{x^2+x+1}=\frac{x^{3m}-1+x^{3n+1}-x+x^{3p+2}-x^2+(x^2+x+1)}{x^2+x+1}=\\
\frac{(x^{3m}-1)+x(x^{3n}-1)+x^2(x^{3p}-1)+(x^2+x+1)}{x^2+x+1}.$$
And:
$$x^{3k}-1 \equiv 0 \pmod{x^2+x+1}, k\in \mathbb N.$$


TITLE: Process of finding the canonical form of conic section
QUESTION [0 upvotes]: I have a problem with understanding the process of transforming a conic section into canonical form.
My conic is:
$11x^2-24xy+4y^2+2x+16y-11=0$
Can someone explain me the complete process?

REPLY [0 votes]: In maxima, first find the center
solve([diff(11*x^2-24*x*y+4*y^2+2*x+16*y-11,x)=0,diff(11*x^2-24*x*y+4*y^2+2*x+16*y-11,y)=0],[x,y]);
(%o1) [[x = 1, y = 1]]

Then transform the equation into one centered on the origin:
expand(11*(x+1)^2-24*(x+1)*(y+1)+4*(y+1)^2+2*(x+1)+16*(y+1)-11);
(%02) 4*y^2-24*x*y+11*x^2-2

Resulting in $11(x-1)^2-24(x-1)(y-1)+4(y-1)^2=2$ or the translated $$11x'^2-24x'y'+4y'^2=2.$$
Now we rotate:
$M=\begin{pmatrix}11&-12\\-12&4\end{pmatrix}$
eigenvectors(M);
                                             4           3
(%o4)            [[[- 5, 20], [1, 1]], [[[1, -]], [[1, - -]]]]
                                             3           4

So, normalising the eigenvectors, $P=\begin{pmatrix}3/5&4/5\\4/5&-3/5\end{pmatrix}$ should work:
transpose(P).M.P;
                                  [ - 5  0  ]
(%o6)                             [         ]
                                  [  0   20 ]

So $-5u^2+20v^2=2$ or $$-\frac52 u^2+10 v^2=1.$$


TITLE: Proof of equidistribution theorem for exponential coefficients
QUESTION [0 upvotes]: Can anyone provide a proof of the equidistribution theorem using Weyl's criterion for the case of $c*a \,\,\, \text{(mod 1)}$ where $c=2^n: \,\,\, \forall n \in N_0$ for irrational algebraic $a$?  The original equidistribution theorem is for the case that $c = n: \,\,\, \forall n \in N_0$.  Much thanks in advance.

REPLY [3 votes]: You are asking whether a number $a$ is a normal number in base 2 or not. This is an extremely complicated problem, and very far from being solved. Generally speaking, there exist several constructions of normal numbers, but we don't have any tool to determine whether a given number is normal or not. It is (sometimes) conjectured that all algebraic irrationals are normal, but a proof of such a statement is very, very, very, very, very far away.
You may be interested in the paper
Bailey, D.; Crandall, R.E.: On the random character of fundamental constant expansions. Experiment. Math. 10 (2001), no. 2, 175–190.
where this topic is discussed.
(By the way, the topic has a connection with ergodic theory. You are looking at the orbit of $a$ under the map $T:~x \mapsto 2x \mod 1$.)


TITLE: find the 2002th term of a binary sequence
QUESTION [5 upvotes]: Source: 2002 UofT Math Competition, problem 9.

A sequence whose entries are 0 and 1 has the property that, if each 0 is replaced by 01 and each 1 by 001, the sequence remains unchanged. What is the 2002th term of the sequence?

Let $S_1 = 0$ and for $k\ge 2$ let $S_k$ be obtained from $S_{k-1}$ by replacing every 0 in $S_{k-1}$ with $01$ and every $1$ with $001$. The crucial claim is that $S_k = S_{k-1}S_{k-2}S_{k-1}$ for all $k\ge 3$.

I'm not sure if this can be shown using some form of induction.

The claim clearly justifies that the given sequence is well-defined, since $S_{k-1}$ is a prefix of $S_k$.
From the claim, the answer is easy: The length of the $S_k$'s for $1\leq k\leq 10$ are $1,2,5,12,29,70,169,408,985,2378.$ Let $S_{k,i}$ denote the ith symbol of $S_k$ where $i$ is at most the length of $S_k$. Then $S_{10,2002} = S_{9, 2002-985-408} = S_{9,609} = S_{8,609-408-169} = S_{8,32} = S_{7,32}=S_{6,32} = S_{4,32-29} = S_{3,3} = 0.$

REPLY [2 votes]: As Greg Martin indicated, you can use induction to prove the crucial claim. When I have encountered this kind of claim and its proof for the first time, I was like, uh, so easy and obvious?!

Let $\phi$ map strings to strings by replacing each $0$ with $01$ and each $1$ by $001$.
It is straightforward to see that $\phi$ preserves string concatenation, i.e., $\phi(st)=\phi(s)\phi(t)$, where $s$ and $t$ are two arbitrary strings, $st$ is the concatenation of $s$ and $t$.
You have constructed $S_k$ by $S_{k+1}=\phi(S_k)$.
Suppose $S_k = S_{k-1}S_{k-2}S_{k-1}$. Then
$$S_{k+1} = \phi(S_{k-1}S_{k-2}S_{k-1})=\phi(S_{k-1})\phi(S_{k-2})\phi(S_{k-1})=S_{k}S_{k-1}S_{k}$$
The base case is $S_3=01001=(01)0(01)=S_2S_1S_2$.


TITLE: Central Motion with a Logarithmic potential
QUESTION [0 upvotes]: A particle is moving under the influence of central force given by the potential:
$$V(r)=\alpha \log(r)$$
I have to calculate the radius of corresponding orbit in terms of L(angular momentum) and whether it is stable.
Assuming that alpha is positive and I differentiate the potential to get an extremum, it turns out there is none of real $L$. What am i doing wrong ?

REPLY [3 votes]: If you minimize the effective potential then you can force a circular orbit.
The effective potential only equals the potential when the angular momentum is zero.
What you have just shown is that there are no circular orbits when the angular momentum is zero.
You would get the same result if you used a $1/r$ central force potential.


TITLE: Is a closed map from X to T1 proper?
QUESTION [2 upvotes]: $f:X→Y$ is a closed map.
Suppose that the inverse image of each point in $Y$ is a compact subset of $X.$
Show that $f^{-1}(K)$ is compact in $X$ whenever $K$ is compact in $Y.$
This is my homework and I understand $Y$ is a $T_1$ space. But what should I do next?

REPLY [1 votes]: A map $f:X\to Y$ which is continuous and closed and has the property that all fibers (the preimages of points) are compact is called perfect. A map is called proper if the preimages of compact sets are compact. So you want to show that each perfect map is proper.
Note that for every $B\subset Y$, the restriction $f':f^{-1}(B)\to B$ is again proper. In particular, if $K\subseteq Y$ is compact, then $f':f^{-1}(K)\to K$ is a perfect map with a compact target space. For that case, you can find a proof at Perfect Map $p:\ X\to Y$, $Y$ compact implies $X$ compact which also shows that the continuity is not needed.


TITLE: How does static friction increase with increase in the applied force?
QUESTION [1 upvotes]: My book says that 'Between surface of contact between two bodies there are contact forces applied between each other (they are equal in magnitude but opposite in direction,obeying Newton's 3rd law) due to electromagnetic interaction. The  perpendicular component of the contact force is the Normal force while the parallel component  is the frictional force. Friction is caused due to the coldwelding ie. formation of bond between the micro contact areas. When they distort due to external force,they create vibration waves which become damped releasing heat.' But it didn't tell about the mechanism of static friction, how it happens in the microscopic level. As the external force on the body increases, static frictional force also increases its value (upto a level). How does the force of static friction increase ? What is the cause of it at the microscopic level? Does the bond become more strong? How does this increase in static friction happen?

REPLY [4 votes]: I am not an expert in such fields, but I'll give you an overview of how I've learnt it.
The main point to realize is that, on a microscopic scale, the surfaces we initially thought of as "smooth" contain actually a great many irregular protuberances.
Coming back to the surface area between the two objects, one must carefully distinguish between the microscopic area of contact and the macroscopic upon which the friction force is independent, meaning they can be lying on top of each other with their larger cross sections or their smaller parts, it will not matter. Of course this seems surprising at first because friction results from adhesion, so one might expect the friction force to be greater when objects slide on their larger sides, because the contact area is larger. However, what determines the amount of adhesion is not the macroscopic contact area, but the microscopic contact area, and the latter is pretty much independent of whether the objects lie on a large face or on a small face.
Key idea is that the normal force puts pressure on the protuberances of one surface against those of the other which causes the protuberances/junctions to undergo a certain flattening (elastic deformation e.g.), and this increases the effective area of contact between the "rough" parts (before, you can imagine that only the tip points where actually bonding), as illustrated in these two pictures:

Second picture: larger effective area of contact or in other words higher number of contact points between the protuberances, also as pointed out by Jim.
To conclude, we now can tell that for large macroscopic contact areas, the number of protuberances in contact is larger but since the normal force is distributed over all of them, their deformations are less important (smaller effective microscopic area), whereas the opposite will hold for smaller macroscopic surfaces, where the deformations are very strong and maximize the contact between the junctions, but their numbers is comparatively lower. All of which explains why macroscopic areas don't matter.
As for larger normal forces, it will increase the deformation of junctions and make the coupling between the surfaces stronger.


TITLE: $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
QUESTION [7 upvotes]: I'm trying to show that $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian, this is another exercise in Hatcher.
I'm not sure but I thought I can use the exact sequence
$$ \cdots 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} H_0(\mathbb{R}) \xrightarrow{i} \cdots$$
and then
$$ 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} \operatorname{ker} (i) \xrightarrow{i}0$$
My idea is to use that an exact sequence $0 \rightarrow A \rightarrow B \rightarrow C  \rightarrow$ splits if $C$ is free. So I thought because $\operatorname{ker}(i)$ is a subgroup of $\mathbb{Z}$, it's free. If this is correct, all I need to finish is $H_0(\mathbb{Q})$. I think it should be a countable infinite direct sum of $\mathbb{Z}$'s but I don't know how to express that. How about $\mathbb{Z}^\mathbb{N}$?
Then I'd have $H_0(\mathbb{Q}) = H_1(\mathbb{Q}, \mathbb{R}) \oplus \operatorname{ker}(i)$ and therefore $H_1(\mathbb{Q}, \mathbb{R}) \cong H_0(\mathbb{Q})  / \operatorname{ker}(i) \cong H_0(\mathbb{Q})$.
Now I'm supposed to find a basis for that.
Can someone tell me if my calculations are correct and if yes give me a hint on how to find a basis for $\mathbb{Z}^\mathbb{N}$? Many thanks for your help!
Edit
I just found out we have the following exact sequence:
$$0 \rightarrow H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{f} H_0(\mathbb{Q}) \xrightarrow{g} H_0(\mathbb{R}) \xrightarrow{h} 0$$
Using the previous exercise, 16 (a), I know that $H_0(\mathbb{R}, \mathbb{Q}) = 0$ because $\mathbb{Q}$ meets every path-component of $\mathbb{R}$. I also have $H_0(\mathbb{R}) = \mathbb{Z}$.
Edit 2
OK, I think I have it:
$$ H_0( \mathbb{Q} )  = \oplus_{q \in \mathbb{Q}} \mathbb{Z} $$
$$ g: \oplus_{q \in \mathbb{Q}} \mathbb{Z}  \rightarrow \mathbb{Z}$$
$$ \operatorname{im}{g} = \mathbb{Z} \implies \operatorname{ker}{g} = \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$$
Here $\mathbb{Q} - \ast$ should be the rationals minus a point.
$$ H_1(\mathbb{R}, \mathbb{Q}) / 0 = H_1(\mathbb{R}, \mathbb{Q}) = \operatorname{im}{f} =  \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$$
Can you tell me if this is correct?
What is a basis for this?
Many thanks for your help!

REPLY [6 votes]: A subgroup of a free abelian group is free abelian, by a well known theorem of Kaplanski.
That fact and the long exact sequence do the trick.
Later. $\newcommand\QQ{\mathbb{Q}}\newcommand\RR{\mathbb{R}}\newcommand\ZZ{\mathbb{Z}}$Let's "compute" $H_1(\RR,\QQ)$.
We know that the path-connected components of $\QQ$ are the points. It follows from the general description of $H_0$ that if we denote, for each $q\in\QQ$, $[q]_\QQ$ the homology class in $H_0(\QQ)$ of the point $q$ and $\ZZ[q]_\QQ$ the subgroup it generates in $H_0(\QQ)$, then there is an isomorphism $$H_0(\QQ)\cong\bigoplus_{q\in\QQ}\ZZ[q]_\QQ$$
We also know that $H_0(\RR)$ is free abelian of rank one, generated by the homology class of any point; moreover, we know that if $x$ and $y$ are points in $\RR$ and we denote $[x]_\RR$ and $[y]_\RR$ their homology classes in $H_0(\RR)$, then $[x]_\RR=[y]_\RR$.
Now, the map $g$ is induced by the inclusion $\QQ\to\RR$. In other words, for each $q\in\QQ$ we have $$g([q]_\QQ)=[q]_\RR.$$ In fact, since the set $\{[q]_\QQ:q\in\QQ\}$ is a basis of the abelian group $H_0(\QQ)$, this information completely determines the map $g$.
Next, and as you observed, there is a short exact sequence $$0\to H_1(\RR,\QQ)\xrightarrow fH_0(\QQ)\xrightarrow gH_0(\RR)$$ coming from the long exact sequence inhomology for the pair $(\RR,\QQ)$. We therefore obtain a description of $H_1(\RR,\QQ)$ as an abelian group isomorphic to the kernel of the map $g$.
Can you construct a basis of $\ker g$?


TITLE: Why $\mathbb{N} \subset V_\omega$ and $Seq(\mathbb{N}) \subset V_\omega$?
QUESTION [0 upvotes]: Let $V_\omega$ denote the set of all hereditarily finite sets.
Let $Seq(\mathbb{N})$ denote the set of all finite sequences from $\mathbb{N}$. That is, $Seq(\mathbb{N}) = \cup\{^n\mathbb{N}|n\in\mathbb{N}\}$ where $^n\mathbb{N}$ is the set of all functions from $n$ to $\mathbb{N}$.
Why $\mathbb{N} \subset V_\omega$ and $Seq(\mathbb{N}) \subset V_\omega$?
I just started to learn some basic set theory, the set $V_\omega$ is quite confusing for me.

REPLY [1 votes]: Note that $V_\omega$ is an inductive set. If $x\in V_\omega$ then $x$ is hereditarily finite, and so $x\cup\{x\}$ is also finite, and its transitive closure is finite (it is the transitive closure of $x$ with the addition $\{x\}$).
Since $\varnothing\in V_\omega$, it means that every integer is in $V_\omega$. Therefore $\Bbb N$, or $\omega$, is a subset of $V_\omega$.
As for the finite sequences, note that if $f$ is a finite sequence of integers then for some $n\in\Bbb N$ we have that $f$ is a sequence of integers smaller than $n$. It is not hard to show, then, that $f$ is also hereditarily finite and therefore in $V_\omega$.


TITLE: Orthogonal complement and subspaces
QUESTION [1 upvotes]: I have the following multiple response question here.

Suppose $V$ is a subspace of $\mathbb{R}^n$ and $S$ is a set of vectors in $V$ whose span is the subspace $W$ of $V$. Select the correct statements.

The orthogonal complement of $S$ in $V$ is the same as the orthogonal complement $W$ in $V$.
Every vector of $S$ belongs to the orthogonal complement of $S$ in $V$.
If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}=\mathbf{0}$.
If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}$ belongs to $S$.
If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $|u|=1$.
If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at least $3$.
If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at most $3$.
If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is always $3$.

The correct answer should be $1$,$3$ and $6$ but could someone explain why?
I understand $3$ but I don't understand why $1$ and $6$ are true. Could someone explain why the other statements are false?
For instance, I thought $8$ would be right instead of $6$ since I thought by the rank-nullity theorem, the nullity must ALWAYS be $3$.

REPLY [2 votes]: By definition $S^\perp=\{\bar{x} \in V | \bar{x}\cdot \bar{s} = 0, \forall \bar{s} \in S\}$, where I'm assuming $(\cdot)$ is the regular dot product.
Let $S$ be a set of vectors in $V$.
Then $\bar{x}\in S^\perp$ implies $\bar{x}\cdot\bar{s} =0$ for all $\bar{s}\in S$, and since $W=\text{span}(S)$, $W$ is the set of all linear combinations of the vectors in $S$, thus for $\bar{w} \in W$, $\bar{w} = a_1\bar{s_1} +\cdots + a_n\bar{s_n}$ for $a_1,...,a_n \in \mathbb{R}$, and $\bar{x}\cdot\bar{w} =a_1(\bar{x}\cdot\bar{s_1})+\cdots+a_n(\bar{x}\cdot\bar{s_n}) = a_1(0)+\cdots+a_n(0)=0$.
Similarly $W^\perp = \{\bar{x} \in V|\bar{x}\cdot\bar{w}=0, \forall \bar{w} \in W\}$, thus $\bar{x}\in W^\perp$ because every $\bar{w} \in W$ is a linear combination of the vectors in $S$ and $\bar{x}\cdot\bar{s} =0$ for all $\bar{s}\in S$. $S^\perp \subseteq W^\perp$.
Every vector $\bar{s} \in S$ is an element of $W$ thus if $\bar{x} \in W^\perp$ then in particular $\bar{x}\cdot\bar{s} =0$ for all $\bar{s} \in S$, thus $\bar{x}\in S^\perp$. $W^\perp \subseteq S^\perp$.
In conclusion $S^\perp = W^\perp$. As for $\#6,7,8$ $\dim(S^\perp)+\dim(S) = \dim(V) =10$, however we cannot be certain that $\dim(S)=7$ because the vectors in $S$ may not be linearly independent, thus at best we know that $\dim(S) \leq 7$, and so $\dim(S^\perp)\geq 10-7=3$. For example say $S$ had $7$ vectors but when finding a basis for $S$ we find that the basis only has $5$ vectors, then $\dim(S) =5\leq 7$ and $\dim(S^\perp)=5 \geq 3$.


TITLE: How to adjust a ratio to create desired proportion
QUESTION [2 upvotes]: Let's say I have a bunch of apples and oranges. Specifically 500 apples and 760 oranges.
I need to create a big basket of fruit. For every 7 apples I include, I need to include 3 oranges.
I want to make the biggest basket possible using as much of my fruit as I can while honoring the desired apple to orange ratio.
I'm not mathematically inclined and I've been just taking bunches of 7 and 3 at a time until I run out. There has to be some kind of formula I can use to tell me upfront the max number of each that I should pull.
Thanks in advance.

REPLY [1 votes]: HINT:
$a\colon o = 7 \colon 3$ and the fraction $7\colon 3$ is in lowest terms so $a = 7 q$, $o = 3 q$ with the conditions $7 q \le 500$ and $3 q \le 760$ so $q \le \frac{497}{7} = 71$ and $q \le \frac{759}{3} = 253$, so the maximal amounts are given by the maximal $q$, $71$ and are
$$a = 7 \cdot 71 = 497 \\
o = 3 \cdot 71 = 213$$
${\bf Added:}$ In general, if the proportion of apples and oranges is given
$a\colon o = m\colon n$ a fraction in lowest terms and the available quantities of fruits are $A$ and $O$, then the largest amounts are $a = m q$, $o = n q$ where
$$q = \min (\left[ \frac{A}{m}\right], \left[ \frac{O}{n}\right])$$
the largest integer $\le$ both $\frac{A}{m}$ and $\frac{O}{n}$.


TITLE: Do modern Formula One cars produce enough down-force to drive upside-down?
QUESTION [12 upvotes]: For example, if they were driving at top speed through a long tunnel, could they transition to and stay on the ceiling?

REPLY [12 votes]: Apparently that is possible.
From http://en.wikipedia.org/wiki/Formula_One_car:

Indycars, for example, produce downforce equal to their weight (that is, a downforce:weight ratio of 1:1) at 190 km/h (118 mph), while an F1 car achieves the same at 125 to 130 km/h (78 to 81 mph), and at 190 km/h (118 mph) the ratio is roughly 2:1.

From http://www.formula1.com/inside_f1/understanding_the_sport/5281.html:

A modern Formula One car is capable of developing 3.5 g lateral cornering force (three and a half times its own weight) thanks to aerodynamic downforce. That means that, theoretically, at high speeds they could drive upside down.


TITLE: If an integer divides $(x-a)$ and $f(x)$ then it divides $f(a)$
QUESTION [1 upvotes]: Prove that if $n\, |\, (x-a)$ and $n \, | \, f(x)$ then $n \, | \, f(a)$ for a polynomial $f$.

It is true if $\operatorname{deg}(f)=1$ or $2$, but what for greater degree of $f(x)$?
I don't know how to continue or if it's any other solution. If possible I would like to prove this without induction.

REPLY [2 votes]: In general, for polynomial division of $f(x)$ by $g(x)$ we have unique polynomials $q(x),r(x)$ such that
$$f(x) = g(x)q(x) + r(x)$$
Where $\deg(r(x)) < \deg(g(x))$. Letting $g(x)=x-a$ we get
$$f(x) = (x-a)q(x) + c$$
for some constant $c$. Since only a constant has lesser degree than $x-a$. The Remainder Theorem tells us this constant must be $f(a)$. So
$$\color{blue}{f(x)} = \color{red}{(x-a)}q(x) + f(a)$$
So if $n$ divides $\color{blue}{f(x)}$ and $\color{red}{(x-a)}$, it must divide $f(a)$.


TITLE: Find all entire functions in the complex plane such that $|f(z)| \leq 2 |z|^{3/2} - 1$ for $|z| \geq 2$ and $f(0)=1$.
QUESTION [2 upvotes]: Find all entire functions in the complex plane such that $|f(z)| \leq 2 |z|^{3/2} - 1$ for $|z| \geq 2$ and $f(0)=1$.

I know how to do this when $a$ in $|z|^a$ is an integer. For that case, we can use Liouville's Theorem. Is this case similar and how do we proceed?

REPLY [3 votes]: You can argue that if the analytic function is bounded by $2|z|^{(3/2)}$ then for sufficieny large $|z|$ it's bounded by $c|z|^2$ for any positive constant $c$.  An entire function having such a bound must either be nonsingular or have only a first-order pole at infinity; thus it must have the form $f(z)=1+bz$.  Add in the absolute value requirement at the circle $|z|=2$ where $|f|\leq(2(2^{3/2})-1)$ and you can place bounds on $b$.


TITLE: How is $\text{End}(M)$ a ring?
QUESTION [1 upvotes]: Let $G$ be an abelian group. I am told that $\text{End}(M)$ forms a ring.
I don't see how that is. The property that I am having difficulty proving is that if $f,g\in \text{End}(M)$, then $f+g\in \text{End}(M)$. Any help would be much appreciated.

REPLY [1 votes]: Write $h=f+g$. What you need to show is that $h:M\to M$ is a group homomorphism, that is, you need to show that for all $m,m'$ in $M$
$$h(m+m')=h(m)+h(m')$$
Just expand both sides separately using the fact that $f$ and $g$ are by definition group homomorphisms, and rearrange the sums using the fact that $M$ is abelian, and you should get the desired equality.
Here are the details:

The first term is \begin{array}{rcl}h(m+m')&=&(f+g)(m+m')\\&=&f(m+m')+g(m+m')\\&=&f(m)+f(m')+g(m)+g(m')\end{array} and \begin{array}{rcl}h(m)+h(m')&=&(f+g)(m)+(f+g)(m')\\&=&f(m)+g(m)+f(m')+g(m')\end{array} The two expressions are equal because $(M,+)$ is abelian.


TITLE: Another real quadratic integer ring, Euclidean but not norm-Euclidean, with norm function needing only two adjustments?
QUESTION [4 upvotes]: I have come to know about $\mathcal O_K$ with $K = \mathbb Q(\sqrt{69})$. The norm function needs to be adjusted to absolute value, as is the case with other real rings, but it also needs to be adjusted for primes lying over $23$.
Is there another real quadratic integer ring that is Euclidean but not norm-Euclidean, but for which the norm function can, with two adjustments (absolute value and something else), be used as the Euclidean function? And if so, what is it?

REPLY [4 votes]: The answer certainly is not known to be ${\mathbb Z}[\sqrt{14}]$: this ring is Euclidean, as was shown by Harper, and it is not norm-Euclidean, as has been known for a long time. Whether it is Euclidean for a weighted norm is not known (see Prop. 4.11
here), and to the best of my knowledge nobody so far has managed to deal with weighted norms at two prime ideals.


TITLE: Find an expression for $(\lambda\textbf{1}-a)^{-1}$.
QUESTION [1 upvotes]: Let $A$ be a unital Banach algebra, and suppose $a\in A$ has the property that $a^2=1$, but that $a\neq \textbf{1}$ and $a\neq -\textbf{1}$.
By the spectral mapping theorem, $\sigma(a^2) = \{\lambda^2 : \lambda \in \sigma(a) \}.$
Therefore $\sigma(a^2)= \sigma(\textbf{1})= \{ 1 \}$, i.e. $\sigma(a)\subseteq \{-1,1\}.$
It can easily be shown (by an argument of contradiction) that equality holds, i.e. $\sigma(a)=\{-1,1\}$.
Now suppose that $\lambda \not\in \sigma(a)$, then $(\lambda\textbf{1}-a)^{-1}$ exists.

I am trying to find an expression for $(\lambda\textbf{1}-a)^{-1}$ in terms of $a$ and $\lambda$ which holds for all $\lambda \not\in \sigma(a)$.

Can anyone please assist? I am currently self-studying some work on spectral theory, and this is one of the problems I am trying to solve as part of the process.

REPLY [1 votes]: Solve for real values $\alpha,\beta,$ $(\lambda I-a)(\alpha I+\beta a)=I.$ Since $a^2=I,$ the equation becomes: $$\lambda \alpha-\beta=1,\\\lambda\beta-\alpha=0.$$
Solving this gives: $$\beta=\frac{1}{\lambda^2-1}\\\alpha=\frac{\lambda}{\lambda^2-1}.$$

In general, if $a\in A$ has a polynomial $p(x),$ with $p(a)=0,$ and if $\lambda\in\mathbb C$ is not a root of $p(x),$ then we can divide $p(x)$ by $x-\lambda$ to get:
$$p(x)=(x-\lambda)q(x)+r,$$ where $q(x)$ is a polynomial of one lower degree and $r=p(\lambda)\in\mathbb C\setminus\{0\}.$ Then, substituting $x=a,$ you get $$(a-\lambda I)q(a)+p(\lambda)I=0,$$ or $$(\lambda I-a)^{-1}=\frac1{p(\lambda)}q(a).$$
In your case, $p(x)=x^2-1,$ and $$p(x)=p((x-\lambda)+\lambda)=(x-\lambda)^2+2\lambda(x-\lambda)+(\lambda^2-1)$$ so $q(x)=x-\lambda+2\lambda=x+\lambda,$ and you get: $$\frac{1}{\lambda^2-1}\left(a+\lambda I\right),$$ again.

Even more generally, if $p_1,p_2$ are relatively prime polynomials (so, they share no roots) we can apply the Euclidean algorithm and solve for polynomials $q_1,q_2$ such that:
$$p_1(x)q_1(x)+p_2(x)q_2(x)=1.$$
Then if $p_1(a)=0,$ then $p_2(a)^{-1}=q_2(a).$