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Let $X$ be smooth manifold of dimension $n$, let $E$ be a vector
bundle over $X$, and let $A$ be a connection on $E$.  WE define the
Sobolev Space $L^p_{k,A}(X,E)$ to be the completion of
$C^\infty(X,E)$ with respect to the norm

$$||s||^p_{p,k,A}=\sum_{i=0}^k\int_{X}|\nabla_A\overset{i \mathrm{
times}}{\circ\cdots\circ}\nabla_A s|^p\mathrm{ d\mathrm{vol}}$$

If no vector bundle is specified, then the trivial bundle, i.e. the
set of smooth functions on $X$, is implied.  Other notations for
this space are $W^p_{k,A}$ and in the case $p=2$, $H_{k,A}$.  We
would like to know when $L^p_k(X)\subset L^q_l(X)$ holds.  Define
the conformal weight of $L^p_k$, to be $k-n/p$.  Then for $f\in
L^p_k$, $||f(tx)||_{p,k}\simeq t^{k-n/p}||f(x)||$, so for such an
inclusion to hold, we must have $k-n/p\leq l-n/q$.  If $X$ is
compact, then this, and the condition that $k\geq l$ are sufficient
for $L^p_k(X)\subset L^q_l(X)$.  Furthermore, Rellich's theorem
tells us that if both inequalities are strict, this inclusion is a
compact embedding,

Let $C^{l,\alpha}(X)$ be the space of functions with $l$
derivatives, such that the $l^{th}$ derivative is Holder continuous
with exponent $\alpha$.  If $0<\alpha<1$, then $L^p_k(X)\subset
C^{l,\alpha}\iff k\geq l$, and $k-n/p\geq l+\alpha$.  The case
$\alpha=0$ is much more subtle.  For example, if $\Sigma$ is a
Riemann surface, $L^2_1(\Sigma)\not\subset C^0(\Sigma)$, and if $X$
is a 4-manifold, $L^4_1(X)\not\subset C^0(X)$.

\section{Sobolev Multiplication}

Multiplication gives a map $L^p_k\times L^q_l\rightarrow L^r_m$ if
$$\mathrm{a)\,\,\, }(k-n/p)+(l-n/q)>m-n/r\,\,\mathrm{  and  }\,\,k,l\geq m,\mathrm{
or}$$
$$\mathrm{b)\,\,\, }k-n/p>0\,\,\mathrm{  and  }\,\,k-n/p,l-n/q\geq m-n/r$$

The proofs of these two results follow from the embedding theorem,
and Holder's inequality.  For a), write $\nabla^m(fg)=\sum_{i=0}^{m}c_i\nabla^if\nabla^{m-i}g$ where the $c_i$ only depend on $m$.  By Holder's inequality, we have $||\nabla^if\nabla^{m-i}g||_r\leq||\nabla^if||_a||\nabla^{m-i}g||_b$ when $1/a+1/b\leq1/r$.  This is equivalent to $m-n/r\leq(i-n/a)+(m-i-n/b)$.  Since $k\geq m\geq i$, $l\geq m\geq m-i$, we can pick $a$ so that $k-n/p=i-n/a$ if $k-n/p\leq i$ or $a\geq 2r$ if $k-n/p>i$ (And similarly for b), and apply Sobolev Embedding term by term.

For b), note that the hypotheses imply the hypotheses of a).

\section{Differentiation on Banach Spaces}

Let $X,Y$ be Banach spaces, $U=subset X$ an open subset, and
$f:U\rightarrow Y$ a map.  We say that $f$ is differentiable at
$x\in U$ if there is $C>0$ and a bounded linear operator
$A:X\rightarrow Y$ such that $||f(x+h)-f(x)-A\dot h||_Y\leq
C||h||^2_X$  (this exponent 2 here is not quite the correct
definition, but it will do for our purposes). In this case, we write
$d_x f$ for $A$, and call this the derivative of $f$ at $x$.

We say that $f$ is continuously differentiable if $x\mapsto
A_x:U\rightarrow \mathrm{Hom}(X,Y)$ is a continuous map.  We say
that $f$ is twice differentiable at $x$ is it is continuously
differentiable at $x$, and the map $x'\mapsto d_{x'}f$ is
differentiable at $x$.  Iterating this definition allows us to
define higher derivatives.  In particular, $f$ is $C^\infty$ if it
has derivatives of all orders.

It turns out that if $(f,g)\mapsto \mu(f,g)=f\cdot g:L^p_k\times
L^q_l\rightarrow L^r_m$ is continuous, then it is $C^\infty$.  In
fact, if we let $\delta$ denote a small variation, and letting
$d_{\mu(f,g)}(\delta f,\delta g)=f\delta g+g\delta f$, we have

$$||\mu(f+\delta f,g+\delta g)-\mu(f,g)-d_{\mu(f,g,)}(\delta
f,\delta g)||_{r.m}$$

$$=||\delta f\cdot\delta g||_{r,m}\leq C(||\delta f||_{p,k}||\delta
g||_{l,q})^2$$

where the last inequality follows from the multiplication theorem.
The derivative $d_{\mu(f,g)}(\delta f,\delta g)$ only depends on the
product $f\cdot g$, so the higher derivatives all vanish.



\section{Application to the Space of Connections}

Let $\pi:P\rightarrow X$ be a principal $G$-bundle, and let
$\mathfrak{g}$ be the Lie algebra of $G$.  Let $A$ denote a
connection on $P$, $F_A\in\Omega^2(X,\mathrm{ad }P)$ its curvature.
If, $\tau$ is a local trivialization, we let $a^\tau$ denote the
one-form associated to the connection, $A\mapsto F_A$ corresponds to
$a^\tau\mapsto da^\tau+[a^\tau\wedge a^\tau]$.  Recall the following
definition:

$$\mathcal{A}^p_k=\{\mathrm{connections }A|\mathrm{ if }\tau:\pi^{-1}(U)\rightarrow
U\times P\mathrm{ is a }C^\infty\mathrm{ trivialization such that if
}$$
$$\phi\in C^\infty_0(U)\mathrm{ we have } a_\tau\in\Omega^1(U,\mathfrak{g}),
\phi a_\tau\in L^p_k(U,T^pU\otimes\mathfrak{g})\}$$

Let $\mathcal{F}:\mathcal{A}^p_k\rightarrow
L^q_l(X,\Lambda^2\otimes\mathrm{ad }P)$ be the map sending a
connection to its curvature.  We would like to know when
$\mathcal{F}$ is continuous.  Clearly, $a^\tau\mapsto da^\tau$ is
continuous into $L^p_{k-1}$.  From the multiplication theorem,
$a^\tau\mapsto [a^\tau\wedge a^\tau]$ is continuous when
$(k-n/p)+(k-n/p)>(k-1)-n/p$, i.e. $k+1\geq n/p$.  In fact, in this
case it is $C^\infty$.  For example, for $n=4$, $(p,k)=(2,1)$,
$(q,l)=(4,0)$, $a^\tau\mapsto [a^\tau\wedge a^\tau]$ is continuous.

We specialize to the the situation where $G=O(n),U(n),Sp(n)$, amd
where $P$ is the frame bundle of some vector bundle $E$.  A smooth
gauge transformation $u$ is a smooth bundle endomorphism of $E$ such
that $u$ is an orthogonal (resp. unitary or symplectic) isomorphism
on each fiber.  In other words, $u\in C^\infty(X,\mathrm{End}(E))$,
and $u^*u=1$, where $u^*$ denotes the transpose if $G=O(n)$, the
conjugate transpose if $G=U(n)$, etc.  We define

$$\mathcal{G}^p_k=\{u\in L^p_k(X,\mathrm{End}(E))|u^*u=1\mathrm{ almost everywhere}\}$$

We would like $\mathcal{G}^p_k$ to be a group, and ideally, a Banach
Lie group.

\begin{thm}
If $(k-n/p)>0$, then $\mathcal{G}^p_k$ is a Banach Lie group, and
$L^p_k(X,\mathrm{End}(E))$ is a Banach space and algebra
\end{thm}

\begin{proof}
Define $\Phi:L^p_k(X,\mathrm{End}(E))\rightarrow
L^p_k(X,\mathrm{End}(E))$ by $\Phi(s)=s^*s$. Then $\Phi$ is a
$C^\infty$ map, and $d_s\Phi(t)=s^*t+t^*s$.  Using $s^*s=1$, once
easily sees that $d_s\Phi(sw/2)=w$, so that $d_s\Phi$ is surjective
for all $s\in\mathcal{G}^p_k$. It is also clear that
$\mathrm{ker}(d_s\Phi)=\{sr|r=-r^*\}$.  These two facts allows us to
apply the implicit function theorem to conclude that
$\Phi^{-1}(1)\subset\mathcal{G}^p_k$ is a Banach Lie group.
\end{proof}

We would next like to determine when $\mathcal{G}^p_k$ acts on
$\mathcal{A}^q_l$. The multiplication theorem implies that
$L^p_{k-1}$ is an $L^p_k$ module if $(k-n/p)$>0.  The argument at
the end of section 3 shows that this multiplication is smooth, so
that $\mathcal{G}^p_k\times\mathcal{A}^p_{k-1}\rightarrow
\mathcal{A}^p_{k-1}$ is a smooth.

Let $\mathcal{B}^p_{k-1}=\mathcal{A}^p_{k-1}/\mathcal{G}^p_k$.  We
want to know when this is a Hausdorff space.  Recall the following
criterion:  If $G$ is a topological group, that acts continuously on
a topological space $X$, then $X/G$ is Hausdorff if and only if the
graph $\Gamma=\{(x,xg)|x\in X,g\in G\}$ is closed in $X\times X$.

\begin{thm}
If $(k-n/p)>0$, then $\mathcal{B}^p_{k-1}$ is Hausdorff
\end{thm}

\begin{proof}
Suppose we have a sequence $(A_i,u_iA_i)$ that converges to $(A,A')$
in $L^p_{k-1}$.  Then $a^\tau_i$ converges to $a^\tau$ in $L^p_k$,
and $u^{-1}_idu_i+u^{-1}_ia^\tau_iu_i\rightarrow a'^\tau$.
Rearranging this gives $du_i=u_ia'^\tau-a^\tau_iu_i$.  Suppose
$k=1$, $p=n+\epsilon$. Then since $a^\tau_i$ converges, and $u_i$
are all in $\mathcal{G}^p_k$, this implies that $du_i$ is uniformly
bounded. If we pass to a subsequence where $u_i$ converges, say to
$u$, in $C^0$, then $du_i$ converges in $L^{n+\epsilon}$, so that
$u_i$ converges to $u$ in $L^{n+\epsilon}_1$.  For $k>1$, one
applies a bootstrap argument.
\end{proof}