09-08 Notes

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Let $X$ be smooth manifold of dimension $n$, let $E$ be a vector
 
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bundle over $X$, and let $A$ be a connection on $E$.  WE define the
\begin{document}
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Sobolev Space $L^p_{k,A}(X,E)$ to be the completion of
 
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\title{Lecture 2}
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\newcommand{\e}{\epsilon}
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\newcommand{\ra}{\rightarrow}
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\newcommand{\g}{\mathfrak{g}}
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\newcommand{\M}{\mathcal{M}}
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\newcommand{\al}{\alpha}
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\newcommand{\PR}{\mathcal{P}}
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\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\G}{\mathcal{G}}
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\newcommand{\F}{\mathcal{F}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\A}{\mathcal{A}}
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\newcommand{\B}{\mathcal{B}}
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\newcommand{\dvol}{\text{ d\textit{vol}}}
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\newtheorem*{thm}{Proposition}
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\newtheorem*{thm1}{Claim}
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\maketitle
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\section{Sobolev Embedding}
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  Let $X$ be smooth manifold of dimension $n$, $\dvol$, let $E$ be a
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vector bundle over $X$, and let $A$ be a connection on $E$.  WE
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define the Sobolev Space $L^p_{k,A}(X,E)$ to be the completion of
+
 
$C^\infty(X,E)$ with respect to the norm
 
$C^\infty(X,E)$ with respect to the norm
  
$$||s||^p_{p,k,A}=\sum_{i=0}^k\int_{X}|\nabla_A\overset{i \text{
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$$||s||^p_{p,k,A}=\sum_{i=0}^k\int_{X}|\nabla_A\overset{i \mathrm{
times}}{\circ\cdots\circ}\nabla_A s|^p\dvol$$
+
times}}{\circ\cdots\circ}\nabla_A s|^p\mathrm{ d\mathrm{vol}}$$
  
 
If no vector bundle is specified, then the trivial bundle, i.e. the
 
If no vector bundle is specified, then the trivial bundle, i.e. the
 
set of smooth functions on $X$, is implied.  Other notations for
 
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
 
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
+
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
 
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
 
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
 
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
 
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
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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
 
tells us that if both inequalities are strict, this inclusion is a
 
compact embedding,
 
compact embedding,
  
Let $C^{l,\al}(X)$ be the space of functions with $l$ derivatives,
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Let $C^{l,\alpha}(X)$ be the space of functions with $l$
such that the $l^{th}$ derivative is Holder continuous with exponent
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derivatives, such that the $l^{th}$ derivative is Holder continuous
$\al$.  If $0\al<1$, then $L^p_k(X)=subset C^{l,\al}\iff k\geq l$,
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with exponent $\alpha$.  If $0<\alpha<1$, then $L^p_k(X)\subset
and $k-n/p\geq l+\al$.  The case $\al=0$ is much more subtle.  For
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C^{l,\alpha}\iff k\geq l$, and $k-n/p\geq l+\alpha$.  The case
example, if $\Sigma$ is a Riemann surface, $L^2_1(\Sigma)\not=subset
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$\alpha=0$ is much more subtle.  For example, if $\Sigma$ is a
C^0(\Sigma)$, and if $X$ is a 4-manifold, $L^4_1(X)\not=subset
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Riemann surface, $L^2_1(\Sigma)\not\subset C^0(\Sigma)$, and if $X$
C^0(X)$.
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is a 4-manifold, $L^4_1(X)\not\subset C^0(X)$.
  
 
\section{Sobolev Multiplication}
 
\section{Sobolev Multiplication}
  
Multiplication gives a map $L^p_k\times L^q_l\ra L^r_m$ if
+
Multiplication gives a map $L^p_k\times L^q_l\rightarrow L^r_m$ if
$$\text{a) }(k-n/p)+(l-n/q)>m-n/r\text{  and  }k,l\geq m,\text{
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$$\mathrm{a)\,\,\, }(k-n/p)+(l-n/q)>m-n/r\,\,\mathrm{  and  }\,\,k,l\geq m,\mathrm{
 
or}$$
 
or}$$
$$\text{b) }k-n/p>0\text{  and  }k-n/p,l-n/q\geq m-n/r$$
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$$\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,
 
The proofs of these two results follow from the embedding theorem,
and Holder's inequality.
+
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}
 
\section{Differentiation on Banach Spaces}
  
Let $X,Y$ be Banach spaces, $U=subset X$ an open subset, and $f:U\ra
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Let $X,Y$ be Banach spaces, $U=subset X$ an open subset, and
Y$ a map.  We say that $f$ is differentiable at $x\in U$ if there is
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$f:U\rightarrow Y$ a map.  We say that $f$ is differentiable at
$C>0$ and a bounded linear operator $A:X\ra Y$ such that
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$x\in U$ if there is $C>0$ and a bounded linear operator
$||f(x+h)-f(x)-A\dot h||_Y\leq C||h||^2_X$  (this exponent 2 here is
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$A:X\rightarrow Y$ such that $||f(x+h)-f(x)-A\dot h||_Y\leq
not quite the correct definition, but it will do for our purposes).
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C||h||^2_X$  (this exponent 2 here is not quite the correct
In this case, we write $d_x f$ for $A$, and call this the derivative
+
definition, but it will do for our purposes). In this case, we write
of $f$ at $x$.
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$d_x f$ for $A$, and call this the derivative of $f$ at $x$.
  
We say that $f$ is continuously differentiable if $x\mapsto A_x:U\ra
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We say that $f$ is continuously differentiable if $x\mapsto
\text{Hom}(X,Y)$ is a continuous map.  We say that $f$ is twice
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A_x:U\rightarrow \mathrm{Hom}(X,Y)$ is a continuous map.  We say
differentiable at $x$ is it is continuously differentiable at $x$,
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that $f$ is twice differentiable at $x$ is it is continuously
and the map $x'\mapsto d_{x'}f$ is differentiable at $x$.  Iterating
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differentiable at $x$, and the map $x'\mapsto d_{x'}f$ is
this definition allows us to define higher derivatives.  In
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differentiable at $x$.  Iterating this definition allows us to
particular, $f$ is $C^\infty$ if it has derivatives of all orders.
+
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
 
It turns out that if $(f,g)\mapsto \mu(f,g)=f\cdot g:L^p_k\times
L^q_l\ra L^r_m$ is continuous, then it is $C^\infty$.  In fact, if
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L^q_l\rightarrow L^r_m$ is continuous, then it is $C^\infty$.  In
we let $\delta$ denote a small variation, and letting
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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
 
$d_{\mu(f,g)}(\delta f,\delta g)=f\delta g+g\delta f$, we have
  
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\section{Application to the Space of Connections}
 
\section{Application to the Space of Connections}
  
Let $\pi:P\ra X$ be a principal $G$-bundle, and let $\g$ be the Lie
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Let $\pi:P\rightarrow X$ be a principal $G$-bundle, and let
algebra of $G$.  Let $A$ denote a connection on $P$,
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$\mathfrak{g}$ be the Lie algebra of $G$.  Let $A$ denote a
$F_A\in\Omega^2(X,\text{ad }P)$ its curvature. If, $\tau$ is a
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connection on $P$, $F_A\in\Omega^2(X,\mathrm{ad }P)$ its curvature.
local trivialization, we let $a^\tau$ denote the one-form associated
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If, $\tau$ is a local trivialization, we let $a^\tau$ denote the
to the connection, $A\mapsto F_A$ corresponds to $a^\tau\mapsto
+
one-form associated to the connection, $A\mapsto F_A$ corresponds to
da^\tau+[a^\tau\wedge a^\tau]$.  Recall the following definition:
+
$a^\tau\mapsto da^\tau+[a^\tau\wedge a^\tau]$.  Recall the following
 +
definition:
  
$$\A^p_k=\{\text{connections }A|\text{ if }\tau:\pi^{-1}(U)\ra
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$$\mathcal{A}^p_k=\{\mathrm{connections }A|\mathrm{ if }\tau:\pi^{-1}(U)\rightarrow
U\times P\text{ is a }C^\infty\text{ trivialization such that if }$$
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U\times P\mathrm{ is a }C^\infty\mathrm{ trivialization such that if
$$\phi\in C^\infty_0(U)\text{ we have } a_\tau\in\Omega^1(U,\g),
+
}$$
\phi a_\tau\in L^p_k(U,T^pU\otimes\g)\}$$
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$$\phi\in C^\infty_0(U)\mathrm{ we have } a_\tau\in\Omega^1(U,\mathfrak{g}),
 +
\phi a_\tau\in L^p_k(U,T^pU\otimes\mathfrak{g})\}$$
  
Let $\F:\A^p_k\ra L^q_l(X,\Lambda^2\otimes\text{ad }P)$ be the map
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Let $\mathcal{F}:\mathcal{A}^p_k\rightarrow
sending a connection to its curvature.  We would like to know when
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L^q_l(X,\Lambda^2\otimes\mathrm{ad }P)$ be the map sending a
$\F$ is continuous.  Clearly, $a^\tau\mapsto da^\tau$ is continuous
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connection to its curvature.  We would like to know when
into $L^p_{k-1}$.  From the multiplication theorem, $a^\tau\mapsto
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$\mathcal{F}$ is continuous.  Clearly, $a^\tau\mapsto da^\tau$ is
[a^\tau\wedge a^\tau]$ is continuous when
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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
 
$(k-n/p)+(k-n/p)>(k-1)-n/p$, i.e. $k+1\geq n/p$.  In fact, in this
 
case it is $C^\infty$.  For example, for $n=4$, $(p,k)=(2,1)$,
 
case it is $C^\infty$.  For example, for $n=4$, $(p,k)=(2,1)$,
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gauge transformation $u$ is a smooth bundle endomorphism of $E$ such
 
gauge transformation $u$ is a smooth bundle endomorphism of $E$ such
 
that $u$ is an orthogonal (resp. unitary or symplectic) isomorphism
 
that $u$ is an orthogonal (resp. unitary or symplectic) isomorphism
on each fiber.  In other words, $u\in C^\infty(X,\text{End}(E))$,
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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
 
and $u^*u=1$, where $u^*$ denotes the transpose if $G=O(n)$, the
 
conjugate transpose if $G=U(n)$, etc.  We define
 
conjugate transpose if $G=U(n)$, etc.  We define
  
$$\G^p_k=\{u\in L^p_k(X,\text{End}(E))|u^*u=1\text{ almost everywhere}\}$$
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$$\mathcal{G}^p_k=\{u\in L^p_k(X,\mathrm{End}(E))|u^*u=1\mathrm{ almost everywhere}\}$$
  
We would like $\G^p_k$ to be a group, and ideally, a Banach Lie
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We would like $\mathcal{G}^p_k$ to be a group, and ideally, a Banach
group.
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Lie group.
  
 
\begin{thm}
 
\begin{thm}
If $(k-n/p)>0$, then $\G^p_k$ is a Banach Lie group, and
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If $(k-n/p)>0$, then $\mathcal{G}^p_k$ is a Banach Lie group, and
$L^p_k(X,\text{End}(E))$ is a Banach space and algebra
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$L^p_k(X,\mathrm{End}(E))$ is a Banach space and algebra
 
\end{thm}
 
\end{thm}
  
 
\begin{proof}
 
\begin{proof}
Define $\Phi:L^p_k(X,\text{End}(E))\ra L^p_k(X,\text{End}(E))$ by
+
Define $\Phi:L^p_k(X,\mathrm{End}(E))\rightarrow
$\Phi(s)=s^*s$. Then $\Phi$ is a $C^\infty$ map, and
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L^p_k(X,\mathrm{End}(E))$ by $\Phi(s)=s^*s$. Then $\Phi$ is a
$d_s\Phi(t)=s^*t+t^*s$.  Using $s^*s=1$, once easily sees that
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$C^\infty$ map, and $d_s\Phi(t)=s^*t+t^*s$.  Using $s^*s=1$, once
$d_s\Phi(sw/2)=w$, so that $d_s\Phi$ is surjective for all
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easily sees that $d_s\Phi(sw/2)=w$, so that $d_s\Phi$ is surjective
$s\in\G^p_k$. It is also clear that
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for all $s\in\mathcal{G}^p_k$. It is also clear that
$\text{ker}(d_s\Phi)=\{sr|r=-r^*\}$.  These two facts allows us to
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$\mathrm{ker}(d_s\Phi)=\{sr|r=-r^*\}$.  These two facts allows us to
 
apply the implicit function theorem to conclude that
 
apply the implicit function theorem to conclude that
$\Phi^{-1}(1)=subset\G^p_k$ is a Banach Lie group.
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$\Phi^{-1}(1)\subset\mathcal{G}^p_k$ is a Banach Lie group.
 
\end{proof}
 
\end{proof}
  
We would next like to determine when $\G^p_k$ acts on $\A^q_l$. The
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We would next like to determine when $\mathcal{G}^p_k$ acts on
multiplication theorem implies that $L^p_{k-1}$ is an $L^p_k$ module
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$\mathcal{A}^q_l$. The multiplication theorem implies that
if $(k-n/p)$>0.  The argument at the end of section 3 shows that
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$L^p_{k-1}$ is an $L^p_k$ module if $(k-n/p)$>0.  The argument at
this multiplication is smooth, so that $\G^p_k\times\A^p_{k-1}\ra
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the end of section 3 shows that this multiplication is smooth, so
\A^p_{k-1}$ is a smooth.
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that $\mathcal{G}^p_k\times\mathcal{A}^p_{k-1}\rightarrow
 +
\mathcal{A}^p_{k-1}$ is a smooth.
  
Let $\B^p_{k-1}=\A^p_{k-1}/\G^p_k$.  We want to know when this is a
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Let $\mathcal{B}^p_{k-1}=\mathcal{A}^p_{k-1}/\mathcal{G}^p_k$.  We
Hausdorff space.  Recall the following criterion:  If $G$ is a
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want to know when this is a Hausdorff space.  Recall the following
topological group, that acts continuously on a topological space
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criterion:  If $G$ is a topological group, that acts continuously on
$X$, then $X/G$ is Hausdorff if and only if the graph
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a topological space $X$, then $X/G$ is Hausdorff if and only if the
$\Gamma=\{(x,xg)|x\in X,g\in G\}$ is closed in $X\times X$.
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graph $\Gamma=\{(x,xg)|x\in X,g\in G\}$ is closed in $X\times X$.
  
\begin{thm1}
+
\begin{thm}
If $(k-n/p)>0$, then $\B^p_{k-1}$ is Hausdorff
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If $(k-n/p)>0$, then $\mathcal{B}^p_{k-1}$ is Hausdorff
\end{thm1}
+
\end{thm}
  
 
\begin{proof}
 
\begin{proof}
 
Suppose we have a sequence $(A_i,u_iA_i)$ that converges to $(A,A')$
 
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 $L^p_k$,
 
in $L^p_{k-1}$.  Then $a^\tau_i$ converges to $a^\tau$ in $L^p_k$,
and $u^{-1}_idu_i+u^{-1}_ia^\tau_iu_i\ra a'^\tau$. Rearranging this
+
and $u^{-1}_idu_i+u^{-1}_ia^\tau_iu_i\rightarrow a'^\tau$.
gives $du_i=u_ia'^\tau-a^\tau_iu_i$.  Suppose $k=1$, $p=n+\e$. Then
+
Rearranging this gives $du_i=u_ia'^\tau-a^\tau_iu_i$.  Suppose
since $a^\tau_i$ converges, and $u_i$ are all in $\G^p_k$, this
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$k=1$, $p=n+\epsilon$. Then since $a^\tau_i$ converges, and $u_i$
implies that $du_i$ is uniformly bounded. If we pass to a
+
are all in $\mathcal{G}^p_k$, this implies that $du_i$ is uniformly
subsequence where $u_i$ converges, say to $u$, in $C^0$, then $du_i$
+
bounded. If we pass to a subsequence where $u_i$ converges, say to
converges in $L^{n+\e}$, so that $u_i$ converges to $u$ in
+
$u$, in $C^0$, then $du_i$ converges in $L^{n+\epsilon}$, so that
$L^{n+\e}_1$.  For $k>1$, one applies a bootstrap argument.
+
$u_i$ converges to $u$ in $L^{n+\epsilon}_1$.  For $k>1$, one
 +
applies a bootstrap argument.
 
\end{proof}
 
\end{proof}
 
 
\end{document}
 

Latest revision as of 01:12, 9 December 2008

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}