09-08 Notes

Revision as of 22:21, 12 November 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.

\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{thm1} If $(k-n/p)>0$, then $\mathcal{B}^p_{k-1}$ is Hausdorff \end{thm1}

\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}