# 09-10 Notes

(Difference between revisions)

\begin{document}

\title{MIT Geometry Seminar \\ Lecture 3 \\ 9/10/2008}

\maketitle

Let $X$ be a compact Riemanian manifold, and $P$ be a principal bundle over $X$ with structure group $G$, associated to a vector bundle $E \rightarrow X$. We denote the space of $L_k^p$ connections by $\mathcal{A}_k^p$, more precisely, $\mathcal{A}_k^p$ = $\{$ A connection $|$ if $\tau: \pi^{-1}(U) \rightarrow U \times P$ is a $C^\infty$ trivialization, $\varphi \in C_o^\infty(U)$, and $a_\tau \in \Omega^1(U,\mathfrak{g})$, then $\varphi\alpha_\tau \in L_k^p(U, T^*U \otimes \mathfrak{g}) \}$. We denote the $L_{k+1}^p$ gauge transformations by $\mathcal{G}_{k+1}^p$, more precisely $\mathcal{G}_{k+1}^p$ = $\{ u \in L_{k+1}^p$($X$, End $E$) $| u^*u = 1$ a.e. $\}$. Last time, we noted that we have a smooth action $\mathcal{G}_{k+1}^p \times \mathcal{A}_k^p \rightarrow \mathcal{A}_k^p$ provided $p(k+1) > n$, we denote the quotient of this action by $\mathcal{B}_k^p$ = $\mathcal{A}_k^p / \mathcal{G}_{k+1}^p$.

In light of the Sobolev Embedding theorems, we can ask the following question: does an embedding $\mathcal{A}_k^p \hookrightarrow \mathcal{A}_l^q$ yield a map $\mathcal{B}_k^p \hookrightarrow \mathcal{B}_l^q$?

\begin{prop}

Suppose $u \in \mathcal{G}_0^\infty = \{ u \in L^\infty(X, End E) | u^*u = 1$  a.e. $\}$ and $\exists A, \tilde{A} \in \mathcal{A}_k^p$ such that $u\cdot A = \tilde{A}$, then $u \in \mathcal{G}_{k+1}^p$.


\end{prop}

\proof We assume $u \in \mathcal{G}_0^\infty$ because that is the weakest possible restriction. In a local trivialization, we have $$u\cdot A = \tilde{A} \Rightarrow u^{-1}du + u^{-1}a^{\tau}u = \tilde{a}^\tau \Rightarrow du = u\tilde{a}^\tau - a^\tau u$$ If $\tilde{a}^\tau, a^\tau \in L^p$ then the above implies that $$du \in L^p \Rightarrow u \in L_1^p .$$ Next, if we have $\tilde{a}^\tau, a^\tau \in L_1^p$ then this implies, by the Leibnitz rule, $$\nabla du = u \nabla \tilde{a}^\tau - \nabla a^\tau u + \nabla u \tilde{a}^\tau - a^\tau \nabla u$$ The first two terms in the above, $u \nabla \tilde{a}^\tau, \nabla a^\tau u \in L^p$. For the third and fourth terms, we know that $\nabla u \in L^p$. We also have $\tilde{a}^\tau, a^\tau \in L_1^p,$ but by the Sobolev embedding theorems, we have $L_1^p \hookrightarrow L^q$ for some q. This implies that $\nabla u \tilde{a}^\tau, a^\tau \nabla u \in L^n,$ where $\frac{1}{n} = \frac{1}{p} + \frac{1}{q}.$ If $p \leq n$, then $L^n \hookrightarrow L^p$, which means $\nabla du \in L^p$ and we are done. If $p > n$, then $L^p \hookrightarrow L^n$, and so $\nabla du \in L^n .$ Then: $$\nabla du \in L^n \Rightarrow u \in L_2^n \Rightarrow \nabla u \in L_1^n \hookrightarrow L^r \forall r \Rightarrow \nabla u \tilde{a}^\tau, a^\tau \nabla u \in L^p \Rightarrow \nabla du \in L^p .$$ We repeat this "bootstrapping" argument and obtain $u \in \mathcal{G}^p_{k+1}$, for any k we desire.

Recall that if $p(k+1) > n$ then $\mathcal{B}_k^p$ is Hausdorff. Consider the borderline case, with $\mathcal{A}_0^n, \mathcal{G}_1^n,$ in this case $\mathcal{G}_1^n$ is a group, but not a topological group.

\begin{prop}

The $G_1^n$ orbits are weakly (hence strogly) closed in $A_0^n.$


\end{prop}

\proof Suppose $A \in \mathcal{A}_0^n$ and $u_i \in \mathcal{G}_1^n$ such that $u_i \cdot A \rightharpoonup \tilde{A}$ in $L^n$, then we want to show that $\exists u \in \mathcal{G}_1^n$ such that $u \cdot A = \tilde{A}$. (Note: $\rightharpoonup$ denotes weak convergence.) Locally, for a trivialization $\tau$, we have: $u_i^{-1}du_i + u_1^{-1}a^{\tau}u_i = \tilde{a}_i^{\tau}$ and by assumption $\tilde{a}_i^{\tau} \stackrel{L^n}\rightharpoonup a^{\tau}$. Then we have $du_i = u_i\tilde{a}_i^{\tau} - a^{\tau}u_i$, and weak convergence in $L^n$ implies that the right hand side is uniformly bounded in $L^n$, so in particular, $u_i$ is uniformly bounded in $L_1^n$. This implies that we can find a subsequence such that $u_{i_k} \stackrel{L^n_1}\rightharpoonup u$. From the Sobolev Embedding Theorem, we have that $L_1^n \hookrightarrow L^p$ for all $p$, so we have a subsequence $u_i \stackrel{L^p}\rightarrow u$, which implies $u_i \rightarrow u$ a.e., thus $u^*u = 1$ a.e. and so $u \in \mathcal{G}_1^n$. Now, in the equation $du_i = u_i\tilde{a}_i^{\tau} - a^{\tau}u_i$, we have $du_i \rightharpoonup du$, and $a^{\tau}u_i \rightharpoonup a^{\tau}u$ (because $u_i \rightharpoonup u$). We want to show that $u\tilde{a}^{\tau} = \tilde{\tilde{a}}$ in $L^{n-\epsilon}$ for small $\epsilon$, where $\tilde{\tilde{a}}$ is such that we have a subsequence $u_{i}\tilde{a}_i^{\tau} \rightharpoonup \tilde{\tilde{a}}$. Since we have $u_i \stackrel{L^p}\rightarrow u$ for all $p$, we can find a large $N$ for which $u_i \in L^N$ and $\tilde{a}_i^{\tau} \in L^n$, which would imply by the Lemma below that $u_i\tilde{a}_i^{\tau} \rightharpoonup u\tilde{a}^{\tau} = \tilde{\tilde{a}}$ in $L^{n-\epsilon}$.

\begin{lemma}

If $a_i \in L^p, b_i \in L^q, a_i \stackrel{L^p}\rightharpoonup a, b_i \stackrel{L^q}\rightharpoonup b$, then $a_ib_i \stackrel{L^r}\rightharpoonup ab$ where $\frac{1}{r}=\frac{1}{p} + \frac{1}{q}$.


\end{lemma}

We will now use the above proposition to prove the following:

\begin{prop}

$\mathcal{B}_0^n$ is a metric space (and hence Hausdorff).

\end{prop}

\proof Define a metric $$d([A],[\tilde{A}])= \inf_{u, \tilde{u} \in \mathcal{G}_1^n} ||uA - \tilde{u}\tilde{A}||_{L^n(X)}$$

Since we can rewrite $uA-\tilde{u}\tilde{A} = u(A-u^{-1}\tilde{u}A)$, so we have the following $\norm{uA - \tilde{u}\tilde{A}}_{L^n(X)} = ||u(A-u^{-1}\tilde{u}A)||_{L^n(X)} = ||A-u^{-1}\tilde{u}A||_{L^n(X)}$. Hence we can rewrite the metric as follows: $$d([A],[\tilde{A}])= \inf_{u \in \mathcal{G}_1^n} ||A - u\tilde{A}||_{L^n(X)}$$

We check that $d([A],[\tilde{A}])$ satisfies the conditions of a metric space:

\begin{enumerate}

\item
Symmetry. Clearly, $$d([A],[\tilde{A}]) = \inf_{u \in \mathcal{G}_1^n} ||A - u\tilde{A}||_{L^n(X)} = \inf_{u \in \mathcal{G}_1^n} ||\tilde{A} - uA||_{L^n(X)} = d([\tilde{A}],[A])$$

\item
Triangle inequality. In the $L^n$ metric, we have:
$$||A-u\tilde{A}||_{L^n} \leq ||A-v\tilde{\tilde{A}}||_{L^n} + ||v\tilde{\tilde{A}}-u\tilde{A}||_{L^n}$$
We take $\inf_{u \in \mathcal{G}_1^n}$ of both sides, and get;
$$d([A],[\tilde{A}]) \leq d([A], [\tilde{\tilde{A}}]) + d([\tilde{\tilde{A}}],[A])$$
\item
$d([A],[\tilde{A}]) = 0 \Rightarrow [A]= [\tilde{A}]$.
If $d([A],[\tilde{A}]) = 0$, then there exists a sequence $u_i$ such that $\lim_{i \rightarrow \infty} ||u_i A - \tilde{A}||_{L^n} = 0$. So the points $u-i A$ lie in some $L^n$ ball about $\tilde{A}$, this implies that, after possibly passing to some subsequence, $u_i A \rightharpoonup \tilde{\tilde{A}}$ for some $\tilde{\tilde{A}}$. However, by the previous proposition, $\mathcal{G}_i^n$ orbits are weakly closed in $\mathcal{A}_0^n$, so $\tilde{\tilde{A}} = u\tilde{A}$ for some $u \in \mathcal{G}_i^n$. Since we have a subsequence $u_i$ for which $\lim_{i \rightarrow \infty} ||u_i A - \tilde{A}||_{L^n} = 0$, we have $\tilde{\tilde{A}} = \tilde{A} = uA$. Hence, $A$ and $\tilde{A}$ are in same $\mathcal{G}_1^n$ orbit, so $[A]=[\tilde{A}]$.


\end{enumerate}