09-10 Notes

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


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 $\mathcal{B}_k^p \hookrightarrow \mathcal{B}_l^q$?


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 $u \in \mathcal{G}_{k+1}^p$.


\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.


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


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